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14800 | https://www.youtube.com/watch?v=5z1mfbvYXMs | Sum of Arithmetic Series Using the Gauss's Method
Mathema Teach
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Posted: 24 Apr 2021
this video will show how to determine the sum of arithmetic series using the method developed by Johann Carl Friedrich Gauss
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Transcript:
Introduction [Music] hello everyone in this video we are going to determine the sum of arithmetic series using gauss's method before we jump into this example we have right here let's have a short life story of gauss and what prompted him to develop a method of determining the sum of arithmetic series Life Story of Carl Gauss [Music] gauss was born as johan carl frederick gauss a german mathematician and physicist legend has it that long time ago when famous mathematician carl gause was a young student his teacher tried to keep his class busy by giving them a task of computing the sum of the first 100 integer that means they were asked to get the sum of 1 plus 2 plus 3 plus 4 plus 5 plus 97 plus 98 plus 99 plus 100 so pretty much adding all the numbers and that really keeps them busy he stunned his teacher and his classmates by calculating the sum quickly without assistance now how did gause do it now let's go back to the video and see how gauss developed a method in adding some of arithmetic series [Music] Sum of Arithmetic Series okay so we are supposed to use gauss's method to calculate the sum of this given arithmetic series that is we're adding the first um 15 numbers that means that's 1 plus 2 plus 3 plus 4 all the way to plus 50. so what is the sum for all of this first 50 um numbers there are steps that we need to follow in order that we can determine the sum of an arithmetic series using gauss's method the first step is to write out the given series so this is the given series that we have right here so we can go ahead and write rewrite it so we can have this as our s s stands for the sum and so i am going to rewrite all of this okay so that's the first step the second step is to add the series to itself but reverse the order of the terms so we're gonna go ahead and uh start with 50 and then all the way to one so i'm just gonna go ahead and label that as again as s so we're now on step two so we start with 50 because we're reversing it so i'm gonna have 50 plus 49 so i'll do it backwards plus 48 plus 47 plus 46 plus and then all the way to five all the way to one i'm just going to write it up here now we're ready to move on to the third step the third step is to add these two equations that we have here so we're gonna add these two so i put a plus there so that we can go ahead and say that if we are going to add this this will be s plus s will be 2s and that is equal to 1 plus 50 is 51 and then 2 plus 49 is 51 and then 3 plus 48 is 51 and then 4 plus 47 is 51 and then 5 plus 46 is 51 plus and then we have 46 plus 5 is 51 plus 47 plus 4 is 51. 48 plus 3 is 51. 49 plus 2 is 51. 50 plus 1 is 51. notice that this right side of the equation that we have here would all be 51 and part of this step is to solve for the sum s so please notice that these are all 51 so that we can go ahead and say that this is 2s is equal to 51 now how many 51s are there that's the number that we multiply in here now in order that we can determine the number that we put in here we are going to go back to the lesson on slope intercept form so i'm just going to go ahead and show the work on the side right here so we are going to create a table so in this table we are going to um name this as so i'm just going to go ahead and write the table right here so we will have this as our n and this will be our t of n so our n here represents the term numbers i'm just going to label that up here while our t of n is the term okay so then we create the table okay so we're gonna start with a term number we're gonna start with one two three four and five so that's the term number this means that is the first term so in the given series that we have here this is our first term second term third term fourth term fifth term so our first term is one our second term is two our third term is three our fourth term is four and our fifth term is five notice that this term right here is growing by a fixed number that is adding one each time so that's a plus one and plus one and then also plus one so if the growth factor is the same we can go ahead and say that this table represents a linear function we can go ahead and use the slope intercept form of a line which is y equals m x plus b so that we go ahead and say that our m for this um uh table we have here again m represents the slope it's the growth factor that's one and our b or the y-intercept or this is otherwise called as the zeroth term that means we go back one time from the first term we go back words one time to get the zero term so i'm just gonna go ahead and put a broken line right there so the zeroth term means we subtract one so this is zero so our zero term right here is actually zero and that is our y-intercept b which is zero so that we can go ahead and change this equation that we have right here using this notation that we used there so our y was actually t of n so which is the term t of n is equal to our slope here is one and our x is the term number n and our b is zero so that's a plus zero so we can go ahead and say that our t of n is actually equal to n so that is the equation that would work for this so if we want to know the term number for 50 because we don't know uh what term number is this we know that this is first term second term third term fourth term fifth term what term number is 50 we are going to solve for n so if our term is 50 we're gonna solve for the term of the last one so that's 50 is equal to n and so we can go ahead and say that our n is actually 50. we can go ahead and say that the total number of terms for this is 50 and this is the one that we put into the parentheses this means that there are 50 50 ones in this series so i'm just gonna go ahead and write 50 right here so if we multiply 51 times 50 so that would equal to 2005 now we are again supposed to solve for s which is the sum so we're gonna divide both sides by two i divide this by two so that i am left with um the sum s we can go ahead and say that the value for s is 1275 if we divide those two this means that this is the sum of the first 50 integers that means we're adding 1 plus 2 plus 3 all the way to 50 the sum for all these would be 1275. okay at this time i would encourage you to pause this video and try this problem out on your own and when you're done and pause it and check your answer okay so we go over the problem here again the first step is to write out the given series so i'm just gonna go ahead and label this as s and then we are going to rewrite this okay now we move on to the second step the second step states that we're supposed to add the series to itself but reverse the order of the terms so i'm just gonna go ahead and label this as s and that is equal to we start with 129 so that's 129 plus uh 122 plus we have 115 plus we have 108 plus and until we get all the way to three okay now we're ready to move on to the third step the third step is to add the two equations and solve for the sum s so that means we're going to add these two equations that we have here that's the first one and this is the second one so we're gonna add them up together so this is how it's gonna look like okay so what i did was i added these two equations and it came out to be 2s and that they're all equal to 132. so that we can go ahead and rewrite this as 2s is actually equal to 132 times our problem right now is to determine how many 132 are there so just by looking at this we have one two three four five six seven eight but please remember there are still these three dots that we have here is telling us that there are still more um 132 in between these two right here so how do we determine the um the rest of this 132 that's in between these two here so that we go ahead and use the um slope intercept form so i go ahead and show the work on the side here okay so i've labeled the term number for n this means this is the first term so we look at the list our first term is three so i'm gonna write three down here that is the term and then for the second term that is 10 so i'm going to write 10 right here and then the third term is 17 so i'm going to write 17 right here and the fourth term is 24 so i'm gonna write 24 right here now looking at this we can determine how much does it grow by so that would be from three to ten it was added seven each time so 10 to 17 that's plus seven so um 17 to 24 that's another plus seven so then again this is a linear function again we are going to use the uh slope intercept form y equals m x plus b where in this equation here our m or our slope is the growth factor which is seven and our b is the zero term that means we go backwards one time so from one we go backwards right there so that value would be from we that would be zero right here so we go back one time and then we go back one time on here as well so then um three minus seven would be negative four so this is negative four right here so this is where our um y-intercept or the zero term is going to come from so this is negative four right here because that's the value of the tn now we can go ahead and rewrite this in terms of this notation that we got right there so our y is actually the t of n so i'm gonna write t of n is equal to our slope is seven our x is n and our y intercept is negative 4. so this is our equation or this is the equation that works for this given series that we have right there now our t of n our term is 129 so that's the last term so we say that this is term number one term number two term number three term number four that's n is equal to four now what n is this so how many terms total are there so we need to solve for n so again our t of n is 129 that's the last one so i plug it in here so 129 is equal to 7 n minus 4. again we're solving for n that's the number of terms total for this given arithmetic series so we can go ahead and solve for n here okay our n value is 19 this means that there are 19 numbers all together and you can get all the way to 129 so pretty much there are still numbers in here so when we count them this is n is equal to 1 uh 2 3 4 all the way to 17 18 and 19. so that's the 19th term so this 19 that we have here will be multiplied to 132 so that is multiplied by 19. this means that there are 19 132 on the right side of the equation so there are 19 of them although it shows that there are only eight here so there's still more in between that and that's how we got it here so that we can go ahead and um solve for s so that would be 2s is equal to 132 times 19 is 2500 so that's 2508 and then again we want to solve for s or the sum we're going to divide 2 from both sides divide this by 2 so we are left with s is equal to 1 1254 so this is the sum of this given arithmetic series that we have here it means that when we add 3 plus 10 plus 17 plus 24 all the way to 129 which is the 19th term it will give us 1254. did you get the same answers as this good perfect if you found this video helpful hit like and subscribe for more math videos see ya |
14801 | https://www.geeksforgeeks.org/maths/derivative-of-tan2x/ | Derivative of tan^2x - GeeksforGeeks
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Derivative of tan^2x
Last Updated : 23 Jul, 2025
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In calculus, finding the derivative of functions involving the trigonometric functions is a common task. One such function is tan 2(x). This article will guide we through the process of differentiating tan 2(x) including the step-by-step calculations and practical applications.
What is Tangent Function?
The tangent function, written as tan(x) is a trigonometric function that relates the angle x in a right triangle to the ratio of the opposite side to the adjacent side. Specifically, for an angle x in a right triangle:
tan(x)=Opposite Side / Adjacent Side
In terms of the unit circle, tan(x) is defined as the ratio of the sine of the angle to the cosine of the angle:
tan(x) = sin(x)/cos(x)
Derivative of tan 2 x
The derivative of tan2 x with respect to x can be found using the chain rule.
The derivative of tan2(x) is 2tan(x) sec2(x).
How to Differentiate tan 2 x?
The chain rule states that if a function f(x) can be expressed as g(h(x)) then its derivative is given by:
d d x[g(h(x))]=g′(h(x))⋅h′(x)\frac{d}{dx}[g(h(x))] = g'(h(x)) \cdot h'(x) d x d[g(h(x))]=g′(h(x))⋅h′(x)
In our case:
Let g(u) = u 2, where u = tan(x). Thus, f(x) = g(tan(x)).
The outer function is g(u) = u 2. The derivative of the u 2 with the respect to u is:
d d u[u 2]=2 u\frac{d}{du}[u^2] = 2u d u d[u 2]=2 u
The inner function is h(x) = tan(x). The derivative of tan(x) with the respect to x is:
d d x[tan(x)]=sec2(x)\frac{d}{dx}[\tan(x)] = \sec^2(x) d x d[tan(x)]=sec 2(x)
Applying the chain rule:
d d x[tan2(x)]=2 tan(x)⋅sec2(x)\frac{d}{dx}[\tan^2(x)] = 2 \tan(x) \cdot \sec^2(x) d x d[tan 2(x)]=2 tan(x)⋅sec 2(x)
Combining these results, the derivative of tan 2(x) is:
d d x[tan2(x)]=2 tan(x)⋅sec2(x)\frac{d}{dx}[\tan^2(x)] = 2 \tan(x) \cdot \sec^2(x) d x d[tan 2(x)]=2 tan(x)⋅sec 2(x)
Read More,
Derivatives
What are Derivatives and How it Works?
Higher Order Derivatives
First and Second Order Derivatives
Left Hand And Right Hand Derivatives
Solved Examples on Derivative of tan 2 x
Example 1: Find the derivative of f(x) = tan2x.
Solution:
We will use the chain rule to the differentiate tan 2(x).
Let u(x) = tan(x). Then, f(x) = u 2(x).
The derivative of u 2(x) with respect to the x is:
d d x[u 2(x)]=2 u(x)⋅d u(x)d x \frac{d}{dx} \left[u^2(x)\right] = 2u(x) \cdot \frac{du(x)}{dx} d x d[u 2(x)]=2 u(x)⋅d x d u(x)
The derivative of u(x) =tan(x) is:
d u(x)d x=sec2(x) \frac{du(x)}{dx} = \sec^2(x) d x d u(x)=sec 2(x)
Substituting u(x) = tan(x) and d u(x)d x=sec2(x)\frac{du(x)}{dx} = \sec^2(x)d x d u(x)=sec 2(x) into the chain rule formula:
d d x[tan2(x)]=2 tan(x)⋅sec2(x) \frac{d}{dx} \left[\tan^2(x)\right] = 2\tan(x) \cdot \sec^2(x) d x d[tan 2(x)]=2 tan(x)⋅sec 2(x)
Example 2: Find the value ofd d x[tan2(x)]\frac{d}{dx} \left[\tan^2(x)\right]d x d[tan 2(x)]atx=π 4 x = \frac{\pi}{4}x=4 π.
Solution:
From Example 1 we have:
d d x[tan2(x)]=2 tan(x)⋅sec2(x) \frac{d}{dx} \left[\tan^2(x)\right] = 2\tan(x) \cdot \sec^2(x) d x d[tan 2(x)]=2 tan(x)⋅sec 2(x)
Evaluate tan(π 4)\tan\left(\frac{\pi}{4}\right)tan(4 π) and sec2(π 4)\sec^2\left(\frac{\pi}{4}\right)sec 2(4 π) :
tan(π 4)=1 and sec2(π 4)=2\tan\left(\frac{\pi}{4}\right) = 1 \quad \text{and} \quad \sec^2\left(\frac{\pi}{4}\right) = 2 tan(4 π)=1 and sec 2(4 π)=2
Substitute these values into the derivative expression:
d d x[tan2(x)]∣x=π 4=2⋅1⋅2=4\frac{d}{dx} \left[\tan^2(x)\right] \bigg|_{x = \frac{\pi}{4}} = 2 \cdot 1 \cdot 2 = 4 d x d[tan 2(x)]∣∣x=4 π=2⋅1⋅2=4
Example 3: Differentiate g(x) = tan2(x) + sin(x).
Solution:
The derivative of a sum of the functions is the sum of their derivatives:
d d x[tan2(x)+sin(x)]=d d x[tan2(x)]+d d x[sin(x)]\frac{d}{dx} \left[\tan^2(x) + \sin(x)\right] = \frac{d}{dx} \left[\tan^2(x)\right] + \frac{d}{dx} \left[\sin(x)\right]d x d[tan 2(x)+sin(x)]=d x d[tan 2(x)]+d x d[sin(x)]
From Example 1 we know:
d d x[tan2(x)]=2 tan(x)⋅sec2(x)\frac{d}{dx} \left[\tan^2(x)\right] = 2\tan(x) \cdot \sec^2(x)d x d[tan 2(x)]=2 tan(x)⋅sec 2(x)
The derivative of sin(x) is:
d d x[sin(x)]=cos(x)\frac{d}{dx} \left[\sin(x)\right] = \cos(x)d x d[sin(x)]=cos(x)
∴d d x[tan2(x)+sin(x)]=2 tan(x)⋅sec2(x)+cos(x)\frac{d}{dx} \left[\tan^2(x) + \sin(x)\right] = 2\tan(x) \cdot \sec^2(x) + \cos(x)d x d[tan 2(x)+sin(x)]=2 tan(x)⋅sec 2(x)+cos(x)
Example 4: Evaluate at x = 0. Find the derivative of f(x) = tan2(x) + sin(x) at x = 0.
Solution:
From Example 3 we have:
d d x\frac{d}{dx}d x d[tan 2(_x_) + sin(_x_)] = 2tan(_x_) ⋅ sec 2(_x_) + cos(_x_)
Evaluate tan(0), sec 2(0) and cos(0):
tan(0) = 0, sec2(0) = 1, cos(0) = 1
Substitute these values into the derivative expression:
d d x[tan2(x)+sin(x)]∣x=0=2⋅0⋅1+1=1\frac{d}{dx} \left[\tan^2(x) + \sin(x)\right] \bigg|_{x = 0} = 2 \cdot 0 \cdot 1 + 1 = 1 d x d[tan 2(x)+sin(x)]∣∣x=0=2⋅0⋅1+1=1
Example 5: Derivative of h(x) = tan2(x).cos(x)
Solution:
We will use the product rule to differentiate h(x) = tan 2(x) . cos(x).
The product rule states:d d x[u(x)⋅v(x)]=d u(x)d x⋅v(x)+u(x)⋅d v(x)d x \frac{d}{dx} \left[u(x) \cdot v(x)\right] = \frac{du(x)}{dx} \cdot v(x) + u(x) \cdot \frac{dv(x)}{dx} d x d[u(x)⋅v(x)]=d x d u(x)⋅v(x)+u(x)⋅d x d v(x)
Let u(x) = tan 2(x) and v(x) = cos(x).
From Example 1 we have:d u(x)d x \frac{du(x)}{dx} d x d u(x)= 2tan(_x_)⋅sec 2(_x_)
The derivative of v(x) = cos(x) is:d v(x)d x \frac{dv(x)}{dx} d x d v(x)= -sin(x)
By applying the product rule we get,
d d x\frac{d}{dx} d x d tan 2(_x_)⋅cos(_x_)= 2tan(_x_)⋅sec 2(_x_)⋅cos(_x_) - tan 2(_x_)⋅sin(_x_)
Practice Questions on Derivative of tan 2 x
Q1. Find d d x\frac{d}{dx}d x d tan 2(_x_)
Q2. Evaluate d d x\frac{d}{dx}d x d tan 2(_x_) at x= _π_/6 .
Q3. Differentiate f(x) = tan 2(x) - cos(x).
Q4. Find the derivative of f(x)= tan 2(_x_)/ x.
Q5. Evaluate the derivative of f(x) = tan 2(x) + e x at x = 0.
Q6. Differentiate f(x) = tan 2(x) . sin(x).
Q7. Compute d d x\frac{d}{dx}d x d tan 2(_x_) when x= _π_/3.
Q8. Find the derivative of f(x) = tan 2(_x_) / sin(x).
Q9. Differentiate f(x) = tan 2(x).ln(x) .
Q10. Evaluate d d x\frac{d}{dx} d x d tan 2(_x_) at x= _π_/2.
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14802 | https://cbhphilly.org/wp-content/uploads/2024/02/CBH_CPG_Pharma-ADHD-Treatment_2024-02.pdf | Clinical Guidelines: Pharmacologic Treatment of Attention Deficit and Hyperactivity Disorder (ADHD) in Children and Adolescents Updated February 12, 2024 PHARMACOLOGIC TREATMENT OF ADHD IN CHILDREN AND ADOLESCENTS Updated February 2024 2 TABLE OF CONTENTS 1. Background ............................................................................................................................................................................................. 3 2. Purpose .................................................................................................................................................................................................... 3 3. Practice Guidelines (Adapted from AAP Guidelines) ............................................................................................................................ 4 3.1. Assessment .................................................................................................................................................................................. 4 3.2. Cultural and Social Determinants of Health ............................................................................................................................... 4 3.3. Prescribing & Treatment Guidelines .......................................................................................................................................... 5 3.4. Medication Side Effect Monitoring ............................................................................................................................................ 5 3.5. Coordination of Care and Engagement ....................................................................................................................................... 6 3.6. Safety and Diversion ................................................................................................................................................................... 7 4. Monitoring .............................................................................................................................................................................................. 7 5. Appendices .............................................................................................................................................................................................. 8 5.1. ADHD and Psychiatric Co-morbidities ...................................................................................................................................... 8 5.2. Special Populations: Transition Age Youth ................................................................................................................................ 8 5.3. Approved Medications for ADHD.............................................................................................................................................. 9 5.3.1. Stimulants ....................................................................................................................................................................... 9 5.3.2. Non-Stimulants ............................................................................................................................................................. 11 5.4. Additional Resources ................................................................................................................................................................ 12 5.5. References ................................................................................................................................................................................. 12 PHARMACOLOGIC TREATMENT OF ADHD IN CHILDREN AND ADOLESCENTS Updated February 2024 3 1. BACKGROUND Community Behavioral Health (CBH) is committed to working with our provider partners to continuously improve the quality of behavioral health care for our shared population. Whenever possible, this is best accomplished through the implementation of evidence-based practices, as well as those informed by nationally recognized treatment guidelines, while respecting the need for individualized treatment. The following are medication prescribing standards, adapted for the CBH Network from national treatment guidelines. They are intended to guide providers in aligning their practices with the best available scientific evidence to help members with ADHD access state-of-the-art care. To assess quality of care, CBH will be collecting several standardized metrics. These metrics come either from the Healthcare Effectiveness Data and Information Set (HEDIS), a set of measures used by many major health care organizations for quality improvement or are measures of clear clinical priority in our network. While CBH will be collecting specific data related only to guidelines that have been issued to the network thus far, the use of empirical guidelines and practice parameters is encouraged in all prescribing. CBH expects providers to follow these guidelines in addition to all other relevant CBH, state, and federal regulations and standards, including CBH prescribing Bulletins (e.g., Provider Bulletin 07-07: Screening for and Treatment of the Components of Metabolic Syndrome), the Department of Behavioral Health and Intellectual disAbility Services (DBHIDS) Practice Guidelines for Resiliency and Recovery-Oriented Treatment, and the DBHIDS Network Inclusion Criteria (NIC) Standards for Excellence. Note further that the following are guidelines for the pharmacologic treatment of ADHD. CBH and DBHIDS encourage a biopsychosocial and recovery- and resiliency-based approach to treatment; in each case, these guidelines for medication treatment should be but one part of a robust, multidisciplinary treatment approach that involves high-quality psychosocial treatment, collaboration with physical health providers, and inclusion of families and other supports. 2. PURPOSE CBH has updated its guidelines for the treatment of ADHD in children and adolescents to reflect the most recently published evidence-based practice parameters available: those of the American Academy of Pediatrics (AAP), issued in 20191. The current guidelines include references to the Diagnostic and Statistical Manual of Mental Disorders, Fifth Edition (DSM-5-TR), published since the last iteration of the guidelines in 2011. The update includes the addition of a key action statement regarding the treatment of comorbid conditions. Also notable is the addition of a supplemental article on systemic barriers to the care of children 1 Wolraich ML, Hagan JF, Allan C, et al. AAP Subcommittee on Children and Adolescents with Attention Deficit/Hyperactive Disorder. Clinical Practice Guideline for the Diagnosis, Evaluation, and Treatment of Attention-Deficit/Hyperactivity Disorder in Children and Adolescents. Pediatrics. 2019;144(4):e20192528. PHARMACOLOGIC TREATMENT OF ADHD IN CHILDREN AND ADOLESCENTS Updated February 2024 4 and adolescents with ADHD. The AAP subcommittee included representatives from the American Academy of Child and Adolescent Psychiatry (AACAP). AACAP Clinical Practice Guidelines for ADHD are pending and will replace the AACAP Practice Parameters for ADHD2. CBH encourages its network providers to remain current with the state of evidence-based practice parameters and to incorporate these into the clinical care offered. These guidelines reflect the best scientific evidence available to guide treatment delivery and should be considered the standard of care in the CBH Network. Resources including further details on behavioral treatments related to these guidelines for providers may be accessed in the AAP Clinical Practice Guidelines. 3. PRACTICE GUIDELINES (ADAPTED FROM AAP GUIDELINES) 3.1. Assessment Any child four through 18 years of age who presents with academic or behavioral problems and symptoms of inattention, hyperactivity, or impulsivity should receive an evaluation for ADHD. To make a diagnosis of ADHD, the prescribing physician should confirm and document that DSM-5-TR criteria have been met (including documentation of impairment in more than one major setting). This diagnosis should be informed primarily by reports from parents or guardians, teachers, and other school and mental health clinicians involved in the child’s care. Parent and teacher behavior rating scales (e.g., the ADHD Rating Scale-IV, the Child Behavior Checklist, the Vanderbilt ADHD diagnostic scales, and the Conners rating scales) may be helpful in clarifying the diagnosis. In the evaluation of a child for ADHD, assessment should be completed for comorbid conditions, including emotional or behavioral (e.g., anxiety, depressive, oppositional defiant, conduct, or trauma-related disorders, substance use), developmental (e.g., learning and language disorder, autism spectrum disorders), and physical (e.g., tics, sleep apnea) conditions. These evaluations must be done by appropriately trained and licensed personnel. The presence of a comorbid condition may alter the treatment of ADHD in some cases. 3.2. Cultural and Social Determinants of Health Recent evidence suggests that African American and Latinx children are less likely to have ADHD diagnosed and are less likely to be treated for ADHD. Special attention should be given to these populations when assessing comorbidities related to ADHD and during ADHD treatment. 2 Pliszka, Steven. AACAP Workgroup on Quality Issues. AACAP Official Action, Practice Parameter for the Assessment and Treatment of Children and Adolescents with Attention-Deficit/Hyperactivity Disorder. Journal of the American Academy of Child and Adolescent Psychiatry. 2007;46(7): 894-921. PHARMACOLOGIC TREATMENT OF ADHD IN CHILDREN AND ADOLESCENTS Updated February 2024 5 3.3. Prescribing & Treatment Guidelines Recommendations for treatment of children and adolescents with ADHD vary by age: 1. For preschool-aged children (defined by the AAP as children 4-5 years of age), the AAP recommends implementing evidence-based parent- and/or teacher-administered behavior therapy (e.g., Parent Child Interaction Therapy) as the first line of treatment. The AAP recommends methylphenidate as the first line medication if behavioral interventions do not provide significant improvement and there is moderate to severe functional impairment. The AAP notes that methylphenidate is not FDA-approved for the treatment of ADHD in preschool-aged children but has the most robust evidence to support its use in this population. Short-acting amphetamine/dextroamphetamine and dextroamphetamine are FDA-approved to treat ADHD in children as young as three years of age. The AAP does not recommend amphetamines as first line medication for ADHD in this population as they report that the criteria for FDA approval at the time does not meet current standards for approval. The AAP does not recommend diagnosing or treating children younger than four years of age with ADHD based on insufficient data, except for parent training in behavior management for ADHD-like symptoms with significant impairment. In areas where evidence-based behavioral treatments are not available, the clinician needs to weigh the risks of starting medication at an early age against the harm of delaying diagnosis and treatment. 2. For elementary school-aged children (6-11 years of age), the clinician should prescribe FDA-approved medications for ADHD (see Table 1 in Section 5.) and/or evidence-based parent and/or teacher-administered behavior therapy as treatment for ADHD, preferably both. The evidence is particularly strong for stimulant medications and sufficient but less strong for atomoxetine, extended-release guanfacine, and extended-release clonidine (in that order). The school environment, placement, and supports should be a part of a comprehensive treatment plan and often includes an Individualized Education Program (IEP) or Section 504 plan. c. For adolescents (12-18 years of age), clinicians should prescribe FDA-approved medications for ADHD (see Table 1) with the assent of the adolescent and may prescribe behavior therapy as treatment for ADHD, preferably both. Stimulant medications are highly effective in reducing core ADHD symptoms in adolescents. The school environment, program, or placement should be a part of a comprehensive treatment plan and often includes an IEP) or Section 504 plan. The prescriber should initiate and titrate the doses of medication for ADHD to achieve the maximum benefit with minimum adverse effects. 3.4. Medication Side Effect Monitoring Prescribers are encouraged to monitor heart rate (HR) and blood pressure (BP) in youth taking stimulant medication. Non-stimulant ADHD medications may also impact HR and BP. Clinicians are advised to obtain the youth’s and family’s cardiac history prior to initiation of ADHD medications. The AAP recommends PHARMACOLOGIC TREATMENT OF ADHD IN CHILDREN AND ADOLESCENTS Updated February 2024 6 further cardiac evaluation prior to initiation of treatment if risk factors are present. Prescribing guidelines for stimulants and atomoxetine include monitoring height and weight at baseline and periodically during treatment. Prescribing of medications for ADHD that are not FDA-approved for this indication is generally discouraged and must proceed according to CBH Provider Bulletin 10-03: Use of Psychotropic Medications in Children and Adolescents (FDA-Approved and Off-Label). To ensure appropriate titration to optimize symptom control, for appropriate reevaluation of symptoms and functional impact, and to monitor for the emergence of adverse effects, these follow-up intervals are required after the initiation of an ADHD medication: 1. One medication management follow-up visit no more than 30 days after the prescription is initiated 2. At least two more such visits after the initial 30-day period of medication 3.5. Coordination of Care and Engagement The public children’s behavioral health system in Pennsylvania is based on the principles developed through the Department of Human Services (DHS) Child and Adolescent Service System Program (CASSP). The principles are defined as follows: child-centered, family-focused, community-based, multi-system, culturally competent, and least intrusive/restrictive. Moreover, Philadelphia’s System of Care initiative, whose mission is to support the delivery of quality mental health services to children and young adults with complex mental health needs, promotes care that is youth and family-driven, community based, and culturally and linguistically competent3. Treatment for ADHD, including pharmacologic interventions, should align with these principles. Youth diagnosed with ADHD should receive evidence-based and holistic care, to include access to resources and services that promote recovery, resilience and thriving. Specifically, care should be characterized by access to behavioral/psychotherapeutic interventions, integration with physical health systems, effective collaboration with child-serving systems including schools and child welfare, peer support, and case management. Achieving improved follow-up care and the provision of evidence-based care are contingent upon successful engagement of youth and families in care. Providers are encouraged to develop and implement processes that promote engagement and strengthen the therapeutic alliance. The National Alliance on Mental Illness (NAMI) has published guidance on promoting a culture of engagement in the mental health system; salient features include principles of shared-decision making, adopting a strengths-based approach, and including natural supports4. 3 Philadelphia System of Care, System of Care Framework. 4 National Alliance on Mental Illness (NAMI), Engagement: A New Standard for Mental Health Care. PHARMACOLOGIC TREATMENT OF ADHD IN CHILDREN AND ADOLESCENTS Updated February 2024 7 3.6. Safety and Diversion As with any medication, providers should educate parents and caregivers regarding safe storage practices for ADHD medications. Children should only have access to medications under the supervision of a responsible adult. To minimize the risk of diversion when prescribing stimulant medications, which are classified as Schedule II medications by the Drug Enforcement Agency (DEA), providers should adhere to established state and federal guidelines including querying prescription drug monitoring programs. In situations where the risk of diversion needs to be minimized, providers should consider prescribing long-acting stimulants with a lower risk of abuse potential. 4. MONITORING CBH encourages its providers to maintain robust internal quality management programs to ensure treatment of CBH members adheres to these and other applicable guidelines. In addition to “as needed” reviews of medical records when quality issues arise, CBH will be tracking and sharing the following performance metrics with providers: HEDIS – ADD: Follow-up care for children prescribed ADHD Medication, tracked via the National Committee for Quality Assurance (NCQA) HEDIS measure5) » Initiation Phase: Assesses children between age 6-12 who were diagnosed with ADHD and had one follow-up visit with a practitioner with prescribing authority within 30 days of their first prescription of ADHD medication » Continuation Phase: Assesses children between age 6-12 who had a prescription for ADHD medication and remained on the medication for at least 210 days and had at least 2 follow-up visits with a practitioner in the 9 months following Initiation Phase Appropriate use of medication for children and adolescents diagnosed with ADHD (will be tracked via claims data to generate percentages of members with ADHD prescribed FDA-approved medications, other medications, and no medication). In addition, providers should maintain documentation of all evaluations and interventions described in these guidelines, whether delivered by the provider or an outside practitioner. CBH and the DBHIDS Network Improvement and Accountability Collaborative (NIAC) will continue to monitor treatment provided to assure that care is consistent with the DBHIDS NIC Standards for Excellence. 5 National Committee for Quality Assurance (NCQA), Follow-Up Care for Children Prescribed ADHD Medication (ADD, ADD-E). PHARMACOLOGIC TREATMENT OF ADHD IN CHILDREN AND ADOLESCENTS Updated February 2024 8 5. APPENDICES 5.1. ADHD and Psychiatric Co-morbidities Special consideration should be given for the complexities in managing ADHD with co-morbid psychiatric diagnoses. ADHD is often comorbid with Oppositional Defiant Disorder, Conduct Disorder, Mood Disorders, Anxiety Disorders, Substance Abuse disorders, Autism Spectrum Disorder, Intellectual disability, language and learning disorders, and Tic Disorders6. Children and adolescents with ADHD and coexisting conditions are at risk for more adverse outcomes. Youth/caregiver preference, symptomatology, the degree of functional impairment, medical/family history, and psychosocial factors should inform treatment planning in this clinical context. Based on the above factors, decisions regarding medication, therapy, and behavioral interventions will vary and should be evidence-based and characterized by psychoeducation, informed consent, and shared decision making. Treating ADHD with psychiatric comorbidities may include using combined therapy (medication with behavioral/therapeutic interventions) as first-line treatment, optimizing drug delivery mechanisms (e.g., prescribing prodrug or long-acting stimulants to reduce the risk of abuse in youth with substance abuse disorders), and prioritizing symptom reduction for the primary diagnosis7. 5.2. Special Populations: Transition Age Youth The term transitional age youth (TAY) generally refers to the cohort of individuals spanning the age demographic of older adolescence to early adulthood, approximately 16-25 years of age. This developmental period is characterized by unique challenges and milestones, including separation-individuation, identity formation, vocational and educational transitions, and achieving intimacy8. Youth with mental health conditions, including ADHD, are particularly vulnerable during this time given the increased rate of emerging psychiatric illnesses, functional impairment, and significant psychosocial stressors for youth who identify as ethnic and/or sexual minorities, are economically disadvantaged, aging out of foster care, or have multiple system involvement. This cohort of youth must also navigate transitioning from child to adult serving systems of care, including mental and physical health care. According to the AAP and AACAP, a significant percentage of children diagnosed with ADHD will continue to experience symptoms and impairment into adulthood. TAY receiving treatment for ADHD may be at risk of disengaging from treatment and suffering adverse health, educational, and socioeconomic outcomes. Strategies and interventions that may facilitate 6 Barbaresi WJ, Campell L, Diekroger EA, et al. Society for Developmental and Behavioral Pediatrics, Clinical Practice Guideline for the Assessment and Treatment of Children and Adolescents with Complex Attention-Deficit/Hyperactivity Disorder, Journal of Developmental & Behavioral Pediatrics. 2020; 41: S35-S57 doi: 10.1097/DBP.0000000000000770. 7 Austerman, J. ADHD and Behavioral Disorders: Assessment, Management and an update from DSM-5, Cleveland Clinic Journal of Medicine. 2015; 82(1): S2-S7. 8 Martel A & Fuchs CD. Transitional Age Youth and Mental Illness – Influences on Young Adult Outcomes, Child and Adolescent Psychiatric Clinics. 2017;26(2): PXIII-XVII. PHARMACOLOGIC TREATMENT OF ADHD IN CHILDREN AND ADOLESCENTS Updated February 2024 9 continuity of care for TAY include motivational interviewing, peer support, mentorship, case management, shared decision making, and strengthening natural supports9. 5.3. Approved Medications for ADHD10 5.3.1. Stimulants Brand Name Generic Name Duration Available Dosage Strengths Class: Methylphenidate Adhansia XR™ methylphenidate hydrochloride – extended-release (capsule) 16 hours 25mg 35mg 45mg 55mg 70mg 85mg AzstarysTM serdexmethylphenidate and dexmethylphenidate (capsule) 10+ hours 26.1mg/5.2mg 39.2mg/7.8mg 52.3mg/10.4mg Aptensio XR™ methylphenidate hydrochloride – extended-release (capsule) 12 hours 10mg 15mg 20mg 30mg 40mg 50mg 60mg Concerta® methylphenidate hydrochloride – extended-release (tablet) 10-12 hours 18mg 27mg 36mg 54mg 72mg Cotempla™XR-ODT methylphenidate extended-release (orally disintegrating tablet) 8-12 hours 8.6mg 17.3mg 25.9mg Daytrana® methylphenidate (transdermal patch) 10-16 hours 10mg 15mg 20mg 30mg Focalin® dexmethylphenidate hydrochloride (tablet) 3-5 hours 2.5mg 5mg 10mg Focalin XR® dexmethylphenidate hydrochloride – extended-release (capsule) 12 hours 5mg 10mg 15m 20mg 25m 30mg 35mg 40mg Jornay PM™ methylphenidate hydrochloride – extended-release (capsule) 12+ hours 20mg 40mg 60mg 80mg 100mg Metadate CD® methylphenidate hydrochloride – extended-release (capsule) 8 hours 10mg 20mg 30mg 40mg 50mg 9 Buitelaar J.K. Optimising treatment strategies for ADHD in adolescence to minimise “lost in transition” to adulthood, Epidemiology and Psychiatric Sciences. 2017;26: 448–452 doi:10.1017/S2045796017000154 10 ADHD Medications Approved by the US FDA. Prepared by Children and Adults with Attention-Deficit/Hyperactivity Disorder (CHADD), 2023. PHARMACOLOGIC TREATMENT OF ADHD IN CHILDREN AND ADOLESCENTS Updated February 2024 10 Brand Name Generic Name Duration Available Dosage Strengths Metadate® ER methylphenidate hydrochloride – extended-release (tablet) 8-12 hours 20mg Methylin® ER methylphenidate hydrochloride – extended-release (tablet) 8 hours 10mg 20mg Methylin® Oral Solution methylphenidate hydrochloride (liquid) 3-5 hours 5mg/5ml and 10mg/5ml QuilliChew ER™ methylphenidate hydrochloride – extended-release (chewable tablet) 8-12 hours 20mg 30mg 40mg Quillivant XR® methylphenidate hydrochloride – extended-release (liquid) 8, 10, and 12 hours 25mg/5ml (5mg/ml) Ritalin® methylphenidate hydrochloride (tablet) 3-5 hours 5mg 10mg 20mg Ritalin LA® methylphenidate hydrochloride – extended-release (capsule) 8 hours 10mg 20mg 30mg 40mg Class: Amphetamine Adderall® Amphetamine and dextroamphetamine mixed salts (tablet) 4-8 hours 5mg 7.5mg 10mg 12.5mg 15mg 20mg 30mg Adderall XR® Amphetamine and dextroamphetamine mixed salts – extended-release (capsule) 8-12 hours 5mg 10mg 15mg 20mg 25mg 30mg Adzenys ER amphetamine extended-release oral suspension (liquid) 9-12 hours 3.1mg/2.5ml 6.3mg/5ml 9.4mg/7.5ml 12.5mg/10ml 15.7mg/12.5ml 18.8mg/15ml Adzenys XR-ODT™ amphetamine extended-release (orally disintegrating tablet) 9-12 hours 3.1mg 6.3mg 9.4mg 12.5mg 15.7mg 18.8mg Desoxyn® methamphetamine hydrochloride (tablet) 4-8 hours 5mg Dexedrine® Dextroamphetamine-sulfate-(tablet) 4-6 hours 2.5mg 5mg 7.5mg 10mg 15 mg 20mg 30mg Dexedrine® dextroamphetamine sulfate – extended-release (tablet) 6-9 hours 15mg Dexedrine Spansule® dextroamphetamine sulfate – extended release (capsule) 8-12 hours 15mg PHARMACOLOGIC TREATMENT OF ADHD IN CHILDREN AND ADOLESCENTS Updated February 2024 11 Brand Name Generic Name Duration Available Dosage Strengths Dyanavel® XR amphetamine extended-release tablet 8–12 hours 2.5mg 5mg 10mg 15mg 20mg Dyanavel® XR amphetamine extended-release oral suspension (liquid) 8-12 hours 2.5mg/ml, 12.5mg/tsp Evekeo® amphetamine sulfate (tablet) 4-6 hours 5mg 10mg Evekeo ODT™ amphetamine sulfate – orally disintegrating (tablet) 4-6 hours 5mg 10mg 15mg 20mg Mydayis™ mixed salts of a single-entity amphetamine product – extended-release (capsule) 16 hours 12.5mg 25mg 37.5mg 50mg ProCentra® dextroamphetamine sulfate (liquid) 4-8 hours 5mg/5ml Vyvanse® lisdexamfetamine dimesylate (chewable tablet) 8-12 hours 10mg 20mg 30mg 40mg 50mg 60mg Vyvanse® lisdexamfetamine dimesylate (capsule) 10-12 hours 10mg 20mg 30mg 40mg 50mg 60mg 70mg Xelstrym™ dextroamphetamine (transdermal patch) 9 hours 10mg 15mg 20mg 30mg Zenzedi® dextroamphetamine sulfate (tablet) 4-8 hours 2.5mg 5mg 7.5mg 10mg 15mg 20mg 30mg 5.3.2. Non-Stimulants Brand Name Generic Name Duration Available Dosage Strengths Class: Norepinephrine Reuptake Inhibitor Strattera® atomoxetine hydrochloride (capsule) 24 hours 10mg 18mg 25mg 40mg 60mg 80mg 100mg Qelbree™ viloxazine extended-release (capsule) 24 hours 100mg 150mg 200mg Class: Alpha Agonist Kapvay® clonidine hydrochloride – extended-release (tablet) 12-24 hours 0.lmg 0.2mg PHARMACOLOGIC TREATMENT OF ADHD IN CHILDREN AND ADOLESCENTS Updated February 2024 12 Brand Name Generic Name Duration Available Dosage Strengths Intuniv® guanfacine hydrochloride – extended-release (tablet) 12-24 hours lmg 2mg 3mg 4mg 5.4. Additional Resources FDA Medication Guides ADHD Medication Guide (Recommended by the AAP) CMS Pediatric Stimulant and Related Medication Factsheet CMS Pediatric Dosing Chart AACAP ADHD: Parents’ Medication Guide 5.5. References Community Behavioral Health (CBH), Provider Bulletin 07-07: Screening for and Treatment of the Components of Metabolic Syndrome. Department of Behavioral Health and Intellectual disAbility Services (DBHIDS), Philadelphia Behavioral Health Practice Guidelines, 2013, or latest version. Department of Behavioral Health and Intellectual disAbility Services (DBHIDS), Network Inclusion Criteria Standards of Excellence, February 2019, or latest version. Wolraich ML, Hagan JF, Allan C, et al. AAP Subcommittee on Children and Adolescents with Attention Deficit/Hyperactive Disorder. Clinical Practice Guideline for the Diagnosis, Evaluation, and Treatment of Attention-Deficit/Hyperactivity Disorder in Children and Adolescents. Pediatrics. 2019;144(4):e20192528. Pliszka, Steven. AACAP Workgroup on Quality Issues. AACAP Official Action, Practice Parameter for the Assessment and Treatment of Children and Adolescents with Attention-Deficit/Hyperactivity Disorder. Journal of the American Academy of Child and Adolescent Psychiatry. 2007;46(7): 2007;46(7):894-921. Community Behavioral Health (CBH), Provider Bulletin 10-03: Use of Psychotropic Medications in Children and Adolescents (FDA-Approved and Off-Label). National Committee for Quality Assurance (NCQA), Follow-Up Care for Children Prescribed ADHD Medication (ADD, ADD-E). PHARMACOLOGIC TREATMENT OF ADHD IN CHILDREN AND ADOLESCENTS Updated February 2024 13 PA Department of Human Services Child and Adolescent Social Service Program (CASSP). Philadelphia System of Care, System of Care Framework. National Alliance on Mental Illness (NAMI), Engagement: A New Standard for Mental Health Care. Barbaresi WJ, Campell L, Diekroger EA, et al. Society for Developmental and Behavioral Pediatrics, Clinical Practice Guideline for the Assessment and Treatment of Children and Adolescents with Complex Attention-Deficit/Hyperactivity Disorder, Journal of Developmental & Behavioral Pediatrics. 2020; 41: S35-S57 doi: 10.1097/DBP.0000000000000770. Austerman, J. ADHD and Behavioral Disorders: Assessment, Management and an update from DSM-5, Cleveland Clinic Journal of Medicine. 2015; 82(1): S2-S7. Martel A & Fuchs CD. Transitional Age Youth and Mental Illness – Influences on Young Adult Outcomes, Child and Adolescent Psychiatric Clinics. 2017;26(2): PXIII-XVII. Buitelaar J.K. Optimising treatment strategies for ADHD in adolescence to minimise ‘lost in transition’ to adulthood, Epidemiology and Psychiatric Sciences. 2017;26: 448–452 doi:10.1017/S2045796017000154 |
14803 | https://www.testing.com/tests/apo-b/ | At a Glance
Why Get Tested?
To help evaluate your risk of developing cardiovascular disease (CVD); sometimes to help monitor treatment for high cholesterol or to help diagnose a rare inherited apolipoprotein B (apo B) deficiency
When To Get Tested?
When you have a personal or family history of heart disease and/or high cholesterol and triglyceride levels and your healthcare provider is trying to determine your risk of developing CVD; sometimes on a regular basis when you are being treated for high cholesterol; rarely when your health care practitioner suspects that you have an inherited apo B deficiency
Sample Required?
A blood sample drawn from a vein in your arm
Test Preparation Needed?
No special preparation is needed for an apo B test. However, since this test is often ordered at the same time as other tests that do require fasting, such as LDL-C, HDL-C and triglycerides, fasting for at least 12 hours may be required.
Best Home Apo B Test
LabCorp Complete Heart Health Test (with Apo B)
Price:$169
Type:Self-collection
Sample:Blood
Tests for:Apo B, Diabetes Risk Index, GlyCa, cholesterol, lipids
Results timeline:1-3 days after the sample is received
We’re highlighting the Complete Heart Health Test (with Apo B) from LabCorp because it’s convenient, comprehensive, and relatively affordable. The test measures key indicators for cardiovascular disease and Type 2 diabetes. In addition to Apo B, the test also measures VLDL, LDL, and HDL cholesterol, triglycerides, and GlyCa.
A kit for self-collecting your blood sample is included with your order. You should receive your results within three business days of the sample arriving at one of LabCorp’s CLIA-certified laboratories.
What is being tested?
Apolipoprotein B-100 (also called apolipoprotein B or apo B) is a protein that is involved in the metabolism of lipids and is the main protein constituent of lipoproteins such as very low-density lipoprotein (VLDL) and low-density lipoprotein (LDL, the “bad cholesterol”). This test measures the amount of apo B in the blood.
Apolipoproteins combine with lipids to transport them throughout the bloodstream. Apolipoproteins provide structural integrity to lipoproteins and shield the water-repellent (hydrophobic) lipids at their center. Most lipoproteins are cholesterol- or triglyceride-rich and carry lipids through the body for uptake by cells.
Chylomicrons are the lipoprotein particles that carry dietary lipids from the digestive tract, via the bloodstream, to tissue – mainly the liver. In the liver, the body repackages these dietary lipids and combines them with apo B-100 to form triglyceride-rich VLDL. This combination is like a taxi full of passengers with apo B-100 as the taxi driver. In the bloodstream, the taxi moves from place to place, releasing one passenger at a time.
An enzyme called lipoprotein lipase (LPL) removes triglycerides from VLDL to produce intermediate density lipoproteins (IDL) first and then LDL. Each VLDL particle contains one molecule of apo B-100, which is retained as VLDL loses triglycerides and shrinks to become the more cholesterol-rich LDL. Apo B-100 is recognized by receptors found on the surface of many of the body’s cells. These receptors promote the uptake of cholesterol into the cells.
The cholesterol that LDL and apo B-100 transport is vital for cell membrane integrity, sex hormone production, and steroid production. In excess, however, LDL can lead to fatty deposits (plaques) in artery walls and lead to hardening and scarring of the blood vessels. These fatty depositions narrow the vessels in a process termed atherosclerosis. The atherosclerotic process increases the risk of heart attack.
Apo B-100 levels tend to mirror LDL-C levels, a test routinely ordered as part of a lipid profile. Many experts think that apo B levels may eventually prove to be a better indicator of risk of cardiovascular disease (CVD) than LDL-C. Some recommend the measurement of apo B to help with risk prediction when a person has multiple risk factors. Other experts disagree; they feel that apo B is only a marginally better alternative and do not recommend its routine use. The clinical utility of apo B and that of other emerging cardiac risk markers such as apo A-I, Lp(a), and hs-CRP has yet to be fully established.
Common Questions
How is the test used?
The apolipoprotein B (apo B) test is used, along with other lipid tests, to help determine an individual’s risk of developing cardiovascular disease (CVD).
This test is not used as a general population screen but may be ordered if a person has a family history of heart disease and/or high cholesterol and triglycerides (hyperlipidemia). It may be performed, along with other tests, to help diagnose the cause of abnormal lipid levels, especially when someone has elevated triglyceride levels.
A healthcare practitioner may order both an apo A-I (associated with high-density lipoprotein (HDL), the “good” cholesterol) and an apo B to determine an apo B/apo A-I ratio. This ratio is sometimes used as an alternative to a total cholesterol/HDL ratio to evaluate risk for developing CVD.
Apo B levels may be ordered to monitor the effectiveness of lipid treatment as an alternative to non-HDL-C (non-HDL-C is the total cholesterol concentration minus the amount of HDL).
In rare cases, an apo B test may be ordered to help diagnose a genetic problem that causes over- or under-production of apo B.
When is it ordered?
Apo B may be measured, along with an apo A-I or other lipid tests, when a healthcare practitioner is trying to evaluate someone’s risk of developing CVD and when a person has a personal or family history of heart disease and/or abnormal lipid levels, especially when the person has significantly elevated triglyceride levels.
Sometimes apo B is ordered to monitor a person who is undergoing treatment for high cholesterol.
What does the test result mean?
Elevated levels of apo B correspond to elevated levels of LDL-C and to non-HDL-C and are associated with an increased risk of cardiovascular disease (CVD). Elevations may be due to a high-fat diet and/or decreased clearing of LDL from the blood.
Some genetic disorders are the direct (primary) cause of abnormal levels of apo B. For example, familial combined hyperlipidemia is an inherited disorder causing high blood levels of cholesterol and triglycerides. Abetalipoproteinemia, also called Apolipoprotein B deficiency or Bassen-Kornzweig syndrome, is a very rare genetic condition that can cause abnormally low levels of apo B.
Abnormal levels of apo B can also be caused by underlying conditions or other factors (secondary causes). Increased levels of apo B are seen, for example, in:
Diabetes
Use of drugs such as: androgens, beta blockers, diuretics, progestins (synthetic progesterones)
Hypothyroidism
Nephrotic syndrome (a kidney disease)
Pregnancy (levels increase temporarily and decrease again after delivery)
Apo B levels may be decreased with any condition that affects lipoprotein production or affects its synthesis and packaging in the liver. Lower levels are seen with secondary causes such as:
Use of drugs such as: estrogen (in post-menopausal women), lovastatin, simvastatin, niacin, and thyroxine
Hyperthyroidism
Malnutrition
Reye syndrome
Weight reduction
Severe illness
Surgery
Cirrhosis
An increased ratio of apo B to apo A-I may indicate a higher risk of developing CVD.
Is there anything else I should know?
Chylomicrons, the lipoprotein particles that carry dietary lipids to the liver, contain a lipoprotein called apolipoprotein B-48. It is about half the size of apo B-100 and is structurally related to apo B-100. It is not considered a risk factor for atherosclerosis and is not measured as part of the apo-B test. The apo B test is specific for apo B-100.
What can I do to lower my apo B?
Diet and exercise changes that lower LDL levels (and increase HDL, the “good” cholesterol) will lower your apo B levels and decrease your risk of heart disease.
What could cause apoB and LDL levels to stay high despite lifestyle changes?
Some elevations of apo B-100 (and LDL-C) are due to mutations in the APOB gene that cause it to produce apo B-100 that is not recognized as easily by LDL receptors. Others are in the LDL receptor system of the liver cell that recognizes apo B-100. These genetic defects impede the clearing of LDL from the blood and result in accumulations of LDL in the circulation, increasing the risk of heart disease.
Can an apo B test be performed in my doctor's office or at home?
No, the apo B test requires specialized equipment and is not offered by every laboratory. Your blood may need to be sent to a reference laboratory for testing.
Related Tests
LDL Cholesterol Blood Test (LDL-C)
Learn More
Triglycerides Testing
Learn More
HDL Cholesterol Testing
Learn More
Lipoprotein (a) Test
Learn More
hs-CRP Test (C-Reactive Protein High-Sensitivity)
Learn More
Homocysteine Test
Learn More
Cardiac Risk Assessment
Learn More
Lipid Panel Test
Learn More
Apo A-I
Learn More
Resources
American Heart Association: Understand Your Risks to Prevent a Heart Attack
Learn More
NHLBI: Ischemic Heart Disease
Learn More
MedlinePlus Medical Encyclopedia: Familial hypercholesterolemia
Learn More
MedlinePlus Medical Encyclopedia: Familial dysbetalipoproteinemia
Learn More
MedlinePlus Medical Encyclopedia: Bassen-Kornzweig syndrome
Learn More
Sources
Sources Used in Current Review
Meeusen, J. (2018 May). Apolipoprotein B or Low-Density Lipoprotein Cholesterol: Is It Time for a Twenty-First-Century Lipid Marker? Clin Chem 2018 64 (6), p. 984. Available online at Accessed on 9/08/18.
Rosenso, R. et. al. (2016 January 19). Integrated Measure for Atherogenic Lipoproteins in the Modern Era: Risk Assessment Based on Apolipoprotein B. JACC v67(2). 202-204. Available online at Accessed on 9/08/18.
Delgado, J. et. al. (2018 August, Updated). Atherosclerotic Cardiovascular Disease (ASCVD) Nontraditional Risk Markers – Cardiovascular Disease Risk Markers (Nontraditional). ARUP Consult. Available online at Accessed on 9/08/18.
(© 1995– 2018). Apolipoprotein B, Plasma. Mayo Clinic Mayo Medical Laboratories. Available online at Accessed on 9/08/18.
Morita, S. (2016). Metabolism and Modification of Apolipoprotein B-Containing Lipoproteins Involved in Dyslipidemia and Atherosclerosis. Biol Pharm Bull 39, 1-24 (2016). Available online at Accessed on 9/08/18.
Szternel, L. et. al. (2018 June 21). Non-fasting lipid profile determination in presumably healthy children: Impact on the assessment of lipid abnormalities. PLoS One. 2018 Jun 21;13(6):e0198433. Available online at Accessed on 9/08/18.
Kim, S. et. al. (2017 September 28). The association between the apolipoprotein B/A-I ratio and coronary calcification may differ depending on kidney function in a healthy population. PLoS One. 2017 Sep 28;12(9):e0185522. Available online at Accessed on 9/08/18.
Sources Used in Previous Reviews
Thomas, Clayton L., Editor (1997). Taber’s Cyclopedic Medical Dictionary. F.A. Davis Company, Philadelphia, PA [18th Edition].
Pagana, Kathleen D. & Pagana, Timothy J. (2001). Mosby’s Diagnostic and Laboratory Test Reference 5th Edition: Mosby, Inc., Saint Louis, MO.
Apolipoprotein B Mutation Detection. ARUP’s Guide to Clinical Laboratory Testing (CLT) [On-line information]. Available online at
Angelo, S., Reviewed (2001 November 21, Reviewed). Apolipoprotein B100. University of Pennsylvania Health System Encyclopedia [On-line information]. Available online at
Gianturco, S. & Bradley, W. (1999). Pathophysiology of Triglyceride-Rich Lipoproteins in Atherothrombosis: Cellular Aspects. Clin. Cardiol. 22, (Suppl. 11) 11-7-11-14 [On-line Journal]. PDF available for download at
Blanchear, M. (2000 April 22). Lipoprotein Interconversions. University of Manitoba, Tutorials in Biochemistry [On-line information]. Available online through
(2001 October). Apolipoprotein B, Detecting Familial Hypercholesterolemia. ARUP [On-line Technical Bulletin]. Available online at
(2002 October 25, Updated). Apolipoprotein B100. MedlinePlus Health Information [On-line information]. Available online at
(2001 October 30). Bassen-Kornzweig Syndrome. MedlinePlus Health Information [On-line information]. Available online at
Pagana, Kathleen D. & Pagana, Timothy J. (© 2007). Mosby’s Diagnostic and Laboratory Test Reference 8th Edition: Mosby, Inc., Saint Louis, MO., Pp 110-114.
Clarke, W. and Dufour, D. R., Editors (2006). Contemporary Practice in Clinical Chemistry, AACC Press, Washington, DC. Winter, W. and Harris, N. Chapter 21: Lipoprotein Disorders, Pp 251-259.
(2006 August). Abetalipoproteinemia. Genetics Home Reference [On-line information]. Accessed on: 4/29/07 Available online at
Gandelman, G. (2007 February 7, Updated). Apolipoprotein B100. MedlinePlus Medical Encyclopedia [On-line information]. Available online at Accessed on 4/29/07
Sondheimer, N. (2005 April 20, Updated). Bassen-Kornzweig syndrome. MedlinePlus Medical Encyclopedia [On-line information]. Available online at Accessed on 4/29/07
Kasper D, et al, eds. (2005). Harrision’s Principles of Internal Medicine 16th edition, McGraw Hill.
Zieve, D. and Dugdale, D. (Updated 2010 May 23). Apolipoprotein B100. MedlinePlus Medical Encyclopedia [On-line information]. Available online at Accessed July 2010.
Ganda, O. (2009 August 31). Refining Lipoprotein Assessment in Diabetes: Apolipoprotein B Makes Sense. Medscape from Endocrine Practice. 2009;15(4):370-376 [On-line information]. Available online at Accessed July 2010.
Benderly, M. et. al. (2009 February 16). Apolipoproteins and Long-Term Prognosis in Coronary Heart Disease Patients. Medscape from American Heart Journal. 2009;157(1):103-110 [On-line information]. Available online at Accessed July 2010.
Delgado, J. et. al. (Updated 2010 April). Cardiovascular Disease (Non-traditional Risk Markers) – Risk Markers – CVD (Non-traditional). ARUP Consult [On-line information]. Available online at . Accessed July 2010.
(© 1995–2010). Unit Code 80308: Apolipoprotein B, Plasma. Mayo Clinic, Mayo Medical Laboratories [On-line information]. Available online at Accessed July 2010.
Myers, G. Editor (2009). Emerging Biomarkers for Primary Prevention of Cardiovascular Disease and Stroke. The National Academy of Clinical Biochemistry Laboratory Medicine Practice Guidelines [On-line information]. PDF available for download at Accessed July 2010.
Pagana, K. D. & Pagana, T. J. (© 2007). Mosby’s Diagnostic and Laboratory Test Reference 8th Edition: Mosby, Inc., Saint Louis, MO. Pp 110-114.
Wu, A. (© 2006). Tietz Clinical Guide to Laboratory Tests, 4th Edition: Saunders Elsevier, St. Louis, MO. Pp 146-149.
Contois J, et al. Apolipoprotein B and Cardiovascular Disease Risk: Position Statement from the AACC Lipoproteins and Vascular Diseases Division Working Group on Best Practices. Clin Chem 55(3):407 (2009). Available online at Accessed Sept 2010.
Tietz Textbook of Clinical Chemistry and Molecular Diagnostics. Burtis CA, Ashwood ER, Bruns DE, eds. St. Louis: Elsevier Saunders; 2006 Pp 916-917, 928-934.
Zieve, D. (Updated 2012 June 4). Apolipoprotein B100. MedlinePlus Medical Encyclopedia [On-line information]. Available online at Accessed March 2014.
Zieve, D. (Updated 2012 June 4). Familial combined hyperlipidemia. MedlinePlus Medical Encyclopedia [On-line information]. Available online at [On-line information]. Available online at Accessed March 2014.
(© 1995–2014). Apolipoprotein B, Plasma. Mayo Clinic Mayo Medical Laboratories [On-line information]. Available online at Accessed March 2014.
Delgado, J. et. al. (Updated 2014 February). Cardiovascular Disease (Non-traditional Risk Markers) – Risk Markers – CVD (Non-traditional) ARUP Consult [On-line information]. Available online at Accessed March 2014.
Ford, E. et. al. (2013). Temporal Changes in Concentrations of Lipids and Apolipoprotein B Among Adults With Diagnosed and Undiagnosed Diabetes, Prediabetes, and Normoglycemia Findings From the National Health and Nutrition Examination Survey 1988-1991 to 2005-2008. Medscape Multispecialty from Cardiovasc Diabetol. 2013;12(26) [On-line information]. Available online at Accessed March 2014.
Elhomsy, G and Griffing, G. (Updated 2012 July 10). Apolipoprotein B. Medscape Reference [On-line information]. Available online at Accessed March 2014.
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·Se il presente indicativo(1° persona singolare)è irregolare, il presente subjuntivoconserva la stessa irregolarità:
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·Se il futuro imperfectoè irregolare, il condicionalha la stessa irregolarità:
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14806 | https://mathworld.wolfram.com/UniformPolyhedron.html | Uniform Polyhedron -- from Wolfram MathWorld
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Uniform Polyhedron
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The uniform polyhedra are polyhedra consisting of regular (possibly polygrammic) faces of equal edge length whose polyhedron vertices are all symmetrically equivalent. The uniform polyhedra include the Platonic solids (consisting of equal convex regular polygon faces), Archimedean soldis (consisting of convex regular faces of more than one type). Unlike these special cases, the uniform polyhedra need not enclose a volume and in general have self-intersections between faces. For example, the Kepler-Poinsot polyhedra (consisting of equal concave regular polygon or polygram faces) are uniform polyhedra whose outer hulls enclose a volume but which contain interior faces corresponding to parts of the faces that are not part of the hull. Badoureau discovered 37 such nonconvex uniform polyhedra in the late nineteenth century, many previously unknown (Wenninger 1983, p.55).
Coxeter et al.(1954) conjectured that there are 75 uniform polyhedra in which only two faces are allowed to meet at an polyhedron edge, and this surmise was subsequently proven. The five pentagonal prisms can also be considered uniform polyhedra, bringing the total to 80. In addition, there are two other polyhedra in which four faces meet at an edge, the great complex icosidodecahedron and small complex icosidodecahedron (both of which are compounds of two uniform polyhedra).
The polyhedron vertices of a uniform polyhedron all lie on a circumsphere whose center is their geometric centroid (Coxeter et al. 1954, Coxeter 1973, p.44). The polyhedron vertices joined to another polyhedron vertex lie on a circle (Coxeter et al. 1954).
Not-necessarily circumscriptable versions of uniform polyhedra with exactified numeric vertices and polygrammic faces sometimes split into separate polygons are implemented in the Wolfram Language as UniformPolyhedron["name"] or UniformPolyhedron["Uniform", n] (cf. Garcia 2019). The full exact, equilateral, circumscriptable uniform polyhedra are implemented in the Wolfram Language as PolyhedronData["name"] or PolyhedronData["Uniform", n].
Except for a single non-Wythoffian case, uniform polyhedra can be generated by Wythoff's kaleidoscopic method of construction. In this construction, an initial vertex inside a special spherical triangle is mapped to all the other vertices by repeated reflections across the three planar sides of this triangle. Similarly, and its kaleidoscopic images must cover the sphere an integral number of times which is referred to as the density of . The density is dependent on the choice of angles , , at , , respectively, where , , are reduced rational numbers greater than one. Such a spherical triangle is called a Schwarz triangle, conveniently denoted . Except for the infinite dihedral family of for , 3, 4, ..., there are only 44 kinds of Schwarz triangles (Coxeter et al. 1954, Coxeter 1973). It has been shown that the numerators of , , are limited to 2, 3, 4, 5 (4 and 5 cannot occur together) and so the nine choices for rational numbers are: 2, 3, 3/2, 4, 4/3, 5, 5/2, 5/3, 5/4 (Messer 2002).
The names of the 75 uniform polyhedra were first formalized in Wenninger (1983, first printed in 1971), based on a list prepared by N.Johnson a few years earlier, as slightly modified by D.Luke. Johnson also suggested a few modifications in the original nomenclature to incorporate some additional thoughts, as well as to undo some of Luke's less felicitous changes. The "List of polyhedra and dual models" in Wenninger (1983) gives revised names for several of the uniform polyhedra. The names of the five pentagonal prisms appeared in Har'El (1993).
The following table gives the names of the uniform polyhedra and their duals as given in Wenninger (1983) and Har'El (1993) and with the numberings of Maeder (1997), Wenninger (1971), Coxeter et al.(1954), and Har'El (1993). Coxeter et al.(1954) give many properties of the uniform solids, and Coxeter et al.(1954), Johnson (2000), and Messer (2002) give the quartic equation for determining the central angle subtending half an edge. The single non-Wythoffian case is the great dirhombicosidodecahedron with Maeder index 75 which has pseudo-Wythoff symbol.
Maeder index Wenninger index Coxeter index Har'El indexWythoff symbolnamedual polyhedron
1 1 15 6regular tetrahedrontetrahedron
2 6 16 7truncated tetrahedrontriakis tetrahedron
3 68 37 8octahemioctahedronoctahemioctacron
4 67 36 9tetrahemihexahedrontetrahemihexacron
5 2 17 10regular octahedroncube
6 3 18 11cubeoctahedron
7 11 19 12cuboctahedronrhombic dodecahedron
8 7 20 13truncated octahedrontetrakis hexahedron
9 8 21 14truncated cubesmall triakis octahedron
10 13 22 15small rhombicuboctahedron (rhombicuboctahedron)deltoidal icositetrahedron
11 15 23 16great rhombicuboctahedron (truncated cuboctahedron)disdyakis dodecahedron
12 17 24 17snub cubepentagonal icositetrahedron
13 69 38 18small cubicuboctahedronsmall hexacronic icositetrahedron
14 77 50 19great cubicuboctahedrongreat hexacronic icositetrahedron
15 78 51 20cubohemioctahedronhexahemioctacron
16 79 52 21cubitruncated cuboctahedrontetradyakis hexahedron
17 85 59 22quasirhombicuboctahedron (great rhombicuboctahedron)great deltoidal icositetrahedron
18 86 60 23small rhombihexahedronsmall rhombihexacron
19 92 66 24stellated truncated hexahedron (quasitruncated hexahedron)great triakis octahedron
20 93 67 25great truncated cuboctahedron (quasitruncated cuboctahedron)great disdyakis dodecahedron
21 103 82 26great rhombihexahedrongreat rhombihexacron
22 4 25 27regular icosahedrondodecahedron
23 5 26 28regular dodecahedronicosahedron
24 12 28 29icosidodecahedronrhombic triacontahedron
25 9 27 30truncated icosahedronpentakis dodecahedron
26 10 29 31truncated dodecahedrontriakis icosahedron
27 14 30 32small rhombicosidodecahedron (rhombicosidodecahedron)deltoidal hexecontahedron
28 16 31 33great rhombicosidodecahedron (truncated icosidodechedon)disdyakis triacontahedron
29 18 32 34snub dodecahedronpentagonal hexecontahedron
30 70 39 35small ditrigonal icosidodecahedronsmall triambic icosahedron
31 71 40 36small icosicosidodecahedronsmall icosacronic hexecontahedron
32 110 41 37small snub icosicosidodecahedronsmall hexagonal hexecontahedron
33 72 42 38small dodecicosidodecahedronsmall dodecacronic hexecontahedron
34 20 43 39small stellated dodecahedrongreat dodecahedron
35 21 44 40great dodecahedronsmall stellated dodecahedron
36 73 45 41dodecadodecahedronmedial rhombic triacontahedron
37 75 47 42truncated great dodecahedronsmall stellapentakis dodecahedron
38 76 48 42rhombidodecadodecahedronmedial deltoidal hexecontahedron
39 74 46 44small rhombidodecahedronsmall rhombidodecacron
40 111 49 45snub dodecadodecahedronmedial pentagonal hexecontahedron
41 80 53 46ditrigonal dodecadodecahedronmedial triambic icosahedron
42 81 54 47great ditrigonal dodecicosidodecahedrongreat ditrigonal dodecacronic hexecontahedron
43 82 55 48small ditrigonal dodecicosidodecahedronsmall ditrigonal dodecacronic hexecontahedron
44 83 56 49icosidodecadodecahedronmedial icosacronic hexecontahedron
45 84 57 50icositruncated dodecadodecahedrontridyakis icosahedron
46 112 58 51snub icosidodecadodecahedronmedial hexagonal hexecontahedron
47 87 61 52great ditrigonal icosidodecahedrongreat triambic icosahedron
48 88 62 53great icosicosidodecahedrongreat icosacronic hexecontahedron
49 89 63 54small icosihemidodecahedronsmall icosihemidodecacron
50 90 64 55small dodecicosahedronsmall dodecicosacron
51 91 65 56small dodecahemidodecahedronsmall dodecahemidodecacron
52 22 68 57great stellated dodecahedrongreat icosahedron
53 41 69 58great icosahedrongreat stellated dodecahedron
54 94 70 59great icosidodecahedrongreat rhombic triacontahedron
55 95 71 60great truncated icosahedrongreat stellapentakis dodecahedron
56 96 72 61rhombicosahedronrhombicosacron
57 116 88 62great snub icosidodecahedrongreat pentagonal hexecontahedron
58 97 74 63small stellated truncated dodecahedrongreat pentakis dodecahedron
59 98 75 64truncated dodecadodecahedronmedial disdyakis triacontahedron
60 114 76 65inverted snub dodecadodecahedronmedial inverted pentagonal hexecontahedron
61 99 77 66great dodecicosidodecahedrongreat dodecacronic hexecontahedron
62 100 78 67small dodecahemicosahedronsmall dodecahemicosacron
63 101 79 68great dodecicosahedrongreat dodecicosacron
64 115 80 69great snub dodecicosidodecahedrongreat hexagonal hexecontahedron
65 102 81 70great dodecahemicosahedrongreat dodecahemicosacron
66 104 83 71great stellated truncated dodecahedrongreat triakis icosahedron
67 105 84 72quasirhombicosidodecahedron (great rhombicosidodecahedron)great deltoidal hexecontahedron
68 108 87 73great truncated icosidodecahedrongreat disdyakis triacontahedron
69 113 73 74great inverted snub icosidodecahedrongreat inverted pentagonal hexecontahedron
70 107 86 75great dodecahemidodecahedrongreat dodecahemidodecacron
71 106 85 76great icosihemidodecahedrongreat icosihemidodecacron
72 118 91 77small retrosnub icosicosidodecahedronsmall hexagrammic hexecontahedron
73 109 89 78great rhombidodecahedrongreat rhombidodecacron
74 117 90 79great retrosnub icosidodecahedrongreat pentagrammic hexecontahedron
75 119 92 80great dirhombicosidodecahedrongreat dirhombicosidodecacron
76 1pentagonal prismpentagonal dipyramid
77 2pentagonal antiprismpentagonal trapezohedron
78 33 3pentagrammic prismpentagrammic dipyramid
79 34 4pentagrammic antiprismpentagrammic deltohedron
80 35 5pentagrammic crossed antiprismpentagrammic concave deltohedron
Johnson (2000) proposed a further revision of the "official" names of the uniform polyhedra and their duals and, at the same time, devised a literal symbol for each uniform polyhedron. For each uniform polyhedron, Johnson (2000) gives its number in Wenninger (1989), a modified Schläfli symbol (following Coxeter), a literal symbol, and its new designated name. Not every uniform polyhedron has a dual that is free from anomalies like coincident vertices or faces extending to infinity. For those that do, Johnson gives the name of the dual polyhedron. In Johnson's new system, the uniform polyhedra are classified as follows:
Regular (regular polygonal vertex figures),
Quasi-regular (rectangular or ditrigonal vertex figures),
Versi-regular (orthodiagonal vertex figures),
Truncated regular (isosceles triangular vertex figures),
Quasi-quasi-regular (trapezoidal vertex figures),
Versi-quasi-regular (dipteroidal vertex figures),
Truncated quasi-regular (scalene triangular vertex figures),
Snub quasi-regular (pentagonal, hexagonal, or octagonal vertex figures),
Prisms (truncated hosohedra),
Antiprisms and crossed antiprisms (snub dihedra)
Here is a brief description of Johnson's symbols for the uniform polyhedra (Johnson). The star operator appended to "D" or "E" replaces pentagons by pentagrams . The bar operator indicates the removal from a related figure of a set (or sets) of faces, leaving "holes" so that a different set of faces takes their place. Thus, CO is obtained from the cuboctahedron CO by replacing the eight triangles by four hexagons. In like manner, rR'CO has the twelve squares of the rhombicuboctahedron rCO and the six octagons of the small cubicuboctahedron R'CO but has holes in place of their six squares and eight triangles. The operator "r" stands for "rectified": a polyhedron is truncated to the midpoints of the edges. Operators "a", "b", and "c" in the Schläfli symbols for the ditrigonary (i.e., having ditrigonal vertex figures) polyhedra stand for "altered," "blended," and "converted." The operator "o" stands for "ossified" (after S.L. van Oss). Operators "s" and "t" stand for "simiated" (snub) and "truncated."
Primes and capital letters are used for certain operators analogous to those just mentioned. For instance, rXY is the "rhombi-XY," with the faces of the quasi-regular XY supplemented by a set of square "rhombical" faces. The isomorphic r'XY has a crossed vertex figure. The operators "R" and "R'" denote a supplementary set of faces of a different kind--hexagons, octagons or octagrams, decagons or decagrams. Likewise, the operators "T" and "S" indicate the presence of faces other than, or in addition to, those produced by the simpler operators "t" and "s." The vertex figure of s'XY, the "vertisnub XY," is a crossed polygon, and that of sXY, the "retrosnub XY," has density 2 relative to its circumcenter.
Regular polyhedra:
1T tetrahedron tetrahedron
2O octahedron cube
3C cube octahedron
4I icosahedron dodecahedron
5D dodecahedron icosahedron
20Dsmall stellated dodecahedron great dodecahedron
21E great dodecahedron small stellated dodecahedron
22Egreat stellated dodecahedron great icosahedron
41J great icosahedron great stellated dodecahedron
Quasi-regular polyhedra:
11 rCO cuboctahedron rhombic dodecahedron
12 rID icosidodecahedron rhombic triacontahedron
73 rEDdodecadodecahedron middle rhombic triacontahedron
94 rJEgreat icosidodecahedron great rhombic triacontahedron
70 aIDsmall ditrigonary icosidodecahedron small triambic icosahedron
80 bDEditrigonary dodecadodecahedron middle triambic icosahedron
87 cJE great ditrigonary icosidodecahedron great triambic icosahedron
Versi-regular polyhedra:
67 oTT tetrahemihexahedron no dual
78 oCO cubohemioctahedron no dual
68 oOC octahemioctahedron no dual
91 oDI small dodecahemidodecahedron no dual
89 oID small icosahemidodecahedron no dual
102 oEDsmall dodecahemiicosahedron no dual
100 oDE great dodecahemiicosahedron no dual
106 oJEgreat icosahemidodecahedron no dual
107 oEJ great dodecahemidodecahedron no dual
Truncated regular polyhedra:
6 ttT truncated tetrahedron triakis tetrahedron
7 ttO truncated octahedron tetrakis hexahedron
8 ttC truncated cube triakis octahedron
92 t't'C stellatruncated cube great triakis octahedron
9 ttI truncated icosahedron pentakis dodecahedron
10 ttD truncated dodecahedron triakis icosahedron
97 t't'Dsmall stellatruncated dodecahedron great pentakis dodecahedron
75 ttE great truncated dodecahedron small stellapentakis dodecahedron
104 t't'Egreat stellatruncated dodecahedron great triakis icosahedron
95 ttJ great truncated icosahedron great stellapentakis dodecahedron
Quasi-quasi-regular polyhedra: and
13 rrrCO rhombicuboctahedron strombic disdodecahedron
69 R'rR'CO small cubicuboctahedron small sagittal disdodecahedron
77 RrRCO great cubicuboctahedron great strombic disdodecahedron
85 r'rr'CO great rhombicuboctahedron great sagittal disdodecahedron
14 rrrID rhombicosidodecahedron strombic hexecontahedron
72 R'rR'ID small dodekicosidodecahedron small sagittal hexecontahedron
71 rarIDsmall icosified icosidodecahedron small strombic trisicosahedron
82 R'aR'IDsmall dodekified icosidodecahedron small sagittal trisicosahedron
76 rrrEDrhombidodecadodecahedron middle strombic trisicosahedron
83 R'rR'EDicosified dodecadodecahedron middle sagittal trisicosahedron
81 RcRJE great dodekified icosidodecahedron great strombic trisicosahedron
88 r'cr'JE great icosified icosidodecahedron great sagittal trisicosahedron
99 RrRJEgreat dodekicosidodecahedron great strombic hexecontahedron
105 r'rr'JEgreat rhombicosidodecahedron great sagittal hexecontahedron
Versi-quasi-regular polyhedra:
86 orrR'CO small rhombicube small dipteral disdodecahedron
103 OrRr'CO great rhombicube great dipteral disdodecahedron
74 orrR'ID small rhombidodecahedron small dipteral hexecontahedron
90 oarR'IDsmall dodekicosahedron small dipteral trisicosahedron
96 orrR'EDrhombicosahedron middle dipteral trisicosahedron
101 OcRr'JE great dodekicosahedron great dipteral trisicosahedron
109 OrRr'JEgreat rhombidodecahedron great dipteral hexecontahedron
Truncated quasi-regular polyhedra:
15 trtCO truncated cuboctahedron disdyakis dodecahedron
93 t'rt'CO stellatruncated cuboctahedron great disdyakis dodecahedron
79 TrTCO cubitruncated cuboctahedron trisdyakis octahedron
16 trtID truncated icosidodecahedron disdyakis triacontahedron
98 t'rt'EDstellatruncated dodecadodecahedron middle disdyakis triacontahedron
84 T'rT'EDicositruncated dodecadodecahedron trisdyakis icosahedron
108 t'rt'JEstellatruncated icosidodecahedron great disdyakis triacontahedron
Snub quasi-regular polyhedra: or
17 srsCO snub cuboctahedron petaloidal disdodecahedron
18 srsID snub icosidodecahedron petaloidal hexecontahedron
110 sasIDsnub disicosidodecahedron no dual
118 sasIDretrosnub disicosidodecahedron no dual
111 srsEDsnub dodecadodecahedron petaloidal trisicosahedron
114 s'rs'EDvertisnub dodecadodecahedron vertipetaloidal trisicosahedron
112 S'rS'EDsnub icosidodecadodecahedron hexaloidal trisicosahedron
113 srsJEgreat snub icosidodecahedron great petaloidal hexecontahedron
116 s'rs'JEgreat vertisnub icosidodecahedron great vertipetaloidal hexecontahedron
117 srsJEgreat retrosnub icosidodecahedron great retropetaloidal hexecontahedron
Snub quasi-regular polyhedron:
119 SSrSSJEgreat disnub disicosidisdodecahedron no dual
Prisms:
P(p)-gonal prism, , 5, 6, ...-gonal bipyramid
P(p/d)-fold -gonal prism, -fold -gonal bipyramid
Antiprisms and crossed antiprisms:
shQ(p)-gonal antiprism, , 5, 6, ...-gonal antibipyramid
shQ(p/d)-fold -gonal antiprism, -fold -gonal antibipyramid
s'hQ'(p/d)-fold -gonal crossed antiprism, -fold -gonal crossed antibipyramid
See also
Archimedean Solid, Augmented Polyhedron, Dual Polyhedron, Johnson Solid, Kepler-Poinsot Polyhedron, Möbius Triangles, Platonic Solid, Polyhedron, Schwarz Triangle, Uniform Polychoron, Vertex Figure, Wythoff Symbol
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References
Ball, W.W.R. and Coxeter, H.S.M. Mathematical Recreations and Essays, 13th ed. New York: Dover, p.136, 1987.Brückner, M. Vielecke under Vielflache. Leipzig, Germany: Teubner, 1900.Bulatov, V. "Compounds of Uniform Polyhedra." V. "Dual Uniform Polyhedra." V. "Uniform Polyhedra." H.S.M. Regular Polytopes, 3rd ed. New York: Dover, 1973.Coxeter, H.S.M.; Longuet-Higgins, M.S.; and Miller, J.C.P. "Uniform Polyhedra." Phil. Trans. Roy. Soc. London Ser. A246, 401-450, 1954.Garcia, K. "Building Uniform Polyhedra for Version 12." July 25, 2019. Z. "Uniform Solution for Uniform Polyhedra." Geometriae Dedicata47, 57-110, 1993.Har'El, Z. "Kaleido." Z. "Eighty Dual Polyhedra Generated by Kaleido." Z. "Eighty Uniform Polyhedra Generated by Kaleido." Z. "Uniform Solution for Uniform Polyhedra." Geom. Dedicata47, 57-110, 1993. A. "Exact Descriptions of Regular and Semi-Regular Polyhedra and Their Duals." Computing Science Tech.Rept. No. 130. Murray Hill, NJ: AT&T Bell Lab., 1986.Johnson, N.W. "Convex Polyhedra with Regular Faces." Canad. J. Math.18, 169-200, 1966.Johnson, N.W. Uniform Polytopes. Cambridge, England: Cambridge University Press, 2000. Maeder, R.E. "Uniform Polyhedra." Mathematica J.3, 48-57, 1993. Maeder, R.E. Polyhedra.m and PolyhedraExamples Mathematica notebooks. R.E. "Visual Index of All Uniform Polyhedra." P.W. "Problem 1094." Crux Math.11, 325, 1985.Messer, P.W. "Solution to Problem 1094." Crux Math.13, 133, 1987.Messer, P.W. "Closed-Form Expressions for Uniform Polyhedra and Their Duals." Disc. Comput. Geom.27, 353-375, 2002.Sandia National Laboratories. "Polyhedron Database." J. "The Complete Set of Uniform Polyhedron." Phil. Trans. Roy. Soc. London, Ser. A278, 111-136, 1975.Skilling, J. "Uniform Compounds of Uniform Polyhedra." Math. Proc. Cambridge Philos. Soc.79, 447-457, 1976.Smith, A. "Uniform Compounds and the Group ." Proc. Cambridge Philos. Soc.75, 115-117, 1974.Sopov, S.P. "Proof of the Completeness of the Enumeration of Uniform Polyhedra." Ukrain. Geom. Sbornik8, 139-156, 1970.Virtual Image. The Uniform Polyhedra CD-ROM. 1997. R. "Uniform/Dual Polyhedra." R. "Stellated Polyhedra." M.J. Dual Models. Cambridge, England: Cambridge University Press, 1983.Wenninger, M.J. Polyhedron Models. New York: Cambridge University Press, pp.1-10 and 98, 1989.Zalgaller, V. Convex Polyhedra with Regular Faces. New York: Consultants Bureau, 1969.Ziegler, G.M. Lectures on Polytopes. Berlin: Springer-Verlag, 1995.
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14807 | https://pubmed.ncbi.nlm.nih.gov/33570647/ | ASH ISTH NHF WFH 2021 guidelines on the management of von Willebrand disease - PubMed
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. 2021 Jan 12;5(1):301-325.
doi: 10.1182/bloodadvances.2020003264.
ASH ISTH NHF WFH 2021 guidelines on the management of von Willebrand disease
Nathan T Connell1,Veronica H Flood2,Romina Brignardello-Petersen3,Rezan Abdul-Kadir4,Alice Arapshian5,Susie Couper6,Jean M Grow7,Peter Kouides8,Michael Laffan9,Michelle Lavin10,Frank W G Leebeek11,Sarah H O'Brien12,Margareth C Ozelo13,Alberto Tosetto14,Angela C Weyand15,Paula D James16,Mohamad A Kalot17,Nedaa Husainat17,Reem A Mustafa17
Affiliations Expand
Affiliations
1 Hematology Division, Department of Medicine, Brigham and Women's Hospital, Harvard Medical School, Boston, MA.
2 Versiti Blood Research Institute, Medical College of Wisconsin, Milwaukee, WI.
3 Department of Health Research Methods, Evidence, and Impact, McMaster University, Hamilton, ON, Canada.
4 Department of Obstetrics and Gynaecology and Katharine Dormandy Haemophilia and Thrombosis Centre, Royal Free Foundation Hospital and Institute for Women's Health, University College London, London, United Kingdom.
5 Middle Village, NY.
6 Maylands, WA, Australia.
7 Department of Strategic Communication, Marquette University, Milwaukee, WI.
8 Mary M. Gooley Hemophilia Treatment Center, University of Rochester, Rochester, NY.
9 Centre for Haematology, Imperial College London, London, United Kingdom.
10 Irish Centre for Vascular Biology, Royal College of Surgeons in Ireland and National Coagulation Centre, St James's Hospital, Dublin, Ireland.
11 Department of Hematology, Erasmus University Medical Center, Rotterdam, The Netherlands.
12 Division of Hematology/Oncology, Department of Pediatrics, Nationwide Children's Hospital, The Ohio State University College of Medicine, Columbus, OH.
13 Hemocentro UNICAMP, University of Campinas, Campinas, Brazil.
14 Hemophilia and Thrombosis Center, Hematology Department, S. Bortolo Hospital, Vicenza, Italy.
15 Department of Pediatrics, University of Michigan Medical School, Ann Arbor, MI.
16 Department of Medicine, Queen's University, Kingston, ON, Canada; and.
17 Outcomes and Implementation Research Unit, Division of Nephrology and Hypertension, Department of Internal Medicine, University of Kansas Medical Center, Kansas City, KS.
PMID: 33570647
PMCID: PMC7805326
DOI: 10.1182/bloodadvances.2020003264
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ASH ISTH NHF WFH 2021 guidelines on the management of von Willebrand disease
Nathan T Connell et al. Blood Adv.2021.
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. 2021 Jan 12;5(1):301-325.
doi: 10.1182/bloodadvances.2020003264.
Authors
Nathan T Connell1,Veronica H Flood2,Romina Brignardello-Petersen3,Rezan Abdul-Kadir4,Alice Arapshian5,Susie Couper6,Jean M Grow7,Peter Kouides8,Michael Laffan9,Michelle Lavin10,Frank W G Leebeek11,Sarah H O'Brien12,Margareth C Ozelo13,Alberto Tosetto14,Angela C Weyand15,Paula D James16,Mohamad A Kalot17,Nedaa Husainat17,Reem A Mustafa17
Affiliations
1 Hematology Division, Department of Medicine, Brigham and Women's Hospital, Harvard Medical School, Boston, MA.
2 Versiti Blood Research Institute, Medical College of Wisconsin, Milwaukee, WI.
3 Department of Health Research Methods, Evidence, and Impact, McMaster University, Hamilton, ON, Canada.
4 Department of Obstetrics and Gynaecology and Katharine Dormandy Haemophilia and Thrombosis Centre, Royal Free Foundation Hospital and Institute for Women's Health, University College London, London, United Kingdom.
5 Middle Village, NY.
6 Maylands, WA, Australia.
7 Department of Strategic Communication, Marquette University, Milwaukee, WI.
8 Mary M. Gooley Hemophilia Treatment Center, University of Rochester, Rochester, NY.
9 Centre for Haematology, Imperial College London, London, United Kingdom.
10 Irish Centre for Vascular Biology, Royal College of Surgeons in Ireland and National Coagulation Centre, St James's Hospital, Dublin, Ireland.
11 Department of Hematology, Erasmus University Medical Center, Rotterdam, The Netherlands.
12 Division of Hematology/Oncology, Department of Pediatrics, Nationwide Children's Hospital, The Ohio State University College of Medicine, Columbus, OH.
13 Hemocentro UNICAMP, University of Campinas, Campinas, Brazil.
14 Hemophilia and Thrombosis Center, Hematology Department, S. Bortolo Hospital, Vicenza, Italy.
15 Department of Pediatrics, University of Michigan Medical School, Ann Arbor, MI.
16 Department of Medicine, Queen's University, Kingston, ON, Canada; and.
17 Outcomes and Implementation Research Unit, Division of Nephrology and Hypertension, Department of Internal Medicine, University of Kansas Medical Center, Kansas City, KS.
PMID: 33570647
PMCID: PMC7805326
DOI: 10.1182/bloodadvances.2020003264
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Abstract
Background: von Willebrand disease (VWD) is a common inherited bleeding disorder. Significant variability exists in management options offered to patients.
Objective: These evidence-based guidelines from the American Society of Hematology (ASH), the International Society on Thrombosis and Haemostasis (ISTH), the National Hemophilia Foundation (NHF), and the World Federation of Hemophilia (WFH) are intended to support patients, clinicians, and health care professionals in their decisions about management of VWD.
Methods: ASH, ISTH, NHF, and WFH formed a multidisciplinary guideline panel. Three patient representatives were included. The panel was balanced to minimize potential bias from conflicts of interest. The University of Kansas Outcomes and Implementation Research Unit and the McMaster Grading of Recommendations Assessment, Development and Evaluation (GRADE) Centre supported the guideline development process, including performing and updating systematic evidence reviews (through November 2019). The panel prioritized clinical questions and outcomes according to their importance to clinicians and patients. The panel used the GRADE approach, including GRADE Evidence-to-Decision frameworks, to assess evidence and make recommendations, which were subject to public comment.
Results: The panel agreed on 12 recommendations and outlined future research priorities.
Conclusions: These guidelines make key recommendations regarding prophylaxis for frequent recurrent bleeding, desmopressin trials to determine therapy, use of antiplatelet agents and anticoagulant therapy, target VWF and factor VIII activity levels for major surgery, strategies to reduce bleeding during minor surgery or invasive procedures, management options for heavy menstrual bleeding, management of VWD in the context of neuraxial anesthesia during labor and delivery, and management in the postpartum setting.
© 2021 by The American Society of Hematology.
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Conflict of interest statement
Conflict-of-interest disclosure; and all authors were members of the guideline panel or members of the systematic review team or both. As such, they completed a disclosure-of-interest form, which was reviewed by ASH and is available as supplemental Files 2 and 3.
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Comprehensive comparison of global coagulation assays to differentiate lupus anticoagulant from acquired hemophilia A in patients with prolonged APTT.Chikasawa Y, Amano K, Shinozawa K, Bingo M, Miyashita R, Yamaguchi T, Mitsuhashi A, Inaba H, Hagiwara T, Kinai E.Chikasawa Y, et al.Int J Hematol. 2023 Nov;118(5):577-588. doi: 10.1007/s12185-023-03659-y. Epub 2023 Sep 26.Int J Hematol. 2023.PMID: 37751038
Laboratory Testing for von Willebrand Factor: Factor VIII Binding for the Diagnosis or Exclusion of Type 2N von Willebrand Disease: An Update.Favaloro EJ, Mohammed S, Vong R, Pasalic L.Favaloro EJ, et al.Methods Mol Biol. 2023;2663:679-691. doi: 10.1007/978-1-0716-3175-1_45.Methods Mol Biol. 2023.PMID: 37204745
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Tension
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AP Physics 1 › Tension
Questions 1 - 10
1
Consider the following system:
If the mass is and , what is the tension, ? Assume no frictional forces.
Explanation
Since there is no friction between the mass and slope, there are only two relevant forces acting on the mass: gravity and tension. Furthermore, since the block is not in motion, we know that these forces are equal to each other. Therefore:
Substituting in an expression for the force of gravity, we get:
We know all of these values, allowing us to solve for the tension:
2
A toy is held to a hook on the ceiling with fishing line, determine the tension in the line.
None of these
Explanation
3
Earth mass:
Earth's distance from the Sun:
Velocity of the Earth:
Imagine that, instead of gravity, the earth was attached to the sun with a giant, unbreakable rope. Determine what the tension would be in the rope.
Explanation
The velocity of the earth is
Convert to :
Plug in values:
4
Suppose that a object is lifted upwards while hanging to a rope. If the object accelerates upwards at a rate of , what is the tension in the rope?
Explanation
For this question, we're presented with a scenario in which an object of a given mass is hanging to a rope. That rope is being pulled on, causing the object to accelerate upwards at a certain rate. We're then asked to calculate the tension that results in the rope.
In order to answer this question, it's best if we approach this by examining a force diagram. Since there are no pertinent forces occurring in the x direction, the only concern we have is in the y direction.
One of the forces acting on the object is the downward force of gravity. Another force is the upward tension caused by the rope. Hence, we know what force components contribute to the net force. Furthermore, we're told that as it is pulled up, the object accelerates at a rate of . Since this is the net acceleration in the y direction, we can determine the net force in the y direction.
Next, we can rearrange the above terms in order to isolate the term for tension.
Finally, if we plug in the values that we know, we can calculate our answer.
5
Two objects of equivalent mass are attached with a very strong rope that goes through a pulley. The masses are left to hang. What will happen?
The masses will remain motionless.
One mass will fall and the other will rise.
Both masses will rise.
Both masses will fall.
None of these
Explanation
The masses will each provide an equal force to each other through the rope. Thus, there will be no net force and no net acceleration.
6
Suppose that a person is pulling a box tied to a string across the ground, as shown in the diagram below.
If the string is situated at an angle of with respect to the horizontal and the coefficient of kinetic friction of the box with respect to the ground is , what tension in the string is necessary so that the box moves at a constant speed?
There is not enough information given to answer this question
Explanation
To answer this question, we need to separately consider the component forces acting in the horizontal and vertical direction.
First, let's consider the vertical component. Since the box is only moving in the horizontal direction, we know that there are no net forces acting in the vertical direction. Consequently, the net force in the y-direction is zero.
Next, let's look at the forces acting in the horizontal direction. Since we need to figure out the force in the string necessary to make the box move at a constant speed, we're looking for a situation in which the box is not accelerating in the horizontal direction. Thus, we're looking for a case in which the net force in the x-direction is zero.
Now, let's rewrite the expression for the force of kinetic friction.
Now, plugging in the expression for the normal force, we obtain the following.
Next, let's go ahead and plug in the values given to us in the question stem.
7
A helicopter is lifting a box of mass with a rope. The helicopter and box are accelerating upward at . Determine the tension in the rope.
Explanation
Plug in values:
Solve for
8
An elevator accelerates upward for a short period of time at a rate of . A mass of is hung by a rope from the top of the elevator.
During the period when the elevator is accelerating, what is the magnitude of the tension in the rope?
None of the other answers
Explanation
Here is a (simplistic) diagram of the elevator. There are three forces acting on this object. There is it's own weight due to gravity ; the tension in the rope holding it up, which we will call ; and there is an external force due to the fact that the elevator is accelerating upward. Since the mass is fixed, it has zero velocity with respect to the elevator, and therefore the net force on the mass is 0 by Newton's 2nd law. The net force is given by . Solving this equation for and plugging in the values gives us:
Since we want the magnitude of the tension, and not the value of the vector, we omit the minus sign. Therefore the tension in the rope is , as desired.
9
What is the tension force on a wire holding a 10kg ball 20ft above the ground, if the ball is not moving at that height?
Explanation
Since the gravitational force must be cancelled by the tension force, as the ball is experiencing no acceleration, and no other forces are being applied to it:
10
A 12kg block is sliding down a incline with an acceleration of as shown in the diagram. If the coefficient of kinetic friction of block 1 on the ramp is 0.18, what is the mass of block 2?
Explanation
In order to find the mass of block 2, we're going to need to calculate a few other things, such as the tension in the rope.
To begin with, we'll need to identify the various forces on our free-body diagram. To do this, we will begin with block 1 and use a rotated coordinate system to simplify things. In such a system, the x-axis will run parallel to the surface of the ramp, while the y-axis will be perpendicular to the ramp's surface, as shown below:
Now we can identify the forces acting on block 1. Along the rotated y-axis, the force of gravity acting on the block is equal to , and the force of the ramp on the block is just the normal force, . Since block 1 is not moving in the y direction, we can set these two forces equal to each other.
Now, considering the forces acting along the rotated x-axis, we have a force pointing downwards equal to . Pointing upwards, we have the tension force and we also have the frictional force, .
The formula for calculating the force due to kinetic friction is:
Since we have already determined what the normal force is, we can substitute that expression into the above equation to obtain:
Now, we can write an expression for the net force acting upon block 1 in the x direction:
Rearrange the above expression to solve for tension.
So far, we have only been looking at block 1. Now let's turn our attention to block 2 and see what forces are acting on it. In the downward direction we have the weight of the block due to gravity, which is equal to . In the upward direction, as we can see in the diagram, we have the tension of the rope, . We need to write an expression that tells us the net force acting upon block 2.
Since we calculated the expression for tension from the information regarding block 1, we can plug that expression into the above equation in order to obtain:
Now rearrange to solve for the mass of block 2.
Then plugging in values, we can finally calculate block 2's mass:
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College Algebra 2e
8.4 Rotation of Axes
College Algebra 2e8.4 Rotation of Axes
Contents Contents
Highlights
Table of contents
Preface
1 Prerequisites
2 Equations and Inequalities
3 Functions
4 Linear Functions
5 Polynomial and Rational Functions
6 Exponential and Logarithmic Functions
7 Systems of Equations and Inequalities
8 Analytic Geometry
Introduction to Analytic Geometry
8.1 The Ellipse
8.2 The Hyperbola
8.3 The Parabola
8.4 Rotation of Axes
8.5 Conic Sections in Polar Coordinates
Chapter Review
Exercises
9 Sequences, Probability, and Counting Theory
Answer Key
Index
Search for key terms or text.
Close
Learning Objectives
In this section, you will:
Identify nondegenerate conic sections given their general form equations.
Use rotation of axes formulas.
Write equations of rotated conics in standard form.
Identify conics without rotating axes.
As we have seen, conic sections are formed when a plane intersects two right circular cones aligned tip to tip and extending infinitely far in opposite directions, which we also call a cone. The way in which we slice the cone will determine the type of conic section formed at the intersection. A circle is formed by slicing a cone with a plane perpendicular to the axis of symmetry of the cone. An ellipse is formed by slicing a single cone with a slanted plane not perpendicular to the axis of symmetry. A parabola is formed by slicing the plane through the top or bottom of the double-cone, whereas a hyperbola is formed when the plane slices both the top and bottom of the cone. See Figure 1.
Figure 1 The nondegenerate conic sections
Ellipses, circles, hyperbolas, and parabolas are sometimes called the nondegenerate conic sections, in contrast to the degenerate conic sections, which are shown in Figure 2. A degenerate conic results when a plane intersects the double cone and passes through the apex. Depending on the angle of the plane, three types of degenerate conic sections are possible: a point, a line, or two intersecting lines.
Figure 2 Degenerate conic sections
Identifying Nondegenerate Conics in General Form
In previous sections of this chapter, we have focused on the standard form equations for nondegenerate conic sections. In this section, we will shift our focus to the general form equation, which can be used for any conic. The general form is set equal to zero, and the terms and coefficients are given in a particular order, as shown below.
A x 2+B x y+C y 2+D x+E y+F=0 A x 2+B x y+C y 2+D x+E y+F=0 A x 2+B x y+C y 2+D x+E y+F=0
where A,B,A,B,A,B, and C C C are not all zero. We can use the values of the coefficients to identify which type conic is represented by a given equation.
You may notice that the general form equation has an x y x y x y term that we have not seen in any of the standard form equations. As we will discuss later, the x y x y x y term rotates the conic whenever B B B is not equal to zero.
| Conic Sections | Example |
--- |
| ellipse | 4 x 2+9 y 2=1 4 x 2+9 y 2=1 4 x 2+9 y 2=1 |
| circle | 4 x 2+4 y 2=1 4 x 2+4 y 2=1 4 x 2+4 y 2=1 |
| hyperbola | 4 x 2−9 y 2=1 4 x 2−9 y 2=1 4 x 2−9 y 2=1 |
| parabola | 4 x 2=9 y or 4 y 2=9 x 4 x 2=9 y or 4 y 2=9 x 4 x 2=9 y or 4 y 2=9 x |
| one line | 4 x+9 y=1 4 x+9 y=1 4 x+9 y=1 |
| intersecting lines | (x−4)(y+4)=0(x−4)(y+4)=0(x−4)(y+4)=0 |
| parallel lines | (x−4)(x−9)=0(x−4)(x−9)=0(x−4)(x−9)=0 |
| a point | 4 x 2+4 y 2=0 4 x 2+4 y 2=0 4 x 2+4 y 2=0 |
| no graph | 4 x 2+4 y 2=−1 4 x 2+4 y 2=−1 4 x 2+4 y 2=−1 |
Table 1
General Form of Conic Sections
A conic section has the general form
A x 2+B x y+C y 2+D x+E y+F=0 A x 2+B x y+C y 2+D x+E y+F=0 A x 2+B x y+C y 2+D x+E y+F=0
where A,B,A,B,A,B, and C C C are not all zero.
Table 2 summarizes the different conic sections where B=0,B=0,B=0, and A A A and C C C are nonzero real numbers. This indicates that the conic has not been rotated.
ellipseA x 2+C y 2+D x+E y+F=0,A≠C and A C>0 A x 2+C y 2+D x+E y+F=0,A≠C and A C>0 A x 2+C y 2+D x+E y+F=0,A≠C and A C>0
circleA x 2+C y 2+D x+E y+F=0,A=C A x 2+C y 2+D x+E y+F=0,A=C A x 2+C y 2+D x+E y+F=0,A=C
hyperbolaA x 2−C y 2+D x+E y+F=0 or−A x 2+C y 2+D x+E y+F=0,A x 2−C y 2+D x+E y+F=0 or−A x 2+C y 2+D x+E y+F=0,A x 2−C y 2+D x+E y+F=0 or−A x 2+C y 2+D x+E y+F=0, where A A A and C C C are positive
parabolaA x 2+D x+E y+F=0 or C y 2+D x+E y+F=0 A x 2+D x+E y+F=0 or C y 2+D x+E y+F=0 A x 2+D x+E y+F=0 or C y 2+D x+E y+F=0
Table 2
How To
Given the equation of a conic, identify the type of conic.
Rewrite the equation in the general form, A x 2+B x y+C y 2+D x+E y+F=0.A x 2+B x y+C y 2+D x+E y+F=0.A x 2+B x y+C y 2+D x+E y+F=0.
Identify the values of A A A and C C C from the general form.
If A A A and C C C are nonzero, have the same sign, and are not equal to each other, then the graph may be an ellipse.
If A A A and C C C are equal and nonzero and have the same sign, then the graph may be a circle.
If A A A and C C C are nonzero and have opposite signs, then the graph may be a hyperbola.
If either A A A or C C C is zero, then the graph may be a parabola.
If B = 0, the conic section will have a vertical and/or horizontal axes. If B does not equal 0, as shown below, the conic section is rotated. Notice the phrase “may be” in the definitions. That is because the equation may not represent a conic section at all, depending on the values of A, B, C, D, E, and F. For example, the degenerate case of a circle or an ellipse is a point:
A x 2+B y 2=0,A x 2+B y 2=0,A x 2+B y 2=0, when A and B have the same sign.
The degenerate case of a hyperbola is two intersecting straight lines: A x 2+B y 2=0,A x 2+B y 2=0,A x 2+B y 2=0, when A and B have opposite signs.
On the other hand, the equation, A x 2+B y 2+1=0,A x 2+B y 2+1=0,A x 2+B y 2+1=0, when A and B are positive does not represent a graph at all, since there are no real ordered pairs which satisfy it.
Example 1
Identifying a Conic from Its General Form
Identify the graph of each of the following nondegenerate conic sections.
ⓐ4 x 2−9 y 2+36 x+36 y−125=0 4 x 2−9 y 2+36 x+36 y−125=0 4 x 2−9 y 2+36 x+36 y−125=0
ⓑ9 y 2+16 x+36 y−10=0 9 y 2+16 x+36 y−10=0 9 y 2+16 x+36 y−10=0
ⓒ3 x 2+3 y 2−2 x−6 y−4=0 3 x 2+3 y 2−2 x−6 y−4=0 3 x 2+3 y 2−2 x−6 y−4=0
ⓓ−25 x 2−4 y 2+100 x+16 y+20=0−25 x 2−4 y 2+100 x+16 y+20=0−25 x 2−4 y 2+100 x+16 y+20=0
Solution
ⓐ Rewriting the general form, we have
A=4 A=4 A=4 and C=−9,C=−9,C=−9, so we observe that A A A and C C C have opposite signs. The graph of this equation is a hyperbola.
ⓑ Rewriting the general form, we have
A=0 A=0 A=0 and C=9.C=9.C=9. We can determine that the equation is a parabola, since A A A is zero.
ⓒ Rewriting the general form, we have
A=3 A=3 A=3 and C=3.C=3.C=3. Because A=C,A=C,A=C, the graph of this equation is a circle.
ⓓ Rewriting the general form, we have
A=−25 A=−25 A=−25 and C=−4.C=−4.C=−4. Because A C>0 A C>0 A C>0 and A≠C,A≠C,A≠C, the graph of this equation is an ellipse.
Try It #1
Identify the graph of each of the following nondegenerate conic sections.
ⓐ16 y 2−x 2+x−4 y−9=0 16 y 2−x 2+x−4 y−9=0 16 y 2−x 2+x−4 y−9=0
ⓑ16 x 2+4 y 2+16 x+49 y−81=0 16 x 2+4 y 2+16 x+49 y−81=0 16 x 2+4 y 2+16 x+49 y−81=0
Finding a New Representation of the Given Equation after Rotating through a Given Angle
Until now, we have looked at equations of conic sections without an x y x y x y term, which aligns the graphs with the x- and y-axes. When we add an x y x y x y term, we are rotating the conic about the origin. If the x- and y-axes are rotated through an angle, say θ,θ,θ, then every point on the plane may be thought of as having two representations: (x,y)(x,y)(x,y) on the Cartesian plane with the original x-axis and y-axis, and (x′,y′)(x′,y′)(x′,y′) on the new plane defined by the new, rotated axes, called the x'-axis and y'-axis. See Figure 3.
Figure 3 The graph of the rotated ellipse x 2+y 2–x y–15=0 x 2+y 2–x y–15=0 x 2+y 2–x y–15=0
We will find the relationships between x x x and y y y on the Cartesian plane with x′x′x′ and y′y′y′ on the new rotated plane. See Figure 4.
Figure 4 The Cartesian plane with x- and y-axes and the resulting x′− and y′−axes formed by a rotation by an angle θ.θ.θ.
The original coordinate x- and y-axes have unit vectors i i i and j.j.j. The rotated coordinate axes have unit vectors i′i′i′ and j′.j′.j′. The angle θ θ θ is known as the angle of rotation. See Figure 5. We may write the new unit vectors in terms of the original ones.
i′=cos θ i+sin θ j j′=−sin θ i+cos θ j i′=cos θ i+sin θ j j′=−sin θ i+cos θ j i′=cos θ i+sin θ j j′=−sin θ i+cos θ j
Figure 5 Relationship between the old and new coordinate planes.
Consider a vectoru u uin the new coordinate plane. It may be represented in terms of its coordinate axes.
u=x′i′+y′j′u=x′(i cos θ+j sin θ)+y′(−i sin θ+j cos θ)u=i x'cos θ+j x'sin θ−i y'sin θ+j y'cos θ u=i x'cos θ−i y'sin θ+j x'sin θ+j y'cos θ u=(x'cos θ−y'sin θ)i+(x'sin θ+y'cos θ)j Substitute.Distribute.Apply commutative property.Factor by grouping.u=x′i′+y′j′u=x′(i cos θ+j sin θ)+y′(−i sin θ+j cos θ)Substitute.u=i x'cos θ+j x'sin θ−i y'sin θ+j y'cos θ Distribute.u=i x'cos θ−i y'sin θ+j x'sin θ+j y'cos θ Apply commutative property.u=(x'cos θ−y'sin θ)i+(x'sin θ+y'cos θ)j Factor by grouping.u=x′i′+y′j′u=x′(i cos θ+j sin θ)+y′(−i sin θ+j cos θ)Substitute.u=i x'cos θ+j x'sin θ−i y'sin θ+j y'cos θ Distribute.u=i x'cos θ−i y'sin θ+j x'sin θ+j y'cos θ Apply commutative property.u=(x'cos θ−y'sin θ)i+(x'sin θ+y'cos θ)j Factor by grouping.
Because u=x′i′+y′j′,u=x′i′+y′j′,u=x′i′+y′j′, we have representations of x x x and y y y in terms of the new coordinate system.
x=x′cos θ−y′sin θ and y=x′sin θ+y′cos θ x=x′cos θ−y′sin θ and y=x′sin θ+y′cos θ x=x′cos θ−y′sin θ and y=x′sin θ+y′cos θ
Equations of Rotation
If a point (x,y)(x,y)(x,y) on the Cartesian plane is represented on a new coordinate plane where the axes of rotation are formed by rotating an angle θ θ θ from the positive x-axis, then the coordinates of the point with respect to the new axes are (x′,y′).(x′,y′).(x′,y′). We can use the following equations of rotation to define the relationship between (x,y)(x,y)(x,y) and (x′,y′):(x′,y′):(x′,y′):
x=x′cos θ−y′sin θ x=x′cos θ−y′sin θ x=x′cos θ−y′sin θ
and
y=x′sin θ+y′cos θ y=x′sin θ+y′cos θ y=x′sin θ+y′cos θ
How To
Given the equation of a conic, find a new representation after rotating through an angle.
Find x x x and y y y where x=x′cos θ−y′sin θ x=x′cos θ−y′sin θ x=x′cos θ−y′sin θ and y=x′sin θ+y′cos θ.y=x′sin θ+y′cos θ.y=x′sin θ+y′cos θ.
Substitute the expression for x x x and y y y into in the given equation, then simplify.
Write the equations with x′x′x′ and y′y′y′ in standard form.
Example 2
Finding a New Representation of an Equation after Rotating through a Given Angle
Find a new representation of the equation 2 x 2−x y+2 y 2−30=0 2 x 2−x y+2 y 2−30=0 2 x 2−x y+2 y 2−30=0 after rotating through an angle of θ=45°.θ=45°.θ=45°.
Solution
Find x x x and y,y,y, where x=x′cos θ−y′sin θ x=x′cos θ−y′sin θ x=x′cos θ−y′sin θ and y=x′sin θ+y′cos θ.y=x′sin θ+y′cos θ.y=x′sin θ+y′cos θ.
Because θ=45°,θ=45°,θ=45°,
x=x′cos(45°)−y′sin(45°)x=x′(1 2√)−y′(1 2√)x=x′−y′2√x=x′cos(45°)−y′sin(45°)x=x′(1 2)−y′(1 2)x=x′−y′2 x=x′cos(45°)−y′sin(45°)x=x′(1 2)−y′(1 2)x=x′−y′2
and
y=x′sin(45°)+y′cos(45°)y=x′(1 2√)+y′(1 2√)y=x′+y′2√y=x′sin(45°)+y′cos(45°)y=x′(1 2)+y′(1 2)y=x′+y′2 y=x′sin(45°)+y′cos(45°)y=x′(1 2)+y′(1 2)y=x′+y′2
Substitute x=x′cos θ−y′sin θ x=x′cos θ−y′sin θ x=x′cos θ−y′sin θ and y=x′sin θ+y′cos θ y=x′sin θ+y′cos θ y=x′sin θ+y′cos θ into 2 x 2−x y+2 y 2−30=0.2 x 2−x y+2 y 2−30=0.2 x 2−x y+2 y 2−30=0.
2(x′−y′2–√)2−(x′−y′2–√)(x′+y′2–√)+2(x′+y′2–√)2−30=0 2(x′−y′2)2−(x′−y′2)(x′+y′2)+2(x′+y′2)2−30=0 2(x′−y′2)2−(x′−y′2)(x′+y′2)+2(x′+y′2)2−30=0
Simplify.
2(x′−y′)(x′−y′)2−(x′−y′)(x′+y′)2+2(x′+y′)(x′+y′)2−30=0 x′2−2 x′y′+y′2−(x′2−y′2)2+x′2+2 x′y′+y′2−30=0 2 x′2+2 y′2−(x′2−y′2)2=30 2(2 x′2+2 y′2−(x′2−y′2)2)=2(30)4 x′2+4 y′2−(x′2−y′2)=60 4 x′2+4 y′2−x′2+y′2=60 3 x′2 60+5 y′2 60=60 60 FOIL method Combine like terms.Combine like terms.Multiply both sides by 2.Simplify.Distribute.Set equal to 1.2(x′−y′)(x′−y′)2−(x′−y′)(x′+y′)2+2(x′+y′)(x′+y′)2−30=0 FOIL method x′2−2 x′y′+y′2−(x′2−y′2)2+x′2+2 x′y′+y′2−30=0 Combine like terms.2 x′2+2 y′2−(x′2−y′2)2=30 Combine like terms.2(2 x′2+2 y′2−(x′2−y′2)2)=2(30)Multiply both sides by 2.4 x′2+4 y′2−(x′2−y′2)=60 Simplify.4 x′2+4 y′2−x′2+y′2=60 Distribute.3 x′2 60+5 y′2 60=60 60 Set equal to 1.2(x′−y′)(x′−y′)2−(x′−y′)(x′+y′)2+2(x′+y′)(x′+y′)2−30=0 FOIL method x′2−2 x′y′+y′2−(x′2−y′2)2+x′2+2 x′y′+y′2−30=0 Combine like terms.2 x′2+2 y′2−(x′2−y′2)2=30 Combine like terms.2(2 x′2+2 y′2−(x′2−y′2)2)=2(30)Multiply both sides by 2.4 x′2+4 y′2−(x′2−y′2)=60 Simplify.4 x′2+4 y′2−x′2+y′2=60 Distribute.3 x′2 60+5 y′2 60=60 60 Set equal to 1.
Write the equations with x′x′x′ and y′y′y′ in the standard form.
x′2 20+y′2 12=1 x′2 20+y′2 12=1 x′2 20+y′2 12=1
This equation is an ellipse. Figure 6 shows the graph.
Figure 6
Writing Equations of Rotated Conics in Standard Form
Now that we can find the standard form of a conic when we are given an angle of rotation, we will learn how to transform the equation of a conic given in the form A x 2+B x y+C y 2+D x+E y+F=0 A x 2+B x y+C y 2+D x+E y+F=0 A x 2+B x y+C y 2+D x+E y+F=0 into standard form by rotating the axes. To do so, we will rewrite the general form as an equation in the x′x′x′ and y′y′y′ coordinate system without the x′y′x′y′x′y′ term, by rotating the axes by a measure of θ θ θ that satisfies
cot(2 θ)=A−C B cot(2 θ)=A−C B cot(2 θ)=A−C B
We have learned already that any conic may be represented by the second degree equation
A x 2+B x y+C y 2+D x+E y+F=0 A x 2+B x y+C y 2+D x+E y+F=0 A x 2+B x y+C y 2+D x+E y+F=0
where A,B,A,B,A,B, and C C C are not all zero. However, if B≠0,B≠0,B≠0, then we have an x y x y x y term that prevents us from rewriting the equation in standard form. To eliminate it, we can rotate the axes by an acute angle θ θ θ where cot(2 θ)=A−C B.cot(2 θ)=A−C B.cot(2 θ)=A−C B.
If cot(2 θ)>0,cot(2 θ)>0,cot(2 θ)>0, then 2 θ 2 θ 2 θ is in the first quadrant, and θ θ θ is between (0°,45°).(0°,45°).(0°,45°).
If cot(2 θ)<0,cot(2 θ)<0,cot(2 θ)<0, then 2 θ 2 θ 2 θ is in the second quadrant, and θ θ θ is between (45°,90°).(45°,90°).(45°,90°).
If A=C,A=C,A=C, then θ=45°.θ=45°.θ=45°.
How To
Given an equation for a conic in the x′y′x′y′x′y′ system, rewrite the equation without the x′y′x′y′x′y′ term in terms of x′x′x′ and y′,y′,y′, where the x′x′x′ and y′y′y′ axes are rotations of the standard axes by θ θ θ degrees.
Find cot(2 θ).cot(2 θ).cot(2 θ).
Find sin θ sin θ sin θ and cos θ.cos θ.cos θ.
Substitute sin θ sin θ sin θ and cos θ cos θ cos θ into x=x′cos θ−y′sin θ x=x′cos θ−y′sin θ x=x′cos θ−y′sin θ and y=x′sin θ+y′cos θ.y=x′sin θ+y′cos θ.y=x′sin θ+y′cos θ.
Substitute the expression for x x x and y y y into in the given equation, and then simplify.
Write the equations with x′x′x′ and y′y′y′ in the standard form with respect to the rotated axes.
Example 3
Rewriting an Equation with respect to the x′ and y′ axes without the x′y′ Term
Rewrite the equation 8 x 2−12 x y+17 y 2=20 8 x 2−12 x y+17 y 2=20 8 x 2−12 x y+17 y 2=20 in the x′y′x′y′x′y′ system without an x′y′x′y′x′y′ term.
Solution
First, we find cot(2 θ).cot(2 θ).cot(2 θ). See Figure 7.
8 x 2−12 x y+17 y 2=20⇒A=8,B=−12 and C=17 cot(2 θ)=A−C B=8−17−12 cot(2 θ)=−9−12=3 4 8 x 2−12 x y+17 y 2=20⇒A=8,B=−12 and C=17 cot(2 θ)=A−C B=8−17−12 cot(2 θ)=−9−12=3 4 8 x 2−12 x y+17 y 2=20⇒A=8,B=−12 and C=17 cot(2 θ)=A−C B=8−17−12 cot(2 θ)=−9−12=3 4
Figure 7
cot(2 θ)=3 4=adjacent opposite cot(2 θ)=3 4=adjacent opposite cot(2 θ)=3 4=adjacent opposite
So the hypotenuse is
3 2+4 2=h 2 9+16=h 2 25=h 2 h=5 3 2+4 2=h 2 9+16=h 2 25=h 2 h=5 3 2+4 2=h 2 9+16=h 2 25=h 2 h=5
Next, we find sin θ sin θ sin θ and cos θ.cos θ.cos θ.
sin θ=1−cos(2 θ)2−−−−−−−√=1−3 5 2−−−−√=5 5−3 5 2−−−−√=5−3 5⋅1 2−−−−−−√=2 10−−√=1 5−−√sin θ=1 5√cos θ=1+cos(2 θ)2−−−−−−−√=1+3 5 2−−−−√=5 5+3 5 2−−−−√=5+3 5⋅1 2−−−−−−√=8 10−−√=4 5−−√cos θ=2 5√sin θ=1−cos(2 θ)2=1−3 5 2=5 5−3 5 2=5−3 5⋅1 2=2 10=1 5 sin θ=1 5 cos θ=1+cos(2 θ)2=1+3 5 2=5 5+3 5 2=5+3 5⋅1 2=8 10=4 5 cos θ=2 5 sin θ=1−cos(2 θ)2=1−3 5 2=5 5−3 5 2=5−3 5⋅1 2=2 10=1 5 sin θ=1 5 cos θ=1+cos(2 θ)2=1+3 5 2=5 5+3 5 2=5+3 5⋅1 2=8 10=4 5 cos θ=2 5
Substitute the values of sin θ sin θ sin θ and cos θ cos θ cos θ into x=x′cos θ−y′sin θ x=x′cos θ−y′sin θ x=x′cos θ−y′sin θ and y=x′sin θ+y′cos θ.y=x′sin θ+y′cos θ.y=x′sin θ+y′cos θ.
x=x′cos θ−y′sin θ x=x′(2 5√)−y′(1 5√)x=2 x′−y′5√x=x′cos θ−y′sin θ x=x′(2 5)−y′(1 5)x=2 x′−y′5 x=x′cos θ−y′sin θ x=x′(2 5)−y′(1 5)x=2 x′−y′5
and
y=x′sin θ+y′cos θ y=x′(1 5√)+y′(2 5√)y=x′+2 y′5√y=x′sin θ+y′cos θ y=x′(1 5)+y′(2 5)y=x′+2 y′5 y=x′sin θ+y′cos θ y=x′(1 5)+y′(2 5)y=x′+2 y′5
Substitute the expressions for x x x and y y y into in the given equation, and then simplify.
8(2 x′−y′5√)2−12(2 x′−y′5√)(x′+2 y′5√)+17(x′+2 y′5√)2=20 8((2 x′−y′)(2 x′−y′)5)−12((2 x′−y′)(x′+2 y′)5)+17((x′+2 y′)(x′+2 y′)5)=20 8(4 x′2−4 x′y′+y′2)−12(2 x′2+3 x′y′−2 y′2)+17(x′2+4 x′y′+4 y′2)=100 32 x′2−32 x′y′+8 y′2−24 x′2−36 x′y′+24 y′2+17 x′2+68 x′y′+68 y′2=100 25 x′2+100 y′2=100 25 100 x′2+100 100 y′2=100 100 8(2 x′−y′5)2−12(2 x′−y′5)(x′+2 y′5)+17(x′+2 y′5)2=20 8((2 x′−y′)(2 x′−y′)5)−12((2 x′−y′)(x′+2 y′)5)+17((x′+2 y′)(x′+2 y′)5)=20 8(4 x′2−4 x′y′+y′2)−12(2 x′2+3 x′y′−2 y′2)+17(x′2+4 x′y′+4 y′2)=100 32 x′2−32 x′y′+8 y′2−24 x′2−36 x′y′+24 y′2+17 x′2+68 x′y′+68 y′2=100 25 x′2+100 y′2=100 25 100 x′2+100 100 y′2=100 100 8(2 x′−y′5)2−12(2 x′−y′5)(x′+2 y′5)+17(x′+2 y′5)2=20 8((2 x′−y′)(2 x′−y′)5)−12((2 x′−y′)(x′+2 y′)5)+17((x′+2 y′)(x′+2 y′)5)=20 8(4 x′2−4 x′y′+y′2)−12(2 x′2+3 x′y′−2 y′2)+17(x′2+4 x′y′+4 y′2)=100 32 x′2−32 x′y′+8 y′2−24 x′2−36 x′y′+24 y′2+17 x′2+68 x′y′+68 y′2=100 25 x′2+100 y′2=100 25 100 x′2+100 100 y′2=100 100
Write the equations with x′x′x′ and y′y′y′ in the standard form with respect to the new coordinate system.
x′2 4+y′2 1=1 x′2 4+y′2 1=1 x′2 4+y′2 1=1
Figure 8 shows the graph of the ellipse.
Figure 8
Try It #2
Rewrite the 13 x 2−6 3–√x y+7 y 2=16 13 x 2−6 3 x y+7 y 2=16 13 x 2−6 3 x y+7 y 2=16 in the x′y′x′y′x′y′ system without the x′y′x′y′x′y′ term.
Example 4
Graphing an Equation That Has No x′y′ Terms
Graph the following equation relative to the x′y′x′y′x′y′ system:
x 2+12 x y−4 y 2=30 x 2+12 x y−4 y 2=30 x 2+12 x y−4 y 2=30
Solution
First, we find cot(2 θ).cot(2 θ).cot(2 θ).
x 2+12 x y−4 y 2=20⇒A=1,B=12,and C=−4 x 2+12 x y−4 y 2=20⇒A=1,B=12,and C=−4 x 2+12 x y−4 y 2=20⇒A=1,B=12,and C=−4
cot(2 θ)=A−C B cot(2 θ)=1−(−4)12 cot(2 θ)=5 12 cot(2 θ)=A−C B cot(2 θ)=1−(−4)12 cot(2 θ)=5 12 cot(2 θ)=A−C B cot(2 θ)=1−(−4)12 cot(2 θ)=5 12
Because cot(2 θ)=5 12,cot(2 θ)=5 12,cot(2 θ)=5 12, we can draw a reference triangle as in Figure 9.
Figure 9
cot(2 θ)=5 12=adjacent opposite cot(2 θ)=5 12=adjacent opposite cot(2 θ)=5 12=adjacent opposite
Thus, the hypotenuse is
5 2+12 2=h 2 25+144=h 2 169=h 2 h=13 5 2+12 2=h 2 25+144=h 2 169=h 2 h=13 5 2+12 2=h 2 25+144=h 2 169=h 2 h=13
Next, we find sin θ sin θ sin θ and cos θ.cos θ.cos θ. We will use half-angle identities.
sin θ=1−cos(2 θ)2−−−−−−−√=1−5 13 2−−−−√=13 13−5 13 2−−−−−√=8 13⋅1 2−−−−−√=2 13√cos θ=1+cos(2 θ)2−−−−−−−√=1+5 13 2−−−−√=13 13+5 13 2−−−−−√=18 13⋅1 2−−−−−√=3 13√sin θ=1−cos(2 θ)2=1−5 13 2=13 13−5 13 2=8 13⋅1 2=2 13 cos θ=1+cos(2 θ)2=1+5 13 2=13 13+5 13 2=18 13⋅1 2=3 13 sin θ=1−cos(2 θ)2=1−5 13 2=13 13−5 13 2=8 13⋅1 2=2 13 cos θ=1+cos(2 θ)2=1+5 13 2=13 13+5 13 2=18 13⋅1 2=3 13
Now we find x x x and y. y. y.
x=x′cos θ−y′sin θ x=x′(3 13√)−y′(2 13√)x=3 x′−2 y′13√x=x′cos θ−y′sin θ x=x′(3 13)−y′(2 13)x=3 x′−2 y′13 x=x′cos θ−y′sin θ x=x′(3 13)−y′(2 13)x=3 x′−2 y′13
and
y=x′sin θ+y′cos θ y=x′(2 13√)+y′(3 13√)y=2 x′+3 y′13√y=x′sin θ+y′cos θ y=x′(2 13)+y′(3 13)y=2 x′+3 y′13 y=x′sin θ+y′cos θ y=x′(2 13)+y′(3 13)y=2 x′+3 y′13
Now we substitute x=3 x′−2 y′13√x=3 x′−2 y′13 x=3 x′−2 y′13 and y=2 x′+3 y′13√y=2 x′+3 y′13 y=2 x′+3 y′13 into x 2+12 x y−4 y 2=30.x 2+12 x y−4 y 2=30.x 2+12 x y−4 y 2=30.
(3 x′−2 y′13√)2+12(3 x′−2 y′13√)(2 x′+3 y′13√)−4(2 x′+3 y′13√)2=30(1 13)[(3 x′−2 y′)2+12(3 x′−2 y′)(2 x′+3 y′)−4(2 x′+3 y′)2]=30(1 13)[9 x′2−12 x′y′+4 y′2+12(6 x′2+5 x′y′−6 y′2)−4(4 x′2+12 x′y′+9 y′2)]=30(1 13)[9 x′2−12 x′y′+4 y′2+72 x′2+60 x′y′−72 y′2−16 x′2−48 x′y′−36 y′2]=30(1 13)[65 x′2−104 y′2]=30 65 x′2−104 y′2=390 x′2 6−4 y′2 15=1 Factor.Multiply.Distribute.Combine like terms.Multiply.Divide by 390.(3 x′−2 y′13)2+12(3 x′−2 y′13)(2 x′+3 y′13)−4(2 x′+3 y′13)2=30(1 13)[(3 x′−2 y′)2+12(3 x′−2 y′)(2 x′+3 y′)−4(2 x′+3 y′)2]=30 Factor.(1 13)[9 x′2−12 x′y′+4 y′2+12(6 x′2+5 x′y′−6 y′2)−4(4 x′2+12 x′y′+9 y′2)]=30 Multiply.(1 13)[9 x′2−12 x′y′+4 y′2+72 x′2+60 x′y′−72 y′2−16 x′2−48 x′y′−36 y′2]=30 Distribute.(1 13)[65 x′2−104 y′2]=30 Combine like terms.65 x′2−104 y′2=390 Multiply.x′2 6−4 y′2 15=1 Divide by 390.(3 x′−2 y′13)2+12(3 x′−2 y′13)(2 x′+3 y′13)−4(2 x′+3 y′13)2=30(1 13)[(3 x′−2 y′)2+12(3 x′−2 y′)(2 x′+3 y′)−4(2 x′+3 y′)2]=30 Factor.(1 13)[9 x′2−12 x′y′+4 y′2+12(6 x′2+5 x′y′−6 y′2)−4(4 x′2+12 x′y′+9 y′2)]=30 Multiply.(1 13)[9 x′2−12 x′y′+4 y′2+72 x′2+60 x′y′−72 y′2−16 x′2−48 x′y′−36 y′2]=30 Distribute.(1 13)[65 x′2−104 y′2]=30 Combine like terms.65 x′2−104 y′2=390 Multiply.x′2 6−4 y′2 15=1 Divide by 390.
Figure 10 shows the graph of the hyperbola x′2 6−4 y′2 15=1.x′2 6−4 y′2 15=1.x′2 6−4 y′2 15=1.
Figure 10
Identifying Conics without Rotating Axes
Now we have come full circle. How do we identify the type of conic described by an equation? What happens when the axes are rotated? Recall, the general form of a conic is
A x 2+B x y+C y 2+D x+E y+F=0 A x 2+B x y+C y 2+D x+E y+F=0 A x 2+B x y+C y 2+D x+E y+F=0
If we apply the rotation formulas to this equation we get the form
A′x′2+B′x′y′+C′y′2+D′x′+E′y′+F′=0 A′x′2+B′x′y′+C′y′2+D′x′+E′y′+F′=0 A′x′2+B′x′y′+C′y′2+D′x′+E′y′+F′=0
It may be shown that B 2−4 A C=B′2−4 A′C′.B 2−4 A C=B′2−4 A′C′.B 2−4 A C=B′2−4 A′C′. The expression does not vary after rotation, so we call the expression invariant. The discriminant, B 2−4 A C,B 2−4 A C,B 2−4 A C, is invariant and remains unchanged after rotation. Because the discriminant remains unchanged, observing the discriminant enables us to identify the conic section.
Using the Discriminant to Identify a Conic
If the equation A x 2+B x y+C y 2+D x+E y+F=0 A x 2+B x y+C y 2+D x+E y+F=0 A x 2+B x y+C y 2+D x+E y+F=0 is transformed by rotating axes into the equation A′x′2+B′x′y′+C′y′2+D′x′+E′y′+F′=0,A′x′2+B′x′y′+C′y′2+D′x′+E′y′+F′=0,A′x′2+B′x′y′+C′y′2+D′x′+E′y′+F′=0, then B 2−4 A C=B′2−4 A′C′.B 2−4 A C=B′2−4 A′C′.B 2−4 A C=B′2−4 A′C′.
The equation A x 2+B x y+C y 2+D x+E y+F=0 A x 2+B x y+C y 2+D x+E y+F=0 A x 2+B x y+C y 2+D x+E y+F=0 is an ellipse, a parabola, or a hyperbola, or a degenerate case of one of these.
If the discriminant, B 2−4 A C,B 2−4 A C,B 2−4 A C, is
<0,<0,<0, the conic section is an ellipse
=0,=0,=0, the conic section is a parabola
0,>0,>0, the conic section is a hyperbola
Example 5
Identifying the Conic without Rotating Axes
Identify the conic for each of the following without rotating axes.
ⓐ5 x 2+2 3–√x y+2 y 2−5=0 5 x 2+2 3 x y+2 y 2−5=0 5 x 2+2 3 x y+2 y 2−5=0
ⓑ5 x 2+2 3–√x y+12 y 2−5=0 5 x 2+2 3 x y+12 y 2−5=0 5 x 2+2 3 x y+12 y 2−5=0
Solution
ⓐ Let’s begin by determining A,B,A,B,A,B, and C.C.C.5A x 2+2 3–√B x y+2C y 2−5=0 5︸A x 2+2 3︸B x y+2︸C y 2−5=0 5︸A x 2+2 3︸B x y+2︸C y 2−5=0
Now, we find the discriminant.
B 2−4 A C=(2 3–√)2−4(5)(2)=4(3)−40=12−40=−28<0 B 2−4 A C=(2 3)2−4(5)(2)=4(3)−40=12−40=−28<0 B 2−4 A C=(2 3)2−4(5)(2)=4(3)−40=12−40=−28<0
Therefore, 5 x 2+2 3–√x y+2 y 2−5=0 5 x 2+2 3 x y+2 y 2−5=0 5 x 2+2 3 x y+2 y 2−5=0 represents an ellipse.
ⓑ Again, let’s begin by determining A,B,A,B,A,B, and C.C.C.5A x 2+2 3–√B x y+12C y 2−5=0 5︸A x 2+2 3︸B x y+12︸C y 2−5=0 5︸A x 2+2 3︸B x y+12︸C y 2−5=0
Now, we find the discriminant.
B 2−4 A C=(2 3–√)2−4(5)(12)=4(3)−240=12−240=−228<0 B 2−4 A C=(2 3)2−4(5)(12)=4(3)−240=12−240=−228<0 B 2−4 A C=(2 3)2−4(5)(12)=4(3)−240=12−240=−228<0
Therefore, 5 x 2+2 3–√x y+12 y 2−5=0 5 x 2+2 3 x y+12 y 2−5=0 5 x 2+2 3 x y+12 y 2−5=0 represents an ellipse.
Try It #3
Identify the conic for each of the following without rotating axes.
ⓐx 2−9 x y+3 y 2−12=0 x 2−9 x y+3 y 2−12=0 x 2−9 x y+3 y 2−12=0
ⓑ10 x 2−9 x y+4 y 2−4=0 10 x 2−9 x y+4 y 2−4=0 10 x 2−9 x y+4 y 2−4=0
Media
Access this online resource for additional instruction and practice with conic sections and rotation of axes.
Introduction to Conic Sections
8.4 Section Exercises
Verbal
1.
What effect does the x y x y x y term have on the graph of a conic section?
If the equation of a conic section is written in the form A x 2+B y 2+C x+D y+E=0 A x 2+B y 2+C x+D y+E=0 A x 2+B y 2+C x+D y+E=0 and A B=0,A B=0,A B=0, what can we conclude?
3.
If the equation of a conic section is written in the form A x 2+B x y+C y 2+D x+E y+F=0,A x 2+B x y+C y 2+D x+E y+F=0,A x 2+B x y+C y 2+D x+E y+F=0, and B 2−4 A C>0,B 2−4 A C>0,B 2−4 A C>0, what can we conclude?
Given the equation a x 2+4 x+3 y 2−12=0,a x 2+4 x+3 y 2−12=0,a x 2+4 x+3 y 2−12=0, what can we conclude if a>0?a>0?a>0?
5.
For the equation A x 2+B x y+C y 2+D x+E y+F=0,A x 2+B x y+C y 2+D x+E y+F=0,A x 2+B x y+C y 2+D x+E y+F=0, the value of θ θ θ that satisfies cot(2 θ)=A−C B cot(2 θ)=A−C B cot(2 θ)=A−C B gives us what information?
Algebraic
For the following exercises, determine which conic section is represented based on the given equation.
9 x 2+4 y 2+72 x+36 y−500=0 9 x 2+4 y 2+72 x+36 y−500=0 9 x 2+4 y 2+72 x+36 y−500=0
7.
x 2−10 x+4 y−10=0 x 2−10 x+4 y−10=0 x 2−10 x+4 y−10=0
2 x 2−2 y 2+4 x−6 y−2=0 2 x 2−2 y 2+4 x−6 y−2=0 2 x 2−2 y 2+4 x−6 y−2=0
9.
4 x 2−y 2+8 x−1=0 4 x 2−y 2+8 x−1=0 4 x 2−y 2+8 x−1=0
4 y 2−5 x+9 y+1=0 4 y 2−5 x+9 y+1=0 4 y 2−5 x+9 y+1=0
11.
2 x 2+3 y 2−8 x−12 y+2=0 2 x 2+3 y 2−8 x−12 y+2=0 2 x 2+3 y 2−8 x−12 y+2=0
4 x 2+9 x y+4 y 2−36 y−125=0 4 x 2+9 x y+4 y 2−36 y−125=0 4 x 2+9 x y+4 y 2−36 y−125=0
13.
3 x 2+6 x y+3 y 2−36 y−125=0 3 x 2+6 x y+3 y 2−36 y−125=0 3 x 2+6 x y+3 y 2−36 y−125=0
−3 x 2+3 3–√x y−4 y 2+9=0−3 x 2+3 3 x y−4 y 2+9=0−3 x 2+3 3 x y−4 y 2+9=0
15.
2 x 2+4 3–√x y+6 y 2−6 x−3=0 2 x 2+4 3 x y+6 y 2−6 x−3=0 2 x 2+4 3 x y+6 y 2−6 x−3=0
−x 2+4 2–√x y+2 y 2−2 y+1=0−x 2+4 2 x y+2 y 2−2 y+1=0−x 2+4 2 x y+2 y 2−2 y+1=0
17.
8 x 2+4 2–√x y+4 y 2−10 x+1=0 8 x 2+4 2 x y+4 y 2−10 x+1=0 8 x 2+4 2 x y+4 y 2−10 x+1=0
For the following exercises, find a new representation of the given equation after rotating through the given angle.
3 x 2+x y+3 y 2−5=0,θ=45°3 x 2+x y+3 y 2−5=0,θ=45°3 x 2+x y+3 y 2−5=0,θ=45°
19.
4 x 2−x y+4 y 2−2=0,θ=45°4 x 2−x y+4 y 2−2=0,θ=45°4 x 2−x y+4 y 2−2=0,θ=45°
2 x 2+8 x y−1=0,θ=30°2 x 2+8 x y−1=0,θ=30°2 x 2+8 x y−1=0,θ=30°
21.
−2 x 2+8 x y+1=0,θ=45°−2 x 2+8 x y+1=0,θ=45°−2 x 2+8 x y+1=0,θ=45°
4 x 2+2–√x y+4 y 2+y+2=0,θ=45°4 x 2+2 x y+4 y 2+y+2=0,θ=45°4 x 2+2 x y+4 y 2+y+2=0,θ=45°
For the following exercises, determine the angle θ θ θ that will eliminate the x y x y x y term and write the corresponding equation without the x y x y x y term.
23.
x 2+3 3–√x y+4 y 2+y−2=0 x 2+3 3 x y+4 y 2+y−2=0 x 2+3 3 x y+4 y 2+y−2=0
4 x 2+2 3–√x y+6 y 2+y−2=0 4 x 2+2 3 x y+6 y 2+y−2=0 4 x 2+2 3 x y+6 y 2+y−2=0
25.
9 x 2−3 3–√x y+6 y 2+4 y−3=0 9 x 2−3 3 x y+6 y 2+4 y−3=0 9 x 2−3 3 x y+6 y 2+4 y−3=0
−3 x 2−3–√x y−2 y 2−x=0−3 x 2−3 x y−2 y 2−x=0−3 x 2−3 x y−2 y 2−x=0
27.
16 x 2+24 x y+9 y 2+6 x−6 y+2=0 16 x 2+24 x y+9 y 2+6 x−6 y+2=0 16 x 2+24 x y+9 y 2+6 x−6 y+2=0
x 2+4 x y+4 y 2+3 x−2=0 x 2+4 x y+4 y 2+3 x−2=0 x 2+4 x y+4 y 2+3 x−2=0
29.
x 2+4 x y+y 2−2 x+1=0 x 2+4 x y+y 2−2 x+1=0 x 2+4 x y+y 2−2 x+1=0
4 x 2−2 3–√x y+6 y 2−1=0 4 x 2−2 3 x y+6 y 2−1=0 4 x 2−2 3 x y+6 y 2−1=0
Graphical
For the following exercises, rotate through the given angle based on the given equation. Give the new equation and graph the original and rotated equation.
31.
y=−x 2,θ=−45∘y=−x 2,θ=−45∘y=−x 2,θ=−45∘
x=y 2,θ=45∘x=y 2,θ=45∘x=y 2,θ=45∘
33.
x 2 4+y 2 1=1,θ=45∘x 2 4+y 2 1=1,θ=45∘x 2 4+y 2 1=1,θ=45∘
y 2 16+x 2 9=1,θ=45∘y 2 16+x 2 9=1,θ=45∘y 2 16+x 2 9=1,θ=45∘
35.
y 2−x 2=1,θ=45∘y 2−x 2=1,θ=45∘y 2−x 2=1,θ=45∘
y=x 2 2,θ=30∘y=x 2 2,θ=30∘y=x 2 2,θ=30∘
37.
x=(y−1)2,θ=30∘x=(y−1)2,θ=30∘x=(y−1)2,θ=30∘
x 2 9+y 2 4=1,θ=30∘x 2 9+y 2 4=1,θ=30∘x 2 9+y 2 4=1,θ=30∘
For the following exercises, graph the equation relative to the x′y′x′y′x′y′ system in which the equation has no x′y′x′y′x′y′ term.
39.
x y=9 x y=9 x y=9
x 2+10 x y+y 2−6=0 x 2+10 x y+y 2−6=0 x 2+10 x y+y 2−6=0
41.
x 2−10 x y+y 2−24=0 x 2−10 x y+y 2−24=0 x 2−10 x y+y 2−24=0
4 x 2−3 3–√x y+y 2−22=0 4 x 2−3 3 x y+y 2−22=0 4 x 2−3 3 x y+y 2−22=0
43.
6 x 2+2 3–√x y+4 y 2−21=0 6 x 2+2 3 x y+4 y 2−21=0 6 x 2+2 3 x y+4 y 2−21=0
11 x 2+10 3–√x y+y 2−64=0 11 x 2+10 3 x y+y 2−64=0 11 x 2+10 3 x y+y 2−64=0
45.
21 x 2+2 3–√x y+19 y 2−18=0 21 x 2+2 3 x y+19 y 2−18=0 21 x 2+2 3 x y+19 y 2−18=0
16 x 2+24 x y+9 y 2−130 x+90 y=0 16 x 2+24 x y+9 y 2−130 x+90 y=0 16 x 2+24 x y+9 y 2−130 x+90 y=0
47.
16 x 2+24 x y+9 y 2−60 x+80 y=0 16 x 2+24 x y+9 y 2−60 x+80 y=0 16 x 2+24 x y+9 y 2−60 x+80 y=0
13 x 2−6 3–√x y+7 y 2−16=0 13 x 2−6 3 x y+7 y 2−16=0 13 x 2−6 3 x y+7 y 2−16=0
49.
4 x 2−4 x y+y 2−8 5–√x−16 5–√y=0 4 x 2−4 x y+y 2−8 5 x−16 5 y=0 4 x 2−4 x y+y 2−8 5 x−16 5 y=0
For the following exercises, determine the angle of rotation in order to eliminate the x y x y x y term. Then graph the new set of axes.
6 x 2−5 3–√x y+y 2+10 x−12 y=0 6 x 2−5 3 x y+y 2+10 x−12 y=0 6 x 2−5 3 x y+y 2+10 x−12 y=0
51.
6 x 2−5 x y+6 y 2+20 x−y=0 6 x 2−5 x y+6 y 2+20 x−y=0 6 x 2−5 x y+6 y 2+20 x−y=0
6 x 2−8 3–√x y+14 y 2+10 x−3 y=0 6 x 2−8 3 x y+14 y 2+10 x−3 y=0 6 x 2−8 3 x y+14 y 2+10 x−3 y=0
53.
4 x 2+6 3–√x y+10 y 2+20 x−40 y=0 4 x 2+6 3 x y+10 y 2+20 x−40 y=0 4 x 2+6 3 x y+10 y 2+20 x−40 y=0
8 x 2+3 x y+4 y 2+2 x−4=0 8 x 2+3 x y+4 y 2+2 x−4=0 8 x 2+3 x y+4 y 2+2 x−4=0
55.
16 x 2+24 x y+9 y 2+20 x−44 y=0 16 x 2+24 x y+9 y 2+20 x−44 y=0 16 x 2+24 x y+9 y 2+20 x−44 y=0
For the following exercises, determine the value of k k k based on the given equation.
Given 4 x 2+k x y+16 y 2+8 x+24 y−48=0,4 x 2+k x y+16 y 2+8 x+24 y−48=0,4 x 2+k x y+16 y 2+8 x+24 y−48=0, find k k k for the graph to be a parabola.
57.
Given 2 x 2+k x y+12 y 2+10 x−16 y+28=0,2 x 2+k x y+12 y 2+10 x−16 y+28=0,2 x 2+k x y+12 y 2+10 x−16 y+28=0, find k k k for the graph to be an ellipse.
Given 3 x 2+k x y+4 y 2−6 x+20 y+128=0,3 x 2+k x y+4 y 2−6 x+20 y+128=0,3 x 2+k x y+4 y 2−6 x+20 y+128=0, find k k k for the graph to be a hyperbola.
59.
Given k x 2+8 x y+8 y 2−12 x+16 y+18=0,k x 2+8 x y+8 y 2−12 x+16 y+18=0,k x 2+8 x y+8 y 2−12 x+16 y+18=0, find k k k for the graph to be a parabola.
Given 6 x 2+12 x y+k y 2+16 x+10 y+4=0,6 x 2+12 x y+k y 2+16 x+10 y+4=0,6 x 2+12 x y+k y 2+16 x+10 y+4=0, find k k k for the graph to be an ellipse.
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14811 | https://direct.physicsclassroom.com/mop/Circular-and-Satellite-Motion/Centripetal-Force-Requirement/QG8help | Physics Classroom - Home
Circular and Satellite Motion - Mission CG4 Detailed Help
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| | A bucket of water is held by a rope and twirled in a vertical circle. The bucket is whirled more rapidly such that the speeds at both the top and the bottom of the circle are increased. As this increase in speed occurs, the _____. List all that apply ... . |
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| | Centripetal Force Requirement: Circular motion requires an inward force. To travel along the curved path of a circle, there must be a force directed centripetally. Any object or thing could supply the force as long as it is directed toward the center of the circular path. |
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| | The acceleration (a) of an object moving in a circle is dependent upon the speed (v) of the object and the radius (R) of the circle. The relationship is expressed by the following equation: a = v2/R |
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| | Circular motion requires an inward force to cause the inward acceleration. The acceleration that is needed is dependent in part upon the speed of the bucket. Consistent with the equation in the Formula Frenzy section, an increase in speed will cause the acceleration to increase. As would be reasoned from Newton's second law (Fnet = m•a), an increase in acceleration would require an increased net force. The net force is the vector sum of the two individual forces - gravity and tension. The force of gravity is dependent upon the object's mass and unaffected by a change in speed. Thus, the increased net force must result from an increase in the magnitude of the tension force. |
| | |
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| | What is the direction of the net force for an object moving in a circle? How do I know when a specific type of force is acting on an object? What principles govern the relative size of individual forces for objects moving through roller coaster loops? | |
14812 | https://www.youtube.com/watch?v=N2PpRnFqnqY | Derivative as a concept | Derivatives introduction | AP Calculus AB | Khan Academy
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Transcript:
you are likely already familiar with the idea of a slope of a line if you're not I encourage you to review it on Con Academy but all it is it's describing the rate of change of a vertical variable with respect to a horizontal variable so for example here I have our classic y AIS in the vertical Direction and xaxis in the horizontal Direction and if I wanted to figure out the slope of this line I could pick two points say that point and that point I could say okay from this point to this point what is my change in X well my change in X would be this distance right over here change in X the Greek letter Delta this triangle here it's just shorthand for change so change in X and I could also calculate the change in y so this point going up to that point our change in y would be this right over here our change in y and then we would Define slope or we have defined slope as change in y over change in X so slope is equal to the rate of change of our vertical variable over the rate of change of our horizontal variable it's sometimes described as rise over run and for any line It's associated with a slope because it has a constant rate of change if you took any two points on this line no matter how far apart or no matter how close together anywhere they sit on the line if you were to do this calculation you would get the same slope that's what makes it a line but what's fascinating about calculus is we're going to build the tools so that we can think about the rate of change not just of a line which we've called slope in the past we can think about the rate of change the instantaneous rate of change of a curve of something whose rate of change is possibly constantly changing so for example here's a curve where the rate of change of Y with respect to X is constantly changing even if we wanted to use our traditional tools if we said okay we can calculate the average rate of change let's say between this point and this point well what would it be well the average rate of change between this point and this point would be the slope of the line that connects them so it be the slope of this line of the secant line but if we pick two different points if we pick this point and this point the average rate of change between those points all of a sudden looks quite different it looks like it has a higher slope so even when we take the slopes between two points on the line the secant lines you can see that those slopes are changing but what if we wanted to ask ourselves an even more interesting question what is the instantaneous rate of change at a point so for example how fast is y changing with respect to X exactly at that point exactly when X X is equal to that value let's call it X1 well one way you could think about it is what if we could draw a tangent line to this point a line that just touches the graph right over there and we can calculate the slope of that line well that should be the rate of change at that point the instantaneous rate of change so in this case the tangent line might look something like that if we know the slope of this well then we could say that that's the instantaneous rate of change at that point why do I say instantaneous rate of change well think about the video on the sprinters the usin bolt example if we wanted to figure out the speed of Usain Bolt at a given instant well maybe this describes his position with respect to time if y was position and X is time usually you would see T is time but let's say x is time so then if we're talking about right at this time we're talking about the instant instantaneous rate and this idea is the central idea of differential calculus and it's known as a derivative the slope of the tangent line which you could also view as the instantaneous rate of change I'm putting exclamation mark because it's so conceptually important here so how can we denote a derivative one way is known as Liv Net's notation and livet is one of the fathers of calculus along with Isaac Newton and his notation you would denote the slope of the tangent line as equaling dy over DX now why do I like this notation because it really comes from this idea of a slope which is change in y over change in X as you'll see in future videos one way to think about the slope of the tangent line is well let's calculate the slope of secant lines let's say between that point and that point but then let's get even closer say that point and that point and then let's gets even closer and that point and that point and then let's get even closer and let's see what happens as the change in X approaches zero and so using these D's instead of Deltas this was liet's way of saying hey what happens if my changes in say x become close to zero so this idea this is known as sometimes differential notation liances notation is instead of just change in y over change in x super small changes in y for a super small change in X especially as the change in X approaches zero and as you will see that is how we will calculate the derivative now there's other notations if this curve is described as Y is equal to F ofx the slope of the tangent line at that point could be denoted as equaling F Prime of X1 so this notation it takes a little bit of time getting used to the lrange notation it's saying F Prime is representing the derivative it's telling us the slope of the tangent line for a given point so if you input an X into this function into F you're getting the corresponding y value if you input an X into F Prime you're getting the slope of the tangent line at that point now another notation that you'll see less likely in a Calculus class but you might see in a physics class is the notation y with a DOT over it so you could write this as y with a DOT over it which also denotes the derivative you might also see y Prime this would be more common in a math class now as we March forward in our calculus Adventure we will build the tools to actually calculate these things and if you're already familiar with limits they will be very useful as you can imagine because we're really going to be taking the limit of our change in y over change in X as our change in X approaches zero and we're not just going to be able to figure it out for a point we're going to be able to figure out General equations that describe the derivative for any given point so be very very excited |
14813 | https://brainly.in/question/60916973 | find the equationof linehavinginclination.45°and y-intercept. - Brainly.in
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sumalurehman29
01.08.2024
Math
Secondary School
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Find the equation
of line
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inclination.45°and
y-intercept.
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riya1tiwari2005
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Step-by-step explanation:
To find the equation of a line with a given inclination angle and y-intercept, follow these steps:
Determine the Slope:
The inclination angle theta is 45°. The slope m of the line is given by m = tantheta)
For ( \theta = 45°, tan45^° = 1. So, the slope m = 1
Use the Slope-Intercept Form:
The slope-intercept form of a line is y = mx + c where m is the slope and c is the y-intercept.
If the y-intercept is given as c then the equation of the line is:
y = 1 \cdot x + c
y = x + c
So, the equation of the line with an inclination of 45° and y-intercept c is:
y = x + c
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14814 | https://patents.google.com/patent/US8357827B2/en | US8357827B2 - Process for the continuous preparation of nitrobenzene - Google Patents
US8357827B2 - Process for the continuous preparation of nitrobenzene - Google Patents
Process for the continuous preparation of nitrobenzene Download PDF
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Publication number US8357827B2 US8357827B2 US13/019,350 US201113019350A US8357827B2 US 8357827 B2 US8357827 B2 US 8357827B2 US 201113019350 A US201113019350 A US 201113019350A US 8357827 B2 US8357827 B2 US 8357827B2 Authority US United States Prior art keywords reactor pressure benzene mixed acid acid Prior art date 2010-02-05 Legal status (The legal status is an assumption and is not a legal conclusion. Google has not performed a legal analysis and makes no representation as to the accuracy of the status listed.)Active, expires 2031-03-10 Application number US13/019,350 Other versionsUS20110196177A1 (enInventor Jurgen Munnig Bernd Pennemann Andreas Karl Rausch Current Assignee (The listed assignees may be inaccurate. Google has not performed a legal analysis and makes no representation or warranty as to the accuracy of the list.) Covestro Deutschland AG Original Assignee Bayer MaterialScience AG Priority date (The priority date is an assumption and is not a legal conclusion. Google has not performed a legal analysis and makes no representation as to the accuracy of the date listed.)2010-02-05 Filing date 2011-02-02 Publication date 2013-01-22 Family has litigation First worldwide family litigation filed litigation Critical patent litigation dataset” by Darts-ip is licensed under a Creative Commons Attribution 4.0 International License.2011-02-02 Application filed by Bayer MaterialScience AG filed Critical Bayer MaterialScience AG 2011-02-02 Assigned to BAYER MATERIALSCIENCE AG reassignment BAYER MATERIALSCIENCE AG ASSIGNMENT OF ASSIGNORS INTEREST (SEE DOCUMENT FOR DETAILS).Assignors: RAUSCH, ANDREAS KARL, MUNNIG, JURGEN, PENNEMANN, BERND 2011-08-11 Publication of US20110196177A1 publication Critical patent/US20110196177A1/en 2013-01-22 Application granted granted Critical 2013-01-22 Publication of US8357827B2 publication Critical patent/US8357827B2/en 2016-03-30 Assigned to COVESTRO DEUTSCHLAND AG reassignment COVESTRO DEUTSCHLAND AG CHANGE OF NAME (SEE DOCUMENT FOR DETAILS).Assignors: COVESTRO DEUTSCHLAND AG 2016-06-13 Assigned to COVESTRO DEUTSCHLAND AG reassignment COVESTRO DEUTSCHLAND AG CORRECTIVE ASSIGNMENT TO CORRECT THE ASSIGNOR NAME PREVIOUSLY RECORDED ON REEL 038301 FRAME 0001. ASSIGNOR(S) HEREBY CONFIRMS THE CHANGE OF NAME FROM BAYER MATERIALSCIENCE AG TO COVESTRO DEUTSCHLAND AG.Assignors: BAYER MATERIALSCIENCE AG Status Active legal-status Critical Current 2031-03-10 Adjusted expiration legal-status Critical
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Classifications
C—CHEMISTRY; METALLURGY
C07—ORGANIC CHEMISTRY
C07C—ACYCLIC OR CARBOCYCLIC COMPOUNDS
C07C201/00—Preparation of esters of nitric or nitrous acid or of compounds containing nitro or nitroso groups bound to a carbon skeleton
C07C201/06—Preparation of nitro compounds
C07C201/08—Preparation of nitro compounds by substitution of hydrogen atoms by nitro groups
C—CHEMISTRY; METALLURGY
C07—ORGANIC CHEMISTRY
C07C—ACYCLIC OR CARBOCYCLIC COMPOUNDS
C07C205/00—Compounds containing nitro groups bound to a carbon skeleton
C07C205/06—Compounds containing nitro groups bound to a carbon skeleton having nitro groups bound to carbon atoms of six-membered aromatic rings
Definitions
the present invention relates to a process for the continuous preparation of nitrobenzene by nitration of benzene with mixed acid.
the pressure upstream of the nitration reactor is from 14 bar to 40 bar above the pressure in the gas phase of a phase separation apparatus for separating crude nitrobenzene and waste acid.
the continuous process for the preparation of nitrobenzene of the present invention is based upon the concept of adiabatic nitration of benzene with a mixture of sulfuric acid and nitric acid (so-called “mixed acid”).
An adiabatic nitration process was claimed for the first time in U.S. Pat. No. 2,256,999. More current embodiments of an adiabatic nitration process are described, for example, in EP 0 436 443 B1; EP 0 771 783 B1; and U.S. Pat. No. 6,562,247 B2. Processes in which the reaction is carried out adiabatically are distinguished by the fact that no technical measures are taken to supply heat to the reaction mixture or to remove heat from the reaction mixture.
a common feature of the known adiabatic processes is that the benzene and nitric acid starting materials are reacted in the presence of a large excess of sulfuric acid.
the sulfuric acid takes up the heat of reaction that is liberated and the water formed in the reaction.
nitric acid and sulfuric acid are generally mixed to form the so-called mixed acid (also called nitrating acid), and benzene is metered into the mixed acid.
the product obtained reacts with the nitric acid or with “nitronium ions” formed in the mixed acid substantially to form water and nitrobenzene.
Benzene is used in at least the stoichiometric amount—based on the molar amount of nitric acid—but preferably in a 2% to 10% excess compared with the stoichiometrically required amount of benzene.
the most important criterion for describing the quality of an adiabatic process for the nitration of aromatic hydrocarbons is the content of undesirable by-products in the product. Such by-products are formed by repeated nitration or oxidation of the aromatic hydrocarbon or of the nitroaromatic compound.
by-products are formed by repeated nitration or oxidation of the aromatic hydrocarbon or of the nitroaromatic compound.
dinitrobenzene and of nitrophenols, in particular trinitrophenol (picric acid), which is rated as explosive is always discussed.
pressure upstream of the reactor must be at least equal to the sum of the pressure losses of all the dispersing elements in the reactor. Other factors may also have to be taken into account. Such factors include the static pressure of the liquid column in the reactor and the pressure in the phase separation apparatus.
pressure upstream of the reactor is understood as being the absolute pressure that the liquid starting materials are under immediately before and also on entry into the reactor.
the total pressure loss over the reactor and, where appropriate, further apparatuses connected downstream of the reactor (and accordingly also the pressure upstream of the reactor)is to be kept as low as possible. See, for example, EP 1 291 078 A2, paragraph . Another example is described in EP 2 070 907 A1, where it is disclosed that an increase in the absolute pressure upstream of the reactor from 13.5 bar to 14.5 bar as a result of deposits of metal sulfates in the dispersing elements leads to a reduction of about 18% in the throughput of sulfuric acid (Example 1).
the prior art therefore teaches that high pressure losses, and accordingly, high absolute pressures, upstream of the reactor are to be avoided.
the possible lower limit for the pressure upstream of the reactor is additionally established by the fact that the benzene should be in liquid form at the reactor inlet under the given conditions (U.S. Pat. No. 4,091,042, column 2, lines 14 to 17).
the pressure loss per static dispersing element is kept as low as possible because, in order to overcome a higher pressure loss, for example, a pump having a higher power is required, which in turn leads to higher costs for the process as a whole (EP 1 291 078 A2, paragraph ).
the pressure inside the reactor is also limited by the material used to construct the tubular reactor.
High-alloy steels can likewise be used, in particular when the sulfuric acid always contains a residual amount of nitric acid, becaue nitric acid has a passivating effect on the high-alloy steel.
steel pipes enamelled with glass are especially used for the adiabatic nitration of benzene.
Nominal pressure level PN25 is permissible only in the case of pipe diameters up to a nominal width of DN150 (nominal pressure levels according to EN1333, nominal width according to DIN EN ISO 6708).
the permissible operating pressure is not identical to the nominal pressure level but must be calculated in view of the temperature and material being used. At higher temperatures, the permissible operating pressure is correspondingly lower due to the reduction in the permissible material parameters.
fittings are required in addition to apparatuses and pipes, which fittings are in turn subject to their own standards.
the result of these high requirements is that the skilled person building large-scale nitration installations must be concerned with keeping the pressure within the installation, particularly the pressure upstream of the reactor, low, as long as he/she does not know that a significant advantage is obtained thereby.
STY space-time yield
the space-time yield is calculated as the quotient of the production of nitrobenzene in metric tonnes per hour and the volume of the reaction space.
the reaction space is defined as the space which begins with the first dispersion of benzene and mixed acid and within which the reaction is completed to a degree of conversion of nitric acid of at least 99%.
the reaction space is arranged in a technical device for carrying out chemical reactions, the reactor. In the simplest case, the reaction space is identical with the inside volume of the reactor.
the first dispersion means the first intensive mixing of benzene and mixed acid. This generally takes place either in a mixing nozzle or in a static mixing element. Simply combining a benzene stream and a mixed acid stream in a common feed pipe leading to the reactor, without taking particular measures to intensively mix the two streams, is not regarded as the first dispersion required in the present invention.
the residence time of the reaction mixture, consisting of the aromatic compound and the mixed acid, within the reaction space is the reaction time.
a high space-time yield is advantageous for the industrial application of a process because it makes it possible to construct compact reaction devices which are distinguished by a low investment volume per capacity.
the present invention is directed to a process for the preparation of nitrobenzene by adiabatic nitration of benzene with mixed acid containing sulfuric acid and nitric acid.
benzene and mixed acid are introduced into a reactor either a) separately from one another or b) together, after they have been brought into contact with one another (i.e., in combination).
benzene and the mixed acid are introduced separately into the reactor, at least one of the benzene or mixed acid reactants is under a pressure, p1, on entry into the reactor.
the benzene and mixed acid have been brought into contact with one another prior to introduction into the reactor, the combined benzene and mixed acid are under a pressure, p1, on entry into the reactor.
the benzene and mixed acid are then dispersed in one another in the reactor in from 1 to 30, preferably from 2 to 20, most preferably from 6 to 15, dispersing elements. Where more than one dispersing element is employed, the dispersing elements are arranged one behind the other.
the benzene and nitric acid present in the mixed acid react to form nitrobenzene and nitrobenzene-containing reaction mixture is removed from the reactor.
the nitrobenzene-containing reaction mixture is subjected to a phase separation in a phase separation apparatus.
a pressure, p2 prevails.
the pressure difference, i.e., p1 ⁇ p2 is from 14 bar to 40 bar, preferably from 15 bar to 30 bar, more preferably from 15 bar to 25 bar, most preferably from 20 bar to 25 bar.
This process of the present invention makes it is possible to prepare large amounts of nitrobenzene in compact reactors with outstanding yields and selectivities.
the reactants i.e., benzene and mixed acid
the reactants are introduced into the reactor either a) separately from one another via different feed pipes or—preferably—b) after they have been brought into contact with one another, i.e., via a common feed pipe.
intensive mixing of the two streams does not yet take place during this first contact between benzene and mixed acid in the common feed pipe.
the pressure p1 on entry into the reactor is preferably measured in one of the respective feed pipes (benzene or mixed acid or common feed pipe for both) leading to the reactor.
Reactors according to the invention are preferably stirred tank, loop or tubular reactors. These can be arranged in series or in parallel. Combinations of different reactor types are also conceivable. Tubular reactors can be cylindrical or conical.
Dispersing elements useful in the process of the present invention are preferably sieve trays or perforated metal sheets.
the crude, liquid reaction product is preferably fed to a phase separation apparatus in which two liquid phases form, one being referred to as crude nitrobenzene (nitrobenzene and impurities) and the other as waste acid (substantially water and sulfuric acid).
the crude nitrobenzene and the waste acid are worked up as described in greater detail below.
gases escape from the liquid phase in the phase separation apparatus, so that the phase separation apparatus also has a third, gaseous phase.
the gas phase of the phase separation apparatus substantially contains nitrogen oxides as well as water vapor and benzene vapor. These gases are generally fed to a waste gas system. The pressure p2 is measured in this gas phase.
benzene is nitrated with mixed acid.
the mixed acid used contains preferably from 64% by mass to 71% by mass sulfuric acid and from 2% by mass to 8% by mass nitric acid; most preferably from 66% by mass to 69% by mass sulfuric acid and from 3% by mass to 5% by mass nitric acid, the remainder to 100% by mass preferably being water and the percentages by mass being based on the total mass of the mixed acid.
the concentration of the sulfuric acid used is preferably from 65% by mass to 80% by mass and that of the nitric acid is preferably from 62% by mass to 70% by mass, in each case based on the total mass of the acid in question.
benzene and mixed acid can be introduced into the nitration reactor separately from one another. It is preferred, however, for benzene to be metered into the mixed acid beforehand and for the two reactants to be passed into the reactor together.
the ratio of the mixed acid stream (in mass of mixed acid added per hour) to the benzene stream (in mass of benzene added per hour)is also referred to as the phase ratio and is preferably from 12:1 to 30:1, more preferably from 12:1 to 18:1.
the pressure p1 is preferably measured in the common feed pipe leading to the reactor, preferably at a point immediately upstream of the reactor.
a pressure p1a can also be measured in the feed pipe for the mixed acid, before the benzene stream is combined with the mixed acid stream.
the pressure p1a therein is identical with the pressure p1 on entry into the reactor, provided the nature of the metering of benzene into the mixed acid stream does not result in a pressure loss for the mixed acid stream. This is the case, for example, when benzene is introduced into the mixed acid stream using a lance or a mixing nozzle that preferably occupies only a small portion (preferably less than 10%) of the cross-sectional area of the mixed acid pipe.
the pressure in the benzene feed pipe (p1b) upstream of the lance or mixing nozzle is preferably greater than the pressure p1a in the mixed acid feed pipe, more preferably from 0.5 bar to 10 bar greater.
the position of the relevant measuring site for determining the pressure p1 should be chosen so that the pressure of the relevant material stream (e.g., the process product obtained by bringing benzene and mixed acid into contact with one another or—in the case of separate feed—the mixed acid) on entry into the reactor can be determined correctly (i.e., there is either no significant pressure loss between the measurement site and the point at which the relevant material stream enters the reactor, or the pressure loss is known and can be taken into account in the calculation).
the relevant material stream e.g., the process product obtained by bringing benzene and mixed acid into contact with one another or—in the case of separate feed—the mixed acid
the pressure measurement is conducted with any of the pressure measuring devices known to those skilled in the art, preferably using digital pressure transducers with a membrane manometer as sensor.
the pressure difference between p1 and the pressure in the gas phase of the phase separation apparatus, p2 is from 14 bar to 40 bar, preferably from 15 bar to 30 bar, more preferably from 15 bar to 25 bar, most preferably from 20 bar to 25 bar, and is accordingly higher than hitherto conventional in the prior art.
the nitration reaction takes place in the reactor under adiabatic conditions, i.e., no technical measures are taken to supply heat to the reaction mixture or to remove heat from the reaction mixture.
An important feature of the adiabatic nitration of aromatic hydrocarbons is that the temperature of the reaction mixture increases proportionally to the progress of the reaction, i.e., proportionally to the nitric acid conversion. A temperature difference is thereby obtained between the temperature of the mixed reactants, the aromatic compound and the mixed acid, before the start of the reaction (which is determined by thermodynamic calculations known to the person skilled in the art and is referred to below as the “start temperature”) and the temperature of the reaction mixture after at least 99% nitric acid conversion (referred to below as the “reaction end temperature”).
the value referred to here as the start temperature is generally advantageously calculated as the combined temperature of the mixed acid and benzene streams, and the value referred to here as the reaction end temperature is preferably measured in the inlet of the phase separation apparatus.
the difference between the start temperature and the reaction end temperature(the adiabatic temperature difference, also referred to below as ⁇ T adiabatic ) depends on the nature of the nitrated hydrocarbon and on the ratio in which the mixed acid and the aromatic hydrocarbon are used.
phase ratio gives a high adiabatic temperature difference and has the advantage that a large amount of the aromatic hydrocarbon is converted per unit time.
the values for ⁇ T adiabatic are preferably from 25 K to 60 K, most preferably from 30 K to 45 K. Despite the high adiabatic temperature difference, very good selectivities are achieved in the process of the present invention, which was not to be expected in view of the prior art.
reaction end temperatures in the reactor are preferably from 120° C. to 160° C., most preferably from 130 to 140° C.
the process of the present invention is preferably carried out in a tubular reactor having a plurality of dispersing elements distributed over the length of the tubular reactor.
These dispersing elements ensure intensive mixing and re-dispersion of benzene, nitric acid and sulfuric acid and water.
Such a reactor, and the form of dispersing elements which can be used,are described, for example, in EP 0 708 076 B1 (FIG. 2) and EP 1 291 078 A2 (FIG. 1). The corresponding portions of these documents are hereby incorporated into the disclosure of the present invention.
EP 1 291 078 A2 A configuration for the tubular reactor as is described in EP 1 291 078 A2 (FIG. 1, paragraphs to ) is most preferred. The corresponding portions of this document are hereby incorporated into the disclosure of the present invention.
EP 1 291 078 A2 from 3 to 11 dispersing elements made of tantalum are used (corresponding to 4 to 12 chambers; see paragraph ). Each of these dispersing elements produces a pressure loss of from 0.5 bar to 4 bar and exhibits from 10 to 25 openings for a mass flow rate of 1 t/h.
the openings can be slots, holes or bores.
the dispersing elements are designed so that the mean droplet diameter is less than 200 ⁇ m, most preferably less than 120 ⁇ M, and, as can be calculated by the person skilled in the art from the examples, a pressure of up to 10 bar is achieved in the enamelled steel reactor.
the dispersing elements are to be designed so that the absolute pressure upstream of the reactor is at least 15 bar, provided that the absolute pressure in the gas phase of the phase separation apparatus is less than or equal to 1.0 bar.
the dispersing elements preferably exhibit decreasing pressure losses in the direction of flow of the reactants.
reactors having permissible absolute operating pressures of preferably at least 18 bar, most preferably at least 21 bar, are used.
These reactors can be made, for example, of high-alloy steels.
the resistance of suitable stainless steels under nitrating conditions is the result in particular of the fact that the nitric acid used for the nitration has a passivating effect.
These reactors can also be made of enamelled steel. A combination of high-alloy and enamelled steels is also possible.
the liquid flowing through the dispersing elements causes the pressure of the liquid within the reactor to fall.
This pressure drop allows modification of the configuration of the reactor so that the front section is constructed to withstand high operating pressures while the section in which the pressure is markedly lower can be constructed to withstand lower operating pressures.
Suitable gaskets preferably cover the majority of the flange face and can optionally be centered between the flange faces by means of a binding. It is also possible to use gaskets which have been reinforced at the surfaces and the outside edges, for example, by means of glass fiber fabric on the surfaces or support rings on the outside edges.
phase separation apparatus After passing through the nitration reactor, the crude, liquid reaction product is fed to a phase separation apparatus. Any phase separation apparatus known to the person skilled in the art can be used. Preferably, the separation is carried out in a gravity separator. The liquid phases, crude nitrobenzene and waste acid, obtained in the phase separation apparatus are preferably worked up as described below.
the waste acid is usually fed to a flash evaporator in which, during decompression of the waste acid to a reduced pressure range, water vaporizes and the waste acid is thus cooled and concentrated.
the adiabatic procedure for the nitration of benzene with mixed acid has the advantage that the heat of reaction of the exothermic reaction is used to heat the waste acid so greatly that at the same time the concentration and the temperature that the sulfuric acid exhibited prior to mixing with nitric acid and benzene can be established again in the flash evaporator.
the crude nitrobenzene obtained in the phase separation apparatus still contains sulfuric acid, water, benzene as well as nitrophenols and dinitrobenzene as impurities. These impurities are separated off by suitable working-up processes, e.g., washing and distillation steps.
suitable working-up processes e.g., washing and distillation steps.
a possible form of this working up is described in EP 1 816 117 A1 (paragraph ).
the corresponding sections of EP 1 816 117 A1 are hereby incorporated into the disclosure of the present invention. Other forms are also possible, however.
the gases formed in the phase separation apparatus are preferably fed to a waste gas system.
a space-time yield of preferably more than 7.0 t of nitrobenzene per cubic meter volume of the reaction space per hour is achievable with a very low content of by-products.
the sulfuric acid always had a concentration of 71% by mass and the nitric acid always had a concentration of 69.2% by mass.
the mixed acid always had a temperature of 96° C., and the benzene was preheated and always had a temperature of 80° C. The purity of the benzene was always more than 99% by mass.
the crude reaction product was introduced into a static phase separation apparatus operated without pressure.
the amount of benzene indicated in Table 1 was reacted with the mixture of the indicated amounts of nitric acid and sulfuric acid.
the adiabatic end temperature was measured and the adiabatic temperature difference ⁇ T adiabatic was calculated by means of the combined temperature of the mixed acid and benzene streams.
the nitrobenzene separated from acid in the phase separation apparatus was analyzed to determine the dinitrobenzene and nitrophenols contents. From the group of nitrophenols, picric acid can be identified separately. In all the tests, the nitric acid used was reacted completely (>99.99% according to ion chromatography analysis of the waste acid for nitrate).
Example 4 it was possible with a pressure difference of 15.8 bar to achieve a space-time yield of more than 7.0 t nitrobenzene /(m 3 reaction space ⁇ h) and, despite the high adiabatic temperature difference, the lowest contents of by-products.
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Abstract
Nitrobenzene is continuously produced by nitration of benzene with mixed acid under adiabatic conditions. In this process, the pressure upstream of the nitration reactor is from 14 bar to 40 bar above the pressure in the gas phase of the phase separation apparatus used to separate crude nitrobenzene and waste acid.
Description
BACKGROUND OF THE INVENTION
The present invention relates to a process for the continuous preparation of nitrobenzene by nitration of benzene with mixed acid. In this process, the pressure upstream of the nitration reactor is from 14 bar to 40 bar above the pressure in the gas phase of a phase separation apparatus for separating crude nitrobenzene and waste acid.
The continuous process for the preparation of nitrobenzene of the present invention is based upon the concept of adiabatic nitration of benzene with a mixture of sulfuric acid and nitric acid (so-called “mixed acid”). An adiabatic nitration process was claimed for the first time in U.S. Pat. No. 2,256,999. More current embodiments of an adiabatic nitration process are described, for example, in EP 0 436 443 B1; EP 0 771 783 B1; and U.S. Pat. No. 6,562,247 B2. Processes in which the reaction is carried out adiabatically are distinguished by the fact that no technical measures are taken to supply heat to the reaction mixture or to remove heat from the reaction mixture.
A common feature of the known adiabatic processes is that the benzene and nitric acid starting materials are reacted in the presence of a large excess of sulfuric acid. The sulfuric acid takes up the heat of reaction that is liberated and the water formed in the reaction.
In order to carry out the reaction, nitric acid and sulfuric acid are generally mixed to form the so-called mixed acid (also called nitrating acid), and benzene is metered into the mixed acid. The product obtained reacts with the nitric acid or with “nitronium ions” formed in the mixed acid substantially to form water and nitrobenzene. Benzene is used in at least the stoichiometric amount—based on the molar amount of nitric acid—but preferably in a 2% to 10% excess compared with the stoichiometrically required amount of benzene.
The most important criterion for describing the quality of an adiabatic process for the nitration of aromatic hydrocarbons is the content of undesirable by-products in the product. Such by-products are formed by repeated nitration or oxidation of the aromatic hydrocarbon or of the nitroaromatic compound. In nitration of benzene, the content of dinitrobenzene and of nitrophenols, in particular trinitrophenol (picric acid), which is rated as explosive, is always discussed.
In order to obtain nitrobenzene with particularly high selectivity, the nature of the mixed acid to be used has been specified in detail. (See, e.g., EP 0 373 966 B1; EP 0 436 443 B1; and EP 0 771 783 B1). It has been noted that the content of by-products is determined by the maximum temperature attained by the reaction mixture (EP 0 436 443 B1, column 15, I. 22-25). It is also known that a high initial conversion is advantageous for high selectivity, and that this high initial conversion may be achieved with optimal mixing at the beginning of the reaction (EP 0 771 783 B1, paragraph ).
The inexpensive and efficient configuration of the initial mixing (dispersion) and the repeated mixing (re-dispersion) of aromatic compounds in the mixed acid is the subject of numerous studies. As a result, use of mixing nozzles (EP 0 373 966 B1; EP 0 771 783 B1) and specially formed static dispersing elements (EP 0 489 211 B1; EP 0 779 270 B1; EP 1 291 078 A1; and U.S. Pat. No. 6,562,247 B2) has been proposed. It is also possible to combine the two concepts.
If static mixing elements (dispersing elements) are used for the mixing, the pressure loss at these static mixing elements is critical for the quality of the mixing. The pressure upstream of the reactor must be at least equal to the sum of the pressure losses of all the dispersing elements in the reactor. Other factors may also have to be taken into account. Such factors include the static pressure of the liquid column in the reactor and the pressure in the phase separation apparatus. Within the scope of this invention, “pressure upstream of the reactor” is understood as being the absolute pressure that the liquid starting materials are under immediately before and also on entry into the reactor.
According to the teaching of the prior art, the total pressure loss over the reactor and, where appropriate, further apparatuses connected downstream of the reactor (and accordingly also the pressure upstream of the reactor) is to be kept as low as possible. See, for example, EP 1 291 078 A2, paragraph . Another example is described in EP 2 070 907 A1, where it is disclosed that an increase in the absolute pressure upstream of the reactor from 13.5 bar to 14.5 bar as a result of deposits of metal sulfates in the dispersing elements leads to a reduction of about 18% in the throughput of sulfuric acid (Example 1). The prior art therefore teaches that high pressure losses, and accordingly, high absolute pressures, upstream of the reactor are to be avoided.
An example of the mixing of aromatic compound and mixed acid by means of a suitable nozzle without static dispersing elements is found in EP 0 373 966 B1. Here, a range of from 0.689 bar to 13.79 bar is given as a suitable range for the working pressure. “Back pressure” equals counter-pressure in the nozzle, i.e., the pressure of the liquid starting material stream (aromatic compound or mixed acid) on entry into the reactor, which is equivalent to the pressure of the starting material stream in question upstream of the reactor. (p. 5, I. 12 to 13) This disclosure also teaches that, under normal conditions, a pressure higher than 11.03 bar is not expected to be necessary (p. 5, I. 15 to 16).
The possible lower limit for the pressure upstream of the reactor is additionally established by the fact that the benzene should be in liquid form at the reactor inlet under the given conditions (U.S. Pat. No. 4,091,042, column 2, lines 14 to 17). Regarding the possible upper limit, it is to be noted that, according to the prior art, the pressure loss per static dispersing element is kept as low as possible because, in order to overcome a higher pressure loss, for example, a pump having a higher power is required, which in turn leads to higher costs for the process as a whole (EP 1 291 078 A2, paragraph ). Also, attempts are generally made to keep the number and thickness (stability) of the dispersing elements preferably as low as possible and thus minimize the cost of the dispersing elements which are often produced from special tantalum material. (EP 1 291 078 A1, paragraph )
The pressure inside the reactor is also limited by the material used to construct the tubular reactor. Under generally conventional conditions for the adiabatic nitration of benzene at from 80° C. to 150° C. using sulfuric acid having a concentration of from 65% by mass to 80% by mass, only tantalum, Teflon and glass are permanently resistant. High-alloy steels can likewise be used, in particular when the sulfuric acid always contains a residual amount of nitric acid, becaue nitric acid has a passivating effect on the high-alloy steel. On an industrial scale, steel pipes enamelled with glass are especially used for the adiabatic nitration of benzene. Steel enamel pipe segments are to be manufactured in accordance with DIN standard 2873 of June 2002 for nominal pressure level PN10 and at most for nominal pressure level PN25. Nominal pressure level PN25 is permissible only in the case of pipe diameters up to a nominal width of DN150 (nominal pressure levels according to EN1333, nominal width according to DIN EN ISO 6708). As is known to the person skilled in the art, the permissible operating pressure is not identical to the nominal pressure level but must be calculated in view of the temperature and material being used. At higher temperatures, the permissible operating pressure is correspondingly lower due to the reduction in the permissible material parameters. In the construction of chemical installations, fittings (valves, slides, etc.) are required in addition to apparatuses and pipes, which fittings are in turn subject to their own standards. The result of these high requirements is that the skilled person building large-scale nitration installations must be concerned with keeping the pressure within the installation, particularly the pressure upstream of the reactor, low, as long as he/she does not know that a significant advantage is obtained thereby.
Although processes described in the prior art permit the preparation of a crude nitrobenzene which has a low content of by-products, i.e., from 100 ppm to 300 ppm dinitrobenzene and from 1500 ppm to 2500 ppm nitrophenols of which picric acid can account for from 10% by mass to 50% by mass of the nitrophenols, a critically important factor for industrial production, apart from the purity of the crude nitrobenzene, is that the preparation of the nitroaromatic compounds be carried out in reaction devices that are as compact as possible. This is a particular concern in view of the constantly rising demand for nitroaromatic compounds, especially for the preparation of aromatic amines and aromatic isocyanates.
An important parameter for describing the relationship between the amount of product that can be produced and the size of the reaction device is the space-time yield (STY). STY is calculated as the quotient of the amount of the target compound that can be produced per unit time and the volume of the reaction device.
STY[t nitrobenzene(m 3 reaction space ·h)]=amount produced[t nitrobenzene /h]/reaction space[m 3]
In the case of the nitration of benzene, the space-time yield is calculated as the quotient of the production of nitrobenzene in metric tonnes per hour and the volume of the reaction space. The reaction space is defined as the space which begins with the first dispersion of benzene and mixed acid and within which the reaction is completed to a degree of conversion of nitric acid of at least 99%. The reaction space is arranged in a technical device for carrying out chemical reactions, the reactor. In the simplest case, the reaction space is identical with the inside volume of the reactor. In this connection, the first dispersion means the first intensive mixing of benzene and mixed acid. This generally takes place either in a mixing nozzle or in a static mixing element. Simply combining a benzene stream and a mixed acid stream in a common feed pipe leading to the reactor, without taking particular measures to intensively mix the two streams, is not regarded as the first dispersion required in the present invention.
The residence time of the reaction mixture, consisting of the aromatic compound and the mixed acid, within the reaction space is the reaction time.
A high space-time yield is advantageous for the industrial application of a process because it makes it possible to construct compact reaction devices which are distinguished by a low investment volume per capacity.
With regard to the space-time yield of aromatic compound nitration, there is still a marked need for improvement over the prior art.
However, high space-time yields when carrying out a nitration adiabatically (in particular with a constant residence time in the reactor) inevitably lead to high temperature differences (adiabatic temperature jumps) between the start temperature (the temperature of the mixed starting materials before the start of the reaction, determined by calculating the combined temperature of the individual streams) and the reaction end temperature (the temperature after conversion of substantially all the nitric acid); and, as is clear from the prior art, high reaction end temperatures lead to an impairment of the selectivity. (See, e.g., EP 0 436 443 B1, column 15, I. 22-25).
SUMMARY OF THE INVENTION
It is an object of the present invention to provide a process for the adiabatic nitration of aromatic compounds, in which a high space-time yield is achieved without impairing the product quality.
It is also an object of the present invention to provide a continuous process for the production of nitrobenzene in compact reaction devices in which the nitrobenzene is obtained with high selectivity and in outstanding yields. These and other objects which will be apparent to those skilled in the art were achieved by the process for the preparation of nitrobenzene by adiabatic nitration of benzene with mixed acid containing sulfuric acid and nitric acid described more fully below.
DETAILED DESCRIPTION OF THE PRESENT INVENTION
The present invention is directed to a process for the preparation of nitrobenzene by adiabatic nitration of benzene with mixed acid containing sulfuric acid and nitric acid. In this process, benzene and mixed acid are introduced into a reactor either a) separately from one another or b) together, after they have been brought into contact with one another (i.e., in combination). When benzene and the mixed acid are introduced separately into the reactor, at least one of the benzene or mixed acid reactants is under a pressure, p1, on entry into the reactor. When the benzene and mixed acid have been brought into contact with one another prior to introduction into the reactor, the combined benzene and mixed acid are under a pressure, p1, on entry into the reactor. The benzene and mixed acid are then dispersed in one another in the reactor in from 1 to 30, preferably from 2 to 20, most preferably from 6 to 15, dispersing elements. Where more than one dispersing element is employed, the dispersing elements are arranged one behind the other. The benzene and nitric acid present in the mixed acid react to form nitrobenzene and nitrobenzene-containing reaction mixture is removed from the reactor. After leaving the reactor, the nitrobenzene-containing reaction mixture is subjected to a phase separation in a phase separation apparatus. In the gas phase of the phase separation apparatus, a pressure, p2, prevails. The pressure difference, i.e., p1−p2 is from 14 bar to 40 bar, preferably from 15 bar to 30 bar, more preferably from 15 bar to 25 bar, most preferably from 20 bar to 25 bar.
This process of the present invention makes it is possible to prepare large amounts of nitrobenzene in compact reactors with outstanding yields and selectivities.
In the first step of the process of the present invention, the reactants (i.e., benzene and mixed acid) are introduced into the reactor either a) separately from one another via different feed pipes or—preferably—b) after they have been brought into contact with one another, i.e., via a common feed pipe. In embodiment b), intensive mixing of the two streams does not yet take place during this first contact between benzene and mixed acid in the common feed pipe.
The pressure p1 on entry into the reactor is preferably measured in one of the respective feed pipes (benzene or mixed acid or common feed pipe for both) leading to the reactor.
Reactors according to the invention are preferably stirred tank, loop or tubular reactors. These can be arranged in series or in parallel. Combinations of different reactor types are also conceivable. Tubular reactors can be cylindrical or conical.
Dispersing elements useful in the process of the present invention are preferably sieve trays or perforated metal sheets.
Following the nitration reactor, the crude, liquid reaction product is preferably fed to a phase separation apparatus in which two liquid phases form, one being referred to as crude nitrobenzene (nitrobenzene and impurities) and the other as waste acid (substantially water and sulfuric acid). The crude nitrobenzene and the waste acid are worked up as described in greater detail below. At the same time as the two liquid phases form, gases escape from the liquid phase in the phase separation apparatus, so that the phase separation apparatus also has a third, gaseous phase.
The gas phase of the phase separation apparatus substantially contains nitrogen oxides as well as water vapor and benzene vapor. These gases are generally fed to a waste gas system. The pressure p2 is measured in this gas phase.
The present invention is explained in greater detail below.
According to the invention, benzene is nitrated with mixed acid. The mixed acid used contains preferably from 64% by mass to 71% by mass sulfuric acid and from 2% by mass to 8% by mass nitric acid; most preferably from 66% by mass to 69% by mass sulfuric acid and from 3% by mass to 5% by mass nitric acid, the remainder to 100% by mass preferably being water and the percentages by mass being based on the total mass of the mixed acid. The concentration of the sulfuric acid used is preferably from 65% by mass to 80% by mass and that of the nitric acid is preferably from 62% by mass to 70% by mass, in each case based on the total mass of the acid in question.
In the process of the present invention, benzene and mixed acid can be introduced into the nitration reactor separately from one another. It is preferred, however, for benzene to be metered into the mixed acid beforehand and for the two reactants to be passed into the reactor together. The ratio of the mixed acid stream (in mass of mixed acid added per hour) to the benzene stream (in mass of benzene added per hour) is also referred to as the phase ratio and is preferably from 12:1 to 30:1, more preferably from 12:1 to 18:1.
When benzene and mixed acid are metered into the reactor together, the pressure p1 is preferably measured in the common feed pipe leading to the reactor, preferably at a point immediately upstream of the reactor.
Alternatively, a pressure p1a can also be measured in the feed pipe for the mixed acid, before the benzene stream is combined with the mixed acid stream. The pressure p1a therein is identical with the pressure p1 on entry into the reactor, provided the nature of the metering of benzene into the mixed acid stream does not result in a pressure loss for the mixed acid stream. This is the case, for example, when benzene is introduced into the mixed acid stream using a lance or a mixing nozzle that preferably occupies only a small portion (preferably less than 10%) of the cross-sectional area of the mixed acid pipe. The pressure in the benzene feed pipe (p1b) upstream of the lance or mixing nozzle is preferably greater than the pressure p1a in the mixed acid feed pipe, more preferably from 0.5 bar to 10 bar greater.
If the manner of metering of the benzene into the mixed acid stream causes a significant pressure loss in the mixed acid stream, then it is preferred to measure the pressure p1 in the common feed pipe.
When benzene and mixed acid are fed separately into the reactor, it is preferred within the scope of this invention to measure the pressure p1 in the feed pipe for the mixed acid, preferably immediately upstream of the reactor, because this is more meaningful, as a result of the high phase ratio used for the adiabatic procedure, than a measurement in the feed pipe for the benzene stream.
Regardless of the precise manner in which benzene and mixed acid are fed (separately or together, with or without a mixing nozzle), the position of the relevant measuring site for determining the pressure p1 should be chosen so that the pressure of the relevant material stream (e.g., the process product obtained by bringing benzene and mixed acid into contact with one another or—in the case of separate feed—the mixed acid) on entry into the reactor can be determined correctly (i.e., there is either no significant pressure loss between the measurement site and the point at which the relevant material stream enters the reactor, or the pressure loss is known and can be taken into account in the calculation).
In the process according to the invention, the pressure measurement is conducted with any of the pressure measuring devices known to those skilled in the art, preferably using digital pressure transducers with a membrane manometer as sensor.
In the process according to the invention, the pressure difference between p1 and the pressure in the gas phase of the phase separation apparatus, p2, is from 14 bar to 40 bar, preferably from 15 bar to 30 bar, more preferably from 15 bar to 25 bar, most preferably from 20 bar to 25 bar, and is accordingly higher than hitherto conventional in the prior art.
In the present invention, the nitration reaction takes place in the reactor under adiabatic conditions, i.e., no technical measures are taken to supply heat to the reaction mixture or to remove heat from the reaction mixture. An important feature of the adiabatic nitration of aromatic hydrocarbons is that the temperature of the reaction mixture increases proportionally to the progress of the reaction, i.e., proportionally to the nitric acid conversion. A temperature difference is thereby obtained between the temperature of the mixed reactants, the aromatic compound and the mixed acid, before the start of the reaction (which is determined by thermodynamic calculations known to the person skilled in the art and is referred to below as the “start temperature”) and the temperature of the reaction mixture after at least 99% nitric acid conversion (referred to below as the “reaction end temperature”). It is known to the person skilled in the art that the value referred to here as the start temperature is generally advantageously calculated as the combined temperature of the mixed acid and benzene streams, and the value referred to here as the reaction end temperature is preferably measured in the inlet of the phase separation apparatus. The difference between the start temperature and the reaction end temperature (the adiabatic temperature difference, also referred to below as ΔT adiabatic) depends on the nature of the nitrated hydrocarbon and on the ratio in which the mixed acid and the aromatic hydrocarbon are used. A low ratio of mixed acid and aromatic hydrocarbon (so-called phase ratio) gives a high adiabatic temperature difference and has the advantage that a large amount of the aromatic hydrocarbon is converted per unit time.
Under otherwise identical conditions, a higher value for ΔT adiabatic indicates a more complete conversion. In the process of the present invention, the values for ΔT adiabatic are preferably from 25 K to 60 K, most preferably from 30 K to 45 K. Despite the high adiabatic temperature difference, very good selectivities are achieved in the process of the present invention, which was not to be expected in view of the prior art.
The reaction end temperatures in the reactor are preferably from 120° C. to 160° C., most preferably from 130 to 140° C.
The process of the present invention is preferably carried out in a tubular reactor having a plurality of dispersing elements distributed over the length of the tubular reactor. These dispersing elements ensure intensive mixing and re-dispersion of benzene, nitric acid and sulfuric acid and water. Such a reactor, and the form of dispersing elements which can be used, are described, for example, in EP 0 708 076 B1 (FIG. 2) and EP 1 291 078 A2 (FIG. 1). The corresponding portions of these documents are hereby incorporated into the disclosure of the present invention.
A configuration for the tubular reactor as is described in EP 1 291 078 A2 (FIG. 1, paragraphs to ) is most preferred. The corresponding portions of this document are hereby incorporated into the disclosure of the present invention. In EP 1 291 078 A2, from 3 to 11 dispersing elements made of tantalum are used (corresponding to 4 to 12 chambers; see paragraph ). Each of these dispersing elements produces a pressure loss of from 0.5 bar to 4 bar and exhibits from 10 to 25 openings for a mass flow rate of 1 t/h. The openings can be slots, holes or bores. These parameters can likewise be implemented in the process of the present invention in order to avoid coalescence of the phases and keep the diameter of the organic droplets in the acid phase small. In EP 1 291 078 A2, the dispersing elements are designed so that the mean droplet diameter is less than 200 μm, most preferably less than 120 μM, and, as can be calculated by the person skilled in the art from the examples, a pressure of up to 10 bar is achieved in the enamelled steel reactor. In the process of the present invention, when a reactor configuration equivalent to that of EP 1 291 078 A2 is used, the dispersing elements are to be designed so that the absolute pressure upstream of the reactor is at least 15 bar, provided that the absolute pressure in the gas phase of the phase separation apparatus is less than or equal to 1.0 bar.
The dispersing elements preferably exhibit decreasing pressure losses in the direction of flow of the reactants. Most preferably, the second and every further dispersing element—in the direction of flow of the reactants—exhibits at most 80% of the pressure loss of the preceding dispersing element in each case.
In carrying out the process of the present invention, reactors having permissible absolute operating pressures of preferably at least 18 bar, most preferably at least 21 bar, are used. These reactors can be made, for example, of high-alloy steels. The resistance of suitable stainless steels under nitrating conditions is the result in particular of the fact that the nitric acid used for the nitration has a passivating effect. These reactors can also be made of enamelled steel. A combination of high-alloy and enamelled steels is also possible.
The liquid flowing through the dispersing elements causes the pressure of the liquid within the reactor to fall. This pressure drop allows modification of the configuration of the reactor so that the front section is constructed to withstand high operating pressures while the section in which the pressure is markedly lower can be constructed to withstand lower operating pressures.
Because steel enamel is a less expensive material compared with high-alloy steels, the use of steel enamel pipe segments for the whole of the structure of the tubular reactor is preferred, since this embodiment in particular achieves the aim of low investment costs. In order to be able to achieve absolute operating pressures of more than 15 bar in a reactor of enamelled steel, it may be necessary to make particular requirements of the quality of manufacture of the enamelled pipe segments that are to be used. Thus, for example, particular attention should be paid to the perpendicularity between the flange and the pipe. It may also be necessary to make particular demands of the enamelled flange face, in particular to remove uneven areas. An optimum contact surface for the gaskets can be obtained by grinding or polishing the enamelled flange face. In addition, uniform contact pressure can be achieved by suitable adjustment of the flange. In order to obtain the required permitted pressure, suitable gaskets should also be chosen. Suitable gaskets preferably cover the majority of the flange face and can optionally be centered between the flange faces by means of a binding. It is also possible to use gaskets which have been reinforced at the surfaces and the outside edges, for example, by means of glass fiber fabric on the surfaces or support rings on the outside edges.
After passing through the nitration reactor, the crude, liquid reaction product is fed to a phase separation apparatus. Any phase separation apparatus known to the person skilled in the art can be used. Preferably, the separation is carried out in a gravity separator. The liquid phases, crude nitrobenzene and waste acid, obtained in the phase separation apparatus are preferably worked up as described below.
The waste acid is usually fed to a flash evaporator in which, during decompression of the waste acid to a reduced pressure range, water vaporizes and the waste acid is thus cooled and concentrated. The adiabatic procedure for the nitration of benzene with mixed acid has the advantage that the heat of reaction of the exothermic reaction is used to heat the waste acid so greatly that at the same time the concentration and the temperature that the sulfuric acid exhibited prior to mixing with nitric acid and benzene can be established again in the flash evaporator.
The crude nitrobenzene obtained in the phase separation apparatus still contains sulfuric acid, water, benzene as well as nitrophenols and dinitrobenzene as impurities. These impurities are separated off by suitable working-up processes, e.g., washing and distillation steps. A possible form of this working up is described in EP 1 816 117 A1 (paragraph ). The corresponding sections of EP 1 816 117 A1 are hereby incorporated into the disclosure of the present invention. Other forms are also possible, however.
The gases formed in the phase separation apparatus are preferably fed to a waste gas system.
By the process according to the invention, a space-time yield of preferably more than 7.0 t of nitrobenzene per cubic meter volume of the reaction space per hour is achievable with a very low content of by-products. The low content of by-products despite large adiabatic temperature differences—as a result of the high space-time yields—is made possible by the use of high pressure (i.e., from 14 to 40 bar above the pressure in the gas phase of the phase separation apparatus upstream of the reactor).
The process according to the invention has been described herein using the example of nitrobenzene. However, the person skilled in the art can readily extend the invention to the preparation of other nitoaromatic compounds; for example, to the preparation of dinitrotoluene by nitration of toluene.
Having thus described the invention, the following Examples are given as being illustrative thereof.
EXAMPLES Examples 1 to 3 (not According to the Invention), Example 4 (According to the Invention)
The examples below were carried out under the following conditions:
All the tests were carried out in a tubular reactor of enamelled steel having an inside volume of 455 ml. The tubular reactor was mounted vertically, and flow was from the bottom. The tubular reactor contained 10 dispersing elements made of tantalum, which exhibited decreasing pressure losses from top to bottom. The pressure prevailing in the mixed acid (p1a) was measured upstream of the tubular reactor using a membrane manometer, before benzene was introduced into the mixed acid by means of a low-pressure-loss lance and the resulting stream was introduced into the reactor. In the present test arrangement, p1a can be equated with the pressure p1 of the process product obtained by bringing benzene and mixed acid into contact on entry into the reactor. The sulfuric acid always had a concentration of 71% by mass and the nitric acid always had a concentration of 69.2% by mass. The mixed acid always had a temperature of 96° C., and the benzene was preheated and always had a temperature of 80° C. The purity of the benzene was always more than 99% by mass. After passing through the reactor, the crude reaction product was introduced into a static phase separation apparatus operated without pressure.
The amount of benzene indicated in Table 1 was reacted with the mixture of the indicated amounts of nitric acid and sulfuric acid. The benzene excess, based on the product nitrobenzene, was in each case 6% by mass.
At the end of the tubular reactor, the adiabatic end temperature was measured and the adiabatic temperature difference ΔT adiabatic was calculated by means of the combined temperature of the mixed acid and benzene streams. The nitrobenzene separated from acid in the phase separation apparatus was analyzed to determine the dinitrobenzene and nitrophenols contents. From the group of nitrophenols, picric acid can be identified separately. In all the tests, the nitric acid used was reacted completely (>99.99% according to ion chromatography analysis of the waste acid for nitrate). In Example 4 (according to the invention), it was possible with a pressure difference of 15.8 bar to achieve a space-time yield of more than 7.0 t nitrobenzene/(m 3 reaction space·h) and, despite the high adiabatic temperature difference, the lowest contents of by-products.
TABLE 1
Parameters and results of Examples 1 to 4.
Example:1234
Space-time yield 1 5.5 6.2 6.4 7.2
HNO 3 stream [g/h]1864 2089 2169 2410
H 2 SO 4 stream [g/h]30000 34000 34000 38000
Benzene stream [g/h]1760 1970 2050 2275
Absolute pressure on 10.7 13.4 13.6 16.8
entry into the reactor
(p1) [bar]
Absolute pressure in 1.0 1.0 1.0 1.0
the gas phase of the
phase separation
apparatus (p2) [bar]
Pressure difference 9.7 12.4 12.6 15.8
p1 − p2 [bar]
ΔT adiabatic [K]42.3 41.7 44.4 44.1
Nitric acid conversion>99.99>99.99>99.99>99.99
[%]
Dinitrobenzene content 293 211 213 176
[ppm by mass]
Nitrophenol content 2153 2169 2055 1973
(total) [ppm by mass]
Picric acid content 171 136 122 89
[ppm by mass]
Comparative Example
1 Space-time yield = [t nitrobenzene/(m 3 reaction space · h)]
Although the invention has been described in detail in the foregoing for the purpose of illustration, it is to be understood that such detail is solely for that purpose and that variations can be made therein by those skilled in the art without departing from the spirit and scope of the invention except as it may be limited by the claims.
Claims (5)
A process for the production of nitrobenzene by adiabatic nitration of benzene with a mixed acid comprising sulfuric acid and nitric acid, the process comprising:
(i) introducing the benzene and the mixed acid into a reactor either
a) separately with at least one of the benzene or the mixed acid being under a pressure p1 on entry into the reactor
or
b) together, after being brought into contact with one another, under a pressure p1 on entry into the reactor,
(ii) dispersing the benzene and the mixed acid in one another in the reactor with from 1 to 30 dispersing elements, provided that when more than one dispersing element is present, the dispersing elements are arranged one behind the other,
(iii) removing reaction product from the reactor,
(iv) subjecting the reaction product removed in (iii) to a phase separation in a phase separation apparatus, in which a gas phase pressure p2 prevails
in which
p1−p2=from 14 bar to 40 bar.
The process of claim 1 in which the reactor temperature after at least 99% conversion of nitric acid minus the combined benzene and mixed acid reactants' temperature before start of the nitration reaction is from 25 K to 60 K.
The process of claim 1 in which from 2 to 20 dispersing elements are used in (ii).
The process of claim 3 in which each dispersing element exhibits a pressure loss, and the pressure losses of the dispersing elements decrease in the direction of flow of the benzene and mixed acid reactants.
The process of claim 4 in which the second and each subsequent dispersing element in the direction of flow of the reactants exhibits at most 80% of the pressure loss of the preceding dispersing element.
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| KR (1) | KR101822193B1 (en) |
| CN (1) | CN102153476B (en) |
| DE (1) | DE102010006984A1 (en) |
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--- ---
| US9260377B2 (en) | 2012-07-27 | 2016-02-16 | Bayer Materialscience Ag | Method for producing nitrobenzene by adiabatic nitriding |
| US9284255B2 (en) | 2012-07-27 | 2016-03-15 | Bayer Materialscience Ag | Method for producing nitrobenzene by adiabatic nitriding |
| US9302978B1 (en) | 2013-04-29 | 2016-04-05 | Covestro Deutschland Ag | Process for the preparation of nitrobenzene by adiabatic nitration |
| US10023524B2 (en) | 2014-06-24 | 2018-07-17 | Covestro Deutschland Ag | Process for producing nitrobenzene |
| WO2020021323A1 (en) | 2018-07-24 | 2020-01-30 | Noram International Limited | Nitration reactor and method |
| RU2828259C1 (en) | 2023-10-09 | 2024-10-08 | Акционерное общество "Российский научный центр "Прикладная химия (ГИПХ)" | Method of producing nitrobenzene |
| US12180135B2 (en) | 2019-04-17 | 2024-12-31 | Covestro Deutschland Ag | Process for the continuous production of nitrobenzene |
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| EP2877442B1 (en) | 2012-07-27 | 2016-11-16 | Covestro Deutschland AG | Method for producing nitrobenzene by adiabatic nitriding |
| CN103304421A (en) | 2013-06-26 | 2013-09-18 | 南京理工大学 | Method for preparing dinitrotoluene by nitrifying ortho-nitrotoluene with fuming sulfuric acid-nitric acid system |
| DE102017110084B4 (en) | 2017-02-03 | 2019-07-04 | Josef Meissner Gmbh & Co. Kg | Process and plant for the adiabatic nitration of aromatics |
| CN110573491B (en) | 2017-03-07 | 2022-11-11 | 科思创德国股份有限公司 | Method for preparing nitrobenzene |
| WO2020011817A1 (en) | 2018-07-12 | 2020-01-16 | Covestro Deutschland Ag | Method for producing nitrobenzene by adiabatic nitration of benzene |
| EP3956290A1 (en) | 2019-04-17 | 2022-02-23 | Covestro Deutschland AG | Process and device for producing nitrobenzene |
| WO2025045698A1 (en) | 2023-08-25 | 2025-03-06 | Covestro Deutschland Ag | Reactor for the nitration of aromatic compounds, production plant comprising said reactor and nitration process using said reactor |
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| US9260377B2 (en) | 2012-07-27 | 2016-02-16 | Bayer Materialscience Ag | Method for producing nitrobenzene by adiabatic nitriding |
| US9284255B2 (en) | 2012-07-27 | 2016-03-15 | Bayer Materialscience Ag | Method for producing nitrobenzene by adiabatic nitriding |
| US9302978B1 (en) | 2013-04-29 | 2016-04-05 | Covestro Deutschland Ag | Process for the preparation of nitrobenzene by adiabatic nitration |
| US10023524B2 (en) | 2014-06-24 | 2018-07-17 | Covestro Deutschland Ag | Process for producing nitrobenzene |
| US11136285B2 (en) | 2014-06-24 | 2021-10-05 | Covestro Deutschland Ag | Process for producing nitrobenzene |
| WO2020021323A1 (en) | 2018-07-24 | 2020-01-30 | Noram International Limited | Nitration reactor and method |
| US11819819B2 (en) | 2018-07-24 | 2023-11-21 | Noram International Limited | Nitration reactor and method |
| US12180135B2 (en) | 2019-04-17 | 2024-12-31 | Covestro Deutschland Ag | Process for the continuous production of nitrobenzene |
| RU2828259C1 (en) | 2023-10-09 | 2024-10-08 | Акционерное общество "Российский научный центр "Прикладная химия (ГИПХ)" | Method of producing nitrobenzene |
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| ES2524397T3 (en) | 2014-12-09 |
| KR20110091480A (en) | 2011-08-11 |
| DE102010006984A1 (en) | 2011-08-11 |
| EP2354117B1 (en) | 2014-09-24 |
| US20110196177A1 (en) | 2011-08-11 |
| PT2354117E (en) | 2014-11-28 |
| JP5791912B2 (en) | 2015-10-07 |
| EP2354117A1 (en) | 2011-08-10 |
| CN102153476A (en) | 2011-08-17 |
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14815 | https://dmtlsd.tripod.com/Involutionmathematics.html | Involution(mathematics)
Involution(mathematics)
In mathematics, an (anti-)involution, or an involutary function, is a function f that is its own inverse: f(f(x)) = x for all x in the domain of f. General properties Any involution is a bijection. The identity map is a trivial example of an involution. Common examples in mathematics of more detailed involutions include multiplication by −1 in arithmetic, the taking of reciprocals,complementation in set theory and complex conjugation. Other examples include circle inversion, the ROT13 transformation, and the Beaufort polyalphabetic cipher. The number of involutions on a set with n = 0, 1, 2, ... elements is given by the recurrence relation: a0 = a1 = 1; an = an− 1 + (n − 1)an− 2, for n > 1. The first few terms of this sequence are 1, 1, 2, 4, 10, 26, 76, 232 (sequence A000085 in OEIS). Involution throughout the fields of mathematics Euclidean geometry A simple example of an involution of the three-dimensional Euclidean space is reflection against a plane. Performing a reflection twice brings us back where we started. Another is the so-called reflection through the origin; this is an abuse of language, as it is an involution, but not a reflection. These transformations are examples of affine involutions. [edit]Linear algebra In linear algebra, an involution is a linear operator T such that T2 = I. Except for in characteristic 2, such operators are diagonalizable with 1's and -1's on the diagonal. If the operator is orthogonal (an orthogonal involution), it is orthonormally diagonalizable. The transpose of a square matrix is an involution because the transpose of the transpose of a matrix is itself. Involutions are related to idempotents; if 2 is invertible, (in a field of characteristic other than 2), then they are equivalent. Quaternion algebra In a Quaternion algebra, an (anti-)involution is defined by following axioms: if we consider a transformation then a involution is § f(f(x)) = x. An involution is its own inverse § An involution is linear: f(x1 + x2) = f(x1) + f(x2) and f(λx) = λf(x) § f(x1x2) = f(x2)f(x1) An anti-involution does not obey the last axiom but instead § f(x1x2) = f(x1)f(x2) [edit]Ring theory In ring theory, the word involution is customarily taken to mean an antihomomorphism that is its own inverse function. Examples of involutions in common rings: § complex conjugation on the complex plane § multiplication by j in the split-complex numbers § taking the transpose in a matrix ring Group theory In group theory, an element of a group is an involution if it has order 2; i.e. an involution is an element a such that a ≠ e and a2 = e, where e is the identity element. Originally, this definition differed not at all from the first definition above, since members of groups were always bijections from a set into itself, i.e., group was taken to mean permutation group. By the end of the 19th century, group was defined more broadly, and accordingly so was involution. The group of bijections generated by an involution through composition, is isomorphic with cyclic groupC2. A permutation is an involution precisely if it can be written as a product of one or more non-overlapping transpositions. The involutions of a group have a large impact on the group's structure. The study of involutions was instrumental in the classification of finite simple groups. Coxeter groups are groups generated by involutions with the relations determined only by relations given for pairs of the generating involutions. Coxeter groups can be used, among other things, to describe the possible regular polyhedra and their generalizations to higher dimensions. Mathematical logic The operation of complement in Boolean algebras is an involution. Accordingly, negation in classical logic satisfies the law of double negation: ¬¬A is equivalent to A. Generally in non-classical logics, negation which satisfies the law of double negation is called involutive. In algebraic semantics, such a negation is realized as an involution on the algebra of truth values. Examples of logics which have involutive negation are, e.g., Kleene and Bochvar three-valued logics, Łukasiewicz many-valued logic, fuzzy logic IMTL, etc. Involutive negation is sometimes added as an additional connective to logics with non-involutive negation; this is usual e.g. in t-norm fuzzy logics. The involutiveness of negation is an important characterization property for logics and the corresponding varieties of algebras. For instance, involutive negation characterizes Boolean algebras among Heyting algebras. Correspondingly, classical Boolean logic arises by adding the law of double negation to intuitionistic logic. The same relationship holds also betweenMV-algebras and BL-algebras (and so correspondingly between Łukasiewicz logic and fuzzy logic BL), IMTL and MTL, and other pairs of important varieties of algebras (resp. corresponding logics).
In mathematics,
an (anti-)involution, or
an involutary function, is
a function f that
is its own inverse:
f(f(x))
= x for all x in the domain of f.
General
properties
Any involution is a bijection.
The identity map is
a trivial example of an involution. Common examples in mathematics of more
detailed involutions include multiplication by
−1 in arithmetic,
the taking of reciprocals,complementation in set theory and complex conjugation. Other examples include circle inversion,
the ROT13 transformation,
and the Beaufort polyalphabetic
cipher.
The number of involutions on a set with n =
0, 1, 2, ... elements is given by the recurrence
relation:
a0 = a1 =
1;
an = an−
1 + (n − 1)an− 2,
for n > 1.
The first few terms of this sequence are 1, 1, 2, 4, 10, 26, 76, 232 (sequence A000085 in OEIS).
Involution
throughout the fields of mathematics
Euclidean geometry
A simple example of an involution of the
three-dimensional Euclidean
space is reflection against a plane.
Performing a reflection twice brings us back where we started.
Another is the so-called reflection through the origin; this is
an abuse
of language, as it is an involution, but not a reflection.
These transformations are examples of affine involutions.
[edit]Linear algebra
In linear algebra, an involution is a linear
operator T such that T2 = I.
Except for in characteristic 2, such operators are diagonalizable with 1's and
-1's on the diagonal. If the operator is orthogonal (an orthogonal involution), it is
orthonormally diagonalizable.
The transpose of a square matrix is an involution
because the transpose of the transpose of a matrix is itself.
Involutions
are related to idempotents; if 2 is invertible, (in a field
of characteristic other than 2), then they are equivalent.
Quaternion algebra
In a Quaternion algebra, an (anti-)involution is
defined by following axioms: if we consider a transformation then a
involution is
§ f(f(x)) = x. An
involution is its own inverse
§ An involution is linear: f(x1 + x2)
= f(x1) + f(x2) and f(λx)
= λf(x)
§ f(x1x2)
= f(x2)f(x1)
An anti-involution does not obey the last axiom but
instead
§ f(x1x2)
= f(x1)f(x2)
[edit]Ring theory
In ring theory,
the word involution is customarily taken to mean an antihomomorphism that
is its own inverse function. Examples of involutions in common rings:
§ complex
conjugation on the complex plane
§ multiplication by j in the split-complex
numbers
§ taking the transpose in
a matrix ring
Group theory
In group theory,
an element of a group is
an involution if it has order 2;
i.e. an involution is an element a such that a ≠ e and a2 = e,
where e is the identity element.
Originally, this definition differed not at all from the first definition
above, since members of groups were always bijections from a set into itself,
i.e., group was taken to mean permutation
group. By the end of the 19th century, group was
defined more broadly, and accordingly so was involution. The group
of bijections generated by an involution through composition, is isomorphic
with cyclic
groupC2.
A permutation is
an involution precisely if it can be written as a product of one or more
non-overlapping transpositions.
The involutions of a group have a large impact on
the group's structure. The study of involutions was instrumental in the classification of finite simple groups.
Coxeter
groups are groups
generated by involutions with the relations determined only by relations given
for pairs of the generating involutions. Coxeter groups can be used, among
other things, to describe the possible regular polyhedra and
their generalizations
to higher dimensions.
Mathematical logic
The operation of complement in Boolean algebras is an involution.
Accordingly, negation in
classical logic satisfies the law of double negation: ¬¬A is
equivalent to A.
Generally in non-classical logics, negation which
satisfies the law of double negation is called involutive. In
algebraic semantics, such a negation is realized as an involution on the
algebra of truth
values. Examples of logics which have involutive negation
are, e.g., Kleene and Bochvar three-valued
logics, Łukasiewicz
many-valued logic, fuzzy logic IMTL,
etc. Involutive negation is sometimes added as an additional connective to
logics with non-involutive negation; this is usual e.g. in t-norm
fuzzy logics.
The involutiveness of negation is an important
characterization property for logics and the corresponding varieties of algebras. For instance,
involutive negation characterizes Boolean algebras among Heyting algebras.
Correspondingly, classical Boolean logic arises
by adding the law of double negation to intuitionistic
logic. The same relationship holds also betweenMV-algebras and BL-algebras (and so correspondingly
between Łukasiewicz
logic and fuzzy logic BL), IMTL and MTL,
and other pairs of important varieties of algebras (resp. corresponding
logics). |
14816 | https://ciaaw.org/chromium.htm | Atomic Weight of Chromium | Commission on Isotopic Abundances and Atomic Weights
Home
Atomic Weights
Isotopic Abundances
Reference Materials
The Commission
FAQ's
Other elements
Chromium
| Isotope | Atomic mass (Da) | Isotopic abundance (amount fraction) |
---
| 50 Cr | 49.946 041(3) | 0.043 45(13) |
| 52 Cr | 51.940 505(3) | 0.837 89(18) |
| 53 Cr | 52.940 647(3) | 0.095 01(17) |
| 54 Cr | 53.938 878(3) | 0.023 65(7) |
In 1983 the Commission recommended the standard atomic weight of chromium to four decimal places, A r(Cr) = 51.9961(6). Meanwhile, report isotope fractionation of chromium during chromate reduction, resulting in δ 53 Cr SRM979 values in groundwater samples as high as +5.8 ‰ or A r(Cr) = 51.9982, which is outside the current range of uncertainty of the standard atomic weight. Measurements of n(53 Cr)/n(52 Cr) can be expressed as δ 53 Cr values with respect to NIST SRM 979.
SOURCEAtomic weights of the elements: Review 2000 by John R de Laeter et al. Pure Appl. Chem. 2003 (75) 683-800
© IUPAC 2003
CIAAW
Chromium
A r(Cr) = 51.9961(6) since 1983
The name derives from the Greek chroma for "colour", from the many coloured compounds of chromium. It was discovered in 1797 by the French chemist and pharmacist Nicolas-Louis Vauquelin, who also isolated chromium in 1798.
Isotopic reference materials of chromium.
Privacy policy|Impressum|Contact|© CIAAW, 2007-2015 |
14817 | https://www.reddit.com/r/probabilitytheory/comments/6x919z/help_with_probability_notation_1_what_does_the/ | Help with probability notation: 1) what does the overbar that's over the A bar union B bar mean? 2) what do the () indicate? 3) what's the difference between A bar union B bar and (AB bar)? Sorry I know it's basic but I have to start somewhere... : r/probabilitytheory
Skip to main contentHelp with probability notation: 1) what does the overbar that's over the A bar union B bar mean? 2) what do the () indicate? 3) what's the difference between A bar union B bar and (AB bar)? Sorry I know it's basic but I have to start somewhere... : r/probabilitytheory
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Help with probability notation: 1) what does the overbar that's over the A bar union B bar mean? 2) what do the () indicate? 3) what's the difference between A bar union B bar and (AB bar)? Sorry I know it's basic but I have to start somewhere...
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14818 | https://www.quora.com/How-do-you-determine-where-a-quadratic-graph-has-only-positive-real-roots | Something went wrong. Wait a moment and try again.
Quadratic Graphs
Roots (mathematics)
Positive Numbers
Graphing (functions)
Basic Algebra
Quadratic Formula
Real Numbers
Solving Quadratic Equatio...
5
How do you determine where a quadratic graph has only positive real roots?
Gary Russell
Former Professor at University of Iowa (1996–2025)
·
Author has 6K answers and 3.1M answer views
·
3y
Note that a quadratic is shaped like a parabola — facing upwards or downwards. The graph of the quadratic can do one the following:
Crosses the x-axis at two points.
Just touches the x-axis and bounces back.
Never crosses the x-axis.
If scenario 3 occurs, then there are no real roots. Otherwise, both roots are real.
The reason for this is that if f(x) is the quadratic function, then you are looking for a real value x such that f(x) = 0. If x exists, then f(x) must touch the x-axis at x.
I see that you are interested in only positive real roots. First, compute the discriminant of the quadra
Note that a quadratic is shaped like a parabola — facing upwards or downwards. The graph of the quadratic can do one the following:
Crosses the x-axis at two points.
Just touches the x-axis and bounces back.
Never crosses the x-axis.
If scenario 3 occurs, then there are no real roots. Otherwise, both roots are real.
The reason for this is that if f(x) is the quadratic function, then you are looking for a real value x such that f(x) = 0. If x exists, then f(x) must touch the x-axis at x.
UPDATE
I see that you are interested in only positive real roots. First, compute the discriminant of the quadratic (D = b^2 - 4ac). If D is 0 or positive, then both roots are real.
If both are real, then using the Quadratic Formula, you need to also verify that
(-b - sqrt(D))/2a > 0
where sqrt(D) is written as a positive number.
Related questions
How do you tell if a quadratic function is positive or negative?
How do I determine where a quadratic graph has only positive roots using the discriminant (delta)?
How do I determine for which values of x a quadratic (parabola) graph has only positive roots?
What is the proof that any quadratic equation with real roots has at least one positive root?
How do you determine the shape of a quadratic graph?
Sandeep Kumar
Maths Faculty at Rao IIT Academy
·
3y
If any quadratic has only positive real roots then it must satisfy the following conditions:
(I) D greater than or equal to 0 (as it has real roots)
(II) Sum of roots is positive i.e. -b/a > 0
(III) Product of roots is positive i.e. c/a > 0
Gregory Schoenmakers
Engineer and former high school maths teacher.
·
Author has 4.4K answers and 8.5M answer views
·
Updated 3y
The roots of are
For the roots to be real, we require that
For the roots to be positive, we require that and
ie
Assistant
Bot
·
1y
To determine where a quadratic graph has only positive real roots, we can analyze the quadratic function given in the standard form:
where , , and are real coefficients, and .
Steps to Determine Conditions for Positive Real Roots
Discriminant Condition: The roots of the quadratic equation can be found using the quadratic formula:
For the roots to be real, the discriminant must be non-negative:
This ensures that the quadratic has either two distinct real roots or one repeated real root.
Positivity Condition:
To determine where a quadratic graph has only positive real roots, we can analyze the quadratic function given in the standard form:
where , , and are real coefficients, and .
Steps to Determine Conditions for Positive Real Roots
Discriminant Condition: The roots of the quadratic equation can be found using the quadratic formula:
For the roots to be real, the discriminant must be non-negative:
This ensures that the quadratic has either two distinct real roots or one repeated real root.
Positivity Condition: For the roots to be positive, we need to ensure the following:
The sum of the roots must be positive.
The product of the roots must be positive.
Using Vieta's formulas, we have:
The sum of the roots:
The product of the roots:
Summary of Conditions
For the quadratic to have only positive real roots, the following conditions must hold:
The discriminant must be non-negative:
The coefficient must be negative:
The coefficient must be positive:
If all these conditions are satisfied, the quadratic graph will intersect the x-axis at two points (both positive) or at one point (a double root at a positive x-value).
Related questions
What is the nature of a quadratic equation that has no real roots (positive or negative)?
How do you prove that a quadratic equation has two real roots when its discriminant is positive?
What is the condition for a quadratic equation to have three real roots?
How do you show that a quadratic equation has no real roots?
What is the relationship between the roots of a quadratic equation and its graph?
Ak Murthy
Former Retired as Prof. of mathematics 30 yrs of service
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Author has 1.2K answers and 800.7K answer views
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3y
The simplest Test is provided by Mr. Sandeep Kumar. ax² + bx + c = 0 hasb
non- negative real roots iff
(1) D ≥ 0 (2) —( b/a) > 0 (3) c/a > 0
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Mark Duffin
Mathematics Teacher (2003–present)
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3y
Originally Answered: How do I determine if a parabola has only positive real roots?
·
The discriminant is the part of the quadratic formula that is under the square root sign - b^2–4ac
If the discriminant is <0 then there are no roots - it is entirely above or below the x axis
If the discriminant is =0 then the two roots are the same, there is only one root - this is when the turning point (maximum or minimum) sits on the x axis
If the discriminant is >0 then there are two roots- the graph crosses the x axis in two places, both negative values, both positive values or one of each
Dave Neary
B.Sc. in Mathematics, National University of Ireland, Galway (Graduated 1996)
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Author has 2.9K answers and 1.2M answer views
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3y
Originally Answered: How do I determine if a parabola has only positive real roots?
·
A parabola [math]y = ax^2+bx+c[/math] has real roots if [math]b^2-4ac\geq 0[/math]. The equation can be re-written in terms of its roots as [math]y=a(x-r_1)(x-r_2)[/math] - which we can also state as: [math]r_1+r_2 = -\frac{b}{a}[/math], [math]r_1r_2 = \frac{c}{a}[/math].
[math]r_1 \geq 0[/math] and [math]r_2 \geq 0[/math] means that [math]r_1r_2 \geq 0, -(r_1+r_2) \leq 0[/math] - so once you have a positive determinant, you can just check whether [math]\frac{b}{a}\leq0, \frac{c}{a}\geq0[/math] to determine if the roots are positive..
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Annette Sheldon
Former Retired Scientist
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Author has 69 answers and 42.1K answer views
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3y
Related
How do I determine for which values of x a quadratic (parabola) graph has only positive roots?
Sorry, your question appears to have been slightly misstated. A parabolic equation (assuming in the form y = ax^2 + bx +c ) has either 0, 1, or 2 real roots. Calculate the values of (-b+-(b^2–4ac)^0.5)/2a , THOSE ARE the roots. If they are positive, so be it!
I think I answered the question you wanted to have asked.
By the way, if the parabola doesn’t touch the x-axis, then if you flipped it vertically around its vertex the x-axis intercepts would give you info on the COMPLEX roots of the equation. The complex part(s) together add up (so to speak, as +- respectively) to the distance laterally be
Sorry, your question appears to have been slightly misstated. A parabolic equation (assuming in the form y = ax^2 + bx +c ) has either 0, 1, or 2 real roots. Calculate the values of (-b+-(b^2–4ac)^0.5)/2a , THOSE ARE the roots. If they are positive, so be it!
I think I answered the question you wanted to have asked.
By the way, if the parabola doesn’t touch the x-axis, then if you flipped it vertically around its vertex the x-axis intercepts would give you info on the COMPLEX roots of the equation. The complex part(s) together add up (so to speak, as +- respectively) to the distance laterally between the intercepts in units of i; the real part is the x-value of the vertex.
—Cheers, —Mr. d.
Kevin Solari
Author has 7K answers and 1.8M answer views
·
3y
Related
How do I determine for which values of x a quadratic (parabola) graph has only positive roots?
It's not the values of x that determine the roots it's the values of the coefficients.
Typically ax² + bx + c =0
And the solutions in roots are
x = [ -b +/- sqrt(b² - 4ac) ] / 2a
So we need 1) b² >= 4ac (to ensure roots are resl)
And
(-b -sqrt(b² - 4ac))/a> 0 (for positive roots)
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Allen Ries
Math Major University of Alberta
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Author has 25.1K answers and 9.7M answer views
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3y
Related
How do I determine for which values of x a quadratic (parabola) graph has only positive roots?
ax²+bx+c=0 is the general quadratic formula. To check find the positive roots you calculate b²-4ac. If b²-4ac ≥0 then both roots will be positive.
David Vanderschel
PhD in Mathematics & Physics, Rice (Houston neighborhood) (Graduated 1970)
·
Author has 37.6K answers and 50.1M answer views
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3y
Related
How do I determine for which values of x a quadratic (parabola) graph has only positive roots?
A2A: Think about the coefficients you get when you multiply
math(x-q)[/math]
when both
[math]p[/math]
and
[math]q[/math]
are positive.
Satya Parkash Sud
Former Professor at Himachal Pradesh University Shimla (1986–2002)
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Author has 8.1K answers and 27.4M answer views
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4y
Related
How do you show that a quadratic equation has no real roots?
The equation of the form
ax² + b x + c = 0, —————(1),
where a ≠ 0, a, b, c belong to complex numbers is called a quadratic equation. The numbers a, b , c are called the coefficients of the equation.
A root of the quadratic equation (1) is comlex number p such that
a p² + b p + c = 0.
The quantity D = b² — 4 a c is known as the discriminant of the equation (1).
Nature of Roots:
If a, b, c belong to R and a ≠ 0, then the following applies to the roots;
(a) The equation (1) has real and distinct roots if and only if D > 0.
(b) The equation (1) has real and equal roots (also called repeated roots) if D = 0
The equation of the form
ax² + b x + c = 0, —————(1),
where a ≠ 0, a, b, c belong to complex numbers is called a quadratic equation. The numbers a, b , c are called the coefficients of the equation.
A root of the quadratic equation (1) is comlex number p such that
a p² + b p + c = 0.
The quantity D = b² — 4 a c is known as the discriminant of the equation (1).
Nature of Roots:
If a, b, c belong to R and a ≠ 0, then the following applies to the roots;
(a) The equation (1) has real and distinct roots if and only if D > 0.
(b) The equation (1) has real and equal roots (also called repeated roots) if D = 0.
(c) The equation (1) has complex roots with non-zero imaginary parts if and only if D < 0.
(d) p + i q (p, q belong to R, q≠ 0) is a root of (1) if and only if p — i q is a root of (1). This means that complex roots occur in conjugate pairs .
To show that the given quadratic equation does not have any real roots, find its discriminant. Show that D< 0. In case D is negative, that is sufficient to imply that there are no real roots of the quadratic equation.
The god of Mathematics
Math Problem Solver at fiverr.com/marcel_lus
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Author has 77 answers and 640.1K answer views
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2y
Related
How can we find the number of positive and negative real roots for a quadratic function?
Hello, I will be glad to help you out.
We can find the number of positive and negative real roots of a Quadratic by using what is known as Descartes’ rule of signs
The rule states that; the number of positive real roots in any polynomial function f(x) is either equal to the number of sign variations in the function, or less than that by an even number.
Similarly, the number of negative roots in the function follows the same definition, but for 𝑓(-𝑥).
By definition, a quadratic is a type of polynomial.
So for example let’s consider the quadratic function 𝑓(𝑥) = -2x^2 + 1
Counting the number of s
Hello, I will be glad to help you out.
We can find the number of positive and negative real roots of a Quadratic by using what is known as Descartes’ rule of signs
The rule states that; the number of positive real roots in any polynomial function f(x) is either equal to the number of sign variations in the function, or less than that by an even number.
Similarly, the number of negative roots in the function follows the same definition, but for 𝑓(-𝑥).
By definition, a quadratic is a type of polynomial.
So for example let’s consider the quadratic function 𝑓(𝑥) = -2x^2 + 1
Counting the number of sign variations in the function, we have 2.
Therefore, there are either 2, or 2 – 2 = 0 positive zeros in the function.
Now 𝑓(-𝑥) = –2(-x)^2 + 1
= –2x^2 + 1
Here there are 2 sign variations too.
Therefore, there are either 2, or 2 – 2 = 0 negative zeros in the function too.
Related questions
How do you tell if a quadratic function is positive or negative?
How do I determine where a quadratic graph has only positive roots using the discriminant (delta)?
How do I determine for which values of x a quadratic (parabola) graph has only positive roots?
What is the proof that any quadratic equation with real roots has at least one positive root?
How do you determine the shape of a quadratic graph?
What is the nature of a quadratic equation that has no real roots (positive or negative)?
How do you prove that a quadratic equation has two real roots when its discriminant is positive?
What is the condition for a quadratic equation to have three real roots?
How do you show that a quadratic equation has no real roots?
What is the relationship between the roots of a quadratic equation and its graph?
How do you show quadratic equations having common roots?
Why are quadratic graphs always symmetrical?
What do imaginary roots in quadratic equations look like in a graph?
How many real roots does a quadratic equation have if its discriminant is negative? How many real roots does a quadratic equation have if its discriminant is positive?
How many times will the graph of a quadratic equation cross the x-axis if it has three real roots?
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14819 | https://www.youtube.com/watch?v=P_8j1jQSI8w | Solving Modulus Equations and Inequalities
Tulla Maths
8220 subscribers
1403 likes
Description
141438 views
Posted: 29 Nov 2022
How to solve Modulus inequalities and equations. Also know as solving absolute value inequalities. Modulus inequalities revision questions / handout. Absolute Value Equations. Leaving Cert Maths and a levels.
106 comments
Transcript:
in this video we're looking at modulus inequalities we're going to look at four examples the first two are using an equal signs that are a little bit easy in nature and then they get a little bit harder in example three and four just a little bit of revision of what modulus is so these parallel lines are standing for a modulus so if I'm looking for the modulus of say um five it basically turns your value into its absolute value so it's positive value so the modulus of five is five now that could have also been the case the modulus of negative five which basically means you turn your value into a positive value so five so it's usually got to do with maybe areas of triangles you might have seen the modulus or the values for lengths of um coordinate geometry so that's what the modulus is doing for us so if we had another example here the modulus of X is equal to 3 technically the values for X there are negative 3 because the modulus of that is three but I could also use x as positive 3 because the modulus of positive 3 obviously has to be tree so there's two values there for X now the easiest way to find the values for X here would be to square both sides so what I mean by that is if we keep our basic example here of the modulus of X is equal to 3. so in order to find those two values of X the easiest way is to just come along and square both sides because when we Square the X we get x squared and we square root a 3 and we get nine and the square root of nine is plus or minus 3. so that's how we're going to solve these one two three four examples of modulus we're going to square both sides and solve for x basically so let's have a look now at example one so it's asking us to solve the inequality for X an element of the real number so real numbers are positive and negative uh whole or rational number so fractions and decimals so the method of doing this is to square both sides so I'm going to go x minus 2 all to be squared is equal to 4 to be squared multiplying x minus 2 by itself is my next step because the square means you multiply it by itself and 4 squared there will give you 16. so first term by my second bracket and second term by my second bracket is equal to 16. and multiplying in my bracket stand so first term by second term and so on is going to give me x squared minus 2x minus 2X and negative 2 by negative 2 will give me a plus 4 is equal to 16. if I move that 16 over or subtract 16 from both sides I'm getting x squared minus 4X um plus 4 minus 16 is equal to zero so my quadratic is x squared minus 4X minus 12 is equal to 0. so I'm trying to find the values of X so I need to solve my quadratic here so you can use your guide number trial and error using your two brackets or your minus B formula your quadratic formula I'm going to just do the two brackets here I think it should be straightforward enough um so it's equal to zero so the multiples of minus 12 to get to minus 4 are going to be minus six and a plus 2. so there are my factors and when I solve those I get X is equal to 6 and X is equal to minus 2. so therefore my two values for X are positive six and negative two and you can always check these by subbing them back in to your modules so that's example one let's have a look there at example two so solve the following inequality for X an element of r x minus five modulus is equal to the modulus of X plus one so the process once again is to square both sides so it's x minus five to be squared is equal to X plus one two B squared squaring something means you multiplied by itself so x minus five multiplied by x minus five is equal to X plus one times X plus one so multiplying out your bracket so first term by the second bracket and second term by the second bracket on both sides of the equal so it's a little bit Ultra heavy here already so just take your time as you multiply out your brackets so multiplying out the first two you get x squared minus five x minus five x minus five by minus five is a positive 25 and on the right hand side of the equals we get One X Plus One X plus one by one which is one tidying it up a little bit x squared minus 10x plus 25 is equal to x squared plus 2X plus 1. so to solve a quadratic we need to move everything to one side so I'm basically just going to tidy this up a little bit and move everything to the left of my equals so I have x squared minus 10x plus 25 and then I'm moving over the x squared to become a negative x squared negative 2x plus 1 will change to a negative one and that is now going to equal to zero on the right hand side hopefully you can spot here that the x squared minus x squared will cancel out so that's leaving me with minus 10 x minus 2x is minus 12x plus 25 minus 1 is plus 24 is equal to zero I need to solve that linear equation now so let me just rewrite it up here so I'm going to have x's on one side and numbers on the other so subtract 24 from both sides we'll leave him with minus 12x is equal to negative 24 and dividing across by negative 12 will give me a value of x as positive 2. so there's only one value for x on example two and that's X is equal to 2. looking at example tree next solved following inequality for X an element of or so it's now getting into a greater than symbol rather than the equals but we carry these out in the same fashion we Square both sides so it's x minus four all to be squared greater than 3 to be squared squaring something once again means you multiply it by itself so x minus 4 by x minus 4 is greater than 3 trees which is nine so first term by my second bracket second term by my second bracket greater than 9 which is giving me x squared minus four x minus 4X plus 16 is greater than 9. so just tidying up this now a little bit is giving me minus eight X plus 16 and if I subtract the 9 from both sides or move over that nine that's leaving me with negative 9 greater than zero because we can only solve our quadratic equations when it is um greater than or less than or equal to zero so that's my quadratic equation now I'm going to come up and solve my quadratic so again guide number minus B formula two brackets whichever way you want so what we're trying to do first of all now is find the roots of the equation so I'm going to let it equal to zero here just to solve the roots because I'm looking for where it crosses the x-axis so when that's equal to zero so my two brackets x and x and equal to zero and the multiples of seven are seven and one and a minus and a minus because minus by minus gives me plus so the roots are X is equal to seven and X is equal to positive one so what does that mean so if we come back if we come down here just do a little sketch of this and it's crossing the x-axis at number one and number seven and it's a positive quadratic so maybe it looks something like that so we're looking for where it's greater than zero so greater than zero is above the x-axis so it's this portion of my graph here and this portion of my graph here so the values for X here are going to be X greater than seven so I could use seven eight nine ten eleven all the way to Infinity so anything basically in this yellow portion here but I could also go down to anything below one so X less than one because I can include zero minus one minus two minus three all these values here the only values I can't use are anything between 1 and 7. um so that's example three Dawn let's have a look now at example four solved following inequality for X an element of the real numbers so again we have an inequality here modulus on both sides so we need to square both sides um to deal with the positive and negative of that inequality so that's going to give me X Plus 1 on the left hand side and I'm going to square it out and that's going to be greater than uh 2 times X plus 3 all to be squared so what you're doing here is you'll have to square the 2 and then you'll have to square the bracket so we can write that out if it makes it easier so it's X Plus 1 to be squared greater than at 2 to be squared and the X plus 3 to be squared so I'm going to do them separately so on the left hand side that'll give me X Plus 1 times X Plus 1 and on the right hand side two twos are 4 and then I need to multiply out my two brackets so I'm just going to use square brackets here just to keep them separately and I need to times them by each other as well so first term by my second bracket and then second term by my second bracket greater than 4 times the x times the second bracket and then the plus 3 by the X Plus 3. so just take your time with your multiplication here now I'm just going to multiply in my bracket here so uh use the arrows if it helps so that's going to give me X by X is x squared plus 1X Plus 1X Plus 1. greater than 4 times x by X is x squared 3x 3 by X is 3x and 3 by 3 is 9. just timesing in that 4 now into this bracket multiplying it in is going to give me uh greater than or equal to 4 x squared well that 3 plus 3 is 6 and 6 by 4 is 24. X and four nines is 36. and on the left I have my x squared plus 2X Plus 1. tidying that up now I'm going to move everything to the left hand side so uh minus all of this quadratic in other words so that's going to give me x squared plus 2X plus 1 and as I move them over let me just change the colors here so that'll be minus 4x squared minus 24x minus 36 and that will all be greater than zero so grouping terms I have x squared minus 4x squared will give me a negative 3x squared I have a 2X minus a 24 is a minus 22x and then I have my 1 minus 35 which or 36 is minus 35 greater than or equal to zero so make it a little bit easier I'm going to multiply across by negative 1 to change the signs so that's going to give me positive 3x squared positive 22x and a positive 35 but remember when you multiply across or divide by a minus you have to change the direction of your inequality sign so my inequality sign there is flipping so now I'm looking to factorize this so using your minus B formula if it's a little bit easier or you can use your um your two brackets uh either way uh let's see what we get so the factors of 35 that are multiplying to 35 and adding to 22 we could have a 5 and a 7 and positive and positive so that would give me 15 plus 7 is 22. yeah so there are my Solutions so they're my factors solving these now I get 3x is equal to minus seven X is equal to minus seven over three or two and one-thirds and my second root is going to be X is equal to negative five so there are my roots where it crosses the axis so technically there if when I'm solving that I'm going to have that equal to a zero so let's see the question again wants us to solve the inequality so there's a range of values here so I'm just going to sketch this out once again and see how it looks so minus 5 would be down here minus two and a third probably around here so minus seven over three and it's a positive quadratic so just sketching it looks something like this and if I come back up to the question here the quadratic that I was solving was less than zero so I'm looking at this portion here where it's below my x-axis so the values of X below the x axis would be X um greater than minus five and less than minus seven over three so a combined inequality there would be between minus seven over three and negative five so that's my final solution there so that's example for them so that's four examples there of the modulus inequality ranging in difficulty |
14820 | https://www.ias.edu/sites/default/files/math/csdm/13-14/HuangSu.pdf | The minimum number of nonnegative edges in hypergraphs Hao Huang∗ Benny Sudakov † Abstract An r-unform n-vertex hypergraph H is said to have the Manickam-Mikl´ os-Singhi (MMS) property if for every assignment of weights to its vertices with nonnegative sum, the number of edges whose total weight is nonnegative is at least the minimum degree of H. In this paper we show that for n > 10r3, every r-uniform n-vertex hypergraph with equal codegrees has the MMS property, and the bound on n is essentially tight up to a constant factor. This result has two immediate corollaries. First it shows that every set of n > 10k3 real numbers with nonnegative sum has at least n−1 k−1 nonnegative k-sums, verifying the Manickam-Mikl´ os-Singhi conjecture for this range.
More importantly, it implies the vector space Manickam-Mikl´ os-Singhi conjecture which states that for n ≥4k and any weighting on the 1-dimensional subspaces of Fn q with nonnegative sum, the number of nonnegative k-dimensional subspaces is at least n−1 k−1 q. We also discuss two additional generalizations, which can be regarded as analogues of the Erd˝ os-Ko-Rado theorem on k-intersecting families.
1 Introduction Given an r-uniform n-vertex hypergraph H with minimum degree δ(H), suppose every vertex has a weight wi such that w1 + · · · + wn ≥0. How many nonnegative edges must H have? An edge of H is nonnegative if the sum of the weights on its vertices is ≥0. Let e+(H) be the number of such edges.
By assigning weight n−1 to the vertex with minimum degree, and −1 to the remaining vertices, it is easy to see that the number of nonnegative edges can be at most δ(H). It is a very natural question to determine when this easy upper bound is tight, which leads us to the following definition.
Definition 1.1. A hypergraph H with minimum degree δ(H) has the MMS property if for every weighting w : V (H) →R satisfying P x∈v(H) w(x) ≥0, the number of nonnegative edges is at least δ(H).
The question, which hypergraphs have MMS property, was motivated by two old conjectures of Manickam, Mikl´ os, and Singhi [8, 9], both of which were raised in their study of so-called first distribution invariant of certain association schemes.
Conjecture 1.2. Suppose n ≥4k, and we have n real numbers w1, · · · , wn such that w1+· · ·+wn ≥0, then there are at least n−1 k−1 subsets A of size k satisfying P wi∈A wi ≥0.
∗Institute for Advanced Study, Princeton, NJ 08540 and DIMACS at Rutgers University.
Email: huanghao@math.ias.edu. Research supported in part by NSF grant DMS-1128155.
†Department of Mathematics, ETH, 8092 Zurich, Switzerland and Department of Mathematics, UCLA, Los Angeles, CA 90095. Email: bsudakov@math.ucla.edu. Research supported in part by NSF grant DMS-1101185 and by a USA-Israel BSF grant.
1 The second conjecture is an analogue of Conjecture 1.2 for vector spaces. Let V be a n-dimensional vector space over a finite field Fq. Denote by V k the family of k-dimensional subspaces of V , and the q-Gaussian binomial coefficient n k q is defined as Q 0≤i 33k2.
Later, Frankl gave a shorter proof for a cubic range n ≥3 2k3. A linear bound n ≥1046k was obtained by Pokrovskiy . He reduced the conjecture to finding a k-uniform hypergraph on n vertices satisfying the MMS property (similar techniques were also employed earlier in ). The second conjecture, Conjecture 1.3, was very recently proved by Chowdhury, Sarkis, and Shahriari .
We observe that both conjectures can be reduced to proving that certain hypergraph has the MMS property. For the first conjecture, simply let the hypergraph H1 be the complete k-uniform hypergraph on n vertices. For the second conjecture, one can take the k 1 q-uniform hypergraph H2 with vertex set V 1 and let edges correspond to k-dimensional subspaces. Both hypergraphs are regular, and moreover the codegree of every pair of vertices is the same. It is tempting to conjecture that all such hypergraphs satisfy the MMS property. The requirement that all the codegrees are equal may not be dropped. For instance, the tight Hamiltonian cycle (the edges are consecutive r-tuples modulo n) when n ≡1 (mod r) is not MMS. This can be seen by choosing the weights w(xr + 1) = n for x = 0, · · · , n/r and all the other weights to be −n+r r−1 , which results in only r −1 nonnegative edges, as opposed to the fact that the degree is r. Our main theorem indeed confirms that equal codegrees imply the MMS property.
Theorem 1.4. Let H be an r-uniform n-vertex hypergraph with n > 10r3 and all the codegrees equal to λ. Then for every weighting w : V (H) →R with P v wv ≥0, we have e+(H) ≥δ(H). Moreover in the case of equality, all nonnegative edges form a star, i.e., contain a fixed vertex of H.
The lower bound on n in this theorem is tight up to a constant factor. Our result immediately implies two corollaries. First it verifies Conjecture 1.2 for a weaker range n ≥Ω(k3). Moreover it also provides a proof of Conjecture 1.3.
As mentioned earlier, there are some subtle connections between Manickam-Mikl´ os-Singhi con-jecture and the Erd˝ os-Ko-Rado theorem on intersecting families. In , Erd˝ os, Ko and Rado also initiated the study of k-intersecting families (any two subsets have at least k common elements).
2 They show that for k < t, there exists an integer n0(k, t) such that for all n ≥n0(k, t) the largest k-intersecting family of t-sets are the k-stars, which are of size n−k t−k . This result is equivalent to saying that in the t k -uniform hypergraph H whose vertices are k-subsets of [n] and edges corre-spond to t-subsets of [n], the maximum intersecting sub-hypergraph has size n−k t−k . The following theorem says that for large n, this hypergraph has the MMS property. Note that this is not implied by Theorem 1.4, because the codegree of two vertices (as k-subsets) depends on the size of their intersection.
Theorem 1.5. Let k, t be positive integers with t > k, n > Ct3k+3 for sufficiently large C and let {wX}X∈([n] k ) be a weight assignment with P X∈([n] k ) wX ≥0. Then there are always at least n−k t−k subsets T of size t such that P X⊂T wX ≥0.
This result can be regarded as an analogue of the k-intersecting version of the Erd˝ os-Ko-Rado theorem. Moreover, the Manickam-Mikl´ os-Singhi conjecture is a special case of this theorem corre-sponding to k = 1 and t = r. Using a similar proof one can also obtain a generalization of the vector space version of Manickam-Mikl´ os-Singhi conjecture.
Theorem 1.6. Let k, t be positive integers with t > k, n > Ck(t−k) for sufficiently large C, V be the n-dimensional vector space over Fq and let {wX}X∈[V k] be a weight assignment with P X∈[V k] wX ≥0.
Then there are always at least n−k t−k q t-dimensional subspaces T such that P X∈[V k],X⊂T wX ≥0.
The rest of the paper is organized as follows. In Section 2 we prove Theorem 1.4 and deduce Conjecture 1.3 as a corollary. In Section 3 we will have two constructions, showing that the n > Ω(k3) bound for Theorem 1.4 is essentially tight. The ideas presented in Section 2 are not enough to prove Theorem 1.5. Hence, in Section 4 we develop more sophisticated techniques to prove this theorem.
We also sketch the lemmas needed to obtain Theorem 1.6 and leave the proof details to the appendix.
The final section contains some open problems and further research directions.
2 Equal codegrees and MMS property In this section we prove Theorem 1.4. Without loss of generality, we may assume that V (H) = [n], and the weights are 1 = w1 ≥w2 ≥· · · ≥wn, such that Pn i=1 wi = 0. Suppose the number of edges in H is e. By double counting, we have that H is d-regular with d = n−1 r−1 λ and that dn = re. Consider the 2r-th largest weight w2r, we will verify Theorem 1.4 for the following three cases respectively: (i) w2r ≤ 1 2r2 ; (ii) w2r ≥1 2r; and (iii) 1 2r2 ≤w2r ≤1 2r.
Lemma 2.1. If w2r ≤ 1 2r2 , then e+(H) ≥d.
Proof. First we show that among the d edges containing w1, the number of negative edges is at most 5d 6r. Denote the negative edges by e1, · · · , em and the nonnegative edges by em+1, · · · , ed. By the definition of a negative edge, for every 1 ≤i ≤m we have X j∈ei{1} wj < −w1 = −1.
3 Summing these inequalities, we get m X i=1 X j∈ei{1} wj < −m.
Now we consider the sum Pd i=m+1 P j∈ei{1} wj and rewrite it as P j αjwj. The sum of coefficients αj’s is equal to (d −m)(r −1). Note that in this sum w2, · · · , w2r each appears at most λ times (their codegree with {1}) and they are bounded by 1, so in total they contribute no more than 2rλ.
The remaining variables w2r+1, · · · , wn contribute less than (d −m)(r −1)w2r < d−m 2r . Combining these three estimates, we obtain that X 1∈e X j∈e{1} wj < −m + 2rλ + d −m 2r .
By double counting, the left hand side is equal to λ(w2 + · · · + wn) = −λ. Comparing these two quantities and doing simple calculations we get m < 2rλ + d 2r + 1 < d 5r + d 2r + 1 < 5d 6r .
Here we used that n > 10r3 and λ = r−1 n−1d < d/(10r2). Therefore we may assume there are at least (1 −5 6r)d nonnegative edges through w1.
If every edge through w1 is positive, then we are already done; so we assume that there exists a negative edge e through w1. We claim that in this case there are many nonnegative edges through w2 which are disjoint from e. Consider the set S of r-tuples consisting of all the edges through w2 which are disjoint from e. Since each of the r vertices of e has at most λ common neighbors with w2, we have |S| ≥d −rλ. Denote by S−the set of negative edges in S and consider the sum P f∈S− P j∈f{2} wj. Obviously it is at most −w2|S−|. Rewrite this sum as λ·(P j̸∈e∪{2} αjwj). Since all codegrees are λ, αj ∈[0, 1] and P j̸∈e∪{2} αj = (r−1)|S−|/λ, which implies that Pn j̸∈e∪{2}(1−αj) = (n −r −1) −(r −1)|S−|/λ. Therefore −w2 < X j̸∈e∪{2} wj = X j̸∈e∪{2} αjwj + X j̸∈e∪{2} (1 −αj)wj < −|S−| λ · w2 + (n −r −1 −(r −1)|S−|/λ) · w2, The first inequality uses that the sum of all the weights is zero and that e is a negative edge, so P j∈e wj < 0. To see the second inequality, just observe that wj ≤w2 for every j ̸∈e ∪{2}.
By simplifying the last inequality we get |S−| λ < n−r r . Therefore the number of nonnegative edges containing w2 that are disjoint from e is at least |S| −|S−| > d −rλ −n −r r λ = n −(r3 −2r2 + 2r) r(n −1) d, which is greater than 5d 6r if n > 10r3. These nonnegative edges, together with the (1−5 6r)d nonnegative edges through w1, already give more than d nonnegative edges.
4 Lemma 2.2. If w2r ≥1 2r, then e+(H) ≥d.
Proof. First we claim for any 1 ≤i ≤2r, there are at least 3 5rd nonnegative edges containing wi. Let Si be the set of negative edges containing wi, then for any edge e ∈Si, P j∈e{i} wj < −wi. Summing up these inequalities, we have X e∈Si X j∈e{i} wj < −|Si|wi.
Like the previous case, suppose the left hand side can be rewritten as λ · P j̸=i αjwj, then αj ∈[0, 1], and P j̸=i αj = (r−1)|Si|/λ, which implies P j̸=i(1−αj) = n−1−(r−1)|Si|/λ. Since P j̸=i wj = −wi, we have −wi = X j̸=i αjwj + X j̸=i (1 −αj)wj < −|Si| λ wi + X 1≤j≤2r wj + wi · X j>2r (1 −αj) ≤−|Si| λ wi + 2rw1 + n −1 −(r −1)|Si| λ wi Substituting λ = r−1 n−1d and wi ≥w2r ≥1 2r, gives |Si| ≤(n + 4r2)λ r = (r −1)(n + 4r2) r(n −1) d, which is less than (1 −3 5r)d when n > 10r3. So there are at least 3 5rd nonnegative edges containing wi. Note that for 1 ≤i < j ≤2r, wi and wj are simultaneously contained in at most λ edges.
Therefore the total number of nonnegative edges is at least 2r · 3 5rd − 2r 2 · λ ≥6 5d −2r2(r −1) n −1 d.
When n > 10r3, this gives more than d nonnegative edges.
Lemma 2.3. If 1 2r2 ≤w2r ≤1 2r, then e+(H) ≥d.
Proof. Let t be the index such that wt ≥2rw2r and wt+1 < 2rw2r. Since w1 = 1 ≥2rw2r such t exists and is between 1 and 2r. For arbitrary 1 ≤i ≤t, let Ti be the set of negative edges containing wi. Similarly as before, we assume X e∈Ti X j∈e{i} wj = λ · X j̸=i αjwj.
Then P j̸=i αjwj < −wi|Ti|/λ, and P j̸=i(1 −αj) = n −1 −(r −1)|Ti|/λ. We also have −wi = X j̸=i wj = X j̸=i αjwj + X j̸=i (1 −αj)wj < −|Ti| λ wi + X 1≤j≤t (1 −αj)wj + X t2r (1 −αj)wj ≤−|Ti| λ wi + t + (2r −t)wt + (n −1 −(r −1)|Ti|/λ)w2r ≤−|Ti| λ wi + t + (2r −t)wt + (n −1 −(r −1)|Ti|/λ)wt 2r .
5 Suppose |Ti|/λ ≥1. Since wi ≥wt, we then have |Ti| λ −1 −(2r −t) −n −1 −(r −1)|Ti|/λ 2r wt ≤t ≤t · 2r2w2r ≤rtwt.
Using that n > 10r3, t ≤2r and λ = r−1 n−1d and rearranging the last inequality gives |Ti| ≤ 2r 3r −1 n −1 2r + (r −1)t + 2r + 1 λ ≤ 2r 3r −1 n −1 2r + 2r2 + 1 r −1 n −1d < 7 15d.
Since λ < d/(10r2), the above inequality also holds when |Ti| < λ. Therefore there are at least 8 15d nonnegative edges through wi. This completes the proof for t ≥2, as the number of nonnegative edges through w1 and w2 is already at least 16 15d −λ > d (when n > 10r3).
If t = 1, it means that w2 < 2rw2r. For 2 ≤i ≤2r, as before denote by Ui the set of all the negative edges through wi. Then similarly we define X f∈Ui X j∈f{i} wj = λ X j̸=i αjwj.
Note that αi ∈[0, 1], P j̸=i αjwj < −wi|Ui|/λ and P j̸=i(1 −αj) = n −1 −(r −1)|Ui|/λ. So −wi = X j̸=i wj = X j̸=i αjwj + X j̸=i (1 −αj)wj ≤−|Ui| λ wi + 2 X i=1 rwi + X j>2r (1 −αj)wj ≤−|Ui| λ wi + 1 + 2rw2 + ((n −1 −(r −1)|Ui|)/λ)wi We have r|Ui| λ −n wi ≤1 + 2rw2 ≤2r2w2r + 2r · 2rw2r = 6r2w2r ≤6r2wi.
Therefore when n > 10r3, |Ui| ≤6r2 + n r · r −1 n −1 · d ≤ 1 −2 5r d.
Hence there are at least 2 5rd nonnegative edges through every wi when 2 ≤i ≤2r, together with the 8 15d nonnegative edges through w1. Thus the total number of nonnegative edges is at least 8 15d + 2 5rd(2r −1) − 2r 2 λ = 4 3 −2 5r −2r3 −3r2 + r n −1 d > d, where we used that λ = r−1 n−1d and n > 10r3.
Combining Lemma 2.1, Lemma 2.2 and Lemma 2.3 we show that e+(H) ≥d. From the proofs it is not hard to see that when n > 10r3 the only way to achieve the inequality is when the nonnegative edges form a star, i.e. contain a fixed vertex of H. This concludes the proof of Theorem 1.4.
□ Next we use Theorem 1.4 to prove the vector space analogue of Manickam-Mikl´ os-Singhi conjec-ture.
6 Proof of Conjecture 1.3: Let H be the hypergraph such that the vertex set V (H) consists of all the 1-dimensional subspaces of V = Fn q . Obviously the number of vertices is equal to n 1 q. Every k-dimensional subspace of V defines an edge of H which contains exactly k 1 q vertices. Therefore H is an r-uniform hypergraph on n′ vertices with r = k 1 q and n′ = n 1 q. Since every two 1-dimensional subspaces span a unique 2-dimensional subspace, so the codegree of any two vertices in H is equal to n−2 k−2 q. Applying Theorem 1.4, as long as n′ > 10r3, the minimum number of nonnegative edges in H is at least equal to its degree, which is equal to n−1 k−1 q. Actually the condition that n′ > 10r3 is equivalent to qn −1 q −1 > 10 qk −1 q −1 3 .
Since n ≥4k, we have qn −1 ≥q4k −1. Moreover (q4k −1)/(qk −1)3 = (q3k + q2k + qk + 1)/(q2k − 2qk + 1) ≥qk. From k ≥2, we also have (q −1)2qk ≥q2(q −1)2 > 10 if q ≥3. Therefore qn −1 q −1 ≥(qk −1)3qk q −1 > 10 qk −1 q −1 3 .
For q = 2, it is not hard to verify that the inequality is still satisfied when k ≥3. The only remaining case is when (q, k) = (2, 2). Again it is easy to check that the inequality holds when n ≥9. The case n = 8 was resolved by Manickam and Singhi , who proved their conjecture when k divides n.
Remark. The statement of Conjecture 1.3 is known to be false only for n < 2k. Hence, it would be interesting to determine the minimal n = n(k) which implies this conjecture. Note that Theorem 1.4 can be used to prove the assertion of Conjecture 1.3 also for n < 4k. For example it shows that this conjecture holds for n ≥3k and q ≥5, n ≥3k + 1 and q ≥3 or n ≥3k + 2 and all q. For large q the proof will work already starting with n = 3k −1.
3 The tightness of Theorem 1.4 In the previous section, we show that for every r-uniform n-vertex hypergraph with equal codegrees and n > 10r3, the minimum number of nonnegative edges is always achieved by the stars. Here we discuss the tightness of this result. As a warm-up example, recall that a finite projective plane has N2 + N + 1 points and N2 + N + 1 lines such that every line contains N + 1 points. Moreover every two points determine a unique line, and every two lines intersect at a unique point. If we regard points as vertices and lines as edges, this naturally corresponds to a (N +1)-uniform (N +1)-regular hypergraph with all codegrees equal to 1. Let us assign weights 1 to the N + 1 points on a fixed line l, and weights −N+1 N2 to the other points. Obviously the sum is nonnegative. On the other hand every line other than l contains at most one point with positive weight, thus its sum of weight is at most 1 −N · N+1 N2 < 0. Therefore there is only one nonnegative edge. This already gives us a hypergraph with n ∼r2 that is not MMS.
The next theorem provides an example of a hypergraph with n ∼r3 for which there is a con-figuration of edges, different from a star, that also achieves the minimum number of nonnegative edges.
7 Theorem 3.1. For infinitely many r there is an r-uniform hypergraph H with equal codegrees on (r3 −2r2 + 2r) vertices, and a weighting w : V (H) →R with nonnegative sum, such that there are δ(H) nonnegative edges that do not form a star.
Proof. Let r = q + 1, where q is a prime power. Denote by Fq the finite field with q elements. Define a hypergraph H with the vertex set V (H) consisting of points from the 3-dimensional projective space PG(3, Fq). Here PG(n, Fq) = (Fn+1 q {0})/ ∼, with the equivalence relation (x0, · · · , xn) ∼ (σx0, · · · , σxn), where σ is an arbitrary number from Fq.
It is easy to see that n = |V (H)| = q3 + q2 + q + 1 = r3 −2r2 + 2r. Every 1-dimensional subspace of PG(3, Fq) defines an edge of H with q + 1 = r elements. It is not hard to check that H is d-regular for d = q2 + q + 1, and every pair of vertices has codegree 1.
Now we assign the weights to V (H) in the following way. Let S be the set of points of a 2-dimensional projective subspace of V (H), then |S| = q2 + q +1. Every vertex from S receives weight 1, and every vertex outside S has weight −q2+q+1 q3 , such that the total weight is zero. Note that every edge has size q + 1, so if it contains at most one vertex from S, its total weight is at most 1 −q · q2+q+1 q3 < 0. Therefore every nonnegative edge must contain at least two vertices from S.
Since S is a subspace, the lines containing 2 points from S are completely contained in S. There are precisely q2 + q + 1 = d lines in S (these are all the nonnegative edges in H) and they do not form a star.
Finally, we give an example which shows that one might find hypergraphs with n ∼r3 and weights such that the number of nonnegative edges is strictly smaller than the vertex degree.
Theorem 3.2. If q and q + 1 are both prime powers, then there exists a (q + 1)-uniform (q + 1)2-regular hypergraph H on (q3 + 2q2 + q + 1) vertices with all codegrees equal to 1, and an assignment of weights with nonnegative sum such that there are strictly less than (q + 1)2 nonnegative edges in H. In particular if there are infinitely many Mersenne primes, then we obtain infinitely many such hypergraphs.
Proof. Let V (H) = V1 ∪V2, such that |V1| = q2 + q + 1, and |V2| = q2(q + 1). We first take H1 to be the projective plane PG(2, Fq) on V1 with edges corresponding to the projective lines. In other words H1 is a (q + 1)-uniform hypergraph with degree q + 1 and codegree 1. The hypergraph H2 consists of some (q + 1)-tuples that intersect V1 in exactly one vertex and intersect V2 in q vertices, such that e(H2) = (q2 + q + 1)(q2 + q). H3 is a (q + 1)-uniform hypergraph on V2 with q3 edges. We will carefully define the edges of H2 and H3 soon.
We hope H2 and H3 to satisfy the following properties: (i) for every pair of vertices u ∈V1 and v ∈V2, their codegree in H2 is equal to 1; (ii) note that every edge in H2 naturally induces a clique of size q in V2 2 ; while every edge in H3 induces a clique of size q + 1 in V2 2 . We hope these cliques form an edge partition of the complete graph K|V2| = Kq2(q+1). It is not hard to see that if (i), (ii) are both satisfied, then the hypergraph H = H1 ∪H2 ∪H3 has codegree δ = 1. And H is a regular hypergraph with degrees equal to δ · n −1 r −1 = (q3 + 2q2 + q + 1) −1 (q + 1) −1 = (q + 1)2.
Now we assign weights to V (H), such that every vertex in V2 receives a weight −1, while every vertex in V1 receives a weight q2(q+1) q2+q+1, so the total weight is zero. If an edge is nonnegative, it must 8 contain at least two vertices from V1, since q2(q+1) q2+q+1 + (−1) · q < 0. Such edge can only come from H1.
However we have e(H1) = q2 + q + 1, which is strictly smaller than the degree (q + 1)2. Therefore what remains is to show the existence of H2 and H3 satisfying (i), (ii). In other words, we need to find a clique partition in (ii) with Kq2(q+1) = q3 · Kq+1 ∪(q2 + q + 1)(q2 + q) · Kq.
Moreover, condition (i) requires that the family of Kq’s can be partitioned into Kq-factors.
A natural idea is to partition [q2(q + 1)] = S1 ∪· · · ∪Sq with |Si| = q(q + 1). Observe that the projective plane PG(2, Fq) defines a clique partition Kq2+q+1 = (q2 + q + 1)Kq+1. By removing one vertex from it, we obtain a partition Kq2+q = (q + 1) · Kq ∪q2 · Kq+1. By doing this for every Si, we get the q3 copies of Kq+1 we want, and (q + 1)q copies of Kq, which clearly form a Kq-factor, since the (q + 1) copies of Kq from each Si are pairwise disjoint. We still need to find an edge partition of the balanced complete q-partite graph Kq2+q,··· ,q2+q into (q2 + q)2 copies of Kq, so that they also can be grouped into q2 + q disjoint Kq-factors.
Suppose we know that q + 1 is also a prime power. Label the vertices in Kq2+q,··· ,q2+q by (x, y, z) where x ∈Fq, y ∈Fq, and z ∈Fq+1. Two vertices (x, y, z) and (x′, y′, z′) are adjacent iffx ̸= x′. Let u and v be the primitive elements of Fq and Fq+1 respectively. Now we define (q2+q)2 cliques Ci,j,k,l’s for i, k ∈Fq and j, l ∈Fq+1. The clique Ci,j,k,l consists of vertices in the form of (ut, i + kut, j + lvt) for t = 0, 1, · · · , q −1. Suppose (x, y, z) and (x′, y′, z′) with x ̸= x′ are both contained in the clique Ci,j,k,l, denote by logu x the unique number t in {0, 1, · · · , q −1} such that ut = x, then we have i + kx = y i + kx′ = y′ j + lvlogu x = z j + lvlogu x′ = z′.
Since x ̸= x′, the first two equations uniquely determine i and k. Moreover, logu x and logu x′ are different numbers between 0 and q−1, so vlogu x ̸= vlogu x′, and thus j, l are also uniquely determined.
Therefore {Ci,j,k,l} forms a Kq-partition of the edges of Kq2+q,··· ,q2+q, and it is not hard to see that they can be partitioned into Kq-factors by fixing k and l.
By the above discussions, if we have both of q and q + 1 to be powers of prime, in particular when q = 2n −1 is a Mersenne prime, then we can explicitly construct the hypergraph.
4 Two additional generalizations In the next two subsections we discuss generalizations of the two Manickam-Mikl´ os-Singhi conjectures and prove Theorem 1.5 and Theorem 1.6. For k = 1 they follow from Theorem 1.4, thus we can assume that k ≥2. In that case, as we mentioned earlier in the introduction, these two theorems are not direct consequences of Theorem 1.4 because the codegrees in the corresponding hypergraphs are not equal.
4.1 Generalization of MMS In this subsection we will prove Theorem 1.5. This requires some new ideas and techniques since direct adaptation of the proof of Theorem 1.4 does not work.
Indeed, it is easy to construct a 9 weighting such that there is no nonnegative edge through the vertex (a k-set) of maximal weight.
For example say k = 2, one can take w{1,2} = 1, the weights of all the 2n −4 pairs containing 1 or 2 to be −n−3 10 , and the rest to have weights roughly 2 5. For sufficiently large n, no t-set containing {1, 2} has nonnegative total weights.
First we prove a simple lemma from linear algebra.
Lemma 4.1. Suppose the s × s lower triangular matrix β = {βi,j} satisfies that βi,i > 0 and for every j < k ≤i, 0 ≤βi,j ≤βi,k. Then for a given vector ⃗ b = (b1, · · · , bs) such that b1 ≥· · · ≥bs ≥0, the equation ⃗ b = ⃗ γ · β has a unique solution ⃗ γ = (γ1, · · · , γs) and moreover 0 ≤γi ≤bi/βi,i.
Proof. The existence and uniqueness of ⃗ γ follow from the fact that β is invertible. Next we inductively prove 0 ≤γi ≤bi/βi,i. We start from γs, from the equation we know bs = γsβs,s. So γs = bs/βs,s and the inductive hypothesis is true. Suppose 0 ≤γi ≤bi/βi,i for every i > k. Now from the linear equation, we have bk = βk,kγk + βk+1,kγk+1 · · · + βs,kγs. Since γi, i > k and βi,j are nonnegative, we have γk ≤bk/βk,k. Note that βi,j is increasing in j, so for every k+1 ≤i ≤s, βi,k ≤βi,k+1. Therefore bk ≤βk,kγk + Ps i=k+1 βi,k+1γi = βk,kγk + bk+1. Since 0 ≤bk+1 ≤bk, we know that γk ≥0.
Without loss of generality, we may assume that P X⊂([n] k ) wX = 0; and w{1,··· ,k}, or alternatively written as w[k], has the largest positive weight. We may let w[k] = 1, then wX ≤1 for every k-set X.
Throughout this section, we also assume that n > Ct3k+3, here C is some sufficiently large constant.
The next lemma shows that if the sum of weights of certain edges is very negative, then we already have enough nonnegative edges.
Lemma 4.2. If for some subset |L| = k, X L⊂Y,|Y |=t X L̸=X⊂Y wX ≤− 1 13t2k n −k t −k , and P X̸=L wX ≥−1, then there are more than n−k t−k nonnegative edges in H.
Proof. We may rewrite the left hand side of the inequality as n −2k t −2k X |X∩L|=0 wX + n −2k + 1 t −2k + 1 X |X∩L|=1 wX + · · · + n −k −1 t −k −1 X |X∩L|=k−1 wX = n −k −1 t −k −1 X |X∩L|≤k−1 wX − k−2 X j=0 bj · X |X∩L|=j wX .
Here we let bj = n−k+1 t−k+1 − n−2k+j t−2k+j . Note that P |X∩L|≤k−1 wX = P X̸=L wX ≥−1. Since n > Ct3k+3, this implies k−2 X j=0 bj · X |X∩L|=j wX ≥ 1 13t2k n −k t −k − n −k −1 t −k −1 ≥ 1 14t2k n −k t −k .
(1) For a fixed integer 0 ≤y ≤k −1, denote by Dy the number of nonnegative t-sets Z with |Z ∩L| = y. If Dy > n−k t−k then we are done. Otherwise assume Dy ≤ n−k t−k for every y. We estimate the following sum: X |Z∩L|=y,|Z|=t X X⊂Z wX.
10 Since every nonnegative t-set contributes to the sum at most t k , it is at most t k Dy ≤ t k n−k t−k . By double counting, the above sum also equals Py j=0(βy,j · P |X∩L|=j wX), where βy,j = k−j y−j n−2k+j t−k−y+j , note that βy,j = 0 when j < k + y −t. When j ≥k + y −t, since n ≫t, for fixed y, βy,j is increasing in j. Also note that bj is decreasing in j. Let ⃗ γ = (γ0, · · · , γk−2) be the unique solution of the system of equations ⃗ b = ⃗ γ · β, then by Lemma 4.1 k−2 X j=0 bj · X |X∩L|=j wX = k−2 X j=0 k−2 X y=j βy,jγy X |X∩L|=j wX = k−2 X y=0 γy · y X j=0 βy,j · X |X∩L|=j wX ≤ t k n −k t −k k−2 X y=0 γy ≤ t k n −k t −k k−2 X y=0 by βy,y .
Since by/βy,y = n−k−1 t−k−1 − n−2k+y t−2k+y / n−2k+y t−k ≤ n−k−1 t−k−1 / n−2k t−k . We have k−2 X j=0 bj · X |X∩L|=j wX ≤ t k n −k t −k · (k −1) · n−k−1 t−k−1 n−2k t−k ≤tk+1 n n −k t −k .
For n > Ct3k+3 this contradicts (1).
We now assume that the tk-th largest weight in H is wP and consider several cases.
Lemma 4.3. If wP > 1 t2k , there are more than n−k t−k nonnegative edges in the hypergraph H.
Proof. We will show that every vertex whose weight is larger than wP is contained in at least 3 2tk n−k t−k nonnegative edges, otherwise there are already > n−k t−k nonnegative edges. For simplicity we just need to prove this statement for wP itself. Suppose there are S negative edges containing wP , which are denoted by e1, · · · , eS (as t-subsets). And eS+1, · · · , e(n−k t−k) are the other (thus nonnnegative) edges containing wP . We have (n−k t−k) X i=1 X P̸=X⊂ei wX = S X i=1 X P̸=X⊂ei wX + (n−k t−k) X i=S+1 X P̸=X⊂ei wX ≤−wP · S + wP · n −k t −k −S · t k −1 + tk · n −k −1 t −k −1 (2) Here we used that there are at most tk sets X whose weight is larger than wP (but always ≤1), and the number of times every such set appears in the sum is at most n−k−1 t−k−1 . If S ≥(1 − 3 2tk ) n−k t−k , then the above expression is at most − n −k t −k 1 −3 2tk wP −3 2tk · t k · wP −tk+1 n , which can be further bounded by − n −k t −k 1 −3 2tk − 3 2 · k!
wP −tk+1 n < − n −k t −k · 1 13wP ≤− n −k t −k · 1 13t2k .
11 The last inequality uses that n > Ct3k+3, t > k ≥2 and therefore 1− 3 2tk − 3 2·k! ≥1− 3 2·32 − 3 2·2! = 1 12.
Since we also have P X̸=P wX = −wP ≥−1, Lemma 4.2 for L = P immediately gives > n−k t−k nonnegative edges.
Therefore we can assume that for the tk sets with largest weights, the number of nonnegative edges containing each such set is at least 3 2tk n−k t−k . Using the union bound, the number of nonnegative edges is at least tk · 3 2tk n −k t −k − tk 2 n −k −1 t −k −1 , which is also larger than n−k t−k .
The next lemma covers the case when wP is smaller than 1 t2k , and there are significant number of negative edges containing {1, · · · , k}.
Lemma 4.4. If the tk-th largest weight wP is smaller than 1/t2k, and there are less than (1−1 tk ) n−k t−k nonnegative edges containing {1, · · · , k}, then there are at least n−k t−k nonnegative edges in H.
Proof. We consider all the t-tuples containing {1, · · · , k}, similarly as before suppose there are S ≥ 1 tk n−k t−k negative edges e1, · · · , eS and nonnegative edges eS+1, · · · , e(n−k t−k), we get X [k]⊂Z,|Z|=t X X⊂Z,X̸=[k] wX = S X i=1 X X⊂ei,X̸=[k] wX + (n−k t−k) X i=S+1 X X⊂ei,X̸=[k] wX ≤−S + 1 t2k · n −k t −k −S · t k −1 + tk · n −k −1 t −k −1 ≤− S − 1 k! · tk n −k t −k −S −tk n −k −1 t −k −1 The first inequality is by bounding the tk largest weights in the second sum by 1 and the rest by 1 t2k . It also uses the fact that two sets are contained in at most n−k−1 t−k−1 edges. Since S ≥ 1 tk n−k t−k , we have X [k]⊂Z,|Z|=t X X⊂Z,X̸=[k] wX ≤− 1 2tk n −k t −k −tk n −k −1 t −k −1 ≤− 1 2tk −tk+1 n n −k t −k .
For large n the right hand side is at most −1 3tk n−k t−k . We also have P X̸=[k] wX = −w[k] = −1. Now we once again can apply Lemma 4.2 for L = {1, · · · , k} to show the existence of > n−k t−k nonnegative edges.
It remains to prove the case when {1, · · · , k} is contained in at least (1 −1 tk ) n−k t−k nonnegative edges.
Lemma 4.5. If {1, · · · , k} is contained in at least (1 −1 tk ) n−k t−k nonnegative edges, then there are at least n−k t−k nonnegative edges in H.
12 Proof. Note that if every edge containing {1, · · · , k} is nonnegative, this already gives n−k t−k nonneg-ative edges and the lemma is proved. So we may assume that there is a negative edge f (as t-subset) through {1, · · · , k} with P X⊂f wX < 0. Suppose the largest weight outside the edge f is wQ, where |Q ∩f| ≤k −1. Now we define new weights w′, such that w′ X = ( − t k if X ⊂f wX/wQ otherwise.
Then for every X ̸⊂f, w′ X ≤1 and w′ Q = 1. Now we consider all the n−k t−k t-tuples containing the k-set Q. As usual, assume that S of them has negative sum according to w′. If S ≥(1 − 3 2tk ) n−k t−k , we have the following estimate: X Q⊂Y,|Y |=t X X⊂Y,X̸=Q w′ X ≤−S + n −k t −k −S · t k ≤− n −k t −k · 1 −3 2tk −3 2tk t k ≤− n −k t −k · 1 −3 2tk −3 2k!
≤−1 12 n −k t −k .
Note that since P X⊂f wX < 0, we have X X̸=Q w′ X = X X̸=Q wX/wQ − X X⊂f wX/wQ + t k ≥−1 − t k 2 ≥−t2k.
If we apply lemma 4.2 for L = Q and the weight w′′ X = w′ X/t2k, we get > n−k t−k nonnegative edges for the new weight function w′. Note that every such nonnegative edge can not share with f a common k-subset, otherwise its total weight is at most ( t k −1) − t k < 0. Hence these nonnegative edges are also nonnegative edges for the original weight function w.
By the above discussion, it remains to consider the case S < (1 − 3 2tk ) n−k t−k . Then there are at least 3 2tk n−k t−k nonnegative edges containing Q, together with the (1 −1 tk ) n−k t−k nonnegative edges containing {1, · · · , k}. Since {1, · · · , k} and wQ have codegree at most n−k−1 t−k−1 < 1 2tk n−k t−k , we have in total more than n−k t−k nonnegative edges.
4.2 Generalization of vector MMS Our techniques from the previous section also allow us to prove a generalization of the vector space version of Manickam-Mikl´ os-Singhi conjecture. Since the proof of this result is very similar to that of Theorem 1.5 we only state the appropriate variants of the lemmas involved. The detailed proofs of these lemmas can be found in the appendix of this paper. The proof of Theorem 1.6 follows immediately from combining these lemmas. As before, we define the hypergraph H to have the vertex set V k and every edge corresponds to a t-dimensional subspace. It is easy to check that the hypergraph is t k q-uniform on n k q vertices. Like the previous section, we also assume that [k] is the k-dimensional subspace with w[k] = 1 and for every X, wX ≤1. All the following lemmas are proven under the assumption that n > C(t −k)k for sufficiently large constant C.
13 Lemma 4.6. If for some k-dimensional subspace L, X L⊂Y,Y ∈[V t] X L̸=X⊂Y wX ≤− 1 24 t k 2 q n −k t −k q , and P X̸=L wX ≥−1, then there are more than n−k t−k q nonnegative edges in H.
We now assume that the 3 t k q-th largest weight in H is wP , and consider the following several cases.
Lemma 4.7. If wP > 1/(4 t k 2 q), there are more than n−k t−k q nonnegative edges in H.
Lemma 4.8. If wP ≤1/(4 t k 2 q), and there are less than (1− 1 2[t k]q ) n−k t−k q nonnegative edges containing [k], then there are at least n−k t−k q nonnegative edges in H.
Lemma 4.9. If [k] is contained in at least (1− 1 2[t k]q ) n−k t−k q nonnegative edges, then there are at least n−k t−k q nonnegative edges in H.
5 Concluding Remarks A r–(n, t, λ) block design is a collection of t-subsets of [n] such that every r elements are contained in exactly λ subsets. In , Rands proved the following generalization of Erd˝ os-Ko-Rado theorem: given a r–(n, t, λ) block design H and 0 < s < r, then there exists a function f(t, r, s) such that if H has an s-intersecting subhypergraph H′, then if n > f(t, r, s), the number of edges in H′ is at most bs, which is the number of blocks through s vertices. Note that Erd˝ os-Ko-Rado theorem corresponds to the very special case when H = [n] t and s = 1. Moreover, when (s, r) = (1, 2), this is an analogue of our Theorem 1.4, and when the block design is complete, it is similar to Theorem 1.5. Using tools developed in the previous section, we can prove the following generalization of Manickam-Mikl´ os-Singhi cojecture to designs. Given an r–(n, t, λ) design H, for j = 1, · · · , t, let dj be the number of blocks containing a fixed set of j elements. Obviously dr = λ, and by double counting, dj = (n−j r−j) (t−j r−j) λ.
Theorem 5.1. Let k, r, t be positive integers with t ≥r ≥2k, n > Ct3k+3 for sufficiently large C and let {wX}X∈([n] k ) be a weight assignment with P X∈([n] k ) wX ≥0. Then for a given r–(n, t, λ) design H, the number of blocks B with P X⊂B,X∈([n] k ) wX ≥0 is at least dk = (n−k r−k) (t−k r−k) λ.
It would be interesting if one can remove the condition t ≥r ≥2k in this statement. This will give a general result unifying our Theorems 1.4 and 1.5. The only additional ingredient needed to prove the above theorem is the following fact. For two disjoint vertex subsets |A| = a and |B| = b of a r–(n, t, λ) design, the number of edges containing every vertex from A while not containing any vertex in B is equal to (n−r−b t−r ) (n−r t−r) (n−a−b r−a ) (t−a r−a) · λ. We will omit any further details here and will return to this problem in the future.
14 In Section 3, we gave an example of infinitely many r-uniform n-vertex hypergraphs with equal codegrees and n ∼r3 not having the MMS property, based on the assumption that there are infinitely many Mersenne primes. Since the largest known Mersenne number has more than ten million digits, our example already gives (unconditionally) a huge hypergraph with n cubic in r. Still it would be interesting to construct infinitely many such hypergraphs directly, without relying on the existence of Mersenne primes?
Acknowledgment. We would like to thank Ameerah Chowdhury for bringing to our attention a Manikam-Miklos-Sighi conjecture for vector spaces and for sharing with us her preprint on this topic.
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15 A Missing proofs from Section 4.2 Throughout this section we use that for a > b, q(a−b)b ≤ a b q ≤q(a−b)b+b.
Proof of Lemma 4.6: We may rewrite the left hand side of the inequality as n −2k t −2k q X dim(X∩L)=0 wX + n −2k + 1 t −2k + 1 q X dim(X∩L)=1 wX + · · · + n −k −1 t −k −1 q X dim(X∩L)=k−1 wX = n −k −1 t −k −1 q X dim(X∩L)≤k−1 wX − k−2 X j=0 bj · X dim(X∩L)=j wX .
Here we let bj = n−k+1 t−k+1 q − n−2k+j t−2k+j q. Note that P dim(X∩L)≤k−1 wX = P X̸=L wX ≥−1. Since n > Ck(t −k), this implies k−2 X j=0 bj · X |X∩L|=j wX ≥ 1 24 t k 2 q n −k t −k q − n −k −1 t −k −1 q ≥ 1 25 t k 2 q n −k t −k q .
(3) For a fixed integer 0 ≤y ≤k −1, denote by Dy the number of nonnegative t-dimensional subspaces Z with dim(Z ∩L) = y. If Dy > n−k t−k q then we are done. Otherwise assume Dy ≤ n−k t−k q for every y. We estimate the following sum: X dim(Z∩L)=y,dim Z=t X X⊂Z wX.
Since every nonnegative t-dimensional subspace contributes to the sum at most t k q, it is at most t k qDy ≤ t k q n−k t−k q. By double counting, the above sum also equals Py j=0(βy,j · P dim(X∩L)=j wX).
Here for a k-dimensional subspace X with dim(X ∩L) = j, βy,j denotes the number of t-dimensional subspaces Z such that X ⊂Z and dim(Z ∩L) = y. There are (qk−qj)···(qk−qy−1) (qy−qj)···(qy−qy−1) ways to extend X ∩L to Z ∩L. Let Q = span{X, Z ∩L}, and R = span{X, L}. Then dim Q = k + y −j, dim R = 2k −j, and Q ⊂R. The next step is to extend Q to Z such that Z ∩R = Q. The number of ways is equal to (qn−q2k−j)···(qn−qt+k−y−1) (qt−qk+y−j)···(qt−qt−1) . Note that this is only nonzero for j ≥k + y −t, in this case βy,j is the product of these two expressions, which is roughly q(k−y)(y−j)+(n−t)(t−k+j−y). Since t−k +j −y ≥0, it is increasing in j for large n. Also note that bj is decreasing in j. Let ⃗ γ = (γ0, · · · , γk−2) be the unique solution of the system of equations ⃗ b = ⃗ γ · β, then by Lemma 4.1 k−2 X j=0 bj · X dim(X∩L)=j wX = k−2 X j=0 k−2 X y=j βy,jγy X dim(X∩L)=j wX = k−2 X y=0 γy · y X j=0 βy,j · X dim(X∩L)=j wX ≤ t k q n −k t −k q k−2 X y=0 γy ≤ t k q n −k t −k q k−2 X y=0 by βy,y .
It is easy to check that βy,y ≥q(n−t)(t−k) and so by/βy,y = ( n−k−1 t−k−1 q − n−2k+y t−2k+y q)/q(n−t)(t−k) ≤ n−k−1 t−k−1 q/q(n−t)(t−k). Therefore k−2 X j=0 bj · X |X∩L|=j wX ≤ t k q n −k t −k q · (k −1) · n−k−1 t−k−1 q q(n−t)(t−k) ≤(k −1) t k q qt−k−(n−t) n −k t −k q , 16 that for n > Ck(t −k) contradicts (3).
Proof of Lemma 4.7: We will show that every k-subspace whose weight is larger than wP is contained in at least 1 2[t k]q n−k t−k q nonnegative edges, otherwise there are already more than n−k t−k q nonnegative edges. For simplicity we just need to prove this statement for wP itself. Suppose there are S negative edges containing wP , which are denoted by e1, · · · , eS (as t-dimensional subspaces).
And eS+1, · · · , e[n−k t−k]q are the other (thus nonnnegative) edges containing wP . We have [n−k t−k]q X i=1 X P̸=X⊂ei wX = S X i=1 X P̸=X⊂ei wX + [n−k t−k]q X i=S+1 X P̸=X⊂ei wX ≤−wP · S + wP · n −k t −k q −S !
· t k q −1 !
+ 3 t k q · n −k −1 t −k −1 q (4) Here we used that there are at most 3 t k q vertices X whose weight is larger than wP (but always ≤1), and the number of times every such weight appear in the sum is at most n−k−1 t−k−1 q. If S ≥ 1 − 1 2[t k]q n−k t−k q, then the above expression is at most − n −k t −k q 1 − 1 2 t k q wP − 1 2 t k q · t k q · wP −3 t k q · qt−k −1 qn−k −1 !
, which can be further bounded by − n −k t −k q 1 2 − 1 2 t k q wP −3 t k q · qt−k −1 qn−k −1 !
< − n −k t −k q · 1 3wP ≤− n −k t −k q · 1 12 t k 2 q .
The first inequality is because t > k ≥2 and q ≥2, so t k q ≥7, and also because n > Ck(t −k) for large C. Since we also have P X̸=P wX = −wP ≥−1. Lemma 4.6 for L = P immediately gives > n−k t−k q nonnegative edges.
Therefore we can assume that for the 3 t k q vertices with largest weights, the number of nonneg-ative edges containing each such vertex is at least 1 2[t k]q n−k t−k q. Using the union bound, the number of nonnegative edges is at least 3 t k q · 1 2 t k q n −k t −k q − 3 t k q 2 n −k −1 t −k −1 q ≥3 2 1 − 3 t k 2 q qn−t n −k t −k q , which is also larger than n−k t−k q when n > Ck(t −k).
Proof of Lemma 4.8: We consider all the t-dimensional subspaces containing [k], similarly as before suppose there are S ≥ 1 2[t k]q n−k t−k q negative edges e1, · · · , eS and nonnegative edges eS+1, · · · , e[n−k t−k]q, 17 we get X [k]⊂Z,dim Z=t X X⊂Z,X̸=[k] wX = S X i=1 X X⊂ei,X̸=[k] wX + [n−k t−k]q X i=S+1 X X⊂ei,X̸=[k] wX ≤−S + 1 4 t k 2 q · n −k t −k q −S !
· t k q −1 !
+ 3 t k q · n −k −1 t −k −1 q ≤− n −k t −k q S n−k t−k q − 1 4 t k q −3 t k q · qt−k −1 qn−k −1 ≤− n −k t −k q 1 2 t k q − 1 4 t k q −3 t k q q−(n−t) !
The first inequality is by bounding the 3 t k q largest weights in the second sum by 1 and the rest by 1 4[t k] 2 q . It also uses the fact that two k-dimensional subspaces are contained in at most n−k−1 t−k−1 q t-dimensional subspaces. For n > Ck(t −k), we have X [k]⊂Z,dim Z=t X X⊂Z,X̸=[k] wX ≤− n −k t −k q · 1 5 t k q .
We also have P X̸=[k] wX = −w[k] = −1. Now we once again can apply Lemma 4.6 for L = [k] to show the existence of > n−k t−k q nonnegative edges.
Proof of Lemma 4.9: Note that if every t-dimensional subspaces containing [k] is nonnegative, this already gives n−k t−k q nonnegative edges and the lemma is proved. So we may assume that there is a negative edge f (as t-dimensional subspace) containing [k] with P X⊂f wX < 0. Suppose the largest weight outside the edge f is wQ, where dim(Q ∩f) ≤k −1. Now we define new weights w′, such that w′ X = ( − t k q if X ⊂f wX/wQ otherwise.
Then for every X ̸⊂f, w′ X ≤1 and w′ Q = 1.
Now we consider all the n−k t−k q t-dimensional subspaces containing Q. As usual, assume that S of them has negative sum according to w′. If S ≥ 1 − 2 3[t k]q n−k t−k q, we have the following estimate: X Q⊂Y,dim Y =t X X⊂Y,X̸=Q w′ X ≤ −S + n −k t −k q −S !
· t k q ≤− n −k t −k q · 1 − 2 3 t k q −2 3 !
≤ −1 12 n −k t −k q .
18 Note that since P X⊂f wX < 0, we have X X̸=Q w′ X = X X̸=Q wX/wQ − X X⊂f wX/wQ + t k q !
≥−1 − t k 2 q .
If we apply Lemma 4.6 for L = Q and the new weighting w′′ X = w′ X/(2 t k 2 q), we get > n−k t−k q nonnegative edges for weight w′. Note that every such nonnegative edge cannot share with f a common k-dimensional subspace, otherwise its total weight is at most ( t k q −1) − t k q < 0. Hence these nonnegative edges are also nonnegative edges for the original weighting w.
By the above discussion, it remains to consider the case S < 1 − 2 3[t k]q n−k t−k q. Then there are at least 2 3[t k]q n−k t−k q nonnegative edges containing Q, together with the 1− 1 2[t k]q n−k t−k q nonnegative edges containing [k]. Since [k] and wQ have codegree at most n−k−1 t−k−1 q ≤ 1 6[t k]q n−k t−k q, we have in total more than n−k t−k q nonnegative edges.
19 |
14821 | https://stats.libretexts.org/Courses/Taft_College/PSYC_2200%3A_Elementary_Statistics_for_Behavioral_and_Social_Sciences_(Oja)/01%3A_Description/03%3A_Descriptive_Statistics/3.06%3A_Introduction_to_Standard_Deviations_and_Calculations | Skip to main content
3.6: Introduction to Standard Deviations and Calculations
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: May 12, 2022
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3.5: Introduction to Measures of Variability
3.7: Practice SD Formula and Interpretation
Page ID
: 22032
Michelle Oja
Taft College
( \newcommand{\kernel}{\mathrm{null}\,})
Sum of Squares
Variability can also be defined in terms of how close the scores in the
distribution
are to the middle of the
distribution
. Using the
mean
as the measure of the middle of the
distribution
, we can see how far, on average, each data point is from the center. The data from a fake Quiz 1 are shown in Table . The
mean
score is 7.0:
Therefore, the column “ ” contains deviations (how far each score deviates from the
mean
), here calculated as the score minus 7. The column “” has the “Squared Deviations” and is simply the previous column squared.
There are a few things to note about how Table is formatted, as this is the format you will use to calculate standard deviation. The raw data scores () are always placed in the left-most column. This column is then summed at the bottom to facilitate calculating the
mean
(simply divided this number by the number of scores in the table). Once you have the
mean
, you can easily work your way down the middle column calculating the deviation scores. This column is also summed and has a very important property: it will always sum to 0 (or close to zero if you have rounding error due to many decimal places). This step is used as a check on your math to make sure you haven’t made a mistake. THIS IS VERY IMPORTANT. When you mis-calculate an equation, it is often because you did some simple math (adding or subtracting) incorrectly. It is very useful when equations have these self-checking points in them, so I encourage you to use them. If this column sums to 0, you can move on to filling in the third column of squared deviations. This column is summed as well and has its own name: the Sum of Squares (abbreviated as and given the formula ). As we will see, the Sum of Squares appears again and again in different formulas – it is a very important value, and this table makes it simple to calculate without error.
Table : Calculation of Variance for Quiz 1 scores.
| | | |
| 9 | 2 | 4 |
| 9 | 2 | 4 |
| 9 | 2 | 4 |
| 8 | 1 | 1 |
| 8 | 1 | 1 |
| 8 | 1 | 1 |
| 8 | 1 | 1 |
| 7 | 0 | 0 |
| 7 | 0 | 0 |
| 7 | 0 | 0 |
| 7 | 0 | 0 |
| 7 | 0 | 0 |
| 6 | -1 | 1 |
| 6 | -1 | 1 |
| 6 | -1 | 1 |
| 6 | -1 | 1 |
| 6 | -1 | 1 |
| 6 | -1 | 1 |
| 5 | -2 | 4 |
| 5 | -2 | 4 |
| | | |
The calculations in Table can be done by hand, but it is also very easy to set up the data in any spreadsheet program and learn the simple commands to make the spreadsheet do the simple math. As long as you tell it what to do with the correct numbers, then your results will be correct. You can also use the memory function in graphing calculators to save the the data set, and run some of the more common mathematical functions. Using spreadsheets and your graphing calculator to do the math also saves problems with rounding since the devices keep all of the decimals so you only have to round your final result. This statistics textbook will not go into explanations on how to use software (like spreadsheets, calculators, or more sophisticated statistical programs), but much that is easily accessible (like spreadsheets on Excel or Google) are relatively easy to learn to use
Variance (of a Sample )
Now that we have the Sum of Squares calculated, we can use it to compute our formal measure of average distance from the
mean
, the variance. The variance is defined as the average squared difference of the scores from the
mean
. We square the deviation scores because, as we saw in the Sum of Squares table, the sum of raw deviations is always 0, and there’s nothing we can do mathematically without changing that.
The
population
parameter
for variance is (“sigma-squared”) and is calculated as:
Notice that the numerator that formula is identical to the formula for Sum of Squares presented above with replaced by . Thus, we can use the Sum of Squares table to easily calculate the numerator then simply divide that value by to get variance. If we assume that the values in Table represent the full
population
, then we can take our value of Sum of Squares and divide it by to get our
population
variance:
So, on average, scores in this
population
are 1.5 squared units away from the
mean
. This measure of spread is much more
robust
(a term used by statisticians to
mean
resilient or resistant to) outliers than the
range
, so it is a much more useful value to compute.
But we won't do much with variance of a
population
. Instead, we'll focus on variance of a
sample
. The
sample
statistic
used to estimate the variance is (“s-squared”):
This formula is very similar to the formula for the
population
variance with one change: we now divide by instead of . The value has a special name: the degrees of freedom (abbreviated as). You don’t need to understand in depth what degrees of freedom are (essentially they account for the fact that we have to use a
sample
statistic
to estimate the
mean
() before we estimate the variance) in order to calculate variance, but knowing that the denominator is called provides a nice shorthand for the variance formula: .
Going back to the values in Table and treating those scores as a
sample
, we can estimate the
sample
variance as:
[s^{2}=\dfrac{30}{20-1}=1.58 \n\nonumber ]
Notice that this value is slightly larger than the one we calculated when we assumed these scores were the full
population
. This is because our value in the denominator is slightly smaller, making the final value larger. In general, as your
sample
size gets bigger, the effect of subtracting 1 becomes less and less. Comparing a
sample
size of 10 to a
sample
size of 1000; 10 – 1 = 9, or 90% of the original value, whereas 1000 – 1 = 999, or 99.9% of the original value. Thus, larger
sample
sizes will bring the estimate of the
sample
variance closer to that of the
population
variance. This is a key idea and principle in statistics that we will see over and over again: larger
sample
sizes better reflect the
population
.
The Big Finish: Standard Deviation
The standard deviation is simply the square root of the variance. This is a useful and interpretable
statistic
because taking the square root of the variance (recalling that variance is the average squared difference) puts the standard deviation back into the original units of the measure we used. Thus, when reporting
descriptive statistics
in a study, scientists virtually always report
mean
and standard deviation. Standard deviation is therefore the most commonly used measure of spread for our purposes.
The
sample
statistic
follows the same conventions and is given as :
The
sample
standard deviation from Table is:
We'll practice calculating standard deviations, then interpreting what the numbers
mean
. Because in behavioral statistics, it's not about the numbers. We never end with a number, we end with a conclusion (which can be as simple as a sentence, or can be several paragraphs). Social scientists want to know what the numbers mean because we use
statistical analyses
to answer real questions.
Contributors
Foster et al. (University of Missouri-St. Louis, Rice University, & University of Houston, Downtown Campus)
Dr. MO (Taft College)
3.5: Introduction to Measures of Variability
3.7: Practice SD Formula and Interpretation |
14822 | https://math.stackexchange.com/questions/1647691/length-of-a-chord-of-a-circle | Skip to main content
Length of a Chord of a circle
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I was wondering about the possible values that the length of a chord of a circle can take. The Length of a chord is always greater than or equal to 0 and smaller than or equal to the diameter (sayd). Is it possible to draw a chord of length 'x', where x could be any real number? In other words, does the chord of circle take all the possible real values between 0andd?
Any help/comments would be much appreciated.
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asked Feb 9, 2016 at 15:51
jimojimo
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There always is such a chord. "Is it possible to draw" is more complicated. If the drawing is to be done with straightedge and compass, the answer is no.
– André Nicolas
Commented
Feb 9, 2016 at 15:58
1
To make the question more accurate you need to specify that 0<d<somevalue(sayr) for example.
– NoChance
Commented
Feb 9, 2016 at 16:02
1
Related perhaps tangentially: Ptolemy's table of chords (See also my answer below.)
– Michael Hardy
Commented
Feb 9, 2016 at 16:03
5
@AndréNicolas: Constructing length x may not be possible (it's not even very well-posed), but if you have such a length, then constructing a chord of that length is trivial (you just select a point A on the circle, draw a circle of radius x centered at A, and draw a line segment from A to one of the two points where the two circles intersect).
– ruakh
Commented
Feb 9, 2016 at 17:26
1
@ruakh: You are certainly right, if x is given we can construct the chord.
– André Nicolas
Commented
Feb 9, 2016 at 17:46
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This comes down to the intermediate value theorem.
On the circle x2+y2=1, the chord from (1,0) to (cosθ,sinθ) has length 2sinθ2. You can see that by means of the usual "distance formula".
As θ goes from 0 to π (or from 0∘ to 180∘ if you like), the chord goes from 0 to 2 and the chord is a continuous function of θ. The fact that it's continuous means you can apply the intermediate value theorem and see that it assumes all intermediate values.
If you don't like transcendental functions (perhaps because proving continuity of those takes a lot of work), you can also do it like this: the point
(1−t21+t2,2t1−t2)(1)
goes around the circle from (−1,0) back to (−1,0) as t goes from −∞ to +∞. The length of the chord from (1,0) to the point in (1) can also be found via the distance formula and the same kind of argument can be used.
Tangentially (no pun intended) related is this: Ptolemy's table of chords
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answered Feb 9, 2016 at 16:03
Michael HardyMichael Hardy
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Open a compass with the desired length x, place the needle on the circumference and draw an arc that intersects the circle. Can you imagine that you'll never meet it ?
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answered Feb 9, 2016 at 16:00
user65203user65203
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Hint:
chord length =2rsin(θ2)
where r is the radius and θ is the angle subtended at the center by the chord. Note the continuity of the RHS, now use the Intermediate Value Theorem.
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edited Feb 9, 2016 at 15:59
answered Feb 9, 2016 at 15:57
foshofosho
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Yes. There are several ways to prove it.
From the axioms for Euclidean geometry: draw a circle with radius x and center on your circle; see where the circles intersect. (Euclid's axioms are a little incomplete - they may not actually guarantee the intersection.)
An argument from continuity: start at one end of the diameter of your circle and move around to the other end. The length of the chord varies continuously, so takes on every value between d and 0. This argument depends on understanding enough of the structure of the real numbers.
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edited Feb 9, 2016 at 16:03
answered Feb 9, 2016 at 16:00
Ethan BolkerEthan Bolker
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You actually need radius x - the two intersection points won't form a diameter of the (second) circle, so they won't be distance (x/2)+(x/2)=x apart.
– Steven Stadnicki
Commented
Feb 9, 2016 at 16:02
@StevenStadnicki Fixed thanks.
– Ethan Bolker
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The answer depends on which axioms you have. If you have any axioms that give you a mesure such that for every real positive number exist a line of that lenght the answer is yes. You can see Hilbert geometry (moder versione of euclidean geometry) and his axioms about continuity and mesure
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answered Feb 9, 2016 at 15:59
Giovanni SiclariGiovanni Siclari
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Consider a point P on the circle and the diameter going through this point. Let N the other end of the diameter. Let M=N. The angle between PN→ and PM→ is zero and the length of the chord is d. now if you increase this angle continuously from 0 to π2 by moving M along the circle, the chord will go from d to 0 and it'll take all the values between 0 and d because the length of the chord is a continuous function of the angle.
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answered Feb 9, 2016 at 16:03
AugustinAugustin
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Let O be the center of the circle. Take a point B at distance b from O, with 0<b<r=d/2.
Let the line thru B, perpendicular to the segment OB, meet the circle at points A and C.
We have AO=CO=r=d/2. The triangles ABO,CBO are right-angled at B. So the length of the chord AC is
AB+BC=r2−b2−−−−−−√+r2−b2−−−−−−√=2r2−b2−−−−−−√=d1−(b/d)2−−−−−−−−√.
The number 1−(b/d)2−−−−−−−−√ can be any value between 0 and 1, depending on b, so the length AC can be anything between 0 and d.
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answered Feb 9, 2016 at 23:35
DanielWainfleetDanielWainfleet
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14823 | https://arxiv.org/pdf/2007.03064 | Maximizing five-cycles in Kr-free graphs
Bernard Lidick´ y∗ Kyle Murphy †
May 12, 2021
Abstract
The Pentagon Problem of Erd˝ os problem asks to find an n-vertex triangle-free graph that is maximizing the number of 5-cycles. The problem was solved using flag algebras by Grzesik and independently by Hatami, Hladk´ y, Kr´ al’, Norin, and Razborov. Recently, Palmer suggested a more general problem of maximizing the number of 5-cycles in Kk+1 -free graphs. Using flag algebras, we show that every Kk+1 -free graph of order n contains at most 110 k4 (k4 − 5k3 + 10 k2 − 10 k + 4) n5 + o(n5)copies of C5 for any k ≥ 3, with the Tur´ an graph being the extremal graph for large enough n.
1 Introduction
All graphs in this paper are simple. Let G, H, and F be graphs. We define ν(H, G ) as the number of (possibly non-induced) subgraphs of G isomorphic to H. If G does not contain any subgraph isomorphic to F , then we say that G is F -free . Let ex( n, H, F ) denote the maximum value of ν(H, G ) among all F -free graphs G on n vertices. The function ex( n, H, F )is well-studied when H is an edge. As such, it is convention when H = K2 to let ex( n, F )denote ex( n, K 2, F ). The value of ex( n, F ) for any graph F is called the Tur´ an number of
F .One of the first results in extremal graph theory was Mantel’s Theorem which states that for all n ≥ 3, ex( n, K 3) ≤ b n2
4
c. When k ≥ 3, the value of ex( n, K k+1 ) was determined by Tur´ an.
Theorem 1.1 (Tur´ an’s Theorem ) For all k ≥ 3, and all n,ex (n, K k+1 ) ≤
(
1 − 1
k
) n2
2 .
∗
Department of Mathematics, Iowa State University. Ames, IA, USA. E-mail: lidicky@iastate.edu .Supported in part by NSF grant DMS-1855653.
†
Department of Mathematics, Iowa State University. Ames, IA, USA. E-mail: kylem2@iastate.edu
1
arXiv:2007.03064v3 [math.CO] 11 May 2021
Moreover, the Tur´ an graph Tk(n), which is the complete balanced (k − 1) -partite graph on n
vertices, is the unique Kk+1 -free graph on n vertices which contains the maximum possible number of edges.
The Erd˝ os-Stone-Simonovits Theorem determined the asymptotic value of ex( n, F )when F is not a complete graph. Let χ(F ) denote the chromatic number of F . Then for all
F for which χ(F ) ≥ 2, ex( n, F ) = χ(F ) − 22 ( χ(F ) − 1) n2 + o(n2).
The systematic study of the function ex( n, H, F ) was initiated by Alon and Shikhel-man , although there were some prior results. When t < k + 1, Zykov showed that the Tur´ an graph Tk(n) is also the unique graph with the maximum number of Kt subgraphs among all Kk+1 -free graphs.
Theorem 1.2 (Zykov ) Let k and t be integers such that t < k + 1 . Then for all n,the Tur´ an graph Tk(n) is the unique Kk+1 -free graph on n vertices containing the maximum number of Kt subgraphs.
We will need the following corollary in our calculations.
Corollary 1.3 Let G be a Kk+1 -free graph on n vertices. Then
ν(K5, G ) ≤ k4 − 10 k3 + 35 k2 − 50 k + 24
k4 n5 + o(n5).
Alon and Shikhelman proved the following analogue of the K˝ ov´ ari-S´ os-Tur´ an Theorem: ex( n, K 3, K s,t ) = O(n3−3/s ).
They also proved that for fixed integers t < k , if F is a k-chromatic graph: ex( n, K t, F ) =
(k − 1
t
) ( nk − 1
)t
o(nt).
In , Gy˝ ori, Pach, and Simonovits studied a handful of cases where F = Kr. The order of magnitude of ex( n, C k, C ) is known for all ≥ 3 and k ≥ 3, see Gishboliner and Shapira . The asymptotic value of ex( n, C k, C 4) was determined by Gerbner, Gy˝ ori, Methuku, and Vizer . They proved a variety of results on ex( n, F, H ) when F and H
were both cycles. This includes showing that ex( n, C 2, C 2k) = Θ( n) for k, ` ≥ 2 and ex( n, C 4, C 2k) = (1 + o(1)) (k − 1)( k − 2) 4 n2
for k ≥ 2. In , Gerbner and Palmer provided more general bounds on ex( n, H, F ). In particular, they showed that if H and F are graphs and χ(F ) = k, then ex( n, H, F ) ≤ ex( n, H, K k) + o (n|H|) .
2Additionally, they extended the result of Gishboliner and Shapira to show that for all k and
t,ex( n, C k, K 2,t ) =
( 12k + o(1)
)
(t − 1) k/ 2nk/ 2,
and ex( n, P k, K 2,t ) =
(12 + o(1)
)
(t − 1) (k−1) /2n(k+1) /2,
where Pk is a path on k vertices. In , Cutler, Nir, and Radcliffe determined the asymptotic value of ex( n, S t, K k+1 ), where St is the star with t leaves. In particular, they showed that while the extremal graph must be complete multi-partite, it is not always isomorphic to the Tur´ an graph Tk(n). The study of the function ex( n, K 3, H ) has seen recent attention as well. In particular, the function ex( n, K 3, C 5) was studied in [2, 8, 13]. In , Mubayi and Mukherjee studied the function ex( n, K 3, H ) for a handful of other 3-chromatic graphs H.In , Gerbner and Palmer found a handful of cases where the value of ex( n, H, F ) is achieved by the Tur´ an graph and in , Gerbner studied the function ex( n, H, F ) when H
and F each have at most 4 vertices. Recently, the authors of studied the problem of maximizing the number of copies of a graph H in some graph G embedded in a particular surface. In 1984, Erd˝ os conjectured that the balanced blow-up of C5 on n vertices maximizes the number of five-cycles among all triangle-free graphs of order n. If G is a graph on m
vertices, then the balanced blow-up of G on n vertices is the graph G(n) obtained from G
by replacing each vertex of G with an independent set of size ⌊ nm
⌋ or ⌈ nm
⌉, and replacing each edge in G with a complete bipartite graph on the corresponding sets. The problem of determining ex( n, C 5, K 3) was known as the Pentagon Problem of Erd˝ os. In a sense, a graph with ex( n, C 5, K 3) five-cycles is the “least bipartite” triangle-free graph on n vertices when measured by the number of 5-cycles. In posing this question, Erd˝ os also proposed the following two measures of “non-bipartiteness” . 1. The minimal possible number of edges in a subgraph spanned by half the vertices. 2. The minimal possible number of edges that have to be removed to make the graph bipartite, which is equivalent to the problem of max cut. In 1989, Gy˝ ori showed that a triangle-free graph on n vertices contains at most 1 .03 (n
5
)5
five-cycles. In 2012, Grzesik and independently in 2013, Hatami Hladk´ y, Kr´ al’, Norin, and Razborov showed that a triangle-free graph on n vertices contains at most (n
5
)5 +
o(n5) five cycles. Moreover, a matching lower bound is given by the balanced blow-up of
C5 when n is divisible by 5. The authors of also proved that for large enough n, the balanced blow-up of a C5 on n vertices is the unique extremal graph. In 2018, Lidick´ y and Pfender proved that a balanced C5 blow-up is the unique extremal construction for all
n, with the exception of n = 8. This observation was made by Michael , who showed that the M¨ obius ladder on 8 vertices contains the same number of five cycles as the balanced
C5 blow-up. Palmer suggested a generalization to the Pentagon Problem of Erd˝ os: maximizing the number of five-cycles in Kk+1 -free graphs for k ≥ 3. Observe that in the more general 3case, the problem of maximizing the number of non-induced C5 subgraphs is different from maximizing the number of induced C5 subgraphs. In this paper, we will discuss the non-induced case. Let H and G be graphs on n1 and
n2 vertices, respectively. The density d(H, G ) of H in G is given by
d(H, G ) = ν(H, G )
(n2
n1
)−1
.
Normally, n−n1
2
would be used as the scaling factor for defining the density of H in G. We will use (n2
n1
)−1, since this is more natural in proofs involving the flag algebra method. Let OPT k(C5) = lim
n→∞
max
Gn∈F kn
d(C5, G n), (1) where Fkn is the set of all Kk+1 -free graphs on n vertices. Note that since d(C5, G ) measures the density of non-induced C5 subgraphs in a graph G, this parameter will often have a value greater than one. For example, d(C5, K ) = 12 for all ≥ 5. Our main goal is to prove the following theorem.
Theorem 1.4 Let k ≥ 3 be an integer. Then (i) OPT k(C5) = 1
k4
(12 k4 − 60 k3 + 120 k2 − 120 k + 48) .
(ii) If n is sufficiently large, then Tk(n) is the unique Kk+1 -free graph on n vertices for which ν(C5, T k(n)) = ex (n, C 5, K k+1 ).
Since our result forbids ( k +1)-cliques, Tur´ an’s Theorem implies that the number of edges in an extremal graph cannot be more than in Tk(n). Interestingly, the authors of proved that if G is a graph with at least k−1
k
(n
2
) edges, then the Tur´ an graph provides a lower bound on the number of five-cycles contained in G.The proof of Theorem 1.4(i) uses flag algebras to calculate the upper bound for OPT k(C5). The second part is done by stability and exact structure arguments. Unlike typical appli-cations of the flag algebra method, our result does not need computer assistance for the calculations involving flag algebras. However, it is still convenient to use a computer for the purpose of multiplying and expanding rational functions. In the next section, we will give a brief overview of the flag algebra method. Section 2 contains the proof of Theorem 1.4(i). Then we prove a stability lemma in Section 3, and use it to prove Theorem 1.4(ii) in Section 4. We will end with some concluding remarks and conjectures concerning the general behavior of the function ex( n, H, F ).
1.1 The Flag Algebra Method
Introduced by Razborov , the flag algebra method provides a framework for computa-tionally solving problems in extremal combinatorics. Flag algebras have been used to solve problems on hypergraphs [4, 14, 20, 31], permutations , graph decomposition problems , and oriented graphs among many other applications. Here we will give a brief introduc-tion and description of the notation and theory we will need for our result. We will not prove 4any claims since they have already been proven by Razborov . Another overview of flag algebras can be found in . Let H and G be graphs on n1 and n2 vertices, respectively, such that n1 ≤ n2. If
X ⊆ V (G), we will denote the induced subgraph of G on the vertices of X by G[X]. Let a subset X be selected uniformly at random from V (G) such that |X| = n1. Then P (H, G ) is the probability that G[X] is isomorphic to H.A sequence of graphs ( Gn)n≥1 of increasing order is said to be convergent if for every finite graph H, the following limit converges: lim
n→∞
P (H, G n).
Let F denote the set of all graphs up to isomorphism, and let Fdenote the set of all graphs on vertices up to isomorphism. Let RF denote the set of all formal finite linear combinations of graphs in F. A type of size k is a graph σ on k labelled vertices labeled by [ k] = {1, . . . , k }. If σ is a type of size k and F is a graph on at least k vertices, then an embedding of σ into F is an injective function θ : [ k] → V (F ), such that θ gives an isomorphism between σ and F [im( θ)]. A σ-flag is a pair ( F, θ ) where F is a graph and θ
is an injective function from [ k] to V (F ) that defines a graph isomorphism of F [im( θ)] and
σ. In this way, σ can be thought of as a labelled subgraph of F . Two σ-flags F and G are isomorphic if there exists a graph isomorphism between F and G that preserves the labelled subgraph σ.Let Fσ denote the set of all σ-flags and Fσdenote the set of all σ-flags on ver-tices. Observe that if σ is the empty graph, then Fσ = F. For two σ-flags F and G
with |V (F )| ≤ | V (G)|, let P (F, G ) denote the probability that an injective map from V (F )to V (G) that fixes the labeled graph σ induces a copy of F in G. Razborov showed that there exists an algebra Aσ after some factorization of RFσ. In doing so, he defined addition and multiplication on the elements of RFσ. Addition can be defined in the natural way, by simply adding coefficients of the elements in RFσ. We will now describe how to define multiplication of elements in Aσ.Let ( G, θ ) ∈ F σ be a σ-flag on n vertices. Let ( F1, θ 1), (F2, θ 2) ∈ F σ be two σ-flags for which |V (F1)| + |V (F2)| ≤ n + |V (σ)|. Let X1 and X2 be two disjoint sets of sizes |V (F1)| − |V (σ)| and |V (F2)| − | V (σ)| respectively, selected uniformly at random from V (G) \ im( θ). We will define the density of F1 and F2 in G, denoted P (F1, F 2; G) as the probability that (G[X1 ∪ im( θ)] , θ ) is isomorphic to ( F1, θ 1) and ( G[X2 ∪ im( θ)] , θ ) is isomorphic to ( F2, θ 2). It can be shown that as n grows, then the density of F1 and F2 is approximately equal to the product of their individual densities:
|P (F1, F 2; G) − P (F1, G )P (F2, G )| ≤ O(n−1). (2) Given this fact, if |V (F1)| + |V (F2)| = ` − | V (σ)| we could ideally define multiplication in
Aσ by
F1 · F2 = ∑
F∈F σ`
P (F1; F2; F )F. (3) The issue with this, however, is that the product F1 · F2 could be also written as a linear combination of elements in Fσ′ for any′ > ` . Hence, before defining Aσ we factor out all 5expressions of the form
F − ∑
F′∈F σ`
P (F, F ′)F ′ (4) from RFσ. Note that (4) corresponds to the law of total probability and hence it should behave as 0 when added to another linear combination. Let Kσ be the linear subspace of
RFσ containing all expressions of the form (4). We define Aσ to be RFσ factorized by Kσ,and we define multiplication in Aσ by naturally extending (3). Returning to the idea of convergent sequences of graphs, let Hom +(Aσ, R) be the set of all homomorphisms from Aσ to R such that for each φ ∈ Hom +(Aσ, R) and H ∈ F σ, φ(H) ≥ 0. If σ has order 0, we omit it in the notation. Razborov showed that each homomorphism in Hom +(A, R) corresponds to some convergent graph sequence and vice versa . In any fixed graph G, we can express d(C5, G ) as the sum of induced densities in the following way:
d(C5, G ) = ∑
Fi∈F 5
cC5
Fi
P (Fi, G ), (5) where cC5
Fi
= ν(C5, F i). Hence, for any sequence of unlabelled graphs ( Gn)n≥1 and its corre-sponding homomorphism φ ∈ Hom +(A, R), lim
n→∞
d(C5, G n) = ∑
Fi∈F 5
cC5
Fi
φ(Fi). (6) Quite often in our computations to simplify notation, we will drop the function notation and simply write Fi or draw the graph Fi in place of φ(Fi). Under this notation equation (6) would be lim
n→∞
d(C5, G n) = ∑
Fi∈F 5
cC5
Fi
Fi.
Finally, while we will often work with σ-flags where σ is not empty, flag algebras are often applied to questions concerning unlabelled graphs. In order to translate information from Hom +(Aσ, R) to Hom +(A, R) Razborov defined the unlabelling operator which is a linear operator
J·Kσ : RFσ → RF,
such that for any σ-flag F = ( H, θ ), JF Kσ = qσ(F )H, where qσ(F ) is equal to the probability (H, θ ′) is isomorphic to F where θ′ is a randomly chosen injective mapping θ′ from [ k] to
V (H). It can be shown that for any a ∈ A σ and any φ ∈ Hom +(A, R),
φ (Ja · aKσ) ≥ 0. (7) We will frequently make use of this fact in our computations. If a flag algebra calculation has a constant number of terms and operations, then it can be interpreted as a calculation in a graph of order n with an error term O(n−1) coming from (2). 62 Proof of Theorem 1.4(i)
In this section we will prove Theorem 1.4(i). First we will provide a lower bound by counting the number of five cycles in the Tur´ an graph. Next, using the flag algebra method, we will provide a matching upper bound. The proof of the upper bound when k = 3 is slightly different than the proof when k ≥ 4.
Proof of Theorem 1.4(i). First we will count the number of five-cycles in Tk(n), which will give an asymptotic lower bound. Observe that the only induced subgraphs of Tk(n) on five vertices containing a five-cycle are , , and . There are (k
5
) ( nk
)5 copies of in
Tk(n), with every such graph containing 12 distinct C5 subgraphs. There are 4 (k
4
)( n/k
2
) ( nk
)3
copies of in Tk(n), with every such graph containing 6 distinct C5 subgraphs. Finally, there are 3 (k
3
)( n/k
2
)2 nk copies of in Tk(n), with every such graph containing 4 distinct C5
subgraphs. Thus,
ν(C5, T k(n)) = 12
(k
5
) (nk
)5
24
(k
4
)( n/k
2
) (nk
)3
12
(k
3
)( n/k
2
)2 nk + o(n5),
where the error term o(n5) accounts for the cases where n is not divisible by k. This implies that for all n,
d(C5, T k(n)) ≥ ν(C5, T k(n))
(n
5
)−1
= 1
k4 (12 k4 − 60 k3 + 120 k2 − 120 k + 48) + o(1) .
Now we will calculate an asymptotic upper bound. Unless it is stated otherwise, assume that k ≥ 3. Let F5 = {F0, . . . , F 33 = K5} be the set of unlabeled graphs (up to isomorphism) on five vertices. Each of these graphs is pictured in Table 1 in the Appendix. After removing each cC5
Fi
for which cC5
Fi
= 0 from (5) we get lim
n→∞
d(C5, G n) = + + · + 2 · + 2 · + 4 · + 6 · + 12 · . (8) Since F5 contains all graphs on five vertices (up to isomorphism) and ∑33
i=0
Fi = 1, lim
n→∞
d(C5, G n) ≤ max {cC5
Fi
: Fi ∈ F 5}.
Therefore, OPT k(C5) ≤ max {cC5
Fi
: Fi ∈ F 5}.
Given this fact, our goal is to find appropriate constants cFi so that lim
n→∞
d(C5, G n) ≤ ∑
Fi∈F 5
cFi Fi,
and max {cFi : Fi ∈ F 5} is as small as possible. To do so, we can take advantage of properties that we know must be true of all φ ∈ Hom +(A, R) which correspond to Kk+1 -free convergent 7sequences of graphs. Additionally, using labeled flags, we can derive nonnegative expressions of unlabeled graphs in F5. We define
σ1 =1
23
σ2 =1
23
σ3 =1
23
so that Fσ1
4
, Fσ2
4
, and Fσ3
4
denote three sets of labeled flags on four vertices. By (7), the following expressions are nonnegative for all k ≥ 3. 1. P1(k) = 10 ·
uv
(
(k − 1) 1
23
−1
23
)2}~
σ1
=(10 k2 − 20 k + 10) + ( k2 − 2k + 1) + ( −k + 1) + ( −4k + 4) + +
P2(k) = 30 ·
uv
(
(k − 2) 1
23
−1
23
)2}~
σ2
=(3 k2 − 12 k + 12) + ( k2 − 6k + 8) + ( −4k + 10) + 3
P3(k) = 30 ·
uv
(
(k − 2) 1
23
−1
23
)2}~
σ2
=(6 k2 −24 k+24) +( k2 −4k+4) +( −k+2) +( −6k+12) +2 +3
P4(k) = 30 ·
uv
(1
23
−1
23
)2}~
σ3
=6 − + 2 − 4
P5(k) = 30 ·
uv
(
(k − 3) 1
23
( k − 3) 1
23
− 21
23
)2}~
σ3
=(6 k2 − 36 k + 54) + (2 k2 − 20 k + 42) + (4 k2 − 24 k + 36) + ( −24 k + 84) +120
8Additionally, we can apply Corollary 1.3, which states that for any Kk+1 -free convergent sequence of graphs,
≤ k4 − 10 k3 + 35 k2 − 50 k + 24
k4 .
At this point we will split the proof into the two cases where k ≥ 4 and k = 3. To gain some intuition as to why this is necessary, we can consider the previous inequality. When
k = 3 or k = 4, the previous bound implies that = 0. The issue is that it does not give any information about the density of K4, which is also equal to zero when k = 3. Thus, two slightly different proofs are required for k = 3 and k ≥ 4.
Case 1: Suppose that k ≥ 4. Since ∑33
i=0
Fi = 1,
≤
33
∑
i=0
Fi
(k4 − 10 k3 + 35 k2 − 50 k + 24
k4
)
. (9) After rearranging the terms from (9) we obtain the following constraint on the elements of
F5:0 ≤
32
∑
i=0
Fi · k4 − 10 k3 + 35 k2 − 50 k + 24
k4 + · −10 k3 + 35 k2 − 50 k + 24
k4 . (10) Let
Z(k) =
32
∑
i=0
Fi · k4 − 10 k3 + 35 k2 − 50 k + 24
k4 + · −10 k3 + 35 k2 − 50 k + 24
k4 .
It is straightforward to verify that the following rational functions are nonnegative for all k ≥ 4.
p1(k) = 3( k5−8k4+22 k3−24 k2+8 k)5k7−35 k6+75 k5−48 k4 p2(k) = 10 k5−60 k4+109 k3−76 k2+18 k
5k7−35 k6+75 k5−48 k4
p3(k) = 5k5−28 k4+45 k3−28 k2+6 k
5k7−35 k6+75 k5−48 k4
p4(k) = 5k7−30 k6+53 k5−52 k4+94 k3−96 k2+24 k
4(5 k7−35 k6+75 k5−48 k4)
p5(k) = 15 k5−60 k4+78 k3−40 k2+8 k
4(5 k7−35 k6+75 k5−48 k4)
z(k) = 6(5 k3−20 k2+30 k−16) 5k3−35 k2+75 k−48
Thus, for any convergent sequence ( Gn)n≥1 of Kk+1 -free graphs with k ≥ 4, lim
n→∞
d(C5, G n) ≤
33
∑
i=0
cC5
Fi
Fi + z(k)Z(k) +
5
∑
j=1
pj (k)Pj (k) =
33
∑
i=0
cFi Fi, (11) where cFi is the coefficient of Fi after all expansions. Then OPT k(C5) ≤ max {cFi : Fi ∈ F 5}.
The values of cFi for each Fi ∈ F 5 are listed below.
• C1(k) = c = c = c = c
= c = c = c = 60 k7−720 k6+3600 k5−9876 k4+16320 k3−16440 k2+9360 k−2304 5k7−35 k6+75 k5−48 k4 .
9• C2(k) = c = 33 k7−450 k6+2547 k5−7824 k4+14214 k3−15360 k2+9144 k−2304 5k7−35 k6+75 k5−48 k4 .
• C3(k) = c = c = c = c = c = c = c = c = c = c = c =
c = c
= c = c = c = c = c = 30 k7−420 k6+2430 k5−7596 k4+13980 k3−15240 k2+9120 k−2304 5k7−35 k6+75 k5−48 k4 .
• C4(k) = c = 30 k7−423 k6+2457 k5−7686 k4+14118 k3−15336 k2+9144 k−2304 5k7−35 k6+75 k5−48 k4 .
• C5(k) = c = 35 k7−468 k6+2607 k5−7916 k4+14278 k3−15367 k2+9144 k−2304 5k7−35 k6+75 k5−48 k4 .
• C6(k) = c = 30 k7−425 k6+2468 k5−7697 k4+14098 k3−15302 k2+9132 k−2304 5k7−35 k6+75 k5−48 k4 .
• C7(k) = c = c = 35 k7−455 k6+2505 k5−7644 k4+13980 k3−15240 k2+9120 k−2304 5k7−35 k6+75 k5−48 k4 .
• C8(k) = c = (135 /4) k7−(895 /2) k6+(9967 /4) k5−7631 k4+(27913 /2) k3−15216 k2+9114 k−2304 5k7−35 k6+75 k5−48 k4 .
• C9(k) = c = 50 k7−610 k6+3129 k5−8902 k4+15326 k3−15956 k2+9264 k−2304 5k7−35 k6+75 k5−48 k4 .
• C10 (k) = c = 50 k7−610 k6+3103 k5−8758 k4+15050 k3−15748 k2+9216 k−2304 5k7−35 k6+75 k5−48 k4 .
Claim 2.1 For all i = 1 , . . . , 10 and k ≥ 4, C1(k) ≥ Ci(k).
Proof. Observe that for all i = 1 , . . . , 10, each polynomial Ci(k) has the same denominator of 5 k7 − 35 k5 + 75 k4 − 48 k3. It is straightforward to verify that 5 k7 − 35 k5 + 75 k4 − 48 k3
is positive for all k ≥ 4. By examining the leading coefficients in the numerator of each polynomial, it is straightforward to check that C1(k) is the largest for k > 1000. For 4 ≤ k ≤ 1000, we have provided Sage code used to verify the claim in Appendix 6.1. By factoring C1(k) it follows that OPT k(C5) ≤ C1(k) = 1
k4 (12 k4 − 60 k3 + 120 k2 − 120 k + 48) ,
completing the proof of Theorem 1.4(i) when k ≥ 4.
Case 2: Suppose that k = 3. Assume that ( Gn)n≥1 is a K4-free convergent sequence of graphs. Each graph in the set H given below has a limit density equal to zero, and therefore can be removed from our calculations.
H =
{
, , , ,
}
.In this case, we will use the same polynomials Pi(k) for i = 1 , 2, 3, 4 that were provided earlier in the proof. We will define one new polynomial P6, which is nonnegative by (7). 10 P6 =
uv
(1
23
+1
23
−1
23
)2}~
σ3
=
2 · − − 2 · + + 6 · − 4 · ≥ 0Now suppose that
p1 = 1 /27 p2 = 13 /27 p3 = 8 /27
p4 = 2 /9 p6 = 17 /54 .
Then
d(C5) = lim
n→∞
d(C5, G n) ≤ ∑
F∈F 5\H
ν(C5, F )F +
4
∑
j=1
pj Pj (3) + p6P6.
Let cF denote the coefficient of each graph F after combining each of the two sums. We provide the values of cF for each graph in F5 \ H for which cF 6 = 0.
• c = c = c = c = c = 40 27
• c = 427
• c = − 227
• c = 11 18
• c = 154
• c = −17 54
• c = c = 1
• c = 79
• c = 89
SageMath code to verify this calculation can be found in Appendix 6.1. It is straightforward to verify that
d(C5) ≤ max {cF : F ∈ F 5 \ H} = 40 27 .
Furthermore, the set T3 of graphs for which cFi = 40 27 is given below.
T3 =
{
, , , ,
}
(12) This completes the proof of Theorem 1.4 (i). 11 2.1 Finding the Optimal Bound
We will now give a short description on how we found the functions z(k)Z(k) and pi(k)Pi(k)that were used in the proof of Theorem 1.4(i). If Fj ∈ F 5 is a graph for which cFj =OPT k(C5), then we call Fj a tight subgraph. In our proof of Theorem 1.4(i), the set T given below contains the tight subgraphs when k ≥ 4.
T =
{
, , , , , ,
}
The set T3, defined in (12), contains the tight subgraphs when k = 3. Note that T3 are the K4-free graphs from T . The following lemma, which appears as Lemma 2.4.3 in , states that any graph appearing with positive probability in the limit of ( Gn)n≥1 must be tight if Gn are extremal graphs.
Lemma 2.2 () Given (Gn)n≥1 a convergent sequence of Kk+1 -free graphs of increasing order, such that d(C5, G n) → OPT k(C5). Let d(H, G ∞) be the value of lim
n→∞
d(H, G n). Then for any exact solution, d(H, G ∞) > 0 implies that H must be a tight subgraph.
Using semidefinite programming, we verified that the conjectured upper bound of OPT k(C5)was correct for small values of k. In doing so, we were able to guess the correct types and labelled flags to use. It was a greedy process and there may be simpler solutions. This cor-responds to the polynomials Pi for i = 1 , . . . , 6. Note that each labelled flag is a four-vertex graph appearing in the Tur´ an graph. Next, Lemma 2.2 implies that each Fi ∈ F 5 that is a subgraph of the Tur´ an graph must have the property that cFi = OPT k(C5). Given this fact, we used SageMath to solve for the correct polynomials pi(k) and z(k). These agreed with the values calculated by the semidefinite program for small k.
3 Stability
In this section we will prove a stability lemma which states that for any Kk+1 -free graph G
on a sufficient number of vertices, if G contains “close” to the extremal number of five-cycles, then G can be made isomorphic to Tk(n) by adding or deleting a small number of edges.
Proposition 3.1 For two positive integers x1 and x2, if x1 ≥ x2 + 2 , then 1. x1x2 < (x1 − 1)( x2 + 1)
x1
(x2
2
) < (x2 + 1) (x1−12
).
Proof. The first inequality is clear from the equation below: (x1 − 1)( x2 + 1) = x1x2 + ( x1 − x2) − 1 ≥ x1x2 + 1 .
Since x2 − 1 < x 1 − 2, the second inequality follows immediately from the first inequality. The next proposition follows immediately from Lemma 3.3, which we will prove next. 12 Proposition 3.2 For any complete k-partite graph H on n vertices, ν(C5, H ) is maximized when the sizes of the partite sets are as equal as possible.
The following lemma will show that if H is a complete k-partite graph with unbalanced partite sets, then we can always increase the number of five-cycles in H by moving the vertices as to make H more balanced. Throughout the proof, we will assume for each i = 1 , . . . , k
that |Xi| = xi. For a graph G and a vertex v ∈ V (G), let νG(v, C 5) denote the number of five-cycles in G that contain v.
Lemma 3.3 Let H be a complete k-partite graph with partite sets X1, . . . , X k. Suppose that for two integers i and j,
xi ≥ xj + 2 .
Let H′ be the graph obtained from H by deleting a vertex in Xi and duplicating a vertex in
Xj . Then
ν(C5, H ′) > ν (C5, H ).
Proof. By symmetry we may assume that i = 1 and j = 2. Let H′ be obtained from H
by removing some vertex v ∈ X1 and adding a new vertex v′ to X2, where v′ is a duplicate of some vertex in X2. Letting X′
1
, . . . , X ′
k
denote the new partite sets in H′, |X′
1
| = x1 − 1,
|X′
2
| = x2 + 1, and |X′
q
| = xq for each q ∈ { 3, . . . , k }.Since the only five-cycles that have been deleted from H are those containing v, we only need to show that νH′ (v′, C 5) > ν H (v, C 5). Additionally, there is a one-to-one correspondence between the five cycles in H containing v and no other vertices in X1 ∪ X2 and the five cycles in H′ containing v′ and no other vertices in X′
1
∪ X′
2
. Because of this, we can focus only on those five-cycles which contained v and at least one other vertex in X1 ∪ X2.Let c(v, n 1, n 2) denote the number of five cycles in H containing v along with n1 and
n2 vertices in X1 and X2, respectively. We define c′(v′, n 1, n 2) in an identical manner, but pertaining to v′ and H′. In order to show that ν(C5, H ′) > ν (C5, H ), it suffices to show the following, 1. c′(v′, 1, 0) + c′(v′, 2, 0) + c′(v′, 0, 1) > c (v, 0, 1) + c(v, 0, 2) + c(v, 1, 0), and 2. c′(v′, 1, 1) + c′(v′, 2, 1) > c (v, 1, 1) + c(v, 1, 2) .
We will prove each of these inequalities as two separate claims. Throughout the proof we will assume that I = {3, . . . , k }.
Claim 3.4 c′(v′, 1, 0) + c′(v′, 2, 0) + c′(v′, 0, 1) > c (v, 0, 1) + c(v, 0, 2) + c(v, 1, 0) .
Proof. Since H is a complete k-partite graph,
c(v, 0, 1) = 6 x2 · ∑
{i,j }∈
(I
2
)
[( xi
2
)
xj +
(xj
2
)
xi
]
12 x2 · ∑
{i,j,h }∈
(I
3
)
xixj xh,c(v, 0, 2) = 4
(x2
2
)
· ∑
i∈I
(xi
2
)
6
(x2
2
)
· ∑
{i,j }∈
(I
2
)
xixj , and
c(v, 1, 0) = 4( x1 − 1) · ∑
{i,j }∈
(I
2
)
[( xi
2
)
xj +
(xj
2
)
xi
]
6 x1 · ∑
{i,j,h }∈
(I
3
)
xixj xh.
13 By counting in similar way in H′,
c′(v′, 1, 0) = 6( x1 − 1) · ∑
{i,j }∈
(I
2
)
[( xi
2
)
xj +
(xj
2
)
xi
]
12( x1 − 1) · ∑
{i,j,h }∈
(I
3
)
xixj xh,c′(v′, 2, 0) = 4
(x1 − 12
)
· ∑
i∈I
(xi
2
)
6
(x1 − 12
)
· ∑
{i,j }∈
(I
2
)
xixj , and
c′(v′, 0, 1) = 4 x2 · ∑
{i,j }∈ I
[( xi
2
)
xj +
(xj
2
)
xi
]
6 x2 · ∑
{i,j,h }∈ I
xixj xh.
Since x1 ≥ x2 + 2, it follows that c′(v′, 2, 0) > c ′(v′, 0, 2). Thus, it suffices to show that
c(v, 0, 1) + c(v, 1, 0) ≤ c′(v′, 0, 1) + c′(v′, 1, 0) .
It is straightforward to verify that
6· ∑
{i,j }∈
(I
2
)
[( xi
2
)
xj +
(xj
2
)
xi
]
+12 · ∑
{i,j,h }∈
(I
3
)
xixj xh ≥ 4· ∑
{i,j }∈
(I
2
)
[( xi
2
)
xj +
(xj
2
)
xi
]
+6 · ∑
{i,j,h }∈
(I
3
)
xixj xh.
This immediately implies that
c(v, 0, 1) − c(v, 0, 1) ≤ c′(v′, 1, 0) − c′(v′, 1, 0) ,
which proves the claim.
Claim 3.5 c′(v′, 1, 1) + c′(v′, 2, 1) > c (v, 1, 1) + c(v, 1, 2) .
Proof. For convenience, we will count c(v, 1, 1) + c(v, 1, 2) in the following way:
c(v, 1, 1) + c(v, 1, 2) = x1x2f11 +
(x2
2
)
x1f21 , (13) where fpq is a function independent of the values x1 and x2 used to count the number of five cycles containing v, p vertices from X1, and q vertices from X2. Using the same method to count c′(v′, 1, 1) + c′(v′, 2, 1), we get
c′(v′, 1, 1) + c′(v′, 2, 1) = ( x1 − 1)( x2 + 1) f11 +
(x1 − 12
)
(x2 + 1) f12 . (14) By Proposition 3.1, (x1 − 1)( x2 + 1) f11 > x 1x2f11 .
Moreover, since the sizes of each set Xj for all j ∈ I have not changed, f12 = f21 . Therefore,
(x1 − 12
)
(x2 + 1) f12 >
(x2
2
)
x1f21
by Proposition 3.1, completing the proof of the claim.
As each of Claims 3.4 and 3.5 are true, it follows that ν(C5, H ′) > ν (C5, H ), completing the proof of Lemma 3.3. For two graphs G and H of the same order, let Dist( G, H ) equal the minimum number of adjacencies that one needs to change in G in order to obtain a graph isomorphic to H.The parameter Dist( G, H ) is commonly known as the edit distance between G and H. Our main goal of this section is to prove the following lemma. 14 Lemma 3.6 (Stability Lemma) For every ε > 0, there exists an n0 and εF > 0 such that for every Kk+1 -free graph G of order n ≥ n0 with d(C5, G ) ≥ OP T k(C5) − εF , the edit distance between G and Tk(n) is at most εn 2.
The proof of Lemma 3.6 requires the following two lemmas along with Lemma 2.2. For a family of graphs F, we say that a graph G is F-free if G does not contain any member of
F as an induced subgraph.
Lemma 3.7 (Induced Removal Lemma ) Let F be a set of graphs. For each ε > 0,there exist n0 ≥ 0 and δ > 0 such that for every graph G of order n0 ≥ n, if G contains at most δn |V (H)| induced copies of H for every H ∈ F , then G can be made F-free by removing or adding at most εn 2 edges from G.
Let ( Gn)n≥1 be a convergent sequence of Kk+1 -free graphs. In the proof of Theorem 1.4(i), we found constants cFi for each Fi ∈ F 5 such that
d(C5, G n) ≤
33
∑
i=0
cFi Fi ≤ max {cFi : Fi ∈ F 5}
and max {cFi : Fi ∈ F 5} = OPT k(C5) = 1
k4 (12 k4 − 60 k3 + 120 k2 − 120 k + 48) .
Let P3 be the three vertex graph with exactly one edge; see Figure 1. The goal of Lemma 3.8 is to prove that if lim n→∞ (C5, G n) = OPT k(C5), then lim n→∞ (P3, G n) = 0. Figure 1: P3
Lemma 3.8 For each δF > 0, there exists εF > 0 and n0 = n0(δF ) such that any Kk+1 -free graph G on n ≥ n0 vertices with d(C5, G ) > OP T k(C5) − εF contains at most δF n3 induced copies of P3.
Proof. Let ( Gn)n≥1 be a convergent sequence of Kk+1 -free graphs maximizing the number of five-cycles. Let T be the set of tight subgraphs in F5 given by the proof of Theorem 1.4(i). This is the same set T provided at the end of Section 2.
T =
{
, , , , , ,
}
.Observe that for each graph F ∈ T ,
P (P3, F ) = 0 .
15 Since T contains the set of tight graphs, the following is a consequence of (4) for the sequence (Gn)n≥1,
P3 =
33
∑
i=0
P (P3, F i)Fi = 0 .
It follows that for the sequence ( Gn)n≥1,lim
n→∞
d(P3, G n) = 0 ,
which completes the proof of Lemma 3.8.
Proof of Lemma 3.6. Let εI > 0 and εF > 0, which we will determine later. By Lemma 3.7, there exists a δF > 0 and an n0 such that any Kk+1 -free graph on n ≥ n0
vertices containing at most δF n3 copies of P3 can be made P3-free after changing at most
εI n2 adjacencies. Assume that G is a graph on n ≥ n0 vertices such that
d(C5, G ) > OP T k(C5) − εF ,
where n0 is large enough to satisfy the conditions of Lemmas 3.7 and 3.8 so that G contains at most δF n3 copies of P3. Moreover, for sufficiently small εI ,
d(C5, G ) > OP T k−1(C5) + 2 · 5! · εI .
Using Lemma 3.7, let G′ be a P3-free graph obtained from G by changing at most εI n2
edges. Since each edge that was removed in this way was contained in at most n3 copies of
C5, ν(C5, G ′) ≥ ν(C5, G ) − εI n5. Therefore, 1. d(C5, G ′) > OP T k(C5) − 5! · εI − εF ,2. d(C5, G ′) > OP T k−1(C5) + 5! · εI .Using the previous two inequalities, along with the fact that G′ is P3-free, we will now show that G′ must be a complete k-partite graph.
Claim 3.9 G′ is a complete k-partite graph.
Proof. Since G′ does not contain any induced copies of P3 as a subgraph, each pair of non-adjacent vertices must have an identical neighborhood. Therefore, we can partition V (G′)into independent sets X1, . . . , X such that for all distinct i, j ∈ [], each vertex in Xi is adjacent to each vertex in Xj . Hence, G′ is a complete `-partite graph. Since
OP T k−1(C5) = lim
n→∞
d(C5, T k−1(n)) and d(C5, G ′) > OP T k−1(C5) + 5! · εI , Proposition 3.2 implies that G′ must be k-partite if n
is sufficiently large. At this point, we know that G′ only differs from Tk(n) in the sizes of the partite sets
X1, X 2, . . . , X k. The next claim will show that we can impose that the partite sets in G′
must be reasonably close to being balanced. 16 Claim 3.10 Let G′ be a complete k-partite graph with partite sets X1, X 2, . . . , X k. Then for any εT > 0, there exists δ > 0 such that if
d(C5, G ′) > OP T k(C5) − δ,
then for each i = 1 , . . . , k n(1 − εT )
k ≤ | Xi| ≤ n(1 + εT )
k .
Proof. For each i = 1 , . . . , k let xi = |Xi|. Let ε′(k − 1) > ε T and assume by symmetry that x1 = 1+ ε′(k−1)
k
n. If we picked x1 = 1+ ε′
k
n, we would get less pleasant expressions in what follows. We want to calculate an upper bound on d(C5, G ′). By Lemma 3.3, d(C5, G ′) is maximized if all remaining parts are balanced. That is, xi = 1−ε′
k
n for i = 2 , . . . , k . With knowing the sizes of all Xi, the following is a straightforward calculation,
d(C5, G ′) ≤ OP T k(C5) − 60 ε′2
(
1 − 6
k + 15
k2 − 18
k3 + 8
k4
)
60 ε′3
(
1 − 8
k + 25
k2 − 34
k3 + 16
k4
)
180 ε′4
( 1
k − 5
k2 + 8
k3 − 4
k4
)
− 12 ε′5
(
1 − 15
k2 + 30
k3 − 16
k4
)
o(1) ,
see Appendix 6.2 for a code in SageMath. For all k ≥ 3, the term 1 − 6
k
15
k2
− 18
k3
8
k4
is positive with minimum 881 at k = 3. For sufficiently small ε′ and large n, we get
d(C5, G ′) ≤ OP T k(C5) − 5ε′2.
This implies the statement of the claim. Returning to the proof of Lemma 3.6, suppose that ε > 0, and let εT = ε/ 2. Next, choose an εI ≤ ε/ 2 small enough so that εF and δF are sufficiently small. In particular, we must select εI , εF , and δF so that any k-partite graph G′ satisfying d(C5, G ′) > OP T k(C5) − 5! εI −
εF , must have partite sets X1, . . . , X k that satisfy
n(1 − ε/ 2)
k ≤ | Xi| ≤ n(1 + ε/ 2)
k
for all i = 1 , . . . , k . Then by changing at most ( εI + εT )n2 pairs we can obtain Tk(n) from the original graph G, which completes the proof of Lemma 3.6.
4 Exact Result
In this section we will prove Theorem 1.4(ii). First we will give a brief outline. As we have shown, if G is a Kk+1 -free graph on n vertices for large enough n that contains close to the extremal number of five-cycles, then the edit distance between G and Tk(n) is very small. Given such a graph G, the process of deleting and adding the necessary edges to transform
G into the Tur´ an graph actually increases the number of five-cycles. This will prove that
Tk(n) is the unique extremal graph for large enough n.17 Proof of Theorem 1.4(ii). Suppose that k ≥ 3. By Lemma 3.6, there exists an ε > 0and an integer n0 = n(k, ε ) so that for any Kk+1 -free graph G on n ≥ n0 vertices satisfying
d(C5, G ) > OP T k(C5) − ε,
we have that Dist( G, T k(n)) ≤ 12k10 n2.Let G be a graph on n vertices, where n is sufficiently large. In particular, n ≥
max {n(k, ε ), 2k5 + 1 } and G satisfies
d(C5, G ) > OP T k(C5) − ε′,
where ε′ ≤ min {ε, 1
k10
}. Lemma 3.6 gives a partition of V (G) into k sets X1, X 2, . . . , X k,where bnk c ≤ | Xi| ≤ d nk e for all i = 1 , . . . , k , so that by changing at most 12k10 n2 pairs uv for
u, v ∈ V (G), we can construct a new graph G′ from G so that G′ is isomorphic to Tk(n) and the partite sets of G′ are X1, X 2, . . . , X k.Call each edge that is removed in this process a surplus edge and call each edge that is added in this process a missing edge . For each vertex v ∈ V (G), let fv denote the sum of the total number of surplus edges and missing edges incident to v. Define the set X0 to contain each vertex v with fv > 1
k6
n. We will refer to each vertex in X0 as a bad vertex .
Claim 4.1 |X0| ≤ 1
k4
n.
Proof. Since fv > 1
k6
n for each vertex v ∈ X0 and the combined total of surplus edges and missing edges in G is at most 12k10 n2, it follows that 1
k6 n|X0| ≤ 1
k10 n2,
which proves Claim 4.1. For a graph G and a vertex v ∈ V (G) let NG(v) denote the neighborhood of v in G. For all v ∈ V (G), let di(v) denote the size of the set NG(v) ∩ (Xi \ X0). Let
d∗(v) =
k
∑
i=1
di(v).
By Claim 4.1, ⌊nk
⌋
− 1
k4 n ≤ | Xi \ X0| ≤
⌈nk
⌉
,
for all i = 1 , . . . , k . Thus, for each vertex v not contained in X0,
d∗(v) ≥
(k − 1
k − 1
k4 − 1
k6
)
n.
For two vertices u and v in a graph G, let NG(u, v ) denote the common neighborhood of u
and v, which is the set of all vertices in G adjacent to both u and v.
Claim 4.2 There are no surplus edges in G − X0.
18 Proof. Assume by way of contradiction that G−X0 contains a surplus edge uv . Our goal is to show that it would be in Kk+1 . Since uv is removed in the the process of transforming
G into the Tur´ an graph, we may assume by symmetry that u and v are contained in the same set X1. Since neither vertex is contained in X0,min {dj (v), d j (u)} ≥
( 1
k − 1
k4 − 1
k6
)
n
for each j = 2 , . . . , k . Therefore,
|NG(u, v ) ∩ (X2 \ X0)| ≥
( 1
k − 1
k4 − 2
k6
)
n > 0.
Pick one vertex w2 contained in NG(u, v ) ∩ (X2 \ X0). Since w2 is not contained in X0,
|NG(w2) ∩ NG(u, v ) ∩ (X3 \ X0)| ≥
( 1
k − 1
k4 − 3
k6
)
n > 0.
This implies that we can find some common neighbor, say w3, of u, v, and w2, where
w3 ∈ X3 \ X0. We continue the process of a selecting a vertex wj ∈ Xj \ X0 in the common neighborhood of the set {u, v, w 1, . . . , w j−1} for all j = 4 , . . . , k . This is possible because after selecting wj−1, the common neighborhood of the set {u, v, w 2, . . . , w j−1} contains at least ( 1
k − 1
k4 − jk6
)
n > 0vertices in Xj \ X0 for all j = 4 , . . . , k . This implies, however, that the set {u, v, w 2, . . . , w k}
obtained by selecting a vertex in this way from each partite set X2, . . . , X k induces a copy of Kk+1 in G, which is a contradiction. An immediate consequence of Claim 4.2 is that every surplus edge in G is incident to at least one vertex in X0, implying that G − X0 is a k-partite graph, albeit not necessarily complete k-partite. We will split the vertices of X0 into two classes. For each vertex v ∈ X0,one of the following holds. 1. There exists some index i ∈ { 1, 2, . . . , k } such that di(v) = 0. In this case we will call
v a type 1 vertex, or 2. di(v) > 0 for all i = 1 , . . . , n . In this case we will call v a type 2 vertex. As we are trying to show that every extremal graph is a complete balanced k-partite graph, we will now prove that G cannot contain any type 2 vertices. First in Claim 4.3, we will prove that if v is a type 2 vertex, then d∗(v) must be relatively small. In Claim 4.4, we will prove a lower bound on the number of five-cycles containing a vertex v. Finally, in Claim 4.5, we will show that a type 2 vertex cannot be contained in enough five-cycles to justify the claim that G is an extremal graph.
Claim 4.3 Let v ∈ X0 be a type 2 vertex. Then there exist distinct integers i and j where
1 ≤ i, j ≤ k such that
1 ≤ di(v) ≤ dj (v) ≤ 1
k5 n.
19 Proof. By symmetry, assume that 1 ≤ d1(v) and d1(v) ≤ dq(v) for all q = 2 , . . . , k . For contradiction, assume dq(v) > 1
k5
n for all q = 2 , . . . , k . Let w1 ∈ X1 \ X0 be adjacent to v in
G. Since w1 /∈ X0,
|NG(v, w 1) ∩ (X2 \ X0)| ≥ 1
k5 n − 1
k6 n,
implying that there exists a vertex w2 ∈ X2 \ X0 for which the set {v, w 1, w 2} induces a triangle in G. If we continue selecting vertices in this way, then for all q = 3 , . . . , k , there are at least 1
k5 n − q − 1
k6 n > 0vertices in Xq \ X0 that are adjacent to all of the previously selected vertices v, w 1, . . . , w q−1.This implies that we can select k vertices w1, . . . , w k so that the set {v, w 1, . . . , w k} induces a copy of Kk+1 in G, which is a contradiction. Therefore, there exists an index j ∈ { 2, . . . , k }
for which 1 ≤ d1(v) ≤ dj (v) ≤ 1
k5
n, completing the proof of Claim 4.3.
Claim 4.4 For all k ≥ 3, and v ∈ V (G), νG(v, C 5) ≥ (OP T k(C5) − 1
k10
)(n
4
) − 1
k5
n4.
Proof. Suppose by way of contradiction that there exists some vertex v for which
νG(v, C 5) <
(
OP T k(C5) − 1
k10
) ( n
4
)
− 1
k5 n4.
Since d(C5, G ) > OPT k(C5) − 1
k10
, it follows by averaging that there exists some vertex
u ∈ V (G) for which
νG(u, C 5) ≥
(
OP T k(C5) − 1
k10
) ( n
4
)
.
Let νG({u, v }, C 5) denote the number of five-cycles containing both u and v. Then
νG({u, v }, C 5) ≤ 2n3.
Let G′ be the graph obtained from G by deleting v and replacing it with a copy u′ of u.Since there is no edge between u′ and u, G′ is also Kk+1 -free. As there were previously
ν({u, v }, C 5) five-cycles containing u and v,
ν(C5, G ′) − ν(C5, G ) ≥ νG(u, C 5) − νG(v, C 5) − νG({u, v }, C 5) ≥ 1
k5 n4 − 2n3 > 0since n > 2k5. This, however, contradicts the assumption that G is an extremal graph as
ν(C5, G ′) > ν (C5, G ). Therefore, if n is sufficiently large it follows that for each v ∈ V (G),
νG(v, C 5) ≥
(
OP T k(C5) − 1
k10
) ( n
4
)
− 1
k5 n4,
which completes the proof of Claim 4.4. In Claims 4.5 and 4.7 we count the number of 5-cycles containing a particular vertex
v ∈ X0. We use the following argument repeatedly. We want to count the number of 5-cycles vu 1u2u3u4v, where v is in X0, u1 ∈ Xi, u4 ∈ Xj and u2, u 3 ∈ V (G) \ X0. Assume we already picked u1 and u4 and want to count the number of choices for u2 and u3. We distinguish two cases. 20 1. i = j: First u2 can be in any of the remaining k − 1 parts. Then u3 has k − 2 choices for a part to complete the 5-cycle as it needs to avoid the parts containing u2 and u4
and these are distinct. After multiplying by n2/k 2, the number of choices for u2 and
u3 in each of the selected parts, we get (k − 1)( k − 2)
k2 n2
choices for u2 and u3 together. 2. i 6 = j: We further distinguish two cases. If u2 /∈ Xj , then there are k − 2 parts which could contain u2 and k − 2 parts which could contain u3. If u2 ∈ Xj , then there are
k − 1 parts which could contain u3. After including the number of choices in each part, we get ((k − 2) 2
k2 + k − 1
k2
)
n2
choices for u2 and u3 together.
Claim 4.5 G does not contain any type 2 vertices.
Proof. Assume for contradiction that v ∈ X0 is a type 2 vertex. Then by Claim 4.3 there are two sets, say X1 and X2, such that 1 ≤ d1(v) ≤ d2(v) ≤ 1
k5 n.
We will now provide an upper bound on the value of νG(v, C 5). We will count the maximum number of such five-cycles of the form vu 1u2u3u4v based on the locations of u1 and u4 as follows: 1. If u1, u 4 ∈ X1 \ X0 or u1, u 4 ∈ X2 \ X0:2
( nk5
2
)(k − 1)( k − 2)
k2 n2. (15) 2. u1 ∈ X1 \ X0 and u4 ∈ X2 \ X0:
( nk5
)2 ((k − 2) 2
k2 + k − 1
k2
)
n2. (16) 3. u1 ∈ (X1 \ X0) ∪ (X2 \ X0) and u4 /∈ X1 ∪ X2:2nk5 · nk
((k − 2) 2
k2 + k − 1
k2
)
n2. (17) 4. u1, u 4 /∈ X1 ∪ X2:(k − 2) ·
(nk
2
)
· (k − 1)( k − 2)
k2 n2 +
(k − 22
)
· n2
k2 ·
((k − 2) 2
k2 + k − 1
k2
)
n2. (18) 21 Finally, there are at most 2n4
k4
five-cycles containing v and at least one other vertex in X0.Combining this, along with the upper bounds obtained in equations (15)–(18),
νG(v, C 5) ≤ n4
24
(
12 − 84
k + 228
k2 − 300
k3 + 216
k4 + 48
k6 − 144
k7 + 144
k8 + 48
k10 − 144
k11 + 120
k12
)
.
The SageMath code for verifying this fact can be found in Appendix 6.3. This implies that for large enough n,
(
OP T k(C5) − 1
k10
) ( n
4
)
− νG(v, C 5) ≥ 1
k5 n4.
Using SageMath, we verified that this was true for 3 ≤ k ≤ 1000. After that, it is straight-forward to check the coefficients in order to verify this fact. This contradicts Claim 4.4 since
G was assumed to be an extremal graph. Therefore, G does not contain any type 2 vertices. Since G does not contain any type 2 vertices, we can place each vertex v ∈ X0 into the set Xi for which di(v) = 0. In order to show that G is a complete k-partite graph, we must show that any pair of vertices u and v that were in X0 and go to the same Xi cannot be adjacent. The next claim will provide an upper bound on the “good degree” of at least one of these adjacent vertices.
Claim 4.6 Suppose that u and v are two adjacent type 1 vertices such that dj (u) = dj (v) = 0
for some index j ∈ { 1, . . . , k }. Then there exists some index i ∈ { 1, . . . , k } such that i 6 = j
and
di(u) ≤ k2 + 1 2k3 n or di(v) ≤ k2 + 1 2k3 n.
Proof. By symmetry we may assume that j = 1. Assume for contradiction that
|NG(u, v ) ∩ (Xi \ X0)| > 1
k3 n
for all i = 2 , . . . , k . Using an identical argument to the one made in the proof of Claim 4.3, there exists a set {w2, . . . , w k} such that wi ∈ (Xi \X0) and the set {u, v, w 2, . . . , w k} induces a Kk+1 in G, which is a contradiction. This implies that for at least one index i,
|NG(u, v ) ∩ (Xi \ X0)| ≤ 1
k3 n.
Without loss of generality assume that di(u) ≤ di(v). Then
di(u) ≤ n
2
( 1
k − 1
k3
)
nk3 = k2 + 1 2k3 n,
which completes the proof of Claim 4.6. We will now show that the vertex u of low degree described in the previous claim cannot be contained in enough five-cycles to justify the assumption that G is an extremal graph. Unlike Claim 4.5, we will only show that the two vertices u and v from Claim 4.6 cannot be adjacent. 22 Claim 4.7 Suppose that u and v are type 1 vertices such that dj (u) = dj (v) = 0 for some
j = 1 , . . . , k . Then u and v are not adjacent.
Proof. By symmetry we may assume that d1(u) = d1(v) = 0. Assume for contradiction that u and v are adjacent. By symmetry and Claim 4.6, we may assume that d1(u) = 0 and
d2(u) ≤ k2 + 1 2k3 n.
In a similar manner as in Claim 4.5, we will count the number of five-cycles of the form
uv 1v2v3v4u incident to u by considering the possibilities for the locations of v1 and v4 as follows: 1. v1, v 4 ∈ X2 \ X0: (k2+1 2k3 n
2
)(k − 1)( k − 2)
k2 n2. (19) 2. v1 ∈ X2 \ X0 and v4 /∈ X2:
k2 + 1 2k3 · k − 2
k
((k − 2) 2
k2 + k − 1
k2
)
n4. (20) 3. v1, v 4 /∈ X2:(k − 2) ·
(nk
2
)
· (k − 1)( k − 2)
k2 n2 +
(k − 22
)
· n2
k2 ·
((k − 2) 2
k2 + k − 1
k2
)
n2. (21) There are at most 2
k4
n4 five-cycles containing u and at least one other vertex in X0. Com-bining this along with equations (19)–(21),
νG(u, C 5) ≤
(
12 − 72
k + 171
k2 − 189
k3 + 96
k4 + 90
k5 + 57
k6 − 9
k7 + 6
k8
) n4
24 .
For k > 1000 it is clear that 12 − 72
k + 171
k2 − 189
k3 + 96
k4 + 90
k5 + 57
k6 − 9
k7 + 6
k8 ≤ OPT k(C5) − 1
k10 .
The SageMath code for verifying that this is also true for 3 ≤ k ≤ 1000 found in Ap-pendix 6.4. Given this fact, it is straightforward to verify that
(
OPT k(C5) − 1
k10
) ( n
4
)
− νG(u, C 5) ≥ 1
k5 n4
for large enough n. This, however contradicts Claim 4.4, which implies that u and v are not adjacent. Claim 4.7 implies that if u and v are type 1 vertices for which di(v) = di(u) = 0, then u
and v cannot be adjacent. This means that we can place each type 1 vertex v into the set
Xi for which di = 0. Since G does not contain any type 2 vertices, this implies that G is a k-partite graph. Since G maximizes the number of (possibly non-induced) C5 subgraphs, it follows that G must be a compelte k-partite graph. Finally, Proposition 3.2 implies that
G is isomorphic to Tk(n), implying that for large enough n, the Tur´ an graph Tk(n) is the unique extremal graph maximizing the number of C5 subgraphs. 23 5 Conclusion
In , Palmer and Gerbner showed that if H is a graph and F is a graph with chromatic number k + 1, then ex( n, H, F ) ≤ ex( n, H, K k+1 ) + o(n|H|).
Since the Tur´ an graph Tk(n) does not contain any ( k + 1)-chromatic graph as a subgraph, this immediately implies that for any ( k + 1)-chromatic graph F ,lim
n→∞
d(C5, F ) = 1
k4 (12 k4 − 60 k3 + 120 k2 − 120 k + 48) ,
which closely resembles the Erd˝ os-Stone-Simonovits theorem. Let G be a graph with chromatic number k. Then for any r ≥ k, the Tur´ an graph
Tr(n) contains G as a subgraph. When trying to maximize the copies of G among Kr+1 -free graphs, evidence seems to suggest that Tr(n) is extremal, as we have shown to be the case with five-cycles. While a complete r-partite graph seems to frequently be the best option, it is not always optimal to balance the partite sets. Let St be a star with t leaves, also known as K1,t . Is it easy to see ex( n, S 4, K 3) is achieved by an unbalanced bipartite graph. For more detailed treatment of stars, see Cutler, Nir, and Radcliffe . It seems very likely that while the Tur´ an graph is not always extremal, that some complete r-partite graph will be best possible.
Conjecture 5.1 Let G be a graph and let k > χ (G) be an integer. Then for all r ≥ k,ex (n, G, K r) is realized by a complete (r − 1) -partite graph.
While an unbalanced r-partite graph might best possible in some cases, we believe that for large enough r ≥ χ(G), the value of ex( n, G, K r+1 ) is realized by the Tur´ an graph. As r
increases, any G-subgraph in Tr(n) can be taken from an increasing number of partite sets. Thus, as r grows larger, the effect of G being unbalanced becomes minimized. The following conjecture also appears in .
Conjecture 5.2 Let G be a graph and let r > |V (G)| be an integer. Then ex (n, G, K r) is realized by the Tur´ an graph Tr−1(n).
Acknowledgement
We would like to thank the anonymous referees for their comments that greatly improved the presentation of the paper.
References
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SageMath code on next pages can be also obtained at .
F0 = F1 = F2 = F3 =
F4 = F5 = F6 = F7 =
F8 = F9 = F10 = F11 =
F12 = F13 = F14 = F15 =
F16 = F17 = F18 = F19 =
F20 = F21 = F22 = F23 =
F24 = F25 = F26 = F27 =
F28 = F29 = F30 = F31 =
F32 = F33 =
Table 1: Graphs on 5 vertices up to isomorphism.
Hσ1
4,1
=1
23 Hσ1
4,2
=1
23
Hσ2
4,3
=1
23 Hσ2
4,4
=1
23 Hσ2
4,5
=1
23
Hσ3
4,6
=1
23 Hσ3
4,7
=1
23 Hσ3
4,8
=1
23
Table 2: Labeled graphs on four vertices. 27 6.1 Proof of Claim 2.1
SageMath code f o r Claim 2 . 1
var ( ’ k ’ )
Vector c o n t a i n i n g c o e f f i c i e n t s f o r each graph i n F 5
c F i = [ 0 ] ∗ 3 4
Z(K)
Zk = [ ( k ∗k∗k∗k−10 ∗k∗k∗k+35 ∗k∗k−50 ∗k+24)/k ˆ 4 ] ∗ 3 4 Zk [ 3 3 ] = ( −10 ∗k∗k∗k+35 ∗k∗k−50 ∗k+24)/kˆ4
P i (K) f i r s t i n i t w i t h 0
P1k = [ 0 ] ∗ 3 4 P2k = [ 0 ] ∗ 3 4 P3k = [ 0 ] ∗ 3 4 P4k = [ 0 ] ∗ 3 4 P5k = [ 0 ] ∗ 3 4
P 1 (K)
P1k [ 0 ] = 10 ∗ kˆ2 − 20 ∗ k + 10 P1k [ 1 ] = kˆ2 −2∗k +1 P1k [ 3 ] = −k + 1 P1k [ 4 ] = −4∗k + 4 P1k [ 1 8 ] = 1 P1k [ 1 9 ] = 1
P 2 (K)
P2k [ 1 8 ] = 3 ∗kˆ2 − 12 ∗ k + 12 P2k [ 2 9 ] = kˆ2 − 6∗k + 8 P2k [ 3 1 ] = −4∗k+10 P2k [ 3 2 ] = 3
P 3 (K)
P3k [ 4 ] = 6 ∗kˆ2 − 24 ∗ k + 24 P3k [ 1 1 ] = kˆ2 − 4∗k + 4 P3k [ 1 7 ] = −k+2 P3k [ 1 9 ] = −6∗k+12 P3k [ 3 1 ] = 2 P3k [ 3 2 ] = 3
P 4 (K)
P4k [ 1 9 ] = 6P4k [ 2 8 ] = −128 P4k [ 3 0 ] = 2P4k [ 3 1 ] = −4
P 5 (K)
P5k [ 1 9 ] = 6∗kˆ2 −36 ∗k+54 P5k [ 3 0 ] = 2∗kˆ2 −20 ∗k+42 P5k [ 3 1 ] = 4∗kˆ2 − 24 ∗ k + 36 P5k [ 3 2 ] = −24 ∗k + 84 P5k [ 3 3 ] = 120
S c a l i n g f u n c t i o n s z ( k ) and p i ( k )
zk = 6 ∗(5 ∗ kˆ3 − 20 ∗ kˆ2 + 30 ∗ k − 1 6 ) / ( 5 ∗ kˆ3 − 35 ∗ kˆ2 + 75 ∗ k − 48) p1k = 3∗( kˆ5 − 8∗kˆ4 + 22 ∗ kˆ3 − 24 ∗ kˆ2 + 8 ∗k )/ \
(5 ∗ kˆ7 − 35 ∗ kˆ6 + 75 ∗ kˆ5 − 48 ∗ k ˆ4) p2k = (10 ∗ kˆ5 − 60 ∗ kˆ4 + 109 ∗ kˆ3 − 76 ∗ kˆ2 + 18 ∗ k )/ \
(5 ∗ kˆ7 − 35 ∗ kˆ6 + 75 ∗ kˆ5 − 48 ∗ k ˆ4) p3k = (5 ∗ kˆ5 − 28 ∗ kˆ4 + 45 ∗ kˆ3 − 28 ∗ kˆ2 + 6 ∗k )/ \
(5 ∗ kˆ7 − 35 ∗ kˆ6 + 75 ∗ kˆ5 − 48 ∗ k ˆ4) p4k = ( 1 / 4 ) ∗ ( 5 ∗ kˆ7 − 30 ∗ kˆ6 + 53 ∗ kˆ5 − 52 ∗ kˆ4 + 94 ∗ kˆ3 − 96 ∗ kˆ2 + 24 ∗ k )/ \
(5 ∗ kˆ7 − 35 ∗ kˆ6 + 75 ∗ kˆ5 − 48 ∗ k ˆ4) p5k = ( 1 / 4 ) ∗ ( 1 5 ∗ kˆ5 − 60 ∗ kˆ4 + 78 ∗ kˆ3 − 40 ∗ kˆ2 + 8 ∗k )/ \
(5 ∗ kˆ7 − 35 ∗ kˆ6 + 75 ∗ kˆ5 − 48 ∗ k ˆ4)
Number o f C 5s i n each o f t h e 34 g r a p h s .
f i v e c y c l e s = [ 0 ] ∗ 2 6 + [ 1 , 1 , 1 , 2 , 2 , 4 , 6 , 1 2 ]
def t e s t p o s i t i v e f o r s m a l l k ( x ) :
for r in [ 4 . . 1 0 0 0 ] :
i f x . s u b s t i t u t e ( k=r ) <= 0 :
print ( ”ERROR: ” , x , ” i s not p o s i t i v e f o r k=” , r )
return
Denominator f o r a l l cFi i n t h e r e s u l t
den = 5 ∗kˆ7 − 35 ∗ kˆ6 + 75 ∗ kˆ5 − 48 ∗ kˆ4
Test i f denominator i s p o s i t i v e f o r s m a l l k
t e s t p o s i t i v e f o r s m a l l k ( den )
for i in [ 0 . . 3 3 ] : c F i [ i ] = expand ( f a c t o r ( ( zk ∗( Zk [ i ] ) + p1k ∗P1k [ i ] + p2k ∗P2k [ i ] +\
p3k ∗P3k [ i ] + p4k ∗P4k [ i ] + p5k ∗P5k [ i ] + f i v e c y c l e s [ i ] )∗ den ) )
This i s t h e p o l y n o m i a l f o r each o f t h e t i g h t g r a p h s ( when m u l t i p l i e d # by den ) # I t i s t h e c o e f f i c i e n t a t c { F 0 } = cFi [ 0 ]
29 o p t p o l = c F i [ 0 ]
#o p t p o l = 60 ∗ k ˆ7 − 720 ∗ k ˆ6 + 3600 ∗ k ˆ5 − 9876 ∗ k ˆ4 + 16320 ∗ k ˆ3 \
− 16440 ∗ k ˆ2 + 9360 ∗ k − 2304 # P r i n t i n g o f a l l c o e f f i c i e n t s c { F i }
print ( ” P r i n t i n g the r e s u l t i n g c o e f f i c i e n t s ” )
for i in [ 0 . . 3 3 ] :
print ( ”cF {} = {} / {} ” . format ( i , c F i [ i ] , den ) )
Test t h a t cFi [ 0 ] == optimum v a l u e
optimum = −60/k + 120/ kˆ2 − 120/ kˆ3 + 48/ kˆ4 + 12
##### S t a r t o f Claim 2 . 1 # C a l c u l a t i n g t h e d i f f e r e n c e s C 1 − C i f o r i = 2 , . . , 1 0 . # We do i t by c h e c k i n g c { F 0 } − c { F i } f o r a l l i
c F 0 c F i = [ 0 ] ∗ 3 4
for i in [ 0 . . 3 3 ] : c F 0 c F i [ i ] = expand ( c F i [ i ] − o p t p o l )
Showing t h a t c F0 i s l a r g e s t by d i s p l a y i n g t h e d i f f e r e n c e . # The l e a d i n g c o e f f i c i e n t a t k ˆ7 i s n e g a t i v e # and t e s t s f o r s m a l l v a l u e s o f k by e v a l u a t i o n
def t e s t n o t p o s i t i v e f o r s m a l l k ( x ) :
for r in [ 4 . . 1 0 0 0 ] :
i f x . s u b s t i t u t e ( k=r ) > 0 :
print ( ”ERROR: ” , x , ” i s p o s i t i v e f o r k=” , r )
return print ( )
print ( ” Showing t h a t c F0 i s l a r g e s t ” )
print ( ” − s e e d i f f e r e n c e i s 0 o r l e a d i n g c o e f f i c i e n t n e g a t i v e ” )
for i in [ 0 . . 3 3 ] :
print ( ” numerator ( c F0 − c F {} )=” . format ( i ) , c F 0 c F i [ i ] ) t e s t n o t p o s i t i v e f o r s m a l l k ( c F 0 c F i [ i ] )
print ( ” a l l done ” ) 30 # SageMath code f o r Claim 2 . 1 when k=3 # This code c a l c u l a t e s t h e c o e f f i c i e n t s cFi #f o r t h e c a s e o f k = 3 i n Claim 2 . 1 . # In \ mathcal {F} 5 : F20 , F24 , F30 , F32 , and F33 a l l c o n t a i n a K4 . # Here we have removed t h o s e g r a p h s and re −i n d e x e d t h e remaining g r a p h s .
c F i = [ 0 ] ∗ 2 9
Counting t h e number o f C5s i n each graph .
c o n s t a n t s = [ 0 ] ∗ 2 9 c o n s t a n t s [ 2 4 ] = 1 c o n s t a n t s [ 2 5 ] = 1 c o n s t a n t s [ 2 6 ] = 1 c o n s t a n t s [ 2 7 ] = 2 c o n s t a n t s [ 2 8 ] = 4
Pi (K) f i r s t i n i t w i t h 0
P1k = [ 0 ] ∗ 3 4 P2k = [ 0 ] ∗ 3 4 P3k = [ 0 ] ∗ 3 4 P4k = [ 0 ] ∗ 3 4 P6k = [ 0 ] ∗ 3 4 P1k [ 0 ] = 40 P1k [ 1 ] = 4 P1k [ 3 ] = −2P1k [ 4 ] = −8P1k [ 1 8 ] = 1 P1k [ 1 9 ] = 1 P2k [ 1 8 ] = 3 P2k [ 2 7 ] = −1P2k [ 2 8 ] = −2P3k [ 4 ] = 6 P3k [ 1 1 ] = 1 P3k [ 1 7 ] = −1P3k [ 1 9 ] = −6P3k [ 2 8 ] = 2 31 P4k [ 1 9 ] = 6 P4k [ 2 6 ] = −1P4k [ 2 8 ] = −4P6k [ 1 1 ] = 1 P6k [ 2 3 ] = 2 P6k [ 2 2 ] = −1P6k [ 2 7 ] = −2P6k [ 1 7 ] = 1 P6k [ 1 9 ] = 6 P6k [ 2 8 ] = −4
The s c a l i n g c o e f f i c i e n t s f o r k = 3
p1k = 1/27 p2k = 13/27 p3k = 8/27 p4k = 2/9 p6k = 17/54
This c a l c u l a t e s cFi
for i in [ 0 . . 2 8 ] : c F i [ i ] = p1k ∗P1k [ i ] + p2k ∗P2k [ i ] + p3k ∗P3k [ i ]
+ p4k ∗P4k [ i ] + p6k ∗P6k [ i ] + c o n s t a n t s [ i ]
This p r i n t s t h e v a l u e o f cFi a l o n g w i t h t h e ( p o s s i b l y ) re −i n d e x e d graph .
for i in [ 0 . . 2 8 ] :
print ( ’ c o e f f i c i e n t o f F ’ , i , ’= ’ , c F i [ i ] ) 32 6.2 SageMath code for Claim 3.10
SageMath code Claim 3 . 1 0
var ( ’ k , e ’ )
t h i s i s t h e s i z e o f t h e s e t s . # We s t a r t by u s i n g e p s i l o n ∗( k −1) f o r e a s i e r c o u n t i n g .
x = (1 + e ∗( k −1))/ k y = (1 − e )/ k
These count t h e number o f f i v e c y c l e s . # The f i r s t i s t h e one we use . # The second i s a s a n i t y c h e c k .
def f i v e c y c l e c o u n t ( x , y ) :
h e r e we count by p i c k i n g one v e r t e x i n x , # t h e n c o u n t i n g t h e number o f p o s s i b l e f i v e c y c l e s .
n e i g h b o r s i n s a m e s e t s = \
x ∗( k −1) ∗( y ˆ 2 / 2 ) ∗ ( y ˆ2 ∗( k −2) ∗(k −3) + x ∗( k −2) ∗y ∗2) n e i g h b o r s i n d i f f s e t s = \
x ∗( ( y ∗( k − 1 ) ) ∗ ( ( k −2) ∗y ) ) / 2 ∗ ( ( k −3) ∗(k −3) ∗yˆ2 + ( k −2) ∗yˆ2 + x ∗( k −3) ∗y )
This i s c o u n t i n g t h e number o f f i v e −c y c l e s not i n X 1 . # Note t h a t i t i s e q u a l t o t h e s a n i t y c h e c k b u t w i t h k −1.
nobadset twosame = ( y ˆ3 ∗( k −1) ∗(k −2)/2) ∗( y ˆ2 ∗( k −2) ∗(k −3)) nobadset nosame = \
( y ˆ3 ∗( k −1) ∗(k −2) ∗(k −3)/2) ∗( ( k −3) ∗(k −3) ∗yˆ2 + ( k −2) ∗y ˆ2)
return 120 ∗( n e i g h b o r s i n d i f f s e t s + \
n e i g h b o r s i n s a m e s e t s ) + \
24 ∗( nobadset twosame + nobadset nosame )
def s a n i t y ( y ) : nobadset twosame = ( y ˆ3 ∗ k ∗( k −1)/2) ∗( y ˆ2 ∗( k −1) ∗(k −2)) nobadset nosame = \
( y ˆ3 ∗ k ∗( k −1) ∗(k −2)/2) ∗( ( k −2) ∗(k −2) ∗yˆ2 + ( k −1) ∗y ˆ2)
return 24 ∗( nobadset twosame + nobadset nosame ) 33 f = f i v e c y c l e c o u n t ( x , y )
t o c h e c k our count i s c o r r e c t , # n o t i c e t h a t t h e non −e p s i l o n terms e q u a l OPT.
view ( f . c o l l e c t ( e ) ) view ( expand ( s a n i t y (1/ k ) ) ) 34 6.3 SageMath code for Claim 4.5
Sage code f o r c l a i m 4 . 5 − showing t h e r e are no t y p e 2 v e r t i c e s
var ( ’ k ’ )
This f u n c t i o n c o u n t s t h e number o f f i v e c y c l e s u s i n g e q u a t i o n s ( 1 5 ) − ( 1 8 )
def f i v e c y c l e c o u n t ( k ) : o n e b a d s e t t w o b a d v e r t i c e s = 2∗( 1/(2 ∗ k ˆ10) ) ∗ ( ( k −1) ∗(k −2)/kˆ2 )t w o b a d s e t t w o b a d v e r t i c e s = \
(1/ k ˆ10 + 2/k ˆ 6 ) ∗ ( ( k −2)ˆ2/kˆ2 + ( k −1)/kˆ2 )nobadset twosame = ( ( k −2)/(2 ∗ k ˆ2) ) ∗ ( ( k −1) ∗(k −2)/kˆ2 )nobadset nosame = \
( ( k −2) ∗(k −3)/(2 ∗ k ˆ2) ) ∗ ( ( k −2)ˆ2/kˆ2 + ( k −1)/kˆ2 )with X0 = 2/kˆ4
return 24 ∗( o n e b a d s e t t w o b a d v e r t i c e s \
t w o b a d s e t t w o b a d v e r t i c e s + nobadset twosame \
nobadset nosame + with X0 )
This g i v e s t h e sum o f e q u a t i o n s ( 1 5 ) − ( 1 8 ) f a c t o r e d i n a n i c e way .
e x p a n d e d f i r s t c h e c k = expand ( f i v e c y c l e c o u n t ( k ) )
print ( ’ f i v e c y c l e s c o n t a i n i n g a type 2 v e r t e x : \ n ’ , e x p a n d e d f i r s t c h e c k )
a c t u a l upper bound once we ac co u nt f o r t h e v e r t i c e s i n X 0
def bad ub ( k ) :
return f i v e c y c l e c o u n t ( k )
The a v e r a g e ” d e n s i t y ” o f f i v e c y c l e s c o n t a i n i n g a p a r t i c u l a r v e r t e x
def good ub ( k ) :
return −60/k + 120/ kˆ2 − 120/ kˆ3 + 48/ kˆ4 + 12 − 1/k ˆ10
The d i f f e r e n c e between t h e o p t i m a l and t h e count f o r t y p e 2 .
def e p s i l o n ( k ) :
return good ub ( k ) − bad ub ( k )
print ( ’ d i f f e r e n c e : ’ , f a c t o r ( e p s i l o n ( k ) ) )
This v e r i f i e s t h a t f o r s m a l l v a l u e s o f k , count ( r ) g r e a t e r than 1/ k ˆ5
def c o u n t c h e c k ( a , b ) :
for i in [ a . . b ] :
i f e p s i l o n ( i ) < 1/( i ˆ 5 ) : 35 return ” the d i f f e r e n c e i s l e s s than 1/kˆ5 f o r k =” , i
return ” ’ d i f f e r e n c e ’ i s g r e a t e r than 1/kˆ5 f o r a l l v a l u e s ”+ \
” o f ”+ s t r ( a)+” <= k up t o ”+ s t r ( b )
print ( c o u n t c h e c k ( 3 , 1 0 0 0 ) ) 36 6.4 SageMath code for Claim 4.7
SageMath code f o r Claim 4 . 7
var ( ’ k ’ )
This f u n c t i o n c o u n t s t h e number o f f i v e −c y c l e s i n e q u a t i o n s ( 1 9 ) − ( 2 1 )
def f i v e c y c l e c o u n t ( k ) : o n e b a d s e t t w o b a d v e r t i c e s = \
( ( 1 / 2 ) ∗ ( ( k ˆ2+1)/(2 ∗ k ˆ3) )ˆ2 ) ∗ ( ( k −1) ∗(k −2)/kˆ2 )o n e b a d s e t o n e b a d v e r t i c e s = \
( ( k ˆ2+1)/(2 ∗ k ˆ3) ) ∗ ( ( k −2)ˆ3/kˆ3 + ( k −1) ∗(k −2)/kˆ3 )nobadset twosame = \
( ( k −2)/(2 ∗ k ˆ2) ) ∗ ( ( k −1) ∗(k −2)/kˆ2 )nobadset nosame = \
( ( k −2) ∗(k −3)/(2 ∗ k ˆ2) ) ∗ ( ( k −2)ˆ2/kˆ2 + ( k −1)/kˆ2 )with X0 = 2/k ˆ ( 4 )
return 24 ∗( o n e b a d s e t t w o b a d v e r t i c e s \
o n e b a d s e t o n e b a d v e r t i c e s + nobadset twosame \
nobadset nosame+with X0 )
This g i v e s t h e sum o f e q u a t i o n s ( 1 9 ) − ( 2 1 ) f a c t o r e d i n a n i c e way
e x p a n d e d f i r s t c h e c k = expand ( f i v e c y c l e c o u n t ( k ) )
print ( ’ f i v e c y c l e s c o n t a i n i n g a t y p e 1 v e r t e x : \ n ’ , e x p a n d e d f i r s t c h e c k )
The upper bound on f i v e c y c l e s c o n t a i n i n g a s u b o p t i m a l t y p e 1 v e r t e x
def bad ub ( k ) :
return f i v e c y c l e c o u n t ( k )
def g o o d l b ( k ) :
return −60/k + 120/ kˆ2 − 120/ kˆ3 + 48/ kˆ4 + 12 − 1/k ˆ10
d i f f e r e n c e from o p t i m a l v a l u e
def e p s i l o n ( k ) :
return g o o d l b ( k ) − bad ub ( k )
print ( ’ d i f f e r e n c e : ’ , f a c t o r ( e p s i l o n ( k ) ) )
This v e r i f i e s t h a t f o r s m a l l v a l u e s o f k , e p s i l o n ( k ) i s g r e a t e r than 1/ k ˆ5
def c o u n t c h e c k ( a , b ) : 37 for i in [ a . . b ] :
i f e p s i l o n ( i ) < 1/( i ˆ ( 5 ) ) :
return ” the d i f f e r e n c e i s l e s s than 1/kˆ5 f o r k =” , i
return ” ’ d i f f e r e n c e ’ i s g r e a t e r than 1/kˆ5 f o r a l l v a l u e s ”+ \
” o f ”+ s t r ( a)+” <= k up t o ”+ s t r ( b )
print ( c o u n t c h e c k ( 3 , 1 0 0 0 ) ) 38 |
14824 | https://www.vedantu.com/maths/collinear-vectors | Courses for Kids
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How to Prove Three Vectors or Points Are Collinear in Geometry
Collinear Vectors are essential for solving many Class 11 and 12 Maths and Physics questions, especially when checking whether points or forces lie in a straight line. Learning this concept makes it easier to approach exam problems involving vector direction, parallelism, and geometry confidently.
Formula Used in Collinear Vectors
The standard formula is: ( \vec{a} = k\vec{b} ), where “k” is a scalar. Alternatively, for vectors ( \vec{a} = (a_1,a_2,a_3) ) and ( \vec{b} = (b_1,b_2,b_3) ), check collinearity with ( \frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} ) (as long as none are zero).
Here’s a helpful table to understand Collinear Vectors more clearly:
Collinear Vectors Table
| Condition | Description | Applies To |
---
| Scalar Multiple | ( \vec{a} = k\vec{b} ) | All Vectors |
| Equal Ratio of Components | ( \frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} ) | 3D Vectors |
| Zero Cross Product | ( \vec{a} \times \vec{b} = \vec{0} ) | 3D Vectors |
This table shows the main ways to check for collinear vectors in real problems, especially in board exams or when solving geometry questions.
Worked Example – Solving a Problem
Let’s check if the vectors ( \vec{P}=(3,4,5) ) and ( \vec{Q}=(6,8,10) ) are collinear:
Write the vectors clearly: ( \vec{P}=(3,4,5) ), ( \vec{Q}=(6,8,10) \ ) 2. Find the ratios of each corresponding component:
( \frac{3}{6} = \frac{1}{2} ) ( \frac{4}{8} = \frac{1}{2} ) ( \frac{5}{10} = \frac{1}{2} )
Since all ratios are equal, vectors ( \vec{P} ) and ( \vec{Q} ) are collinear.
Alternatively, you can use the cross product test for 3D vectors. For more on cross products, see Vector Cross Product on Vedantu.
Practice Problems
Are the vectors ( \vec{a} = (2, 6, -4) ) and ( \vec{b} = (1, 3, -2) ) collinear?
Show that the vectors joining the points A(1,2), B(3,6), and C(5,10) are collinear.
If ( \vec{m} = (k, 8, 12) ) and ( \vec{n} = (3, 12, 18) ) are collinear, find the value of k.
Which of the following are not collinear vectors: ( (1,2,3), (2,4,6), (5,7,8) )?
Common Mistakes to Avoid
Confusing collinear vectors with coplanar or just parallel vectors. (For the difference, refer to Coplanar Vectors and Parallel Lines.)
Using the ratio method when any vector component is zero, which can cause division errors.
Real-World Applications
Checking for collinear vectors is useful in analysing forces in engineering, alignment of objects in physics, and determining if points lie on a straight road or path in navigation. For deeper geometric problems, try studying Vector Equations or Vector Joining Two Points on Vedantu.
We explored the idea of Collinear Vectors, their formulae, stepwise problem-solving, and real uses. Keep practising to master these checks, and visit Vedantu’s Vector Algebra for more detailed explanations and related vector concepts.
FAQs on Understanding Collinear Vectors: Conditions, Formulas & Proofs
What are collinear vectors?
Collinear vectors are vectors that lie along the same straight line or on parallel lines, meaning the direction of one vector is either the same as or exactly opposite to the other. In other words, two or more vectors are collinear if they have the same or exactly opposite direction irrespective of their magnitudes.
How do you know if vectors are collinear?
Two vectors are collinear if one is a scalar multiple of the other. Mathematically, vectors a and b are collinear if a = k × b, where k is a scalar. Alternatively, their cross product will be zero (for three-dimensional vectors).
What is the formula for collinear vectors?
The condition for collinear vectors is that two vectors a = (a1, a2, a3) and b = (b1, b2, b3) are collinear if: (a1/b1) = (a2/b2) = (a3/b3), provided denominators are not zero. Alternatively, for 3D vectors, a × b = 0.
What is the condition for two vectors to be collinear?
The condition for two vectors to be collinear is that one vector must be a scalar multiple of the other. If vector a = λ × vector b, where λ is a real number (scalar), the vectors are collinear. Also, the cross product a × b = 0 for collinear vectors.
How to prove that three points are collinear?
To prove that three points (A, B, C) are collinear, check if the vectors AB and AC form a straight line:1. Find AB = B - A and AC = C - A2. The points are collinear if AB and AC are collinear vectors (i.e., AB = k × AC for some scalar k), or if the area of triangle ABC is zero.
How do you visually represent collinear vectors?
On a diagram, collinear vectors are shown as arrows lying along the same straight line, either pointing in the same or exactly opposite directions with respect to a common origin or anywhere on that line. They do not form angles other than 0° or 180° between each other.
What is the difference between collinear vectors and parallel vectors?
While both collinear and parallel vectors point in the same or opposite directions, collinear vectors must lie on the same line (or be scalar multiples of each other). Parallel vectors can lie on different lines but maintain a constant angle (0° or 180°) with each other. All collinear vectors are parallel, but not all parallel vectors are strictly collinear.
What are the important properties of collinear vectors?
Properties of collinear vectors include:• They have the same or opposite direction.• They can be represented as scalar multiples of each other.• The angle between them is 0° or 180°.• Their cross product equals zero.• They are linearly dependent.
What are non-collinear vectors?
Non-collinear vectors are vectors that do not lie on the same straight line and cannot be expressed as scalar multiples of each other. The angle between non-collinear vectors is not 0° or 180°, and their cross product is not zero. Three non-collinear points or vectors form a triangle and do not lie on the same straight line.
What is an example of collinear vectors?
For example, let vector a = (2, 4, 6) and vector b = (1, 2, 3). Here, a = 2 × b, so both are collinear vectors since one is a scalar multiple of the other.
How are collinear vectors used in physics?
In physics, collinear vectors often describe forces acting along the same straight line, like tension in a rope or opposite electric charges on a single axis. Identifying collinear vectors helps in analyzing equilibrium and net force calculations.
How do you solve questions on collinear vectors for exams?
To solve collinear vectors questions in exams:• Check if vectors are scalar multiples.• Verify the cross product (for 3D vectors) is zero.• For points, ensure the area of the formed triangle is zero.• Substitute given values and solve for the unknown scalar if required.
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Breaking Down Matrices — Part III: Vector Products and Determinants
We know how to add vectors. Can we multiply them? Matrices transform vectors. What does the determinant of a matrix mean? Where does the formula for computing determinant comes from? Let us connect them all together!
Prasanna Sethuraman
12 min read
·
Mar 1, 2025
Given two vectors from the same vector space, a = a1e1 + a2e2 + … + anen and b = b1e1 + b2e2 + … + bnen, we defined the dot product of vectors to be a·b = a1b1 + a2b2 + … + anbn in part I. Note that this is true only when the basis vectors e1, e2, …, en are orthogonal unit vectors. For non-orthogonal basis, we would have a·b = ∑∑ ak bℓg(k,ℓ)where g(k,ℓ) = ek·eℓ is the metric tensor that plays a crucial role in Einstein’s general relativity.
The dot product is the scaled projection of the vector a onto vector b and is a·b = |a||b|cos(θ) in 2D geometry (image below) and is a scalar value. The projected vector of a in the direction of b is then given by (a·hat(b))hat(b) where hat(b) = b/|b| is the unit vector in the direction of b. See that this is the same as ((a·b)/(b·b))b.
The two vectors form the sides of a parallelogram, and as we can see from the following image, the area of this parallelogram, which is base times the…
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## Written by Prasanna Sethuraman
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It has been two decades of building Wireless Systems with Signal Processing, but the learning never stops!
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14826 | https://www.youtube.com/watch?v=Q2UgMYwWiO4 | Base 10 to Base 2 (Decimal to Binary) Conversion
ElectronX Lab
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Posted: 6 Jan 2014
A demonstration of the repeated divide by 2 method for converting numbers from base 10 (or decimal) into base 2 (or binary) form
68 comments
Transcript:
Intro in this video I'm going to show you how to do conversion from base 10 or decimal numbers into base 2 or binary numbers there are a couple of different methods for doing this by hand the method I'm going to show you is the repeated divide by two method I think the easiest way to show Example how to do this repeated divide by two method is just to go with an example and so here I'm going to convert 45 base 10 into binary form and with the repeated divide by two method this is the way that I like to set it up so I take my number and I divide it by two and now what I'm going to create over here are two different columns the First Column is going to have the whole number part of this Division and the second part of the second column is going to have the remainder of this Division and just as a as a refresher to to remind you what I mean by the remainder just going back to doing long division I'm doing two divid 45 45 so 2 into 4 is 2 2 2 is 4 0 bring down the five 2 into 5 also goes 2 2 2 is 4 one one is the left over from 5 - 4 and so now that I've reached the end of the whole number part what I'm left over with here is the one and so what I have is the the one is the remainder and the nice thing about dividing by two is that you're going to either have a remainder of zero if the number is an even number or one if the number is an odd number so so 45 divid 2 is going to give me 22 with a remainder of 1 and then the reason that it's called repeated divide by two is I'm going to take that whole number part that I've just calculated the result of the that previous Division and I'm going to divide that by two so 22 / 2 is going to give me an 11 with a remainder of zero bring the whole number part down 11 / 2 is 5 with the remainder of one bring the whole number part down again 5 / two is two with a remainder of one 2/ two is one with a remainder of 0o and finally 1 / 2 is equal to 0 with a remainder of 1 so this repeated divide by two is I keep dividing the result of the previous operation by two until my whole number part that I've got left is zero and then once that is complete my my answer or my binary equivalent of the decimal number I was trying to convert is created by building the number back up from from the remainder parts from the bottom to the top with the bottom being the most significant bit and the top being the least significant bit so the answer of converting 45 into binary is going to be 1 Z 1 1 0 1 so 45 base 10 is equal to 1 1 1 0 01 base Check 2 now as a check I can convert this binary back into decimal form to see if I've if I've done everything correctly so I can do the the sum of the weights here to convert from the binary back into decimal so this will be equal to 2 0 + 0 plus this one this column here is 2^ 2ar plus 2 the 3 2 the 4th is zero plus 2 to the 5th so this is 32 + 8 + 4 + 1 which is equal to 45 so I did the conversion correctly all right let's do another Second Conversion conversion this time I'm going to convert 321 into binary so there's the number I'm trying to convert 321 and then I'm going to do the repeated divide by two so divide 321 by 2 that's going to give me 160 so my result I'll write in one column and the remainder I will write in a second column and 160 and in this case the remainder is going to be one so now I need to divide 160 by 2 160 divid by 2 is 80 with a remainder of zero bring the 80 down and divide it by two 80 divid by two is 40 also with a remainder of zero and then 40 divided two which is 20 with a remainder of zero 20 over 2 is 10 with a remainder of zero bring the 10 down 10 / two is five with a remainder of zero and then 5 / 2 is two with a remainder of one 2 / 2 is 1 with a remainder of zero and then 1/ 2 is zero with a remainder of one and now that I've got my whole number part down to zero I'm finished and now I can build my number back up from the most significant bit at the bottom to the least significant bit here at the top so 321 in base 10 is equal to 1 1 0 000000 0 so five zeros and then a one and once again I can do a check to make sure I did this conversion correctly and looking at converting this base two number back into the base 10 number so this is the 2 to the 0o column and then 1 2 3 4 5 6 2 to the 6 column 7 8 2 the e8th column so this is going to be equal to 2 8th + 2 6 + 2 0 which is 256 + 64 + 1 which is equal to 321 so yes I did the conversion correctly and this conversion method can be used for any whole decimal number doesn't matter the size it's just if it's a really big number you're going to have to do a lot of dividing by two so I hope you learned a little bit in this video and I'll see you in the next one |
14827 | https://math.libretexts.org/Bookshelves/Algebra/Algebra_and_Trigonometry_1e_(OpenStax)/01%3A_Prerequisites/1.01%3A_Real_Numbers_-_Algebra_Essentials | Skip to main content
1.1: Real Numbers - Algebra Essentials
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: Dec 26, 2024
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1.0: Prelude to Prerequisites
1.1E: Real Numbers - Algebra Essentials (Exercises)
Page ID
: 1266
OpenStax
OpenStax
( \newcommand{\kernel}{\mathrm{null}\,})
Learning Objectives
Classify a real number as a natural, whole, integer, rational, or irrational number.
Perform calculations using order of operations.
Use the following properties of real numbers: commutative, associative, distributive, inverse, and identity.
Evaluate algebraic expressions.
Simplify algebraic expressions.
It is often said that mathematics is the language of science. If this is true, then the language of mathematics is numbers. The earliest use of numbers occurred centuries ago in the Middle East to count, or enumerate items. Farmers, cattlemen, and tradesmen used tokens, stones, or markers to signify a single quantity—a sheaf of grain, a head of livestock, or a fixed length of cloth, for example. Doing so made commerce possible, leading to improved communications and the spread of civilization.
Three to four thousand years ago, Egyptians introduced fractions. They first used them to show reciprocals. Later, they used them to represent the amount when a quantity was divided into equal parts.
But what if there were no cattle to trade or an entire crop of grain was lost in a flood? How could someone indicate the existence of nothing? From earliest times, people had thought of a “base state” while counting and used various symbols to represent this null condition. However, it was not until about the fifth century A.D. in India that zero was added to the number system and used as a numeral in calculations.
Clearly, there was also a need for numbers to represent loss or debt. In India, in the seventh century A.D., negative numbers were used as solutions to mathematical equations and commercial debts. The opposites of the counting numbers expanded the number system even further.
Because of the evolution of the number system, we can now perform complex calculations using these and other categories of real numbers. In this section, we will explore sets of numbers, calculations with different kinds of numbers, and the use of numbers in expressions.
Classifying a Real Number
The numbers we use for counting, or enumerating items, are the natural numbers: and so on. We describe them in set notation as where the ellipsis indicates that the numbers continue to infinity. The natural numbers are, of course, also called the counting numbers. Any time we enumerate the members of a team, count the coins in a collection, or tally the trees in a grove, we are using the set of natural numbers. The set of whole numbers is the set of natural numbers plus zero: .
The set of integers adds the opposites of the natural numbers to the set of whole numbers: . It is useful to note that the set of integers is made up of three distinct subsets: negative integers, zero, and positive integers. In this sense, the positive integers are just the natural numbers. Another way to think about it is that the natural numbers are a subset of the integers.
The set of rational numbers is written as .Notice from the definition that rational numbers are fractions (or quotients) containing integers in both the numerator and the denominator, and the denominator is never . We can also see that every natural number, whole number, and integer is a rational number with a denominator of .
Because they are fractions, any rational number can also be expressed in decimal form. Any rational number can be represented as either:
a terminating decimal: , or
a repeating decimal:
We use a line drawn over the repeating block of numbers instead of writing the group multiple times.
Example : Writing Integers as Rational Numbers
Write each of the following as a rational number. Write a fraction with the integer in the numerator and in the denominator.
Solution
a.
b.
c.
Exercise
Write each of the following as a rational number.
Answer
Example : Identifying Rational Numbers
Write each of the following rational numbers as either a terminating or repeating decimal.
Solution
a. a repeating decimal
b. (or ), a terminating decimal
c. , a terminating decimal
Exercise
Write each of the following rational numbers as either a terminating or repeating decimal.
Answer
: 1. (or ), terminating
2. , repeating
3. , terminating
Irrational Numbers
At some point in the ancient past, someone discovered that not all numbers are rational numbers. A builder, for instance, may have found that the diagonal of a square with unit sides was not or even , but was something else. Or a garment maker might have observed that the ratio of the circumference to the diameter of a roll of cloth was a little bit more than , but still not a rational number. Such numbers are said to be irrational because they cannot be written as fractions. These numbers make up the set of irrational numbers. Irrational numbers cannot be expressed as a fraction of two integers. It is impossible to describe this set of numbers by a single rule except to say that a number is irrational if it is not rational. So we write this as shown.
Example : Differentiating Rational and Irrational Numbers
Determine whether each of the following numbers is rational or irrational. If it is rational, determine whether it is a terminating or repeating decimal.
Solution
: This can be simplified as . Therefore,is rational.
: Because it is a fraction, is a rational number. Next, simplify and divide. So, is rational and a repeating decimal.
: This cannot be simplified any further. Therefore, is an irrational number.
: Because it is a fraction, is a rational number. Simplify and divide. So, is rational and a terminating decimal.
is not a terminating decimal. Also note that there is no repeating pattern because the group of increases each time. Therefore it is neither a terminating nor a repeating decimal and, hence, not a rational number. It is an irrational number.
Exercise
Determine whether each of the following numbers is rational or irrational. If it is rational, determine whether it is a terminating or repeating decimal.
Answer
: 1. rational and repeating;
2. rational and terminating;
3. irrational;
4. rational and terminating;
5. irrational
Real Numbers
Given any number , we know that is either rational or irrational. It cannot be both. The sets of rational and irrational numbers together make up the set of real numbers. As we saw with integers, the real numbers can be divided into three subsets: negative real numbers, zero, and positive real numbers. Each subset includes fractions, decimals, and irrational numbers according to their algebraic sign (+ or –). Zero is considered neither positive nor negative.
The real numbers can be visualized on a horizontal number line with an arbitrary point chosen as , with negative numbers to the left of and positive numbers to the right of . A fixed unit distance is then used to mark off each integer (or other basic value) on either side of . Any real number corresponds to a unique position on the number line.The converse is also true: Each location on the number line corresponds to exactly one real number. This is known as a one-to-one correspondence. We refer to this as the real number line as shown in Figure (.
Example : Classifying Real Numbers
Classify each number as either positive or negative and as either rational or irrational. Does the number lie to the left or the right of on the number line?
Solution
is negative and rational. It lies to the left of on the number line.
is positive and irrational. It lies to the right of .
is negative and rational. It lies to the left of .
is negative and irrational. It lies to the left of .
is a repeating decimal so it is rational and positive. It lies to the right of .
Exercise
Classify each number as either positive or negative and as either rational or irrational. Does the number lie to the left or the right of on the number line?
Answer
: 1. positive, irrational; right
2. negative, rational; left
3. positive, rational; right
4. negative, irrational; left
5. positive, rational; right
Sets of Numbers as Subsets
Beginning with the natural numbers, we have expanded each set to form a larger set, meaning that there is a subset relationship between the sets of numbers we have encountered so far. These relationships become more obvious when seen as a diagram, such as Figure().
SETS OF NUMBERS
The set of natural numbers includes the numbers used for counting: .
The set of whole numbers is the set of natural numbers plus zero: .
The set of integers adds the negative natural numbers to the set of whole numbers: .
The set of rational numbers includes fractions written as .
The set of irrational numbers is the set of numbers that are not rational, are nonrepeating, and are nonterminating: .
Example : Differentiating the Sets of Numbers
Classify each number as being a natural number (N), whole number (W), integer (I), rational number (Q), and/or irrational number (Q′).
Solution
| | N | W | I | Q | Q' |
--- --- --- |
| a. | X | X | X | X | |
| b. | | | | X | |
| c. | | | | | X |
| d. | | | X | X | |
| e. | | | | | X |
Exercise
Classify each number as being a natural number (N), whole number (W), integer (I), rational number (Q), and/or irrational number (Q′).
Answer
: | | N | W | I | Q | Q' |
--- --- --- |
| a. | | | X | X | |
| b. | | X | X | X | |
| c. | X | X | X | X | |
| d. | | | | | X |
| e. | | | | X | |
Performing Calculations Using the Order of Operations
When we multiply a number by itself, we square it or raise it to a power of . For example, . We can raise any number to any power. In general, the exponential notation an means that the number or variable is used as a factor times.
In this notation, is read as the power of , where is called the base and is called the exponent. A term in exponential notation may be part of a mathematical expression, which is a combination of numbers and operations. For example, is a mathematical expression.
To evaluate a mathematical expression, we perform the various operations. However, we do not perform them in any random order. We use the order of operations. This is a sequence of rules for evaluating such expressions.
Recall that in mathematics we use parentheses ( ), brackets [ ], and braces { } to group numbers and expressions so that anything appearing within the symbols is treated as a unit. Additionally, fraction bars, radicals, and absolute value bars are treated as grouping symbols. When evaluating a mathematical expression, begin by simplifying expressions within grouping symbols.
The next step is to address any exponents or radicals. Afterward, perform multiplication and division from left to right and finally addition and subtraction from left to right.
Let’s take a look at the expression provided.
There are no grouping symbols, so we move on to exponents or radicals. The number is raised to a power of , so simplify as .
Next, perform multiplication or division, left to right.
Lastly, perform addition or subtraction, left to right.
Therefore,
For some complicated expressions, several passes through the order of operations will be needed. For instance, there may be a radical expression inside parentheses that must be simplified before the parentheses are evaluated. Following the order of operations ensures that anyone simplifying the same mathematical expression will get the same result.
ORDER OF OPERATIONS
Operations in mathematical expressions must be evaluated in a systematic order, which can be simplified using the acronym PEMDAS:
P(arentheses)
E(xponents)
M(ultiplication) and D(ivision)
A(ddition) and S(ubtraction)
HOW TO: Given a mathematical expression, simplify it using the order of operations.
Simplify any expressions within grouping symbols.
Simplify any expressions containing exponents or radicals.
Perform any multiplication and division in order, from left to right.
Perform any addition and subtraction in order, from left to right.
Example : Using the Order of Operations
Use the order of operations to evaluate each of the following expressions.
Solution
Note that in the first step, the radical is treated as a grouping symbol, like parentheses. Also, in the third step, the fraction bar is considered a grouping symbol so the numerator is considered to be grouped.
In this example, the fraction bar separates the numerator and denominator, which we simplify separately until the last step.
Exercise
Use the order of operations to evaluate each of the following expressions.
Answer
Using Properties of Real Numbers
For some activities we perform, the order of certain operations does not matter, but the order of other operations does. For example, it does not make a difference if we put on the right shoe before the left or vice-versa. However, it does matter whether we put on shoes or socks first. The same thing is true for operations in mathematics.
Commutative Properties
The commutative property of addition states that numbers may be added in any order without affecting the sum.
We can better see this relationship when using real numbers.
Similarly, the commutative property of multiplication states that numbers may be multiplied in any order without affecting the product.
Again, consider an example with real numbers.
and
It is important to note that neither subtraction nor division is commutative. For example, is not the same as . Similarly, .
Associative Properties
The associative property of multiplication tells us that it does not matter how we group numbers when multiplying. We can move the grouping symbols to make the calculation easier, and the product remains the same.
Consider this example.
The associative property of addition tells us that numbers may be grouped differently without affecting the sum.
This property can be especially helpful when dealing with negative integers. Consider this example.
Are subtraction and division associative? Review these examples.
As we can see, neither subtraction nor division is associative.
Distributive Property
The distributive property states that the product of a factor times a sum is the sum of the factor times each term in the sum.
This property combines both addition and multiplication (and is the only property to do so). Let us consider an example.
Note that is outside the grouping symbols, so we distribute the by multiplying it by , multiplying it by , and adding the products.
Example
To be more precise when describing this property, we say that multiplication distributes over addition. The reverse is not true, as we can see in this example.
Multiplication does not distribute over subtraction, and division distributes over neither addition nor subtraction.
A special case of the distributive property occurs when a sum of terms is subtracted.
For example, consider the difference . We can rewrite the difference of the two terms and by turning the subtraction expression into addition of the opposite. So instead of subtracting , we add the opposite.
Now, distribute and simplify the result.
This seems like a lot of trouble for a simple sum, but it illustrates a powerful result that will be useful once we introduce algebraic terms. To subtract a sum of terms, change the sign of each term and add the results. With this in mind, we can rewrite the last example.
Identity Properties
The identity property of addition states that there is a unique number, called the additive identity that, when added to a number, results in the original number.
The identity property of multiplication states that there is a unique number, called the multiplicative identity that, when multiplied by a number, results in the original number.
For example, we have and . There are no exceptions for these properties; they work for every real number, including and .
Inverse Properties
The inverse property of addition states that, for every real number a, there is a unique number, called the additive inverse (or opposite), denoted , that, when added to the original number, results in the additive identity, .
For example, if , the additive inverse is , since .
The inverse property of multiplication holds for all real numbers except because the reciprocal of is not defined. The property states that, for every real number , there is a unique number, called the multiplicative inverse (or reciprocal), denoted , that, when multiplied by the original number, results in the multiplicative identity, .
For example, if , the reciprocal, denoted , is because
PROPERTIES OF REAL NUMBERS
The following properties hold for real numbers , , and .
Table
| | Addition | Multiplication |
| Commutative Property | | |
| Associative Property | | |
| Distributive Property | | |
| Identity Property | There exists a unique real number called the additive identity, 0, such that, for any real number a |
There exists a unique real number called the multiplicative identity, 1, such that, for any real number a
Inverse Property
Every real number a has an additive inverse, or opposite, denoted –a, such that
Every nonzero real number a has a multiplicative inverse, or reciprocal, denoted 1a , such that
Example : Using Properties of Real Numbers
Use the properties of real numbers to rewrite and simplify each expression. State which properties apply.
Solution
Exercise
Use the properties of real numbers to rewrite and simplify each expression. State which properties apply.
Answer
: 1. , commutative property of multiplication
2. , distributive property
3. , distributive property
4. , commutative property of addition, associative property of addition, inverse property of addition, identity property of addition
5. , distributive property, inverse property of addition, identity property of addition
Evaluating Algebraic Expressions
So far, the mathematical expressions we have seen have involved real numbers only. In mathematics, we may see expressions such as , , or . In the expression , is called a constant because it does not vary and is called a variable because it does. (In naming the variable, ignore any exponents or radicals containing the variable.) An algebraic expression is a collection of constants and variables joined together by the algebraic operations of addition, subtraction, multiplication, and division.
We have already seen some real number examples of exponential notation, a shorthand method of writing products of the same factor. When variables are used, the constants and variables are treated the same way.
In each case, the exponent tells us how many factors of the base to use, whether the base consists of constants or variables.
Any variable in an algebraic expression may take on or be assigned different values. When that happens, the value of the algebraic expression changes. To evaluate an algebraic expression means to determine the value of the expression for a given value of each variable in the expression. Replace each variable in the expression with the given value, then simplify the resulting expression using the order of operations. If the algebraic expression contains more than one variable, replace each variable with its assigned value and simplify the expression as before.
Example : Describing Algebraic Expressions
List the constants and variables for each algebraic expression.
Solution
| | Constants | Variables |
---
| a. | | |
| b. | , | |
| c. | | , |
Exercise
List the constants and variables for each algebraic expression.
Answer
: | | Constants | Variables |
---
| a. | , | , |
| b. | | , |
| c. | | |
Example : Evaluating an Algebraic Expression at Different Values
Evaluate the expression for each value for .
Solution
Substitute for .
Substitute for .
Substitute for .
Substitute for .
Exercise
Evaluate the expression for each value for .
Answer
Example : Evaluating Algebraic Expressions
Evaluate each expression for the given values.
for
for
for
for ,
for ,
Solution
Substitute for .
Substitute for .
Substitute for .
Substitute for and for .
Substitute for and for .
Exercise
Evaluate each expression for the given values.
for
for
for
for ,
for
Answer
Formulas
An equation is a mathematical statement indicating that two expressions are equal. The expressions can be numerical or algebraic. The equation is not inherently true or false, but only a proposition. The values that make the equation true, the solutions, are found using the properties of real numbers and other results. For example, the equation has the unique solution of because when we substitute for in the equation, we obtain the true statement .
A formula is an equation expressing a relationship between constant and variable quantities. Very often, the equation is a means of finding the value of one quantity (often a single variable) in terms of another or other quantities. One of the most common examples is the formula for finding the area of a circle in terms of the radius of the circle: . For any value of , the area can be found by evaluating the expression .
Example : Using a Formula
A right circular cylinder with radius and height has the surface area (in square units) given by the formula . See Figure . Find the surface area of a cylinder with radius in. and height in. Leave the answer in terms of .
Evaluate the expression for and .
Solution
The surface area is square inches.
Exercise
A photograph with length and width is placed in a matte of width centimeters (cm). The area of the matte (in square centimeters, or is found to be ⋅W.See Figure . Find the area of a matte for a photograph with length cm and width cm.
Answer
Simplifying Algebraic Expressions
Sometimes we can simplify an algebraic expression to make it easier to evaluate or to use in some other way. To do so, we use the properties of real numbers. We can use the same properties in formulas because they contain algebraic expressions.
Example : Simplifying Algebraic Expressions
Simplify each algebraic expression.
Solution
Exercise
Simplify each algebraic expression.
Answer
: 1. or
Example : Simplifying a Formula
A rectangle with length and width has a perimeter given by . Simplify this expression.
Solution
Exercise
If the amount is deposited into an account paying simple interest for time , the total value of the deposit is given by . Simplify the expression. (This formula will be explored in more detail later in the course.)
Answer
Access these online resources for additional instruction and practice with real numbers.
Simplify an Expression
Evaluate an Expression1
Evaluate an Expression2
Key Concepts
Rational numbers may be written as fractions or terminating or repeating decimals. See Example and Example.
Determine whether a number is rational or irrational by writing it as a decimal. See Example.
The rational numbers and irrational numbers make up the set of real numbers. See Example. A number can be classified as natural, whole, integer, rational, or irrational. See Example.
The order of operations is used to evaluate expressions. See Example.
The real numbers under the operations of addition and multiplication obey basic rules, known as the properties of real numbers. These are the commutative properties, the associative properties, the distributive property, the identity properties, and the inverse properties. See Example.
Algebraic expressions are composed of constants and variables that are combined using addition, subtraction, multiplication, and division. See Example. They take on a numerical value when evaluated by replacing variables with constants. See Example,Example, and Example
Formulas are equations in which one quantity is represented in terms of other quantities. They may be simplified or evaluated as any mathematical expression. See Example and Example.
Glossary
algebraic expression
: constants and variables combined using addition, subtraction, multiplication, and division
associative property of addition
: the sum of three numbers may be grouped differently without affecting the result; in symbols,a+(b+c)=(a+b)+c
associative property of multiplication
: the product of three numbers may be grouped differently without affecting the result; in symbols,a⋅(b⋅c)=(a⋅b)⋅c
base
: in exponential notation, the expression that is being multiplied
commutative property of addition
: two numbers may be added in either order without affecting the result; in symbols,a+b=b+a
commutative property of multiplication
: two numbers may be multiplied in any order without affecting the result; in symbols,a⋅b=b⋅a
constant
: a quantity that does not change value
distributive property
: the product of a factor times a sum is the sum of the factor times each term in the sum; in symbols,a⋅(b+c)=a⋅b+a⋅c
equation
: a mathematical statement indicating that two expressions are equal
exponent
: in exponential notation, the raised number or variable that indicates how many times the base is being multiplied
exponential notation
: a shorthand method of writing products of the same factor
formula
: an equation expressing a relationship between constant and variable quantities
identity property of addition
: there is a unique number, called the additive identity, 0, which, when added to a number, results in the original number; in symbols,a+0=a
identity property of multiplication
: there is a unique number, called the multiplicative identity, 1, which, when multiplied by a number, results in the original number; in symbols,a⋅1=a
integers
: the set consisting of the natural numbers, their opposites, and 0:{…,−3,−2,−1,0,1,2,3,…}
inverse property of addition
: for every real numbera,there is a unique number, called the additive inverse (or opposite), denoted−a,which, when added to the original number, results in the additive identity, 0; in symbols,a+(−a)=0
inverse property of multiplication
: for every non-zero real numbera,there is a unique number, called the multiplicative inverse (or reciprocal), denoted1a,which, when multiplied by the original number, results in the multiplicative identity, 1; in symbols,a⋅1a=1
irrational numbers
: the set of all numbers that are not rational; they cannot be written as either a terminating or repeating decimal; they cannot be expressed as a fraction of two integers
natural numbers
: the set of counting numbers:{1,2,3,…}
order of operations
: a set of rules governing how mathematical expressions are to be evaluated, assigning priorities to operations
rational numbers
: the set of all numbers of the formmn,wheremandnare integers andn≠0.Any rational number may be written as a fraction or a terminating or repeating decimal.
real number line
: a horizontal line used to represent the real numbers. An arbitrary fixed point is chosen to represent 0; positive numbers lie to the right of 0 and negative numbers to the left.
real numbers
: the sets of rational numbers and irrational numbers taken together
variable
: a quantity that may change value
whole numbers
: the set consisting of 0 plus the natural numbers:{0,1,2,3,…}
1.0: Prelude to Prerequisites
1.1E: Real Numbers - Algebra Essentials (Exercises) |
14828 | https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Group/Group_01%3A_Hydrogen_and_the_Alkali_Metals/Z001_Chemistry_of_Hydrogen_(Z1) | Chemistry of Hydrogen (Z=1) - Chemistry LibreTexts
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Group 1: Hydrogen and the Alkali Metals
Elements Organized by Group
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Fri, 30 Jun 2023 23:46:45 GMT
Chemistry of Hydrogen (Z=1)
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[ "article:topic", "hydrogenation", "deuterium", "tritium", "isotopes", "combustion", "hydrogen", "showtoc:no", "Cavendish", "water producer", "Antoine Lavoisier", "fusion reactions", "Protium", "license:ccbyncsa", "licenseversion:40" ]
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4. Supplemental Modules and Websites (Inorganic Chemistry)
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7. Group 1: Hydrogen and the Alkali Metals
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Chemistry of Hydrogen (Z=1)
Last updated Jun 30, 2023
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Group 1: Reactivity of Alkali Metals
Chemistry of Lithium (Z=3)
Page ID 92188
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Hydrogen Facts
2. History of Hydrogen
3. Properties of Hydrogen
4. Reactions of Hydrogen
1. Reactions of Hydrogen with Active Metals
1. Group 1 metals
2. Group 2 metals
2. Reactions of Hydrogen with Nonmetals/Descriptive_Chemistry/Elements_Organized_by_Group/Group_01:_Hydrogen_and_the_Alkali_Metals/Z001_Chemistry_of_Hydrogen_(Z1)#Reactions_of_Hydrogen_with_Nonmetals)
3. Reactions with Transition Metals/Descriptive_Chemistry/Elements_Organized_by_Group/Group_01:_Hydrogen_and_the_Alkali_Metals/Z001_Chemistry_of_Hydrogen_(Z1)#Reactions_with_Transition_Metals)
Uses & Application
Natural Occurrence & Other Sources
Naturally Occurring Hydrogen
Other Sources of Hydrogen
Isotopes
References
Outside Links
Problems
Contributors and Attributions
Hydrogen is a colorless, odorless and tasteless gas that is the most abundant element in the known universe. It is also the lightest (in terms of atomic mass) and the simplest, having only one proton and one electron (and no neutrons in its most common isotope). It is all around us. It is a component of water (H 2 O), fats, petroleum, table sugar (C 6 H 12 O 6), ammonia (NH 3), and hydrogen peroxide (H 2 O 2)—things essential to life, as we know it.
Hydrogen Facts
Atomic Number: 1
Atomic Symbol: H
Atomic Weight: 1.0079
Electronic Configuration: 1s 1
Oxidation States: 1, -1
Atomic Radius: 78 pm
Melting Point: -259.34°C
Boiling Point: -252.87° C
Elemental Classification: Non-Metal
At Room Temperature: Colorless & Odorless Diatomic Gas
History of Hydrogen
Hydrogen comes from Greek meaning “water producer” (“hydro” =water and “gennao”=to make). First isolated and identified as an element by Cavendish in 1766, hydrogen was believed to be many different things. Cavendish himself thought that it was "inflammable air from metals", owing to its production by the action of acids on metals. Before that, Robert Boyle and Paracelsus both used reactions of iron and acids to produce hydrogen gas and Antoine Lavoisier gave hydrogen its name because it produced water when ignited in air. Others thought it was pure phlogiston because of its flammability. Hydrogen is among the ten most abundant elements on the planet, but very little is found in elemental form due to its low density and reactivity. Much of the terrestrial hydrogen is locked up in water molecules and organic compounds like hydrocarbons.
Properties of Hydrogen
Hydrogen is a nonmetal and is placed above group in the periodic table because it has ns 1 electron configuration like the alkali metals. However, it varies greatly from the alkali metals as it forms cations (H+) more reluctantly than the other alkali metals. Hydrogen‘s ionization energy is 1312 kJ/mol, while lithium (the alkali metal with the highest ionization energy) has an ionization energy of 520 kJ/mol.
Because hydrogen is a nonmetal and forms H- (hydride anions), it is sometimes placed above the halogens in the periodic table. Hydrogen also forms H 2 dihydrogen like halogens. However, hydrogen is very different from the halogens. Hydrogen has a much smaller electron affinity than the halogens.
H 2 dihydrogen or molecular hydrogen is non-polar with two electrons. There are weak attractive forces between H 2 molecules, resulting in low boiling and melting points. However, H 2 has very strong intramolecular forces; H 2 reactions are generally slow at room temperature due to strong H—H bond. H 2 is easily activated by heat, irradiation, or catalysis. Activated hydrogen gas reacts very quickly and exothermically with many substances.
Hydrogen also has an ability to form covalent bonds with a large variety of substances. Because it makes strong O—H bonds, it is a good reducing agent for metal oxides. Example: CuO(s) + H 2(g) → Cu(s) + H 2 O(g) H 2(g) passes over CuO(s) to reduce the Cu 2+ to Cu(s), while getting oxidized itself.
Reactions of Hydrogen
Hydrogen's low ionization energy makes it act like an alkali metal:
H(g)→H(g)++e−
However, it half-filled valence shell (with a 1s 1 configuration) with one e− also causes hydrogen to act like a halogen non-metal to gain noble gas configuration by adding an additional electron
H(g)+e−→H(g)−
Reactions of Hydrogen with Active Metals
Hydrogen accepts e- from an active metal to form ionic hydrides like LiH. By forming an ion with -1 charge, the hydrogen behaves like a halogen.
Group 1 metals
2M(s)+H 2(g)→2MH(s)
with M representing Group 1 Alkali metals
Examples:
2K(s)+H 2(g)→2KH(s)
2K(s)+Cl 2(g)→2KCl(s)
Group 2 metals
M(s)+H 2(g)→MH 2(s)
with M representing Group 2 Alkaline Earth metals
Example:
Ca(s)+H 2(g)→CaH 2(s)
Ca(s)+Cl 2(g)→CaCl 2(s)
Reactions of Hydrogen with Nonmetals
Unlike metals forming ionic bonds with nonmetals, hydrogen forms polar covalent bonds. Despite being electropositive like the active metals that form ionic bonds with nonmetals, hydrogen is much less electropositive than the active metals, and forms covalent bonds.
Hydrogen + Halogen → Hydrogen Halide
H 2(g)+Cl 2(g)→HCl(g)
Hydrogen gas reacting with oxygen to produce water and a large amount of heat: Hydrogen + Oxygen → Water
(H 2(g)+O 2(g)→H 2O(g)
Reactions with Transition Metals
Reactions of hydrogen with Transition metals (Group 3-12) form metallic hydrides. There is no fixed ratio of hydrogen atom to metal because the hydrogen atoms fill holes between metal atoms in the crystalline structure.
Uses & Application
The vast majority of hydrogen produced industrially today is made either from treatment of methane gas with steam or in the production of "water gas" from the reaction of coal with steam. Most of this hydrogen is used in the Haber process to manufacture ammonia.
Hydrogen is also used for hydrogenation vegetable oils, turning them into margarine and shortening, and some is used for liquid rocket fuel. Liquid hydrogen (combined with liquid oxygen) is a major component of rocket fuel (as mentioned above combination of hydrogen and oxygen relapses a huge amount of energy). Because hydrogen is a good reducing agent, it is used to produce metals like iron, copper, nickel, and cobalt from their ores.
Because one cubic feet of hydrogen can lift about 0.07 lbs, hydrogen lifted airships or Zeppelins became very common in the early 1900s.However, the use of hydrogen for this purpose was largely discontinued around World War II after the explosion of The Hindenburg; this prompted greater use of inert helium, rather than flammable hydrogen for air travel.
Video Showing the explosion of The Hindenburg. (Video from Youtube)
Recently, due to the fear of fossil fuels running out, extensive research is being done on hydrogen as a source of energy.Because of their moderately high energy densities liquid hydrogen and compressed hydrogen gas are possible fuels for the future.A huge advantage in using them is that their combustion only produces water (it burns “clean”). However, it is very costly, and not economically feasible with current technology.
Combustion of fuel produces energy that can be converted into electrical energy when energy in the steam turns a turbine to drive a generator.However, this is not very efficient because a great deal of energy is lost as heat.The production of electricity using voltaic cell can yield more electricity (a form of usable energy).Voltaic cells that transform chemical energy in fuels (like H 2 and CH 4) are called fuel cells.These are not self-contained and so are not considered batteries.The hydrogen cell is a type of fuel cell involving the reaction between H 2(g) with O 2(g) to form liquid water; this cell is twice as efficient as the best internal combustion engine.In the cell (in basic conditions), the oxygen is reduced at the cathode, while the hydrogen is oxidized at the anode.
Reduction: O 2(g)+2H 2 O(l)+4e- → 4OH-(aq)
Oxidation: H 2(g) + 2OH-(aq) → 2H 2 O(l) + 2e-
Overall: 2H 2(g) + O 2(g) → 2H 2 O(l)
E°cell= Reduction- Oxidation= E°O 2/OH- - E°H2O/H2 = 0.401V – (-0.828V) = +1.23
However, this technology is far from being used in everyday life due to its great costs.
Image of A Hydrogen Fuel Cell. (Image made by Ridhi Sachdev)
Natural Occurrence & Other Sources
Naturally Occurring Hydrogen
Hydrogen is the fuel for reactions of the Sun and other stars (fusion reactions). Hydrogen is the lightest and most abundant element in the universe. About 70%- 75% of the universe is composed of hydrogen by mass. All stars are essentially large masses of hydrogen gas that produce enormous amounts of energy through the fusion of hydrogen atoms at their dense cores. In smaller stars, hydrogen atoms collided and fused to form helium and other light elements like nitrogen and carbon(essential for life). In the larger stars, fusion produces the lighter and heavier elements like calcium, oxygen, and silicon.
On Earth, hydrogen is mostly found in association with oxygen; its most abundant form being water (H 2 O). Hydrogen is only .9% by mass and 15% by volume abundant on the earth, despite water covering about 70% of the planet. Because hydrogen is so light, there is only 0.5 ppm (parts per million) in the atmosphere, which is a good thing considering it is EXTREMELY flammable.
Other Sources of Hydrogen
Hydrogen gas can be prepared by reacting a dilute strong acid like hydrochloric acids with an active metal. The metal becomes oxides, while the H+ (from the acid) is reduced to hydrogen gas. This method is only practical for producing small amounts of hydrogen in the lab, but is much too costly for industrial production:
Zn(s)+2H(aq)+→Zn(aq)2++H 2(g)
The purest form of H 2(g) can come from electrolysis of H 2 O(l), the most common hydrogen compound on this plant. This method is also not commercially viable because it requires a significant amount of energy (ΔH=572 kJ):
2H 2O(l)→2H 2(g)+O 2(g)
(H _2 O) is the most abundant form of hydrogen on the planet, so it seems logical to try to extract hydrogen from water without electrolysis of water. To do so, we must reduce the hydrogen with +1 oxidation state to hydrogen with 0 oxidation state (in hydrogen gas). Three commonly used reducing agents are carbon (in coke or coal), carbon monoxide, and methane. These react with water vapor form H 2(g):
C(s)+2H 2O(g)→CO(g)+H 2(g)
CO(g)+2H 2O(g)→CO2+H 2(g)
Reforming of Methane:
CH 4(g)+H 2O(g)→CO(g)+3H 2(g)
These three methods are most industrially feasible (cost effective) methods of producing H 2(g).
Isotopes
There are two important isotopes of hydrogen. Deuterium (2 H) has an abundance of 0.015% of terrestrial hydrogen and the nucleus of the isotope contains one neutron.
Figure : Three Hydrogen Isotopes (Image Made by of Ridhi Sachdev)
Protium (1 H) is the most common isotope, consisting of 99.98% of naturally occurring hydrogen. It is a nucleus containing a single proton.
Deuterium (2 H) is another an isotope containing a proton and neutron, consisting of only 0.0156% of the naturally occurring hydrogen. Commonly indicated with symbol D and sometimes called heavy hydrogen, deuterium is separated by the fractional distillation of liquid hydrogen but it can also be produced by the prolonged electrolysis of ordinary water. Approximately 100,000 gallons of water will produce a single gallon of D 2 O, "heavy water". This special kind of water has a higher density, melting point, and boiling point than regular water and used as a moderator in some fission power reactors. Deuterium fuel is used in experimental fusion reactors. Replacing protium with deuterium has important uses for exploring reaction mechanisms via the kinetic isotope effect.
Tritium (3 H) contains two neutrons in its nucleus and is radioactive with a 12.3-year half-life, which is continuously formed in the upper atmosphere due to cosmic rays. It is can also be made in a lab from Lithium-6 in a nuclear reactor. Tritium is also used in hydrogen bombs. It is very rare (about 1 in every 1,018 atoms) and is formed in the environment by cosmic ray bombardment. Most tritium is manufactured by bombarding Li with neutrons. Tritium is used in thermonuclear weapons and experimental fusion reactors.
References
Shultz, M., Kelly, M., Paritsky, L., Wagner, J. A Theme-Based Course: Hydrogen as the Fuel of the Future. Journal of Chemical Education200986 (9), 105.
Rigden, John. Hydrogen: The Essential Element. The President and Fellows of Harvard College. 2003.
Banks, Alton. Hydrogen. Journal of Chemical Education198966(10), 801.
Petrucci, Ralph H. General Chemistry. 9th ed. Upper Saddle River: Prentice Hall, 2007. Print
Sadava, Heller, Orians, Purves, Hillis. Life The Science of Biology. 8th ed. Sunderland, MA: W.H. Freeman, 2008.
Dinga, G. Hydrogen:The ultimate fuel and energy carrier.Journal of Chemical Education198865 (8), 688.
Outside Links
Problems
Write the reaction of Na(s) with H 2(g).
What is the name of the radioactive isotope of hydrogen?
What characteristics of alkali metals does hydrogen display?
What characteristics of halogens does hydrogen display?
How does the electronegativity of hydrogen compare to that of the halogens?
What is the electron configuration of a neutral hydrogen atom.
Answers
2Na(s) + H 2(g)→ 2NaH(s)
Tritium
Hydrogen is placed above group in the periodic table because it has ns 1 electron configuration like the alkali metals. However, it varies greatly from the alkali metals as it forms cations (H+) more reluctantly than the other alkali metals. Hydrogen‘s ionization energy is 1312 kJ/mol, while lithium (the alkali metal with the highest ionization energy) has an ionization energy of 520 kJ/mol.
Because hydrogen is a nonmetal and forms H- (hydride anions), it is sometimes placed above the halogens in the periodic table. Hydrogen also forms H 2 dihydrogen like halogens. However, hydrogen is very different from the halogens. Hydrogen has a much smaller electron affinity than the halogens.
Hydrogen is less electronegative than the halogens.
1s 1
Contributors and Attributions
Ridhi Sachdev (UC Davis)
Stephen R. Marsden
Chemistry of Hydrogen (Z=1) is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.
Chemistry of Hydrogen (Z=1) is licensed CC BY-NC-SA 4.0.
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Group 1: Reactivity of Alkali Metals
Chemistry of Lithium (Z=3)
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14829 | https://www.geeksforgeeks.org/maths/multiplicative-inverse/ | Multiplicative Inverse
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23 Jul, 2025
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Multiplicative inverse of a number is another number that, when multiplied by the original number, results in the identity element for multiplication, which is 1. In other words, for a non-zero number a, its multiplicative inverse is denoted as a−1, and it satisfies the equation: a⋅a-1 = 1. We can also define multiplicative inverse as the reciprocal of a number. A number when multiplied with its own multiplicative inverse(reciprocal), then we get 1. In this article, we will learn about multiplicative inverse their definition, multiplicative inverse of natural numbers, fraction, unit fraction, mixed fraction, and complex numbers.
Table of Content
What is Multiplicative Inverse?
Multiplicative Inverse of Natural Number
Multiplicative Inverse of Fraction
Multiplicative inverse of Mixed Fraction
Multiplicative Inverse of Complex Numbers
What is Multiplicative Inverse?
Multiplicative inverse meaning is a reciprocal of a number which is denoted as a-1. Let there is a number X its multiplicative inverse is represented as 1/X or X-1. The English meaning of "inverse" is "opposite". It's important to note that the multiplicative inverse is not defined for zero because there is no number you can multiply by 0 to get 1. Let us understand the examples of multiplicative inverse.
Examples for multiplicative inverse:
2-1 or 1/2 is a multiplicative inverse because 2 × 1/2 = 1.
5-1 or 1/5 is a multiplicative inverse because 5 × 1/5 = 1.
4-1 or 1/4 is a multiplicative inverse because 4 × 1/4 = 1.
Multiplicative Inverse Definition
Multiplicative inverse is defined as the reciprocal of a number and is represented as 1/n or n-1 where n is a number.
Multiplicative Inverse Property
The identity of multiplicative inverse states that every nonzero number a has a unique multiplicative inverse, denoted as a-1, such that the product of a and its multiplicative inverse is equal to the multiplicative identity that is 1.
Mathematically it is written as:
a . a-1 = 1
Multiplicative Inverse of Natural Number
The multiplicative inverse of a natural number N i.e. {1,2, 3, ..., n} will be 1/N. Example: The multiplicative inverse of 1 is 1. One important thing that we have to remember is that we cannot find multiplicative inverse of zero (0).
Learn more about Natural Number
Multiplicative Inverse of Fraction
We can also find the multiplicative inverse of a fraction. We have to find the reciprocal of a fraction in order to get its multiplicative inverse. Let there is a fraction i.e. X/Y then its multiplicative inverse of X/Y will be Y/X because when Y/X is multiplies by X/Y then the result will be 1.
Examples of Multiplicative Inverse of Fraction are:
The multiplicative inverse of 2/3 is 3/2 because when we multiply 2/3 with 3/2 we get 1.
The multiplicative inverse of 5/6 is 6/5 because when we multiply 5/6 with 6/5 we get 1.
The multiplicative inverse of 7/6 is 6/7 because when we multiply 7/6 with 6/7 we get 1.
Learn more about Fraction
Multiplication Inverse of Unit Fraction
The multiplicative inverse of unit fraction can also be find out by reciprocal. Unit fraction includes numbers like 1/2, 1/3, 1/5, etc. The examples of multiplicative inverse of a unit fraction are:
The multiplicative inverse of 1/5 will be 5.
The multiplicative inverse of 1/3 will be 3.
The multiplicative inverse of 1/7 will be 7.
Multiplicative inverse of Mixed Fraction
In order to find multiplicative inverse of mixed fraction we have to solve the mixed fraction into simple fraction and then we reciprocal the simple fraction.
Examples of multiplicative inverse of Mixed Fraction are as follows:
The simple fraction of mixed fraction 4(3/2) will be 11/2 and its multiplicative inverse will be 2/11.
The simple fraction of mixed fraction 2(1/2) will be 5/2 and its multiplicative inverse will be 2/5.
The simple fraction of mixed fraction 4(3/2) will be 11/2 and its multiplicative inverse will be 2/11.
Learn more about Mixed Fraction
Multiplicative Inverse of Complex Numbers
The multiplicative inverse of a complex number a + bi, where a and b are real numbers and i is the imaginary unit (i2 = -1), is given by (a + bi)-1 or 1/ (a + bi).
To find the multiplicative inverse, you often multiply the numerator and denominator by the conjugate of the complex number. The conjugate of a + bi is a − bi. So, the multiplicative inverse becomes 1/ (a + bi) × (a-bi)/ (a + bi) = (a-bi)/ (a2 + b2).
Following are the examples of Multiplicative Inverse of Complex Numbers:
Multiplicative inverse of 2 + 3i will be 1/ (2 + 3i) × (2 - 3i) / (2 - 3i) = 2-3i/13.
Multiplicative inverse of -1 + 2i will be 1/ (-1 + 2i) × (-1-2i)/(-1-2i) = -1-2i/5.
Learn more about
Complex numbers
Multiplicative Inverse of 0
Reciprocals
Multiplicative Inverse - Solved Examples
Example 1: What is the Multiplicative Inverse of 7?
Solution:
The multiplicative Inverse of 7 is 1/7
Example 2: What is the Multiplicative Inverse of 2-3i?
Solution:
Multiplicative inverse of 2-3i is 1/(2-3i)
z-1 = 1/(2-3i) × (2 + 3i)/(2 + 3i)
Solving Above we get
z-1 = (2 + 3i)/13
Example 3: Find the Multiplicative Inverse of -2/7.
Solution:
The Multiplicative Inverse of -2/7 is -7/2
Multiplicative Inverse - Practice Questions
1. Find the multiplicative inverse of the following:
(i) Multiplicative Inverse of 2
(ii) Multiplicative Inverse of 5
(iii) Multiplicative Inverse of 2 + 3i
2. Find the multiplicative inverse of 3/7.
3. Find the multiplicative inverse of 4+3i, where i is the imaginary unit.
4. Calculate the multiplicative inverse of -8.
5. Find the multiplicative inverse of 1/17.
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14830 | https://www.tiger-algebra.com/drill/-x~2-4x-5=0/ | Copyright Ⓒ 2013-2025
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Solution - Quadratic equations
Other Ways to Solve
Step by Step Solution
Step by step solution :
Step 1 :
Step 2 :
Pulling out like terms :
2.1 Pull out like factors :
-x2 - 4x - 5 = -1 • (x2 + 4x + 5)
Trying to factor by splitting the middle term
2.2 Factoring x2 + 4x + 5
The first term is, x2 its coefficient is 1 .
The middle term is, +4x its coefficient is 4 .
The last term, "the constant", is +5
Step-1 : Multiply the coefficient of the first term by the constant 1 • 5 = 5
Step-2 : Find two factors of 5 whose sum equals the coefficient of the middle term, which is 4 .
| | | | | | | |
--- --- ---
| | -5 | + | -1 | = | -6 | |
| | -1 | + | -5 | = | -6 | |
| | 1 | + | 5 | = | 6 | |
| | 5 | + | 1 | = | 6 | |
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Equation at the end of step 2 :
Step 3 :
Parabola, Finding the Vertex :
3.1 Find the Vertex of y = -x2-4x-5
Parabolas have a highest or a lowest point called the Vertex . Our parabola opens down and accordingly has a highest point (AKA absolute maximum) . We know this even before plotting "y" because the coefficient of the first term, -1 , is negative (smaller than zero).
Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.
Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.
For any parabola,Ax2+Bx+C,the x -coordinate of the vertex is given by -B/(2A) . In our case the x coordinate is -2.0000
Plugging into the parabola formula -2.0000 for x we can calculate the y -coordinate :
y = -1.0 -2.00 -2.00 - 4.0 -2.00 - 5.0
or y = -1.000
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = -x2-4x-5
Axis of Symmetry (dashed) {x}={-2.00}
Vertex at {x,y} = {-2.00,-1.00}
Function has no real roots
Solve Quadratic Equation by Completing The Square
3.2 Solving -x2-4x-5 = 0 by Completing The Square .
Multiply both sides of the equation by (-1) to obtain positive coefficient for the first term:
x2+4x+5 = 0 Subtract 5 from both side of the equation :
x2+4x = -5
Now the clever bit: Take the coefficient of x , which is 4 , divide by two, giving 2 , and finally square it giving 4
Add 4 to both sides of the equation :
On the right hand side we have :
-5 + 4 or, (-5/1)+(4/1)
The common denominator of the two fractions is 1 Adding (-5/1)+(4/1) gives -1/1
So adding to both sides we finally get :
x2+4x+4 = -1
Adding 4 has completed the left hand side into a perfect square :
x2+4x+4 =
(x+2) • (x+2) =
(x+2)2
Things which are equal to the same thing are also equal to one another. Since
x2+4x+4 = -1 and
x2+4x+4 = (x+2)2
then, according to the law of transitivity,
(x+2)2 = -1
We'll refer to this Equation as Eq. #3.2.1
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
(x+2)2 is
(x+2)2/2 =
(x+2)1 =
x+2
Now, applying the Square Root Principle to Eq. #3.2.1 we get:
x+2 = √ -1
Subtract 2 from both sides to obtain:
x = -2 + √ -1
In Math, i is called the imaginary unit. It satisfies i2 =-1. Both i and -i are the square roots of -1
Since a square root has two values, one positive and the other negative
x2 + 4x + 5 = 0
has two solutions:
x = -2 + √ 1 • i
or
x = -2 - √ 1 • i
Solve Quadratic Equation using the Quadratic Formula
3.3 Solving -x2-4x-5 = 0 by the Quadratic Formula .
According to the Quadratic Formula, x , the solution for Ax2+Bx+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
x = ————————
2A
In our case, A = -1
B = -4
C = -5
Accordingly, B2 - 4AC =
16 - 20 =
-4
Applying the quadratic formula :
4 ± √ -4
x = —————
-2
In the set of real numbers, negative numbers do not have square roots. A new set of numbers, called complex, was invented so that negative numbers would have a square root. These numbers are written (a+bi)
Both i and -i are the square roots of minus 1
Accordingly,√ -4 =
√ 4 • (-1) =
√ 4 • √ -1 =
± √ 4 • i
Can √ 4 be simplified ?
Yes! The prime factorization of 4 is
2•2
To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).
√ 4 = √ 2•2 =
± 2 • √ 1 =
± 2
So now we are looking at:
x = ( 4 ± 2i ) / -2
Two imaginary solutions :
Two solutions were found :
How did we do?
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14831 | https://www.ccjm.org/content/ccjom/44/1/41.full.pdf | Benign hepatic tumors and cysts in women using oral contraceptive agents Computed tomography as a diagnostic aid Thomas R. Havrilla, M.D. Richard G. Pepe, D.O. Ralph J. Alfidi, M.D. John R. Haaga, M.D. Department of Diagnostic Radiology Benign hepatic tumors in women taking oral contraceptive agents have been reported in in-creasing numbers. Considering the number of women taking contraceptive agents, the inci-dence of these tumors is low. Previously, a de-finitive diagnosis of a benign hepatic tumor or cyst could be made only by laparotomy and open biopsy. However, with the advent of whole body computed tomography (CT) an earlier di-agnosis may be possible and laparotomy for di-agnostic purposes may be unnecessary. Two cases are reported and a review of the literature is included. Method and equipment The CT scanner used in the study was the whole body Delta unit (Delta 50, Ohio Nuclear Corporation, Solon, Ohio). This scanner ob-tains two simultaneous I3-mm transverse slices in approximately 21h minutes and scans in air through an arc of 180 degrees. CRT and digital display are shown on a 256-square matrix. The Delta scanner is also equipped with a color dis-play. Other technical features have previously been described.1 Respiratory motion was not limited. The patients were examined in the su-pine and right lateral decubitus positions follow-41 uses require permission. on September 28, 2025. For personal use only. All other www.ccjm.org Downloaded from 42 Cleveland Clinic Quarterly Vol. 44, No. 1 ing the intravenous administration of 100 cc Renografin 60. Early experi-ence with contrast injection has con-firmed the enhancement of the dif-ferential attenuation coefficients of normal and abnormal tissues in cer-tain patients. In addition, a "cursor circle" can measure Delta attenuation numbers of any specific area on the CRT display. In this way, the mean density of the contents of the circle can be obtained. This is valuable in discriminating between solid and cystic components of lesions under evaluation. Case reports C a s e 1. A 28-year-old woman was re-ferred to the Cleveland Clinic because of a large liver mass found 2 months prior to admission. The patient was asymptomatic except for an increase in girth and weight. On physical examination, a large nontender mass was noted in the upper right quadrant extending to the iliac crest. The remainder of the physical ex-amination was unremarkable and the re-sults of laboratory studies including liver function tests were normal. The history was noncontributory except for the in-take of OvuIen-21 for 21 months. B-mode ultrasound examination of the abdomen at another hospital revealed a large, echo-free volume even at high gain compatible with a liver cyst. Liver scan revealed a large filling defect in the right lobe. After admission to the Cleveland Clinic Hospi-tal, abdominal angiography was per-formed and a 35 X 22-cm avascular mass was observed in the upper right quadrant displacing the right kidney to the left of midline (Fig. 1A-D). A CT scan of the abdomen depicted a huge mass in the upper right quadrant replacing the right lobe of the liver (Fig. 2).The specific den-sity of this mass measured 0 Delta units, equivalent to water density and consistent with a fluid-filled cyst. At laparotomy, a cyst involving the entire right lobe of the liver was found and contained 3000 cc of clear fluid. A portion of the cyst wall was left attached to the liver and opened to the peritoneal cavity for drainage. Patho-logic examination revealed several focal nodules in the cyst wall. On section, these nodules contained thin-walled cysts with clear, yellow fluid. The diagnosis was cyst-adenoma of the liver (hepatic cyst) aris-ing from bile ducts. C a s e 2. A 27-year-old woman was ad-mitted with a history of hepatomegaly and chronic cholecystitis. Pertinent drug history included the intake of Ovulen-21 for 7 years. Prior to admission, signs and symptoms of acute cholecystitis devel-oped. Results of physical examination were normal except for a palpable liver 7 cm below the right costal margin. Labora-tory findings were normal except for ele-vated liver enzyme levels. Plain film of the abdomen showed an upper right quadrant mass (Fig. 3). The oral cholecys-togram revealed four large partially ra-diolucent stones. On the radionuclide liver scan, moderate hepatomegaly of the right lobe and the anterior left lobe was present with multiple areas of decreased activity consistent with multiple space-oc-cupying masses (Fig. 4). The CT scan displayed a bulbous mass in the right he-patic lobe consistent with a normal var-iant or a well-differentiated neoplasm (Fig. 5): The tumor was of the same den-sity as the liver and measured 50 Delta units without contrast and 90 Delta units with contrast enhancement. On explora-tory laparotomy, a large hepatic adenoma measuring 20 x 10 x 10 cm was present in the right lobe. A second mass 15 x 10 x 7 cm was also present in the left hepatic lobe (in retrospect, this tumor was also seen on the CT scan). Diagnosis by frozen section was hepatic adenoma. The tumor was not resected. Cholecystectomy was performed. An intraoperative cholangio-gram again demonstrated the liver mass (Fig. 6). Because of bleeding from the biopsy sites, a hepatic artery was ligated. Microscopic examination of the tissue segment showed a loss of the normal lob-uses require permission. on September 28, 2025. For personal use only. All other www.ccjm.org Downloaded from Spring 1977 Hepatic tumors, cysts, oral contraceptives 43 Fig. 1. A, On the selective right renal arteriogram, the right renal artery (arrow) and the right kidney (arrow head) are displaced to the left of midline by a large upper right quadrant mass. B, Selective superior mesenteric arteriogram. On the arterial phase, the superior mesenteric artery (arrow) and its peripheral branches to the colon and j e j u n u m are deviated to the left of midline by a large u p p e r right quadrant mass. C, Selective superior mesenteric arteriogram. On the venous phase, the large u p p e r quadrant density is displacing the colon and the small intestine inferiorly and to the left. D, Selective celiac arteriogram by transfemoral percutaneous catheterization again reveals the main celiac trunk and the splenic and hepatic vessels displaced to the left. T h e right hepatic and gastroduodenal arteries (arrows) are also displaced upward and to the left by a large right avascular hepatic mass. u l a r p a t t e r n c o m p a t i b l e with a h e p a t o c e l -lular a d e n o m a . P o s t o p e r a t i v e l y , t h e pa-tient h a d f e v e r a n d u p p e r right q u a d r a n t p a i n . B e c a u s e of a s u s p e c t e d liver abscess, a r e p e a t C T scan was p e r f o r m e d w h i c h s h o w e d a l a r g e a r e a of d e c r e a s e d d e n s i t y in t h e lower right lobe of t h e liver m e a s -u r i n g a p p r o x i m a t e l y 30 Delta u n i t s (Fig. uses require permission. on September 28, 2025. For personal use only. All other www.ccjm.org Downloaded from 44 Cleveland Clinic Quarterly Vol. 44, No. 1 Fig. 2. Computed axial tomographic scan taken at level of the liver is compatible with a large cystic lesion of water density replacing almost entirely the righl lobe of the liver (ar-row). Reproduced by permission of Alfidi RJ, et al: T h e effects of biological motion on C T resolution. Am J Roentgenol 127:11-15, 1976. Fig. 3. Plain film of the abdomen reveals dis-placement of the righl hepatic flexure and transverse colon by an enlarged right lobe of the liver or right upper quadrant mass. 7). This area of decreased density re-placed the previously described bulbous area of the right lobe and resembled an abscess or necrotic hematoma. At reoper-ation, a large hematoma with necrosis of the hepatic adenoma was demonstrated and confirmed by biopsy. A drainage procedure was performed; postopera-A n t . \ / Fig. 4. Liver scan was obtained using 4mCi of Tc99m sulfur colloid. Moderate hepatomegaly was evident involving the lower aspect of the right lobe and also the left lobe anteriorly. These areas had discrete margins and were compatible with space-occupying masses (ar-rows). Fig. 5. C T scan of the liver in the right lateral decubitus position was compatible with bul-bous enlargement of the right lower lobe of the liver. This was a nonspecific finding and could be a normal variant or a secondary to a low grade hepatoma or adenoma. T h e Delta num-bers were consistent with normal liver paren-chyma. uses require permission. on September 28, 2025. For personal use only. All other www.ccjm.org Downloaded from Spring 1977 Hepatic tumors, cysts, oral contraceptives 45 Fig. 6. On the operative cholangiogram, the biliary tree is visualized through the hepatic mass and is patent without evidence of calculi or obstruction. Contrast was seen within lhe d u o d e n u m on subsequent films. T h e large he-patic mass is well outlined as a large ovoid density (arrow) because of an air-tissue inter-face about the mass in the operative field. T h e mass itself is not filled with contrast. tively, t h e p a t i e n t d i d well a n d was dis-c h a r g e d f r o m t h e hospital 11 clays later. contraceptives, and more study is needed. Liver adenomas and cystadenomas are rare. Until 1972, 76 cases of be-nign lesions of the extrahepatic ducts, excluding the ampulla of Va-ter, were reported. 5 In a study done by Malt et al,6 at the Massachusetts General Hospital between 1947 and 1968, there were 26 cases of benign tumors of the liver. Of these 26 cases, four hepatoadenomas and seven cysts were present. Hepatic cysts are rare and are more often found in children than in adults. Causes of acute symptoms are secondary to intracystic hemorrhage, intraperitoneal rupture, or abscess formation. A cystic hamartoma of the liver in an adult can be an acute ab-dominal emergency simulating a per-forated duodenal ulcer.7 Until recently, a diagnosis of cystic liver could only be made definitively by laparotomy, which also had the advantage of affording immediate treatment. Some investigators advo-Discussion Several reports of benign hepatic tumors in women taking oral contra-ceptives have been published in re-cent years. Horvath et al,2 in 1972, reported a case of well-differentiated hepatoma in a 28-year-old woman who had taken oral contraceptives for 7 years. In addition, seven cases of hepatic adenoma were also re-ported in 1973.3 All these women were taking oral contraceptives. The possible relationship of hepatomas in women who use an oral contraceptive was suggested in a report by the com-mittee on Safety of Medicine.4 Many variables, however, can affect the as-sessment of the carcinogenesis of oral Fig. 7. C T scan done postoperatively at the level of the liver now reveals a large area of decreased density present in the lower right lobe of the liver in [he previously described region of the adenoma (arrows). T his finding was consistent with an abscess or necrotic he-matoma with a density measurement of 30 Delta units. uses require permission. on September 28, 2025. For personal use only. All other www.ccjm.org Downloaded from 46 Cleveland Clinic Quarterly Vol. 44, No. 1 cate aspiration biopsy and others per-itoneoscopy.8 Also at one time, Tho-rotrast given intravenously was sug-gested as a means of diagnosis. At angiography, hepatic cysts are usu-ally avascular structures with vascular displacement of normal vessels.9 These vessels form a smooth curve about the cyst. The venous hepato-gram usually shows a large radiolu-cent defect. CT scanning is accurate in cyst ver-sus solid tumor identification and dif-ferentiation. In a comparison study between the detection of lesions by CT scan and radionuclide liver scans, CT was found to be equally accu-rate.10 Density determinations pro-vide an appraisal of the tumor sub-stance without surgical intervention, because the density values are very precise in differentiating between pa-renchymal masses and cysts. With utilization of the CT cursor, cystic lesions usually measure 0 to 10 Delta units and rarely exceed 20 Delta units (0 Delta units equal the density of water). In contrast, tumor masses usually measure 30 to 50 Delta units after contrast enhancement while normal liver parenchyma with con-trast enhancement measures approx-imately 90 Delta units. Therefore, CT scanning offers a noninvasive source of valuable data of clinical im-portance. CT may be limited in diag-nosing a small, solid hepatic tumor of the same density as liver paren-chyma. Through the years, different types of treatment have been suggested for benign hepatic tumors and cysts. Pri-mary resection has been advocated because of the previously described complications. If a cystic structure is noted, marsupialization or irrigation with coagulating fluids is possible. Excessive bleeding at operation may be difficult to control, and a hepatic lobectomy may be necessary.8, 11 Gal-loway et al12 reported four cases of "minimal deviation hepatoma," a tu-mor of atypical cells with a relatively benign course. One patient was tak-ing oral contraceptives. The tumor was classified as an intermediate stage between adenoma and hepa-toma. Because 4 of 19 previous pa-tients died after attempted radical re-section, limited resection combined with chemotherapy was suggested to reduce operative mortality. It is hoped that with some of the sophisticated CT refinements antici-pated, nonsurgical diagnosis will soon become a reality. CT guided needle aspiration may lessen the need for laparotomy as a primary means of definitive diagnosis in se-lected cases. The advantage of accu-rate needle localization should facili-tate drainage of the fluid when a cyst is found. As previously mentioned, irrigation with a coagulating fluid could be performed to help eliminate recurrence of the cysts. CT is an accurate method for per-forming percutaneous biopsies and aspiration procedures and is the im-aging modality of choice.8 With aspi-ration, a definitive preoperative cyto-logic diagnosis can be made. The ac-curacy is directly related to the locali-zation of the needle path and the pre-cise location of the needle tip. Since the scan thickness is only 13 mm, an exact three dimensional position of the needle tip can be obtained. To date, a cystic liver was aspirated by CT guidance in only one patient. In this patient, clear cyst fluid was ob-tained from several cysts within the liver; and the diagnosis of polycystic renal and liver disease was con-uses require permission. on September 28, 2025. For personal use only. All other www.ccjm.org Downloaded from Spring 1977 Hepatic tumors, cysts, oral contraceptives 47 firmed. CT guided biopsies have also been performed on retroperitoneal masses, pulmonary and pleural le-sions, renal cysts and tumors, liver abscesses, hematomas, and metas-tasis. In addition, percutaneous transhepatic cholangiograms, aspira-tion of a urinoma, biopsies of bone and pancreas have been performed. Therefore, if a cystic lesion is present and the density approximates water density, the diagnosis is almost path-ognomonic. This information in con-junction with a CT guided aspiration will confirm the diagnosis. Summary Primary benign adenomas and cystadenomas of the liver are rare, but the number of reported cases is increasing. The majority of patients are women taking oral contracep-tives, and the incidence seems to be more than coincidental. The use of CT scanning has been helpful in pre-liminary studies to differentiate cysts from solid tumors. CT coupled with aspiration may alleviate the need for a laparotomy for a definitive diagno-sis in certain cases. References 1. Alfidi RJ, Haaga J R , Meaney T F , et al: C o m p u t e d tomography of the thorax and abdomen; a preliminary report. Radiology 117: 257-264, 1975. 2. Horvath E, Kovacs K, Ross RC: Ultra-structural findings in a well-differentiated hepatoma. Digestion 7: 74-82, 1972. 3. Baum J K , Bookstein J J , Holtz F, et al: Possible association between benign hepa-tomas and oral contraceptives. Lancet 2: 926-929, 1973. 4. Lingeman CH: Letter: Liver-cell neo-plasms and oral contraceptives. Lancet 1: 64, 1974. 5. Hossack KF, H e r r o n JJ: Benign tumors of the common bile duct; report of a case and review of the literature. Aust NZ J Surg 142: 22-25, 1972. 6. Malt RA, Hershberg RA, Miller WL: Ex-perience with benign tumors of the liver. Surg Gynecol Obstet 130: 285-291, 1970. 7. Gough AS, Pike C: Cystic hamartoma of the liver in an adult presenting as an ab-dominal emergency. Br J Surg 50: 342-343, 1962. 8. Grime RT, Moore T , Nicholson A, et al: Cystic hamartomas and polycystic disease of the liver. Br J Surg 47: 307-313,1959. 9. Alfidi RJ, Rastogi H, Buonocore E, et al: Hepatic arteriography. Radiology 90: 1136-1142, 1968. 10. Alfidi RJ, Haaga J R , Havrilla T R , et al: Computed tomography of the liver. Am J Roentgenol 127: 69-74, 1976. 11. Larsson-Cohn U, Stenram U: Liver ultra-structure and function in icteric and non-icteric women using oral contraceptive agents. Acta Med Scand 181: 257-264, 1967. 12. Galloway SJ, Casarella WJ, Lattes R, et al: Minimal deviation hepatoma; a new entity. Am J Roentgenol 125: 184-192, 1975. uses require permission. on September 28, 2025. For personal use only. All other www.ccjm.org Downloaded from |
14832 | http://www.wallace.ccfaculty.org/book/6.5%20Factor%20Special%20Products.pdf | 6.5 Factoring - Factoring Special Products Objective: Identify and factor special products including a difference of squares, perfect squares, and sum and difference of cubes.
When factoring there are a few special products that, if we can recognize them, can help us factor polynomials. The first is one we have seen before. When multi-plying special products we found that a sum and a difference could multiply to a difference of squares. Here we will use this special product to help us factor Difference of Squares: a2 −b2 = (a + b)(a −b) If we are subtracting two perfect squares then it will always factor to the sum and difference of the square roots.
Example 1.
x2 −16 Subtracting two perfect squares, the square roots are x and 4 (x + 4)(x −4) Our Solution Example 2.
9a2 −25b2 Subtracting two perfect squares, the square roots are 3a and 5b (3a + 5b)(3a −5b) Our Solution It is important to note, that a sum of squares will never factor. It is always prime. This can be seen if we try to use the ac method to factor x2 + 36.
Example 3.
x2 + 36 No bx term, we use 0x.
x2 + 0x + 36 Multiply to 36, add to 0 1 · 36, 2 · 18, 3 · 12, 4 · 9, 6 · 6 No combinations that multiply to 36 add to 0 Prime, cannot factor Our Solution It turns out that a sum of squares is always prime.
Sum of Squares: a2 + b2 = Prime 1 A great example where we see a sum of squares comes from factoring a difference of 4th powers. Because the square root of a fourth power is a square ( a4 √ = a2), we can factor a difference of fourth powers just like we factor a difference of squares, to a sum and difference of the square roots. This will give us two factors, one which will be a prime sum of squares, and a second which will be a difference of squares which we can factor again. This is shown in the following examples.
Example 4.
a4 −b4 Difference of squares with roots a2 and b2 (a2 + b2)(a2 −b2) The first factor is prime, the second is a difference of squares!
(a2 + b2)(a + b)(a −b) Our Solution Example 5.
x4 −16 Difference of squares with roots x2 and 4 (x2 + 4)(x2 −4) The first factor is prime, the second is a difference of squares!
(x2 + 4)(x + 2)(x −2) Our Solution Another factoring shortcut is the perfect square. We had a shortcut for multi-plying a perfect square which can be reversed to help us factor a perfect square Perfect Square: a2 + 2ab + b2 = (a + b)2 A perfect square can be difficult to recognize at first glance, but if we use the ac method and get two of the same numbers we know we have a perfect square.
Then we can just factor using the square roots of the first and last terms and the sign from the middle. This is shown in the following examples.
Example 6.
x2 −6x + 9 Multiply to 9, add to −6 The numbers are −3 and −3, the same! Perfect square (x −3)2 Use square roots from first and last terms and sign from the middle Example 7.
4x2 + 20xy + 25y2 Multiply to 100, add to 20 The numbers are 10 and 10, the same! Perfect square (2x + 5y)2 Usesquarerootsfromfirstandlasttermsandsignfromthemiddle 2 World View Note: The first known record of work with polynomials comes from the Chinese around 200 BC. Problems would be written as “three sheafs of a good crop, two sheafs of a mediocre crop, and one sheaf of a bad crop sold for 29 dou. This would be the polynomial (trinomial) 3x + 2y + z = 29.
Another factoring shortcut has cubes. With cubes we can either do a sum or a difference of cubes. Both sum and difference of cubes have very similar factoring formulas Sum of Cubes: a3 + b3 = (a + b)(a2 −ab + b2) Difference of Cubes: a3 −b3 = (a −b)(a2 + ab + b2) Comparing the formulas you may notice that the only difference is the signs in between the terms. One way to keep these two formulas straight is to think of SOAP. S stands for Same sign as the problem. If we have a sum of cubes, we add first, a difference of cubes we subtract first. O stands for Opposite sign. If we have a sum, then subtraction is the second sign, a difference would have addition for the second sign. Finally, AP stands for Always Positive. Both formulas end with addition. The following examples show factoring with cubes.
Example 8.
m3 −27 We have cube roots m and 3 (m 3)(m2 3m 9) Use formula, use SOAP to fill in signs (m −3)(m2 + 3m + 9) Our Solution Example 9.
125p3 + 8r3 We have cube roots 5p and 2r (5p 2r)(25p2 10r 4r2) Use formula, use SOAP to fill in signs (5p + 2r)(25p2 −10r + 4r2) Our Solution The previous example illustrates an important point. When we fill in the trino-mial’s first and last terms we square the cube roots 5p and 2r. Often students forget to square the number in addition to the variable. Notice that when done correctly, both get cubed.
Often after factoring a sum or difference of cubes, students want to factor the second factor, the trinomial further. As a general rule, this factor will always be prime (unless there is a GCF which should have been factored out before using cubes rule).
3 The following table sumarizes all of the shortcuts that we can use to factor special products Factoring Special Products Difference of Squares a2 −b2 = (a + b)(a −b) Sum of Squares a2 + b2 = Prime Perfect Square a2 + 2ab + b2 = (a + b)2 Sum of Cubes a3 + b3 = (a + b)(a2 −ab + b2) Difference of Cubes a3 −b3 = (a −b)(a2 + ab + b2) As always, when factoring special products it is important to check for a GCF first. Only after checking for a GCF should we be using the special products.
This is shown in the following examples Example 10.
72x2 −2 GCF is 2 2(36x2 −1) Difference of Squares, square roots are 6x and 1 2(6x + 1)(6x −1) Our Solution Example 11.
48x2y −24xy + 3y GCF is 3y 3y(16x2 −8x + 1) Multiply to 16 add to 8 The numbers are 4 and 4, the same! Perfect Square 3y(4x −1)2 Our Solution Example 12.
128a4b2 + 54ab5 GCF is 2ab2 2ab2(64a3 + 27b3) Sum of cubes! Cube roots are 4a and 3b 2ab2(4a + 3b)(16a2 −12ab + 9b2) Our Solution Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative Commons Attribution 3.0 Unported License. ( 4 6.5 Practice - Factoring Special Products Factor each completely.
1) r2 −16 3) v2 −25 5) p2 −4 7) 9k2 −4 9) 3x2 −27 11) 16x2 −36 13) 18a2 −50b2 15) a2 −2a + 1 17) x2 + 6x + 9 19) x2 −6x + 9 21) 25p2 −10p + 1 23) 25a2 + 30ab + 9b2 25) 4a2 −20ab + 25b2 27) 8x2 −24xy + 18y2 29) 8 −m3 31) x3 −64 33) 216 −u3 35) 125a3 −64 37) 64x3 + 27y3 39) 54x3 + 250y3 41) a4 −81 43) 16 −z4 45) x4 −y4 47) m4 −81b4 2) x2 −9 4) x2 −1 6) 4v2 −1 8) 9a2 −1 10) 5n2 −20 12) 125x2 + 45y2 14) 4m2 + 64n2 16) k2 + 4k + 4 18) n2 −8n + 16 20) k2 −4k + 4 22) x2 + 2x + 1 24) x2 + 8xy + 16y2 26) 18m2 −24mn + 8n2 28) 20x2 + 20xy + 5y2 30) x3 + 64 32) x3 + 8 34) 125x3 −216 36) 64x3 −27 38) 32m3 −108n3 40) 375m3 + 648n3 42) x4 −256 44) n4 −1 46) 16a4 −b4 48) 81c4 −16d4 Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative Commons Attribution 3.0 Unported License. ( 5 6.5 Answers - Factoring Special Products 1) (r + 4)(r −4) 2) (x + 3)(x −3) 3) (v + 5)(v −5) 4) (x + 1)(x −1) 5) (p + 2)(p −2) 6) (2v + 1)(2v −1) 7) (3k + 2)(3k −2) 8) (3a + 1)(3a −1) 9) 3(x + 3)(x −3) 10) 5(n + 2)(n −2) 11) 4(2x + 3)(2x −3) 12) 5(25x2 + 9y2) 13) 2(3a + 5b)(3a −5b) 14) 4(m2 + 16n2) 15) (a −1)2 16) (k + 2)2 17) (x + 3)2 18) (n −4)2 19) (x −3)2 20) (k −2)2 21) (5p −1)2 22) (x + 1)2 23) (5a + 3b)2 24) (x + 4y)2 25) (2a −5b)2 26) 2(3m −2n)2 27) 2(2x −3y)2 28) 5(2x + y)2 29) (2 −m)(4 + 2m + m2) 30) (x + 4)(x2 −4x + 16) 31) (x −4)(x2 + 4x + 16) 32) (x + 2)(x2 −2x + 4) 33) (6 −u)(36 + 6u + u2) 34) (5x −6)(25x2 + 30x + 36) 35) (5a −4)(25a2 + 20a + 16) 36) (4x −3)(16x2 + 12x + 9) 37) (4x + 3y)(16x2 −12xy + 9y2) 38) 4(2m −3n)(4m2 + 6mn + 9n2) 39) 2(3x + 5y)(9x2 −15xy + 25y2) 40) 3(5m + 6n)(25m2 −30mn + 36n2) 41) (a2 + 9)(a + 3)(a −3) 42) (x2 + 16)(x + 4)(x −4) 43) (4 + z2)(2 + z)(2 −z) 44) (n2 + 1)(n + 1)(n −1) 45) (x2 + y2)(x + y)(x −y) 46) (4a2 + b2)(2a + b)(2a −b) 47) (m2 + 9b2)(m + 3b)(m −3b) 48) (9c2 + 4d2)(3c + 2d)(3c −2d) Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative Commons Attribution 3.0 Unported License. ( 6 |
14833 | https://jessicaweinkle.substack.com/p/rapid-intensification | Rapid Intensification - by Jessica Weinkle - Conflicted
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Rapid Intensification
An ode to the nation's hurricane hunters
Jessica Weinkle
Oct 09, 2024
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1944 Navy hurricane reconnaissance team flying WWII Bombers B-25 to gather hurricane data
As I write, Florida is bracing for the impacts of Hurricane Milton, a fierce storm that will be costly and memorable. My thoughts are with the people of this colorful state.
Prelude to Milton’s landfall is the media storm whirled up by the appropriate effort to notify the public of the coming threat, and newscasters working their superbowl moment.
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Captivating the public is Hurricane Milton’s rapid intensification. Here is Climate.gov recounting early bouts of Milton’s rapid intensification:
Milton formed as a tropical depression over the southwestern Gulf of Mexico on Saturday, October 5, 2024. Just a couple hours after its formation the National Hurricane Center (NHC) deemed it a tropical storm. By Sunday afternoon, just 24-hours after becoming a named storm, Milton had rapidly intensified into a Category 1 hurricane. Rapid intensification is defined as a 35 mph increase in wind speed in 24 hours.
From 1:00 p.m. CDT on Sunday to 1:00 p.m. CDT on Monday (a 24-hour period), Milton increased an additional 95 mph, more than doubling the requirement for rapid intensification. Based on these early numbers, this explosive amount of rapid intensification is only eclipsed by Wilma 2005 and Felix 2007 according to NHC records. Maximum sustained winds peaked at 180 mph on Monday afternoon (a strong Category 5) and the hurricane’s pressure bottomed out at 897 mb.
The media was awash with loud claims linking rapid intensification to fossil fuel emissions and anthropogenic climate change.
NBC’s National Climate Reporter
On this matter, the IPCC is more reserved than NBC attention grabbing reporters:
The global frequency of TC rapid intensification events has likely increased over the past four decades.
The statement deserves some context, and a proper tribute to the nation’s hurricane hunters and forecasters.
NOAA’s Hurricane Hunter’s capturing data to support forecasting
A characteristic of the chatter about Milton’s forecasting efforts is its emphasis on the work of the Hurricane Hunters. These are brave teams under NOAA and Air Force Reserve that fly into hurricanes to collect data in support of NHC forecasting efforts.
The practice is referred to as hurricane reconnaissance and it began in 1944. The photo at the top of the page is the first team (then, with the Army Air Force) to have made this effort. The reconnaissance team made their trip in a B-25, a popular WWII bomber.
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Hurricane reconnaissance was a game changer for forecasting and observation capabilities prior to the development of satellites in the 1960s. By comparison, meteorologists recounting the experience forecasting the 1926 Great Miami Hurricane note that a challenge in getting a hurricane warning out to Miami was in part due to a lack of data about the storm as it moved west past the Bahamas into open water:
The notable feature of this table [ie observation data] is the small number of reports from oceanic areas and this is strikingly manifest for the 17th-the critical date. It may be that the advance notice of the presence of the storm deterred vessel masters from entering the storm area; any event the absence of reports from oceanic areas a critical times must be a serious handicap to any organization that attempts to forecast the coming of these destructive storms.
So, the development of aircraft reconnaissance was a big deal for hurricane forecasting.
Even still what hurricane hunters offer today is heaps more than they could in earlier days when they were limited not just by meteorological technology but by aircraft capability.
A 2012 paper by Andrew Hagan and Chris Landsea, cautions against about making adamant statements about trends regarding strong hurricanes in recent decades. They find that historic methods would not have captured characteristics of the strongest hurricanes with which we have recent experience:
The results suggest that intensity estimates for extreme tropical cyclones prior to the satellite era are unreliable for trend and variability analysis
There are specific mentions of the inability to capture rapid intensification of storms. This excerpt notes that in the 1940s, meteorologists would not have captured rapid intensification of Hurricane Katrina (2005):
The rapid intensification and subsequent rapid weakening that occurred in the Gulf of Mexico would not have been captured with the observational platforms of the late 1940s/early 1950s, and Katrina very likely would not have been listed as a Category 5. It would have been assumed that Katrina slowly intensified until reaching its peak intensity at landfall in Louisiana.
Should rapid intensification occur at night reconnaissance efforts defiantly would not have captured it because they did not fly at night nor did they fly into storms this strong. Here for Hurricane Wilma (2005):
The next aircraft flight occurred at night, and no intensity information was available during the late 1940s/early 1950s at night. That is the night when Wilma underwent its extreme rapid intensification. The first fix during daylight occurred at 1806 UTC 19 October. Aircraft in the late 1940s and early 1950s would not have been able to penetrate the center since the central pressure was below 940 mb.
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What Hurricane Hunters are able to do today is a technological feat in comparison to what they could do in the 1940s. Hagan and coauthors (2012) give a careful account of the first decade of aircraft reconnaissance noting basic limitations:
Aircraft central pressures were only reported during daylight hours due to the need to visually see the ocean surface and primarily in tropical storms and minor hurricanes. Beginning in 1950 penetrations were generally attempted more often and for somewhat stronger hurricanes compared with the late 1940s (roughly a Saffir– Simpson category stronger on average). Nevertheless, it was still a common occurrence in the 1950s for a plane to attempt a penetration and have to abort before the RMW or even the inner core was reached due to extreme turbulence causing the plane to become uncontrollable
In my own scan of the literature the earliest mention of rapid intensification occurs in discussion of 1964 Hurricane Cleo written in the Monthly Weather Review.
MWR in 1966
The analysis presents brief discussion of these two images.
Which -goodness me- is a far cry from the analytical data and imagery we have today. Check out Ryan Maue at
Weather Trader
for his collection of data, imagery, and analyses.
Of course, as data collection capabilities increased so too did publications on ‘rapid intensification.’ Below is the annual count of papers in Web of Science mentioning rapid intensification and hurricane or tropical cyclone.
Web of Science
Today, NOAA boasts the ability of their Hurricane Hunters to use their aircraft to get “an MRI-like look at the storm” with planes that were acquired in the 1970s (though receiving substantial updates since).
Just last month NOAA announced an order in for new planes. Imagine the capabilities then!
Left: the original reconnaissance plane Right: the plane used today
So, when viewing claims about rapid intensification and climate change keep the above context in mind- which, is not to say that anthropogenic climate change is not relevant for the study of rapid intensification.
It is more the point that our fascination with rapid intensification is guided by our more recent ability to closely observe it.
Good luck to Florida and a big Thank You to our Hurricane Hunters.
Nine weather reconnaissance missions. Nine, y’all. Nine.
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Andy @Revkin
Oct 9, 2024
And now Musk-enabled dissemblers are attacking Hurricane Hunters.
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14834 | https://www.youtube.com/watch?v=xjNW6LgPJVY | Sodium-potassium ATPase: Active Transport
AMBOSS: Medical Knowledge Distilled
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Posted: 12 Nov 2020
Sodium-potassium ATPase, also known as the sodium-potassium pump, is a membrane-bound transport protein. It’s an ion pump, which functions by working against a concentration gradient and requires a direct energy source, ATP. This gives rise to the alternate name for this type of transport, primary active transport.
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Sodium-potassium ATPase Sodium-potassium ATPase, also
known as the sodium-potassium pump, is a membrane-bound transport protein.
It’s an ion pump, which functions by working against a concentration gradient
and requires a direct energy source, ATP. This gives rise to the alternate name for
this type of transport, primary active transport. Sodium-potassium ATPase has an intracellular
ATP binding site. If ATP binds here, the protein's conformation opens up to the
cytosol, eventually taking in three sodium ions. The protein changes its conformation and closes at
the intracellular side. Now, autophosphorylation occurs, which releases ADP while phosphate
still remains bound to the protein. Next, the protein opens towards the
extracellular side, releasing three sodium ions into the extracellular space. In turn,
the binding pocket becomes available again, and two extracellular potassium ions can bind.
A new conformational change of the protein causes the binding pocket to close again. The
intracellularly bound phosphate is cleaved hydrolytically, that is, by a water molecule.
Through the subsequent binding of another ATP molecule, a new conformational change of the ion
pump occurs that opens it to the intracellular side. The potassium ions are released into
the cytosol. Three sodium ions now bind to the empty protein for transport into the
extracellular space, starting a new cycle. Sodium-potassium ATPase is one of the most
important transporters for maintaining the vital concentration gradients of sodium
and potassium. This concentration gradient generated by ATP consumption is fundamental
for regulating water content in cells and electrochemical cell communication. It can
also drive secondary active transport processes. |
14835 | https://www.youtube.com/watch?v=oSCaqdi0SY0 | How to build a 3X3 and 4X4 Magic Square easily
IPM Leap
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Posted: 24 Nov 2021
One of the most fascinating tasks in Mathematics is to Create a Magic Square and this video illustrates how to build a 3X3 and 4X4 Magic Square easily from scratch without remembering any complicated algorithms.
What are Magic Squares?
They are square grids, most commonly 3×3 or 4×4, which are filled with numbers. The numbers in every row, every column, and across diagonals add up to the same value which is known as the Magic constant.
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Transcript:
Introduction [Music] hi and welcome to our channel ipm leap one of the most fascinating tasks in mathematics is to create a magic square and today i am going to show you how to build a 3 cross 3 and 4 cross 4 magic square i am going to teach you a super easy method to create these grids and i am sure once you learn it you will never forget it so what are magic squares in simple words these are square grids most commonly 3 cross 3 or 4 cross 4 which are filled with numbers the numbers in every row every column and across diagonals add up to the same value which is known as the magic constant in fact you may verify for yourself in this 3 cross 3 grid whether you take the first row the second row or the third row the three numbers will always add up to the value 15. same holds true for numbers in the columns whether you take the first column the second or the third numbers added to the value 15 also numbers across the two diagonals add up to 15. it's simply magical so now that you know why it is called a magic square let's discuss how to create one How to create a 3X3 Magic Square okay the objective is clear we have to arrange the numbers 1 2 3 up till 9 in a 3 cross 3 square grid in such a way that each number occurs once and the entries along each column each row and each of the two diagonals added to the same value so let's discuss how to place those numbers so that we are able to create a 3 cross 3 magic square we will use a very simple and effective memory trick first we write the numbers back to back from 1 to nine in a square one two and three in the first row then four five and six and then we write seven eight and nine next we have to draw lines diagonally to separate out these numbers so first line second line parallel to the first third line and fourth line similarly now draw lines in the opposite direction first line second line parallel to that third line and fourth line all the numbers are now separated and we have grid made by these two sets of parallel lines isn't it in this grid we can see that exactly four spaces are vacant or empty and we can also see that exactly four out of these nine numbers are lying outside this grid so the next step is to create this magic square is to put these four numbers inside this grid okay let's do that let's replicate the grid so that you may see properly what's happening so first we will shift 3 from this position to a diagonally opposite position in the grid which is an empty space similarly we will shift the number 7 from this position to the diagonally opposite position in the grid shown by this arrow next we will shift number one to this diagonally opposite position in the grid which is empty and finally we will shift the number nine to this only remaining position in the grid that's it we have got a grid which is numbers as in a magic square you may write this properly and our magic square will look like this [Music] How to calculate the Magic Constant now comes the magic constant in any magic square the entries along any row any column and both diagonals add up to the same value known as the magic constant even though we already have the magic square in front of us for calculating the magic constant we don't need to look at the numbers in this square it's pure logic since a 3 cross 3 magic square uses numbers from 1 to 9 we first sum up these 9 numbers the sum is equal to 45 now there are three rows so we divide this sum 45 equally over the three rows so each row sum will be equal to 45 divided by 3 which is 15. easy isn't it ok so just to appreciate the beauty of a magic square let's sum up the numbers along each row each column and both diagonals to see if they add up to the magic constant fifteen row one four plus nine plus two is equal to fifteen row 2 3 plus 5 plus 7 is equal to 15. row 3 8 plus 1 plus 6 is equal to 15. the columns now column 1 4 plus 3 plus eight is equal to fifteen column two nine plus five plus one is equal to fifteen and column three two plus seven plus six is equal to fifteen now the two diagonals diagonal one four plus five plus six is equal to fifteen and diagonal two eight plus five plus two is also equal to fifteen so that was the three cross three magic square for you Rotational Variance now there are eight variations of the same three cross three square that i am going to explain first we need to account for rotational variance so let's bring each of the four edges of our original square on top one by one first a quarter spin clockwise this medic square would look like then a half spin 180 degree clockwise this one would look like and then a spin by three quarters or 270 degrees this one would look like the following [Music] and next we reflect these four patterns about a vertical axis so there could be eight variations of the same three cross three magic square they are all on your screen [Laughter] [Music] Magic Square 4X4 let's move on to a 4 cross 4 magic square we know the objective so first things first how do we create a 4 cross 4 magic square effortlessly well just like we did in the previous case we write all numbers from 1 to 16 sequentially along a square grid so one two three and four in first row five six seven and eight in second row nine ten eleven and twelve in third row and 13 14 15 and 16 in the fourth row next we need to shuffle a few of these numbers while keeping the others at the original position so numbers which stay back are these two the middle ones along all edges we copy 2 and 3 8 and 12 stay back we copy 8 in 12 14 and 15 stay back we copy 14 and 15 5 and 9 stay back so we copy 5 and 9. next step is to interchange these four corner numbers diagonally one gets interchanged with 16 we write 16 here and one here and 13 gets interchanged with the number four we write 13 here and four here now we just need to fill numbers in the inner square these original numbers in the inner square get interchanged diagonally for our magic square so 6 and 11 get interchanged we write 11 here and 6 here and 7 and 10 also get interchanged we write 10 here and 7 here that completes our magic square this is what our 4 cross 4 magic square looks like Magic Constant just like the 3 cross 3 square in a 4 cross 4 square each number from 1 to 16 occurs once and the entries along each row and each column add up to the same value the magic constant let's calculate it for a 4 cross 4 grid so we have filled numbers from 1 to 16 in this grid the sum of these numbers is 136 divided equally over 4 rows each row sum will be 136 divided by 4 which is equal to 34 our magic constant is 34 for a 4 cross 4 magic square let's verify row 1 16 plus 2 plus 3 plus 13 is equal to 34 row 2 5 plus 11 plus 10 plus 8 is equal to 34 row 3 9 plus 7 plus 6 plus 12 is equal to 34 row 4. 4 plus 14 plus 15 plus 1 is equal to 34 also along the columns column 1 16 plus 5 plus 9 plus 4 is equal to 34 column 2 2 plus 11 plus 7 plus 14 is equal to 34 column 3 3 plus 10 plus 6 plus 15 is equal to 34 column 4 13 plus 8 plus 12 plus 1 is also equal to 34. additionally in a four cross four square the entries along each of the two diagonals the four quadrants and the four corner squares also add up to the magic constant 34 would you like to see the two diagonals 16 plus 11 plus 6 plus 1 is equal to 34 and 4 plus 7 plus 10 plus 13 is equal to 34. the four quadrants quadrant one sixteen plus two plus five plus eleven is equal to thirty-four quadrant two three plus thirteen plus ten plus eight is equal to thirty-four quadrant three nine plus seven plus four plus fourteen is equal to thirty-four and quadrant four six plus twelve plus fifteen plus one is equal to thirty-four and finally the four corners 16 plus 13 plus 4 plus 1 is equal to 34 the magic constant [Music] that brings us to the end of this video if you enjoy this lesson don't forget to subscribe our channel and like and share our videos for any doubts or queries drop in a comment on this video page below we'll be glad to address them hope you all stay safe happy and healthy always take care and happy learning |
14836 | https://www.kenhub.com/en/library/anatomy/recurrent-branch-of-median-nerve | Recurrent median nerve: anatomy, pathway and supply | Kenhub
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ArticlesAnatomyUpper limbNerves and vesselsRecurrent branch of median nerve
Recurrent branch of median nerve
Author: Christina Loukopoulou, MSc • Reviewer: Roberto Grujičić, MD
Last reviewed: October 30, 2023
Reading time: 4 minutes
Recommended video: Median nerve [01:57]
Anatomy, distribution and function of the median nerve.
Recurrent branch of median nerve
Ramus recurrens nervi mediani
1/3
Synonyms:Thenar muscular branch of median nerve, Ramus muscularis thenaris nervi mediani
The recurrent branch of median nerve (C8, T1) is a small muscular branch given off by the median nerve (C6-T1) to supply the thenar eminence in the hand. It is a purely motor nerve. While recurrent branch of median nerve is the official anatomical term, variant terminology can be found in textbooks. Synonyms include recurrent motor branch of median nerve, recurrent median nerve, thenar muscular branch and million dollar nerve.
As a nerve arising from the brachial plexus, the median recurrent nerve reaches the hand by traveling through the carpal tunnel. Also known as the transverse carpal ligament, the carpal tunnel lies deep to the flexor retinaculum and acts as a passageway for the forearm structures to reach the hand. These structures include the median nerve along with the nine tendons of the forearm muscles: more specifically four tendons of flexor digitorum superficialis, four tendons of flexor digitorum profundus and a single tendon of flexor pollicis longus muscle.
Anatomical variations in the branching pattern of the recurrent branch of median nerve are fairly common. The Poisel classification is often used to describe the course and variations of the nerve’s branches as well as their relationship to the transverse carpal ligament. This classification divided the variations into three types:
Extraligamentous type: recurrent branch of median nerve arises close to the distal end of the transverse carpal ligament and continues a retrograde course to reach the thenar eminence.
Subligamentous type: recurrent branch of median nerve arises within the carpal tunnel and maintains its course deep to the transverse carpal ligament to reach the thenar eminence.
Transligamentous type: recurrent branch of median nerve arises within the carpal tunnel and proceeds to pierce the transverse carpal ligament to reach the thenar eminence.
Owing to its subcutaneous location, the recurrent branch of median nerve can be easily severed when this area is lacerated resulting in paralysis of the thenar muscles. Colloquially, the nerve is also known as the ‘million dollar nerve’. Iatrogenic injury to the nerve during surgery can lead to a ‘million dollar lawsuit’ due to loss of basic hand function.
Mnemonic
Terminal branches of the recurrent branch of median nerve supply the thenar eminence, the fleshy prominence of the thumb. Specific muscles innervated by the nerve are the abductor pollicis brevis, opponens pollicis and the larger superficial head of the flexor pollicis brevis.
In order to remember the muscles innervated by the recurrent branch, you can use the mnemonic LOAF to recall the motor innervation of the median nerve. This stands for:
L: lateral two lumbricals
O: opponens pollicis
A: abductor pollicis brevis
F: flexor pollicis brevis (superficial head)
Terminology (variations)Recurrent branch of median nerve
Variations: recurrent motor branch of median nerve, recurrent median nerve, thenar muscular branch, million dollar nerve
Definition The median recurrent nerve (C8, T1) is a small motor branch given off by the median nerve (C6-T1).
Field of innervation Abductor pollicis brevis muscle, opponens pollicis muscle and superficial head of flexor pollicis brevis muscle
Function Motor innervation of thenar muscles
Learn more about the intrinsic muscles of the hand with this study unit:
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Sources
All content published on Kenhub is reviewed by medical and anatomy experts. The information we provide is grounded on academic literature and peer-reviewed research. Kenhub does not provide medical advice. You can learn more about our content creation and review standards by reading our content quality guidelines.
References:
Al-Qattan, M. (2010). Variations in the course of the thenar motor branch of the median nerve and their relationship to the hypertrophic muscle overlying the transverse carpal ligament. The Journal of Hand Surgery, 35(11), pp.1820-1824. DOI: 10.1016/j.jhsa.2010.08.011
Henry, B., Zwinczewska, H., Roy, J., Vikse, J., Ramakrishnan, P., Walocha, J. and Tomaszewski, K. (2015). The prevalence of anatomical variations of the median nerve in the carpal tunnel: A systematic review and meta-analysis. PLOS ONE, 10(8), p.e0136477. DOI: 10.1371/journal.pone.0136477
Moore, K. L., Dalley, A. F., & Agur, A. (2017). Clinically oriented anatomy (8th ed.). Lippincott Williams and Wilkins.
Standring, S. (2016). Gray's Anatomy (41st ed.). Edinburgh: Elsevier Churchill Livingstone.
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14837 | https://harperjames.co.uk/article/contract-formation-capacity/ | Capacity of a company to contract in UK law – Commercial Contracts
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What is capacity in contract law?
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Article
10 mins read
Updated on 16 June 2022
Our subject expert
David Sant
Senior Solicitor - Commercial Technology & Data Protection
Before you sign any commercial contract, you need to be sure the other party has the legal capacity to agree. If they don’t, you risk ending up with an unenforceable deal.
Our commercial law solicitors regularly advise on whether the legal and structural conditions for contract formation are met, including director authority, corporate power, and contractual capacity across sectors. We can help you structure your agreements so they are watertight, enforceable, and structured correctly from the start.
Jump to:
What is capacity to contract?
Contracts – the basics
Contract capacity and corporations
Contract capacity and contracting with trustees
The capacity of an individual to contract
Contracting with an under 18
Contracting and mental capacity
Contracting capacity and intoxication
Practical tips on contracts and capacity
Eliminate doubt around your contractual relationships
What is capacity to contract?
To form a contract, a party must have the legal capacity to contract. This is a basic of contract law. Without capacity the contract may not be valid and, if so, won't be enforceable.
Not checking capacity to contract is the sort of mistake that an entrepreneur or salesperson is likely to only make once, but it can be a very expensive error to make.
Capacity to contract means the person or business you are contracting with has the legal ability to enter into a contractual relationship with you. Capacity depends on whether you are contracting with:
A corporation
An individual acting in a particular role, such as a trustee
An individual
Once the status of the person you are contracting with is established, then their capacity to contract is assessed using a combination of statute and common law, assisted by case law.
Contracts – the basics
It isn’t just a person’s capacity to contract that’s important when you are entering into a contract. If a person is acting on behalf of another, such as their employer, an association, a partnership or company, they need contractual authority.
You also need to make sure that there is:
An offer
An acceptance
An intention to create legal relations
Without the basics covered, you may not have a contract. That could be disastrous for you or for your fledgling business if you have entered into a contract to sell goods to A on the assumption that you have already entered into a valid contract to buy the goods from B, with a decent profit margin on sale, only to find that you have no valid contract with B to buy but you are bound by a contract to sell to A what you don’t have and can't buy on the same terms as you thought you had validly contracted with B for.
Contract capacity and corporations
The capacity of corporations to enter into contracts depends on the type of corporation. The topic is closely allied to the issues surrounding authority to contract. In summary, the position is:
Registered companies - a registered company has a legal personality and has capacity to contract. Any person authorised by the company can enter into a contract on behalf of the company. Every director and the company secretary are authorised signatories to a contract. Provided that the other party to the contract is acting in good faith, they are not required to check whether the powers of the directors are limited in any way.
Limited liability partnerships (LLPs) - an LLP, like a company, is a body corporate with its own legal personality. It can therefore legally contract in its own right. Each member can bind the LLP although the LLP agreement between the members may include restrictions on the powers of individual members. In usual business dealings, a third party may assume that any member of the LLP is authorised to act on behalf of, and bind, the LLP, unless they know to the contrary.
General partnerships - a general partnership does not have a legal personality, and so, liability rests with each of the partners. The agreement which governs the relationship between the partners and any limitations on the authority of those partners is private. Every partner acts as an agent of the partnership and of the other partners.
Limited partnerships- a limited partnership does not have its own legal personality. Therefore, liability rests with the general partner. The general partner runs the day-to-day business of a limited partnership and acts as its agent. The limited partners have no active role and lose their limited liability status if they breach this. The general partner has a wide scope to bind the partnership and the other partners, but the private partnership agreement may set out restrictions on authority. It is important to consider the legal capacity of the general partner, for example, whether they are an individual, body corporate etc.
Overseas companies - the position of an overseas company is governed by local law and the company constitution. It is therefore important when entering into a contract with an overseas company to obtain a legal opinion from local counsel that the company can enter into the contract and has done so correctly. This does not guarantee the enforceability of the contract but provides a little more protection than taking the word of the overseas company.
If a corporation acts outside of its powers, the actions are deemed ultra vires. An ultra vires contract will be void. It is possible for a company to ratify an ultra vires act by a special resolution, but this will not affect the liability of the directors or anyone else which has arisen as a result of the ultra vires act.
Contract capacity and contracting with trustees
The ability of a trustee to enter into a contract comes from statute and the trust instrument. Although individual trustees can enter into contracts it is crucial to review the provisions of the trust instrument as it may affect the powers of the trustees. Whilst the trustees may be personally liable by entering the contract this won't assist your business if the trustees don’t have the personal assets to compensate for loss because they lacked capacity to bind the trust.
The capacity of an individual to contract
Individuals are presumed to have capacity to enter a contract . If an individual later says they didn’t have capacity it is up to them to prove it. If they can show they lacked capacity the contract is voidable at the option of the person who lacked capacity.
An individual’s capacity can be questioned where the party claiming incapacity is:
A minor under the age of 18
Mentally incapacitated
Was intoxicated at the time of the contract
Whether a minor or a mentally incapacitated or intoxicated person has the capacity to contract isn’t a simple yes or no answer as the law on whether they have the capacity to contract depends on the nature of the contract.
Contracting with an under 18
For some businesses, whose sales are targeted at the youth market, it is crucial to understand the capacity to contract of under 18s so your business understands its risks.
With under 18s and contract capacity , it is all about the nature of the contract. That’s important to know if you are selling apps, games, tech stuff or trainers etc.
There are three relevant types of contract:
Contract for necessaries – under 18s are bound by a contract for necessaries. The Sale of Goods Act 1979 defines them as ‘goods suitable to the condition in life of the minor and to his or her actual requirements at the time of the sale and delivery’. Necessaries includes food, clothes, accommodation and other essentials and the law says a minor who purchases such necessaries must pay a reasonable price for them.
Contract of apprenticeship, education and service – under 18s are bound by this type of contract if the minor will benefit from it.
Any other contract entered into by an under 18 – if the contract doesn’t fall into the above two categories it is voidable at the minor’s option. This means the contract can be avoided by the under 18 before they reach 18 years of age or within a reasonable time after. What is deemed a ‘reasonable time’ depends on the circumstances of each case. Frustratingly for businesses, there is no requirement for the under 18 to explain or justify the reason for voiding the contract . However, if the contract is voided, the under 18 should return any goods acquired under the contract. If the under 18 has paid money under a contract, they are unlikely to be reimbursed unless they can prove the contract did not benefit them. When a minor reaches 18 years of age, they can choose to ratify the contract and so make it valid and enforceable.
Contracting and mental capacity
Contracting and mental capacity is becoming a hot topic as businesses, and their employees, can't always recognise that a person has an invisible condition that may affect their capacity to contract. Trying to hold a person to a contract can result in reputational damage and adverse publicity but the reality is that it can be hard to identify if someone doesn’t have the mental capacity to contract.
The law under the common law and the Mental Capacity Act 2005 (MCA) says a person can assume the other contracting party has mental capacity and the onus is on the other party, or their advocate, to prove that they didn’t have capacity when they entered the contract.
With mental capacity and contracts, the type of contract is relevant. Under the MCA, a mentally incapacitated person must pay a reasonable price for necessary goods or services supplied or provided to them. ‘Necessary’ means suitable to the person's condition in life and to their actual requirements at the time when the goods or services were supplied.
Any other contract will be valid and binding unless the person challenging the contract on the basis of mental incapacity can prove that:
When they entered into the contract, they did not have the mental capacity to do so and
The other party to the contract knew, or should have known, that they didn’t have capacity
There is case law on what amounts to mental capacity and knowledge but each case turns on its facts. A person who was mentally incapacitated at the time of the contract can choose to rescind the contract. This places the parties in the position they were in prior to the contract.
Contracting capacity and intoxication
According to an 1811 case (Pitt v Smith (1811) 3 Camp 33) an intoxicated person doesn’t have capacity to enter into a contract. However, an intoxicated person will be bound by a contract for necessaries and is required to pay a reasonable price for any such necessaries purchased by virtue of the Sale of Goods Act.
The problem with contract capacity and intoxication is that there are degrees of drunkenness. To determine whether a person was so drunk as to lack contractual capacity, the courts have said the following two questions are relevant:
When the intoxicated person entered into the contract, were they too drunk to understand and
Did the other party have, or should have had, knowledge of the intoxication
If a person is successful in claiming that they didn’t have capacity through intoxication they can either rescind or ratify the contract. Rescinding the contract puts the parties in the position they were in prior to the contract. Ratifying the contract makes the contract binding and enforceable.
Practical tips on contracts and capacity
Reading an article on contract capacity can really bring it home to sole traders, start-ups and big businesses that you have to have your wits about you when you enter into a contract. In some ways that’s easier when you are a founder and it is just you, and maybe a trusted partner, negotiating all your contracts. That’s not possible as your business grows and you scale up so here are some practical tips:
Carry out a risk profile for your business -are most of your sales to overseas companies or to youngsters under the age of 18? Is your product likely to attract those who may not have individual capacity? These are the types of questions you need to be asking to assess your business risk profile.
Consider if your products are necessities -if you are in the food and hospitality industry selling burgers you will be OK but maybe not if you are high end retail or if you market top of the range technology and digital products.
Do you carry out risk checks when contracting with individuals - for example, if you are selling phones on a contract, do you carry out an age check?
Checklists on contracting – even if you and your team are highly experienced, a checklist can be very helpful, not least because without it you may not remember the difference in contractual capacity between a LLP and general partnership.
Training– where you have staff negotiating contracts on your behalf it is sensible to not only have a thorough induction process but ongoing training to remind of the importance of contract basics, such as contract capacity, contract authority and whether you are contracting without realising it through exchange of email, which in some cases can be enforceable.
Eliminate doubt around your contractual relationships
If there's uncertainty about the capacity of another party– or your internal authority to contract– our commercial law solicitors can help you assess and verify the necessary legal standing. We advise on director and board powers, enforceability issues, and best practices for avoiding void agreements. With us, your business can contract with full legal confidence.
About our expert
David Sant
Senior Solicitor - Commercial Technology & Data Protection
David is a senior solicitor at Harper James, specialising in technology and data protection.
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Expanded Form With Decimals – Definition with Examples
Home » Math Vocabulary » Expanded Form With Decimals – Definition with Examples
What is Decimals?
Expanded Form
Expanded Form of Decimal Numbers
Standard Form
Application of Expanded Form with Decimals
Solved Examples
Practice Problems
Frequently Asked Questions
The expanded form with decimal numbers is the mathematical expression to show the sum of the values of each digit in the number. Let’s learn more about decimals, place value, and expanded forms below.
What is Decimals?
A decimal number is a number that consists of a whole number and a fractional part. The decimal point separates the whole number from the fractional part.
In a decimal number, the digits to the left of the decimal point represent a whole number. The digits to the right of the decimal represent the parts.
As we move towards right after the decimal point, the place value of the digits becomes 10 times smaller.
The first digit on the right of the decimal point means tenths i.e. . The next place becomes ten times smaller and is called the hundredths i.e. and so on.
For example,
In 31.356, 31 is the whole number part, 3 is in tens place and its place value is 30,
1 is in ones place, and its place value is 1.
There are three digits to the right of the decimal point,
3 is in the tenths place, and its place value is 0.3 or
5 is in the hundredths place, and its place value is 0.05 or
6 is in the thousandths place, and its place value is 0.006 or
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Expanded Form
In the expanded form. We break up a number according to its place value and expand it to show the value of each digit.
For example, the expanded form of 254 is given.
Thus, the expanded form of the number is .
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Expanded Form of Decimal Numbers
The expanded form with decimals is the mathematical expression to show the sum of the values of each digit in the number. Writing decimals in an expanded form simply means writing each number according to its place value. Before the decimal point, the expanded form is the same as it is for whole numbers. After the decimal point, it is different.
When you write a decimal number in its expanded form, you are expressing the number separated into its composite individual place values and decimal values if necessary in the form of an expression.
Likewise, the expanded form of a number with a decimal or a fraction is written with a base 10-multiple denominator, represented by the power of 10.
For example, the number 3.482 in its expanded form is written as:
The expanded form of the number 3.482 is:
4⁄10 + 8⁄100 + 42⁄1000 ,or
How to Write a Decimal in an Expanded Form?
Writing decimals in expanded form can be achieved by following the steps given below.
Suppose we want to expand 1.234.
Step 1: We start from the left hand side first. We have ones place, i.e., 1.
Step 2: Next, we have the first decimal place, the tenths. We take 2 and multiply it by fraction .
Step 3: Then, we have the hundreds place. We move to a higher multiple of 10 for the denominator. In other words, we multiply 3 by fraction 1100.
Step 4: Finally, we have the thousands place. Add another 0 in the denominator. We multiply 4 by .
Hence, the expanded form of is .
Standard Form
Writing a number in standard form (or numeric form) is the opposite or reverse of writing a number in expanded form.
For example, the standard form for is .
Expanded form and standard form are opposites of each other.
Application of Expanded Form with Decimals
The scope of decimal numbers is in the areas where accuracy and precision are required. The concept of the expanded form is useful in comprehending the value of a number as well as the numeric value of a quantity.
Fun Facts
The weights in the ratios of 1⁄20 , 1⁄10 , 1⁄5 , 1⁄2 and multiples of 10 were used in the Indus Valley civilization for the purpose of trade.
Rod calculus in ancient China used bamboo strips with the decimal system for math operations like multiplication during 305 BCE.
Mathematician Archimedes invented a decimal positional system for Sand Reckoner to represent large numbers that were multiples of 10.
Solved Examples
Example 1. Write 8.12 in its expanded form.
Solution: The expanded form of
Example 2. Write in standard form.
Solution: The standard form is 7.893.
Example 3. What will be the place value of 4 in 56.924?
Solution: The place value of 4 in 56.924 will be .
Practice Problems
Expanded Form With Decimals - Definition with Examples
Attend this quiz & Test your knowledge.
1
The standard form of is
58.765
587.65
5876.5
5.8765
CorrectIncorrect
Correct answer is: 58.765
The standard form of
.
2
The expanded form of 123.45 is
CorrectIncorrect
Correct answer is:
The expanded form of 123.45 is
.
3
The place value of 6 in 987.65 is
CorrectIncorrect
Correct answer is: 0.6
The expanded form of 987.65 is
.
The place value of 6 in 987.65 is 0.6.
Conclusion
The expanded form is nothing but a technique of rewriting a number by including the values of the digits. Given above was everything you needed to know about the expanded form.
Frequently Asked Questions
What is the difference between expanded form and standard form?
How is expanded form with decimals used?
Is expanded form the same as expanded notation? |
14839 | https://www.amaehealth.com/blog/what-is-petulant-bpd | min read
What is Petulant BPD? Understanding the Symptoms, Causes, and Management
By
Leslie Kolb
|
February 25, 2025
Summarize with AI
Table of Contents
Have you ever met someone whose moods seem to switch like a light? They might have petulant BPD. This article will explore this lesser-known subtype of Borderline Personality Disorder (BPD) and provide insights into its complexities.
What is Petulant BPD?
Borderline Personality Disorder is a complex mental health condition affecting emotions, behavior, and relationships. Petulant BPD is a specific subtype characterized by intense emotional volatility and controlling behaviors.
People with petulant borderline often struggle with expressing their feelings in healthy ways. They may experience frequent mood swings and have difficulty maintaining stable relationships. Their emotions can feel like a rollercoaster, with highs and lows happening rapidly.
The irritable mood associated with this subtype can lead to explosive anger and irritability. These individuals might seem constantly dissatisfied or complaining, hence the term "petulant." It's as if they're perpetually frustrated with the world and those around them.
Understanding petulant BPD is crucial because it helps differentiate it from other BPD subtypes. This knowledge can lead to more targeted and effective treatment approaches. It's important to note that while these traits can be challenging, they're often a response to deep-seated pain and fear.
Key Symptoms of Petulant BPD
Common Symptoms of Petulant BPD
Understanding the symptoms of petulant BPD is crucial for recognizing and addressing this condition. Here are the primary symptoms:
Let's consider Sarah, a 28-year-old with petulant borderline. She often feels misunderstood and struggles to express her needs. When frustrated, Sarah's anger explodes, pushing away those closest to her. Her relationships are a series of intense beginnings and dramatic endings.
The petulant mood in BPD can manifest as chronic feelings of bitterness or resentment. Sarah often feels the world is against her, leading to frequent complaints and dissatisfaction. She might lash out at friends for perceived slights or become furious over minor inconveniences.
It's important to remember that these symptoms are not a choice. They stem from deep-seated emotional pain and learned coping mechanisms. With proper treatment and support, individuals with petulant BPD can learn healthier ways of managing their emotions and relationships.
How Petulant BPD Impacts Relationships and Daily Life
Petulant BPD can significantly affect a person's relationships and everyday functioning. The emotional instability and fear of abandonment often create a push-pull dynamic in relationships.
For instance, Tom, diagnosed with a petulant borderline, constantly fears his girlfriend will leave him. He alternates between clinging to her and pushing her away when he feels vulnerable. This behavior strains their relationship and leaves both parties exhausted.
The petulant mood associated with this subtype can make maintaining connections challenging in friendships. Frequent irritability and complaints may drive others away, reinforcing the person's fear of abandonment. A friend might cancel plans due to illness, but someone with petulant BPD might interpret this as rejection, leading to an angry outburst.
Daily life with such a problem can feel like an emotional minefield. Simple disagreements might trigger intense reactions. Work relationships can suffer due to difficulty managing emotions in professional settings. For example, constructive criticism from a boss might be perceived as a personal attack, leading to defensive or aggressive behavior.
The controlling behaviors characteristic of the petulant borderline can also impact family dynamics. A parent with this condition might become overly protective or demanding of their children, straining the parent-child relationship.
It's crucial to note that despite these challenges, individuals with petulant BPD are often deeply caring and sensitive people. Their behaviors stem from intense emotions and a fear of rejection. Many can learn to manage their symptoms and build healthier relationships with proper treatment and understanding.
Amae Health recognizes the complex nature of petulant BPD and its impact on daily life. Our integrated care approach addresses the emotional and practical challenges individuals face with this condition. Providing comprehensive support, we help patients navigate the complexities of relationships and daily functioning.
Causes and Risk Factors of Petulant BPD
What Causes Petulant BPD?
The exact cause of petulant BPD isn't fully understood, but several factors contribute to its development:
Risk Factors for Petulant BPD
Several factors can increase the likelihood of developing petulant borderline:
Understanding these factors helps in early identification and intervention. Amae Health emphasizes considering biological and environmental factors in treatment planning. It's important to note that having one or more risk factors doesn't guarantee someone will develop petulant BPD. Conversely, some individuals may develop the condition without obvious risk factors. Each person's journey is unique, and treatment should be tailored accordingly.
Treatment Options for Petulant BPD
Managing Symptoms of Petulant BPD
Effective treatment for petulant BPD often involves a combination of approaches. Here's an overview of available treatments:
Psychotherapy:
Cognitive Behavioral Therapy (CBT) helps identify and change negative thought patterns and behaviors. CBT can be particularly useful in managing the petulant mood swings associated with this subtype.
Medication:
While there's no specific medication for BPD, certain drugs can help manage symptoms:
Holistic Approaches:
Mindfulness practices can help individuals stay grounded in the present moment. It can be particularly useful in managing the intense emotions of petulant BPD.
Amae Health offers a comprehensive treatment approach for petulant BPD. We have programs that combine evidence-based therapies with holistic care to address all aspects of the condition. They recognize that recovery is a journey and provide ongoing support to help individuals manage their symptoms long-term.
It's important to note that treatment for petulant BPD is not one-size-fits-all. What works for one person may not work for another. Patience and persistence are key, as finding the right combination of treatments may take time.
Living with Petulant BPD: Personal Accounts and Coping Strategies
Living with a petulant borderline can be challenging, but many individuals find ways to manage their symptoms effectively. Here are some personal accounts and coping strategies:
Emily, diagnosed with petulant BPD at 25, shares: "Learning to recognize my triggers was a game-changer. I now use deep breathing when I feel my anger rising." Emily found that keeping a mood diary helped her identify patterns in her emotions and behaviors.
Coping strategies that many find helpful include:
John, another individual with petulant mood swings, found journaling helpful: "Writing down my thoughts helps me understand my emotions better and communicate more effectively." John also uses art to express emotions that are difficult to put into words.
Sarah, who has been managing her petulant borderline for several years, emphasizes the importance of self-compassion: "I've learned to be kinder to myself. My emotions are intense, but they don't define me." Sarah practices positive self-talk and celebrates small victories in her recovery journey.
It's important to remember that recovery is not linear. There may be setbacks along the way, but each challenge overcome is a step towards better management of petulant BPD.
Conclusion
Understanding the nuances of petulant borderline can help both individuals with the condition and their loved ones navigate the challenges it presents. It's important to remember that behind the angry outbursts and controlling behaviors is often a person struggling with intense emotions and a deep fear of abandonment.
Remember, if you or someone you know is struggling with symptoms of petulant borderline, help is available. Organizations like Amae Health provide comprehensive, personalized care for individuals with BPD.
In the end, it's important to remember that individuals with petulant BPD are not defined by their diagnosis. They are complex, valuable human beings capable of growth, love, and positive change. They can thrive and lead fulfilling lives with the right support and resources.
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min read
The Difference Between Learning Disabilities and ADHD
By
Sonia Garcia
|
April 18, 2025
Picture a classroom where two students are struggling with their reading assignment. Michael reads slowly, mixing up letters despite his best efforts to focus. Meanwhile, Olivia reads fluently but can't sit long enough to finish a page. Michael has a learning disability called dyslexia, while Olivia has ADHD. Though both students face challenges, their underlying difficulties are quite different. Understanding these differences is crucial for parents, teachers, and healthcare providers. This guide explores how these conditions differ and, most importantly, how to support each unique situation.
What Are Learning Disabilities (LD) and ADHD?
Defining Learning Disabilities (LD)
A learning disability affects how the brain processes information. Think of it as a unique wiring system in the brain. People with learning disabilities often have average or above-average intelligence. Their challenges lie in specific areas of learning.
Common types of learning disorders include:
Understanding ADHD
ADHD creates unique processing challenges in the brain. It affects how people manage daily tasks and activities. The brain struggles with focus and impulse control throughout various situations. It impacts school, work, and social interactions.
Executive functions work differently in people with ADHD. Simple tasks might feel overwhelming. Many everyday activities require extra mental effort. It makes daily routines more challenging than they appear.
Three distinct types of learning disorders present different challenges:
Everyone experiences symptoms differently depending on their environment and daily schedule. Some find mornings most challenging. Others struggle more during quiet afternoon activities. Understanding these patterns helps create better support strategies.
Key Differences Between Learning Disabilities and ADHD
Focus on Specific vs. Global Skills
Learning disabilities target specific academic skills. Meanwhile, ADHD affects overall attention and behavior management. This fundamental difference shapes how each condition impacts daily life. The impact varies significantly:
| Area of Impact | Learning Disabilities (LD) | ADHD |
---
| Academic Skills | Struggles with specific subjects while excelling in others | Performance varies across all subjects based on interest and focus |
| Reading Ability | May have specific reading difficulties (dyslexia) | Can read well but loses focus during reading tasks |
| Math Skills | Might struggle specifically with calculations (dyscalculia) | Can understand math but makes careless errors due to inattention |
| Writing Tasks | May have trouble forming letters or expressing ideas in writing | Writing is rushed, disorganized, or left incomplete |
| Attention Span | Generally able to focus but struggles with specific tasks | Difficulty maintaining attention across all activities |
| Organization | Usually capable of keeping materials and spaces organized | Consistently struggles with organization in all areas |
| Task Completion | Completes tasks but may take longer in specific areas | Starts many tasks but has trouble finishing them |
| Social Skills | Social abilities typically unaffected by the disability | May struggle with turn-taking and reading social cues |
| Memory | Specific memory challenges related to disability area | General difficulties with working memory and recall |
| Following Instructions | Can follow directions but may need help in specific areas | Trouble remembering and following multi-step instructions |
| Test Performance | Consistent difficulties in specific subject areas | Variable performance depending on attention level |
| Project Planning | Generally able to plan and execute projects | Struggles with project planning and time management |
Impact on Executive Functions
Executive functions play a crucial role in daily activities. These brain-based skills affect how people manage tasks and behaviors.
People with learning disabilities usually maintain strong executive function skills. They can:
Those with ADHD often struggle with executive function skills. They experience challenges with:
How Are They Diagnosed?
Diagnostic Criteria for LD
Learning disability diagnosis requires comprehensive evaluation. The process includes multiple steps and assessments. Key components of LD diagnosis:
ADHD Diagnosis
ADHD diagnosis follows a different path. It focuses on behavioral patterns across various settings. Essential elements of ADHD diagnosis:
Challenges in Diagnosis
Several factors can complicate accurate diagnosis:
Treatment and Support for LD and ADHD
Managing Learning Disabilities
Effective support for learning disabilities requires targeted intervention strategies. Key support elements include:
Strategies for ADHD
ADHD management requires a comprehensive approach. Different strategies address various aspects of the condition. Essential management components:
Support for Co-occurring Conditions
Many individuals experience both conditions simultaneously. It requires carefully coordinated support approaches. Combined support strategies include:
Why Understanding the Difference Matters
Tailoring Interventions
Understanding differences between conditions leads to more effective support. Proper identification helps create targeted assistance plans. Important considerations include:
Supporting Children and Families
Proper understanding enables better support from all involved parties. This knowledge helps create effective support networks. Support network components include:
Conclusion: Empowering Success Through Understanding
Understanding the differences between learning disabilities and ADHD enables better support. This knowledge helps create effective intervention strategies. Success becomes possible through appropriate understanding and targeted assistance.
Key takeaway points include:
Individuals with these conditions can achieve significant success with proper support and understanding. Recognition of differences leads to more effective assistance strategies. This understanding helps create pathways to achievement and growth.
Personalized Support at Amae Health
At Amae Health, we recognize the unique challenges individuals face when living with learning disabilities or ADHD. Our team of compassionate clinicians and mental health experts is dedicated to providing personalized, evidence-based care that addresses each person’s specific needs. Whether you're seeking clarity through diagnosis or ongoing support for cognitive, emotional, or behavioral health, Amae Health offers a safe and supportive environment where healing and growth can begin.
min read
Understanding High-Functioning ADHD in Women
By
Sonia Garcia
|
May 15, 2025
High-functioning ADHD in women is often misunderstood, masked by coping strategies and societal expectations to appear organized and composed. A study published in BMC Psychiatry suggests that the gap in ADHD diagnoses between males and females is largely due to under-recognition and referral bias, with females often showing subtler symptoms. Unlike traditional ADHD, which is more visibly disruptive, high-functioning ADHD manifests through procrastination, emotional overwhelm, and struggles with focus.
Women with ADHD may excel professionally and academically, but this success often comes at a cost — hidden exhaustion, burnout, and self-doubt. Their tendency to internalize symptoms makes diagnosis challenging, leading to misdiagnoses such as anxiety or depression. Understanding these unique manifestations is essential for providing the right support and treatment. This article explores the symptoms, challenges, and strategies for managing high-functioning ADHD in women.
What is High-Functioning ADHD?
High-functioning ADHD refers to a presentation of attention-deficit/hyperactivity disorder where symptoms are less obvious or more effectively masked, particularly by women who often develop advanced coping mechanisms. Unlike traditional ADHD, which is characterized by visibly disruptive behaviors and severe impairments, high-functioning ADHD allows individuals to maintain a semblance of control in daily life. However, this comes at a cost.
Women with ADHD may excel academically or professionally, but they often do so by exerting tremendous effort to counteract symptoms such as inattentiveness, impulsivity, and time management issues. This hidden struggle can lead to chronic stress, burnout, and mental health challenges like anxiety and depression. Because the symptoms appear less severe, high-functioning ADHD frequently goes undiagnosed, especially in women who are culturally conditioned to internalize difficulties and maintain outward composure.
Recognizing the unique manifestations of high-functioning ADHD is crucial for proper diagnosis and support, helping those affected to access appropriate treatment and improve their quality of life.
Symptoms of High-Functioning ADHD in Women
Inattentiveness & Distractibility
Women with high-functioning ADHD often struggle with maintaining focus, especially during tasks that require prolonged attention or lack immediate rewards. These ADHD symptoms frequently manifest as minds wandering, making it difficult to complete assignments or follow through on conversations. This inattentiveness can manifest as forgetting details, losing track of tasks, or zoning out during meetings. Despite being capable and intelligent, these women often expend significant mental energy to appear attentive and organized.
Common signs of inattentiveness include:
Impulsivity Manifestations
Impulsivity in women with ADHD might not always present as overtly risky behavior. Instead, it can appear as interrupting conversations, making snap decisions without fully thinking them through, or struggling to resist distractions like online shopping or social media. In professional settings, this impulsivity can lead to speaking out of turn or overcommitting to tasks, while in personal relationships, it might cause difficulty with boundaries or emotional outbursts.
Examples of impulsivity manifestations:
Emotional Regulation Difficulties
High-functioning ADHD in women is often accompanied by challenges in managing emotions. Heightened sensitivity and a tendency to feel emotions intensely can lead to sudden mood swings, irritability, or feelings of being overwhelmed. This heightened emotional response is sometimes mistaken for mood disorders, making it harder to diagnose ADHD accurately. Emotional dysregulation can impact relationships and contribute to chronic stress.
Key challenges with emotional regulation:
Organizational & Time Management Challenges
Women with high-functioning ADHD frequently battle with staying organized and managing time effectively. Procrastination, difficulty prioritizing tasks, and a tendency to underestimate how long activities will take are common challenges. This can result in missed deadlines, cluttered workspaces, and a constant sense of falling behind, despite considerable effort to stay on top of responsibilities.
Other challenges with organization and time management:
Differences in ADHD Symptoms Between Genders
ADHD symptoms can manifest quite differently in males and females, leading to widespread misconceptions and often causing ADHD in women to go unnoticed. While males typically exhibit more visible signs of hyperactivity and impulsivity, females are more likely to internalize their symptoms, presenting as inattentiveness, anxiety, or perfectionism. Recognizing these differences is crucial for accurate diagnosis and effective treatment.
The table below highlights some of the key differences in how ADHD symptoms present in males versus females:
| Symptom | Females | Males |
---
| Hyperactivity | Internal restlessness; often appears distracted or daydreamy, masking the need to move or fidget. | Overt physical hyperactivity, fidgeting, and running around. |
| Impulsivity | Emotional impulsivity, like sudden outbursts, oversharing, or impulsive spending as a form of coping. | Impulsive actions, such as interrupting conversations or engaging in risky behaviors without considering consequences. |
| Inattentiveness | Difficulty focusing on tasks, often masked by over-preparation or perfectionism to compensate. | Visible forgetfulness, difficulty following instructions, and frequent mistakes in school or work. |
| Emotional Regulation | Tendency to internalize emotions, leading to chronic anxiety, self-blame, and perfectionist tendencies. | Outward expressions of irritability, frustration, or impatience, often perceived as aggressive. |
Why Do Women with High-Functioning ADHD Go Unnoticed?
A systematic review published in BMC Psychiatry highlighted that many women with ADHD remain undiagnosed well into adulthood due to a lack of awareness and diagnostic biases. Plus, social conditioning teaches many women to appear organized, attentive, and emotionally stable, even when they are struggling internally. As a result, they may excel academically or professionally at the cost of significant mental and emotional exhaustion.
Additionally, the diagnostic criteria for ADHD were historically based on male presentations of the disorder, which tend to be more externally disruptive. Women, however, are more likely to experience internalized symptoms such as inattentiveness, low self-esteem, and anxiety. This mismatch between symptoms and diagnostic criteria leads to many women being misdiagnosed with anxiety or depression instead of ADHD. Recognizing these patterns is essential for improving diagnosis rates and access to treatment for women with high-functioning ADHD.
The Impact of High-Functioning ADHD on Daily Life
Professional Life
Women with ADHD often face unique challenges in the workplace. While they may excel in their roles, maintaining focus and managing time efficiently can be a constant struggle. Procrastination, difficulty prioritizing tasks, and impulsive decision-making can lead to missed deadlines and burnout.
Additionally, the pressure to appear competent and composed can result in overworking and perfectionism, which exacerbates stress. Seeking accommodations like flexible deadlines, noise-canceling headphones, or utilizing project management tools can significantly improve productivity and reduce overwhelm.
Personal Relationships
High-functioning ADHD can also impact personal relationships, making it difficult for women to balance social obligations and emotional needs. Forgetfulness, distraction, and challenges with emotional regulation can cause misunderstandings with partners, friends, and family members. Women with ADHD may also struggle with maintaining consistent communication or remembering important dates, which can lead to feelings of guilt or inadequacy.
Despite these challenges, many women with ADHD use their empathy and creativity to form deep and meaningful connections. Being open about their struggles and setting clear expectations with loved ones can help in building stronger, more supportive relationships.
Physical Health & Lifestyle Habits
High-functioning ADHD in women can also influence physical health and lifestyle habits, often making it challenging to maintain a balanced routine. Women with ADHD may struggle with regular exercise due to difficulties with planning and motivation, leading to a more sedentary lifestyle.
Additionally, impulsivity can contribute to irregular eating patterns, such as binge eating or skipping meals. Sleep disturbances are also common, with many women experiencing difficulty falling asleep or maintaining a consistent sleep schedule.
Mental Health Implications
The continuous effort to mask symptoms and maintain a facade of normalcy can take a toll on mental health. ADHD in women makes them more prone to anxiety, depression, and low self-esteem due to feelings of inadequacy and chronic stress. The fear of being perceived as lazy or incompetent can further fuel perfectionism and self-criticism, creating a vicious cycle.
Without proper diagnosis and support, these mental health struggles can intensify over time, leading to burnout and a diminished quality of life. Addressing these underlying mental health challenges with therapy, support groups, and self-compassion practices is essential for improving overall well-being and quality of life.
Coping Strategies & Management for Women with ADHD
Managing high-functioning ADHD requires a combination of self-awareness, structure, and support. While the challenges can be significant, adopting effective coping strategies can help adult women manage symptoms and improve their quality of life. Here are some practical tips:
Experimenting with different approaches can help women find what works best for them, ensuring that ADHD symptoms are managed effectively.
You’re Not Alone: Get Help for Your ADHD
Living with high-functioning ADHD can be exhausting, especially when symptoms go unnoticed or are misunderstood. Many women struggle in silence, masking their challenges and battling chronic stress and burnout. Understanding that you’re not alone in this experience is the first step towards finding effective support.
At Amae Health, we provide comprehensive psychiatric care for individuals facing complex mental health challenges which may include ADHD. If ADHD symptoms are impacting your daily life, our expert team can help assess your needs and develop a personalized approach to improve focus, emotional regulation, and overall well-being.
If you’re ready to take the first step towards better managing your ADHD symptoms, reach out to Amae Health for a consultation today. You don’t have to navigate this journey alone.
min read
Psychosis vs. Schizophrenia: What’s the Difference?
By
Sonia Garcia
|
June 16, 2025
Psychosis is a group of symptoms where a person loses touch with reality, experiencing hallucinations or delusions. Schizophrenia is a chronic mental health disorder that affects thinking, emotions, and behavior. Psychosis is one of the main symptoms of schizophrenia, but it can also occur on its own in other conditions. Understanding the difference between psychosis vs schizophrenia is key to recognizing symptoms early and finding the right support.
While psychosis can be short-term and triggered by factors like stress or substance use, schizophrenia involves persistent, long-term challenges. In this guide, we’ll explore how these conditions differ, their causes, symptoms, and how professional care can support recovery.
What Is Psychosis?
Psychosis is a mental state where a person loses touch with reality. According to the National Institute of Mental Health (NIMH), it involves a disruption of thoughts and perceptions, making it difficult to recognize what is real and what is not. These episodes can cause significant confusion and may impact how a person communicates or makes decisions.
Importantly, psychosis is not a standalone diagnosis but a symptom found in several mental health conditions. Among these, psychosis and schizophrenia are closely linked, but psychosis can also arise from bipolar disorder, severe depression, medical conditions, or substance use. Treatment for psychosis focuses on identifying the underlying cause and providing compassionate, tailored support to help manage symptoms effectively. Early intervention is crucial, as it can significantly improve outcomes and reduce the impact of future episodes.
What Is Schizophrenia?
Schizophrenia is a chronic mental health condition that affects how a person thinks, feels, and behaves. Because psychosis and schizophrenia are closely linked, understanding their differences helps tailor effective treatment strategies. While psychosis is a key feature of schizophrenia, it represents only one part of a broader set of challenges. This condition involves disruptions in thought processes, emotional regulation, and behavior, often impacting daily life and relationships.
Unlike isolated episodes of psychosis, schizophrenia is a long-term condition that requires ongoing treatment and support. In professional clinics, such as Amae Health, specialists take a comprehensive approach that addresses not only acute symptoms but also the cognitive and emotional aspects of the disorder. With the right combination of medication, therapy, and community support, many people with schizophrenia can manage their condition and lead fulfilling lives. Early diagnosis and personalized care are key to improving quality of life.
Psychosis vs. Schizophrenia: Key Symptoms
Common Indicators of Psychosis
Psychosis is marked by a range of experiences that distort perception and thinking. Typical signs include:
These symptoms are shared with schizophrenia, but unlike schizophrenia, psychosis can also occur briefly in response to specific triggers such as trauma or substance use. Comparing psychosis vs schizophrenia highlights these differences in symptom duration and underlying causes.
Signs of Schizophrenia
Schizophrenia includes the signs of psychosis but extends beyond them. According to the National Institute of Mental Health, symptoms of schizophrenia include:
While psychosis is one component, schizophrenia encompasses ongoing disruptions in thought, emotion, and behavior, making comprehensive, long-term care essential for effective management.
Psychosis vs. Schizophrenia: Causes
Understanding the causes of psychosis and schizophrenia helps clarify the differences between these conditions and supports accurate diagnosis. Psychosis often arises from immediate external factors or short-term internal stressors, while schizophrenia typically develops from a combination of long-term biological and environmental influences. Recognizing these distinctions helps clinicians tailor treatment approaches and support recovery.
What Triggers Psychosis
Psychosis can be triggered by intense short-term stressors or health-related issues. Common triggers include:
When looking at psychosis vs schizophrenia, the causes of each condition highlight their fundamental differences.
Why Schizophrenia Develops
Schizophrenia tends to emerge from deeper, long-term factors. According to the World Health Organization (WHO), the condition likely results from a combination of genetic factors and environmental influences, such as early-life stress or heavy cannabis use. Contributing influences include:
Understanding these factors enables professionals to create personalized treatment plans aimed at managing the condition over time.
Diagnosis and Evaluation
Accurate diagnosis is essential when comparing psychosis vs schizophrenia, as the conditions share similarities but have distinct causes and treatments. While both share overlapping symptoms, their underlying causes and long-term outlooks differ. Here’s how clinicians typically differentiate the two:
| Criteria | Psychosis | Schizophrenia |
---
| Nature | A symptom, not a diagnosis | A chronic mental health condition |
| Duration | Often short-term, episode-based | Long-term, persistent |
| Triggers | Trauma, substances, sleep deprivation, medical issues | Genetic, neurobiological, and environmental factors |
| Assessment | Focus on identifying the immediate cause | Comprehensive evaluation of persistent patterns and cognitive/emotional symptoms |
| Treatment focus | Addressing the trigger and stabilizing symptoms | Long-term management with medication, therapy, and support systems |
In professional clinics, such as Amae Health, clinicians combine thorough evaluations with compassionate care to ensure accurate diagnosis and effective treatment planning. Early identification of either condition significantly improves the chances of recovery and long-term stability.
Treatment Options for Psychosis and Schizophrenia
Treatment Approaches for Psychotic Episodes
Short-term psychotic episodes often require a focused, immediate treatment plan. Common approaches include:
In cases linked to substance use or sleep deprivation, addressing the root cause is essential. Timely intervention helps manage symptoms effectively and prevents recurrence.
Long-Term Management of Schizophrenia
Managing both psychosis and schizophrenia requires a sustained, multifaceted approach. But the most effective strategies for schizophrenia include:
NIMH recommends coordinated specialty care as the standard for early schizophrenia treatment, focusing on shared decision-making and a recovery-oriented approach. Integrated care plans, like those offered by clinics such as Amae Health, provide the continuity needed to support recovery and improve quality of life over time.
When to Seek Help
Recognizing the right moment to seek professional help is vital. Red-flag symptoms include:
If you or someone you know is experiencing these signs, early consultation with a mental health professional is crucial. Prompt intervention can prevent psychosis and schizophrenia symptoms from worsening and open the door to effective treatment options. In professional settings compassionate specialists provide thorough assessments and personalized care plans, helping individuals regain stability and improve their quality of life. Remember, seeking help early makes a significant difference.
Recognizing the Difference Is the First Step Toward Healing
Understanding the distinction between psychosis and schizophrenia is essential for timely support. While psychosis can be a temporary reaction to stress or substance use, schizophrenia involves ongoing challenges that require long-term care. Recognizing early warning signs and consulting a professional helps prevent complications.
In expert clinics specialists offer thorough evaluations and individualized treatment plans. Whether addressing short-term psychotic episodes or managing schizophrenia, early and tailored care greatly improves outcomes.
If you notice concerning symptoms, don’t wait — reach out Amae Health for professional support today.
All clinical services are provided by licensed physicians and clinicians affiliated with independently owned and operated professional practices. For patients in New York, this is known as ‘Scott C. Fears M.D. P.C.’ For patients in California, this is known as ‘Amae Health Medical Associates, P.C.’ For patients in North Carolina, this is known as ‘Amae Medical Associates North Carolina P.C.’ Amae Health Services, LLC provides administrative and technology services to the Amae Health Medical Associates practices it supports and does not provide any professional medical services itself. |
14840 | https://www.quora.com/Can-you-make-a-3X3-magic-square-of-integers-whose-row-column-and-diagonal-sums-are-not-a-multiple-of-3-and-all-equal | Something went wrong. Wait a moment and try again.
Magic Squares
Logical Math Puzzles
Algorithms, Combinatorics
Positive Integers
Number Theory
Combinatorial Math
Scientific Puzzles
5
Can you make a 3X3 magic square of integers whose row, column, and diagonal sums are not a multiple of 3 and all equal?
David Shaffer
Bachelors degree in Maths and Physics from a long time ago
·
Author has 2K answers and 2.4M answer views
·
3y
Suppose the middle square is m, the row total is t and the whole grid adds to 3t. Then adding together the middle column, middle row and two diagonals gives the whole grid plus 3 extra copies of the middle square:
4t=3t+3m
t=3m
and so the row total must be a multiple of 3.
Related questions
How do I create a 3x3 magic square with integers?
How can I fill a matrix whose column sum, row sum, and diagonals sum are the same?
In a magic square, each row column and diagonal have the same some check. Which of the following is the magic square 1, - 10, 0, -4, -3, -2, -6, 4, -7?
How do I solve a 3×3 magic square with diagonals 7,7,-22?
In a magic square, the sum of the number in each row in each column and along the diagonal is the same. Is this a magic squared?
Kiran Kumar
I write Code
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Author has 116 answers and 5M answer views
·
Updated 7y
Related
How can I fill a matrix whose column sum, row sum, and diagonals sum are the same?
How about a python code which can auto-fill required matrix for you !
Is this what you want ?
(made with Pencil tool)
sum of each individual row is 4+3+8=15 ; 9+5+1=15; 2+7+6=15
sum of each individual column is 4+9+2=15; 3+5+7=15; 8+1+6=15
sum of each individual diagonal is 4+5+6=15; 8+5+2=15;
These type of matrices popularly known as magic squares.
I’ll tell you a simple way to fill the matrix. Assume every box has directions like
Every time you have to place the new number to the south-east of previous number.
Always place the first number at middle row last column
Now 2 has to be placed to the south-
How about a python code which can auto-fill required matrix for you !
Is this what you want ?
(made with Pencil tool)
sum of each individual row is 4+3+8=15 ; 9+5+1=15; 2+7+6=15
sum of each individual column is 4+9+2=15; 3+5+7=15; 8+1+6=15
sum of each individual diagonal is 4+5+6=15; 8+5+2=15;
These type of matrices popularly known as magic squares.
I’ll tell you a simple way to fill the matrix. Assume every box has directions like
Every time you have to place the new number to the south-east of previous number.
Always place the first number at middle row last column
Now 2 has to be placed to the south-east of 1
But that box is out of the matrix so you have to place the number opposite to that box
Now 3 has to be placed south-east to the 2 but it will be again out of the matrix so place number opposite to that box
Now 4 has to be put to the south-east to 3 but it was already occupied by 1 so place 4 to the left side of 3
Now 5,6 can placed to south-east because you won’t get out of the box or already fill box
Now 7 has to be placed south-east but it will out of matrix and opposite to the out of box will also be out of the box
In this case you have to fill the box which is left to the previous number
like this fill 8 & 9.
Coding will be very easy once you get the concept.
table[M][n]=1;for i in range(2,k+1): try: M=M+1 n=n+1; if(table[M][n]==0): table[M][n]=i else: M=M-1 n=n-2 table[M][n]=i; except: try: if(table[M]==0): n=0; table[M][n]=i else: table[M][n-1]=i except: try: M=0; table[M][n]=i; except: M=M-1 n=n-1; table[M][n-1]=i print(table)
I’ve used try -exceptions because every time I’m trying to access the box which is out of the matrix will ultimately lead to Index out of bound error.
I hope I’ve helped you.
Here you go Kiran-G1/myscripts the code.
Gregorio Morales
Math Teacher
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Author has 357 answers and 374.9K answer views
·
3y
Nope, nobody can. The magic constant of a magic square is always three times the middle cell.
The proof is easy, and I will not deprive you of the pleasure of proving it yourself.
Your response is private
Was this worth your time?
This helps us sort answers on the page.
Absolutely not
Definitely yes
Alberto Cid
M.S.E. in Telecommunications Engineering & Data Transmission, Technical University of Madrid (Graduated 2008)
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Author has 2K answers and 3.8M answer views
·
5y
Related
How do I create a 3x3 magic square with integers?
I think it’s not very difficult.
It’s just logic and trial and error…
First, to make things simpler I will assume you use numbers from 1 to 9, since the 3x3 square has 33 = 9 cells.
In a magic square the sums in every column and row must be the same.
That “constant” sum is usually called the “magic sum”.
It’s usually also the sum in the diagonals.
What’s the value of that magic sum?
Well, since the numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9 the sum of all the 9 numbers is:
45, and, so, the sum in every column or row must be 15.
If you include the requirement for the diagonals then the number in
I think it’s not very difficult.
It’s just logic and trial and error…
First, to make things simpler I will assume you use numbers from 1 to 9, since the 3x3 square has 33 = 9 cells.
In a magic square the sums in every column and row must be the same.
That “constant” sum is usually called the “magic sum”.
It’s usually also the sum in the diagonals.
What’s the value of that magic sum?
Well, since the numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9 the sum of all the 9 numbers is:
45, and, so, the sum in every column or row must be 15.
If you include the requirement for the diagonals then the number in the middle must be the 5…
Let’s use letters for the 9 cells:
a b c
d e f
g h i
The sums that include the cell in the middle are:
d+e+f = 15
b+e+h = 15
a+e+i = 15
c+e+g = 15
Notice all the 9 different letters are in those equations.
Let’s sum it all:
(a+b+c+d+e+f+g+h+i ) + 3e = 154 = 60
And the sum from a to i is 45.
So, 3e = 15 … so e = 5.
Now, just try.
What if a = 1? → i = 9
And what if we also force the b=2 ?
Then c = 15 - 2- 1 = 12 … Not possible.
Then b must be 5 or higher, but 5 is in the middle, so b must be 6 or 7 or 8.
Try 6:
1 6 8
5
9
The last column exceeds 15…
Try b = 8
c = 6 and also the last column exceeds 15
So, no corner can be 1.
Then, the only possible place for the 1 is in b, or d or f or h.
If we try b=1 a = 6 → c = 8 and i = 4
6 1 8
5
4
Then:
6 1 8
7 5 3
2 9 4
You can check it.
Other solutions are rotations or images in a mirror of rotations, then a total of 8 solutions (all symmetric, either specular symmetry or rotational symmetry).
Notice the corners have the four even numbers… and 5, the middle (or average) number, is in the middle.
Promoted by Cash Canvas
Ethan Anderson
Senior Writer at CashCanvas (2024–present)
·
Updated Apr 21
What are the biggest missed opportunities for building wealth that most people don’t know about?
Overpaying on Auto Insurance
Believe it or not, the average American family still overspends by $461/year¹ on car insurance.
Sometimes it’s even worse: I switched carriers last year and saved literally $1,300/year.
Here’s how to quickly see how much you’re being overcharged (takes maybe a couple of minutes):
Pull up Coverage.com – it’s a free site that will compare offers for you
Answer the questions on the page
It’ll spit out a bunch of insurance offers for you.
That’s literally it. You’ll likely save yourself a bunch of money.
Overlook how much you can save when shopping online
Many people over
Overpaying on Auto Insurance
Believe it or not, the average American family still overspends by $461/year¹ on car insurance.
Sometimes it’s even worse: I switched carriers last year and saved literally $1,300/year.
Here’s how to quickly see how much you’re being overcharged (takes maybe a couple of minutes):
Pull up Coverage.com – it’s a free site that will compare offers for you
Answer the questions on the page
It’ll spit out a bunch of insurance offers for you.
That’s literally it. You’ll likely save yourself a bunch of money.
Overlook how much you can save when shopping online
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Not Investing in Real Estate (Starting at Just $20)
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Wasting Time on Unproductive Habits
As a rule of thumb, I’d ignore most sites that claim to pay for surveys, but a few legitimate ones actually offer decent payouts.
I usually use Survey Junkie. You basically just get paid to give your opinions on different products/services, etc. Perfect for multitasking while watching TV!
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Paying off credit card debt on your own
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Overspending on Mortgages
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Ignoring Home Equity
Your home can be one of your most valuable financial assets, yet many homeowners miss out on opportunities to leverage its equity. Bankrate’s Best Home Equity Options helps you find the right loan for renovations, debt consolidation, or unexpected expenses.
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Missing Out on Smart Investing
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Related questions
Can two different magic squares have the same sum for each row, column, and diagonal? If so, can you provide an example?
What is a 33 magic square whose sum is 72?
Using the negative integers from -11 to -19, can you complete the magic square 3 by 3 such that all the rows, columns, and diagonals are -45?
Can you make a three by three magic square in which the product of each row column and diagonal is 1000?
How will we make , ( 4 × 4 ) magic square in which the sum of each row or column is 32?
Vijaykumar N. Joshi
Former Prof and Head , Dept. Of Computer Science , RETD (1980–2002)
·
Author has 1.9K answers and 1.7M answer views
·
5y
Related
How do I create a 3x3 magic square with integers?
CMA K S Narayanan
Cost Accountant having 24 years Telecom Experience
·
Author has 7K answers and 13.4M answer views
·
5y
Related
How do I create a 3x3 magic square with negative and positive numbers?
Draw a 3 x 3 Grid.
Extend the grid as shown and write integers from 1 to 9 in pattern as shown.
Move numbers lying outside the main 3 x 3 grid as shown to the vacant cell opposite to it.
You now have a magic square that totals 15 across all rows, columns and diagonals.
Notes :
You can extend this method to any ODD x ODD square matrix.
You can use consecutive terms of any arithmetic progression to fill in the grids to get new magic squares.
:-)
Draw a 3 x 3 Grid.
Extend the grid as shown and write integers from 1 to 9 in pattern as shown.
Move numbers lying outside the main 3 x 3 grid as shown to the vacant cell opposite to it.
You now have a magic square that totals 15 across all rows, columns and diagonals.
Notes :
You can extend this method to any ODD x ODD square matrix.
You can use consecutive terms of any arithmetic progression to fill in the grids to get new magic squares.
:-)
Promoted by The Penny Hoarder
Lisa Dawson
Finance Writer at The Penny Hoarder
·
Updated Jul 31
What's some brutally honest advice that everyone should know?
Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included.
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Cancel Your Car Insurance
You might not even realize it, but your car insurance company is probably overcharging you. In fact, they’re kind of counting on you not noticing. Luckily, this problem is easy to fix.
Don’t waste your time
Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included.
And honestly? Putting them to use was way easier than I expected. I bet you can knock out at least three or four of these right now—yes, even from your phone. Don’t wait like I did.
Cancel Your Car Insurance
You might not even realize it, but your car insurance company is probably overcharging you. In fact, they’re kind of counting on you not noticing. Luckily, this problem is easy to fix.
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Get Paid $225/Month While Watching Movie Previews
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Earn $1000/Month by Reviewing Games and Products You Love
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Mathivanan Palraj
Author of GMAT:Problem Solving Techniques for Top Score
·
Author has 1.7K answers and 3.8M answer views
·
5y
Related
How do I create a 3x3 magic square with integers?
I assume that you use integers from 1 to 9 and no integer is repeated.
First find the sum from 1 to 9. 910/2 = 45
Since each row, column, and diagonal has three cells/integers, divide 45 by 3. You get 15. This is the sum of each row, column, and diagonal.
The next step is find triplets that give 15 as sum.
I start with 9. You can start with any number.
9+5+1; 9+4+2
8+4+3; 8+6+1; 8+5+2;
7+6+2; 7+5+3;
6+5+4
We have got all 8 triplets(3 rows, 3 columns, and 2 diagonals)
There is a definite math logic in placing the integers in the cells.
The most repeated integer: 5 (four times). 5 is the candidate at the
I assume that you use integers from 1 to 9 and no integer is repeated.
First find the sum from 1 to 9. 910/2 = 45
Since each row, column, and diagonal has three cells/integers, divide 45 by 3. You get 15. This is the sum of each row, column, and diagonal.
The next step is find triplets that give 15 as sum.
I start with 9. You can start with any number.
9+5+1; 9+4+2
8+4+3; 8+6+1; 8+5+2;
7+6+2; 7+5+3;
6+5+4
We have got all 8 triplets(3 rows, 3 columns, and 2 diagonals)
There is a definite math logic in placing the integers in the cells.
The most repeated integer: 5 (four times). 5 is the candidate at the center.
Three times repeated: 8, 6, 4, 2. (These are the candidates for the four corners) All corners have 1 row, 1 column, and 1 diagonal.
The other integers are two times repeated that will be placed at the middle of row, column.
8 x 6
x 5 x
4 x 2
I think the remaining integers can easily be placed by anyone.
Hope you got the math logic.
Neil Morrison
I do all my own differentiation!
·
Author has 9.4K answers and 27.4M answer views
·
3y
Related
A magic square of order 4 is created by putting the integers 1 to 16 into a 4 by 4 square grid so that the sum of the numbers in each row, column and main diagonal is the same. What is the sum of the integers in any row or column?
Others have already covered how to calculate the ‘Magic Sum’, so let me show you a neat way to place the numbers so as to achieve that sum.
Draw diagonals on the 4×4 square. (In your mind at least)
Start at the top left square and work right and down while counting from 1 to 16 - but only write-in the number if you are on a diagonal square. So Write 1, miss 2 and 3, write 4, miss 5, write 6 and 7, etc. Once you reach 16, you should have Diagram A
Start at the bottom right and work left and up while counting from 1 to 16 - but only write-in the number if you are
not
on a diagonal square. (i.e. if
Others have already covered how to calculate the ‘Magic Sum’, so let me show you a neat way to place the numbers so as to achieve that sum.
Draw diagonals on the 4×4 square. (In your mind at least)
Start at the top left square and work right and down while counting from 1 to 16 - but only write-in the number if you are on a diagonal square. So Write 1, miss 2 and 3, write 4, miss 5, write 6 and 7, etc. Once you reach 16, you should have Diagram A
Start at the bottom right and work left and up while counting from 1 to 16 - but only write-in the number if you are not on a diagonal square. (i.e. if the square isn’t already filled). So miss 1, write 2 and 3, miss 4, write 5, etc. Once you reach 16, you should have Diagram B
See how many 34s you can find? It isn’t just the rows, columns, and diagonals. e.g.
1+15+2+16=34
12+6+11+5=34
8+10+7+9=34
…and a whole lot of similarly symmetrical patterns. (Don’t forget to check the 2×2 sub squares)
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Mario Skrtic
IT at Public Libraries (2003–present)
·
Upvoted by
Bernard Montaron
, PhD Mathematics & Discrete Mathematics, Université Pierre Et Marie Curie Paris VI (1980) · Author has 4.4K answers and 1.1M answer views
·
1y
Related
Can you find all 3 x 3 matrices similar to
3
6
7
9
5
3
7
3
9
where the horizontal numbers (left to right), the vertical numbers (top to down), and the diagonal numbers are all prime? Can you find all such matrices with constant sum of lines, columns, and diagonals?
First question example: 3 digit primes 367,953,739 and vertical 397,653,739 and diagonal 359,737
I wasn’t sure about second question:
Sum of lines are sum of digits in each line 3+6+7=9+5+3=7+3+9=3+9+7=6+5+3=7+3+9=3+5+9=7+3+7 (my answer) or sum of primes 367+953+739=397+653+739=359+737
Brute force in PariGP:
{
\ a,b,c are horizontal lines
s1=0;
s2=0;
k=vector(8);
forprime(a=100,999,
da=digits(a);
k=sumdigits(a);
forprime(b=100,999,
db=digits(b);
k=sumdigits(b);
forprime(c=100,999,
dc=digits(c);
k=sumdigits(c);
d=da100+db10+dc;
k=sumdigits(d);
if (isprime(d),
e=da100+db10+dc;
k=
First question example: 3 digit primes 367,953,739 and vertical 397,653,739 and diagonal 359,737
I wasn’t sure about second question:
Sum of lines are sum of digits in each line 3+6+7=9+5+3=7+3+9=3+9+7=6+5+3=7+3+9=3+5+9=7+3+7 (my answer) or sum of primes 367+953+739=397+653+739=359+737
Brute force in PariGP:
{
\ a,b,c are horizontal lines
s1=0;
s2=0;
k=vector(8);
forprime(a=100,999,
da=digits(a);
k=sumdigits(a);
forprime(b=100,999,
db=digits(b);
k=sumdigits(b);
forprime(c=100,999,
dc=digits(c);
k=sumdigits(c);
d=da100+db10+dc;
k=sumdigits(d);
if (isprime(d),
e=da100+db10+dc;
k=sumdigits(e);
if (isprime(e),
f=da100+db10+dc;
k=sumdigits(f);
if (isprime(f),
g=da100+db10+dc;
k=sumdigits(g);
if (isprime(g),
h=da100+db10+dc;
k=sumdigits(h);
if (isprime(h),
s1+=1;
\ print(a);
\ print(b);
\ print(c);
\ print("");
t=1;
for (i=2,8,
if(k[i]!=k,t=0);
);
if (t==1,
s2+=1;
print(a);
print(b);
print(c);
print("");
);
);
);
);
);
);
);
);
);
print("First question: There are ",s1," such matrices");
print("Second question: There are ",s2," such matrices");
}
First question: There are 8007 such matrices
Second question: There are 0 such matrices
Mark Chandler
Related
Can you find a magic square with squared numbers?
Yes - there is a 7x7 magic square made up of the squares of the numbers from 0 to 48. Each row, column and diagonal adds up to 5432. To my mind this is the most magical of all magic squares, and I didn’t believe it existed until I checked it myself! I didn’t discover this myself and am afraid that I don’t know who deserves the credit - if I find out I will mention it (I saw an incorrect version in print, but managed to work out what it should have showed):
Yes - there is a 7x7 magic square made up of the squares of the numbers from 0 to 48. Each row, column and diagonal adds up to 5432. To my mind this is the most magical of all magic squares, and I didn’t believe it existed until I checked it myself! I didn’t discover this myself and am afraid that I don’t know who deserves the credit - if I find out I will mention it (I saw an incorrect version in print, but managed to work out what it should have showed):
Frank Abbing
Former Pensioner at Philips (Electronics Company) (1965–1990)
·
Author has 1K answers and 223.2K answer views
·
Updated 2y
Related
How many different ways can you arrange the numbers 1-9 in a 3x3 grid such that each row, column, and diagonal sum to the same number?
My computer finds eight solutions:
The first one is OK. The other seven are just transformations of this first one
```
include #define gotoxy(lin, col) printf("\033[%d;%dH", lin, col) #define SUM 15 int list; // all horizontal tripletsint arr; void box(int l, int c){ l = 4; c = 8; for(int i = 0; i<3; i++){ gotoxy(i+l+1,c+1); for(int j = 0; j<3; j++) printf(" %d",arr[i][j]); }} int test_arr(){ if((arr+arr+arr)!=SUM) return 0; // vertical if((arr+arr+arr)!=SUM) return 0; if((arr+arr+arr)!=SUM) return 0; // diagonal if((ar
```
My computer finds eight solutions:
The first one is OK. The other seven are just transformations of this first one
```
include #define gotoxy(lin, col) printf("\033[%d;%dH", lin, col) #define SUM 15 int list; // all horizontal tripletsint arr; void box(int l, int c){ l = 4; c = 8; for(int i = 0; i<3; i++){ gotoxy(i+l+1,c+1); for(int j = 0; j<3; j++) printf(" %d",arr[i][j]); }} int test_arr(){ if((arr+arr+arr)!=SUM) return 0; // vertical if((arr+arr+arr)!=SUM) return 0; if((arr+arr+arr)!=SUM) return 0; // diagonal if((arr+arr+arr)!=SUM) return 0; int n, i, j; for(j = 0; j<3; j++){ // duplicated digits n = arr[j]; for(i = 0; i<3; i++) if(arr[i]==n || arr[i]==n) return 0; } return 1; // arr is valid} int main(){ int a, b, c, l0, l1, l2, bl = 0, bc = 0, x = 0; for(a = 1; a<=9; a++) for(b = 1; b<=9; b++) for(c = 1; c<=9; c++) if((a+b+c)==SUM && a!=b && a!=c && b!=c){ list[x] = a; list[x] = b; list[x] = c; x++; } for(l0 = 0; l0<48; l0++) for(l1 = 0; l1<48; l1++) for(l2 = 0; l2<48; l2++) if(l0!=l1 && l0!=l2 && l1!=l2){ for(x = 0; x<3; x++){ arr[x] = list[l0][x]; arr[x] = list[l1][x]; arr[x] = list[l2][x]; } if(test_arr()){ box(bl, bc); bc++; if(bc>3){bc = 0; bl++;} } }}
```
There was an earlier program, this one is much faster!
John K WilliamsSon
Accredited (MS Educ) nerd who loves talking about math
·
Author has 9K answers and 23.2M answer views
·
3y
Related
Can you make a three by three magic square in which the product of each row column and diagonal is 1000?
One we remember that 1000 is equal to 10³, it becomes relatively easy to make a magic square with nine different numbers where the product of each row, each column, and each diagonal is equal to 2³×3³.
There may be other ways to do it, like brute force computer programs, but if we think about it logically, we can think through the process on paper.
To do this, we need to create two different exponent magic squares, one for 2ⁿ and one for 5ⁿ.
Each magic square must contain all the numbers 0, 1 and 2, three times each,
111 is the / diagonal of one square. 111 is the \ diagonal of the other square. (
One we remember that 1000 is equal to 10³, it becomes relatively easy to make a magic square with nine different numbers where the product of each row, each column, and each diagonal is equal to 2³×3³.
There may be other ways to do it, like brute force computer programs, but if we think about it logically, we can think through the process on paper.
To do this, we need to create two different exponent magic squares, one for 2ⁿ and one for 5ⁿ.
Each magic square must contain all the numbers 0, 1 and 2, three times each,
111 is the / diagonal of one square. 111 is the \ diagonal of the other square. (We can’t have the same diagonal in both squares equal to 111.)
The non-111 diagonal of each square will either be 012 or 210.
We will calculate all the other numbers in each possible exponent magic square by making the top, bottom, left and right edges add up to 3.
Here are the only four possible exponent magic squares
0 2 1 1 2 0 1 0 2 2 0 1 2 1 0 0 1 2 2 1 0 0 1 2 1 0 2 2 0 1 0 2 1 1 2 0
Using these exponents, there are four possible TWOⁿ magic squares
/ \ \ / 2⁰ 2² 2¹ 2¹ 2² 2⁰ 2¹ 2⁰ 2² 2² 2⁰ 2¹ 2² 2¹ 2⁰ 2⁰ 2¹ 2² 2² 2¹ 2⁰ 2⁰ 2¹ 2² 2¹ 2⁰ 2² 2² 2⁰ 2¹ 2⁰ 2² 2¹ 2¹ 2² 2⁰ / \ \ /
and four possible FIVEⁿ magic squares
/ \ \ / 5⁰ 5² 5¹ 5¹ 5² 2⁰ 5¹ 5⁰ 5² 5² 5⁰ 5¹ 5² 5¹ 5⁰ 5⁰ 5¹ 2² 5² 5¹ 5⁰ 5⁰ 5¹ 5² 5¹ 5⁰ 5² 5² 5⁰ 2¹ 5⁰ 5² 5¹ 5¹ 5² 5⁰ / \ \ /
The / and \ mark the diagonals that must be different. If we use the / diagonal for our TWOⁿ square, we must use the \ diagonal for the FIVEⁿ and vice versa.
Here is what our TWOⁿ squares look like:
/ \ \ / 1 4 2 2 4 1 2 1 4 4 1 2 4 2 1 1 2 4 4 2 1 1 2 4 2 1 4 4 1 2 1 4 2 2 4 1 / \ \ /
Your turn: Make a similar set of four grids for the FIVES (I’m not going to do all your work for you). I’ll get you started:
1 25 5 25 5 1 5 1 25
For each of your four FIVEⁿ squares, there are two TWOⁿ that you can multiply the numbers by to get a magic product square:
Since I showed you one of the /5/ FIVEⁿ squares, you will use one of the \2\ TWO² squares.
+----+-----+----+ 1 25 5 2 4 1 | 2 | 100 | 5 | 25 5 1 × 1 2 4 = | 25 | 10 | 4 | 5 1 25 4 1 2 | 20 | 1 | 50 | +----+-----+----+
+----+-----+----+ 1 25 5 2 1 4 | | | | ← your 25 5 1 × 4 2 1 = | | | | ← turn 5 1 25 1 4 2 | | | | ← now +----+-----+----+
There are still SIX more magic product squares for you to find.
Shubham Kumar
B.Tech from Indian Institute of Technology, Kharagpur (IIT KGP) (Graduated 2024)
·
8y
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Ancient History
Agamemnon and His Family: The Cycle of Blood
Agamemnon, the mighty king of the Greeks, and his family left a trail of blood in Greek myth. Read on to discover the harrowing tales of Agamemnon and his family.
Published: Jun 5, 2022
written by Bethany Williams, BA Classics and English, MA Literature
Published: Jun 5, 2022written by Bethany Williams, BA Classics and English, MA Literature
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Williams, Bethany. "Agamemnon and His Family: The Cycle of Blood" TheCollector.com, June 5, 2022,
While Greek myth is fascinating and compelling to a variety of audiences, a lot of this fascination comes from its shocking horror and violence. One of the most compelling Greek myths is the story of Agamemnon. His life was full of war, deceit, blood, and death. But it didn’t stop with him: his family continued the cycle of blood and revenge. Clytemnestra, Orestes, and Iphigenia — members of Agamemnon’s family — all had their fair share of blood spilled on them or were the ones to spill it.
Agamemnon’s Lineage
Agamemnon was a descendant of a cursed line, stemming from the legendary Pelops. Pelops’ sons had brought about a blood curse on their descendants because they murdered their step-brother. Their descendants were cursed to continually murder — through accidents and revenge — members of their own family tree. The family line became one of the greatest complexities in Greek myth.
Atreus was the father of Agamemnon and the King of Mycenae. The curse caused Atreus to kill one of his sons by accident, not knowing his identity. In a complex to and fro of revenge sequences, Atreus was eventually deposed by his own brother, Thyestes, who then took the Mycenaean throne.
At the time of Atreus’ death, he had two surviving sons: Agamemnon and Menelaus. The sons of Atreus are collectively referred to and known as the Atredai in ancient Greek, or Atreides in English. When their father was murdered, they took refuge with King Tyndareus of Sparta.
Refuge and Marriages
Under the protection of King Tyndareus, the brothers fared well. King Tyndareus married two of his daughters to the brothers. Agamemnon was married to the strong-willed and remarkable Clytemnestra, who had a fierce soul to match Agamemnon’s. Meanwhile, Menelaus was lucky to win the contest for Helen’s hand.
Helen was so famous throughout Greece for her beauty that she had around 45 suitors vying for her. Her father Tyndareus arranged that all the suitors should take an oath to protect whoever was the victor. In the end, Menelaus proved the victor, and so he was married to Helen. The Oath of Tyndareus would later be enacted, causing the Trojan War.
After the marriages, Agamemnon decided on returning to his hometown and taking it back from his uncle. Like many of his peers, Agamemnon was driven by the prospect of glory and fame. He longed for prestige and renown in war. As the son of the famous Atreus, he had a family reputation to live up to. Taking back his home would empower him and his new wife, giving them the titles of King and Queen.
Agamemnon, with the help of Tyndareus (who proved to be a trusted ally after all this time), attacked Mycenae and defeated the usurper. Agamemnon’s rightful throne was restored, and with it came many powers including a huge army and a great amount of land. Around the same time, Menelaus succeeded Tyndareus as King of Sparta.
Back in Mycenae
After returning to Mycenae, Clytemnestra and Agamemnon appeared to have a happy marriage at the beginning. They had many children together: Orestes was their only son, and they had three daughters named Iphigenia, Electra, and Chrysothemis.
Agamemnon expanded his kingdom to an even greater size by successively attacking and conquering nearby lands and cities. He became one of the most powerful kings in Greece, and this later led to his title of “King of Kings” among the Greeks. His symbol became the lion, and above the gate to his kingdom, two lions were carved. While Agamemnon’s strength and power were growing in Mycenae, his brother Menelaus in Sparta was approaching one of the greatest events of Greek history.
Menelaus had invited the Trojans to Sparta for talks on trade and peace. The young prince Paris, however, fell in love with Helen. The Trojan Prince rashly decided to take her with him back to Troy, to become his own wife. When the betrayal was realized, Menelaus asked his brother for help, and also invoked the Oath of Tyndareus. This oath was a promise that the suitors would help Menelaus, as the husband of Helen, in a time of need. He summoned the Kings and Princes of Greece along with their armies to attack Troy.
Before sailing to Troy, the armies of Greece gathered together at the port of Aulis. The port of Aulis was situated on the Eastern coast of Greece, and faced the Aegean sea. On the other side of the Aegean sea, was the land of the Trojans.
The Sacrifice of Iphigenia
Now, when the fleet of a thousand ships had gathered, they could not yet set sail. The goddess Artemis was angry at Agamemnon, and so had caused the wind to stop. Without the wind, the ships could not sail to Troy. Agamemnon had killed a deer that was sacred to Artemis, and so Artemis in retaliation demanded that Agamemnon sacrifice one of his dearest daughters to him.
“CHORUS: But you, Iphigeneia, on yourlovely hair the Argives will seta wreath, as on the browsof a spotted heifer, led downfrom caves in the mountainsto the sacrifice,and the knife will open the throatand let the blood of a girl.
…
Oh where is the noble faceof modesty, or the strength of virtue, nowthat blasphemy is in powerand men have put justicebehind them, and there is no law but lawlessness,and none join in fear of the gods?”(Euripides, Iphigenia at Aulis)
In the version of this myth envisioned by Euripides, Agamemnon is emotionally distraught. His ambition for power and glory warred with his love for his daughter. However, his ambition eventually won and he summoned Iphigenia to Aulis.
Agamemnon did not tell Iphigenia nor her accompanying mother of her fate, but rather he invited Iphigenia to Aulis under the pretence of marrying her to the greatest warrior of their generation: Achilles. Iphigenia and Clytemnestra were thrilled at the match, and excitedly traveled in a wonderful bridal train to Aulis. Iphigenia was brought to her father, in her wedding gown, and then sacrificed with a knife across the throat.
Clytemnestra’s Revenge
Clytemnestra never forgot the horrific, violent betrayal. For years and years, she plotted Agamemnon’s demise, by her own hand, in revenge for the murder of her daughter. She claimed that the curse on Atreus’ house, as well as her right to vengeance, justified the murder of Agamemnon:
“And what of the doom of craft that firstHe planted, making the House accurst?What of the blossom, from this root riven,Iphigenîa, the unforgiven?Even as the wrong was, so is the pain:He shall not laugh in the House of the slain,When the count is scored;He hath but spoiled and paid againThe due of the sword.” — Clytemnestra on her right to kill Agamemnon (Aeschylus, Agamemnon)
Clytemnestra arranged that bonfire torches should be placed all along the way from Mycenae to Troy. They would be lit when Agamemnon began his return home. Agamemnon interpreted this as a loving wife wishing to know when he was going to return, but for Clytemnestra, it was a warning for his return so that her plan should be put into action.
Clytemnestra’s Revenge: II
Agamemnon was away at Troy for about ten years. He did not know how life in Mycenae had changed so greatly in that time. Clytemnestra had been Queen and sole ruler for a long time, and she had taken up a new lover named Aegisthus who helped her plot her revenge.
When Agamemnon finally returned, Clytemnestra laid a blood-red carpet from the entrance of the great city walls to the doors of their home. Stepping on this carpet was taboo because it indicated pride. The very same pride which had led him to the murder of Agamemnon’s own daughter.
Clytemnestra invited her husband to take a bath after his long journey, and he readily agreed. When relaxed and soaking in the bath, Clytemnestra laid heavy robes down on top of him so that he could not move under the weight. As the robes collected water, it was even harder to move. Clytemnestra then used an ax to murder him.
The Concubine and Agamemnon’s Pride
To add insult to injury, Agamemnon had brought a new concubine from Troy. Clytemnestra’s love and attraction to Agamemnon had long since dissipated, but the insult could not be left unpunished. And so, Cassandra was killed, too.
In some versions, not just Cassandra, but Agamemnon’s whole returning party were killed by Clytemnestra and her accomplices in a gigantic show of revenge and justice for the murder of Iphigenia. Only through this did Clytemnestra believe her beloved daughter could be avenged.
All of Agamemnon’s efforts to win the war and attain glory, stemming from the sacrifice of his daughter, were completely eradicated. Agamemnon was never able to taste the fruits of his victory in his hometown after ten years at war, and so his sacrifice became his greatest mistake. Pride or hubris had led Agamemnon to his doom.
Agamemnon’s Progeny
Many playwrights have recaptured the myth about Agamemnon’s family, but the Oresteia by Aeschylus is a trilogy of plays that runs from the murder of Agamemnon to the trial of Orestes. Agamemnon vividly brings to life Clytemnestra’s plot and murder of Agamemnon; The Libation Bearers is the next episode, in which Orestes, the son of Clytemnestra and Agamemnon, plots and carries out the death of his mother in revenge for this father’s murder. The final play of the three, called the Eumenides, is about Orestes being hunted by the Furies, who are the underworld goddesses of justice.
Orestes and Electra were horrified at the murder of their father, and less sympathetic to their mother’s case. Together they plotted the death of Clytemnestra and Aegisthus. The gods were even more enraged by this return to violence and bloodshed within the family. The Furies were sent to haunt Orestes for killing his mother.
The play ends with the formation of the first court of justice, heralded by the Goddess of Wisdom, Athena. Orestes was put to trial and found not guilty. The curse of Atreus’ House was brought to an end, and the justice system of Athens was founded.
The myths of Agamemnon’s family draw attention to the terms and classifications of justice and revenge. Who was in the right? And who was in the wrong? Are things ever so black and white?
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The Curse Of Atreus and the House of Atreides in Greek Mythology
Bethany Williams
BA Classics and English, MA Literature
Bethany is a Masters student, currently studying the adaptation of Greek myth in modern literature. She is a graduate of Classics and English (BA), during which she studied Ancient Greek language, classical reception within its own time and throughout history, as well as Greek and Roman history. Apart from her studies, she has an appreciation for art, philosophy, and travel. She may be based in England, but her heart is in Greece.
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Contact UsOrder of Operations ============================================================================ Arithmetic Order of Operations (BEDMAS) ---------------------------------------------------------------------------------------------------------- Addition:a + b Subtraction:a − b Multiplication:a × b or ab Division:a ÷ b or a/b Exponent:a b or a^b Note that −a = 0 − a =( − 1)× a, and a − b=a +( −b ) Order of Operations ------------------------------------------------------------------------------------------ One need only think of a toddler with two operations to do, eat lunch and wash face, to realise that the order in which the operations are done makes a tremendous difference to the result. When evaluating numerical or algebraic expressions, we need to know the order in which addition, subtraction, multiplication, division and exponents are carried out. For all numerical or algebraic expressions, the order of evaluation is ( BEDMAS ): B rackets and Parentheses First Priority Exponents Second Priority Division Third Priority Multiplication Third Priority A ddition Fourth Priority Subtraction Fourth Priority If an expression involves two or more operations at the same level of priority, those operations are done from left to right. ### Example 1A. Click on the question marks to see the following examples done step-by-step. Keep the following order of operations in mind when studying these examples: 1.) 2.) 3.) 4.) 5.) 6.) 7.) 8.) More Examples ### Example 1B. When brackets occur within brackets, solve the expression inside the inner-most brackets first. [4 + 2(3 + 2 × 4)] ÷ 2 = [4 + 2(3 + 8)]÷ 2 =[4 + 2(11)] ÷ 2 = [4 + 22] ÷ 2 = 26 ÷ 2 = 13 ### Example 1C. When a quotient is written as a ratio the numerator and the denominator are evaluated first. That is, we treat the numerator and denominator as if they were inside brackets. For example, ### Example 1D. To calculate -a 2, first find the square of a, a 2. Then take its negative. So -a 2 is the negative of a 2. To calculate (-a)2, square -a. For example, -3 2 = -9. But (-3)2 = 9. (While the above is the correct order of operations, note that Excel evaluates -a 2 as (-a)2.) ### Exercise 1. Now try some of these exercises: Compute: = If an expression involves functions such as ,e x, ln(x), sin(x) or cos(x) they have to be evaluated first, before they can be raised to a power, multiplied, divided, added or subtracted. They are given First Priority. ### Example 2. More Examples ### Exercise 2. Now try some of these exercises. (The notation sqrt(a) is used for and exp(x) for e x.) Compute, rounding any decimal answers to 2 d.p.: = Worksheet Order of Operations Index | Evaluating Algebraic Expressions>>
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14844 | https://study.com/academy/lesson/euclidean-distance-calculation-formula-examples.html | Euclidean Distance | Calculation, Formula & Examples | Study.com
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Math Courses / General Math Lessons Course
Euclidean Distance | Calculation, Formula & Examples
Instructor Cynthia HelznerShow bio
Cynthia Helzner has tutored middle school through college-level math and science for over 20 years. She has a B.S. in microbiology from The Schreyer Honors College at Penn State and a J.D. from the Dickinson School of Law. She also taught math and test prep classes and volunteered as a MathCounts assistant coach.
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Discover Euclidean distance and comprehend what it represents in math. Explore the Euclidean distance formula and steps on how to calculate Euclidean distance. Updated: 10/05/2023
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Table of Contents
What is Euclidean Distance?
Euclidean Distance Formula
Examples of Euclidean Distances
Lesson Summary
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What is Euclidean Distance? ---------------------------
Euclidean distance is the length of the line segment connecting two points. It is often used in math to find the distance between two points on a coordinate plane, but it can also be used in other contexts, such as finding the distance between two cities or the distance driven between two mile markers on a straight highway.
Euclidean distance is a scalar quantity, meaning that it has a magnitude but not a direction or sign. For example, the distance between two points could be 3 but not -3.
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Euclidean Distance Formula --------------------------
The Euclidean distance formula for one dimension is D=|x 1−x 2|, where D is the Euclidean distance, and x 1 and x 2 are the positions (such as on a number line) of the two points. The absolute value bars in the formula are a result of distance being a scalar quantity, so it cannot be negative.
The formula for Euclidean distance in two dimensions is D=(x 2−x 1)2+(y 2−y 1)2, where D is the Euclidean distance, and (x 1,y 1) and (x 2,y 2) are the Cartesian coordinates of the two points. The squares and the square root in the formula take care of distance being a scalar quantity (squares and square roots cannot be negative), so absolute value bars are not needed in the two-dimensional formula.
The two-dimensional Euclidean distance formula is a consequence of the Pythagorean theorem, a 2+b 2=c 2, in which a and b are the lengths of the two legs of a right triangle, and c is the length of the hypotenuse. Two-dimensional Euclidean distance is the hypotenuse connecting the two points, while (x 2−x 1) and (y 2−y 1) are the horizontal and vertical legs, respectively, of the triangle.
Figure 1: Two-dimensional Euclidean distance
How to Calculate Euclidean Distance
It is important to select the appropriate formula when calculating Euclidian distance. The steps for calculating Euclidean distance are as follows:
Step 1 — Choose the correct formula based on whether the distance is along one or two dimensions. A number line, mile markers on a highway, and positions on a ruler are all examples of one-dimensional problems. Coordinate points, north/south and east/west positions, and map grid locations are examples of two-dimensional problems.
Step 2 — Label the numbers in the problem according to their abbreviations in the formula from step 1 (x 1 and x 2 for one dimension; x 1, x 2, y 1, and y 2 for two dimensions).
Step 3 — Plug the labeled values from step 2 into the formula from step 1.
Step 4 — Perform the mathematical operations in the formula and simplify the result.
Properties of Euclidean Distance
Euclidean distance has various properties that result from its scalar and geometric features. The properties of Euclidean distance include the following:
Euclidean distance is positive. Because Euclidean distance is a scalar quantity, it cannot be negative. If two points are not at the same location, the distance between them will not be 0. So, for two distinct points, the Euclidean distance must be positive. For example, the distance between -2 and 10 on a number line is 12, not -12.
Euclidean distance is symmetric, meaning that the distance from A to B is the same as the distance from B to A. In other words, it does not matter which point is considered point 1 and which is considered point 2. This is a consequence of the fact that Euclidean distance is scalar.
Euclidean distance abides by the triangle inequality property. The triangle inequality property states that any side of a triangle is more than the difference of the other two sides and less than the sum of the other two sides. The first part of the theorem dictates that the Euclidean distance is longer than the difference between (x 2−x 1) and (y 2−y 1). Applying the second part of the theorem to Euclidean distance means that the distance is the shortest path from the starting point to the ending point. Any other path would be the sum of the other two sides of a triangle, which must collectively be longer than the Euclidean distance.
Examples of Euclidean Distances -------------------------------
When learning how to calculate Euclidean distance, it is helpful to work through some examples of Euclidean distances. The formula used in each example depends on the number of dimensions in the problem.
Example One: One Dimension
What is the distance between -20 and 10 on a number line?
A number line is a one-dimensional structure, so the Euclidean distance formula for one dimension is appropriate here. Plugging x 1=−20 and x 2=10 into the Euclidean distance formula for one dimension yields:
D=|x 1−x 2|D=|−20−10|D=|−30|D=30
So, the distance between -20 and 10 is 30.
Example Two: Two Dimensions
Two points on a coordinate plane are located at (-2, 0) and (1, -4). What is the distance between them?
A coordinate plane is a two-dimensional structure, so use the Euclidean distance formula for two dimensions. Plugging x 1=−2, x 2=1, y 1=0, and y 2=−4 into the Euclidean distance formula for two dimensions yields:
D=(1−(−2))2+(−4−0)2 D=3 2+(−4)2 D=9+16 D=25 D=5
Thus, the distance between (-2, 0) and (1, -4) is 5.
Lesson Summary --------------
The Euclidean distance between two points is the length of the line segment connecting those two points. Typically, this is applied on a coordinate plane, but it can be used in other contexts as well. Because Euclidean distance is scalar (i.e., it has a magnitude but not a direction or sign), it cannot be negative. For two distinct points, the distance will not be 0. Thus, the Euclidean distance between two different points is always positive. As a result of Euclidean distance being scalar, it is symmetric, meaning that the distance from one point to the other is the same as if the direction were reversed.
The formula to use for Euclidean distance depends on whether the distance is one- or two-dimensional. Euclidean distance in one dimension is given by D=|x 1−x 2|, where D is the distance, and x 1 and x 2 are the positions of the two points. Euclidean distance in two dimensions is given by D=(x 2−x 1)2+(y 2−y 1)2, where D is the distance, and (x 1,y 1) and (x 2,y 2) are the Cartesian coordinates of the two points. Two-dimensional Euclidean distance abides by the triangle inequality property, which states that any side of a triangle is more than the difference of the other two sides and less than the sum of the other two sides.
Frequently Asked Questions
What is Euclidean distance used to identify?
Euclidean distance is used to identify the shortest distance between two points. Examples of Euclidean distance include the distance between two cities on a map or between two numbers on a number line.
What is the Euclidean distance in simple terms?
Euclidean distance is the length of the line segment connecting two points. It is also the straight-line path connecting the two points and is therefore the shortest path connecting two points.
What is an example of Euclidean distance?
An example of one-dimensional Euclidean distance is the distance between two numbers on a number line. An example of two-dimensional Euclidean distance is the distance between two points on a coordinate plane.
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General Math Lessons
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Ch 1. General Mathematics Lessons
Large Numbers: Usage, Names & Order | What are Big Numbers?
Imperial System of Measurement | Overview, History & Units
Simplex Tableau Procedure & Examples | What is Simplex Method?
Algebraic Properties Definition, Equations & Examples
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Nth Root Definition & Formula
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Absolute Convergence Overview, Proof & Examples
Euclidean Distance | Calculation, Formula & Examples
Next Lesson Test Statistic | Definition, Types & Formula
Parametric Statistics | Tests, Data & Distribution
Rates in Mathematics | Definition & Examples
Vigenre Cipher History, Example & Coding Variants
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Linearity in Algebra | Functions, Equations & Examples
Margin of Error Definition, Calculation & Formula
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Complex Plane Definition, Graph & Examples
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Nonlinear System | Definition, Functions & Equations
Abstract Algebra: Overview & Applications | What is Abstract Algebra?
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Integration Techniques in Calculus
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14845 | https://people.math.wisc.edu/~pmterwil/Htmlfiles/ex3solp1.pdf | Math 475 Text: Brualdi, Introductory Combinatorics 5th Ed.
Prof: Paul Terwilliger Selected solutions for Chapter 3 4, 5, 6. Given integers n ≥1 and k ≥2 suppose that n + 1 distinct elements are chosen from {1, 2, . . . , kn}.
We show that there exist two that differ by less than k.
Partition {1, 2, . . . , nk} = ∪n i=1Si where Si = {ki, ki −1, ki −2, . . . , ki −k + 1}. Among our n + 1 chosen elements, there exist two in the same Si. These two differ by less than k.
9. Consider the set of 10 people. The number of subsets is 210 = 1024. For each subset consider the sum of the ages of its members.
This sum is among 0, 1, . . . , 600.
By the pigeonhole principle the 1024 sums are not distinct. The result follows. Now suppose we consider at set of 9 people. Then the number of subsets is 29 = 512 < 600. Therefore we cannot invoke the pigeonhole principle.
10. For 1 ≤i ≤49 let bi denote the number of hours the child watches TV on day i.
Consider the numbers {b1 + b2 + · · · + bi + 20}48 i=0 ∪{b1 + b2 + · · · + bj}49 j=1.
There are 98 numbers in the list, all among 1, 2, . . . , 96.
By the pigeonhole principle the numbers {b1 + b2 + · · · + bi + 20}48 i=0 ∪{b1 + b2 + · · · + bj}49 j=1. are not distinct. Therefore there exist integers i, j (0 ≤i < j ≤49) such that bi+1 + · · · + bj = 20. During the days i + 1, . . . , j the child watches TV for exactly 20 hours.
14. After n minutes we have removed n pieces of fruit from the bag. Suppose that among the removed fruit there are at most 11 pieces for each of the four kinds. Then our total n must be at most 4 × 11 = 44. After n = 45 minutes we will have picked at least a dozen pieces of fruit of the same kind.
16. Label the people 1, 2, . . . , n. For 1 ≤i ≤n let ai denote the number of people aquainted with person i. By construction 0 ≤ai ≤n −1. Suppose the numbers {ai}n i=1 are mutually distinct. Then for 0 ≤j ≤n −1 there exists a unique integer i (1 ≤i ≤n) such that ai = j.
Taking j = 0 and j = n −1, we see that there exists a person aquainted with nobody else, and a person aquainted with everybody else. These people are distinct since n ≥2. These two people know each other and do not know each other, for a contradiction. Therefore the numbers {ai}n i=1 are not mutually distinct.
18. Divide the 2 × 2 square into four 1 × 1 squares. By the pigeonhole principle there exists a 1 × 1 square that contains at least two of the five points. For these two points the distance apart is at most √ 2.
20. Color the edges of K17 red or blue or green. We show that there exists a K3 subgraph of K17 that is red or blue or green. Pick a vertex x of K17. In K17 there are 16 edges that contain x. By the pigeonhole principle, at least 6 of these are the same color (let us say red).
Pick distinct vertices {xi}6 i=1 of K17 that are connected to x via a red edge. Consider the K6 subgraph with vertices {xi}6 i=1. If this K6 subgraph contains a red edge, then the two vertices involved together with x form the vertex set of a red K3 subgraph. On the other 1 hand, if the K6 subgraph does not contain a red edge, then since r(3, 3) = 6, it contains a K3 subgraph that is blue or green. We have shown that K17 has a K3 subgraph that is red or blue or green.
27. Let s1, s2, . . . , sk denote the subsets in the collection. By assumption these subsets are mutually distinct. Consider their complements s1, s2, . . . , sk. These complements are mutu-ally distinct. Also, none of these complements are in the collection. Therefore s1, s2, . . . , sk, s1, s2, . . . , sk are mutually distinct. Therefore 2k ≤2n so k ≤2n−1. There are at most 2n−1 subsets in the collection.
28. The answer is 1620. Note that 1620 = 81 × 20. First assume that P100 i=1 ai < 1620.
We show that no matter how the dance lists are selected, there exists a group of 20 men that cannot be paired with the 20 women. Let the dance lists be given. Label the women 1, 2, . . . , 20. For 1 ≤j ≤20 let bj denote the number of men among the 100 that listed woman j. Note that P20 j=1 bj = P100 i=1 ai so (P20 j=1 bj)/20 < 81. By the pigeonhole principle there exists an integer j (1 ≤j ≤20) such that bj ≤80. We have 100 −bj ≥20. Therefore there exist at least 20 men that did not list woman j. This group of 20 men cannot be paired with the 20 women.
Consider the following selection of dance lists. For 1 ≤i ≤20 man i lists woman i and no one else. For 21 ≤i ≤100 man i lists all 20 women. Thus ai = 1 for 1 ≤i ≤20 and ai = 20 for 21 ≤i ≤100. Note that P100 i=1 ai = 20 + 80 × 20 = 1620. Note also that every group of 20 men can be paired with the 20 women.
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14846 | https://sboh.wa.gov/sites/default/files/2022-01/HIR-2018-01-SB6003.pdf | Executive Summary: Health Impact Review of SB 6003 Concerning Breakfast After the Bell Programs in Certain Public Schools (2017-2018 Legislative Sessions) For more information contact: (360)-236-4109 | hir@sboh.wa.gov or go to sboh.wa.gov BILL INFORMATION Sponsors: Wellman, Billig, Palumbo, Frockt, Rolfes, Van De Wege, Liias, Keiser, Pedersen, Hunt, Conway, Chase, Saldaña, Kuderer Summary of Bill: Requires high-needs schools that have not reached target participation (70% of free or reduced-price eligible students) in both the School Lunch and Breakfast Programs to offer Breakfast After the Bell and provide adequate time for students to eat. Requires that all breakfasts served under these programs comply with federal meal patterns and nutrition standards. Allows the Office of Superintendent of Public Instruction (OSPI) to administer one-time grants to each high needs school to implement a program, depending on appropriation of funding from the state. Requires OSPI to develop and distribute procedures and guidelines, and to offer training and technical and marketing assistance to schools to implement Breakfast After the Bell. Directs OSPI to partner with nonprofit and philanthropic organizations. Requires the Joint Legislative Audit and Review Committee (JLARC) to conduct an analysis of the programs established in schools. Requires OSPI and the Education Data Center to assist in providing data required to conduct this analysis. HEALTH IMPACT REVIEW Summary of Findings: This Health Impact Review found the following evidence regarding the provisions in SB 6003: Strong evidence that Breakfast After the Bell programs would likely increase the number of low-income students participating in the School Breakfast Program and eating breakfast. Strong evidence that eating breakfast would likely improve health outcomes for these students and decrease health disparities. Strong evidence that eating breakfast would likely improve educational outcomes. Very strong evidence that improving educational outcomes would likely improve educational attainment. Very strong evidence that improving educational attainment would likely improve earning potential. Very strong evidence that improving educational attainment would likely decrease health disparities. Very strong evidence that improving earning potential would likely decrease health disparities. Evidence indicates that SB 6003 has potential to increase the number of low-income students and students of color who eat breakfast, which in turn has potential to improve educational outcomes, to improve earning potential, and to decrease health disparities. Health Impact Review of SB 6003 Concerning Breakfast After the Bell Programs in Certain Public Schools (2017-2018 Legislative Sessions) January 11, 2018 Staff Contact: Lindsay Herendeen Acknowledgement We would like to thank the experts who provided consultation and technical support during this Health Impact Review and previous Health Impact Reviews related to this topic. Contents Introduction and Methods ............................................................................................................... 1 Analysis of SB 6003 and the Scientific Evidence .......................................................................... 2 Logic Model .................................................................................................................................... 5 Summaries of Findings ................................................................................................................... 6 Annotated References ................................................................................................................... 10 1 January 2018- Health Impact Review of SB 6003 Introduction and Methods A Health Impact Review is an analysis of how a proposed legislative or budgetary change will likely impact health and health disparities in Washington State (RCW 43.20.285). For the purpose of this review ‘health disparities’ have been defined as the differences in disease, death, and other adverse health conditions that exist between populations (RCW 43.20.270). This document provides summaries of the evidence analyzed by State Board of Health staff during the Health Impact Review of Senate Bill 6003 (SB 6003) from the 2017-2018 legislative sessions. Staff analyzed the content of SB 6003 and created a logic model depicting possible pathways leading from the provisions of the bill to health outcomes. We consulted with experts and contacted stakeholders with diverse perspectives on the bill. State Board of Health staff can be contacted for more information on which stakeholders were consulted on this review and previous reviews on this topic. We conducted objective reviews of the literature for each pathway using databases including PubMed and Google Scholar. The following pages provide a detailed analysis of the bill including the logic model, summaries of evidence, and annotated references. The logic model is presented both in text and through a flowchart (Figure 1). The logic model includes information on the strength of the evidence for each relationship. The strength-of-evidence has been defined using the following criteria: Not well researched: the literature review yielded few if any studies or only yielded studies that were poorly designed or executed or had high risk of bias. A fair amount of evidence: the literature review yielded several studies supporting the association, but a large body of evidence was not established; or the review yielded a large body of evidence but findings were inconsistent with only a slightly larger percent of the studies supporting the association; or the research did not incorporate the most robust study designs or execution or had a higher than average risk of bias. Strong evidence: the literature review yielded a large body of evidence on the relationship (a vast majority of which supported the association) but the body of evidence did contain some contradictory findings or studies that did not incorporate the most robust study designs or execution or had a higher than average risk of bias; or there were too few studies to reach the rigor of ‘very strong evidence’; or some combination of these. Very strong evidence: the literature review yielded a very large body of robust evidence supporting the association with few if any contradictory findings. The evidence indicates that the scientific community largely accepts the existence of the association. This review was subject to time constraints, which influenced the scope of work for this review. The annotated references are only a representation of the evidence and provide examples of current research. In some cases only a few review articles or meta-analyses are referenced. One article may cite or provide analysis of dozens of other articles. Therefore the number of references included in the bibliography does not necessarily reflect the strength-of-evidence. In addition, some articles provide evidence for more than one research question so they are referenced multiple times. 2 January 2018- Health Impact Review of SB 6003 Analysis of SB 6003 and the Scientific Evidence Summary of relevant background information All public schools, nonprofit private schools, and Residential Child Care Institutions in the United States can participate in the federally funded National School Lunch and School Breakfast Programs.1 Children from families with incomes at or below 130% of the federal poverty level receive free school meals. Children from families with incomes between 130% and 185% of the poverty level receive school meals at a reduced-price.1 Summary of SB 6003 Requires high-needs schools that have not reached target participation (70% of free or reduced-price eligible students) in both the School Lunch and Breakfast Programs to offer Breakfast After the Bell and provide adequate time for students to eat. Requires that all breakfasts served under these programs comply with federal meal patterns and nutrition standards. Allows the Office of Superintendent of Public Instruction (OSPI) to administer one-time grants to each high needs school to implement a program, depending on appropriation of funding from the state. Requires OSPI to develop and distribute procedures and guidelines, and to offer training and technical and marketing assistance to schools to implement Breakfast After the Bell. Directs OSPI to partner with nonprofit and philanthropic organizations. Requires the Joint Legislative Audit and Review Committee (JLARC) to conduct an analysis of the programs established in schools. Requires OSPI and the Education Data Center to assist in providing data required to conduct this analysis. Health impact SB 6003 Evidence indicates that SB 6003 has potential to increase the number of low-income students and students of color who eat breakfast, which in turn has potential to improve educational outcomes, to improve earning potential, and to decrease health disparities. Scope of this Health Impact Review Due to time limitations, this Health Impact Review does not focus on potential pathways leading from the provisions outlined in section 7 of SB 6003. Section 7 relates to grants that may be awarded for activities such as increasing awareness of and participation in school breakfast and lunch programs, improving the nutritional content of food, and promoting programs such as organic gardens that provide produce for school meals. Pathways to health impacts The potential pathways leading from the provisions of SB 6003 to decreased health disparities are depicted in Figure 1. There is strong evidence that Breakfast After the Bell programs increase participation in the School Breakfast Program and increase the number of students who eat breakfast, particularly among low-income students and students of color who are more likely to skip breakfast than 3 January 2018- Health Impact Review of SB 6003 their peers.2-15 There is also strong evidence that eating breakfast is directly associated with improved education7,16,17 and health outcomes.12,18-21 In addition, the literature indicates that increased educational opportunities and outcomes are very strongly linked to increased educational attainment,22-25 which in turn is very strongly associated with both improved health26-38 and increased income27,39 (which is also very strongly linked to improved health).21,26-28,30,35,36,40-45 Because this bill specifically applies to students who are eligible for reduced-price lunch and to schools with large populations of students eligible for free and reduced-price meals it would likely have the greatest impact among low-income students. In addition, because schools in Washington with high percentages of low-income students also have higher percentages of students of color,9 this bill could help increase the number of students of color who participate in school breakfast and lunch programs as well. Low-income students and students of color are more likely to experience health disparities9,46-49 and therefore, increasing participation in school meals among these students has potential to decrease these disparities. Due to time limitations we only researched the most direct connections between the provisions of the bill and decreased health disparities and did not explore the evidence for all possible pathways. For example, potential pathways that were not researched include: Evidence of how breakfast consumption impacts student behavior and discipline incidence and how discipline incidence relates to educational outcomes. Evidence for how eating breakfast at school impacts a family’s financial stability. Magnitude of Impact- Breakfast After the Bell This legislation could impact a large number of students in Washington as more than 20% of 6th grade students, more than 30% of 8th grade students, more than 37% of 10th grade students, and nearly 39% of 12th grade students reported skipping breakfast that morning on the 2012 Healthy Youth Survey.11 In addition, evidence indicates that Breakfast After the Bell models have increased the participation of students in the School Breakfast Program, with the level of increase varying by the type of program and level of implementation. For example, one school which added a mobile breakfast cart and extended the morning cafeteria hours saw the average daily School Breakfast Participation increase by nearly 400% and the number of students who skipped breakfast at least once per week decrease by 8.7% following implementation of these programs.10 4 January 2018- Health Impact Review of SB 6003 Logic Model Key Not Well Researched A Fair Amount of Evidence Strong Evidence Very Strong Evidence Breakfast After the Bell programs implemented More low - income students and students of color participate in the School Breakfast Program and eat breakfast Improved educational outcomes Improved health outcomes and decreased health disparities Improved earning potential Improved educational attainment Figure 1 Concerning Breakfast After the Bell Programs SB 6003 5 January 2018- Health Impact Review of SB 6003 Summaries of Findings Will implementing Breakfast After the Bell programs increase the number of low-income students and students of color who participate in the School Breakfast Program and who eat breakfast in Washington State? There is strong evidence that Breakfast After the Bell models such as Grab-and-Go, extended breakfast hours, and Breakfast in the Classroom increase participation in the School Breakfast Program and decrease the number of students who skip breakfast.2,4-8,10,12,13,15 The language of SB 6003 specifically requires that schools with large populations of students eligible for free and reduced-price meals and with low percentages of these students participating in school meals offer Breakfast After the Bell, therefore this bill is more likely to increase breakfast participation among low-income students. Some evidence indicates that Breakfast After the Bell programs are more likely to increase breakfast participation for low-income students than for their higher income peers even within the same school,8,14 so this positive effect may be enhanced for low-income students in the schools where these programs are implemented. Similarly, a study evaluating school breakfast programs in New York City found that Breakfast in the Classroom programs increased breakfast participation and that schools with a higher percentage of students eligible for free or reduced-price meals, students of color, and students with Limited English Proficiency were more likely to implement the program.15 Therefore, increases in breakfast participation most directly impacted students of greatest disadvantage. In addition, because schools in Washington with high percentages of low-income students also have higher percentages of students of color,3,9 this bill could help decrease the number of students of color who skip breakfast as well. This is important because Washington state data indicate that students of color and American Indian/Alaska Native students were more likely than their white peers to report having skipped breakfast on the day of the Healthy Youth Survey.11 Will increasing the number of low-income students who participate in the School Breakfast Program and eat breakfast in Washington improve health outcomes and decrease health disparities? There is strong evidence that eating breakfast is directly associated with improved health outcomes. We found evidence that eating breakfast is positively associated with a number of beneficial health outcomes such as healthy weight, lower fasting insulin, lower cholesterol, increased daily nutrient intake, and decreased hunger related health symptoms.12,18-20 One review article found that the literature overall supports an association between eating breakfast and decreased rates of obesity, but that the literature yields inconsistent results.18 In contrast the evidence shows a strong consistent link between eating breakfast and other health indicators such as adequate daily nutrient intake.19,20 Recent studies have also evaluated the impact of Breakfast in the Classroom policies on student energy intake and weight gain, citing concerns that some students may eat “double breakfasts” (one at home and one at school) at schools with alternative breakfast policies and may therefore be at greater risk for overweight or obesity. Overall, these studies have found no evidence to support the concern that students participating in Breakfast in the Classroom consume excess calories or are at risk for excessive weight gain.50,51 Due to time limitations we focused this literature search specifically on the impacts of eating breakfast on health outcomes and did not expand the scope to include evidence on the impacts of food insecurity in general on health outcomes. Including publications on these additional aspects would likely further bolster the evidence for the link between eating breakfast and positive health 6 January 2018- Health Impact Review of SB 6003 impacts. Because the same students who experience disparities in access to breakfast are also experiencing health disparities,21 improved health outcomes for these students has potential to decrease health disparities. Will increasing the number of low-income students who participate in the School Breakfast Program and eat breakfast in Washington improve educational outcomes? There is strong evidence that eating breakfast is associated with improved educational and cognitive outcomes such as better grades, achievement test scores, memory, punctuality, readiness to learn, classroom behavior, reading scores, fewer errors on attention tasks, and decreased dropout rates.2,16,17 The literature also indicates that these positive effects are more apparent in nutritionally vulnerable children, indicating that increasing access to breakfast may have a greater positive impact on low-income students. Access to breakfast may therefore work to narrow the opportunity gaps experienced by low-income students and students of color. There is also evidence that School Breakfast Programs are associated with improved attendance which may at least partially account for improvements in academic outcomes.7,16,17 Will improving educational outcomes improve educational attainment? There is very strong evidence that improved educational outcomes such as those linked to eating breakfast and lunch (e.g. higher grades and increased readiness to learn) are associated with higher educational attainment.22-25 For example, one study found that low grades during primary school were predictive of not having completed a secondary education by age 20 or 21.25 These links are well documented and because this connection is widely accepted, less time was dedicated to researching this relationship. In addition several measures of educational outcomes are innately indicative of education attainment (e.g. specific grades are required as a prerequisite for high school graduation—one measure of educational attainment) further supporting the strength-of-evidence for this relationship. Will improving educational attainment improve earning potential? There is very strong evidence for the connections between increasing educational attainment and increasing income as well as decreasing rates of unemployment. These links are well documented globally, and data indicate that these trends do exist in Washington state as well.27,39 Because this connection is widely accepted, less time was dedicated to researching this relationship. Will improving earning potential improve health outcomes and decrease health disparities? There is very strong evidence that improving earning potential will improve health outcomes and decrease health disparities. There is a large body of robust evidence that supports the association between income, or socioeconomic position, and health.21,26-28,30,35,36,40-45 Significant correlations exist between lower income and a number of health indicators including worse overall self-reported health, depression, stress, asthma, arthritis, stroke, oral health, tobacco use, women's health indicators, health screening rates, physical activity, and diabetes.27,28,40,43,45 Further, 2015 data indicate that age-adjusted death rates were higher in Washington census tracks with higher poverty rates.35 Household income was also the strongest predictor of self-reported health status in Washington in 2016, even after accounting for age, education, and race/ethnicity.36 Among children, evidence indicates that low socioeconomic status in the first five years of life has negative health outcomes in later childhood and adolescence including activity-limiting illness, parent-reported poor health status, acute and recurrent infections, increasing body mass index 7 January 2018- Health Impact Review of SB 6003 (BMI), dental caries, and higher rates of hospitalization.44 Finally, financial stress is also associated with adverse outcomes for families such as problem behavior in adolescents, interparental conflict, and parental depression.42 Will improving educational attainment improve health and decrease health disparities? There is very strong evidence that higher educational attainment is associated with better health. Data collected nationally and in Washington State indicate a correlation between higher educational attainment and positive health outcomes such as decreased rates of diabetes, oral health problems, tobacco use, inactivity, obesity, depression, and coronary heart disease. The correlation between health and education is observed even after controlling for income, which can also serve as a mediating factor.26-38 8 January 2018- Health Impact Review of SB 6003 Annotated References 1. Levin Madeleine, Hewins Jessie. Universal Free School Meals: Ensuring that All Children Are Able to Learn. Sargent Shriver National Center on Poverty Law 2014. Levin et al. present an overview of the school meal participation, the importance of nutrition on academic achievement, and success stories from districts and states across the country that have implemented alternative meal delivery programs. The authors discuss research that indicates that nourished children are not only better test-takers and participants in school but they are also more likely to arrive on time, behave, have better attendance and be alert in class. Further, hungry children and teens are more likely to have lower math scores, repeat a grade, be suspended from school, and have altercations with their peers. School meal participants are more likely to consume fruit, vegetables, and milk and when meals are offered at no charge to students, a growing body of evidence indicates that school meal participation increases. The report continues on to discuss the various ways that schools have been able to implement universal free meals to their students, including through the community eligibility provision, and the success these schools have seen in increased participation. The authors conclude that school meal programs often help families stretch their limited resources further to ensure that their children can succeed in school and further efforts are needed to expand the uptake of free meal models and the use of community eligibility. 2. Anzman-Frasca Stephanie, Djang Holly Carmichael, Halmo Megan M., et al. Estimating Impacts of a Breakfast in the Classroom Program on School Outcomes. JAMA Pediatrics. 2015;169(1):71. Anzman-Frasca et al. conducted a quasi-experimental study to estimate the impact of Breakfast in the Classroom (BIC) programs on School Breakfast Program participation, school attendance, and academic outcomes. The authors analyzed data from 257 schools implementing BIC programs and 189 schools without BIC programs within one large urban school district in the United States. Over 80% of students in the district were eligible for free or reduced-price school meals, and more than 70% of the students were Hispanic/Latino. After controlling for potential confounding factors, the authors found that schools with BIC programs had significantly higher rates of participation/higher meal counts in the School Breakfast Program and greater overall student attendance than schools without these programs. On average 73.3% of students in BIC schools participated in the School Breakfast Program compared to 42.9% of students in non-BIC schools. BIC programs were implemented in different months in each school with some schools having their programs operational by November and other schools having their programs running by March. As more BIC programs were implemented participation in the breakfast program increased, with 41.9% participation in the BIC schools in August 2012 increasing to 94.6% participation in May 2013. The respective rates for non-BIC schools were 46.4% and 43.4%. BIC schools also had attendance rates of 95.5% versus rates of 95.3% for non-BIC schools. This difference represents an additional 76 attended days per grade per month. The data did not reveal significantly different outcomes for BIC versus non-BIC schools in relation to achievement test scores. The measure of test scores may not have been timely as some of the schools did not implement their BIC programs until March. 9 January 2018- Health Impact Review of SB 6003 3. Aud S, Fox MA, KewalRamani A. Status of Trends in the Education of Racial and Ethnic Groups. Washington DC: National Center for Education Statistics; U.S. Department of Education; 2010. This report compiled national level school data and survey data from 2007, and was analyzed by US Department of Education staff. They found that nationally, Hispanic, Native American/Alaska Native, Native Hawaiian and other Pacific Islander, and African American children under 18 are more likely to be living in poverty than White, Asian, or mixed race children. 4. Bartfeld J., Kim M. Participation in the School Breakfast Program: new evidence from the ECLS-K. The Social service review. 2010;84(4):541-562. Bartfeld and Kim analyzed data from the Early Childhood Longitudinal Study—Kindergarten Cohort in order to determine predictors of participation in the School Breakfast Program. They analyzed parent survey data from wave 5 (third grade) collected in 2002. The primary analysis includes 6,680 children who attended schools with a School Breakfast Program. The authors found that the probability of participating in the School Breakfast Program increases if breakfast is served in the classroom or if the duration of the breakfast period increases. They also found that for students who ride the bus, as the length of time increases between when their bus arrives at school and their first class, participation in the School Breakfast Program also increases. The authors conclude that the “results strongly support the hypothesis that increasing the convenience of the School Breakfast Program leads to greater participation.” 5. Bernstein LS, McLaughlin JE, Crepinsek MK, et al. Evaluation of the school breakfast program pilot project: Final report. Special nutrition programs. Report number cn-04-sbp. Nutrition assistance program report series. Alexandria, VA USDA, Food and Nutrition Service; 2004. In 1998 Congress authorized a three year (2000-2003) School Breakfast Program Pilot Project to evaluate the effects of providing universal free school breakfast in six districts across the United States. Control schools continued to offer the regular School Breakfast Program. Bernstein et al. found that schools offering universal free school breakfast saw an increase in participation in the School Breakfast Program in the first year of the pilot. However, schools that also offered breakfast in the classroom saw significantly larger increases in participation (from 27% in the base year to 66% the following year) compared to schools that did not offer breakfast in the classroom (from 17% in the base year to 28% the following year). The rate of participation in the control schools stayed relatively constant increasing from 20% to 21%. 6. Morris Chad T., Courtney Anita, Bryant Carol A., et al. Grab 'N' Go Breakfast at School: Observations from a Pilot Program. Journal of Nutrition Education & Behavior. 2010;42(3). Morris et al. evaluated a two week Grab ‘N’ Go pilot program in a middle school where 48% of the student body qualified for free or reduced-price lunches. For this program school nutrition staff prepared sack breakfasts that were available in the cafeteria prior to the first class of the day. The first week these were provided free to all students while in the second week they were priced at the regular breakfast prices for free, reduced ($0.30), and full price breakfast ($1.10). The evaluation included field notes, focus groups, interviews with school staff, and student and 10 January 2018- Health Impact Review of SB 6003 teacher surveys. Forty-nine percent of the students surveyed who ‘rarely ate breakfast’ indicated that they had participate in the Grab ‘N’ Go meals. Sixty-six percent of surveyed students reported participating in the program, with 27% participating for all 10 days. 7. Mosehauer K et al. The Future of School Breakfast: An Analysis of Evidence-Based Practices to Improve School Breakfast Participation in Washington State. Washington Appleseed; 2013. Mosehauer et al. analyzed 2013 Washington student School Breakfast Program participation data and found that districts with some schools using Breakfast after the Bell models had higher average participation rates among the target population (students eligible for free and reduced-price lunch) in the School Breakfast Program than districts only using traditional cafeteria models. For example, the districts using traditional models had participation rates for the target population of 49.85% while districts with some schools using traditional cafeteria models and some using Second Chance Breakfast had rates of 50.97%, districts with some schools using traditional models and some using Second Chance Breakfast or Grab 'N' Go had rates of 55.46%, and districts with some schools using traditional models and some using Second Chance Breakfast, Grab 'N' Go, and Universal Breakfast had rates of 77.47%. The report does not indicate if these rates are statistically significantly different. These data also indicate that low-income students missed more days of school than their high-income counterparts, but that schools with higher participation in the School Breakfast Program by students eligible for free and reduced-price meals had smaller disparities in attendance between low and high-income students. Schools meeting national breakfast participation goals saw 40% fewer absences for low-income students than schools not meeting these goals. For example, in Auburn School District, low-income students at schools serving breakfast to 70% or more of the students eligible for free or reduced-price meals missed 1.7 more days of school each year than their higher-income peers, while free and reduced-price eligible students at schools serving less than 70% of the target population missed 3.62 days more than their peers. These are descriptive data only and do not account for potential confounding factors, however the authors did exclude students who were not enrolled in the same school for the entire year which reduces the risk of cross-over. This report also indicates that in Washington, low-income students who experience greater food insecurity missed 1 to 5 more days of school per year than their counterparts. 8. Nanney Marilyn S., Olaleye Temitope M., Wang Qi, et al. A pilot study to expand the school breakfast program in one middle school. Behav. Med. Pract. Policy Res. Translational Behavioral Medicine : Practice, Policy, Research. 2011;1(3):436-442. Nanney et al. conducted a cohort study to evaluate the impact of a Grab 'N' Go School Breakfast Program meal that was delivered in the hallway and eaten in the classroom. The program was implemented in a classroom in Minneapolis, Minnesota with sixth grade students (n=239) for six weeks in 2010. The researchers analyzed School Breakfast Program participation data and pre-intervention BMI as well as data from post intervention student and teacher surveys and researcher observations. The response rate for the student survey was 83.9% and the teacher survey response rate was not reported. One hundred percent of teacher respondents indicated that the meals were not messy or disruptive and that the student behavior was overall excellent or good. The authors found a significant increase in School Breakfast Program participation during the intervention—with an average increase in participation of 0.47 days per week. This increase 11 January 2018- Health Impact Review of SB 6003 was stronger for students who received fee and reduced price lunches (increase of 0.63 days per week) compared to full paid students (increase of 0.29 days per week). 9. Student Enrollment Demographic Data 2014; Available at: Accessed November 13, 2016. These recent Washington state data indicate that school districts that serve high percentages of students eligible for free and reduced-price lunch also tend to serve high percentages of students of color. Although the relationship is not true for every district, the trend is apparent when looking at data for all of the districts combined. 10. Olsta Julia. Bringing Breakfast to Our Students: A Program to Increase School Breakfast Participation. Journal of School Nursing. 2013;29(4):263-270. Olsta evaluated the impact of a program extending breakfast cafeteria hours and providing a mobile cart that served a complete school breakfast during students’ morning study hall classes. This study was conducted in a public high school in a Midwestern suburban area in the 2010-2011 school year. This school held a morning study hall in the cafeteria. The student body was made up of approximately 28% low-income students and 43% students of color. Before implementing the program the school’s Wellness Team collected a baseline survey as a needs assessment from students enrolled in physical education (PE) classes (n=1,405) which had a response rate of 86%. Following implementation of both the food cart and the extended breakfast hours the school saw an increase in average daily school breakfast participation of nearly 400% by the end of the school year. The school Wellness Team conducted a follow-up survey in the spring of 2011 with current students enrolled in PE classes (n=826) with a response rate of 49%. These were not the same students who took the baseline survey. The survey data indicated that the percentage of students who reported that they had skipped breakfast at least once per week was 8.7% lower when compared to the baseline survey. This research did not include a comparison group. This study was partially supported by a grant from the Midwest Dairy Council so there is potential for a conflict of interest. 11. QxQ Analysis. 2012; Available at: Accessed. Washington State Healthy Youth Survey data from 2012 indicate that among 6th , 8th, 10th, and 12th grade respondents students of color and American Indian/Alaska Native (AI/AN) students were more likely to report having gone without breakfast on the day of the survey than their white peers. Although these disparities did not reach significance in every grade for each race/ethnicity classification, the rates of skipping breakfast were significantly higher than those for white students for the following: 6th grade AI/AN and Hispanic students as well as those that self-reported their race/ethnicity as “multiple” or “other”; 8th AI/AN, black, and Hispanic students as well as those reporting their race/ethnicity as “multiple” or “other”; and 10th grade AI/AN and Hispanic students; and 12th grade black and Hispanic students. High percentages of students from every grade reported skipping breakfast. These rates reached as high as 49.5% (95% CI 39.5-59.5%) for AI/AN 10th graders. The percentage of students who skipped breakfast in 6th, 8th, 10th, and 12th grade (all racial ethnic groups combined) were 20.2% ± 2.0%, 31.7% ± 2.0%, 37.1%± 2.3%, and 38.5% ± 2.5% respectively. 12 January 2018- Health Impact Review of SB 6003 12. Sweeney Nancy M., Tucker Joanne, Reynosa Brenda, et al. Reducing Hunger-Associated Symptoms: The Midmorning Nutrition Break. The Journal of School Nursing. 2006;22(1):32-39. Sweeney et al. evaluated the impact of a 9:00 AM Nutritional Break one academic year after its implementation in an inner-city high school. A baseline survey of the high school students (n=846) indicated that 57% of the respondents had not eaten breakfast on the day of the survey despite the universal free breakfast program offered at the school (response rate not noted). Following the intervention researchers collected 590 student surveys (71% response rate) and 46 staff surveys (33% response rate). Students reported a 36% participation rate in the before-school on campus breakfast while 69% of students reported participating at least one day per week in the Nutrition Break. The authors also found that as participation in the Nutrition Break increased, students reported a significant decrease in several huger-related symptoms (i.e. inability to focus, tiredness, stomachache, and midmorning hunger), although participation in the Nutrition Break only accounted for small percentages in the variation. Seventy-four percent of the staff members who responded to the survey indicated that the Nutrition Breaks had a positive effect on students such as improved performance, students being more alert, energetic, motivated and refreshed after the Break and that hunger was no longer an issue during class. Thirty-five percent of staff responded that the Break had negative impacts such as sugar rushes following the Break and sugar crashes later in the day. 13. Van Wye G., Seoh H., Adjoian T., et al. Evaluation of the New York City breakfast in the classroom program. American journal of public health. 2013;103(10):59-64. Van Wye et al. conducted a cross-sectional survey of third through fifth grade students attending schools in New York City in nine schools providing Breakfast in the Classroom (BIC) programs and in seven geographically and demographically matched randomly selected control schools. All students in the BIC and control classrooms were surveyed on what and where they had eaten that morning (n=2,289) with a 98% response rate. Both control and intervention schools offered universal free breakfast. Students in BIC classrooms were significantly less likely to report having skipped breakfast (8.7%) compared to students in the control classrooms (15.0%). While BIC students were more likely to report eating in more than one location (e.g. home and in the classroom) than control students, there were no significant differences between the two groups in consumption of fruits and vegetables, candy, doughnuts, chips, or sugary drinks. BIC students ate an estimated 95 calories more in the morning compared to control students. Among students offered BIC, those actually eating BIC consumed an estimated 151 more calories than students in control schools. The authors indicate that while the data support that BIC decreased the number of students skipping breakfast, it was also associated with an increase in caloric consumption that may be above what is needed. They also note that because the BIC programs were relatively new it is possible that as they are normalized students will be less likely to eat breakfast at more than one location. 14. Larson N., Wang Q., Grannon K., et al. A Low-Cost, Grab-and-Go Breakfast Intervention for Rural High School Students: Changes in School Breakfast Program Participation Among At-Risk Students in Minnesota. J Nutr Educ Behav. 2017. Larson, et. al. (2017) evaluated the impact of a Grab-And-Go program on School Breakfast Program participation at eight high schools in Minnesota. Data were based on outcomes from Project BreakFAST, a group-randomized trial to increase School Breakfast Participation in 13 January 2018- Health Impact Review of SB 6003 Minnesota high schools. For this study, 16 schools were selected and randomized to implement the full Project BreakFAST intervention or just the Grab-And-Go component. The authors concluded that the Grab-And-Go component alone increased participation by 13.9% to 30.7% among students eligible for free or reduced-price meals, and by 4.3% to 17.2% for non-elgibile students. Overall, participation in the School Breakfast Program due to Grab-And-Go increased across all student groups, regardless of eligiblity for free and reduced-price meals, race, or ethnicity. 15. Corcoran Sean P., Elbel Brian, Schwartz Amy Ellen. The Effect of Breakfast in the Classroom on Obesity and Academic Performance: Evidence from New York City. Journal of Policy Analysis and Management. 2016;35(3):509-532. Corcoran, et. al. (2016) looked at the impacts of a Breakfast in the Classroom program in New York City. They examined the impact of Breakfast in the Classroom on participation in the School Breakfast Program, academic performance, engagement in the classroom, and body mass index. They used the staggered implemenetation of Breakfast in the Classroom programs in New York City schools between 2007 and 2012 to complete a longitudinal study of students. Using data from the New York City Department of Education, they compared students in schools with and without Breakfast in the Classroom both before and after implementation. They found that School Breakfast Program participation rates increased between 4.4 and 30.2 percentage points after the implementation of Breakfast in the Classroom. Schools with a higher percentage of students eligible for free-or-reduced-price meals, students of color, and students with Limited English Proficiency were more likely to implement Breakfast in the Classroom programs. Therefore, increases in breakfast participation most directly impacted students of greatest disadvantage. They did not find an increase in BMI or the incidence of obesity among participating students. They also did not find an impact on school attendance. They also found a small, but not significant impact on reading and math acheivement. 16. Adolphus K., Lawton C. L., Dye L. The effects of breakfast on behavior and academic performance in children and adolescents. Frontiers in human neuroscience. 2013;7. Adolphus et al. conducted a systematic review of the literature to determine the effects of breakfast on school behavior and on academic performance. The authors included articles and reviews published between 1950 and 2013. Thirty-six studies met their inclusion criteria with 14 including behavior measures, 17 including academic performance measures, and five studies including both behavior and academic outcomes. Of the 19 studies which evaluated the effects of breakfast on behavior 11 found a positive effect on behavior such as being on-task in the classroom. This effect was found in both well-nourished and undernourished children as well as those with low socioeconomic position. Most of these studies are evaluations of School Breakfast Programs, which both lack scientific rigor and are looking at the connection between having a School Breakfast Program and behavior irrespective of how many students were actually participating in the program or eating breakfast. The authors indicate that in order for the School Breakfast Program to impact behavior barriers to accessing the program must be minimized. The review identified 21 studies evaluating the effect of habitual breakfast and School Breakfast Programs on academic performance. The authors found that increased frequency of habitual breakfast was consistently and positively associated with improved school performance such as school grades and achievement test scores. The evidence also supports that 14 January 2018- Health Impact Review of SB 6003 School Breakfast Programs have positive effects on school performance, particularly math grades and arithmetic scores. The evidence indicates that these positive effects may be more pronounced in schools with more undernourished children and worse academic outcomes— where studies carried out with these populations showed consistent positive effects of breakfast on school performance. The researchers highlight studies which have found School Breakfast Programs to be associated with increased attendance, punctuality, readiness to learn, decreased dropout rates, and better classroom behavior. 17. Hoyland A., Dye L., Lawton C. L. A systematic review of the effect of breakfast on the cognitive performance of children and adolescents. Nutrition research reviews. 2009;22(2):220-243. Hoyland et al. conducted a systematic review of the literature on the effects of breakfast on cognitive function in children and adolescents. The authors included studies published between 1950 and 2009. Forty-two articles consisting of 45 studies met their inclusion criteria. The authors provided a quality rating score for each study using pre-defined criteria. The researchers conclude that overall the evidence supports that eating breakfast has positive effects on cognitive performance compared to skipping breakfast. This finding was apparent in both acute studies and evaluations of longer-term School Breakfast Programs. They also found that these positive effects were more apparent in nutritionally vulnerable children, indicating that increasing access to breakfast may have a greater positive impact on low-income students. The authors also highlight evidence that School Breakfast Programs are associated with improved attendance and indicate that this may at least partially account for the benefits observed as part of the breakfast programs. They cite studies which found that breakfast was associated with improvements in, for example, memory, fewer errors on attention tasks, achievement tests, math grades, and reading scores. Although there was not complete consensus among the 45 studies, and some publications found no association between breakfast and some measure of cognition, six of the seven highest quality studies (receiving at least a 16 out of 18 possible points for quality), found a positive association between breakfast and at least one measure of cognitive performance. One of the authors of this study was supported by a grant from Kellogg Company UK, which could introduce a conflict of interest. The authors also note that many of the "studies reviewed were sponsored in whole or in part by industry." 18. Mesas A. E., Muñoz-Pareja M., López-García E., et al. Selected eating behaviours and excess body weight: a systematic review. Obesity reviews : an official journal of the International Association for the Study of Obesity. 2012;13(2):106-135. Mesas et al. conducted a systematic review of the literature on the connection between skipping breakfast and obesity. They included studies in English, Spanish, and Portuguese that had been published through the end of 2010. The authors excluded studies that only included subjects with excess weight. This exclusion may mask any effects primarily or exclusively impacting individuals with high BMI. The authors identified 63 cross sectional and 10 longitudinal studies that fit their inclusion criteria. Of the publications focusing on children and/or adolescents 35 of the 48 cross sectional studies found that skipping breakfast was associated with overweight or obesity. This association between skipping breakfast and obesity was observed even by studies that controlled for confounding factors. Seven of the longitudinal studies focused on children and/or adolescents and these found conflicting results. In three of these seven studies skipping breakfast was associated with excess weight. The cross-sectional studies on adults were 15 January 2018- Health Impact Review of SB 6003 conflicting, though a slightly higher percentage of the studies found that skipping breakfast was associated with increased BMI. However, the two longitudinal studies in adults that controlled for confounders (higher quality studies) found that skipping breakfast was associated with an increase in BMI of at least 5% after one year of follow-up. 19. O'Neil C. E., Nicklas T. A., Fulgoni V. L., et al. Nutrient Intake, Diet Quality, and Weight/Adiposity Parameters in Breakfast Patterns Compared with No Breakfast in Adults: National Health and Nutrition Examination Survey 2001-2008. Journal of the Academy of Nutrition and Dietetics. 2014;114(12):S27-S43. O’Neil et al. cite eleven studies in their introduction which indicate that breakfast consumption among adults is associated with improved daily nutrient intake, food group selection, dietary adequacy, diet quality, and intake of micronutrients. The authors also cite evidence that breakfast consumption is associated with positive measures of BMI and cardiovascular risk factors but note that the literature addressing these measures has been inconsistent. O’Neil et al. analyzed 2001-2008 National Health and Nutrition Examination Survey data for respondents 19 years of age or older (N=18,988). They found eleven different meal patterns among respondents. The data indicate that individuals who ate quality breakfasts (e.g. grains, 100% fruit juice, lower-fat milk, whole fruit) had higher daily intakes of nutrients and lower BMI and waist circumference than breakfast skippers. Note that individuals with meal patterns that included, for example, doughnuts, muffins, etc., did not have higher nutrient intake or lower BMI than breakfast skippers. O’Neil was a member of the Kellogg’s Breakfast Council at the time this article was written, and publication of this article was supported by an unrestricted education grant from the Kellogg Company, both of which may introduce a conflict of interest. 20. Smith K. J., Gall S. L., McNaughton S. A., et al. Skipping breakfast: longitudinal associations with cardiometabolic risk factors in the Childhood Determinants of Adult Health Study. The American journal of clinical nutrition. 2010;92(6):1316-1325. Smith et al. cite evidence in their introduction that skipping breakfast is associated with a poorer overall diet; lower physical activity levels; higher intakes of fat, cholesterol, and energy; and lower intakes of fiber, vitamins and minerals. The authors conducted a longitudinal cohort study in Australia in order to determine the effects of skipping breakfast in both childhood and adulthood. The national cohort of 9-15 year olds were first surveyed in 1985 (67.5% response rate). Between 2004 and 2006, participants (n=2,248) filled out a questionnaire and just over 1,720 participants had their diet quality assessed, their waists measured, and their fasting blood drawn by researchers. The data indicate that those who skipped breakfast as a child and as an adult had the least healthy eating habits. The authors also found that, after controlling for lifestyle factors, age, sex, and sociodemographic factors, participants who skipped breakfast both as a child and as an adult had significantly larger waist circumferences, higher fasting insulin, higher total cholesterol, and higher low-density lipoproteins (LDL or ‘bad’) cholesterol than participants who ate breakfast both as a child and in adulthood. Lifetime breakfast skippers had an average waist circumference of 4.63cm larger than their counterparts. This study was partially funded by an unrestricted grant from the Kellogg Company and support from the Kellogg’s Corporate Citizenship Fund which has potential to introduce a conflict of interest. 21. VanEenwyk J. Health of Washington State Report: Socioeconomic Position in Washington. Washington State Department of Health; 2014. 16 January 2018- Health Impact Review of SB 6003 VanEenwyk presents data about socioeconomic position in Washington State including differences within the state as well as statewide differences compared to national data. Data indicate that compared to the United States as a whole, fewer Washington residents are living in poverty and a higher percentage of residents ages 25 and older have college degrees. However, these economic resources are not evenly distributed among all Washington residents. Females in Washington were more likely to be living in poverty than males and were also more likely to have lower wages. Further, American Indian and Alaska Native, Hispanic, and black residents had higher percentages of living in poverty and lower median household incomes compared to other groups. Data also indicated that counties in eastern Washington were more likely to have high poverty rates and high rates of unemployment than counties in western Washington. 22. Lucio R, Hunt E, M Bornovalova. Identifying the necessary and sufficient number of risk factors for predicting academic failure. Developmental psychology. 2012;48(2):422-428. Lucio et al. analyzed data from the Educational Longitudinal Study: 2002 which includes a national sample of 14,796 students. The authors used a 5-step process to identify which factors contribute to academic ‘failure’—a grade point average (GPA) of less than 2.0 which is the minimum GPA needed to graduate from high school. They found that a number of academic outcomes impact a student’s GPA and therefore their ability to attain a high school diploma. Many of these are academic outcomes that other research has found to be impacted by skipping breakfast such as academic engagement, grade retention, and behavior among students. The authors also found that the odds of passing decreased with each additional risk factor: “For each risk factor that is added, there is a 47% increased likelihood of failing.” 23. Melby J. N., Conger R. D., Fang S. A., et al. Adolescent family experiences and educational attainment during early adulthood. Developmental psychology. 2008;44(6):1519-1536. Melby et al. analyzed data from a longitudinal study of two-biological-parent intact families in Iowa. They had a sample size of 451 families. The researchers conducted modeling to determine what factors impact educational attainment and found level of academic engagement was strongly correlated with later educational attainment. 24. Ou Suh-Ruu, Reynolds Arthur J. Predictors of educational attainment in the Chicago Longitudinal Study. School Psychology Quarterly. 2008;23(2):199-229. Ou and Reynolds analyzed data from the Chicago Longitudinal Study, using a sample size of 1,286 youth in order to investigate predictors of high school completion and total educational attainment. They found that, among other factors, school absences, grade retention, and youth’s educational expectations all influenced educational attainment. 25. Winding T. N., Nohr E. A., Labriola M., et al. Personal predictors of educational attainment after compulsory school: influence of measures of vulnerability, health, and school performance. Scandinavian journal of public health. 2013;41(1):92-101. Winding et al. analyzed data from a 2004 questionnaire completed by a cohort of adolescents born in 1989 (n=3053) in Denmark (83% response rate) and linked 2010 educational attainment data from Statistics Denmark. This allowed for a follow-up of 6.5 years. The authors found that 17 January 2018- Health Impact Review of SB 6003 low grades during primary school was predictive of not having completed a secondary education by age 20/21 (odds ratios between 1.7 and 2.5). For students with low math grades this association was even stronger. The authors cite two additional studies which have also found an association between school performance and later educational attainment. 26. Health of Washington State: Mental Health. Washington State Department of Health; 2008. Washington Behavioral Risk Factor Surveillance System (BRFSS) data from 2004-2006 indicate that American Indians and Alaska Natives and non-Hispanic black individuals reported significantly higher rates of poor mental health compared to other groups. These relationships persisted after adjusting for additional factors such as age, income, and education. Washington BRFSS data also show an association between lower annual household income and poor mental health, a relationship that was also shown with education. It is well understood that mental health is also closely related to other areas such as employment opportunities, physical health, substance abuse. This report also highlights a Washington state study from 2002 that reveal that 16% of individuals in the state who were receiving publicly funded mental health services had at least one felony conviction, a rate over twice that of the general population. 27. Centers for Disease Control and Prevention. Behavioral Risk Factor Surveillance System Prevalence And Trends Data: Washington-2014. 2014; Available at: Accessed August 16, 2016. Behavioral Risk Factor Surveillance System (BRFSS) 2014 data from Washington state show significant correlations between lower income and a number of health indicators including: worse overall self-reported health, depression, asthma, arthritis, stroke, oral health, tobacco use, women's health indicators, health screening rates, physical activity, and diabetes. Data also show that as educational attainment increases income level also increases. 28. Christensen Trevor, Weisser Justin. Health of Washington State Report: Tobacco Use. Washington State Department of Health; 2015. Christensen et al. report Washington state Behavioral Risk Factor Surveillance System (BRFSS) data from 2012 to 2014 indicate that prevalence of smoking decreases as income and levels of education increase. Further, American Indians and Alaska Natives (AI/AN) and Native Hawaiian/Other Pacific Islander populations have significantly higher smoking rates than white, black, Hispanic, and Asian populations. 29. Kandel Denise B., Griesler Pamela C., Schaffran Christine. Educational attainment and smoking among women: Risk factors and consequences for offspring. Drug and Alcohol Dependence. 2009;104:S24-S33. Researchers examined United States data from four national data sets and found that, among women, lower levels of education are associated with greater risk of being a current smoker, smoking daily, smoking heavily, being nicotine dependent, starting to smoke at an early age, having higher levels of circulating cotinine (a metabolite of nicotine) per cigarettes smoked, and continuing to smoke in pregnancy. In addition, lower levels of maternal education were linked to increased risk of antisocial behavior among offspring. 18 January 2018- Health Impact Review of SB 6003 30. Kemple Angela. Health of Washington State Report: Diabetes. Washington State Department of Health; 2016. Kemple presents data from Washington regarding diabetes in the state. Washington data from the Behavioral Risk Factor Surveillance System (BRFSS) from 2012-2014 show that among adults, the percentage of persons with diabetes increased as household income decreased. This relationship was also true for education. Further, BRFSS data also show that age-adjusted diabetes prevalence is highest among those who are Hispanic, American Indian/Alaska Native, and black. 31. McCarty C. A., Mason W. A., Kosterman R., et al. Adolescent school failure predicts later depression among girls. Journal of Adolescent Health. 2008;43(2):180-187. McCarty et al. conducted a prospective longitudinal cohort study with a sample of 808 youth followed from ages 10 to 21. The researchers discovered that adolescent school ‘failure’ (meaning being suspended, expelled, or dropping out of high school early) predisposed girls to depression in early adulthood. 32. McLaren L. Socioeconomic status and obesity. Epidemiologic reviews. 2007;29:29-48. McLaren et al. conducted a meta-analysis exploring the relationship between obesity and SES among adults. A total of 333 studies published internationally met the inclusion criteria. In highly developed countries, the majority of the studies found higher body weights among women with lower education attainment. Nearly 50% of the studies in highly developed countries found the same relationship for men. 33. Mersky JP, AJ Reynolds. Educational success and adult health: Findings from the Chicago longitudinal study. Prevention Science. 2009;10(2):175-195. Mersky and Reynolds analyzed data from a Chicago prospective cohort study that followed 1,539 individuals. Results indicate that high school completion was significantly and inversely associated with tobacco smoking, frequent substance use, depression, and no health insurance coverage. In addition, middle school reading performance was inversely related to depression and student’s expectation to attend college was negatively associated with frequent drug use. 34. Mezuk B, Eaton WW, Golden SH, et al. The influence of educational attainment on depression and risk of type 2 diabetes. American journal of public health. 2011;98(8):1480. Researchers analyzed adult survey data collected in the Baltimore Epidemiological Catchment Area and then conducted follow-up interviews with the survey cohort. Mezuk et al. found a statistically significant association between type 2 diabetes and lower educational attainment. In addition, the data indicate that depression was associated with type 2 diabetes, but each year of education attained decreased the risk of type 2 diabetes for those experiencing depression. 35. Poel A. Health of Washington State Report: Mortality and Life Expectancy. Data Update 2015. Washington State Department of Health; 2015. 19 January 2018- Health Impact Review of SB 6003 Poel presents Washington state data on mortality and life expectancy. The data show that age-adjusted death rates were higher in Washington census tracks with higher poverty rates. The state data also show that American Indian/Alaska Natives, Native Hawaiian/Other Pacific Islanders, and black residents had the highest age-adjusted death rate and shortest life expectancy at birth compared to other groups in the state. 36. Serafin M. Health of Washington State Report: Self-reported Health Status. Data Update 2016. Washington State Department of Health; 2016. Serafin presents data from Washington state on self-reported health status. The data show that after accounting for age, education, race and ethnicity, household income was a strong predictor of self-reported health status. Health status varied by race and ethnicity, with close to 35% of Hispanics, 30% of American Indian/Alaska Natives, and 20% of Native Hawaiian/Other Pacific Islanders reporting fair or poor health. Washington Behavioral Risk Factor Surveillance System (BRFSS) data from 2012-2014 also show that education was a strong predictor of self-reported fair or poor health after adjusting for age. 37. Skodova Z., Nagyova I., van Dijk J. P., et al. Socioeconomic differences in psychosocial factors contributing to coronary heart disease: a review. Journal of clinical psychology in medical settings. 2008;15(3):204-213. Skodova et al. conducted a meta-analysis of the literature addressing the relationships between SES, coronary heart disease (CHD), and psychosocial factors contributing to coronary heart disease. Researchers identified 12 studies that met their inclusion criteria. They found that higher levels of education are associated with lower rates of CHD, and that decreasing education is associated with factors that are linked to CHD such as depression, anxiety, hostility, and a lack of social supports. 38. Steptoe A., Hamer M., Butcher L., et al. Educational attainment but not measures of current socioeconomic circumstances are associated with leukocyte telomere length in healthy older men and women. Brain, behavior, and immunity. 2011;25(7):1292-1298. Steptoe et al. analyzed data collected from 543 male and female London-based civil servants of white European origin. All participants were between the ages of 53 and 76 and healthy. Researchers looked at blood samples to determine telomere length and telomerase activity. Telomere shortening is associated with aging. Short telomeres are also associated with increased risk of premature heart attack and mortality. Researchers found that lower educational attainment was associated with shorter telomere length after controlling for biological and behavioral covariates. This association remained significant even after adjusting for current SES. Researchers speculated that low educational attainment may be an indicator of long-term lower SES, and may be associated with accumulated stress resulting in telomere shortening. They also postulate that education may promote problem-solving skills leading to reduced responses to stress, thereby impacting aging. 39. Bureau of Labor Statistics website. Employment projections: Earnings and unemployment rates by educational attainment. Last Updated March 15, 2016; Available at: Accessed November 1, 2016. 20 January 2018- Health Impact Review of SB 6003 National data from 2015 indicate that as educational attainment increases median weekly earnings also increase and unemployment rates decrease. 40. Ellings Amy. Health of Washington State Report: Obesity and Overweight. Washington State Department of Health; 2015. Ellings reports Washington state Behavioral Risk Factor Surveillance System (BRFSS) data from 2002-2014, which shows that obesity rates are the highest among low income families and that as income increases, rates of obesity decrease. Further, individuals that graduated college or attended some college had lower rates of obesity than those who had a high school education or less. Black, American Indian and Alaska Native, and Hispanic Washington residents had higher rates of obesity even after accounting for gender, income, education, and age. 41. Paul Karsten I., Moser Klaus. Unemployment impairs mental health: Meta-analyses. Journal of Vocational Behavior. 2009;74(3):264-282. Paul et al. conducted a meta-analysis of 237 cross-sectional and 87 longitudinal studies that examined the relationship between mental health and unemployment. The meta-analysis of cross-sectional data revealed that unemployed persons showed significantly more symptoms of distress and impaired well-being than did employed persons. The meta-analyses of longitudinal studies and natural experiments supported the concept that unemployment is not only correlated to distress but also causes it. 42. Ponnet K. Financial stress, parent functioning and adolescent problem behavior: an actor-partner interdependence approach to family stress processes in low-, middle-, and high-income families. Journal of youth and adolescence. 2014;43(10):1752-1769. Ponnet cites extensive evidence on the relationship between financial hardship and emotional problems among youth and adults, family conflict, problem behavior among adolescents, and psychological distress. The author analyzed data from a subsample of two-parent families with children between 11 and 17 years of age from the Relationship between Mothers, Fathers and Children study drawn from the Dutch-speaking part of Belgium (n= 1,596 individuals from 798 families). Analysis showed that parents in low-income groups had significantly more financial stress than those in middle-income and high-income groups. The author found that the association between financial stress and problem behavior in adolescents is mediated by depressive symptoms, interparental conflict, and positive parenting. They also found that financial stress had more detrimental impacts on depressive feelings for mothers with low incomes than for those with higher incomes. 43. Prause J., Dooley D., Huh J. Income volatility and psychological depression. American journal of community psychology. 2009;43(1-2):57-70. Prause et al. analyzed a sample (n = 4,493) from the National Longitudinal Survey of Youth. Researchers found that income volatility was significantly associated with depression; and downward volatility (frequent losses in income) was significantly associated with depression even after controlling for baseline depression. High income appeared to act as a buffer, so those with lower incomes were more vulnerable to the adverse effects of downward volatility. 21 January 2018- Health Impact Review of SB 6003 44. Spencer N., Thanh T. M., Louise S. Low income/socio-economic status in early childhood and physical health in later childhood/adolescence: a systematic review. Maternal and child health journal. 2013;17(3):424-431. Spencer et al. conducted a meta-analysis of studies examining the relationship between low socioeconomic status in the first five years of life and physical health outcomes in later childhood and adolescence. Nine studies met the researchers’ strict inclusion criteria. The studies indicated significant associations between early childhood low-income status and a number of adverse health outcomes including: activity-limiting illness, parent-reported poor health status, acute and recurrent infections, increasing body mass index (BMI), dental caries, and higher rates of hospitalization. 45. Subramanyam M., Kawachi I., Berkman L., et al. Relative deprivation in income and self-rated health in the United States. Social science & medicine. 2009;69(3):327-334. Subramanyam et al. analyzed data from the 2002, 2004, and 2006 Current Population Surveys conducted by the United States Census Bureau. Researchers found that individuals from the lowest income category were over five times more likely to report being in poor health than participants from the highest income category. In addition, they found that relative deprivation (the differences in incomes between an individual and others who have higher incomes than that individual [one measure of income inequality]) appeared to explain a large part of this association. 46. Child Weight and Physical Activity. Washington State Department of Health; 2013. The authors present Washington state data on child weight and physical activity. The data show that in 2012, around 10% of Washington students in grades 8, 10, and 12 were obese and another 13-14% were overweight. Among 10th grade students, American Indian/Alaska Natives, Blacks, Hispanics, and Pacific Islanders were more likely than their white counterparts to be overweight or obese. Nationally, the authors indicate that the percentage of children and adolescents who were defined as overweight has doubled since the early 1970's and in 2012, around 42% of Washington students in grades 8, 10, and 12 reported that they were trying to lose weight. 47. Food Research & Action Center. Relationship Between Poverty and Obesity. 2015; Available at: Accessed November 14, 2016. Overview of studies from the United States that present research on the relationship between obesity and poverty. Provides relevant study conclusions for both adult and child populations. 48. Ebbeling Cara B., Pawlak Dorota B., Ludwig David S. Childhood obesity: public-health crisis, common sense cure. The Lancet. 2002;360(9331):473-482. Ebbeling et al. present a global literature review on the scope of the childhood obesity problem and developments in the establishment of a cause, prevention, and treatment for obesity. Rates of childhood obesity have grown across the globe, with a nearly 2 to 3 fold increase in the rates in the United States over the last 25 years. Most relevant to this review, the authors examined extensive literature that demonstrates the association between childhood obesity and hypertension, high blood lipids, chronic inflammation, increased blood clotting tendency, 22 January 2018- Health Impact Review of SB 6003 endothelial dysfunction, hyperinsulinaemia, sleep apnea, asthma, and type 2 diabetes. Further, type 2 diabetes presents additional risks such as heart disease, stroke, kidney failure, and blindness. In addition to the physical health risks from childhood obesity, many studies have indicated substantial psychosocial consequences such as negative self-image and stereotyping. The authors note that Black and Hispanic youth in the United States are at a greater risk for type 2 diabetes and cardiovascular disease than their white counterparts. 49. Jay Shubrook Jr. Childhood Obesity and the Risk of Diabetes in Minority Populations American Osteopathic Association Health Watch; 2011. Shubrook presents data on childhood obesity and diabetes among children in the United States. Data shows that childhood obesity increases the risk of adult obesity with estimates indicating that obese children as young as age 6 have a 50% chance of being obese as an adult. Further data indicates that childhood obesity increases the risk of coronary heart disease and mortality as an adult. Data from the National Health and Nutrition Examination Survey (NHANES) show that Hispanic and non-Hispanic black children have the highest rates for childhood obesity in the United States. There also appears to be a disproportionately higher incidence of type 2 diabetes among minority children with the highest incidence found among Navajo Indian females (38.42 cases per 100,000 people compared to 3.7 cases per 100,000 white females). Shubrook concludes that the burden of obesity is of great concern, particularly among minority populations in the U.S. and this increased risk needs to be acknowledged in order to address the problem effectively. 50. Ritchie L. D., Rosen N. J., Fenton K., et al. School Breakfast Policy Is Associated with Dietary Intake of Fourth- and Fifth-Grade Students. J Acad Nutr Diet. 2016;116(3):449-457. In response to recent concerns that Breakfast in the Classroom programs may increase the risk of excessive energy intake and weight gain, Ritchie et.al. (2016) looked at energy intake and diet quality for fourth-grade students in schools in California with alternative breakfast program policies. They compared energy intake and diet quality of fourth-grade students that do not eat breakfast, that eat breakfast either at home or at school, and that eat two breakfasts (at home and at school). Of the 43 schools they evaluated, 20 served breakfast in the cafeteria before school, 17 has Breakfast in the Classroom programs, and 6 had "Second Chance" breakfast programs which offered breakfast in the cafeteria during the first recess break. The authors concluded that Breakfast in the Classroom programs were associated with more students eating breakfast and more students eating two breakfasts (one at home and one at school). However, these students did not have higher mean energy intakes or higher daily energy intakes than other students. In addition, students participating in the Breakfast in the Classroom program had an overall higher dietary quality than students in other groups. Overall, the authors concluded that there was no evidience to support the concern that students participating in Breakfast in the Classroom consume excess calories. 51. Wang S., Schwartz M. B., Shebl F. M., et al. School breakfast and body mass index: a longitudinal observational study of middle school students. Pediatr Obes. 2017;12(3):213-220. 23 January 2018- Health Impact Review of SB 6003 Wang et.al. (2017) randomly selected twelve out of 27 schools from a medium-sized urban school district. They surveyed students enrolled in fifth grade during the 2011-2012 school year, and followed these students to seventh grade . They collected information about breakfast location and consumption, height and weight measurements, and demographics. Students were grouped into six categories related to breakfast location and consumption: frequent skippers, inconsistent school eaters, inconsistent home eaters, regular home eaters, regular school eaters, and double breakfast eaters (one at home, one at school). The authors found that the odds of being overweight or obese was significantly higher for students that skipped breakfast than for students that ate double breakfasts. Simiarly, weight changes over time for students that ate double breakfasts were the same as changes for students in other breakfast consumption categories. Overall, the study concluded that there is no evidence to support the concern that students who eat two breakfasts (one at home and one at school) are at increased risk for excessive weight gain. |
14847 | https://pmc.ncbi.nlm.nih.gov/articles/PMC4262697/ | Retinal Detachment Associated with AIDS-Related Cytomegalovirus Retinitis: Risk Factors in a Resource-Limited Setting - PMC
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Am J Ophthalmol
. Author manuscript; available in PMC: 2016 Jan 1.
Published in final edited form as: Am J Ophthalmol. 2014 Oct 22;159(1):185–192. doi: 10.1016/j.ajo.2014.10.014
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Retinal Detachment Associated with AIDS-Related Cytomegalovirus Retinitis: Risk Factors in a Resource-Limited Setting
Michael Yen
Michael Yen
1 Francis I. Proctor Foundation, University of California, San Francisco, San Francisco, California
2 Icahn School of Medicine at Mount Sinai, New York, New York
Find articles by Michael Yen
1,2, Jenny Chen
Jenny Chen
1 Francis I. Proctor Foundation, University of California, San Francisco, San Francisco, California
3 Ocular Inflammatory Disease Center, Jules Stein Eye Institute and Department of Ophthalmology, David Geffen School of Medicine at UCLA, Los Angeles, CA
Find articles by Jenny Chen
1,3, Somsanguan Ausayakhun
Somsanguan Ausayakhun
4 Department of Ophthalmology, Faculty of Medicine, Chiang Mai University, Chiang Mai, Thailand
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4, Paradee Kunavisarut
Paradee Kunavisarut
4 Department of Ophthalmology, Faculty of Medicine, Chiang Mai University, Chiang Mai, Thailand
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4, Pornpattana Vichitvejpaisal
Pornpattana Vichitvejpaisal
4 Department of Ophthalmology, Faculty of Medicine, Chiang Mai University, Chiang Mai, Thailand
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4, Sakarin Ausayakhun
Sakarin Ausayakhun
4 Department of Ophthalmology, Faculty of Medicine, Chiang Mai University, Chiang Mai, Thailand
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4, Choeng Jirawison
Choeng Jirawison
5 Department of Ophthalmology, Nakornping Hospital, Chiang Mai, Thailand
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5, Jessica Shantha
Jessica Shantha
1 Francis I. Proctor Foundation, University of California, San Francisco, San Francisco, California
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1, Gary N Holland
Gary N Holland
3 Ocular Inflammatory Disease Center, Jules Stein Eye Institute and Department of Ophthalmology, David Geffen School of Medicine at UCLA, Los Angeles, CA
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3, David Heiden
David Heiden
6 Department of Ophthalmology and Pacific Vision Foundation, California Pacific Medical Center, San Francisco, CA
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6, Todd P Margolis
Todd P Margolis
1 Francis I. Proctor Foundation, University of California, San Francisco, San Francisco, California
7 Department of Ophthalmology, University of California, San Francisco, San Francisco, California
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1,7, Jeremy D Keenan
Jeremy D Keenan
1 Francis I. Proctor Foundation, University of California, San Francisco, San Francisco, California
7 Department of Ophthalmology, University of California, San Francisco, San Francisco, California
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1,7
Author information
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Copyright and License information
1 Francis I. Proctor Foundation, University of California, San Francisco, San Francisco, California
2 Icahn School of Medicine at Mount Sinai, New York, New York
3 Ocular Inflammatory Disease Center, Jules Stein Eye Institute and Department of Ophthalmology, David Geffen School of Medicine at UCLA, Los Angeles, CA
4 Department of Ophthalmology, Faculty of Medicine, Chiang Mai University, Chiang Mai, Thailand
5 Department of Ophthalmology, Nakornping Hospital, Chiang Mai, Thailand
6 Department of Ophthalmology and Pacific Vision Foundation, California Pacific Medical Center, San Francisco, CA
7 Department of Ophthalmology, University of California, San Francisco, San Francisco, California
Todd Margolis’ current affiliation is: Department of Ophthalmology and Visual Sciences, Washington University School of Medicine, St. Louis, Missouri
✉
Corresponding Author: Jeremy D Keenan, MD MPH Medical Sciences Building S334, Box 0412 513 Parnassus Ave San Francisco, CA 94143-0412 jeremy.keenan@ucsf.edu Telephone: (415) 476-1442 Fax: (415) 476-0527
Issue date 2015 Jan.
PMC Copyright notice
PMCID: PMC4262697 NIHMSID: NIHMS636827 PMID: 25448999
The publisher's version of this article is available at Am J Ophthalmol
Abstract
Purpose
To determine risk factors predictive of retinal detachment in patients with cytomegalovirus (CMV) retinitis in a setting with limited access to ophthalmic care.
Design
Case-control study.
Methods
Sixty-four patients with CMV retinitis and retinal detachment were identified from the Ocular Infectious Diseases and Retina Clinics at Chiang Mai University. Three control patients with CMV retinitis but no retinal detachment were selected for each case, matched by calendar date. The medical records of each patient were reviewed, with patient-level and eye-level features recorded for the clinic visit used to match cases and controls, and also for the initial clinic visit at which CMV retinitis was diagnosed. Risk factors for retinal detachment were assessed separately for each of these time points using multivariate conditional logistic regression models that included 1 eye from each patient.
Results
Patients with a retinal detachment were more likely than controls to have low visual acuity (OR, 1.24 per line of worse vision on the logMAR scale; 95%CI, 1.16-1.33) and bilateral disease (OR, 2.12; 95%CI, 0.92-4.90). Features present at the time of the initial diagnosis of CMV retinitis that predicted subsequent retinal detachment included bilateral disease (OR, 2.68; 95%CI, 1.18-6.08) and lesion size (OR, 2.64 per 10% increase in lesion size; 95%CI, 1.41-4.94).
Conclusion
Bilateral CMV retinitis and larger lesion sizes, each of which is a marker of advanced disease, were associated with subsequent retinal detachment. Earlier detection and treatment may reduce the likelihood that patients with CMV retinitis develop a retinal detachment.
Keywords: CMV Retinitis, HIV/AIDS, International Ophthalmology, Retinal Detachment
Introduction
Cytomegalovirus (CMV) retinitis is the most common ocular opportunistic infection and the leading cause of blindness among patients with the acquired immunodeficiency syndrome (AIDS).1, 2 Although it is now uncommon in the United States with the widespread use of highly active antiretroviral therapy (HAART),3-6 CMV retinitis continues to be the most common ocular opportunistic infections in AIDS patients and an important cause of blindness in many developing countries, especially in Asia.7-13 For example, at a tertiary ophthalmology center in Thailand, CMV retinitis affected 33% of AIDS patients and was the second leading cause of blindness among all patients seen at the clinic.9, 11 Of the major reasons for visual acuity loss associated with CMV retinitis, retinal detachments are perhaps the most devastating.14 Patients often have poor long-term visual outcomes even with surgical intervention, leading to permanent loss of vision.15-18 Because it can be a devastating complication of CMV retinitis, it is critical to understand the clinical characteristics that are associated with retinal detachment.
Risk factors for retinal detachment have been described for CMV retinitis patients in the United States.19-23 However, little information exists on CMV-related retinal detachment in the developing world, where the vast majority of CMV retinitis currently occurs and where retinal detachment is a common complication.13, 17 The circumstances in such resource-limited settings are very different from those seen in the United States. For example, patients at risk for CMV retinitis in Southeast Asia are not routinely screened, and the disease is often at an advanced stage at the time of initial diagnosis.10 Patients are often diagnosed several months after starting HAART, raising the possibility of preexisting CMV retinitis that becomes symptomatic due to intra-ocular inflammation from immune recovery. Furthermore, CMV retinitis in the developing world is treated almost exclusively with intravitreal ganciclovir injections, presenting additional risk for complications. Because the disease course and treatment regimen differ significantly in resource-limited settings, the risk factors for retinal detachment may be different than those found in the United States. In this study, we describe the results from a case-control study conducted in Thailand that investigated the risk factors for CMV retinitis-related retinal detachment.
Methods
Ethical approval was obtained from the Committee on Human Research at the University of California, San Francisco and the Chiang Mai University Faculty of Medicine Research Ethics Committee prior to conducting this case-control study. This study adhered to the tenets of the Declaration of Helsinki.
This study was conducted at the Ocular Infectious Diseases Clinic at Chiang Mai University, a tertiary referral ophthalmology clinic in northern Thailand. The clinic generally examines at-risk HIV patients with visual symptoms. We identified cases of CMV retinitis-related retinal detachment that were diagnosed between December 2004 and August 2010 by reviewing the medical records of all patients with CMV retinitis seen in the Ocular Infectious Diseases Clinic and all patients from the Retina Clinic logbook who had a procedure performed during this time. Cases were defined as any patient for whom retinal detachment was documented in the medical record. We selected 3 controls per case using incidence density sampling. Specifically, we reviewed the Ocular Infectious Diseases Clinic CMV retinitis logbook and enumerated all patients with a visit for CMV retinitis ± 3 months from the date the retinal detachment was initially diagnosed. We used a random number generator to select 3 patients without retinal detachment from this list to serve as controls. We sampled controls without replacement; each patient could serve as a control to only one case of retinal detachment. If a case had sequential bilateral retinal detachments, the date of the first retinal detachment was used to identify controls.
We collected eye-level and patient-level clinical and demographic information from the visit that the retinal detachment was first diagnosed for cases, and from the corresponding matched visit for controls. We also collected the same information from the initial clinic visit at which the diagnosis of CMV retinitis was initially made, in order to determine whether features that were present at baseline were predictive of future retinal detachment.
The medical record included detailed retinal drawings for all patients, drawn on a template that included the optic disc, fovea and vascular arcades, as well as the three zones traditionally used for CMV retinitis (zone 1 included the area within 3000 μm of the fovea or 1500 μm of the margin of the optic disc, zone 2 extended from zone 1 to the vortex veins, and zone 3 was the remaining retinal area that extended to the ora serrata).24 We noted the zone(s) with CMV retinitis and then grouped them based on the most anterior zone that was involved. In addition, we assessed 1) the area of retinitis as a proportion of the retinal surface area, 2) the most anterior extent of the retinitis, expressed as the proportion of the retinal radius as measured from the fovea to the ora serrata, and 3) the most posterior extent of the retinitis, also expressed as the proportion of the retinal radius. Measurements of the size and location of retinitis were performed by one of two graders using ImageJ software (Bethesda, MD),25 masked to whether the eye was a case or control. Lesion size measurements had good inter-rater reliability (intraclass correlation coefficient [ICC] between the two graders on a random set of 50 eyes was 0.88, 95%CI 0.81-0.94). Based on the charted retinal drawings it was often difficult to clearly distinguish detached retina from areas with CMV retinitis. Therefore, for this study we included lesion size and location only from eyes at the time of initial CMV retinitis diagnosis, before a retinal detachment had developed.
We constructed univariate and multivariate conditional logistic regression models to determine risk factors for retinal detachment, using the eye as the unit of analysis. Separate models were constructed for the matched visit (visit at which the retinal detachment was diagnosed or the time-matched visit for controls) and the visit at which CMV retinitis was first diagnosed. Because eyes from the same person are not independent, we selected only a single eye from each patient for analyses: for cases with sequential bilateral retinal detachments, we selected the first eye that developed a detachment; for cases with bilateral retinal detachments initially diagnosed at the same visit, one eye was randomly selected; and for controls with bilateral CMV retinitis, we randomly selected one of the eyes. Continuous predictors were tested for specification error (linktest in Stata) and transformations applied as needed. We explored the use of cubic splines for continuous variables but these did not significantly improve model fit and were therefore not used in the final model. All factors with P<0.05 in univariate models were included in the multivariate model, with a backwards stepwise selection algorithm employed until all covariates had a P<0.10 using the likelihood ratio test. All statistics were performed using Stata version 13 (Statacorp, College Station, Texas, USA).
Results
This study included 64 eyes from 64 patients with CMV retinitis-related retinal detachment and 192 matched control eyes from 192 patients with CMV retinitis but no retinal detachment. Greater than 90% of patients with a retinal detachment were on antiretroviral therapy, and the median CD4 count was 200 cells/μl (IQR 30 to 378) at the time of detachment (mean 226 cells/μl, 95%CI 172-279). Most retinal detachments occurred relatively soon after diagnosis of CMV retinitis (median 1.15 months, IQR 0 to 9.78; mean 8.2 months, 95%CI 4.5-11.9). Visual acuity was generally poor at the time of retinal detachment (median logMAR 1.8, IQR 1.0-1.8 [Snellen equivalent: hand motions, IQR 20/200 to Hand Motions]; mean logMAR 1.46, 95%CI 1.32 to 1.60).
Univariate analyses indicated that at the time the retinal detachment was diagnosed, cases with retinal detachment were more likely than controls to have CMV retinitis in the contralateral eye (67.2% vs. 52.6%; P=0.04), decreased visual acuity (mean logMAR 1.46 vs. 0.60 [Snellen equivalent 20/600 vs. 20/80]; P<0.001), and a shorter period of time since diagnosis of CMV retinitis (mean 8.2 months vs. 12.9 months; P=0.045; Table 1). Patients with retinal detachment were also more likely than controls to have received their initial CMV retinitis diagnosis at the time of the matched visit (35.9% vs. 14.1%; P<0.001), and therefore less likely to have started management for CMV retinitis before the time of detachment. In multivariate analysis, only decreased visual acuity was strongly associated with retinal detachment (OR 1.24 per line of logMAR acuity, 95%CI 1.16-1.33), although CMV retinitis in the contralateral eye was also included in the model (OR 2.12, 95%CI 0.92-4.90; Table 2).
Table 1.
Characteristics of patients with the acquired immunodeficiency syndrome (AIDS) at the time of diagnosis for cytomegalovirus retinitis and at the time of retinal detachment
| | Initial CMV Retinitis Visita | Retinal Detachment Visitb |
:---:
| | Mean (95%CI) or N(%) | Mean (95%CI) or N(%) |
| Characteristic | Cases (N=41) | Controls (N=123) | OR (95% CI) | Cases (N=64) | Controls (N=192) | OR (95% CI) |
| PATIENT CHARACTERISTICS | | | | | | |
| Age (years) | 37 (34-41) | 36 (31-40) | 1.02 (0.97-1.07) | 38 (34-42) | 38.5 (33-44) | 0.99 (0.96-1.03) |
| Female | 23 (56.1%) | 71 (57.7%) | 0.93 (0.46-1.92) | 30 (46.9%) | 109 (56.8%) | 0.68 (0.39-1.19) |
| CD4 count (cells/μm 3)c | 83.0 (37-129) | 78.2 (56-111) | 1.00 (0.98-1.03) | 226 (172-279) | 189 (163-215) | 1.01 (0.99-1.02) |
| Taking antiretroviral therapyd | 26 (92.9%) | 90 (91.8%) | 1.08 (0.16-7.55) | 46 (95.8%) | 163 (95.3%) | 1.24 (0.24-6.33) |
| CMV retinitis in contralateral eye | 26 (63.4%) | 49 (39.8%) | 2.74 (1.25-6.03) | 43 (67.2%) | 101 (52.6%) | 1.91 (1.03-3.57) |
| No prior CMV retinitis diagnosis | 41 (100%) | 123 (100%) 23 (35.9%) | 27 (14.1%) | 3.16 (1.67-6.01) |
| Months since CMV diagnosise - 8.2 (4.5-11.9) | 12.9 (10.3-15.6) | 0.98 (0.96-1.00) |
| EYE CHARACTERISTICS | | | | | | |
| Visual acuity (logMAR)f | 0.87 (0.68-1.06) | 0.72 (0.60-0.80) | 1.04 (0.98-1.10) | 1.46 (1.32-1.60) | 0.60 (0.50-0.70) | 1.24 (1.16-1.33) |
| Lesion active | 30 (73.2%) | 93 (75.6%) | 0.87 (0.38-2.00) | 18 (28.1%) | 52 (27.1%) | 1.06 (0.55-2.03) |
| Lesion size (% retinal surface)g | 12.8% (9.5-16.1%) | 8.2% (6.7-9.6%) | 2.67 (1.45-4.91) -
| Most posterior extent of retintiish | 28.3% (22.1-34.4%) | 27.9% (24.8-31.0%) | 1.01 (0.82-1.25) -
| Most anterior extent of retinitisi | 78.1% (72.9-83.3%) | 72.5% (69.4-75.5%) | 1.26 (0.99-1.60) -
| Number of lesions | 1.6 (1.3-1.9) | 1.3 (1.2-1.4) | 1.68 (1.03-2.74) -
| Vitreous haze present | 6 (14.6%) | 6 (4.9%) | 4.37 (1.06-18.11) | 4 (6.3%) | 3 (1.6%) | 4.00 (0.90-17.87) |
| Retinal hemorrhage present | 18 (43.9%) | 61 (49.6%) | 0.78 (0.37-1.64) | 13 (20.3%) | 43 (22.4%) | 0.89 (0.45-1.75) |
| Frosted branch angiitis present | 6 (14.6%) | 6 (4.9%) | 3.29 (0.99-10.92) | 2 (3.1%) | 1 (0.5%) | 6.00 (0.54-66.17) |
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Bold values indicate a likelihood ratio p-value <0.05. CMV= cytomegalovirus; CD4= cluster of differentiation 4; logMAR= logarithm of the minimum angle of resolution
a
Longitudinal analysis of the risk factors at the time of initial CMV retinitis diagnosis that predict subsequent retinal detachment; excludes matched groups in which the case had a retinal detachment at the time of initial CMV retinitis diagnosis.
b
Cross-sectional analysis of the characteristics present at the time of retinal detachment diagnosis and their association with retinal detachment; includes all matched groups.
c
Data not available for all patients. 32 (78.0%) cases and 108 (87.8%) controls had data at the initial visit; 59 (92.2%) cases and 181 (94.3%) controls had data at the matched visit. Odds ratio indicates odds of retinal detachment per 10 cells/μm 3.
d
Data not available for all patients. 28 (68.3%) cases and 98 (79.7%) controls had data at the initial visit; 48 (75.0%) cases and 171 (89.1%) controls had data at the matched visit.
e
Time since diagnosis of CMV retinitis. Numbers in table include all cases and controls; if the 23 cases with a retinal detachment at the time of CMV retinitis diagnosis and their 69 corresponding controls are excluded, the mean time since CMV retinitis diagnosis was 12.8 months (95%CI 7.5 to 18.1) for cases and 13.4 (95%CI 9.9 to 16.9) for controls.
f
Odds ratio indicates odds of retinal detachment per line of logMAR visual acuity.
g
Percent of the total retinal area occupied by the lesion by assessment of retinal drawings; odds ratio indicates odds of retinal detachment per 10% total retinal surface area
h
Distance from the fovea to the most proximal lesion border, calculated as a percent of the distance from the fovea to the ora serrata; odds ratio indicates odds of retinal detachment per 10% increase in the distance from the fovea to the most posterior border.
i
Distance from the fovea to the most distal lesion border, calculated as a percent of the distance from the fovea to the ora serrata; odds ratio indicates odds of retinal detachment per 10% increase in the distance from the fovea to the most anterior border.
Table 2.
Risk factors for retinal detachment in patients with the acquired immunodeficiency syndrome (AIDS) and cytomegalovirus retinitis: multivariate analyses
| | Initial CMV Retinitis Visita | Retinal Detachment Visitb |
:---:
| Characteristic | OR (95% CI)c | OR (95% CI)c |
| Lesion Size, per 10% total retinal surface area | 2.64 (1.41-4.94)
| CMV retinitis in contralateral eye | 2.68 (1.18-6.08) | 2.12 (0.92-4.90) |
| Decreased visual acuity (per logMAR line) 1.24 (1.16-1.33) |
Open in a new tab
CMV= cytomegalovirus; logMAR= logarithm of the minimum angle of resolution
a
Longitudinal analysis of the risk factors present at the time of initial CMV retinitis diagnosis that are predictive of subsequent retinal detachment; excludes matched groups in which the case had a retinal detachment at the time of initial CMV retinitis diagnosis.
b
Cross-sectional analysis of the characteristics present at the time of retinal detachment diagnosis and their association with retinal detachment; includes all matched groups.
c
Initial multivariate models included factors with P<0.05 in univariate analyses (bold items in Table 1); the final multivariate models shown above were selected with a backwards stepwise algorithm.
In order to assess for risk factors that might predict a subsequent retinal detachment, we performed similar analyses using data from the initial visit at which CMV retinitis was diagnosed. For this analysis, we excluded the 23 cases that had a retinal detachment at the time of initial CMV retinitis diagnosis and their corresponding 69 controls, leaving 41 patients with retinal detachment and 123 controls. In univariate analyses, cases were more likely than controls to have a larger lesion size (mean 12.8% of the total retinal surface area was involved with retinitis, vs. 8.2%, P<0.001), more peripheral retinitis (the most anterior border of the lesion was on average 78.1% of the total distance from the fovea to the ora serrata, vs. 72.5%; P=0.049), a higher number of retinal lesions (mean 1.6 vs. 1.3; P=0.04), CMV retinitis in the contralateral eye (63.4% vs. 39.8%, P=0.008), and vitreous haze (14.6% vs. 4.9%; P=0.04; Table 1). In the multivariate analysis, only lesion size (OR 2.64 per 10% increase in lesion size measured as a percentage of the total retinal surface area, 95%CI 1.41-4.94) and retinitis in the contralateral eye (OR 2.68, 95%CI 1.18-6.08) remained significant (Table 2).
As noted in the Figure, cases with retinal detachment tended to have more anterior disease at the time of CMV retinitis diagnosis than did controls (46.3% of cases vs. 29.3% of controls had disease in zone 3; OR 2.01, 95%CI 0.99 to 4.09). However, we could not demonstrate an association between zone 3 disease and retinal detachment after adjusting for lesion size (OR 1.54, 95%CI 0.72 to 3.30).
Figure. Risk of retinal detachment in patients with the acquired immunodeficiency syndrome (AIDS) and cytomegalovirus retinitis, stratified by most anterior zone of retinal involvement.
Open in a new tab
Patients are grouped by the most anterior zone occupied by the lesion, with cases of retinal detachment (+RD) on the left and controls without retinal detachment (-RD) on the right. The zones represented by each pair of columns are shown in the key above the figure, with each colored box corresponding to the color of the bar.
Because a large number (35.9%) of cases presented with retinal detachment at the time of initial CMV retinitis diagnosis and were therefore ineligible for treatment before the retinal detachment occurred, we analyzed the effect of prior intravitreal injections only for the cases who did not have a retinal detachment at their initial visit (N=41) and their matched controls (N=123). At the time of retinal detachment or the matched control visit, there was no significant difference in the number of prior injections between cases (mean 6, 95%CI 4 to 9) and controls (mean 6, 95%CI 5 to 8; P=0.86). There was also no difference in the amount of time since the most recent injection (mean 30.4 weeks [95%CI 13.7 to 47.0] for cases, and 37.3 weeks [95%CI 23.7 to 50.9] for controls; P=0.57).
Discussion
Although most retinal detachments at this tertiary ophthalmology clinic in Thailand occurred approximately 1 month after diagnosis of CMV retinitis, 25% were diagnosed more than 9 months after the initial CMV retinitis diagnosis, suggesting that patients in this setting continued to be at risk for retinal detachments long after they had been diagnosed with and treated for CMV retinitis. CMV retinitis patients with a retinal detachment were more likely than controls to have bilateral disease and low visual acuity. At the time of initial CMV retinitis diagnosis, larger lesion size and bilateral disease were risk factors that significantly predicted a future retinal detachment. Neither the number of prior intravitreal injections nor the time since the most recent injection were associated with retinal detachment.
We found that development of a retinal detachment was strongly associated with large retinitis lesion size, a finding that has been described before.19-23 The lesion sizes reported here are somewhat smaller than those typically reported in previous studies. This discrepancy may be due to differences in the way lesion size was measured, with most prior studies using a categorical measurement from the clinical examination to estimate lesion size, whereas we used a continuous measure from tracings of retinal drawings. Our methodology is analogous to using retinal photographs to assess lesion size—a technique which has been shown to result in smaller lesion measurements compared with clinician assessment.26
Anterior lesions may predispose to retinal detachment because the anterior retina is thinner and more susceptible to tears, and also because it underlies the vitreous base, making it more vulnerable to the effects of vitreoretinal traction. Other studies have found zone 3 (i.e. more anterior) lesions to be significantly associated with retinal detachment.19, 20, 23 We expanded upon these findings by showing in univariate analyses that having a more anterior retinitis lesion at the time of initial CMV retinitis diagnosis was associated with subsequent retinal detachment. However, lesion location may be confounded by lesion size because large lesions are more likely to encompass more than one zone.27 The association between anterior lesion location and retinal detachment found in this study did not persist after adjusting for lesion size, suggesting that the size of a retinitis lesion is a stronger risk factor for retinal detachment than is anterior extent of the lesion, at least in this population.
In this study bilateral disease at the time of CMV retinitis diagnosis was a strong risk factor for future retinal detachment. Previous studies in the United States have not found retinitis in the contralateral eye to be an independent predictor of retinal detachment.19-23 In contrast, our study found that an eye was at increased risk for a retinal detachment if the contralateral eye also had CMV retinitis. In this setting with limited capacity for screening, bilateral disease is likely a marker of the duration of untreated CMV retinitis, as well as an indication of the amount of time that a patient has been severely immunocompromised.27 Although we did not find the time between diagnosis of CMV retinitis and retinal detachment to be predictive of retinal detachment in the multivariate analysis, this variable does not capture how long the patient had CMV retinitis prior to being diagnosed. In this Thai setting, patients likely develop CMV retinitis several months before they present to the ophthalmologist.10 Our finding that bilateral disease is a significant predictor of subsequent retinal detachment suggests that advanced disease due to delayed diagnosis plays a major role in the development of retinal detachment.
Intravitreal ganciclovir injections may prevent retinal detachments by limiting the progression of CMV retinitis.19 However, retinal detachment is also a known complication of intravitreal injections. 28, 29 It is difficult to determine whether intravitreal ganciclovir injections were a risk factor for retinal detachment in this population since this was the only treatment modality available to patients in the study. We did not find an association between the number of intravitreal injections and retinal detachment nor the time since the most recent injection, although this may be confounded by other factors such as the length of time in treatment. The overall effect is particularly difficult to determine given the retrospective design of this study, and could best be evaluated with a randomized controlled trial.
We found that patients had a higher CD4 count at the time of retinal detachment compared to the time of diagnosis of CMV retinitis, suggesting a possible role of inflammation due to immune reconstitution. However, when we compared cases to controls we did not find either CD4 count or HAART status to be a risk factor. Some studies carried out in the United States found that a low CD4 count was a risk factor for retinal detachment.21, 22 This discrepancy is likely due to differences in the patient population. Patients in the United States who developed CMV retinitis in the HAART era were often non-compliant, intolerant of, or failing HAART. 4, 30, 31 In contrast, almost all newly diagnosed CMV retinitis patients in our study clinic in Thailand started HAART within 2 months prior to the diagnosis of CMV retinitis.10 Thus, patients with CMV retinitis and a low CD4 count in our study population were likely in the early stages of immune reconstitution, in contrast to those in the studies carried out in the United States. Previous research has also identified HAART to be a protective factor against CMV retinitis-related retinal detachment.23 Because almost all cases and controls in our study were on three-drug antiretroviral therapy at the time of diagnosis, we were underpowered to detect a difference between those who were and were not receiving HAART. Nonetheless, HAART is clearly important for both the prevention and treatment of CMV retinitis and improving access to HAART in resource-poor settings should be a priority.5, 32-34
This study is novel in that it reports risk factors for CMV retinitis-related retinal detachment in a developing county setting in which almost all patients were treated with HAART and intravitreal ganciclovir injections. Previous studies have usually been limited to patients in the pre-HAART area treated systemically for CMV retinitis,19-21 although two studies did contain a small number of patients with retinal detachment who were treated with HAART.22, 23 In general, we found similar risk factors to those reported before, with the exception that anterior lesion location was less important than lesion size, and retinitis in the contralateral eye was found to be a strong risk factor for subsequent retinal detachment. Large retinitis lesion size and bilateral disease are likely signs of longstanding disease, suggesting that earlier detection of CMV retinitis could result in fewer retinal detachments and less blindness. Routine screening would result in earlier diagnosis and treatment, which could limit lesion growth and therefore reduce the chance of a retinal detachment.19, 29, 35 In addition, routine screening may detect patients who have signs that are concerning for a retinal detachment, such as a retinal tear or hole. Ophthalmologists could then pursue laser photocoagulation, which some hypothesize limits progression of a retinal detachment for lesions that are small and inactive.36-39 Adequately screening at-risk individuals can be challenging in resource-poor areas due to the relative lack of ophthalmologists.40, 41 Alternate methods could be explored to increase diagnostic capacity, including training primary care providers in indirect ophthalmoscopy42 and remote diagnosis via telemedicine.43-45
We acknowledge several limitations to this study. Besides the limitations inherent in all retrospective studies, this study was conducted at a tertiary center and is likely subject to referral bias. The case-control study design prevented us from determining the prevalence or rate of retinal detachments at this clinic. We used ophthalmologist drawings to determine the size and location of retinitis lesion sizes. Although drawings could introduce imprecision and misclassification error, their use should not introduce bias since the drawings were performed before the onset of retinal detachment.26 Moreover, in an attempt to reduce bias, all retinal lesion measurements were made by an individual masked to retinal detachment status. Finally, although this study included a much larger number of post-HAART era CMV retinitis-related retinal detachments than previous studies, the number was still relatively small. We improved the statistical power of the study by including 3 controls per case, but we may have nonetheless been underpowered to establish an association between certain risk factors and retinal detachment.
In conclusion, in this northern Thai population larger lesion size and bilateral disease at the time of diagnosis of CMV retinitis, both of which indicate late stage disease, were predictive of a retinal detachment. Efforts to expand the diagnostic capacity for CMV retinitis may lead to earlier detection and treatment, thereby reducing retinal detachments and blindness in these patients.
Acknowledgements and Disclosures
A. Funding/Support: This study was supported by the Gladstone Institute of Virology and Immunology Center for AIDS Research (San Francisco, California), the University of California, San Francisco Research Evaluation and Allocation Committee (San Francisco, California), That Man May See, (San Francisco, California), grant K23EY019071 from the National Eye Institute (National Institutes of Health, Bethesda, Maryland), the Littlefield Trust (San Francisco, CA), the Peierls Foundation (Austin, Texas), and the Doris Duke Charitable Foundation (New York, New York) through a grant supporting the Doris Duke International Clinical Research Fellows Program at the University of California, San Francisco. Michael Yen is a Doris Duke International Clinical Research Fellow.
D. Other Acknowledgements: None
Footnotes
C. Contributions of Authors:
Study Design and Conduct (JC, SoA, TPM, JDK)
Data Collection, Management, Analysis and Interpretation (MY, JC, SoA, PK, PV, SaA, CJ, JS, GNH, DH, TPM, JDK)
Preparation, Review or Approval of the Manuscript (MY, JC, SoA, PK, PV, SaA, CJ, JS, GNH, DH, TPM, JDK)
Disclosures
B. Financial Disclosures: Dr. Holland has served on Advisory Boards for the following companies: Novartis International AG (Basel, Switzerland); Santen, Incorporated (Emeryville, California); and Xoma Corporation (Berkeley, California).
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March 2024 - Volume 96 - Issue 3
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LETTERS TO THE EDITOR
Diagnostic peritoneal lavage is dead: Long live diagnostic peritoneal aspiration
Rafaqat, Wardah MD; Lagazzi, Emanuele MD; Abiad, May MD; Velmahos, George C. MD, PhD
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Division of Trauma, Emergency General Surgery and Surgical Critical Care, Massachusetts General Hospital, Boston, MA
Journal of Trauma and Acute Care Surgery 96(3):p e24, March 2024. | DOI: 10.1097/TA.0000000000004211
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Diagnostic Peritoneal Lavage has become redundant but Diagnostic Peritoneal Aspiration has a role in hemodynamically unstable patients with inconclusive FAST U/S.
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Journal of Trauma and Acute Care Surgery96(3):e24, March 2024.
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14849 | https://projecteuclid.org/journals/electronic-journal-of-probability/volume-30/issue-none/Exact-threshold-for-approximate-ellipsoid-fitting-of-random-points/10.1214/25-EJP1378.pdf | E l e c t r o n i c J o u r n a l o f P r o b a b i l i t y Electron. J. Probab. 30 (2025), article no. 126, 1–46.
ISSN: 1083-6489 Exact threshold for approximate ellipsoid fitting of random points Afonso S. Bandeira Antoine Maillard† Abstract We consider the problem (P) of exactly fitting an ellipsoid (centered at 0) to n standard Gaussian random vectors in Rd, as n, d →∞with n/d2 →α > 0. This problem is conjectured to undergo a sharp transition: with high probability, (P) has a solution if α < 1/4, while (P) has no solutions if α > 1/4. So far, only a trivial bound α > 1/2 is known to imply the absence of solutions, while the sharpest results on the positive side assume α ≤η (for η > 0 a small constant) to prove that (P) is solvable. In this work we show a universality property for the minimal fitting error achievable by ellipsoids: we show that, to leading order, it coincides with the minimal error in a so-called “Gaussian equivalent” problem, for which the satisfiability transition can be rigorously analyzed. Our main results follow from this finding, and they are twofold.
On the positive side, we prove that if α < 1/4, there exists an ellipsoid fitting all the points up to a small error, and that the lengths of its principal axes are bounded above and below. On the other hand, for α > 1/4, we show that achieving small fitting error is not possible if the length of the ellipsoid’s shortest axis does not approach 0 as d →∞(and in particular there does not exist any ellipsoid fit whose shortest axis length is bounded away from 0 as d →∞). To the best of our knowledge, our work is the first rigorous result characterizing the expected phase transition in ellipsoid fitting at α = 1/4. In a companion non-rigorous work, the second author and D. Kunisky give a general analysis of ellipsoid fitting using the replica method of statistical physics, which inspired the present work.
Keywords: ellipsoid fitting; high-dimensional probability; statistical physics.
MSC2020 subject classifications: 60D05; 52A22; 90C22.
Submitted to EJP on September 5, 2024, final version accepted on June 25, 2025.
Supersedes arXiv:2310.05787.
Department of Mathematics, ETH Zürich (Switzerland). E-mail: bandeira@math.ethz.ch †Department of Mathematics, ETH Zürich (Switzerland) and Inria Paris & DI ENS, PSL University, Paris (France). E-mail: antoine.maillard@inria.fr. To whom correspondence shall be sent.
Exact threshold for approximate ellipsoid fitting of random points 1 Introduction and main results 1.1 The ellipsoid fitting conjecture We consider the random ellipsoid fitting problem: given n random standard Gaussian vectors in dimension d, when do they all lie on the boundary of a (centered) ellipsoid?
Formally, we define an ellipsoid fit using the set Sd of d × d real symmetric matrices, as follows.
Definition 1.1 (Ellipsoid fit). Let x1, · · · , xn ∈Rd. We say that S ∈Sd is an ellipsoid fit for (xµ)n µ=1 if it satisfies: (P) : ( x⊺ µSxµ = d for all µ ∈{1, · · · , n}, S ⪰0.
(1.1) In Definition 1.1, the matrix S ⪰0 defines the ellipsoid Σ := {x ∈Rd : x⊺Sx = d}.
Geometrically speaking, the eigenvectors of S give the directions of the principal axes of the ellipsoid, while its eigenvalues (λi)d i=1 are related to the lengths (ri)d i=1 of its principal (semi-)axes by ri = √ dλ−1/2 i .
Scaling – In what follows, we will rather refer to the rescaled quantities r′ i = ri/ √ d as the lengths of the ellipsoid axes, effectively rescaling distances so that the sphere of radius √ d (with S = Id) has all (semi-)axes of length 1. In particular, the lengths of the ellipsoid’s longest and shortest axis are then respectively λmin(S)−1/2 and λmax(S)−1/2.
We are interested in finding an ellipsoid fit to a set of random points x1, · · · , xn i.i.d.
∼ N(0, Id). The question of the existence of such an ellipsoid arose first in [31, 29, 30], which conjectured the following (see e.g. Conjecture 2 in or Conjecture 1.1 in ).
Conjecture 1.2 (The ellipsoid fitting conjecture). Let n, d ≥1, and x1, · · · , xn be drawn i.i.d. from N(0, Id). Let p(n, d) := P[∃S ∈Sd an ellipsoid fit for (xµ)n µ=1]. For any ε > 0, the following holds: lim sup d→∞ n d2 ≤1 −ε 4 ⇒lim d→∞p(n, d) = 1, (1.2) lim inf d→∞ n d2 ≥1 + ε 4 ⇒lim d→∞p(n, d) = 0.
(1.3) Informally, Conjecture 1.2 predicts a sharp transition for the existence of an ellipsoid fit in the regime n/d2 →α > 0 exactly at α = 1/4.
1.2 Related works Conjecture 1.2 was first stated and studied in the series of works [31, 29, 30], where it arose as being connected to the decomposition of a (random) data matrix M as M = L + D, with L ⪰0 being low-rank, and D a diagonal matrix. Connections to other problems throughout theoretical computer science have since then been unveiled, such as certifying a lower bound on the discrepancy of a random matrix using a canonical semidefinite relaxation [29, 28], overcomplete independent component analysis , or Sum-of-Squares lower bound for the Sherrington-Kirkpatrick Hamiltonian . We refer the reader to the detailed expositions of [28, 21] on the connections of ellipsoid fitting to theoretical computer science and machine learning.
Interestingly, Conjecture 1.2 arose both from numerical evidence1 and the remark that d2/4 is known to be the statistical dimension (or squared Gaussian width) of S+ d , the 1Given (xµ)n µ=1, eq. (1.1) is a convex problem (it is an example of a semidefinite program, or SDP), which is efficiently solvable when solutions exist.
EJP 30 (2025), paper 126.
Page 2/46 Exact threshold for approximate ellipsoid fitting of random points n d6/5−ε d3/2−ε d2/polylog(d) [28, 19] d2/C [5, 16, 36] d2/4 d2/2 (Trivial) SAT (rigorous) SAT (conjecture) UNSAT (conjecture) UNSAT (rigorous) Figure 1: A summary of the current state of the ellipsoid fitting conjecture. In red, we show regions for which ellipsoid fitting is rigorously known to be unsatisfiable (UNSAT), and in orange regions which are conjectured to be. Similarly, we show in green regions rigorously known to be satisfiable (SAT), and in yellow regions which are conjectured to be so. Figure is taken from .
set of positive semidefinite matrices [7, 2]. As such, if one replaces eq. (1.1) by (PGauss.) : ( Tr[SGµ] = d for all µ ∈{1, · · · , n}, S ⪰0, (1.4) in which (Gµ)n µ=1 are (independent) standard Gaussian matrices, Conjecture 1.2 provably holds for (PGauss.). The crucial property of (PGauss.) behind this result is that the affine subspace {S ∈Sd : (Tr[SGµ] = d)n µ=1} is randomly oriented, uniformly in all directions.
Although this motivation for the conjecture was known, our work is (to the best of our knowledge) the first mathematically rigorous approach to leverage the connection between (P) and (PGauss.).
Indeed, previous progress on Conjecture 1.2 has mostly focused on proving the existence of a fitting ellipsoid using an ansatz solution: the first line of eq. (1.1) defines an affine subspace V of symmetric matrices of codimension n, so one can study a well-chosen S⋆∈V , and argue that for small enough n it satisfies S⋆⪰0 with high probability as d →∞. Various such constructions have been used, and we summarize in Fig. 1 the current rigorous progress on the ellipsoid fitting conjecture that arose from these approaches. Presently, the best rigorous results on Conjecture 1.2 are due to the recent works [5, 16, 36] and can be summarized as follows: Theorem 1.3 ([5, 16, 36]). Let n, d ≥1, and x1, · · · , xn i.i.d.
∼ N(0, Id). Let p(n, d) := P[∃S ∈Sd an ellipsoid fit for (xµ)n µ=1]. There exists a (small) universal constant η > 0 such that: lim sup d→∞ n d2 ≤η ⇒lim d→∞p(n, d) = 1, Moreover, if n > d(d + 1)/2, then p(n, d) = 0.
Note that the bound n > d(d + 1)/2 in Theorem 1.3 arises from a simple dimension counting argument, as d(d + 1)/2 is the dimension of the space of symmetric matrices: for such values of n, not only does there not exist a solution to eq. (1.1), there does not exist any solution even without the constraint S ⪰0!
Statistical physics approaches: heuristic and rigorous – In this work, we tackle Conjecture 1.2 using techniques inspired by the statistical physics of disordered systems.
While analytical methods developed in this field were originally designed to study models known as spin glasses [4, 22], they have seen in the past decades a great number of applications in high dimensional statistics, theoretical computer science, and machine learning. Moreover, despite these techniques often being non-rigorous, a growing line of mathematics literature has emerged establishing many of their predictions. Notably, ellipsoid fitting is an example of a semidefinite program (SDP) 2 with random linear 2i.e. a combination of linear equations with a positivity constraint S ⪰0.
EJP 30 (2025), paper 126.
Page 3/46 Exact threshold for approximate ellipsoid fitting of random points constraints, and some such SDPs have been previously analyzed with tools of statistical physics [24, 18], although the methods of these works fall short for analyzing the satisfiability transition in random ellipsoid fitting . We refer the interested reader to the recent book that compiles many (sometimes surprising) applications of the theory of disordered systems, as well as mathematically rigorous approaches to it.
Notably, in the companion work to our manuscript , non-rigorous methods of statistical physics are employed to provide a detailed picture of the satisfiability transition in random ellipsoid fitting. Besides predicting a threshold for n ∼d2/4, this work gives analytical formulas for the typical shape of ellipsoid fits in the satisfiable phase (i.e. the spectral density of S), generalizes these predictions for non-Gaussian but rotationally-invariant vectors {xµ}n µ=1, and also studies the performance of different explicit solutions, notably ones used in the previous literature (see Fig. 1). We emphasize that the present paper is, in contrast, mathematically rigorous.
Inspiration of our approach – Importantly, the non-rigorous analysis of suggests that a quantity known as the free entropy (or free energy) in statistical physics, is universal: its value is (with high probability) the same for (P) and a variant of (PGauss.), as d →∞.
Such a universality property would have major consequences, as the free entropy carries deep information about the structure of the space of solutions to the problem. Remarkably, similar phenomena have been studied numerically and theoretically in statistical learning models, in which one can effectively replace an arbitrary (and possibly complicated) data distribution by its “Gaussian equivalent”.
Investigating this Gaussian equivalence phenomenon is the object of a recent and very active line of work, with consequences on the theory of empirical risk minimization and beyond [13, 17, 23, 11, 1, 9, 20, 10, 32]. Inspired by these works (in particular ) we provide a rigorous proof of the universality conjectured in , using an interpolation argument. We then leverage tools of the theory of random convex programs [7, 2], such as Gordon’s min-max inequality , to sharply characterize the space of solutions to (PGauss.). Using the aforementioned universality allows to transfer many of these conclusions to the original problem (P), yielding our main results.
1.3 Main results We now state our main results, separating the conjecturally satisfiable (α = n/d2 < 1/4) and unsatisfiable (α > 1/4) regimes.
Notation – f = od(g) (respectively f = Od(g)) means that f/g →0 as d →∞(re-spectively f/g is bounded as d →∞). We also use g ≳f to denote f = Od(g). We denote Sd the set of d × d real symmetric matrices, while Sd−1(r) refers to the Euclidean sphere of radius r in Rd. For S ∈Sd, Sp(S) = {λi}d i=1 is the set of eigenvalues of S.
For γ ∈[1, ∞], ∥S∥Sγ := (P i |λi|γ)1/γ stands for the Schatten-γ norm. Bγ(S, δ) is the Schatten-γ ball of radius δ centered in S, and Bγ(δ) the ball centered at S = 0. We denote by ∥S∥op := ∥S∥S∞the operator norm, and ∥S∥F := ∥S∥S2 the Frobenius norm.
For a function ψ : R →R, we write ∥ψ∥L to denote its Lipschitz constant. Finally, we use a generic nomenclature C, c > 0 to denote positive constants (not depending on the dimension), that may vary from line to line. If necessary, we will make explicit the dependency of these constants on parameters of the problems. Finally, we use in the text the abbreviation “w.h.p.”, short for “with high probability”, to refer to events that have probability 1 −od(1) as d →∞.
1.3.1 The satisfiable phase: α < 1/4 Our main result on the “positive” side of the ellipsoid fitting conjecture can be stated as follows.
EJP 30 (2025), paper 126.
Page 4/46 Exact threshold for approximate ellipsoid fitting of random points Theorem 1.4 (Satisfiable regime). Assume α := lim sup(n/d2) < 1/4 and let r ∈[1, 4/3).
There exist 0 < λ−≤λ+, depending only on α, such that the following holds. Let Γr(ε) := ( S ∈Sd : Sp(S) ⊆[λ−, λ+] and 1 n n X µ=1 √ d x⊺ µSxµ d −1 r ≤ε ) .
Then for any ε > 0, if x1, · · · , xn i.i.d.
∼N(0, Id), P[Γr(ε) ̸= ∅] →1 as n, d →∞.
Let us make a series of remarks on Theorem 1.4. First, one can alternatively formulate its conclusion as: p-lim d→∞ min Sp(S)⊆[λ−,λ+] 1 n n X µ=1 √ d x⊺ µSxµ d −1 r = 0, (1.5) where p-lim denotes limit in probability. Secondly, while our current proof limits the choice of r ∈[1, 4/3), it might be possible to refine our arguments to reach the same result for any r ∈[1, 2], see our discussion in Section 1.4. Moreover, note that by standard concentration arguments, we expect Gaussian points to be close to the sphere Sd−1( √ d), i.e. the ellipsoid defined by S = Id. A detailed analysis yields, for any r ∈[1, 2]: p-lim d→∞ 1 n n X µ=1 √ d ∥xµ∥2 d −1 r = E[|Z|r] > 0, (1.6) where Z ∼N(0, 2). This follows from classical concentration arguments, see Appendix B for a detailed derivation. Theorem 1.4 therefore shows that there exists an ellipsoid whose “fitting error” improves by an arbitrary factor over the one achieved by the unit sphere, as long as α < 1/4. On the other hand, we will see that this is not possible for α > 1/4, strongly suggesting that our results capture the phenomenon responsible for the conjectured satisfiability transition of ellipsoid fitting. In Section 1.4 we will consider possible future directions that could allow to improve our results to the existence of fitting ellipsoids with exactly zero error, i.e. the conjecture of eq. (1.2).
Finally, we notice that Theorem 1.4 is coherent with the non-rigorous analysis of , which predicts that for any α < 1/4 typical solutions to ellipsoid fitting have spectral density contained in an interval of the type [λ−, λ+] depending only on α.
1.3.2 The unsatisfiable phase: α > 1/4 Our main result towards proving the non-existence of fitting ellipsoids for α > 1/4 is the following.
Theorem 1.5 (Unsatisfiable regime). Assume α := lim inf(n/d2) > 1/4. Let φ : R+ →R+ be a non-decreasing differentiable function, with φ(0) = 0, and such that φ has a unique global minimum in 0. For any ε > 0 and M > 0 we let: Γ(ε, M) := ( S ∈Sd : Sp(S) ⊆[0, M] and 1 n n X µ=1 φ √ d x⊺ µSxµ d −1 ≤ε ) .
There exists ε = ε(α, φ) > 0 such that for all M > 0, if x1, · · · , xn i.i.d.
∼N(0, Id), lim d→∞P [Γ(ε, M) ̸= ∅] = 0.
As stated below, a direct corollary of Theorem 1.5 is that, when α > 1/4, ellipsoid fitting admits (w.h.p.) no solutions with spectrum bounded above as d →∞(i.e. an ellipsoid whose smallest axis has length bounded away from zero).
EJP 30 (2025), paper 126.
Page 5/46 Exact threshold for approximate ellipsoid fitting of random points Corollary 1.6 (Non-existence of fitting ellipsoids with bounded spectrum). Let α > 1/4.
Let n, d →∞with n/d2 →α > 0, and x1, · · · , xn i.i.d.
∼N(0, Id). We denote Γ the set of ellipsoid fits for (xµ)n µ=1. Then, for all c > 0, lim d→∞P[∃S ∈Γ : ∥S∥op ≤c] = 0.
In particular, our results imply that the negative side of the ellipsoid fitting conjecture, i.e. eq. (1.3), would follow from either: H.1 a proof that (for α > 1/4 and with high probability) the set of ellipsoid fits is bounded in spectral norm as d →∞, or H.2 a proof that, if S ∈Γ is an ellipsoid fit, there exists a (possibly different) ellipsoid fit ˆ S ∈Γ that has bounded spectral norm as d →∞.
As a consequence of Corollary 1.6, proving either H.1 or H.2 would yield the negative side of the ellipsoid fitting conjecture. Notice further that Theorem 1.5 implies the non-existence of bounded ellipsoid fits even allowing a small “fitting error”, so that it would be sufficient to prove H.2 considering an ellipsoid ˆ S that fits (xµ)n µ=1 only up to a small enough error ε > 0 (in the sense of Theorem 1.5).
1.4 Discussion and consequences The combination of our two main results (Theorem 1.4 and Theorem 1.5) provides strong evidence for the original ellipsoid fitting conjecture (Conjecture 1.2). Our con-clusions are attained through the study of a “Gaussian equivalent” problem, which partly motivated Conjecture 1.2. Informally, we show that an approximate version of the ellipsoid fitting (i.e. by allowing infinitesimally small error) undergoes a sharp satis-fiability transition at α = n/d2 = 1/4. Moreover, we also show in our proof that in the Gaussian equivalent problem, the satisfiability transition for this “approximate” version corresponds to the one of the exact fitting problem (i.e. not allowing for any non-zero error). This strongly suggests that our method is indeed capturing the phenomenon responsible for the ellipsoid fitting transition.
Our results are an example of a universality phenomenon in high-dimensional stochas-tic geometry: we show that the statistical dimension (or the square of the Gaus-sian width) of the set of positive semidefinite matrices determines the satisfiability of – a modified version of – random ellipsoid fitting, even though the affine subset {S ∈Sd : (x⊺ µSxµ = d)n µ=1} is not randomly oriented uniformly in all directions. In general, understanding the conditions under which universality holds in such problems of high-dimensional random geometry is an important open question. We mention , which proves universality between a model in which the random subspace is given by the kernel of a random i.i.d. Gaussian matrix, and a second model where the subspace is the kernel of a matrix with independent elements (not necessarily Gaussian).
1.4.1 Towards Conjecture 1.2 Unfortunately, while our main theorems characterize a satisfiability transition for ellip-soid fitting at the expected threshold, they do not formally imply Conjecture 1.2. We discuss briefly some improvements of our results that could potentially allow to bridge this gap.
On the “positive” side of the conjecture (i.e. the regime α = n/d2 < 1/4), Theorem 1.4 shows the existence of bounded ellipsoids that can achieve arbitrarily small error ε EJP 30 (2025), paper 126.
Page 6/46 Exact threshold for approximate ellipsoid fitting of random points (where the error is taken to 0 after d →∞). On the other hand, a proof of eq. (1.2) would require to invert these limits, and take ε →0 before d →∞. In this regard, an important strengthening of Theorem 1.4 would be to obtain non-asymptotic bounds on P[Γr(ε) = ∅], that depend on ε.
Another potential for improvement stems from geometrical considerations: denoting V := {S ∈Sd : x⊺ µSxµ = d for all µ ∈[n]}, one may use Theorem 1.4 to bound the distance of the set Γr(ε) to the affine subspace V .
Since λmin(S) ≥λ−(α) for any S ∈Γr(ε), it would suffice to show that dop(Γr(ε), V ) ≤λ−(α) to deduce eq. (1.2), the first part of Conjecture 1.2. We perform in Appendix C a naive analysis of necessary conditions for this conclusion to follow from Theorem 1.4. Unfortunately, we find that (among other considerations) these conditions would require a significantly stronger form of Theorem 1.4, by proving the conclusion for larger values of r ∈[1, 2] and/or a better scaling with d of the minimal error achievable (i.e. proving the conclusion of Theorem 1.4 for Γr(εd) with εd →0 as d →∞).
Moreover, let us emphasize that a critical difficulty in improving our proof techniques would be to quantitatively sharpen the universality arguments we carry out, and in particular to strengthen Proposition 2.2, which shows the universality of the minimal fitting error, or “ground state” energy, for ellipsoid fitting and a simpler “Gaussian equivalent” problem. While the present form of Proposition 2.2 shows universality of this error up to a od(1) difference, this estimate would likely have to be improved in order to carry out the aforementioned approaches. This part of our proof is greatly inspired by a recent literature on similar universality phenomena [13, 17, 23, 11, 1, 9, 20, 10, 32], and we are not aware of the existence of such universality results at a finer scale (or even predictions/conjectures of conditions under which they should hold).
Finally, on the “negative” side of the conjecture (i.e. for α > 1/4), as emphasized in Corollary 1.6 and the discussion thereafter, Theorem 1.5 reduces the second part of Conjecture 1.2 (eq. (1.3)) to proving either H.1 or H.2. If such a proof were to become available, our results would imply the regime α > 1/4 of Conjecture 1.2.
1.4.2 Further directions Our proof method that leverages universality of the minimal fitting error is quite versatile, and we end our discussion by mentioning a few further directions and generalized results that stem from our analysis.
The dual program – First, as a semidefinite program, ellipsoid fitting admits a dual formulation, as written e.g. in .
While the limitations of Theorems 1.4 and 1.5, discussed above, prevent us from directly drawing conclusions on the dual, it might be possible to directly apply to it a similar universality approach. Such an application might allow to overcome some current limitations of our results, and we leave this investigation for future work.
Beyond Gaussian vectors – Secondly, while we perform our analysis for x1, · · · , xn ∼ N(0, Id), it is clear from our proof that our results (both Theorems 1.4 and 1.5) hold for any i.i.d. (xµ)n µ=1 such that the matrices Wµ := (xµx⊺ µ−Id)/ √ d satisfy a uniform pointwise normality (or uniform one-dimensional CLT) assumption, as defined in Definition 4.5, and proven for the case of Gaussian vectors in Lemma 4.9. An interesting example of a non-Gaussian distribution is given by the case of rotationally-invariant vectors with fluctuating norm, of the form xµ d = √rµωµ, with rµ and ωµ independent, and ωµ ∼Unif(Sd−1( √ d)). Letting τ := limd→∞[ √ dVar(r1)], conjectures that the ellipsoid fitting transition point for this model is located at EJP 30 (2025), paper 126.
Page 7/46 Exact threshold for approximate ellipsoid fitting of random points n/d2 = αc(τ) ∈(0, 1/2), and gives an exact expression of αc(τ) (see Fig. 5 of ), showing that ellipsoid fitting becomes harder as the fluctuations of the norm increase.
While pointwise normality may not hold in this setting, it is conceivable that our proof techniques can be adapted to handle these distributions, by following the calculations of , to obtain results akin to Theorems 1.4 and 1.5. More generally, while it is clear that some distributions can not satisfy uniform pointwise normality (see the discussion below Lemma 4.9 for examples), a more thorough investigation of the class of distributions of xµ’s for which pointwise normality holds (and thus our proof applies) is, in our opinion, an interesting direction to explore.
A minimal nuclear norm estimator – Let us conclude by mentioning a different approach to a possible solution of the first part of Conjecture 1.2. It is conjectured in (through non-rigorous methods) that the minimal nuclear norm solution, i.e.
ˆ SNN := arg min S∈Sd {x⊺ µSxµ=d}n µ=1 ∥S∥S1 = arg min S∈Sd {x⊺ µSxµ=d}n µ=1 d X i=1 |λi(S)|, satisfies ˆ SNN ⪰0 with high probability for any α < 1/4. Analyzing ˆ SNN, whether through the techniques of the present paper or with different methods, is another promising approach to prove eq. (1.2), the “positive” part of Conjecture 1.2.
1.5 Structure of the paper In Section 2 we present the proof of our main results. The proof of some intermediate results is postponed to later sections: in Section 3 we study in detail the “Gaussian equivalent” problem to random ellipsoid fitting, and in Section 4 we prove a crucial universality property for the minimal fitting error between ellipsoid fitting and its “Gaussian equivalent”.
2 Proof of the main results We prove here Theorems 1.4 and 1.5. The core idea of our proof can be sketched as follows: (i) Using rigorous methods inspired by statistical physics, we prove that a quantity known as the asymptotic free entropy is universal for the ellipsoid fitting problem of eq. (1.1) and a variant of its Gaussian counterpart of eq. (1.4), for any value of α = n/d2. The main technique we use is an interpolation method. In a suitable limit (known as the low-temperature limit in statistical physics), this implies the universality of the minimal “fitting error”.
(ii) We study the Gaussian equivalent problem using methods of random convex geome-try [7, 2], leveraging in particular Gordon’s min-max inequality . When α < 1/4 we show that not only a zero error is achievable, but that one can achieve it by a matrix whose spectrum is contained in an interval of the type [λ−, λ+], i.e. the axis of the corresponding ellipsoid have lengths bounded above and below. On the other hand, for α > 1/4, we prove that not only is the Gaussian equivalent problem not satisfiable, but one can lower bound the minimal fitting error as long as the set of candidate matrices is contained in an operator norm ball Bop(M) (for any constant M > 0).
(iii) We prove that the conclusions of (ii) transfer to the original ellipsoid fitting problem, using the universality shown in (i).
EJP 30 (2025), paper 126.
Page 8/46 Exact threshold for approximate ellipsoid fitting of random points Our proof of (i) leverages an important line of work on free entropy universality [17, 23, 11], and a part of it closely follows the proof of , which we will point out in relevant places. Nevertheless, as our setting does not satisfy all the hypotheses of this work, and for completeness of our exposition, we chose to write the whole proof in a self-contained manner.
2.1 Reduction of the problem and Gaussian equivalent For any 0 ≤λ−≤λ+, any function φ : R →R and any ε > 0 we define the set Γ(φ, λ−, λ+, ε) := ( S ∈Sd : Sp(S) ⊆[λ−, λ+] and 1 n n X µ=1 φ √ d x⊺ µSxµ d −1 ≤ε ) .
(2.1) If one thinks of φ as an error (or loss) function, then Γ(φ, λ−, λ+, ε) represents the set of matrices with spectrum in [λ−, λ+] that solve (P) up to an approximation error ε. Notice that for any S ∈Sd: √ d x⊺ µSxµ d −1 = Tr[WµS] −d −Tr[S] √ d , (2.2) with Wµ := (xµx⊺ µ −Id)/ √ d. Moreover, Wµ has the same first two moments as a Gaussian matrix. Formally, we define: Definition 2.1 (Matrix ensembles). Let d ≥1. We say that a random symmetric W ∈Sd is generated according to3: • W ∼GOE(d) if Wij i.i.d.
∼N(0, [1 + δij]/d) for i ≤j.
• W ∼Ellipse(d) if W d = (xx⊺−Id)/ √ d for x ∼N(0, Id).
One checks easily that EEllipse(d)[WijWkl] = EGOE(d)[WijWkl] for any i ≤j and k ≤l.
This remark and eq. (2.2) lead to consider the following modified problem, with Wµ := (xµx⊺ µ −Id)/ √ d and b ∈R: Γb(φ, λ−, λ+, ε) := ( S ∈Sd : Sp(S) ⊆[λ−, λ+] and 1 n n X µ=1 φ (Tr[WµS] −b) ≤ε ) .
(2.3) In the rest of the proof we will focus on studying the set Γb of eq. (2.3) with4 b ∈R, for both Wµ ∼Ellipse(d) and Wµ ∼GOE(d) (which we call the “Gaussian equivalent” problem). At the end of our proof, we will transfer our conclusions on Γb back to the original solution set Γ of eq. (2.1).
2.2 Universality of the minimal error We can now state the main result concerning on the universality of the minimal error (or “ground state energy” in statistical physics jargon). This result is inspired by a rich line of work on universality of empirical risk minimization [17, 23, 11, 9].
Proposition 2.2 (Ground state universality). Let φ : R →R+ and ψ : R →R two bounded differentiable functions with bounded derivatives, and assume furthermore ∥ψ′∥L < ∞. Let n, d ≥1 and n, d →∞with α1d2 ≤n ≤α2d2 for some 0 < α1 ≤α2, and B ⊆Sd a closed set such that B ⊆Bop(C0) for some C0 > 0 (not depending on d). For 3GOE(d) stands for Gaussian Orthogonal Ensemble.
4Furthermore, by rescaling S (and up to a change in λ−, λ+, φ) we will reduce to the case b ∈{−1, 0, 1}.
EJP 30 (2025), paper 126.
Page 9/46 Exact threshold for approximate ellipsoid fitting of random points X1, · · · , Xn ∈Sd we define the ground state energy: GSd({Xµ}) := inf S∈B 1 d2 n X µ=1 φ(Tr[XµS]).
(2.4) Then we have: lim d→∞ E{Wµ} i.i.d.
∼Ellipse(d)ψ[GSd({Wµ})] −E{Gµ} i.i.d.
∼GOE(d)ψ[GSd({Gµ})] = 0.
(2.5) Therefore, for any ρ ≥0 and δ > 0: lim d→∞P[GSd({Wµ}) ≥ρ + δ] ≤lim d→∞P[GSd({Gµ}) ≥ρ], lim d→∞P[GSd({Wµ}) ≤ρ −δ] ≤lim d→∞P[GSd({Gµ}) ≤ρ].
(2.6) A word on the proof – The main proof technique we use is Gaussian interpolation: namely we define an interpolating Uµ(t) such that Uµ(0) = Gµ and Uµ(1) = Wµ, and show that GSd({Uµ(t)}) is constant (up to negligible terms) along the interpolation path.
Note that while Proposition 2.2 is very close to the results of , there is a technical difference with the setup of this work: for any fixed S, the random variable Tr[WS] for W ∼Ellipse(d) is not sub-Gaussian but only sub-exponential. As a consequence, we can not achieve a good control of the Lipschitz constant of the error (or “energy” function) of eq. (2.4) with respect to the Frobenius norm of S, as is required in . We bypass this difficulty by controlling instead the Lipschitz constant with respect to the operator norm (see Lemma 4.1), using important empirical process bounds over the operator norm ball (see Lemma 4.3): this leads to the limitation B ⊆Bop(C0). Interestingly, improving these bounds would also allow to relax the limitation r ∈[1, 4/3) in Theorem 1.4, as we discuss after. Having dealt with this difficulty, the rest of the interpolation argument is very similar to . We show Proposition 2.2 in Section 4, deferring some arguments to Appendix D.
2.3 The Gaussian equivalent problem We now study the Gaussian equivalent problem. We will later transfer our analysis to the original ellipsoid fitting case using Proposition 2.2. Our results are stated separately for the satisfiable and unsatisfiable regimes.
Proposition 2.3 (Regular solutions in the satisfiable regime). Let n, d →∞with n/d2 → α < 1/4, and let {Gµ}n µ=1 i.i.d.
∼GOE(d). Let V := {S ∈Sd : ∀µ ∈[n], Tr[GµS] = 1}.
There exist 0 < λ−≤λ+ (depending only on α) such that: lim d→∞P{∃S ∈V s.t. Sp(S) ⊆[λ−, λ+]} = 1.
Proposition 2.3 shows that for α < 1/4, there exist with high probability ellipsoids satisfying the “Gaussian equivalent” to random ellipsoid fitting, and that such solutions might also be assumed to have their axes’ lengths bounded above and below as d →∞.
In the unsatisfiable regime α > 1/4, we show on the other hand that with high probability there are no solutions to the Gaussian equivalent problem, even allowing for some error when fitting the random points.
Proposition 2.4 (No approximate solution in the unsatisfiable regime). Let n, d →∞ with n/d2 →α > 1/4, and let {Gµ}n µ=1 i.i.d.
∼ GOE(d). Let b ∈R and denote C(b) µ (S) := |Tr(GµS) −b|. Define the affine subspace: Vb := {S ∈Sd : ∀µ ∈[n], C(b) µ (S) = 0}.
EJP 30 (2025), paper 126.
Page 10/46 Exact threshold for approximate ellipsoid fitting of random points (i) Assume that b ̸= 0. Then there exist c = c(α, b) > 0 and η = η(α, b) ∈(0, 1) such that lim d→∞P{∀S ⪰0 : #{µ ∈[n] : C(b) µ (S) > c} ≥ηn} = 1.
(ii) Assume that b = 0. Then there exist c = c(α) > 0 and η = η(α) ∈(0, 1) such that, with probability 1 −od(1), the following holds for all τ ≥0: sup S⪰0 #{µ∈[n] : C(0) µ (S)>cτ}<ηn ∥S∥F ≤τ √ d.
Propositions 2.3 and 2.4 are proven in Section 3. Our proof follows a standard approach in random geometry problems involving Gaussian distributions, by leveraging Gordon’s min-max inequality and its sharpness in convex settings [35, 34].
It strengthens for our setting the results obtained for general random convex programs in [7, 2] (using either Gordon’s inequality or tools of integral geometry).
2.4 The satisfiable regime: Proof of Theorem 1.4 Propositions 2.2 and 2.3 have the following consequence, taking B := {S : λ−Id ⪯ S ⪯λ+Id}, with (λ−, λ+) given by Proposition 2.3.
Corollary 2.5. Let n, d →∞with n/d2 →α < 1/4, and W1, · · · , Wn i.i.d.
∼ Ellipse(d).
There exist λ−, λ+ > 0 depending only on α such that the following holds. If we have φ : R →R+ with ∥φ∥∞, ∥φ′∥∞< ∞and such that φ(0) = 0, then p-lim d→∞ min Sp(S)⊆[λ−,λ+] 1 n n X µ=1 φ(Tr[WµS] −1) = 0.
The proof of Corollary 2.5 is immediate by combining Propositions 2.2 and 2.3. We can furthermore relax some of the assumptions on φ in Corollary 2.5, as we now show.
Lemma 2.6. Corollary 2.5 holds for φ(x) = |x|r, for any 1 ≤r < 4/3.
Note that the limitation r < 4/3 is a consequence of a limitation on the control of an empirical process that is done in Lemma 4.3 (see also the discussion in Section 4.1).
Proof of Lemma 2.6. Let us first assume that r > 1, so that φ(x) = |x|r is continuously differentiable in x = 0. Let ε > 0 and A > 0, and let us denote uA : R+ →[0, 1] a C∞function such that uA(x) = 1 if x ≤A and uA(x) = 0 if x ≥A + 1. We denote φA(z) := |z|ruA(|z|). Then φA is bounded, with bounded derivative. Moreover, we have for any x ∈R: |x|r = φA(x) + |x|r(1 −uA(|x|)) ≤φA(x) + |x|r1{|x| ≥A}.
By Corollary 2.5, under an event of probability 1−od(1) we can fix S with Sp(S) ∈[λ−, λ+] and such that Pn µ=1 φA(Tr[WµS] −1) ≤nε/2. We pick γ > 1 such that γr ≤4/3, and condition on the 1 −od(1) probability event, thanks to Lemma 4.3: max ∥R∥op=1 n X µ=1 |Tr(WµR)|γr ≤Cn (2.7) EJP 30 (2025), paper 126.
Page 11/46 Exact threshold for approximate ellipsoid fitting of random points We have, with probability 1 −od(1): 1 n n X µ=1 |Tr[WµS] −1|r ≤ε 2 + 1 n n X µ=1 |Tr[WµS] −1|r1{|Tr[WµS] −1| ≥A}, (a) ≤ε 2 + Ar(1−γ) n n X µ=1 |Tr[WµS] −1|γr, (b) ≤ε 2 + Ar(1−γ)2γr−1 n [n + Cλγr + n], ≤ε 2 + C(γ, r, α)Ar(1−γ).
We used in (a) the following inequality, for a positive random variable X, t ≥0, and any γ > 1: E[X1{X ≥t}] = tE X t 1 X t ≥1 ≤t1−γE[Xγ].
In (b) we used eq. (2.7) and |a + b|r ≤2r−1(|a|r + |b|r). We pick A = ε 2C(γ, r, α) 1/[r(1−γ)] .
We have then, with probability 1 −od(1): min Sp(S)⊆[λ−,λ+] 1 n n X µ=1 |Tr[WµS] −1|r ≤ε, which ends the proof.
We now tackle the case r = 1. For η > 0, we let vη : R+ →[0, 1] a C∞function such that vη(x) = 1 for x ≥η and vη(x) = 0 for x ≤η/2. A transposition of the argument above shows that Corollary 2.5 applies to φη(x) := |x|vη(|x|), which is continuously differentiable everywhere. Notice that for any x ∈R: |x| = φη(x) + |x|(1 −vη(|x|)) ≤φη(x) + |x|1{|x| ≤η} ≤φη(x) + η.
So, for any S ∈Sd: 1 n n X µ=1 |Tr[WµS] −1| ≤1 n n X µ=1 φη(Tr[WµS] −1) + η.
(2.8) Fixing now any ε > 0, and letting η := ε/2, we get from Corollary 2.5 applied to φη that with probability 1 −od(1): min Sp(S)⊆[λ−,λ+] 1 n n X µ=1 φη(Tr[WµS] −1) ≤ε 2.
Combining this result with eq. (2.8), we get that with probability 1 −od(1): min Sp(S)⊆[λ−,λ+] 1 n n X µ=1 |Tr[WµS] −1| ≤ε, which ends the proof.
Proof of Theorem 1.4 – Notice that Lemma 2.6 precisely shows that, for φ(x) = |x|r and ε > 0, the set Γ1 of eq. (2.3) is non-empty with high probability. We now use the following remark (recall the definition of Γ in eq. (2.1)): EJP 30 (2025), paper 126.
Page 12/46 Exact threshold for approximate ellipsoid fitting of random points Lemma 2.7. For any (xµ)n µ=1 and λ−, λ+, ε > 0, r ≥1, if S ∈Γ1(| · |r, λ−, λ+, ε), then ˆ S ∈Γ(| · |r, λ′ −, λ′ +, ε′), with ˆ S = dS √ d + Tr[S] , λ′ −= λ− λ+ + d−1/2 , λ′ + = λ+ λ−+ d−1/2 , ε′ = ε λ−+ d−1/2r .
Since λ′ + ≤λ+/λ−, ε′ ≤ε/(λ−)r, and λ′ −≥λ−/(2λ+) for d large enough, combining Lemmas 2.6 and 2.7 imply that for any r ∈[1, 4/3) and ε > 0: P[Γ(| · |r, a, b, ε) ̸= ∅] →d→∞1, for some 0 < a ≤b depending only on α, which ends the proof of Theorem 1.4.
Proof of Lemma 2.7. Let S ∈Γ1(| · |r, λ−, λ+, ε). Defining ˆ S = dS/( √ d + Tr[S]), we have by eq. (2.2): √ d " x⊺ µ ˆ Sxµ d −1 # r = d √ d + Tr[S] r |Tr[SWµ] −1|r , ≤ 1 λ−+ d−1/2r |Tr[SWµ] −1|r .
2.5 The unsatisfiable regime: Proof of Theorem 1.5 Propositions 2.2 and 2.4 have the following corollary.
Corollary 2.8. Let n, d →∞with n/d2 →α > 1/4, and W1, · · · , Wn i.i.d.
∼Ellipse(d). Let φ : R+ →R+ be a non-decreasing differentiable function, with φ(0) = 0, and such that φ has a unique global minimum in 0. Then: (i) Let b ∈{−1, 1}. There exists ε = ε(α, φ) > 0 such that for all M > 0: lim d→∞P " min Sp(S)⊆[0,M] 1 n n X µ=1 φ(|Tr[WµS] −b|) ≥ε # = 1.
(ii) Let b = 0. For all τ > 0, there exists ε = ε(τ, α, φ) > 0 such that for all M > 0: lim d→∞P min Sp(S)⊆[0,M] ∥S∥F ≥τ √ d 1 n n X µ=1 φ(|Tr[WµS]|) ≥ε = 1.
Proof of Corollary 2.8. Note that we can assume that φ is bounded with bounded deriva-tive and φ′(0) = 0: if not it is always possible to lower bound φ by such a function.
x 7→φ(|x|) is then a bounded function on R with bounded derivative. We start with (i).
By Proposition 2.4, there exist cα, ηα > 0 such that lim d→∞P{∀S ⪰0 : #{µ ∈[n] : |Tr(GµS) −b| ≤cα} ≤(1 −ηα)n} = 1.
Conditioning on this event, we have inf S⪰0 n X µ=1 φ(|Tr[GµS] −b|) ≥nηαφ(cα).
Using Proposition 2.2 with B = {S : Sp(S) ⊆[0, M]} we reach that, for all M > 0, with probability 1 −od(1): inf Sp(S)⊆[0,M] 1 n n X µ=1 φ(|Tr[WµS] −b|) ≥1 2ηαφ(cα).
EJP 30 (2025), paper 126.
Page 13/46 Exact threshold for approximate ellipsoid fitting of random points We now turn to (ii). Again by Proposition 2.4, we fix cα, ηα > 0 such that for all τ ≥0: lim d→∞P sup S⪰0 #{µ∈[n] : |Tr(GµS)|>cατ}<ηαn ∥S∥F ≤τ 2 √ d = 1.
Stated differently: lim d→∞P{∀S ⪰0 : ∥S∥F ≤τ 2 √ d or #{µ ∈[n] : |Tr(GµS)| > cατ} ≥ηαn} = 1.
Conditioning on this event and since φ is non-decreasing on R+: inf S⪰0 ∥S∥F ≥τ √ d n X µ=1 φ(|Tr[GµS]|) ≥nηαφ(cατ).
Using Proposition 2.2 with B = {S : 0 ⪯S ⪯MId and ∥S∥F ≥τ √ d} we reach that, for all τ, M, with probability 1 −od(1): inf Sp(S)⊆[0,M] ∥S∥F ≥τ √ d 1 n n X µ=1 φ(|Tr[WµS]|) ≥1 2ηαφ(cατ), which ends the proof.
We now turn to the proof of Theorem 1.5.
Proof of Theorem 1.5 –. Let M > 0. As in the proof of Corollary 2.8, we can assume without loss of generality that φ has bounded derivative: if it does not, it is always possible to lower bound φ by such a function.
Let δ ∈(0, 1), and S with Sp(S) ⊆[0, M] and |Tr[S] −d| ≥δ √ d. Notice that, defining S′ := √ dS/|d −Tr[S]| ⪰0, we have with b := sign(d −Tr[S]) ∈{±1}: Tr[S′Wµ] −b = x⊺ µSxµ −d |TrS −d| , and so since φ is non-decreasing: φ √ d x⊺ µSxµ d −1 ≥φ(δ|Tr(WµS′) −b|).
Moreover, Sp(S′) ⊆[0, M/δ]. Since this argument is valid for any S with Sp(S) ⊆[0, M] we get: min Sp(S)⊆[0,M] |Tr[S]−d|≥δ √ d 1 n n X µ=1 φ √ d x⊺ µSxµ d −1 ≥ min b∈{±1} min Sp(S)⊆[0,M/δ] 1 n n X µ=1 φ(δ|Tr[WµS] −b|).
Using Corollary 2.8 applied to x 7→φ(δx) there exists therefore c = c(α, δ, φ) > 0 such that with probability 1 −od(1): min Sp(S)⊆[0,M] |Tr[S]−d|≥δ √ d 1 n n X µ=1 φ √ d x⊺ µSxµ d −1 ≥c(α, δ, φ).
(2.9) EJP 30 (2025), paper 126.
Page 14/46 Exact threshold for approximate ellipsoid fitting of random points Let now S ∈Sd with Sp(S) ⊆[0, M] and |Tr[S] −d| ≤δ √ d. Then: Tr[WµS] = √ d x⊺ µSxµ d −1 + d −Tr[S] √ d | {z } |·|≤δ , so that min Sp(S)⊆[0,M] |Tr[S]−d|≤δ √ d 1 n n X µ=1 φ √ d x⊺ µSxµ d −1 ≥ min Sp(S)⊆[0,M] |Tr[S]−d|≤δ √ d 1 n n X µ=1 φ(|Tr[WµS]|) −∥φ′∥∞δ.
(2.10) Notice that since δ < 1, for large enough d we have |Tr[S] −d| ≤δ √ d ⇒Tr[S] ≥d/2. If moreover S ⪰0, by Cauchy-Schwarz we have ∥S∥F ≥Tr[S]/ √ d ≥ √ d/2. This implies: min Sp(S)⊆[0,M] |Tr[S]−d|≤δ √ d 1 n n X µ=1 φ(|Tr[WµS]|) ≥ min Sp(S)⊆[0,M] ∥S∥F ≥ √ d/2 1 n n X µ=1 φ(|Tr[WµS]|).
(2.11) Using Corollary 2.8 we can obtain ε = ε(α, φ) > 0 such that, with probability 1 −od(1), we have: min Sp(S)⊆[0,M] ∥S∥F ≥ √ d/2 1 n n X µ=1 φ(|Tr[WµS]|) ≥ε.
(2.12) Combining eq. (2.9) with all three equations (2.10), (2.11), (2.12), we get that for any δ > 0, with probability 1 −od(1): min Sp(S)⊆[0,M] 1 n n X µ=1 φ √ d x⊺ µSxµ d −1 ≥min[c(α, δ, φ), ε(α, φ) −δ∥φ′∥∞].
Taking δ := min (1, ε(α, φ)/(2∥φ′∥∞)) > 0 ends the proof.
3 The Gaussian equivalent problem 3.1 The satisfiable regime: Proof of Proposition 2.3 3.1.1 Gordon’s min-max theorem We will use the Gaussian min-max theorem of Gordon , as stated in [35, 34]: Proposition 3.1 (Gaussian min-max theorem [14, 35, 34]). Let n, p ≥1, W ∈Rn×p an i.i.d. standard normal matrix, and g ∈Rn, h ∈Rp two independent vectors with i.i.d.
N(0, 1) coordinates. Let Sv, Su be two compact subsets respectively of Rp and Rn, and let ψ : Sv × Su →R a continuous function. We define the two optimization problems: C(W) := min v∈Sv max u∈Su {u⊺Wv + ψ(v, u)} , C(g, h) := min v∈Sv max u∈Su {∥u∥2h⊺v + ∥v∥2g⊺u + ψ(v, u)} .
Then: (i) For all t ∈R, one has P[C(W) < t] ≤2P[C(g, h) ≤t].
(ii) Assume that Sv, Su are convex and that ψ is convex-concave on Sv × Su. Then for all t ∈R: P[C(W) > t] ≤2P[C(g, h) ≥t].
EJP 30 (2025), paper 126.
Page 15/46 Exact threshold for approximate ellipsoid fitting of random points 3.1.2 Gordon’s min-max inequality and random geometry We first introduce the notion of Gaussian width of a convex cone.
Definition 3.2 (Gaussian width). For p ≥1, and a closed convex cone K ⊆Rp, we define its Gaussian width as ω(K) := E max x∈K∩Sp−1⟨g, x⟩, for g ∼N(0, Ip).
We show now a general result leveraging Gordon’s min-max inequality to prove the existence of a solution to a general type of random geometry problem. Such applications are classical, and we show here that one can assume furthermore that the solution is bounded.
Lemma 3.3. Let n, p ≥1, and (gµ)n µ=1 i.i.d.
∼ N(0, Ip). We define V := {x ∈Rp : ∀µ ∈ [n], ⟨gµ, x⟩= 1}. Let K ⊆Rp be a closed convex cone, with Gaussian width ω(K). Assume that there exists ε ∈(0, 1) such that n ≤(1 −ε) ω(K)2 as n →∞. Then: lim n→∞P ∃x ∈K ∩V s.t. ∥x∥2 ≤2 √ε = 1.
(3.1) Proof of Lemma 3.3. Let us denote, for A > 0: Pn(A) := P {∃x ∈K ∩V s.t. ∥x∥2 ≤A} , and define G ∈Rn×p as the Gaussian matrix with gµ as its µ-th row. By elementary compactness and duality arguments, we have: 1 −Pn(A) = P h min x∈K ∥x∥2≤A ∥Gx −1n∥2 > 0 i = P h min x∈K ∥x∥2≤A max ∥λ∥2≤1{−λ⊺1n + λ⊺Gx} > 0 i , for G ∈Rn×p with i.i.d. N(0, 1) elements. By dominated convergence we have then 1 −Pn(A) = lim η→0 P h min x∈K ∥x∥2≤A max ∥λ∥2≤1{−λ⊺1n + λ⊺Gx} > η i .
(3.2) Note that in eq. (3.2), both x and λ belong to a convex and compact set (since K is closed and convex), and ψ(x, λ) = −λ⊺1n is clearly convex-concave. We can thus apply item (ii) of Proposition 3.1: 1 −Pn(A) ≤2 lim η→0 P h min x∈K ∥x∥2≤A max ∥λ∥2≤1{−λ⊺1n + ∥λ∥2g⊺x + ∥x∥2λ⊺h} ≥η i , (3.3) with g ∼N(0, Ip) and h ∼N(0, In). We then control the right-hand-side of the last equation, using that K is a cone: min x∈K ∥x∥2≤A max ∥λ∥2≤1{−λ⊺1n + ∥λ∥2g⊺x + ∥x∥2λ⊺h} = min x∈K ∥x∥2≤A max{0, ∥∥x∥2h −1n∥2 + g⊺x}, = max n 0, min x∈K ∥x∥2≤A ∥∥x∥2h −1n∥2 + g⊺x o , = max n 0, min v∈[0,A] ∥vh −1n∥2 + v min x∈K∩Sp−1 g⊺x o .
(3.4) Note that g →maxx∈K∩Sp−1[g⊺x] is 1-Lipschitz, and in particular concentrates on its average, which by definition is the Gaussian width ω(K). We use the classical result (see e.g. Theorem 3.25 of ): EJP 30 (2025), paper 126.
Page 16/46 Exact threshold for approximate ellipsoid fitting of random points Theorem 3.4. Let X1, · · · , Xn i.i.d.
∼N(0, 1). Let f : Rn →R a Lipschitz function. Then for all t ≥0: P[f(X1, · · · , Xn) −Ef(X1, · · · , Xn) ≥t] ≤exp n − t2 2∥f∥2 L o .
Therefore, for δ ∈(0, ω(K)), we have5: P n min x∈K∩Sp−1[g⊺x] ≥−ω(K) + δ o ≤e−δ2/2.
(3.5) From eqs. (3.3), (3.4) and (3.5) we have: 1 −Pn(A) ≤2 lim η→0 P h max n 0, min v∈[0,A] ∥vh −1n∥2 + v(−ω(K) + δ) o ≥η i + 2e−δ2/2. (3.6) Recall that we assumed n ≤(1 −ε) ω(K)2 and n →∞. Therefore, for n ≥n0(ε, δ) large enough we can assume n ≤(1 −ε/2) [ω(K) −δ]2. Let v⋆= v⋆(ε) = p (4 −ε)/ε such that: 1 −ε 4 EX∼N (0,1)[(v⋆X −1)2] = (v⋆)2.
Thus, for A = v⋆(ε), we have from eq. (3.6): 1 −Pn(v⋆) ≤2 lim η→0 P h max n 0, ∥v⋆h −1n∥2 + v⋆(−ω(K) + δ) o ≥η i + 2e−δ2/2.
(3.7) By the law of large numbers we have ( p →denotes convergence in probability): 1 √n∥v⋆h −1n∥2 p → p E[(v⋆X −1)2] = v⋆ p1 −ε 4 < v⋆ p1 −ε 3 .
In particular, with probability 1 −on(1), we have ∥v⋆h −1n∥2 + v⋆(−ω(K) + δ) ≤v⋆√n 1 −ε 3 −1/2 − 1 −ε 2 −1/2 < 0.
Therefore, we have by eq. (3.7): 1 −Pn(v⋆) ≤on(1) + 2e−δ2/2.
Taking the limit n →∞and then δ →∞finishes the proof6 (notice that v⋆≤2/√ε).
3.1.3 The cone of positive matrices with bounded condition number We study here the Gaussian width of the convex cone of positive semidefinite matrices with bounded condition number. We start with a classical result on the Gaussian width of S+ d [7, 2], we refer the reader to Proposition 10.2 of for a proof.
Proposition 3.5 (Gaussian width of S+ d ). The Gaussian width of S+ d satisfies: r d(d + 1) 4 −1 ≤ω(S+ d ) ≤ r d(d + 1) 4 .
In particular, notice that ω(S+ d ) ∼d/2 as d →∞. We generalize this result by asking that the matrices have bounded condition number.
5Since g d = −g, minx∈K∩Sp−1[g⊺x] d = −maxx∈K∩Sp−1[g⊺x].
6Recall that δ < ω(K) but ω(K) →∞as n →∞by hypothesis.
EJP 30 (2025), paper 126.
Page 17/46 Exact threshold for approximate ellipsoid fitting of random points Lemma 3.6 (Gaussian width of PSD matrices with bounded condition number). For any κ ≥1, define Kκ := {S ∈S+ d : λmax(S) ≤κλmin(S)}. Then Kκ is a closed convex cone, and its Gaussian width satisfies lim inf d→∞ 2ω(Kκ) d = f(κ), where κ →f(κ) ∈[0, 1] is non-decreasing, with limκ→∞f(κ) = 1.
Proof of Lemma 3.6. The fact that Kκ is a closed convex cone is easy to verify. Moreover, for any κ ≤κ′ we have Kκ ⊆Kκ′ ⊆S+ d , and ω(S+ d ) ∼d/2 by Proposition 3.5, so κ 7→f(κ) ∈[0, 1], and f is non-decreasing. To finish the proof, we show that f(κ) →1 as κ →∞. We let κ > 1. To identify Sd with Rd(d+1)/2, we use the matrix flattening function, for M ∈Sd: vec(M) := (( √ 2Mab)1≤a<b≤d, (Maa)d a=1) ∈Rd(d+1)/2, (3.8) = ((2 −δab)1/2Mab)a≤b.
It is an isometry (⟨vec(M), vec(N)⟩= Tr[MN]), and if Z ∼GOE(d) then the entries of vec(Z) satisfy vec(Z)k i.i.d.
∼N(0, 2/d). We thus reach: ω(Kκ) := E max S∈Kκ ∥S∥2 F=2d 1 2Tr[ZS] , in which Z ∼GOE(d), cf. Definition 2.1. Letting z1 ≥· · · ≥zd be the eigenvalues of Z, by Wigner’s theorem (1/d) P i δzi weakly converges as d →∞(a.s.) to σs.c.(dx) = (2π)−1√ 4 −x21{|x| ≤2}dx. Moreover, we have (taking S having same eigenvectors as Z)7: ω(Kκ) ≥E max λ1≥···λd≥0 P λ2 i ≤2d λ1≤κλd ( 1 2 d X i=1 λizi ) .
(3.9) Let us now sketch the end of the proof. We define γ(κ) := s 2 R 2 2κ−1 x2 σs.c.(dx) + 4 κ2 R 2κ−1 −2 σs.c.(dx) , λ⋆ i := γ(κ) h zi1{zi ≥2κ−1} + 2 κ1{zi < 2κ−1} i .
(3.10) Let ε > 0. Since one can show that z1 →2 a.s. as d →∞, with high probability we have λ⋆ 1 ≤κ(1 + ε)λ⋆ d. Moreover, one checks easily that d−1 Pd i=1[λ⋆ i ]2 p →2 as d →∞.
Letting µi := λ⋆ i /√1 + ε, we can therefore use {µi} to lower bound ω(Kκ) as d →∞, with κε := κ(1 + ε). This yields that, for any ε > 0: f(κε) = lim inf d→∞ 2ω(Kκε) d ≥ 1 √1 + ε v u u u t2 hR 2 2κ−1 x2 σs.c.(dx) + 2 κ R 2κ−1 −2 x σs.c.(dx) i2 R 2 2κ−1 x2 σs.c.(dx) + 4 κ2 R 2κ−1 −2 σs.c.(dx) .
(3.11) Taking the limits κ →∞and ε →0 in eq. (3.11) yields limκ→∞f(κ) ≥1, which ends the proof.
7Notice that we replaced ∥S∥2 F = 2d by ∥S∥2 F ≤2d since w.h.p. one can always find S ∈Kκ such that Tr[SZ] > 0.
EJP 30 (2025), paper 126.
Page 18/46 Exact threshold for approximate ellipsoid fitting of random points 3.1.4 Proof of Proposition 2.3 We can now complete the proof of Proposition 2.3. Recall that V = {S ∈Sd : ∀µ ∈[n], Tr[GµS] = 1}.
Since α = n/d2 < 1/4, by Lemma 3.6, we can find κ = κ(α) and ε = ε(α) such that n ≤(1−ε)ω(Kκ)2 for n large enough. Therefore, by Lemma 3.3 there exists A = A(α) > 0 such that: lim d→∞P ∃S ∈V s.t. S ⪰0 and Tr[S2] ≤Ad and λmax(S) ≤κλmin(S) = 1.
(3.12) Notice that H := n−1/2 Pn µ=1 Gµ ∼GOE(d), and thus P[∥H∥op ≤3] = 1 −od(1) .
Conditioning on this event, let S ∈V . Then Tr[HS] = √n = √αd, and thus by duality of ∥· ∥op and ∥· ∥S1: Tr|S| ≥ 1 ∥H∥op |Tr[HS]| ≥ √α 3 d.
The proof of Proposition 2.3 is then ended by noticing that: Sp(S) ⊆[λ−, λ+] ⇐ S ⪰0, Tr[S2] ≤A(α)d, λmax(S) ≤κ(α)λmin(S), Tr[S] ≥B(α)d, for some 0 < λ−≤λ+ depending only on α.
3.2 The unsatisfiable regime: Proof of Proposition 2.4 We will show the following general result on the unsatisfiability of approximate versions of a general class of random geometry problems.
Proposition 3.7. Let b ∈R, p, n →∞, K ⊆Rp a closed convex cone, with Gaussian width ω(K), (aµ)n µ=1 i.i.d.
∼N(0, Ip), and denote C(b) µ (x) := |a⊺ µx −b| the µ-th “constraint”.
We denote Vb := {x ∈Rp : ∀µ ∈[n], C(b) µ (x) = 0} a randomly-oriented affine subspace. Then we have the following: (i) Assume that b ̸= 0. Then (a) If there exists ε > 0 such that n ≤(1 −ε) ω(K)2 as n →∞, then P[Vb ∩K ̸= ∅] →n→∞1.
(b) If there exists ε > 0 such that n ≥(1 + ε) ω(K)2 as n →∞, then there exist c = c(ε, b) > 0 and η = η(ε, b) ∈(0, 1) such that lim n→∞P{∀x ∈K : #{µ ∈[n] : C(b) µ (x) > c} ≥ηn} = 1.
(ii) Assume that b = 0. Note that V0 is a linear subspace, and V0 ∩K is a closed convex cone.
(a) Assume that n ≤(1 −ε) ω(K)2 for some ε > 0. Then as n →∞, P[V0 ∩K ∩ Sp−1 ̸= ∅] →1.
(b) Assume that n ≥(1 + ε) ω(K)2 for some ε > 0. Then there exist c = c(ε, b) > 0 and η = η(ε, b) ∈(0, 1) such that, with probability 1 −on(1), the following holds for all τ ≥0: max x∈K #{µ∈[n] : C(0) µ (x)>cτ}<ηn ∥x∥2 ≤τ.
EJP 30 (2025), paper 126.
Page 19/46 Exact threshold for approximate ellipsoid fitting of random points It is clear that Proposition 3.7 ends the proof of Proposition 2.4. Indeed, seen as an element of Rd(d+1)/2, p d/2Gµ i.i.d.
∼N(0, Id(d+1)/2), see the canonical embedding in eq. (3.8). By Proposition 3.5, we have ω(S+ d )2 = d2/4 + o(d2), and thus we can apply Proposition 3.7 with p = d(d + 1)/2. The difference √ d in normalization in point (ii) of Proposition 2.4 comes from the additional √ d factor that arises when relating Gµ to a standard Gaussian.
We now focus on proving Proposition 3.7.
3.2.1 Proof of Proposition 3.7 We first show (i), assuming b ̸= 0. Results of [7, 2] show that Vb ∩K ̸= ∅with high probability if n ≤(1 −ϵ)ω(K)2 (see e.g. Theorem 8.1 of ), i.e. point (a). While the converse is also shown in these works for n ≥(1 + ε)ω(K)2, here we wish to prove the stronger statement (b). Assume therefore that n ≥(1 + ε)ω(K)2. By the union bound, for all c ≥0 and η ∈(0, 1): P(∃x ∈K : #{µ ∈[n] : C(b) µ (x) > c} < ηn) = P(∃x ∈K, ∃S ⊆[n] : |S| > (1 −η)n and ∀µ ∈S, C(b) µ (x) ≤c) ≤ X S⊆[n] |S|>(1−η)n P(∃x ∈K : ∀µ ∈S, C(b) µ (x) ≤c), (a) ≤ X k<η·n n k P(∃x ∈K : ∥{a⊺ µx −b}(1−η)n µ=1 ∥∞≤c), (b) ≤exp ηn log e η P(∃x ∈K : ∥{a⊺ µx −b}(1−η)n µ=1 ∥∞≤c).
(3.13) We used that aµ are i.i.d. in (a), and the bound Pk i=0 n i ≤(en/k)k in (b). We now make use of the following lemma (proven later on).
Lemma 3.8. Recall that n ≥(1+ε)ω(K)2. There exist c1, c2 > 0 and η0 ∈(0, 1) depending only on ε such that for any η ∈(0, η0): P(∃x ∈K : ∥{a⊺ µx −b}(1−η)n µ=1 ∥∞≤c1 · b) ≤2 exp{−nc2}.
Applying Lemma 3.8 in eq. (3.13), we can consider η = η(ε, b) ∈(0, 1/2) small enough, such that η < η0 and η log(e/η) ≤c2/2. This yields then that P(∃x ∈K : #{µ ∈[n] : C(b) µ (x) > c1 · b} < ηn) ≤2 exp{−nc2/2} →0, and ends the proof. We now prove (ii) of Proposition 3.7, assuming b = 0. (a) is here a simple consequence of the usual Gordon’s “escape through a mesh” theorem , so we focus on (b), assuming n ≥(1 + ε)ω(K)2. Note that since K is a cone, we have for all c ≥0, η ∈(0, 1): sup x∈K #{µ∈[n] : C(0) µ (x)>c}<ηn ∥x∥2 = sup{v ≥0 : ∃x ∈K ∩Sp−1 s.t. #{µ ∈[n] : C(0) µ (x) > c/v} < ηn}.
(3.14) We now use the following counterpart to Lemma 3.8 in the case b = 0, also proven later: Lemma 3.9. Recall that n ≥(1+ε)ω(K)2. There exist c1, c2 > 0 and η0 ∈(0, 1) depending only on ε such that for any η ∈(0, η0): P(∃x ∈K ∩Sp−1 : ∥{a⊺ µx}(1−η)n µ=1 ∥∞≤c1) ≤2 exp{−nc2}.
EJP 30 (2025), paper 126.
Page 20/46 Exact threshold for approximate ellipsoid fitting of random points Repeating the same reasoning as in the b ̸= 0 case, we can then find c = c(ε) > 0 and η = η(ε) ∈(0, 1/2) such that with probability 1 −on(1): ∀x ∈K ∩Sp−1 : #{µ ∈[n] : C(0) µ (x) > c} ≥ηn.
Plugging this in eq. (3.14) yields that with probability 1 −on(1), for all τ ≥0: max x∈K #{µ∈[n] : C(0) µ (x)>cτ}<ηn ∥x∥2 ≤τ, which ends the proof.
We now prove the two Lemmas 3.8 and 3.9.
3.2.2 Proofs of Lemma 3.8 and 3.9 Proof of Lemma 3.8. Note that for all t ≥0 and η ∈(0, 1): P ∃x ∈K : ∥{a⊺ µx −b}(1−η)n µ=1 ∥∞≤t = P [∃x ∈K : ∥Gx −b1m∥∞≤t] , (3.15) with m := n(1 −η), G ∈Rm×p an i.i.d. N(0, 1) matrix, and 1m the all-ones vector. We thus have: P(∃x ∈K : ∥{a⊺ µx −b}(1−η)n µ=1 ∥∞≤t) (a) = lim A→∞P [∃x ∈K : ∥x∥2 ≤A and ∥Gx −b1m∥∞≤t] , (b) = lim A→∞P h min x∈K ∥x∥2≤A ∥Gx −b1m∥∞≤t i , (3.16) where (a) follows from dominated convergence, and (b) uses that the minimum is now over a compact set since K is closed. Since ∥· ∥∞and ∥· ∥1 are dual norms, we have for all x ∈K: ∥Gx −b1m∥∞= max λ∈Rm ∥λ∥1≤1 [−bλ⊺1m + λ⊺Gx] .
We can now use the Gaussian min-max inequality (Proposition 3.1), which, together with eq. (3.16), implies: P(∃x ∈K : ∥{a⊺ µx −b}(1−η)n µ=1 ∥∞≤t) ≤lim A→∞2P[γA(g, h) ≤t] (a) ≤2P[γ(g, h) ≤t].
Here we defined: γ(g, h) := inf x∈K max λ∈Rm ∥λ∥1≤1 [−bλ⊺1m + ∥λ∥2g⊺x + ∥x∥2h⊺λ] , (3.17) and γA is defined by restricting the infimum to ∥x∥2 ≤A. Moreover, g ∼N(0, Ip), h ∼ N(0, Im). The inequality (a) holds since γ(g, h) ≤γA(g, h) for all A > 0. To conclude the proof, it therefore suffices to show: P[γ(g, h) ≤c1b] ≤2 exp{−nc2}, (3.18) for c1, c2 small enough (depending on ε, b). We use again that g →maxx∈K∩Sp−1[g⊺x] is 1-Lipschitz, and in particular concentrates on the Gaussian width by Theorem 3.4. Using that g is distributed as −g, this implies that for any u > 0: P min x∈K∩Sp−1[g⊺x] ≤−ω(K) −u ≤e−u2/2.
EJP 30 (2025), paper 126.
Page 21/46 Exact threshold for approximate ellipsoid fitting of random points Since ω(K) ≤ p n/(1 + ε) by hypothesis, we can fix δ = δ(ε) > 0 and η0 = η0(ε) > 0 such that for n large enough we have for η < η0: ω(K) + δ√m ≤ p m/(1 + ε/2) (recall that m = (1 −η)n). Thus, since K is a cone, and using the max-min inequality, we have with probability at least 1 −e−n(1−η0)δ2/2: γ(g, h) ≥inf v≥0 max λ∈Rm ∥λ∥1≤1 −bλ⊺1m + v h⊺λ −∥λ∥2(ω(K) + δ√m) .
(3.19) Let us now assume that b > 0, and let X ∼N(0, 1), and D := E[|X|] = p 2/π. We pick σ = σ(ε) ∈(0, 1) (its choice will be constrained later on), and define Aε by: E[X21{X ≤Aε}] = σ(ε).
(3.20) Finally, we define λ⋆= λ⋆(h) ∈Rm by λ⋆ µ := 1 Dmhµ1{hµ ≤Aε}.
(3.21) It is a simple exercise based on Hoeffding’s and Bernstein’s inequalities to check that for any u > 0 (recall that m = (1 −η)n ≥(1 −η0(ε))n): P[∥λ⋆∥1 ≤1] ≥1 −exp{−C1n}, P h⊺λ⋆−σ(ε)2 D ≤−u ≤exp{−C2n min(u2, u)}, P ∥λ⋆∥2 2 −σ(ε)2 mD2 ≥u n ≤exp{−C3n min(u2, u)}, P [1⊺ mλ⋆≥−C4] ≤exp{−C5n}, (3.22) for some positive constants (Ca)5 a=1, all depending on ε. In particular, the first three lines of eq. (3.22) imply that for all u ∈(0, 1), with probability at least 1 −3 exp{−C(ε)nu2}: ∥λ⋆∥1 ≤1 and h⊺λ⋆ ∥λ⋆∥2 1 ω(K) + δ√m ≥ σ(ε)2/D −u p σ(ε)2/D2 + u r 1 + ε 2, (3.23) in which we used that ω(K) + δ√m ≤√m(1 + ε/2)−1/2, and m/n ≤1. We can choose σ(ε) ∈(0, 1) sufficiently close to 1, and u(ε) ∈(0, 1) sufficiently close to 0 such that the right-hand side of eq. (3.23) is greater than 1. Combining it with the last equation of eq. (3.22) and the lower bound of eq. (3.19), we get that (with new constants c1, c2 depending on ε), with probability at least 1 −2 exp{−c2(ε)n}: γ(g, h) ≥bc1(ε), which implies eq. (3.18) and ends the proof. The case b < 0 is treated similarly, constrain-ing hµ ≥−Aε rather than hµ ≤Aε in eq. (3.21).
Proof of Lemma 3.9. Let t ≥0.
Repeating the same arguments as in the proof of Lemma 3.8 one obtains that P(∃x ∈K ∩Sp−1 : ∥{a⊺ µx}(1−η)n µ=1 ∥∞≤t) ≤2P " min x∈K∩Sp−1 max λ∈Rm ∥λ∥1≤1 {∥λ∥2g⊺x + h⊺λ} ≤t | {z } =:γ(g,h) # .
Again, we can fix η0(ε) > 0 and δ(ε) > 0 such that for η < η0 and n large enough we have ω(K) + δ√m ≤√m[1 + ε/2]−1/2. By the max-min inequality and the concentration of the Gaussian width, this implies that with probability at least 1 −e−n(1−η0)δ2: γ(g, h) ≥max λ∈Rm ∥λ∥1≤1 h⊺λ −∥λ∥2(ω(K) + δ√m) .
(3.24) EJP 30 (2025), paper 126.
Page 22/46 Exact threshold for approximate ellipsoid fitting of random points Defining again D := p 2/π so that D = E[|X|] for X ∼N(0, 1), we now define λ⋆as: λ⋆ µ := 1 Dmhµ.
We have the counterpart to eq. (3.22) for this case: for any u > 0 and τ > 0, P[∥λ⋆∥1 ≤1 + τ] ≥1 −exp{−C1nτ 2}, P h⊺λ⋆−1 D ≤−u ≤exp{−C2n min(u2, u)}, P ∥λ⋆∥2 2 − 1 mD2 ≥u n ≤exp{−C3n min(u2, u)}, (3.25) for some (Ca)3 a=1 depending on ε. We can fix u = u(ε) > 0 such that D−1 −u − s u + D−2 1 + ε/2 =: 2c1(ε) > 0, since the limit as u →0 of the left-hand side is strictly positive. Letting τ = 1, we finally get that with probability at least 1 −3 exp(−c2(ε)n) we can lower bound (using λ = λ⋆/2 such that ∥λ∥1 ≤1) γ(g, h) ≥D−1 −u 2 −1 2 r u n + 1 mD2 (ω(K) + δ√m), ≥D−1 −u 2 −1 2 s u + D−2 1 + ε/2 , ≥c1(ε).
This ends the proof.
4 Universality: Proof of Proposition 2.2 This section is devoted to the proof of Proposition 2.2. We first show in Section 4.1 a critical result on the Lipschitz constant of the “error” function appearing in eq. (2.4).
This requires controlling a random process on the operator norm sphere, which is also useful in the proof of Theorem 1.4, see Section 2.4. We leverage this control to show in Section 4.2 a general result on the universality of a quantity known as the asymptotic free entropy of the model, both for matrices arising from ellipsoid fitting and its Gaussian equivalent. This result follows from an interpolation argument. Finally, we apply these results in the so-called “low-temperature” limit in Section 4.3 to deduce Proposition 2.2.
As we mentioned, while we can not directly apply the results of , parts of Sections 4.2 and 4.3.2 closely follows their approach. We defer to Appendix D some technicalities, as well as some parts of the proof that more directly follow the arguments of .
4.1 Lipschitz constant of the energy function, and bounding random processes We show here the following result on the behavior of the error (or “energy”) function, under both models Xµ ∼Ellipse(d) and Xµ ∼GOE(d).
Lemma 4.1 (Lipschitz constant of the energy). Let n, d ≥1 and n, d →∞with α1d2 ≤n ≤ α2d2 for some 0 < α1 < α2. Let φ : R →R+ such that ∥φ′∥∞< ∞, and X1, · · · , Xn ∈Sd be generated i.i.d. according to either GOE(d) or Ellipse(d). For S ∈Sd, we define the energy: E{Xµ}(S) := 1 d2 n X µ=1 φ[Tr(XµS)].
EJP 30 (2025), paper 126.
Page 23/46 Exact threshold for approximate ellipsoid fitting of random points Then the following holds for some C > 0 (depending only on α): P sup S1,S2∈Sd |E{Xµ}(S1) −E{Xµ}(S2)| ∥S1 −S2∥op ≤C∥φ′∥∞ ≥1 −2e−n.
In other words, the energy function has a bounded Lipschitz constant (as d →∞) with respect to the operator norm. Note that this is strictly stronger than what a naive use of the triangular inequality and of the duality ∥· ∥op ↔∥· ∥S1 yields: |E{Xµ}(S1) −E{Xµ}(S2)| ∥S1 −S2∥op ≤∥φ′∥∞ d2 n X µ=1 Tr|Xµ|, since Tr|Xµ| ≳d for Xµ ∼GOE(d), and Tr|Xµ| ≳ √ d for Xµ ∼Ellipse(d). Instead, the proof of Lemma 4.1 is based on the following bounds for random processes, for which we separate the GOE(d) and Ellipse(d) setting. Lemma 4.2 is a consequence of elementary concentration results, and is proven in Appendix D.1, while Lemma 4.3 is proven in the following.
Lemma 4.2 (Bounding random processes, GOE(d) setting). Let (Gµ)n µ=1 i.i.d.
∼GOE(d).
Let r ∈[1, 2]. We assume that n ≥α1d2, for some α1 > 0. There exists C = C(α1) > 0 such that for all t > 0: P " max ∥S∥2 F =d n X µ=1 |Tr[GµS]|r !1/r ≥(C + t)n1/r # ≤exp −nt2/2 .
Lemma 4.3 (Bounding random processes, Ellipse(d) setting). Let W1, · · · , Wn be drawn i.i.d. from Ellipse(d).
Let r ∈[1, 4/3].
We assume that α1d2 ≤n ≤α2d2, for some 0 < α1 < α2. There are constants C1, C2 > 0 (that might depend on α1, α2) such that for all t > 0: P h max ∥S∥op=1 n X µ=1 |Tr(WµS)|r ≥n(C1 + t) i ≤2 exp n −C2 min(nt 2 r , n 1 4 + 1 r t 1 r ) o .
Proof of Lemma 4.1. We finish here the proof, assuming Lemmas 4.2 and 4.3. By the mean value theorem and the duality ∥· ∥op ↔∥· ∥S1: |E{Xµ}(S1) −E{Xµ}(S2)| ≤ sup S∈Sd ⟨∇SE{Xµ}(S), S1 −S2⟩ , ≤∥S1 −S2∥op sup S∈Sd ∥∇SE{Xµ}(S)∥S1.
Again using the duality ∥· ∥op ↔∥· ∥S1: ∥∇SE{Xµ}(S)∥S1 = 1 d2 n X µ=1 Xµφ′[Tr(XµS)] S1 , = 1 d2 sup ∥R∥op=1 n X µ=1 Tr[XµR]φ′[Tr(XµS)], ≤∥φ′∥∞ d2 sup ∥R∥op=1 n X µ=1 |Tr[XµR]|.
Using Lemmas 4.2 and 4.3 in the case r = 1, we reach the sought statement.
EJP 30 (2025), paper 126.
Page 24/46 Exact threshold for approximate ellipsoid fitting of random points Remark I – Note that a naive argument using that Tr[WµS] is a sub-exponential random variable yields Lemma 4.3 for r = 1, which is already enough to deduce Lemma 4.1.
However, Lemma 4.3 is also used later in the proof of Theorem 1.4, see Section 2.4. Since Tr[WµS] is the sum of many independent random variables, we can leverage its two-tailed behavior (by Bernstein’s inequality) to prove Lemma 4.3 for all r ≤4/3, yielding the limitation r < 4/3 in Theorem 1.4. It is possible that a finer analysis could lead to a proof of Lemma 4.3 for the case 4/3 ≤r ≤2, which would in turn imply Theorem 1.4 for r < 2.
Remark II – We give an informal argument as to why we can not hope to extend Lemma 4.2 nor 4.3 for r > 2. Indeed, in the GOE(d) setting, the choice S = √ dGµ/∥Gµ∥F (for some µ ∈[n]) yields that the objective function is at least √ d∥Gµ∥F ≳√n. In the Ellipse(d) setting, assume that r > 2. If 1/q + 1/r = 1, then by the dualities ℓp ↔ℓq and ∥· ∥op ↔∥· ∥S1: max ∥S∥op=1 n X µ=1 |Tr(WµS)|r !1/r = max ∥λ∥q=1 n X µ=1 λµWµ S1 .
(4.1) Let us lower bound the right-hand side of eq. (4.1). Let β ∈(0, 1), p = βn, and λ1 = · · · = λp = p−1/q > λp+1 = · · · = λn = 0. Then ∥λ∥q = 1. Moreover, n X µ=1 λµWµ S1 = p1−1/qd−1/2 1 p p X µ=1 xµx⊺ µ −Id S1 .
By classical results of concentration of Wishart matrices , we know that since p ≲d2 then 1 p p X µ=1 xµx⊺ µ −Id S1 ≳d3/2 √p ≳ s d β , where ≳might hide constants that depend on α. We then reach: n X µ=1 λµWµ S1 ≳β1/2−1/qn1−1/q.
Since r > 2, one has 1/2 −1/q < 0. Letting β going to 0, this shows that any bound of the type max ∥λ∥q=1 n X µ=1 λµWµ S1 ≤C(α)n1−1/q can not hold.
Proof of Lemma 4.3. Throughout this proof, constants might depend on α1, α2. Let us define, for any S ∈Sd: Y (S) := n X µ=1 |Tr(WµS)|r, X(S) := Y (S) −EY (S).
(4.2) For a set E ⊆Sd, a norm ∥· ∥on Sd, and a value ε > 0, an ε-net of E is a set A ⊆E such that every point of E is at distance at most ε from A (for the distance induced by ∥· ∥).
We define the covering number N(E, ∥· ∥, ε) as the smallest possible cardinality of an ε-net of E. We fix ε ∈(0, 1). It follows from classical covering number bounds that if T := {S ∈Sd : ∥S∥op = 1}, then log N(T, ∥· ∥op, ε) ≤d(d + 1) 2 log 3 ε.
(4.3) EJP 30 (2025), paper 126.
Page 25/46 Exact threshold for approximate ellipsoid fitting of random points Let us fix N an ε-net of T for ∥· ∥op, of minimal cardinality.
If we let S⋆:= arg max∥S∥op=1 Y (S), and S0 ∈N with ∥S⋆−S0∥op ≤ε, then we have by Minkowski’s inequality (recall r ≥1): ∥{Tr[WµS⋆]}n µ=1∥r −∥{Tr[WµS0]}n µ=1∥r ≤∥{Tr[Wµ(S0 −S⋆)]}n µ=1∥r ≤ε max ∥S∥op=1 ∥{Tr[WµS]}n µ=1∥r.
This implies max ∥S∥op=1 n X µ=1 |Tr(WµS)|r ≤ 1 (1 −ε)r max S∈N n X µ=1 |Tr(WµS)|r.
(4.4) We combine the covering number upper bound of eq. (4.3) and the relation of eq. (4.4) with the following lemma, proven later on, which bounds the deviation probability of the process for a given S.
Lemma 4.4 (Tail bound at a fixed point). With the notations of eq. (4.2), we have, for any S ∈T : (i) E [Y (S)] ≤C1n.
(ii) For all t ≥0: P[X(S) ≥nt] ≤2 exp n −C2 min(nt2, nt 2 r , n 1 4 + 1 r t 1 r ) o .
Note that the above constants may depend on r.
Picking ε = 1/2 and performing a union bound over N, we reach using eqs. (4.3), (4.4) and Lemma 4.4: P[sup S∈T Y (S) ≥n(C1 + t)] ≤2 exp n C3n −C2 min(nt2, nt 2 r , n 1 4 + 1 r t 1 r ) o .
We thus have for any t ≥1: P[sup S∈T Y (S) ≥n(C1 + t)] ≤2 exp n C3n −C2nt 2 r o if t ≤n1−3r 4 , exp n C3n −C2n 1 4 + 1 r t 1 r o if t ≥n1−3r 4 .
Note that n1/4+1/r ≥n since r ≤4/3. Therefore, for (new) constants C1, C2 we have for all t > 0: P[sup S∈T Y (S) ≥n(C1 + t)] ≤2 exp n −C2 min(nt 2 r , n 1 4 + 1 r t 1 r ) o , which ends the proof.
We now tackle Lemma 4.4.
Proof of Lemma 4.4. We start with (i). One has E[Y (S)] = nE[|Tr(W1S)|r]. Let Z := Tr(W1S) d = (x⊺Sx −Tr[S])/ √ d, with x ∼N(0, Id). Since ∥S∥op = 1, we have by Hanson-Wright’s inequality : P[|Z| ≥t] ≤2 exp n −C min dt2 ∥S∥2 F , √ dt o , ≤2 exp n −C min t2, √ dt o , (4.5) EJP 30 (2025), paper 126.
Page 26/46 Exact threshold for approximate ellipsoid fitting of random points where C > 0 is an absolute constant, and we used that ∥S∥F ≤ √ d∥S∥op. Separating the sub-Gaussian and the sub-exponential parts of the tail we have, we have for all p ≥1: E[|Z|p] = p Z ∞ 0 du up−1 P[|Z| ≥u], ≤2p " Z √ d 0 du up−1 e−Cu2 + Z ∞ √ d du up−1 e−C √ du # , ≤2p " Z ∞ 0 du up−1 e−Cu2 + e−Cd Z ∞ 0 du (u + √ d)p−1 e−C √ du # , (a) ≤2p " 1 2Cp/2 Γ(p/2) + e−Cd max(1, 2p−2) Z ∞ 0 du [up−1 + d(p−1)/2] e−C √ du # , ≤2p " Γ(p/2) 2Cp/2 + e−Cd max(1, 2p−2) Γ(p) Cpdp/2 + d(p−2)/2 C # .
We used in (a) that (a + b)x ≤max(1, 2x−1)(ax + bx) for all a, b, x ≥0. Using Minkowski’s inequality, we reach that for all p ≥1: E[|Z|p]1/p ≤C1 √p + C2e−C3d p p √ d + d 1 2 −1 p , (4.6) for some positive constants (Ca)3 a=1 independent of p and d. Informally, the sub-Gaussian tail dominates the first moments of Z since the sub-exponential tail only kicks in at the scale O( √ d). Eq. (4.6) implies claim (i) of Lemma 4.4 by taking p = r (since the second term goes to 0 as d →∞for any fixed p).
We turn to (ii). We make use of classical tail bounds for sub-Weibull random variables, recalled in Lemma A.1. Denoting Zµ := Tr(WµS), we have X(S) = Pn µ=1{|Zµ|r−E[|Zµ|r]}.
We decompose X(S) in two parts, i.e. X(S) = X1(S) + X2(S), with X1(S) := n X µ=1 h min(|Zµ|, √ d)r −E{min(|Zµ|, √ d)r} i , X2(S) := n X µ=1 [|Zµ|r −dr/2]1{|Zµ| > √ d} −E n [|Zµ|r −dr/2]1{|Zµ| > √ d} o .
(4.7) We will successively bound X1(S), X2(S). To lighten the notations, we do not write their dependency on S in what follows. Observe that (Zµ)n µ=1 are i.i.d. random variables, and that they satisfy the tail bound of eq. (4.5).
Bounding X1 – Denoting Tµ := min(|Zµ|, √ d), we have P[Tµ ≥t] ≤2 exp{−C2t2} by the tail bound of eq. (4.5). Moreover, E[T r µ] ≤E[|Zµ|r] ≤C1 (depending only on r) by eq. (4.6). Therefore, for every t > C1, we have P[|T r µ −E[T r µ]| ≥t] = P[T r µ ≥E[T r µ] + t] ≤2e−C2(t+E[T r µ])2/r ≤2e−C2t2/r.
This implies that P[|T r µ −E[T r µ]| ≥t] ≤2e−Ct2/r for all t ≥0 and some (new) constant C > 0, depending only on r. We can thus apply (i) of Lemma A.1 for q = 2/r ∈[1, 2] and ai = 1/n (so ∥a∥2 2 = ∥a∥q q⋆= n−1), which yields that for all t ≥0 P[|X1| ≥nt] = P " 1 n n X µ=1 {T r µ −E[T r µ]} ≥t # ≤2 exp n −Cn min(t2, t2/r) o .
(4.8) EJP 30 (2025), paper 126.
Page 27/46 Exact threshold for approximate ellipsoid fitting of random points Bounding X2 – We proceed similarly, using (ii) of Lemma A.1. Letting Uµ := dr/2[|Zµ|r −dr/2]1{|Zµ| > √ d}, then Uµ ≥0, and moreover, by the Cauchy-Schwarz inequality: E[Uµ] ≤dr/2q E|Zµ|2r q P[|Zµ| > √ d], ≤C1e−C2d, for some C1, C2 > 0 depending only on r, using the moments and tail bound of eqs. (4.5) and (4.6). Repeating the argument used on Tµ above (using this time the second part of the tail of eq. (4.5)), we then reach that for all t ≥0: P[|Uµ −E[Uµ]| ≥t] ≤2e−Ct1/r.
We can then apply (ii) of Lemma A.1 with q = 1/r ∈[1/2, 1] to reach: P[|X2| ≥nt] = P " 1 n n X µ=1 {Uµ −E[Uµ]} ≥tdr/2 # ≤2 exp n −C min(ndrt2, d1/2(nt)1/r) o , ≤2 exp n −C min(n1+r/2t2, n1/4+1/rt1/r) o , (4.9) using that α1d2 ≤n ≤α2d2.
We conclude the proof of Lemma 4.4 by combining eqs. (4.8) and eq. (4.9), along with the union bound P[|X| ≥nt] ≤P[|X1| ≥nt/2] + P[|X2| ≥nt/2].
4.2 Free entropy universality for matrix models In this section we state and prove a general universality theorem for the asymptotic free entropy in a large class of matrix models, under a “uniform one-dimensional central limit theorem” assumption (or pointwise normality). We first need to define such an assumption.
Definition 4.5 (Uniform pointwise normality). Let d ≥1, and ρ a probability distribution on Sd. We say that ρ satisfies a one-dimensional CLT with respect to the set Ad ⊆Sd if: (i) The mean and covariance of ρ are matching the GOE(d) distribution, i.e. for W ∼ρ and G ∼GOE(d), we have E[W] = E[G] = 0 and for all i ≤j and k ≤l: E[WijWkl] = E[GijGkl] = δikδjl(1 + δijkl)/d.
(ii) For any bounded Lipschitz function ϕ, we have: lim d→∞sup S∈Ad EW ∼ρ ϕ Tr[WS] −EG∼GOE(d) ϕ Tr[GS] = 0.
(4.10) We can now state the universality theorem for the free entropy. Its proof is in great part an adaptation of the proof arguments for Theorem 1 and Lemma 1 in (see also [17, 11, 9]). We sketch the ideas of its proof in the following, deferring some technicalities and adaptations of the arguments of to appendices.
Theorem 4.6 (Free entropy universality for matrix models). Let n, d ≥1 and n, d →∞ with α1d2 ≤n ≤α2d2 for some 0 < α1 ≤α2. We are given: (i) P0 a probability distribution on Sd, such that supp(P0) ⊆B2(C0 √ d), for a constant C0 > 0.
(ii) φ : R →R+ a bounded differentiable function with bounded derivative.
EJP 30 (2025), paper 126.
Page 28/46 Exact threshold for approximate ellipsoid fitting of random points (iii) A series of symmetric convex sets Ad such that supp(P0) ⊆Ad.
(iv) ρ a probability distribution on Sd, which satisfies a one-dimensional CLT with respect to Ad as per Definition 4.5.
For W1, · · · , Wn ∈Sd we define the free entropy: Fd({Wµ}) := 1 d2 log Z P0(dS) exp n − n X µ=1 φ (Tr[WµS]) o .
(4.11) Then for any bounded differentiable function ψ with bounded Lipschitz derivative we have lim d→∞ E{Wµ} i.i.d.
∼ρψ[Fd({Wµ})] −E{Gµ} i.i.d.
∼GOE(d)ψ[Fd({Gµ})] = 0.
(4.12) Remark I – One could straightforwardly weaken the hypothesis supp(P0) ⊆A in Theo-rem 4.6 to the weaker condition d−2 log P0(Ac) →−∞as d →∞.
Remark II – Note that our setup differs slightly from the one of , as we consider distributions P0 with possibly continuous support, and (more importantly) for a fixed S ∈Sd, the projections {Tr[WµS]}n µ=1 are not sub-Gaussian when Wµ ∼Ellipse(d), but only sub-exponential. Nevertheless, we will see that the approach of can in large part be adapted to prove Theorem 4.6, thanks to the results we showed in Section 4.1.
Sketch of proof of Theorem 4.6 – Since supp(P0) ⊆Ad, the integral in eq. (4.11) can be restricted to S ∈Ad. We make use of an interpolation argument to show the universality of the free entropy. We define, for t ∈[0, π/2] and µ ∈[n]: Uµ(t) := cos(t)Wµ + sin(t)Gµ, e Uµ(t) := ∂Uµ(t) ∂t = −sin(t)Wµ + cos(t)Gµ.
(4.13) Note that {Uµ}n µ=1 are still i.i.d., and are smooth functions of t. Moreover, if Wµ was also a GOE(d) matrix, then Uµ(t), e Uµ(t) would be independent GOE(d) matrices. By the fundamental theorem of calculus: |Eψ[Fd(W)] −Eψ[Fd(G)]| = Z π/2 0 ∂ ∂t{E ψ[Fd(U(t))]}dt (a) ≤ Z π/2 0 E∂ψ[Fd(U(t))] ∂t dt, (4.14) where (a) follows by dominated convergence since ψ[Fd(U(t))] is continuously differen-tiable on [0, π/2], and the triangular inequality. We will deduce Theorem 4.6 if we can show the following two lemmas: Lemma 4.7 (Domination). Under the hypotheses of Theorem 4.6: Z π/2 0 sup d≥1 E∂ψ[Fd(U(t))] ∂t dt < ∞.
Lemma 4.8 (Pointwise limit). Under the hypotheses of Theorem 4.6, for any t ∈(0, π/2): lim d→∞E ∂ψ[Fd(U(t))] ∂t = 0.
Indeed, plugging Lemmas 4.7 and 4.8 in eq. (4.14) and taking the d →∞limit using dominated convergence ends the proof of Theorem 4.6. We therefore focus on proving these two lemmas in the following.
EJP 30 (2025), paper 126.
Page 29/46 Exact threshold for approximate ellipsoid fitting of random points As it will be useful, we state the result of the elementary computation of the derivative: ∂ψ[Fd(U(t))] ∂t (4.15) = −ψ′[Fd(U(t))] d2 n X µ=1 R P0(dS) e−P ν φ(Tr[Uν(t)S]) Tr[S e Uµ(t)] φ′(Tr[Uµ(t)S]) R P0(dS) e−P ν φ(Tr[Uν(t)S]) .
Because {Gµ, Wµ} are i.i.d. we get further: E ∂ψ[Fd(U(t))] ∂t (4.16) = −n d2 E " ψ′[Fd(U(t))] R P0(dS) e−P ν φ(Tr[Uν(t)S]) Tr[S e U1(t)] φ′(Tr[U1(t)S]) R P0(dS) e−P ν φ(Tr[Uν(t)S]) # .
Note that if Wµ was also a GOE(d) matrix, for any t, Uµ(t) and e Uµ(t) would be independent GOE(d) matrices. The main idea behind the interpolation is that the matrix Wµ will only appear through some one-dimensional projection with a matrix S. We will then use Definition 4.5 to argue that one can effectively replace Wµ by a GOE(d) matrix, which by the argument above would mean that we can consider the case of independent GOE(d) matrices Uµ(t) and e Uµ(t). In this case, the RHS of eq. (4.16) would be 0, since there is only a single term involving e U1(t), which has zero mean: this crucial idea is the intuition behind Lemma 4.8.
The details of the proofs of Lemmas 4.7 and 4.8 are fairly technical and substan-tially follow the ones of their counterparts in . For this reason, we defer them to Appendix D.2.
4.3 Proof of Proposition 2.2 4.3.1 Consequences of universality for ellipsoid fitting We investigate here the consequences of Theorem 4.6 for the ellipsoid fitting prob-lem. It follows by the Berry-Esseen central limit theorem that the distribution Ellipse(d) satisfies uniform pointwise normality on a large set of matrices (in the sense of Definition 4.5).
Lemma 4.9 (One-dimensional CLT for the ellipse problem). Let d ≥1 and W ∼Ellipse(d).
Fix any η ∈(0, 1/2) Let Ad := {S ∈Sd : Tr[|S|3] ≤d3/2−η}. Then Ad is convex and symmetric, and the law of W satisfies a one-dimensional CLT with respect to Ad, in the sense of Definition 4.5.
Remark – This lemma makes crucial use of the Gaussian nature of the vectors, and more specifically it relies on their rotation invariance and the first moments of their norm, as is clear from the proof. On the other hand, for vectors sampled from other distributions, such as x ∼Unif({±1}d) or x ∼Unif(Sd−1), it is easy to see that Lemma 4.9 can not hold: indeed, S = Id is such that Tr[SW] = 0 deterministically, while Tr[SG] = Tr[G] ∼N(0, 2) for G ∼GOE(d). This is consistent, as in the example of these two distributions there always exists an ellipsoid fit, which is the sphere itself, and therefore Theorem 1.4 can not possibly hold.
Proof of Lemma 4.9. Note that Ad is a centered ball for the S3-norm, and is therefore con-vex and symmetric. The proof of the first and second moments condition of Definition 4.5 is immediate via a simple calculation. We focus on proving condition (ii) of Definition 4.5.
EJP 30 (2025), paper 126.
Page 30/46 Exact threshold for approximate ellipsoid fitting of random points Fix S ∈Ad, with eigenvalues (λi)d i=1. With W ∼Ellipse(d) and G ∼GOE(d), let ( X := Tr[SW], Y := Tr[SG].
It is trivial to see that Y ∼N(0, 2Tr[S2]/d), so that Y d = d−1/2 Pd i=1 λizi for zi i.i.d.
∼N(0, 2).
Moreover, X d = 1 √ d d X i=1 λi(x2 i −1), with xi i.i.d.
∼N(0, 1). We use the Berry-Esseen central limit theorem, and in particular the formulation of Chapter 11 of – itself a simple consequence of the Lindeberg exchange method.
Lemma 4.10 (Corollary 11.59 of ). There exists a universal constant C > 0 such that the following holds. Let p ≥1 and X1, · · · , Xp and Y1, · · · , Yp be independent random variables, such that E[Xi] = E[Yi] and E[X2 i ] = E[Y 2 i ] for all i ∈[p]. Let ϕ : R →R a Lipschitz function with Lipschitz constant ∥ϕ∥L. Then E ϕ p X i=1 Xi !
−E ϕ p X i=1 Yi ! ≤C∥ϕ∥L " p X i=1 E |Xi|3 + E |Yi|3 #1/3 .
Lemma 4.10 yields: |E ϕ(X) −E ϕ(Y )| ≤C∥ϕ∥L B Tr[|S|3] d3/2 1/3 , with B = E[|z2 −1|3] + 23/2E|z|3 for z ∼N(0, 1). Using the definition of Ad, we reach: sup S∈Ad |E ϕ(X) −E ϕ(Y )| ≤C∥ϕ∥L sup S∈Ad 1 √ d Tr|S|3 d 1/3 ≤C∥ϕ∥L d−η/3 →0.
This ends the proof.
We can now state the main result of this section, a corollary of Theorem 4.6 and Lemma 4.9.
Corollary 4.11 (Universality for ellipsoid fitting). Let n, d ≥1 and n, d →∞with α1d2 ≤ n ≤α2d2 for some 0 < α1 ≤α2. Let P0 be a probability distribution such that supp(P0) ⊆ Bop(C0) for some constant C0 > 0. Let φ : R →R+ a bounded differentiable function with bounded derivative. For X1, · · · , Xn ∈Sd we define the free entropy: Fd({Xµ}) := 1 d2 log Z P0(dS) exp n − n X µ=1 φ (Tr[XµS]) o .
Then for any ψ such that ∥ψ∥∞, ∥ψ′∥∞, ∥ψ′∥L < ∞we have lim d→∞ E{Wµ} i.i.d.
∼Ellipse(d)ψ[Fd({Wµ})] −E{Gµ} i.i.d.
∼GOE(d)ψ[Fd({Gµ})] = 0.
(4.17) We notice that the only requirement for Corollary 4.11 to hold is supp(P0) ⊆ BF (C √ d) ∩B3(d3/2−η) for some C > 0 and η > 0, a weaker requirement than supp(P0) ⊆ Bop(C0).
Proof of Corollary 4.11. Since Bop(C0) ⊆B2(C0 √ d), hypothesis (i) of Theorem 4.6 is satisfied.
Condition (ii) is satisfied by hypothesis.
Since Bop(C0) ⊆B3(C0d1/3) ⊆ B3(d1/2−η) for any η ∈(0, 1/6), condition (iii) of Theorem 4.6 is satisfied with Ad = B3(d1/2−η). Finally, Lemma 4.9 verifies condition (iv) for this choice of Ad. All in all we can apply Theorem 4.6, from which the conclusion follows.
EJP 30 (2025), paper 126.
Page 31/46 Exact threshold for approximate ellipsoid fitting of random points 4.3.2 Proof of Proposition 2.2 We are now ready to prove Proposition 2.2, taking a “small-temperature” limit. Such arguments are classical in rigorous statistical mechanics, see e.g. Appendix A of .
Notice that the restriction B ⊆Bop(C0) will be critical because we proved an upper bound on the Lipschitz constant of the energy for the operator norm, cf. Lemma 4.1.
Recall the definition of the energy function: E{Xµ}(S) := 1 d2 n X µ=1 φ[Tr(XµS)].
We fix η ∈(0, 1), and Nη ⊆B a minimal η-net of B for ∥· ∥op. Since B ⊆Bop(C0) and dim(Sd) = d(d + 1)/2, it follows by standard covering number upper bounds [38, 37] that log |Nη| = log N(B, ∥· ∥op, η) ≤log N Bop(C0), ∥· ∥op, η 2 ≤d2 log K η , (4.18) for some K > 0 depending on C0. Recall the definition of GSd({Xµ}) in eq. (2.4). We define: GSd(η, {Xµ}) := inf S∈Nη E{Xµ}(S).
We will show the two lemmas: Lemma 4.12. For any η > 0 and any ψ such that ∥ψ∥∞, ∥ψ′∥∞, ∥ψ′∥L < ∞: lim d→∞ E{Wµ} i.i.d.
∼Ellipse(d)ψ[GSd(η, {Wµ})] −E{Gµ} i.i.d.
∼GOE(d)ψ[GSd(η, {Gµ})] = 0.
Lemma 4.13. Let X1, · · · , Xn i.i.d.
∼ρ, with ρ ∈{GOE(d), Ellipse(d)}. Then, with probabil-ity at least 1 −2e−n: |GSd(η, {Xµ}) −GSd({Xµ})| ≤C∥φ′∥∞· η.
These results are proven in the following, let us first see how they end the proof of Proposition 2.2. We fix η ∈(0, 1). We have |Eψ[GSd({Wµ})] −Eψ[GSd({Gµ})]| ≤|Eψ[GSd(η, {Wµ})] −Eψ[GSd(η, {Gµ})]| + ∥ψ∥L X X∈{W,G} E|GSd(η, {Xµ}) −GSd({Xµ})|.
The first term goes to 0 as d →∞by Lemma 4.12. Finally, using Lemma 4.13 and the Cauchy-Schwarz inequality, we have: E|GSd(η, {Xµ}) −GSd({Xµ})| ≤e−n/2q 2E|GSd(η, {Xµ}) −GSd({Xµ})|2 + C∥φ′∥∞η, (a) ≤4α∥φ∥∞e−n/2 + C∥φ′∥∞η, using in (a) that |E(S)| ≤α∥φ∥∞. Letting d →∞, we get lim d→∞|Eψ[GSd({Wµ})] −Eψ[GSd({Gµ})]| ≤C∥φ′∥∞∥ψ∥L · η.
Taking the limit η →0 ends the proof of eq. (2.5). The claim of eq. (2.6) can be obtained easily by picking ψ approximating an indicator function, see e.g. Section A.1.3 of for a detail of this argument.
EJP 30 (2025), paper 126.
Page 32/46 Exact threshold for approximate ellipsoid fitting of random points Proof of Lemma 4.12. We define, for β > 0: Fd(η, β, {Xµ}) := 1 d2β log 1 |Nη| X S∈Nη exp n −β n X µ=1 φ (Tr[XµS]) o .
Using Corollary 4.11 with P0 being the uniform distribution over Nη, we have, for any β > 0: lim d→∞|Eψ[Fd(η, β, {Wµ})] −Eψ[Fd(η, β, {Gµ})]| = 0.
(4.19) Moreover, for any fixed d, η, we have GSd(η, {Xµ}) = limβ→∞Fd(η, β, {Xµ}). Thus: |GSd(η, {Xµ}) −Fd(η, β, {Xµ})| ≤ Z ∞ β ∂Fd(η, s, {Xµ}) ∂s ds.
(4.20) Defining the “Gibbs” measure for S ∈Nη: Pβ(S) := exp n −β Pn µ=1 φ (Tr[XµS]) o P S′∈Nη exp n −β Pn µ=1 φ (Tr[XµS′]) o, it is easy to check that ∂Fd(η, s, {Xµ}) ∂s = 1 s2d2 X S∈Nη Ps(S) log Ps(S) + log |Nη| , (a) ≤ 1 s2d2 log |Nη|, (b) ≤1 s2 log K η , where (a) follows from the fact that, the uniform distribution over Nη maximizes the entropy, and (b) is eq. (4.18). Plugging the result back in eq. (4.20) we get: |GSd(η, {Xµ}) −Fd(η, β, {Xµ})| ≤log K η Z ∞ β ds s2 , ≤1 β log K η .
(4.21) Combining eqs. (4.19) and (4.21) we get, for any β > 0: lim sup d→∞ |Eψ[GSd(η, {Wµ})] −Eψ[GSd(η, {Gµ})]| ≤∥ψ∥L lim sup d→∞ X X∈{W,G} E|GSd(η, {Xµ}) −Fd(η, β, {Xµ})|, ≤2∥ψ∥L β log K η .
Taking the limit β →∞ends the proof of Lemma 4.12.
Proof of Lemma 4.13. Note that GSd(η, {Xµ}) ≥GSd({Xµ}) since Nη ⊆B. The other side of this inequality is a direct consequence of Lemma 4.1. Indeed, assuming that E(S) is C∥φ′∥∞-Lipschitz with respect to the operator norm, let us fix S⋆∈B such EJP 30 (2025), paper 126.
Page 33/46 Exact threshold for approximate ellipsoid fitting of random points that E(S⋆) = GSd({Xµ}) (since B is closed and bounded it is compact, therefore this minimizer exists). Letting S ∈Nη such that ∥S⋆−S∥op ≤η, we have GSd({Xµ}) = E(S⋆), ≥E(S) −|E(S⋆) −E(S)|, ≥GSd(η, {Xµ}) −C∥φ′∥∞· η, which ends the proof.
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Roman Vershynin, High-dimensional probability: An introduction with applications in data science, vol. 47, Cambridge university press, 2018. MR3837109 Acknowledgments. The authors are grateful to March Boedihardjo, Tim Kunisky, Petar Nizi´ c-Nikolac, and Joel Tropp for insightful discussions and suggestions.
A A classical concentration inequality We will make use of the following elementary concentration inequality, a generaliza-tion of Bernstein’s inequality for ψq tails [33, 15, 1].
Lemma A.1 (Tail of sum of independent sub-Weibull random variables –). Let q ∈(0, 2], and W1, · · · , Wn be i.i.d. centered random variables satisfying P[|W1| ≥t] ≤2e−Ctq.
(i) If q ∈[1, 2], then for all a ∈Rn and all t ≥0: P " n X µ=1 aµWµ ≥t # ≤2 exp n −c min t2 ∥a∥2 2 , tq ∥a∥q q⋆ o , where q⋆∈[2, +∞] with 1/q + 1/q⋆= 1.
(ii) If q ∈[1/2, 1], then for all t > 0 P " 1 n n X µ=1 Wµ ≥t # ≤2 exp −cq min(nt2, (nt)q) .
This lemma is stated in , see Lemmas 3.6 and 3.7 – and eq. (3.7) – and is a consequence of the same result for symmetric Weibull random variables .
B Fitting error of the sphere We show here eq. (1.6). By Bernstein’s inequality , we have for all µ ∈[n] and u ≥0: P ∥xµ∥2 d −1 ≥u ≤2 exp −Cd min(u, u2) .
As a consequence, if Xµ := √ d(∥xµ∥2/d−1), then8 ∥Xµ∥ψ1 ≤C. Let Yµ := |Xµ|r −E[|Xr µ|].
By the central limit theorem, Xµ d →N(0, 2) as d →∞. One shows easily that (e.g. for ε = 2): sup d≥1 E[|Xµ|2+ε] < ∞, and thus (since r ≤2) |Xµ|r is uniformly integrable as d →∞.
This implies (cf.
Theorem 3.5 of ) that E|Xµ|r →E|Z|r with Z ∼N(0, 2).
Notice that E[|Z|r] = 2rΓ([r + 1]/2)/√π.
8Where for q ∈[1, ∞), we defined the Orlicz norm of a random variable X as ∥X∥ψq := inf{t > 0 : E exp(|X|q/tq) ≤2}. In particular if ∥X∥ψ1 < ∞then X is said to be a sub-exponential random variable. We refer to for more details on these classical definitions.
EJP 30 (2025), paper 126.
Page 36/46 Exact threshold for approximate ellipsoid fitting of random points Let q := 1/r ∈[1/2, 1]. Since ∥Xµ∥ψ1 ≤C, we have ∥|Xµ|r∥ψq ≤C, and thus ∥Yµ∥ψq ≤ C′. We can then use Lemma A.1, and we get: P " 1 n n X µ=1 (|Xµ|r −E[|Xµ|r]) ≥t # ≤2 exp n −cr min(nt2, (nt)1/r) o .
Combining the above, we get that for any ε > 0, we have with probability 1 −od(1): E[|Z|r] −ε ≤1 n n X µ=1 √ d ∥xµ∥2 d −1 r ≤E[|Z|r] + ε.
C Towards exact ellipsoid fitting We show here the following proposition.
Proposition C.1 (From approximate to exact ellipsoid fitting). Let n, d →∞with n/d2 → α ∈(0, 1/2). Let {Xµ}n µ=1 be symmetric random matrices, and Hµν := Tr[XµXν]. Assume that: (i) With probability 1 −od(1), ∥H−1∥op ≤Cn−1/2 for some C = C(α) > 0.
(ii) There exists λ−> 0 (depending only on α) such that p-lim n→∞ min S⪰λ−Id 1 √n n X µ=1 |Tr(XµS) −1|2 = 0.
(C.1) Then, with probability 1 −od(1) there exists S ⪰0 such that Tr[XµS] = 1 for all µ ∈[n].
Let us emphasize that given our current proof of Theorem 1.4 (cf. Section 2.4), if the assumptions of Proposition C.1 hold for Xµ ∼Ellipse(d) and α < 1/4, then the first part of Conjecture 1.2 will hold.
However, the proof of Proposition C.1 is rather naive, as it crudely bounds the operator norm distance of the minimizer of eq. (C.1) to a subspace V (the affine subspace of solutions to the linear constraints Tr[XµS] = 1) by its distance in Frobenius norm. For these reasons, the assumptions of Proposition C.1 may be far from being optimal.
Remark I – Note that the condition (i) is clearly satisfied if Xµ i.i.d.
∼ GOE(d) and α ∈(0, 1/2), since H is then distributed as a Wishart matrix. On the other hand, while we expect it to hold as well for Xµ i.i.d.
∼Ellipse(d), this condition is (to the best of our knowledge) not known unless α is small enough: interestingly, this was one of the limitations in the recent works [5, 36, 16] that proved that ellipsoid fitting is feasible for α sufficiently small.
Remark II – Note that it is sufficient for eq. (C.1) to hold that there exists r ∈[1, 2] such that: p-lim n→∞ min S⪰λ−Id 1 nr/4 n X µ=1 |Tr(XµS) −1|r = 0.
At the moment, our proof of Theorem 1.4 (cf. Lemma 2.6) only implies (for r < 4/3) a similar statement with a prefactor 1/n rather than the required 1/nr/4. It would thus need to be improved to show that eq. (C.1) holds for the ellipsoid fitting setting.
Proof of Proposition C.1. Let V := {S ∈Sd : Tr[XµS] = 1, ∀µ ∈[n]} be the affine space of solutions to the constraints. Let ε > 0, and ˆ S ⪰λ−Id such that 1 √n n X µ=1 |Tr(Xµ ˆ S) −1|2 ≤ε.
EJP 30 (2025), paper 126.
Page 37/46 Exact threshold for approximate ellipsoid fitting of random points Note that for all M ∈Sd, if ∥M∥op ≤λ−, then ˆ S + M ⪰0. In particular, ∥M∥F ≤λ−⇒ ˆ S + M ⪰0. In order to conclude it thus suffices to show that dF( ˆ S, V ) ≤λ−. The following lemma is an elementary geometrical result: Lemma C.2 (Euclidean distance to an affine subspace –). Let d ≥1 and 1 ≤r ≤d two integers. Let (ak, bk)r k=1 ∈(Rd × R)r, with (ak)r k=1 linearly independent. We define G := {x ∈Rd : a⊺ kx + bk = 0, ∀k ∈[r]}. Then, for any y ∈Rd: d2(y, G)2 = v⊺H−1v, in which vk := a⊺ ky + bk, and Hkk′ = ⟨ak, ak′⟩is the Gram matrix of the {ak}.
We now condition on the event of condition (i). For any ε > 0, applying Lemma C.2 yields (with probability 1 −od(1)): dF( ˆ S, V )2 ≤C(α)ε, so that picking ε ≤λ2 −/C(α) implies the result.
D Additional proofs for Proposition 2.2 D.1 Proof of Lemma 4.2 The lemma is a direct corollary of the following elementary result.
Proposition D.1 (Bounding a Gaussian process on the sphere). Let n, p ≥1. Let G ∈ Rn×p with Gij i.i.d.
∼N(0, 1). There is A > 0 such that for all δ > 0 and r ≥1: P h max ∥x∥2=1 ∥Gx∥r ≥ A(√n + √p) + δ√n nmax(1/r−1/2,0)i ≤exp n −nδ2 2 o .
Proof of Proposition D.1. Notice that G →max∥x∥2=1 ∥Gx∥r is Lipschitz: max ∥x∥2=1 ∥G1x∥r −max ∥x∥2=1 ∥G2x∥r ≤max ∥x∥2=1 ∥(G1 −G2)x∥r, (a) ≤nmax( 1 r −1 2 ,0) max ∥x∥2=1 ∥(G1 −G2)x∥2, ≤nmax( 1 r −1 2 ,0)∥G1 −G2∥F, where we used Hölder’s inequality in the form ∥y∥r ≤n 1 r −1 2 ∥y∥2 for r ≤2, and for r ≥2 the fact that ∥y∥r ≤∥y∥2, alongside with ∥G∥op ≤∥G∥F . By Theorem 3.4 we reach: P h max ∥x∥2=1 ∥Gx∥r ≥E max ∥x∥2=1 ∥Gx∥r + δnmax(1/r,1/2)] ≤exp n −nδ2 2 o .
The proof is complete if we can show that E max∥x∥2=1 ∥Gx∥r = O(nmax(1/r−1/2,0)) · (√n + √p). This follows by the same inequality as above: E max ∥x∥2=1 ∥Gx∥r ≤nmax( 1 r −1 2 ,0)E max ∥x∥2=1 ∥Gx∥2.
The bound E max∥x∥2=1 ∥Gx∥2 = O(√n + √p) is well-known, see e.g. .
D.2 Proof of Theorem 4.6 As we have seen, it suffices to prove Lemmas 4.7 and 4.8.
We introduce some additional notations.
EJP 30 (2025), paper 126.
Page 38/46 Exact threshold for approximate ellipsoid fitting of random points • For any µ ∈[n], we denote ⟨·⟩µ := R P0(dS) e−P ν(̸=µ) φ[Tr(Uν(t)S)] · R P0(dS) e−P ν(̸=µ) φ[Tr(Uν(t)S)] .
(D.1) We do not write explicitly the t dependency of this average as it will be clear from context.
• For any µ, we denote E(µ) the expectation conditioned on {Gµ, Wµ}, i.e. the expec-tation over {Gν, Wν}ν(̸=µ). Note that this notation is different from , for which E(µ) was the expectation with respect to (Gµ, Wµ). On the other hand, we denote without parenthesis the expectation with respect to these variables: e.g. E(1) is conditioned on {G1, W1}, but EG1,W1 is the expectation w.r.t. G1, W1.
D.2.1 Proof of Lemma 4.7 We fix t ∈(0, π/2), and we start from eq. (4.16), which we can rewrite using eq. (D.1) as: E ∂ψ[Fd(U(t))] ∂t = −n d2 E " ψ′[Fd(U(t))] D e−φ(Tr[U1(t)S]) Tr[S e U1(t)] φ′(Tr[U1(t)S]) E 1 D e−φ(Tr[U1(t)S]) E 1 # .
Since ∥ψ′∥∞< ∞and n/d2 ≤α2, using the triangular inequality the proof of Lemma 4.7 is complete if one can show the following bound, which will also be useful afterwards: Lemma D.2. There exists a universal constant C > 0 such that: sup d≥1 sup t∈(0,π/2) sup {Wµ,Gµ}n µ=2 EW1,G1 D e−φ(Tr[U1S]) Tr[S e U1] φ′(Tr[U1S]) E 1 D e−φ(Tr[U1S]) E 1 ≤C.
Proof of Lemma D.2. Note that EW1,G1 D e−φ(Tr[U1S]) Tr[S e U1] φ′(Tr[U1S]) E 1 D e−φ(Tr[U1S]) E 1 ≤∥φ′∥∞EW1,G1 "D e−φ(Tr[U1S]) Tr[S e U1(t)] E 1 D e−φ[Tr(U1S)] E 1 # , ≤e∥φ∥∞∥φ′∥∞EW1,G1 hD Tr[S e U1(t)] E 1 i , (D.2) in which we used the positivity and boundedness of φ in the last inequality. To control the last term in eq. (D.2) we write: EG1,W1 hD Tr[S e U1(t)] E 1 i (a) = D EG1,W1 Tr[S e U1(t)] E 1, (b) ≤ Dn EG1,W1 Tr[S e U1(t)]2o1/2E 1, (c) ≤ √ 2 Dn1 dTr[S2] o1/2E 1, (D.3) (d) ≤ √ 2C0, (D.4) where (a) uses that ⟨·⟩1 is independent of {W1, G1}, (b) is from the Cauchy-Schwarz inequality, in (c) we use the hypothesis on the first two moments of ρ matching the ones of GOE(d), and in (d) that supp(P0) ⊆B2(C √ d).
EJP 30 (2025), paper 126.
Page 39/46 Exact threshold for approximate ellipsoid fitting of random points D.2.2 Proof of Lemma 4.8 We fix t ∈(0, π/2) for the rest of the proof, and we write U, e U for U(t), e U(t). We follow the ideas of Appendix A.3 of , and start again from eq. (4.16): E ∂ψ[Fd(U(t))] ∂t ≤α2(I1 + I2), (D.5) with: I1 = E " n ψ′[Fd(U)] −ψ′[Fd(U (1))] o R P0(dS) e−P ν φ(Tr[UνS]) Tr[S e U1] φ′(Tr[U1S]) R P0(dS) e−P ν φ(Tr[UνS]) # , I2 = E " ψ′[Fd(U (1))] R P0(dS) e−P ν φ(Tr[UνS]) Tr[S e U1] φ′(Tr[U1S]) R P0(dS) e−P ν φ(Tr[UνS]) # , with U (µ) obtained from U by setting Uµ = 0. We show successively I1 →0 and I2 →0.
Since ψ′ is assumed to be Lipschitz, we have: |ψ′[Fd(U)] −ψ′[Fd(U (1))]| ≤∥ψ′∥Lip d2 log R P0(dS)e−Pn ν=1 φ(Tr[UνS]) R P0(dS)e−Pn ν=2 φ(Tr[UνS]) , ≤∥ψ′∥Lip d2 log D e−φ(Tr[U1S])E 1 , ≤−∥ψ′∥Lip d2 log D e−φ(Tr[U1S])E 1, ≤∥ψ′∥Lip∥φ∥∞ d2 .
Therefore, I1 ≤Od 1 d2 !
× E R P0(dS) e−P ν φ(Tr[UνS]) Tr[S e U1] φ′(Tr[U1S]) R P0(dS) e−P ν φ(Tr[UνS]) .
(D.6) The second term in eq. (D.6) is bounded by Lemma D.2, uniformly in t. Therefore, we reach that I1 →0 as d →∞.
We now tackle I2. Note that since U (1) is independent of U1, we can rewrite it as: I2 = E(1) " ψ′[Fd(U (1))]EG1,W1 "D e−φ(Tr[U1S]) Tr[S e U1] φ′(Tr[U1S]) E 1 D e−φ(Tr[U1S]) E 1 ## , ≤∥ψ′∥∞E(1) EG1,W1 "e−φ(Tr[U1S]) Tr[S e U1] φ′(Tr[U1S]) D e−φ(Tr[U1S]) E 1 #+ 1 .
(D.7) We focus on bounding the right-hand side of eq. (D.7). We will show the following lemma: Lemma D.3. Uniformly over all t ∈[0, π/2] and {Wµ, Gµ}n µ=2, and under the hypotheses of Theorem 4.6: lim d→∞ EG1,W1 "e−φ(Tr[U1S]) Tr[S e U1] φ′(Tr[U1S]) D e−φ(Tr[U1S]) E 1 #+ 1 = 0.
(D.8) EJP 30 (2025), paper 126.
Page 40/46 Exact threshold for approximate ellipsoid fitting of random points One directly concludes that I2 →0 from using the dominated convergence theorem in eq. (D.7) (the pointwise limit is given by Lemma D.3 and the domination hypothesis by Lemma D.2). This ends the proof of Lemma 4.8.
We thus focus on the proof of Lemma D.3. Following , the sketch of the proof is the following: (i) Show that the denominator appearing in eq. (D.8) can be moved to the numerator by using the expansion of 1/x in power series around 1.
This transforms the quantity to control to a sum of terms of the type ⟨EG1,W1[f(S1, · · · , Sk)]⟩1, with S1, · · · , Sk independent samples under ⟨·⟩1.
(ii) Extend the one-dimensional CLT of Definition 4.5 to k-dimensional projections of G and W (with k = Od(1)), and to square-integrable locally-Lipschitz functions. This allows to apply it to the terms appearing in (i), and write (uniformly in S1, · · · , Sk) that EG1,W1[f(S1, · · · , Sk)] ≃EG1, e G1[f(S1, · · · , Sk)], with e G1 an independent GOE(d) matrix.
(iii) For the case of Gaussian matrices, as explained above we have e U1 independent of U1. Using the form of the function f that appears in eq. (D.8) this implies that EG1, e G1[f(S1, · · · , Sk)] = 0 and concludes the proof.
Let us perform this strategy in detail. The following lemma is proven in Section D.3.
Lemma D.4 (Polynomial approximation to the fraction –). For all δ > 0, there exists a real polynomial Q (depending only on δ) such that for all d ≥1, all t ∈(0, π/2) and all {Wµ, Gµ}n µ=2: EG1,W1 "e−φ(Tr[U1S]) Tr[S e U1] φ′(Tr[U1S]) D e−φ(Tr[U1S]) E 1 #+ 1 ≤ EG1,W1 nD e−φ(Tr[U1S]) Tr[S e U1] φ′(Tr[U1S]) E 1Q D e−φ(Tr[U1S])E 1 o + δ.
We fix δ > 0, and denote Q(X) = PK k=0 akXk the polynomial of Lemma D.4. Therefore, uniformly in d, t, and {Wµ, Gµ}: EG1,W1 "e−φ(Tr[U1S]) Tr[S e U1] φ′(Tr[U1S]) D e−φ(Tr[U1S]) E 1 #+ 1 ≤ K X k=0 |ak| D EG1,W1 n e−Pk a=0 φ(Tr[U1Sa]) Tr[S0 e U1] φ′(Tr[U1S0]) oE 1 + δ, (D.9) with {Sa}k a=0 i.i.d. samples from ⟨·⟩1.
We then extend the one-dimensional CLT of Definition 4.5 to finite-dimensional projections, similarly to Lemmas 29 and 30 of .
The proof of this lemma is deferred to Section D.3.
Lemma D.5 (Extension of the CLT to finite-dimensional projections –). Let R ≥1 an integer, and e G ∼GOE(d), independent of everything else. Let ϕ : R2R →R a locally Lipschitz function such that for both X ∈{W, e G}: sup d≥1 sup {Wµ,Gµ}n µ=2 sup t∈(0,π/2) D EX,G h ϕ {Tr(XSa)}R a=1, {Tr(GSa)}R a=1 2iE 1 < ∞.
(D.10) EJP 30 (2025), paper 126.
Page 41/46 Exact threshold for approximate ellipsoid fitting of random points It is understood there that {Sa} i.i.d.
∼⟨·⟩1. Then lim d→∞ sup {Wµ,Gµ}n µ=2 sup t∈(0,π/2) D EW,G ϕ({Tr(WSa)}, {Tr(GSa)}) −E e G,G ϕ({Tr( e GSa)}, {Tr(GSa)}) E 1 = 0.
We wish to apply Lemma D.5 to eq. (D.9), i.e. to ϕ({Tr(W1Sa)}, {Tr(G1Sa)}) := e−Pk a=0 φ(Tr[U1Sa]) Tr[S0 e U1] φ′(Tr[U1S0]) .
ϕ is locally Lipschitz by our hypotheses on φ. Moreover, note that for X ∈{W, e G}, and {Sa} ∈supp(P0) (using that W has the same two first moments as e G): EX,G h ϕ {Tr(XSa)}, {Tr(GSa)} 2i ≤2∥φ′∥2 ∞Tr[S2 0]/d ≤2C2 0∥φ′∥2 ∞< ∞.
This allows to apply Lemma D.5 in eq. (D.9), and to reach that, uniformly in {Wµ, Gµ}n µ=2 and t ∈(0, π/2) we have: lim sup d→∞ EG1,W1 "e−φ(Tr[U1S]) Tr[S e U1] φ′(Tr[U1S]) D e−φ(Tr[U1S]) E 1 #+ 1 ≤δ + K X k=0 |ak| lim sup d→∞ D EG1,W1 ϕ({Tr(W1Sa)}, {Tr(G1Sa)}) E 1, ≤δ + K X k=0 |ak| lim sup d→∞ D EG1, e G1 ϕ({Tr( e G1Sa)}, {Tr(G1Sa)}) E 1 + K X k=0 |ak| lim sup d→∞ D EG1,W1 ϕ({Tr(W1Sa)}, {Tr(G1Sa)}) −EG1, e G1 ϕ({Tr( e G1Sa)}, {Tr(G1Sa)}) E 1, (a) ≤δ + K X k=0 |ak| lim sup d→∞ D EG1, e G1 n e−Pk a=0 φ(Tr[V1Sa]) Tr[S0 e V1] φ′(Tr[V1S0]) oE 1 , where we used Lemma D.5 in (a), and with V1 = cos(t) e G1+sin(t)G1, and e V1 = −sin(t) e G1+ cos(t)G1. Since G1, e G1 are gaussians, so are V1 and e V1, and one verifies easily that they are independent since their covariance is zero. Therefore, we have: EG1, e G1 n e−Pk a=0 φ(Tr[V1Sa]) Tr[S0 e V1] φ′(Tr[V1S0]) o = EV1 n e−Pk a=0 φ(Tr[V1Sa]) h Ee V1Tr[S0 e V1] i | {z } =0 φ′(Tr[V1S0]) o = 0.
Thus, we reach, for any δ > 0: lim sup d→∞ sup t∈(0,π/2) sup {Wµ,Gµ}n µ=2 EG1,W1 "e−φ(Tr[U1S]) Tr[S e U1] φ′(Tr[U1S]) D e−φ(Tr[U1S]) E 1 #+ 1 ≤δ.
Letting δ →0 finishes the proof of Lemma D.3.
EJP 30 (2025), paper 126.
Page 42/46 Exact threshold for approximate ellipsoid fitting of random points D.3 Additional proofs D.3.1 Proof of Lemma D.4 We use the power expansion of x 7→1/x around x = 1, defining, for M ≥1: QM(x) := M X k=0 (1 −x)k, RM(x) := 1 x −QM(x).
We make use of Lemma 27 of : Lemma D.6 (). For any integer M ≥1 we have • For all x ̸= 0, RM(x) = (1 −x)M+1/x.
• x 7→RM(x)2 is convex on (0, 1].
• For any s ∈(0, 1) and η > 0, there exists M ≥1 such that supt∈[s,1] |RM(t)| ≤η.
Since e−∥φ∥∞≤e−φ ≤1, we have that for all η > 0 there exists Mη ≥1 such that for all M ≥Mη, all matrices {Wµ, Gµ}n µ=1, all t ∈(0, π/2): RM ⟨e−φ(Tr[SU1])⟩1 ≤η.
(D.11) Thus, since 1/x = QM(x) + RM(x): EG1,W1 "e−φ(Tr[U1S]) Tr[S e U1] φ′(Tr[U1S]) D e−φ(Tr[U1S]) E 1 #+ 1 ≤ D EG1,W1 h e−φ(Tr[U1S]) Tr[S e U1] φ′(Tr[U1S]) QM D e−φ(Tr[U1S])E 1 iE 1 + D EG1,W1 h e−φ(Tr[U1S]) Tr[S e U1] φ′(Tr[U1S]) RM D e−φ(Tr[U1S])E 1 iE 1 , (a) ≤ D EG1,W1 h e−φ(Tr[U1S]) Tr[S e U1] φ′(Tr[U1S]) QM D e−φ(Tr[U1S])E 1 iE 1 + η D EG1,W1 h e−φ(Tr[U1S]) Tr[S e U1] φ′(Tr[U1S]) iE 1, ≤ D EG1,W1 h e−φ(Tr[U1S]) Tr[S e U1] φ′(Tr[U1S]) QM D e−φ(Tr[U1S])E 1 iE 1 + η∥φ′∥∞⟨EG1,W1|Tr[S e U1]|⟩1, (b) ≤ D EG1,W1 h e−φ(Tr[U1S]) Tr[S e U1] φ′(Tr[U1S]) QM D e−φ(Tr[U1S])E 1 iE 1 + Cη∥φ′∥∞, using eq. (D.11) in (a), and the Cauchy-Schwarz inequality (cf. the proof of Lemma D.2) in (b). We emphasize that this bound is uniform in t and {Wµ, Gµ}. Choosing η = δ/(C∥φ′∥∞) ends the proof of Lemma D.4.
D.4 Proof of Lemma D.5 We first prove that the conclusion of Lemma D.5 holds uniformly over all S1, · · · , SR ∈ Ad when the function ϕ is assumed to be continuous with compact support: Lemma D.7. Let R ∈N⋆, and e G ∼GOE(d), independent of everything else.
Let ϕ : R2R →R be Lipschitz and compactly supported. Recall that W ∼ρ, a measure who is assumed to satisfy a one-dimensional CLT with respect to a set Ad ⊆Sd, see Definition 4.5. We assume that Ad is convex and symmetric. Then: lim d→∞ sup S1,··· ,SR∈Ad EW,G ϕ {Tr(WSa)}, {Tr(GSa)} −E e G,G ϕ {Tr( e GSa)}, {Tr(GSa)} = 0.
EJP 30 (2025), paper 126.
Page 43/46 Exact threshold for approximate ellipsoid fitting of random points Proof of Lemma D.7. Recall that we use the matrix flattening function of eq. (3.8). Let us denote: H := vec(S1) 0 vec(S2) 0 · · · vec(SR) 0 0 vec(S1) 0 vec(S2) · · · 0 vec(SR) ∈Rd(d+1)×2R, v := (vec(W)⊺, vec(G)⊺)⊺∈Rd(d+1), h := (vec( e G)⊺, vec(G)⊺)⊺∈Rd(d+1).
(D.12) Using these notations, we have: {Tr(WSa)}R a=1, {Tr(GSa)}R a=1 = H⊺v ∈R2R, {Tr( e GSa)}R a=1, {Tr(GSa)}R a=1 = H⊺h ∈R2R.
(D.13) We add a small Gaussian noise to help us deal with characteristic functions later on. Let δ > 0, and Z ∼N(0, δ2I2R). For all S1, · · · , SR we have: |E ϕ(H⊺v) −E ϕ(H⊺h)| ≤|E ϕ(H⊺v) −E ϕ(H⊺v + Z)| + |E ϕ(H⊺h) −E ϕ(H⊺h + Z)| + |E ϕ(H⊺v + Z) −E ϕ(H⊺h + Z)|, ≤2∥ϕ∥LE∥Z∥2 + |E ϕ(H⊺v + Z) −E ϕ(H⊺h + Z)|, ≤CR1/2∥ϕ∥Lδ + |E ϕ(H⊺v + Z) −E ϕ(H⊺h + Z)|.
(D.14) We now control the last term of eq. (D.14). For X ∈R2R a random variable, we define its characteristic function as φX(u) := E e−iu⊺X. We have then 1 (2π)2R Z R2R ϕ(t) Z R2R eit⊺u−δ2 2 ∥u∥2φX(u)du dt = 1 (2πδ2)R EX Z R2R ϕ(t)e−∥t−X∥2 2δ2 dt, = E ϕ(X + Z).
Coming back to eq. (D.14) we get: |E ϕ(H⊺v + Z) −E ϕ(H⊺h + Z)| ≤ 1 (2π)2R Z R2R |ϕ(t)| Z R2R eit⊺u−δ2 2 ∥u∥2[φH⊺v(u) −φH⊺h(u)]du dt, ≤∥ϕ∥L1 (2π)2R Z R2R e−δ2 2 ∥u∥2|φH⊺v(u) −φH⊺h(u)|du.
(D.15) Here ∥ϕ∥L1 := R |ϕ(t)|dt. We will show that for any u ∈R2R: lim d→∞ sup S1,··· ,SR∈Ad |φH⊺v(u) −φH⊺h(u)| = 0.
(D.16) Combining eq. (D.16) with the dominated convergence theorem applied in eq. (D.15), we get: lim d→∞ sup S1,··· ,SR∈Ad |E ϕ(H⊺v + Z) −E ϕ(H⊺h + Z)| = 0.
Plugging this back into eq. (D.14), we get that for any δ > 0: lim sup d→∞ sup S1,··· ,SR∈Ad |E ϕ(H⊺v) −E ϕ(H⊺h)| ≤CR1/2∥ϕ∥Lδ.
EJP 30 (2025), paper 126.
Page 44/46 Exact threshold for approximate ellipsoid fitting of random points Letting δ →0 ends the proof of Lemma D.7. There remains to prove eq. (D.16). Let u = (u(1), u(2)) ∈R2R, with u(1), u(2) ∈RR, and let us fix S1, · · · , SR ∈Ad. We have |φH⊺v(u) −φH⊺h(u)| = E e−i Pr a=1 u(1) a Tr[W Sa]−i Pr a=1 u(2) a Tr[GSa] −E e−i Pr a=1 u(1) a Tr[ e GSa]−i Pr a=1 u(2) a Tr[GSa] , (a) = E exp n −i Tr h W r X a=1 u(1) a Sa io −E exp n −i Tr h e G r X a=1 u(1) a Sa io , (D.17) using in (a) that G is independent of W, e G. We can assume that u(1) ̸= 0, otherwise the result of eq. (D.16) is clear. Since Ad is symmetric and convex we have ˆ S := 1 ∥u(1)∥1 r X a=1 u(1) a Sa ∈Ad.
Therefore, letting ϕu(x) := e−i∥u∥1x, we have by eq. (D.17): |φH⊺v(u) −φH⊺h(u)| = |E ϕu(Tr[ ˆ SW]) −E ϕu(Tr[ ˆ S e G])| (D.18) Moreover, |ϕu(x) −ϕu(y)| = ∥u∥1 Z y−x 0 ei∥u∥1tdt ≤∥u∥1|y −x|, so that ∥ϕu∥L ≤∥u∥1. We can then therefore apply the one-dimensional CLT of Defini-tion 4.5 in eq. (D.18), and we get that sup S1,··· ,SR∈Ad |φH⊺v(u) −φH⊺h(u)| ≤sup S∈Ad |E ϕu(Tr[SW]) −E ϕu(Tr[S e G])| →d→∞0.
This ends the proof of eq. (D.16).
We then deduce the full statement of Lemma D.5 by a truncation argument.
End of the proof of Lemma D.5. ϕ is now only assumed to be locally Lipschitz and square integrable, in the sense of eq. (D.10). We take the same notations as in the proof of Lemma D.7, see in particular eq. (D.12), so that EW,G ϕ {Tr(WSa)}, {Tr(GSa)} −E e G,G ϕ {Tr( e GSa)}, {Tr(GSa)} = E ϕ(H⊺v) −E ϕ(H⊺h).
Let B > 0, and let us denote uB : R+ →[0, 1] a C∞function such that uB(x) = 1 if x ≤B and uB(x) = 0 if x ≥B + 1. We denote ϕB(z) := ϕ(z)uB(∥z∥). One can then check easily that ϕB is Lipschitz (because ϕ is locally-Lipschitz), and compactly supported. Moreover, we have: ⟨|E ϕ(H⊺v) −E ϕ(H⊺h)|⟩1 ≤⟨|E ϕB(H⊺v) −E ϕB(H⊺h)|⟩1 + X z∈{h,v} ⟨E |ϕ(H⊺z)|(1 −uB(∥H⊺z∥))⟩1.
(D.19) We now control the different terms in eq. (D.19) successively. Notice that ⟨|E ϕB(H⊺v) −E ϕB(H⊺h)|⟩1 ≤ sup S1,··· ,SR∈Ad |E ϕB(H⊺v) −E ϕB(H⊺h)|, EJP 30 (2025), paper 126.
Page 45/46 Exact threshold for approximate ellipsoid fitting of random points so that by Lemma D.7: lim d→∞ sup {Wµ,Gµ}n µ=2 sup t∈(0,π/2) ⟨|E ϕB(H⊺v) −E ϕB(H⊺h)|⟩1 = 0.
(D.20) We now tackle the remaining terms in eq. (D.19). Let z ∈{h, v}. Using the Cauchy-Schwarz inequality twice we get: ⟨E |ϕ(H⊺z)|(1 −uB(∥H⊺z∥))⟩1 ≤⟨E |ϕ(H⊺z)|1{∥H⊺z∥2 ≥B}⟩1, ≤⟨(Ez[ϕ(H⊺z)2])1/2Pz{∥H⊺z∥2 ≥B}1/2⟩1, ≤⟨(Ez[ϕ(H⊺z)2])⟩1/2 1 · ⟨Pz{∥H⊺z∥2 ≥B}⟩1/2 1 .
(D.21) The first term in eq. (D.21) is bounded by the square integrability assumption, cf.
eq. (D.10). To bound the second term, we use Markov’s inequality: ⟨Pz{∥H⊺z∥2 ≥B}⟩1 ≤1 B2 ⟨Ez ∥H⊺z∥2 2⟩1.
From eq. (D.13) and the matching of the first two moments of ρ with GOE(d), we have: Ez ∥H⊺z∥2 2 = 4 d R X a=1 Tr[S2 a].
Thus, we get: ⟨Pz{∥H⊺z∥2 ≥B}⟩1 ≤4R B2 TrS2 d + 1 ≤4RC0 B2 .
All in all, we get: sup d≥1 sup {Wµ,Gµ}n µ=2 sup t∈(0,π/2) ⟨E |ϕ(H⊺z)|(1 −uB(∥H⊺z∥))⟩1 ≤C(R, ϕ) B .
(D.22) Combining eqs. (D.20) and eq. (D.22) into eq. (D.19), and taking B →∞after d →∞, we conclude the proof of Lemma D.5.
EJP 30 (2025), paper 126.
Page 46/46 |
14850 | https://pubmed.ncbi.nlm.nih.gov/2493116/ | Thrombolytic therapy in acute myocardial infarction: Part II--rt-PA - PubMed
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. 1989 Mar;73(2):387-407.
doi: 10.1016/s0025-7125(16)30679-4.
Thrombolytic therapy in acute myocardial infarction: Part II--rt-PA
H S Mueller1,R Roberts,S L Teichman,B E Sobel
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1 Albert Einstein College of Medicine, Bronx, New York.
PMID: 2493116
DOI: 10.1016/s0025-7125(16)30679-4
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Thrombolytic therapy in acute myocardial infarction: Part II--rt-PA
H S Mueller et al. Med Clin North Am.1989 Mar.
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. 1989 Mar;73(2):387-407.
doi: 10.1016/s0025-7125(16)30679-4.
Authors
H S Mueller1,R Roberts,S L Teichman,B E Sobel
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1 Albert Einstein College of Medicine, Bronx, New York.
PMID: 2493116
DOI: 10.1016/s0025-7125(16)30679-4
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Thrombolysis with pharmacologic agents is a valuable modality for treatment of acute myocardial infarction. The results of several clinical studies indicate that early recanalization can be elicited with intravenous agents and that it is associated with substantial reductions of infarct size, improvement of ventricular function, and reduction in mortality. The recently introduced fibrin-selective agent, rt-PA, appears to represent a significant pharmacologic advance. Its use intravenously elicits high recanalization rates without marked derangements of the coagulation system, reflecting its relative fibrin selectivity. The efficacy of thrombolysis with any drug given by any route, unfortunately, is not 100 per cent. Bleeding remains an important risk. The optimal approach to management of residual stenosis after thrombolysis has not yet been delineated. Currently available information appears to justify the following conclusions: 1. Transmural myocardial infarction usually is caused by an acute obstructing coronary thrombus superimposed on a chronic atherosclerotic lesion. Myocardial necrosis following interruption of blood flow generally is complete within several hours. 2. The thrombus can be lysed and blood flow can be restored with intravenous agents that activate plasminogen. Intravenous rt-PA, a relatively fibrin-specific agent, elicits recanalization in 70 to 75 per cent of infarct-related arteries. 3. Recombinant t-PA evokes only modest depletion of fibrinogen (16 to 36 per cent reduction from baseline). 4. Early reperfusion preserves myocardium and ventricular function and reduces mortality. 5. The extent of benefit after pharmacologic reperfusion is correlated strongly with the brevity of myocardial ischemia prior to initiation of therapy. The greatest benefit is realized in patients treated within the first few hours of onset of acute myocardial infarction. 6. The incidence and optimal means of prevention of reocclusion and reinfarction following successful pharmacologic reperfusion are not yet entirely clear. Mechanical recanalization with PTCA in conjunction with thrombolysis is promising, but its routine immediate use on an emergency basis does not appear to be beneficial. Vigorous educational efforts are needed to heighten the awareness of prospective patients and all members of the health care team to the value of prompt diagnosis of incipient or evolving infarction so that prompt implementation of thrombolysis in appropriate candidates can be facilitated.
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14851 | https://www.youtube.com/watch?v=8NhmfmaXlLo | How to Multiply Fractions Using Cancellation | Multiplying Fractions | Math with Mr. J
Math with Mr. J
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Welcome to How to Multiply Fractions Using Cancellation (aka Cross Cancellation) with Mr. J! Need help with multiplying fractions using cancellation? You're in the right place!
Whether you're just starting out, or need a quick refresher, this is the video for you if you're looking for help with multiplying fractions using cancellation. Mr. J will go through examples of multiplying fractions and explain the steps of how to multiply fractions using cancellation.
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Introduction Welcome to Math with Mr. J. In this video, I'm going to cover how to use cancellation, also called cross cancellation when multiplying two fractions. Cancellation is a way to simplify a problem before we multiply. Now remember, when we multiply fractions, we multiply straight across. So multiply the numerators, the top numbers, and then multiply the denominators, Examples the bottom numbers. We use cancellation before we multiply straight across. This strategy gives us smaller numbers in value to work with and easier numbers to work with. Let's jump into our examples, starting with number one, where we have 7/10 times 5/6. When we use cancellation, again, we need to look to simplify the problem before we multiply. Look for common factors other than 1 diagonally. That way we can divide those numbers by that common factor in order to simplify the problem before multiplying. Think of it like simplifying fractions, but we are looking diagonally. So let's look at 7 and 6 first. Are there any common factors between 7 and 6 other than 1? No, the only common factor is 1, so we can't use cancellation with the 7 and the 6. Now let's take a look at the 10 and the 5. Are there any common factors other than 1 between 10 and 5? Yes, 5 is a common factor and it happens to be the greatest common factor. So let's divide 10 and 5 by 5. So 10 divided by 5 gives us 2. 5 divided by 5 gives us 1. Now we have a simpler problem. We have smaller numbers in value to work with. So let's multiply straight across now, starting with the numerators. 7 times 1 is 7. Now for the denominators. 2 times 6 gives us 12. 7/12 is our final answer and it is in simplest form, the only common factor between 7 and 12 is 1, so we are done 7/12. Examples without using cancellation Now let's go through that problem without using cancellation in order to see the difference. So we have 7/10 times 5/6. So let's multiply straight across, again, without using cancellation. We will start with the numerators, so 7 times 5 is 35. Now for the denominators. 10 times 6 is 60. So we get 35/60. Which is different than when we used cancellation. Well, 35/60 isn't in simplest form, so let's simplify. Are there any common factors between 35 and 60 other than one? Yes, 5 and 5 is the greatest common factor. So let's divide 35 and 60 by 5 in order to simplify. 35 divided by 5 is 7. 60 divided by 5 is 12. So our final simplified answer 7/12. The only common factor between 7 and 12 is 1, so that is in simplest form and we are done. So we get 7/12 that way as well. So do we have to use cancellation in order to multiply fractions? No, but it is a good strategy to be familiar with and use when possible. It can be helpful. Let's move on to number two, where we have 15/22 times 11/12. Let's look to use cancellation and again, we look diagonally. So let's take a look at 15 and 12 first. Do we have any common factors other than 1? Yes. 3 is a common factor and the greatest common factor. So let's divide them both by 3. 15 divided by 3 is 5. 12 divided by 3 is 4. Now let's take a look at 22 and 11. Are there any common factors other than 1 between 22 and 11? Yes, 11 and 11 is the greatest common factor. So let's divide 22 by 11, that gives us 2. And then 11 divided by 11, that gives us 1. Now we can multiply straight across and you can see that we have much simpler and easier numbers to work with when we multiply straight across. So let's start with the numerators. 5 times 1 gives us 5. Now for the denominators. 2 times 4 gives us 8. So we get 5/8. The only common factor between 5 and 8 is 1, so we are in simplest form and we are done. For number two, that problem was much simpler to work through after we used cancellation. We had 5 times 1 and 2 times 4 instead of 15 times 11 and 22 times 12. Now did we have to use cancellation? No, we could have Conclusion gone straight across and done 15 times 11 and then 22 times 12 and then simplified, but the cancellation made it easier and simpler as far as multiplying straight across. So understanding cancellation and how to use it can be helpful when multiplying fractions. So there you have it. There's how to use cancellation when multiplying two fractions. I hope that helped. Thanks so much for watching. Until next time. Peace. |
14852 | https://www.math.ucla.edu/~dam/220b.1.11w/220notes5.pdf | CHAPTER 5 INTRODUCTION TO COMPUTABILITY THEORY The class of recursive functions was originally introduced as a tool for estab-lishing undecidability results (via the Church-Turing Thesis); but it is a very interesting class, it has been studied extensively since the 1930s, and its theory has found important applications in many mathematical areas. Here we will give only a brief introduction to some of its aspects.
5A. Semirecursive relations It is convenient to introduce the additional notation {e}(⃗ x) = ϕn e (⃗ x) for the recursive n-ary partial function with code e, as in the Normal Form Theorem 4F.1, which puts the “program” e and the “data” ⃗ x on equal footing and eliminates the need for double and triple subscripts in the formulas to follow.
We start with a Corollary to the proof of Theorem 4F.1, which gives some additional information about the coding of recursive partial functions and whose significance will become apparent in the sequel.
Theorem 5A.1 (Sm n -Theorem, Kleene). For all m, n ≥1, there is a one-to-one, m+1-ary primitive recursive function Sm n (e, y1, . . . , ym), such that for all ⃗ y = y1, . . . , ym, ⃗ x = x1, . . . , xn, ϕSm n (e,⃗ y)(⃗ x) = ϕe(⃗ y, ⃗ x), i.e., {Sm n (e, ⃗ y)}(⃗ x) = {e}(⃗ y, ⃗ x).
Proof. For each sequence of numbers e, ⃗ y = e, y1, . . . , ym, let θ′ ≡∆e = 0 & ∆y1 = 0 & · · · & ∆ym = 0 & 0 = 1, and for each full extended formula ψ ≡ψ(v0, . . . , vm−1, vm, . . . , vm+n) 183 184 5. Introduction to computability theory (as in the definition of Tn+m(e, ⃗ y, ⃗ x, z) in Theorem 4F.1) let θ ≡φ(v0, . . . , vn) ≡ψ(∆y1, . . . , ∆ym, v0, . . . , vn) and put Sm n (e, ⃗ y) = ( the code of θ, if e is the code of some ψ as above the code of θ′, otherwise.
It is clear that each Sm n (e, ⃗ y) is a primitive recursive function, and it is also one-to-one, because the value Sm n (e, ⃗ y) codes all the numbers e, y1, . . . , ym— this was the reason for introducing the extra restriction on the variables in the definition of the T predicate. Moreover: Tm+n(e, ⃗ y, ⃗ x, z) ⇐ ⇒e is the code of some ψ as above, and (z)1 is the code of a proof in Q of φ(∆y1, . . . , ∆ym, ∆x1, . . . , ∆xn, ∆(z)0) ⇐ ⇒Sm n (e, ⃗ y) is the code of the associated θ and (z)1 is the code of a proof in Q of θ(∆x1, . . . , ∆xn, ∆(z)0) ⇐ ⇒Tn(Sm n (e, ⃗ y), ⃗ x, z).
To see this, check first the implications in the direction = ⇒, which are all immediate—with the crucial, middle implication holding because (literally) θ(∆x1, . . . , ∆xn, ∆(z)0) ≡ψ(∆y1, . . . , ∆ym, ∆x1, . . . , ∆xn, ∆(z)0).
For the implications in the direction ⇐ =, notice that if Tn(Sm n (e, ⃗ y), ⃗ x, z) holds, then Sm n (e, ⃗ y) is the code of a true sentence, since (z)0 is the code of a proof of it in Q, and so it cannot be the code of θ′, which is false; so it is the code of θ, which means that e is the code of some φ as above, and then the argument runs exactly as in the direction = ⇒.
From this we get immediately, by the definitions, that {Sm n (e, ⃗ y)}(⃗ x) = {e}(⃗ y, ⃗ x).
⊣ Example 5A.2. The class of recursive partial functions is “uniformly” closed for composition, for example there is a primitive recursive function un(e, m1, m2) such that for all ⃗ x = (x1, . . . , xn), {un(e, m1, m2)}(⃗ x) = {e}({m1}(⃗ x), {m2}(⃗ x)).
5A. Semirecursive relations 185 Proof. The partial function f(e, m1, m2, ⃗ x) = {e}({m1}(⃗ x), {m2}(⃗ x)) is recursive, and so for some number b f and by Theorem 5A.1, f(e, m1, m2, ⃗ x) = { b f}(e, m1, m2, ⃗ x) = {S3 n( b f, e, m1, m2)}(⃗ x), and it is enough to set un(e, m1, m2) = S3 n( b f, e, m1, m2).
⊣ This is, obviously, a special case of a general fact which follows from the Sm n -Theorem, in slogan form: if the class of recursive partial function is closed for some operation, it is then closed uniformly (in the codes) for the same operation.
To simplify the statements of several definitions and results in the sequel, we recall here and name the basic, “logical” operations on relations: P(⃗ x) ⇐ ⇒¬Q(⃗ x) (¬) P(⃗ x) ⇐ ⇒Q(⃗ x) & R(⃗ x) (&) P(⃗ x) ⇐ ⇒Q(⃗ x) ∨R(⃗ x) (∨) P(⃗ x) ⇐ ⇒Q(⃗ x) = ⇒R(⃗ x) (⇒) P(⃗ x) ⇐ ⇒(∃y)Q(⃗ x, y) (∃N) P(z, ⃗ x) ⇐ ⇒(∃i ≤z)Q(⃗ x, i) (∃≤) P(⃗ x) ⇐ ⇒(∀y)Q(⃗ x, y) (∀N) P(z, ⃗ x) ⇐ ⇒(∀i ≤z)Q(⃗ x, i) (∀≤) P(⃗ x) ⇐ ⇒Q(f1(⃗ x), . . . , fm(⃗ x)) (replacement) For example, we have already shown that the class of primitive recursive re-lations is closed under all these operations (with primitive recursive fi(⃗ x)), except for the (unbounded) quantifiers ∃N, ∀N, under which it is not closed by Theorem 4F.2.
Proposition 5A.3. The class of recursive relations is closed under the propositional operations ¬, & , ∨, ⇒, the bounded quantifiers ∃≤, ∀≤, and sub-stitution of (total) recursive functions, but it is not closed under the (un-bounded) quantifiers ∃, ∀.
186 5. Introduction to computability theory Definition 5A.4. (1) A relation P(⃗ x) is semirecursive if it is the domain of some recursive partial function f(⃗ x), i.e., P(⃗ x) ⇐ ⇒f(⃗ x)↓.
(2) A relation P(⃗ x) is Σ0 1 if there is some recursive relation Q(⃗ x, y), such that P(⃗ x) ⇐ ⇒(∃y)Q(⃗ x, y).
Proposition 5A.5. The following are equivalent, for an arbitrary relation P(⃗ x): (1) P(⃗ x) is semirecursive.
(2) P(⃗ x) is Σ0 1.
(3) P(⃗ x) satisfies the equivalence P(⃗ x) ⇐ ⇒(∃y)Q(⃗ x, y) with some primitive recursive Q(⃗ x, y).
Proof. (1) = ⇒(3) by the Normal Form Theorem; (3) = ⇒(2) trivially; and for (2) = ⇒(1) we set f(⃗ x) = µyQ(⃗ x, y), so that (∃y)Q(⃗ x, y) ⇐ ⇒f(⃗ x)↓.
⊣ Proposition 5A.6 (Kleene’s Theorem). A relation P(⃗ x) is recursive if and only if both P(⃗ x) and its negation ¬P(⃗ x) are semirecursive.
Proof. If P(⃗ x) is recursive, then the relations Q(⃗ x, y) ⇐ ⇒P(⃗ x), R(⃗ x, y) ⇐ ⇒¬P(⃗ x) are both recursive, and (trivially) P(⃗ x) ⇐ ⇒(∃y)Q(⃗ x, y) ¬P(⃗ x) ⇐ ⇒(∃y)R(⃗ x, y).
For the other direction, if P(⃗ x) ⇐ ⇒(∃y)Q(⃗ x, y) ¬P(⃗ x) ⇐ ⇒(∃y)R(⃗ x, y) with recursive Q and R, then the function f(⃗ x) = µy[Q(⃗ x, y) ∨R(⃗ x, y)] 5A. Semirecursive relations 187 is total and recursive, and P(⃗ x) ⇐ ⇒Q(⃗ x, f(⃗ x)).
⊣ Proposition 5A.7. The class of semirecursive relations is closed under the “positive” propositional operations &, ∨, under the bounded quantifiers ∃≤, ∀≤, and under the existential quantifier ∃N; it is not closed under negation ¬ and under the universal quantifier ∀N.
Proof. Closure under (total) recursive substitutions is trivial, and the fol-lowing transformations show the remaining positive claims of the proposi-tion: (∃y)Q(⃗ x, y) ∨(∃y)R(⃗ x, y) ⇐ ⇒(∃u)[Q(⃗ x, u) ∨R(⃗ x, u)] (∃y)Q(⃗ x, y) & (∃y)R(⃗ x, y) ⇐ ⇒(∃u)[Q(⃗ x, (u)0) & R(⃗ x, (u)1)] (∃z)(∃y)Q(⃗ x, y, z) ⇐ ⇒(∃u)R(⃗ x, (u)0, (u)1) (∃i ≤z)(∃y)Q(⃗ x, y, i) ⇐ ⇒(∃y)(∃i ≤z)Q(⃗ x, y, i) (∀i ≤z)(∃y)Q(⃗ x, y, i) ⇐ ⇒(∃u)(∀i ≤z)Q(⃗ x, (u)i, i).
On the other hand, the class of semirecursive relations is not closed under ¬ or ∀N, otherwise the basic Halting relation H(e, x) ⇐ ⇒(∃y)T1(e, x, y) would have a semirecursive negation and so would be recursive by 5A.6, which it is not.
⊣ The graph of a partial function f(⃗ x) is the relation Gf(⃗ x, w) ⇐ ⇒f(⃗ x) = w, (111) and the next restatement of Theorem 4E.10 often gives (with the closure prop-erties of Σ0 1) simple proofs of recursiveness for partial functions: Proposition 5A.8 (The Σ0 1-Graph Lemma). A partial function f(⃗ x) is re-cursive if and only if its graph Gf(⃗ x, w) is a semirecursive relation.
Proof. If f(⃗ x) is recursive with code b f, then Gf(⃗ x, w) ⇐ ⇒(∃y)[Tn( b f, ⃗ x, y) & U(y) = w], so that Gf(⃗ x, w) is semirecursive; and if f(⃗ x) = w ⇐ ⇒(∃u)R(⃗ x, w, u) with some recursive R(⃗ x, w, u), then f(⃗ x) = µuR(⃗ x, (u)0, (u)1) 0, 188 5. Introduction to computability theory so that f(⃗ x) is recursive.
⊣ The next Lemma is also very simple, but it simplifies many proofs.
Proposition 5A.9 (The Σ0 1-Selection Lemma). For each semirecursive re-lation R(⃗ x, w), there is a recursive partial function f(⃗ x) = νwR(⃗ x, w) such that for all ⃗ x, (∃w)R(⃗ x, w) ⇐ ⇒f(⃗ x)↓ (∃w)R(⃗ x, w) = ⇒R(⃗ x, f(⃗ x)).
Proof. By the hypothesis, there is a recursive relation P(⃗ x, w, y) such that R(⃗ x, w) ⇐ ⇒(∃y)P(⃗ x, w, y), and the conclusion of the lemma follows if we just set f(⃗ x) = µuP(⃗ x, (u)0, (u)1) 0.
⊣ Problems for Section 5A Problem 5A.1. Prove that if R(x1, . . . , xn) is a recursive relation, then there exists a formula R(v1, . . . , vn) in the language of arithmetic such that R(x1, . . . , xn) = ⇒Q ⊢R(∆x1 . . . , ∆xn) and ¬R(x1, . . . , xn) = ⇒Q ⊢¬R(∆x1 . . . , ∆xn).
Problem 5A.2. Prove that every axiomatizable, complete theory is decid-able.
Problem 5A.3. Show that the class of recursive partial functions is uni-formly closed under definition by primitive recursion in the following, precise sense: there is a primitive recursive function un(e, m), such that if f(y, ⃗ x) is defined by the primitive recursion f(0, ⃗ x) = ϕe(⃗ x), f(y + 1, ⃗ x) = ϕm(f(y, ⃗ x), y, ⃗ x), then f(y, ⃗ x) = {un(e, m)}(y, ⃗ x).
Problem 5A.4. Define a total, recursive, one-to-one function un(e, i), such that for all e, i, ⃗ x, {un(e, i)}(⃗ x) = {e}(⃗ x).
(In particular, each recursive partial function has, effectively, an infinite num-ber of distinct codes.) 5B. Recursively enumerable sets 189 Problem 5A.5. Show that the partial function f(e, u) = ⟨ϕe((u)0), ϕe((u)1), . . . , ϕe((u)lh(u) − · 1)⟩ is recursive.
Problem 5A.6 (Craig’s Lemma). Show that a theory T has a primitive recursive set of axioms if and only if it has a semirecursive set of axioms.
Problem 5A.7. Prove that there is a recursive relation R(x) which is not primitive recursive.
Problem 5A.8. Suppose R(⃗ x, w) is a semirecursive relation such that for each ⃗ x there exist at least two distinct numbers w1 ̸= w2 such that R(⃗ x, w1) and R(⃗ x, w2). Prove that there exist two, total recursive functions g(⃗ x) and h(⃗ x), such tht for all ⃗ x, R(⃗ x, g(⃗ x)) & R(⃗ x, h(⃗ x)) & g(⃗ x ̸= h(⃗ x).
Problem 5A.9. Suppose R(⃗ x, w) is a semirecursive relation such that for each ⃗ x, there exists at least one w such that R(⃗ x, w).
(a) Prove that there is a total recursive function f(n, ⃗ x) such that R(⃗ x, w) ⇐ ⇒(∃n)[w = f(n, ⃗ x)].
(∗) (b) Prove that if (in addition), for each ⃗ x, there exist infinitely many w such that R(⃗ x, w), then there exists a total, recursive f(n, ⃗ x) which satisfies (∗) and which is one-to-one, i.e., m ̸= n = ⇒f(m, ⃗ x) ̸= f(n, ⃗ x).
5B. Recursively enumerable sets Some of the properties of semirecursive relations are easier to identify when we view unary relations as sets: Definition 5B.1 (R.e. sets). A set A ⊆N is recursively or computably enumerable if either A = ∅, or some total, recursive function enumerates it, i.e., A = {f(0), f(1), . . . , }.
(112) The term “recursively enumerable” is unwieldy and it is always abbreviated by the initials “r.e.” or ”c.e.”.
190 5. Introduction to computability theory Proposition 5B.2. The following are equivalent for any A ⊆N: (1) A is r.e.
(2) The relation x ∈A is semirecursive.
(3) A is finite, or there exists a one-to-one recursive function f : N ↣N which enumerates it.
Proof. The implication (3) = ⇒(1) is trivial, and (1) = ⇒(2) follows from the equivalence x ∈A ⇐ ⇒(∃n)[f(n) = x] which holds for all non-empty r.e. sets A. To show (2) = ⇒(3), we suppose that A is infinite and x ∈A ⇐ ⇒(∃y)R(x, y) with a recursive R(x, y), and set B = {u | R((u)0, (u)1) & (∀v < u)[R((v)0, (v)1) = ⇒(v)0 ̸= (u)0]}.
It is clear that B is a recursive set, that u ∈B = ⇒(u)0 ∈A, and that if x ∈A and we let t = (µu)[R((u)0, (u)1) & (u)0 = x], then (directly from the definition of B), t ∈B & (∀u)[(u ∈B & (u)0 = x) ⇐ ⇒u = t]; it follows that the projection π(u) = (u)0 is a one-to-one correspondence of B with A, and hence B is infinite. Now B is enumerated without repetitions by the recursive function g(0) = (µu)[u ∈B] g(n + 1) = (µu)[u > g(n) & u ∈B], and the composition f(n) = (g(n))0 enumerates A without repetitions.
⊣ The next fact shows that we cannot go any further in producing “nice” enumerations of arbitrary r.e. sets.
5B. Recursively enumerable sets 191 Proposition 5B.3. A set A ⊆N is recursive if and only if it is finite, or there exists an increasing, total recursive function which enumerates it, A = {f(0) < f(1) < . . . , }.
Proof. A function f : N →N is increasing if f(n) < f(n + 1) (n ∈N), from which it follows (by an easy induction) that for all n n ≤f(n); thus, if some increasing, recursive f enumerates A, then x ∈A ⇐ ⇒(∃n ≤x)[x = f(n)], and A is recursive. For the opposite direction, if A is recursive and infinite, then the function f(0) = (µx)[x ∈A] f(n + 1) = (µx)[x > f(n) & x ∈A] is recursive, increasing and enumerates A.
⊣ The simplest example of an r.e. non-recursive set is the “diagonal” set K = {x | (∃y)T1(x, x, y)} = {x | {x}(x)↓}, (113) and the next Proposition shows that (in some sense) K is the “most complex” r.e. set.
Proposition 5B.4. For each r.e. set A, there is a one-to-one recursive function f : N ↣N such that x ∈A ⇐ ⇒f(x) ∈K.
(114) Proof. By hypothesis, A = {x | g(x)↓} for some recursive partial function g(x). We set h(x, y) = g(x) and we choose some code b h of h, so that for any y, x ∈A ⇐ ⇒h(x, y)↓ ⇐ ⇒{b h}(x, y)↓ ⇐ ⇒{S1 1(b h, x)}(y)↓; in particular, this holds for y = S1 1(b h, x) and it yields in that case x ∈A ⇐ ⇒{S1 1(b h, x)}(S1 1(b h, x))↓ ⇐ ⇒S1 1(b h, x) ∈K, 192 5. Introduction to computability theory so that (114) holds with f(x) = S1 1(b h, x).
⊣ Definition 5B.5 (Reducibilities). A reduction of a set A to another set B is any (total) recursive function f, such that x ∈A ⇐ ⇒f(x) ∈B, (115) and we set: A ≤m B ⇐ ⇒there exists a reduction of A to B, A ≤1 B ⇐ ⇒there exists a one-to-one reduction of A to B, A ≡B ⇐ ⇒there exists a reduction f of A to B which is a permutation, where a permutation f : N ↣ →N is any one-to-one correspondence of N onto N. Clearly A ≡B = ⇒A ≤1 B = ⇒A ≤m B.
Proposition 5B.6. For all sets A, B, C, A ≤m A and [A ≤m B & B ≤m C] = ⇒A ≤m C, and the same holds for the stronger reductions ≤1 and ≡; in addition, the relation ≡of recursive isomorphism is symmetric, A ≡B ⇐ ⇒B ≡A.
Definition 5B.7. A set B is r.e. complete if it is r.e., and every r.e. set A is one-one reducible to B, A ≤1 B.
Proposition 5B.4 expresses precisely the r.e. completeness of K, and the next, basic result shows that up to recursive isomorphism, there is only one r.e. complete set.
Theorem 5B.8 (John Myhill). For any two sets A and B, A ≤1 B & B ≤1 A = ⇒A ≡B.
Proof. The argument is s constructive version of the classical Schr¨ oder-Bernstein in set theory, and it is based on the next Lemma, in which a sequence of pairs W = (x0, y0), (x1, y1), . . . , (xn, yn) (116) is called good (as an approximation to an isomorphism) for A and B if i ̸= j = ⇒[xi ̸= xj and yi ̸= yj], xi ∈A ⇐ ⇒yi ∈B (i ≤n).
5B. Recursively enumerable sets 193 For any good sequence, we set X = {x0, x1, . . . , xn}, Y = {y0, y1, . . . , yn}.
Lemma X. If there is a recursive one-to-one function f : N ↣N such that x ∈A ⇐ ⇒f(x) ∈B, then for every good sequence (116) and each x / ∈X, we can find some y / ∈Y such that the extension W ′ = (x0, y0), (x1, y1), . . . , (xn, yn), (x, y) (117) is also good, i.e., x ∈A ⇐ ⇒y ∈B, Proof of Lemma X. We set z0 = f(x) zi+1 = ( zi if zi / ∈Y, f(xj) otherwise, if zi = yj, and we verify two basic properties of the sequence z0, z1, . . . .
(1) For each i, x ∈A ⇐ ⇒zi ∈B.
Proof. For i = 0, x ∈A ⇐ ⇒ f(x) = z0 ∈B, by the hypothesis on f.
Inductively, if zi / ∈Y , then x ∈A ⇐ ⇒zi+1 = zi ∈B by the induction hypothesis, and if zi ∈Y , then x ∈A ⇐ ⇒zi = yj ∈B (by the induction hypothesis) ⇐ ⇒xj ∈A (because the given sequence is good) ⇐ ⇒f(xj) = zi+1 ∈B.
(2) For every i, zi ∈Y = ⇒(∀k < i)[zi ̸= zk].
Proof.
The proposition is trivially true if z0 = f(x) / ∈Y , since, in this case, zi = f(x) / ∈Y for every i, by the definition. The proposition is also trivially true for i = 0, and, inductively, we assume that it holds for i and that zi+1 ∈Y . Notice that zi ∈Y , otherwise (by the definition) zi+1 = zi / ∈Y ; hence, by the definition again, for some j, zi = yj, zi+1 = f(xj).
(118) 194 5. Introduction to computability theory Towards a contradiction, let k be the least counterexample for zi+1, i.e., zi+1 = zk & (∀l < k)[zi+1 ̸= zl].
Notice that k ̸= 0, since z0 = f(x), zi+1 = f(xj), and hence, zi+1 = z0 = ⇒f(xj) = f(x) = ⇒xj = x, which is absurd, since x / ∈X while xj ∈X; hence k = t + 1 for some t < i, and by the selection of k, zt ∈Y (otherwise zt+1 = zt and t + 1 would not be a counterexample), and hence, for some s, zt = ys, zt+1 = f(xs).
(119) We compute: zi+1 = zt+1 = ⇒f(xj) = f(xs)(from (118) and (119)) = ⇒xj = xs (because f is one-to-one) = ⇒yj = ys (because the sequence is good) = ⇒zi = zt (by (118) and (119)).
It follows from the inductive hypothesis that t ≥i, hence t + 1 ≥i + 1, and this contradicts the assumption t + 1 < i + 1.
Now (2) implies that for some j < n+2, zj / ∈Y (since Y has n+1 members), and the Lemma holds if we choose y = zj, W ′ = W, (x, y).
⊣(Lemma X) The symmetric Lemma Y gives us for each good sequence W and each y / ∈Y some x / ∈X such that the extension (117) W ′ = W, (x, y) is good, and the construction of the required recursive permutation proceeds by successive application of these two Lemmas starting with the good sequence W0 = ⟨0, f(0)⟩, X0 = {0}, Y0 = {f(0)}.
Odd step 2n + 1. Let y = min(N \ Y2n) and extend W2n by applying Lemma Y, so that y ∈Y2n+1.
Even step 2n + 2. Let x = min(N \ X2n+1) and extend W2n+1 by applying Lemma X so that x ∈X2n+2.
In the end, the union S n Wn is the graph of a permutation h : N ↣ →N which reduces A to B, x ∈A ⇐ ⇒h(x) ∈B.
The recursiveness of h follows from the construction and completes the proof that A ≡B.
⊣ Definition 5B.9 (Codes for r.e. sets). For each e ∈N, let We = {x | ϕe(x)↓}, 5B. Recursively enumerable sets 195 so that the relation x ∈We is semirecursive and the sequence W0, W1, . . .
enumerates all the r.e. sets.
Proposition 5B.10. If A ≤m B and B is recursive, then A is also re-cursive; hence, if A ≤m B and A is not recursive, then B is not recursive either.
With the r.e. completeness of K, this simple fact is the basic tool for proving non-recursiveness for sets and relations: because if we verify that K ≤m B, then B is not recursive.
Example 5B.11. The set A = {e | We ̸= ∅} is r.e. but not recursive.
Proof. The set A is r.e. because the relation e ∈A ⇐ ⇒(∃x)[x ∈We] is Σ0 1. To show that K ≤m A, we let g(e, x) = µyT1(e, e, y), so that the value g(e, x) is independent of x, i.e., g(e, x) = ( µyT1(e, e, y) if (∃y)T1(e, e, y) undefined otherwise.
It follows that for all e and x, e ∈K ⇐ ⇒g(e, x)↓, so that e ∈K ⇐ ⇒(∃x)g(e, x)↓; and so, if b g is any code of g(x, y), e ∈K ⇐ ⇒(∃x)[{b g}(e, x)↓] ⇐ ⇒(∃x)[{S1 1(b g, e)}(x)↓] ⇐ ⇒WS1 1(b g,e) ̸= ∅ ⇐ ⇒S1 1(b g, e) ∈A, so that K ≤1 A and A is not recursive.
⊣ 196 5. Introduction to computability theory Notice that with this construction, e ∈K ⇐ ⇒WS1 1(b g,e) = N ⇐ ⇒WS1 1(b g,e) has at least 2 members, so that the sets B = {e | We = N}, C = {e | We has at least 2 members} are also not recursive.
Problems for Section 5B Problem 5B.1. Prove that a relation P(⃗ x) is Σ0 1 if and only if it is definable by a Σ1 formula, in the sense of Definition 4C.12.
Problem 5B.2. Show that there is a recursive, partial function f(e) such that We ̸= ∅= ⇒[f(e)↓& f(e) ∈We].
Problem 5B.3. Prove or give a counterexample to each of the following propositions: (a) There is a total recursive function u1(e, m) such that for all e, m, Wu1(e,m) = We ∪Wm.
(b) There is a total recursive function u2(e, m) such that for all e, m, Wu2(e,m) = We ∩Wm.
(c) There is a total recursive function u3(e, m) such that for all e, m, Wu3(e,m) = We \ Wm.
Problem 5B.4. Prove or give a counterexample to each of the following propositions, where f : N →N is a total recursive function and f[A] = {f(x) | x ∈A}, f−1[A] = {x | f(x) ∈A}.
(a) If A is recursive, then f[A] is also recursive.
(b) If A is r.e., then f[A] is also r.e.
(c) If A is recursive, then f−1[A] is also recursive.
(d) If A is r.e., then f−1[A] is also r.e.
Problem 5B.5. Prove that there is a total recursive function u(e, m) such that Wu(e,m) = {x + y | x ∈We & y ∈Wm}.
5C. Productive, creative and simple sets 197 Problem 5B.6. Prove that every infinite r.e. set A has an infinite recursive subset.
Problem 5B.7. (The Reduction Property of r.e. sets.) Prove that for every two r.e. sets A, B, there exist r.e. sets A∗, B∗with the following properties: A∗⊆A, B∗⊆B, A ∪B = A∗∪B∗, A∗∩B∗= ∅.
Problem 5B.8. (The Separation Property for r.,e. complements.) Prove that if A and B are disjoint sets whose complements are r.e., then there exists a recursive set C such that A ⊆C, C ∩B = ∅.
Problem 5B.9. (Recursively inseparable r.e. sets.) Prove that there exist two disjoint r.e. sets A and B such that there is no recursive set C satisfying A ⊆C, C ∩B = ∅.
Problem 5B.10. Prove that for every two r.e. sets A, B, there is a formula φ(x) so that whenever x ∈(A ∪B), Q ⊢φ(x) = ⇒x ∈A, (120) Q ⊢¬φ(x) = ⇒x ∈B, (121) Q ⊢φ(x) or Q ⊢¬φ(x) (122) 5C. Productive, creative and simple sets Up until now, the only r.e non-recursive sets we have seen are r.e. complete, and the question arises whether that is all there is.
The next sequence of definitions and results (due to Emil Post) shows that this simplistic picture of the class of r.e. sets is far from the truth.
Definition 5C.1. A function p : N ↣N is a productive function for a set B if it is recursive, one-to-one, and such that We ⊆B = ⇒p(e) ∈B \ We; and a set B is productive if it has a productive function.
A set A is creative if it is r.e. and its complement Ac = {x ∈N | x / ∈A} is productive.
198 5. Introduction to computability theory Proposition 5C.2. The complete set K is creative, with productive func-tion for its complement the identity p(e) = e.
Proof. We must show that We ⊆Kc = ⇒e ∈Kc \ We, i.e., (∀t)[t ∈We = ⇒t / ∈K] = ⇒[e / ∈We & e / ∈K].
Spelling out the hypothesis of the required implication: (∀t)[{e}(t)↓= ⇒{t}(t) ↑]; and the conclusion simply says that {e}(e) ↑, because e / ∈We ⇐ ⇒e / ∈K ⇐ ⇒{e}(e) ↑.
Finally, the hypothesis implies the conclusion because if {e}(e)↓, then, setting t = e in the hypothesis we get {e}(e) ↑, which is contradictory.
⊣ Corollary 5C.3. Every r.e. complete is creative.
Proof. It is enough to show that if A is productive and A ≤1 B, then B is also productive, and then apply this to the complement Xc of the given, r.e.
complete set X for which we have Kc ≤1 Xc (because K ≤1 X). So suppose that x ∈A ⇐ ⇒f(x) ∈B with f(x) recursive and 1 - 1, and that p(e) is a productive function for A.
Choose u(e) (by appealing to the Sm n -Theorem) such that it is recursive, 1 - 1, and for each e, Wu(e) = f−1[We], and let q(e) = f(p(u(e))).
To verify that q(e) is a productive function for B, we compute: We ⊆B = ⇒Wu(e) = f−1[We] ⊆A = ⇒p(u(e)) ∈A \ f−1[We] = ⇒q(e) = f(p(u(e))) ∈B \ We.
⊣ Proposition 5C.4. Every productive set B has an infinite r.e. subset.
5C. Productive, creative and simple sets 199 Proof. The idea is to define a function f : N →N by the recursion f(0) = e0, where We0 = ∅ f(x + 1) = some code of Wf(x) ∪{p(f(x))}, where p(e) is a given productive function for B. If we manage this, then a simple induction will show that, for every x, Wf(x) ⊊Wf(x+1) ⊆B, so that the set A = Wf(0) ∪Wf(1) ∪. . . = {y | (∃x)[y ∈Wf(x)} is an infinite, r.e. subset of B. For the computation of the required function h(w, x) such that f(x + 1) = h(f(x), x), let first R(e, y, x) ⇐ ⇒x ∈We ∨x = y and notice that this is a semirecursive relation, so that for some b g, x ∈We ∪{y} ⇐ ⇒{b g}(e, y, x)↓ ⇐ ⇒{S2 1(b g, e, y)}(x)↓.
This means that if we set u(e, y) = S2 1(b g, e, y), then Wu(e,y) = We ∪{y}.
Finally, set h(w, x) = u(w, p(w)), and in the definition of f, f(x + 1) = h(f(x), x) = u(f(x), p(f(x))), so that Wf(x+1) = Wf(x) ∪{p(f(x))} as required.
⊣ Definition 5C.5. A set A is simple if it is r.e., and its complement Ac is infinite and has no infinite r.e. subset, i.e., We ∩A = ∅= ⇒We is finite.
Theorem 5C.6 (Emil Post). There exists a simple set.
200 5. Introduction to computability theory Proof. The relation R(x, y) ⇐ ⇒y ∈Wx & y > 2x is semirecursive, so that by the Σ0 1-Selection Lemma 5A.9, there is a recursive partial function f(x) such that (∃y)[y ∈Wx & y > 2x] ⇐ ⇒f(x)↓ ⇐ ⇒f(x)↓& f(x) ∈Wx & f(x) > 2x.
The required set is the image of f, A = {f(x) | f(x)↓} = {y | (∃x)[f(x) = y]} = {y | (∃x)[f(x) = y & 2x < y]}, (123) where the last, basic equality follows from the definition of the relation R(x, y).
(1) A is r.e., from its definition, because the graph of f(x) is Σ0 1.
(2) The complement Ac of A is infinite, because y ∈A & y ≤2z = ⇒(∃x)[y = f(x) & 2x < y ≤2z] = ⇒(∃x)[y = f(x) & x < z], so that at most z of the 2z + 1 numbers ≤2z belong to A; it follows that some y ≥z belongs to the complement Ac, and since this holds for every z, the set Ac is infinite.
(3) For every infinite We, We ∩A ̸= ∅, because We is infinite = ⇒(∃y)[y ∈We & y > 2e] = ⇒f(e)↓& f(e) ∈We = ⇒f(e) ∈We ∩A.
⊣ Corollary 5C.7. Simple sets are neither recursive nor r.e. complete, and so there exists an r.e., non-recursive set which is not r.e. complete.
Proof. A simple set cannot be recursive, because its (infinite, by definition) complement is a witness against its simplicity; and it cannot be r.e. complete, because it is not creative by Proposition 5C.4.
⊣ Problems for Section 5C Problem 5C.1. Show that if A is simple and B is infinite r.e., then the intersection A ∩B is infinite.
5C. Productive, creative and simple sets 201 Problem 5C.2. Show that the intersection of two simple sets is simple.
Problem 5C.3. Prove or give a counterexample: (a) For each infinite r.e. set A, there is a total, recursive function f(x) such that for each x, f(x) > x & f(x) ∈A.
(b) (a) For each r.e. set A with infinite complement, there is a total, recursive function f(x) such that for each x, f(x) > x & f(x) / ∈A.
Problem 5C.4. Prove that if Q is interpretable in a consistent, axiomati-zable theory T, then the set #T = {#θ | θ is a sentence and T ⊢θ} of (codes of) theorems of T is creative.
Solution.
We will show how to define for each m some m such that Wm ∩#T = ∅= ⇒m / ∈#T & m / ∈Wm, skipping the (easy) uniformity argument, that, in fact, m = f(m) for some recursive, total f. The idea is to imitate the proof of the Rosser Theorem 4C.4, replacing the refutation relation by membership in Wm. So put Rm(e, y) ⇐ ⇒e codes a sentence φ of PA and T1(m, #π(φ), y).
We let R(e, y) be a formula which numeralwise expresses this relation in Q, and choose by the Fixed Point Lemma 4B.14 a sentence σ of PA such that Q ⊢σ ↔(∀y)[Proof π,T (⌜σ⌝, y) →(∃u ≤y)R(⌜σ⌝, u)].
We will show that m = #π(σ) has the required property—and it can be computed recursively from m because the proof of the Fixed Point Lemma is effective.
So assume that Wm ∩#T = ∅.
(a) T ̸⊢π(σ).
The argument is almost exactly like that in the proof of Rosser’s Theorem: if there were a proof in T of π(σ) with code y, then we would know that Q ⊢Proof π,T (⌜σ⌝, ∆y); and since (by the hypothesis) #π(σ) / ∈Wm, we know that for each u, Q ⊢¬R(∆m, ⌜σ⌝, ∆u); and using properties of Q as in the proof of the Rosser Theorem, we show that Q ⊢¬σ so that T ⊢¬π(σ) contradicting the consistency of T.
202 5. Introduction to computability theory (b) #π(σ) / ∈Wm. Suppose #π(σ) ∈Wm. This means that there is a u such that R(#σ, u), and so Q ⊢R(⌜σ⌝, ∆u). On the other hand, no y codes a proof of π(σ) in T by (a), and so for every y, Q ⊢¬Proof π,T (⌜σ⌝, ∆y); and then the usual arguments about Q (as in the proof of the Rosser Theorem) show that Q ⊢σ, which contradicts (a).
5D. The Second Recursion Theorem In this section we will prove a very simple result of Kleene, which has sur-prisingly strong and unexpected consequences in many parts of definability theory, and even in analysis and set theory. Here we will prove just one, sub-stantial application of the Second Recursion Theorem, but we will also use it later in the theory of recursive functionals and effective operations.
Theorem 5D.1 (Kleene). For each recursive partial function f(z, ⃗ x), there is a number z∗, such that for all ⃗ x, ϕz∗(⃗ x) = {z∗}(⃗ x) = f(z∗, ⃗ x).
(124) In fact, for each n, there is a primitive recursive function hn(e), such that if f = ϕn+1 e , is n + 1-ary, then equation (124) holds with z∗= hn(e), i.e., ϕhn(e)(⃗ x) = {hn(e)}(⃗ x) = ϕe(hn(e), ⃗ x).
(125) The theorem gives immediately several simple propositions which show that the coding of recursive partial functions has many unexpected (and even weird) properties.
Example 5D.2. There exist numbers z1 – z4, such that ϕz1(x) = z1 ϕz2(x) = z2 + x Wz3 = {z3} Wz4 = {0, . . . , z4}.
Proof. For z1, we apply the Second Recursion Theorem to the function f(z, x) = z and we set z1 = z∗; it follows that ϕz1(x) = f(z1, x) = z1.
The rest are similar and equally easy.
⊣ 5D. The Second Recursion Theorem 203 Proof of the Second Recursion Theorem 5D.1. The partial function g(z, ⃗ x) = f(S1 n(z, z), ⃗ x) is recursive, and so there some number b g such that {S1 n(b g, z)}(⃗ x) = {b g}(z, ⃗ x) = f(S1 n(z, z), ⃗ x); the result follows from this equation if we set z∗= S1 n(b g, b g).
For the stronger (uniform) version (125), let d be a number such that ϕd(e, z, ⃗ x) = ϕe(S1 n(z, z), ⃗ x); it follows that b g = S1 n+1(d, e) is a code of ϕe(S1 n(z, z), ⃗ x), and the required function is h(e) = S1 n(b g, b g) = S1 n(S1 n+1(d, e), S1 n+1(d, e)).
⊣ For a (much more significant) example of the strength of the Second Recur-sion Theorem, we show here the converse of 5C.3, that every creative set is r.e. complete (and a bit more).
Theorem 5D.3 (John Myhill). The following are equivalent for every r.e.
set A.
(1) There is a recursive partial function p(e) such that We ∩A = ∅= ⇒[p(e)↓& p(e) ∈Ac \ We].
(2) There is a total recursive function q(e) such that We ∩A = ∅= ⇒q(e) ∈Ac \ We.
(126) (3) A is creative, i.e., (126) holds with a one-to-one recursive function q(e).
(4) A is r.e. complete.
In particular, an r.e. set is complete if and only if it is creative.
Proof. (1) ⇒(2).
For the given, recursive partial function p(e), there exists (by the Second Recursion Theorem) some number z such that {S1 1(z, e)}(t) = ϕz(e, t) = ( ϕe(t), if p(S1 1(z, e))↓, ↑, otherwise.
204 5. Introduction to computability theory We set q(e) = p(S1 1(z, e)) with this z, and we observe that q(e) is a total function, because q(e) = p(S1 1(z, e)) ↑= ⇒WS1 1(z,e) = ∅by the definition = ⇒p(S1 1(z, e))↓.
In addition, since q(e)↓, WS1 1(z,e) = We, and hence We ∩A = ∅= ⇒q(e) = p(S1 1(z, e)) ∈Ac \ WS1 1(z,e) = Ac \ We which is what we needed to show.
(2) ⇒(3) (This implication does not use the Second recursion Theorem, and could have been given in Section 5A.) For the given q(e) which satisfies (126), we observe first that there is a recursive function h(e) such that Wh(e) = We ∪{q(e)}; and then we set, by primitive recursion, g(0, e) = e g(i + 1, e) = h(g(i, e)), so that (easily, by induction on i), Wg(i+1,e) = We ∪{q(g(0, e)), q(g(1, e)), . . . , q(g(i, e))}.
It follows that for each i > 0, (127) We ∩A = ∅ = ⇒q(g(i, e)) ∈Ac \ (We ∪{q(g(0, e)), q(g(1, e)), . . . , q(g(i −1, e))}), and, more specifically, We ∩A = ∅= ⇒(∀j < i)[q(g(i, e)) ̸= q(g(j, e))].
(128) Finally, we set f(0) = q(0), and for the (recursive) definition of f(e + 1), we compute first, in sequence, the values q(g(0, e + 1)), . . . , q(g(e + 1, e + 1)) and we distinguish two cases.
Case 1. If these values are all distinct, then one of them is different from the values f(0), . . . , f(e), and we just set j = (µi ≤(e + 1))(∀y ≤e)[q(g(i, e + 1)) ̸= f(y)] f(e + 1) = q(g(j, e + 1)).
5D. The Second Recursion Theorem 205 Case 2. There exist i, j ≤e+1, i ̸= j, such that q(g(i, e+1)) = q(g(j, e+1)).
In this case we set f(e + 1) = max{f(0), . . . , f(e)} + 1.
It is clear that f(e) is recursive and one-to-one, and that it is a productive function for Ac follows immediately from (128) and (127).
(3) ⇒(4). If q(e) is a productive function for the complement Ac and B is any r.e. set, then (by the Second Recursion Theorem) there is some number z such that ϕz(x, t) = ( 1 if x ∈B & t = q(S1 1(z, x)) ↑ otherwise; the function f(x) = q(S1 1(z, x)) is one-to-one (as a composition of one-to-one functions), and it reduces B to A, as follows.
If x ∈B, then WS1 1(z,x) = {q(S1 1(z, x)} = {f(x)}, and f(x) / ∈A = ⇒WS1 1(z,x) ∩A = ∅ = ⇒q(S1 1(z, x)) ∈Ac \ WS1 1(z,x) = ⇒f(x) ∈Ac \ {f(x)}, which is a contradiction; hence f(x) ∈A. On the other hand, if x / ∈B, then WS1 1(z,x) = ∅⊆Ac, hence f(x) = q(S1 1(z, x)) ∈Ac.
⊣ Problems for Section 5D Problem 5D.1. Prove that there is some number z such that Wz = {z, z + 1, . . . } = {x | x ≥z}.
Problem 5D.2. Prove that for some number t and all x, ϕt(x) = t + x.
Problem 5D.3. Prove that for each total, recursive function f(x) one of the following holds: (a) There is a number z such that f(z) is odd and for all x, ϕz(x) = f(z + x); or (b) there is a number w such that f(w) is even and for all x, ϕw(x) = f(2w + x + 1).
206 5. Introduction to computability theory Problem 5D.4. Prove or give a counterexample: for each total, recursive function f(x), there is some z such that Wf(z) = Wz.
Problem 5D.5. Prove or give a counterexample: for every total, recursive function f(x), there is a number z such that for all t, ϕf(z)(t) = ϕz(t).
Problem 5D.6. (a) Prove that for every total, recursive function f(x), there is a number z such that Wz = {f(z)}.
(b) Prove that there is some number z such that ϕz(z)↓and Wz = {ϕz(z)}.
Problem 5D.7. Suppose A ⊆N, x ⪯y is a wellordering of A, F : (N ⇀N) × N →N is a function(al) defined on partial functions on N, and g : A →N is defined by the transfinite recursion g(x) = F(g↾{y ∈A | y ≺x}, x).
Suppose in addition, that there is a recursive partial function ψ(e, x) satisfying the following, for every e, every x, and every partial function p: if (∀y ≺x)[p(y) = ϕe(y)], then F(p, x) = ψ(e, x).
Prove that there is a recursive partial function g∗such that for all x ∈A, g(x) = g∗(x).
Note: It is not assumed that A is a recursive set, or that x ⪯y is a recursive relation; the result holds for completely arbitrary A and x ⪯y.
5E. The arithmetical hierarchy The semirecursive (Σ0 1) relations are of the form (∃y)Q(⃗ x, y) where Q(⃗ x, y) is recursive, and so they are just one existential quantifier “away” from the recursive relations in complexity. The next definition gives us a useful tool for the classification of complex, undecidable relations.
5E. The arithmetical hierarchy 207 Definition 5E.1. The classes (sets) of relations Σ0 k, Π0 k, ∆0 k are defined recursively, as follows: Σ0 1 : the semirecursive relations Π0 k = ¬Σ0 k : the negations (complements) of relations in Σ0 k Σ0 k+1 = ∃NΠ0 k : the relations which satisfy an equivalence P(⃗ x) ⇐ ⇒(∃y)Q(⃗ x, y), where Q(⃗ x, y) is Π0 k ∆0 k = Σ0 k ∩Π0 k : the relations which are both Σ0 k and Π0 k.
A set A is in one of these classes Γ if the relation x ∈A is in Γ.
5E.2. Canonical forms. These classes of the arithmetical hierarchy are (obviously) characterized by the following “canonical forms”, in the sense that a given relation P(⃗ x) is in a class Γ if it is equivalent with the canonical form for Γ, with some recursive Q: Σ0 1 : (∃y)Q(⃗ x, y) Π0 1 : (∀y)Q(⃗ x, y) Σ0 2 : (∃y1)(∀y2)Q(⃗ x, y1, y2) Π0 2 : (∀y1)(∃y2)Q(⃗ x, y1, y2) Σ0 3 : (∃y1)(∀y2)(∃y3)Q(⃗ x, y1, y2, y3) .
.
.
A trivial corollary of these canonical forms is that: Proposition 5E.3. The relations which belong to some Σ0 k or some Π0 k are precisely the arithmetical relations.
Proof. Each primitive recursive relation is arithmetical, by Theorem 4B.13 and Lemma 4B.2, and then (inductively) every Σ0 k and every Π0 k relation is arithmetical, because the class of arithmetical relations is closed under nega-tion and quantification on N.
For the other direction, we notice that re-lations defined by quantifier-free formulas are (trivially) recursive, and that every arithmetical relation is defined by some formula in prenex form with quantifier-free matrix; and by introducing dummy quantifiers, if necessary, we may assume that the quantifiers in the prefix are alternating and start with an ∃, so that the relation defined by each formula is in some Σ0 k.
⊣ Theorem 5E.4. (1) For each k ≥1, the classes Σ0 k, Π0 k, and ∆0 k are closed for (total) recursive substitutions and for the operations &, ∨, ∃≤and ∀≤. In addition: • Each ∆0 k is closed for negation ¬.
208 5. Introduction to computability theory • Each Σ0 k is closed for ∃N, existential quantification over N.
• Each Π0 k is closed for ∀N, universal quantification over N.
(2) For each k ≥1, Σ0 k ⊆∆0 k+1, (129) and hence the arithmetical classes satisfy the following diagram of inclusions: Σ0 1 Σ0 2 Σ0 3 ⊆ ⊆ ⊆ ⊆ ⊆ ⊆ ∆0 1 ∆0 2 ∆0 3 · · · ⊆ ⊆ ⊆ ⊆ ⊆ ⊆ Π0 1 Π0 2 Π0 3 Proof. First we verify the closure of all the arithmetical classes for recursive substitutions, by induction on k; the proposition is known for k = 1 by 5A.7, and (inductively), for the case of Σ0 k+1, we compute: P(⃗ x) ⇐ ⇒R(f1(⃗ x), . . . , fn(⃗ x)) ⇐ ⇒(∃y)Q(f1(⃗ x), . . . , fn(⃗ x), y) where Q ∈Π0 k, by definition ⇐ ⇒(∃y)Q′(⃗ x, y) where Q′ ∈Π0 k by the induction hypothesis.
The remaining parts of (1) are shown directly (with no induction) using the transformations in the proof of 5A.7.
We show (2) by induction on k, where, in the basis, if P(⃗ x) ⇐ ⇒(∃y)Q(⃗ x, y) with a recursive Q, then P is surely Σ0 2, since each recursive relation is Π0 1; but a semirecursive relation is also Π0 2, since, obviously, P(⃗ x) ⇐ ⇒(∀z)(∃y)Q(⃗ x, y) and the relation Q1(⃗ x, z, y) ⇐ ⇒Q(⃗ x, y) is recursive. The induction step of the proof is practically identical, and the inclusions in the diagram follow easily from (129) and simple computations.⊣ More interesting is the next theorem which justifies the appellation “hierar-chy” for the classes Σ0 k, Π0 k: 5E. The arithmetical hierarchy 209 Theorem 5E.5 (Kleene).
(1) (Enumeration for Σ0 k) For each k ≥1 and each n ≥1, there is an n + 1-ary relation e S0 k,n(e, ⃗ x) in the class Σ0 k which enumerates all the n-ary, Σ0 k relations, i.e., P(⃗ x) is Σ0 k if and only if for some e, P(⃗ x) ⇐ ⇒e S0 k,n(e, ⃗ x).
(2) (Enumeration for Π0 k) For each k ≥1 and each n ≥1, there is an n + 1-ary relation e P 0 k,n(e, ⃗ x) in Π0 k which enumerates all the n-ary, Π0 k relations, i.e., P(⃗ x) is Π0 k if and only if, for some e, P(⃗ x) ⇐ ⇒e P 0 k,n(e, ⃗ x).
(3) (Hierarchy Theorem) The inclusions in the Diagram of Proposition 5E.4 are all strict, i.e., Σ0 1 Σ0 2 Σ0 3 ⊊ ⊊ ⊊ ⊊ ⊊ ⊊ ∆0 1 ∆0 2 ∆0 3 · · · ⊊ ⊊ ⊊ ⊊ ⊊ ⊊ Π0 1 Π0 2 Π0 3 Proof. For (1) and (2) we set, recursively, e S0 1,n(e, ⃗ x) ⇐ ⇒(∃y)Tn(e, ⃗ x, y) e P 0 k,n(e, ⃗ x) ⇐ ⇒¬e S0 k,n(e, ⃗ x) e S0 k+1,n(e, ⃗ x) ⇐ ⇒(∃y) e P 0 k,n+1(e, ⃗ x, y), and the proofs are easy, with induction on k. For (3), we observe that the “diagonal” relation Dk(x) ⇐ ⇒e S0 k,1(x, x) is Σ0 k but cannot be Π0 k, because, if it were, then for some e we would have ¬e S0 k,1(x, x, ) ⇐ ⇒e S0 k,1(e, x) which is absurd when x = e. It follows that for each k, there exist relations which are Σ0 k but not Π0 k, and from this follows easily the strictness of all the inclusions in the diagram.
⊣ Theorem 5E.5 gives an alternative proof—and a better understanding— of Tarscki’s Theorem 4A.5, that the truth set of arithmetic TruthN is not arithmetical, cf. Problem 5E.1.
210 5. Introduction to computability theory Definition 5E.6 (Classifications). A (complete) classification of a rela-tion P(⃗ x) (in the arithmetical hierarchy) is the determination of “the least” arithmetical class to which P(⃗ x) belongs, i.e., the proof of a proposition of the form P ∈Σ0 k \ Π0 k, P ∈Π0 k \ Σ0 k, or P ∈∆0 k+1 \ (Σ0 k ∪Π0 k); for example, in 5B.11 we showed that {e | We ̸= ∅} ∈Σ0 1 \ Π0 1.
The complete classification of a relation P is sometimes very difficult, and we are often satisfied with the computation of some “upper bound”, i.e., some k such that P ∈Σ0 k or P ∈Π0 k. The basic method for the computation of a “lower bound,” when this can be done, is to show that the given relation is complete in some class Σ0 k or Π0 k as in the next result.
Proposition 5E.7. (1) The set F = {e | ϕe is total } is Π0 2 but it is not Σ0 2.
(2) The set Fin = {e | We is finite} is Σ0 2 \ Π0 2.
Proof. (1) The upper bound is obvious, since e ∈F ⇐ ⇒(∀x)(∃y)T1(e, x, y).
To show (by contradiction) that F is not Σ0 2, suppose P(x) is any Π0 2 relation, so that P(x) ⇐ ⇒(∀u)(∃v)Q(x, u, v) with a recursive Q(x, u, v), and set f(x, u) = µvQ(x, u, v).
If b f is a code of this (recursive) partial function f(x, u), then P(x) ⇐ ⇒(∀u)[f(x, u)↓] ⇐ ⇒(∀u)[{S1 1( b f, x)}(u)↓] ⇐ ⇒S1 1( b f, x) ∈F; it follows that if F were Σ0 2, then every Π0 2 would be Σ0 2, which contradicts the Hierarchy Theorem 5E.5 (3).
(2) The upper bound is again trivial, e ∈Fin ⇐ ⇒(∃k)(∀x)[x ∈We = ⇒x ≤k].
For the lower bound, let P(x) be any Σ0 2 relation, so that P(x) ⇐ ⇒(∃u)(∀v)Q(x, u, v) 5E. The arithmetical hierarchy 211 with a recursive Q. We set g(x, u) = µy(∀i ≤u)¬Q(x, i, (y)i), so that if b g is a code of g, then (∃u)(∀v)Q(x, u, v) ⇐ ⇒{u | g(x, u)↓} is finite ⇐ ⇒{u | {b g}(x, u)↓} is finite ⇐ ⇒{u | {S1 1(b g, x)}(u)↓} is finite, i.e., P(x) ⇐ ⇒S1 1(b g, x) ∈Fin; but this implies that Fin is not Π0 2, because, if it were, then every Σ0 2 relation would be Π0 2, which it is not.
⊣ Problems for Section 5E Problem 5E.1. Prove that for every arithmetical relation P(⃗ x), there is a 1-1, total recursive function f : Nn →N such that P(⃗ x) ⇐ ⇒f(⃗ x) ∈TruthN; infer Tarski’s Theorem 4A.5, that TruthN is not arithmetical.
Problem 5E.2. Classify in the arithmetical hierarchy the set A = {e | We ⊆{0, 1}}.
Problem 5E.3. Classify in the arithmetical hierarchy the set A = {e | We is a singleton}.
Problem 5E.4. Classify in the arithmetical hierarchy the set A = {e | We is finite and non-empty}.
Problem 5E.5. Classify in the arithmetical hierarchy the set B = {x | there are infinitely many twin primes p ≥x}, where p is a twin prime if both p and p + 2 are prime numbers.
Problem 5E.6. Classify in the arithmetical hierarchy the relation Q(e, m) ⇐ ⇒ϕe ⊑ϕm ⇐ ⇒(∀x)(∀w)[ϕe(x) = w = ⇒ϕm(x) = w].
Problem 5E.7. Classify in the arithmetical hierarchy the set A = {e | We has at least e members}.
212 5. Introduction to computability theory Problem 5E.8. Classify in the arithmetical hierarchy the set B = {e | for some w and all x, if ϕe(x)↓, then ϕe(x) ≤w}.
(This is the set of codes of bounded recursive partial functions.) Problem 5E.9. For a fixed, unary, total recursive function f, classify in the arithmetical hierarchy the set of all the codes of f, Cf = {e | ϕe = f}.
Problem 5E.10. Let A be some recursive set with non-empty comple-ments, i.e., A ⊊N. Classify in the arithmetical hierarchy the set B = {e | We ⊆A}.
Problem 5E.11. Show that the graph Gf(⃗ x, w) ⇐ ⇒f(⃗ x) = w of a total function is Σ0 k if and only if it is ∆0 k.
Is this also true of partial functions?
Problem 5E.12. A total function f : Nn →N is limiting recursive if there is a total, recursive function g : Nn+1 →N such that for all ⃗ x, f(⃗ x) = lim m→∞g(m, ⃗ x).
Prove that a total f(⃗ x) is limiting recursive if and only if the graph Gf of f(⃗ x) is ∆0 2.
5F. Relativization The notions of µ-recursiveness in 4E.5 and reckonability in 4E.9 “relativize” naturally to a “given” partial function as follows.
Definition 5F.1. For a fixed partial function p : Nm ⇀N: (1) A µ-recursive derivation from (or relative to) p is a sequence of partial functions on N f1, f2, . . . , fk, where each fi is S, or a constant Cn q or a projection P n i , or p, or is defined by composition, primitive recursion or minimalization from functions before it in the sequence; and a partial function f : Nn ⇀N is µ-recursive in p if it occurs in a µ-recursive derivation from p.
5F. Relativization 213 (2) For each partial function p, let Qp be Robinson’s system in the extension of the language of arithmetic with a single m-ary function symbol p and with the additional axioms Dp = {p(∆x1, . . . , ∆xm) = ∆w | p(x1, . . . , xm) = w}, which express formally the graph of p. A partial function f is reckonable in p, if there is a formula F(v1, . . . , vn, y, p) of Qp, such that for all ⃗ x, w, f(⃗ x) = w ⇐ ⇒Qp ⊢F(∆x1, . . . , ∆xn, ∆w, p).
These notions express two different ways in which we can compute a function f given access to the values of p, and they behave best when the “given” p is total, in which case they coincide: Proposition 5F.2. If p : Nm →N is a total function, then, for every (possibly partial) f, f is µ-recursive in p ⇐ ⇒f is reckonable in p.
Proof is a simple modification of the proof of 4E.10.
⊣ We will just say f is recursive in p or f is Turing-recursive in p for this notion of reduction of a partial function to a total one, the reference to “Turing” coming from a third equivalent definition which involves “Turing machines with oracles”.
The Enumeration Theorem The Enumeration Theorem (4F.1) generalizes easily to the present context.
For simplicity, we restrict ourselves to the case that p is a function of one variable.
Define T p n(e, x1, . . . , xn, y) exactly as Tn(e, x1, . . . , xn, y) was defined, except use Qp in place of Q.
If T p n(e, ⃗ x, y) and the number of p(∆x) = ∆w is ((y)1)i for some i < lh(y), then all of x and w are smaller than y. This allows us to define a single relation that does the work of all the T p n.
Set ˜ Tn(a, e, ⃗ x, y) ⇔(∃p)(p(y) = a ∧T p n(e, ⃗ x, y)).
Notice that (∃p) could be replaced by (∀p) without affecting the relation de-fined. Define ϕp,n e (⃗ x) = U(µy ˜ Tn(p(y), e, ⃗ x, y))).
214 5. Introduction to computability theory Theorem 5F.3. (1) Each relation ˜ Tn(a, e, ⃗ x, y) is primitive recursive.
(2) Each ϕp,n e (⃗ x) is a partial function recursive in p, and so is the partial function which “enumerates” all these, ϕp,n(e, ⃗ x) = ϕp,n e (⃗ x).
(3) For each partial function f(x1, . . . , xn) of n arguments that is recursive in p, there exists some e (a code of f) such that f(⃗ x) = ϕp,n e (⃗ x) = U(µy ˜ Tn(p(y), e, x1, . . . , xn, y)).
(130) 5F.4. Turing reducibility and Turing degrees.
For any two total functions f, g : N →N, we set f ≤T g ⇐ ⇒f is recursive in (or Turing reducible to) g We also set f ≡T g ⇐ ⇒f ≤T g & g ≤T f, and we assign to each set g its degree (of unsolvability) deg(f) = {g | f ≡T g}.
We set deg(f) ≤deg(g) ⇔f ≤T g.
We extend these definitions to subsets of N by, for example, letting A ≤T B just in case χA ≤T χB, where χA, χB are the characteristic functions of A and B.
Proposition 5F.5. (1) A ≤m B = ⇒A ≤T B, but the converse is not al-ways true.
(2) If f ≤T g and g ≤T h, then f ≤T h.
(3) If g is recursive, then, for every f, f ≤T g ⇐ ⇒f is recursive, and so deg(∅) = deg(N) = {f | g is recursive}.
Proposition 5F.6. Any two Turing degrees have a least upper bound.
Proof. Given f and g, let h(2n) = f(n) and h(2n + 1) = g(n).
⊣ Proposition 5F.7. For every Turing degree, there is a greater one.
5F. Relativization 215 Proof. Since only countably many functions are recursive in f, there is a g that is not recursive in f. The least upper bound of deg(f) and deg(g) is greater than deg(f).
Another proof: It is easy to check that {e | ϕf e(e)↓} is is not recursive in f.
(In fact, it has greater degree than deg(f).) ⊣ Theorem 5F.8 (The Kleene-Post Theorem). There exist Turing incompa-rable functions f, g : N →N, i.e., f ̸≤T g and g ̸≤T f.
(131) Proof. There are numerous ways to prove the theorem.
The following proof is essentially the original one, and it gives f and g that are recursive in the set K.
We define by recursion se, te s′ e, and t′ e for each e ∈N. Each will be a finite sequence of natural numbers. For each e, we shall have se ⊆s′ e ⊊se+1 and te ⊊t′ e ⊆te+1.
We will then let f extend all the se and g extend all the te.
Let s0 = t0 = the empty sequence.
Given se and te, let ae = µa(a codes s(a) ⊇se & (∃y ≤lh(a) ˜ T1(a↾y, e, lh(te), y)) if it exists; #se otherwise.
Let s′ e be the sequence coded by ae. Let lh(t′ e) = lh(te) + 1. If ae is defined by the first clause, let t′ e(lh(te)) = U(µy ˜ T1(ae↾y, e, lh(te), y)) + 1.
If ae is defined by the “otherwise” clause, let t′ e(lh(te)) = 0. Note that the definitions assure that ϕf e(lh(te)) is either undefined or defined and different from g(lh(te)).
Define te+1 and se+1 from t′ e and s′ e as s′ e and t′ e were defined from se and te. This definition guarantees that f(lh(te)) is different from ϕg e.(lh(s′ e)).
⊣ A set is recursively enumerable (r.e.) in f if it is the domain of a partial function recursive in f.
Proposition 5F.9. A set is recursive in f just in case both it and its com-plement are r.e. in f.
For each f, W f e be the domain of ϕf e. Let Kf = {e | e ∈W f e }.
Proposition 5F.10. Kf is r.e. in f but not recursive in f.
216 5. Introduction to computability theory 5F.11. Remark. Actually deg(f) < deg(Kf), and every set r.e. in f is ≤1 Kf. Moreover f ≤T g ⇔Kf ≤1 Kg.
A degree of unsolvability is r.e. if it is the degree of an r.e. set.
Theorem 5F.12. There is a simple set A such that KA ≤T K. (Hence the remark above implies that deg(KA) = deg(K).) Corollary 5F.13. There is an r.e. degree d such that 0 < deg(d) < deg(K).
(Here 0 is the degree of recursive functions and sets.) Proof. To prove the theorem, we will define by recursion a total recursive function g(e, x) and a recursive relation A(s, x). We let A be the r.e. set {x | (∃s)A(s, x)}.
Let As = {x | A(s, x)}. We will arrange that (s < s′ & x ∈As) ⇒x ∈As′.
Let A0 = ∅.
Let as be the characteristic function of As and let a be the characteristic function of A. For each s, let g(s, e) = µy ≤s ˜ T1( ¯ as(y), e, e, y) if (∃y ≤s) ˜ T1( ¯ as(y), e, e, y); 0 otherwise.
Note that if g(s, e) ̸= 0 then (i) e ∈KAs and (ii) if ¯ a(g(s, e)) = ¯ as(g(s, e)) then e ∈KA.
Let W s e = {x | (∃y ≤s)T1(e, x, y)}.
Let x ∈As+1 just in case x ∈As or there is an e ≤s such that the following conditions are satisfied: (1) W s e ∩As = ∅; (2) x ∈W s e ; (3) x > 2e; (4) (∀e′ < e)g(s, e′) ≤x; (5) x is the smallest number satisfying (1)-(4).
Lemma 5F.14. Ac is infinite.
Proof. Suppose conditions (1)-(5) are satisfied by x for e and s. By con-dition (5), no x′ ̸= x satisfies (1)-(5) for e and s. Since x ∈W s+1 e ∩As+1 ̸= ∅, no x′ can satisfy condition (1) for e and any s′ > s. Thus for each e there is at most one x that satisfies (1)-(5) for e and any s. By condition (3), for any e′ there are at most e′ + 1 numbers x ≤2e′ that satisfy (1)-(5) for any e and s. Thus A has at most e′ + 1 members that are ≤2e′.
⊣ 5F. Relativization 217 Lemma 5F.15. For each e, lims g(s, e) exists.
Proof. Fix e. Let s0 be large enough that (∀e′ ≤e)(We ∩A ̸= ∅⇒W s e ∩A ̸= 0).
We show that there is no s ≥s0 such that g(s, e) ̸= 0 and g(s + 1, e) ̸= g(s, e).
Assume that there is such an s. Some x < g(s, e) must satisfy (1)-(5) for some e′ and s. By condition (4), e′ must be ≤e. Since s ≥s0, this is impossible. ⊣ Lemma 5F.16. A is simple.
Proof. By Lemma 5F.14, it is enough to prove that A ∩We ̸= ∅for every e such that We is infinite. Fix e and assume that We is infinite. Let s0 be such that (∀e′ < e)(∀s ≥s0)g(s, e′) = g(s0, e′).
Let x0 = max{g(s0, e′) | e′ < e}. Since We is infinite, it has a member ≥x0 and > 2e. Let x be the least such member. Let s ≥s0 be such that x ∈W s e .
Either W s e ∩As ̸= ∅or else conditions (1)-(5) are satisfied by x for e and s.
Whichever is the case, W s+1 e ∩As+1 ̸= ∅.
⊣ Lemma 5F.17. KA ≤T K.
Proof. Define a function f by f(e) = µs(∀s′ ≥s)g(s′, e) = g(s, e).
Since e ∈KA ⇔g(f(e), e) > 0, KA ≤T f. The relation R(s, e) ⇔(∀s′ ≥s)g(s′, e) = g(s, e) is Π0 1, and so it is recursive in K. Since f is recursive in R, KA ≤T f ≤T K.
⊣ This completes the proof of the theorem.
⊣ Definition 5F.18 (Functionals). A (partial) functional is any partial func-tion α(x1, . . . , xn, p1, . . . , pm) of n number arguments and m partial function arguments, such that for i = 1, . . . , m, pi ranges over the ki-ary partial func-tions on N, and such that (when it takes a value), α(x1, . . . , xn, p1, . . . , pm) ∈ N.
We view every partial function f(⃗ x) as a functional 0 partial function arguments. More interesting examples include: α1(x, p) = p(x + 1) α2(x, p, q) = if x = 0 then p(x) else q(x, p(x −1)) α3(p) = ( 1 if p(0)↓or p(1)↓ ↑ otherwise 218 5. Introduction to computability theory α4(p) = ( 1 if p(0)↓ 0 otherwise α5(p) = ( 1 if (∀x)[p(x)↓] ↑ otherwise From these examples, we might say that α1 and α2 are “recursive”, in the sense that we can see a direct method for computing their values if we have access to a “oracles” who can respond to questions of the form What is p(x)?
What is q(x, y)?
for specific x. To compute α2(x, p), for example, if x = 0 we request of the oracle the value p(0, x) and give it as output, while, if x > 0, then we first request the value v = p(x −1, 0), and then we request and give as output the value p(x, v). On the other hand, there is no obvious way to compute the values of α4 and α5 in this way, unless we can ask the oracle questions about the domain of convergence of p, a conception which does not yield a natural and useful notion of computability. Finally, α3(p) is a borderline case, which appears to be recursive if we can ask the oracle “non-deterministic” questions of the form what is p(0) or p(1)?, which looks iffy—or, at the least, suggests on a different notion of “non-deterministic computability” for functionals.
Definition 5F.19 (Recursive functionals). We make these two notions of functional computability precise, using the relativization process.
(1) A µ-recursive (functional) derivation (in one, m-ary partial func-tion variable) is a sequence of functionals α1(⃗ x1, p), . . . , αm(⃗ xm, p) in which each αi is S, Cn q or P n j (not depending on p); an evaluation functional evm(x1, . . . , xm, p) = p(x1, . . . , xm) (132) which introduces dependence on p; or it is defined from previously listed func-tionals by composition, primitive recursion or minimalization, which are de-fined as before, e.g, αi(⃗ x, p) = µy[αj(⃗ x, y, p) = 0] (j < i).
A functional is µ-recursive, or just recursive, if it occurs in some µ-recursive derivation.
5F. Relativization 219 (2) A functional α(⃗ x, p) is reckonable (or non-deterministically recur-sive) if there is a formula F(v1, . . . , vn, y, p) in the system Qp introduced in (2) of 5F.1, such that for all ⃗ x, w and p, α(⃗ x, p) = w ⇐ ⇒Qp ⊢F(∆x1, . . . , ∆xn, ∆w, p).
(133) 5F.20. Remark. There is no formula F(v1, . . . , vn, y, p) such that, for all ⃗ x, w and p p(⃗ x) = w ⇐ ⇒F(∆x1, . . . , ∆xn, ∆w) (A) Qp ⊢(∃!y)F(∆x1, . . . , ∆xn, y), (B) simply because if (A) holds for all p, then the formula on the right of (B) is not true, whenever p is not a totally defined function. This means that the simplest evaluation functional (132) is not “numeralwise representable” in Q, in the most natural extension of this notion to functionals, which is why we have not introduced it.
Proposition 5F.21. Every recursive functional is reckonable.
Proof is a minor modification of the argument for (1) = ⇒(2) of Theo-rem 4E.10 (skipping the argument for the characteristic property of numeral-wise representability which does not hold here), and we will skip it.
⊣ To separate recursiveness from reckonability for functionals, we need to in-troduce some basic notions, all of them depending on the following, partial ordering of partial functions of the same arity.
Definition 5F.22. For any two, m-ary partial functions p and q, we set p ≤q ⇐ ⇒(∀⃗ x, w)[p(⃗ x) = w = ⇒q(⃗ x) = w], i.e., if the domain of convergence of p is a subset of the domain of convergence of q, and q agrees with p whenever they are both defined. For example, if ∅ is the nowhere-defined m-ary partial function, then, for every m-ary q, ∅≤q; and, at the other extreme, (∀⃗ x)p(⃗ x)↓& p ≤q = ⇒p = q.
Proposition 5F.23. For each m, ≤is a partial ordering of the set of all m-ary partial functions, i.e., p ≤p, [p ≤q & q ≤r] = ⇒p ≤r, [p ≤q & q ≤p] = ⇒p = q.
Proof is simple and we will skip it.
⊣ Definition 5F.24. A functional α(⃗ x, p) is: 220 5. Introduction to computability theory 1. monotonic (or monotone), if for all partial functions p, q, and all ⃗ x, w, [α(⃗ x, p) = w & p ≤q] = ⇒α(⃗ x, q) = w; 2. continuous, if for each p and al ⃗ x, w, α(⃗ x, p) = w = ⇒(∃r)[r ≤p & α(⃗ x, r) = w & r is finite], where a partial function is finite if its domain of convergence is finite; and 3. deterministic, if for each p and all ⃗ x, w, α(⃗ x, p) = w = ⇒(∃!r ≤p)[α(⃗ x, r) = w & (∀r′ ≤r)[α(⃗ x, r′)↓= ⇒r′ = r]].
5F.25. Exercise.
Give counterexamples to show that no two of these properties imply the third.
Theorem 5F.26. (1) Every reckonable functional is monotonic and con-tinuous.
(2) Every recursive functional is monotonic, continuous and deterministic.
(3) There are reckonable functionals which are not deterministic.
Proof. (1) is immediate, using the (corresponding) properties of proofs: for example, if Qp ⊢F(∆x1, . . . , ∆xn, ∆w, p) for some p, ⃗ x and w, then the proof can only use a finite number of the axioms in Qp, which “fix” p only on a finite set of arguments—and if r is the (finite) restriction of p to this set, then Qr ⊢F(∆x1, . . . , ∆xn, ∆w, r), so that α(⃗ x, r) = w.
(2) is proved by induction on a given µ-recursive derivation.
There are several cases to consider, but the arguments are simple and we will skip them.
(3) The standard example is α3(p) = ( 1 if p(0)↓or p(1)↓ ↑ otherwise as above, which is not deterministic because if p(0) = p(1) = 0 and p(x) ↑for all x > 1, then there is no least r ≤p which determines the value α3(p) = 1.⊣ Part (1) of this theorem yields a simple normal form for reckonable func-tionals which characterizes them without reference to any formal systems. We need another coding.
5F. Relativization 221 5F.27. Coding of finite partial functions and sets. For each a ∈N and each m ≥1, d(a, x) = ( (a)x − · 1 if x < lh(a) & (a)x > 0, ↑ otherwise da(x) = d(a, x), Dx = {i | dx(i)↓} dm a (⃗ x) = dm(a, ⃗ x) = d(a, ⟨⃗ x⟩) (⃗ x = x1, . . . , xm).
Note that, easily, each partial function dm(a, ⃗ x) is primitive recursive; the sequence dm 0 , dm 1 , . . .
enumerates all finite partial functions of m arguments; and the sequence D0, D1, . . .
enumerates all finite sets, so that the binary relation of membership i ∈Dx ⇐ ⇒i < lh(x) & (x)i > 0, is primitive recursive.
Theorem 5F.28 (Normal form for reckonable functionals). A functional α(⃗ x, p) is reckonable if and only if there exists a semirecursive relation R(⃗ x, w, a), such that for all ⃗ x, w and p, α(⃗ x, p) = w ⇐ ⇒(∃a)[dm a ≤p & R(⃗ x, w, a)].
(134) Proof. Suppose first that α(⃗ x, p) is reckonable, and compute: α(⃗ x, p) = w ⇐ ⇒(∃finite r ≤p)[α(⃗ x, r) = w] (by 5F.26) ⇐ ⇒(∃a)[dm a ≤p & α(⃗ x, dm a ) = w)].
Thus, it is enough to prove that the relation R(⃗ x, w, a) ⇐ ⇒α(⃗ x, dm a ) = w is semirecursive; but if (133) holds with some formula F(v1, . . . , vm, y, p), then R(⃗ x, w, a) ⇐ ⇒Qdm a ⊢F(∆x1, . . . , ∆xn, ∆w, p) ⇐ ⇒Q ⊢σm,a,p →F(∆x1, . . . , ∆xn, ∆w, p) where σm,a,p is the finite conjunction of equations p(∆u1, . . . , ∆um) = ∆dm a (u1, . . . , um), 222 5. Introduction to computability theory one for each u1, . . . , um in the domain of dm a . A code of the sentence on the right can be computed primitive recursively from ⃗ x, w, a, so that R(⃗ x, w, a) is reducible to the relation of provability in Q and hence semirecursive.
For the converse, we observe that with the same σm,a,p we just used and for any m-ary p, dm a ≤p ⇐ ⇒Qp ⊢σm,a,p, and that, with some care, this σm,a,p can be converted to a formula σ∗(a, p) with the free variable a, in which bounded quantification replaces the blunt, finite conjunction so that dm a ≤p ⇐ ⇒Qp ⊢σ∗(∆a, p).
(135) Assume now that α(⃗ x, p) satisfies (134), choose a primitive recursive P(⃗ x, w, a, z) such that R(⃗ x, w, a) ⇐ ⇒(∃z)P(⃗ x, w, a, z), choose a formula P(v1, . . . , vn, vn+1, vn+2, z) which numeralwise expresses P in Q, and set F(v1, . . . , vn, vn+1, vn+2)) ≡(∃a)[σ∗(a, p) & (∃z)P(v1, . . . , vn, vn+1, a, z)]].
Now, Qp ⊢F(∆x1, . . . , ∆xn, ∆w, p) ⇐ ⇒Qp ⊢(∃a)[σ∗(a, p) & (∃z)[P(∆x1, . . . , ∆xn, ∆w, a, z)]] ⇐ ⇒α(⃗ x, p) = w, with the last equivalence easy to verify, using the soundness of Qp.
⊣ There is no simple normal form of this type for recursive functionals, because predicate logic is not well suitable for expressing “determinism”.
5G. Effective operations Intuitively, a functional α(⃗ x, p) is recursive in either of the two ways that we made precise, if its values can be computed effectively and uniformly for all partial functions p, given access only to specific values of p—which sim-ply means that the evaluation functional (132) is declared recursive. In many cases, however, we are interested in the values α(⃗ x, p) only for recursive partial functions p, and then we might make available to the computation procedure some code of p, from which (perhaps) more than the values of p can be ex-tracted.
5G. Effective operations 223 Definition 5G.1. The associate of a functional α(⃗ x, p) is the partial func-tion fα(⃗ x, e) = α(⃗ x, ϕe), (136) and we call α(⃗ x, p) an effective operation if its associate is recursive.
Note that this imposes no restriction on the values α(⃗ x, p) for non-recursive p, and so, properly speaking, we should think of effective operations as (par-tial) functions on recursive partial functions, not on all partial functions—this is why the term “operation” is used. For purposes of comparison with recur-sive functionals, however, it is convenient to consider effective operations as functionals, with arbitrary values on non-recursive arguments, as we did in the precise definition.
Proposition 5G.2. A recursive partial function f(⃗ x, e) is the associate of some effective operation if and only if it satisfies the invariance condition ϕe = ϕm = ⇒f(⃗ x, e) = f(⃗ x, m); (137) and if f satisfies this condition, then it is the associate of the effective operation α(⃗ x, ϕe) = f(⃗ x, e), (with α(⃗ x, p) defined arbitrarily when p is not recursive).
Proof is immediate.
⊣ Theorem 5G.3. Every reckonable functional (and hence every recursive functional) is an effective operation.
Proof. This is immediate from the Normal Form Theorem for reckonable functionals 5F.28; because if f(⃗ x, e) is the associate of α(⃗ x, p), then f(⃗ x, e) = w ⇐ ⇒(∃a)[dm a ≤ϕe & R(⃗ x, w, a)] with a semirecursive R(⃗ x, w, a) by 5F.28, and so the graph of f is semirecursive and f is recursive.
⊣ Definition 5G.4. A functional α(⃗ x, p) is operative if ⃗ x = x1, . . . , xn varies over n-tuples and p over n-ary partial functions, for the same n, so that the fixed point equation p(⃗ x) = α(⃗ x, p) (138) makes sense. Solutions of this equation are called fixed points of α.
224 5. Introduction to computability theory Theorem 5G.5 (The Fixed Point Lemma). Every operative effective oper-ation α has a recursive fixed point, i.e., there exists a recursive partial function p such that, for all ⃗ x, p(⃗ x) = α(⃗ x, p).
Proof. The partial function f(z, ⃗ x) = α(⃗ x, ϕz) is recursive, and so, by the Second Recursion Theorem, there is some z∗such that ϕz∗(⃗ x) = f(z∗, ⃗ x) = α(⃗ x, ϕz∗); and so p = ϕz∗is a fixed point of α.
⊣ 5G.6. Remark. By an elaboration of these methods (or different argu-ments), it can be shown that every effective operation has a recursive least fixed point: i.e., that for some recursive partial function p, the fixed point equation (138) holds, and in addition, for all q, (∀⃗ x)[q(⃗ x) = α(⃗ x, q)] = ⇒p ≤q.
The Fixed Point Lemma applies to all reckonable operative functionals, and it is a powerful tool for showing easily the recursiveness of partial functions defined by very general recursive definitions, for example by double recursion: 5G.7. Example. If g1, g2, g3, π1, π2 are total recursive functions and f(x, y, z) is defined by the double recursion f(0, y, z) = g1(y, z) f(x + 1, 0, z) = g2(f(x, π1(x, y, z), z), x, y, z) f(x + 1, y + 1) = g3(f(x + 1, y, π2(x, y, z)), x, y, z), then f(x, y, z) is recursive.
Proof. The functional h(x, y, z, p) = g1(y, z) if x = 0 g2(p(x − · 1, π1(x − · 1, 0, z), z), x − · 1, 0, z) otherwise, if y = 0 g3(p(x, y − · 1, Π0 2(x, y − · 1, z)), x, y − · 1, z) otherwise is recursive, and so it has a recursive fixed point f(x, y, z), which, easily, satisfies the required equations. It remains to show that f(x, y, z) is a total function, and we do this by showing by an induction on x that (∀x)f(x, y, z)↓; both the basis case and the induction step require separate inductions on y.⊣ 5G. Effective operations 225 The converse of Theorem 5G.3 depends on the following, basic result.
Lemma 5G.8. Every effective operation is monotonic and continuous on recursive partial arguments.
Proof. To simplify the argument we consider only effective operations of the form α(p), with no numerical argument and a unary partial function ar-gument, but the proof for the general case is only notationally more complex.
To show monotonicity, suppose p ≤q, where p = ϕe and q = ϕm, and let b f be a code of the associate of α, so that for every z, α(ϕz) = { b f}(z).
Suppose also that α(ϕe) = w; we must show that α(ϕm) = w.
The relation R(z, x, v) ⇐ ⇒ϕe(x) = v or [{ b f}(z) = w & ϕm(x) = v] is semirecursive; the hypothesis ϕe ≤ϕm implies that R(z, x, v) = ⇒ϕm(x) = v; hence R(z, x, v) is the graph of some recursive partial function g(z, x); and so, by the Second recursion Theorem, there is some number z∗such that ϕz∗(x) = g(z∗, x), so that ϕz∗(x) = v ⇐ ⇒ϕe(x) = v or [{ b f}(z∗) = w & ϕm(x) = v].
(139) We now observe that: (1a) α(ϕz∗) = { b f}(z∗) = w; because, if not, then ϕz∗= ϕe from (139), and so α(ϕz∗) = α(ϕe) = w.
(1b) ϕz∗= ϕm, directly from the hypothesis ϕe ≤ϕm and (1a).
It follows that α(ϕm) = α(ϕz∗) = w.
The construction for the proof of continuity is a small variation, as follows.
First, we find using the Second recursion Theorem some z∗such that ϕz∗(x) = v ⇐ ⇒(∀u ≤x)¬[T1( b f, z∗, u) & U(u) = w] & ϕe(x) = v, (140) and we observe: (2a) α(ϕz∗) = w. Because, if not, then (∀u)¬[T1( b f, z∗, u) & U(u) = w], 226 5. Introduction to computability theory and hence, for every x, (∀u ≤x)¬[T1( b f, z∗, u) & U(u) = w], and so, from (140), ϕz∗= ϕe and α(ϕz∗) = α(ϕe) = w.
(2b) ϕz∗≤ϕe, directly from (140).
(2c) The partial function ϕz∗is finite, because it converges only when x < (µu)[T1( b f, z∗, u) & U(u) = w].
(141) ⊣ Theorem 5G.9 (Myhill-Shepherdson). For each effective operation α(⃗ x, p), there is a reckonable functional α∗(⃗ x, p) such that for all recursive partial func-tions p, α(⃗ x, p) = α∗(⃗ x, p).
(142) Proof. By the Lemma, α(⃗ x, ϕe) = w ⇐ ⇒(∃a)[dm a ≤ϕe & α(⃗ x, dm a ) = w], and so (142) holds with α∗(⃗ x, p) = w ⇐ ⇒(∃a)[dm a ≤p & α(⃗ x, dm a ) = w].
(143) To show that this α∗is reckonable, note that (by an easy application of the Sm n -Theorem) there is a primitive recursive u(a) such that dm a = ϕu(a), and so the partial function α(⃗ x, dm a ) = fα(⃗ x, u(a)) is recursive, its graph is semirecursive, and (143) with Theorem 5F.28 imply that α∗is reckonable.
⊣ 5G.10. Remark. It is natural to think of a functional α(⃗ x, p) as inter-preting a program A, which computes some function f(⃗ x) but requires for the computations some unspecified partial function p—and hence, A must be “given” p in addition to the arguments ⃗ x. Now if p could be any partial func-tion whatsoever, then the only reasonable way by which A can be “given” p is through its values: we imagine that A can look up a table or ask an “oracle” for p(u), for any specific u, during the computation. We generally refer to this manner of “accessing” a partial function by a program as call-by-value, and it is modeled mathematically by recursive or reckonable functionals, depending on whether the program A is deterministic or not. On the other hand, if it is known that p = ϕe is a recursive partial function, then some code e of it 5H. Computability on Baire space 227 may be given to A, at the start of the computation, so that A can compute any p(u) that it wishes, but also (perhaps) infer general properties of p from e, and use these properties in its computations; this manner of accessing a recursive partial function is (one version of) call-by-name, and it is modeled mathematically by effective operations.
One might suspect that given access to a code of p, one might be able to com-pute effectively partial functions (depending on p) which cannot be computed when access to p is restricted in call-by-value fashion. The Myhill-Shepherdson Theorem tells us that, for non-deterministic programs, this cannot happen— knowledge of a code of p does not enlarge the class of partial functions which can be non-deterministically computed from it. Note that this is certainly false for deterministic computations, because of the basic example α3(p) in 5F.18, which is reckonable but not recursive.
5H. Computability on Baire space We will extend here the basic results about recursive partial functions and relations on N, to partial functions and relations which can also take arguments in Baire space, the set N = (N →N) = {α | α : N →N} of all total, unary functions on the natural numbers. For example, the total functions f(α) = α(0), g(x, α, β, y) = α(β(x)) + y, will be deemed recursive by the definitions we will give, and so will the partial function h(α) = µt[α(t) = 0], which is defined only if α(t) = 0 for some t. The relation R(α) ⇐ ⇒(∃t)[α(t) = 0] will be semirecursive but not recursive.
5H.1. Notation.
More precisely, in this section we will study partial functions f : Nn × N ν ⇀N, with n = 0 or ν = 0 allowed, so that the partial functions on N we have been studying are included. To avoid “too many dots”, we set once and for all 228 5. Introduction to computability theory boldface abbreviations for vectors, x = (x1, . . . , xn), y = (y1, . . . , ym), α = (α1, . . . , αν), β = (β1, . . . , βµ), so that the values of our partial functions will be denoted compactly by ex-pression like f(x, α), g(t, x, y, α, γ, β), etc.
In defining specific functions we will sometimes mix the number with the Baire arguments, as above, the “official” reading always being the one in which all number arguments precede all the Baire ones, e.g., g(x, α, β, y) = g(x, y, α, β).
Codings of initial segments.
Recall from Lemma 3I.13 the following notation, which will now be very useful: α(0) = 1, α(t) = ⟨α(0), α(1), . . . , α(t − · 1)⟩∈N and for vectors of Baire points, α(t) = (α1(t), . . . , αν(t)) ∈Nν.
For any sequence code u and any s, put u↾s = ( ⟨(u)0, . . . , (u)s−1⟩, if s ≤lh(u), u, otherwise, so that lh(u↾s) = min{lh(u), s}, and s ≤t = ⇒α(s) = α(t)↾s.
Similarly, for any tuple of sequence codes ⃗ u = (u1, . . . , uν) and any s, put ⃗ u↾s = (u1↾s, . . . , uν ↾s), so that if s ≤t, then α(s) = α(t)↾s.
A relation R(x, ⃗ u) is monotone in ⃗ u if, for every α and every s, if R(x, α(s)) and s ≤t, then R(x, α(t)).
λ-abstraction. We will also find useful Church’s λ operation, by which, for any partial function f : Nn+1 × N ν ⇀N, λ(t)f(x, t, α) = gx,α : N ⇀N where gx,α(t) = f(x, t, α), 5H. Computability on Baire space 229 For example, λ(t)(xt + t2) is that function gx : N →N such that for all t, gx(t) = xt + t2, and (closer to the way we will use this), λ(x)U(µyT1(e, x, y)) = ϕe.
(144) Definition 5H.2 (Semirecursive relations on N). A relation P(x, α) ⇐ ⇒P(x1, . . . , xn, α1, . . . , αν) is semirecursive or Σ0 1, if there is a semirecursive relation R(x, ⃗ u) ⇐ ⇒R(x1, . . . , xn, u1, . . . , uν) on N, such that P(x, α) ⇐ ⇒(∃t)R(x, α(t)).
(145) A relation P(x, α) is recursive or ∆0 1 if both P(x, α) and its negation ¬P(x, α) are semirecursive.
Notice that these definitions agree with the old ones for relations R(x) which have no Baire arguments.
Lemma 5H.3. The class of semirecursive relations with arguments in N and N is closed under permutations and identifications of variables: i.e., if π : {1, . . . , n} →{1, . . . , m}, ρ : {1, . . . , ν} →{1, . . . , µ} are any functions and P(y1, . . . , ym, β1, . . . , βµ) is semirecursive, then so is the relation P ′(x1, . . . , xn, α1, . . . , αν) ⇐ ⇒P(xπ(1), . . . , xπ(n), αρ(1), . . . , αρ(n)) This justifies “explicit” definitions of the form P ′(x, y, α, β) ⇐ ⇒P(y, x, x, β, β, β) within Σ0 1, and it is immediate from the definition.
Lemma 5H.4. (1) If P(x, α) ⇐ ⇒ (∃t)R(x, t, α) with a semirecursive R(x, t, ⃗ u), then P(x, α) is semirecursive.
(2) If P(x, α) is semirecursive and ν ≥1, then it satisfies (145) with some recursive relation R(x, ⃗ u) which is monotone in ⃗ u.
Proof. (1) The claim is obvious when ν = 0, since the assumed equivalence implies immediately that R(x) is a semirecursive relation on N.
If ν ≥1, so that P(x, α) has at least one Baire argument, we set R′(x, ⃗ u) ⇐ ⇒(∃s ≤lh(u1))R(x, s, ⃗ u↾s), 230 5. Introduction to computability theory and compute: P(x, α) ⇐ ⇒(∃t)R(x, t, α(t)) ⇐ ⇒(∃t)(∃s ≤t)R(x, s, α(s)) ⇐ ⇒(∃t)(∃s ≤lh(α1(t)))R(x, s, α(t)↾s) ⇐ ⇒(∃t)R′(x, α(t)).
(2) Assume that (145) holds with R(x, ⃗ u) ⇐ ⇒(∃y)Q(x, ⃗ u, y) where Q(x, ⃗ u, y) is recursive, and let R′(x, ⃗ u) ⇐ ⇒(for i = 1, . . . , ν)[lh(ui) = lh(u1)] & (∃s ≤lh(u1))(∃y ≤lh(u1))Q(x, ⃗ u↾s, y).
Now R′(x, ⃗ u) is clearly recursive and monotone in ⃗ u, and P(x, α) ⇐ ⇒(∃s)R(x, α(s)) ⇐ ⇒(∃s)(∃y)Q(x, α(s), y) ⇐ ⇒(∃t)(∃s ≤t)(∃y ≤t)Q(x, α(s), y) ⇐ ⇒(∃t)R′(x, α(t)).
⊣ Lemma 5H.5 (Closure properties of Σ0 1 and ∆0 1). (1) The class of semire-cursive relations with arguments in N and N is closed under substitutions of total, recursive functions f : Nn →N; the positive propositional operations & and ∨; bounded number quantification of both kinds; existential number quan-tification (∃x); and also existential quantification over N, (∃α).
(2) The class of recursive relations with arguments in N and N is closed under substitutions of total, recursive functions f : Nn →N; the propositional operations ¬, & and ∨; and bounded number quantification of both kinds.
Proof. We may assume in the proofs that ν ≥1 (i.e., Baire arguments are present), since otherwise these results are known.
For conjunction, assume that P1 and P2 satisfy (145) with a recursive, mono-tone matrix, by (2) of Lemma 5H.4, and compute: P1(x, α) & P2(x, α) ⇐ ⇒(∃t)R1(x, α(t)) & (∃t)R2(x, α(t)) ⇐ ⇒(∃t)[R1(x, α(t)) & R2(x, t, α(t))], the last equivalence by the monotonicity.
5H. Computability on Baire space 231 For existential quantification over N: P(x, α) ⇐ ⇒(∃β)P1(x, α, β) ⇐ ⇒(∃β)(∃t)R1(x, α(t), β(t)) ⇐ ⇒(∃t)(∃β)R1(x, α(t), β(t)) ⇐ ⇒(∃t)(∃v)[Seq(v) & lh(v) = t & R1(x, α(t), v)]; the crucial “quantifier-drop” equivalence (∃β)R1(x, α(t), β(t)) ⇐ ⇒(∃v)[Seq(v) & lh(v) = t & R1(x, α(t), v)] is proved from left-to-right by setting v = β(t), and from right-to-left by taking β to be an arbitrary, infinite extension of the sequence v. Now set R′(x, t, ⃗ u) ⇐ ⇒(∃v)[Seq(v) & lh(v) = t & R(x, ⃗ u, v)]; this is a semirecursive relation, P(x, α) ⇐ ⇒(∃β)P1(x, α, β) ⇐ ⇒(∃t)R′(x, t, α(t)), and so P(x, α) is semirecursive by (1) of Lemma 5H.4.
The remaining arguments are similar.
⊣ Theorem 5H.6. (1) For every n and every ν, there exists a Σ0 1 relation e S0 n,ν(e, x, α) ⇐ ⇒e S0 n,ν(e, x1, . . . , xn, α1, . . . , αν) such that an arbitrary relation R(x, α) is semirecursive if and only if there is some number e such that R(x, α) ⇐ ⇒e S0 n,ν(e, x, α).
In fact, if ν ≥1, then e S0 n,ν(e, x, α) ⇐ ⇒(∃t)T r n,ν(e, x, α(t)), (146) where T r n,µ(e, x, ⃗ u) is a primitive recursive and monotone in ⃗ u relation on N.
(2) If n + ν > 0, then there exists a semirecursive relation P(x, α) which is not recursive.
(3) For every m, every n and every ν, there exists a primitive recursive function Sr,m n,ν (e, y) such that for all y, x, α, e S0 m+n,ν(e, y, x, α) ⇐ ⇒e S0 n,ν(Sr,m n,ν (e, y), x, α).
232 5. Introduction to computability theory Proof. (1) The result is known when ν = 0, so we assume ν ≥1, and with the notation of the Normal Form and Enumeration Theorem 4F.1, we let T r n,ν(e, x, ⃗ u) ⇐ ⇒(for i = 1, . . . , ν)[Seq(ui) & lh(ui) = lh(u1)] & (∃s ≤lh(u1))(∃y ≤lh(u1))Tn+ν(e, x, ⃗ u↾s, y).
This is clearly primitive recursive and monotone in ⃗ u, and if e S0 n,ν is defined from it by (146), then it is semirecursive. For the converse, suppose P(x, α) satis-fies (145) with a semirecursive R(x, ⃗ u). By the Enumeration Theorem 4F.1, there is some e such that R(x, ⃗ u) ⇐ ⇒(∃y)Tn+ν(e, x, ⃗ u, y), and then we compute: P(x, α) ⇐ ⇒(∃t)R(x, α(t)) ⇐ ⇒(∃t)(∃s ≤t)R(x, α(s)) ⇐ ⇒(∃t)(∃s ≤t)(∃y)Tn+ν(e, x, α(s), y) ⇐ ⇒(∃t)(∃y ≤t)(∃s ≤t)Tn+ν(e, x, α(s), y) ⇐ ⇒(∃t)T r n,ν(e, x, α(t)).
(2) follows from (1) by the usual, diagonal method, and (3) follows from the Sm n theorem for recursive partial functions on N by setting Sr,m n,ν (e, y) = Sm n+ν(e, y) and chasing the definitions.
⊣ Definition 5H.7 (Recursive partial functions on N to N). A partial func-tion f(x, α) with values in N is recursive if its graph Gf(x, α, w) ⇐ ⇒f(x, α) = w is semirecursive. For example, the (total) evaluation function ev(x, α) = α(x) is recursive, because α(x) = w ⇐ ⇒(∃t)[t > x & (α(t))x = w].
Lemma 5H.8 (Closure properties for recursive partial functions into N). The class of recursive partial functions on Baire space with values in N contains all recursive partial functions f : Nn ⇀N; it is closed under permutations and identifications of variables, i.e., if f(x1, . . . , xn, α1, . . . , αν) = g(xπ(1), . . . , xπ(n), αρ(1), . . . , αρ(n)) 5H. Computability on Baire space 233 with π, ρ as in Lemma 5H.3 and g(y, β) is recursive, then so is f(x, α); and it is also closed under substitution (in its number arguments), primitive recur-sion, and minimalization.
Proof. These claims all follow easily from the closure properties of the class of semirecursive relations, and we will consider just two of them, as examples.
For substitution (in a simple case), we are given that f(x, α) = g(h1(x, α), x, α), where g(y, x, α) and h(x, α) are recursive, and we compute the graph of f(x, α): f(x, α) = w ⇐ ⇒(∃y)[h1(x, α) = y & g(y, x, α) = w]; the result follows from the closure properties of Σ0 1 in Lemma 5H.5.
For primitive recursion, we are given that f(0, x, α) = g(x, α), f(y + 1, x, α) = h(f(y, x, α), y, x, α).
Hence, f(y, x, α) = w ⇐ ⇒ (∃u)[(u)0 = g(x, α) & (∀i ≤y)[(u)i+1 = h((u)i, x, α)] & (u)y = w, and so the graph of f(y, x ¯, α) is semirecursive.
⊣ Theorem 5H.9 (Normal Form and Enumeration). (1) For every n and ev-ery ν ≥1, there is a (primitive) recursive and monotone in ⃗ u relation T 1 n,ν(e, x, ⃗ u) on N, such that a partial function f(x, α) into N is recursive if and only if there is a number e such that f(x, α) = {e}(x, α) = U(µtT 1 n,ν(e, x, α(t))), (147) with U(t) = (t)0.
(2) For every m, every n and every ν, there exists a primitive recursive function Sm n,ν(e, y) such that for all y, x, α, {e}(y, x, α) = {(Sm n,ν(e, y)}(x, α).
Proof. (1) Every partial function defined by (147) with a recursive T 1 n,ν is recursive, by the closure properties. To define a suitable T 1 n,ν, we note that by the definitions and Theorem 5H.6, for each recursive f(x, α) = w, there is some e such that f(x, α) = w ⇐ ⇒(∃t)T r n+1,ν(e, x, w, α(t)).
(148) 234 5. Introduction to computability theory We set T 1 n,ν(e, x, ⃗ u) ⇐ ⇒(for i = 1, . . . , ν)[Seq(ui) & lh(ui) = lh(u1)] & (∃s ≤lh(u1))T r n+1,ν(e, x, (s)0, ⃗ u↾s).
This is obviously primitive recursive and monotone in ⃗ u, and to complete the proof, we need only show that if (148) holds for some e, then f(x, α) = µtT 1 n,ν(e, x, α(t)) 0.
(149) So fix x, α, and first check that s = µtT 1 n,ν(e, x, α(t)) = ⇒f(x, α) = (s)0; (150) this holds because the hypothesis implies T r n+1,ν(e, x, (s)0, α(s)), which by (148) yields the conclusion. Conversely, if f(x, α) = w, then T r n+1,ν(e, x, w, α(t)) holds for some t, and then taking s = ⟨w, t⟩> t and using the monotonicity of T r n+1,ν(e, x, w, ⃗ u), we have T r n+1,ν(e, x, (s)0, α(s)); so (∃t)T 1 n,ν(e, x, α(t)), and then (150) gives f(x, α) = w = ⇒ µtT 1 n,ν(e, x, α(t)) 0 = w, (151) which together with (150) yield (149).
(2) follows from (3) of Theorem 5H.6 by setting Sm n,ν(e, y) = Sr,m n+1,ν(e, y) and chasing the definitions.
⊣ Definition 5H.10 (Recursive partial functions with values in N). A partial function f : Nn × N ν ⇀N is recursive, if the following, associated, unfolding partial function f∗: Nn+1 × N ⇀N is recursive: f∗(x, α, t) = f(x, α)(t); or, equivalently, if f(x, α) = λ(t)f∗(x, α, t), with some recursive f∗: Nn+1 × N ν ⇀N. Thus f(x, α) = β ⇐ ⇒(∀t)[f∗(x, α, t) = β(t)], f(x, α)↓⇐ ⇒(∀t)[f∗(x, α, t)↓], 5H. Computability on Baire space 235 which suggests that neither the graph nor the domain of convergence of a recursive partial functions with values in N need be semirecursive. In fact, if we view (144) as a definition of a partial function h : N ⇀N, then h(e) = λ(x)U(µyT1(e, x, y))↓⇐ ⇒(∀x)(∃y)T1(e, x, y), so that by Proposition 5E.7, the domain of convergence of h is Π0 2 \ Σ0 2.
Lemma 5H.11. The class of recursive partial functions with arguments in N is not closed under substitution.
Proof. If g(α) = 0, h(e) = λ(x)U(µyT1(e, x, y)), and f(e) = g(h(e)), then f : N ⇀N, f(e) = w ⇐ ⇒(∀t)[{e}(t)↓] & w = 0, and the graph of f is not semirecursive, so that f is not recursive.
⊣ However, f(e) agrees with a recursive partial function (the constant 0) for values of e for which h(e)↓; this is a general and useful fact: Theorem 5H.12. (1) Suppose g : Nn×N ×N ν ⇀N and h : Nn×N ν ⇀N are recursive partial functions, and let f(x, α) = g(x, h(x, α), α); then there exists a recursive partial function e f : Nn × N ν ⇀N such that if h(x, α)↓, then f(x, α) = e f(x, α).
(2) If g(z1, . . . , zm), h1(x, α), . . . , hm(x, α) are recursive partial functions such that for i = 1, . . . , m, if hi : Nn × N ν ⇀N then hi is total, then the substitution f(x, α) = g(h1(x, α), . . . , hm(x, α)) is a recursive partial function.
Proof. (1) By the Normal Form Theorem 5H.9, g(x, β, α) = U(µtR(x, β(t), α(t))) with a recursive relation R(x, v, ⃗ u). Let also e h(t, x, α) = ⟨h(x, α)(0), . . . , h(x, α)(t −1)⟩; this is a recursive partial function, such that if h(x, α) = β, then e h(t, x, α) = β(t).
Finally, put e f(x, α) = U(µtR(x,e h(t, x, α), x, α)), α(t))).
236 5. Introduction to computability theory Now this is a recursive partial function, and if h(x, α) = β, then e f(x, α) = U(µtR(x, β(t), x, α)) = g(x, h(x, α), α), as claimed.
(2) follows immediately from (1).
⊣ Corollary 5H.13. The classes of semirecursive and recursive relations with arguments in Baire space are closed under substitution of total, recursive func-tions with values in N.
Among the most useful such substitutions are those which use the following codings of finite and infinite tuples of Baire points: Definition 5H.14 (Sequence codings for Baire space). We set ⟨α0, . . . , αν−1⟩= λ(t) ( αi(s), if t = ⟨i, s⟩for some i < ν and some s, 0, otherwise, and similarly for an infinite sequence of Baire points, ⟨α0, α1, . . .⟩= λ(t) ( αi(s), if t = ⟨i, s⟩for some i and some s, 0, otherwise.
We also set, (α)i = λ(s)α(⟨i, s⟩).
Lemma 5H.15. (1) The function (α0, . . . , αν−1) 7→⟨α0, . . . , αν−1⟩, is recursive and one-to-one on N ν →N; and the function (α, i) 7→(α)i is recursive and an inverse of the tuple functions, in the sense that (⟨α0, . . . , αn−1⟩)i = αi (i < n).
(2) The function (α0, α1, . . . ) 7→⟨α0, α1, . . .⟩ is one-to-one on N ∞→N.
5H.16. The arithmetical hierarchy with arguments in Baire space.
We can now define and establish the basic properties of the arithmetical hi-erarchy for relations with arguments in Baire space, exactly as we did for relations with arguments in N in Section 5E, starting with the Σ0 1 relations on 5H. Computability on Baire space 237 Baire space and using recursive functions with arguments and values in Baire space. We comment briefly on the changes that must be made, which involve only the results needed to justify the theorems.
The basic definition of the classes Σ0 k, Π0 k and ∆0 k is exactly that in 5E.1, starting with the definition 5H.2 of semirecursive relations on Baire space.
The canonical forms of these classes are those in 5E.2, whose Table we repeat to emphasize the form of the dependence on the Baire arguments when these are present, i.e., with ν ≥1: Σ0 1 : (∃y)Q(x, α(y)) Π0 1 : (∀y)Q(x, α(y)) Σ0 2 : (∃y1)(∀y2)Q(x, α(y2), y1) Π0 2 : (∀y1)(∃y2)Q(x, , α(y2), y1) Σ0 3 : (∃y1)(∀y2)(∃y3)Q(x, α(y3), y2, y3) .
.
.
The closure properties are again those in Theorem 5A.7, with the closure under substitutions of total recursive functions with values either in N or in N depending on Corollary 5H.13, and the quantifier contractions justified by the sequence codings in 5H.14 and 5H.15. Finally, the Enumeration and Hierarchy Theorems 5E.5 are proved as before, starting with Theorem 5H.6 now, and the proper inclusions diagram in that theorem still holds, with the “properness” witnessed by the same relations on N.
5H.17. Baire codes of subsets of N and relations on N. With each Baire point γ, we associate the set of natural numbers Aγ = {s ∈N | γ(s) = 1}, (152) and for each n ≥2, the n-ary relation Rn γ(x1, . . . , xn) ⇐ ⇒γ(⟨x1, . . . , xn⟩) = 1; (153) we say that γ is a code of A if Aγ = A, and a code of R ⊆Nn if R = Rn γ. If n = 2, we often use “infix notation”, xRγy ⇐ ⇒Rγ(x, y) ⇐ ⇒γ(⟨x, y⟩) = 1.
These trivial codings allow us to classify subsets of N and relations on N in the arithmetical hierarchy, as in the following, simple example.
Lemma 5H.18. The set of codes of linear orderings LO = {γ | γ is a code of a linear ordering (of a subset of N)} (154) is Π0 1 \ Σ0 1.
238 5. Introduction to computability theory Proof. To simplify notation, set x ≤γ y ⇐ ⇒γ(⟨x, y⟩) = 1, x ∈Dγ ⇐ ⇒x ≤γ x, and compute: γ ∈LO ⇐ ⇒≤γ is a linear ordering of Dγ ⇐ ⇒(∀x, y)[x ≤γ y = ⇒[x ∈Dγ & y ∈Dγ]] & (∀x, y)[[x ≤γ y & y ≤γ x] = ⇒x = y] & (∀x, y, z)[[x ≤γ y & y ≤γ z] = ⇒x ≤γ z] & (∀x, y)[[x ∈Dγ & y ∈Dγ] = ⇒x ≤γ y ∨y ≤γ x].
For the converse, suppose that P(x) ⇐ ⇒(∀u)R(x, u) with R(x, u) recursive, and let f∗(x, t) = 1, if R(x, min((t)0, (t)1)) & (t)0 ≤(t)1, 0, if R(x, min((t)0, (t)1)) & (t)0 > (t)1, 1, if ¬R(x, min((t)0, (t)1)).
The function f∗(x, t) is recursive and total, and hence so is the function f(x) = λ(t)f∗(x, t); moreover, if (∀t)R(x, t), then f(x)(⟨u, v⟩) = f∗(x, ⟨u, v⟩) = ( 1, if u ≤v, 0, otherwise, so that f(x) ∈LO, in fact f(x) is a code of the natural ordering on N. On the other hand, if, for some t, ¬R(x, t), then f(x)(⟨t, t + 1⟩) = f(x)(⟨t + 1, t⟩) = 1, i.e., t ≤f(x) (t + 1) & (t + 1) ≤f(x) t, so that f(x) / ∈LO. This establishes the reduction.
P(x) ⇐ ⇒(∀u)R(x, u) ⇐ ⇒f(x) ∈LO.
It follows that if LO were Σ0 1, then every Π0 1 relation on N would be in Σ0 1, which it is not, and hence LO is not Σ0 1.
⊣ Problems for Section 5H Problem 5H.1. Prove that the class of recursive partial functions on Baire space to N is closed under minimalization.
5I. The analytical hierarchy 239 Problem 5H.2. Prove that a relation P(x, α) is Σ0 1 if and only if it is the domain of some recursive f : Nn × N ν ⇀N, P(x, α) ⇐ ⇒f(x, α)↓.
Problem 5H.3. (Σ0 1-Selection). Prove that for every semirecursive relation P(x, y, α), there is a recursive partial function f(x, α) such that: (1) f(x, α)↓⇐ ⇒(∃y)P(x, y, α).
(2) If (∃y)P(x, y, α), then P(x, f(x, α), α).
Problem 5H.4. Prove that a relation P(x, α) is recursive if and only if its characteristic (total) function χR(x, α) is recursive.
Problem 5H.5. Prove that the set A = {α ∈N | for infinitely many x, α(x) = 0} is in Π0 2 \ Σ0 2.
5I. The analytical hierarchy Once we have relations with arguments in Baire space, we can apply quan-tification over N on them to define more complex (and more interesting) relations. The resulting analytical hierarchy resembles in structure the arith-metical structure, which it extends, but it contains many of the fundamental relations of analysis and set theory.
Definition 5I.1. The Kleene classes of relations Σ1 k, Π1 k, ∆1 k with argu-ments in N and N are defined recursively, for k ≥0, as follows: Π1 0 = Π0 1 : the negations of semirecursive relations Σ1 k+1 = ∃N Π1 k : the relations which satisfy an equivalence of the form P(x, α) ⇐ ⇒(∃β)Q(x, α, β), where Q(x, α, β) is Π1 k Π1 k = ¬Σ1 k : the negations (complements) of relations in Σ1 k ∆1 k = Σ1 k ∩Π1 k : the relations which are both Σ1 k and Π1 k.
A set of numbers A ⊆N or of Baire points A ⊆N is in one of these classes Γ if the relation x ∈A or α ∈A is in Γ.
5I.2. Canonical forms. Using the canonical form for Σ0 1 relations in (2) of Lemma 5H.4, we obtain immediately the following canonical forms for the 240 5. Introduction to computability theory Kleene classes, (with k ≥1), where R(x, ⃗ u, v) is recursive on N and monotone in ⃗ u, v: Π1 1 : (∀β)(∃t)R(x, α(t), β(t)) Σ1 1 : (∃β)(∀t)R(x, α(t), β(t)) Π1 2 : (∀β1)(∃β2)(∀t)R(x, α(t), β1(t), β2(t)) Σ1 2 : (∃β1)(∀β2)(∃t)R(x, α(t), β1(t), β2(t)) Π1 3 : (∀β1)(∃β2)(∀β3)(∃t)R(x, α(t), β1(t), β2(t), β3(t)) .
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.
It is worth singling out the form of Π1 1 subsets of N and N which include some of the most interesting examples: (Π1 1) x ∈A ⇐ ⇒(∀β)(∃t)R(x, β(t)), α ∈A ⇐ ⇒(∀β)(∃t)R(α(t), β(t)), Theorem 5I.3. (1) For each k ≥1, the classes Σ1 k, Π1 k, and ∆1 k are closed for (total) recursive substitutions with values in N or N, and for the operations &, ∨, ∃≤, ∀≤, ∃N and ∀N. In addition: • Each ∆1 k is closed for negation ¬.
• Each Σ1 k is closed for ∃N , existential quantification over N.
• Each Π1 k is closed for ∀N , universal quantification over N.
(2) Every arithmetical relation is ∆1 1.
(3) For each k ≥1, Σ1 k ⊊∆1 k+1, (155) and hence the Kleene classes satisfy the following diagram of proper inclusions: Σ1 1 Σ1 2 Σ1 3 ⊊ ⊊ ⊊ ⊊ ⊊ ⊊ S k Σ0 k ⊆∆1 1 ∆1 2 ∆1 3 · · · ⊊ ⊊ ⊊ ⊊ ⊊ ⊊ Π1 1 Π1 2 Π1 3 Proof. (1) The closure of all Kleene classes under total, recursive substi-tutions follows from the canonical forms and the closure of Σ0 1 and Π0 1 under total recursive substitutions, Corollary 5H.13. The remaining closure proper-ties are proved by induction on k ≥1, using the recursiveness of the projection 5I. The analytical hierarchy 241 functions and the function γ 7→γ′ = λ(t)γ(t); the known closure properties of Π0 1; and, to contract quantifiers, the following equivalences and their duals, where we abbreviate z = x, α.
(∃β)P(z, β) ∨(∃β)Q(z, β) ⇐ ⇒(∃β)[P(z, β) ∨Q(z, β)] (E1) (∃β)P(z, β) & (∃β)Q(z, β) ⇐ ⇒(∃γ)[P(z, (γ)0) & Q(z, (γ)1)] (E2) (∃s ≤t)(∃β)P(z, s, β) ⇐ ⇒(∃β)(∃s ≤t)P(z, s, β) (E3) (∀s ≤t)(∃β)P(z, s, β) ⇐ ⇒(∃γ)(∀s ≤t)P(z, s, (γ)s) (E4) (∃s)(∃β)P(z, s, β) ⇐ ⇒(∃β)(∃s)P(z, s, β) (E5) (∀s)(∃β)P(z, s, β) ⇐ ⇒(∃γ)(∀s)P(z, s, (γ)s) (E6) (∃δ)(∃β)P(z, δ, β) ⇐ ⇒(∃γ)P(z, (γ)0, (γ)1) (E7) These are all either trivial, or direct expressions of the countable Axiom of Choice.
In some more detail: (1a) The closure properties of Σ1 1 follow from those of Π0 1 and (E1)-(E7). To show closure under ∀N, for example, suppose P(z, t) ⇐ ⇒(∃β)Q(z, β, t) with Q ∈Π0 1, and compute: (∀t)P(z, t) ⇐ ⇒(∀t)(∃β)Q(z, β, t) ⇐ ⇒(∃γ)(∀t)Q(z, (γ)t, t)by (E6).
This is enough, because Q(z, (γ)t, t) is Π0 1 by the closure of this class under recursive substitutions.
(1b) The closure properties of Π1 k follow from those of Σ1 k taking negations and pushing the negation operator through the quantifier prefix.
(1c) The closure properties of Σ1 k+1 follow from those of Π1 k using (E1)-(E7).
(2) follows from (1), since ∆0 1 ⊆∆1 1 and ∆1 1 is closed under both num-ber quantifiers. (And we will see in 5I.5 that the arithmetical relations are contained properly in ∆1 1.) (3) We notice first that the non-strict version of the diagram (with ⊆in place of ⊊) is trivial, using dummy quantification, and closure under recursive substitutions, e.g., (∃β)P(z, β) ⇐ ⇒(∃β)(∀α)P(z, β) ⇐ ⇒(∀α)(∃β)P(z, β).
242 5. Introduction to computability theory To show that the inclusions are strict, we need to define enumerating (univer-sal) sets e S1 k,n,ν for Σ1 k and e P 1 k,n,ν for Π1 k.
We start with the fact that the relation e P 0 n,ν(e, x, α) ⇐ ⇒(∀t)¬T r n,ν(e, x, α(t)) (156) enumerates all Π0 1 relations with arguments x, α, by (146) in Theorem 5H.6, and set recursively: e S1 1,n,ν(e, x, α) ⇐ ⇒(∃β)(∀t)¬T r n,ν+1(e, x, α(t), β(t)), e P 1 k,n,ν(e, x, α) ⇐ ⇒¬e S1 k,n,ν(e, x, α) e S1 k+1,n,ν(e, x, α) ⇐ ⇒(∃β) e P 1 k,n,ν+1(e, x, α, β).
It is easy to verify (chasing the definitions) that a relation R(x, α) is Σ1 k if and only if there is some e such that R(x, α) ⇐ ⇒e S1 k,n,ν(e, x, α), and then, as usual, the diagonal relation D(x) ⇐ ⇒e S1 k,1,0(x, x) is in Σ1 k \ Π1 k.
⊣ We now place in the analytical hierarchy some important relations on Baire space, starting with the basic satisfaction relation on (codes of) structures.
5I.4. Codes of structures. Consider (for simplicity) FOL(τ), where the signature τ has K relation symbols P1, . . . , PK where Pi is ni-ary. We code τ by its characteristic u = ⟨n1, . . . , nK⟩.
With each α ∈N and each u, we associate the tuple A(u, α) = (Aα, R1,α, . . . , RK,α), where Aα = {n ∈N | (α)0(n) = 1}, Ri,α(x1, . . . , xni) ⇐ ⇒x1, . . . , xni ∈A & (α)i(⟨x1, . . . , xni⟩) = 1.
This is a structure when Aα ̸= ∅, and so the semirecursive relation (u, α) ⇐ ⇒Aα ̸= ∅⇐ ⇒(∃n)[(α)0(n) = 1] 5I. The analytical hierarchy 243 determines which Baire points code structures. Let Assgn(u, x, α) ⇐ ⇒x codes an (ultimately 0) assignment into Aα ⇐ ⇒(u, α) & (∀i < lh(x))[(x)i ∈Aα]; this, too, is a semirecursive relation. Finally, using the codings of Section 4A, we set Sat(u, α, m, x) ⇐ ⇒(u, α) (157) & m codes a formula φ of FOL(τ) & A(u, α), (x)0, (x)1, . . . | = φ This relation codes the basic satisfaction relation between structures (on sub-sets of N), assignments and formulas.
Theorem 5I.5. (1) The relation Sat(u, α, m, x) is ∆1 1.
(2) The set Truth of codes of true arithmetical sentences is ∆1 1, and so the arithmetical relations are properly contained in ∆1 1.
Proof. (1) Let P(u, α, β) ⇐ ⇒(∀m, x)[β(⟨m, x⟩) ≤1 & β(⟨m, x⟩) = 1 ⇐ ⇒Sat(u, α, m, x)]; the equivalence determines β completely, once u and α are fixed, and so Sat(u, α, m, x) ⇐ ⇒(∃β)[P(u, α, β) & β(⟨m, x⟩) = 1] ⇐ ⇒(∀β)[P(u, α, β) = ⇒β(⟨m, x⟩) = 1].
Thus the proof will be complete once we show that P(u, α, β) is arithmetical, which is not too hard to do, since for it to hold, β must satisfy the “Tarski conditions” for satisfaction: β(⟨m, x⟩) must give the correct value when m codes a prime formula, and for complex formulas, the correct value of β(⟨m, x⟩) can be computed in terms of β(⟨s, y⟩) for codes s of shorter formulas.
(2) follows by first translating the formulas of number theory to FOL(τ) with a signature τ which has only relation symbols (for the graphs of 0, S, + and ·), and then reformulating questions of truth to quries about satisfaction—which for sentences are one and the same thing.) ⊣ Next we turn to the classification of relation on countable ordinal numbers via their Baire codes. We set WO = {α ∈LO |≤α is a well ordering}, (158) 244 5. Introduction to computability theory where the set LO of codes of linear orderings is defined in (154), and for each α ∈WO, we set |α| = the ordinal similar with ≤α (α ∈WO).
(159) Each |α| is a countable ordinal, and each countable ordinal is |α| with some (in fact many) α ∈WO.
Theorem 5I.6. The set WO of ordinal codes is Π1 1. Moreover, there are relations ≤Π, ≤Σ in Π1 1 and Σ1 1 respectively, such that if β ∈WO, then for all α, α ≤Π β ⇐ ⇒α ≤Σ β ⇐ ⇒[α ∈WO & |α| ≤|β|].
Proof. To see that WO is Π1 1, we compute: α ∈WO ⇐ ⇒α ∈LO & (∀β) h (∀t)[β(t + 1) ≤α β(t)] = ⇒(∃t)[β(t + 1) = β(t)] i .
To prove the second assertion, take first α ≤Σ β ⇐ ⇒α ∈LO & (∃γ)[γ maps ≤α into ≤β in a one-to-one order-preserving manner] ⇐ ⇒α ∈LO & (∃γ)(∀n)(∀m)[n <α m = ⇒γ(n) <β γ(m)].
It is immediate that ≤Σ is Σ1 1 and for β ∈WO, α ≤Σ β ⇐ ⇒[α ∈WO & |α| ≤|β|].
For the relation ≤Π, take α ≤Π β ⇐ ⇒α ∈WO & there is no order-preserving map of ≤β onto a proper initial segment of ≤α ⇐ ⇒α ∈WO & (∀γ)¬(∃k)(∀n)(∀m) n ≤β m ⇐ ⇒[γ(n) ≤α γ(m) <α k] , where of course we abbreviate s <α t ⇐ ⇒s ≤α t & s ̸= t.
⊣ That WO is not Σ1 1 is a consequence of the following, basic result of defin-ability theory, established (in various forms) by Lusin-Sierpinski and Kleene.
5I. The analytical hierarchy 245 Theorem 5I.7 (The Basic Representation Theorem for Π1 1). If P(x, α) is Π1 1, then there exists a total recursive function f : Nn × N ν →N, that for all x, α, f(x, α) ∈LO, and P(x, α) ⇐ ⇒f(x, α) ∈WO.
(160) Proof. We set z = x, β to save some typing, and by 5I.2 we choose a recursive and monotone in u relation R(z, u) such that P(z) ⇐ ⇒(∀β)(∃t)R z, β(t) .
For each z, put T(z) = {(u0, . . . , ut−1) | ¬R(z, ⟨u0, . . . , ut−1⟩)} so that T(z) is a tree on N (by the monotonicity or R) and clearly P(z) ⇐ ⇒T(z) is wellfounded.
What we must do is replace T(z) by a linear ordering on a subset of N which will be wellfounded precisely when T(z) is. Put (v0, . . . ,vs−1) >z (u0, . . . , ut−1) ⇐ ⇒(v0, . . . , vs−1), (u0, . . . , ut−1) ∈T(z) & h v0 > u0 ∨[v0 = u0 & v1 > u1] ∨[v0 = u0 & v1 = u1 & v2 > u2] ∨· · · ∨[v0 = u0 & v1 = u1 & · · · & vs−1 = us−1 & s < t] i where > on the right is the usual “greater than” in N.
It is immediate that if (v0, . . . , vs−1), (u0, . . . , ut−1) are both in T(z) and (v0, . . . , vs−1) is an initial segment of (u0, . . . , ut−1), then (v0, . . . , vs−1) >z (u0, . . . , ut−1); thus if T(z) has an infinite branch, then >z has an infinite descending chain.
Assume now that >z has an infinite descending chain, say v0 >z v1 >z v2 >z · · · , where vi = (vi 0, vi 1, . . . , vi si−1), 246 5. Introduction to computability theory and consider the following array: v0 = (v0 0, v0 1, . . . , v0 s0−1) v1 = (v1 0, v1 1, . . . , v1 s1−1) · · · · · · vi = (vi 0, vi 1, . . . , vi si−1) · · · · · · The definition of >z implies immediately that v0 0 ≥v1 0 ≥v2 0 ≥· · · , i.e., the first column is a non-increasing sequence of natural numbers. Hence after a while they all are the same, say vi 0 = k0 for i ≥i0.
Now the second column is non-increasing below level i0, so that for some pair i1, k1, vi 1 = k1 for i ≥i1.
Proceeding in the same way we find an infinite sequence k0, k1, . . .
such that for each s, (k0, . . . , ks−1) ∈T(z), so T(z) is not wellfounded. Thus we have shown, P(z) ⇐ ⇒T(z) is wellfounded ⇐ ⇒>z has no infinite descending chains.
Finally put u ≤z v ⇐ ⇒(∃t ≤u)(∃s ≤v) h Seq(u) & lh(u) = t & Seq(v) & lh(v) = s & u = v ∨ (v)0, . . . , (v)s−1 >z (u)0, . . . , (u)t−1 i and notice that ≤z is always a linear ordering, and P(z) ⇐ ⇒≤z is a wellordering.
Moreover, the relation P(z, u, v) ⇐ ⇒u ≤z v is easily recursive. The proof is completed by taking f(z)(n) = ( 1, if (n)0 ≤z (n)1, 0, otherwise.
⊣ 5I. The analytical hierarchy 247 Corollary 5I.8. The set WO of ordinal codes is not Σ1 1. |
14853 | https://influentialpoints.com/Training/unequal_variance_anova.htm | Welch's test
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your browser cannot display this list of links.Unequal variance ANOVA ====================== ### Background Although conventional parametric (and non-parametric) ANOVA require the assumption of equal variances, Welch (1951) developed an unequal variance one-way analysis of variance. This test has been little used to date, although its provision in R (as oneway.test) probably means it will be more heavily used in future. Subsequent multiple comparison of means should be done using pairwise.t.test with a non-pooled standard deviation. This calculates pairwise comparisons between group levels with Bonferroni-type corrections for multiple testing. ### How to do it #### Worked example We will analyse the data from Johnston et al.(2001) using R: ### One-way unequal variance ANOVA Gives something like this: One-way analysis of means (not assuming equal variances) data: con and trt F = 5.9763, num df = 2.000, denom df = 42.893, p-value = 0.005134 Pairwise comparisons using t tests with non-pooled SD data: con and trt A B B 0.0596 - C 0.0044 0.1280 P value adjustment method: holm Except where otherwise specified, all text and images on this page are copyright InfluentialPoints, all rights reserved. Images not copyright InfluentialPoints credit their source on web-pages attached via hypertext links from those images. |
14854 | https://brainly.in/question/7060160 | Is microfilament in plant and animal cells? - Brainly.in
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14855 | https://www.ck12.org/flexi/cbse-math/reciprocal-identities/lesspgreaterthe-numbers-b-and-b-are-their-own-reciprocal-less-by-pgreater/ | Flexi answers - The numbers ___ and ___ are their own reciprocal. | CK-12 Foundation
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14856 | https://oeis.org/wiki/Sums_of_squares | This site is supported by donations to The OEIS Foundation.
Sums of squares
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All positive integers can be expressed as sums of squares. Some can be expressed as the sum of two or three squares, some can be expressed as the sum of a million squares. And some can be expressed expressed as sums of squares in multiple ways. For example, 338350 is the sum of the first hundred nonzero squares. It can also be represented as 5802 + 432 + 102 + 12.
Since the square of a negative number is a positive number, we will not bother to distinguish between the squares of negative integers and the squares of positive integers in this article. However, we will distinguish between sums of squares that may include instances of 02 and those that must consist solely of the squares of nonzero integers; the latter will be referred to as "nonzero squares" here.
Among numbers that can be represented as the sum of a given number of squares, we may distinguish between those that can be represented with fewer squares and those that can't. For example, 25 = 42 + 32, but also 52, whereas 29 = 52 + 22 but we can't use (that is inelegant at best and cheating at worst).
| | | | | |
--- ---
| | Numbers that can be represented with squares | | Numbers that can't be represented with fewer than squares | |
| 1 | 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, ... | A000290 |
| 2 | 2, 5, 8, 10, 13, 17, 18, 20, 25, 26, 29, 32, ... | A000404 | 2, 5, 8, 10, 13, 17, 18, 20, 26, 29, 32, 34, ... | A000415 |
| 3 | 3, 6, 9, 11, 12, 14, 17, 18, 19, 21, 24, 26, 27, 30, ... | A000408 | 3, 6, 11, 12, 14, 19, 21, 22, 24, 27, 30, 33, 35, 38, ... | A000419 |
| 4 | 4, 7, 10, 12, 13, 15, 16, 18, 19, 20, 21, 22, 23, 25, ... | A000414 | 7, 15, 23, 31, 39, 47, 55, 63, 71, 79, 87, 95, 103, ... | A004771 |
| 5 | 5, 8, 11, 13, 14, 16, 17, 19, 20, 21, 22, 23, 24, 25, ... | A047700 | None | |
| 6 | 6, 9, 12, 14, 15, 16, 17, 20, 21, 22, ... | |
| 7 | 7, 10, 13, 15, 16, 18, 21, 22, ... | |
Sums of four squares
The answer to Waring's problem for squares is that all integers have at least one representation as the sum of at most four squares.
: Theorem SQS4. Every positive integer can be expressed as the sum of at most four nonzero squares. Or we can say that all positive integers are the sum of four squares, some, but not all, of which may be zero: that is, , with .
: Proof. PROOF GOES HERE. ENDOFPROOFMARK
When , the number requires four nonzero squares. Obviously prime numbers require at least two nonzero squares for their Waring representation. See Theorem P2SQ in the Gaussian integers article for a result regarding which primes can be expressed as the sum of two squares.
: Corollary to Theorem SQS4. If is the sum of four equal nonzero squares, then it has at least two representations as sum of squares, at least one of which includes zeroes. The stricture may have given the impression that the theorem does not cover the sums of equal squares, but this would be a false impression, as a little rewriting will show. If , then (or , etc.). We can rewrite . Reassigning and 'resetting' , and to 0 gives us a representation of satisfying the inequality given in the theorem.
Sums of five squares
With just a few exceptions (see A047701), all positive integers can be expressed as the sum of five nonzero squares. For example, 255 = 92 + 82 + 72 + 62 + 52.
: Theorem SQS5. Every integer can be expressed as the sum of five nonzero squares.
: Proof. The cases are examined one by one in the table below, demonstrating that each of them has at least one representation as the sum of five nonzero squares. For , we will need the 4-square representation, which may include zeroes, which Theorem SQS4 tells us exists for all . What is special about 169 is not that it is a square, but that it can also be represented as the sum of five, four, three or two squares. Solve the equation observing the stricture . If , then . If , then . If , then . And if only , then we have . □
Of course some integers have more than one representation as a sum of five nonzero squares, and the proof of Theorem SQS5 might not always give us the most "interesting" representation. Returning to our example of 255, the method outlined in the proof would give us 255 = (122 + 52) + (92 + 22 + 12).
| | | | | | |
--- --- --- |
| 34 | 42 + 32 + 22 + 22 + 12 | 35 | 52 + 22 + 22 + 12 + 12 | 36 | 42 + 32 + 32 + 12 + 12 |
| 37 | 52 + 32 + 12 + 12 + 12 | 38 | 42 + 42 + 22 + 12 + 12 | 39 | 42 + 32 + 32 + 22 + 12 |
| 40 | 62 + 12 + 12 + 12 + 12 | 41 | 52 + 32 + 22 + 12 + 12 | 42 | 42 + 32 + 32 + 22 + 22 |
| 43 | 52 + 32 + 22 + 22 + 12 | 44 | 52 + 42 + 12 + 12 + 12 | 45 | 52 + 32 + 32 + 12 + 12 |
| 46 | 42 + 42 + 32 + 22 + 12 | 47 | 52 + 42 + 22 + 12 + 12 | 48 | 52 + 32 + 32 + 22 + 12 |
| 49 | 62 + 22 + 22 + 22 + 12 | 50 | 52 + 42 + 22 + 22 + 12 | 51 | 62 + 32 + 22 + 12 + 12 |
| 52 | 52 + 42 + 32 + 12 + 12 | 53 | 42 + 42 + 42 + 22 + 12 | 54 | 62 + 32 + 22 + 22 + 12 |
| 55 | 52 + 42 + 32 + 22 + 12 | 56 | 52 + 52 + 22 + 12 + 12 | 57 | 62 + 32 + 22 + 22 + 22 |
| 58 | 62 + 42 + 22 + 12 + 12 | 59 | 62 + 32 + 32 + 22 + 12 | 60 | 52 + 42 + 32 + 32 + 12 |
| 61 | 62 + 42 + 22 + 22 + 12 | 62 | 52 + 42 + 42 + 22 + 12 | 63 | 62 + 42 + 32 + 12 + 12 |
| 64 | 72 + 32 + 22 + 12 + 12 | 65 | 42 + 42 + 42 + 42 + 12 | 66 | 62 + 42 + 32 + 22 + 12 |
| 67 | 62 + 52 + 22 + 12 + 12 | 68 | 52 + 52 + 42 + 12 + 12 | 69 | 72 + 32 + 32 + 12 + 12 |
| 70 | 62 + 52 + 22 + 22 + 12 | 71 | 72 + 42 + 22 + 12 + 12 | 72 | 62 + 52 + 32 + 12 + 12 |
| 73 | 62 + 52 + 22 + 22 + 22 | 74 | 72 + 42 + 22 + 22 + 12 | 75 | 62 + 52 + 32 + 22 + 12 |
| 76 | 72 + 42 + 32 + 12 + 12 | 77 | 72 + 52 + 12 + 12 + 12 | 78 | 62 + 42 + 42 + 32 + 12 |
| 79 | 72 + 42 + 32 + 22 + 12 | 80 | 72 + 52 + 22 + 12 + 12 | 81 | 62 + 62 + 22 + 22 + 12 |
| 82 | 62 + 52 + 42 + 22 + 12 | 83 | 72 + 52 + 22 + 22 + 12 | 84 | 72 + 42 + 32 + 32 + 12 |
| 85 | 72 + 52 + 32 + 12 + 12 | 86 | 82 + 42 + 22 + 12 + 12 | 87 | 62 + 52 + 42 + 32 + 12 |
| 88 | 72 + 52 + 32 + 22 + 12 | 89 | 82 + 42 + 22 + 22 + 12 | 90 | 62 + 62 + 42 + 12 + 12 |
| 91 | 72 + 62 + 22 + 12 + 12 | 92 | 72 + 52 + 42 + 12 + 12 | 93 | 72 + 52 + 32 + 32 + 12 |
| 94 | 82 + 42 + 32 + 22 + 12 | 95 | 72 + 52 + 42 + 22 + 12 | 96 | 92 + 32 + 22 + 12 + 12 |
| 97 | 72 + 62 + 22 + 22 + 22 | 98 | 82 + 52 + 22 + 22 + 12 | 99 | 72 + 62 + 32 + 22 + 12 |
| 100 | 72 + 52 + 42 + 32 + 12 | 101 | 82 + 42 + 42 + 22 + 12 | 102 | 62 + 62 + 52 + 22 + 12 |
| 103 | 82 + 52 + 32 + 22 + 12 | 104 | 92 + 32 + 32 + 22 + 12 | 105 | 62 + 62 + 42 + 42 + 12 |
| 106 | 72 + 62 + 42 + 22 + 12 | 107 | 62 + 62 + 52 + 32 + 12 | 108 | 92 + 42 + 32 + 12 + 12 |
| 109 | 82 + 62 + 22 + 22 + 12 | 110 | 82 + 52 + 42 + 22 + 12 | 111 | 92 + 42 + 32 + 22 + 12 |
| 112 | 92 + 52 + 22 + 12 + 12 | 113 | 102 + 22 + 22 + 22 + 12 | 114 | 82 + 62 + 32 + 22 + 12 |
| 115 | 72 + 62 + 52 + 22 + 12 | 116 | 92 + 42 + 32 + 32 + 12 | 117 | 92 + 52 + 32 + 12 + 12 |
| 118 | 82 + 62 + 42 + 12 + 12 | 119 | 82 + 72 + 22 + 12 + 12 | 120 | 92 + 52 + 32 + 22 + 12 |
| 121 | 82 + 62 + 42 + 22 + 12 | 122 | 102 + 42 + 22 + 12 + 12 | 123 | 92 + 62 + 22 + 12 + 12 |
| 124 | 82 + 72 + 32 + 12 + 12 | 125 | 102 + 42 + 22 + 22 + 12 | 126 | 82 + 62 + 42 + 32 + 12 |
| 127 | 82 + 72 + 32 + 22 + 12 | 128 | 92 + 62 + 32 + 12 + 12 | 129 | 92 + 62 + 22 + 22 + 22 |
| 130 | 82 + 62 + 52 + 22 + 12 | 131 | 92 + 62 + 32 + 22 + 12 | 132 | 92 + 52 + 42 + 32 + 12 |
| 133 | 72 + 72 + 52 + 32 + 12 | 134 | 82 + 72 + 42 + 22 + 12 | 135 | 82 + 62 + 52 + 32 + 12 |
| 136 | 112 + 32 + 22 + 12 + 12 | 137 | 102 + 42 + 42 + 22 + 12 | 138 | 92 + 62 + 42 + 22 + 12 |
| 139 | 102 + 52 + 32 + 22 + 12 | 140 | 82 + 72 + 52 + 12 + 12 | 141 | 92 + 72 + 32 + 12 + 12 |
| 142 | 82 + 62 + 52 + 42 + 12 | 143 | 82 + 72 + 52 + 22 + 12 | 144 | 92 + 72 + 32 + 22 + 12 |
| 145 | 102 + 62 + 22 + 22 + 12 | 146 | 102 + 52 + 42 + 22 + 12 | 147 | 92 + 62 + 52 + 22 + 12 |
| 148 | 82 + 72 + 52 + 32 + 12 | 149 | 82 + 82 + 42 + 22 + 12 | 150 | 102 + 62 + 32 + 22 + 12 |
| 151 | 112 + 42 + 32 + 22 + 12 | 152 | 92 + 62 + 52 + 32 + 12 | 153 | 82 + 62 + 62 + 42 + 12 |
| 154 | 82 + 72 + 62 + 32 + 12 | 155 | 82 + 72 + 52 + 42 + 12 | 156 | 92 + 72 + 42 + 32 + 12 |
| 157 | 102 + 62 + 42 + 22 + 12 | 158 | 102 + 72 + 22 + 22 + 12 | 159 | 92 + 82 + 32 + 22 + 12 |
| 160 | 112 + 52 + 32 + 22 + 12 | 161 | 102 + 72 + 22 + 22 + 22 | 162 | 102 + 62 + 42 + 32 + 12 |
| 163 | 102 + 72 + 32 + 22 + 12 | 164 | 112 + 52 + 42 + 12 + 12 | 165 | 92 + 72 + 52 + 32 + 12 |
| 166 | 92 + 82 + 42 + 22 + 12 | 167 | 112 + 52 + 42 + 22 + 12 | 168 | 112 + 62 + 32 + 12 + 12 |
| 169 | 122 + 42 + 22 + 22 + 12 | | | | |
References
Jump up ↑ This is part of Theorem 5.6 in Niven & Zuckerman (1980), on pages 144 and 145. In that book, they also prove that there infinitely many integers which are not the sum of four nonzero squares.
Jump up ↑ This is pretty much the same as the proof given in Niven & Zuckerman (1980), p. 145, but with the statements regarding numbers that are not the sums of four nonzero squares left out.
Ivan Niven & Herbert S. Zuckerman, An Introduction to the Theory of Numbers, New York: John Wiley (1980).
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14857 | https://math.stackexchange.com/questions/2912507/what-does-it-mean-to-have-an-absolute-value-equal-an-absolute-value | What does it mean to have an absolute value equal an absolute value? - Mathematics Stack Exchange
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What does it mean to have an absolute value equal an absolute value?
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I have no problem reading absolute value equations such as |x−2|=2|x−2|=2.
I know this means that the distance of some real number is 2 2 away from the origin. Because the origin splits the number line into a negative side and positive side then the numbers inside the absolute value symbol will be 2 2 and −2−2, since those are the only two numbers 2 2 units away from the origin. Then, it's just a matter of finding the values of x x which will give 2 2 and −2−2 inside the absolute value.
Therefore, |x−2|=2|x−2|=2 which is
x−2=2 x−2=2
or
x−2=−2 x−2=−2
And the solutions are {0,4}{0,4}
But when I see |3 x−1|=|x+5||3 x−1|=|x+5|, I have no idea know what this means. I know how to solve it, but I don't know how this relates to the distance from the origin or how to interpret this on a number line. My initial interpretation is to say, "the absolute value of some unknown number is the absolute value of some unknown number," but that doesn't tell me the distance from 0 0.
My Algebra textbook gave the following definition:
If |u|=|v||u|=|v|, then u=v u=v or u=−v u=−v.
But I can't really tell why this is the case.
absolute-value
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edited Sep 11, 2018 at 6:44
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2 It might help to plot the equation y=|3 x−1|=3|x−1/3|y=|3 x−1|=3|x−1/3|, and then plot the equation y=|x+5|y=|x+5| on the same graph. Your equation is valid at the intersection point(s) of those two plots.Andy Walls –Andy Walls 2018-09-10 23:25:55 +00:00 Commented Sep 10, 2018 at 23:25
Maybe this will help: |u|=|v||u|=|v| is equivalent to u=v u=v or u=−v u=−v or −u=v−u=v or −u=−v−u=−v. But of course, the last two of these are redundant being equivalent to the first two.MasB –MasB 2018-09-10 23:43:33 +00:00 Commented Sep 10, 2018 at 23:43
1 The algebraic d e f i n i t i o n d e f i n i t i o n of |x||x| is that |x|=x|x|=x if x≥0 x≥0 and |x|=−x|x|=−x if x<0 x<0. In all cases, |x||x| is the non-negative member of {x,−x}.{x,−x}. So if |x|=|y||x|=|y| then one member of {x,−x}{x,−x} is equal to one member of {y,−y},{y,−y}, which implies (x=y∨x=−y∨−x=y∨−x=−y)(x=y∨x=−y∨−x=y∨−x=−y), which is equivalent to (x=y∨x=−y)(x=y∨x=−y). Conversely , if x=±y,x=±y, then the non-negative member of {x,−x},{x,−x}, which is |x|,|x|, must equal the non-negative member of {y,−y},{y,−y}, which is |y|.|y|.DanielWainfleet –DanielWainfleet 2018-09-11 00:10:09 +00:00 Commented Sep 11, 2018 at 0:10
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Your interpretation is good.
Any valuev v is at distance|v||v| from origin. Sometimes we are given the distance and are asked to find original value. When something (∈R∈R) is at distance|w||w| from origin, it has a value either w w or −w−w.
As you said, |x−2|=2|x−2|=2 means that (x−2)(x−2) is at distance 2 2 from origin.
The same goes for the example that confuses you;
|3 x−1|=|x+5||3 x−1|=|x+5|
means that (3 x−1)(3 x−1) is at distance|x+5||x+5| from origin.
And what can we conclude from this? That the value of (3 x−1)(3 x−1) is either (x+5)(x+5) or −(x+5)−(x+5), and that is what your textbook says using u u and v v.
[Also, you can flip it and say that (x+5)(x+5) is at distance |3 x−1||3 x−1| from origin, and those are the redundant cases you see mentioned in other answers.]
And to directly refer to your title question; Having two absolute values equal means, in terms of distance from origin, that they are both equaly far from origin. So
|u|=|v||u|=|v|
means that u u and v v are equaly far from origin. How far specificaly? Exactly |u||u| (or |v||v|, because they are equal).
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edited Sep 11, 2018 at 0:56
answered Sep 10, 2018 at 23:45
Sandro LovničkiSandro Lovnički
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Thanks for your answer, it's cleared up most of my confusion, just one more stumbling block. The solution set, {0, 4}, for |x - 2| = 2 makes sense to me since, when plugged back in, they give |2| = 2 and |-2| = 2 which are the absolute values that are two away from the origin. The solution set for |3x - 1| = |x + 5| is {-1, 3}. When I plug these values back into the equation I get |8| = |8| and |-4| = |4|. I don't get how there could be two different absolute values. Does this mean that the points -4, 4, and 8 are the absolute values, or the points filled on the number line?Slecker –Slecker 2018-09-11 01:48:51 +00:00 Commented Sep 11, 2018 at 1:48
1 @Slecker As opposed to the simpler equation of the form |u|=2|u|=2 that asks you “which values of x x cause u u to be 2 away from the origin”, you have an equation |u|=|v||u|=|v| and the question is “which values of x x cause u u to be as far away from the origin as v v is”. x=−1 x=−1 causes u=−4 u=−4 and v=4 v=4 which fits (both are 4 away from the origin, on different sides), and x=3 x=3 results in u=v=8 u=v=8 (both are obviously 8 away from the origin on the same side, on the same exact point).Roman Odaisky –Roman Odaisky 2018-09-11 02:48:00 +00:00 Commented Sep 11, 2018 at 2:48
@Roman Aha! Thank you very much for that last bit of insight: “|u|=|v| and the question is 'which values of x cause u to be as far away from the origin as v is.'" It helps me tremendously fit the last conceptual puzzle piece in my head!Slecker –Slecker 2018-09-11 03:04:52 +00:00 Commented Sep 11, 2018 at 3:04
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Maybe trying to answer the following question can help you..
In real line, when do two points u u and v v have the same absolute value (the same distance away from the origin)?
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answered Sep 10, 2018 at 23:40
Rodrigo DiasRodrigo Dias
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If you’re having trouble imagining what the equation might represent, consider two objects moving along the number line. Let the coordinate of one of those change with time as 3 t−1 3 t−1 and the other one as t+5 t+5 (this would imply that at the moment of time t=0 t=0, presumably when you started observing the situation, one was at −1−1 and was moving at a constant speed of 3 units of distance per unit of time in the positive direction, while the second one was at the point 5 and its speed was 1).
Now you’re asked the question, at what time one was as far away from the origin as the other one? (Maybe you’re a spy trying to determine the timestamp of some event, but all you know is that the signal from two ships or planes or whatever was equally strong at that point.) That’s what your equation
|3 t−1|=|t+5||3 t−1|=|t+5|
says. Intuitively, because the first object’s velocity is greater but at t=0 t=0 it’s to the left of the second object, this would happen two times: once in the past, when the origin was exactly between the objects, and once in the future, when one object overtakes the other. The equation 3 t−1=−(t+5)3 t−1=−(t+5) corresponds to the former case (coordinates are −4−4 and 4 4 respectively), and 3 t−1=t+5 3 t−1=t+5 to the latter (both objects at 8 8).
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answered Sep 11, 2018 at 2:36
Roman OdaiskyRoman Odaisky
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That's actually a good way of conceptualizing it, and I never could think of a way to apply absolute values to real life, so your illustration with the spy is quite interesting and helpful.Slecker –Slecker 2018-09-11 03:19:39 +00:00 Commented Sep 11, 2018 at 3:19
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if |u|=|v||u|=|v| there are actually four possibilities, but they are two redundant pairs.
The four possibilities are.
A) u≥0;v≥0 u≥0;v≥0 and therefore u=|u|;v=|v|=|u|=u u=|u|;v=|v|=|u|=u and RESULT 1) u=v u=v.
B) u<0;v<0 u<0;v<0 and therefore u=−|u|;v=−|v|=−|u|=u u=−|u|;v=−|v|=−|u|=u and RESULT 1) u=v u=v. (that's redudant.)
C) u≥0;v<0 u≥0;v<0 and therefore u=|u|;v=−|v|=−|u|=−u u=|u|;v=−|v|=−|u|=−u and RESULT 2) u=−v u=−v
D) u<0;v≥0 u<0;v≥0 and therefore u=−|u|;v=|v|=|u|=−u u=−|u|;v=|v|=|u|=−u and RESULT 2) u=−v u=−v. (that's reducant).
Now if |3 x−1|=|x+5||3 x−1|=|x+5| we could solve by doing all four cases but that is unnecessary as we will get redundant and contradictory results.
Let's do it to see what happens and see if we can learn from it:
A) 3 x−1≥0 3 x−1≥0 and x+5≥0 x+5≥0 and therefore 3 x−1=x+5 3 x−1=x+5.
In other words 3 x≥1 3 x≥1 and x≥−5 x≥−5 and 2 x=6 2 x=6
In other words x≥1 3 x≥1 3 and x≥−5 x≥−5 and x=3 x=3.
Or in other words x=3 x=3.
B) 3 x−1<0 3 x−1<0 and x+5<0 x+5<0 and therefore 3 x−1=x+5 3 x−1=x+5.
Or x<1 3 x<1 3 and x<−5 x<−5 and x=3 x=3.
That's a contradiction. Notice there was no reason to consider the two case whether 3 x−1 3 x−1 and x+5 x+5 are greater or less than 0 0. That was just a waste of time. It would have been just as well to only consider that 3 x−1=x+5⟹x=3 3 x−1=x+5⟹x=3. That's all we had to do.
C) 3 x−1<0 3 x−1<0 and x+5≥0 x+5≥0 and therefore 3 x−1=−x−5 3 x−1=−x−5
Or x<1 3 x<1 3 and x≥−5 x≥−5 and 4 x=−4⟹x=−1 4 x=−4⟹x=−1. So x=−1 x=−1 is possible.
D) 3 x−1≥0 3 x−1≥0 and x+5<0 x+5<0 and therefore 3 x−1=−x−5 3 x−1=−x−5
Or x≥1 3 x≥1 3 and x<−5 x<−5 and 4 x=−4⟹x=−1 4 x=−4⟹x=−1. That's a contradiction.
But again we didn't have to do both C) and D). That was redundant.
It's enough to know that if |3 x−1|=|x+5||3 x−1|=|x+5| then either 3 x−1=x+5 3 x−1=x+5 (and we don't care if they are both positive or both negative-- we can figure that out later) or 3 x−1=−x−5 3 x−1=−x−5 (ditto).
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answered Sep 11, 2018 at 0:11
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If |u|=|v||u|=|v|, then u=v u=v or u=−v u=−v.
An easy way to understand why this is true is to square both sides of the equation. The statement |u|=|v||u|=|v| is the same statement as u 2=v 2 u 2=v 2 (since x 2−−√=|x|x 2=|x| for any x x). Rearrange this to u 2−v 2=0 u 2−v 2=0, then factor the LHS to obtain (u−v)(u+v)=0(u−v)(u+v)=0. Conclude that the original statement |u|=|v||u|=|v| is equivalent to the statement u−v=0 u−v=0 or u+v=0 u+v=0.
You can similarly solve |3 x−1|=|x+5||3 x−1|=|x+5| by squaring both sides.
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answered Sep 11, 2018 at 5:32
grand_chatgrand_chat
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Red Tide and PSP
Dinoflagellate Blooms, “Red Tides” “Rusty Tides, PSP, and Harmful Algal Blooms
Related Pages:Shellfish Bed closures |Beach Information |Seafood
Overview
Excessive growth of microalgae in coastal waters may result in adverse impacts to the ecosystem of human health, and are often broadly defined as harmful algal blooms (HAB). “Red Tides” are a long used lay term referring to blooms of the dinoflagellate algae Alexandrium fundyense (a type of microscopic phytoplankton). An algal “bloom” is a population explosion of microalgae where cell densities may exceed millions of cells per liter of water, and can even result in the discoloration of the water. However, Alexandrium blooms may not occur at high enough concentrations to discolor the water red, yet still have adverse effects. This algae can contaminate filter-feeding shellfish with a toxin that results in a syndrome known as paralytic shellfish poisoning (PSP) when the shellfish are consumed. Shellfish are monitored by the Massachusetts Division of Marine Fisheries for this PSP toxin contamination. When the PSP toxin is elevated in mussels (the indicator species used by DMF), they will enact a PSP shellfish bed closure in a region. The press typically report these as “red tide” closures. Since 2009, there have been no shellfish bed PSP closures in Buzzards Bay, however, in most years, other areas of Massachusetts have had PSP closures, particularly in waters north of Cape Cod. Go to the DMF PSP website for the latest information. You should always consult with your local shellfish constable for the status of shellfish bed closures your community.
Since 2005, the Buzzards Bay Coalition received reports of “Rusty Tides” caused by the non-toxic dinoflagellate Cochlodinium polykrikoides. Go to the bottom of this page to learn more about that species of microalgae.
The most widespread PSP closures in Massachusetts in recent years occurred during the summers of 2005 and 2006. Coincidentally, because of exceptionally heavy rains in early June of 2006 (6 to 10 inches of rain), on June 7, 2006 the state also closed all of state waters to shellfishing, including Buzzards Bay. This occurred in addition to the PSP closures. The first closure of any part of Buzzards Bay due to PSP occurred in 2005, when the northernmost end of the bay was closed, near the canal. That year, large portions of the rest of state suffered from prolonged PSP closures that had a disastrous impact to the incomes of commercial shellfishermen, and the state even sought federal disaster relief funds.
On October 7, 2016, Buzzards Bay was closed to shellfishing for the first time due to a bloom of the diatom Pseudo-nitzschia. The Massachusetts Division of Marine Fisheries (DMF) has banned the harvesting of shellfish in all of Buzzards Bay, Mount Hope Bay and Lackeys Bay off Naushon Island on Vineyard Sound. The ban later extended to Nantucket Sound. Certain Pseudo-nitzschia species can produce a toxin called domoic acid. This biotoxin can concentrate in filter feeding shellfish. High concentrations of domoic acid in shellfish can cause Amnesic Shellfish Poisoning (ASP) with symptoms that include vomiting, cramps, diarrhea and incapacitating headaches followed by confusion, disorientation, permanent loss of short-term memory, and in severe cases, seizures and coma.
Thismap (pdf file)shows the maximum extent of the 2005 PSP closures.
Information about the Massachusetts 2005 Red Tide and PSP closures in Buzzards Bay
Beginning on April 27, 2005 the Massachusetts Division of Marine Fisheries initiated a prohibition on the taking of shellfish in Nauset Harbor, in Orleans because of elevated levels of paralytic shellfish poison (PSP) toxin in shellfish. A wider scale closure was initiated at Cape Ann to New Hampshire on May 19. This closure expanded southward, until on May 27, the Division of Marine Fisheries announced a widespread prohibition on the taking of shellfish in Massachusetts Bay, Cape Cod Bay, and upper Buzzards Bay. On the outer Cape, the closure continued to extend south to Monomoy Island on June 2, and further south to Nantucket Island on June 3. By June 6, the PSP closure extended southwest to Edgartown on Martha’s Vineyard.
On June 9, 2005, Governor Mitt Romney declared an economic disaster in Massachusetts, allowing the state to seek federal disaster aid. On June 10, DMF closed Federal Waters adjacent to Massachusetts to the taking and landing of shellfish (read the DMF notice). On June 16, closures extended to the entire south coast of Martha’s Vineyard closed. In response to the Governor’s declaration, the Small Business Administration announced it would issue low-interest loans to qualified individuals on June 21. However, on June 24, the federal government decided the Red Tide outbreak and resulting PSP shellfish bed closures did not qualify for federal disaster relief.
Alexandrium fundyense
PSP closures are enacted in response to a bloom of a type of a type of planktonic algae, a dinoflagellate of the genus Alexandrium. Blooms of Alexandrium are often called “Red Tide” because the water can have a pinkish tinge from the density of algae in the water, although it is possible to have PSP closures without any discoloration of the water, so Red Tide has become synonymous elevated Alexandrium populations in the water that result in PSP shellfish closures. PSP contamination of shellfish is a threat to human health because the algae produce PSP toxins, specifically one called saxitoxin, that can accumulate in shellfish during blooms. PSP toxin is a threat to human health, and can even cause death.
While PSP closures occur more regularly in the cold waters north of Cape Cod, in 2005 for the first time ever, the PSP closure extended through the Cape Cod Canal and south into Buzzards Bay to Wings Neck in Bourne and Stony Point Dike in Wareham. This area includes Buttermilk Bay, Onset Bay, and Phinneys Harbor. The inclusion of the north end of Buzzards Bay reflects the historically high levels of PSP toxin observed at the end of May in Cape Cod Bay, that were transported southward, through the canal. The exceptionally high levels of Alexandrium in Cape Cod Bay were believed to be the result of heavy spring rains and unusual May Nor’easters that drove in the offshore phytoplankton bloom.
Click to read the May 27, 2005 New Hampshire to Buzzards Bay PSP closure notice.
On July 1, 2005, DMF reopened the closed areas of Buzzards Bay and few selected areas of Cape Cod. Openings continued gradually through July and August, but a few areas remained closed as shown in the map above. Concerns remain about the possible long-term implications of future outbreaks of Red Tide south of Cape Cod, and in Buzzards Bay, and whether the dinoflagellate cysts that remain in the bottom sediments from this outbreak, will make these areas more prone to future Red Tides and likely PSP closures. It is presumed that cysts deposited in one season provided the inoculums for new outbreaks in future years. Some studies have shown that Alexandrium cysts may survive in sediments under natural conditions for at least several years.
Cochlodinium Blooms in Buzzards Bay (Rusty Tides)
A boater’s view of the “rust colored” water observed off North Falmouth in September 2005 caused by Cochlodinium. Photo taken by Larry Soule, Baywatcher for the Buzzards Bay Coalition.
Image 4: Cochlodinium dinoflagellate”>
Warm waters and high nutrients are thought to contribute to blooms of Cochlodinium. In Peconic Bay on Long Island, fertilizer runoff and surface and groundwater discharges of wastewater is believed to have contributed to some recent “Brown Tides” and Cochlodinium blooms observed in that estuary in 2004 and 2005. The Peconic Bay National Estuary Program has prepared and easy-to-understand <a class=
Microscopic view of _Cochlodinium_ spp. dinoflagellate.
In September 2005, we received the first report of a _Cochlodinium_ Bloom in Buzzards Bay, sometimes referred to as “Rusty Tide”, to distinguish it from the PSP related “Red Tides.” _Cochlodinium_ blooms may have occurred previously in Buzzards Bay, but if so, they were not widely reported. We have received reports of these Rusty or Rust Tides in every subsequent year. For example, in August 2012, _Cochlodinium_ blooms were widely reported in the northern half of Buzzards Bay, both in bays and offhsore, in late August. We also received a report from the south central portion of the bay from Dartmouth Health Agent Wendy Henderson who observed a _Cochlodinium_ Bloom at Anthony Beach near Clarks Cove, exposed coast areas near Round Hill and Salters Point.
The species of dinoflagellate is _Cochlodinium polykrikoides_ (synonym= _C. heterolobatum_). The 2005 bloom in Buzzards Bay had counts that exceeded 7.6 million cells per liter, and Rusty patches of water were observed primarily off Falmouth and Bourne as illustrated by the photograph to the right. The 2005 bloom subsided a few weeks later after waters cooled.
Unlike the _Alexandrium_ dinoflagellate, _Cochlodinium_ does not contain neurotoxins that affect people, but like _Alexandrium_, it contains reddish pigments, so that in large enough concentrations, _Cochlodinium_ can also discolor coastal waters with a rusty hue if it appears in great enough concentration. _Cochlodinium_ had been considered a Harmful Algal Bloom (HAB) species simply because in high concentrations in contained systems, it caused the death of fish by consuming all the oxygen in the water (plants generally produce excess oxygen in the day, but like animals, they are always consuming oxygen too). However, more recent studies have concluded that _Cochlodinium polykrikoides_ in fact exhibit a pronounced chemical toxicity (Gobler et. al, 2008; Harmful Algae 7(3):293-307). Tanga and Gobler (2008, Harmful Algae, 8(3):454-462) documented that juvenile bay scallops experienced “100% mortality during 3 days exposure to cultures at cell densities an order of magnitude lower than raw bloom water.” These authors proposed that the algae releases reactive oxygen compounds as the mechanism of mortality, but a subsequent study by Kim et. al (2009; Biosci Biotechnol Biochem. 73(3):613-8.) suggested that other compounds might be the cause of observed mortality, which cause the observed neurotoxic, hemolytic, and hemagglutinative effects. Large concentrations of _Cochlodinium_ may also affect the growth, survival, and development of some zooplankton larvae of more desirable species, like oysters.
Warm waters and high nutrients are thought to contribute to blooms of _Cochlodinium_. In Peconic Bay on Long Island, fertilizer runoff and surface and groundwater discharges of wastewater is believed to have contributed to some recent “Brown Tides” and _Cochlodinium_ blooms observed in that estuary in 2004 and 2005. The Peconic Bay National Estuary Program has prepared and easy-to-understand _Cochlodinium_ Page and Fact Sheet that contains much useful information about the species.
What are Dinoflagellates?
Dinoflagellates are a broad group of single celled organisms generally classified as a Class (Dinoflagellata) in the phylum Protista. Protists are a catchall group that includes plant-like organisms and animal-like organisms. Dinoflagellates are typically found floating in the water, and are thus known as plankton. Some dinoflagellates have chlorophyll (and make their own food), and are thus considered “algae” or phytoplankton. Other dinoflagellates eat small plants or animals, and are therefore considered zooplankton. Some dinoflagellates are mixotrophic; that is they both have chlorophyll, and they eat other organisms. _Cochlodinium_ is mixotrophic.
Some dinoflagellates are even parasitic on other animals
The distinguishing features of dinoflagellates that separate them from other Protists include the fact that they have armored plates made of cellulose, they may contain chlorophylls a and c and fucoxanthin (along with other pigments that give them a reddish color), and they have two flagella for swimming through the water, one of which is contained in a spiral groove on the body.
Besides the notoriety of the Red Tide dinoflagellates, many beach goers are familiar with another dinoflagellate know as “sea sparkles.” These are in fact a large (at least for a single-cell organism – 1mm) plankton-eating species of dinoflagellate called _Noctiluca_ that can be seen scintillating in the water as bright sparkles of light at night. [Ctenophore jellies are also bioluminescent, and also cause flashes during nighttime swimming in Buzzards Bay, but they are another story.].
Brown Tides
In recent years, brown tides have been frequently reported in a broad region from Rhode Island to New Jersey, with many occurrences in Long Island embayments, The term “brown tide” is somewhat generic, and can refer to several different phytoplankton species that color the water brown at high concentrations. In Long Island waters, the most common species is _Aureococcus anophagefferens_. This species is not a dinoflagellate, but is in a diverse class of single celled and multi-celled yellow-green and brown algae that includes kelps and diatoms. Unlike _Alexandrium_ spp., as described above, this organism produces a non-toxic bloom and is harmless to humans.
Newspaper and Magazine Articles, and information pages
Woods Hole Oceanographic Institution Northeast PSP page
2008 WHOI Oceanus Magazine Article on Red Tide
Mass.gov page on red tide monitoring
Mass.gov page on Red Tide poisoning.
Woods Hole Oceanographic Institution Harmful Algal Bloom page
NOAA Harmful algal blooms
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Steiner Triple System
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$\begingroup$
A Steiner Triple System, denoted by $STS(v),$ is a pair $(S,T)$ consisting of a set $S$ with $v$ elements, and a set $T$ consisting of triples of $S$ such that every pair of elements of $S$ appear together in a unique triple of $T$.
Now, my book goes on to say that the number of triples of a $STS(n)$ disjoint from a given triple is $(n-3)(n-7)/6$ but I am not sure how they got that result?
I know that there are $n(n-1)/6$ triples altogether where each point of a triple lies in $(n-1)/2$ triples but I am not sure how they got that $(n-3)(n-7)/6$.
combinatorics
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edited Apr 26, 2015 at 6:37
LuffyLuffy
asked Apr 26, 2015 at 5:47
LuffyLuffy
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$\endgroup$
5
1
$\begingroup$ $(n3)(n7)/6=n(n-1)/6-3(n-1)/2+3-1$ $\endgroup$
Alexey Burdin
– Alexey Burdin
2015-04-26 07:36:39 +00:00
Commented Apr 26, 2015 at 7:36
$\begingroup$ @AlexeyBurdin: you should post your comment as an answer. I managed to tie myself in knots with the inclusion-exclusion argument. Yours is much simpler. $\endgroup$
Will Orrick
– Will Orrick
2015-04-26 07:42:43 +00:00
Commented Apr 26, 2015 at 7:42
$\begingroup$ @WillOrrick: sorry I'm not sure about that "$+3$" -- the equation is a trial-and-error product actually... $\endgroup$
Alexey Burdin
– Alexey Burdin
2015-04-26 07:52:36 +00:00
Commented Apr 26, 2015 at 7:52
$\begingroup$ @AlexeyBurdin Can you please expand on your results because I would like to see another approach to this problem? $\endgroup$
Luffy
– Luffy
2015-04-26 08:30:52 +00:00
Commented Apr 26, 2015 at 8:30
$\begingroup$ @WillOrrick: finally did, thanks much to both of you for attention. $\endgroup$
Alexey Burdin
– Alexey Burdin
2015-04-26 09:13:22 +00:00
Commented Apr 26, 2015 at 9:13
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3 Answers 3
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$\begingroup$
Consider
there are $n(n1)/6$ triples altogether
so we take a tuple(=triple) $T$ to exclude
each point of a triple lies in $(n1)/2$ triples
call these tuples a "tuple line". Each "tuple line" containing $T$ does not intersect with an other (if any 2 did, say in $T'$ , we would have, say $a,b\in T$, and $T'$ in a "$a$-tuple line" and "$b$-tuple line", so $a,b\in T'$ that's impossible by definition since $T\neq T'$).So we exclude $3(n-1)/2$ tuples, but we excluded tuple $T$ thrice, while we need to do it only once, so we get$$n(n1)/63(n1)/2+31=(n3)(n7)/6$$
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answered Apr 26, 2015 at 9:11
Alexey BurdinAlexey Burdin
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$\begingroup$ Alexey, can you elaborate on how we excluded tuple $T$ thrice? $\endgroup$
Luffy
– Luffy
2015-04-29 01:34:20 +00:00
Commented Apr 29, 2015 at 1:34
2
$\begingroup$ @Luffy : we excluded $3$ "tuple lines" $(n-1)/2$ tuples each and each containing $T$. $\endgroup$
Alexey Burdin
– Alexey Burdin
2015-04-29 02:42:01 +00:00
Commented Apr 29, 2015 at 2:42
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2
$\begingroup$
The inclusion-exclusion argument given here should probably be ignored. Alexey Burdin gives a much cleaner answer.
There are $\binom{n}{2}$ pairs that can be formed from elements of $S$, and each pair is in exactly one triple of $T$. But each triple contains three pairs. So there are $\frac{1}{3}\binom{n}{2}=\frac{n(n-1)}{6}$ triples in $T$.
Let ${a,b,c}\in T$. Now there are $n-1$ pairs containing $a$, each of which appears in one triple of $T$. Since there are two elements apart from $a$ in each such triple, there are $\frac{n-1}{2}$ triples containing $a$ and $\frac{n(n-1)}{6}-\frac{n-1}{2}=\frac{(n-1)(n-3)}{6}$ triples of $T$ that do not contain $a$. The same number do not contain $b$; likewise the same number do not contain $c$.
Let $T_s$ be the set of triples in $T$ that do not have the set $s$ as a subset. We want $\lvert T_{{a}}\cap T_{{b}}\cap T_{{c}}\rvert$. By inclusion-exclusion, this is $$ \begin{aligned} \lvert T_{{a}}\cap T_{{b}}\cap T_{{c}}\rvert=&\lvert T_{{a}}\rvert+\lvert T_{{b}}\rvert+\lvert T_{{c}}\rvert-\lvert T_{{a}}\cup T_{{b}}\rvert-\lvert T_{{a}}\cup T_{{c}}\rvert-\lvert T_{{b}}\cup T_{{c}}\rvert\ &+\lvert T_{{a}}\cup T_{{b}}\cup T_{{c}}\rvert\ =&\lvert T_{{a}}\rvert+\lvert T_{{b}}\rvert+\lvert T_{{c}}\rvert-\lvert T_{{a,b}}\rvert-\lvert T_{{a,c}}\rvert-\lvert T_{{b,c}}\rvert+\lvert T_{{a,b,c}}\rvert. \end{aligned} $$ But $\lvert T_{{a,b}}\rvert=\lvert T_{{a,c}}\rvert=\lvert T_{{b,c}}\rvert=\lvert T_{{a,b,c}}\rvert=\frac{n(n-1)}{6}-1=\frac{(n+2)(n-3)}{6}$ since the only triple containing two or more of $a$, $b$, $c$ is ${a,b,c}$. So the number is $$ 3\cdot\frac{(n-1)(n-3)}{6}-3\frac{(n+2)(n-3)}{6}+\frac{(n+2)(n-3)}{6}=\frac{n-3}{6}(3(n-1)-2(n+2))=\frac{(n-3)(n-7)}{6}. $$
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edited Apr 26, 2015 at 15:39
community wiki
5 revsWill Orrick
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4
$\begingroup$ Sorry for the confusion! I got off on the wrong foot with my explanation by defining $T_s$ as the set of triples NOT containing the elements of $s$. I think this made things hopelessly confusing, essentially because it required thinking in terms of double negatives, and because it required an unusual form of inclusion-exclusion (intersections on the left, unions on the right, rather than the reverse). It also required a bit more algebra than necessary. $\endgroup$
Will Orrick
– Will Orrick
2015-04-26 14:44:27 +00:00
Commented Apr 26, 2015 at 14:44
1
$\begingroup$ Much more straightforward to do things Alexey Burdin's way. I should have defined $T_s$ to be the set of triples that DO contain the elements of $s$. With this definition of $T_s$ you want to compute $$\lvert(T_{{a}}\cup T_{{b}}\cup T_{{c}})'\rvert=\lvert T\rvert-\lvert T_{{a}}\rvert-\ldots+\lvert T_{{a,b}}\rvert+\ldots-\lvert T_{{a,b,c}}\rvert.$$ This leads directly to Alexey Burdin's much simpler answer. $\endgroup$
Will Orrick
– Will Orrick
2015-04-26 14:45:03 +00:00
Commented Apr 26, 2015 at 14:45
$\begingroup$ My original derivation of the statement that the number of pairs not containing $a$ is $\frac{n-1}{2}$ was flawed. I've corrected this now. Sorry for all the mess. You really should focus on understanding Alexey Burdin's answer rather than mine, but I leave this here for the time being since you've already spent some time on it. In partial recompense, I offer a completely different approach, not using inclusion-exclusion, as a separate answer. $\endgroup$
Will Orrick
– Will Orrick
2015-04-26 15:39:38 +00:00
Commented Apr 26, 2015 at 15:39
$\begingroup$ Thank you very much! It is more clear now. $\endgroup$
Luffy
– Luffy
2015-04-26 19:01:48 +00:00
Commented Apr 26, 2015 at 19:01
Add a comment |
2
$\begingroup$
Here's a derivation that explains the factors $n-3$ and $n-7$ directly. Consider the triple ${a,b,c}\in T$. There are $n-3$ elements in $S\setminus{a,b,c}$. Each such element $d$ appears in one triple with $a$, in one triple with $b$, and in one triple with $c$. These three triples are distinct, since the only triple containing two or more of $a$, $b$, $c$ is ${a,b,c}$. Let these three triples be ${d,a,e}$, ${d,b,f}$, and ${d,c,g}$. The elements $e$, $f$, and $g$ are distinct since, for example, $e$ can appear in only one triple with $d$. Hence there are $n-7$ elements in $S\setminus{a,b,c,d,e,f,g}$.
Each pair consisting of $d$ and one element $h$ of $S\setminus{a,b,c,d,e,f,g}$ appears in exactly one triple. The triple containing $d$ and $h$ cannot contain $a$, $b$, $c$, $e$, $f$, or $g$ since if it contained, say, $a$, it would have to be the triple ${d,a,e}$, but $e\ne h$.
Hence a triple not containing $a$, $b$, or $c$ is specified by making one of $n-3$ choices for the element $d$, and one of $n-7$ choices for the element $h$. But such a triple is not uniquely specified by this procedure: any of its three elements can play the role of $d$, and any of the two remaining elements can play the role of $h$. So the number of triples is $$ \frac{1}{3\cdot2}(n-3)(n-7). $$
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answered Apr 26, 2015 at 15:40
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14860 | https://www.cut-the-knot.org/Curriculum/Geometry/ApollonianCircle.shtml | Typesetting math: 100%
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Apollonian Circles Theorem
Given two points A and B and a number r. What is the locus of points P such that AP/BP=r? The answer is a circle. The circle which is known as the Apollonian Circle. For every positive r there is a different one.
This problem has been treated elsewhere. Here we present a different solution based on the inversion transform. Along the way we show that the whole family of Apollonian circles can be inverted into a family of concentric circles.
Solution
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Copyright © 1996-2018 Alexander Bogomolny
Apollonian Circles Theorem
For two fixed points A and B and a real r>0, the locus of points P such that AP/BP=r is a circle. The circle is known as the Apollonian Circle.
Consider a circle of radius R centered at A. Let t be the inversion in this circle. Set t(B)=B′ and t(P)=P′. The calculations are simplified by taking R=1, but, in principle, any circle with center at A will do. So, assume R=1.
This means, in particular, that
(1)
AP⋅AP′=AB⋅AB′=1.
Triangles AB′P′ and APB, in which the sides satisfy (1), also share the angle at A. They are, therefore, similar. It then follows that
(2)
BP/B′P′=AB/AP′.
From here,
(3)
B′P′=BP⋅AP′/AB.
Combining (1) and (3) gives
(4)
B′P′=BP/(AP⋅AB)=1/rAB.
This tells us that P′ lies on a circle of radius 1/rAB centered at B′. In other words, the inversive image of the locus of points P is a circle centered at B′. To repeat, the Apollonian circle defined by the point circles A and B and r>0 is mapped by t to a circle centered at t(B). This is true for any Apollonian circle defined by A and B, so that the whole family of them is mapped on the family of concentric circles with center t(B).
In establishing Steiner's Porism we showed that any two non-intersecting circles can be inverted into a pair of concentric circles. The above strengthens this assertion with a more direct proof.
The Apollonian circles defined by two point circles are said to be coaxal. This is one of the three varieties of coaxal families.
References
D. A. Brannan et al, Geometry, Cambridge University Press, 2002
Inversion - Introduction
Angle Preservation Property
Apollonian Circles Theorem
Archimedes' Twin Circles and a Brother
Bisectal Circle
Chain of Inscribed Circles
Circle Inscribed in a Circular Segment
Circle Inversion: Reflection in a Circle
Circle Inversion Tool
Feuerbach's Theorem: a Proof
Four Touching Circles
Hart's Inversor
Inversion in the Incircle
Inversion with a Negative Power
Miquel's Theorem for Circles
Peaucellier Linkage
Polar Circle
Poles and Polars
Ptolemy by Inversion
Radical Axis of Circles Inscribed in a Circular Segment
Steiner's porism
Stereographic Projection and Inversion
Tangent Circles and an Isosceles Triangle
Tangent Circles and an Isosceles Triangle II
Three Tangents, Three Secants
Viviani by Inversion
Simultaneous Diameters in Concurrent Circles
An Euclidean Construction with Inversion
Construction and Properties of Mixtilinear Incircles
Two Quadruplets of Concyclic Points
Seven and the Eighth Circle Theorem
Invert Two Circles Into Equal Ones
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Copyright © 1996-2018 Alexander Bogomolny |
14861 | https://www.expii.com/t/discriminant-to-find-nature-of-roots-4856 | Expii
Discriminant to Find Nature of Roots - Expii
If the discriminant of a quadratic is negative, then the quadratic has no real solutions. But it still has complex number solutions!.
Explanations (7)
Have you ever wondered what type of solutions a quadratic equation will have without actually finding the solutions?
The discriminant does just that! It is the part of the quadratic formula that appears directly under the radical sign. Take a look at the photo shown below:
The Quadratic Formula
Based on what the value of the discriminant is we can tell what types of solutions we will have.
For instance, imagine the discriminant is a negative number. In this scenario, we will have a negative number under the radical sign. This would indicate 2 imaginary solutions.
Now let's imagine another scenario. If the discriminant equals zero we will now have zero under the radical sign. Well the square root of zero is zero so we would essentially be adding zero to negative b in the numerator. Because adding or subtracting zero yields the same response, having a discriminant value of zero would indicate only one real solution. This is often called a "double solution" or "two equal solutions"
The last scenario is when the discriminant is positive. In this case, we will have two real solutions. The solutions cannot be imaginary because the number under the radical sign is not negative. If the positive discriminant value is a perfect square, we will have two rational solutions. If the discriminant value is a non perfect square, we will have two irrational solutions.
This chart below should summarize the information about the discriminant.
Discriminat Chart
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Quadratic Equations from 3 Points
Completing the Square to Find Complex Roots
Converting Between Different Forms of a Quadratic
Using the Quadratic Formula to Find Complex Roots
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Vinjai Vale
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Now that we know what quadratics are, a pressing question arises. Just as we were able to find solutions to linear equations like 2x+3=5, can we find the solutions to a quadratic like 5x2−12x+5=1?
Let's start simple.
Think about the solutions to x2−9=0.
We know that the solutions are x=3 and x=−3. The equation x2−a=0 always has two solutions for x: √a and −√a. The only exception is a=0.
Now, think about what the solutions are to (x−1)2=9.
We have two cases: x−1=3 or x−1=−3. This gives the solutions −2 and 4.
Diving In
Think about the solutions to x2+12x+20=48.
We can start by trying to make a perfect square using x. Note that x2+12x+36=(x+6)2; thus, if we add 16 to both sides of our equation, we get (x+6)2=64. That's great -- we know how to solve that! Since 64=82, we have x+6=8 or x+6=−8. In the first case we get x=2, and in the second we get x=−14. Hence those are our two solutions.
Let's move things up a notch.
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Michael Livshits
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Let our equation be of the form x2+bx+c=0 and let x1 and x2 be the roots of the quadratic.
Then x2+bx+c=(x−x1)(x−x2) =x2−(x1+x2)x+x1x2 and therefore x1+x2=−b and x1x2=c. Now, if we knew d=x1−x2, we could figure out x1=(−b+d)/2 and x2=(−b−d)/2.
The trouble is that our equation does not know which of the roots is x1 and which one is x2, so we can know d only up to a sign, so we can know d2. Let us try to figure it out in terms of b and c. Can you do it?
Here comes a little magic: d2=x21−2x1x2+x22 and b2=x21+2x1x2+x22, they differ just by 4x1x2=4c. Ta-taam!
Now we get d2=b2−4c and finally we get our quadratic formula: x1,2=(−b±√b2−4c)/2
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Anusha Rahman
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The Discriminant
The discriminant of a quadratic equation tells you the nature of the roots. That means you can use the discriminant to find out whether a quadratic function has two solutions, one solution, or imaginary solutions.
The discriminant is found within the quadratic equation, the value under the radicand.
Image source: by Anusha Rahman
When we solve the discriminant, if our value is:
d>0, the quadratic has 2 real solutions.
d=0, the quadratic has 1 real solution.
d<0, the quadratic has 1 imaginary solution.
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Sarah Walker
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What is the Quadratic Formula?
When given an equation in the form ax2+bx+c, usually the simplest way to solve for the value of x is to factor the quadratic, set each factor equal to zero, and then solve for each factor. However, the quadratic can sometimes be too difficult to solve by factoring. At this point, you may want to throw in the towel, give up on math, and run far away. Never fear! You can count on the quadratic formula to find the solution for you!
When using the quadratic formula, you will need to have an ”a”,“b”, and ”c” term from ax2+bx+c, where ”a”,“b”, and ”c” are numbers. These are considered numerical coefficients of the quadratic equation that you are asked to solve.
You may be asking, “but what IS the quadratic formula?” Well, here it is, derived from the process of completing the square (which you can refresh on here!)
x=−b±√b2−4ac2a
This beautiful equation will be engrained into your mind forever. I learned it five years ago, and I still remember it to this day!
Rules for the Quadratic Formula
1) In order for the quadratic formula to work, you must have the equation arranged so that x=−b±√b2−4ac2a is set equal to 0.
2) Make sure that you are careful not to drop the square root, or the ± before the square root. Because you have the ±, this means you will have 2 roots!
3) Our a term can NEVER be 0. This will cause the denominator of the quadratic formula to be 0, and it will void the equation since you can never EVER divide by 0. If you are given an equation in the form of bx+c=0, you will have to solve for x by subtracting both sides by c, then dividing both sides by b.
4) The b2 under the square root means that you square ALL of b, including a negative sign if applicable. So, the b2 term will never be negative!
5) Also, take note that you will multiply your b by a negative sign at the beginning of the numerator.
All in all, make sure you take note of your signs! If you try to take shortcuts, or try to shorthand the problem, you will mess it up! Take the quadratic formula slow, and I promise you will become a pro!
Example of Quadratic Formula
Solve x2+7x–4
Since this is a quadratic equation in the form ax2+bx+c, we can say that our a term is 1, since the coefficient of x2 is 1 (even if you can’t see it!) Our b term is 7, and our c term is −4. Make sure that you include the negative sign in your c term, or you’ll get the wrong answer!
With a=1, b=7, and c=−4 we can plug these values into the quadratic formula to get:
x=−7±√72−4(1)(−4)2(1)=0
We can simplify what is being multiplied, to get:
x=−7±√49+162=0
NOTE that under the square root, we now have a summation of the numbers rather than a subtraction, because our c term is negative! (and we know that two negative numbers multiplied together give us a positive!)
We can further simplify this to:
x=−7±√652=0
Now, since 65 is not a perfect square, and it does not contain a perfect square, we can leave our answer in this form!
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How is it possible to minimize the maximum of 3 functions?
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Anonymous
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Originally Answered: How is it possible to minimize the maximum of 3 functions ? ·
Conceptually you define/compute first a new function P(x,y) as max(f,g,h) (it is the upper envelope) and then you look for the minimum as with any other function.
Numerically you can compute f,g,h over a grid and then you get P() over a grid and look for the minimum.
If you would like to use other methods, like the Newton method, the problem you will approach is that it may be tedious to compute P() analytically even if you have f,g,h analytically and they are "nice" functions (for example, f,g,h are polynomials). The reason is that the upper envelope (which is what I would call P()) consists
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Conceptually you define/compute first a new function P(x,y) as max(f,g,h) (it is the upper envelope) and then you look for the minimum as with any other function.
Numerically you can compute f,g,h over a grid and then you get P() over a grid and look for the minimum.
If you would like to use other methods, like the Newton method, the problem you will approach is that it may be tedious to compute P() analytically even if you have f,g,h analytically and they are "nice" functions (for example, f,g,h are polynomials). The reason is that the upper envelope (which is what I would call P()) consists of "intervals" where it follows one of the three functions and kinks at the borders of these "intervals."
If you want to build intuition, start in R1 (i.e. f(x), g(x), h(x)). Then you can see the problems visually: even if all f,g,h, are differentiable, P will not - it will have kink points when it will switch between these functions. Draw a graph when f,g,h are all linear to see the pattern...
Minimum of P will be either at a local minimum of f,g,h or at one of these "kink points" which means that whatever you do, finding the kink points will be important and whatever method you use has to be able to deal with non-differentiabilities at the kinks.
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What is the maximum value of the function f(x) = |x|+3, x€ [-3,2]?
What is Minimize Minimize?
How does one minimize this function of four variables?
How do you minimize and maximize a function?
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Justin Rising
consumer of optimization theory · Author has 12.1K answers and 26.5M answer views
·9y
Originally Answered: How is it possible to minimize the maximum of 3 functions ? ·
If all of your functions are convex, then their maximum is as well, and you can use standard convex optimization techniques. Otherwise it's just a nonlinear optimization problem, and you can use your favorite algorithms for that class of problems.
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Minimizing the maximum of three functions involves finding a point where the largest value among the three functions is as small as possible. This is often referred to as solving a "minimax" problem. Here’s a structured approach to understand and tackle this problem:
Problem Statement
Given three functions f 1(x)f 1(x), f 2(x)f 2(x), and f 3(x)f 3(x), the goal is to minimize:
M(x)=max(f 1(x),f 2(x),f 3(x))M(x)=max(f 1(x),f 2(x),f 3(x))
Steps to Minimize the Maximum
Define the Domain: Identify the range of x x values over which you want to minimize M(x)M(x).
Find Critical Points:
Calculate the derivatives of the functions to find critical poi
Continue Reading
Minimizing the maximum of three functions involves finding a point where the largest value among the three functions is as small as possible. This is often referred to as solving a "minimax" problem. Here’s a structured approach to understand and tackle this problem:
Problem Statement
Given three functions f 1(x)f 1(x), f 2(x)f 2(x), and f 3(x)f 3(x), the goal is to minimize:
M(x)=max(f 1(x),f 2(x),f 3(x))M(x)=max(f 1(x),f 2(x),f 3(x))
Steps to Minimize the Maximum
Define the Domain: Identify the range of x x values over which you want to minimize M(x)M(x).
Find Critical Points:
Calculate the derivatives of the functions to find critical points where f 1(x)=f 2(x)f 1(x)=f 2(x), f 2(x)=f 3(x)f 2(x)=f 3(x), and f 1(x)=f 3(x)f 1(x)=f 3(x).
This helps in identifying points where the maximum function might change.
Evaluate the Functions:
At each critical point, compute the values of f 1(x)f 1(x), f 2(x)f 2(x), and f 3(x)f 3(x).
Determine M(x)M(x) at these points.
Consider Endpoints: If the domain is closed, evaluate M(x)M(x) at the endpoints of the interval as well.
Compare Values: Find the minimum value of M(x)M(x) from the critical points and endpoints. This minimum value will be your optimal solution.
Example
Suppose we have the following functions:
f 1(x)=x 2 f 1(x)=x 2
f 2(x)=(x−2)2 f 2(x)=(x−2)2
f 3(x)=(x+1)2 f 3(x)=(x+1)2
Find critical points where the functions intersect:
Set f 1(x)=f 2(x)f 1(x)=f 2(x): x 2=(x−2)2 x 2=(x−2)2
Set f 2(x)=f 3(x)f 2(x)=f 3(x): (x−2)2=(x+1)2(x−2)2=(x+1)2
Set f 1(x)=f 3(x)f 1(x)=f 3(x): x 2=(x+1)2 x 2=(x+1)2
Solve these equations to find the critical points.
Evaluate M(x)M(x) at the critical points and endpoints of the domain.
Select the minimum value of M(x)M(x).
Numerical Methods
For more complex functions or higher dimensions, numerical optimization techniques such as gradient descent, genetic algorithms, or other optimization methods might be employed to find the minimum of the maximum efficiently.
Conclusion
Minimizing the maximum of multiple functions can be a powerful technique in optimization problems, especially in fields like operations research, economics, and engineering. By following the structured approach above, you can systematically find the optimal solution.
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Anupreet Choudhary
Engineer,IITian, Trying to become good in mathematics · Author has 71 answers and 572.5K answer views
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How do I find maximum and minimum value of a function?
There are several methods of doing this. The easiest way is to use differential calculus and try to plot the curve. Tarun Desu has already stated the steps involved but it is only true for functions in single variable like y= f(x).
For functions in multi-variables, like z=f(x,y) & z=f(x,y,z......) things become more complex and it is harder to visualize these functions. Anyway, the concept remains the same.
So, here is the method for two-variable functions:
Find partial differentials of function f(x,y) w.r.t x and y and equate them to zero. There will be two equations - f x=0 f x=0 and f y=0 f y=0
Solve t
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There are several methods of doing this. The easiest way is to use differential calculus and try to plot the curve. Tarun Desu has already stated the steps involved but it is only true for functions in single variable like y= f(x).
For functions in multi-variables, like z=f(x,y) & z=f(x,y,z......) things become more complex and it is harder to visualize these functions. Anyway, the concept remains the same.
So, here is the method for two-variable functions:
Find partial differentials of function f(x,y) w.r.t x and y and equate them to zero. There will be two equations - f x=0 f x=0 and f y=0 f y=0
Solve these two equations simultaneously in x and y. Let (a,b) (c,d)... be the pair of values
Calculate: r=∂2 f∂x 2 r=∂2 f∂x 2 , s=∂2 f∂y∂x s=∂2 f∂y∂x and t=∂2 f∂y 2 t=∂2 f∂y 2 for (a,b) (c,d)...
Now if r t−s 2>0 r t−s 2>0 and r<0 r<0 , then f(a,b) is maximum. If r t−s 2>0 r t−s 2>0 and r>0 r>0 then f(a,b) is minimum. If r t−s 2<0 r t−s 2<0 then f(a,b) is a saddle point
You can apply the above steps for the following example -
f(x,y)=x 4+y 4−2 x 2+4 x y−2 y 2 f(x,y)=x 4+y 4−2 x 2+4 x y−2 y 2
The graph of the above function is
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How do you minimize X = ((A'B'C')' + (A'B)')'?
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Huseyin Tugrul Buyukisik
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Justin Rising
, MSE in CS, PhD in Statistics · Author has 3.2K answers and 5.4M answer views
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Is the fitness function in genetic algorithm minimized or maximized?
Depends on how you sort the DNAs. If you sort them in increasing order and take first DNA for best choice, then it is minimization. If you sort them in decreasing order and pick first one as best one, then it is maximization.
Minimization example:
Drawing a Mona Lisa painting by triangles
RMS pixel error = minimized
Maximization example:
Designing a leg for a robot to jump high
Jump height = maximized
I see more of robot/artificial life design in videos than Mona Lisa painting so I guess maximization is more popular.
Also its easy to change it. Just put a “-” sign before the return value. Minimization
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Depends on how you sort the DNAs. If you sort them in increasing order and take first DNA for best choice, then it is minimization. If you sort them in decreasing order and pick first one as best one, then it is maximization.
Minimization example:
Drawing a Mona Lisa painting by triangles
RMS pixel error = minimized
Maximization example:
Designing a leg for a robot to jump high
Jump height = maximized
I see more of robot/artificial life design in videos than Mona Lisa painting so I guess maximization is more popular.
Also its easy to change it. Just put a “-” sign before the return value. Minimization becomes maximization. Maximization becomes minimization.
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Tarun Desu
Undergrad Electronics student, Aiming infinity and beyond
·9y
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How do I find maximum and minimum value of a function?
There are Various methods in order to find maximum or minimum value of a function.
One of the conventional methods is:
Find the derivative of the function and equate it to zero.
Find the roots of the differentiated equation.
Do double differentiation of original function and substitute the values of roots in the 2nd differentiated expression.
If the value comes out to be negative, At the particular value of the root Maximum occurs. Then substitute the value in original expression to get Maximum of the function.
If the value of double derivative after substituting the root is positive, Minimum occur
Continue Reading
There are Various methods in order to find maximum or minimum value of a function.
One of the conventional methods is:
Find the derivative of the function and equate it to zero.
Find the roots of the differentiated equation.
Do double differentiation of original function and substitute the values of roots in the 2nd differentiated expression.
If the value comes out to be negative, At the particular value of the root Maximum occurs. Then substitute the value in original expression to get Maximum of the function.
If the value of double derivative after substituting the root is positive, Minimum occurs. Then substitute the value in original equation to get Minimum value of the function.
If the Second derivative is Zero: Then go for higher derivatives of the function & substitute the value of the root in the nth order derivative expression. If it's positive it would give the Maximum of the function at the particular root.
Hope the answer Helps.
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Ted Horton
25+ years experience teaching physics and math. · Upvoted by
Terry Moore
, M.Sc. Mathematics, University of Southampton (1968) · Author has 401 answers and 688.7K answer views
·5y
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How do you minimize a function with two variables?
As Terry Moore explained in his answer, you start by taking both partial derivatives and set them to 0 (calling the function f and the variables x and y):
i. ∂f∂x=0∂f∂x=0
ii. ∂f∂y=0∂f∂y=0
This will give you 2 equations for x and y. Hopefully, you can find unique solutions. It is possible that you will find a curve along which the equations are satisfied, instead of unique points. If you combine these as much as possible, you will at least have a constraint to use in the next part.
There is also something analogous to the second
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As Terry Moore explained in his answer, you start by taking both partial derivatives and set them to 0 (calling the function f and the variables x and y):
i. ∂f∂x=0∂f∂x=0
ii. ∂f∂y=0∂f∂y=0
This will give you 2 equations for x and y. Hopefully, you can find unique solutions. It is possible that you will find a curve along which the equations are satisfied, instead of unique points. If you combine these as much as possible, you will at least have a constraint to use in the next part.
There is also something analogous to the second derivative test that you need to apply next. First some notation for the second order partial derivatives, in case you are not familiar:
f x x=∂2 f∂x 2 f x x=∂2 f∂x 2
f y y=∂2 f∂y 2 f y y=∂2 f∂y 2
f x y=f y x=∂2 f∂x∂y f x y=f y x=∂2 f∂x∂y
Keep in mind that order of differentiation does not effect the result of this last one.
Here are the steps:
(1) First take all the derivatives (first and second order). Do not use any constraints before you do this.
(2) Try solving i. and ii. to get a solution/constraint.
(3) Now set up this quantity:
D=f x x f y y−f 2 x y D=f x x f y y−f x y 2
If you want to plug in your result from i. and ii., you can. Here is what the second derivative test looks like:
Local minimum: D>0 D>0 , and f x x>0 f x x>0
Local maximum: D>0 D>0 , and f x x<0 f x x<0
Saddle point: D<0 D<0
If D=0 , there is supposed to be a third order test. I don’t know that one. All of this is related to the Taylor series expansion of a function of two variables.
Quick edit, in case you get a result here, you should also check the boundary of your domain for x and y. If the function has a limited domain, then there should be a curve along the boundary. It is possible that your critical points are on that boundary, even without satisfying the conditions laid out above.
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How do I minimize quadratic function?
To minimize a quadratic function, you need to find the value of the independent variable (usually denoted as 'x') that corresponds to the minimum value of the function. The process for doing this involves finding the vertex of the quadratic function.
Here are the steps to minimize a quadratic function:
Write the quadratic function in standard form: f(x) = ax^2 + bx + c, where a, b, and c are constants.
Find the x-coordinate of the vertex: x = -b/(2a).
Substitute the value of x found in step 2 into the quadratic function to find the minimum value of the function: f(x) = a(x)^2 + b(x) + c.
The value
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To minimize a quadratic function, you need to find the value of the independent variable (usually denoted as 'x') that corresponds to the minimum value of the function. The process for doing this involves finding the vertex of the quadratic function.
Here are the steps to minimize a quadratic function:
Write the quadratic function in standard form: f(x) = ax^2 + bx + c, where a, b, and c are constants.
Find the x-coordinate of the vertex: x = -b/(2a).
Substitute the value of x found in step 2 into the quadratic function to find the minimum value of the function: f(x) = a(x)^2 + b(x) + c.
The value of f(x) found in step 3 is the minimum value of the quadratic function.
So, the minimum value of the quadratic function occurs at the vertex of the parabola. If the coefficient of x^2 (a) is positive, the vertex corresponds to a minimum point, and if a is negative, the vertex corresponds to a maximum point.
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Ray Rodgers
B.S. Physics, B.S. Chemical Eng, M.S. Electrical Eng · Author has 255 answers and 458.9K answer views
·7y
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How do I minimize this function: y=∑N i=1 X i p i where∑N i=1 p i=1 y=∑i=1 N X i p i where∑i=1 N p i=1? X is fixed, and I’m trying to find all values of p i p i to minimize that function.
You’re trying to minimize the following function:
f(p 1,p 2,p 3,...,p N)=N∑i=1 x i p i f(p 1,p 2,p 3,...,p N)=∑i=1 N x i p i
Subject to the constraint:
g(p 1,p 2,p 3,...,p N)=N∑i=1 p i−1=0 g(p 1,p 2,p 3,...,p N)=∑i=1 N p i−1=0
This is best solved using Lagrange multipliers, or just one Lagrange multiplier in your case.
Note that each term in the summations includes only one variable and each variable only appears in one term at most. This means you can drop the summations and solve your equations with respect to p i p i rather than p 1 p 1, p 2 p 2, p 3 p 3, etc. without any loss of generality. Specifically, you’re minimizing the
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You’re trying to minimize the following function:
f(p 1,p 2,p 3,...,p N)=N∑i=1 x i p i f(p 1,p 2,p 3,...,p N)=∑i=1 N x i p i
Subject to the constraint:
g(p 1,p 2,p 3,...,p N)=N∑i=1 p i−1=0 g(p 1,p 2,p 3,...,p N)=∑i=1 N p i−1=0
This is best solved using Lagrange multipliers, or just one Lagrange multiplier in your case.
Note that each term in the summations includes only one variable and each variable only appears in one term at most. This means you can drop the summations and solve your equations with respect to p i p i rather than p 1 p 1, p 2 p 2, p 3 p 3, etc. without any loss of generality. Specifically, you’re minimizing the following function (i.e. the “Lagrangian”):
L=f+λ g L=f+λ g
With respect to p i p i
∂L∂p i=∂f∂p i+λ∂g∂p i=0∂L∂p i=∂f∂p i+λ∂g∂p i=0
Which generates the set of equations:
−x i p 2 i+λ=0−x i p i 2+λ=0
This generates N equations, one for each x i x i and p i p i. You have solved for p i p i in terms of x i x i:
p i=√x i λ p i=x i λ
All that remains is to determine λ λ which is found by substituting p i p i in your constraint equation:
N∑i=1 p i−1=N∑i=1√x i λ−1=0∑i=1 N p i−1=∑i=1 N x i λ−1=0
And solving for λ λ (you can separate the λ λ into its own square root and pull it out of the summation):
1√λ∗N∑i=1√x i=1 1 λ∗∑i=1 N x i=1
λ=(N∑i=1√x i)2 λ=(∑i=1 N x i)2
Thus the selection of p i p i that minimizes your function is:
p i= ⎷x i(∑N i=1√x i)2 p i=x i(∑i=1 N x i)2
The constraints are generally all set to zero when working with Lagrange multipliers.
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Gary Ward
MaEd in Education&Mathematics, Austin Peay State University (Graduated 1997) · Author has 4.9K answers and 7.6M answer views
·1y
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How can you determine if a function has more than one maximum without using calculus or graphing software?
How can you determine if a function has more than one maximum without using calculus or graphing software?
An even function, -x²ⁿ where 2n is the greatest exponent of the function, has one maximum. It may have relative maximums if the exponent is greater than 2, but still even, but it may not.
Thanks to Desmos, we can see two negative even functions in red and orange compared to an odd function in green. This odd function has a relative maximum, but y=x 3 y=x 3 does not.
Continue Reading
How can you determine if a function has more than one maximum without using calculus or graphing software?
An even function, -x²ⁿ where 2n is the greatest exponent of the function, has one maximum. It may have relative maximums if the exponent is greater than 2, but still even, but it may not.
Thanks to Desmos, we can see two negative even functions in red and orange compared to an odd function in green. This odd function has a relative maximum, but y=x 3 y=x 3 does not.
Upvote ·
9 5
Simon Tsai
Teaching Assistant at National Taiwan University (2024–present) · Author has 4.9K answers and 2M answer views
·Updated 4y
Related
How do you maximize the function y=(x-2) ^3×(x-3) ^2?
I will take you to mean that you want a local maximum. In that case, intuitively, we begin with differentiation:
y=(x−a)b(x−b)a y=(x−a)b(x−b)a
⟹ln(y)=b ln(x−a)+a ln(x−b)⟹ln(y)=b ln(x−a)+a ln(x−b)
⟹∂y y=(b x−a+a x−b)∂x⟹∂y y=(b x−a+a x−b)∂x
From the above, we find
∂y∂x=0⟺x=a 2+b 2 a+b∂y∂x=0⟺x=a 2+b 2 a+b
At this point,
\displa\displa
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I will take you to mean that you want a local maximum. In that case, intuitively, we begin with differentiation:
y=(x−a)b(x−b)a y=(x−a)b(x−b)a
⟹ln(y)=b ln(x−a)+a ln(x−b)⟹ln(y)=b ln(x−a)+a ln(x−b)
⟹∂y y=(b x−a+a x−b)∂x⟹∂y y=(b x−a+a x−b)∂x
From the above, we find
∂y∂x=0⟺x=a 2+b 2 a+b∂y∂x=0⟺x=a 2+b 2 a+b
At this point,
y=k a a b b∣∣∣a−b a+b∣∣∣a+b y=k a a b b|a−b a+b|a+b
k=(−1)s−2⌊s/2⌋k=(−1)s−2⌊s/2⌋
s=min(a,b)s=min(a,b)
If k=+1 k=+1, it is a local maximum; if k=−1 k=−1, it is a local minimum. In your case, a=2 a=2 and b=3 b=3, so it is positive, and what we obtained is a local maximum.
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Marek Čtrnáct
Translator From English · Author has 821 answers and 938.4K answer views
·2y
Related
Can a function have a local minimum but no global minimum/maximum?
Sure! A good example is y=x sin(x)y=x sin(x):
Each “wave” is bigger than the one before. There are infinitely many minimums and maximums, but none of them is global.
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Sure! A good example is y=x sin(x)y=x sin(x):
Each “wave” is bigger than the one before. There are infinitely many minimums and maximums, but none of them is global.
Upvote ·
9 2
9 2
Jörg Straube
M.Sc. in Computer Science, ETH Zurich (Graduated 1987) · Author has 6.2K answers and 1.7M answer views
·10mo
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How do I minimize f(x) =5x1^2+x2_4 subject to x2_ 4>=_4x1 ; _x2+3<=2x1?
Let x = x1 and y = x2
f(x, y) = 5x^2 + y - 4 → minimal
y - 4 >= -4x → y >= 4 - 4x
-y + 3 <= 2x → y >= 3 - 2x
Both lines intersect at (1, 0)
x < 1 → border of constrained area: y = 4 - 4x
f(x, y) = 5x^2 - 4x
f‘ = 10x - 4 = 0 → x = 2/5 → y = 12/5
f(2/5, 12/5) = 20/25 + 60/25 - 100/25 = -4/5
x > 1 → border of constrained area: y = 3 - 2x
f(x, y) = 5x^2 -2x - 1
f‘ = 10x - 2 = 0 → x = 1/5 (not >1) → x = 1 → y = 0
f(1, 0) = 5 + 0 - 4 = 1
f(x, y) is minimal at (2/5, 12/5)
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9 1
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14863 | https://www.talkorigins.org/faqs/faq-intro-to-biology.html | Introduction to Evolutionary Biology Version 2 Copyright © 1996-1997 by Chris Colby [Last Update: January 7, 1996]
volution is the cornerstone of modern biology. It unites all the fields of biology under one theoretical umbrella. It is not a difficult concept, but very few people -- the majority of biologists included -- have a satisfactory grasp of it. One common mistake is believing that species can be arranged on an evolutionary ladder from bacteria through "lower" animals, to "higher" animals and, finally, up to man. Mistakes permeate popular science expositions of evolutionary biology. Mistakes even filter into biology journals and texts. For example, Lodish, et. al., in their cell biology text, proclaim, "It was Charles Darwin's great insight that organisms are all related in a great chain of being..." In fact, the idea of a great chain of being, which traces to Linnaeus, was overturned by Darwin's idea of common descent.
Misunderstandings about evolution are damaging to the study of evolution and biology as a whole. People who have a general interest in science are likely to dismiss evolution as a soft science after absorbing the pop science nonsense that abounds. The impression of it being a soft science is reinforced when biologists in unrelated fields speculate publicly about evolution.
This is a brief introduction to evolutionary biology. I attempt to explain basics of the theory of evolution and correct many of the misconceptions.
What is Evolution?
Evolution is a change in the gene pool of a population over time. A gene is a hereditary unit that can be passed on unaltered for many generations. The gene pool is the set of all genes in a species or population.
The English moth, Biston betularia, is a frequently cited example of observed evolution. [evolution: a change in the gene pool] In this moth there are two color morphs, light and dark. H. B. D. Kettlewell found that dark moths constituted less than 2% of the population prior to 1848. The frequency of the dark morph increased in the years following. By 1898, the 95% of the moths in Manchester and other highly industrialized areas were of the dark type. Their frequency was less in rural areas. The moth population changed from mostly light colored moths to mostly dark colored moths. The moths' color was primarily determined by a single gene. [gene: a hereditary unit] So, the change in frequency of dark colored moths represented a change in the gene pool. [gene pool: the set all of genes in a population] This change was, by definition, evolution.
The increase in relative abundance of the dark type was due to natural selection. The late eighteen hundreds was the time of England's industrial revolution. Soot from factories darkened the birch trees the moths landed on. Against a sooty background, birds could see the lighter colored moths better and ate more of them. As a result, more dark moths survived until reproductive age and left offspring. The greater number of offspring left by dark moths is what caused their increase in frequency. This is an example of natural selection.
Populations evolve. [evolution: a change in the gene pool] In order to understand evolution, it is necessary to view populations as a collection of individuals, each harboring a different set of traits. A single organism is never typical of an entire population unless there is no variation within that population. Individual organisms do not evolve, they retain the same genes throughout their life. When a population is evolving, the ratio of different genetic types is changing -- each individual organism within a population does not change. For example, in the previous example, the frequency of black moths increased; the moths did not turn from light to gray to dark in concert. The process of evolution can be summarized in three sentences: Genes mutate. [gene: a hereditary unit] Individuals are selected. Populations evolve.
Evolution can be divided into microevolution and macroevolution. The kind of evolution documented above is microevolution. Larger changes, such as when a new species is formed, are called macroevolution. Some biologists feel the mechanisms of macroevolution are different from those of microevolutionary change. Others think the distinction between the two is arbitrary -- macroevolution is cumulative microevolution.
The word evolution has a variety of meanings. The fact that all organisms are linked via descent to a common ancestor is often called evolution. The theory of how the first living organisms appeared is often called evolution. This should be called abiogenesis. And frequently, people use the word evolution when they really mean natural selection -- one of the many mechanisms of evolution.
Common Misconceptions about Evolution
Evolution can occur without morphological change; and morphological change can occur without evolution. Humans are larger now than in the recent past, a result of better diet and medicine. Phenotypic changes, like this, induced solely by changes in environment do not count as evolution because they are not heritable; in other words the change is not passed on to the organism's offspring. Phenotype is the morphological, physiological, biochemical, behavioral and other properties exhibited by a living organism. An organism's phenotype is determined by its genes and its environment. Most changes due to environment are fairly subtle, for example size differences. Large scale phenotypic changes are obviously due to genetic changes, and therefore are evolution.
Evolution is not progress. Populations simply adapt to their current surroundings. They do not necessarily become better in any absolute sense over time. A trait or strategy that is successful at one time may be unsuccessful at another. Paquin and Adams demonstrated this experimentally. They founded a yeast culture and maintained it for many generations. Occasionally, a mutation would arise that allowed its bearer to reproduce better than its contemporaries. These mutant strains would crowd out the formerly dominant strains. Samples of the most successful strains from the culture were taken at a variety of times. In later competition experiments, each strain would outcompete the immediately previously dominant type in a culture. However, some earlier isolates could outcompete strains that arose late in the experiment. Competitive ability of a strain was always better than its previous type, but competitiveness in a general sense was not increasing. Any organism's success depends on the behavior of its contemporaries. For most traits or behaviors there is likely no optimal design or strategy, only contingent ones. Evolution can be like a game of paper/scissors/rock.
Organisms are not passive targets of their environment. Each species modifies its own environment. At the least, organisms remove nutrients from and add waste to their surroundings. Often, waste products benefit other species. Animal dung is fertilizer for plants. Conversely, the oxygen we breathe is a waste product of plants. Species do not simply change to fit their environment; they modify their environment to suit them as well. Beavers build a dam to create a pond suitable to sustain them and raise young. Alternately, when the environment changes, species can migrate to suitable climes or seek out microenvironments to which they are adapted.
Genetic Variation
Evolution requires genetic variation. If there were no dark moths, the population could not have evolved from mostly light to mostly dark. In order for continuing evolution there must be mechanisms to increase or create genetic variation and mechanisms to decrease it. Mutation is a change in a gene. These changes are the source of new genetic variation. Natural selection operates on this variation.
Genetic variation has two components: allelic diversity and non- random associations of alleles. Alleles are different versions of the same gene. For example, humans can have A, B or O alleles that determine one aspect of their blood type. Most animals, including humans, are diploid -- they contain two alleles for every gene at every locus, one inherited from their mother and one inherited from their father. Locus is the location of a gene on a chromosome. Humans can be AA, AB, AO, BB, BO or OO at the blood group locus. If the two alleles at a locus are the same type (for instance two A alleles) the individual would be called homozygous. An individual with two different alleles at a locus (for example, an AB individual) is called heterozygous. At any locus there can be many different alleles in a population, more alleles than any single organism can possess. For example, no single human can have an A, B and an O allele.
Considerable variation is present in natural populations. At 45 percent of loci in plants there is more than one allele in the gene pool. [allele: alternate version of a gene (created by mutation)] Any given plant is likely to be heterozygous at about 15 percent of its loci. Levels of genetic variation in animals range from roughly 15% of loci having more than one allele (polymorphic) in birds, to over 50% of loci being polymorphic in insects. Mammals and reptiles are polymorphic at about 20% of their loci - - amphibians and fish are polymorphic at around 30% of their loci. In most populations, there are enough loci and enough different alleles that every individual, identical twins excepted, has a unique combination of alleles.
Linkage disequilibrium is a measure of association between alleles of two different genes. [allele: alternate version of a gene] If two alleles were found together in organisms more often than would be expected, the alleles are in linkage disequilibrium. If there two loci in an organism (A and B) and two alleles at each of these loci (A1, A2, B1 and B2) linkage disequilibrium (D) is calculated as D = f(A1B1) f(A2B2) - f(A1B2) f(A2B1) (where f(X) is the frequency of X in the population). [Loci (plural of locus): location of a gene on a chromosome] D varies between -1/4 and 1/4; the greater the deviation from zero, the greater the linkage. The sign is simply a consequence of how the alleles are numbered. Linkage disequilibrium can be the result of physical proximity of the genes. Or, it can be maintained by natural selection if some combinations of alleles work better as a team.
Natural selection maintains the linkage disequilibrium between color and pattern alleles in Papilio memnon. [linkage disequilibrium: association between alleles at different loci] In this moth species, there is a gene that determines wing morphology. One allele at this locus leads to a moth that has a tail; the other allele codes for a untailed moth. There is another gene that determines if the wing is brightly or darkly colored. There are thus four possible types of moths: brightly colored moths with and without tails, and dark moths with and without tails. All four can be produced when moths are brought into the lab and bred. However, only two of these types of moths are found in the wild: brightly colored moths with tails and darkly colored moths without tails. The non-random association is maintained by natural selection. Bright, tailed moths mimic the pattern of an unpalatable species. The dark morph is cryptic. The other two combinations are neither mimetic nor cryptic and are quickly eaten by birds.
Assortative mating causes a non-random distribution of alleles at a single locus. [locus: location of a gene on a chromosome] If there are two alleles (A and a) at a locus with frequencies p and q, the frequency of the three possible genotypes (AA, Aa and aa) will be p2, 2pq and q2, respectively. For example, if the frequency of A is 0.9 and the frequency of a is 0.1, the frequencies of AA, Aa and aa individuals are: 0.81, 0.18 and 0.01. This distribution is called the Hardy-Weinberg equilibrium.
Non-random mating results in a deviation from the Hardy-Weinberg distribution. Humans mate assortatively according to race; we are more likely to mate with someone of own race than another. In populations that mate this way, fewer heterozygotes are found than would be predicted under random mating. [heterozygote: an organism that has two different alleles at a locus] A decrease in heterozygotes can be the result of mate choice, or simply the result of population subdivision. Most organisms have a limited dispersal capability, so their mate will be chosen from the local population.
Evolution within a Lineage
In order for continuing evolution there must be mechanisms to increase or create genetic variation and mechanisms to decrease it. The mechanisms of evolution are mutation, natural selection, genetic drift, recombination and gene flow. I have grouped them into two classes -- those that decrease genetic variation and those that increase it.
Mechanisms that Decrease Genetic Variation
Natural Selection
Some types of organisms within a population leave more offspring than others. Over time, the frequency of the more prolific type will increase. The difference in reproductive capability is called natural selection. Natural selection is the only mechanism of adaptive evolution; it is defined as differential reproductive success of pre- existing classes of genetic variants in the gene pool.
The most common action of natural selection is to remove unfit variants as they arise via mutation. [natural selection: differential reproductive success of genotypes] In other words, natural selection usually prevents new alleles from increasing in frequency. This led a famous evolutionist, George Williams, to say "Evolution proceeds in spite of natural selection."
Natural selection can maintain or deplete genetic variation depending on how it acts. When selection acts to weed out deleterious alleles, or causes an allele to sweep to fixation, it depletes genetic variation. When heterozygotes are more fit than either of the homozygotes, however, selection causes genetic variation to be maintained. [heterozygote: an organism that has two different alleles at a locus. | homozygote: an organism that has two identical alleles at a locus] This is called balancing selection. An example of this is the maintenance of sickle-cell alleles in human populations subject to malaria. Variation at a single locus determines whether red blood cells are shaped normally or sickled. If a human has two alleles for sickle-cell, he/she develops anemia -- the shape of sickle-cells precludes them carrying normal levels of oxygen. However, heterozygotes who have one copy of the sickle-cell allele, coupled with one normal allele enjoy some resistance to malaria -- the shape of sickled cells make it harder for the plasmodia (malaria causing agents) to enter the cell. Thus, individuals homozygous for the normal allele suffer more malaria than heterozygotes. Individuals homozygous for the sickle- cell are anemic. Heterozygotes have the highest fitness of these three types. Heterozygotes pass on both sickle-cell and normal alleles to the next generation. Thus, neither allele can be eliminated from the gene pool. The sickle-cell allele is at its highest frequency in regions of Africa where malaria is most pervasive.
Balancing selection is rare in natural populations. [balancing selection: selection favoring heterozygotes] Only a handful of other cases beside the sickle-cell example have been found. At one time population geneticists thought balancing selection could be a general explanation for the levels of genetic variation found in natural populations. That is no longer the case. Balancing selection is only rarely found in natural populations. And, there are theoretical reasons why natural selection cannot maintain polymorphisms at several loci via balancing selection.
Individuals are selected. The example I gave earlier was an example of evolution via natural selection. [natural selection: differential reproductive success of genotypes] Dark colored moths had a higher reproductive success because light colored moths suffered a higher predation rate. The decline of light colored alleles was caused by light colored individuals being removed from the gene pool (selected against). Individual organisms either reproduce or fail to reproduce and are hence the unit of selection. One way alleles can change in frequency is to be housed in organisms with different reproductive rates. Genes are not the unit of selection (because their success depends on the organism's other genes as well); neither are groups of organisms a unit of selection. There are some exceptions to this "rule," but it is a good generalization.
Organisms do not perform any behaviors that are for the good of their species. An individual organism competes primarily with others of it own species for its reproductive success. Natural selection favors selfish behavior because any truly altruistic act increases the recipient's reproductive success while lowering the donors. Altruists would disappear from a population as the non- altruists would reap the benefits, but not pay the costs, of altruistic acts. Many behaviors appear altruistic. Biologists, however, can demonstrate that these behaviors are only apparently altruistic. Cooperating with or helping other organisms is often the most selfish strategy for an animal. This is called reciprocal altruism. A good example of this is blood sharing in vampire bats. In these bats, those lucky enough to find a meal will often share part of it with an unsuccessful bat by regurgitating some blood into the other's mouth. Biologists have found that these bats form bonds with partners and help each other out when the other is needy. If a bat is found to be a "cheater," (he accepts blood when starving, but does not donate when his partner is) his partner will abandon him. The bats are thus not helping each other altruistically; they form pacts that are mutually beneficial.
Helping closely related organisms can appear altruistic; but this is also a selfish behavior. Reproductive success (fitness) has two components; direct fitness and indirect fitness. Direct fitness is a measure of how many alleles, on average, a genotype contributes to the subsequent generation's gene pool by reproducing. Indirect fitness is a measure of how many alleles identical to its own it helps to enter the gene pool. Direct fitness plus indirect fitness is inclusive fitness. J. B. S. Haldane once remarked he would gladly drown, if by doing so he saved two siblings or eight cousins. Each of his siblings would share one half his alleles; his cousins, one eighth. They could potentially add as many of his alleles to the gene pool as he could.
Natural selection favors traits or behaviors that increase a genotype's inclusive fitness. Closely related organisms share many of the same alleles. In diploid species, siblings share on average at least 50% of their alleles. The percentage is higher if the parents are related. So, helping close relatives to reproduce gets an organism's own alleles better represented in the gene pool. The benefit of helping relatives increases dramatically in highly inbred species. In some cases, organisms will completely forgo reproducing and only help their relatives reproduce. Ants, and other eusocial insects, have sterile castes that only serve the queen and assist her reproductive efforts. The sterile workers are reproducing by proxy.
The words selfish and altruistic have connotations in everyday use that biologists do not intend. Selfish simply means behaving in such a way that one's own inclusive fitness is maximized; altruistic means behaving in such a way that another's fitness is increased at the expense of ones' own. Use of the words selfish and altruistic is not meant to imply that organisms consciously understand their motives.
The opportunity for natural selection to operate does not induce genetic variation to appear -- selection only distinguishes between existing variants. Variation is not possible along every imaginable axis, so all possible adaptive solutions are not open to populations. To pick a somewhat ridiculous example, a steel shelled turtle might be an improvement over regular turtles. Turtles are killed quite a bit by cars these days because when confronted with danger, they retreat into their shells -- this is not a great strategy against a two ton automobile. However, there is no variation in metal content of shells, so it would not be possible to select for a steel shelled turtle.
Here is a second example of natural selection. Geospiza fortis lives on the Galapagos islands along with fourteen other finch species. It feeds on the seeds of the plant Tribulus cistoides, specializing on the smaller seeds. Another species, G. Magnirostris, has a larger beak and specializes on the larger seeds. The health of these bird populations depends on seed production. Seed production, in turn, depends on the arrival of wet season. In 1977, there was a drought. Rainfall was well below normal and fewer seeds were produced. As the season progressed, the G. fortis population depleted the supply of small seeds. Eventually, only larger seeds remained. Most of the finches starved; the population plummeted from about twelve hundred birds to less than two hundred. Peter Grant, who had been studying these finches, noted that larger beaked birds fared better than smaller beaked ones. These larger birds had offspring with correspondingly large beaks. Thus, there was an increase in the proportion of large beaked birds in the population the next generation. To prove that the change in bill size in Geospiza fortis was an evolutionary change, Grant had to show that differences in bill size were at least partially genetically based. He did so by crossing finches of various beak sizes and showing that a finch's beak size was influenced by its parent's genes. Large beaked birds had large beaked offspring; beak size was not due to environmental differences (in parental care, for example).
Natural selection may not lead a population to have the optimal set of traits. In any population, there would be a certain combination of possible alleles that would produce the optimal set of traits (the global optimum); but there are other sets of alleles that would yield a population almost as adapted (local optima). Transition from a local optimum to the global optimum may be hindered or forbidden because the population would have to pass through less adaptive states to make the transition. Natural selection only works to bring populations to the nearest optimal point. This idea is Sewall Wright's adaptive landscape. This is one of the most influential models that shape how evolutionary biologists view evolution.
Natural selection does not have any foresight. It only allows organisms to adapt to their current environment. Structures or behaviors do not evolve for future utility. An organism adapts to its environment at each stage of its evolution. As the environment changes, new traits may be selected for. Large changes in populations are the result of cumulative natural selection. Changes are introduced into the population by mutation; the small minority of these changes that result in a greater reproductive output of their bearers are amplified in frequency by selection.
Complex traits must evolve through viable intermediates. For many traits, it initially seems unlikely that intermediates would be viable. What good is half a wing? Half a wing may be no good for flying, but it may be useful in other ways. Feathers are thought to have evolved as insulation (ever worn a down jacket?) and/or as a way to trap insects. Later, proto-birds may have learned to glide when leaping from tree to tree. Eventually, the feathers that originally served as insulation now became co-opted for use in flight. A trait's current utility is not always indicative of its past utility. It can evolve for one purpose, and be used later for another. A trait evolved for its current utility is an adaptation; one that evolved for another utility is an exaptation. An example of an exaptation is a penguin's wing. Penguins evolved from flying ancestors; now they are flightless and use their wings for swimming.
Common Misconceptions about Selection
Selection is not a force in the sense that gravity or the strong nuclear force is. However, for the sake of brevity, biologists sometimes refer to it that way. This often leads to some confusion when biologists speak of selection "pressures." This implies that the environment "pushes" a population to more adapted state. This is not the case. Selection merely favors beneficial genetic changes when they occur by chance -- it does not contribute to their appearance. The potential for selection to act may long precede the appearance of selectable genetic variation. When selection is spoken of as a force, it often seems that it is has a mind of its own; or as if it was nature personified. This most often occurs when biologists are waxing poetic about selection. This has no place in scientific discussions of evolution. Selection is not a guided or cognizant entity; it is simply an effect.
A related pitfall in discussing selection is anthropomorphizing on behalf of living things. Often conscious motives are seemingly imputed to organisms, or even genes, when discussing evolution. This happens most frequently when discussing animal behavior. Animals are often said to perform some behavior because selection will favor it. This could more accurately worded as "animals that, due to their genetic composition, perform this behavior tend to be favored by natural selection relative to those who, due to their genetic composition, don't." Such wording is cumbersome. To avoid this, biologists often anthropomorphize. This is unfortunate because it often makes evolutionary arguments sound silly. Keep in mind this is only for convenience of expression.
The phrase "survival of the fittest" is often used synonymously with natural selection. The phrase is both incomplete and misleading. For one thing, survival is only one component of selection -- and perhaps one of the less important ones in many populations. For example, in polygynous species, a number of males survive to reproductive age, but only a few ever mate. Males may differ little in their ability to survive, but greatly in their ability to attract mates -- the difference in reproductive success stems mainly from the latter consideration. Also, the word fit is often confused with physically fit. Fitness, in an evolutionary sense, is the average reproductive output of a class of genetic variants in a gene pool. Fit does not necessarily mean biggest, fastest or strongest.
Sexual Selection
In many species, males develop prominent secondary sexual characteristics. A few oft cited examples are the peacock's tail, coloring and patterns in male birds in general, voice calls in frogs and flashes in fireflies. Many of these traits are a liability from the standpoint of survival. Any ostentatious trait or noisy, attention getting behavior will alert predators as well as potential mates. How then could natural selection favor these traits?
Natural selection can be broken down into many components, of which survival is only one. Sexual attractiveness is a very important component of selection, so much so that biologists use the term sexual selection when they talk about this subset of natural selection.
Sexual selection is natural selection operating on factors that contribute to an organism's mating success. Traits that are a liability to survival can evolve when the sexual attractiveness of a trait outweighs the liability incurred for survival. A male who lives a short time, but produces many offspring is much more successful than a long lived one that produces few. The former's genes will eventually dominate the gene pool of his species. In many species, especially polygynous species where only a few males monopolize all the females, sexual selection has caused pronounced sexual dimorphism. In these species males compete against other males for mates. The competition can be either direct or mediated by female choice. In species where females choose, males compete by displaying striking phenotypic characteristics and/or performing elaborate courtship behaviors. The females then mate with the males that most interest them, usually the ones with the most outlandish displays. There are many competing theories as to why females are attracted to these displays.
The good genes model states that the display indicates some component of male fitness. A good genes advocate would say that bright coloring in male birds indicates a lack of parasites. The females are cueing on some signal that is correlated with some other component of viability.
Selection for good genes can be seen in sticklebacks. In these fish, males have red coloration on their sides. Milinski and Bakker showed that intensity of color was correlated to both parasite load and sexual attractiveness. Females preferred redder males. The redness indicated that he was carrying fewer parasites.
Evolution can get stuck in a positive feedback loop. Another model to explain secondary sexual characteristics is called the runaway sexual selection model. R. A. Fisher proposed that females may have an innate preference for some male trait before it appears in a population. Females would then mate with male carriers when the trait appears. The offspring of these matings have the genes for both the trait and the preference for the trait. As a result, the process snowballs until natural selection brings it into check. Suppose that female birds prefer males with longer than average tail feathers. Mutant males with longer than average feathers will produce more offspring than the short feathered males. In the next generation, average tail length will increase. As the generations progress, feather length will increase because females do not prefer a specific length tail, but a longer than average tail. Eventually tail length will increase to the point were the liability to survival is matched by the sexual attractiveness of the trait and an equilibrium will be established. Note that in many exotic birds male plumage is often very showy and many species do in fact have males with greatly elongated feathers. In some cases these feathers are shed after the breeding season.
None of the above models are mutually exclusive. There are millions of sexually dimorphic species on this planet and the forms of sexual selection probably vary amongst them.
Genetic Drift
Allele frequencies can change due to chance alone. This is called genetic drift. Drift is a binomial sampling error of the gene pool. What this means is, the alleles that form the next generation's gene pool are a sample of the alleles from the current generation. When sampled from a population, the frequency of alleles differs slightly due to chance alone.
Alleles can increase or decrease in frequency due to drift. The average expected change in allele frequency is zero, since increasing or decreasing in frequency is equally probable. A small percentage of alleles may continually change frequency in a single direction for several generations just as flipping a fair coin may, on occasion, result in a string of heads or tails. A very few new mutant alleles can drift to fixation in this manner.
In small populations, the variance in the rate of change of allele frequencies is greater than in large populations. However, the overall rate of genetic drift (measured in substitutions per generation) is independent of population size. [genetic drift: a random change in allele frequencies] If the mutation rate is constant, large and small populations lose alleles to drift at the same rate. This is because large populations will have more alleles in the gene pool, but they will lose them more slowly. Smaller populations will have fewer alleles, but these will quickly cycle through. This assumes that mutation is constantly adding new alleles to the gene pool and selection is not operating on any of these alleles.
Sharp drops in population size can change allele frequencies substantially. When a population crashes, the alleles in the surviving sample may not be representative of the precrash gene pool. This change in the gene pool is called the founder effect, because small populations of organisms that invade a new territory (founders) are subject to this. Many biologists feel the genetic changes brought about by founder effects may contribute to isolated populations developing reproductive isolation from their parent populations. In sufficiently small populations, genetic drift can counteract selection. [genetic drift: a random change in allele frequencies] Mildly deleterious alleles may drift to fixation.
Wright and Fisher disagreed on the importance of drift. Fisher thought populations were sufficiently large that drift could be neglected. Wright argued that populations were often divided into smaller subpopulations. Drift could cause allele frequency differences between subpopulations if gene flow was small enough. If a subpopulation was small enough, the population could even drift through fitness valleys in the adaptive landscape. Then, the subpopulation could climb a larger fitness hill. Gene flow out of this subpopulation could contribute to the population as a whole adapting. This is Wright's Shifting Balance theory of evolution.
Both natural selection and genetic drift decrease genetic variation. If they were the only mechanisms of evolution, populations would eventually become homogeneous and further evolution would be impossible. There are, however, mechanisms that replace variation depleted by selection and drift. These are discussed below.
Mechanisms that Increase Genetic Variation
Mutation
The cellular machinery that copies DNA sometimes makes mistakes. These mistakes alter the sequence of a gene. This is called a mutation. There are many kinds of mutations. A point mutation is a mutation in which one "letter" of the genetic code is changed to another. Lengths of DNA can also be deleted or inserted in a gene; these are also mutations. Finally, genes or parts of genes can become inverted or duplicated. Typical rates of mutation are between 10-10 and 10-12 mutations per base pair of DNA per generation.
Most mutations are thought to be neutral with regards to fitness. (Kimura defines neutral as |s| < 1/2Ne, where s is the selective coefficient and Ne is the effective population size.) Only a small portion of the genome of eukaryotes contains coding segments. And, although some non-coding DNA is involved in gene regulation or other cellular functions, it is probable that most base changes would have no fitness consequence.
Most mutations that have any phenotypic effect are deleterious. Mutations that result in amino acid substitutions can change the shape of a protein, potentially changing or eliminating its function. This can lead to inadequacies in biochemical pathways or interfere with the process of development. Organisms are sufficiently integrated that most random changes will not produce a fitness benefit. Only a very small percentage of mutations are beneficial. The ratio of neutral to deleterious to beneficial mutations is unknown and probably varies with respect to details of the locus in question and environment.
Mutation limits the rate of evolution. The rate of evolution can be expressed in terms of nucleotide substitutions in a lineage per generation. Substitution is the replacement of an allele by another in a population. This is a two step process: First a mutation occurs in an individual, creating a new allele. This allele subsequently increases in frequency to fixation in the population. The rate of evolution is k = 2Nvu (in diploids) where k is nucleotide substitutions, N is the effective population size, v is the rate of mutation and u is the proportion of mutants that eventually fix in the population.
Mutation need not be limiting over short time spans. The rate of evolution expressed above is given as a steady state equation; it assumes the system is at equilibrium. Given the time frames for a single mutant to fix, it is unclear if populations are ever at equilibrium. A change in environment can cause previously neutral alleles to have selective values; in the short term evolution can run on "stored" variation and thus is independent of mutation rate. Other mechanisms can also contribute selectable variation. Recombination creates new combinations of alleles (or new alleles) by joining sequences with separate microevolutionary histories within a population. Gene flow can also supply the gene pool with variants. Of course, the ultimate source of these variants is mutation.
The Fate of Mutant Alleles
Mutation creates new alleles. Each new allele enters the gene pool as a single copy amongst many. Most are lost from the gene pool, the organism carrying them fails to reproduce, or reproduces but does not pass on that particular allele. A mutant's fate is shared with the genetic background it appears in. A new allele will initially be linked to other loci in its genetic background, even loci on other chromosomes. If the allele increases in frequency in the population, initially it will be paired with other alleles at that locus -- the new allele will primarily be carried in individuals heterozygous for that locus. The chance of it being paired with itself is low until it reaches intermediate frequency. If the allele is recessive, its effect won't be seen in any individual until a homozygote is formed. The eventual fate of the allele depends on whether it is neutral, deleterious or beneficial.
Neutral alleles
Most neutral alleles are lost soon after they appear. The average time (in generations) until loss of a neutral allele is 2(Ne/N) ln(2N) where N is the effective population size (the number of individuals contributing to the next generation's gene pool) and N is the total population size. Only a small percentage of alleles fix. Fixation is the process of an allele increasing to a frequency at or near one. The probability of a neutral allele fixing in a population is equal to its frequency. For a new mutant in a diploid population, this frequency is 1/2N.
If mutations are neutral with respect to fitness, the rate of substitution (k) is equal to the rate of mutation(v). This does not mean every new mutant eventually reaches fixation. Alleles are added to the gene pool by mutation at the same rate they are lost to drift. For neutral alleles that do fix, it takes an average of 4N generations to do so. However, at equilibrium there are multiple alleles segregating in the population. In small populations, few mutations appear each generation. The ones that fix do so quickly relative to large populations. In large populations, more mutants appear over the generations. But, the ones that fix take much longer to do so. Thus, the rate of neutral evolution (in substitutions per generation) is independent of population size.
The rate of mutation determines the level of heterozygosity at a locus according to the neutral theory. Heterozygosity is simply the proportion of the population that is heterozygous. Equilibrium heterozygosity is given as H = 4Nv/[4Nv+1] (for diploid populations). H can vary from a very small number to almost one. In small populations, H is small (because the equation is approximately a very small number divided by one). In (biologically unrealistically) large populations, heterozygosity approaches one (because the equation is approximately a large number divided by itself). Directly testing this model is difficult because N and v can only be estimated for most natural populations. But, heterozygosities are believed to be too low to be described by a strictly neutral model. Solutions offered by neutralists for this discrepancy include hypothesizing that natural populations may not be at equilibrium.
At equilibrium there should be a few alleles at intermediate frequency and many at very low frequencies. This is the Ewens- Watterson distribution. New alleles enter a population every generation, most remain at low frequency until they are lost. A few drift to intermediate frequencies, a very few drift all the way to fixation. In Drosophila pseudoobscura, the protein Xanthine dehydrogenase (Xdh) has many variants. In a single population, Keith, et. al., found that 59 of 96 proteins were of one type, two others were represented ten and nine times and nine other types were present singly or in low numbers.
Deleterious alleles
Deleterious mutants are selected against but remain at low frequency in the gene pool. In diploids, a deleterious recessive mutant may increase in frequency due to drift. Selection cannot see it when it is masked by a dominant allele. Many disease causing alleles remain at low frequency for this reason. People who are carriers do not suffer the negative effect of the allele. Unless they mate with another carrier, the allele may simply continue to be passed on. Deleterious alleles also remain in populations at a low frequency due to a balance between recurrent mutation and selection. This is called the mutation load.
Beneficial alleles
Most new mutants are lost, even beneficial ones. Wright calculated that the probability of fixation of a beneficial allele is 2s. (This assumes a large population size, a small fitness benefit, and that heterozygotes have an intermediate fitness. A benefit of 2s yields an overall rate of evolution: k=4Nvs where v is the mutation rate to beneficial alleles) An allele that conferred a one percent increase in fitness only has a two percent chance of fixing. The probability of fixation of beneficial type of mutant is boosted by recurrent mutation. The beneficial mutant may be lost several times, but eventually it will arise and stick in a population. (Recall that even deleterious mutants recur in a population.)
Directional selection depletes genetic variation at the selected locus as the fitter allele sweeps to fixation. Sequences linked to the selected allele also increase in frequency due to hitchhiking. The lower the rate of recombination, the larger the window of sequence that hitchhikes. Begun and Aquadro compared the level of nucleotide polymorphism within and between species with the rate of recombination at a locus. Low levels of nucleotide polymorphism within species coincided with low rates of recombination. This could be explained by molecular mechanisms if recombination itself was mutagenic. In this case, recombination with also be correlated with nucleotide divergence between species. But, the level of sequence divergence did not correlate with the rate of recombination. Thus, they inferred that selection was the cause. The correlation between recombination and nucleotide polymorphism leaves the conclusion that selective sweeps occur often enough to leave an imprint on the level of genetic variation in natural populations.
One example of a beneficial mutation comes from the mosquito Culex pipiens. In this organism, a gene that was involved with breaking down organophosphates - common insecticide ingredients -became duplicated. Progeny of the organism with this mutation quickly swept across the worldwide mosquito population. There are numerous examples of insects developing resistance to chemicals, especially DDT which was once heavily used in this country. And, most importantly, even though "good" mutations happen much less frequently than "bad" ones, organisms with "good" mutations thrive while organisms with "bad" ones die out.
If beneficial mutants arise infrequently, the only fitness differences in a population will be due to new deleterious mutants and the deleterious recessives. Selection will simply be weeding out unfit variants. Only occasionally will a beneficial allele be sweeping through a population. The general lack of large fitness differences segregating in natural populations argues that beneficial mutants do indeed arise infrequently. However, the impact of a beneficial mutant on the level of variation at a locus can be large and lasting. It takes many generations for a locus to regain appreciable levels of heterozygosity following a selective sweep.
Recombination
Each chromosome in our sperm or egg cells is a mixture of genes from our mother and our father. Recombination can be thought of as gene shuffling. Most organisms have linear chromosomes and their genes lie at specific location (loci) along them. Bacteria have circular chromosomes. In most sexually reproducing organisms, there are two of each chromosome type in every cell. For instance in humans, every chromosome is paired, one inherited from the mother, the other inherited from the father. When an organism produces gametes, the gametes end up with only one of each chromosome per cell. Haploid gametes are produced from diploid cells by a process called meiosis.
In meiosis, homologous chromosomes line up. The DNA of the chromosome is broken on both chromosomes in several places and rejoined with the other strand. Later, the two homologous chromosomes are split into two separate cells that divide and become gametes. But, because of recombination, both of the chromosomes are a mix of alleles from the mother and father.
Recombination creates new combinations of alleles. Alleles that arose at different times and different places can be brought together. Recombination can occur not only between genes, but within genes as well. Recombination within a gene can form a new allele. Recombination is a mechanism of evolution because it adds new alleles and combinations of alleles to the gene pool.
Gene Flow
New organisms may enter a population by migration from another population. If they mate within the population, they can bring new alleles to the local gene pool. This is called gene flow. In some closely related species, fertile hybrids can result from interspecific matings. These hybrids can vector genes from species to species.
Gene flow between more distantly related species occurs infrequently. This is called horizontal transfer. One interesting case of this involves genetic elements called P elements. Margaret Kidwell found that P elements were transferred from some species in the Drosophila willistoni group to Drosophila melanogaster. These two species of fruit flies are distantly related and hybrids do not form. Their ranges do, however, overlap. The P elements were vectored into D. melanogaster via a parasitic mite that targets both these species. This mite punctures the exoskeleton of the flies and feeds on the "juices". Material, including DNA, from one fly can be transferred to another when the mite feeds. Since P elements actively move in the genome (they are themselves parasites of DNA), one incorporated itself into the genome of a melanogaster fly and subsequently spread through the species. Laboratory stocks of melanogaster caught prior to the 1940's lack of P elements. All natural populations today harbor them.
Overview of Evolution within a Lineage
Evolution is a change in the gene pool of a population over time; it can occur due to several factors. Three mechanisms add new alleles to the gene pool: mutation, recombination and gene flow. Two mechanisms remove alleles, genetic drift and natural selection. Drift removes alleles randomly from the gene pool. Selection removes deleterious alleles from the gene pool. The amount of genetic variation found in a population is the balance between the actions of these mechanisms.
Natural selection can also increase the frequency of an allele. Selection that weeds out harmful alleles is called negative selection. Selection that increases the frequency of helpful alleles is called positive, or sometimes positive Darwinian, selection. A new allele can also drift to high frequency. But, since the change in frequency of an allele each generation is random, nobody speaks of positive or negative drift.
Except in rare cases of high gene flow, new alleles enter the gene pool as a single copy. Most new alleles added to the gene pool are lost almost immediately due to drift or selection; only a small percent ever reach a high frequency in the population. Even most moderately beneficial alleles are lost due to drift when they appear. But, a mutation can reappear numerous times.
The fate of any new allele depends a great deal on the organism it appears in. This allele will be linked to the other alleles near it for many generations. A mutant allele can increase in frequency simply because it is linked to a beneficial allele at a nearby locus. This can occur even if the mutant allele is deleterious, although it must not be so deleterious as to offset the benefit of the other allele. Likewise a potentially beneficial new allele can be eliminated from the gene pool because it was linked to deleterious alleles when it first arose. An allele "riding on the coat tails" of a beneficial allele is called a hitchhiker. Eventually, recombination will bring the two loci to linkage equilibrium. But, the more closely linked two alleles are, the longer the hitchhiking will last.
The effects of selection and drift are coupled. Drift is intensified as selection pressures increase. This is because increased selection (i.e. a greater difference in reproductive success among organisms in a population) reduces the effective population size, the number of individuals contributing alleles to the next generation.
Adaptation is brought about by cumulative natural selection, the repeated sifting of mutations by natural selection. Small changes, favored by selection, can be the stepping-stone to further changes. The summation of large numbers of these changes is macroevolution.
The Development of Evolutionary Theory
Biology came of age as a science when Charles Darwin published "On the Origin of Species." But, the idea of evolution wasn't new to Darwin. Lamarck published a theory of evolution in 1809. Lamarck thought that species arose continually from nonliving sources. These species were initially very primitive, but increased in complexity over time due to some inherent tendency. This type of evolution is called orthogenesis. Lamarck proposed that an organism's acclimation to the environment could be passed on to its offspring. For example, he thought proto-giraffes stretched their necks to reach higher twigs. This caused their offspring to be born with longer necks. This proposed mechanism of evolution is called the inheritance of acquired characteristics. Lamarck also believed species never went extinct, although they may change into newer forms. All three of these ideas are now known to be wrong.
Darwin's contributions include hypothesizing the pattern of common descent and proposing a mechanism for evolution -- natural selection. In Darwin's theory of natural selection, new variants arise continually within populations. A small percentage of these variants cause their bearers to produce more offspring than others. These variants thrive and supplant their less productive competitors. The effect of numerous instances of selection would lead to a species being modified over time.
Darwin's theory did not accord with older theories of genetics. In Darwin's time, biologists held to the theory of blending inheritance -- an offspring was an average of its parents. If an individual had one short parent and one tall parent, it would be of medium height. And, the offspring would pass on genes for medium sized offspring. If this was the case, new genetic variations would quickly be diluted out of a population. They could not accumulate as the theory of evolution required. We now know that the idea of blending inheritance is wrong.
Darwin didn't know that the true mode of inheritance was discovered in his lifetime. Gregor Mendel, in his experiments on hybrid peas, showed that genes from a mother and father do not blend. An offspring from a short and a tall parent may be medium sized; but it carries genes for shortness and tallness. The genes remain distinct and can be passed on to subsequent generations. Mendel mailed his paper to Darwin, but Darwin never opened it.
It was a long time until Mendel's ideas were accepted. One group of biologists, called biometricians, thought Mendel's laws only held for a few traits. Most traits, they claimed, were governed by blending inheritance. Mendel studied discrete traits. Two of the traits in his famous experiments were smooth versus wrinkled coat on peas. This trait did not vary continuously. In other words, peas are either wrinkled or smooth -- intermediates are not found. Biometricians considered these traits aberrations. They studied continuously varying traits like size and believed most traits showed blending inheritance. ## Incorporating Genetics into Evolutionary Theory
The discrete genes Mendel discovered would exist at some frequency in natural populations. Biologists wondered how and if these frequencies would change. Many thought that the more common versions of genes would increase in frequency simply because they were already at high frequency.
Hardy and Weinberg independently showed that the frequency of an allele would not change over time simply due to its being rare or common. Their model had several assumptions -- that all alleles reproduced at the same rate, that the population size was very large and that alleles did not change in form. Later, R. A. Fisher showed that Mendel's laws could explain continuous traits if the expression of these traits were due to the action of many genes. After this, geneticists accepted Mendel's Laws as the basic rules of genetics. From this basis, Fisher, Sewall Wright and J. B. S.. Haldane founded the field of population genetics. Population genetics is a field of biology that attempts to measure and explain the levels of genetic variation in populations.
R. A. Fisher studied the effect of natural selection on large populations. He demonstrated that even very small selective differences amongst alleles could cause appreciable changes in allele frequencies over time. He also showed that the rate of adaptive change in a population is proportional to the amount of genetic variation present. This is called Fisher's Fundamental Theorem of Natural Selection. Although it is called the fundamental theorem, it does not hold in all cases. The rate at which natural selection brings about adaptation depends on the details of how selection is working. In some rare cases, natural selection can actually cause a decline in the mean relative fitness of a population.
Sewall Wright was more concerned with drift. He stressed that large populations are often subdivided into many subpopulations. In his theory, genetic drift played a more important role compared to selection. Differentiation between subpopulations, followed by migration among them, could contribute to adaptations amongst populations. Wright also came up with the idea of the adaptive landscape -- an idea that remains influential to this day. Its influence remains even though P. A. P. Moran has shown that, mathematically, adaptive landscapes don't exist as Wright envisioned them. Wright extended his results of one-locus models to a two-locus case in proposing the adaptive landscape. But, unbeknownst to him, the general conclusions of the one-locus model don't extend to the two-locus case.
J. B. S. Haldane developed many of the mathematical models of natural and artificial selection. He showed that selection and mutation could oppose each other, that deleterious mutations could remain in a population due to recurrent mutation. He also demonstrated that there was a cost to natural selection, placing a limit on the amount of adaptive substitutions a population could undergo in a given time frame.
For a long time, population genetics developed as a theoretical field. But, gathering the data needed to test the theories was nearly impossible. Prior to the advent of molecular biology, estimates of genetic variability could only be inferred from levels of morphological differences in populations. Lewontin and Hubby were the first to get a good estimate of genetic variation in a population. Using the then new technique of protein electrophoresis, they showed that 30% of the loci in a population of Drosophila pseudoobscura were polymorphic. They also showed that it was likely that they could not detect all the variation that was present. Upon finding this level of variation, the question became -- was this maintained by natural selection, or simply the result of genetic drift? This level of variation was too high to be explained by balancing selection.
Motoo Kimura theorized that most variation found in populations was selectively equivalent (neutral). Multiple alleles at a locus differed in sequence, but their fitnesses were the same. Kimura's neutral theory described rates of evolution and levels of polymorphism solely in terms of mutation and genetic drift. The neutral theory did not deny that natural selection acted on natural populations; but it claimed that the majority of natural variation was transient polymorphisms of neutral alleles. Selection did not act frequently or strongly enough to influence rates of evolution or levels of polymorphism.
Initially, a wide variety of observations seemed to be consistent with the neutral theory. Eventually, however, several lines of evidence toppled it. There is less variation in natural populations than the neutral theory predicts. Also, there is too much variance in rates of substitutions in different lineages to be explained by mutation and drift alone. Finally, selection itself has been shown to have an impact on levels of nucleotide variation. Currently, there is no comprehensive mathematical theory of evolution that accurately predicts rates of evolution and levels of heterozygosity in natural populations.
Evolution Among Lineages
The Pattern of Macroevolution
Evolution is not progress. The popular notion that evolution can be represented as a series of improvements from simple cells, through more complex life forms, to humans (the pinnacle of evolution), can be traced to the concept of the scale of nature. This view is incorrect.
All species have descended from a common ancestor. As time went on, different lineages of organisms were modified with descent to adapt to their environments. Thus, evolution is best viewed as a branching tree or bush, with the tips of each branch representing currently living species. No living organisms today are our ancestors. Every living species is as fully modern as we are with its own unique evolutionary history. No extant species are "lower life forms," atavistic stepping stones paving the road to humanity.
A related, and common, fallacy about evolution is that humans evolved from some living species of ape. This is not the case -- humans and apes share a common ancestor. Both humans and living apes are fully modern species; the ancestor we evolved from was an ape, but it is now extinct and was not the same as present day apes (or humans for that matter). If it were not for the vanity of human beings, we would be classified as an ape. Our closest relatives are, collectively, the chimpanzee and the pygmy chimp. Our next nearest relative is the gorilla. ## Evidence for Common Descent and Macroevolution
Microevolution can be studied directly. Macroevolution cannot. Macroevolution is studied by examining patterns in biological populations and groups of related organisms and inferring process from pattern. Given the observation of microevolution and the knowledge that the earth is billions of years old -- macroevolution could be postulated. But this extrapolation, in and of itself, does not provide a compelling explanation of the patterns of biological diversity we see today. Evidence for macroevolution, or common ancestry and modification with descent, comes from several other fields of study. These include: comparative biochemical and genetic studies, comparative developmental biology, patterns of biogeography, comparative morphology and anatomy and the fossil record.
Closely related species (as determined by morphologists) have similar gene sequences. Overall sequence similarity is not the whole story, however. The pattern of differences we see in closely related genomes is worth examining.
All living organisms use DNA as their genetic material, although some viruses use RNA. DNA is composed of strings of nucleotides. There are four different kinds of nucleotides: adenine (A), guanine (G), cytosine (C) and thymine (T). Genes are sequences of nucleotides that code for proteins. Within a gene, each block of three nucleotides is called a codon. Each codon designates an amino acid (the subunits of proteins).
The three letter code is the same for all organisms (with a few exceptions). There are 64 codons, but only 20 amino acids to code for; so, most amino acids are coded for by several codons. In many cases the first two nucleotides in the codon designate the amino acid. The third position can have any of the four nucleotides and not effect how the code is translated.
A gene, when in use, is transcribed into RNA -- a nucleic acid similar to DNA. (RNA, like DNA, is made up of nucleotides although t he nucleotide uracil (U) is used in place of thymine (T).) The RNA transcribed from a gene is called messenger RNA. Messenger RNA is then translated via cellular machinery called ribosomes into a string of amino acids -- a protein. Some proteins function as enzymes, catalysts that speed the chemical reactions in cells. Others are structural or involved in regulating development.
Gene sequences in closely related species are very similar. Often, the same codon specifies a given amino acid in two related species, even though alternate codons could serve functionally as well. But, some differences do exist in gene sequences. Most often, differences are in third codon positions, where changes in the DNA sequence would not disrupt the sequence of the protein.
There are other sites in the genome where nucleotide differences do not effect protein sequences. The genome of eukaryotes is loaded with 'dead genes' called pseudogenes. Pseudogenes are copies of working genes that have been inactivated by mutation. Most pseudogenes do not produce full proteins. They may be transcribed, but not translated. Or, they may be translated, but only a truncated protein is produced. Pseudogenes evolve much faster than their working counterparts. Mutations in them do not get incorporated into proteins, so they have no effect on the fitness of an organism.
Introns are sequences of DNA that interrupt a gene, but do not code for anything. The coding portions of a gene are called exons. Introns are spliced out of the messenger RNA prior to translation, so they do not contribute information needed to make the protein. They are sometimes, however, involved in regulation of the gene. Like pseudogenes, introns (in general) evolve faster than coding portions of a gene.
Nucleotide positions that can be changed without changing the sequence of a protein are called silent sites. Sites where changes result in an amino acid substitution are called replacement sites. Silent sites are expected to be more polymorphic within a population and show more differences between populations. Although both silent and replacement sites receive the same amount of mutations, natural selection only infrequently allows changes at replacement sites. Silent sites, however, are not as constrained.
Kreitman was the first demonstrate that silent sites were more variable than coding sites. Shortly after the methods of DNA sequencing were discovered, he sequenced 11 alleles of the enzyme alcohol dehydrogenase (AdH). Of the 43 polymorphic nucleotide sites he found, only one resulted in a change in the amino acid sequence of the protein.
Silent sites may not be entirely selectively neutral. Some DNA sequences are involved with regulation of genes, changes in these sites may be deleterious. Likewise, although several codons code for a single amino acid, an organism may have a preferred codon for each amino acid. This is called codon bias.
If two species shared a recent common ancestor one would expect genetic information, even information such as redundant nucleotides and the position of introns or pseudogenes, to be similar. Both species would have inherited this information from their common ancestor.
The degree of similarity in nucleotide sequence is a function of divergence time. If two populations had recently separated, few differences would have built up between them. If they separated long ago, each population would have evolved numerous differences from their common ancestor (and each other). The degree of similarity would also be a function of silent versus replacement sites. Li and Graur, in their molecular evolution text, give the rates of evolution for silent vs. replacement rates. The rates were estimated from sequence comparisons of 30 genes from humans and rodents, which diverged about 80 million years ago. Silent sites evolved at an average rate of 4.61 nucleotide substitution per site per 109 years. Replacement sites evolved much slower at an average rate of 0.85 nucleotide substitutions per site per 109 years.
Groups of related organisms are 'variations on a theme' -- the same set of bones are used to construct all vertebrates. The bones of the human hand grow out of the same tissue as the bones of a bat's wing or a whale's flipper; and, they share many identifying features such as muscle insertion points and ridges. The only difference is that they are scaled differently. Evolutionary biologists say this indicates that all mammals are modified descendants of a common ancestor which had the same set of bones.
Closely related organisms share similar developmental pathways. The differences in development are most evident at the end. As organisms evolve, their developmental pathway gets modified. An alteration near the end of a developmental pathway is less likely to be deleterious than changes in early development. Changes early on may have a cascading effect. Thus most evolutionary changes in development are expected to take place at the periphery of development, or in early aspects of development that have no later repercussions. For a change in early development to be propagated, the benefit of the early alteration must outweigh the consequences to later development.
Because they have evolved this way, organisms pass through the early stages of development that their ancestors passed through up to the point of divergence. So, an organism's development mimics its ancestors although it doesn't recreate it exactly. Development of the flatfish, Pleuronectes, illustrates this point. Early on, Pleuronectes develops a tail that comes to a point. In the next developmental stage, the top lobe of the tail is larger than the bottom lobe (as in sharks). When development is complete, the upper and lower lobes are equally sized. This developmental pattern mirrors the evolutionary transitions it has undergone.
Natural selection can modify any stage of a life cycle, so some differences are seen in early development. Thus, evolution does not always recapitulate ancestral forms -- butterflies did not evolve from ancestral caterpillars, for example. There are differences in the appearance of early vertebrate embryos. Amphibians rapidly form a ball of cells in early development. Birds, reptiles and mammals form a disk. The shape of the early embryo is a result of different yolk concentrations in the eggs. Birds' and reptiles' eggs are heavily yolked. Their eggs develop similarly to amphibians except the yolk has deformed the shape of the embryo. The ball is stretched out and lying atop the yolk. Mammals have no yolk, but still form a disk early. This is because they have descended from reptiles. Mammals lost their yolky eggs, but retained the early pattern of development. In all these vertebrates, the pattern of cell movements is similar despite superficial differences in appearance. In addition, all types quickly converge upon a primitive, fish-like stage within a few days. From there, development diverges.
Traces of an organism's ancestry sometimes remain even when an organism's development is complete. These are called vestigial structures. Many snakes have rudimentary pelvic bones retained from their walking ancestors. Vestigial does not mean useless, it means the structure is clearly a vestige of an structure inherited from ancestral organisms. Vestigial structures may acquire new functions. In humans, the appendix now houses some immune system cells.
Closely related organisms are usually found in close geographic proximity; this is especially true of organisms with limited dispersal opportunities. The mammalian fauna of Australia is often cited as an example of this; marsupial mammals fill most of the equivalent niches that placentals fill in other ecosystems. If all organisms descended from a common ancestor, species distribution across the planet would be a function of site of origination, potential for dispersal, distribution of suitable habitat, and time since origination. In the case of Australian mammals, their physical separation from sources of placentals means potential niches were filled by a marsupial radiation rather than a placental radiation or invasion.
Natural selection can only mold available genetically based variation. In addition, natural selection provides no mechanism for advance planning. If selection can only tinker with the available genetic variation, we should expect to see examples of jury-rigged design in living species. This is indeed the case. In lizards of the genus Cnemodophorus, females reproduce parthenogenetically. Fertility in these lizards is increased when a female mounts another female and simulates copulation. These lizards evolved from sexual lizards whose hormones were aroused by sexual behavior. Now, although the sexual mode of reproduction has been lost, the means of getting aroused (and hence fertile) has been retained.
Fossils show hard structures of organisms less and less similar to modern organisms in progressively older rocks. In addition, patterns of biogeography apply to fossils as well as extant organisms. When combined with plate tectonics, fossils provide evidence of distributions and dispersals of ancient species. For example, South America had a very distinct marsupial mammalian fauna until the land bridge formed between North and South America. After that marsupials started disappearing and placentals took their place. This is commonly interpreted as the placentals wiping out the marsupials, but this may be an over simplification.
Transitional fossils between groups have been found. One of the most impressive transitional series is the ancient reptile to modern mammal transition. Mammals and reptiles differ in skeletal details, especially in their skulls. Reptilian jaws have four bones. The foremost is called the dentary. In mammals, the dentary bone is the only bone in the lower jaw. The other bones are part of the middle ear. Reptiles have a weak jaw and a mouthful of undifferentiated teeth. Their jaw is closed by three muscles: the external, posterior and internal adductor. Each reptile tooth is single cusped. Mammals have powerful jaws with differentiated teeth. Many of these teeth, such as the molars, are multi-cusped. The temporalis and masseter muscles, derived from the external adductor, close the mammalian jaw. Mammals have a secondary palate, a bony structure separating their nostril passages and throat, so most can swallow and breathe simultaneously. Reptiles lack this.
The evolution of these traits can be seen in a series of fossils. Procynosuchus shows an increase in size of the dentary bone and the beginnings of a palate. Thrinaxodon has a reduced number of incisors, a precursor to tooth differentiation. Cynognathus (a doglike carnivore) shows a further increase in size of the dentary bone. The other three bones are located inside the back portion of the jaw. Some teeth are multicusped and the teeth fit together tightly. Diademodon (a plant eater) shows a more advanced degree of occlusion (teeth fitting tightly). Probelesodon has developed a double joint in the jaw. The jaw could hinge off two points with the upper skull. The front hinge was probably the actual hinge while the rear hinge was an alignment guide. The forward movement of a hinge point allowed for the precursor to the modern masseter muscle to anchor further forward in the jaw. This allowed for a more powerful bite. The first true mammal was Morgonucudon, a rodent-like insectivore from the late Triassic. It had all the traits common to modern mammals. These species were not from a single, unbranched lineage. Each is an example from a group of organisms along the main line of mammalian ancestry.
The strongest evidence for macroevolution comes from the fact that suites of traits in biological entities fall into a nested pattern. For example, plants can be divided into two broad categories, non- vascular (ex. mosses) and vascular. Vascular plants can be divided into seedless (ex. ferns) and seeded. Vascular seeded plants can be divided into gymnosperms (ex. pines) and flowering plants (angiosperms). Angiosperms can be divided into monocots and dicots. Each of these types of plants have several characters that distinguish them from other plants. Traits are not mixed and matched in groups of organisms. For example, flowers are only seen in plants that carry several other characters that distinguish them as angiosperms. This is the expected pattern of common descent. All the species in a group will share traits they inherited from their common ancestor. But, each subgroup will have evolved unique traits of its own. Similarities bind groups together. Differences show how they are subdivided.
The real test of any scientific theory is its ability to generate testable predictions and, of course, have the predictions borne out. Evolution easily meets this criterion. In several of the above examples I stated, closely related organisms share X. If I define closely related as sharing X, this is an empty statement. It does however, provide a prediction. If two organisms share a similar anatomy, one would then predict that their gene sequences would be more similar than a morphologically distinct organism. This has been spectacularly borne out by the recent flood of gene sequences -- the correspondence to trees drawn by morphological data is very high. The discrepancies are never too great and usually confined to cases where the pattern of relationship was debated.
Mechanisms of Macroevolution
The following deals with mechanisms of evolution above the species level.
Speciation -- Increasing Biological Diversity
Speciation is the process of a single species becoming two or more species. Many biologists think speciation is key to understanding evolution. Some would argue that certain evolutionary phenomena apply only at speciation and macroevolutionary change cannot occur without speciation. Other biologists think major evolutionary change can occur without speciation. Changes between lineages are only an extension of the changes within each lineage. In general, paleontologists fall into the former category and geneticists in the latter.
Modes of Speciation
Biologists recognize two types of speciation: allopatric and sympatric speciation. The two differ in geographical distribution of the populations in question. Allopatric speciation is thought to be the most common form of speciation. It occurs when a population is split into two (or more) geographically isolated subdivisions that organisms cannot bridge. Eventually, the two populations' gene pools change independently until they could not interbreed even if they were brought back together. In other words, they have speciated.
Sympatric speciation occurs when two subpopulations become reproductively isolated without first becoming geographically isolated. Insects that live on a single host plant provide a model for sympatric speciation. If a group of insects switched host plants they would not breed with other members of their species still living on their former host plant. The two subpopulations could diverge and speciate. Agricultural records show that a strain of the apple maggot fly Rhagolettis pomenella began infesting apples in the 1860's. Formerly it had only infested hawthorn fruit. Feder, Chilcote and Bush have shown that two races of Rhagolettis pomenella have become behaviorally isolated. Allele frequencies at six loci (aconitase 2, malic enzyme, mannose phosphate isomerase, aspartate amino-transferase, NADH-diaphorase-2, and beta-hydroxy acid dehydrogenase) are diverging. Significant amounts of linkage disequilibrium have been found at these loci, indicating that they may all be hitchhiking on some allele under selection. Some biologists call sympatric speciation microallopatric speciation to emphasize that the subpopulations are still physically separate at an ecological level.
Biologists know little about the genetic mechanisms of speciation. Some think a series of small changes in each subdivision gradually lead to speciation. The founder effect could set the stage for relatively rapid speciation, a genetic revolution in Ernst Mayr's terms. Alan Templeton hypothesized that a few key genes could change and confer reproductive isolation. He called this a genetic transilience. Lynn Margulis thinks most speciation events are caused by changes in internal symbionts. Populations of organisms are very complicated. It is likely that there are many ways speciation can occur. Thus, all of the above ideas may be correct, each in different circumstances. Darwin's book was titled "The Origin of Species" despite the fact that he did not really address this question; over one hundred and fifty years later, how species originate is still largely a mystery.
Observed Speciations
Speciation has been observed. In the plant genus Tragopogon, two new species have evolved within the past 50-60 years. They are T. mirus and T. miscellus. The new species were formed when one diploid species fertilized a different diploid species and produced a tetraploid offspring. This tetraploid offspring could not fertilize or be fertilized by either of its two parent species types. It is reproductively isolated, the definition of a species.
Extinction -- Decreasing Biological Diversity
Ordinary Extinction
Extinction is the ultimate fate of all species. The reasons for extinction are numerous. A species can be competitively excluded by a closely related species, the habitat a species lives in can disappear and/or the organisms that the species exploits could come up with an unbeatable defense.
Some species enjoy a long tenure on the planet while others are short- lived. Some biologists believe species are programmed to go extinct in a manner analogous to organisms being destined to die. The majority, however, believe that if the environment stays fairly constant, a well adapted species could continue to survive indefinitely.
Mass Extinction
Mass extinctions shape the overall pattern of macroevolution. If you view evolution as a branching tree, it's best to picture it as one that has been severely pruned a few times in its life. The history of life on this earth includes many episodes of mass extinction in which many groups of organisms were wiped off the face of the planet. Mass extinctions are followed by periods of radiation where new species evolve to fill the empty niches left behind. It is probable that surviving a mass extinction is largely a function of luck. Thus, contingency plays a large role in patterns of macroevolution.
The largest mass extinction came at the end of the Permian, about 250 million years ago. This coincides with the formation of Pangaea II, when all the world's continents were brought together by plate tectonics. A worldwide drop in sea level also occurred at this time.
The most well-known extinction occurred at the boundary between the Cretaceous and Tertiary Periods. This called the K/T Boundary and is dated at around 65 million years ago. This extinction eradicated the dinosaurs. The K/T event was probably caused by environmental disruption brought on by a large impact of an asteroid with the earth. Following this extinction the mammalian radiation occurred. Mammals coexisted for a long time with the dinosaurs but were confined mostly to nocturnal insectivore niches. With the eradication of the dinosaurs, mammals radiated to fill the vacant niches.
Currently, human alteration of the ecosphere is causing a global mass extinction. ### Punctuated Equilibrium
The theory of punctuated equilibrium is an inference about the process of macroevolution from the pattern of species documented in the fossil record. In the fossil record, transition from one species to another is usually abrupt in most geographic locales -- no transitional forms are found. In short, it appears that species remain unchanged for long stretches of time and then are quickly replaced by new species. However, if wide ranges are searched, transitional forms that bridge the gap between the two species are sometimes found in small, localized areas. For example, in Jurassic brachiopods of the genus Kutchithyris, K. acutiplicata appears below another species, K. euryptycha. Both species were common and covered a wide geographical area. They differ enough that some have argued they should be in a different genera. In just one small locality an approximately 1.25m sedimentary layer with these fossils is found. In the narrow (10 cm) layer that separates the two species, both species are found along with transitional forms. In other localities there is a sharp transition.
Eldredge and Gould proposed that most major morphological change occurs (relatively) quickly in small peripheral population at the time of speciation. New forms will then invade the range of their ancestral species. Thus, at most locations that fossils are found, transition from one species to another will be abrupt. This abrupt change will reflect replacement by migration however, not evolution. In order to find the transitional fossils, the area of speciation must be found.
There has been considerable confusion about the theory. Some popular accounts give the impression that abrupt changes in the fossil record are due to blindingly fast evolution; this is not a part of the theory.
Punctuated equilibrium has been presented as a hierarchical theory of evolution. Proponents of punctuated equilibrium see speciation as analogous to mutation and the replacement of one species by another as analogous to natural selection. This is called species selection. Speciation adds new species to the species pool just as mutation adds new alleles to the gene pool. Species selection favors one species over another just as natural selection can favor one allele over another. Evolutionary trends within a group would be the result of selection among species, not natural selection acting within species. This is the most controversial part of the theory. Many biologists agree with the pattern of macroevolution these paleontologists posit, but believe species selection is not even theoretically likely to occur.
Critics would argue that species selection is not analogous to natural selection and therefore evolution is not hierarchical. Also, the number of species produced over time is far less than the amount of different alleles that enter gene pools over time. So, the amount of adaptive evolution produced by species selection (if it did occur) would have to be orders of magnitude less than adaptive evolution within populations by natural selection.
Tests of punctuated equilibrium have been equivocal. It has been known for a long time that rates of evolution vary over time, that is not controversial. However, phylogenetic studies conflict as to whether there is a clear association between speciation and morphological change. In addition, there are major polymorphisms within some species. For example, bluegill sunfish have two male morphs. One is a large, long-lived, mate-protecting male; the other is a smaller, shorter-lived male who sneaks matings from females guarded by large males. The existence of within species polymorphisms demonstrates that speciation is not a requirement for major morphological change. # A Brief History of Life
Biologists studying evolution do a variety of things: population geneticists study the process as it is occurring; systematists seek to determine relationships between species and paleontologists seek to uncover details of the unfolding of life in the past. Discerning these details is often difficult, but hypotheses can be made and tested as new evidence comes to light. This section should be viewed as the best hypothesis scientists have as to the history of the planet. The material here ranges from some issues that are fairly certain to some topics that are nothing more than informed speculation. For some points there are opposing hypotheses -- I have tried to compile a consensus picture. In general, the more remote the time, the more likely the story is incomplete or in error.
Life evolved in the sea. It stayed there for the majority of the history of earth.
The first replicating molecules were most likely RNA. RNA is a nucleic acid similar to DNA. In laboratory studies it has been shown that some RNA sequences have catalytic capabilities. Most importantly, certain RNA sequences act as polymerases -- enzymes that form strands of RNA from its monomers. This process of self replication is the crucial step in the formation of life. This is called the RNA world hypothesis.
The common ancestor of all life probably used RNA as its genetic material. This ancestor gave rise to three major lineages of life. These are: the prokaryotes ("ordinary" bacteria), archaebacteria (thermophilic, methanogenic and halophilic bacteria) and eukaryotes. Eukaryotes include protists (single celled organisms like amoebas and diatoms and a few multicellular forms such as kelp), fungi (including mushrooms and yeast), plants and animals. Eukaryotes and archaebacteria are the two most closely related of the three. The process of translation (making protein from the instructions on a messenger RNA template) is similar in these lineages, but the organization of the genome and transcription (making messenger RNA from a DNA template) is very different in prokaryotes than in eukaryotes and archaebacteria. Scientists interpret this to mean that the common ancestor was RNA based; it gave rise to two lineages that independently formed a DNA genome and hence independently evolved mechanisms to transcribe DNA into RNA.
The first cells must have been anaerobic because there was no oxygen in the atmosphere. In addition, they were probably thermophilic ("heat-loving") and fermentative. Rocks as old as 3.5 billion years old have yielded prokaryotic fossils. Specifically, some rocks from Australia called the Warrawoona series give evidence of bacterial communities organized into structures called stromatolites. Fossils like these have subsequently been found all over the world. These mats of bacteria still form today in a few locales (for example, Shark Bay Australia). Bacteria are the only life forms found in the rocks for a long, long time --eukaryotes (protists) appear about 1.5 billion years ago and fungi-like things appear about 900 million years ago (0.9 billion years ago).
Photosynthesis evolved around 3.4 billion years ago. Photosynthesis is a process that allows organisms to harness sunlight to manufacture sugar from simpler precursors. The first photosystem to evolve, PSI, uses light to convert carbon dioxide (CO2) and hydrogen sulfide (H2S) to glucose. This process releases sulfur as a waste product. About a billion years later, a second photosystem (PS) evolved, probably from a duplication of the first photosystem. Organisms with PSII use both photosystems in conjunction to convert carbon dioxide (CO2) and water (H2O) into glucose. This process releases oxygen as a waste product. Anoxygenic (or H2S) photosynthesis, using PSI, is seen in living purple and green bacteria. Oxygenic (or H2O) photosynthesis, using PSI and PSII, takes place in cyanobacteria. Cyanobacteria are closely related to and hence probably evolved from purple bacterial ancestors. Green bacteria are an outgroup. Since oxygenic bacteria are a lineage within a cluster of anoxygenic lineages, scientists infer that PSI evolved first. This also corroborates with geological evidence.
Green plants and algae also use both photosystems. In these organisms, photosynthesis occurs in organelles (membrane bound structures within the cell) called chloroplasts. These organelles originated as free living bacteria related to the cyanobacteria that were engulfed by ur-eukaryotes and eventually entered into an endosymbiotic relationship. This endosymbiotic theory of eukaryotic organelles was championed by Lynn Margulis. Originally controversial, this theory is now accepted. One key line of evidence in support of this idea came when the DNA inside chloroplasts was sequenced -- the gene sequences were more similar to free-living cyanobacteria sequences than to sequences from the plants the chloroplasts resided in.
After the advent of photosystem II, oxygen levels increased. Dissolved oxygen in the oceans increased as well as atmospheric oxygen. This is sometimes called the oxygen holocaust. Oxygen is a very good electron acceptor and can be very damaging to living organisms. Many bacteria are anaerobic and die almost immediately in the presence of oxygen. Other organisms, like animals, have special ways to avoid cellular damage due to this element (and in fact require it to live.) Initially, when oxygen began building up in the environment, it was neutralized by materials already present. Iron, which existed in high concentrations in the sea was oxidized and precipitated. Evidence of this can be seen in banded iron formations from this time, layers of iron deposited on the sea floor. As one geologist put it, "the world rusted." Eventually, it grew to high enough concentrations to be dangerous to living things. In response, many species went extinct, some continued (and still continue) to thrive in anaerobic microenvironments and several lineages independently evolved oxygen respiration.
The purple bacteria evolved oxygen respiration by reversing the flow of molecules through their carbon fixing pathways and modifying their electron transport chains. Purple bacteria also enabled the eukaryotic lineage to become aerobic. Eukaryotic cells have membrane bound organelles called mitochondria that take care of respiration for the cell. These are endosymbionts like chloroplasts. Mitochondria formed this symbiotic relationship very early in eukaryotic history, all but a few groups of eukaryotic cells have mitochondria. Later, a few lineages picked up chloroplasts. Chloroplasts have multiple origins. Red algae picked up ur-chloroplasts from the cyanobacterial lineage. Green algae, the group plants evolved from, picked up different urchloroplasts from a prochlorophyte, a lineage closely related to cyanobacteria.
Animals start appearing prior to the Cambrian, about 600 million years ago. The first animals dating from just before the Cambrian were found in rocks near Adelaide, Australia. They are called the Ediacarian fauna and have subsequently been found in other locales as well. It is unclear if these forms have any surviving descendants. Some look a bit like Cnidarians (jellyfish, sea anemones and the like); others resemble annelids (earthworms). All the phyla (the second highest taxonomic category) of animals appeared around the Cambrian. The Cambrian 'explosion' may have been a result of higher oxygen concentrations enabling larger organisms with higher metabolisms to evolve. Or it might be due to the spreading of shallow seas at that time providing a variety of new niches. In any case, the radiation produced a wide variety of animals.
Some paleontologists think more animal phyla were present then than now. The animals of the Burgess shale are an example of Cambrian animal fossils. These fossils, from Canada, show a bizarre array of creatures, some which appear to have unique body plans unlike those seen in any living animals.
The extent of the Cambrian explosion is often overstated. Although quick, the Cambrian explosion is not instantaneous in geologic time. Also, there is evidence of animal life prior to the Cambrian. In addition, although all the phyla of animals came into being, these were not the modern forms we see today. Our own phylum (which we share with other mammals, reptiles, birds, amphibians and fish) was represented by a small, sliver-like thing called Pikaia. Plants were not yet present. Photosynthetic protists and algae were the bottom of the food chain. Following the Cambrian, the number of marine families leveled off at a little less than 200.
The Ordovician explosion, around 500 million years ago, followed. This 'explosion', larger than the Cambrian, introduced numerous families of the Paleozoic fauna (including crinoids, articulate brachiopods, cephalopods and corals). The Cambrian fauna, (trilobites, inarticulate brachiopods, etc.) declined slowly during this time. By the end of the Ordovician, the Cambrian fauna had mostly given way to the Paleozoic fauna and the number of marine families was just over 400. It stayed at this level until the end of the Permian period.
Plants evolved from ancient green algae over 400 million years ago. Both groups use chlorophyll a and b as photosynthetic pigments. In addition, plants and green algae are the only groups to store starch in their chloroplasts. Plants and fungi (in symbiosis) invaded the land about 400 million years ago. The first plants were moss-like and required moist environments to survive. Later, evolutionary developments such as a waxy cuticle allowed some plants to exploit more inland environments. Still mosses lack true vascular tissue to transport fluids and nutrients. This limits their size since these must diffuse through the plant. Vascular plants evolved from mosses. The first vascular land plant known is Cooksonia, a spiky, branching, leafless structure. At the same time, or shortly thereafter, arthropods followed plants onto the land. The first land animals known are myriapods -- centipedes and millipedes.
Vertebrates moved onto the land by the Devonian period, about 380 million years ago. Ichthyostega, an amphibian, is the among the first known land vertebrates. It was found in Greenland and was derived from lobe-finned fishes called Rhipidistians. Amphibians gave rise to reptiles. Reptiles had evolved scales to decrease water loss and a shelled egg permitting young to be hatched on land. Among the earliest well preserved reptiles is Hylonomus, from rocks in Nova Scotia.
The Permian extinction was the largest extinction in history. It happened about 250 million years ago. The last of the Cambrian Fauna went extinct. The Paleozoic fauna took a nose dive from about 300 families to about 50. It is estimated that 96% of all species (50% of all Families) met their end. Following this event, the Modern fauna, which had been slowly expanding since the Ordovician, took over.
The Modern fauna includes fish, bivalves, gastropods and crabs. These were barely affected by the Permian extinction. The Modern fauna subsequently increased to over 600 marine families at present. The Paleozoic fauna held steady at about 100 families. A second extinction event shortly following the Permian kept animal diversity low for awhile.
During the Carboniferous (the period just prior to the Permian) and in the Permian the landscape was dominated by ferns and their relatives. After the Permian extinction, gymnosperms (ex. pines) became more abundant. Gymnosperms had evolved seeds, from seedless fern ancestors, which helped their ability to disperse. Gymnosperms also evolved pollen, encased sperm which allowed for more outcrossing.
Dinosaurs evolved from archosaur reptiles, their closest living relatives are crocodiles. One modification that may have been a key to their success was the evolution of an upright stance. Amphibians and reptiles have a splayed stance and walk with an undulating pattern because their limbs are modified from fins. Their gait is modified from the swimming movement of fish. Splay stanced animals cannot sustain continued locomotion because they cannot breathe while they move; their undulating movement compresses their chest cavity. Thus, they must stop every few steps and breath before continuing on their way. Dinosaurs evolved an upright stance similar to the upright stance mammals independently evolved. This allowed for continual locomotion. In addition, dinosaurs evolved to be warm-blooded. Warmbloodedness allows an increase in the vigor of movements in erect organisms. Splay stanced organisms would probably not benefit from warm- bloodedness. Birds evolved from sauriscian dinosaurs. Cladistically, birds are dinosaurs. The transitional fossil Archaeopteryx has a mixture of reptilian and avian features.
Angiosperms evolved from gymnosperms, their closest relatives are Gnetae. Two key adaptations allowed them to displace gymnosperms as the dominant fauna -- fruits and flowers. Fruits (modified plant ovaries) allow for animal-based seed dispersal and deposition with plenty of fertilizer. Flowers evolved to facilitate animal, especially insect, based pollen dispersal. Petals are modified leaves. Angiosperms currently dominate the flora of the world -- over three fourths of all living plants are angiosperms.
Insects evolved from primitive segmented arthropods. The mouth parts of insects are modified legs. Insects are closely related to annelids. Insects dominate the fauna of the world. Over half of all named species are insects. One third of this number are beetles.
The end of the Cretaceous, about 65 million years ago, is marked by a minor mass extinction. This extinction marked the demise of all the lineages of dinosaurs save the birds. Up to this point mammals were confined to nocturnal, insectivorous niches. Once the dinosaurs were out of the picture, they diversified. Morgonucudon , a contemporary of dinosaurs, is an example of one of the first mammals. Mammals evolved from therapsid reptiles. The finback reptile Diametrodon is an example of a therapsid. One of the most successful lineages of mammals is, of course, humans. Humans are neotenous apes. Neoteny is a process which leads to an organism reaching reproductive capacity in its juvenile form. The primary line of evidence for this is the similarities between young apes and adult humans. Louis Bolk compiled a list of 25 features shared between adult humans and juvenile apes, including facial morphology, high relative brain weight, absence of brow ridges and cranial crests.
The earth has been in a state of flux for 4 billion years. Across this time, the abundance of different lineages varies wildly. New lineages evolve and radiate out across the face of the planet, pushing older lineages to extinction, or relictual existence in protected refugia or suitable microhabitats. Organisms modify their environments. This can be disastrous, as in the case of the oxygen holocaust. However, environmental modification can be the impetus for further evolutionary change. Overall, diversity has increased since the beginning of life. This increase is, however, interrupted numerous times by mass extinctions. Diversity appears to have hit an all-time high just prior to the appearance of humans. As the human population has increased, biological diversity has decreased at an ever-increasing pace. The correlation is probably causal.
Scientific Standing of Evolution and its Critics
The theory of evolution and common descent were once controversial in scientific circles. This is no longer the case. Debates continue about how various aspects of evolution work. For example, all the details of patterns of relationships are not fully worked out. However, evolution and common descent are considered fact by the scientific community.
Scientific creationism is 100% crap. So-called "scientific" creationists do not base their objections on scientific reasoning or data. Their ideas are based on religious dogma, and their approach is simply to attack evolution. The types of arguments they use fall into several categories: distortions of scientific principles ( the second law of thermodynamics argument), straw man versions of evolution (the "too improbable to evolve by chance" argument), dishonest selective use of data (the declining speed of light argument) appeals to emotion or wishful thinking ("I don't want to be related to an ape"), appeals to personal incredulity ("I don't see how this could have evolved"), dishonestly quoting scientists out of context (Darwin's comments on the evolution of the eye) and simply fabricating data to suit their arguments (Gish's "bullfrog proteins").
Most importantly, scientific creationists do not have a testable, scientific theory to replace evolution with. Even if evolution turned out to be wrong, it would simply be replaced by another scientific theory. Creationists do not conduct scientific experiments, nor do they seek publication in peer-reviewed scientific journals. Much of their output is "preaching to the choir."
The most persuasive creationist argument is a non-scientific one -- the appeal to fair play. "Shouldn't we present both sides of the argument?," they ask. The answer is no -- the fair thing to do is exclude scientific creationism from public school science courses. Scientists have studied and tested evolution for 150 years. There is voluminous evidence for it. Within the scientific community, there are no competing theories. Until scientific creationists formulate a scientific theory, and submit it for testing, they have no right to demand equal time in science class to present their ideas. Evolution has earned a place in the science curriculum. Creationism has not.
Science is based on an open and honest look at the data. Much of creationism is built on dishonest debating techniques and special pleading for a case the data does not support. Science belongs in science classes. Evolution is science. Creationism is not. It's that simple.
The creationist attack on public school education means that school children are denied the possibility of learning about the most powerful and elegant theory in biology. Politicians are willing to allow the scientifically ignorant, but politically strong, to wreck the educational system in exchange for votes. People interested in evolution, and science education in general, need to closely watch school board elections. Creationist "stealth" candidates have been elected in several regions. Thankfully, many have been voted out once their views became apparent.
The majority of Americans are religious, but only a minority are religious nuts. The version of religion the far right wants to impose on America is as repulsive to most mainstream Christians as it is to members of other religions, atheists and agnostics. Most informed religious people see no reason for biological facts and theories to interfere with their religious beliefs.
The Importance of Evolution in Biology
"Nothing in biology makes sense except in the light of evolution." -- Theodosius Dobzhansky
Evolution has been called the cornerstone of biology, and for good reasons. It is possible to do research in biology with little or no knowledge of evolution. Most biologists do. But, without evolution biology becomes a disparate set of fields. Evolutionary explanations pervade all fields in biology and brings them together under one theoretical umbrella.
We know from microevolutionary theory that natural selection should optimize the existing genetic variation in a population to maximize reproductive success. This provides a framework for interpreting a variety of biological traits and their relative importance. For example, a signal intended to attract a mate could be intercepted by predators. Natural selection has caused a trade- off between attracting mates and getting preyed upon. If you assume something other than reproductive success is optimized, many things in biology would make little sense. Without the theory of evolution, life history strategies would be poorly understood.
Macroevolutionary theory also helps explain many things about how living things work. Organisms are modified over time by cumulative natural selection. The numerous examples of jury- rigged design in nature are a direct result of this. The distribution of genetically based traits across groups is explained by splitting of lineages and the continued production of new traits by mutation. The traits are restricted to the lineages they arise in.
Details of the past also hold explanatory power in biology. Plants obtain their carbon by joining carbon dioxide gas to an organic molecule within their cells. This is called carbon fixation. The enzyme that fixes carbon is RuBP carboxlyase. Plants using C3 photosynthesis lose 1/3 to 1/2 of the carbon dioxide they originally fix. RuBP carboxlyase works well in the absence of oxygen, but poorly in its presence. This is because photosynthesis evolved when there was little gaseous oxygen present. Later, when oxygen became more abundant, the efficiency of photosynthesis decreased. Photosynthetic organisms compensated by making more of the enzyme. RuBP carboxylase is the most abundant protein on the planet partially because it is one of the least efficient.
Ecosystems, species, organisms and their genes all have long histories. A complete explanation of any biological trait must have two components. First, a proximal explanation -- how does it work? And second, an ultimate explanation -- what was it modified from? For centuries humans have asked, "Why are we here?" The answer to that question lies outside the realm of science. Biologists, however, can provide an elegant answer to the question, "How did we get here?"
Some Books about Biology and Evolution
Futuyma, Douglas J. (1997). Evolutionary Biology. Sunderland, Mass.: Sinauer Associates.
Ridley, Mark. (2003). Evolution. Boston: Blackwell Scientific.
Hartl, Daniel L. & Andrew G. Clark. (1997). Principles of Population Genetics. Sunderland, Mass.: Sinauer Associates.
Crow, James F. & Motoo Kimura. (1970). Introduction to Population Genetics Theory. Edina, Minn.: Burgess Publishing Company.
Graur, Dan & Wen-Hsiung Li. (2000). Fundamentals of Molecular Evolution. Sunderland, Mass.: Sinauer Associates.
Lewontin, Richard C. (1974). The Genetic Basis of Evolutionary Change. New York: Columbia Univ. Press.
Gillespie, John H. (1997). The Causes of Molecular Evolution. New York: Oxford Univ. Press.
Golding, Brian, ed. (1994). Non-Neutral Evolution. Boston: Chapman and Hall.
Kimura, Motoo. (1983). The Neutral Theory of Molecular Evolution. Cambridge, U.K.: Cambridge Univ. Press.
Endler, John A. (1986). Natural Selection in the Wild. Princeton, N.J.: Princeton Univ. Press.
Eldredge, Niles. (1989). Macroevolutionary Dynamics. New York: McGraw-Hill.
Cowen, Richard. (2004). History of Life. Boston: Blackwell Scientific.
Dawkins, Richard. (1987). The Blind Watchmaker. New York: W.W. Norton.
Kitcher, Philip. (1982). Abusing Science. Cambridge, Mass.: MIT Press.
Wilson, Edward O. (1992). The Diversity of Life. Cambridge, Mass.: Harvard Belknap.
Darwin, Charles. (1859). On the Origin of Species.
Darwin, Charles. (1871). The Descent of Man.
Haldane, J.B.S. (1932). The Causes of Evolution. Princeton, N.J.: Princeton Univ. Press (reprinted 1990).
Simpson, George G. (1944). Tempo and Mode in Evolution. New York: Columbia Univ. Press.
Mayr, Ernst E. (1982). The Growth of Biological Thought. Cambridge, Mass: Harvard Belknap.
Provine, William B. (2001). The Origins of Theoretical Population Genetics. Chicago: Univ. of Chicago Press.
Appendix
Part one: Geological time
Millions of years ago Preambrian Time Archean Era 4600-2500 Proterozoic Era 2500-570 Phanerozoic Time Paleozoic Era Cambrian Period 570-505 Ordovician Period 505-438 Silurian Period 438-408 Devonian Period 408-360 Carboniferous Period 360-286 Permian Period 286-245 Mesozoic Era Triassic Period 245-208 Jurassic Period 208-144 Cretaceous Period 144-66.4 Cenozoic Era Tertiary Period Paleocene Epoch 66.4-57.8 Eocene Epoch 57.8-38.6 Oligocene Epoch 38.6-23.7 Miocene Epoch 23.7-5.3 Pliocene Epoch 5.3-1.6 Quarternary Period Pleistocene Epoch 1.6-0.01 Holocene Epoch 0.01-0
Part two: Universal phylogeny
green bacteria flavobacteria EUBACTERIA spirochetes gram positive bacteria purple bacteria eukaryotic mitochondria cyanobacteria eukaryotic chloroplasts deinococci thermotagales ARCHAEBACTERIA halophiles methanogens methanogens thermophiles thermophiles choanoflaggelates EUKARYOTES animals fungi plants ciliates cellular slime molds flaggelates microsporidia
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14864 | https://helda.helsinki.fi/bitstreams/91fc6d91-b3ce-4655-8fe6-700e4b590944/download | UNIVERSITY OF HELSINKI REPORT SERIES IN PHYSICS HU-P-D100
STRONG AND ELECTROMAGNETIC TRANSITIONS IN HEAVY FLAVOR MESONS TIMO L ¨ AHDE
Faculty of Science Helsinki Institute of Physics and Department of Physical Sciences, PL 64 University of Helsinki, 00014 Helsinki, Finland
e-mail: talahde@pcu.helsinki.fi
ACADEMIC DISSERTATION To be presented, with the permission of the Faculty of Science of the University of Helsinki, for public criticism in Auditorium E204 of the Department of Physical Sciences, on December 13th, 2002, at noon.
Helsinki 2002 This thesis is dedicated to all those who have shown interest in my work and encouraged me when progress has been slow.
Ut desint vires, tamen est laudanda voluntas.
ISBN 952-10-0563-7 ISSN 0356-0961 Helsinki 2002 Yliopistopaino ISBN 952-10-0564-5 (PDF version) Helsinki 2002 Helsingin yliopiston verkkojulkaisut
Preface
This thesis is a summary of research done at the Department of Physics of the University of Helsinki, and later at the Department of Physical Sciences and the Helsinki Institute of Physics (HIP), during the years 1998 - 2002. This research has mainly been funded by the University of Helsinki, the Academy of Finland and HIP. Fund grants by the V. K. & Y. V¨ ais¨ ala, M. Ehrnrooth and W. von Frenckell foundations are also gratefully acknowledged. First and foremost, the author wishes to express his thanks to the esteemed colleague and supervisor of this thesis, Prof. Dan-Olof Riska, who in spite of numerous other pressing duties, including the directorship of HIP, has provided invaluable guidance, without which the completion of this thesis would have been a formidable task. Prof. Dan-Olof Riska and Doc. Mikko Sainio are also gratefully acknowledged by the author for providing a postgraduate researcher position at HIP. Special thanks are due to the chairman of the Department of Physical Sciences, Prof. Juhani Keinonen, for suggesting the referees for this thesis, and to Profs. Anthony Green and Jukka Maalampi for agreeing to referee the manuscript. The author also wishes to thank all the colleagues at HIP and elsewhere that have in any way contributed to the research presented in this thesis. Special thanks are due to Prof. Anthony Green for his many suggestions and remarks, to Dr. Tomas Lind´ en for assistance with the typesetting of the manuscript, and to Dr. Christina Helminen, with whom the author has had many useful conversations. Instructive discussions with Profs. Keijo Kajantie, Paul Hoyer, Masud Chaichian, Andy Jackson, Carl Carlson, Doc. Claus Montonen, Doc. Jouni Niskanen, Dr. Gunnar Bali and Dr. M.R. Robilotta are also gratefully acknowledged. Several other colleagues have also contributed indirectly, most notably Christer Nyf¨ alt, Krister Henriksson, Lars-Erik Hannelius, Mikko Jahma, Jonna Koponen and Pekko Piirola. The author has also enjoyed many instructive off-topic conversations with his colleagues at the Department of Theoretical Physics (TFO), the Laser Physics and X-ray Research Units and the group of Doc. Kai Nordlund. The thorough nitpicking by Christer Nyf¨ alt, although a time-consuming source of irrita-tion, is gratefully acknowledged by the author and has led to a substantial improvement in the mathematical quality of the manuscript, and in many cases to a better under-standing of the underlying physics. Finally, all colleagues at HIP are acknowledged for the creation of a pleasant working atmosphere. Helsingfors, October 2002
Timo L¨ ahde
IContents
Preface IAbstract IV List of Publications VShort Introduction to the Papers VI 1 Introduction 2
1.1 The quark model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Quantum Chromodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Heavy flavor mesons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4 Transitions in heavy flavor mesons . . . . . . . . . . . . . . . . . . . . . . 51.5 Notation and layout . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2 Models for the Spectra of Q ¯Q and Q¯q Mesons 7
2.1 The BSLT quasipotential reduction . . . . . . . . . . . . . . . . . . . . . . 72.2 The BSLT and Lippmann-Schwinger equations . . . . . . . . . . . . . . . 10 2.3 The Q ¯Q interaction in the BSLT framework . . . . . . . . . . . . . . . . . 11 2.4 Relativistic Q ¯Q potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.5 Spectra of heavy flavor mesons . . . . . . . . . . . . . . . . . . . . . . . . 15
3 Electromagnetic Transitions 19
3.1 Charge density and electric dipole operators . . . . . . . . . . . . . . . . . 20 3.2 Current density and magnetic moment operators . . . . . . . . . . . . . . 23 3.3 Widths for radiative decay . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 II 3.4 E1 and M1 transitions in heavy quarkonia . . . . . . . . . . . . . . . . . . 28 3.4.1 The M1 transition J/ψ → ηc γ . . . . . . . . . . . . . . . . . . . . 28 3.4.2 The M1 transition ψ ′ → ηc γ . . . . . . . . . . . . . . . . . . . . . 28 3.4.3 Other M1 transitions . . . . . . . . . . . . . . . . . . . . . . . . . . 29 3.4.4 The E1 transitions χcJ → J/ψ γ and ψ ′ → χcJ γ . . . . . . . . . . 31 3.4.5 The E1 transitions from the χbJ states . . . . . . . . . . . . . . . . 31 3.4.6 The E1 transitions from the Υ states . . . . . . . . . . . . . . . . . 33 3.4.7 Other E1 transitions . . . . . . . . . . . . . . . . . . . . . . . . . . 33 3.5 M1 transitions in heavy-light mesons . . . . . . . . . . . . . . . . . . . . . 35
4 Pionic Transitions 37
4.1 The amplitude for pion emission . . . . . . . . . . . . . . . . . . . . . . . 38 4.2 The pionic widths of the D mesons . . . . . . . . . . . . . . . . . . . . . . 40 4.3 π0 and γ transitions from the D∗
s
meson . . . . . . . . . . . . . . . . . . . 41 4.4 Estimation of fηN N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
5 Two-pion Transitions 43
5.1 The width for a ππ transition . . . . . . . . . . . . . . . . . . . . . . . . . 43 5.2 ππ transitions in heavy-light mesons . . . . . . . . . . . . . . . . . . . . . 45 5.3 ππ transitions in heavy quarkonia . . . . . . . . . . . . . . . . . . . . . . . 48 5.4 The transitions Υ ′ → Υ ππ and ψ′ → J/ψ ππ . . . . . . . . . . . . . . . . 52
6 Conclusions 55 Svenskspr˚ akigt sammandrag 57 Suomenkielinen tiivistelm¨ a 59 References 61
III Abstract
T. L¨ ahde: Strong and Electromagnetic Transitions in Heavy Flavor Mesons, University of Helsinki, 2002, VII, 65 p. + reprints, University of Helsinki Report Series in Physics, HU-P-D100, ISSN 0356-0961, ISBN 952-10-0563-7, ISBN 952-10-0564-5 (PDF version)
Classification (INSPEC): A1240Q, A1325, A1340F, A1340H, A1440L, A1440N Keywords: charmed mesons, bottom mesons, pionic transitions, electromagnetic transitions, potential models
The electromagnetic and pionic transitions in mesons with heavy flavor charm ( c) or bot-tom ( b) quarks are calculated within the framework of the covariant Blankenbecler-Sugar (BSLT) equation. The magnetic dipole (M1) transitions in the charmonium ( c¯c) system are shown to be sensitive to the relativistic aspect of the spin-flip magnetic moment op-erator, and the Lorentz coupling structure of the Q ¯Q interaction. The observed rate for the M1 transition J/ψ → ηcγ is shown to provide strong evidence for a scalar confining interaction. On the other hand, the electric dipole (E1) transitions are shown to be sensitive to the hyperfine splittings in the Q ¯Q system, and to require a nonperturbative treatment of the hyperfine components in the Q ¯Q interaction. In addition to the spin-flip M1 transitions, the single pion ( π) and dipion ( ππ ) widths are calculated for the heavy-light ( Q¯q) D mesons, by employment of the pseudovector pion-quark coupling suggested by chiral perturbation theory. The pionic transitions
D∗ → Dπ are shown to provide useful and constraining information on the pion-quark axial coupling gqA. It is also shown that axial exchange charge contributions associated with the Q¯q interaction suppress the axial charge amplitude for pion emission by an order of magnitude. The models for π and M1 transitions also make it possible to estimate the
η-nucleon coupling from the transition D∗
s
→ Dsπ0, once the value of the π0 − η mixing angle is known. Finally, the ππ dipion transition rates of the L = 1 D mesons are calculated, and are shown to make up a significant fraction of their total widths for strong decay. The ππ
transitions between S-wave charmonium ( c¯c) and bottomonium ( b¯b) states are modeled in terms of a broad σ meson or a glueball, with derivative couplings to pions. The effects of pion rescattering by the spectator quark are also investigated. IV List of Publications
This thesis consists of two parts. The first and main part of the thesis is a summary and discussion of the results obtained in the published, peer-reviewed research papers listed below. Where available, newer and more accurate results are also presented. The second part consists of reprints of selected papers that have been signed by the author. These papers are based on research done at the Department of Physical Sciences of the University of Helsinki and the Helsinki Institute of Physics in 1998 - 2002.
I T.A. L¨ ahde, C.J. Nyf¨ alt and D.O. Riska,
The Confining Interaction and Radiative Decays of Heavy Quarkonia ,Published in: Nucl.Phys. A645:587-603 (1999), eprint hep-ph/9808438
II K.O.E. Henriksson, T.A. L¨ ahde, C.J. Nyf¨ alt and D.O. Riska,
Pion Decay Widths of D Mesons ,Published in: Nucl.Phys. A686:355-378 (2001), eprint hep-ph/0009095
III T.A. L¨ ahde and D.O. Riska,
Two-Pion Decay Widths of Excited Charm Mesons ,Published in: Nucl.Phys. A693:755-774 (2001), eprint hep-ph/0102039
IV T.A. L¨ ahde and D.O. Riska,
Pion Rescattering in Two-Pion Decay of Heavy Quarkonia ,Published in: Nucl.Phys. A707:425-451 (2002), eprint hep-ph/0112131
V T.A. L¨ ahde and D.O. Riska,
The Coupling of η Mesons to Quarks and Baryons from D∗
s
→ Dsπ0 Decay ,Published in: Nucl.Phys. A710:99-116 (2002), eprint hep-ph/0204230
VI T.A. L¨ ahde,
Exchange Current Operators and Electromagnetic Dipole Transitions in Heavy Quarkonia ,Published in: to be published in Nucl.Phys. A, eprint hep-ph/0208110 VShort Introduction to the Papers
I This paper presents a calculation of the M1 transition rates in the c¯c and b¯b systems within the framework of the nonrelativistic Schr¨ odinger equation. A relativistic version of the single quark spin-flip magnetic moment operator is derived, along with the two-quark exchange current operators for M1 transitions. It is shown that the two-quark operator associated with a scalar confining interaction may provide, together with the relativistic single quark operator, a possible explanation of the empirically measured width of ∼ 1 keV for the transition J/ψ → ηcγ.
II The framework of the covariant Blankenbecler-Sugar (BSLT) equation is used to-gether with the pseudovector pion-quark coupling suggested by chiral perturbation theory to predict the widths for pionic transitions in the heavy-light ( c¯q) D meson systems. It is found that useful and constraining information on the pion-quark axial coupling gqA is provided by the D∗ → Dπ transitions. A satisfactory descrip-tion of the empirically measured pion widths of the L = 1 D∗
2
meson is obtained. Also, the axial charge component of the amplitude for pion emission is shown to be suppressed by axial exchange charge contributions associated with the Q¯q interac-tion.
III The chiral pseudovector Lagrangian, augmented by a Weinberg-Tomozawa term for dipion emission, is used to predict the widths for ππ transitions from the L = 1 D
mesons. It is found that widths of several MeV are expected for these transitions, in analogy with the experimentally well-studied decays of the strange K∗
2
meson. It is thus expected that the ππ modes should constitute a significant fraction of the total widths of the L = 1 D mesons.
IV The dipion transitions between S-wave states in the charmonium ( c¯c) and bot-tomonium ( b¯b) systems are studied using a phenomenological model with deriva-tive couplings to pions. The dipions are modeled in terms of a broad σ meson or a glueball. Effects of pion rescattering by the spectator quark are investigated and shown to be small for 2 S → 1S transitions. The present experimental data on these transitions is shown to constrain the σ meson mass to about 500 MeV. Finally, it is demonstrated that the anomalous double-peaked ππ spectrum of the Υ(3 S) → Υ(1 S) ππ transition may be modeled in terms of a heavier ∼ 1500 MeV scalar meson. VI V The empirically measured branching ratios for D∗
s
→ Ds π0 and D∗
s
→ Ds γ are shown to provide a means of determining the strength of the η coupling to quarks and baryons. This requires that the value of the π0 − η mixing angle is available, along with realistic models for the M1 and pionic transitions in heavy-light mesons. The value thus obtained for the η-nucleon pseudovector coupling fηN N is shown to be much smaller than that suggested by SU (3) symmetry, which is consistent with other recent phenomenological analyses. It is also shown that a significant η-charm coupling, if present, serves to increase the estimated value of fηN N .
VI The two-quark exchange current operators that arise from the elimination of the negative energy components of the Bethe-Salpeter equation in the BSLT quasipo-tential reduction, are calculated for electromagnetic E1 and M1 transitions in heavy quarkonium systems. Although the exchange charge operators that contribute to E1 transitions are shown to be mostly negligible, the corresponding exchange cur-rent operators for M1 transitions are shown to be crucial, if agreement with the empirical width for J/ψ → ηc γ is to be achieved. This requires that the effective confining interaction couples as a Lorentz scalar, since an effective vector interac-tion is shown to yield a spin-flip magnetic moment operator only if the constituent quark masses are unequal. Consequently, in the B±
c
system, the one-gluon exchange interaction also contributes a two-quark spin-flip operator. VII Chapter 1 Introduction
It has been widely accepted, since the beginning of the 20th century, that the visible matter in the universe is composed of protons and neutrons (or baryons), and electrons (or leptons). However, the discovery of the positron in 1933, predicted by Dirac a few years earlier, suggested that short-lived, transient particles may exist alongside the stable protons and electrons. This was confirmed in 1936, when a heavier, unstable electron-like particle, the muon ( μ), was discovered in cosmic ray experiments by Anderson and Neddermeyer. This discovery was followed up in 1947, when the existence of the neutral (π0) and charged ( π±) pions, predicted earlier by Yukawa to be the carriers of the strong nuclear force, was confirmed by a similar experiment. These particles were the first ones of a large number of short-lived, unstable baryons, mesons and leptons which were subsequently produced in copious numbers by accelerator experiments. In particular, the pions were shown to be the lightest members of a new family of particles known as mesons, to denote that they are intermediate in mass between the baryons and leptons.
1.1 The quark model
Around 1960, the number of short-lived baryons (∆, Σ ,Λ ,Ξ ...) and mesons ( π ,K ,ρ ,η ...) that had been discovered by accelerator experiments was overwhelming. This suggested that the hypothesis of Mendeleev could be extended to the baryons and mesons; Rather than being elementary, they might possess substructure and could perhaps be classified according to a ”periodic table” of subatomic particles. This notion, originally put forward by Gell-Mann and Ne’eman among others, became known as the ”Quark Model” and attempts to explain the observed properties (spin, isospin, electric charge, parity) of the mesons and baryons by arranging them into multiplets according to the symmetry group SU (3). It was found that three quark flavors, ”up” ( u), ”down” ( d) and ”strange” (s) with spin 1/2 and fractional electric charges were required to accommodate all of the mesons and baryons known at that time. The experimental discovery of the Ω baryon, which was predicted by the quark model because of a gap in the ”periodic table” of the baryons, soon provided dramatic confir-mation of the quark hypothesis. Although the quarks were at first only thought of as a useful theoretical tool, their actual existence inside the proton was confirmed by 21.2. Quantum Chromodynamics 3
deep inelastic e−p scattering (DIS) experiments at high energies. However, in spite of these remarkable successes, the quark model soon ran into a difficulty of symmetry. The spin-parity quantum numbers of the ∆ resonance were empirically found to be consistent with a combined spin-flavor and configuration space wavefunction which is symmetric. This is inconsistent with Fermi-Dirac statistics, which requires that the total baryon wavefunction should be antisymmetric. This critical problem was finally circumvented by the introduction of a new property for the quarks, ”color”, which allows the wavefunction to be made antisymmetric by means of three color quantum numbers. In order to avoid an undesirable proliferation of unobserved states, a further constraint was placed, namely that the quarks only combine into colorless states (or singlet representations of the color SU (3) C group). This restricts the possible ways of combining quarks and antiquarks to hadrons, the simplest being
q ¯q (mesons), qqq (baryons) and ¯ q ¯q ¯q (antibaryons). Together with the proposal that eight spin 1 gauge fields, ”gluons”, should be associated with the new symmetry group
SU (3) C , these notions were eventually developed into the theory of strong interactions, called Quantum Chromodynamics (QCD) . It was also realized that a fourth quark is required in the theory of weak interactions to explain e.g. the observed rate for the decay K0 → μ+μ−. The fourth quark was eventually discovered in the form of narrow resonances in November 1974 at center-of-mass energies of 3.1 GeV and 3.7 GeV in e+e− annihilation and, independently, in proton-proton collisions. These resonances, named J/ψ and ψ′, were interpreted as mesonic bound states of the new ”charm” quark and its antiquark, c¯c. The charm quark turned out to have a mass of ∼ 1500 MeV, and is thus much more massive than the
∼ 5 MeV u, d quarks and the ∼ 100 MeV s quark. Later, as higher collision energies became available, an unexpected ”bottom” ( b) quark with a mass of ∼ 4800 MeV was similarly discovered in the form of b¯b or Υ mesons. This again raised the question of a possible partner for the b quark, and indeed an extraordinarily heavy ”top” ( t) quark was finally detected in 1995, by the proton-antiproton collider experiments at Fermilab. The t quark turned out to have a mass of 175 GeV, which makes it the most massive elementary particle known, and it is too short-lived for mesonic t¯t bound states to form.
1.2 Quantum Chromodynamics
In the theory of Quantum Chromodynamics (QCD), the interactions between quarks are mediated by eight massless vector bosons called gluons. However, a number of compli-cations effectively prevent the properties of hadrons to be predicted from QCD; First of all, the theory is nonlinear due to gluon self-interactions, and it describes systems that interact strongly enough so that perturbative methods are inapplicable. Only at the very highest energy scales, where quarks become asymptotically free and the coupling between them small, can the predictions of perturbative QCD be compared with exper-imental results. At low energies, the quarks interact strongly, are confined into hadronic bound states and acquire effective masses. These constituent quark masses are for the light u, d quarks of the order ∼ 400 MeV. At present, the only way to analyze QCD at a fundamental level is the method of ”lattice QCD” simulations, where the properties of hadrons are probed by means of numerical Monte Carlo algorithms. Although much progress is being made in the development 4 Chapter 1. Introduction
of more efficient algorithms and the inclusion of dynamical fermions (unquenched lat-tice QCD) into the simulations, the applicability of such methods is still limited by the huge demands on computing power. In such a situation, it is natural to attempt to understand the properties of hadrons by means of effective theories and phenomenolog-ical, QCD-motivated models. The physical motivation of such an approach is that the fundamental degrees of freedom of QCD are quarks and gluons, whereas low-energy ex-periments observe hadrons, which at least at long range interact by Yukawa-type meson exchange. It is, therefore, a reasonable expectation that the low-energy properties of QCD can be described in terms of an effective theory. In the limit of vanishing quark masses, QCD exhibits an invariance under chiral transformations that involve left- and right-handed quark fields separately. This symmetry is only approximate for quarks with a nonzero mass. The absence of parity doublets in the low-energy region of the hadron spectra suggests that this chiral symmetry is spontaneously broken at low energies .
1.3 Heavy flavor mesons
Mesons that contain either two heavy quarks ( c¯c, b¯b, c¯b) or one heavy quark and one light ( c¯q, c¯s, b¯q, b¯s) are special, since their masses lie in a region which is intermediate between the high-energy perturbative regime of QCD and the low-energy regime where the dynamics are governed by chiral symmetry breaking. Thus these heavy flavor mesons are likely to share features that are encountered in these two limits of QCD. One task at hand is then to determine phenomenologically, or from lattice QCD , the functional form, strength and Lorentz structure of the Q¯q and Q ¯Q interaction. Although the nonrelativistic Schr¨ odinger framework can be applied to Q ¯Q systems with some success, a realistic treatment of the Q¯q system has a priori to be relativistic, as the velocity of the confined light constitutent quark is close to that of light. The papers presented in this thesis employ the covariant Blankenbecler-Sugar (BSLT) reduction of the Bethe-Salpeter equation, which has the advantage of formal similarity to the Schr¨ odinger framework. An alternate approach is provided by the Gross quasipotential reduction , which has been shown to yield comparable results for the spectra of
Q ¯Q and Q¯q mesons. However, as the mass spectra of the Q ¯Q and Q¯q mesons are well described by a large number of phenomenological and QCD-motivated models, the spectrum alone cannot discriminate between different assumptions for the Q ¯Q and Q¯q interaction. Fortunately, as will be shown in this thesis, the observed rates for γ and π transitions in heavy flavor mesons may provide useful and constraining information on the quark-antiquark interaction, the quarkonium wave functions, and in particular, on the Lorentz structure of the effective confining interaction. As the negative energy components of the Bethe-Salpeter equation are eliminated in the BSLT (or Schr¨ odinger) quasipotential reduction, two-quark transition operators that depend explicitly on the Lorentz structure of the
Q ¯Q interaction appear as a consequence . In particular, it will be demonstrated in this thesis that a pure scalar confining interaction compares favorably with the current empirical knowledge of M1 transitions in the charmonium system. It is noteworthy, that similar results have been obtained within the instantaneous approximation to the Bethe-Salpeter equation , which treats the negative energy components explicitly, i.e. without two-quark currents. 1.4. Transitions in heavy flavor mesons 5
1.4 Transitions in heavy flavor mesons
The transitions considered in this thesis include the radiative E1 and M1 transitions in the Q ¯Q systems, the M1 transitions in the heavy-light charm ( D) and strange charm ( Ds)mesons, and the π and ππ transitions in the Q ¯Q and Q¯q systems. It is shown in papers I
and VI that a possible solution to the long-standing overprediction by a factor ∼ 3of the width for the M1 transition J/ψ → ηcγ emerges, if the two-quark exchange current operator associated with a scalar confining interaction is included along with a relativistic treatment of the single quark spin-flip operator. On the other hand, the exchange charge contributions to the E1 transition rates are shown in paper VI to be highly suppressed by the large masses of the charm and bottom constituent quarks . Similarly, the nonrelativistic predictions for the spin-flip M1 widths of the Q¯q mesons are shown in paper V to be unrealistic, as the confined light constituent quark requires a relativistic treatment. It is also shown that accidental cancellations in the single quark spin-flip operators render the M1 widths very sensitive to two-quark exchange current contributions. However, as the form of the Q¯q interaction is uncertain, the results are suggestive rather than definite, quantitative predictions. In the heavy-light D mesons, the excited states decay to the ground state predominantly through pion emission. In this thesis, the pionic transitions in the D mesons are described in terms of the chiral pseudovector Lagrangian which couples the pions to the light constituent quarks. It is shown in paper II that the D∗ → Dπ transitions can provide useful and constraining information on the pion-quark axial coupling gqA. Also, the axial charge component of the amplitude for pion emission is shown to be highly affected by two-quark axial exchange charge contributions associated with the Q¯q interaction. The pionic transitions which are driven by the axial charge operator may, therefore, provide information on the Lorentz structure of the Q¯q interaction. In particular, it is shown that a scalar confining interaction has the effect of reducing the widths for such transitions. The chiral Lagrangian may, when augmented with a Weinberg-Tomozawa term for dipion emission, describe the ππ widths of the excited L = 1 D mesons. In this thesis the ππ
widths of the D mesons are shown to be of significant magnitude compared to the widths for single pion emission. This is known to be the case for the strange K∗
2
meson, where the empirical ππ width is ∼ 1/2 of the π width. This model for pseudoscalar emission has also been applied to the D∗
s
→ Dsπ0 transition, which can then be used to extract the coupling of η mesons to quarks and baryons from the empirical branching ratios for those transitions, once an estimate for the π0 − η mixing angle is available. The value for the η-nucleon pseudovector coupling constant fηN N so obtained, is shown to be much smaller than that suggested by naive SU (3) symmetry, but consistent with other recent phenomenological analyses of e.g. photoproduction of η mesons on the nucleon. Whereas the dipion transitions in the D mesons may be modeled in terms of the chiral Lagrangian, the ππ transitions between S-wave c¯c or b¯b states are likely to involve a broad σ meson or a glueball. It is shown, within a model where the coupling of dipions to heavy quarks is mediated by a broad and heavy scalar meson, that the empirical ππ
energy spectra constrain the σ meson mass to ∼ 500 MeV. A possible explanation for the anomalous double-peaked ππ spectrum of the Υ(3 S) → Υ(1 S)ππ transition is obtained, if the ππ emission is described in terms of a heavier ( ∼ 1500 MeV) scalar meson. 6 Chapter 1. Introduction
1.5 Notation and layout
Throughout this thesis, the natural units with ¯ hc = 1 and the δμν metric have been employed. The Euclidean δμν or Pauli metric assigns imaginary time components to four-vectors. The momentum four-vector k is thus of the form k = ( k, ik 0), where the three-vector has been denoted by bold-faced type. However, for typesetting reasons, three-vectors in exponents have been denoted with an arrow, according to k. Also, in obvious cases the modulus |k| has been denoted simply by k. In the Pauli metric the square of a four-vector is of the form
k2 = kμkμ = k2 − k0 2 = −m2, (1.1) and the Dirac γμ matrices are all hermitian with square equal to one. The explicit form of these matrices in the Pauli metric is then γμ = ( γ,γ4) and γ5 = γ1γ2γ3γ4, where
γ =
( 0 −iσ
iσ 0
)
, γ4 =
( 1 00 −1
)
, γ5 =
( 0 −1
−1 0
)
. (1.2) Factors of i are also introduced into the Dirac current and charge density operators to make them real-valued, and for Lagrangians which include a γ5, in order to assure hermiticity. A number of abbreviations that are frequently used in this thesis are OGE (one-gluon ex-change), BSLT (Blankenbecler-Sugar-Logunov-Tavkhelidze), NRIA (non-relativistic im-pulse approximation) and RIA (relativistic impulse approximation). Excited states in the heavy quarkonium systems have been denoted either by the ψ(nJ ) or the primed notation, where the nth excited state is denoted by n primes, e.g. ψ(3 S) ≡ ψ′′ . Note that in the primed notation, the primes refer to radial excitations only. This thesis contains a summary which comprises six chapters, and reprints of selected research papers that have been signed by the author. Chapter 2 of the summary presents the Blankenbecler-Sugar quasipotential reduction, the Q ¯Q and Q¯q Hamiltonian models and the numerical results for the spectra of the heavy flavor mesons. Chapter 3 discusses the calculations of the electromagnetic E1 and M1 widths of papers I, V and VI , while chapter 4 presents the calculation of the pionic transitions in the D mesons of paper II
and the estimation of the η-nucleon pseudovector coupling fηN N from paper V. Chapter 5 deals with the ππ transitions in the D mesons (paper III ) and the model for the ππ
transitions in the Q ¯Q mesons from paper IV . Chapter 6 contains a concluding discussion. Chapter 2 Models for the Spectra of Q ¯Q
and Q¯q Mesons
Although several phenomenological models that employ a nonrelativistic treatment of the heavy quarkonia have succeeded in describing many features of the c¯c and b¯b
systems, a realistic treatment of the heavy-light mesons has a priori to be relativistic as the velocity of the confined light constituent quark is close to that of light. Also in the case of charmonium and bottomonium, the compact size of the Q ¯Q system causes the charm and bottom quarks to move with relativistic velocities, in spite of their large masses. The reason for this is the effective confining interaction, which has a string tension of
∼ 1 GeV/fm and confines the constituent quarks to a region of radius < 0.5 fm. In this situation, a quasipotential reduction of the relativistic Bethe-Salpeter equation suggests itself as a natural framework for a covariant description of the heavy quarkonium systems.
2.1 The BSLT quasipotential reduction
The field-theoretical scattering matrix S may be written in the form
Sf i = δf i − i (2 π)4δ(Pf − Pi) Mf i , (2.1) where the second term on the r.h.s. has been defined, for notational convenience, with a minus sign. The scattering amplitude M is then defined as
Mf i = ¯ u(p′
Q
)¯ u(p′
¯q
) M u (pQ)u(p¯q ), (2.2) where pi and p′
i
denote the initial and final four-momenta of the quarks. Note that the antiquark will be described throughout by positive energy spinors. The Bethe-Salpeter equation for the scattering amplitude M can then be written (schematically) in the form
M = K + K G M, (2.3) or explicitly, for an arbitrary frame, as
M (p′, p, P ) = K(p′, p, P ) + i
∫ d4k
(2 π)4 K(p′, k, P ) G(k, P ) M (k, p, P ), (2.4) 78 Chapter 2. Models for the Spectra of Q ¯Q and Q¯q Mesons
where P is the total four-momentum of the Q¯q system, and p, k and p′ denote the initial, intermediate, and final relative four-momenta of the constituent quarks. In eq. (2.4), K
denotes the interaction kernel of the Bethe-Salpeter equation, which in the nonrelativistic limit corresponds to the potential defined for the Schr¨ odinger equation. This can be seen by comparison of eq. (2.3) and eq. (2.1) in the Born approximation. Also, G denotes the Green’s function of the Bethe-Salpeter equation, which is here taken to be the free fermion propagator. When bound states are considered, the inhomogeneous term in eq. (2.4) is dropped. The second term of the Bethe-Salpeter scattering equation is illustrated, along with the choice of momentum variables for the Blankenbecler-Sugar quasipotential reduction, by Fig. 2.1.
W + pW − p′
W + ∆ + kW − pW + p′
W − ∆ − k
K M
Figure 2.1: Illustration of the choice of frame and variables for the derivation of the Blankenbecler-Sugar (BSLT) reduction of the Bethe-Salpeter scattering equation for unequal quark masses. The upper and lower quark lines are taken to have mass-es mQ and m¯q, respectively. The four-momenta are defined as W = (0 , iP 0/2), ∆ = (0 , i [m 2
Q
− m 2¯q ]/4W0), p = ( p, ip 0) and k = ( k, ik 0). It is instructive, in order to perform the BSLT quasipotential reduction, to introduce the variables presented in Fig. 2.1 and write the Bethe-Salpeter equation, schematically, as two coupled integral equations,
M = U + U g M (2.5)
U = K + K (G − g) U. (2.6) Here the quasipotential U is defined by eq. (2.6) in terms of the Bethe-Salpeter propa-gator G and a three-dimensional propagator g. The propagator g is then constructed so that it has an identical elastic unitarity cut (right hand cut) as G in the physical region. The approximation U K will be employed here, in order to arrive at a major sim-plification of the Bethe-Salpeter problem. The propagators G and g will have identical discontinuities across the right hand cut if Disc G = 2 i Im g. By means of the Cutkosky rules, Im g may then be obtained as Im g = − 2π2
(2 π)4
[ γQ (W + k + ∆) + im Q
] [ γ ¯q (W − k − ∆) + im ¯q
]
δ(+) [ (W + k + ∆) 2 + m 2
Q
] δ(+) [ (W − k − ∆) 2 + m 2¯q
] , (2.7) where it has been indicated that only the positive energy roots of the arguments in the delta functions are to be included. The complete propagator g is then reconstructed by 2.1. The BSLT quasipotential reduction 9
means of the dispersion integral
g = 1
π
∫ ∞
0
dq2
q2 − p2 − i Im g (W ′, k, ∆′), (2.8) where W ′ is defined as q0/2 with q0 on shell. Evaluation of the above integral yields the following form for the BSLT propagator g:
g = − 1
2
δ(k0)
(2 π)3
[γQ
0
EQ(k) − γQ · k − im Q] [ γ ¯q
0
E¯q (k) + γ ¯q · k − im ¯q ]
(EQ(k) + E¯q (k)) ( k2 − p2 − i ) , (2.9) where the delta function ensures the condition k0 = 0 in the resulting three-dimensional integral equation. Note that the (in principle arbitrary) variable ∆ was chosen so that this condition is realized also in the case of unequal constituent quark masses. By introduction of the positive energy projection operators Λ i
+
, the above propagator can be written in the form
g = − δ(k0)
(2 π)3
2mQm¯q
EQ(k) + E¯q (k)ΛQ
+
(k) Λ ¯q
+
(−k)
k2 − p2 − i . (2.10) This form is convenient when matrix elements are taken between positive-energy spinors according to
M, V(p′, p) = ¯uQ(p′)¯ u¯q (−p′) M, U (p′, p) uQ(p)u¯q (−p), (2.11) which, together with eq. (2.5), yields the three-dimensional BSLT scattering equation
M(p′, p) = V(p′, p)−
∫ d3k
(2 π)3 V(p′, k)
( 2mQm¯q
EQ(k) + E¯q (k)
) 1
k2 − p2 − i M(k, p), (2.12) where V denotes the nonlocal interaction operator as obtained from the Feynman rules for Sf i using eq. (2.1) in the Born approximation. The above extension of the original BSLT equation to the case of unequal masses is similar to that of ref. , which has been employed in ref. for the case of Λ N scattering. The elimination of the negative energy components of the Bethe-Salpeter equation in the derivation of eq. (2.12) has been shown to give rise to two-quark exchange cur-rent operators that depend explicitly on the Lorentz structure of the quark-antiquark interaction. These may then contribute significantly to the strong and electromagnetic transition rates in the Q¯q and Q ¯Q systems. In particular, the exchange current oper-ator associated with the scalar confining interaction has been shown to be of decisive importance for the M1 transitions of heavy quarkonia . It should be noted that the appearance of such two-quark operators depends on the type of quasipotential reduction. 10 Chapter 2. Models for the Spectra of Q ¯Q and Q¯q Mesons
Although eq. (2.12) is a widely used quasipotential reduction of the type discussed in this thesis, there is in principle an infinite number of different ways to reduce the Bethe-Salpeter equation to a 3-dimensional form. Another commonly used reduction is the Thompson equation , which differs from the BSLT equation by the choice of the dispersion integral (2.8). These have been shown to produce results that are very close to the full Bethe-Salpeter equation in ref. . There exists also a large variety of quasipotential reductions that differ in the choice of the propagator (2.10), which attempt to include the effects of intermediate negative energy states by various combinations of the negative energy projection operators . It is also noteworthy that the Bethe-Salpeter equation in the ladder approximation has been shown , not to reduce to the desired one-body (Dirac) equation when one of the quarks becomes much heavier than the other. However, a large number of quasipotential reductions (e.g. Gross) are known to be closely related to the Dirac equation. This suggests that such reductions are more appropriate for two-quark systems with a large difference between the constituent masses, while the BSLT equation is ideal for quarkonia such as c¯c and b¯b. As the light constituent quarks in Q¯q mesons have masses that are lighter than those of the heavy quarks by factors of 3 − 10, then the Gross and BSLT reductions are expected to give results of similar quality, which indeed appears to be the case .
2.2 The BSLT and Lippmann-Schwinger equations
As eq. (2.12) is similar to the nonrelativistic Lippmann-Schwinger equation, except for the factor in parentheses, then it can be transformed into such an equation by means of the ”minimal relativity” ansatz
T (p′, p) =
( mQ + m¯q
EQ(p′) + E¯q (p′)
) 1
2
M(p′, p)
( mQ + m¯q
EQ(p) + E¯q (p)
) 1
2
, (2.13)
V (p′, p) =
( mQ + m¯q
EQ(p′) + E¯q (p′)
) 1
2
V(p′, p)
( mQ + m¯q
EQ(p) + E¯q (p)
) 1
2
. (2.14) This yields the equation
T (p′, p) = V (p′, p) −
∫ d3k
(2 π)3 V (p′, k) 2μ
k2 − p2 − i T (k, p), (2.15) which is formally identical to the Lippmann-Schwinger equation. Here μ stands for the usual reduced mass of the two-quark system. The advantage of eq. (2.15) is that it can be transformed to a Schr¨ odinger-type differential equation where the potential is given by eq. (2.14). This transformation gives the differential equation
(
H0 − p2
2μ
)
ψnlm (r) = −V ψ nlm (r), (2.16) where H0 is the kinetic energy operator of the nonrelativistic Schr¨ odinger equation. The factor p2 can be expressed in terms of the total energy of the Q¯q state and the constituent quark masses mQ and m¯q .2.3. The Q ¯Q interaction in the BSLT framework 11
The eigenvalue ε of the BSLT equation is obtained as
ε = p2
2μ =
[E2 − (mQ + m¯q)2] [ E2 − (mQ − m¯q)2]
8μE 2 , (2.17) where E is the mass of the Q¯q state. The BSLT equation can thus be expressed as an eigenvalue equation of the form (H0 + Hint ) ψnlm (r) = ε ψ nlm (r), (2.18) where the interaction Hamiltonian Hint is given in terms of the potential defined in eq. (2.14). The introduction of the quadratic mass operator (2.17) leads to an effective weakening of the repulsive kinetic energy operator, which means that higher excited states will have lower masses in the BSLT equation than they would in the Schr¨ odinger framework. The BSLT eigenvalue ε, expressed in terms of the Schr¨ odinger excitation energy Eex = E − (m1 + m2), is of the form
ε = Eex
8μ
[ (Eex + 2( mQ + m¯q )) ( E2ex + 2 Eex (mQ + m¯q) + 4 mQm¯q )
E2ex + 2 Eex (mQ + m¯q) + ( mQ + m¯q )2
]
, (2.19) where the expression in parentheses tends toward 8 μ when mQ, m ¯q → ∞ . This demon-strates that in the limit of heavy quark masses, or when the quark masses become large compared to the excitation energy Eex , the BSLT equation reduces to the nonrelativistic Schr¨ odinger equation. Although the role of the BSLT potential V as given by eq. (2.14) is equivalent to that of the nonrelativistic, static potential in the Schr¨ odinger framework, the multiplication of the full non-local interaction (in momentum space) by the minimal relativity square root factors is shown in the next section to have important consequences, not only for the numerical treatment of eq. (2.18) but also for the modeling of the strong and electro-magnetic transitions between Q¯q and Q ¯Q states. In particular, the well-known problem of too singular and thus illegal hyperfine operators in the Schr¨ odinger equation is shown to disappear in the BSLT framework.
2.3 The Q ¯Q interaction in the BSLT framework
The interaction between heavy quarks and heavy or light antiquarks is dominated by the (presumably linearly) rising confining interaction. The observed spectra of Q ¯Q mesons also require the presence of a short-range hyperfine interaction that gives rise to e.g. the
J/ψ −ηc splitting. The one-gluon exchange (OGE) interaction of perturbative QCD is a natural candidate for the Q ¯Q systems, whereas the origin of the hyperfine interaction in the Q¯q systems is less obvious. A recently suggested possibility is the pointlike instanton induced interaction proposed by ref. . The interaction Hamiltonians used in this thesis in conjunction with the covariant BSLT equation are, therefore, of the form
Hint = Vconf + VOGE + Vinst , (2.20) with confining, OGE, and instanton induced components, respectively. The effective confining interaction is taken to have scalar Lorentz structure, while the OGE interaction 12 Chapter 2. Models for the Spectra of Q ¯Q and Q¯q Mesons
has vector coupling structure. In the nonrelativistic approximation, the effective linear confining interaction has the form (in the LS -coupling scheme):
Vconf = cr
[
1 − 3
2
P 2
m 2
Q
m 2¯q
( m 2
Q
m 2¯q + mQm¯q
3
)]
c
4mQm¯q r
− c
rm 2
Q
m 2¯q
4m 2
Q
m 2¯q
S · L + c
rm 2
Q
− m 2¯q
8m 2
Q
m 2¯q
(σQ − σ¯q ) · L, (2.21) where the string tension c is of the order ∼ 1 GeV/fm. The above form contains also the momentum dependent terms from eq. (2.14) up to second order in v2/c 2. In the Schr¨ odinger framework (i.e. without the minimal relativity factors), the numerical factor 3/2 in front of the P 2 term would be 1. Likewise, the Darwin-Foldy term in eq. (2.21) would vanish. In addition to the familiar Thomas-precession term, an antisymmetric spin-orbit interaction also appears for unequal quark masses, which mixes the states with L = 1 and J = 1. The interaction components associated with the perturbative OGE interaction are, to order v2/c 2 in the nonrelativistic approximation, of the form
VOGE = − 4
3 αs
[
1
r − 3π
2
( m 2
Q
m 2¯q + mQm¯q
3 m 2
Q
m 2¯q
)
δ(r) + 1
2
P 2
mQm¯q r
]
2
3
αs
r3
( m 2
Q
m 2¯q
2m 2
Q
m 2¯q
2
mQm¯q
)
S · L + αs
6r3
m 2
Q
− m 2¯q
m 2
Q
m 2¯q
(σQ − σ¯q ) · L
8π
9
αs
mQm¯q
δ(r) σQ · σ¯q + αs
3mQm¯qr3 S12 , (2.22) where αs denotes the strong coupling of perturbative QCD, and S12 is the tensor operator
S12 = 3( σQ · ˆr)( σ ¯q · ˆr) − σQ · σ¯q . In the Schr¨ odinger framework, the coefficients for the contact and P 2 terms would be −π and 1, respectively. The instanton induced interaction, considered by ref. for systems with heavy quarks, consists of a spin-independent term as well as a σQ · σ¯q term which contributes to the pseudoscalar-vector splittings in heavy quarkonia. The effective instanton interaction derived in ref. is of the form
Vinst = − ∆MQ ∆M¯q
4n δ(r) + ∆M spin
Q
∆M¯q
4n δ(r) σQ · σ¯q, (2.23) where the factors ∆ MQ and ∆ M¯q denote the mass shifts of the heavy and light con-stituent quarks due to the instanton induced interaction. These shifts are, for light constituent quarks, of the order of the constituent quark mass ( ∼ 400 MeV), and smaller (∼ 100 MeV) for the charm quark. The parameter M spin
Q
controls the strength of the spin-spin interaction, which has the same sign as that from the perturbative OGE in-teraction. The parameter n represents the instanton density, which is typically assigned values around ∼ 1 fm −4. The spin-independent term has scalar coupling for the light constituent quark line and a mixed scalar-γ0 vertex for the heavy quark. 2.4. Relativistic Q ¯Q potentials 13
2.4 Relativistic Q ¯Q potentials
For systems that contain light quarks, the above static interaction Hamiltonians have but qualitative value because of the slow convergence of the asymptotic expansion in
v/c . Even for systems composed of heavy quarks only, the compact size of the wave functions lead to very large matrix elements in first order perturbation theory for the
P 2 terms in eqs. (2.21) and (2.22). Therefore, it was chosen in ref. to employ a local interaction model for the heavy quarkonium systems which takes into account the minimal relativity factors (2.14), as well as the relativistic effects due to the quark spinors and the running coupling of QCD. The central, spin-independent part of the OGE interaction is thus modified to
V 0OGE (r) = − 4
32
π
∫ ∞
0
dk j 0(kr ) WQ¯q
mQm¯q
eQe¯q
αs(k2), (2.24) where the following notation has been introduced for convenience:
eQ =
√
m2
Q
k2
4 , e¯q =
√
m2¯q + k2
4 , WQ¯q =
( mQ + m¯q
eQ + e¯q
)
. (2.25) For the running QCD coupling αs(k2), the parameterization of ref. :
αs(k2) = 12 π
27 ln −1
[ k2 + 4 m2
g
Λ2QCD
]
. (2.26) has been employed. Here the QCD scale parameter Λ QCD and the dynamical gluon mass
mg, which determines the low-momentum cutoff of the inverse logarithmic behavior of
αs have been determined by a fit to the experimental spectra of the Q ¯Q and Q¯q systems. In general, the relativistic effects in eq. (2.24) lead to a strong suppression of the short-range coulombic potential. On the other hand, the running coupling αs, when employed according to eq. (2.26), increases the strength of the OGE interaction for large distances. The end result is, that the OGE interaction, when calculated using eqs. (2.24) and (2.26) bears little or no resemblance to a coulombic potential, even for the heavy c¯c system. This can potentially have serious phenomenological consequences since models that em-ploy a short-range coulombic interaction have, in general, provided good descriptions of the c¯c and b¯b spectra. However, the spin-independent part of the instanton induced interaction (2.23) has been shown in paper VI to provide the necessary short-range at-traction, even if the OGE interaction becomes weak. In principle, the effective confining interaction is also subject to similar relativistic effects, but in view of its long-range nature, their effect will be very small. The hyperfine components of the Q ¯Q interaction, as given by eqs. (2.21) and (2.22) are usually treated as first order perturbations since their behavior for small r is too singular to allow for direct numerical treatment. Modification of those hyperfine components according to eq. (2.14) leads to expressions, which are weaker and more well-behaved, and may consequently be fully taken into account. The employment of individual wave functions for each member in a given hyperfine multiplet was shown, in paper VI , to be important for a realistic description of several electromagnetic E1 and M1 transitions in the heavy quarkonium systems. 14 Chapter 2. Models for the Spectra of Q ¯Q and Q¯q Mesons
The expressions for the local hyperfine components of the Q¯q interaction that take into account the minimal relativity factors and the running QCD coupling are, in configuration space, of the form
V LS OGE = 4
3πr S · L
∫ ∞
0
dk k j 1(kr ) WQ¯q
eQe¯q
[
2 + mQ
e¯q + mq
m¯q
e ¯Q + mQ
]
αs(k2), (2.27)
V LS conf = − 2
πc
r S · L
∫ ∞
0
dk j1(kr )
kWQ¯q
eQe¯q
[ e¯q
eQ + mQ
eQ
e¯q + m¯q
]
, (2.28)
V SS OGE = 4
9π σQ · σ¯q
∫ ∞
0
dk k 2 j0(kr ) WQ¯q
eQe¯q
αs(k2), (2.29)
V TOGE = 2
9π S12
∫ ∞
0
dk k 2 j2(kr ) WQ¯q
eQe¯q
αs(k2), (2.30) for the spin-orbit, spin-spin and tensor components of the OGE interaction, and the spin-orbit (Thomas precession) term from the effective scalar confining interaction. Note that the expression (2.28) for the spin-orbit term associated with the linear scalar confining interaction is obtained by means of the representation − 8πc/ k4 in momentum space. This can be understood as the Fourier transform of a modified linear potential cr e −λr in the limit λ → 0. The integral (2.28) is convergent even if that limit is taken analytically. The above expressions are also free of singularities that require a perturbative treatment. If the QCD coupling αs is taken to be constant, then the hyperfine components, as given by eqs. (2.27)-(2.30), reduce to the static expressions of eqs. (2.21) and (2.22) for large distances. As the instanton induced interaction for Q¯q systems, as given by ref. , consists of delta functions, it has to be treated as a first order perturbation. Such a treatment is very unfortunate here since the repulsive kinetic energy as given by the BSLT quadratic mass operator (2.17) is very sensitive to the ground state energy relative to the sum of the quark masses. A perturbative treatment of a strong attractive interaction component would thus effectively lead to unrealistically small level spacings between the higher excited states. In view of this, the delta function of eq. (2.23) has been treated according to
Vinst = − ∆MQ∆M¯q
4n
∫ ∞
0
dk k 2 j0(kr ) WQ¯q
mQm¯q
eQe¯q
, (2.31) which effectively leads to a smeared-out form of the instanton induced interaction. While the presence of WQ¯q is naturally suggested by the BSLT minimal relativity factors, the
mQ/e Q factors are entirely phenomenological, and have been inserted to allow for better convergence of the above integral. In the limit of very large constituent masses (the static limit), the above equation reduces to the form (2.23). The spin-spin component of the instanton induced interaction was found in ref. to be significant for the heavy-light Q¯q systems, but very weak for the heavy-heavy Q ¯Q mesons. Because of this, the spectra shown in Table 2.1 and Fig. 2.2 do not include that interaction. The calculated
Q¯q spectra employed in paper II , shown for the D meson in Fig. 2.3, do not include the instanton induced interaction, since sufficient attraction was provided there by the OGE interaction, although at the price of an unrealistically large value for the QCD scale parameter Λ QCD .2.5. Spectra of heavy flavor mesons 15
2.5 Spectra of heavy flavor mesons
The c¯c, b¯b and B±
c
spectra that are shown in Table 2.1 and Fig. 2.2 have been obtained by solution of the BSLT equation for a linear scalar confining interaction, and OGE + instanton components modeled according to the expressions given in this section. The hyperfine components have been taken fully into account, so that all the states given in Table 2.1 are represented by different radial wave functions. This model has been employed for the calculations of the electromagnetic transitions in paper VI as well as the dipion transitions in paper IV .Table 2.1: Calculated and experimental c¯c, b¯b and B±
c
states rounded to the nearest MeV, as obtained in papers IV and VI . The states are classified according to excitation number n, total spin S, total orbital angular momentum L and total angular momentum
J. The experimental values are from ref. , except for the recently observed ηc(2 S).
n 2S+1 LJ b¯b Exp( b¯b) c¯c Exp( c¯c) c¯b
1 1S0 9401 – 2997 2980 ± 1.8 6308 2 1S0 10005 – 3640 3654 ± 6 6888 3 1S0 10361 – 4015 – 7229 4 1S0 10634 – 4300 – 7488
1 3S1 9458 9460 3099 3097 6361 2 3S1 10030 10023 3678 3686 6910 3 3S1 10377 10355 4040 4040 ± 10 7244 4 3S1 10648 10580 4319 4159 ± 20 ? 7500
1 1P1 9888 – 3513 – 6754 2 1P1 10266 – 3912 – 7126 3 1P1 10552 – 4211 – 7401
1 3P0 9855 9860 3464 3415 6723 2 3P0 10244 10232 3884 – 7107 3 3P0 10535 – 4192 – 7387
1 3P1 9883 9893 3513 3511 6751 2 3P1 10263 10255 3913 – 7125 3 3P1 10550 – 4213 – 7400
1 3P2 9903 9913 3540 3556 6770 2 3P2 10277 10269 3930 – 7136 3 3P2 10561 – 4226 – 7410
1 3D3 10158 – 3790 – 7009 1 3D2 10149 – 3784 – 7006 1 3D1 10139 – 3768 3770 ± 2.5 6998 Although preliminary, the measured mass of the B±
c
was reported in ref. as 6 .40 ±0.39 GeV, which is about ∼ 100 MeV higher than the predicted 6308 MeV, and most other models give even lower masses for the B±
c
ground state. However, the predicted B±
c
spectrum agrees very well with the QCD-inspired model of ref. . 16 Chapter 2. Models for the Spectra of Q ¯Q and Q¯q Mesons
The quality of the calculated Q ¯Q spectra in Table 2.1 is generally quite satisfactory, as both the ψ′ − J/ψ and J/ψ − ηc splittings are given realistically. In particular, the ηc(2 S)state has recently been reported by the BELLE collaboration with a mass of about 3650 MeV. This suggests that the spin-spin splitting is much smaller for the 2 S states than for the J/ψ and the ηc, a feature which is well described by the present model. The main difficulty is the prediction of the hyperfine splittings in the L = 1 multiplet of charmonium. Table 2.1 and Fig. 2.2 indicate that the splittings are underpredicted for
c¯c but in reasonable agreement with experiment for b¯b. This problem can be traced, in part, to the weakness of the OGE tensor interaction as given by eq. (2.30).
Paper VI Other models
Mb 4885 MeV 4870 MeV
Mc 1500 MeV 1530 MeV
ΛQCD 260 MeV 200-300 MeV
mg 290 MeV mg > ΛQCD
c 890 MeV/fm 912 MeV/fm
(∆ Mc)2
4n
0.084 fm 2 ∼ 0.05 fm 2
(∆ Mb)2
4n
0.004 fm 2 ?Table 2.2: Quark masses and coupling constants used for the calculated spec-tra in Fig. 2.2. The values should be considered as best fits within the BSLT model to the empirical c¯c and b¯b spec-tra. The heavy quark masses are close to those obtained by Roberts et al. in ref. within the framework of the Gross equation. The values of Λ QCD and mg are in line with those suggested in ref. , while the string tension c is somewhat smaller than that suggested by the lattice QCD calculations of ref. . The strength of the instanton induced interaction in the c¯c system is comparable to the estimate given in ref. . In spite of the generally satisfactory results, perfect agreement with experiment had to be sacrificed in the b¯b system in order to obtain an optimal description of the c¯c spectrum with the same set of parameters. As this nevertheless is a small effect, the results indicate that a flavor-independent confining interaction is a reasonable first approximation, in contrast to the instanton induced interaction, the strength of which depends explicitly on the quark flavors involved. The spectra of the heavy-light Q¯q mesons that were obtained in ref. within the framework of the BSLT equation have been used here for calculation of the M1 and pion widths of the Q¯q states. The D meson spectrum so obtained is shown in Fig. 2.3. Al-though quite satisfactory agreement with the empirical Q¯q spectra was achieved, this was only at the price of a very strong OGE interaction and the introduction of a negative constant into the scalar confining interaction, which was treated as a free parameter. The hyperfine components of the OGE and scalar confining interactions were treated as perturbations, while the instanton induced interaction was dropped since the OGE interaction was found to give sufficient attraction to account for the empirical spectra. However, later (unpublished) calculations of the Q¯q spectra have indicated that a non-perturbative treatment of the OGE spin-spin interaction strongly favors the inclusion of the instanton induced spin-spin interaction proposed by ref. , in which case the spin-independent term of that interaction may also lead to a more realistic strength and low-momentum behavior of αs.2.5. Spectra of heavy flavor mesons 17
9400 9500 9600 9700 9800 9900 10000 10100 10200 10300 10400 10500 10600 10700 1S03S11P13P03P13P23D1
ΥΥ’
ηbhbχb0 χb1 χb2 Experimental Calculated 2900 3000 3100 3200 3300 3400 3500 3600 3700 3800 3900 4000 4100 4200 4300 4400 1S03S11P13P03P13P23D1J/ ΨΨ’
Ψ(3770)
ηchcχc0 χc1 χc2 Experimental Calculated
Figure 2.2: Calculated and experimental b¯b and c¯c spectra. All states are given in MeV, and correspond to the data in Table 2.1. The thickness of the lines denoting the experimentally determined states indicates the uncertainty in the mass of the state. Note that the identification of the 4 3S1 state in charmonium is uncertain, and may actually be a 2 3D1 state, or a mixture of the two. The threshold for D ¯D decay is at ∼ 3750 MeV, and for B ¯B decay at ∼ 10500 MeV. The small number of empirically known Q¯q states makes a determination of the quality of a given model difficult. However, the most significant unsolved problem in the case of the Q¯q spectra is the ordering of the L = 1 multiplet, of which only two resonances have been detected so far. These probably correspond to spin-triplet states with J = 2 and J = 1 ( jq = 3 /2 in the heavy quark limit), and are denoted D∗
2
and D1, respectively (note the notational confusion). The empirical fact that the D∗
2
is higher in mass by ∼ 40 MeV then suggests that the ordering of the L = 1 states in the D mesons is similar to that observed for the c¯c and b¯b systems. It is reassuring that this result is consistent with lattice QCD calculations of the spin-orbit splittings in heavy-light mesons although the problem of poor convergence of such calculations is still not solved to satisfaction. In view of these results, it appears that the possibility of spin-orbit inversion in heavy-light mesons is not realized. It should be noted, however, that the spin-orbit splittings as calculated from static expressions like eqs. (2.21) and (2.22) are unrealistic because of the low masses of the light constituent quarks. An unphysical dominance of the Thomas precession associated with the scalar confining interaction may then suggest that the spin-orbit splittings of the L = 1 Q¯q states should be inverted. 18 Chapter 2. Models for the Spectra of Q ¯Q and Q¯q Mesons
1800 2000 2200 2400 2600 2800 3000 3200 3400 1S03S11P13P03P13P21D23D13D23D3Empirical and calculated D meson spectra (State energies in MeV) DDD ?D1D2Experimental Calculated
Figure 2.3: Experimental and calculated D meson spectra from ref. . It should be noted that the ex-cited 2 S state D∗∗ at ∼ 2630 MeV, which was reported by ref. was not detected by a subsequent search and may therefore not exist at that energy. It is instructive to compare the parameters of ref. with the values suggested by the more realistic calculation (paper VI ) of the Q ¯Q spectra in Table 2.2. The heavy quark masses mc = 1580 MeV and mb = 4885 MeV of ref. , though slightly higher, agree quite well with those of Table 2.2. On the other hand, the masses of the light quarks were obtained as mu,d = 450 MeV and ms = 560 MeV, respectively. Although the light quark mass is higher than the usual phenomenological value of ∼ 350 MeV, it is close to the value 420 MeV which has been employed for the instanton induced interaction. It has also been suggested that a natural value of the constituent quark mass is one third of the ∆ mass, rather than the nucleon mass. The parameters Λ QCD and mg were obtained as 280 MeV and 240 MeV in ref. and lead to a much stronger coupling αs than is necessary in paper VI . Most other analyses suggest a much weaker form. The phenomenological consequences of a very strong OGE interaction for light constituent quarks are potentially serious , since incorrect ordering of the positive and negative parity nucleon and Λ states may result. Moreover, the relativistic damping of the short-range part of the attractive coulombic OGE potential in ref. is such that the OGE interaction alone cannot describe e.g. the 1 S − 1P level splittings in the Q¯q spectra. It was therefore necessary to push the other parameters of that model to their limits. In this situation, the instanton induced interaction of ref. suggests itself naturally as it provides both a strong attraction in the S-wave and contributes to the D∗ −D splitting. Also, the OGE potential of eq. (2.24) for a system composed of two light ( ∼ 400 MeV) quarks, becomes depleted for distances equal to the meson radius. The central OGE component may therefore only play a minor role in the dynamics of light constituent quarks, while for the Q¯q mesons, the short-range attraction may turn out to be best described by a combination of OGE and instanton induced components. Chapter 3 Electromagnetic Transitions
The radiative transitions in heavy quarkonium ( c¯c, b¯b, c¯b) systems have drawn much theoretical interest [44, 45], as they can provide direct information on both the heavy quarkonium wave functions and the Q ¯Q interaction. As reasonably reliable empirical data now exists for a number of transitions in both the c¯c and b¯b systems , a fair assessment of the quality of theoretical models is already possible. The measured γ transitions in the charmonium ( c¯c) system include the E1 transitions χcJ → J/ψ γ and ψ ′ → χcJ γ, as well as the spin-flip M1 transitions J/ψ → ηcγ and ψ ′ → ηcγ. The situation concerning the analogous transitions in the bottomonium ( b¯b) system is, however, less satisfactory as the total widths of the χbJ states are not known, and none of the spin-flip M1 transitions observed. Previous calculations of the E1 widths of heavy quarkonia have demonstrated that the E1 approximation leads to overpredictions of most transition rates, and that this over-prediction can be significantly reduced, if not entirely eliminated, by the consideration of relativistic effects. On the other hand, theoretical predictions for the M1 transitions have remained unsatisfactory for a long time as the width for J/ψ → ηcγ has typically been overpredicted by a factor ∼ 3. However, calculations of M1 widths using the non-relativistic Schr¨ odinger equation in ref. and paper I have demonstrated that the M1 transitions in charmonium are sensitive both to the relativistic aspects of the spin-flip operator as well as the Lorentz structure of the Q ¯Q interaction. The results obtained in papers I and VI suggest that a scalar confining interaction may explain the observed width of ∼ 1 keV for J/ψ → ηcγ, provided that an unapproximated single quark spin-flip operator is used. This conclusion is supported by the calculation of ref. , which is based on the instantaneous approximation to the Bethe-Salpeter equation, even though a quantitative understanding of the radiative transitions in charmonium was not achieved. The situation concerning the M1 transitions in the heavy-light Q¯q mesons is more un-certain because of the scarcity of reliable empirical data. Only recently has a first mea-surement of the width of the D∗ state been published, which allows a determination of the partial width for the M1 transition D±∗ → D±γ from its known branching fraction. Even though the total width of the D0∗ is not known, the partial width for D0∗ → D0γ
can be inferred from the measured width of the D±∗ and model calculations of the pionic widths of the D∗ mesons, as the relative branching fractions for π and γ emission are known. 19 20 Chapter 3. Electromagnetic Transitions
As the velocity of the light constituent quark in the Q¯q systems is close to that of light, the relativistic corrections to both single quark and two-quark operators will a priori be large. It is shown in paper V that the large relativistic corrections to the single quark spin-flip operators, that yield unfavorable results for the M1 widths of Q¯q mesons, are in general counteracted by the two-quark operators associated with an interaction Hamiltonian that consists of OGE + scalar confinement components, thus allowing for better agreement with experiment. It is also suggested that the instanton induced interaction for Q¯q
systems of ref. may have a favorable effect on the predictions for the M1 widths of the D∗ mesons.
3.1 Charge density and electric dipole operators
The electromagnetic transition amplitude for a two-quark system, in the impulse approx-imation, is of the form
Tf i = −
∫
d3r1d3r2 ϕ∗
f
(r1, r2) ˆε · [ei q·r1 1(q) + ei q·r2 2(q)] ϕi(r1, r2), (3.1) where q and ˆε denote the momentum and polarization of the emitted photon, respec-tively, while ϕi and ϕf denote the orbital wave functions of the initial and final heavy quarkonium states. In the above equation, 1 and 2 denote the single quark current op-erators of quarks 1 and 2, respectively. By Fourier transformation, the current operators may be rewritten as
(q) =
∫
d3r′ei q·r ′
(r′) (3.2) = −
∫
d3r′r′( · ∇ ) ei q·r ′
−
∫
d3r′ei q·r ′
r′(∇ · ), (3.3) from which the E1 approximation is obtained if the exponentials in eq. (3.3) are dropped (i.e. q → 0). Application of the continuity equation then gives ∇ · = iωρ . For nonzero
q, the second term in eq. (3.3) has to be retained without approximation. Note that
(q) is taken to contain the quantity in square brackets in eq. (3.1). Application of eq. (3.3) together with eq. (3.1) then leads to the following form for the amplitude of a
γ transition,
Tf i = i |q|
∫
d3r1d3r2 ϕ∗
f
(r1, r2) ˆε · d (r1, r2) ϕi(r1, r2). (3.4) The dipole operator d(r1, r2),
d (r1, r2) =
∫
d3r′ei q·r ′
r′ ρ(r′, r1, r2), (3.5) reduces to the E1 approximation in the limit q → 0. In general, the charge density operator ρ(r′) contains, in addition to the single quark contribution ρsq , an exchange part ρex , which arises from the two-quark currents that are illustrated by the diagrams in Fig. 3.1. A necessary constraint is that two-quark contributions to the charge density must have vanishing volume integrals. The dipole operator that corresponds to the single quark charge operator ρsq (r′, r) = ρ1(r′, r1) + ρ2(r′, r2) may be expressed as
d sq (r1, r2) =
∫ d3q
(2 π)3 d3r′ ei qf ·r ′ [
ρ1(q) r′ ei q·(r1−r ′ ) + ρ2(q) r′ ei q·(r2−r ′)]
, (3.6) 3.1. Charge density and electric dipole operators 21
p1
p′
2
pa
qp′
1
VQ ¯Q(k2)
p2
p1
p′
2
pb
VQ ¯Q(k2)
p′
1
qp2
p1
p′
2
pa
qp′
1
VQ ¯Q(k2)
p2
p1
p′
2
pb
VQ ¯Q(k2)
p′
1
qp2
Figure 3.1: Two-quark contributions to the Q ¯Q current and charge density operators. In the decomposition of the Q ¯Qγ vertex, the irreducible two-quark contributions give rise to exchange current operators, the most important of which are illustrated by the upper Born diagrams. In order to obtain the correct two-quark contribution, the positive energy part of the intermediate propagator is subtracted in the lower Born diagrams, since that part is already accounted for by the impulse approximation. Note that similar diagrams describe photon emission by the heavy antiquark. In the case of the Q ¯Q interaction, the scalar confining and vector OGE components have been taken into account. The contributions from the instanton induced interaction have not been considered but are in any case small for the Q ¯Q systems. which upon evaluation yields
d sq (r1, r2) = lim
q→qf
[r1 ei q·r1 ρ1(q) + ei q·r1 i∇q ρ1(q)] + (1 → 2) , (3.7) where qf refers to the physical photon momentum of each transition. The above form reduces to the E1 expression by the substitution qf → 0. The nonrelativistic single quark dipole operator is therefore of the form
d sq (r1, r2) = Q1 r1 ei qf ·r1 + Q2 r2 ei qf ·r2 , (3.8) where Q1 is the charge of the heavy quark, while Q2 denotes that of the heavy antiquark. Insertion of eq. (3.8) into eq. (3.4) yields the single quark dipole operator
d sq (r) =
[ Q1m2 − Q2m1
m1 + m2
]
r ei qf ·r/ 2 . (3.9) 22 Chapter 3. Electromagnetic Transitions
The charge density operator in eq. (3.7) is, to second order in v/c , of the form
ρsq Q1
[
1 − q2
8m2 + iσ1 · p′
1
× p1
4m2
]
(1 → 2) , (3.10) where the second term on the r.h.s. is the relativistic Darwin-Foldy term. The effect of this term is very small because of the large masses of the heavy constituent quarks. The justification of this expansion lies in the small coefficient of that term; It has been shown e.g. in papers I and VI that such an expansion cannot be used for the magnetic moment operator. If the two-quark exchange charge operators from the Born diagrams in Fig. 3.1 are decomposed as ρex (r′, r1, r2) = ρex1 (r′, r1) + ρex2 (r′, r2), then the contribution from quark 1 may be expressed as
ρex1 (r′, r1) =
∫ d3q
(2 π)3 ei q·(r1−r ′)
∫ d3k2
(2 π)3 e−ik2 ·r ρex1 (q, k2). (3.11) Here k2 is the momentum transfered to the heavy antiquark and r is defined as r1 − r2.The exchange charge contribution to the two-quark dipole operator
d ex (r1, r2) =
∫
d3r′ei qf ·r ′
r′ ρex (r′, r1, r2) (3.12) from quark 1 may then be expressed as
d ex1 (r1) = r1 ei qf ·r1
∫ d3k2
(2 π)3 e−ik2 ·r ρex1 (qf , k2)
− lim
q→qf
[
ei q·r1 i∇q
∫ d3k2
(2 π)3 e−ik2 ·r ρex1 (q, k2)
]
, (3.13) which again reduces to the E1 approximation by the substitution qf → 0. The exchange charge density operators that are associated with the Q ¯Q interaction have been extracted in ref. , for different Lorentz invariants for equal-mass systems. When generalized to unequal quark masses, the required operators are obtained as
ρ cex = Q1
4m 31
q2 Vc(k2 ) + Q2
4m 32
q2 Vc(k1 ), (3.14)
ρ gex = Q1
4m 21
[ q · k2
m1
2
3
q · k2 σ1 · σ2
m2
]
Vg (k2 ) + Q2
4m 22
[ q · k1
m2
(3.15) + 2
3
q · k1 σ1 · σ2
m1
]
Vg (k1 ).
In the above expressions, Vc and Vg denote the Fourier transforms of the confining and OGE interactions, respectively. Evaluation of eq. (3.13) thus yields the dipole operators
d Conf ex (r) = q2
f
[ Q1
4m 31
m2
m1 + m2
− Q2
4m 32
m1
m1 + m2
]
r Vc(r) ei qf ·r/ 2 , (3.16)
d Oge ex (r) =
[ Q1
4m 21
( 1
m1
2
3
σ1 · σ2
m2
)
− Q2
4m 22
( 1
m2
2
3
σ1 · σ2
m1
)]
r ei qf ·r/ 2 ∂V g (r)
r ∂r .
Here Vc(r) is the linear confining interaction, while Vg (r) denotes the form of the OGE interaction in configuration space, which is taken to include the effects of the running coupling of QCD. 3.2. Current density and magnetic moment operators 23
3.2 Current density and magnetic moment operators
In the impulse approximation, the spin-flip magnetic moment operator for M1 transitions between S-wave heavy quarkonium states have been derived from the amplitude
Tf i = −(2 π)3 δ3(Pf −Pi −qf )
∫
d3r ϕ ∗
f
(r) ˆε·
[
ei q·r/ 2 1(q) + e−i q·r/ 2 2(q)
]
ϕi(r), (3.17) where r = r1 − r2. Expansion of the exponential in eq. (3.2) according to 1 + iq · r′
then yields the M1 and E2 amplitudes for photon emission. Upon isolation of the M1 contribution, the matrix element for J/ψ → ηcγ and ψ′ → ηcγ may be written in the form
Mf i = i
∫
d3r ϕ ∗
f
(r) q × ˆε · μsf ϕi(r), (3.18) where μsf denotes the spin-flip part of the magnetic moment operator
μ = 1
2
∫
d3r′ r′ × (r′). (3.19) In eq. (3.19), the current operator consists of a single quark contribution sq and a two-quark contribution ex , which arises from the pair terms given in Fig. 3.1. The corresponding single quark magnetic moment operator may be expressed as
μsq = 1
2
∫ d3q
(2 π)3 d3r′ [
r′ × 1(q) ei q·(r/ 2−r ′) + r′ × 2(q) e−i q·(r/ 2+ r ′)]
, (3.20) which yields
μsq = lim
q→0
[
− i
2 ∇q ×
(
ei q·r/ 2 1(q) + e−i q·r/ 2 2(q)
) ]
. (3.21) The magnetic moment operator is given by eq. (3.21) in the nonrelativistic impulse approximation. However, previous work has demonstrated that the static magnetic mo-ment operators of the baryons receive large corrections from the canonical boosts of the constituent quark spinors . Furthermore, it has been shown in paper I that the nonrelativistic impulse approximation does not provide a satisfactory description of the spin-flip magnetic moment operators for Q ¯Q systems, even though the masses of the charm and bottom constituent quarks are large. The matrix element that corresponds to eq. (3.18) in the relativistic impulse approximation is of the form
MRel
f i
= i
∫ d3P
(2 π)3 d3r d 3r′ ei P ·(r ′−r ) ϕ∗
f
(r′) q × ˆε · μ Rel sq (P ) ϕi(r), (3.22) where the final and initial state coordinates r′ and r are defined as r′
1
− r′
2
and r1 − r2
respectively. In eq. (3.22), the momentum variable P is defined as P = ( p′ + p)/2, where
p′ and p are the relative momenta in the representation p1 = P i/2 + p, p2 = P i/2 − p
and p′
1
= P f /2 + p′, p′
2
= P f /2 − p′. The relativistic single quark magnetic moment operator that appears in the matrix element (3.22) is of the form
μ Rel sq = lim
q→0
[
− i
2 ∇q ×
[
ei q·(r ′ +r )/4 (1(q, P ) + 2(q, P ))
] ]
, (3.23) where the single quark current operators i (q, P ) are now treated without approximation. 24 Chapter 3. Electromagnetic Transitions
In the nonrelativistic approximation, the spin-dependent part of the single quark current operator is given by
spin sq = ie
2 (σ1 + σ2) × q
[ Q1
2m1
Q2
2m2
]
ie
2 (σ1 − σ2) × q
[ Q1
2m1
− Q2
2m2
]
, (3.24) where the first term describes the magnetic moment of the two-quark system whereas the second term is the spin-flip operator for an M1 transition in the nonrelativistic impulse approximation (NRIA). In order to obtain the relativistic single quark current operator to be used with eq. (3.23), the nonrelativistic current operator for quark 1 should be replaced according to
1 = e Q 1
2m1
[ p1 + p′
1
iσ1 × (p′
1
− p1) ] (3.25)
−→ e Q 1
√
(E′
1
m1)( E1 + m1)
4E′
1
E1
[ p1
E1 + m1
p′
1
E′
1
m1
iσ1 ×
( p′
1
E′
1
m1
− p1
E1 + m1
) ]
,
and similarly for quark 2. In the above equation, the energy factors of the quarks are defined as E1 = √p 21 + m 21 and E′
1
= √p′21 + m 21 . The spin-flip magnetic moment oper-ator in the non-relativistic impulse approximation (NRIA) may be obtained by insertion of eq. (3.24) into eq. (3.21), giving
μ sq = e
2
[ Q1
2m1
− Q2
2m2
]
(σ1 − σ2). (3.26) The corresponding operator in the relativistic impulse approximation (RIA) has been considered in refs. [38, 48], and may for transitions between S-wave states be expressed as
μ Rel sq = e
2
[ Q1
2m1
f γ
1
− Q2
2m2
f γ
2
]
(σ1 − σ2), (3.27) where the relativistic factors f γi are defined as
f γi = mi
3Ei
[
2 + mi
Ei
]
, (3.28) where Ei denotes the energy factor Ei =
√
P 2 + m 2
i
. This shows that a relativistic treatment will lead to an effective weakening of the NRIA spin-flip operator. In addition to the above single quark current operators, the pair terms in Fig. 3.1 also give large contributions to the magnetic moment operators of mesons and baryons . However, in the case of the magnetic moments of the baryons, additional complications are known to arise from flavor dependent meson exchange interactions which also con-tribute significant exchange current operators . If the exchange current operators of Fig. 3.1 are decomposed as ex (q, k1, k2) = ex1 (q, k2)+ ex2 (q, k1), then the contribution to the two-quark magnetic moment operator may be written in the form
μex = 1
2
∫ d3q d 3k2
(2 π)6 d3r′ [
ei q·(r/ 2−r ′) eik2 ·r r′ × ex1 (q, k2) + (1 → 2)
]
, (3.29) 3.2. Current density and magnetic moment operators 25
where it is again understood that r → − r in the contribution from quark 2. Evaluation of the above equation leads to an expression analogous to eq. (3.21),
μex = lim
q→0
[
− i
2 ∇q ×
(
ei q·r/ 2
∫ d3k2
(2 π)3 e−ik2 ·r ex1 (q, k2)+ e−i q·r/ 2
∫ d3k1
(2 π)3 eik1 ·r ex2 (q, k1)
) ]
. (3.30) As the exchange current operators for most Lorentz invariants do not depend explicitly on the photon momentum q, one notable exception being that for the scalar invariant , then the exchange magnetic moment operators turn out to be difficult to calculate directly from eq. (3.30). It has therefore been shown in paper VI that a convenient way to extract the two-quark magnetic moment operators results, if eq. (3.30) is cast in the form
μ ex = lim
q→0
[
− i
2
∫ d3k
(2 π)3 e−ik·r ∇q ×
{
ex1
( q
2 + k
)
ex2
( q
2 − k
)} ]
, (3.31) which is similar to that obtained in ref. . By means of eq. (3.31), it is now possible to consider the two-quark current operators for the scalar confining and vector OGE interactions, as calculated from the diagrams in Fig. 3.1 in paper V and ref. . The two-quark current operator associated with the scalar confining interaction is then of the form
cex (q, k1, k2) = −e
( Q∗
1
P 1
m 21
Q∗
2
P 2
m 22
i
2 (σ1 + σ2) × q
[ Q∗
1
2m 21
Q∗
2
2m 22
]
i
2 (σ1 − σ2) × q
[ Q∗
1
2m 21
− Q∗
2
2m 22
])
, (3.32) where the variables Q∗
1
and Q∗
2
are defined as Q∗
1
= Vc(k2)Q1 and Q∗
2
= Vc(k1)Q2, respec-tively. The corresponding current operator for the OGE interaction may be expressed as
gex (q, k1, k2) = −e
(
Q∗
1
[ iσ1 × k2
2m 21
2P 2 + iσ2 × k2
2m1m2
]
Q∗
2
[ iσ2 × k1
2m 22
(3.33) + 2P 1 + iσ1 × k1
2m1m2
])
,
with Q∗
1
= Vg (k2)Q1 and Q∗
2
= Vg (k1)Q2. As the above equation depends only on k1 and
k2, the OGE magnetic moment operator is most conveniently calculated using eq. (3.31). The corresponding spin-flip operators for transitions between S-wave quarkonium states have been obtained in paper V as
μ Conf ex = − eV c(r)
4
{[ Q1
m 21
− Q2
m 22
]
(σ1 − σ2) +
[ Q1
m 21
Q2
m 22
]
(σ1 + σ2)
}
(3.34) for the scalar confining interaction, and
μ Oge ex = − eV g(r)
8
{[ Q1
m 21
− Q2
m 22
− Q1 − Q2
m1m2
]
(σ1 − σ2)+
[ Q1
m 21
Q2
m 22
Q1 + Q2
m1m2
]
(σ1 + σ2)
}
(3.35) for the OGE interaction. 26 Chapter 3. Electromagnetic Transitions
For equal constituent quark masses, eqs. (3.34) and (3.35) reduce to the expressions given in ref. . Note that the presence of a spin-flip term in the OGE operator (3.35) is solely a consequence of the difference in mass between the constituent quarks, and will thus not contribute to the M1 widths of the charmonium and bottomonium states. Similarly, the terms that are symmetric in the quark spins vanish for equal mass quarkonia. However, in the case of the B±
c
system, these terms will contribute to the magnetic moment of the
c¯b system. Also the spin-flip M1 transitions in the B±
c
system will receive a contribution from the OGE operator.
3.3 Widths for radiative decay
The widths for E1 dominated transitions of the type χcJ → J/ψ γ or ψ′ → χcJ γ have in paper VI been calculated according to Γ = Sf i
2Jf +1
3 q3α Mf
Mi
[ 4
9 |M 0|2 + 8
9 |M 2|2
]
, (3.36) where Jf is the total angular momentum of the final quarkonium state, and q is the momentum of the emitted photon. The widths for ψ′ → χcJ γ with J = 0 , 1, 2 are then expected to scale as 1 : 3 : 5 respectively, but that result is highly modified by the large hyperfine splittings of the Q ¯Q with L = 1. The statistical factor Sf i is defined as in ref. and assumes the values Sf i = 1 for a triplet-triplet transition and Sf i = 3 for a singlet-singlet transition. On the other hand, in paper VI the widths for transitions between D- and P -wave states were obtained from Γ = 4 Sf i
2Jf +1
27 q3α Mf
Mi
|M 0|2, Sf i = 18
{ 2 1 Jd
Jp 1 1
}2
, (3.37) where Jd and Jp are the total angular momenta of the D- and P -wave states, respec-tively. The values of Sf i are then given by the above Wigner 6 j symbol. Note that the triangularity of the 6 j symbol requires that |Jd − Jp| = 1 or 0. Transitions that change the value of J by more than one unit are thus forbidden. In eqs. (3.36) and (3.37), M0
and M2 denote radial matrix elements for S- and D-wave photon emission, respectively. The radial matrix element for S-wave emission receives contributions not only from the impulse approximation, eq. (3.9), but also from the confinement and OGE operators in eq. (3.16). That matrix element may thus be expressed as
M0 =
∫ ∞
0
dr r u f (r) ui(r) j0
( qr
2
) [
〈Q〉IA + q2 Vc(r) 〈Q〉c +
( ∂V g (r)
r ∂r
)
〈Q〉g
]
, (3.38) where ui and uf are the reduced radial wave functions for the initial and final heavy quarkonium states. Similarly, the matrix element for D-wave emission, which vanishes in the E1 approximation, is of the form
M2 = 〈Q〉ID
∫ ∞
0
dr r u f (r) ui(r) j2
( qr
2
)
. (3.39) The contribution from this matrix element is usually very small as the product qr 1for typical values of the photon momenta and quarkonium radii. Therefore, that matrix element has not been included in eq. (3.37). 3.3. Widths for radiative decay 27
The impulse approximation charge factor 〈Q〉IA , and the exchange charge factors 〈Q〉c
for the scalar confining interaction and 〈Q〉g for the OGE interaction that appear in eqs. (3.38) and (3.39) are of the form
〈Q〉IA =
[
Q1
(
1 − q2
8m 21
) m2
m1 + m2
− Q2
(
1 − q2
8m 22
) m1
m1 + m2
]
(3.40) for the impulse approximation, where the quark charge operators have been multiplied with the Darwin-Foldy terms from eq. (3.10), and
〈Q〉c =
[ Q1
4m 31
m2
m1 + m2
− Q2
4m 32
m1
m1 + m2
]
, (3.41)
〈Q〉g =
[ Q1
4m 21
( 1
m1
2
3
〈Sf |σ1 · σ2|Si〉
m2
)
− Q2
4m 22
( 1
m2
2
3
〈Sf |σ1 · σ2|Si〉
m1
)]
,
for the charge factors that are associated with the scalar confining and OGE interactions, respectively. In the spin dependent terms of eqs. (3.41), Si and Sf denote the total spins of the initial and final quarkonium states. For triplet-triplet and singlet-singlet transitions,
〈Sf |σ1 · σ2|Si〉 = +1 and −3, respectively. The charge factor 〈Q〉ID in eq. (3.39) is defined as 〈Q〉ID = lim q→0 〈Q〉IA . This is permissible since the Darwin-Foldy and exchange charge terms are very small compared to the dominant dipole contribution, which in itself is already insignificant because of the suppression by the j2 function in the matrix element. The expression for the width of a spin-flip M1 transition between S-wave heavy quarko-nium states can be written in the form ΓM1 = 16
2Si +1 q3α Mf
Mi
|M γ |2, (3.42) where Mγ denotes the radial matrix element for an M1 transition and Si is the total spin of the initial state. That matrix element consists of the relativistic impulse approx-imation, scalar confining and OGE components, according to
Mγ = MRIA
γ
MConf
γ
MOge
γ
, (3.43) where the matrix element in the relativistic impulse approximation is defined according to
MRIA
γ
= 2
π
∫ ∞
0
dr dr ′ r r ′uf (r′) ui(r)
∫ ∞
0
dP P 2 1
4
[ Q1
m1
f γ
1
− Q2
m2
f γ
2
]
j0 (r′P ) j0 (rP ) ,
(3.44) where the factors f γi are given by eq. (3.28). The matrix elements associated with the scalar confining and vector OGE interactions, which have been shown to be large in papers I,V and VI are, in the nonrelativistic approximation, of the form
MConf
γ
= −
∫ ∞
0
dr u f (r) ui(r) Vc(r)
4
[ Q1
m 21
− Q2
m 22
]
, (3.45)
MOge
γ
= −
∫ ∞
0
dr u f (r) ui(r) Vg (r)
8
[ Q1
m 21
− Q2
m 22
− Q1 − Q2
m1m2
]
. (3.46) In particular, eq. (3.45) is shown, in the next section, to provide an explanation for the experimental width of the M1 transition J/ψ → ηcγ.28 Chapter 3. Electromagnetic Transitions
3.4 E1 and M1 transitions in heavy quarkonia
A detailed comparison of the numerical results obtained in papers V and VI with exper-imental results and other theoretical calculations is instructive, as there are issues with several of the E1 and M1 transitions that are not readily apparent by casual inspection of the large amount of numerical data presented in those papers. This is even more important as the branching fractions for various transitions are typically better known than the total width of the initial state. With this in mind, the most important ones of the M1 and E1 transitions given in Tables 3.1 and 3.2 are discussed below.
3.4.1 The M1 transition J/ψ → ηc γ
The major importance of this M1 transition, from both experimental and theoretical points of view, has been stated e.g. in the review of ref. . The experimental width of 1.14 ± 0.39 keV has been difficult to explain theoretically, since nonrelativistic calcula-tions overestimate this width by a factor ∼ 3. A possible solution for this overprediction, which was already hinted at in ref. , is presented in Table 3.1, where the exchange current contribution from the scalar confining interaction brings the width down to the desired level. The importance of such negative energy components for the transition
J/ψ → ηc γ has also been demonstrated within the framework of the instantaneous approximation to the Bethe-Salpeter equation in ref. and within the Schr¨ odinger approach in paper I.If the entire Q ¯Q potential had effective vector coupling structure, which has often been suggested in the literature , then no exchange current contributions would arise, as a vector interaction contributes a significant spin-flip operator only if the quark and antiquark masses are unequal, and agreement with experiment would thus be excluded. Furthermore, a large family of effective vector confining interactions have been shown to be inconsistent with QCD by ref. . However, it has also been shown in paper I that an expansion of the RIA spin-flip operator to order v2/c 2 overestimates the relativistic correction to the static (NRIA) result, which originally led to an opposite conclusion concerning the usefulness of a scalar two-quark spin-flip operator. It was also suggested that the charm quark might possess a large anomalous magnetic moment, but that possibility has apparently not been substantiated.
3.4.2 The M1 transition ψ ′ → ηc γ
This nonrelativistically forbidden M1 transition has also proved challenging to explain theoretically, since the (near) orthogonality of the quarkonium wave functions renders the results hypersensitive to small effects. In the recent calculation by ref. , where good agreement with experiment was found for J/ψ → ηc γ, the width for ψ ′ → ηc γ was however overpredicted by almost an order of magnitude. It is shown in Table 3.1 that the M1 model employed in paper VI gives a width of ∼ 1.1 keV for that transition, which is close to the upper uncertainty limit of the current empirical result 0 .84 ± 0.24 keV . That such a favorable result is obtained depends on several factors in the present work, such as the employment of ψ ′ and ηc wave functions that model the spin-spin interaction in the S-wave. 3.4. E1 and M1 transitions in heavy quarkonia 29
The choice of approximation for the M1 matrix element is also important in this respect. The amplitude (3.17) has the additional advantage of allowing the use of a realistic pho-ton momentum in the expression (3.42) for the M1 width. Also, this treatment yields the same spin-flip operators as in the calculation of the exchange magnetic moment operators in ref. , where the rigorous M1 approximation was used. Furthermore, the M1 approx-imation has been taken to affect the entire factor in brackets in eq. (3.17), which leads to the elimination of the photon momentum q from the RIA matrix element (3.44). If the exponentials were separated from the current operators in eq. (3.17), then the width for ψ ′ → ηc γ would be overpredicted by a factor ∼ 4. However, if spin-averaged wave functions were employed, as in paper I, then the conclusion would be exactly the oppo-site; In that case the present treatment would lead to unfavorable results. As seen from Table 3.1, the exchange current operator associated with the scalar confining interaction gives the main contribution to the width for ψ ′ → ηc γ within this calculation. The present treatment of the M1 approximation may be regarded as consistent since it leads to the correct spin-flip operators and simultaneously allows a realistic photon momentum to be used. However, the large photon momentum introduces an additional uncertainty, which involves boosts on the Q ¯Q wavefunction in the final state, an effect which is yet to be considered.
3.4.3 Other M1 transitions
In principle, the width for Υ → ηb γ could be predicted with much better accuracy than the corresponding one in the c¯c system, because of the large mass of the bottom quark. In particular, the exchange current contribution from the scalar confining interaction is much smaller than for c¯c. The largest uncertainty is introduced by the unknown photon momentum for the Υ → ηb γ transition, as the mass of the ηb state is not known empirically. As realistic models of the spin-spin splittings for S-wave quarkonia give an
ηb mass around 9400 MeV, then the width for Υ → ηb γ is likely to be less than 10 eV, as given in Table 3.1. In addition to the M1 transitions discussed above, predictions have also been given in paper VI for M1 transitions between Q ¯Q states below the thresholds for fragmentation. Among these is the transition ψ ′ → η′
c
γ, which is similar to J/ψ → ηc γ. As recent experimental results indicate that the mass of the η′
c
is much higher than previously thought , then the amount of phase space available for ψ ′ → η′
c
γ is also smaller. The predicted width for that transition is thus significantly smaller than the values suggested by previous work . The width for η′
c
→ J/ψ γ is also sensitive to particulars of the model because of cancellations in the matrix element and the large photon momentum involved. The results in Table 3.1 suggest that the width for this transition should be around 2 keV. As the experimental situation concerning the η′
c
continues to improve, then the width for η′
c
→ J/ψ γ may possibly be measured in the near future. In the case of the b¯b system, the number of measurable M1 transitions is larger since the 3S states of bottomonium lie below the threshold for B ¯B fragmentation. These widths are difficult to predict and provide an important test for models of the M1 transitions. The results of paper VI suggest that the widths for transitions which do not change the principal quantum number of the quarkonium state should be highly suppressed, whereas the widths for transitions from excited ηb states to the Υ ground state are predicted to have larger widths of about 100 eV. 30 Chapter 3. Electromagnetic Transitions
Table 3.1: The M1 transitions between low-lying S-wave states in the charmonium ( c¯c), bottomonium ( b¯b) and B±
c (c¯b, ¯cb ) systems. Experimental data is available only for
J/ψ → ηcγ and ψ ′ → ηcγ. The quoted photon momenta qγ have been obtained by combination of the empirical masses of the spin triplet states with splittings given by the Hamiltonian model of paper VI in Table 2.1. Note that the OGE interaction contributes only to M1 transitions between Bc states.
Transition Matrix element [fm] Width
NRIA RIA Exch NRIA RIA RIA+Exch
J/ψ → ηcγ 4.356 · 10 −2 3.762 · 10 −2 −8.724 · 10 −3 2.85 2.12 1.25 keV
qγ : 116 MeV ex: 1 .14 ± 0.39
ψ ′ → ηcγ 3.985 · 10 −3 −5.14 · 10 −4 2.826 · 10 −3 3.35 0.06 1.13 keV
qγ : 639 MeV ex: 0 .84 ± 0.24
ψ ′ → η′
c
γ 4.344 · 10 −2 3.735 · 10 −2 −1.870 · 10 −2 0.18 0.13 0.03 keV
qγ : 46 MeV
η′
c
→ J/ψ γ −4.271 · 10 −3 −7.584 · 10 −3 5.206 · 10 −3 5.89 18.6 1.83 keV
qγ : 502 MeV
Υ → ηbγ −6.71 · 10 −3 −6.39 · 10 −3 2.41 · 10 −4 9.2 8.3 7.7 eV
qγ : 59 MeV
Υ′ → ηbγ −3.94 · 10 −4 −1.44 · 10 −4 −8.76 · 10 −5 31.8 4.3 11.0 eV
qγ : 603 MeV
Υ′ → η′
b
γ −6.70 · 10 −3 −6.39 · 10 −3 5.45 · 10 −4 0.70 0.64 0.53 eV
qγ : 25 MeV
η′
b
→ Υ γ 4.18 · 10 −4 6.30 · 10 −4 −1.31 · 10 −4 71.5 162 102 eV
qγ : 530 MeV
Υ′′ → η′′
b
γ −6.70 · 10 −3 −6.35 · 10 −3 8.02 · 10 −4 0.18 0.16 0.13 eV
qγ : 16 MeV
Υ′′ → η′
b
γ −3.59 · 10 −4 −1.11 · 10 −4 −1.55 · 10 −4 5.3 0.5 2.9 eV
qγ : 350 MeV
Υ′′ → ηb γ −2.10 · 10 −4 −6.59 · 10 −5 −3.77 · 10 −5 30.2 3.0 7.3 eV
qγ : 910 MeV
η′′
b
→ Υ′ γ 3.96 · 10 −4 6.05 · 10 −4 −2.25 · 10 −4 13.7 32.0 12.6 eV
qγ : 311 MeV
η′′
b
→ Υγ 2.05 · 10 −4 3.02 · 10 −4 −4.68 · 10 −5 68.7 149 106 eV
qγ : 842 MeV
B∗
c
→ Bcγ 1.851 · 10 −2 1.496 · 10 −2 0.326 · 10 −3 50.0 32.6 34.0 eV
qγ : 53 MeV
B∗
c′
→ Bcγ 1.015 · 10 −3 −1.437 · 10 −3 2.524 · 10 −3 179 360 206 eV
qγ : 576 MeV
B∗
c′
→ B′
c
γ 1.849 · 10 −2 1.480 · 10 −2 −5.186 · 10 −3 3.61 2.31 0.98 eV
qγ : 22 MeV
B′
c
→ B∗
c
γ −1.067 · 10 −3 −3.089 · 10 −3 1.028 · 10 −2 411 3440 39.5 eV
qγ : 507 MeV 3.4. E1 and M1 transitions in heavy quarkonia 31
3.4.4 The E1 transitions χcJ → J/ψ γ and ψ ′ → χcJ γ
The E1 transitions from the spin-triplet P -wave states are in principle the simplest to predict accurately, as the wave functions involved do not contain any nodes. Although the empirical data from ref. has suggested that the E1 widths are generally over-predicted , that discrepancy is apparently resolved by the new data presented in the 2002 edition of the PDG . However, the results in Table 3.2 indicate that the rigorous E1 approximation, with q = Mi − Mf , overpredicts the widths by a factor ∼ 2. If the E1 approximation is removed, the recoil of the J/ψ can be accounted for, in which case that overprediction is eliminated. Prediction of the widths for γ transitions from the ψ′ state has proved to be difficult, as the E1 approximation typically overpredicts the widths by at least a factor ∼ 2. The present empirical data on the ψ′ suggests that the widths for ψ ′ → χcJ γ should be around 25 keV. As demonstrated in paper VI , the E1 approximation yields widths in excess of 40 keV. This is puzzling, since recoil effects are small and cannot explain this overprediction. Also, it is seen by inspection of Table 3.2 that the predicted relative widths also do not agree well with experiment, although the experimental uncertainties are considerable. Not surprisingly, the matrix elements in Table 3.2 reveal that these transitions are very sensitive to small hyperfine effects in the Q ¯Q wave functions, which has also been demonstrated in ref. . It is therefore conceivable that small modifica-tions of the Q ¯Q wave functions may be sufficient to solve this overprediction. It should also be noted that significant reductions of the E1 widths were achieved in ref. by consideration of closed c¯q − q¯c fragmentation channels.
3.4.5 The E1 transitions from the χbJ states
The calculated widths for the χbJ → Υγ transitions agree rather well with those of the other models presented in Table 3.3, although they appear to be slightly larger. If the calculated E1 widths are used to predict the total widths of the χbJ states, then it is found that the width of the χb2 should be 164 ± 22 keV and that of the χb1 about 93 ± 22 keV. Similarly, the calculated E1 width of the χb0 suggests that the total width of that state is at least ∼ 440 keV. This situation is similar to that observed for c¯c , where the χc2 is wider than the χc1 by about a factor ∼ 2. The E1 transitions from the χbJ (2 P ) states in bottomonium provide a useful test for theoretical models since experimental data is available on all six branching fractions , even though the total widths of the χbJ (2 P ) states are not known. These data indicate that the widths for transitions to the Υ should be about one half of those for transitions to the Υ(2 S), even though much more phase space is available for the former. Indeed, it can be seen from Table 3.3 that spin-averaged wave functions do not provide a good description of the experimental branching fractions even though the hyperfine splittings of the χbJ (2 P ) states are small. On the other hand, much better agreement with experiment is obtained if the hyperfine effects are accounted for by the Q ¯Q wave functions. The calculated widths for χbJ (2 P ) → Υ(2 S) γ may be used to estimate the total widths of the χbJ (2 P ) states from the known branching fractions. The predicted width of the
χb2(2 P ) state is then 100 ± 15 keV, while that of the χb1(2 P ) is 72 ± 14 keV. The χb0(2 P )state appears to be significantly broader, but because of the large errors in the reported E1 branching fractions, only a rough estimate of 267 ± 140 keV is possible. 32 Chapter 3. Electromagnetic Transitions
Table 3.2: The E1 dominated transitions between low-lying states in the charmonium ( c¯c)system, together with the empirical data given by ref. . The column ”IA” contains the matrix element (3.38) in the impulse approximation, while in the column labeled ”DYN”, the exchange charge contributions have been included. Note that a good experimental candidate for the 3D1 state is the ψ (3770) resonance.
Transition M0 [fm] M2 [fm] Width
IA DYN E1 E1 DYN
χc2 → J/ψ γ 0.2389 0.2442 0.2632 7.145 · 10 −3 558 keV 343 keV
qγ : 429 MeV ex: 389 ± 60
χc1 → J/ψ γ 0.2464 0.2519 0.2673 5.729 · 10 −3 422 keV 276 keV
qγ : 390 MeV ex: 290 ± 60
χc0 → J/ψ γ 0.2556 0.2612 0.2701 3.345 · 10 −3 196 keV 144 keV
qγ : 303 MeV exp: 165 ± 40
ψ ′ → χc0 γ −0.2685 −0.2686 −0.2840 −6.106 · 10 −3 44.6 keV 33.1 keV
qγ : 261 MeV ex: 26 .1 ± 4.5
ψ ′ → χc1 γ −0.3126 −0.3126 −0.3202 −3.028 · 10 −3 45.8 keV 38.7 keV
qγ : 171 MeV ex: 25 .2 ± 4.5
ψ ′ → χc2 γ −0.3440 −0.3442 −0.3489 −1.871 · 10 −3 37.1 keV 33.1 keV
qγ : 127 MeV ex: 20 .4 ± 4.0
hc → ηc γ 0.2098 0.2091 0.2289 7.377 · 10 −3 661 keV 370 keV
qγ : 493 MeV
η′
c
→ hc γ −0.3420 −0.3424 −0.3465 −1.618 · 10 −3 61.5 keV 55.0 keV
qγ : 125 MeV
ψ′′ → χc0 γ −0.0456 −0.0450 −0.0199 0.926 · 10 −2 2.69 keV 9.86 keV
qγ : 577 MeV
ψ′′ → χc1 γ −0.0306 −0.0298 −0.0033 1.016 · 10 −2 0.13 keV 9.57 keV
qγ : 494 MeV
ψ′′ → χc2 γ −0.0168 −0.0161 0.0123 1.099 · 10 −2 2.38 keV 5.75 keV
qγ : 455 MeV
ψ′′ → χ′
c0
γ −0.4315 −0.4315 −0.4497 −7.344 · 10 −3 21.3 keV 17.8 keV
qγ : 153 MeV
ψ′′ → χ′
c1
γ −0.4860 −0.4861 −0.4995 −5.399 · 10 −3 42.6 keV 37.3 keV
qγ : 125 MeV
ψ′′ → χ′
c2
γ −0.5280 −0.5283 −0.5391 −4.367 · 10 −3 53.7 keV 48.2 keV
qγ : 109 MeV
3
D3 → χc2 γ 0.4164 0.4194 0.4353 – 243 keV 192 keV
qγ : 227 MeV
3
D2 → χc2 γ 0.4188 0.4219 0.4367 – 56.5 keV 45.2 keV
qγ : 221 MeV
3
D2 → χc1 γ 0.3920 0.3953 0.4145 – 262 keV 198 keV
qγ : 263 MeV
3
D1 → χc2 γ 0.4216 0.4246 0.4372 – 5.06 keV 4.13 keV
qγ : 206 MeV
3
D1 → χc1 γ 0.3963 0.3997 0.4164 – 123 keV 94.9 keV
qγ : 248 MeV
3
D1 → χc0 γ 0.3578 0.3619 0.3889 – 370 keV 251 keV
qγ : 336 MeV 3.4. E1 and M1 transitions in heavy quarkonia 33
3.4.6 The E1 transitions from the Υ states
The experimental situation concerning the Υ(2 S) → χbJ γ transitions has lately become more uncertain since the total width of the Υ(2 S) as reported by ref. , originally given as ∼ 27 keV, has increased over time and now stands at 44 ± 7 keV. This situation is analogous to that for the ψ′, which has undergone a similar, albeit smaller, increase. This has made the model predictions in Table 3.3, which originally fitted the experimental data very well, much less satisfactory. It is therefore very difficult to judge the quality of any given prediction until the experimental situation is stabilized. Still, it is noteworthy that the calculation of paper VI gives slightly better agreement with experiment than the other models in Table 3.3. As the reported total width of the Υ(3 S) state , 26 .3 ± 3.5 keV, is better known than that of the Υ(2 S) state, then it is expected that systematic uncertainties in the reported experimental results for Υ(3 S) → χbJ (2 P ) γ should be smaller than for the analogous Υ(2 S) → χbJ γ transitions. By inspection of Table 3.3, it can be seen that the Υ(3 S) → χbJ (2 P ) γ transitions are generally rather well described by a number of models, although the calculation of ref. , where spin-averaged wave functions were employed, apparently underpredicts the empirical widths. Also, the results of ref. appear to compare slightly more favorably with experiment than those of the calculation in paper VI .While the Υ(3 S) → χbJ (2 P ) γ transitions are relatively well described by different mod-els, the situation concerning the Υ(3 S) → χbJ γ transitions remains unsettled because of a strong cancellation in the E1 matrix element. However, experimental detection of these transitions may be a formidable task since the widths are an order of magnitude smaller than those of any previously measured E1 transition in the b¯b system. Inspection of the results in paper VI reveals that within the dynamical model, the width for Υ(3 S) → χb0 γ
should be the largest and that for Υ(3 S) → χb2 γ the smallest. It is encouraging that the same pattern is also predicted in Table 3.2 for the analogous transitions in the c¯c
system, where the widths are much larger relative to the other E1 transitions.
3.4.7 Other E1 transitions
The widths for E1 transitions from the ψ(3 S) state have also been given in Table 3.2. Those results suggest that the transitions to the (2 P ) states should have widths that are comparable to those for the ψ ′ → χcJ γ transitions. On the other hand, the widths for the ψ(3 S) → χcJ γ transitions are predicted to be smaller by factors 3 − 4. The empirical detection of any of these transitions will probably be difficult since the ψ(3 S) state lies above the D ¯D threshold. However, since the spin singlet hc state is well below threshold, then the photon produced in the hc → ηcγ transition may be detected in the near future. The dynamical model yields a width of 370 keV for this transition, which is then the largest E1 width in the c¯c system, although the difference between the E1 approximation and the dynamical model is large for that transition. The E1 transitions from the 3D1 state are also of particular interest, as that state proba-bly corresponds to the empirical ψ(3770) resonance. The predictions of paper VI suggest that the transitions to the χc1 and χc0 states should be detectable by experiment, whereas that to the χc2 state is highly suppressed by the statistical factor Sf i .34 Chapter 3. Electromagnetic Transitions
Table 3.3: Comparison of the predicted E1 widths in the bottomonium ( b¯b) system with those of other models that use a scalar confining interaction. All widths are given in keV. The experimental widths have been extracted from the branching fractions and total widths reported by ref. .
GS (ref. ) GZ (ref. ) paper VI Exp (ref. )
χb2 → Υ γ 33.0 33.8 36.0 22 ± 3%
χb1 → Υ γ 29.8 30.4 32.5 35 ± 8%
χb0 → Υ γ 25.7 25.3 26.6 < 6 %
Υ′ → χb0 γ 0.73 0.76 1.01 1.7 ± 0.5 keV Υ′ → χb1 γ 1.62 1.37 1.80 3.0 ± 0.7 keV Υ′ → χb2 γ 1.84 1.45 2.03 3.1 ± 0.7 keV
χ′
b2
→ Υ′ γ 12.9 16.2 16.4 16 .4 ± 2.4 %
χ′
b1
→ Υ′ γ 11.9 14.7 15.1 21 ± 4 %
χ′
b0
→ Υ′ γ 10.6 12.3 12.3 4.6 ± 2.1 %
χ′
b2
→ Υ γ 18.2 10.4 11.4 7.1 ± 1.0 %
χ′
b1
→ Υ γ 11.8 7.51 8.40 8.5 ± 1.3 %
χ′
b0
→ Υ γ 6.50 3.57 4.93 0.9 ± 0.6 %
Υ′′ → χb0 γ 0.114 0.029 0.15 –Υ′′ → χb1 γ 0.003 0.095 0.11 –Υ′′ → χb2 γ 0.194 0.248 0.04 –
Υ′′ → χ′
b0
γ 1.09 1.30 1.14 1.4 ± 0.3 keV Υ′′ → χ′
b1
γ 2.15 2.34 2.12 3.0 ± 0.5 keV Υ′′ → χ′
b2
γ 2.29 2.71 2.50 3.0 ± 0.6 keV The determination of the photon momenta for transitions in the bottom-charm B±
c
sys-tem has to rely on model predictions for the masses of the c¯b states. The uncertainty introduced by this is, however, rather small for the E1 transitions, as the model predic-tions for the major level splittings agree with each other to a large extent . Inspec-tion of the results in paper VI reveals that the predicted widths are similar to those obtained by ref. , although differences exist for transitions like B∗
c
(2 S) → B∗
c0
γ and
B∗
c2
(2 P ) → B∗
c
(2 S) γ, where the widths are sensitive to the effects of the hyperfine com-ponents of the Q ¯Q interaction. It is noteworthy that while the predicted widths for the
B∗
cJ
→ B∗
c
γ transitions agree rather well with those from ref. , there is a signifi-cant disagreement for Bc1 → Bc γ. When the somewhat different photon momenta are accounted for, this disagreement amounts to about a factor ∼ 3. An issue not considered in paper VI is the spin mixing of the L = 1 states with J = 1, that is due to the antisymmetric spin-orbit interaction which was not included in the Q ¯Q
interaction Hamiltonian. This mixing, which was considered in ref. , has the effect of allowing ”spin-flip” E1 transitions of the type B∗
c1
→ Bc γ. However, the widths for such ”forbidden” transitions were found in ref. to be typically suppressed by a factor
∼ 100 relative to the ”allowed” ones considered in paper VI .3.5. M1 transitions in heavy-light mesons 35
3.5 M1 transitions in heavy-light mesons
The M1 widths of the spin-flip M1 transition between the vector and pseudoscalar states in the charm mesons have been calculated in paper V, where the Q¯q interaction was modeled as a scalar confining + OGE potential. A comparison of the results presented in Table 3.4 is instructive since the total width of the D±∗ state has recently been measured by the CLEO collaboration . The width is reported as Γ( D±∗ ) = 96 ±4±22 keV, where the latter error represents the systematic uncertainty. The reported branching ratio of 1 .6 ± 0.4% for radiative decay then gives a width of 1 .5 ± 0.6 keV for the M1 transition
D±∗ → D±γ. Here most of the uncertainty can be traced to the systematic errors of the experimental result. Such a comparison shows that the results of Table 3.4 reproduce the empirical width fairly well for a range of values of the light constituent quark mass
mq. The value mq = 450 MeV corresponds to the potential model of ref. , while the value mq = 420 MeV has been suggested in ref. . Indeed, for a light constituent quark mass of 420 MeV, a width for radiative M1 decay which is close to 1.5 keV is reproduced. Values close to that are also favored by the analysis within the framework of the Gross equation by ref. .
D0∗→D0γNRIA RIA RIA + Exch
450 MeV 21.1 keV 8.86 keV 8.95 keV
420 MeV 23.5 keV 9.18 keV 9.89 keV
390 MeV 26.4 keV 9.52 keV 11.1 keV
D±∗ →D±γNRIA RIA RIA + Exch
450 MeV 0.58 keV 9.4·10 −3keV 1.09 keV
420 MeV 0.79 keV 1.5·10 −2keV 1.43 keV
390 MeV 1.07 keV 2.2·10 −2keV 1.90 keV
D±∗
s→D±
sγNRIA RIA RIA + Exch
560 MeV 0.18 keV 2.6·10 −4keV 0.38 keV
530 MeV 0.26 keV 3.9·10 −5keV 0.49 keV
500 MeV 0.36 keV 8.6·10 −4keV 0.64 keV
Table 3.4: Numerical results from paper V for the M1 transi-tions between ground state vec-tor and pseudoscalar mesons in the D and Ds systems, for different values of the light and strange constituent quark mass-es. The charm quark mass of 1580 MeV corresponds to the potential model of ref. . In the right-hand column, the two-quark exchange current contri-butions from the scalar confin-ing and OGE interactions have been added to the RIA result. Even though the total width of the neutral D0∗ meson has not yet been determined , considerable information about the expected width for D0∗ → D0γ may be extracted from the reported branching fraction of 38 .1 ± 2.9% , since the corresponding width for pion emission can be constrained by means of the empirically determined width of the
D±∗ and model calculations of the pionic transitions in D mesons [56, 57]. If one notes that the branching fraction of D±∗ → D0π± is reported as 67 .7 ± 0.5% which implies a width for this transition of ∼ 65 ± 14 keV, then it is found from the model calculation of paper III that this corresponds to a width of ∼ 40 ± 10 keV for D0∗ → D0π0. As the relative branching fractions for π0 and γ emission by the D0∗ are well known , the 36 Chapter 3. Electromagnetic Transitions
best estimate for the width of D0∗ → D0γ is ∼ 25 keV, which is close to that preferred by ref. . There remains, however, a considerable uncertainty of ∼ ± 10 keV from the systematic errors in the empirical measurement of Γ( D±∗ ). The results of paper V in Table 3.4 underpredict this expectation by about a factor ∼ 2, which underlines a basic weakness of the present approach to the M1 transitions in the
D mesons. Firstly, as the two-quark spin-flip operators of eqs. (3.34) and (3.35) are given for the nonrelativistic approximation, a relativistic treatment of those operators will, in general, lead to a weakening of the two-quark contributions to the matrix element for an M1 transition. It is thus entirely possible that a fully relativistic treatment will render the two-quark contributions too weak to account for the experimental data on M1 decay of the D± meson as well. This conclusion is in line with that reached in ref. , where the discrepancy between the model and experiment was parameterized in terms of a large anomalous magnetic moment for the light quark. Another interesting possibility discussed in paper V suggests itself, in view of the prob-lems of fitting the spectra of the Q¯q mesons within the framework of the BSLT equation using a OGE interaction alone, namely that the instanton induced interaction for Q¯q
systems given in ref. may also contribute a significant two-quark current. The inter-action of ref. , which was found to be short-ranged, attractive and with negative sign, has scalar coupling to the light constituent quark. It has been noted in paper V that such an interaction adds up constructively with the OGE contribution, and counteracts that from the scalar confining interaction. An overall favorable effect on the widths for M1 transitions may thus be obtained, which may be inferred from the matrix elements given in paper V.Chapter 4 Pionic Transitions
The pionic transitions in the heavy-light D mesons are instructive, as they can provide direct information on the strength of the coupling between pions and light constituent quarks. Furthermore, as the charm quark in the D mesons does not couple to pions, the decay mechanism is governed by the pion coupling to the light constitutent quark alone. As a first approximation, the pion-light constitutent quark coupling can be taken as being independent of the quark-antiquark interaction in the D meson. It has been shown in paper II , where the pion emission was modeled in terms of the chiral pseudovector Lagrangian, that while this assumption is reasonable for the axial current term, it leads to a large overestimate of the axial charge contribution. In this case two-quark pair terms analogous to those that are required for a realistic description of the M1 transition
J/ψ → ηcγ may reduce the axial charge amplitude to a realistic level. An analogous suppression of the S-wave pion transitions was achieved in ref. within the framework of the Gross quasipotential reduction. The empirical information on the widths and branching fractions of excited charm mesons is still very limited. Absolute values, albeit with large uncertainties, are known for the
D1(2420) and D∗
2
(2460) mesons, and recently the width of the D∗(2010) ±, for which only an upper bound of 0.131 MeV was available earlier, has now been reported as 96 ± 4 ± 22 keV by the CLEO collaboration . This result is shown to be consistent with values of the pion-quark axial coupling gqA that are slightly smaller than 1. The observed widths of the L = 1 D1(2420) and D∗
2
(2460) mesons are also shown to be fairly well described, although a slight underprediction is expected as it has been demonstrated in paper III that two-pion ππ decay may also contribute significantly to the total widths of those mesons. In particular, the analogy with the K∗
2
(1430) strange meson suggests that ππ transitions may account for a significant fraction of the observed total widths. The flavor symmetry breaking decay mode D∗
s
→ Dsπ0, which has been considered in paper V is mainly due to a small isoscalar η meson component in the physical π0 meson, as only the η meson can couple to the strange quark in the Ds meson. The known ratio of the branching fractions for D∗
s
→ Dsπ0 and D∗
s
→ Dsγ may be used to extract the coupling of η mesons to strange quarks, once the value of the π0−η mixing angle is known. Applied to the quark model for the baryons, an ηN N pseudovector coupling constant, small enough to be consistent with the phenomenological analysis of photoproduction of the η on the nucleon and the reaction pp → ppη , has been obtained in paper V.37 38 Chapter 4. Pionic Transitions
4.1 The amplitude for pion emission
In paper II , the emission of pions from a D meson was described by the pseudovector La-grangian, which constitutes the lowest order chiral coupling for pions to light constituent quarks:
Lqqπ = i gqA
2fπ
¯ψq γ5γμ ∂μ πaτa ψq . (4.1) Here gqA denotes the axial coupling constant of pions to light constituent quarks, and fπ
is the pion decay constant, the empirical value of which is 93 MeV. The axial coupling constant is conventionally taken to be equal to, or somewhat less than, unity . The Lagrangian (4.1) yields the following transition amplitude for pion emission:
Tπ = ξ gqA
2fπ
¯uq(p′)γ5γμqμ uq(p), (4.2) where ξ is an isospin factor, the value of which is √2 for π± and 1 for π0 emission. Decomposition of the above amplitude into axial current and axial charge components yields the pion emission amplitude in the impulse approximation,
Tπ = −i ξ gqA
2fπ
√
(E′
q + mq )( Eq + mq)
4E′
q Eq
[
1 − P 2 − q2/4
3( E′
q + mq )( Eq + mq )
]
σq · q
+i ξ gqA
2fπ
2mq + Eq + E′
q
√
4EqE′
q (Eq + mq)( E′
q + mq )
ωπ σq · P , (4.3) where the first (axial current) term gives rise to the P -wave transitions D∗ → Dπ and the D-wave transitions from the L = 1 charm mesons. On the other hand, the axial charge term leads to S-wave transitions from states with L = 1.
p1
p′
2
pa
qp′
1
VQ¯q (k2)
p2
p1
p′
2
pb
VQ¯q (k2)
p′
1
qp2
Figure 4.1: Irreducible two-quark contributions associated with the Q¯q interaction to the axial current and axial charge operators, with four-momentum variables defined as for Fig. 3.1. The pion always interacts with the light constituent quark, as the charm quark does not couple to pions. 4.1. The amplitude for pion emission 39
As demonstrated in paper II , two-quark axial exchange current operators illustrated by Fig. 4.1 give large contributions to several π transitions, in particular to those which involve the axial charge amplitude. In that case the axial exchange charge operator is of equal magnitude as the single quark operator, while the axial exchange current operators typically represent ∼ 10% corrections to the impulse approximation result. If, in the static approximation, the axial current Aμa = ( Aa, iA 0a) of the light constituent quark is expressed as
Aa = −gqAσq τa, (4.4) then the contribution to the axial current from a scalar confining interaction is, to lowest order in v/c , of the form
AConf
a
= − gqA
4m 3
q
Vc(k2)
[
3 σq P 2 − 1
4 σq k22 − 4 P σ q · P + 2 i P × k2
]
τa, (4.5) where Vc(k2) is the Fourier transform of the scalar confining interaction. The above expression does not include the corrections from the canonical boost factors on the single quark spinors that are included in the single quark operator, eq. (4.3). Moreover, a factor
m−2
q
in the axial exchange current operator (4.5) arises as the static approximation to the propagator of the intermediate negative energy quark. Hence a more realistic evaluation requires that those factors are taken into account. For simplicity, the same spinor factors as for the single quark operator were used in paper II . The so obtained results indicate that the static approximation implies a very large overestimate of the axial exchange current contribution. Nevertheless, as shown in Tables 4.1 and 4.2, it serves to increase the calculated widths for pion emission. Acomplete calculation would also require consideration of the axial exchange currents associated with the short-range OGE or instanton induced interactions, which was not attempted in paper II .Both the scalar confining and OGE interactions contribute a two-quark operator to the axial charge amplitude in eq. (4.3). These operators have been calculated in ref. and are of the form
AConf 0a = gqA
m 2
q
Vc(k2) σq · P τa, (4.6) for the scalar confining interaction, and
AOge 0a = gqA
mq M ¯Q
Vg (k2)
[
σq · P ¯Q + 1
2 σq × σ ¯Q · k2
]
τa, (4.7) for the OGE interaction, which, when compared with the single quark axial charge
A0a = − gqA
mq
σq · P τa (4.8) reveals that the scalar confining interaction will tend to cancel out the axial charge component of the amplitude for pion emission. However, the calculations in paper II
show that the OGE contribution is also large, although formally suppressed by a factor
mq/M ¯Q. It has also been shown in paper II that a relativistic treatment of the axial exchange charge operators will weaken them significantly. Although qualitative, these conclusions are in line with the experimentally small width of the L = 1 , J = 1 D1
resonance. 40 Chapter 4. Pionic Transitions
4.2 The pionic widths of the D mesons
The pionic decays of the D∗ mesons presented in Table 4.1 are intriguing since the emitted pions are extremely soft. Due to the very small phase space, the transition
D∗0 → D±π∓ is kinematically forbidden. The results appear to favor the value of
gqA = 0 .87, although that is accidental since only an upper limit on the width of the D±∗
was known when paper II was published. For this value of gqA, the total widths of the L = 1 D mesons in Table 4.2 appear to be underpredicted, although ππ transitions may contribute significantly to these, as proposed in paper III . However, the empirical widths are still rather uncertain , and have decreased over time. The results for the pionic widths of the excited D mesons are rather similar to those of ref. , especially for the transitions D∗ → Dπ , even though the Gross framework was employed in that paper. That calculation was restricted to the transitions allowed by the lowest order selection rules suggested by Heavy Quark Symmetry (HQS) , whereas the present work uses the LS -coupling scheme, which is more appropriate for equal mass quarkonia. The connection between the present calculation and the heavy quark limit remains as yet unexplored. Table 4.1: The calculated and experimental [46, 55] pionic widths in MeV for the D∗
mesons, corresponding to gqA = 0.87. The single quark approximation, with relativistic corrections is denoted RIA, and the result obtained upon addition of the axial exchange current contribution is denoted RIA + EXCH. The net results are also shown for gqA = 1. Transition π mom. RIA RIA + EXCH gqA = 1 Experiment
D∗± → D±π0 38.3 keV 0.026 0.029 0.038 0.029 ± 0.008 MeV
D∗± → D0π± 39.6 keV 0.056 0.064 0.084 0.065 ± 0.017 MeV
D∗0 → D0π0 43.1 keV 0.036 0.041 0.054 < 1.3 MeV Table 4.2: Calculated and empirical pion decay widths of the D1 and D∗
2
mesons driven by the axial current and charge operators respectively, for gqA = 0 .87. The empirical values are total widths , which should mainly be due to pionic transitions to the ground state. The numbers in parentheses are the widths obtained without the axial exchange current contribution. The calculated values are also shown for gqA = 1. Transition Current (RIA) Charge Total gqA = 1 Experiment
D1 → D∗π 4.2 (3.4) 6.1 10.3 MeV 13.6 18 .9+4 .6
−3.5
MeV
D∗
2
→ Dπ 8.1 (6.7) – 8.1 MeV 10.6 –
D∗
2
→ D∗π 3.9 (3.1) – 3.9 MeV 5.1 –
D∗
2
→ Dπ + D∗π 11.9 (9.9) – 11.9 MeV 15.7 25 +8
−7
MeV 4.3. π0 and γ transitions from the D∗
s
meson 41
4.3 π0 and γ transitions from the D∗
s meson
Consideration of the coupling of the octet of light pseudoscalar mesons to the light ( u, d, s )quarks yields, analogously to eq. (4.1) the couplings
Lqqϕ = i gqA
2fϕ
¯ψq γ5γμ∂μ ϕaλa ψq . (4.9) For the pions and the η meson, the empirical decay constants are fπ = 93 MeV and
fη = 112 MeV, respectively, so at least for the decay constants SU (3) flavor symmetry is broken only at the 10% level. Combination of the chiral coupling (4.9) with the representation
ϕaλa = √2
π0
√2
η0
√6
π+ K+
π− − π0
√2
η0
√6
K0
¯K− ¯K0 −
√ 2
3
η0
, ψq =
uds
(4.10) gives the following definitions for the quark-level pseudovector coupling constants fϕqq ,
fπqq = mπ
2fπ
gqA, fηqq = mη
2√3fη
gqA, fηss = − mη
√3fη
gqA. (4.11) The above relations then suggest that the magnitude of the coupling of η mesons to
u, d quarks should be one-half that of the η coupling to strange quarks, independently of the η meson mass. In the static quark model the meson-quark coupling constants of eq. (4.11) are related to the meson-nucleon coupling constants as
fπN N = 5
3 fπqq , fηN N = fηqq . (4.12) Application of the relations (4.11) together with eq. (4.9) yields a coupling of η mesons to strange quarks in terms of fηss . The coupling of the π0 meson to the strange quark may thus be expressed in terms of the ”effective” mixing angle θm, which corresponds to the sum of the π0−η and π0−η′ contributions. The width for the process D∗
s
→ Dsπ0 can then be conveniently obtained from the expression for D±∗ → D±π0 given in paper II
by the replacement gqA/2fπ → fηss θm/m η, giving Γ( D∗
s
→ Dsπ0) = 1
6πMDs
MD∗
s
f 2
ηss
m2
η
θ2
m
q3
π
|M π |2, (4.13) if the π0 emission takes place at the strange quark. Here Mπ is a radial matrix element for pion emission. On the other hand, the width for the radiative M1 transition D∗
s
→ Dsγ
was obtained, in paper V, as Γ( D∗
s
→ Dsγ) = 16
3
MDs
MD∗
s
α q 3
γ
|M γ |2, (4.14) where Mγ is a radial matrix element for M1 decay. By means of eqs. (4.13) and (4.14), the ratio of the π0 and γ widths of the D∗
s
meson is then obtained as Γπ
Γγ
= 8
9πf 2
ηss
θ2
m
m2
η
α
( qπ
qγ
)3 ( |M π |
|M γ |
)2
, (4.15) where the dimension of |M γ | is [MeV] −1.42 Chapter 4. Pionic Transitions
4.4 Estimation of fηN N
Through use of the empirical ratio of pion and photon momenta known to be ap-proximately 139/48 and the η meson mass of 547 MeV one may solve for the coupling constant fηss to get
f 2
ηss
= θ−2
m
Γπ
Γγ
( |M γ |
|M π |
)2
· 4.814 fm −2. (4.16) As the ratio of the π0 and γ decay rates is experimentally known, albeit with quite large errors, to be 0 .062 ± 0.028 , it is, given the rather well known value of θm, possible to obtain an estimate for the coupling constant fηss . However, if the charm quark also couples to π0, which is suggested by the empirically detected transitions ψ′ → J/ψ η
and ψ′ → J/ψ π 0, then eq. (4.16) should be modified. The appropriate modifications are given in paper V, where it was found that a significant coupling of the π0 to the charm quark will increase the value of fηN N . For that calculation, a matrix element for
π0 emission by the charm quark is also required. The matrix elements required for eq. (4.16) were in paper V found to be
Mγ = −1.22 · 10 −2 fm for the M1 transition D∗
s
→ Ds γ, and
Msπ = 0 .794 , Mcπ = 0 .949, for π0 emission by the strange and charm quarks. Assuming that in the π0 emission by the D∗
s
, the pion couples mostly to the strange constituent quark, eq. (4.16) may be used directly together with the above matrix elements for the π0 and γ transitions. Insertion of those matrix elements for a value of the π0 − η mixing angle of θm ∼ 0.012 yields
|fηss | ∼ 0.70. If the uncertainties in the mixing angle and the empirical widths for the
π0 and γ transitions are taken into account, then the best estimate of paper V for the coupling of the η meson to strange constituent quarks is
fηss = − 0.7 +0 .5
−0.3
.In the above result, the negative sign is suggested by the relations in eq. (4.11). The static quark model then implies, through eq. (4.12), that the magnitude of the corresponding pseudovector η-nucleon coupling constant fηN N should be one half of this value. Thus one obtains the following final result for the η-nucleon coupling:
fηN N = 0.35 +0 .15
−0.25
This result should be compared with the value for fηN N or the equivalent pseudoscalar coupling constant gηN N = (2 mN /m η)fηN N , which has been determined by phenomeno-logical model fits to photoproduction of η mesons on the nucleon . The latter value for fηN N is ∼ 0.64. That value has also been found to be realistic in calculations of the cross section for pp → ppη near threshold . Although the result obtained above for
fηN N has quite large uncertainties which are mostly of empirical origin, it still appears to be significantly smaller. A larger value for fηN N could, of course, be obtained by decreasing the mixing angle θm.Chapter 5 Two-pion Transitions
As the orbitally excited L = 1 D1(2420) and D∗
2
(2460) charm meson states lie well above the threshold, not only for single pion but also for two pion decay, then it is likely that a significant fraction of their total widths are made up by ππ transitions to the ground state
D and D∗ mesons. It is thus particularly instructive to obtain theoretical predictions and empirical information on the branching ratios for the latter decay modes. At present, however, the total widths of the D1(2420) and D∗
2
(2460) states are known only within a very wide uncertainty range, and the remaining two members of the L = 1 multiplet have not yet been discovered. Paper III reports a calculation of the ππ decay widths of the excited L = 1 charm meson states, by extending a similar calculation of the widths of their single pion transitions in paper II .Two pion emission from radially excited heavy quarkonium ( Q ¯Q) states empirically con-stitutes a significant fraction of their total decay widths . Indeed, in the case of the
ψ′ (or ψ(2 S)), the branching ratio is empirically as large as ∼ 50%. As the charm quarks themselves do not couple to pions, the ππ coupling to heavy flavor mesons (or quarks) involves at least two gluons if not a glueball. A number of different theoretical approach-es for the coupling of two-pions to heavy mesons have been proposed, from effective field theory descriptions and directly QCD-motivated models to phenomenological models . In ref. , a Lagrangian motivated by chiral perturbation theory has been fitted to experiment. In paper IV , the ππ transitions from excited Q ¯Q states has been described by a derivative Qππ coupling, mediated by a heavy scalar resonance.
5.1 The width for a ππ transition
The ππ width of an excited heavy flavor meson is of the form Γππ = (2 π)4
∫ d3ka
(2 π)3
d3kb
(2 π)3
d3Pf
(2 π)3
Mf Mi
Ef Ei
|Tf i |2
4ωaωb
δ(4) (Pf + ka + kb − Pi), (5.1) where ka and kb are the four-momenta of the emitted pions, Pi and Pf are those of the initial and final state quarkonia, while ωa and ωb denote the energies of the emitted pions, respectively. The factors M/E are normalization factors for the heavy meson states similar to those employed in ref. . 43 44 Chapter 5. Two-pion Transitions
Evaluation of the above expression leads to the following form for the differential width of a ππ transition,
dΓππ
dΩq
= 1
41
(2 π)4
∫ qf
0
dq q 2 (5.2)
∫ 1
−1
dz Q2
f
(q, z )
ωa(q, z ) (Qf + qz
2
) + ωb(q, z ) (Qf − qz
2
) Mf
Ef (q) |Tf i |2,
where the variable z is defined by Q · q = Qqz , and the energies of the pions and the final state quarkonium are given by
ωa =
√
m2
π
Q2
f
q2/4 − Qf qz, (5.3)
ωb =
√
m2
π
Q2
f
q2/4 + Qf qz, (5.4)
Ef =
√
q2 + M 2
f
. (5.5) In the above expressions, the relative momentum of the emitted pions has been fixed by the delta functions in eq. (5.1), and can be expressed as
Q2
f
= (Ef − Mi)4 − (4 m2
π
q2)( Ef − Mi)2
4( Ef − Mi)2 − 4q2z2 . (5.6) The integration limit qf corresponds to the maximal momentum of any one of the final state particles, e.g. the final state quarkonium. Thus qf can be calculated as the q-value of a transition A′ → AX , where X is a particle with mass MX = 2 mπ .The above formalism is adapted for computation of a ππ width using a hadronic matrix element Tf i . However, since experiments generally measure the invariant mass √sππ ,then it is useful to define a dimensionless variable x,
x = mππ − 2mπ
∆M , (5.7) where mππ denotes the invariant mass √sππ of the two-pion system and ∆ M = Mi −
Mf − 2mπ. The relation between q (= |q|) and mππ may then be obtained as
|q| =
[M 2
i
− (Mf + mππ )2] [ M 2
i
− (Mf − mππ )2]
4M 2
i
. (5.8) From this relation, the transformation Jacobian may be obtained as
dq
dx = ∆M
|q|
{[ M 2
i
− (Mf + mππ )2
4 M 2
i
]
(Mf −mππ ) −
[ M 2
i
− (Mf − mππ )2
4 M 2
i
]
(Mf +mππ )
}
,
(5.9) where |q| is given in terms of mππ by eq. (5.8). The width for a ππ transition may thus be calculated as Γππ =
∫ qf
0
dq dΓππ
dq = −
∫ 10
dx dΓππ
dq dq
dx . (5.10) In the above equation, the latter form turns out to be the most useful, since the ex-perimental ππ spectra are presented either in terms of Γ −1dΓ/dx or dΓ/dx versus x.5.2. ππ transitions in heavy-light mesons 45
5.2 ππ transitions in heavy-light mesons
The emission of two pions from the light constituent quarks is modeled in paper III in terms of the conventional chiral pion-quark pseudovector coupling model, augmented by a pointlike Weinberg-Tomozawa term. The only free parameter in this model is the axial coupling gqA of the light constituent quarks, as the quark masses and other parameters of the Hamiltonian model were fitted to the D meson spectrum. In addition to the single quark amplitude for two-pion emission, the exchange current contribution to the Weinberg-Tomozawa interaction was also considered, and found to interfere destructively with the single quark amplitude. The pion-quark pseudovector coupling (4.1) gives rise to Born and crossed Born ampli-tudes of conventional form for the emission of two pions from an interacting constituent quark. The chiral model for the ππ emission amplitude is completed by the Weinberg-Tomozawa (WT) interaction, which is described by the Lagrangian
LWT = − i
4f 2
π
¯ψq γμ τa πa × ∂μπa ψq . (5.11) The general isospin decomposition of the ππ emission amplitude for constituent quarks is, in analogy with that for nucleons,
T = δab T + + 1
2 [τb, τ a]T −, (5.12) while the general expression for the amplitudes T + and T − is
T ± = ¯ u(p′) (A± − iγ μQμB±) u(p). (5.13) In these expressions, Q denotes the relative four-momentum Q = ( kb − ka)/2 of the emit-ted pions. In this notation the Born, crossed Born and Weinberg-Tomozawa amplitudes are, respectively
TB = i
( gqA
2fπ
)2 [
γμQμ − 2im q − 4m2
q
γμQμ
p2
a
m2
q
] (
δba + 1
2 [τb, τ a]
)
, (5.14)
TCB = −i
( gqA
2fπ
)2 [
γμQμ + 2 im q − 4m2
q
γμQμ
p2
b
m2
q
] (
δba − 1
2 [τb, τ a]
)
, (5.15)
TWT = −i γμQμ
4f 2
π
[τb, τ a]. (5.16) Comparison of these amplitudes with eq. (5.13) yields the desired expressions for the sub-amplitudes A± and B±, which are of the form
A+ =
( gqA
2fπ
)2
4mq, (5.17)
A− = 0, (5.18)
B+ = −
( gqA
2fπ
)2
4m2
q
[ 1
s − m2
q
− 1
u − m2
q
]
, (5.19)
B− = −
( gqA
2fπ
)2 (
2 + 4 m2
q
[ 1
s − m2
q
1
u − m2
q
])
1
2f 2
π
. (5.20) 46 Chapter 5. Two-pion Transitions
Here the identities p2
a
= −s and p2
b
= −u, where s and u are the invariant Mandelstam variables, have been used. These results for the A and B amplitudes are formally equiv-alent to the corresponding results for the two-pion emission amplitude for nucleons . Note that in eq. (5.20), the contribution from the Weinberg-Tomozawa interaction tends to cancel the constant term in the B− amplitude that arises from the Born terms. If the axial coupling constant gqA is taken to equal 1, then this cancellation is exact. For calculational purposes, eq. (5.13) was split in paper III into spin-independent and spin-dependent parts according to
T ± = α± + iσq · β±. (5.21) The spin and isospin summed squared amplitude for a given ππ transition is then ex-pressed in the form
|T |2 = |T |2
α+
|T |2
α−
|T |2
β+
|T |2
β−
, (5.22) from which the ππ width is obtained by insertion into eq. (5.2). The explicit expressions for the above squared amplitudes, given in paper III , are different for each transition because of the summation over spin and isospin. The numerical results for the above squared amplitudes are presented in Table 5.1. It is instructive for the determination of the two-quark contribution to the Weinberg-Tomozawa interaction to write the Lagrangian (5.11) in the form of a current-current coupling,
LWT = − 1
4f 2
π
Vμa πa × ∂μπa, (5.23) where Vμa = i ¯ψq γμ ψq τa is the isovector current of the light constituent quark and
πa × ∂μπa is the current of the ππ system. Given this expression, it becomes natural to describe the irreducible two-quark contribution to the ππ production operator by means of two-quark interaction current contributions to the isovector current Vμa .In the non-relativistic limit, the spatial term in the isovector current Vμa = ( V a, iV 0a)of the light constituent quark takes the form
V a =
[ p′
q
pq − iσq × q
2mq
]
τa, (5.24) where mq is the light constituent quark mass and pq and p′
q
the initial and final quark momenta, respectively. The expressions for the exchange current contributions for the scalar confining and OGE interactions have been calculated in ref. , and may be expressed as
V Conf
a
= − Vc(k2)
mq
V a, (5.25)
V Oge
a
= − Vg(k2)
2m2
q
[ mq
MQ
(p′
Q
pQ + iσQ × k2
) + iσq × k2
]
τa. (5.26) In paper III , the above operators contribute only to |T |2
α−
, and were found to reduce the widths for ππ transitions. However, it was also found that the nonrelativistic treatment of the two-quark operators is somewhat unrealistic because of the small mass of the light constituent quark. An approximate relativistic treatment akin to that employed for the single pion transitions in paper II was therefore used in paper III . The results obtained upon addition of the two-quark operators (5.25) and (5.26) are shown in Table 5.2. 5.2. ππ transitions in heavy-light mesons 47
Table 5.1: Numerical results from paper III with single-quark amplitudes and relativistic matrix elements for the ππ widths of the spin triplet D∗
2
, D1, D∗
0
and the spin singlet
D∗
1
mesons. The individual results from the spin independent ( α±) and spin dependent amplitudes ( β±) are shown for gqA = 1.
Transition |T |2
α+
|T |2
α−
|T |2
β−
|T |2
β+
gqA = 1 gqA = 0 .87
D∗
2
→ D∗ππ 0.896 2.864 5.06 · 10 −3 1.17 · 10 −3 3.77 MeV 2.39 MeV
D∗
2
→ Dππ – – 6.45 · 10 −2 7.51 · 10 −3 0.07 MeV 0.05 MeV
D1 → D∗ ππ 0.367 1.291 2.33 · 10 −3 7.85 · 10 −4 1.66 MeV 1.05 MeV
D∗
1
→ D∗ππ – – 6.54 · 10 −4 3.30 · 10 −4 0 MeV 0 MeV
D∗
1
→ Dππ 2.749 7.915 – – 10.7 MeV 6.80 MeV
D∗
0
→ D∗ππ 0.020 0.098 – – 0.12 MeV 0.07 MeV
D∗
0
→ Dππ – – 1.43 · 10 −2 2.86 · 10 −3 0.02 MeV 0.01 MeV
Table 5.2: Numerical results from paper III for the most important ππ transitions, upon addition of the two-quark contributions associated with the scalar confining and OGE interactions to the Weinberg-Tomozawa Lagrangian, for gqA = 1.
|T |2
α−
Total Transition Rel +Conf +OGE gqA = 1 gqA = 0 .87
D∗
2
→ D∗ππ 2.864 2.377 2.144 3.05 MeV 1.82 MeV
D1 → D∗ ππ 1.291 1.076 0.974 1.34 MeV 0.80 MeV
D∗
1
→ Dππ 7.915 6.535 5.872 8.62 MeV 5.17 MeV
The results for the ππ widths of the positive parity charm mesons with L = 1 in Table 5.1 reveal a strong sensitivity to the value of gqA as well as the large hyperfine splittings in the empirical D meson spectrum. Consequently, some transitions are kinematically favored, while others, in particular D∗
0
→ D∗ππ , are strongly inhibited by the small phase space available. The ππ widths of the L = 1 D mesons are, therefore, also very sensitive to the spin-orbit structure of the Q¯q interaction, which was also found to be the case for the analogous single pion transitions in paper II . In the absence of empirical information and definite QCD lattice calculations , the energies of the as yet undiscovered D∗
0
and D∗
1
mesons were in paper III taken to equal those predicted by the calculation of ref. . The predicted total widths of the L = 1 D mesons may be obtained by adding the calculated ππ widths in Table 5.1 to those for the π transitions obtained in paper II .With gqA = 1 the total calculated ππ width in the single quark approximation of the
D∗
2
(2460) is 3.8 MeV and that of the D1(2420) is 1.7 MeV. Such an addition brings the total calculated width of the D∗
2
(2460) to 19.5 MeV, which is well within the uncertainty margin of the empirical value 25 +8
−7
MeV . In the case of the D1(2420) meson, the total calculated width for π and ππ emission comes to 15.3 MeV, which is close to the empirical uncertainty margin of the total width 18 .9+4 .6
−3.5
MeV . However, reduction of the value for gqA to 0.87 brings the calculated widths for π and ππ emission somewhat below the empirical values for the total widths. Likewise, employment of the two-quark contributions associated with the Q¯q interaction has the effect of reducing the calculated
ππ widths, as shown in Table 5.2. 48 Chapter 5. Two-pion Transitions
5.3 ππ transitions in heavy quarkonia
Theoretical work on the ππ decays of excited heavy quarkonia has demonstrated that the empirical energy spectra of the emitted ππ system demands that the pions be derivatively coupled to the heavy quarkonium states. This is consistent with the role of the pions as Goldstone bosons of the spontaneously broken approximate chiral symmetry of QCD. Most models [67, 68, 69, 70] have dealt with the coupling of two-pions to the heavy meson as a whole rather than to its constituent quarks. The satisfactory description obtained suggests that the decay amplitude Tf i at the quark level should be a smooth function of the ππ momentum q, which is dominated by single-quark mechanisms for ππ
emission. However, the pion rescattering or pion exchange term that appears naturally as a consequence of the coupling of two-pions to constituent quarks has been found, in paper IV , to be dominant since it is not suppressed by the orthogonality of the quarkonium wave functions. It was shown in paper IV that an unrealistically large pion rescattering contribution may be avoided if the Qππ vertex involves an intermediate, fairly light and broad σ
meson, in line with the phenomenological resonance model of ref. . An intermediate
σ meson suppresses the contributions from pion rescattering mechanisms, while single quark amplitudes are but slightly affected. Together with a relativistic treatment of the single quark amplitude, this suppression of the pion rescattering contribution has been shown in paper IV to reproduce the expected smooth behavior of the transition amplitude.
¯QQkb
ka
¯QQkb
ka
k
¯QQka
kb
k
¯QQkb
ka
Figure 5.1: Tree-level Feynman diagrams for the emission of two-pions by heavy con-stituent quarks. The two upper diagrams correspond to the single-quark amplitudes
TQ and T ¯Q respectively, while the two lower diagrams describe the pion rescattering amplitudes Tex and Texc which involve a pion exchange between the heavy constituent quarks. 5.3. ππ transitions in heavy quarkonia 49
If the coupling of the pions to the constituent quark does not involve derivatives of the pion field, then agreement with experiment is excluded for the pion invariant mass distri-butions in the decays Υ ′ → Υ ππ and ψ′ → J/ψ ππ [66, 69]. Derivative couplings for the pions are also consistent with the role of pions as Goldstone bosons of the spontaneously broken approximate chiral symmetry of QCD. The effctive Qππ interaction Lagrangian is therefore expected to have the form
LQππ = 4 πλ ¯ψQ ∂μπa ∂μπa ψQ, (5.27) where λ is a coupling constant of dimension [MeV] −3 and ψQ, ¯ψQ denote the heavy quark spinors. The total tree-level amplitude for ππ emission can then be expressed in terms of single quark and pion rescattering terms, as illustrated in Fig. 5.1. The isospin dependence of the dipion-quark coupling then implies that
|T |2
ππ
= 2 |T |2
π+π−
|T |2
π0π0
. (5.28) As a consequence the width for emission of charged pions should be twice that for emission of neutral pions, which is in fair agreement with what is found experimentally [32, 66]. The effect of the strong interaction in the ππ system may be approximately accounted for by inclusion of an intermediate scalar meson ( σ or glueball) resonance in the vertex. This is brought about by modification of the coupling constant λ with a relativistic scalar meson propagator of the Breit-Wigner type:
λ → λ
( M 2
σ
Γ 2
σ
/4
M 2
σ
q2 + Γ 2
σ
/4
)
. (5.29) Here Mσ denotes the pole position mσ −iΓσ/2, and q the four-momentum, of the effective scalar ( σ) meson resonance. The σ resonance appears by infinite iteration of the four-pion vertex in the isospin 0 spin 0 channel. Therefore, as pointed out in ref. , it is natural to describe the strongly interacting ππ state by a broad σ pole rather than by the driving term (4-pion vertex) alone. When the modification (5.29) is taken into account, the expression for the single-quark amplitude becomes
T1q = −16 πλ
( M 2
σ
Γ 2
σ
/4
M 2
σ
q2 + Γ 2
σ
/4
) [
m2
π
− 1
2
((ωa + ωb)2 − q2)]
M1q . (5.30) The nonrelativistic approximation for the matrix element M1q is unreliable, as the radial
S-wave quarkonium wave functions are orthogonal. A relativistic form for the matrix element M1q , where P = |P |, q = |q| and P · q = P qv , may be obtained as
Mrel 1q = 1
π
∫ ∞
0
dr ′ r′ uf (r′)
∫ ∞
0
dr r u i(r)
∫ ∞
0
dP P 2
∫ 1
−1
dv α (P, v, q )
j0
(
r′
√
P 2 + q2
16 − P qv
2
)
j0
(
r
√
P 2 + q2
16 + P qv
2
)
, (5.31) where α(P, v, q ) is a factor which includes the quark spinors in the coupling (5.27),
α(P, v, q ) =
√ (E + MQ)( E′ + MQ)
4EE ′
(
1 − P 2 − q2/4
(E′ + MQ)( E + MQ)
)
. (5.32) 50 Chapter 5. Two-pion Transitions
In principle the quark spinors in the coupling (5.27) also contain a spin dependent part that is proportional to q × P . For the present purposes that contribution turns out to be very tiny and may be safely neglected. It has been shown in paper IV that the relativistic modifications to the single-quark matrix element increase the magnitude of the single quark amplitudes for ππ decay. Figure 5.2: Modeling of the pion rescat-tering term in Fig. 5.1 in terms of inter-mediate σ mesons. The four-momenta of the σ mesons are defined as k1 = −k − ka
and k2 = k − kb. The crossed rescatter-ing diagram in Fig. 5.1 can be obtained by interchanging ka and kb. The physi-cally reasonable approximations k1 ≈ − k,
k2 ≈ k, |k01 | ≈ | k02 | ≈ (ωa + ωb)/2 and
k0 ≈ (ωb−ωa)/2 were made in paper IV , to allow for a simpler treatment of the triple propagators in the pion rescattering ampli-tudes.
σ(k1 )
σ(k2)
π(k)¯QQπ(ka)
π(kb)The pion rescattering amplitude that corresponds to Fig. 5.2 may be expressed as
Tex = −64 π2λ2 (M 2
σ + Γ 2
σ /4) 2 kaμ kμ kbν kν
(k21 + M 2
σ + Γ 2
σ/4)( k2 + m2
π )( k22 + M 2
σ + Γ 2
σ/4) , (5.33) where the momenta are defined as in Fig. 5.2. Upon simplification of the triple propagator according to the recipe of Fig. 5.2, the crossed pion rescattering diagram gives an extra factor of 2, yielding the expression
T2q = −128 π2λ2
{ 1
3
( q2
4 − Q2
f
) [Me1 − A2(Me2 − M e3 )] (5.34) +
( q2z2
4 − 2
3 Q2
f − q2
12
)
Me4 + ωaωb
4 (ωa − ωb)2(Me2 − M e3 )
}
,
where the term which is proportional to the matrix element Me4 represents an amplitude for D-wave ππ emission. The matrix elements in eq. (5.34) may, in the non-relativistic approximation, be expressed as
Me1 =
∫ ∞
0
dr u f (r)ui(r) j0(Qf r) (M 2
σ + Γ 2
σ /4) 2
4π
( e−Xr
2X
)
, (5.35)
Me2 =
∫ ∞
0
dr u f (r)ui(r) j0(Qf r) (M 2
σ + Γ 2
σ /4) 2
4π(X2 − A2)2 A Y 0(Ar ), (5.36)
Me3 =
∫ ∞
0
dr u f (r)ui(r) j0(Qf r) (M 2
σ + Γ 2
σ /4) 2
4π(X2 − A2)2
[
XY 0(Xr )+ (X2 −A2)
2X e−Xr
]
, (5.37)
Me4 =
∫ ∞
0
dr u f (r)ui(r) j2(Qf r) (M 2
σ + Γ 2
σ /4) 2
4π(X2 − A2)2 F2(r). (5.38) In the above matrix elements, X is defined as X = √M 2
σ + Γ 2
σ/4 − (ωa + ωb)2/4, while
Y0(r) denotes the Yukawa function e−r/r . Note that when the value of k20 exceeds m2
π , the 5.3. ππ transitions in heavy quarkonia 51
analytic continuation A → − i √k20 − m2
π
is employed for the matrix element (5.36). Further, uf (r) and ui(r) denote the reduced radial wave functions for the final and initial state heavy quarkonia, respectively. The function F2(r), which is defined in paper IV ,is closely related to, and in the limit mσ → ∞ actually reduces to, a Yukawa Y2 func-tion . It turns out that the matrix element (5.38) is numerically quite insignificant, because of the strong suppression caused by the j2 function for small values of Qr . Also, the smallness of k0 as compared with k precludes any terms proportional to k0 or k20
from playing any major role. In view of the many approximations involved in the above treatment of the pion rescat-tering terms, a check against an unapproximated calculation is desirable. This is possible since the triple propagator in eq. (5.33) may also be considered without any approxima-tion in k1 and k2, at the price of numerically much more cumbersome expressions. By means of the Feynman parameterization 1
ABC = 2
∫ 10
dx x
∫ 10
dy 1
[A(1 − x) + Bxy + Cx (1 − y)] 3 , (5.39) the two-quark amplitudes of Fig. 5.2 may be cast in the form
T2q = −(8 πλ )2
{ 1
3
( q2
4 − Q2
f
) [∫ 10
dx x
∫ 10
dy {MI − A2MII
}]
+
∫ 10
dx x
∫ 10
dy
(
− q2
4 (1 − 2x + xy ) − Q2
f
(1 − xy ) + qQ f z(1 − x)
)
MII
(
− q2
4 (1 − 2x + xy ) + Q2
f
(1 − xy ) + qQ f zx (1 − y)
)
ωaωb k20
∫ 10
dx x
∫ 10
dy MII
}
Texc , (5.40) where the matrix elements are given by
MI = 2
∫ ∞
0
dr u f (r)ui(r) (M 2
σ
Γ 2
σ
/4) 2
8πA e−Ar
j0
(
r
√ q2
4 (1 − 2x + xy )2 + Q2
f
x2y2 + qQ f z(1 − 2x + xy )xy
)
, (5.41)
MII = 2
∫ ∞
0
dr u f (r)ui(r) (M 2
σ
Γ 2
σ
/4) 2
32 πA 3 e−Ar (rA + 1)
j0
(
r
√ q2
4 (1 − 2x + xy )2 + Q2
f
x2y2 + qQ f z(1 − 2x + xy )xy
)
. (5.42) Here the term proportional to k20 is again only of minor importance. Note that in order to obtain the contribution Texc to eq. (5.40), it is necessary to make the substitution
ka ↔ kb, which implies ωa ↔ ωb and Qf → − Qf . In the above matrix elements, the quantity A is now defined as A = √m2
∗
− K20 and involves an effective mass m∗ and an energy transfer variable K0. These are abbreviations of the following expressions: 52 Chapter 5. Two-pion Transitions
m2
∗
=
(
M 2
σ
Γ2
σ
4
)
(1 − xy ) − m 2
π
x (2(1 − x)(1 − y) − xy 2) (5.43) + 2
(q2
4 − Q2
f
)
x(1 − x)(1 − y),K0 = ωa(1 − x) − ωbx(1 − y) + k0, (5.44) where Mσ and k0 are defined as mσ − iΓσ/2 and ( ωb − ωa)/2, respectively. Numerical comparison of the above formalism with the approximate model of paper IV indicates that eq. (5.34) is accurate to within ∼ 3%. All pion rescattering matrix elements have, in paper IV , been considered in the non-relativistic limit, even though it was noted that that limit is not realistic in the case of the single quark amplitudes. This treatment is expected to be permissible in the case of two-quark amplitudes, since the Yukawa functions from the propagators of the exchanged pions and σ mesons cancel the orthogonality of the radial S-wave quarkonium wave func-tions. Relativistic effects thus constitute only a correction to the pion rescattering matrix elements, which is expected to be rather small because of the large constituent masses of the charm and bottom quarks.
5.4 The transitions Υ′ → Υ ππ and ψ′ → J/ψ ππ
If a σ meson lighter than about 1 GeV is employed, together with a relativistic treatment of the single quark amplitudes, then the effects of pion rescattering diagrams may be reduced, and they may become subdominant as compared to the single quark amplitudes, which allows for agreement with experiment. The results of paper IV indicate that a σ
mass of ∼ 500 MeV gives a favorable description of the present experimental data on the
ππ transitions from the 2 S states of heavy quarkonia. The calculated widths and ππ energy distributions were obtained by simultaneously optimizing the results for Υ ′ → Υ π+π− and ψ′ → J/ψ π +π−. The best results from paper IV , which yielded the σ meson parameters mσ = 450 MeV and Γ σ = 550 MeV, are shown in Table 5.3 and Fig. 5.3. A coupling constant close to λ = −0.02 fm 3 was found to provide an optimal description of the ππ widths and energy spectra, where the sensitivity to the sign of λ is due to the consideration of pion rescattering amplitudes. The heavy quark masses correspond to those of the Q ¯Q Hamiltonian model in paper IV .Table 5.3: Experimental data and calculated widths for ππ transitions from the Υ ′
and ψ′ states, for λ = −0.02 fm 3, mσ = 450 MeV and Γ σ = 550 MeV, together with experimental widths, branching fractions, and maximal momenta for each transition.
Transition Γtot % of Γ tot Γexp Γcalc qmax
Υ′→Υπ+π−44 ±7 keV 18.8 ±0.6 % 8.3 ±1.3 keV 5.89 keV 475 MeV Υ′→Υπ0π09.0 ±0.8 % 4.0 ±0.8 keV 3.07 keV 480 MeV
ψ′→J/ψ π +π−277 ±31 keV 31.0 ±2.8 % 86 ±12 keV 53.5 keV 477 MeV
ψ′→J/ψ π 0π018.2 ±2.3 % 50 ±10 keV 27.8 keV 481 MeV
5.4. The transitions Υ′ → Υ ππ and ψ′ → J/ψ ππ 53
00.5 11.5 22.5 300.2 0.4 0.6 0.8 11/ Γ dΓ/dx xTwo-pion decay distribution -Υ(2S) -> Υ(1S) π+π-Experiment Predicted distribution (dimensionless units) 00.5 11.5 22.5 300.2 0.4 0.6 0.8 11/ Γ dΓ/dx xTwo-pion decay distribution -Ψ(2S) -> Ψ(1S) π+π-Experiment Predicted distribution (dimensionless units)
Figure 5.3: Comparison of calculated and experimental ππ energy distributions for Υ ′ → Υπ+π− and ψ′ → J/ψ π +π−, for mσ = 450 MeV, Γ σ = 550 MeV and
λ = −0.02 fm 3. The calculated width for π+π− decay is 5.89 keV for b¯b and 53.5 keV for
c¯c. The scaled ππ invariant mass x is defined in eq. (5.7). The results presented in Fig. 5.3 indicate that the shapes of the experimental ππ spectra for b¯b and c¯c are slightly different. In particular, the peak at high x appears somewhat lower for b¯b, while the tail at low x is more pronounced for b¯b. It was therefore noted in ref. that the resonance model of ref. cannot be simultaneously fitted to both the b¯b and c¯c data. This is because the shape of the ππ energy distribution is rather insensitive to the properties of the σ meson when only amplitudes of the single quark type are considered. Thus widely different masses and widths of the σ meson have to be employed to fit the empirical ππ energy spectra in the c¯c and b¯b systems. 54 Chapter 5. Two-pion Transitions
However, if pion rescattering amplitudes are considered as well, then a σ mass higher than about 450 MeV was shown in paper IV to lead to unrealistically large pion rescat-tering contributions, as the ππ energy spectrum then begins to develop a second peak at low x. As the pion rescattering effects are of a short-ranged character, then it turns out that they are significant only for the ππ spectrum in the bottomonium system. While the pion rescattering contribution is seen from Fig. 5.3 to account for the qualitative differences between the ππ spectra for b¯b and c¯c, it nevertheless appears to be some-what overpredicted. This problem can be traced to the nonrelativistic treatment of the pion rescattering contribution, and may be alleviated if relativistic matrix elements are employed, as discussed in paper IV .It may be seen from Table 5.3 that the ππ widths of the b¯b system are somewhat un-derpredicted. This is because the presence of the pion rescattering terms precludes an independent fit of the width and the ππ spectrum. A larger value of λ, which would improve the ππ widths, would then worsen the description of the ππ spectrum presented in Fig. 5.3. However, in the case of the transition ψ′ → J/ψ π +π−, it may actually be desirable to employ a 20 − 30 % larger value of λ, as was also suggested in ref. .
00.5 11.5 22.5 33.5 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1dΓ/dm ππ (keV/Gev) mππ (GeV) Two-pion decay distribution -Υ(3S) -> Υ(1S) π+π-Exclusive data Inclusive data Best fit to data
Figure 5.4: The empirical double-peaked Υ(3 S) → Υ ππ spectrum, fitted in paper IV
by the parameters mσ = 1400 MeV, Γ σ = 100 MeV, λ = 2 .7 · 10 −3, yielding a width of Γπ+π− = 1.07 keV. An outstanding problem, lately verified by experimental reanalysis , is the double-peaked structure of the empirical Υ(3 S) → Υ ππ spectrum, which cannot be explained by models dominated by single-quark amplitudes, such as the one employed in paper IV .However, it was also shown that single-quark + pion rescattering models do in fact have sufficient freedom to accommodate a double-peaked ππ spectrum, as may be seen from Fig. 5.4. As a much heavier scalar meson is employed, the contributions from the pion rescattering and single quark amplitudes are of equal magnitude. Incidentally, the best fit parameters fall within the range of the empirical scalar resonances f0(1370) and
f0(1500) , both of which possess a strong coupling to ππ .Chapter 6 Conclusions
The electromagnetic and pionic transitions in the heavy Q ¯Q and heavy-light Q¯q mesons have been calculated within the framework of the covariant BSLT equation, with the assumption that the quark-antiquark interaction can be modeled in terms of a long-range confining interaction and a short-ranged OGE or instanton induced interaction. It has been demonstrated that a reasonable description of the empirical heavy meson spectra can be achieved within such an approach, which also yields reasonable values for the confining string tension, constituent quark masses and the parameters Λ QCD and
mg which control the strong coupling αs. However, the question of the effective Lorentz structure of the quark-antiquark interaction cannot be answered by the quarkonium spectra alone, since there are many models which produce equally satisfactory spectra using different assumptions for the Lorentz structure of the effective confining interaction. Numerical lattice QCD is unrevealing in this case, as the different components of the Q¯q
interaction that can be measured on the lattice may also be well fitted by different assumptions for the effective confining interaction. In this situation, a study of electromagnetic and pionic transitions is instructive, as two-quark or negative energy contributions have been shown, within the framework of e.g. the Schr¨ odinger and Gross equations, or the instantaneous approximation to the Bethe-Salpeter equation, to be significant for several of these transitions. As the computation of these effects requires an explicit assumption of the Lorentz structure of the confining interaction, then it is possible that the question may be settled in the future when adequate experimental data on electromagnetic and pionic transitions is available. For the time being, the M1 transitions in the charmonium system are the most instructive, as the two-quark effects for those transitions have been shown to be large. Furthermore, as the OGE interaction does not contribute any two-quark operator for M1 transitions in equal-mass quarkonia, then the transition J/ψ → ηcγ may provide direct insight into the Lorentz structure of the confining interaction. Such a calculation has been described in this thesis, where it is found that the two-quark operator associated with a scalar confining interaction can explain the observed width of about 1 keV. The theoretical importance of the M1 transition J/ψ → ηcγ suggests that a new and more accurate measurement of the width for that transition should be performed as soon as possible. The experimental situation is similar to that for the E1 transitions
χcJ → J/ψ γ , which were thought to be overpredicted by most model calculations for a 55 56 Chapter 6. Conclusions
long time, until the issue was resolved by the new data on these E1 transitions reported in the latest edition of the PDG . Thus, in order to avoid prolonged speculation, a new and independent measurement of J/ψ → ηcγ is desirable. More empirical information is also needed on the total widths of the b¯b states and their E1 branching fractions, as the E1 widths are very sensitive to the Q ¯Q wave functions, if not to the Lorentz structure of the quark-antiquark interaction. Progress has recently been made for the E1 transitions in the b¯b and c¯c systems as well, since models that employ nonperturbative hyperfine interactions have become available. It was found in the calculation reported in this thesis that many of the E1 transitions in the bottomonium system cannot be accurately modeled with spin-averaged wave functions. A realistic description of the heavy-light Q¯q mesons, most notably the charmed D meson, presents much more serious theoretical challenges for a number of reasons, most notably the uncertain composition of the Q¯q interaction, the relativistic nature of the light con-stituent quark and the limited empirical knowledge of the excitation spectra. It has nevertheless been demonstrated in this thesis that the M1 transitions provide an instruc-tive test for the Lorentz structure of the Q¯q interaction Hamiltonian. Promising results have been obtained for a Hamiltonian with scalar confining and vector OGE components, possibly augmented by an instanton induced interaction. The single pion transitions of the D meson are likewise instructive in this sense since the two-quark contributions to the axial charge component of the transition amplitude are large. Consequently, transitions that involve S-wave pion emission are predicted to be suppressed, an effect which has also been observed within the framework of the Gross equation. The recently measured total width of the D∗ vector meson has been shown to provide useful and constraining information on the pion-quark axial coupling constant gqA. Also, the flavor symmetry violating D∗
s
→ Dsπ0 transition can not only provide constraining information on the magnitude of π0 − η mixing, but can also be used to estimate the magnitude of the η-quark, and in particular, the η-nucleon coupling fηN N . In addition to single pion emission, ππ transitions have also been found to contribute significantly to the predicted total widths of the L = 1 D mesons, a situation which is similar to that in the well explored K meson spectrum. These ππ transitions were found to be very sensitive, both to the value of gqA, as well as to the hyperfine splittings in the D meson spectrum. In all likelihood, the ππ transitions will be found to contribute several MeV to the total widths for strong decay. The empirically strong ππ transitions in the charmonium and bottomonium systems have been investigated by means of a phenomenological model, where the Qππ interaction is mediated by a broad scalar σ meson. It has been shown that such a model can explain several features of the empirically observed ππ transitions in heavy quarkonia, although a completely satisfactory description was not achieved. An important conclusion reached in this thesis is that the most instructive test for a given Q ¯Q or Q¯q interaction Hamiltonian is not the excitation spectrum but rather the radiative M1 transitions between the ground state vector and pseudoscalar mesons. Once the question of hyperfine splittings and total widths has been settled by experiment, then the pionic transitions from the L = 1 D mesons will provide a similar testing ground. So far, a pure scalar confining interaction has passed the above tests, although other conceivable forms have not been systematically ruled out. Svenskspr˚ akigt sammandrag
Denna avhandling presenterar en utr¨ akning av elektromagnetiska och starka ¨ overg˚ an-gar i mesoner med en ( Q¯q) eller tv˚ a ( Q ¯Q) tunga kvarkar. Dessa mesoner har beskriv-its med hj¨ alp av den kovarianta Blankenbecler-Sugar (BSLT) ekvationen, under an-tagandet att v¨ axelverkningen mellan kvarkarna kan beskrivas som summan av en ef-fektiv fj¨ attrande v¨ axelverkning med l˚ ang r¨ ackvidd och en kort-r¨ ackviddskomponent, vilken ger upphov till hyperfinstruktur i mesonernas excitationsspektrum. En dylik v¨ axelverkning ger en acceptabel beskrivning av det empiriska mass-spektret, ifall kort-r¨ ackviddskomponenten beskrivs med hj¨ alp av st¨ orningsteoretiskt gluon-utbyte (OGE) eller en instanton-inducerad v¨ axelverkning. Emellertid kan mass-spektret inte ensamt avg¨ ora den relativistiska Lorentz-strukturen hos den fj¨ attrande v¨ axelverkningen, efter-som flera olika ansatser ger ekvivalenta beskrivningar av det empiriska spektret. I denna situation ¨ ar de elektromagnetiska och starka ¨ overg˚ angarna av stor betydelse, eftersom dessa har visats vara k¨ ansliga f¨ or negativ-energi bidrag till ¨ overg˚ angsampli-tuden, vilka i sin tur beror uttryckligen p˚ a v¨ axelverkningens Lorentz-struktur. Dessa effekter har, till dags dato, p˚ avisats inom ett antal teoretiska beskrivningar, d¨ aribland Schr¨ odinger- och Gross-ekvationerna. Eftersom utr¨ akningen av dessa bidrag till s¨ on-derfallsvidderna kr¨ aver en ansats f¨ or kvark-antikvark v¨ axelverkningens Lorentz-struktur, kan de starka och elektromagnetiska s¨ onderfallen utg¨ ora ett test f¨ or olika modeller f¨ or den fj¨ attrande v¨ axelverkningen. F¨ or ¨ ogonblicket ¨ ar den magnetiska (M1) dipol¨ overg˚ angen
J/ψ → ηc γ av st¨ orsta betydelse, eftersom den experimentellt upp¨ atta vidden p˚ a ∼ 1 keV beskrivs d˚ aligt av den icke-relativistiska kvarkmodellen, vilken leder till en tredubbel ¨overuppskattning av detta resultat. Ett av nyckelresultaten i denna avhandling ¨ ar, att en skal¨ art kopplad effektiv fj¨ attrande v¨ axelverkning kan f¨ orklara den uppm¨ atta vidden p˚ a ∼ 1 keV. Emellertid b¨ or en ny experimentell m¨ atning utf¨ oras innan en definitiv slut-sats kan dras av det ovanst˚ aende resultatet. En annan slutsats, presenterad i denna avhandling, ang˚ ar de elektriska (E1) dipols¨ on-derfallen i charmonium ( c¯c) och bottomonium ( b¯b). I detta fall visar det sig att Lorentz-struktren hos kvark-antikvark v¨ axelverkningen i de flesta fall endast har en f¨ orsvinnande liten effekt p˚ a ¨ overg˚ angsamplituden. D¨ aremot ¨ ar E1 ¨ overg˚ angarna k¨ ansliga f¨ or sm˚ a effek-ter i mesonernas v˚ agfunktioner. D¨ armed kan en realistisk beskrivning av ett flertal s¨ on-derfall endast uppn˚ as, ifall kvark-antikvark v¨ axelverkningens hyperfinstruktur behandlas fullst¨ andigt, vilket ¨ ar m¨ ojligt ifall mesonerna beskrivs med hj¨ alp av BSLT-ekvationen. J¨ amf¨ ort med Q ¯Q mesonerna st¨ aller en realistisk beskrivning av Q¯q systemet mycket stora krav p˚ a de teoretiska modellerna, eftersom v¨ axelverkningens form mellan tunga och l¨ atta kvarker ¨ ar os¨ aker, och den l¨ atta kvarken i h¨ og grad relativistisk. Dessutom 57 58 Chapter 6. Conclusions
kompliceras situationen ytterligare av den knapph¨ andiga empiriska kunskapen om Q¯q
mesonernas mass-spektrum. Inte desto mindre har det visats i denna avhandling, att M1 ¨ overg˚ angarna i D mesonerna utg¨ or ett viktigt test f¨ or Q¯q v¨ axelverkningens Lorentz-struktur. Lovande resultat har erh˚ allits f¨ or en skal¨ ar fj¨ attrande + OGE v¨ axelverkning, m¨ ojligen med tillsats av en instanton-inducerad komponent. De starka pion-s¨ onderfallen inom D mesonerna ¨ ar ocks˚ a av stort intresse eftersom negativ-energi bidragen till den axiala laddningsamplituden ¨ ar stora. En f¨ oruts¨ agelse presenterad i denna avhandling ¨ar att pion-¨ overg˚ angar drivna av den axiala laddningsamplituden ¨ ar starkt f¨ orhindrade, vilket nyligen ocks˚ a har observerats med hj¨ alp av Gross-ekvationen. Den nyligen uppm¨ atta totala vidden f¨ or den exciterade D∗ mesonen ¨ ar av stor betydelse, eftersom den kan fixera v¨ ardet p˚ a den axiala kopplingskonstanten gqA f¨ or l¨ atta kvarkar. Dessutom kan det smak-symmetri brytande s¨ onderfallet D∗
s
→ Dsπ0 ge information om styrkan hos η mesonens koppling till kvarkar och baryoner, f¨ orutsatt av storleken hos
η − π0 blandningsvinkeln ¨ ar k¨ and. Ut¨ over s¨ onderfall, i vilka endast en pion emitteras, kan ¨ aven tv˚ a-pion ( ππ ) s¨ onderfall vara av betydelse i D mesonerna. Denna slutsats ¨overensst¨ ammer med den experimentella situationen i de s¨ ara K mesonerna, d¨ ar ππ
s¨ onderfallen ¨ ar v¨ al uppm¨ atta. I denna avhandling befanns ππ s¨ onderfallen vara mycket k¨ ansliga b˚ ade f¨ or v¨ ardet p˚ a gqA och det tillg¨ angliga fasrummet. Det ¨ ar d¨ arigenom sanno-likt, att ππ s¨ onderfallen utg¨ or flera MeV av de totala vidderna f¨ or starkt s¨ onderfall i D
mesonerna. De empiriskt betydande ππ ¨overg˚ angarna i charmonium ( c¯c) och bottomonium ( b¯b) har i denna avhandling unders¨ okts med hj¨ alp av en fenomenologisk modell, vilken beskriver kvark-pion v¨ axelverkningen med hj¨ alp av en skal¨ ar σ resonans. En dylik modell befanns ge en god beskrivning av flera egenskaper hos ππ ¨overg˚ angarna, ¨ aven om en fullst¨ andigt tillfredst¨ allande beskrivning inte uppn˚ addes. Den viktigaste slutsatsen i denna avhandling ¨ ar att det mest betydelsefulla testet f¨ or en given kvark-antikvark v¨ axelverkningsmodell ¨ ar inte mass-spektret, utan snarare de elek-tromagnetiska M1 ¨ overg˚ angarna d¨ ar mesonernas grundtillst˚ and byter spinn. I framtiden, n¨ ar hyperfin-niv˚ aerna och de totala vidderna hos de exciterade D mesonerna ¨ ar exper-imentellt kartlagda, kommer dessa att utg¨ ora ett ytterligare test f¨ or de ovann¨ amnda v¨ axelverkningsmodellerna. Hittills har en skal¨ ar fj¨ attrande v¨ axelverkning klarat dessa test, ¨ aven om andra t¨ ankbara former inte har uteslutits systematiskt. Helsingfors, Oktober 2002
Timo L¨ ahde Suomenkielinen tiivistelm¨ a
T¨ am¨ a v¨ ait¨ oskirja k¨ asittelee s¨ ahk¨ omagneettisia ja vahvoja siirtymi¨ a mesoneissa, jotka koostuvat joko kahdesta raskaasta ( Q ¯Q) tai yhdest¨ a raskaasta ja yhdest¨ a kevyest¨ a ( Q¯q)kvarkista. N¨ ait¨ a mesoneja on kuvattu kovariantin Blankenbecler-Sugar (BSLT) yht¨ al¨ on ratkaisuina, olettaen ett¨ a kvarkkien v¨ alinen vuorovaikutus voidaan esitt¨ a¨ a efektiivisen, pitk¨ an kantaman kvarkkeja kahlitsevan vuorovaikutuksen ja lyhyen kantaman ylihieno-vuorovaikutuksen summana. Mik¨ ali lyhyen kantaman vuorovaikutus oletetaan joko h¨ airi¨ oteoreettiseksi gluonivaihdoksi taikka instantoni-indusoiduksi, on tuloksena laadul-taan hyv¨ aksytt¨ av¨ a malli mesonien kokeellisille viritysspektreille. Valitettavasti viritys-spektrit eiv¨ at yksin¨ a¨ an riit¨ a antamaan ratkaisevaa tietoa kahlitsevan vuorovaikutuksen suhteellisuusteoreettisesta Lorentz-rakenteesta, sill¨ a useampien eri oletusten on todettu johtavan samanlaatuisiin spektreihin. S¨ ahk¨ omagneettisten ja vahvojen vuorovaikutusten merkitys on t¨ ass¨ a tilanteessa erit-t¨ ain suuri, sill¨ a on osoitettu n¨ aiden riippuvan herk¨ asti siirtym¨ aamplitudin negatiivisen energian komponenteista. N¨ am¨ a puolestaan riippuvat eksplisiittisesti kvarkkien v¨ alisen vuorovaikutuksen Lorentz-rakenteesta. T¨ ah¨ an johtop¨ a¨ at¨ okseen on p¨ a¨ adytty aikaisemmin m.m. Schr¨ odingerin ja Grossin yht¨ al¨ oiden kautta. Koska siirtym¨ aamplitudien laskeminen vaatii oletuksen kvarkkien v¨ alisen vuorovaikutuksen Lorentz-rakenteesta, voivat n¨ am¨ asiis toimia testin¨ a, mill¨ a voidaan vertailla eri mallien todenmukaisuutta my¨ os silloin, kun viritysspektrien ennustukset ovat degeneroituneet. T¨ all¨ a hetkell¨ a magneettisen (M1) dipolisiirtym¨ an J/ψ → ηc γ merkitys on hyvin suuri, sill¨ a sen kokeellinen viivaleveys
∼ 1 keV ei ole sopusoinnussa ep¨ arelativistisen ennustuksen kanssa, joka on kolme kertaa suurempi. Er¨ as t¨ am¨ an teoksen t¨ arkeimmist¨ a tuloksista on, ett¨ a skalaarinen kahlitseva vuorovaikutus tarjoaa mahdollisen selityksen yll¨ amainitulle kokeelliselle viivaleveydelle. Lopullinen johtop¨ a¨ at¨ os vaatii kuitenkin kokeellisen tuloksen varmistamisen uudella mit-tauksella. T¨ ass¨ a v¨ ait¨ oskirjassa on my¨ os k¨ asitelty raskaiden mesonien s¨ ahk¨ oisi¨ a (E1) dipolisiirtymi¨ a, jolloin on todettu ett¨ a kvarkkien v¨ alisen vuorovaikutuksen Lorentz-rakenteella on useimmiten vain h¨ avi¨ av¨ an pieni vaikutus teoreettisiin viivaleveyksiin. Toisaalta E1 siir-tym¨ at ovat herkki¨ a pienille muutoksille mesonien aaltofunktioissa. N¨ ain ollen onkin t¨ ass¨ ateoksessa osoitettu, ett¨ a useita E1 siirtymi¨ a voidaan kuvata onnistuneesti vain, mik¨ ali kvarkkien v¨ alinen ylihieno-vuorovaikutus otetaan huomioon k¨ aytt¨ am¨ att¨ a ensimm¨ aisen kertaluvun h¨ airi¨ oteoriaa. Mik¨ ali mesoniin sis¨ altyy kevyt kvarkki, on realistisen mallin rakentaminen heti huo-mattavasti vaikeampaa, koska raskaiden ja kevyiden kvarkkien v¨ alisen vuorovaikutuksen muoto on eritt¨ ain kyseenalainen. Lis¨ aksi kevyt kvarkki on hyvin relativistinen, ja viri-59 60 Chapter 6. Conclusions
tysspektrin kokeellinen tuntemus v¨ ah¨ ainen. T¨ ast¨ a huolimatta on t¨ ass¨ a teoksessa osoitet-tu, ett¨ a D mesonien M1 siirtym¨ at tarjoavat mahdollisuuden tutkia Q¯q-vuorovaikutuksen Lorentz-rakennetta, jolloin lupaavia tuloksia on saatu edell¨ amainituilla vuorovaikutus-malleilla. D mesonien vahvat pionisiirtym¨ at ovat my¨ os t¨ ass¨ a mieless¨ a kiinnostavia, sill¨ apionin ja kevyen kvarkin v¨ alisen vuorovaikutuksen aksiaalinen varauskomponentti riip-puu vahvasti kvarkkien v¨ alisen vuorovaikutuksen Lorentz-rakenteesta. T¨ ass¨ a v¨ ait¨ oskir-jassa on havaittu, ett¨ a aksiaalisesta varauskomponentista riippuvat siirtym¨ at lienev¨ at voimakkaasti estyneit¨ a. Samankaltaiseen johtop¨ a¨ at¨ okseen ollaan p¨ a¨ asty hiljattain my¨ os Grossin yht¨ al¨ on kautta. Eksitoituneen D∗ mesonin viivaleveys, josta hiljattain saatiin ensimm¨ ainen kokeellinen mittaus, on teoreettisesti hyvin t¨ arke¨ a suure, sill¨ a sen avulla voidaan m¨ a¨ ar¨ at¨ a kevyi-den kvarkkien aksiaalinen kytkinvakio gqA. Lis¨ aksi voidaan makusymmetriaa rikkovan
D∗
s
→ Dsπ0 siirtym¨ an avulla tutkia η mesonin ja kvarkkien (tai baryonien) v¨ alist¨ avuorovaikutusta. T¨ am¨ a edellytt¨ a¨ a, ett¨ a neutraalin pionin ja η mesonin v¨ alinen sekoi-tuskulma on tunnettu. Lis¨ aksi on t¨ ass¨ a v¨ ait¨ oskirjassa tutkittu kahden pionin ( ππ ) siir-tymi¨ a, jotka hyvin todenn¨ ak¨ oisesti ovat merkityksellisi¨ a D mesoneissa, mik¨ a johtop¨ a¨ at¨ os on yht¨ apit¨ av¨ a oudosta K mesonista saadun kokeellisen tiedon kanssa. T¨ ass¨ a teoksessa on osoitettu D mesonien ππ siirtymien olevan herkki¨ a gqA:n numeeriselle arvolle sek¨ a k¨ aytet-t¨ aviss¨ a olevalle faasiavaruudelle. T¨ aten ππ siirtymien viivaleveydet ovat todenn¨ ak¨ oisesti muutaman MeV:n suuruisia. Charmonium ( c¯c) ja bottomonium ( b¯b) mesonien kokeellisesti merkitt¨ avi¨ a ππ siirtymi¨ aon t¨ ass¨ a v¨ ait¨ oskirjassa tutkittu fenomenologisella mallilla, jossa raskaiden kvarkkien ja pionien v¨ alist¨ a vuorovaikutusta kuvataan skalaarin σ resonanssin avulla. T¨ all¨ a mallil-la saavutettiin tyydytt¨ av¨ a, joskaan ei t¨ aydellinen, kuvaus raskaiden mesonien ππ siir-tymist¨ a. T¨ am¨ an v¨ ait¨ oskirjan t¨ arkein johtop¨ a¨ at¨ os on, ett¨ a s¨ ahk¨ omagneettiset M1 siirtym¨ at, joissa kvarkkien spin muuttuu, muodostavat viritysspektrist¨ a riippumattoman testin kvarkke-ja kahlitsevan vuorovaikutuksen Lorentz-rakenteelle. Tulevaisuudessa, ylihienorakenteen ja viivaleveyksien ollessa tunnettuja, tulevat D mesonien M1 ja pionisiirtym¨ at tarjoa-maan vastaavan testin. T¨ ah¨ an menness¨ a on skalaarinen yrite kvarkkeja kahlitsevalle vuorovaikutukselle osoittautunut menestyksekk¨ a¨ aksi, vaikka muita mahdollisia muotoja ei ole j¨ arjestelm¨ allisesti eliminoitu. Helsingiss¨ a, Lokakuussa 2002
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14866 | https://math.stackexchange.com/questions/3637442/is-this-an-appropriate-time-to-use-wlog-in-my-proof | Is this an appropriate time to use 'WLOG' in my proof? - Mathematics Stack Exchange
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Is this an appropriate time to use 'WLOG' in my proof?
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A bipartite graph G=(V,E)G=(V,E) is known to be connected given its partitions are of equal size n n, and |E|≥n 2−n+1|E|≥n 2−n+1
G G is necessarily connected if |E|≥n 2−n+1|E|≥n 2−n+1:
1) Based on above |V||V| = 2 n 2 n.
2) WLOG suppose we label the vertices in one partition {x 1,...,x n}{x 1,...,𝑥 n} and those of the other {y 1,...,y n}{y 1,...,y n}.
...and so on. Is this a normal time to use 'WLOG'?
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asked Apr 22, 2020 at 0:12
PlasticCasioPlasticCasio
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1 Not really: it's your choice how you label the vertices. "WLOG" would apply if there was an asymmetry in the conditions that was unimportant in the remaining steps in the proof.Rob Arthan –Rob Arthan 2020-04-22 00:21:07 +00:00 Commented Apr 22, 2020 at 0:21
Thanks, everyone!PlasticCasio –PlasticCasio 2020-04-22 00:31:26 +00:00 Commented Apr 22, 2020 at 0:31
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The appropriate time to use "WLOG" is when you are moving from a general statement to a more specific statement by making a choice that is irrelevant to the matter at hand.
In this case, you are simply giving names to the vertices. You are not introducing any assumptions, so you don't need to argue that you're not actually losing generality.
If, later in the proof, you wanted to make an argument based on the order of the vertices, you might say:
WLOG, suppose that x 1≤x 2≤⋯≤x n x 1≤x 2≤⋯≤x n. Then ...
In this case WLOG is appropriate because you are introducing an additional assumption, but it does not affect the generality of your argument as (if you needed to) you could simply relabel the vertices as you desired.
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answered Apr 22, 2020 at 0:23
NealNeal
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No, this is not an appropriate use of "WLOG". "WLOG" is for when you are making an assumption that might be false, but for some reason (e.g., symmetry), if you prove the result under that assumption then the result without the assumption follows easily. Here you are not making any assumption: you are simply giving names to the vertices.
It would be appropriate to use "WLOG" if the proof had started with a statement like
Let V={x 1,…,x n,y 1,…,y n}V={x 1,…,x n,y 1,…,y n}.
Then, since the x i x i and y i y i have already been defined, it's not necessarily true that the two sets of the partition are {x 1,…,x n}{x 1,…,x n} and {y 1,…,y n}{y 1,…,y n}, which is why "WLOG" is needed. But in your proof, where you have not yet defined the x i x i and y i y i at all, there is no assumption needed to label the two sets in this way, since you are just defining new variables.
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answered Apr 22, 2020 at 0:21
Eric WofseyEric Wofsey
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There is no need for WLOG here. A simpler sentence is this:
Label the vertices in one partition {x 1,...,x n}{x 1,...,𝑥 n} and those in the other {y 1,...,y n}{y 1,...,y n}.
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answered Apr 22, 2020 at 0:16
lhflhf
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14867 | https://pmc.ncbi.nlm.nih.gov/articles/PMC10636317/ | CAMSAP1 role in orchestrating structure and dynamics of manchette microtubule minus-ends impacts male fertility during spermiogenesis - PMC
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CAMSAP1 role in orchestrating structure and dynamics of manchette microtubule minus-ends impacts male fertility during spermiogenesis
Weichang Hu
Weichang Hu
a State Key Laboratory of Molecular Developmental Biology, Institute of Genetics and Developmental Biology, Chinese Academy of Sciences, Beijing 10019, China
b College of Advanced Agricultural Sciences, University of Chinese Academy of Sciences, Beijing 100049, China
Find articles by Weichang Hu
a,b,1, Rui Zhang
Rui Zhang
c National Laboratory of Biomacromolecules, Chinese Academy of Sciences Center for Excellence in Biomacromolecules, Institute of Biophysics, Chinese Academy of Sciences, Beijing 100101, China
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c,1, Honglin Xu
Honglin Xu
a State Key Laboratory of Molecular Developmental Biology, Institute of Genetics and Developmental Biology, Chinese Academy of Sciences, Beijing 10019, China
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a, Yuejia Li
Yuejia Li
a State Key Laboratory of Molecular Developmental Biology, Institute of Genetics and Developmental Biology, Chinese Academy of Sciences, Beijing 10019, China
b College of Advanced Agricultural Sciences, University of Chinese Academy of Sciences, Beijing 100049, China
Find articles by Yuejia Li
a,b, Xiaojuan Yang
Xiaojuan Yang
a State Key Laboratory of Molecular Developmental Biology, Institute of Genetics and Developmental Biology, Chinese Academy of Sciences, Beijing 10019, China
b College of Advanced Agricultural Sciences, University of Chinese Academy of Sciences, Beijing 100049, China
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a,b, Zhengrong Zhou
Zhengrong Zhou
a State Key Laboratory of Molecular Developmental Biology, Institute of Genetics and Developmental Biology, Chinese Academy of Sciences, Beijing 10019, China
b College of Advanced Agricultural Sciences, University of Chinese Academy of Sciences, Beijing 100049, China
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a,b, Xiahe Huang
Xiahe Huang
a State Key Laboratory of Molecular Developmental Biology, Institute of Genetics and Developmental Biology, Chinese Academy of Sciences, Beijing 10019, China
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a, Yingchun Wang
Yingchun Wang
a State Key Laboratory of Molecular Developmental Biology, Institute of Genetics and Developmental Biology, Chinese Academy of Sciences, Beijing 10019, China
b College of Advanced Agricultural Sciences, University of Chinese Academy of Sciences, Beijing 100049, China
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a,b, Wei Ji
Wei Ji
c National Laboratory of Biomacromolecules, Chinese Academy of Sciences Center for Excellence in Biomacromolecules, Institute of Biophysics, Chinese Academy of Sciences, Beijing 100101, China
d College of Life Science, University of Chinese Academy of Sciences, Beijing 100049, China
e Bioland Laboratory (Guangzhou Regenerative Medicine and Health Guangdong Laboratory), Guangzhou, Guangdong 510320, China
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c,d,e, Fei Gao
Fei Gao
d College of Life Science, University of Chinese Academy of Sciences, Beijing 100049, China
f State Key Laboratory of Stem Cell and Reproductive Biology, Institute of Zoology, Chinese Academy of Sciences, Beijing 100101, China
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d,f,2, Wenxiang Meng
Wenxiang Meng
a State Key Laboratory of Molecular Developmental Biology, Institute of Genetics and Developmental Biology, Chinese Academy of Sciences, Beijing 10019, China
b College of Advanced Agricultural Sciences, University of Chinese Academy of Sciences, Beijing 100049, China
g Innovation Academy for Seed Design, Chinese Academy of Sciences, Beijing 100101, China
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a State Key Laboratory of Molecular Developmental Biology, Institute of Genetics and Developmental Biology, Chinese Academy of Sciences, Beijing 10019, China
b College of Advanced Agricultural Sciences, University of Chinese Academy of Sciences, Beijing 100049, China
c National Laboratory of Biomacromolecules, Chinese Academy of Sciences Center for Excellence in Biomacromolecules, Institute of Biophysics, Chinese Academy of Sciences, Beijing 100101, China
d College of Life Science, University of Chinese Academy of Sciences, Beijing 100049, China
e Bioland Laboratory (Guangzhou Regenerative Medicine and Health Guangdong Laboratory), Guangzhou, Guangdong 510320, China
f State Key Laboratory of Stem Cell and Reproductive Biology, Institute of Zoology, Chinese Academy of Sciences, Beijing 100101, China
g Innovation Academy for Seed Design, Chinese Academy of Sciences, Beijing 100101, China
2
To whom correspondence may be addressed. Email: gaof@ioz.ac.cn or wxmeng@genetics.ac.cn.
Edited by Anna Akhmanova, Universiteit Utrecht, Utrecht, Netherlands; received August 23, 2023; accepted September 13, 2023 by Editorial Board Member Rebecca Heald
1 W.H. and R.Z. contributed equally to this work.
Received 2023 Aug 23; Accepted 2023 Sep 13; Issue date 2023 Nov 7.
Copyright © 2023 the Author(s). Published by PNAS.
This article is distributed under Creative Commons Attribution-NonCommercial-NoDerivatives License 4.0 (CC BY-NC-ND).
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PMCID: PMC10636317 PMID: 37903275
Significance
The manchette is a transient structure and plays a crucial role in sperm head shaping and sperm flagellum formation during spermiogenesis. However, the composition of the minus ends of manchette microtubules and their dynamic regulatory mechanisms are not fully understood. In this study, we explored how CAMSAPs (calmodulin-regulated spectrin-associated proteins) affect spermiogenesis and identified the molecular processes that lead to infertility and sperm abnormalities in male mice lacking CAMSAP1. We identified a diverse array of potential manchette proteins by proteomics and elucidating the depolymerization mechanism of the minus end of the manchette microtubules. This highlights the significance of manchette microtubule depolymerization in shaping the morphology of the manchette throughout the late stages of spermiogenesis.
Keywords: CAMSAP1, manchette, microtubule stability, spermiogenesis, male fertility
Abstract
The manchette is a crucial transient structure involved in sperm development, with its composition and regulation still not fully understood. This study focused on investigating the roles of CAMSAP1 and CAMSAP2, microtubule (MT) minus-end binding proteins, in regulating manchette MTs, spermiogenesis, and male fertility. The loss of CAMSAP1, but not CAMSAP2, disrupts the well-orchestrated process of spermiogenesis, leading to abnormal manchette elongation and delayed removal, resulting in deformed sperm nuclei and tails resembling oligoasthenozoospermia symptoms. We investigated the underlying molecular mechanisms by purifying manchette assemblies and comparing them through proteomic analysis, and results showed that the absence of CAMSAP1 disrupted the proper localization of key proteins (CEP170 and KIF2A) at the manchette minus end, compromising its structural integrity and hindering MT depolymerization. These findings highlight the significance of maintaining homeostasis in manchette MT minus-ends for shaping manchette morphology during late spermiogenesis, offering insights into the molecular mechanisms underlying infertility and sperm abnormalities.
Spermiogenesis is a vital stage in male gametogenesis where spermatids are transformed into sperm cells (1). This phase is characterized by forming a distinctive head shape for each species, accomplished through the cooperation of the acrosome–acroplaxome complex and the manchette (2, 3). Any disruptions during this process can result in poorly formed sperm and male infertility (4–6).
The sperm's manchette consists of MT and actin filaments, serving as the fundamental framework and creating a transient intracellular structure. It plays a crucial role in shaping the spermatid head during spermiogenesis (7–10). The manchette emerges at step 8 of spermiogenesis, coinciding with the initiation of sperm nucleus elongation, and disintegrates around steps 13 to 14 when the nucleus completes morphogenesis. Abnormalities in the manchette are closely linked to a failure in shaping the sperm head (9–11).
Manchette MTs, a crucial component of the manchette, have their plus ends stabilized by the perinuclear ring, and their minus ends extend toward the caudal side (7–10). They provide structural support and act as a scaffold for morphological remodeling, contributing to the formation of the sperm head, neck, and tail. They also participate in nuclear condensation, cytoskeletal transport of organelles and proteins, acrosome formation, and flagellum development (12, 13). Despite the critical role played by manchette MTs, our knowledge regarding these structures is still limited, particularly concerning the composition, and the regulation mechanisms by which they are eliminated upon the completion of nuclear remodeling. This knowledge is essential for understanding the processes involved in spermiogenesis and potentially developing treatments for infertility and reproductive disorders.
The calmodulin-regulated spectrin-associated proteins (CAMSAPs), a family of proteins that are characterized by a (calponin homology) CH domain at the N-terminus, two coiled-coil (CC) domains in the middle, and a CC domain and CKK domain at the C-terminus. They interact with the C-terminal CKK domain of α-tubulin, recognize the minus ends of noncentrosomal MTs, and play a critical role in regulating MT networks (14, 15). There are three members in the CAMSAP family of mammals, namely CAMSAP1-3. Despite having similar structures, these proteins have different distribution patterns within intracellular spaces and tissues. Within neurons, CAMSAP1 and CAMSAP3 localize at the axon and contribute to the establishment of axon polarity, whereas CAMSAP2 resides in the dendrite and is responsible for dendrite pruning (16–18). In addition, CAMSAP1 is mainly expressed in nervous tissue, testis, and liver, while CAMSAP2 and CAMSAP3 are more commonly found in epithelial tissue (17, 19, 20). These proteins exhibit distinct spatiotemporal specificities and are involved in regulating diverse physiological processes.
CAMSAPs have been linked to male infertility (21). We investigate the involvement of CAMSAP family proteins in spermiogenesis and show that male mice lacking Camsap 1 are infertile, displaying deformed sperm nuclei and tails resembling those found in oligoasthenozospermia (OAT), a human disease with similar characteristics (22). Our analysis revealed that CAMSAP1 is primarily localized at the caudal end of manchette MTs. In the absence of CAMSAP1, there is a disruption in the composition and dynamics of the minus ends of manchette MTs, resulting in abnormal extension of these structures into the distal cytoplasm and delayed removal. Through proteomic analysis of manchette proteins, we identified CEP170 as a protein localizing to the minus end of manchette MTs, forming globular punctate ring structures resembling the dense plaque observed at the caudal side of the manchette under electron microscopy (23, 24). Notably, the absence of CAMSAP1 led to disarray in these structures. Furthermore, a comparison of purified manchettes from Camsap 1-deficient mice and wild-type (WT) mice revealed significant alterations in the levels of MT depolymerization proteins. Supporting results showed a decreased presence of the depolymerization protein KIF2A in the caudal side of manchette MTs in the absence of CAMSAP1. These findings collectively highlight the crucial role of CAMSAP1 in regulating the structure and dynamics of the minus ends of manchette MTs, thereby influencing spermiogenesis and male fertility.
Results
CAMSAP1 Is Required for Spermiogenesis.
CAMSAPs are a family of proteins that bind to noncentrosomal MT minus ends to regulate the MT network in various tissues and physiological processes (25–27), and mutations in these genes can cause OAT and impact male fertility (21). The CAMSAP family in humans and mice contains three members, CAMSAP1-3; previous studies have shown that the sperm morphology and fertility of Camsap 3 −/− male mice are normal (28), prompting us to investigate the roles of CAMSAP1 and CAMSAP2 in spermiogenesis.
Results revealed that both CAMSAP1 and CAMSAP2 are present in mouse testes and have similar temporal and spatial distributions in developing spermatids (Fig. 1 A and B). Specifically, during steps 1 to 8 of spermiogenesis, CAMSAP1 and CAMSAP2 are absent in round spermatids. However, they become strongly detectable at the caudal end of manchette MTs when spermatids enter step 9 and elongate. By step 16, manchette MTs have disassembled, and CAMSAP1 and CAMSAP2 are no longer detectable. This association between the distribution of CAMSAP1/2 and changes in manchette MTs suggests that these proteins have a specific role in developing elongating spermatids.
Fig. 1.
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CAMSAP1 and CAMSAP2 are associated with manchette MTs in late-stage spermatids. (A) Immunofluorescent co-staining of CAMSAP1 or CAMSAP2 and α-tubulin in adult mouse seminiferous tubule sections. Green, CAMSAP1 or CAMSAP2; red, α-tubulin; blue, DAPI. (B) Immunofluorescent co-staining of CAMSAP1 or CAMSAP2 and α-tubulin in isolated germ cells from adult mice. Green, CAMSAP1 or CAMSAP2; red, α-tubulin; blue, DAPI. [Scale bars: (A), 10 μm; (B), 5 μm.]
We generated Camsap 1 and Camsap 2 null allele mice via gene knockout to investigate their respective roles in spermiogenesis (17) (SI Appendix, Fig. S1 A–C). Testes and caudal epididymis morphology and histology analysis showed no significant differences in spermatozoa morphology or sperm count between Camsap 2 −/− mice and WT mice (SI Appendix, Fig. S1 D–G). However, the spermatozoa of Camsap 1−/− mice exhibited a striking contrast (Fig. 2 A–F). While there was no discernible difference in the shape of the testis compared to WT mice (Fig. 2 A), the counts and motility of spermatozoa obtained from Camsap 1−/− cauda epididymis were drastically reduced (Fig. 2 B and C). Moreover, a significant proportion of the produced sperms displayed abnormal morphology (Fig. 2 D–F). These abnormalities included spermatozoa with irregular shapes, undetached cytoplasm, and deformed spermatozoa with or without bent, coiled, and shortened flagella (Fig. 2 D). Further immunostaining revealed that virtually almost none of the elongating spermatids present in the seminiferous epithelium of CAMSAP1-deficient mice appeared to possess a sperm tail (Fig. 2 G). We then observed the transverse sections of Camsap 1 - null cauda epididymis and found that the remaining small number of spermatozoa showed a large portion of sperm with unremoved fractions such as mitochondria and cytoplasmic vacuoles (Fig. 2 H and I). Note that no developmental defects were observed in Camsap1 heterozygous mice (Camsap 1 +/−). These defects resembled those associated with OAT, a human infertility disorder (22, 29).
Fig. 2.
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CAMSAP1 is essential to spermiogenesis. (A) Representative image of testis of two genotypes of mice. (B) The concentrations of epididymal sperm from WT and Camsap 1 −/− mice. n = 3 for each group. (C) The percentage of motile spermatozoa from WT and Camsap 1 −/− mice. n = 3 for each group. (D) HE-stained sperm smears from WT and Camsap 1 −/− mice. (E) The percentage of spermatozoa with abnormal head. n = 234 spermatozoa from 3 WT mice; n = 171 spermatozoa from 3 Camsap 1 −/− mice. (F) Percentage of spermatozoa with malformed head in epididymis from Camsap 1 −/− and WT mice. n = 409 spermatozoa from 3 WT mice; n = 402 spermatozoa from 3 Camsap 1 −/− mice. (G) Immunofluorescence stained testicular sections from Camsap 1 −/− and WT mice. Green, AQP3; blue, DAPI. (H) Immunofluorescence stained epididymis sections from Camsap 1 −/− and WT mice. Green, acetylated-tubulin; red, COX IV; blue, DAPI. (I) Immunofluorescence stained epididymis sections from Camsap 1 −/− and WT mice. Green, ubiquitin; blue, DAPI. [Scale bars: (A), 1 mm; (D), 5 μm; (G–I), 50 μm.]
Loss of CAMSAP1 Causes the Formation of Sperm with Abnormal Morphology.
We used various techniques to examine the spermatozoa from Camsap 1 −/− mice. Through transmission electron microscopy (TEM) analysis, we found that most spermatozoa in the caudate epididymis had a distorted nucleus surrounded by residual cytoplasm, the translocated MTs, outer dense fibers (ODFs), fiber sheath, and mitochondria were structurally disorganized in Camsap 1-null sperm tails (Fig. 3 A). Sometimes, coiled tails were surrounded by mislocated and poorly condensed mitochondria and connected by coiled ODFs. Additionally, almost all sperm flagella from Camsap 1 −/− mice exhibited a loss of the proper "9+2" MT structure in the axoneme, indicating that MTs were misplaced during their assembly in spermiogenesis (Fig. 3 A). This suggested that the fine structure of abnormal spermatozoa in Camsap 1 −/− mice was impaired.
Fig. 3.
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Loss of CAMSAP1 causes the formation of sperm with abnormal morphology. (A) Sperm section TEM images of WT (Upper) and Camsap1−/− (Bottom) mice. Red arrowheads indicate a distorted nucleus, the red arrow indicates residual cytoplasm, black arrowhead indicates poorly condensed mitochondria, and the black arrow indicates mislocated ODFs in sperm flagella. (B) Stage-matched PAS-stained (Upper) and immunofluorescence-stained (Bottom) tubules of testis sections from WT and Camsap 1 −/− mice. Green, PNA; blue, DAPI. (C) Representative images of spermatids at each step of development, with a schematic of spermatid morphology. [Scale bars: (A), first, third, and fifth column, 0.5 μm; second and fourth column, 0.2 μm; (B) and (C), 10 μm.]
To investigate spermatid developmental stages in Camsap 1−/− and WT testes, we conducted periodic acid-Schiff (PAS) staining analysis and immunostaining for testicular sections, respectively (Fig. 3 B). We found that the most significant defects were present in the Camsap 1-null spermatids during spermiogenesis. A detailed morphological examination of the structure of sperm heads revealed that normal morphology was maintained up to step 8 of spermiogenesis (Fig. 3 C). However, abnormal V-shaped heads were detected in Camsap 1-null spermatids from step 9 onward. This became increasingly apparent from steps 10 to 16, where spermatid heads exhibited a knob-like nuclear morphology as opposed to the normal hook-shaped head in WT elongating spermatids (Fig. 3 C). Indeed, it is crucial to highlight that no significant difference was observed in the protein level of CAMSAP1 in testis protein extracts at different postnatal time points (SI Appendix, Fig. S2 A). These findings strongly indicate that CAMSAP1 plays an irreplaceable role in the later stages of sperm development, specifically after the 8th step (Fig. 3 C).
We note that besides the testis, CAMSAP1 is expressed in the brain and intestine at varying levels (SI Appendix, Fig. S2 B). To determine whether the abnormal spermiogenesis observed is a result of CAMSAP1 absence specifically in spermatids, we generated a germ cell–specific Camsap 1 knockout mouse Camsap 1 flox/−; Stra8-Cre by crossing Camsap 1 flox/flox mice with Stra8-Cre transgenic mice (30) (SI Appendix, Fig. S2 C–F). Our findings demonstrated a phenotype in Camsap 1 flox/−; Stra8-Cre mice that closely resembled that of Camsap 1−/− mice (Fig. 2 A–F and SI Appendix, Fig. S3 A–J). While no notable abnormalities were observed in testis size, cauda epididymis, or the body weight/testis weight ratio when compared to the Camsap 1 flox/+; Stra8-Cre mice, the Camsap 1 flox/−; Stra8-Cre mice exhibited significantly lower motile sperm percentage and concentration (SI Appendix, Fig. S3 F and G). Moreover, significant abnormalities were observed in sperm heads and tails (SI Appendix, Fig. S3 H–J), and Camsap 1 flox/−; Stra8-Cre mice displayed the same infertility phenotype as the Camsap 1−/− mice (SI Appendix, Fig. S3 D and E). These similarities indicate that CAMSAP1 plays a vital role in spermatogenesis in a cell-autonomous manner.
CAMSAP1 Regulates Manchette MT Organization During Spermiogenesis.
Abnormal sperm head shape is often linked to defects in acrosome formation and the manchette (31). These structures are crucial during the elongation phase of spermatids, as they work together (2, 3).
To initiate our investigation, we explored the potential impact of CAMSAP1 on acrosome formation, which undergoes four distinct developmental phases: the Golgi phase, cap phase, acrosome phase, and maturation phase (32). TEM and PNA staining images indicated no noticeable difference in acrosome biogenesis between WT and Camsap 1 −/− mice, and all these four typical developmental phases could be found in Camsap 1-null spermatids (SI Appendix, Fig. S4).
The manchette is a typical noncentrosomal MT-enriched structure (11), and CAMSAP family proteins are known to bind noncentrosomal MT-minus-ends and regulate the MT networks (Fig. 1 B) (26, 27). Based on this knowledge, we hypothesized that CAMSAP1 holds a critical regulatory role in manchette MTs. To test our idea, we performed a comprehensive analysis of manchette MTs using TEM imaging of testis sections as well as immunostaining (Fig. 4 A– E).
Fig. 4.
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CAMSAP1 deficiency leads to abnormally long manchette MTs. (A) The ultrastructure of step 7 to 11 spermatids from WT and Camsap 1 −/− mouse testes. Nu, nucleus. Ma, manchette. (B) Comparison of manchette MT length between WT and Camsap 1 −/− spermatids at different stages in testis sections. Manchette MTs were stained with anti-α-tubulin (red), and the nucleus was stained with DAPI (blue). (C) Quantification of manchette MT length in step 9 to step 12 spermatids from WT and Camsap 1 −/− testis sections. WT testis: n = 71 for steps 8 to 9; n = 68 for step 10; n = 53 for step 11; n = 46 for step 12. Camsap 1 −/− testis: n = 39 for steps 8 to 9; n = 50 for step 10; n = 69 for step 11; n = 44 for step 12. (D) α-tubulin staining (red) of manchette MTs in isolated elongating spermatids from WT and Camsap 1 −/− mice are shown in progressive steps during head elongation. (E) Quantification of manchette MTs length in isolated spermatids from WT and Camsap 1 −/− mice. For the WT group, n = 46; for the Camsap 1 −/− group, n = 50. [Scale bars: (A), 1 μm; (B), 10 μm; (D), 5 μm.]
Our results revealed that the manchette MTs in sperm cells of Camsap 1 -/- testes exhibited normal morphology until step 8 of spermiogenesis. However, during steps 9 to 11, the head structures of the Camsap 1 -/- spermatids became abnormally elongated, and overly elongated manchette MTs surrounded the nuclei (Fig. 4 A). To eliminate the observation bias that the angle of the section might cause, immunofluorescence staining was conducted (Fig. 4 B). Consistent with the observation of TEM analyses, the MTs of manchette in Camsap 1–null spermatids continued to lengthen, producing a long, thin nucleus lacking the characteristic hook shape as spermiogenesis proceeded (Fig. 4 B and C). Additionally, the above defects were confirmed in elongating spermatids isolated from Camsap 1 -/- mice (Fig. 4 D and E).
These findings suggest that CAMSAP1 plays a role in manchette MT network organization.
CAMSAP1 Participates in the Regulation of Manchette MTs Disassembly.
The elongated manchette MTs of Camsap 1 −/− mice indicated a change in MT dynamics. It is known that stable or long-lived MTs tend to accumulate acetylated tubulin due to the presence of more lattice defects, which facilitate the entry of acetyltransferases (33–35). Therefore, we investigated the acetylation modification of manchette MTs in CAMSAP1-deficient mice. Our study revealed that manchette MTs are hyperacetylated in elongating spermatids of Camsap 1 −/− mice during stages IX-XII (Fig. 5 A and B), suggesting that the loss of CAMSAP1 may increase the stability of manchette MTs. However, western blot analysis revealed that the acetylation level of α-tubulin in the testis lysates of Camsap 1 flox/−; Stra8-Cre mice did not exhibit a significant increase compared to Camsap 1 flox/+; Stra8-Cre mice (SI Appendix, Fig. S5 A). CAMSAP1-induced increases in acetylation are therefore likely specific to manchette MTs and do not impact acetylation modifications of intracellular MTs at other stages. Analogously, previous studies have reported that mice lacking HDAC6 (Histone Deacetylase 6) exhibit hyperacetylated microtubules, but it does not affect testis architecture, fertility, and sperm count (36). Our results further support the idea that CAMSAP1 influences the modification level of manchette MTs through the regulation of their dynamics.
Fig. 5.
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CAMSAP1 is involved in the removal and acetylation of manchette MTs. (A) Acetylated α-tubulin staining of testis sections from WT and Camsap 1 −/− mice. (B) Quantification of fluorescence intensity of the acetylated manchette MTs from step 9 to 12 spermatids. WT testis: n = 80 for step 9; n = 96 for step 10; n = 52 for steps 11 to 12. Camsap 1 −/− testis: n = 62 for step 8; n = 88 for step 10; n = 74 for steps 11 to 12. (C) α-tubulin and FITC-PNA co-staining of testis sections from WT and Camsap 1 −/− mice. The images show stages II-III of spermiogenesis. White arrows indicate the manchette MTs stained with α-tubulin (red). The acrosome was stained with PNA (green). (D) Quantification of fluorescence intensity of manchette MTs in steps 13 to 14 spermatids. For the WT group, n = 81; for the Camsap 1 −/− group, n = 86. [Scale bars: (A), (C), 5 μm; enlarged images: 2 μm.]
To further test this hypothesis, we compared manchette MTs in spermatids from WT and Camsap 1 −/− mice at stages 13 to 14, which is when the manchette MTs typically undergo gradual disintegration (8). We found that immunostaining signals of manchette MTs were significantly stronger due to the depletion of CAMSAP1 (Fig. 5 C and D). It should be noted that residual manchette in the epididymal sperm of these mice was not observed (such as Fig. 3 A). Based on our observations, we believe that the manchette is likely disaggregated during steps 15 to 16. These data support the idea that the delayed removal of manchette MTs in Camsap 1 −/− mice is a result of imbalanced dynamics.
Composition and Characterization of Manchette MTs and Associated Proteins.
The manchette is primarily composed of three components: a perinuclear ring, an MT mantle with one end attached to the ring, and dense plaques located at the far end of the mantle (23, 24, 37). Upon observing the spatial distribution of CAMSAP1, we noted a striking similarity to the dense plaques on the manchette. This intriguing finding led us to speculate that CAMSAP1 could be intricately related to these dense plaques.
Understanding the composition of these electron-dense plaques will provide a deeper understanding of the relationship between structural and functional abnormalities in manchette MT. In particular, manchette MT's minus-end structure and protein composition are still fully unknown. For this reason, we purified manchette MTs based on previous studies (37, 38) (Fig. 6 A). The success of our fractionation was confirmed by immunostaining and TEM analysis, as many aggregates in the purified sample closely resemble the manchette MT structure seen in testis sections (Fig. 6 B and C). Meanwhile, western blot analysis showed that the currently reported manchette proteins KIF3A, HOOK1, and α-tubulin were sufficiently enriched in the manchette fraction (SI Appendix, Fig. S5 B).
Fig. 6.
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Mass spectrometry analysis of manchettes enriched using sucrose gradients. (A) Experimental scheme of manchette purification for mass spectrometric analysis. (B) Examination of purified manchette by immunofluorescence staining. Red, α-tubulin; blue, DAPI. (C) Examination of purified manchette by negative-staining electron microscopy. (D) The categorization of candidate proteins for manchette was performed by a Sankey diagram based on the gene ontology enrichment analysis of cellular compartment. We focused on showcasing cellular components related to the manchette. The schematic diagram of elongating spermatids and manchette components is displayed on the left. The protein lists on the right side of the Sankey diagram display the representative proteins in each cellular compartment, the intersection of these candidate proteins and well-known manchette proteins or CAMSAP1 binding proteins. (E) Scatter plot of up- and down-regulated manchette proteins identified by mass spectrometry. (F) Gene ontology annotations of the down-regulated manchette proteins in Camsap 1 −/− mice testes. (B), 5 μm; (C), 1 μm.
Through mass spectrometry analysis of manchette-enriched fractions, we successfully identified a diverse range of proteins (SI Appendix, Table S1). Many of these proteins exhibit close associations with essential cellular components such as MTs, flagellum, mitochondria, and cytoplasmic vesicles (Fig. 6 D). The alignment between these findings and the known function and structure of the manchette strongly indicates that these proteins may encompass a significant pool of potential manchette-related proteins.
To elucidate the mechanism by which CASMAP1 regulates the structure and dynamics of the manchette, we conducted an intersection analysis between the manchette proteins and the CAMSAP1 interacting proteins, based on data from NCBI. We found that 23 proteins were present in both data sets, such as EB1, CEP170, KIF2A, and YWHAs; most of them are microtubule or cilium-related proteins (Fig. 6 D, Right and SI Appendix, Fig. S6). In our investigation to explore the potential involvement of CAMSAP1, we further conducted a comparison of protein levels in manchette-enriched fractions from both WT and Camsap 1−/− mice using mass spectrometry data. The results exhibited significant alterations in 190 proteins, predominantly characterized by down-regulation in the mutant condition (Fig. 6 E). These findings strongly suggest that the absence of CAMSAP1 led to dysfunction in the manchette.
To ensure the appropriate length of the manchette MTs and their timely dismantling, regulatory agencies may be required to participate in dismantling the structure. During late sperm development, the structure of manchette MTs depolymerizes from its caudal side (5), where CAMSAP1 is present (Fig. 1). We hypothesized that CAMSAP1 might play a role in this process. Through bioinformatics analysis, we further identified that the down-regulated proteins in our dataset play essential roles in the regulation of MTs. Specifically, proteins like KIF2A, KIF2C, and KIF27 (members of the kinesin-4 and kinesin-13 families) were found to be affected. These kinesins are well known for their association with MT depolymerization (39–41), suggesting a potential link between CAMSAP1 and the dynamic control of microtubules in the manchette (Fig. 6 E and F).
The Localization of CEP170 and KIF2A at the Minus End of Manchette MTs Is Impaired in Camsap 1 −/− Spermatids.
Previous studies have shown that CEP170, one of the components of the mother centriole subdistal appendage, participates in the regulation of MT organization and assembly (42). We noticed that CEP170 not only appears in the intersection of CAMSAP1 interacting proteins and manchette proteins, but it is also significantly downregulated in the manchette components of Camsap 1−/− mice (Fig. 6 E and SI Appendix, Fig. S6). Interestingly, CEP170 has also been reported to recruit MT-depolymerizing factor KIF2A to promote cilium disassembly (43). A recent study has revealed that CEP170 has the ability to localize to the minus end of noncentrosomal microtubules in cells, and KIF2A is responsible for regulating noncentrosomal microtubules (44). Therefore, we chose to prioritize CEP170 and KIF2A in our subsequent analyses to determine whether the same mechanism applied to spermatozoa.
Analysis of the testis sections revealed that robust CEP170 signals were observed at the caudal side (minus end) of manchette MTs (Fig. 7 A). Our analysis revealed partial overlap between CAMSAP1 and CEP170, with the latter exhibiting a more caudal localization (Fig. 7 A). Therefore, CEP170 might be a potential protein located at the minus end of manchette MTs. Further confirmation of their interaction was obtained through co-immunoprecipitation (Co-IP) analysis (Fig. 7 B).
Fig. 7.
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Impairment of CEP170 and KIF2A localization at the minus end of manchette MTs in Camsap1−/− mice. (A) Representative immunostaining images of testis sections from WT mice. Images show stage IX of spermiogenesis and were obtained through max intensity projection of the z axis. Green, CEP170; red, Camsap1. (B) CEP170 interacts with CAMSAP1 by co-IP assay. (C) Representative immunostaining images of stage IX seminiferous tubule from WT and Camsap 1−/− mice. Images were obtained through max intensity projection of the z axis. Green, CEP170; red, α-tubulin (manchette). (D) The fluorescence intensity of CEP170 in elongating spermatids from WT and Camsap 1 −/− mice. For the WT group, n = 60; for the Camsap1−/− group, n = 61. (E) The number of CEP170 puncta in circles in elongating spermatids from WT and Camsap 1 −/− mice. For the WT group, n = 154; for the Camsap1−/− group, n = 148. (F) Area of the circles formed by CEP170 puncta in elongating spermatids from WT and Camsap 1 −/− mice. For the WT group, n = 122; for the Camsap1−/− group, n = 39. (G) Distance between adjacent CEP170 puncta in elongating spermatids from WT and Camsap 1 −/− mice. For the WT group, n = 370; for Camsap1−/− group, n = 406. (H) Testis sections were stained with antibodies against CEP170 (red) and KIF2A (green). Nuclei were stained with DAPI (blue). (I) CEP170 interacts with KIF2A by co-IP assay. (J) Immunostaining images of testis sections from WT and Camsap 1 −/− mice labeled with antibodies against KIF2A (green) and α-tubulin (red). DAPI stains nuclei (blue). (K) Quantification of KIF2A signal intensity at the minus end of manchette MTs. [Scale bar: (A), (C), 5 μm. (H), (J), 10 μm; enlarged image: 5 μm.]
Next, we examined whether CAMSAP1 deficiency affects the localization of CEP170 in seminiferous tubules at stage IX when spermatids became apparently aberrant in Camsap1-deficient mice. We found that there was a significant decrease in CEP170 distribution at the minus end of manchette MTs in the Camsap 1 −/− testis (Fig. 7 C and D). The number of CEP170 puncta in circles at the minus ends of manchette MTs was significantly reduced in Camsap 1-null spermatids (Fig. 7 C and E). In addition, the area of the circles formed by CEP170 puncta and the distance between adjacent CEP170 puncta in step 9 spermatids from Camsap 1 −/− mice appeared much smaller than that in WT mice (Fig. 7 C, F, and G). Together, these results suggest that CEP170 is one of the potential components of the minus end of manchette MTs, and loss of CAMSAP1 leads to disordered manchette minus-end structures.
We also examined the localization of KIF2A and found it to be located at the minus end of manchette MTs, where it co-localized with CEP170 in elongating spermatids (Fig. 7 H). Their interaction was also confirmed by Co-IP analysis (Fig. 7 I).
Furthermore, the manchette MTs minus end localization of KIF2A was reduced in Camsap 1 −/− mice (Fig. 7 J and K). These findings collectively demonstrate that CAMSAP1 is required for the localization of KIF2A at the minus end of manchette MTs in spermatid differentiation. It is interesting to note that according to biochemical data, CAMSAP1, but not CAMSAP2, has the ability to interact with KIF2A (SI Appendix, Fig. S5 C). This suggests that CAMSAP1 and CAMSAP2 have unique and separate functions.
In conclusion, we have shown that CAMSAP1 plays a critical role in the proper regulation of manchette MTs in spermatids. The deficiency of CAMSAP1 resulted in an altered spatial relationship of the manchette minus-end dense plaques and a reduction in the localization of CEP70 and KIF2A at the minus end of manchette MTs. These findings provide insight into the importance of CAMSAP1 in spermiogenesis and spermatid functions and the potential mechanism by which it regulates MT dynamics (Fig. 8).
Fig. 8.
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Working model of CAMSAP1 on the regulation of manchette morphology during spermiogenesis. CAMSAP1 co-localizes with CEP170 and KIF2A at the minus ends of the manchette microtubules, cooperating to regulate the integrity and dynamics of the manchette microtubules minus ends.
Discussion
The manchette is a transient structure that is essential for mammalian sperm development and plays a vital role in shaping sperm (7–10). Our research reveals that CAMSAP1 is present at the caudal side of manchette MTs and regulates their structure and dynamics throughout spermiogenesis. A deficiency in CAMSAP1 leads to disruption of the acrosome–acroplaxome–manchette complex, resulting in abnormal shaping of the heads and tails of sperm in mice. Ultimately, these abnormalities lead to infertility in the mice.
CAMSAPs and Their Distinct Roles in Sperm Development and Maturation.
The formation and organization of specialized structures in mature sperm is a complex process involving multiple mechanisms (45). A well-orchestrated series of events ensures the correct development of these structures, and any misstep can result in malformed spermatids with head or tail defects, ultimately leading to infertility (7). CAMSAP family proteins are known to have essential roles in diverse cellular processes such as MT organization, cell division, and cell migration. Specifically, previous studies have linked CAMSAP1 and CAMSAP2 to OAT, a condition characterized by abnormal sperm morphology (21). This is consistent with our results suggesting that CAMSAP1 is involved in sperm development and maturation.
However, the fertility and sperm morphology in Camsap 2−/− mice remained unaffected (SI Appendix, Fig. S1 D–G), indicating that CAMSAP2 may serve distinct functions in sperm development across species. The lack of an impaired sperm development phenotype in Camsap 2−/− mice could potentially be attributed to the compensatory role of CAMSAP3 protein. CAMSAP2 and CAMSAP3 share similar functions in regulating the minus end of noncentrosomal MTs and can complement each other in various cell biological processes, while CAMSAP1 exhibits significant differences in its function (26). It is worth noting that similar to Camsap 2−/− mice, no infertility phenotype was observed in CAMSAP3-deficient mice or CAMSAP2-knockdown mice through RNAi (46), likely due to the same compensatory mechanism. This aspect will require further confirmation through future studies. Furthermore, the differences in the interactions between CAMSAP family proteins and CEP170 also suggest that they play diverse regulatory roles in various physiological processes (SI Appendix, Fig. S5 C). These findings collectively highlight the complexity of CAMSAP proteins and their unique contributions to specific cellular functions.
Our conclusions are strongly supported by clinical reports linking CEP170 and KIF2A to OAT (21). Our data point toward CAMSAP1 recruiting KIF2A to the minus ends of manchette MTs. Rescue experiments artificially anchoring KIF2A to the manchette in Camsap 1 −/− mice could not be used to test this hypothesis due to current technological limitations.
The CAMSAP1 Regulates Manchette MTs during Spermiogenesis.
A critical contributor to sperm head morphology is the acrosome–acroplaxome–manchette complex, which plays a vital role in shaping the anterior and posterior portions of the spermatid head (2, 3).
Numerous studies have uncovered extensive knowledge about the plus ends of manchette MTs, specifically the perinuclear ring (6, 9). However, there is a limited understanding of the minus ends of manchette MTs, which is the caudal end. Studies have shown that minus end anchoring and composition of manchette MTs may differ from other noncentrosomal MT arrays (14, 47–49). While γ-tubulin and ninein localize at the minus end of many noncentrosomal MTs (47, 49), they are absent on the caudal portion of manchette MTs. These data suggest that other proteins are involved in the nucleation and anchoring of manchette MTs during spermiogenesis. Our findings demonstrate that CAMSAP1 regulates the structural integrity and stability of the caudal side of manchette MTs, thus filling a crucial gap in this field.
First, it is particularly important to mention that knowledge about the dense plaques of the manchette is still lacking. To identify the components of the manchette, we utilized proteomics and identified several potential manchette proteins. Additionally, we successfully identified two potential dense plaque proteins, CEP170 and KIF2A, distributed in puncta in the distal manchette MTs. This distribution is like the dense plaques observed through electron microscopy (23, 24, 37). This marks identification of dense plaque components, and future in-depth research will be conducted using immuno-electron microscopy.
Next, our investigation proposes a potential mechanism by which CAMSAP1 regulates manchette MTs (Fig. 6 F). CAMSAP1, CEP170, and KIF2A co-localized at the minus ends of manchette MTs (Figs. 1 A and Fig. 7 A and H), and their interaction may be linked with the structural integrity and stability of manchette MT minus ends (Fig. 7 C–J). These results will have great significance for revealing the molecular mechanism of regulation of manchette MTs and even for developing related infertility treatments.
Interestingly, although certain MT-binding proteins like MAP2, TPX2, and MT-cutting enzymes such as katanins are known to be involved in regulating MT length and dynamics (50–52), their levels did not show significant differences between the manchettes of WT and Camsap 1 −/− mice. Nevertheless, it is essential to acknowledge that we cannot rule out the possibility that MAP2, TPX2, and katanin may still play a role in CAMSAP1-induced abnormal elongation of manchette MTs. Further research will be necessary to explore and ascertain their potential involvement in this process.
In conclusion, our findings shed light on the potential roles of CAMSAP1, CEP170, and KIF2A in the regulation of manchette MTs and their contributions to the dynamic process of sperm development. The intricate interplay between these proteins warrants further investigation to comprehend the underlying molecular mechanisms involved fully.
Regulatory Role of CAMSAP1 in Mammalian Sperm Flagella Development.
Our findings indicate that CAMSAP1 is required to develop mammalian sperm flagella and subsequent motility. Previous studies have shown that Wdr47 concentrates CAMSAP1 into the axonemal central lumen to stabilize minus ends of MT seeds, facilitating MT outgrowth (53). However, it remains unclear whether CAMSAP1 affects cilia formation. Defects in the axoneme of sperm (including the "9+2" structure) and secondary systems necessary for flagella function in Camsap 1 −/− mice suggest that CAMSAP1 regulates sperm motility by acting at multiple sites during spermatid differentiation. Specifically, CAMSAP1 is essential for regulating axonemal assembly and translocating proteins to the developing flagellum via intra-manchette transport. Our findings suggest that CAMSAP1 plays a critical role in regulating the localization and function of CEP170 and KIF2A, which are involved in MT dynamics during spermiogenesis. Further studies are necessary to understand the precise mechanisms by which CAMSAP1 regulates MT dynamics in this process and how these regulatory mechanisms may contribute to proper spermatid development and function.
In summary, CAMSAP1 in the CAMSAP1 family is essential for male fertility. Camsap 1 −/− mouse models exhibit a typical clinical phenotype reminiscent of male infertility, characterized by low sperm count, poor motility, and abnormal sperm morphology, or OAT. CAMSAP1 regulates MT depolymerizing activity at the caudal end of the manchette to control manchette length and dissolution, the disruption of which contributes to OAT and male infertility. Our study in the mouse suggests that the human CAMSAP1 could serve as a candidate gene for mutational analysis in infertile individuals with OAT.
Materials and Methods
The animals, primers, antibodies, and reagents used in this study are described in _SI Appendix_, Materials and Methods. The detailed procedures of cell culture and transfection, immunoblot and Co-IP, immunofluorescence staining, fertility test, TEM assay, isolation of manchette, tissue collection and histological analysis, sperm concentration and motility detection, testis smear, and data and statistical analysis are provided in SI Appendix, Materials and Methods.
Supplementary Material
Appendix 01 (PDF)
Click here for additional data file. (37.1MB, pdf)
Acknowledgments
We thank Y. Tian for providing antibodies against COX IV and L. Yang for assistance with TEM microscopy. We thank X. Liang (Tsinghua University) for the advice. This work was funded by the National Natural Science Foundation of China (31930025, 32270736) and National Key Research and Development Program of China (2018YFA0801104, 2021YFA0804802).
Author contributions
W.H., R.Z., F.G., and W.M. designed all experiments, interpreted the results, and prepared the manuscript; W.H., R.Z., Y.L., X.Y., and Z.Z. performed research; X.H. and Y.W. contributed to the mass spectrometric analysis; and H.X., W.J., and F.G. provided reagents and advice.
Competing interests
The authors declare no competing interest.
Footnotes
This article is a PNAS Direct Submission. A.A. is a guest editor invited by the Editorial Board.
Contributor Information
Fei Gao, Email: gaof@ioz.ac.cn.
Wenxiang Meng, Email: wxmeng@genetics.ac.cn.
Data, Materials, and Software Availability
All study data are included in the article and/or SI Appendix. The mass-spectrometry datasets generated in this study are available from the Science Data Bank as entry ( (54).
Supporting Information
References
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Associated Data
This section collects any data citations, data availability statements, or supplementary materials included in this article.
Supplementary Materials
Appendix 01 (PDF)
Click here for additional data file. (37.1MB, pdf)
Data Availability Statement
All study data are included in the article and/or SI Appendix. The mass-spectrometry datasets generated in this study are available from the Science Data Bank as entry ( (54).
Articles from Proceedings of the National Academy of Sciences of the United States of America are provided here courtesy of National Academy of Sciences
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14868 | https://math.libretexts.org/Bookshelves/Mathematical_Logic_and_Proof/Transition_to_Higher_Mathematics_(Dumas_and_McCarthy)/02%3A_New_Page/2.05%3A_New_Page | Skip to main content
2.5: Modular Arithmetic
Last updated
: Jun 9, 2022
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2.4: Constructing Bijections
2.6: Exercises
Page ID
: 99060
Bob Dumas and John E. McCarthy
University of Washington and Washington University in St. Louis
( \newcommand{\kernel}{\mathrm{null}\,})
We define an equivalence relation that will help us derive insights in number theory. DEFINITION. Divides, a∣ba∣b Let aa and bb be integers. Then aa divides bb, written a∣ba∣b, if there is c∈Z such that a⋅c=b. DEFINITION. Congruence, x≡ymodn,≡n Let x,y,n∈Z and n>1. Then x≡ymodn (or x≡ny ) if n∣(x−y). The relation ≡n on Z is called congruence modn.
Theorem 2.25
Congruence mod n is an equivalence relation on Z.
Exercise 2.5
Prove Theorem 2.25.
Definition: Congruence class
The equivalence classes of the relation ≡n are called congruence classes, residue classes, or remainder classes modn. The set of congruence classes modn can be written Zn or Z/nZ.
Of course Zn is a partition of Z. When n=2, the residue classes are called the even and the odd numbers. Many of the facts you know about even and odd numbers generalize if you think of them as residue classes. What are the residue classes for n=3 ?
We leave it as an exercise (Exercise 2.6) to prove that two integers are in the same remainder class modn provided that they have the same remainder when divided by n.
Notation. [a] Fix a natural number n≥2. Let a be in Z. We represent the equivalence class of a with respect to the relation ≡n by [a].
Proposition 2.26. If a≡rmodn and b≡smodn, then (i) a+b≡r+smodn and (ii) ab≡rsmodn. Proof. (i) Assume that a≡rmodn and b≡smodn. Then n∣(a−r) and n∣(b−s). So n∣(a+b−(r+s)) Therefore a+b≡r+smodn proving (i).
To prove (ii), note that there are i,j∈Z such that a=ni+r and b=nj+s Then ab=n2ji+rnj+sni+rs=n(nji+rj+si)+rs Therefore n∣(ab−rs) and ab≡rsmodn Hence the algebraic operations that Zn "inherits" from Z are welldefined. That is, we may define + and ⋅ on Zn by [a]+[b]=[a+b] and [a]⋅[b]=[a⋅b] In mathematics, when you ask whether something is "well-defined", you mean that somewhere in the definition a choice was made, and you want to know whether a different choice would have resulted in the same final result. For example, let X1={−2,2} and let X2={−1,2}. Define y1 by: "Choose x in X1 and let y1=x2." Define y2 by: "Choose x in X2 and let y2=x2." Then y1 is well-defined, and is the number 4 ; but y2 is not well-defined, as different choices of x give rise to different numbers.
In (2.27) and (2.28), the right-hand sides depend a priori on a particular choice of elements from the equivalence classes [a] and [b]. But Proposition 2.26 ensures that sum and product so defined are independent of the choice of representatives of the equivalence classes.
Example 2.29
In Z2 addition and multiplication are defined as follows:
(1) +=
(2) +=+=
(3) +=
(4) ⋅=⋅=⋅=
(5) ⋅=.
Notice that if you read as "even" and as "odd", these are rules that you learned a long time ago.
When working with modular arithmetic we may pick the representatives of remainder classes which best suit our needs. For instance, 79⋅23≡2⋅2≡4mod7. In other words [79⋅23]=⋅=⋅=. ExAMPLE 2.30. You may recall from your early exposure to multiplication tables that multiplication by nine resulted in a product whose digits summed to nine. This generalizes nicely with modular arithmetic. Specifically, if an∈⌜10⌝ for 0≤n≤N then N∑n=0an10n≡N∑n=0anmod9. The remainder of any integer divided by 9 equals the remainder of the sum of the digits of that integer when divided by 9 . Proof. The key observation is that 10≡1mod9. Therefore 102≡1⋅1≡1mod9103≡1⋅1⋅1≡1mod9, and so on for any power of 10 : 10n≡1mod9 for all n∈N. (This induction to all powers of 10 is straightforward, but to prove it formally we shall need the notion of mathematical induction from Chapter 4). Therefore on the left-hand side of (2.31), working mod 9, we can replace all the powers of 10 by 1 , and this gives us the right-hand side.
Example 2.32
The observation that a number’s residue mod 9 is the same as that of the sum of the digits can be used in a technique called "casting out nines" to check arithmetic.
For example, consider the following (incorrect) sum. The number in the penultimate column is the sum of the digits, and the number in the last column is the repeated sum of the digits until reaching a number between 0 and 9 .
1588 22 4
+1805 14 5
3493 19 1
If the addition had been correctly performed, the remainder mod 9 of the sum would equal the sum of the remainders; so we know a mistake was made.
Example 2.33
What is the last digit of 77 ?
We want to know 77mod10. Note that, modulo 10,70≡1,71≡ 7,72≡9,73≡3,74≡1. So 77=7473≡1⋅3≡3, and so 3 is the last digit of 77. EXAMPLE 2.34. What is the last digit of 777 ?
By Example 2.33, we see that the residues of 7nmod10 repeat themselves every time n increases by 4 . Therefore if m≡nmod4, then 7m≡7nmod10.
What is 77mod4 ? Well 71≡3mod4,72≡1mod4, so 77≡ (72)3⋅7≡3mod4. Therefore 777≡73≡3mod10.
2.4: Constructing Bijections
2.6: Exercises |
14869 | https://www.youtube.com/watch?v=Gkzplu4D9rI | Absolute Value (Modulus) of a Complex Number
James Elliott
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Posted: 6 Oct 2020
This video offers a visual representation of determining the absolute value (modulus) of a complex number and also works through several examples.
For more math help and resources, visit www.hsmathsolutions.com.
Transcript:
in this video we're going to talk about the absolute value of a complex number this is also called the modulus of a complex number so here we have our complex plane with our real number axis and our imaginary number axis and let's go ahead and look at the number 2 plus 3i and here's where it lands on the complex plane and remember that absolute value just means the distance a number is from the origin so here's that distance and of course in the plane here we can make a right triangle out of this and the right triangle will have dimensions of 2 and 3 and of course yet again we can use the pythagorean theorem and we have the square root of a squared plus b squared is actually going to be that distance so when we plug the numbers in we'll see that we have the square root of 4 plus 9 which is the square root of 13 which is going to be that distance so as we look at it with our complex number 2 plus 3i the absolute value of that will be the square root of thirteen so in this first example we have the complex number six minus three i and so if you're gonna write this in a plus bi form which it is your a value is 6 and your b value is a negative 3. so to find the absolute value or the modulus we're going to do the square root of a squared plus b squared so my a value is positive six my b value is negative three i'm going to square each one and take the square root so it'll be the square root of 36 plus 9 which is the square root of 45 so 45 over here we can do the prime factorization and this is in an effort to simplify so we see this is the square root of three times three times five and of course here we have a pair of threes because three times three is nine which is a perfect square so that will simplify and so the absolute value or the modulus of 6 minus 3i becomes 3 times the square root of 5. and in our second example we have the complex number 4 times the square root of five plus eight i so again in standard form a plus bi our a value is four times the square root of five and our b value is eight if we're going to put this in our absolute value formula it's the square root of a squared plus b squared which will be the square root of well we have 4 radical 5 squared and then 8 squared okay well how do you square this radical well it's going to be 4 times 4 is 16 and then the square root of 5 squared is just 5. so 16 times 5 is 80 and then of course 8 squared is 64. when i add those i get the square root of 144 well that's nice because that's a perfect square so the value here is going to be 12. so the absolute value or the modulus of 4 radical 5 plus 8i is just 12. |
14870 | https://medcraveonline.com/MOJGG/non-pharmacological-management-for-delirium.html | Non-pharmacological management for delirium - MedCrave online
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MOJGG
Non pharmacological management for delirium
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MOJ
eISSN: 2574-8130
Gerontology & Geriatrics
Case Report Volume 3 Issue 2
Non-pharmacological management for delirium
Natalia Capistran Paramo
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Hospital General de México, México
Correspondence: Natalia Capistrán Páramo, Hospital General de México, México, Tel (044)5534880764
Received: February 21, 2018 | Published: March 7, 2018
Citation:Páramo NC. Non-pharmacological management for delirium. MOJ Gerontol Ger. 2018;3(2):113 – 115. DOI: 10.15406/mojgg.2018.03.00100
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Abstract
Delirium is an acute attention and cognitive disorder, frequently presented in the elderly by multifactorial causes, which has a serious impact on mortality, health costs and quality of life.1 Non-pharmacological measures are the standard of prevention and treatment of delirium. This approaches such as HELP (Hospital Elder Life Program) include cognitive stimulation techniques, however, we have observed in our Hospital, that the “Day Room”, based on Snoezelen sensory stimulation theory, improves delirium severity. This article describes a clinical case of a female geariatric patient with delirium, whom receives both programs (HELP and Snoezelen), where we observed a decrease on delirium severity scales. For the above, we can conclude that “Day Room” might be a very useful non-pharmacological instrument on patients with delirium, so we should do comparative research that sustains clinical efficacy.
Keywords: delirium, acute confusional state, non-pharmacological treatment, “Day Room”, neurocognitive stimulation, sensory stimulation
Abbreviations
HELP, hospital elder life program; MDAS, memorial delirium assessment scale; DRS-R-98, delirium rating scale revised
Introduction
Delirium, an acute decline in attention and cognition, is a common life-threatening, and potentially preventable clinical syndrome among persons who are 65 years of age or older.1 Prevalence is highest among those who are frail or patients who are critically unwell or at the end of life. Over the age of 80 years, more than one third of those in hospital will experience delirium.2 The cause of delirium is typically multifactorial. In fact, the development of delirium involves the complex interrelationship between a vulnerable patient (with a predisposing factor) and exposure to precipitating factors or noxious insults.3 some of the main predisposing factors are: age of 65 or older, dementia, cognitive impairment and functional dependence, visual and hearing impairment and coexisting medical conditions. And precipitating factors such as: Drugs, primary neurologic diseases, infections, severe acute illness, dehydratation, surgery, use of physical restraints, use of catheters.1-3
Pathophysiology
Rather, accumulating evidence suggests that several different sets of interacting biological factor result in disruption of large-scale neuronal networks in the brain, leading to acute cognitive dysfunction.4 Some of the leading hypothesized mechanisms contributing to delirium includes neurotransmitters, inflammation, physiologic stressors, metabolic derangements, electrolyte disorders, and genetic factors.5 The list of potential neurotransmitters involved in delirium is long, but a relative cholinergic deficiency and dopamine excess are the most commonly inferred.6
Clinical features and diagnosis
The hallmark of delirium is an acute impairment of cognition with a fluctuating course. It includes change in the level of consciousness, ranging from a hyperactive state with prominent agitation, hypervigilance and combativeness to a hypoactive state characterized by lethargy, stupor or even coma. Multiple cognitive domains can be affected, with attention being invariably involved, and orientation, memory, language, visuospatial and executive function are also often impaired.7
The current reference standard diagnostic criteria are the 5th edition of American Psychiatric Association’s Diagnostic and Statistical Manual of Mental Disorders (DSM-5) and WHO’s International Classification of Diseases, 10th Revision (ICD-10), but the most widely used instrument for identification of delirium is the Confusion Assessment Method (CAM), which has been validated in high-quality studies with sensitivity of 94%.5 Although there are other validated instruments such as Global Attentiveness Rating (GAR), Clinical Assessment of Confusion (CAC), Delirium Observation Screening Scale (DOSS), Memorial Delirium Assessment Scale (MDAS), Delirium Rating Scale Revised-98 (DRS-R-98).8
Treatment
Primary prevention with non-pharmacological multicomponent approaches is widely accepted as the most effective strategy for delirium.9-11 The Hospital Elder Life Program (HELP), a multicomponent intervention strategy with proven effectiveness and cost-effectiveness in the prevention of delirium and functional decline through targeting of risk factors for delirium is the most widely disseminated approach. The interventions include reorientation, therapeutic activities, reduced use and doses of psychoactive drugs, early mobilization, and promotion of sleep, maintenance of adequate hydration and nutrition, and provision of vision and hearing adaptations. The intervention prevented the initial development of delirium and reduced the total number of days of delirium. Once an initial episode of delirium had occurred, however, the intervention had no significant effect on the severity of delirium or on the likelihood of recurrence.12 HELP is now implemented in more than 200 hospitals worldwide, but adaptations and alternatives may be necessary in some settings.
Other non-pharmacological strategies include proactive old age medicine consultation, multifactorial targeted interventions, staff training, interventions delivered by family members and mobility or rehabilitation interventions, the use of earplugs at night,13 non-pharmacological sleep protocols.14 Delirium rooms,15 and other spaces that provide restraint-free care for patients with delirium, are staffed with specially trained nurses, and promote non-pharmacological management approaches, are an intriguing idea for provision of specialized management for patients with delirium, but have not yet been assessed in a controlled trial.5
The general purpose of this case presentation is to open the discussion, to the probability of implementing improves to the non-pharmacological delirium treatment. Our patient receives cognitive and sensory stimulation (based on Snoezelen stimulation) in “day room”, doing brain stimulation activities, so this way it is planned to identify a positive response on delirium severity.
Case presentation
The patient was born in Mexico City, attended 2 years of primary school, currently unemployed, widowed five years ago, never had any children, cared for by her nephew, she is dependent for basic activities (Katz 1/6 preserves feeding, Lawton brody 0/8). She uses a walker after having suffered a cerebrovascular accident (CVA).
Her medical records indicated hypertension treated with enalapril 10 mg q12 hr and metoprolol 50 mg q12 hr; diabetes mellitus type 2, in treatment with metformin 850 mg q12 hr, paroxystic atrial fibrillation anticoagulated with acenocoumarin 1 mg q24 hr, right middle cerebral artery ischemic stroke on 2013 and 2015, causing as sequels, left-sided hemiparesis and epilepsy in treatment with levetiracetam 1g q12 hr.
Requires a knee prothesis two years ago due to osteoarthritis. She fell a month ago from her own height, causing a right fronto-temporal contusion, after it, she presents clonic movements of left hemibody, with no alteration of wakefulness, managed with midazolam 10 mg IV, single dose with crisis referral, brain tomography is done and shows encephalomalacia in frontal and temporal lobe and frontoparietal chronic subdural hematoma, with no surgery criteria, deciding treatment with carbamazepine and discharged after 72 hours. Two days later, she presents hyporexia, vomit in one occasion and periods of disorientation alternated with periods of lucidity and visual hallucinations, incoherent speech and psychomotor agitation, next day she presents tonic movements of two minutes long with no loss of wakefulness, for what she is taken to ER in which she’s found with blood pressure of 180/120 mmHg, 120 bpm, dehydrated, arrhythmic, with disorientation and psychomotor agitation, handle begins with beta blocker, IECA and antagonist calcium, for every agitation period 5 mg of haloperidol and hospitalization is decided in charge of geriatrics where urinary tract infection is detected, treated with amikacin, severity of delirium is measured with MDAS obtaining 21/30. Next day she presents 160/90 mmHg blood pressure. FC 96 bpm Gluc 156, she receives sensory stimulation based on Snoezelen model in “day Room” for about 3 hours where she paints, cooperating 60% of time, severity of delirium is measured with MDAS obtaining 15/30. The day after she presents 140/80 blood pressure, FC 88 bpm, goes to “Day Room” for sensory and cognitive stimulation, where she plays “lotería” (sort of Bingo), word games, orientation and categories for 4 hours, cooperating 80% of time, severity of delirium is measured with MDAS obtaining 10/30, reducing from severe to mild delirium in 72 hours.
Discussion
As mentioned before, angular treatment of delirium is based on non-pharmacological measures in which many recommendations, with no scientific evidence or clinical trials had been done. However, HELP program was developed which refers that is only useful as prevention, with no significance on severity or recurrence once the first episode is presented.
The HELP program, based on the study of multicomponent intervention, determined the presence of delirium, according to CAM criteria, severity of delirium was measured by adding a numeric value for every of the four designed symptoms: fluctuation, disorganized thinking and consciousness alteration. Assigned punctuation was: missing 0 points, mild 1 point or strong 2 points, except fluctuation which was determined as: missing 0 points or present 1 point. The sum of these values allows to obtain scores from 0-7, the higher the score, the greater the severity.5
Nonetheless, other mind-altered functions were not evaluated, so the use of scales such as MDAS or DRS-R98 is recommended, since they evaluate other affected functions such as consciousness, orientation, short term memory impairment, digit span, attention, disorganized thinking, perceptual disturbance, delusions, psychomotor activity and sleep-wake cycle disturbance (for MDAS 1996).
In this case we base the treatment, according to scientific evidence of Snoezelen sensory stimulation, which report wheter a combined treatment comprised of standard psychiatric inpatient care and a non-pharmacological intervention, multi-sensory behavior therapy, reduces agitation and apathy and improves ADLs in people with dementia on an acute care psychiatric hospital unit compared to standard psychiatric inpatient care alone.16 This, based on the operant paradigm of automatic reinforcement and the physiological model of the relaxation respone.17 Even though this sensory stimulation has implemented only in patients with dementia, we believe is a good alternative for adding to the multicomponent management in patients with acute delirium.
This patient receives sensorial stimulation in two times, with 3 and 4 hours per session, finding an important improvement in severity of delirium based on MDAS scale.
We need to validate this theory in randomized control trials to obtain statistical significance, nevertheless, it could open an alternative for management of patients with delirium, an so, validate the effectiveness in complications such as hospital stay, cognitive impairment and functionality, but also factors that interfere with quality of life, that are not usually measured like collapse of the caregiver, adherence to treatment, hospital cooperation, a recurrence decrease in pharmacological or restrain handle, recalling that the patient medical treatment shall also include psychological and social measures.
Acknowledgment
None.
Conflict of interest
The authors declare no conflict of interest.
References
Sharon SK. Delirium in Older Persons. N Engl J Med. 2006;354(11);1157–65.
Todd OM1, Teale EA. Delirium: a guide for the general physician. Clinical Medicine. 2017;17(1):48–53.
Rolfson D. The causes of delirium. In: Lindesay J, Rockwood K, Macdonald A, editors. Delirium in old age. Oxford, England: Oxford University Press; 2002:101–22.
Watt D, Koziol K, Budding D. Delirium and confusional states. In: Noggle C, Dean R, editors. Disorders in Neuropsychiatry. New York: Springer Publishing Company; 2012.
Inouye SK, Westendorp RG, Saczynski JS. Delirium in elderly people. Lancet. 2014;383(9920):911–922.
Hshieh TT, Fong TG, Marcantonio ER, et al. Cholinergic Deficiency Hypotesis in Delirium: A Synthesis of Current Evidence. J Gerontol A Biol Sci Med Sci. 2012;63(7):764–772.
Vanja CD, Andrew JS. Delirium. Continuum Lifelong Learning Neurol. 2010;16(2):120–134.
Camilla LW, Jayna HL, David LS. Does This Patient Have Delirium?Value of Bedside Instruments. JAMA. 2010;304(7):779–786.
Mahony OR, Murthy L, Akunne A. Synopsis of the national institute for health and clinical excellence guideline for prevention of delirium. Ann Intern Med. 2011;154(11):746–51.
Inouye SK. Delirium in older persons. N Engl J Med. 2006;354:1157–65.
Wei LA, Fearing MA, Sternberg EJ, et al. The confusion assessment method: a systematic review of current usage. J Am Geriatr Soc. 2008;56(5):823–830.
Inouye SK, Bogardus ST, Charpentier PA, et al. A multicomponent intervention to prevent delirium in older patients. N Engl J Med. 1999;340(9):669–76.
Van RB, Elseviers MM, Van DW, et al. The effect of earplugs during the night on the onset of delirium and sleep perception: a randomized controlled trial in intensive care patients. Crit Care. 2012;16(3):R73.
Mc DJA, Mion LC, Lydon TJ, et al. A nonpharmacologic sleep protocol for hospitalized older patients. J Am Geriatr Soc. 1998; 46(6):700–05.
Flaherty JH, Steele DK, Chibnall JT, et al. An ACE unit with a delirium room may improve function and equalize length of stay among older delirious medical inpatients. J Gerontol A Biol Sci Med Sci. 2010;65(12):1387–92.
Staal JA, Sacks A, Matheis R, et al. The effects of Snoezelen (multi-sensory behavior therapy) and psychiatric care on agitation, apathy, and activities of daily living in dementia patients on a short term geriatric psychiatric inpatient unit. Int’l Psychiatry In medicine. 2007;37(4):357–370.
Benson H, Klipper M. The relaxation response. New York: Avon Books; 1975.
©2018 Páramo. This is an open access article distributed under the terms of the, which permits unrestricted use, distribution, and build upon your work non-commercially.
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14871 | https://cemc.uwaterloo.ca/sites/default/files/documents/2023/POTWB-22-N-24-S.html | Solution
Problem of the Week
Problem B and Solution
The Puzzler Returns
Problem
Our superhero The Puzzler is back, seeking your help once more to solve the following number puzzle.
Place each of the numbers 2, 3, 4, 5, 6, 7, 8, 9, and 10 in a different circle in the diagram so that each line of three circled numbers has the same sum.
Can you find more than one possibility for the number that can go in the middle circle?
Not printing this page? You can use our interactive worksheet.
Solution
The key discovery for this puzzle is that the middle circle is on every line of three circled numbers. If we were to remove the number in the middle circle from the diagram, then we would be left with four pairs of numbers that each have the same sum. So after choosing the middle number, it must be possible to pair up the remaining eight numbers so that each pair has the same sum.
If the middle number is 2, the other numbers can be paired as follows: 3+10, 4+9, 5+8, and 6+7. Each of these sums is 13, so the sum of any line of three circled numbers would be 13+2=15.
Similarly, if the middle number is 10, the other numbers can be paired as follows: 2+9, 3+8, 4+7, and 5+6. Each of these sums is 11, so the sum of any line of three circled numbers would be 11+10=21.
We can also choose 6 as the middle number. Then the other numbers can be paired as follows: 2+10, 3+9, 4+8, and 5+7. Each of these sums is 12, so the sum of any line of three circled numbers would be 12+6=18.
It is not possible to choose any other number as the middle number. In each case, if you try pairing up the remaining numbers, you will find that you cannot do so in a way such that each pair has the same sum.
With middle numbers of 2, 6, or 10, there are many possible ways to place the remaining numbers in the diagram. The only condition is that the pairs of numbers with the same sum must be placed on the same line. Some examples are shown. |
14872 | https://www.youtube.com/watch?v=g-E6SqrH4_Y | Flow Visualization
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An introduction to some flow visualization techniques used at MIT. These methods help us understand the fluid dynamics of natural and manmade systems.
I made this video to compile the sweet videos we've taken over the years. Otherwise they would just be sitting there, you know.
For more information visit:
Math Fluids Lab -
Peko Hosoi's Group -
Hatsopoulos Microfluids Lab -
See the original post on MIT TechTV -
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Transcript:
whether it is a swimming fish or a hovering dragonfly flow visualization techniques helped scientists and engineers should like unimportant fluid dynamic phenomena that would otherwise be invisible this is a three link swimmer a machine designed to swim at low Reynolds number around where fluid viscosity dominates microscopic organisms such as bacteria and protozoans swim at low Reynolds number these swimmers must propel themselves not by generating momentum but by fighting viscosity when we place the swimmer in a tank of viscous oil all we can observe is this swimmers position and the fact that it does indeed move forward by placing lines of colored dye into the fluid we are able to chase the path of the fluid as the swimmer makes its way across the tank another way to trace fluid motion is to disperse tiny particles which align themselves with a fluid when it shears when the swimmer is placed in a suspension of these particles we are able to see where the fluid is sharing the most at high Reynolds number animals like the water Strider use fluid momentum to create thrust water Striders are particularly interesting and that they are supported on the water's surface by surface tension for a long time scientists disagreed on how they were able to transfer momentum through the surface without breaking it in 2003 John Bush David who and Brian Chan used dyed water to show that the Striders transfer momentum using vortices rather than ways before this point it was assumed that the waves have the most significance from the Striders mode of locomotion particle image velocimetry is a method where many neutrally buoyant particles are randomly distributed into the fluid a laser sheet illuminates one section of the experiment to be picked up by a camera a computer tracks the motion of the particles and the entire velocity field of that section can be determined here the fluid motion around a flapping fin is analyzed polarized light can be used to visualize flow as well this is a dilute solution of water and tobacco mosaic virus the flow of fluid is made visible using polarized light the virus which is harmless to humans and fish has a long rod shaped structure which aligns to flow much like the particles used the three linked swimmer because the virus particles are microscopic we cannot use them to see the flow using ordinary light but polarized light makes them visible most sources of light emit light ways that have random orientation when the light is passed through special filters it becomes polarized which means the ways are oriented in one direction the particles block or transmit the polarized light depending on their alignment this visualization method enables researchers to see the vortices and the wake created by the fish while it generates thrust these and other techniques have helped scientists to understand a myriad of fluid phenomena and have enabled engineers to develop better fluid machines |
14873 | https://www.aafp.org/pubs/afp/issues/2012/1001/p631.html | Acute Kidney Injury: A Guide to Diagnosis and Management
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MAHBOOB RAHMAN, MD, MS, FARIHA SHAD, MD, AND MICHAEL C. SMITH, MD
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Am Fam Physician. 2012;86(7):631-639
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A more recent article on acute kidney injury is available.
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Author disclosure: No relevant financial affiliations to disclose.
Acute kidney injury is characterized by abrupt deterioration in kidney function, manifested by an increase in serum creatinine level with or without reduced urine output. The spectrum of injury ranges from mild to advanced, sometimes requiring renal replacement therapy. The diagnostic evaluation can be used to classify acute kidney injury as prerenal, intrinsic renal, or postrenal. The initial workup includes a patient history to identify the use of nephrotoxic medications or systemic illnesses that might cause poor renal perfusion or directly impair renal function. Physical examination should assess intravascular volume status and identify skin rashes indicative of systemic illness. The initial laboratory evaluation should include measurement of serum creatinine level, complete blood count, urinalysis, and fractional excretion of sodium. Ultrasonography of the kidneys should be performed in most patients, particularly in older men, to rule out obstruction. Management of acute kidney injury involves fluid resuscitation, avoidance of nephrotoxic medications and contrast media exposure, and correction of electrolyte imbalances. Renal replacement therapy (dialysis) is indicated for refractory hyperkalemia; volume overload; intractable acidosis; uremic encephalopathy, pericarditis, or pleuritis; and removal of certain toxins. Recognition of risk factors (e.g., older age, sepsis, hypovolemia/shock, cardiac surgery, infusion of contrast agents, diabetes mellitus, preexisting chronic kidney disease, cardiac failure, liver failure) is important. Team-based approaches for prevention, early diagnosis, and aggressive management are critical for improving outcomes.
The incidence of acute kidney injury has increased in recent years, both in the community and in hospital settings.1,2 The estimated incidence of acute kidney injury is two to three cases per 1,000 persons.3 Seven percent of hospitalized patients and about two-thirds of patients in intensive care units develop acute kidney injury,2 often as part of the multiple organ dysfunction syndrome.4
SORT: KEY RECOMMENDATIONS FOR PRACTICE
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| Clinical recommendation | Evidence rating | References |
---
| The diagnosis of acute kidney injury is based on serum creatinine levels, urine output, and the need for renal replacement therapy. | C | 8 |
| Renal ultrasonography should be performed in most patients with acute kidney injury to rule out obstruction. | C | 17 |
| Adequate fluid balance should be maintained in patients with acute kidney injury by using isotonic solutions (e.g., normal saline) instead of hyperoncotic solutions (e.g., dextrans, hydroxyethyl starch, albumin). | C | 19 |
| Dopamine use is not recommended for the prevention of acute kidney injury. | A | 21 |
| Diuretics do not improve morbidity, mortality, or renal outcomes, and should not be used to prevent or treat acute kidney injury in the absence of volume overload. | A | 22 |
| Consider therapy with immunosuppressive agents (e.g., cyclophosphamide, prednisone) in patients with rapidly progressive glomerulonephritis. | C | 23 |
A = consistent, good-quality patient-oriented evidence; B = inconsistent or limited-quality patient-oriented evidence; C = consensus, disease-oriented evidence, usual practice, expert opinion, or case series. For information about the SORT evidence rating system, go to
Acute kidney injury is associated with a high rate of adverse outcomes; mortality rates range between 25 and 80 percent, depending on the cause and the clinical status of the patient.5–7 These data highlight the importance of recognition and appropriate management, usually in collaboration with nephrologists and other subspecialists.
Definition
Acute kidney injury is defined as an abrupt (within 48 hours) reduction in kidney function based on an elevation in serum creatinine level, a reduction in urine output, the need for renal replacement therapy (dialysis), or a combination of these factors. It is classified in three stages (Table 1).8 The term acute kidney injury should replace terms such as acute renal failure and acute renal insufficiency, which previously have been used to describe the same clinical condition.
Table 1.
Stages of Acute Kidney Injury
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| Stage | Change in serum creatinine level | Urine output | Other |
--- --- |
| 1 | Increase ≥ 0.3 mg per dL (26.52 μmol per L) or ≥ 1.5- to twofold from baseline | < 0.5 mL per kg per hour for more than six hours | — |
| 2 | Increase > two- to threefold from baseline | < 0.5 mL per kg per hour for more than 12 hours | — |
| 3 | Increase > threefold from baseline or ≥ 4.0 mg per dL (353.60 μmol per L) with an acute rise of at least 0.5 mg per dL (44.20 μmol per L) | < 0.3 mL per kg per hour for 24 hours or anuria for 12 hours | Renal replacement therapy required |
note: Each stage is defined by the change in serum creatinine level, the change in urine output, or the need for renal replacement therapy.
Adapted with permission from Mehta RL, Kellum JA, Shah SV, et al. Acute Kidney Injury Network: report of an initiative to improve outcomes in acute kidney injury. Crit Care. 2007;11(2):R31.
Etiology
The causes of acute kidney injury can be divided into three categories (Table 29 ): prerenal (caused by decreased renal perfusion, often because of volume depletion), intrinsic renal (caused by a process within the kidneys), and postrenal (caused by inadequate drainage of urine distal to the kidneys). In patients who already have underlying chronic kidney disease, any of these factors, but especially volume depletion, may cause acute kidney injury in addition to the chronic impairment of renal function.
Table 2.
Causes of Acute Kidney Injury
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| | |
--- |
| Prerenal | |
| Intrarenal vasoconstriction (hemodynamically mediated) | |
| | Medications: nonsteroidal anti-inflammatory drugs, angiotensin-converting enzyme inhibitors, angiotensin receptor blockers, cyclosporine (Sandimmune), tacrolimus (Prograf) |
| | Cardiorenal syndrome |
| | Hepatorenal syndrome |
| | Abdominal compartment syndrome |
| | Hypercalcemia |
| Systemic vasodilation (e.g., sepsis, neurogenic shock) | |
| Volume depletion | |
| | Renal loss from diuretic overuse, osmotic diuresis (e.g., diabetic ketoacidosis) |
| | Extrarenal loss from vomiting, diarrhea, burns, sweating, blood loss |
| Intrinsic renal | |
| Glomerular (e.g., postinfectious and other glomerulonephritis) | |
| Interstitial | |
| | Medications: penicillin analogues, cephalosporins, sulfonamides, ciprofloxacin (Cipro), acyclovir (Zovirax), rifampin, phenytoin (Dilantin), interferon, proton pump inhibitors, nonsteroidal anti-inflammatory drugs |
| | Infections (e.g., direct infection of renal parenchyma or associated with systemic infections) |
| | Viruses: Epstein-Barr virus, cytomegalovirus, human immunodeficiency virus |
| | Bacteria: Streptococcus species, Legionella species |
| | Fungi: candidiasis, histoplasmosis |
| | Systemic disease: sarcoidosis, lupus |
| Tubular | |
| | Ischemic: prolonged hypotension |
| | Nephrotoxic: exogenous toxins (e.g., radiographic contrast agents, aminoglycosides, cisplatin, methotrexate, ethylene glycol, amphotericin B) and endogenous toxins (e.g., hemolysis and rhabdomyolysis [pigment nephropathy], tumor lysis syndrome, myeloma) |
| Vascular | |
| | Renal vein thrombosis, malignant hypertension, scleroderma renal crisis, renal atheroembolic disease, and renal infarction |
| Postrenal | |
| Extrarenal obstruction: prostate hypertrophy; neurogenic bladder; retroperitoneal fibrosis; bladder, prostate, or cervical cancer | |
| Intrarenal obstruction: stones, crystals (acyclovir, indinavir [Crixivan]), clots, tumors | |
—Most common causes.
Adapted with permission from Holley JL. Clinical approach to the diagnosis of acute renal failure. In: Greenberg A, Cheung AK, eds. Primer on Kidney Diseases. 5th ed. Philadelphia, Pa.: National Kidney Foundation; 2009:278.
PRERENAL CAUSES
Approximately 70 percent of community-acquired cases of acute kidney injury are attributed to prerenal causes.10 In these cases, underlying kidney function may be normal, but decreased renal perfusion associated with intravascular volume depletion (e.g., from vomiting or diarrhea) or decreased arterial pressure (e.g., from heart failure or sepsis) results in a reduced glomerular filtration rate. Autoregulatory mechanisms often can compensate for some degree of reduced renal perfusion in an attempt to maintain the glomerular filtration rate. In patients with preexisting chronic kidney disease, however, these mechanisms are impaired, and the susceptibility to develop acute-on-chronic renal failure is higher.11
Several medications can cause prerenal acute kidney injury. Notably, angiotensin-converting enzyme inhibitors and angiotensin receptor blockers can impair renal perfusion by causing dilation of the efferent arteriole and reduce intraglomerular pressure. Nonsteroidal anti-inflammatory drugs also can decrease the glomerular filtration rate by changing the balance of vasodilatory/vasoconstrictive agents in the renal microcirculation. These drugs and others limit the normal homeostatic responses to volume depletion and can be associated with a decline in renal function. In patients with prerenal acute kidney injury, kidney function typically returns to baseline after adequate volume status is established, the underlying cause is treated, or the offending drug is discontinued.
INTRINSIC RENAL CAUSES
Intrinsic renal causes are also important sources of acute kidney injury and can be categorized by the component of the kidney that is primarily affected (i.e., tubular, glomerular, interstitial, or vascular).
Acute tubular necrosis is the most common type of intrinsic acute kidney injury in hospitalized patients. The cause is usually ischemic (from prolonged hypotension) or nephrotoxic (from an agent that is toxic to the tubular cells). In contrast to a prerenal etiology, acute kidney injury caused by acute tubular necrosis does not improve with adequate repletion of intravascular volume and blood flow to the kidneys. Both ischemic and nephrotoxic acute tubular necrosis can resolve over time, although temporary renal replacement therapy may be required, depending on the degree of renal injury and the presence of preexisting chronic kidney disease.
Glomerular causes of acute kidney injury are the result of acute inflammation of blood vessels and glomeruli. Glomerulonephritis is usually a manifestation of a systemic illness (e.g., systemic lupus erythematosus) or pulmonary renal syndromes (e.g., Goodpasture syndrome, Wegener granulomatosis). History, physical examination, and urinalysis are crucial for diagnosing glomerulonephritis (Table 39 and Figure 112 ). Because management often involves administration of immunosuppressive or cytotoxic medications with potentially severe adverse effects, renal biopsy is often required to confirm the diagnosis before initiating therapy.
Table 3.
History and Physical Examination Findings for Categorizing Acute Kidney Injury
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| Type of acute kidney injury | | History findings | Physical examination findings |
--- --- |
| Prerenal | | Volume loss (e.g., history of vomiting, diarrhea, diuretic overuse, hemorrhage, burns) | Weight loss, orthostatic hypotension and tachycardia |
| Thirst and reduced fluid intake | Poor skin turgor |
| Cardiac disease | Dilated neck veins, S3 heart sound, pulmonary rales, peripheral edema |
| Liver disease | Ascites, caput medusae, spider angiomas |
| Intrinsic renal | | | |
| | Acute tubular necrosis | History of receiving nephrotoxic medications (including over-the-counter, illicit, and herbal), hypotension, trauma or myalgias suggesting rhabdomyolysis, recent exposure to radiographic contrast agents | Muscle tenderness, compartment syndrome, assessment of volume status |
| | Glomerular | Lupus, systemic sclerosis, rash, arthritis, uveitis, weight loss, fatigue, hepatitis C virus infection, human immunodeficiency virus infection, hematuria, foamy urine, cough, sinusitis, hemoptysis | Periorbital, sacral, and lower-extremity edema; rash; oral/nasal ulcers |
| | Interstitial | Medication use (e.g., antibiotics, proton pump inhibitors), rash, arthralgias, fever, infectious illness | Fever, drug-related rash |
| | Vascular | Nephrotic syndrome, trauma, flank pain, anticoagulation (atheroembolic disease), vessel catheterization or vascular surgery | Livedo reticularis, funduscopic examination (showing malignant hypertension), abdominal bruits |
| Postrenal | | Urinary urgency or hesitancy, gross hematuria, polyuria, stones, medications, cancer | Bladder distention, pelvic mass, prostate enlargement |
Adapted with permission from Holley JL. Clinical approach to the diagnosis of acute renal failure. In: Greenberg A, Cheung AK, eds. Primer on Kidney Diseases. 5th ed. Philadelphia, Pa.: National Kidney Foundation; 2009:280.
Figure 1.
Diagnosis and Treatment of Acute Kidney Injury
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Acute interstitial nephritis can be secondary to many conditions, but most cases are related to medication use, making patient history the key to diagnosis. In about one-third of cases, there is a history of maculopapular erythematous rash, fever, arthralgias, or a combination of these symptoms.13 Eosinophiluria may be found in patients with acute interstitial nephritis, but it is not pathognomonic of this disease. A kidney biopsy may be needed to distinguish between allergic interstitial nephritis and other renal causes of acute kidney injury. In addition to discontinuing offending agents, steroids may be beneficial if given early in the course of disease.14
Acute events involving renal arteries or veins can also lead to intrinsic acute kidney injury. Renal atheroembolic disease is the most common cause and is suspected with a recent history of arterial catheterization, the presence of a condition requiring anticoagulation, or after vascular surgery. Physical examination and history provide important clues to the diagnosis (Table 39 ). Vascular causes of acute kidney injury usually require imaging to confirm the diagnosis.
POSTRENAL CAUSES
Postrenal causes typically result from obstruction of urinary flow, and prostatic hypertrophy is the most common cause of obstruction in older men. Prompt diagnosis followed by early relief of obstruction is associated with improvement in renal function in most patients.
Clinical Presentation
Clinical presentation varies with the cause and severity of renal injury, and associated diseases. Most patients with mild to moderate acute kidney injury are asymptomatic and are identified on laboratory testing. Patients with severe cases, however, may be symptomatic and present with listlessness, confusion, fatigue, anorexia, nausea, vomiting, weight gain, or edema.15 Patients can also present with oliguria (urine output less than 400 mL per day), anuria (urine output less than 100 mL per day), or normal volumes of urine (nonoliguric acute kidney injury). Other presentations of acute kidney injury may include development of uremic encephalopathy (manifested by a decline in mental status, asterixis, or other neurologic symptoms), anemia, or bleeding caused by uremic platelet dysfunction.
Diagnosis
A patient history and physical examination, with an emphasis on assessing the patient’s volume status, are crucial for determining the cause of acute kidney injury (Table 39 ). The history should identify use of nephrotoxic medications or systemic illnesses that might cause poor renal perfusion or directly impair renal function. Physical examination should assess intravascular volume status and any skin rashes indicative of systemic illness. The initial laboratory evaluation should include urinalysis, complete blood count, and measurement of serum creatinine level and fractional excretion of sodium (FENa). Imaging studies can help rule out obstruction. Useful tests are summarized in Table 4.16 Figure 1 presents an overview of the diagnosis and management of acute kidney injury.12
Table 4.
Diagnostic Test Results and Corresponding Diseases in Patients with Acute Kidney Injury
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| Test result | When to order | Associated diseases/conditions |
---
| Elevated antineutrophil cytoplasmic antibody, antiglomerular basement membrane antibody | Suspected acute glomerulonephritis, pulmonary renal syndromes | Vasculitis, Goodpasture syndrome |
| Elevated antistreptolysin O titer | Recent infection and clinical picture of acute glomerulonephritis | Poststreptococcal glomerulonephritis |
| Elevated creatine kinase level, elevated myoglobin level, dipstick positive for blood but negative for red blood cells | Recent trauma, muscle injury | Rhabdomyolysis |
| Elevated prostate-specific antigen level | Older men with symptoms suggestive of urinary obstruction | Prostate hypertrophy, prostate cancer |
| Elevated uric acid level | History of rapidly proliferating tumors, recent chemotherapy | Malignancy, tumor lysis syndrome |
| Eosinophiluria | Fever, rash | Allergic interstitial nephritis |
| Evidence of hemolysis (schistocytes on peripheral smear, decreased haptoglobin level, elevated indirect bilirubin level, elevated lactate dehydrogenase level) | Fever, anemia, thrombocytopenia, neurologic signs | Hemolytic uremic syndrome, thrombotic thrombocytopenic purpura, systemic lupus erythematosus, other autoimmune diseases |
| Hydronephrosis on renal ultrasonography | Suspected obstruction | Malignancy, prostate hypertrophy, uterine fibroids, nephrolithiasis, ureterolithiasis |
| Increased anion gap with increased osmolar gap | Suspected poisoning, unresponsive patient | Ethylene glycol or methanol poisoning |
| Low complement level | Suspected acute glomerulonephritis | Systemic lupus erythematosus, endocarditis, postinfectious glomerulonephritis |
| Monoclonal spike on serum protein electrophoresis | Anemia, proteinuria, acute kidney injury in older patients | Multiple myeloma |
| Positive antinuclear antibody, double-stranded DNA antibody | Proteinuria, skin rash, arthritis | Autoimmune diseases, systemic lupus erythematosus |
| Positive blood cultures | Intravenous drug use, recent infection, new cardiac murmur | Endocarditis |
| Positive HIV test | Risk factors for HIV infection | HIV nephropathy |
HIV = human immunodeficiency virus.
—Calculations are as follows:
Anion gap = sodium – (chloride + bicarbonate)
Calculated serum osmolality = 2(sodium [in mEq per L]) + (blood urea nitrogen [in mg per dL] ÷ 2.8) + (glucose [in mg per dL] ÷ 18)
Osmolar gap = measured serum osmolality – calculated serum osmolality.
Adapted with permission from Agrawal M, Swartz R. Acute renal failure [published correction appears in Am Fam Physician. 2001;63(3):445]. Am Fam Physician. 2000;61(7):2081.
SERUM CREATININE LEVEL
It is important to compare the patient’s current serum creatinine level with previous levels to determine the duration and acuity of the disease. The definition of acute kidney injury indicates that a rise in creatinine has occurred within 48 hours, although in the outpatient setting, it may be hard to ascertain when the rise actually happened. A high serum creatinine level in a patient with a previously normal documented level suggests an acute process, whereas a rise over weeks to months represents a subacute or chronic process.
URINALYSIS
Urinalysis is the most important noninvasive test in the initial workup of acute kidney injury. Findings on urinalysis guide the differential diagnosis and direct further workup (Figure 112 ).
COMPLETE BLOOD COUNT
The presence of acute hemolytic anemia with the peripheral smear showing schistocytes in the setting of acute kidney injury should raise the possibility of hemolytic uremic syndrome or thrombotic thrombocytopenic purpura.
URINE ELECTROLYTES
In patients with oliguria, measurement of FENa is helpful in distinguishing prerenal from intrinsic renal causes of acute kidney injury. FENa is defined by the following formula:
Online calculators are also available. A value less than 1 percent indicates a prerenal cause of acute kidney injury, whereas a value greater than 2 percent indicates an intrinsic renal cause. In patients on diuretic therapy, however, a FENa higher than 1 percent may be caused by natriuresis induced by the diuretic, and is a less reliable measure of a prerenal state. In such cases, fractional excretion of urea may be helpful, with values less than 35 percent indicating a prerenal cause. FENa values less than 1 percent are not specific for prerenal causes of acute kidney injury because these values can occur in other conditions, such as contrast nephropathy, rhabdomyolysis, acute glomerulonephritis, and urinary tract obstruction.
IMAGING STUDIES
Renal ultrasonography should be performed in most patients with acute kidney injury, particularly in older men, to rule out obstruction (i.e., a postrenal cause).17,18 The presence of postvoid residual urine greater than 100 mL (determined by a bladder scan or via urethral catheterization if bladder scan is unavailable) suggests postrenal acute kidney injury and requires renal ultrasonography to detect hydronephrosis or outlet obstruction. To diagnose extrarenal causes of obstruction (e.g., pelvic tumors), other imaging modalities, such as computed tomography or magnetic resonance imaging, may be required.
RENAL BIOPSY
Renal biopsy is reserved for patients in whom prerenal and postrenal causes of acute kidney injury have been excluded and the cause of intrinsic renal injury is unclear. Renal biopsy is particularly important when clinical assessment and laboratory investigations suggest a diagnosis that requires confirmation before disease-specific therapy (e.g., immunosuppressive medications) is instituted. Renal biopsy may need to be performed urgently in patients with oliguria who have rapidly worsening acute kidney injury, hematuria, and red blood cell casts. In this setting, in addition to indicating a diagnosis that requires immunosuppressive therapy, the biopsy may support the initiation of special therapies, such as plasmapheresis if Goodpasture syndrome is present.
Management
Optimal management of acute kidney injury requires close collaboration among primary care physicians, nephrologists, hospitalists, and other subspecialists participating in the care of the patient. After acute kidney injury is established, management is primarily supportive.
Patients with acute kidney injury generally should be hospitalized unless the condition is mild and clearly resulting from an easily reversible cause. The key to management is assuring adequate renal perfusion by achieving and maintaining hemodynamic stability and avoiding hypovolemia. In some patients, clinical assessment of intravascular volume status and avoidance of volume overload may be difficult, in which case measurement of central venous pressures in an intensive care setting may be helpful.
If fluid resuscitation is required because of intravascular volume depletion, isotonic solutions (e.g., normal saline) are preferred over hyperoncotic solutions (e.g., dextrans, hydroxyethyl starch, albumin).19 A reasonable goal is a mean arterial pressure greater than 65 mm Hg, which may require the use of vasopressors in patients with persistent hypotension.20 Renal-dose dopamine is associated with poorer outcomes in patients with acute kidney injury; it is no longer recommended.21 Cardiac function can be optimized as needed with positive inotropes, or afterload and preload reduction.
Attention to electrolyte imbalances (e.g., hyperkalemia, hyperphosphatemia, hypermagnesemia, hyponatremia, hypernatremia, metabolic acidosis) is important. Severe hyperkalemia is defined as potassium levels of 6.5 mEq per L (6.5 mmol per L) or greater, or less than 6.5 mEq per L with electrocardiographic changes typical of hyperkalemia (e.g., tall, peaked T waves). In severe hyperkalemia, 5 to 10 units of regular insulin and dextrose 50% given intravenously can shift potassium out of circulation and into the cells. Calcium gluconate (10 mL of 10% solution infused intravenously over five minutes) is also used to stabilize the membrane and reduce the risk of arrhythmias when there are electrocardiographic changes showing hyperkalemia. In patients without electrocardiographic evidence of hyperkalemia, calcium gluconate is not necessary, but sodium polystyrene sulfonate (Kayexalate) can be given to lower potassium levels gradually, and loop diuretics can be used in patients who are responsive to diuretics. Dietary intake of potassium should be restricted.
The main indication for use of diuretics is management of volume overload. Intravenous loop diuretics, as a bolus or continuous infusion, can be helpful for this purpose. However, it is important to note that diuretics do not improve morbidity, mortality, or renal outcomes, and should not be used to prevent or treat acute kidney injury in the absence of volume overload.22
All medications that may potentially affect renal function by direct toxicity or by hemodynamic mechanisms should be discontinued, if possible. For example, metformin (Glucophage) should not be given to patients with diabetes mellitus who develop acute kidney injury. The dosages of essential medications should be adjusted for the lower level of kidney function. Avoidance of iodinated contrast media and gadolinium is important and, if imaging is needed, noncontrast studies are recommended.
Supportive therapies (e.g., antibiotics, maintenance of adequate nutrition, mechanical ventilation, glycemic control, anemia management) should be pursued based on standard management practices. In patients with rapidly progressive glomerulonephritis, treatment with pulse steroids, cytotoxic therapy, or a combination may be considered, often after confirmation of the diagnosis by kidney biopsy.23 In some patients, the metabolic consequences of acute kidney injury cannot be adequately controlled with conservative management, and renal replacement therapy will be required. The indications for initiation of renal replacement therapy include refractory hyperkalemia, volume overload refractory to medical management, uremic pericarditis or pleuritis, uremic encephalopathy, intractable acidosis, and certain poisonings and intoxications (e.g., ethylene glycol, lithium).24
Prognosis
Patients with acute kidney injury are more likely to develop chronic kidney disease in the future. They are also at higher risk of end-stage renal disease and premature death.25–27 Patients who have an episode of acute kidney injury should be monitored for the development or worsening of chronic kidney disease.
Prevention
Because of the morbidity and mortality associated with acute kidney injury, it is important for primary care physicians to identify patients who are at high risk of developing this type of injury and to implement preventive strategies. Those at highest risk include adults older than 75 years; persons with diabetes or preexisting chronic kidney disease; persons with medical problems such as cardiac failure, liver failure, or sepsis; and those who are exposed to contrast agents or who are undergoing cardiac surgery.28 Preventive strategies can be tailored to the clinical circumstances of the individual patient (Table 5).19–21,27,29–31
Table 5.
Preventive Strategies for Patients at High Risk of Acute Kidney Injury
zoom_out_mapEnlargeprintPrint
| Risk factors | Preventive strategies |
--- |
| Cancer chemotherapy with risk of tumor lysis syndrome27 | Hydration and allopurinol (Zyloprim) administration a few days before chemotherapy initiation in patients at high risk of tumor lysis syndrome to prevent uric acid nephropathy |
| Exposure to nephrotoxic medications | Avoid nephrotoxic medications if possible |
| Measure and follow drug levels if available |
| Use appropriate dosing, intervals, and duration of therapy |
| Exposure to radiographic contrast agents29 | Avoid use of intravenous contrast media when risks outweigh benefits |
| If use of contrast media is essential, use iso-osmolar or low-osmolar contrast agent with lowest volume possible |
| Optimize volume status before administration of contrast media; use of isotonic normal saline or sodium bicarbonate may be considered in high-risk patients who are not at risk of volume overload |
| Use of N-acetylcysteine may be considered |
| Hemodynamic instability | Optimal fluid resuscitation; although there is no consensus, a mean arterial pressure goal of > 65 mm Hg is widely used; isotonic solutions (e.g., normal saline) are preferred over hyperoncotic solutions (e.g., albumin)19 |
| Vasopressors are recommended for persistent hypotension (mean arterial pressure < 65 mm Hg) despite fluid resuscitation; choice of vasoactive agent should be tailored to patients’ needs20 |
| Dopamine is not recommended21 |
| Hepatic failure30 | Avoid hypotension and gastrointestinal bleeding |
| Early recognition and treatment of spontaneous bacterial peritonitis; use albumin, 1.5 g per kg at diagnosis and 1 g per kg at 48 hours |
| Early recognition and management of ascites |
| Albumin infusion during large volume paracentesis |
| Avoid nephrotoxic medications |
| Rhabdomyolysis20 | Maintain adequate hydration |
| Alkalinization of the urine with intravenous sodium bicarbonate in select patients (normal calcium, bicarbonate less than 30 mEq per L [30 mmol per L], and arterial pH less than 7.5) |
| Undergoing surgery | Adequate volume resuscitation/prevention of hypotension, sepsis, optimizing cardiac function Consider holding renin-angiotensin system antagonists preoperatively31 |
Information from references 19 through 21, 27, and 29 through 31.
Data Sources: We searched PubMed (also with the Clinical Queries function), the Cochrane Database of Systematic Reviews, and the National Guidelines Clearinghouse using the key words AKI, acute kidney injury, and acute renal failure. Search date: February 2012.
MAHBOOB RAHMAN, MD, MS, is an associate professor of medicine at Case Western Reserve University School of Medicine in Cleveland, Ohio, and a staff nephrologist at University Hospitals Case Medical Center in Cleveland and at Louis Stokes Cleveland VA Medical Center.
FARIHA SHAD, MD, is a nephrologist at Kaiser Permanente in Cleveland. At the time the article was written, Dr. Shad was a fellow at Case Western Reserve University School of Medicine.
MICHAEL C. SMITH, MD, is a professor of medicine at Case Western Reserve University School of Medicine, and a staff nephrologist at University Hospitals Case Medical Center.
Address correspondence to Mahboob Rahman, MD, MS, Case Western Reserve University, 11100 Euclid Ave., Cleveland, OH 44106. Reprints are not available from the authors.
Author disclosure: No relevant financial affiliations to disclose.
Hsu CY, McCulloch CE, Fan D, Ordoñez JD, Chertow GM, Go AS. Community-based incidence of acute renal failure. Kidney Int. 2007;72(2):208-212.
Nash K, Hafeez A, Hou S. Hospital-acquired renal insufficiency. Am J Kidney Dis. 2002;39(5):930-936.
Hoste EA, Schurgers M. Epidemiology of acute kidney injury: how big is the problem?. Crit Care Med. 2008;36(4 suppl):S146-S151.
Hoste EA, Clermont G, Kersten A, et al. RIFLE criteria for acute kidney injury are associated with hospital mortality in critically ill patients: a cohort analysis. Crit Care. 2006;10(3):R73.
Ympa YP, Sakr Y, Reinhart K, Vincent JL. Has mortality from acute renal failure decreased? A systematic review of the literature. Am J Med. 2005;118(8):827-832.
Gruberg L, Weissman NJ, Pichard AD, et al. Impact of renal function on morbidity and mortality after percutaneous aortocoronary saphenous vein graft intervention. Am Heart J. 2003;145(3):529-534.
Uchino S, Kellum JA, Bellomo R, et al.; Beginning and Ending Supportive Therapy for the Kidney (BEST Kidney) Investigators. Acute renal failure in critically ill patients: a multinational, multicenter study. JAMA. 2005;294(7):813-818.
Mehta RL, Kellum JA, Shah SV, et al. Acute Kidney Injury Network: report of an initiative to improve outcomes in acute kidney injury. Crit Care. 2007;11(2):R31.
Holley JL. Clinical approach to the diagnosis of acute renal failure. In: Greenberg A, Cheung AK, eds. Primer on Kidney Diseases. 5th ed. Philadelphia, Pa.: National Kidney Foundation; 2009.
Kaufman J, Dhakal M, Patel B, Hamburger R. Community-acquired acute renal failure. Am J Kidney Dis. 1991;17(2):191-198.
Christensen PK, Hansen HP, Parving HH. Impaired autoregulation of GFR in hypertensive non-insulin dependent diabetic patients. Kidney Int. 1997;52(5):1369-1374.
Smith MC. Acute renal failure. In: Resnick MI, Elder JS, Spirnak JP, eds. Clinical Decisions in Urology. 3rd ed. Hamilton, Ontario, Canada: BC Decker, Inc.; 2004:60–63.
Clarkson MR, Giblin L, O’Connell FP, et al. Acute interstitial nephritis: clinical features and response to corticosteroid therapy. Nephrol Dial Transplant. 2004;19(11):2778-2783.
González E, Gutiérrez E, Galeano C, et al.; Grupo Madrileño De Nefritis Intersticiales. Early steroid treatment improves the recovery of renal function in patients with drug-induced acute interstitial nephritis. Kidney Int. 2008;73(8):940-946.
Meyer TW, Hostetter TH. Uremia. N Engl J Med. 2007;357(13):1316-1325.
Agrawal M, Swartz R. Acute renal failure [published correction appears in Am Fam Physician. 2001;63(3):445]. Am Fam Physician. 2000;61(7):2077-2088.
Lewington A, Kanagasundaram S. Clinical practice guidelines: acute kidney injury. 2011. Accessed September 7, 2012.
O’Neill WC. Sonographic evaluation of renal failure. Am J Kidney Dis. 2000;35(6):1021-1038.
Schortgen F, Lacherade JC, Bruneel F, et al. Effects of hydroxyethylstarch and gelatin on renal function in severe sepsis: a multicentre randomised study. Lancet. 2001;357(9260):911-916.
Brochard L, Abroug F, Brenner M, et al. An Official ATS/ERS/ESICM/SCCM/SRLF Statement: Prevention and Management of Acute Renal Failure in the ICU Patient: an international consensus conference in intensive care medicine. Am J Respir Crit Care Med. 2010;181(10):1128-1155.
Friedrich JO, Adhikari N, Herridge MS, Beyene J. Meta-analysis: low-dose dopamine increases urine output but does not prevent renal dysfunction or death. Ann Intern Med. 2005;142(7):510-524.
Ho KM, Sheridan DJ. Meta-analysis of frusemide to prevent or treat acute renal failure. BMJ. 2006;333(7565):420.
Walters G, Willis NS, Craig JC. Interventions for renal vasculitis in adults. Cochrane Database Syst Rev. ;2008(3):CD003232.
Mehta RL. Indications for dialysis in the ICU: renal replacement vs. renal support. Blood Purif. 2001;19(2):227-232.
Goldberg R, Dennen P. Long-term outcomes of acute kidney injury. Adv Chronic Kidney Dis. 2008;15(3):297-307.
Coca SG, Yusuf B, Shlipak MG, Garg AX, Parikh CR. Long-term risk of mortality and other adverse outcomes after acute kidney injury: a systematic review and meta-analysis. Am J Kidney Dis. 2009;53(6):961-973.
Pession A, Masetti R, Gaidano G, et al. Risk evaluation, prophylaxis, and treatment of tumor lysis syndrome: consensus of an Italian expert panel. Adv Ther. 2011;28(8):684-697.
Leblanc M, Kellum JA, Gibney RT, Lieberthal W, Tumlin J, Mehta R. Risk factors for acute renal failure: inherent and modifiable risks. Curr Opin Crit Care. 2005;11(6):533-536.
Rundback JH, Nahl D, Yoo V. Contrast-induced nephropathy. J Vasc Surg. 2011;54(2):575-579.
Nadim MK, Kellum JA, Davenport A, et al. Hepatorenal syndrome: the 8th international consensus conference of the Acute Dialysis Quality Initiative (ADQI) group. Crit Care. 2012;16(1):R23.
Auron M, Harte B, Kumar A, Michota F. Renin-angiotensin system antagonists in the perioperative setting: clinical consequences and recommendations for practice. Postgrad Med J. 2011;87(1029):472-481.
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14874 | https://www.youtube.com/watch?v=bb1Qdkvwarc | Finding Areas Using the Cross Product (Calculus 3)
Houston Math Prep
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Posted: 25 Jan 2021
This Calculus 3 video explains how to use the magnitude of the cross product to find areas of parallelograms and triangles in 3D space. We show why the magnitude of the cross product is equal to the area of a parallelogram defined by two vectors, and describe why we can use the same idea to find the area of a triangle in 3D space using the cross product. We also work two examples of finding these types of areas.
0:00 Cross product magnitudes
3:18 Areas of parallelograms
4:45 Areas of triangles
5:30 Example 1 - Parallelogram
8:52 Example 2 - Triangle
Houston Math Prep Calculus 3 Playlist:
Houston Math Prep YouTube:
Transcript:
Cross product magnitudes welcome back everyone houston math prep here this video is going to look at using the cross product to find areas in 3d space in particular we're going to be taking a look at the magnitude of the cross product so here based on our formula for the cross product we know that the magnitude of u cross v would be the square of each component added up and then we would take the square root of it all this isn't a formula we need to have memorized remember as long as we get the process of finding a determinant like we did in our cross product intro video so we're going to take a brief journey here you may want to put your pencil down just for the next minute or so but just bear with us so taking a brief journey using this magnitude i promise it'll pay off for us in the long run if this is the magnitude of u cross v here then if we square both sides we get rid of the square root so this is the magnitude squared here you can probably see distributing all of this out and then combining all the like terms and such from all this right hand side is pretty monstrous so we're going to skip all the algebra here for the sake of you staying awake in short after doing all of that work we get something that ends up looking like this and i think this is sort of a nice place in the algebra to continue from so we've skipped a little bit of work and now continuing on from here because we can see some items we might recognize and relate to stuff that we already know right so if we look at this first set of parentheses you might notice this is also a magnitude squared this is the magnitude of u squared and similarly we can probably see that this second set of parentheses is the magnitude of v squared so we've got two magnitude statements in the first half of this formula if we look at the last set of parentheses the inside hopefully looks something familiar to you this is actually a dot product in here so we actually have the dot product of u and v squared right so we get minus u dot v all squared so to make this a little easier to write in one term altogether we'll go ahead and change our dot product into a formula that we've covered that involves cosine so we get magnitudes of u and v times cosine of theta for our u dot v all of that's going to be squared we're going to rewrite our dot product as this other way of writing the dot product remember theta is the angle between these vectors u and v right so now we go ahead and we factor out these common magnitudes squared from each term leaving us with a lovely pythagorean identity right of one minus cosine squared theta which we can then change to sine squared theta of course and so if you look at the top of the screen and the bottom of the screen we really get that the square of the magnitude of the cross product is equal to these magnitudes squared times sine squared theta okay so now this is the point where you can pick your pencil back up and maybe follow along with us so looking at everything being squared here i think we can do a nice simplification here and simplify to the magnitude of u cross v being equal to magnitude u times magnitude v times sine of theta and at this point you may be saying why do we care about any of this but this little nugget right here is what's going to help us in finding areas so we'll just kind of Areas of parallelograms set that formula up in the top right corner here for now and let's look at finding the area of a parallelogram defined by two vectors in space and here it's going to be u and v so when we say defined by two vectors we think of two adjacent sides of the parallelogram being those two vectors with the other two sides just being copies of those vectors as well so we might remember that the formula for the area of a parallelogram same as area of a rectangle really is simply the width times the height of the parallelogram for the width the way i've drawn this one here i think you can probably tell the width is just however long vector u is so we'll say our area is going to be the magnitude of u times whatever the height is when we talk about the height we mean the height perpendicular to the base here not the slant height some basic right triangle trigonometry here is going to tell us that the height here is actually equal to the magnitude of v times sine of theta for our height and that would mean that the area of the parallelogram is equal to our magnitude of u cross v here because magnitude of u times magnitude of v times sine of theta we just found at the beginning of our video is the same as the magnitude of u cross v so the area of a parallelogram is equal to the magnitude of the cross product of those vectors if Areas of triangles the magnitude of the cross product of two vectors is equal to the area of the parallelogram with those vectors as adjacent sides what do we notice if we cut a parallelogram in half like this you can see that we have half our parallelogram on one side and the other half on the other side of the line i've drawn here in other words half a parallelogram cut this way will give us a triangle defined by vectors u and v so that tells us that we can find the area of a triangle defined by two vectors by taking half the magnitude of the cross product since a triangle is going to be half of a parallelogram Example 1 - Parallelogram so we'll go ahead and work through a couple of area examples with you so this one says find the area of the parallelogram defined by vector v which is negative 1 3 2 and vector w which is 4 1 0 and so remember the area of the parallelogram so our area is going to be the magnitude of the cross product of our vector here's the magnitude of v cross w well let's first go ahead and figure out what is v cross w right so vector v cross vector w if you remember from our cross product intro video we set this up as a three by three determinant with our standard unit vectors in the top row and our vectors in the order stated here in the next rows so negative 1 3 2 goes in our second row that's vector v and then w 4 1 0 goes in the bottom here and remember we take an entry in the top row here and we mark out the row and column that that entry is in and multiply that entry by the 2 by 2 determinant containing those entries left over so for i hat's case it would be three two one zero remember it's always minus j hat when we do this second part here so marking out the row and column j hat is in we have negative one two four zero left over negative 1 2 4 0 plus k hat and if we mark out k hat's row and column you'll see we have negative 1 3 4 1. so multiply by the determinant of negative one three four one as a two by two and if we do this we'll get i hat times three times zero would be zero minus one times two would be two minus j hat times our 2 by 2 determinant negative 1 times 0 would be 0 minus 4 times 2 would give us 8 plus k hat times our two by two determinant negative one times one would give us negative one minus four times three would give us twelve and if we simplify here we'll get negative two i hat minus j hat times negative 8 gives us plus 8 j hat k hat times negative 13 here so minus 13 k hat that's our v cross w and so now we want the magnitude of v cross w right so our magnitude of v cross w is going to be the square root of all of these components squared and added up just a magnitude problem here so negative 2 squared plus 8 squared plus negative 13 squared so that will actually give us the square root of 4 plus 64 plus 169 and that would actually give us the square root of 237 for our area of this parallelogram Example 2 - Triangle let's work through one more type of problem with you here we want to find the area of the triangle in space with vertices p 2 1 0 q which is negative 1 0 1 and r which is 0 0 3. now you'll notice these are points in space not vectors these are vertices the corners of the triangle and the idea here the area of the triangle is really going to be finding the area of what would be the parallelogram but dividing by 2. now this area of the parallelogram is going to be based on the cross product of two vectors and we don't have vectors yet so what we'll do is we'll find two vectors with the same initial point so we could find for example vector pq and we could find vector pr and then we could use those and we could find the magnitude of pq cross pr then we'll have the magnitude of a cross product and because we want a triangle not a parallelogram we'll divide by two okay so let's figure out pq first remember that will be terminal point minus initial point so for pq we'll do negative one minus two will be negative three zero minus one will be negative one and then one minus 0 will give us 1. so our vector pq is negative 3 negative 1 positive 1. pr then would be terminal point minus initial point so r minus p here as a vector we will get 0 minus two will be negative two zero minus one will be negative one and three minus zero will be positive three next thing we'll need is the cross product of these right to be part of our formula then we'll find the magnitudes and then we'll divide by two but let's do our cross product next so this will be vector pq cross vector pr remember we'll set up our i hat j hat k hat in the first row since it's the order p q then p r we'll take p q as our first row under the standard unit vectors here negative three negative one one and then p r last negative two negative one three this cross product will have i hat we'll mark out the row and column i hat as in that leaves us negative 1 1 negative 1 3 so i hat times the determinant of those four entries minus j hat times if we mark out the row and column j hat is in we have negative 3 1 negative 2 3. so our 2 by 2 determinant multiplying negative j hat will be negative 3 1 and negative 2 3 plus k hat if we mark out the row and column k hat is in we have negative 3 negative 1 negative 2 negative 1. that will multiply k hat if we do these 2 by 2 determinants we'll have i hat times negative 1 times 3 would be negative 3 minus negative 1 times 1 would be negative 1 minus negative 1 would really be plus 1 there minus j hat negative 3 times 3 would be negative 9 minus negative 2 times 1 would be negative 2 minus negative 2 would be plus 2. plus k hat times negative 3 times negative 1 would be positive 3 minus negative 2 times negative 1 would be positive 2. so here we get the vector negative three plus one would be negative two i hat negative j hat here we get a negative seven in here so this would actually be plus seven j hat plus three minus two would be one so we just get one k hat that is our pq cross pr our cross product and now we need to know the magnitude of this right so we'll take the magnitude and then we'll have to remember to divide by 2. so the magnitude of this will be negative two squared plus seven squared plus one k hat right so this is just one squared all of that over two so we'll get the square root of four plus 49 plus one over two that will actually give us the square root of 54 over 2 and we can actually simplify this if i do it down here so the square root of 54 over 2 i know a perfect square that goes into 54. 9 is a perfect square and that goes into 54. so i pull out the square root of 9. so if i pull out 9 9 times 6 is 54 so that would leave a root 6 left over we actually get 3 root 6 over 2 or the area of our triangle in space with these as vertices all right everyone thanks for watching hopefully this helps you with your areas using the cross product we'll see in the next video |
14875 | https://openstax.org/books/algebra-and-trigonometry-2e/pages/6-key-concepts | Skip to ContentGo to accessibility pageKeyboard shortcuts menu
Algebra and Trigonometry 2e
Key Concepts
Algebra and Trigonometry 2eKey Concepts
Search for key terms or text.
Key Concepts
## 6.1 Exponential Functions
An exponential function is defined as a function with a positive constant other than raised to a variable exponent. See Example 1.
A function is evaluated by solving at a specific value. See Example 2 and Example 3.
An exponential model can be found when the growth rate and initial value are known. See Example 4.
An exponential model can be found when the two data points from the model are known. See Example 5.
An exponential model can be found using two data points from the graph of the model. See Example 6.
An exponential model can be found using two data points from the graph and a calculator. See Example 7.
The value of an account at any time can be calculated using the compound interest formula when the principal, annual interest rate, and compounding periods are known. See Example 8.
The initial investment of an account can be found using the compound interest formula when the value of the account, annual interest rate, compounding periods, and life span of the account are known. See Example 9.
The number is a mathematical constant often used as the base of real world exponential growth and decay models. Its decimal approximation is
Scientific and graphing calculators have the key or for calculating powers of See Example 10.
Continuous growth or decay models are exponential models that use as the base. Continuous growth and decay models can be found when the initial value and growth or decay rate are known. See Example 11 and Example 12.
## 6.2 Graphs of Exponential Functions
The graph of the function has a y-intercept at domain range and horizontal asymptote See Example 1.
If the function is increasing. The left tail of the graph will approach the asymptote and the right tail will increase without bound.
If the function is decreasing. The left tail of the graph will increase without bound, and the right tail will approach the asymptote
The equation represents a vertical shift of the parent function
The equation represents a horizontal shift of the parent function See Example 2.
Approximate solutions of the equation can be found using a graphing calculator. See Example 3.
The equation where represents a vertical stretch if or compression if of the parent function See Example 4.
When the parent function is multiplied by the result, is a reflection about the x-axis. When the input is multiplied by the result, is a reflection about the y-axis. See Example 5.
All translations of the exponential function can be summarized by the general equation See Table 3.
Using the general equation we can write the equation of a function given its description. See Example 6.
## 6.3 Logarithmic Functions
The inverse of an exponential function is a logarithmic function, and the inverse of a logarithmic function is an exponential function.
Logarithmic equations can be written in an equivalent exponential form, using the definition of a logarithm. See Example 1.
Exponential equations can be written in their equivalent logarithmic form using the definition of a logarithm See Example 2.
Logarithmic functions with base can be evaluated mentally using previous knowledge of powers of See Example 3 and Example 4.
Common logarithms can be evaluated mentally using previous knowledge of powers of See Example 5.
When common logarithms cannot be evaluated mentally, a calculator can be used. See Example 6.
Real-world exponential problems with base can be rewritten as a common logarithm and then evaluated using a calculator. See Example 7.
Natural logarithms can be evaluated using a calculator Example 8.
## 6.4 Graphs of Logarithmic Functions
To find the domain of a logarithmic function, set up an inequality showing the argument greater than zero, and solve for See Example 1 and Example 2
The graph of the parent function has an x-intercept at domain range vertical asymptote and
if the function is increasing.
if the function is decreasing.See Example 3.
The equation shifts the parent function horizontally
left units if
right units ifSee Example 4.
The equation shifts the parent function vertically
up units if
down units ifSee Example 5.
For any constant the equation
stretches the parent function vertically by a factor of if
compresses the parent function vertically by a factor of ifSee Example 6 and Example 7.
When the parent function is multiplied by the result is a reflection about the x-axis. When the input is multiplied by the result is a reflection about the y-axis.
The equation represents a reflection of the parent function about the x-axis.
The equation represents a reflection of the parent function about the y-axis.See Example 8.
A graphing calculator may be used to approximate solutions to some logarithmic equations See Example 9.
All translations of the logarithmic function can be summarized by the general equation See Table 4.
Given an equation with the general form we can identify the vertical asymptote for the transformation. See Example 10.
Using the general equation we can write the equation of a logarithmic function given its graph. See Example 11.
## 6.5 Logarithmic Properties
We can use the product rule of logarithms to rewrite the log of a product as a sum of logarithms. See Example 1.
We can use the quotient rule of logarithms to rewrite the log of a quotient as a difference of logarithms. See Example 2.
We can use the power rule for logarithms to rewrite the log of a power as the product of the exponent and the log of its base. See Example 3, Example 4, and Example 5.
We can use the product rule, the quotient rule, and the power rule together to combine or expand a logarithm with a complex input. See Example 6, Example 7, and Example 8.
The rules of logarithms can also be used to condense sums, differences, and products with the same base as a single logarithm. See Example 9, Example 10, Example 11, and Example 12.
We can convert a logarithm with any base to a quotient of logarithms with any other base using the change-of-base formula. See Example 13.
The change-of-base formula is often used to rewrite a logarithm with a base other than 10 and as the quotient of natural or common logs. That way a calculator can be used to evaluate. See Example 14.
## 6.6 Exponential and Logarithmic Equations
We can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then we use the fact that exponential functions are one-to-one to set the exponents equal to one another and solve for the unknown.
When we are given an exponential equation where the bases are explicitly shown as being equal, set the exponents equal to one another and solve for the unknown. See Example 1.
When we are given an exponential equation where the bases are not explicitly shown as being equal, rewrite each side of the equation as powers of the same base, then set the exponents equal to one another and solve for the unknown. See Example 2, Example 3, and Example 4.
When an exponential equation cannot be rewritten with a common base, solve by taking the logarithm of each side. See Example 5.
We can solve exponential equations with base by applying the natural logarithm of both sides because exponential and logarithmic functions are inverses of each other. See Example 6 and Example 7.
After solving an exponential equation, check each solution in the original equation to find and eliminate any extraneous solutions. See Example 8.
When given an equation of the form where is an algebraic expression, we can use the definition of a logarithm to rewrite the equation as the equivalent exponential equation and solve for the unknown. See Example 9 and Example 10.
We can also use graphing to solve equations with the form We graph both equations and on the same coordinate plane and identify the solution as the x-value of the intersecting point. See Example 11.
When given an equation of the form where and are algebraic expressions, we can use the one-to-one property of logarithms to solve the equation for the unknown. See Example 12.
Combining the skills learned in this and previous sections, we can solve equations that model real world situations, whether the unknown is in an exponent or in the argument of a logarithm. See Example 13.
## 6.7 Exponential and Logarithmic Models
The basic exponential function is If we have exponential growth; if we have exponential decay.
We can also write this formula in terms of continuous growth as where is the starting value. If is positive, then we have exponential growth when and exponential decay when See Example 1.
In general, we solve problems involving exponential growth or decay in two steps. First, we set up a model and use the model to find the parameters. Then we use the formula with these parameters to predict growth and decay. See Example 2.
We can find the age, of an organic artifact by measuring the amount, of carbon-14 remaining in the artifact and using the formula to solve for See Example 3.
Given a substance’s doubling time or half-time, we can find a function that represents its exponential growth or decay. See Example 4.
We can use Newton’s Law of Cooling to find how long it will take for a cooling object to reach a desired temperature, or to find what temperature an object will be after a given time. See Example 5.
We can use logistic growth functions to model real-world situations where the rate of growth changes over time, such as population growth, spread of disease, and spread of rumors. See Example 6.
We can use real-world data gathered over time to observe trends. Knowledge of linear, exponential, logarithmic, and logistic graphs help us to develop models that best fit our data. See Example 7.
Any exponential function with the form can be rewritten as an equivalent exponential function with the form where See Example 8.
## 6.8 Fitting Exponential Models to Data
Exponential regression is used to model situations where growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and closer to zero.
We use the command “ExpReg” on a graphing utility to fit function of the form to a set of data points. See Example 1.
Logarithmic regression is used to model situations where growth or decay accelerates rapidly at first and then slows over time.
We use the command “LnReg” on a graphing utility to fit a function of the form to a set of data points. See Example 2.
Logistic regression is used to model situations where growth accelerates rapidly at first and then steadily slows as the function approaches an upper limit.
We use the command “Logistic” on a graphing utility to fit a function of the form to a set of data points. See Example 3.
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14877 | https://www.youtube.com/watch?v=uudB5XiAYOw | Variance of a binomial variable | Random variables | AP Statistics | Khan Academy
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Deriving the variance and standard deviation formulas for binomial random variables.
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what we're going to do in this video is continue our journey trying to understand what the expected value and what the variance of a binomial variable is going to be or what a what the expected value or the variance of a binomial distribution is going to be which is just the distribu distribution of a binomial variable and so like in the last video i have this binomial variable x that's defined in a very general sense it's the number of successes from n trials so it's a finite number of trials where the probability of success is equal to p so the probability is constant across the trials for each of these independent trials so the probability of success in one trial is not dependent on what happened in the other trials and we also talked in that previous video where we talked about the expected value of this binomial variable is we said hey it could be viewed the this binomial variable can be viewed as the sum of n of what you could really consider to be a bernoulli variable here so this variable this random variable y the probability that's equal to one you could view that as a success is equal to p the probability that it's a failure that y is equal to zero is one minus p so you could view y the outcome of y is really the or whether y is one or zero is really whether we had a success or not in each of these trials so if you add up n y's then you are going to get x and we use that information to figure out what the expected value of x is going to be because the expected value of y is pretty straightforward to directly compute expected value of y is just probability weighted outcomes so it's p times 1. plus 1 minus p 1 minus p times 0 times 0. this whole term is going to be 0 and so the expected value of y is really just p and so if you said the expected value of x well that's just going to be we could let me just write it over here this is all review we could say that the expected value of x is just going to be equal to we know from our expected value properties that's going to be equal to the sum of the expected values of these and y's or you could say it is n times the expected value times the expected value of y the expected value of y is p so this is going to be equal to n times p now we're going to do the same idea to figure out what the variance of x is going to be equal to because we could see we know from our variance properties you can't do this with standard deviation but you could do it with variance and then once you figure out the variance you just take the square root for the standard deviation the variance of x is similarly going to be the sum of the variances of these and y's so it's going to be similarly n times the variance and times the variance of y so this all boils down to what is the variance of y going to be equal to so let me scroll over a little bit get a little bit of more real estate and i will figure that out right over right over here all right so we want to figure out the variance of y so variance of y is going to be equal to what well here it's going to be the probability squared distances from the expected value so we have a probability of p where what is going to be our squared distance from the expected value well we have we're going to get a 1 with the probability of p so in that case our distance from the mean or from the expected value we're at one the expected value we already know is equal to p so that's that for that possible outcome the square distance times its probability weight and then we have actually let me scroll over well i'll just do it right over here plus we have a probability of one minus p one minus p for the other possible outcome so in that outcome we are at and the difference between 0 and our expected value well that's just going to be 0 minus p and once again we are going to square that distance and so this is the expression or the square that quantity and so this is the expression for the variance of y and we can simplify it a little bit so this is all going to be equal to so let me just p times 1 minus p squared and then this is just going to be p squared times 1 minus p plus p squared times 1 minus p and let's see we can factor out a p times 1 minus p so what is that going to be left with so if we factor out a p times 1 minus p here we're just going to be left with a 1 minus p and if we factor out a p times 1 minus p here we're just going to have a plus p these two cancel out this is just this whole thing is just a 1. so you're left with p times 1 minus p which is indeed the variance for a binomial variable we actually prove that in other videos i guess it doesn't hurt to see it again but there you have it we know what the variance of y is it is p times 1 minus p and the variance of x is just n times the variance of y so there we go we deserve a little bit of a drum roll the variance of x is equal to n times p times 1 minus p so if we were to take the concrete example of the last video where if i were to take 10 free throws so each trial is a shot is a free throw so if i were to take 10 free throws and my probability of success is 0.3 i have a 30 free throw percentage the variance that i would expect to see so in that case the variance if x is the number of free throws i make after these 10 shots my variance will be 10 times 0.3 0.3 times 1 minus 0.3 so 0.7 and so that would be what this right over here so this would be equal to 10 times 0.3 times 0.7 times 0.21 so my variance in this situation is going to be equal to 2.1 is equal to 2.1 and if i wanted to figure out the standard deviation of this right over here i would just take the square root of this so if you want the standard deviation just take the square root of this expression right over here |
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Singapore Math: Number Sense and Computational Strategies
In this teacher training course, you'll learn Singapore's innovative and practical strategies for solving addition, subtraction, multiplication, and division problems.
Singaporean teachers make math purposeful, interesting, and relevant using a layered curriculum founded on solid number sense and concrete, pictorial, and abstract computational strategies. This course will introduce numerous strategies to create meaningful math lessons of your own.
You will be introduced to what Singapore Math is and how it has become such a powerful and highly regarded math curriculum. Then you discover how number sense and place value instruction are the basis for all Singapore Math. From there, you will learn a variety of computational strategies to make addition, subtraction, multiplication, and division a cinch.
You won't need a passport to discover the curriculum of the world's math leader!
Requirements:
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Instructional Material Requirements:
The instructional materials required for this course are included in enrollment and will be available online.
Learn how to make math purposeful, interesting and relevant. This course will help you discover how to use concrete, pictorial and abstract strategies to make addition, subtraction, multiplication and division a cinch for any student.
Discovering New Computational Strategies
Get ready to learn the revolutionary curriculum from Singapore, one of the world's math leaders! This lesson introduces Singapore Math, the number sense instruction that it revolves around, and how number sense and Singapore computation can help you reach your students.
Get ready to learn the revolutionary curriculum from Singapore, one of the world's math leaders! This lesson introduces Singapore Math, the number sense instruction that it revolves around, and how number sense and Singapore computation can help you reach your students.
Building Number Sense, Part 1
Singapore Math takes number sense to the next level. It integrates number sense into every computation by building a solid foundation on concrete, pictorial, and abstract number sense activities. This lesson explores Singapore's number sense instruction in detail.
Singapore Math takes number sense to the next level. It integrates number sense into every computation by building a solid foundation on concrete, pictorial, and abstract number sense activities. This lesson explores Singapore's number sense instruction in detail.
Building Number Sense, Part 2
Ready to learn how Singapore Math brings place value instruction to life? Place value instruction shows how a number can occupy different places in an equation and represent different quantities in doing so. If your students love games, you're sure to enjoy this concrete way of learning about place value.
Ready to learn how Singapore Math brings place value instruction to life? Place value instruction shows how a number can occupy different places in an equation and represent different quantities in doing so. If your students love games, you're sure to enjoy this concrete way of learning about place value.
Addition Strategies, Part 1
Addition is such a basic part of math, but surprisingly, many students struggle with this operation. In this lesson, you will learn three impressive Singapore Math addition strategies and use place value mats and disks to complete all of your addition problems.
Addition is such a basic part of math, but surprisingly, many students struggle with this operation. In this lesson, you will learn three impressive Singapore Math addition strategies and use place value mats and disks to complete all of your addition problems.
Addition Strategies, Part 2
Imagine that your students have been working with mats and disks for a while, and they're ready for more advanced addition strategies. That's when you will be glad you know about branching, left-to-right addition, and vertical addition, the three strategies you will meet in this lesson.
Imagine that your students have been working with mats and disks for a while, and they're ready for more advanced addition strategies. That's when you will be glad you know about branching, left-to-right addition, and vertical addition, the three strategies you will meet in this lesson.
Subtraction Strategies, Part 1
Once students are comfortable doing addition, it's time to move on to subtraction. It may come as no surprise that this lesson starts with subtraction on your place value mats. As you did with addition, you will start with single-digit problems that don't require regrouping.
Once students are comfortable doing addition, it's time to move on to subtraction. It may come as no surprise that this lesson starts with subtraction on your place value mats. As you did with addition, you will start with single-digit problems that don't require regrouping.
Subtraction Strategies, Part 2
This lesson introduces more advanced subtraction strategies. You will practice subtraction with branching and with the traditional algorithm. You will build on previous knowledge of how branching builds on the work you laid with the mats.
This lesson introduces more advanced subtraction strategies. You will practice subtraction with branching and with the traditional algorithm. You will build on previous knowledge of how branching builds on the work you laid with the mats.
Multiplication Strategies, Part 1
Now that you have mastered two operations using Singaporean computational strategies, it's time to turn your attention to multiplication strategies that double the learning. You guessed it—you will begin by using your place value mats to complete simple multiplication problems.
Now that you have mastered two operations using Singaporean computational strategies, it's time to turn your attention to multiplication strategies that double the learning. You guessed it—you will begin by using your place value mats to complete simple multiplication problems.
Multiplication Strategies, Part 2
While the place value mats are a great start with multiplication, they aren't the only strategy. In this lesson, you will meet model drawing, multiplication through the distributive property, and area model multiplication, three inventive strategies for looking at multiplication.
While the place value mats are a great start with multiplication, they aren't the only strategy. In this lesson, you will meet model drawing, multiplication through the distributive property, and area model multiplication, three inventive strategies for looking at multiplication.
Division Strategies, Part 1
Are you ready to tackle division? Often considered the hardest of the four core mathematical operations, division sometimes gets the short end of the stick. There are a number of ways that you can make this operation click for the struggling learner, so this lesson starts with place value mats and disks.
Are you ready to tackle division? Often considered the hardest of the four core mathematical operations, division sometimes gets the short end of the stick. There are a number of ways that you can make this operation click for the struggling learner, so this lesson starts with place value mats and disks.
Division Strategies, Part 2
This lesson is all about computational strategies, so you will take your division skills to the next level. You will see the distributive property in action, partial quotient division, and short division. Each of these strategies has a Singaporean twist that makes it particularly powerful and innovative.
This lesson is all about computational strategies, so you will take your division skills to the next level. You will see the distributive property in action, partial quotient division, and short division. Each of these strategies has a Singaporean twist that makes it particularly powerful and innovative.
Integrating Computational Strategies
Now that you these new computational strategies at your disposal, you may be wondering how you bring this together in the classroom. This lesson will answer this question. You will create a portable toolkit you can use to share Singapore Math with your students and other teachers at your school.
Now that you these new computational strategies at your disposal, you may be wondering how you bring this together in the classroom. This lesson will answer this question. You will create a portable toolkit you can use to share Singapore Math with your students and other teachers at your school.
What you will learn
How you will benefit
Anni Stipek
Anni Stipek holds bachelor's degree in Elementary Education and a minor in Special Education from University of Puget Sound. She has more than 20 years of experience in the K-12 classroom and worked as a Staff Development for Educators math consultant for five years. She has written and facilitates numerous online courses, including "The Foundations of Singapore Math Model Drawing Grades 1-6" and "The Foundations of Singapore Math Number Sense/Computation Strategies Grades 1-6."
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14879 | https://en.wikipedia.org/wiki/Spectroscopic_notation | Spectroscopic notation - Wikipedia
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1 Ionization states
2 Atomic and molecular orbitalsToggle Atomic and molecular orbitals subsection
2.1 Molecular spectroscopic notation
3 Quarkonium
4 See also
5 References
Spectroscopic notation
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From Wikipedia, the free encyclopedia
Format for notating atoms and molecules
This article is about the notation for atomic and molecular orbitals and is not to be confused with the Russel–Saunders term symbol, a similar notation that describes multi-electron systems, which is often referred to as "spectroscopic notation" as well.
Spectroscopic notation provides a way to specify atomic ionization states, atomic orbitals, and molecular orbitals.
Ionization states
[edit]
Spectroscopists customarily refer to the spectrum arising from a given ionization state of a given element by the element's symbol followed by a Roman numeral. The numeral I is used for spectral lines associated with the neutral element, II for those from the first ionization state, III for those from the second ionization state, and so on. For example, "He I" denotes lines of neutral helium, and "C IV" denotes lines arising from the third ionization state, C 3+, of carbon. This notation is used for example to retrieve data from the NIST Atomic Spectrum Database.
Atomic and molecular orbitals
[edit]
Before atomic orbitals were understood, spectroscopists discovered various distinctive series of spectral lines in atomic spectra, which they identified by letters. These letters were later associated with the azimuthal quantum number, ℓ. The letters, "s", "p", "d", and "f", for the first four values of ℓ were chosen to be the first letters of properties of the spectral series observed in alkali metals. Other letters for subsequent values of ℓ were assigned in alphabetical order, omitting the letter "j" because some languages do not distinguish between the letters "i" and "j":
| letter | name | ℓ |
---
| s | sharp | 0 |
| p | principal | 1 |
| d | diffuse | 2 |
| f | fundamental | 3 |
| g | | 4 |
| h | | 5 |
| i | | 6 |
| k | | 7 |
| l | | 8 |
| m | | 9 |
| n | | 10 |
| o | | 11 |
| q | | 12 |
| r | | 13 |
| t | | 14 |
| u | | 15 |
| v | | 16 |
| ... | | ... |
This notation is used to specify electron configurations and to create the term symbol for the electron states in a multi-electron atom. When writing a term symbol, the above scheme for a single electron's orbital quantum number is applied to the total orbital angular momentum associated to an electron state.
Molecular spectroscopic notation
[edit]
Main article: Molecular term symbol
The spectroscopic notation of molecules uses Greek letters to represent the modulus of the orbital angular momentum along the internuclear axis. The quantum number that represents this angular momentum is Λ.
Λ = 0, 1, 2, 3, ...Symbols: Σ, Π, Δ, Φ
For Σ states, one denotes if there is a reflection in a plane containing the nuclei (symmetric), using the + above. The − is used to indicate that there is not.
For homonuclear diatomic molecules, the index g or u denotes the existence of a center of symmetry (or inversion center) and indicates the symmetry of the vibronic wave function with respect to the point-group inversion operation i. Vibronic states that are symmetric with respect to i are denoted g for gerade (German for "even"), and unsymmetric states are denoted u for ungerade (German for "odd").
Quarkonium
[edit]
For mesons whose constituents are a heavy quark and its own antiquark (quarkonium) the same notation applies as for atomic states. However, uppercase letters are used.
Furthermore, the first number is (as in nuclear physics) n=N+1{\displaystyle n=N+1} where N{\displaystyle N} is the number of nodes in the radial wave function, while in atomic physics n=N+ℓ+1{\displaystyle n=N+\ell +1} is used. Hence, a 1P state in quarkonium corresponds to a 2p state in an atom or positronium.
See also
[edit]
Azimuthal quantum number
Electron configuration
Mnemonics for orbital letters
Principal quantum number
Term symbol
X-ray notation
Molecular symmetry
References
[edit]
^p. 92, Guide to the Sun, Kenneth J. H. Phillips, Cambridge, UK: Cambridge University Press, 1992. ISBN0-521-39788-X.
^§12-7, An Introduction to Quantum Physics, Anthony Philip French and Edwin Floriman Taylor, CRC Press, 1979. ISBN0-7487-4078-3.
^§7.12, Stellar Atmospheres, Jeremy B. Tatum, online book. Accessed on line September 19, 2007.
^ Jump up to: abSpectroscopic notation, web page at accessed on line September 19, 2007.
^ P.Atkins et al. Quanta, Matter, and Change: A Molecular Approach to Physical Chemistry (Oxford University Press) p.106
^W.C. Martin and W.L. Wiese (2002), Atomic, Molecular, and Optical Physics Handbook (version 2.2). [Online] Available: [2021, May 12]. National Institute of Standards and Technology, Gaithersburg, MD.
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14880 | https://www.youtube.com/watch?v=zRLxs35qpJc | Proving Algebraic Inequalities (2 of 3: Using the sign)
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Transcript:
Example so let me give you an example i'm going to word this to you and then i'll write down the symbols can you prove that the average of two square numbers and i'm the sim by the way turning this into symbols is part of the question which is why i'm deliberately not writing that for you the average of two square numbers is always going to be bigger than make sure i get my inequality direction right it's always going to be bigger than the square of the average of the two numbers you started with does that make sense let me say it one more time right you have two numbers two numbers if you squared them both and then averaged whatever you got right you started with two numbers you squared them and then you took the average of whatever the sum was okay that's always going to be bigger than if you took those two numbers you started with find the average first and then square okay can you turn that into symbols can you think about whether one or several of these strategies might help you to actually prove that i'll give you about a minute to see if you can at least state it and then i'll put it up on the board i'm still not going to show you the answer but at least you'll know you're on the right track have a shot turn that into some symbols How not to do this so i'm sensing that feeling in your brain when you're like uh where am i going or i think i have something but is it right let me help you out first by telling you how not to do this question because it's so important as you've seen in this topic counter example can be very powerful so eyes up i promise everything i'm about to write i will leave on the board so you'll have the time to jot it down afterwards but i really want your focus for this part because it's a crucial bit right at the beginning that several of you because i've had a look at your books i know you've missed so still waiting for full attention we're almost there thank you all right let's have a go now i promised i would show you how not to do this before i show you how to do it here's how not to do this this is my algebra i've worked so hard to turn this verbal mess into a symbolic nice thing now i'm going to start doing things with this i'm going to be like oh i know i want to try and do what i can see here right see that getting everything on one side this is what i want to work with and then i start to say well on the right hand side i've got that being greater than 0 because i subtracted it and then off i go and continue right what's the problem with this line of argument why is this an issue in our topic nature of proof before i even go any further i've already made a huge blunder anyone want to tell me How to do this it's solving what do you mean by or what does anyone think that means like i'm i am kind of trying to solve get down to the bottom here why is that a problem i love solving things what's the big deal again you've taken the statement as true which is what you're already trying to believe right in lesson one in this topic right we said statements are the things you're going to be working with right mathematical statements either true or they are false sometimes you don't know which it is sometimes it's conditional but the point of a proof question is i want to establish the fact that this is true that's my job right but everything i've i'm going to continue from here is actually standing on the foundation of this already being true otherwise you can't do anything with it does that make sense so if you have a look and this is why i keep using these same letters over and over again this stands for required to prove i can write this thing down i want it in my head but i don't want to write it down as if i know it to be the case because then you've already started at your logical conclusion and wherever you go at the end you'll be like oh what have i just established answer nothing because that's where you started okay so here's how i'm going to do this instead i can't say this because i'm not using this required to prove line as a foundation there's lots of ways to do this first bit but i think a way that's fairly nice and neat is just to say well this thing having subtracted this from both sides this is the thing that i want you can see if i prove this then it implies this which is what i'm after so all i have to do is say well how do i introduce this thing i'm going to consider this particular mathematical object where did i get it from and the answer is this is the thing that was on the left of the inequality and then this is the thing that was on the right of the inequality that's the object i'm now thinking about and now i can let loose on my all the algebraic manipulation that i want and i can try and prove that this thing is greater than zero which will prove that the original inequality is what i'm after does that make sense now from here actually several of you have some pretty good working you can see for instance this equals like you're going to get this is going to be 2 but that's going to become a 4 when you square it out so i'm going to anticipate i want to collect like terms i'll write this with a denominator of four and then i'll get busy expanding this thing so what do you get on the numerator this is the easy bit yeah this is a squared square sure you can do it whichever order you like that's okay addition's commutative after all okay so i've got this right uh and you can see you've got a whole bunch of like terms because i've already got my common denominators so how many a squareds well i have left on the top y squared and one b squared single a squared a single b squared and then what else do i get on the numerator yeah what's your negative right so minus two a b but hopefully at this point you're like aha this is another object having simplified that i can work on i should forget i shouldn't forget that there's division by four what can you do with this numerator that uses one of these techniques over here what can you do with it you can starts with an f what's this thing called i can factorize this right so this is actually a perfect square namely this perfect square right that's going to be greater than four i can now say hold on a second remembering that right at the beginning of this i said this big umbrella over this is that we're in real number land so this thing here right a minus b if a is real and b is real then what can you tell me about a minus b also real right so i've got a real number being squared the smallest that can possibly be is zero make sense so i can say and this is this is that weird bit going to this whole like we can change what's happening to both sides right i can say that this is greater than or equal to 0 that's a claim i also need a substantiation what's my substantiation a minus b that's a real number because a and b are real numbers does that make sense now it's true there's one thing that i left out it's somewhat usually assumed but a good nice neat question unlike me a human being will always state this explicitly at the beginning when we say a and b we're also not just going to include that a and b are real but we tend to mean when we use different premium rules we also tend to mean that a and b you should write this down because the phrase you're going to see fair bit and b are distinct you might remember this word back from when we were doing uh discriminate quadratics sometimes you got roots that were together but sometimes they were apart which is what we mean by this word so i've so far got that this left hand side minus right hand side is greater than or equal to zero i actually don't want that equal to case if i return back to my original question it's just an inequality without the boundary so how do i get rid of using this fact that i've just reminded you of how do i get rid of that boundary what do you think bran yeah very good and be a distinct i can say that just like i did in this line i can state that algebraically i can say but a is not equal to b yeah since and b are distinct and what that applies is a minus b it can't be zero like i've just subtracted b from both sides are you okay with that flow of logic there this implies this so therefore that's gotten rid of my boundary case so now i can tie this up in a nice neat bow what have i got i promised that everything would stay on the board i think that's just barely still true if i say that i began with my left hand side and right hand side let's write that down a squared plus b squared on two average of the squares and i'm subtracting a plus b on to the square of the averages what have i just established that that's greater than zero i got that it was greater than or equal to zero and then i went a step further to prove that it's greater than zero full stop okay and now i'm right here at this line see how i've gotten there right so now i can get back to the thing that i wanted a squared plus b squared sum or rather average of the squares is greater than the square of the average how do you feel about that is that okay do you see how we had to use these pieces here |
14881 | https://arxiv.org/pdf/2508.13917 | ENUMERATING VECTOR PARKING FUNCTIONS AND THEIR OUTCOMES BASED ON SPECIFIED LUCKY CARS
MELANIE FERRERI, PAMELA E. HARRIS, LUCY MARTINEZ, AND ERIC SWARTZ
Abstract. In a parking function, a car is considered lucky if it is able to park in its preferred spot. Extending work of Harris and Martinez, we enumerate outcomes of parking functions with a fixed set of lucky cars. We then consider a generalization of parking functions known as vector parking functions or
u-parking functions , in which a nonnegative integer capacity is given to each parking spot in the street. With certain restrictions on u, we enumerate outcomes of u-parking functions with a fixed set of lucky cars or with a fixed number of lucky cars. We also count outcomes according to which spots contain lucky cars, and give formulas for enumerating u-parking functions themselves according to their set of lucky cars.
Introduction
Fix n ∈ N = {1, 2, 3, . . . } and let [ n] := {1, 2, . . . , n }. A parking function of length n is a sequence
a = ( a1, a 2, . . . , a n) ∈ [n]n whose weakly increasing rearrangement a↑ = ( a(1) , a (2) , . . . , a (n)) satisfies a(i) ≤ i
for all i ∈ [n]. Parking functions can be described by means of a deterministic parking process as follows. Consider a parking lot with n parking spots on a one-way street (directed from left to right) labeled by [n]. A sequence of n cars enters the street from the left one by one, with car i having a preferred spot ai,which we call its parking preference . For each i ∈ N, car i drives to its preferred spot ai and attempts to park. If the spot is not available, the car continues to drive to the right and parks in the next available spot, if one exists. If there is no available spot, the car exits the lot and is unable to park. We call this the
parking rule , and the set of preference lists α = ( a1, a 2, . . . , a n) is called a parking function if all cars can park under the parking rule. We let PF n denote the set of parking functions of length n, and Konheim and Weiss established that |PF n| = ( n + 1) n−1 . For a parking function α, the order in which the cars park on the street is called the outcome of α and is denoted O(α). If Sn denotes the set of permutations of [ n]written in one-line notation, then the outcome of α is
O(α) = π1π2 · · · πn,
where πi denotes that car πi parked in spot i on the street. Much has been done in studying parking functions, subsets of parking functions, and generalizations. This includes MVP parking functions in which a car bumps an earlier car out of their preference , Naples parking functions in which a car backs up attempting to park in the spot behind their preference whenever they find their preferred spot occupied [9, 13, 16], cases in which a car only tolerates parking in a subset of spots on the street (with some specified order in which they check those spots) [2, 7, 14, 22, 24, 36], and cases in which some spots are unavailable at the start of parking . There has also been work in parking cars of various lengths [12, 25], and others in which cars are only allowed to stay in their parking spot a fixed length of time . Moreover, certain families of parking objects have connections to noncrossing partitions , computing volumes of flow polytopes , counting Boolean intervals in the weak Bruhat order of the symmetric group , ideal states in the Tower of Hanoi game , the faces of the permutohedron , and the Quicksort algorithm . For more on parking functions and other connections, see the survey by Yan , and for open problems in this area see . As with permutations, there is a desire to understand parking objects based on some statistic. Such work includes the discrete statistics of descents, ascents, ties, peaks, valleys, see [10, 17, 34]. One natural statistic of parking functions is the number of cars that park in their preference; such cars are called lucky . Given a parking function α ∈ PF n, we let Lucky (α) = {i ∈ [n] : car i is lucky } and lucky (α) = |Lucky (α)|.
1
arXiv:2508.13917v2 [math.CO] 10 Sep 2025
In the case of parking functions, Gessel and Seo show that
X
α∈PF n
qlucky (α) = q
n−1
Y
i=1
(i + ( n − i + 1) q). (1) A bijective proof of this result was given by Shin, which also demonstrates that parking functions with a given lucky set are in bijection with labeled forests with the same set of leaders , i.e., vertices with the property that they are minimal among their descendants . However, the formula in Equation (1) enumerates parking functions with a fixed number of lucky cars and does not account for which cars are lucky. In this paper, we study lucky sets as “admissible sets” with regard to the lucky statistic for generalized parking functions. For a given permutation statistic, a set S is said to be admissible if there exists a permutation whose instances of that statistic appear exactly at the indices given by S. Admissible sets for permutations were introduced by Billey, Burdzy, and Sagan in studying the peak statistic of permutations . Similarly, admissible sets have been studied for various other permutations statistics. Davis, Nelson, Petersen and Tenner considered admissible pinnacle sets and Gonz´ alez et al. studied the analogous question for signed permutations . Other work includes studying the descent polynomial, which counts permutations according to their descent sets . On the other hand, admissible sets for statistics on parking functions have rarely appeared in the literature. The first few instances include the work on descent sets for parking functions studied by Cruz et al. , and the lucky statistic on Stirling permutations considered as parking functions by Colmenarejo et al. . Such examples demonstrate the potential for further connections between the lucky statistic in parking functions and other combinatorial objects. Motivated by this, Harris and Martinez determined formulas for the number of parking functions with a fixed set of lucky cars . In that work, they observed the close connection between a descent in the outcome permutation and lucky cars. For example, the parking function (1 , 3, 4, 1) has parking outcome 1423. Harris and Martinez show that whenever πi−1 > π i in the outcome permutation, then car πi is lucky, because as car πi parks, the spot to its left was occupied by a car later in the queue, which implies that spot was available upon car πi
attempting to park. Thus car πi parked in their preference. This observation implies that if the set of lucky cars is fixed, then certain parking outcomes are not possible. Moreover, car 1 is always a lucky car as it is the first to park on the street. Our work begins by answering an open problem of Harris and Martinez stated in by giving a formula for the number of parking outcomes given a fixed set of lucky cars.
Theorem 2.2. Let I = {1, i 1, i 2, . . . , i k−1} be a lucky set of PF n with 1 < i 1 < · · · < i k−1 ≤ n. If
On(I) = {O (α) : α ∈ PF n and Lucky (α) = I}, then
|O n(I)| = k! Y
j∈[n]\I
|I ∩ [j]| = 2 i2−i1 3i3−i2 · · · (k − 1) ik−1−ik−2 kn−(ik−1−1) .
Our main contributions consider the lucky sets of cars for a generalization of parking functions called vector parking functions, where each parking spot has a nonnegative integer capacity [32, 38]. To begin, we define vector parking functions and then give a list of our main results. Throughout we label the parking spots by N, and let u = ( u1, . . . , u n), where 1 ≤ u1 ≤ · · · ≤ un. For each 1 ≤ i ≤ n, the capacity of spot
i is the multiplicity of i in the vector u, which we denote by mi(u). In particular, P
i≥1
mi(u) = n. With this notation at hand, a u-parking function can be described via the following parking process . A queue of n cars with parking preferences given by a = ( a1, . . . , a n) ∈ Nn enters the lot, and each car drives to its preferred spot and attempts to park. A car may park at a spot j if there are fewer than mj (u) cars already parked there. Otherwise, it attempts to park in the next spot j + 1, and so on. If there are no available spots past spot j, the car fails to park. We say that a is a u-(vector) parking function if all cars are able to park under this parking rule. We generally denote vector parking functions with bold symbols. For example, let
n = 4, u = (2 , 2, 3, 3) and a = (1 , 3, 3, 1). Then we have m2(u) = 2 , m 3(u) = 2, and m1(u) = 0. Thus, there are 2 non-empty spots with total capacity 4: spots 2, and 3 with capacities 2, and 2, respectively. The cars with parking preferences given by a then park as follows: Car 1 parks in spot 2, car 2 parks in spot 3, car 3 parks in spot 3, and car 4 parks in spot 2. In Figure 1, we show the street after all of the cars have parked. Note that the spot marked with an X has capacity zero, i.e. this spot is unavailable. When u = ( u1, u 2, . . . , u n) is a weakly increasing sequence of positive integers, we let PF( u) denote the set of u-parking functions of length n. A characterization of u-parking functions is as follows. A preference list sequence a = ( a1, a 2, . . . , a n) ∈ Nn, is in PF( u) if the rearrangement into weakly increasing order
2
1 4 2 3
1 2 2 3 3
Figure 1. Vector parking process with u = (2 , 2, 3, 3) and a = (1 , 3, 3, 1).
a↑ = ( a(1) , a (2) , . . . , a (n)) satisfies a(i) ≤ ui for each i ∈ [n]. In this way, when u = (1 , 2, . . . , n ), the set of u
parking functions is precisely the set of classical parking functions. Kung and Yan show that when u is an arithmetic progression, that is ui = a + b(i − 1) for some a, b ∈ N, the number of u-parking functions is
a(a + bn )n−1. For general u, the number of u-parking functions is given by a determinantal formula [32, 38], which is not easy to evaluate. In the classical parking function setting, the parking outcome can be described by a permutation of [ n] in one-line notation π = π1π2 · · · πn, where πi = j means that car j parked in spot i. We introduce the outcome of a u-parking function, which we formally state in Theorem 3.4. Given the capacities of the parking spaces in a u-parking function, whenever multiple cars park in the same numbered spot, we adopt the convention of listing those cars in increasing order. Moreover, whenever there are unavailable parking spots in a u-parking function, we denote those with an X. Namely, if u = ( u1, u 2, . . . , u n), with M = max( u), then the outcome
of a = ( a1, a 2, . . . , a n) is denoted by
Ou(a) = B1B2 . . . B M ,
where Bi contains the set of cars (in increasing order) that parked in spot i, and if a spot i is unavailable (i.e., because mi(u) = 0), then Bi = X, which means no cars can park there. In this way, when ignoring the unavailable spots, i.e. any Bi = X, the outcome of a u-parking function is an ordered set partition of [ n]where the size of the blocks are encoded by positive capacities for the spots described by u. For example, the outcome of the (2 , 2, 3, 3)-parking function (1 , 3, 3, 1), illustrated in Figure 1 is O((1 , 3, 3, 1)) = X{1, 4}{ 2, 3}.In a vector parking function, a car is said to be lucky if it parks in its preferred parking spot. For example, in Figure 1, the only lucky cars are 2 and 3 as they park in their preferred spot 3, while cars 1 and 4 park in spot 2 making them unlucky. We study vector parking functions with a fixed set of lucky cars. We let
Lucky u(a) denote the set of lucky cars of a ∈ PF( u). Unlike the classical parking functions in which the first car is always lucky, this is not always the case for the vector parking functions, as some parking spots might be unavailable at the start of the parking process. For example, if u = (2 , 2, 3, 3) and a = (1 , 3, 3, 1) then Lucky u(a) = {2, 3}. For any subset I ⊆ [n], we define
LuckyPF u(I) := {a ∈ PF( u) : Lucky u(a) = I},
which is the set of u-parking functions of length n with lucky cars in the set I and unlucky cars in the set [n] \ I.Similar to classical parking functions, not every ordered set partition can be the outcome of a u-parking function with a fixed set of lucky cars. For example, if u = (2 , 2, 3, 3), then the outcome O((1 , 3, 3, 1)) =
X{1, 4}{ 2, 3} given in Figure 1 cannot be the outcome of a u-parking function with lucky set I = {3}. This is because car 2 and car 3 parked in spot 3, however when car 2 enters the street, spot 2 is not fully occupied. This means that parking in spot 3 would make car 2 lucky. We let Ou(I) denote the set of outcomes of all u-parking functions with lucky set I. In Section 3.1 we characterize the possible outcomes for u-parking functions given a lucky set I, and enumerate outcomes of vector parking functions with no repetitions and where I has a fixed size.
Theorem 3.7. Fix a lucky set of cars I, let M = max ( u), and let B = B1B2 . . . B M ∈ O u := {O u(a) : a ∈
PF( u)}. Then B ∈ O u(I) if and only if the following hold: (1) if B1̸ = X, then B1 ⊆ I,(2) if there exists a b1 ∈ Bi−1̸ = X and b2 ∈ Bi̸ = X such that b1 > b 2 then b2 ∈ I.In Sections 3.2.1-3.2.3, we enumerate outcomes for a general I in certain special cases for u. In Section 4 we enumerate the outcomes from Theorem 3.7 with a fixed set of lucky spots , defined in that section. Lastly in Section 5, we give a formula for the number of u-parking functions with a fixed set of lucky cars or with
k lucky cars for a general vector u, and discuss related results. We conclude with Section 6 by providing some directions for future work.
3
Counting outcomes of classical parking functions with a fixed set of lucky cars
In this section, we provide a formula for the number of outcomes of parking functions with a fixed lucky set. Harris and Martinez characterized the permutations that are outcomes of parking functions with a fixed lucky set I. We recall this result next.
Theorem 2.1 (, Theorem 2.3) . Fix a lucky set I ⊆ [n] of PF n. Then π = π1π2 · · · πn ∈ Sn is the outcome α ∈ LuckyPF n(I) if and only if (1) π1 ∈ I,(2) if πi ∈ I and πi+1 /∈ I, then πi < π i+1 , and (3) if πi−1 > π i, then πi ∈ I.We now provide a formula that counts the permutations described in Theorem 2.1.
Theorem 2.2. Let I = {1, i 1, i 2, . . . , i k−1} be a lucky set of PF n with 1 < i 1 < · · · < i k−1 ≤ n. If
On(I) = {O (α) : α ∈ PF n and Lucky (α) = I}, then
|O n(I)| = k! Y
j∈[n]\I
|I ∩ [j]| = 2 i2−i1 3i3−i2 · · · (k − 1) ik−1−ik−2 kn−(ik−1−1) .
Proof. We first show that any permutation satisfying condition (3) in Theorem 2.1 also satisfies condition (2). Let π ∈ Sn. Suppose for all i ∈ [n] such that πi−1 > π i, πi ∈ I. Then by reindexing, we have that for all i ∈ [n], if πi > π i+1 , then πi+1 ∈ I. By the contrapositive, we have that if πi+1 ̸ ∈ I, then πi < π i+1 ,which implies that (2) holds for π as well. We proceed to count permutations in Sn satisfying conditions (1) and (3). Given a lucky set I, let k = |I|,and let J = [ n] \ I. Let π be a permutation satisfying the desired conditions. Suppose
π−1(I) = {a1 = 1 , a 2, . . . , a k},
where a1 < a 2 < · · · < a k. Then π1 ∈ I, and π can be partitioned into |I| = k increasing subsequences, where each subsequence begins with some π∗
ai
∈ I and the remaining entries in the subsequence are not in
I. More technically, π (written in one-line notation) is of the form
π = π∗
1
π2 · · · πa2−1 π∗
a2
πa2+1 · · · πa3−1 π∗
a3
πa3+1 · · · πa4−1 · · · π∗
ak
πak +1 · · · πn,
where the starred entries are the elements of I, and the non-starred entries are the elements of [ n] \ I.By condition (3), each underlined subsequence must be increasing. We now count the number of possible permutations π of this form. Let j ∈ J. In the permutation π, j must appear in one of the k subsequences after some π∗
ai
. Since each subsequence is increasing, j must appear in one of the subsequences beginning with a π∗
ai
that is less than j.Define the set Ij = {x ∈ I : x < j } = I ∩ [j]. It follows that |Ij | is the number of possible sequences in which
j can be placed. Once a subsequence is chosen, the position of j within that subsequence is fixed, since each subsequence is increasing. Then the number of possible placements of all j ∈ J into the k subsequences is given by Q
j∈J
|Ij |. Once all of the k underlined subsequences are constructed, they can be rearranged in k!ways. So the total number of permutations of the desired form is k! Q
j∈[n]\I
|Ij |.We check that this product is equivalent to 2 i2−i1 3i3−i2 · · · (k − 1) ik−1−ik−2 kn−(ik−1−1) . Let j ∈ J. Then,
|Ij | = 1 if and only if 1 < j < i − 1, and the number of j ∈ J such that |Ij | = 1 is given by the number of integers between 1 and i1; that is, i1 −2. Now let ℓ ∈ { 2, . . . , k −1}. Then |Ij | = ℓ if and only if iℓ−1 < j < i ℓ.It follows that the number of j ∈ J such that |Ij | = ℓ is given by the number of integers between iℓ−1 and
iℓ, which is iℓ − iℓ−1 − 1. Also, the number of j ∈ J such that |Ij | = k is given by the number of integers in [n] greater than ik−1, which is n − ik−1. Then
Y
j∈[n]\I
|Ij | = (1) |{ j∈[n]\I:|Ij |=1 }| (2) |{ j∈[n]\I:|Ij |=2 }| · · · (k − 1) |{ j∈[n]\I:|Ij |=k−1}| (k)|{ j∈[n]\I:|Ij |=k}|
= (1) i1−2(2) i2−i1−1 · · · (k − 1) ik−1−ik−2−1kn−ik−1 .
4
By substitution, we obtain
k! Y
j∈[n]\I
|Ij | = k!(1) i1−2(2) i2−i1−1 · · · (k − 1) ik−1−ik−2−1kn−ik−1
= ( k)( k − 1) · · · (2)(1)(1) i1−2(2) i2−i1−1 · · · (k − 1) ik−1−ik−2−1kn−ik−1
= (1) i1−1(2) i2−i1 · · · (k − 1) ik−1−ik−2 kn−ik−1+1
= 2 i2−i1 3i3−i2 · · · (k − 1) ik−1−ik−2 kn−(ik−1−1) . □
Example 2.3. We provide an example using the formula in Theorem 2.2 by using the same example in [30, Example 2], which counts the number of outcome permutations for I = {1, 4} when n = 5 to be four. If
I = {1, 4}, then J = {2, 3, 5}. Then I2 = {1}, I3 = {1}, I5 = {1, 4}. So the number of outcomes we obtain is
|I|! · | I2| · | I3| · | I5| = 2!(1)(1)(2) = 4 .
We recall the following result with a formula that counts the number of outcomes when the first k cars are lucky.
Theorem 2.4. [30, Theorem 2.4] Let I = {1, 2, 3, . . . , k } ⊆ [n] be a lucky set of PF n. Then
|O n(I)| = X
J={j1=1 ,j 2,j 3,...,j k}⊆ [n]
k!
n − kj2 − j1 − 1, j 3 − j2 − 1, . . . , j k − jk−1 − 1, n − jk
.
Using our formula in Theorem 2.2, we can obtain the same result, and we remark that this can also be recovered by applying the multinomial theorem to the sum.
Corollary 2.4.1. If I = {1, 2, . . . , k } is a lucky set of PF n with k ≤ n, then
|O n(I)| = k! Y
j∈{ k+1 ,...,n }
|I| = k!( k)n−k.
We provide an example that uses the formula in Corollary 2.4.1 and the formula in next.
Example 2.5. If I = {1, 2, 3} and n = 5, [30, Theorem 2.4] yields
|O n({1, 2, 3})| = X
J={j1=1 ,j 2,j 3,...,j k}⊆ [n]
k!
n − kj2 − j1 − 1, j 3 − j2 − 1, . . . , j k − jk−1 − 1, n − jk
,
where in this case k = 3 and the possible subsets J are {1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, 5}.The sum can be written as
|O n({1, 2, 3})| = 3!
20, 0, 2
3!
20, 1, 1
3!
20, 2, 0
3!
21, 0, 1
3!
21, 1, 0
3!
22, 0, 0
= 3!(1 + 2 + 1 + 2 + 2 + 1) = 3!(9) = 3!(3) 5−3.
In , the authors also considered the case for when there are more spots than cars and gave a char-acterization of the outcomes of parking functions with a fixed lucky set I. We let PF m,n denote the set of ( m, n )-parking functions with m cars and n spots, and we assume that 1 ≤ m ≤ n. We begin with a definition of the outcomes in a parking function with more spots than cars.
Definition 2.6 (, Definition 3.1) . Let Sm,n denote the set of permutations of the multiset {X, . . . , X}∪ [m]with X having multiplicity n−m. Given a parking function α = ( a1, a 2, . . . , a m) ∈ PF m,n , define the outcome
of α by
O(α) = π1π2 · · · πn ∈ Sm,n ,
where πi = j ∈ [m] denotes that car j parked in spot i, and πi = X indicates that spot i remained vacant. For sake of simplicity in our arguments, we assume that X > i for any i ∈ N. Moreover, to every element
π = π1π2 · · · πn ∈ Sm,n we will prepend π0 = X.We recall the characterization of the outcomes of parking functions with more spots than cars with a fixed lucky set I.
5
Theorem 2.7 (, Theorem 3.3) . Fix a lucky set I ⊆ [m] of PF m,n . Then π = π1π2 · · · πn ∈ Sm,n is the outcome of α ∈ LuckyPF n(I) if and only if Xπ = Xπ1π2 . . . π n satisfies the following: (1) if πi−1 = X and πi ∈ [m], then πi ∈ I,(2) if πi ∈ I and πi+1 /∈ I ∪ { X}, then πi < π i+1 ,(3) if πi−1, π i ∈ [m] and πi−1 > π i, then πi ∈ I.We now provide a formula that counts such permutations described in Theorem 2.7.
Theorem 2.8. If I = {1, i 1, i 2, . . . , i k−2, i k−1} ⊆ [m] is a lucky set of PF m,n with 1 < i 1 < i 2 < · · · < i k−1 ≤
n, then
|O m,n (I)| = (k + n − m)! (n − m)!
Y
j∈[m]\I
|I ∩ [j]|
= 2 i2−i1−13i3−i2−14i4−i3−1 · · · (k − 1) ik−1−ik−2−1km−ik−1 (n − m + k)! (n − m)! .
Proof. We observe, similarly as in the proof of Theorem 2.2, that for two consecutive nonempty parking spots πi−1, π i ∈ [m], the last two conditions are equivalent to condition (3). Conditions (2) and (3) in Theorem 2.7 both only concern the case where there are two consecutive entries in π from the set [ m]. (This is directly stated in (3), and in (2), πi ∈ I ⊆ [m] and πi+1 /∈ { X} implies
πi+1 ∈ [m].) So if either πi−1 /∈ [m] or πi /∈ [m], there are no restrictions imposed by conditions (2) and (3). Then we seek to count permutations π ∈ Sm,n satisfying (1) if πi−1 = X and πi ∈ [m], then πi ∈ I, and (2) for πi−1, π i ∈ [m]: if πi̸ ∈ I then πi−1 < π i.Then, similarly as in the proof of Theorem 2.1, a permutation π ∈ Sm,n satisfying these conditions can be partitioned into subsequences. In this case, we have length 1 subsequences containing only the entry
X, and k increasing subsequences (also possibly of length 1), each beginning with some π∗
ai
∈ I, and with the remaining entries being from [ m] \ I. Because each element of [ m] that appears directly after an X is from I, we know that all of the subsequences of elements from [ m] will begin with an element from I. The assumption that π0 = X also guarantees that this holds when π1 ∈ [m]. More technically,
π = X · · · X π∗
a1
πa1+1 · · · πa2−1 X · · · X π∗
a2
πa2+1 · · · πa3−1 X · · · X π∗
a3
πa3+1 · · · πa4−1 · · · π∗
ak
πak +1 · · · πm X · · · X,
where the starred entries are the elements of I, and the non-starred entries are the elements of [ m] \ I. By condition (2), each underlined subsequence must be increasing. The X’s need not appear between every pair of increasing subsequences. We count the number of multiset permutations of this form. We first check the possible ways to sort the elements of [ m] \ I into the subsequences. Similarly as in the proof of Theorem 2.1, each j ∈ [m] \ I can be placed into one of the subsequences beginning with a π∗
ai
that is less than j. Let Ij = {x ∈ I : x < j } = I ∩ [j]. Then |Ij | is the number of possible subsequences in which an element j can be placed. Once a subsequence is chosen, the position of j
within that subsequence is fixed, since the subsequence is increasing. Then, over all j ∈ [m] \ I, the number of possible ways to place them into the k different subsequences of elements from [ m] is Q
j∈[m]\I
|Ij |.Once all of the k subsequences of elements from [ m] are constructed, they may appear in any order. We also must place n − m X’s throughout the permutation, and these can be placed anywhere between the
k subsequences. By construction, this will not contradict condition (1), since each subsequence we have constructed begins with an element of I. Then, to count all multiset permutations of the desired form, we count the number of ways to arrange k distinguishable subsequences and n−m indistinguishable subsequences (i.e., the subsequences of the form “ X”), which is given by (k+n−m)! (n−m)! . It follows that |O m,n (I)| is counted by (k + n − m)! (n − m)!
Y
j∈[m]\I
|Ij |.
We check that this product is equivalent to 2i2−i1−13i3−i2−14i4−i3−1 · · · (k − 1) ik−1−ik−2−1km−ik−1 (k + n − m)! (n − m)! .
6
Let j ∈ [m] \ I. By the same reasoning as in the proof of Theorem 2.1, we have
Y
j∈[m]\I
|Ij | = (1) i1−2(2) i2−i1−1(3) i3−i2−1 · · · (k − 1) ik−1−ik−2−1km−ik−1 .
By substitution, it follows that (k + n − m)! (n − m)!
Y
j∈[m]\I
|Ij | = (k + n − m)! (n − m)! (1) i1−2(2) i2−i1−1(3) i3−i2−1 · · · (k − 1) ik−1−ik−2−1km−ik−1
= (k + n − m)! (n − m)! (2) i2−i1−1(3) i3−i2−1 · · · (k − 1) ik−1−ik−2−1km−ik−1 . □
Using the formula in Theorem 2.8, we obtain a formula for the number of outcomes of parking functions when the first k cars are lucky.
Corollary 2.8.1. If I = {1, 2, 3, . . . , k } ⊆ [m] is a lucky set of PF m,n , then
|O m,n (I)| = km−k (n − m + k)! (n − m)! .
We note that Corollary 2.8.1 gives an alternative expression for the count given in [30, Theorem 3.4]. Next we determine how many outcomes there are such that exactly k cars are lucky. Note that this cannot be obtained by simply summing the formula above over all possible lucky sets of size k, since this would cause us to overcount. See Theorem 2.9.
Example 2.9. If m = n = 5, the parking function (1 , 2, 1, 1, 1) has outcome π = 12345 and lucky set {1, 2}.Also, the parking function (1 , 1, 3, 1, 1) has outcome π′ = 12345 and lucky set {1, 3}. So this single outcome occurs in correspondence with two different lucky sets of the same size.
Lemma 2.10. If S is an outcome of an ( m, n )-parking function in which k cars are lucky, then S is also an outcome for a parking function in which ℓ cars are lucky, for any k < ℓ ≤ m. Consequently the number of outcomes in which at most k cars are lucky is the same as the number of outcomes in which exactly k cars are lucky.
Proof. Suppose that k cars in S are lucky. We may specify ℓ−k unlucky cars in S and alter the corresponding parking function so that those cars’ preferences are now the spots where they parked. This yields a parking function with outcome S in which now ℓ cars are lucky. So any outcome for which k cars are lucky is also an outcome for which ℓ > k cars are lucky. That is, if Om,n (k) denotes the set of outcomes of ( m, n )-parking functions with exactly k lucky cars, then Om,n (1) ⊆ O m,n (2) ⊆ · · · ⊆ O m,n (m). The number of parking functions in which at most k cars are lucky is given by k[
i=1
Om,n (i) = Om,n (k),
since all sets in the union are equal to or contained in Om,n (k). □
Let nk denote the Eulerian number A(n, k ), i.e., the number of permutations of [ n] with exactly k
descents. We now connect this sequence to the number of outcomes of ( m, n )-parking functions.
Theorem 2.11. Let 1 ≤ m ≤ n. The number of outcomes for ( m, n )-parking functions in which exactly k
cars are lucky is given by
[
|I|=k
Om,n (I) =
k−1
X
j=0
"k−j−1X
d=0
m − j − 1
d
n − m + j + 1
j + 1 + d
# mj
.
Proof. As noted earlier, only lucky cars can form descents with the car to their left. Suppose there are a total of k lucky cars, and in the outcome, we have j lucky cars that form descents; that is, j lucky cars which park directly to the right of a car of higher index. Since the car that parks in the leftmost occupied spot is always lucky, it follows that j ≤ k − 1. Also, for n ≥ m, we have n − m unoccupied spots in the outcome. We count the possible outcomes with
j descents and n − m unoccupied spots, for all possible j.
7
Fix j ∈ { 0, 1, . . . , k − 1}. Omitting the unoccupied spots, we may view the outcome as a permutation of [m] with j descents. There are mj such options. Let d be the number of lucky cars that park directly to the right of an unoccupied spot, not including the leftmost car (which may or may not appear to the right of an unoccupied spot). Then d + j ≤ k − 1, since the leftmost car is always lucky and is not included in those sets counted by j and d. Among the m total cars, we have already selected j of them to form descents, and the leftmost one is excluded from the count, so we select d among the remaining m − j − 1 cars to park directly to the right of unoccupied spots. There are m−j−1
d
ways to choose them. Once these d cars are chosen, we know the positions of d unoccupied spots in the outcome. It remains to place the remaining n − m − d unoccupied spots. These cannot be placed to the left of unlucky cars, since this would cause those cars to be lucky. If they were to be placed directly to the left of a lucky car that did not form a descent, other than the leftmost lucky car, they would have already been placed when the d
such cars were selected. It remains that the n − m − d unoccupied spots may be placed either to the left of one of the d existing unoccupied spots, to the left of one of the j descents, at the far left before all occupied spots, or at the far right, after all cars. We may view this as having n − m − d X’s and j + d + 1 contiguous subsequences whose restriction to integers form increasing runs, beginning with either descents, the leftmost car, or the d unoccupied spots placed earlier, and we must choose how to place the X’s among them. These runs appear in a fixed relative order. The number of such arrangements is counted by
n − m − d + j + d + 1
j + d + 1
=
n − m + j + 1
j + d + 1
.
To find the total number of possible outcomes, we apply these counts summing over all possible j and d.This yields k−1X
j=0
mj
"k−j−1X
d=0
m − jd
n − m + j + 1
j + d + 1
#
,
as claimed. □
Simplifying the formula in Theorem 2.11 to the case where m = n, we have
k−1
X
j=0
"k−j−1X
d=0
n − j − 1
d
j + 1
j + 1 + d
# nj
. (2) Now notice j+1
j+1+ d
= 0 for all d ≥ 1, so we can further simplify Equation (2) to
k−1
X
j=0
n − j − 10
j + 1
j + 1
nj
=
k−1
X
j=0
nj
,
which counts the permutations with at most k − 1 descents. We recall that every descent in an outcome forces a lucky car. Moreover, the car parking in the first spot is always lucky, but this is not a descent of the permutation. Hence we see that this gives the count of possible outcomes having exactly k lucky cars. Thus, we have established the following.
Corollary 2.11.1. For any n ≥ 1, the number of outcomes of elements in PF n having exactly k lucky cars is given by Pk−1
j=0
nj
, which counts the permutations with at most k − 1 descents. 3. Counting outcomes for vector parking functions
In this section, we begin by describing all of the possible lucky sets for u-parking functions. Then, we consider counting outcomes for families of u-parking functions for various u. We first recall, as described in [4, pg. 4], that u is a vector that encodes the capacities of the parking spots on the street. If u =(u1, u 2, . . . , u n) with 1 ≤ u1 ≤ u2 ≤ · · · ≤ un and ui ∈ N = {1, 2, . . . } then the capacity of spot i is the multiplicity of i in the vector u.
Definition 3.1. Let a = ( a1, a 2, . . . , a n) be a u-parking function. We say that a has lucky set I if all cars indexed by I can park in their preferred spot, and all other cars park in a spot other than their preference. We denote this by Lucky u(a) = I.
8
The first observation is that in contrast to classical parking functions, u-parking functions need not have 1 appear in their lucky sets. Consider the following example.
Example 3.2. If u = (1 , 1, 3, 3, 3) and a = (2 , 1, 2, 1, 1), then Lucky u(a) = {2, 4} because cars 2 and 4 park in their preferred spot 1 and cars 1 , 3 and 5 park in spot 3.
Lemma 3.3. If m1(u) = m, i.e. the multiplicity of 1 in u is m, then every car with preference 1 is lucky.
Proof. As in the classical case, the car that parks in spot 1 is lucky [30, Theorem 2.3]. Hence, if spot 1 has capacity m, all m cars that park in spot 1 are lucky. □
Next, we formally define the outcome of a u-parking function.
Definition 3.4. Let u = ( u1, u 2, . . . , u n), with M = max( u). Given a u-parking function a = ( a1, a 2, . . . , a n), the outcome of a is denoted by
Ou(a) = B1B2 . . . B M ,
where Bi contains the set of cars (in increasing order) that parked in spot i, and if a spot i is unavailable (i.e., because mi(u) = 0), then Bi = X, which means no cars can park there. We denote by Ou the set of all possible outcomes of u-parking functions.
Example 3.5. If n = 5, u = (1 , 1, 3, 3, 3), and a = (2 , 1, 2, 1, 2), then the outcome is the ordered set partition of {X} ∪ given by B = B1B2B3 = {2, 4}X{1, 3, 5}, because cars 2 and 4 parked in spot 1, cars 1 , 3 and 5 parked in spot 3, and spot 2 is unavailable since it does not appear in the vector u. Furthermore, cars 2 and 4 both park in their preferred spot; namely, spot 1. All other cars park in a spot other than their preferred spot. Hence, the u-parking function a has lucky set I = {2, 4}.Next, we observe that not every ordered set partition can be the outcome of a u-parking function with a fixed set of lucky cars.
Example 3.6. Let a be a u-parking function with u = (2 , 3, 5, 5) and suppose that Lucky u(a) = {3}.Notice that B = X{4}{ 2}X{1, 3} is not a possible outcome since when car 2 entered the street, car 4 had not entered the street and so spot 2 would be empty. This means that in order for car 2 to park in spot 3, it would have to be its preference. Thus, the only way for car 2 to park in spot 3 is if it preferred that spot, in which case car 2 would be lucky, implying 2 ∈ Lucky u(a), which is a contradiction. In the case of the classical parking functions, descents forced certain cars to be lucky. This still holds for
u-parking functions. However, if a car follows an unavailable spot, then that car may or may not be lucky. We now give a characterization of the types of outcomes that arise in a u-parking function with a fixed set of lucky cars. Recall that LuckyPF u(I) denotes the set of u-parking functions that have lucky set of cars I
and that Ou(I) denotes the set of outcomes of all u-parking functions with lucky set I.
Theorem 3.7. Fix a lucky set of cars I, let M = max ( u), and let B = B1B2 . . . B M ∈ O u := {O u(a) : a ∈
PF( u)}. Then B ∈ O u(I) if and only if the following hold: (1) if B1̸ = X, then B1 ⊆ I,(2) if there exists a b1 ∈ Bi−1̸ = X and b2 ∈ Bi̸ = X such that b1 > b 2, then b2 ∈ I.
Proof. Let B = B1B2 . . . B M , where M = max ( u). (⇒) For the forward direction, we verify that when B = B1B2 . . . B M is the outcome of a u-parking function with lucky set of cars I, it satisfies each of the conditions. (1) This follows by Theorem 3.3. (2) Suppose that there exists a b1 ∈ Bi−1̸ = X and b2 ∈ Bi̸ = X such that b1 > b 2. This implies that car b2 parks on the street before car b1 enters the street to park. Thus, when car b2 entered the street to park, spot i − 1 was not completely occupied. Assume for contradiction that car b2 has a preference 1 ≤ ai ≤ i − 1. Then car b2 would park in the first available spot past ai. Among the spots numbered ai, a i + 1 , . . . , i − 1, at least spot i − 1 still has not reached capacity. So car b2 would park in spot x satisfying ai ≤ x ≤ i − 1 < i , contradicting the assumption that car b2 parks in spot i.Thus, car b2 parks in spot i and must have done so by preferring spot i. This makes car b2 a lucky car, and so whenever b1 ∈ Bi−1̸ = X and b2 ∈ Bi̸ = X we have that b2 ∈ I, as claimed.
9
(⇐) Let B = B1B2 . . . B M satisfy conditions (1) and (2) above. We construct a u-parking function whose lucky set is I and outcome is B. Let a = ( a1, . . . , a n) be a vector of preferences, in which car i prefers spot
ai. For each car h, we set ah to be the spot where car h appears in B. It follows that the outcome of a
is B, and all cars are lucky in a. We construct a parking function a′ in which only the cars indexed by I
are lucky. For each car i in I, let a′
i
= ai. Otherwise for j / ∈ I, let a′
j
= aj − 1. Since j / ∈ I, condition (1) implies aj̸ = 1, so a′
j
≥ 1. Hence a′ is a vector consisting of positive integers with a′
h
≤ ah for all h. We know that a is a valid u-parking function since all cars are able to park. Then by the previous inequality,
a′ also satisfies the definition of a u-parking function. Suppose for contradiction that the outcome of a′ is different from the outcome of a. Let j be the car of lowest index whose parking spot differs in the outcomes of a and a′. Only unlucky cars’ preferences are changed. If j were a lucky car which ended up parking in a different spot, then the parking spot of an earlier unlucky car must have been changed. By minimality of j, we have that j is then unlucky. Then if car j
parks in spot d in the outcome of a, its preference is changed to d − 1 in a′, and, since the spot that j parks in changes in the outcome of a′, we have that j is now able to park in spot d − 1 in the outcome of a′.This is only possible if spot d − 1 is available and not entirely occupied when car j attempts to park, which implies that in the outcome of a, there must have been some car f > d which parks in spot Bd−1. However, by property (2), this implies j ∈ I, meaning j was not unlucky, which is a contradiction. Then a′ is a valid
u-parking function with outcome B and lucky set I. So if B satisfies (1) and (2), then B ∈ O u(I). □
3.1. Counting outcomes when k cars are lucky and the vector has no repetition. In this section, we consider removing m spots from a street of length n + m and let n be the number of cars. Moreover, we let u = ( u1, u 2, . . . , u n) where none of the entries of u repeat. Later, we consider the case in which there are repeated entries in the vector u. It turns out that u-parking functions are a generalization of parking completions , in which all spots have capacity 1, and various spots are presumed to be unavailable. Suppose there are m of the n + m spots already taken. Then, a parking completion a = ( a1, a 2, . . . , a n) for a sequence
s = ( s1, s 2, . . . , s m) is a sequence in which all n cars can successfully park where the entries of s denote which spots are unavailable in increasing order . In this context, u is a vector that contains the n spots that are available on the street. More precisely,
u = (1 , 2, . . . , s 1 − 1, s 1 + 1 , s 1 + 2 , . . . , s 2 − 1, s 2 + 1 , . . . , s m−1 − 1, s m−1 + 1 , . . . , s m − 1, s m + 1 , . . . , n + m),
where u is of length n.
Example 3.8. Let n = 3 and u = (2 , 4, 6). Then, s = (1 , 3, 5) and some possible u-parking functions or parking completions are (1 , 1, 1) , (1 , 1, 6) , (1 , 3, 6) , (1 , 5, 2) , (1 , 5, 3) , (1 , 5, 4). Throughout this section, we fix the size of a lucky set I and let |I| = k. We stress that in fixing the size of the lucky set, we are not concerned with the elements in the lucky set. We give a formula in terms of the Eulerian numbers that counts the number of distinct outcomes when |I| = k. We only count distinct outcomes, as two different lucky sets with the same size could yield the same outcome. See Theorem 3.9.
Example 3.9. Consider u = (1 , 2, 3, 5) , k = 2 and consider two lucky sets I = {1, 3} and I′ = {1, 4}.A possible outcome for both lucky sets is Ou(I) = Ou(I′) = {1}{ 2}{ 4}X{3} with a = (1 , 1, 5, 1) and
a′ = (1 , 1, 4, 3), respectively. Following notation from , let δ = ( δ1, . . . , δ k) be a weak composition of n. We denote the multinomial coefficient nδ
:= n!
δ1! δ2! · · · δk! .
The following result can be viewed as a formula for enumerating outcomes of parking completions with k
lucky cars.
Theorem 3.10. Let u = ( u1, u 2, . . . , u n) such that all ui are distinct and in increasing order. Suppose spots
s1, s 2, . . . , s m are unavailable. Let δ1 = s1 − 1, δ2 = s2 − s1 − 1, . . . , δm = sm − sm−1 − 1, δm+1 = n + m − sm.
10
The number of distinct outcomes with k lucky cars is
[
|I|=k
Ou(I) =
nδ
X
(i1, . . . , i m+1 ) nonnegative,
P
p
ip ≤ max {k − δ1, k − 1}
m+1
Y
j=1
δj
ij
,
where δj
ij
is the Eulerian number with initial conditions δj
0
= 1 for any δj ≥ 0.
Proof. Suppose that spots s1, s 2, . . . , s m are forbidden, and there are n cars to park in the n available spots. We construct a (weak) composition δ of n as follows: Let δ1 = s1 −1, δ2 = s2 −s1 −1, . . . , δm = sm −sm−1 −1,
δm+1 = n + m − sm. The nonzero parts of δ are the lengths of contiguous runs of available spots. This construction is illustrated in Figure 2.
. . . . . . . . .
| {z }
δ1
s1 | {z }
δ2
s2 s3 | {z }
δ4
Figure 2. An example where there are 3 unavailable spots: s1, s 2, and s3. The lengths of the contiguous blocks of available spots determine the parts of the weak composition δ. In this example, since s2 and s3 are consecutive unavailable spots, we have s3 −s2 −1 = δ3 = 0. We then distribute the n cars into subsets of sizes δ1, δ 2, . . . , δ m+1 . This determines the block of contiguous available spots in which each car will park. There are nδ
= n!
δ1!··· δm+1 !
ways to do this. Once this distribution is chosen, we then may have up to k − 1 total descents among all blocks (if the first spot is available; i.e. s1̸ = 1), or k descents if the first spot is unavailable. Supposing that the jth subsequence has ij descents, we have that i1 + · · · + im+1 ≤ max {k − δ1, k − 1}. For each choice of ( i1, . . . , i m+1 ) we then have Qm+1
j=1
δj
ij
choices over all subsequences. In total we obtain nδ
X
(i1, . . . , i m+1 ) nonnegative,
P
r
ir ≤ max {k − δ1, k − 1}
m+1
Y
j=1
δj
ij
possible outcomes when there are k lucky cars and spots s1, s 2, . . . , s m are forbidden. □
Remark. For the classical case in which u = (1 , 2, . . . , n ), we have the composition δ = ( n), and the formula in Theorem 3.10 specializes to Corollary 2.8.1:
nn
X
i≤k−1
1
Y
j=1
ni
=
k−1
X
i=1
ni
.
So again we have the ( k − 1)th summation of the Euler triangle, counting the number of permutations of [ n]with at most k − 1 descents. 3.2. Counting outcomes with one unavailable spot. In this subsection, we are interested in enumer-ating outcomes based on which cars are lucky, regardless of which spots they end up choosing, when there is exactly one unavailable spot. 3.2.1. Counting outcomes when the only unavailable spot is spot 1. In this section, we give a formula for the number of outcomes of u-parking functions for u = (2 , 3, . . . , n, n + 1) with lucky set I = {i1, i 2 . . . , i k}
where k ≤ n. In this scenario, there are n + 1 spots on the street with spot 1 unavailable and the remaining
n spots are available, and for any outcome B ∈ O u(I), B1 = X. In this case we can imagine that a lucky car 0 has first come to occupy the unavailable spot 1.
11
Theorem 3.11. Let u = (2 , 3, . . . , n, n + 1). If I = {i1, i 2, . . . , i k} is a lucky set of PF n(u) with 1 ≤ i1 <
· · · < i k ≤ n, then
|O u(I)| = k! Y
j∈[n]\I
(|I ∩ [j]| + 1) = 2 i2−i1 3i3−i2 · · · kik −ik−1 (k + 1) n−ik . (3)
Proof. To account for the assumption that spot 1 is the only unavailable spot, we assume that an imaginary car 0 parks in spot 1, and proceed similarly as in the classical case. Each unlucky car j can park after any lucky car i where i < j , i = 0 included. In this case, the number of options for each unlucky car j is the number of entries in I which are less than j, plus one to account for spot 0. Then the jth car has |I ∩ [j]| + 1 options for which increasing subsequence to park in. The sequence starting with 0 is fixed to appear first, and the remaining k other subsequences may appear in any order. It follows that the number of outcomes in this case is
|O n(I)| = k! Y
j∈[n]\I
(|I ∩ [j]| + 1) .
Next we establish an alternative count similar to the formula appearing in Theorem 2.2, which will give the right-hand-side of (3). The number of unlucky cars which have only one option is the number of unlucky cars with index less than i1. In this case, these cars must park in the sequence following car 0. There are i1 − 1 such cars. The number of unlucky cars which have exactly two options is the number of unlucky cars with index between
i1 and i2. There are i2 − i1 − 1 such cars. Similarly, for ℓ = 3 , . . . , k , the number of unlucky cars which have exactly ℓ options is the number of unlucky cars with index between iℓ−1 and iℓ. There are iℓ − iℓ−1 − 1 such cars. The number of unlucky cars which have k + 1 options is the number of cars with index greater than that of every car indexed by I. There are n − ik such cars. Then, to place the unlucky cars into the k + 1 subsequences, the total number of options is 2i2−i1−13i3−i2−1 · · · kik −ik−1−1(k + 1) n−ik .
We may then arrange the subsequences in any order, as long as the subsequence beginning with car 0 is always first. There are k! such arrangements. In total the number of outcomes is
k! 2 i2−i1−13i3−i2−1 · · · kik −ik−1−1(k + 1) n−ik = 2 i2−i1 3i3−i2 · · · kik −ik−1 (k + 1) n−ik ,
and this completes the proof. □
Remark. Theorem 3.11 could be viewed as a corollary of Theorem 2.2 where we let I′ = {1, i 1 + 1 , i 2 +1, . . . , i k + 1 } be the lucky set. If there are k + 1 lucky cars, the probability that car 1 parks in spot 1 is 1
k+1
,since the way that outcomes are constructed gives equal likelihood for any of the lucky cars to appear first. We can then divide the formula in Theorem 2.2 by k + 1 to obtain the same result.
Corollary 3.11.1. If u = (2 , 3, . . . , n, n + 1) and I = {1, 2, . . . , k }. Then
|O u(I)| = k!( k + 1) n−k.
Proof. We write I = {i1, i 2, . . . , i k} with i1 = 1 , i 2 = 2 , . . . , i k = k. Then by Theorem 3.11,
|O u(I)| = 2 i2−i1 3i3−i2 · · · kik −ik−1 (k + 1) n−ik = 2 131 · · · k1(k + 1) n−k = k!( k + 1) n−k. □
3.2.2. Counting outcomes when the only unavailable spot is spot 2. In this section, we give a formula for the number of outcomes of u-parking functions for u = (1 , 3, 4, . . . , n, n + 1) with lucky set I = {i1, i 2, . . . , i k}
where k ≤ n. In this scenario, there are n + 1 spots on the street with spot 2 unavailable and the remaining
n spots are available, and for any outcome B ∈ O u(I), B2 = X.
Definition 3.12. For j1 < · · · < j ℓ ∈ [n], we define the set I(j1,...,j ℓ) to be the set constructed from
I{ j1, . . . , j ℓ}, in which each element i ∈ I{ j1, . . . , j ℓ} is decreased by |{ j1, . . . , j ℓ} ∩ [i]|, i.e.
I(j1,...,j ℓ) := {i − |{ j1, . . . , j ℓ} ∩ [i]| : i ∈ I{ j1, . . . , j ℓ}} .
Example 3.13. Let I = {1, 4, 5}. Then,
• I(1 ,3) = {2, 3} since I \ { 1, 3} = {4, 5}, and both entries in (1 , 3) are less than 4 , 5, so the entries 4 and 5 are decreased by 2 each.
• I(4) = {1, 4} since I \ { 4} = {1, 5}, and 4 is less than 5, so 5 is decreased by 1.
12
• I(2 ,3) = {1, 2, 3} since I \ { 2, 3} = {1, 4, 5}, and 1, 2, and 3 are less than 4 and 5, so 4 and 5 are decreased by 3.
Definition 3.14. For a given lucky set of cars I = {i1, i 2 . . . , i k} with 1 ≤ i1 < · · · < i k ≤ n, we define
cs(n, I ) to be the number of outcomes for u = (1 , 2, 3, . . . , s − 1, s + 1 , . . . , n + 1) where spot s is unavailable and there are n cars.
Example 3.15. Using the result in Theorem 3.11, we can write c1(n, I ) as follows
c1(n, I ) = 2 i2−i1 3i3−i2 · · · (k + 1) n−ik .
Theorem 3.16. Suppose u = (1 , 3, 4, . . . , n, n + 1). If I = {i1, . . . , i k} with 1 ≤ i1 < · · · < i k ≤ n and
c2(n, I ) denotes the number of outcomes of u-parking functions with lucky set I, then
c2(n, I ) =
k
X
j=1
c1(n − 1, I (ij )).
Proof. Since spot 1 is available, we have that the car parking in that spot is always lucky. Suppose that car
ij parks in spot 1 where 1 ≤ j ≤ k.The rest of the street consists of the unavailable spot 2, followed by spots 3 through n + 1, where the cars parking are indexed by the set [ n + 1] \ { ij } with lucky set I \ { ij }. We may view this as a copy of the parking situation described in Theorem 3.11, in which we have n spots and the first is unavailable, where the cars in [ n] \ { ij } are viewed in correspondence with the set [ n − 1]. The corresponding lucky set in this setting can be formed from I \ { ij } as follows: for each car whose index is higher than ij , we decrease its index by 1. That is, the lucky set is I(ij ). The number of ways for the cars to park in spots 3 , . . . , n + 1 is then given by c1(n − 1, I (ij )). The lucky car ij can be any car from the set I = {i1, . . . , i k}. Then in total, the number of possible outcomes is
c2(n, I ) =
k
X
j=1
c1(n − 1, I (ij )),
as claimed. □
3.2.3. Counting outcomes when the only unavailable spot is spot 3. We now give a formula for the number of outcomes of u-parking functions for u = (1 , 2, 4, . . . , n + 1) with lucky set I = {i1, i 2 . . . , i k} where k ≤ n.In this scenario, there are n + 1 spots on the street with spot 3 unavailable and the remaining n spots are available, and for any outcome B ∈ O u(I), B3 = X. We follow the notation from Theorem 3.14, and let
c3(n, I ) denote the number of possible outcomes in this case.
Theorem 3.17. Suppose u = (1 , 2, 4, . . . , n, n + 1). If I = {i1, . . . , i k} with 1 ≤ i1 < · · · < i k ≤ n and
c3(n, I ) denotes the number of outcomes of u-parking functions with lucky set I, then
c3(n, I ) =
k
X
j=1
c2(n − 1, I (ij )) +
k
X
j=1
X
s∈[n]\I,s>i j
c1(n − 2, I (ij ,s ))
,
where c2(n, I ) is as in Theorem 3.14.
Proof. Similarly as before, since spot 1 is available, the car parking in that spot is always lucky. Suppose this car is car ij . The car that parks in spot 2 may or may not be lucky. If the car in spot 2 is lucky, then for the last n spots {2, 3, . . . , n + 1 } where spot 3 is unavailable, we have a copy of the parking scenario in Theorem 3.16 in which the corresponding lucky set is I(ij ) where we shift the vector u′ = (2 , 4, . . . , n, n + 1) to u = (1 , 3, 4, . . . , n −1, n ). In this case, there are n−1 cars and n−1 available spots with spot 2 unavailable. The possible outcomes in this case are then counted by Pkj=1 c2(n − 1, I (ij )). Otherwise suppose the car that parks in spot 2 is unlucky. In this case, the car in spot 2 must be from the set [ n] \ I, and its preference must have been spot 1. In order for it to have parked in spot 2, it must have parked after car ij , so its index must be higher than ij . Let s be the index of the car in spot 2. Then we have s ∈ [n] \ I, and s > i j .Once car s has been chosen, we count the possible options for how the remaining cars occupy the remaining spots (the forbidden spot 3, and then spots 4 through n + 1). Similarly as in the proof of Theorem 3.16, we
13
can view this as a parking scenario for a street with n − 1 spots in which the first spot is forbidden, and the cars indexed by [ n − 2] are viewed in correspondence with [ n] \ { ij , s } with the same relative order. In this case, the lucky set is I(ij ,s ). Then for a fixed choice of ij and s, we have c1(n − 2, I (ij ,s )) options for the last
n − 1 spots. Over all possible choice of ij and s, we then have Pkj=1
P
s∈[n]\I,s>i j
c1(n − 2, I (ij ,s )) options. Then the total number of possible outcomes is given by
c3(n, I ) =
k
X
j=1
c2(n − 1, I (ij )) +
k
X
j=1
X
s∈[n]\I,s>i j
c1(n − 2, I (ij ,s )). □
Remark. If we keep moving down the street, only removing one spot at a time, we should obtain a recurrence relation that depends on the previous cases. That is, if the only unavailable spot is s then the recurrence should depend on c1(x1, I y1 ), c 2(x2, I y2 ), . . . , c s−1(xs−1, I ys−1 ) where x1, x 2, . . . , x s−1 are some lengths (de-pending on how many spots are available) and y1, y 2, . . . , y s−1 are the corresponding sets of cars depending on which cars we are removing from the lucky set Iyi . We pose this as an open problem in Section 6. 4. Counting outcomes of vector parking functions with a fixed set of lucky spots
In this section, given a vector u, we count outcomes corresponding to a given set L of lucky spots, where a parking spot ℓ is a lucky spot if it is the preference of a lucky car. By construction, L is a subset of the entries in u. Notice that if any car parks in its preferred spot ℓ, that spot is said to be a lucky spot (even if other cars park in spot ℓ that did not prefer ℓ). Throughout we let L be a set containing the indices of lucky spots; i.e. the spots in which lucky cars park.
Definition 4.1. Given a vector u of length n, and L a subset of the entries in u, we associate a composition of n, constructed by separating the entries in u into sub-intervals as follows: (1) Place a bar to the left of the first entry in u.(2) For any i ∈ { 1, 2, . . . , max( u)} such that i ∈ u but i + 1 /∈ u, place a bar to the right of i in u.(3) For each spot ℓ ∈ L where ℓ > 1, place a bar to the left of the first instance of this value in u.Let k denote the total number of contiguous blocks of unavailable spots plus the number of lucky spots. Once the bars have been placed, for all 1 ≤ i ≤ k, let δi be the length of the contiguous set of entries between the
ith and ( i + 1)th bar. Now define the associated composition δu,L = ( δ1, . . . , δ k) to be a weak composition of n which is given via this construction. We illustrate Theorem 4.1 next.
Example 4.2. Let n = 6, u = (1 , 1, 2, 4, 4, 5) and L = {1, 5}. We construct the associate composition by placing the following bars:
| 112 | 44 | 5 | .
The resulting associated composition is δu,L = (3 , 2, 1). If u is the same and instead we consider L′ = {1, 4}, then we would place the following bars:
| 112 || 445 | .
The resulting associated composition is δu,L ′ = (3 , 0, 3).
Definition 4.3. Viewing the outcome of a u-parking function as a permutation of [ n] with X’s inserted to represent unavailable spots between entries, we define the underlying permutation to be the sequence of entries with X’s removed. If multiple cars park in the same spot, we represent them as a sequence in increasing order.
Example 4.4. If the outcome of a u-parking function is {1}X{2, 3, 6}XX {4}{ 7, 8}X{5}, the underlying per-mutation (in one-line notation) is 12364785.
Theorem 4.5. Let u be a vector of length n, and let L be a subset of the entries in u indexing the lucky spots. Let δu,L = ( δ1, . . . , δ k) be the associated composition of n. Then the number of outcomes for
u-parking functions with lucky spots indexed by L is
nδu,L
=
nδ1, . . . , δ k
= n!
δ1! δ2! · · · δk! .
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Proof. Once the lucky spots are chosen for the lucky cars, the unlucky cars may park in the remaining available spots. So, an unlucky car may park to the right of a lucky car who parked prior, or to the right of an unavailable spot. Within the available “unlucky” spots—i.e., spots not indexed by L, the unlucky cars must form increasing subsequences (starting with each lucky car, or unlucky car following an unavailable spot), by the same reasoning as in the proof of Theorem 2.2. Each car that parks in an unlucky spot must have preferred an earlier spot, and it must have been completely occupied or unavailable before that car checked it. So within each lucky spot, the cars parking in that spot must all have smaller indices than each of the unlucky spots in a contiguous run to the right (even if not all cars in the lucky spot preferred this spot). By the discussion above, the underlying permutation of an outcome with lucky spots given by a set L will be constructed of increasing runs, which may be broken at the beginning of lucky spots (where the first entry corresponding to the lucky spot is the starting position of a new increasing subsequence) or at unavailable spots (where in the underlying permutation, a new increasing run may start following where an unavailable spot appeared). By construction, the lengths of these increasing runs are exactly the parts of the associated composition δu,L = ( δ1, . . . , δ k). We may alternatively view these objects as permutations with possible descents either at the beginning of a lucky spot, or where there are “gaps” in which spots are unavailable. To count such permutations, we refer to a result of MacMahon [33, Art. 157], and obtain that the number of permutations of the desired form is given by the multinomial coefficient
nδu,L
= n!
δ1! δ2! · · · δk! . □
Example 4.6. Let u = (1 , 2, 4, 5). Let us count the outcomes in which the lucky spots are given by
L = {1, 5}. To do so, we partition u as follows:
| 12 | 4 | 5 | .
Then the associated composition δu,L = (2 , 1, 1) and the number of outcomes is 42,1,1
= 4! 2!1!1! = 12. One can verify that the possible outcomes are
{1}{ 2}X{3}{ 4}, {1}{ 4}X{2}{ 3}, {2}{ 3}X{1}{ 4}, {2}{ 3}X{4}{ 1},
{1}{ 2}X{4}{ 3}, {1}{ 3}X{4}{ 2}, {2}{ 4}X{1}{ 3}, {2}{ 4}X{3}{ 1},
{1}{ 3}X{2}{ 4}, {1}{ 4}X{3}{ 2}, {3}{ 4}X{1}{ 2}, {3}{ 4}X{2}{ 1}
.
Example 4.7. Let u = (2 , 2, 3, 5). Let us count the outcomes in which the lucky spots are given by
L = {3, 5}. To do so, we partition u as follows:
| 22 | 3 || 5 |
Then the associated composition is δu,L = (2 , 1, 0, 1) and the number of outcomes is again 42,1,0,1
= 4! 2!1!0!1! =12. One can verify that the outcomes are
X{1, 2}{ 3}X{4}, X{1, 4}{ 2}X{3}, X{2, 3}{ 1}X{4}, X{2, 3}{ 4}X{1},
X{1, 2}{ 4}X{3}, X{1, 3}{ 4}X{2}, X{2, 4}{ 1}X{3}, X{2, 4}{ 3}X{1},
X{1, 3}{ 2}X{4}, X{1, 4}{ 3}X{2}, X{3, 4}{ 1}X{2}, X{3, 4}{ 2}X{1}
.
In this case, the two cars that park in spot 2 are both unlucky but still park in ascending order. The cars in spots 3 and 5 are both lucky and can be chosen as any ordered pair among the four possible cars. We remark that this construction counts outcomes based on which spots are occupied by lucky cars. However, the actual cars that end up being lucky are not necessarily restricted. 5. Counting u-parking functions with certain sets of lucky cars
So far, we have looked at counting outcomes for families of certain vector parking functions with certain lucky sets and lucky spots. In this section, we provide a formula for the number of vector parking functions with a fixed set of lucky cars I. We also provide a formula for the number of vector parking functions with a lucky set of fixed size; that is, |I| = k. For sake of simplicity in our arguments, we assume that X < i for any i ∈ N.
15
5.1. Counting vector parking functions with a fixed set of lucky cars. Recall that if there exists some i / ∈ u where i < max( u), then spot i has capacity zero, i.e. spot i is unavailable. We begin with a definition.
Definition 5.1. For an ordered set partition B = B1B2 · · · BM ∈ O u(a) where M = max( u), let bi =max( Bi) and if Bi = X then bi = X. For each b ∈ Bi, let sB (b) be the length of the longest subsequence
bj bj+1 · · · bi−1 with bt < b for all j ≤ t ≤ i − 1. We denote the length of a sequence b1b2 · · · bi by |b1b2 · · · bi|.
Example 5.2. Let u = (1 , 1, 3, 3, 3) and B = {2, 3}X{1, 4, 5} where B1 = {2, 3}, B 2 = X and B3 = {1, 4, 5}.Then b1 = 3 , b 2 = X and b3 = 5. Then, sB (b = 2) = 0 and sB (b = 3) = 0, since 2 , 3 ∈ B1 and there are no bi’s with index i less than 1. Now, for each b ∈ B3, we have the following. For b = 1, since X = b2 < 1and 3 = b1 ≮ 1, we have sB (b = 1) = |b2| = 1. For b = 4, since X = b2 < 4 and 3 = b1 < 4, it follows that
sB (b = 4) = |b1b2| = 2. Similarly, for b = 5, since X = b2 < 5 and 3 = b1 < 5, we have sB (b = 5) = |b1b2| = 2. It follows that sB (b = 1) = 1 , s B (b = 2) = 0 , s B (b = 3) = 0 , s B (b = 4) = 2 and sB (b = 5) = 2. For each b ∈ Bi, the definition sB (b) counts the number of distinct spots where cars arriving before car b
parked contiguously and immediately to the left of car b (if any) and the number of spots that have capacity 0 contiguously and immediately to the left of car b (if any). We stress that sB (b) is the number of spots and not the number of cars. We show (in Theorem 5.5) that in the case that car b is an unlucky car, the number
sB (b) is exactly the number of potential distinct spots car b could prefer and which car b would find either occupied by cars that arrived and parked before car b or spots that have capacity 0. We make this precise next.
Definition 5.3. Fix a lucky set I of PF( u) and an outcome B = B1B2 · · · BM ∈ O u(I) where M = max( u). For each b ∈ Bi, let Pref I (b, u) denote the set of preferences of car b in a u-parking function a of length n
satisfying Lucky u(a) = I and Ou(a) = B.
Example 5.4 (Continuing Theorem 5.2) . If B = {2, 3}X{1, 4, 5} and if car 1 is unlucky, then it could only prefer the unavailable spot to its left and it cannot prefer spot 1 as it would find it empty and park there. This agrees with sB (b = 1) = 1. If car 4 is unlucky, then it could only prefer spot 1, which is occupied by cars 2 and 3, or the unavailable spot 2. This is true because cars 2 and 3 are parked to the left of car 4 and arrived prior to car 4, and spot 2 is unavailable, so car 4 could have preferred this spot and be unlucky. This agrees with sB (b = 4) = 2. The count for the number of preferences of unlucky cars detailed in Theorem 5.4 holds in general.
Lemma 5.5. Fix a lucky set I of PF( u) and fix B = B1B2 · · · BM ∈ O u(I), where M = max( u). Then for each b ∈ Bi,
|Pref I (b, u)| =
(
1 if b ∈ IsB (b) if b / ∈ I,
and the number of possible u-parking functions a with outcome B and lucky set I is equal to
Y
b∈[n]
|Pref I (b, u)| = Y
b / ∈I
sB (b).
Proof. We count possible preferred parking spots for each car b ∈ [n]. If b is a lucky car, its preference must be the spot in which it parked. So for b ∈ I, |Pref I (b, u)| = 1. Otherwise, suppose car b is unlucky. In this case, its preference must have been to the left of the spot in which it parked. We look to the left at all spots which it could have preferred. It could not have preferred a spot that is occupied by a car of higher index, since if that was the case, then it would have parked in that spot first. Similarly, it could not have preferred any spot to the left of a car of higher index. So we look only at consecutive spots to the left, between the spot containing car b and the spot containing the nearest car of higher index. If no car of higher index appears to the left of the spot where car b parked, we consider all spots to the left. Among spots to the left that are either unavailable or occupied completely by cars of lower index, the unlucky car b could have preferred any of these to end up parking where it did. So each of these spots are possible preferences for car b, and they are counted by sB (b). So we have, for b / ∈ I, |Pref I (b, u)| = sB (b).
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Then we may count the total number of parking functions corresponding to this outcome as the product of possible preferences for each of the n cars; that is,
Y
b∈[n]
|Pref I (b, u)|.
Since |Pref I (b, u)| = 1 for all b ∈ I, these terms do not change the product so we may restrict attention to all b / ∈ I. The formula reduces to Y
b / ∈I
sB (b). □
Example 5.6 (Continuing Theorem 5.2) . Suppose that I = {2, 3, 4} and B = {2, 3}X{1, 4, 5}. By Theo-rem 5.5, the number of u-parking functions a satisfying Ou(a) = B and Lucky u(a) = I is given by
Y
b∈[n]
|Pref I (b, u)| = Y
b / ∈I
sB (b) = sB (b = 1) · sB (b = 5) = 2 .
Indeed, if u = (1 , 1, 3, 3, 3) then the only u-parking functions with lucky set I = {2, 3, 4} and outcome
B = {2, 3}X{1, 4, 5} are
a = (2 , 1, 1, 3, 1 ) and a′ = (2 , 1, 1, 3, 2 ) ,
where we box the difference between these u-parking functions, which denote the possible preferences of car 5 being spots 1 and 2. Since cars 2, 3, and 4 are lucky, their preferences are fixed to be the spots where they parked. Since car 1 parks in spot 3 and is unlucky, and since cars 2 and 3 park in spot 1, we have that car 1 may only prefer spot 2. Since car 5 is unlucky, and spot 1 is fully occupied when car 5 goes to park and spot 2 is unavailable, car 5 may prefer either spot 1 or spot 2. We now give a formula for the number of u-parking functions with a fixed set of lucky cars.
Theorem 5.7. Fix a lucky set I of PF( u). Let Ou(I) denote the set of outcomes B = B1B2 . . . B M where
M = max( u) and lucky set I where Bi denotes the set of cars that parked in spot i and if Bi = X then spot
i is unavailable (that is, no cars can park there). Then
|LuckyPF u(I)| = X
B∈O u(I)
Y
b / ∈I
sB (b)
!
.
Proof. Theorem 5.5 established that for a specific outcome and lucky set, the number of u-parking functions yielding that outcome and lucky set is given by Q
b / ∈I
sB (b). Since each u-parking function corresponds to exactly one outcome, we may compute the total number of u-parking functions with a given lucky set I by summing over all possible outcomes with lucky set I, and counting the u-parking functions corresponding to each. The result follows. □
5.2. Counting vector parking functions when k cars are lucky. In this section, we give a formula for the number of u-parking functions with a lucky set of size k. For ease of our arguments, our convention is that 0 0 = 1. In this section, we view an outcome of a vector parking function as usual, but without the unavailable spots X and we reindex the nonempty parts in the outcome. This will make the outcome an ordered set partition of [ n]. See the following for an example.
Example 5.8. If u = (3 , 3, 5, 5) and Ou(a) = XX {2, 3}X{1, 4} for some a, then the outcome becomes
B = {2, 3}{ 1, 4} where B1 = {2, 3} and B2 = {1, 4}.We first set some notation as follows. Let
u = ( u1, u 1, . . . , u 1
| {z }
m1
, u 2, u 2, . . . , u 2
| {z }
m2
, . . . , u r , u r , . . . , u r
| {z }
mr
),
where for each uj ∈ u, uj has multiplicity mj for all j ∈ [r]. Suppose there are k lucky cars. Let B = ( B1, B 2, . . . , B r ) be an ordered set partition of the interval of integers [ Prj=1 mj ] = {1, 2, . . . , m 1 + m2 + · · · + mr } where |Bj | = mj (each Bj contains the cars that will park in spot ij ). Let κ = ( k1, . . . , k r ) be a partition of k such that kj denotes the number of lucky cars in set Bj . Moreover, let Lj be the set that contains the lucky cars in Bj where |Lj | = kj .
17
Definition 5.9. For 0 < t < j , let ℓj,t be the number of unlucky cars in set Bj such that when they park, spots ut through uj−1 are completely occupied. We define ℓj,j = |Bj \ Lj | = mj − kj (total number of unlucky cars in Bj ) and ℓj, 0 = 0.
Lemma 5.10. For 0 < t < j , the count ℓj,t is given by
ℓj,t :=
(
b ∈ Bj : b > max
j−1
[
d=t
Bd
!)
∩ (Bj \ Lj ) .
Proof. Let b be a car included in the set counted by ℓj,t . When this car parks, every spot from ut to uj−1 is filled. So every car included in those spots in the outcome has already parked; therefore they are all smaller in index than car b. That is, b > max
Sj−1
d=t
Bd
. Also, in order for car b to be included in this count, it must be in the set Bj and be unlucky, so it is in Bj \ Lj . This is accounted for with the intersection and the result follows. □
Theorem 5.11. Let u = ( u1, u 1, . . . , u 1
| {z }
m1
, u 2, u 2, . . . , u 2
| {z }
m2
, . . . , u r , u r , . . . , u r
| {z }
mr
) where for each uj ∈ u, uj has multiplicity mj for all j ∈ [r]. Then the number of u-parking functions with k lucky cars is
[
|I|=k
LuckyPF u(I) = X
κ=( k1,k 2,...,k r),Pri=1 ki=k,
0≤ki≤mi
X
B=( B1,B 2,...,B r)∈O u
m1
k1
(u1 − 1) m1−k1 S(B; κ),
where
S(B; κ) := X
(L2,L 3,...,L r),Ld⊆Bd,|Ld|=kd
r
Y
j=2
j
Y
t=1
(uj − ut−1 − 1) ℓj,t −ℓj,t −1 .
Proof. Recall that in this subsection, Bi is no longer the set of cars that parked in spot i, but given the reindexing it is now is the ith nonempty set that appears in the outcome. By assumption, we have r spots with capacities given by m1, . . . , m r . Fix an outcome B = ( B1, B 2, . . . , B r ) ∈ O u, and suppose that there are k lucky cars. Let κ = ( k1, k 2, . . . , k r ) be a weak composition of k, such that the ith spot contains ki lucky cars. Then we have that 0 ≤ ki ≤ mi for all i. We count the number of parking functions corresponding to this outcome. The first available spot is u1. We have that B1 is the set of cars which parked in spot u1, and k1 of them are lucky. First we choose which cars are lucky; there are m1
k1
options. For these cars, their preference is fixed to be u1.Then there are m1 − k1 unlucky cars remaining in B1. We count the possible preferences for the unlucky cars in this set. For each such car, their preference could have been any spot earlier than u1, so they each have ( u1 − 1) options for their preference. Hence, the number of ways to count the preferences of cars in the set B1 is given by m1
k1
(u1 − 1) m1−k1 .We now consider the possible preferences for the cars in spots B2, . . . , B r . Let S(B, κ ) denote the number of possible preferences among the cars parking in the remaining r − 1 available spots. Fix uj to be one of the spots after u1, and let Lj be the set of lucky cars in spot uj (so Lj is a subset of
Bj consisting of kj cars). For the cars in Lj , their preference is fixed to be uj (the spot where they parked). We consider sets of possible preferences for the remaining mj − kj unlucky cars. Fix a spot ut ≤ uj−1 and define u0 = 0. We count the number of unlucky cars c in Bj whose set of possible preferences contains any spot between ut−1 and uj , not including ut−1 (or uj , since c is an unlucky car). When such a car c went to park, its preference was after spot ut−1, so it did not check this spot, and all spots from ut to uj−1 were full. However, spot ut−1 would not have been full since if it was, then car c
could have preferred spot ut−1 as well. The number of such cars is given by ℓj,t − ℓj,t −1. These cars could have preferred any spot in the set
{ut−1 + 1 , , u t−1 + 2 , . . . , u j − 1}, and in particular not any spot ut−1 or earlier, to end up parking in spot uj .For each such car, there are uj − 1 − (ut−1 + 1) + 1 = uj − ut−1 − 1 options for its preference. Then the number of possible preferences among all cars which could have preferred any spot ut ≤ uj−1 is given by
Qj−1
t=1
(uj − ut−1 − 1) ℓj,t −ℓj,t −1 .
18
The unlucky cars that remain to be accounted for are those that preferred a spot between uj−1 and uj , but were unable to park there since it was unavailable. There are uj − uj−1 − 1 unavailable spots between uj−1
and uj . To avoid double-counting cars that were accounted for in the previous product, we only consider cars such that when they parked, spot uj−1 was available; in other words spot uj−1 was not full yet. This guarantees that their preference could not have been uj−1 or earlier (or else they would have parked there instead). The number of unlucky cars such that spot uj−1 was available when they parked is given by mj −kj −ℓj,j −1.It follows that ( uj −uj−1 −1) mj −kj −ℓj,j −1 is the number of possible preferences among unlucky cars preferring a spot after uj−1 and parking in spot uj . In total we have, for a fixed spot uj ,
j−1
Y
t=1
(uj − ut−1 − 1) ℓj,t −ℓj,t −1
!
(uj − uj−1 − 1) mj −kj −ℓj,j −1
possible preferences for the cars in the set Bj . With the convention that ℓj,j = mj −kj , this product simplifies to jY
t=1
(uj − ut−1 − 1) ℓj,t −ℓj,t −1 .
Taking the product over all j = 2 , . . . , r , we have for a given outcome B = ( B1, B 2, . . . , B r ), with lucky cars in each of the spots given by ( L2, . . . , L r ), a total of
r
Y
j=2
j
Y
t=1
(uj − ut−1 − 1) ℓj,t −ℓj,t −1
possible preference lists for the cars in spots 2 through r. Summing over all possible distributions of lucky cars ( L2, . . . , L r ), we have
S(B; κ) = X
(L2,L 3,...,L r),Ld⊆Bd,|Ld|=kd
r
Y
j=2
j
Y
t=1
(uj − ut−1 − 1) ℓj,t −ℓj,t −1
total options for the preferences of cars parking in the last r − 1 spots. Then for a fixed outcome B =(B1, B 2, . . . , B r ) ∈ O (u), the total number of corresponding parking functions is
m1
k1
(u1 − 1) m1−k1 S(B; κ).
Since each parking function corresponds to one outcome, we may enumerate all parking functions with k
lucky cars by summing over all possible outcomes with k lucky cars. Then the number of parking functions of the desired form is
X
κ=( k1,...,k r),
0≤ki≤mi,Pri=1 ki=k
X
B=( B1,B 2,...,B r)∈O (u)
m1
k1
(u1 − 1) m1−k1 S(B; κ). □
With restrictions on the form of u, the formula in Theorem 5.11 can be simplified further.
Corollary 5.11.1. Let u = ( i, i, . . . , i,
| {z }
n−1 times
j). Then the number of u-parking functions with k lucky cars is
[
|I|=k
LuckyPF u(I) =
n − 1
k
(i − 1) n−1−k (( n − 1)( j − i − 1) + ( j − 1)) + n
n − 1
k − 1
(i − 1) n−k.
Proof. We apply the formula in Theorem 5.11, with m1 = n − 1, m2 = 1, u1 = i, and u2 = j. Suppose a is the car that ends up parking in spot j. Then B2 = {a}. Either a is a lucky car or an unlucky car. Hence,
k2 = 0 or k2 = 1. If k2 = 0 then all of the lucky cars park in spot i, and if k2 = 1 then one lucky car parks in spot j and the remaining k − 1 lucky cars park in spot i. The summation is then
19
X
B2={a}
n − 1
k1
(i − 1) n−1−k1 S(B; κ)
| {z }
k1=k,k 2=0
X
B2={a}
n − 1
k1
(i − 1) n−1−k1 S(B; κ)
| {z }
k1=k−1,k 2=1
, (4) and for a composition κ = ( k1, k 2), we have
S(B; κ) = X
(L2,L 3,...,L r ),Ld⊆Bd,|Ld|=kd
rY
j=2
jY
t=1
(uj − ut−1 − 1) ℓj,t −ℓj,t −1
= X
(L2),L2⊆B2,|L2|=k2
2Y
j=2
jY
t=1
(uj − ut−1 − 1) ℓj,t −ℓj,t −1
= X
L2⊆B2,|L2|=k2
2Y
t=1
(u2 − ut−1 − 1) ℓ2,t −ℓ2,t −1
= X
L2⊆B2,|L2|=k2
(u2 − u0 − 1) ℓ2,1−ℓ2,0 (u2 − u1 − 1) ℓ2,2−ℓ2,1
= X
L2⊆B2,|L2|=k2
(j − 1) ℓ2,1 (j − i − 1) ℓ2,2−ℓ2,1 .
Then the summation in (4) becomes
X
B2={a}
n − 1
k
(i − 1) n−1−k X
L2=∅
(j − 1) ℓ2,1 (j − i − 1) ℓ2,2−ℓ2,1
| {z }
k1=k,k 2=0
X
B2={a}
n − 1
k − 1
(i − 1) n−1−k+1 X
L2=B2
(j − 1) ℓ2,1 (j − i − 1) ℓ2,2−ℓ2,1
| {z }
k1=k−1,k 2=1
. (5) The sum P
L2=B2 (j − 1) ℓ2,1 (j − i − 1) ℓ2,2−ℓ2,1 is equal to 1 since there is only one instance when L2 = B2 and in this case, ℓj,t = 0 always since there are no unlucky cars in the set to count. Hence, the two underbraced terms in (5) become
X
B2={a}
n − 1
k
(i − 1) n−1−k X
L2=∅
(j − 1) ℓ2,1 (j − i − 1) ℓ2,2−ℓ2,1
| {z }
k1=k,k 2=0
X
B2={a}
n − 1
k − 1
(i − 1) n−k
| {z }
k1=k−1,k 2=1
= X
B2={a}
n − 1
k
(i − 1) n−1−k X
L2=∅
(j − 1) ℓ2,1 (j − i − 1) ℓ2,2−ℓ2,1
| {z }
k1=k,k 2=0
n
n − 1
k − 1
(i − 1) n−k
| {z }
k1=k−1,k 2=1
. (6) We break up the left sum in (6) and arrive at
X
B2={a}
n − 1
k
(i − 1) n−1−k X
L2=∅
(j − 1) ℓ2,1 (j − i − 1) ℓ2,2−ℓ2,1
=
n − 1
k
(i − 1) n−1−k
X
B2={a},
1≤a≤n−1
X
L2=∅
(j − 1) ℓ2,1 (j − i − 1) ℓ2,2−ℓ2,1 + X
B2={n}
X
L2=∅
(j − 1) ℓ2,1 (j − i − 1) ℓ2,2−ℓ2,1
.
(7)
20 When L2 = ∅ we have ℓ2,2 = 1. If B2 = {a} for 1 ≤ a ≤ n then ℓ2,1 = 0, since spot i will not be completely occupied when car a parks. Otherwise, if B2 = {n} then ℓ2,1 = 1 since the car in B2 is the last car to park. The sum in (7) simplifies to
n − 1
k
(i − 1) n−1−k
X
B2={a},
1≤a≤n−1
X
L2=∅
(j − 1) 0(j − i − 1) 1−0 + X
B2={n}
X
L2=∅
(j − 1) 1(j − i − 1) 1−1
=
n − 1
k
(i − 1) n−1−k
X
B2={a},
1≤a≤n−1
X
L2=∅
(j − i − 1) + X
B2={n}
X
L2=∅
(j − 1)
=
n − 1
k
(i − 1) n−1−k (( n − 1)( j − i − 1) + ( j − 1)) .
In total we obtain
n − 1
k
(i − 1) n−1−k (( n − 1)( j − i − 1) + ( j − 1)) + n
n − 1
k − 1
(i − 1) n−k
u-parking functions with k lucky cars.
□
Remark. Instead of applying Theorem 5.11, the previous result can be obtained from a purely combinatorial argument found by considering possible cases. The resulting count can be written as
n − 1
k
(i − 1) n−1−k(n − 1)( j − i − 1) +
n − 1
k
(i − 1) n−1−k(j − 1) + n
n − 1
k − 1
(i − 1) n−k.
The first term of the above sum counts the possible u-parking functions when all of the k lucky cars park in spot i and the unlucky car in spot j is not the nth car, and the second term counts the possible u-parking functions when all of the k lucky cars park in spot i and the unlucky car in spot j is the nth car. Lastly, the third term counts the possible u-parking functions when k − 1 lucky cars park in spot i and one lucky car parks in spot j.
Remark. In the case of u = (1 , 2, . . . , n ), Theorem 5.11 gives the number of classical parking functions with
k lucky cars, providing an alternative formula to compute the coefficient on qk in (1). 6. Further directions
We conclude with some directions for further study.
• In Section 3.1, we considered the case where the vector u has no repetition and enumerated the possible outcomes that have k lucky cars. It remains an open problem to count the outcomes with k
lucky cars for a general u. A place to start would be to consider cases where the vector u has some repetitions, and count outcomes with a fixed lucky set or a fixed number of lucky cars.
• In Section 3.2, we considered streets with exactly one unavailable spot and no repetition, and we counted outcomes for a fixed set of lucky cars. We give an explicit formula for the case when the first spot is unavailable, and then recursive formulas for the cases when spot 2 or spot 3 is unavailable. Based on those results we remark that for a general unavailable spot i > 3, it should be possible to obtain a recursive formula for the number of outcomes for a fixed set of lucky cars. It remains an open problem to find a closed formula for this count.
• In Section 5, we provide a formula for the number of vector parking functions with a fixed set of lucky cars. It would be of interest to determine closed formulas for the results in this section as well as for the specializations presented. One generalization of parking functions is the set of ( m, n )-parking functions, where there are m cars and
n parking spots, with m ≤ n, and cars follow the standard parking rule to park on the street. It would be of interest to introduce a generalization of vector parking functions in which there is more capacity for cars than cars parking on the street. With such a generalization defined, one direction for further study is to characterize and enumerate the outcomes of these parking functions with a fixed set of lucky cars.
21
Another generalization of parking functions considers cars with various lengths, which are called parking sequences and parking assortments [12, 21]. The difference between sequences and assortments is the parking rule. In parking sequences, cars attempt to park in a contiguous section of the street, and if they fit they park; otherwise they cause a collision and exit the street. In parking assortments, the parking rule allows for the cars to seek forward in the street to find a contiguous segment of parking spots on the street in which the car can park. It would be of interest to define a version of vector parking sequences and vector parking assortments, where spots have a certain capacity and cars have specified lengths and parking preferences. With these generalizations, one could study the lucky statistic on these parking functions, as well as the possible outcomes that arise when a set of lucky cars is specified.
Acknowledgments
L. Martinez gratefully acknowledges William & Mary for hosting her research visit, and acknowledges sup-port by the National Science Foundation Graduate Research Fellowship Program under Grant No. 2233066. This work was also supported by a grant from the Simons Foundation (Travel Support for Mathematicians, P. E. Harris).
References
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Email address : mjferreri@wm.edu
(P. E. Harris) Department of Mathematical Sciences, University of Wisconsin-Milwaukee, Milwaukee, WI 53211
Email address : peharris@uwm.edu
(L. Martinez) Department of Mathematics, Rutgers University, Piscataway, NJ 08854
Email address : lucy.martinez@rutgers.edu
(E. Swartz) Department of Mathematics, William & Mary, Williamsburg, VA 23187
Email address : easwartz@wm.edu |
14882 | https://www.truecable.com/blogs/cable-academy/fiber-optic-splicing-examining-the-factors-that-affect-splice-performance?srsltid=AfmBOopqZdY681CsUwuNiSYHpjC3vkOKPJrl0tjqs1xLg1KRDe2gUmpF | Fiber Optic Splicing: Examining the Factors that Affect Splice Perform
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Fiber Optic Splicing: Examining the Factors that Affect Splice Performance
Written by Ben Hamlitsch, trueCABLE Technical and Product Innovation Manager RCDD, FOI
Are you looking for ways to improve the performance of your fiber optic splices? If so, you've come to the right place. In this blog post, we'll examine the factors that affect splice performance, including intrinsic factors, extrinsic factors, and core diameter mismatch. We'll also discuss the different types of fiber optic splices, such as mechanical splices and fusion splices. Finally, we'll provide some tips on troubleshooting fusion splices.
The performance of a fiber optic splice is determined by a number of factors, including the quality of the fiber, the cleanliness of the splice, and the techniques used to make the splice. Intrinsic factors, such as the refractive index of the fiber, are those that are inherent to the fiber itself. Extrinsic factors, such as the presence of microbends, are those that are external to the fiber. Core diameter mismatch is a type of extrinsic factor that can cause significant loss in a splice.
By understanding the factors that affect splice performance, you can make informed decisions about the type of splice to use and the techniques to employ. This can help you achieve the best possible performance from your fiber optic splices.
What is it that gets spliced onto a fiber optic cable strand or strands? We call it a fiber-optic pigtail. A fiber optic pigtail is a fiber optic cable with one end terminated with a factory-installed connector and the other end unterminated. As a result, the connector side can be connected to equipment, while the other side is fused in the case of fusion splicing and a mechanical connection in the case of mechanical splicing optical fiber cables. Fiber optic pigtails are used to connect fiber optic cables using fusion or mechanical splicing.
We are going to learn
How to determine if the splice loss is due to an intrinsic factor
Determine if the splice loss is from an extrinsic factor
What is a mechanical splice?
What is a fusion splice?
Fusion splicing troubleshooting
Why splice?
Fiber splicing is one way to join two optical fibers together so the light energy from one optical fiber can be transferred to another optical fiber. A fiber splice is the permanent connection of two optical fibers. Once the two optical fibers are joined with a splice, they cannot be taken apart and put back together, as they can if you join them using connectors. Fiber splices are typically employed for one of four reasons: to repair a damaged cable, extend the length of a cable, join two different cable types, or attach a pigtail. We'll talk about fiber pigtails later on in the article
Splice Performance
How well a fiber splice performs depends on many variables. These variables can be broken into two groups: intrinsic factors and extrinsic factors.
An important thing to note and keep in mind is that optical fibers are not perfect, and variations between optical fibers can affect splice performance. These variations are referred to as intrinsic factors.
The performance of a fiber splice can also be affected by alignment and optical fiber mating issues that have nothing to do with the optical fiber. The factors that affect the alignment and mating of the optical fibers are referred to as extrinsic factors.
Intrinsic Factors
Even when fibers are manufactured within specific tolerances, slight variations still exist from one optical fiber to another. These variations can affect the performance of the splice even though the optical fibers are perfectly aligned when mated. The variations between two optical fibers that affect splice performance are referred to as intrinsic factors. Let us now look at the most common types of intrinsic factors.
Numerical Aperture (NA) Mismatch
A numerical aperture (NA) mismatch occurs when the NA of one optical fiber is different from the NA of the other optical fiber. If the NA of the transmitting fiber is larger than the NA of the receiving optical fiber, a loss may occur. However, a loss will not occur if the NA of the transmitting optical fiber is less than the NA of the receiving optical fiber.
Light must enter within a specified range defined by the acceptance cone or cone of acceptance. In order for light to be contained within a fiber, it must stay above the critical angle, or the angle at which it reflects off the boundary between the core and the cladding, rather than penetrating the boundary and refracting through the cladding.
To maintain the critical angle, light must enter within a specified range defined by the acceptance cone (or cone of acceptance). This region is defined by a cone extending outside the optical fiber core. Light entering the core from outside of the cone will either miss the core or enter at an angle that will allow it to pass through the boundary with the cladding and be lost. So the acceptance angle defines the acceptance cone. Light entering the core of the optical fiber at an angle greater than the acceptance angle may not propagate the length of the fiber. For light to propagate the length of the optical fiber, it must enter the core at an angle that does not exceed the acceptance angle.
The exact loss from an NA mismatch is difficult to calculate. Factors such as light source type, light source launch condition, optical fiber length, and bends in the optical fiber all affect the potential loss. It is possible to have an NA mismatch between two optical fibers with no loss from the mismatch.
Core Diameter Mismatch
A core diameter mismatch occurs when there is a difference in the core diameters of the two optical fibers. A core diameter mismatch loss results when the core diameter of the transmitting optical fiber is greater than the core diameter of the receiving optical fiber. A loss occurs when the light at the outer edge of the transmitting optical fiber core falls outside the diameter of the receiving optical fiber core. This light is lost in the cladding of the receiving optical fiber. Core diameter mismatch loss is typically only a concern with multimode optical fiber.
It is not uncommon for two multimode optical fibers with different core diameters to be spliced together. This is because multimode fiber has two common core sizes: 62.5um and 50um.
Mode Field Diameter Mismatch
Mode field diameter is a concern when it comes to singlemode optical fiber. It describes the diameter of the light beam traveling through the core and a portion of the inner cladding. Values for the mode field diameter are defined by the wavelength, with the longer wavelength typically having 8.6 to 9.55um at 1310nm and 8 to 11um at 1550nm.
A mode field diameter mismatch occurs when there is a difference in the mode field diameters of two singlemode optical fibers. A mode field diameter mismatch loss results when the mode field diameter of the transmitting optical fiber is greater than the mode field diameter of the receiving fiber. A loss will occur when the optical fiber with the smaller mode field diameter will not accept all of the light from the optical fiber with the larger mode field diameter.
Cladding Diameter Mismatch
Cladding diameter mismatch occurs when the cladding diameters of the transmit and receive optical fibers are not the same. Cladding diameter mismatch loss occurs when the cores of the optical fiber are not aligned because the cladding diameters are not matching. A cladding diameter mismatch issue can cause the light exiting the core of the transmitting optical fiber to enter the cladding of the receiving optical fiber. The light entering the cladding is lost, causing attenuation.
Concentricity
Ideally, the core in the cladding of the optical fiber are perfectly round and concentric, which means that they share a common geometric center. However, optical fibers are not perfect, and there will be concentricity variations. These concentricity variations can cause the optical fiber cores to misalign, causing a loss when the light exiting the core of the transmitting optical fiber enters the cladding of the receiving optical fiber.
Noncircular
Just as the core and cladding of an optical fiber may not be perfectly concentric, they may also not be perfectly circular. The non-circulatory nature of the core will cause a loss when light from the core of the transmitting optical fiber enters the cladding of the receiving optical fiber.
Cladding that is non-circulatory may cause loss when it causes part of the core of the transmitting optical fiber to align with the cladding of the receiving optical fiber. Any light that injures the cladding of the receiving optical fiber will be lost, causing attenuation.
Note that the amount of loss depends on the alignment of the elliptic curves of the two cores. Maximum loss occurs when the long or major axes of the cores are at right angles to one another, and minimum loss occurs when the axes of the cores are aligned.
Extrinsic Factors
Extrinsic factors that affect optical fiber splice performance are factors related to the condition of the splice itself and external to the optical fiber. In an ideal splice, the optical fibers are identical, and they are aligned so that the cores are perfectly centered on each other and the core axes are perpendicular to the end phase being joined. However, there is no such thing as an ideal splice; only in a real splice do intrinsic and extrinsic factors affect splice performance.
We are going to examine common extrinsic factors that affect splice performance. as we talk about these extrinsic factors, keep in mind that many times they are caused by dirt and contamination. Microscopic particles of dirt can cause the misalignment of one or both optical fibers, creating a high-loss splice.
Lateral Misalignment
Lateral misalignment occurs when two optical fibers are offset. Lateral misalignment loss occurs when light from the core of the transmitting optical fiber enters the cladding of the receiving optical fiber, creating a loss. as the lateral misalignment increases, less light from the core of the transmitting optical fiber makes its way into the core of the receiving optical fiber, increasing the loss of the splice.
End Separation
Even if the optical fibers are perfectly aligned, the splice may still experience loss from end separation. Separation is simply a gap between the transmitting and receiving optical fibers.
Separation can result in two different types of losses. The first is through Fresno reflection, which takes place when light passes from the higher refractive index of the core into the transmitting optical fiber into the lower refractive index of the air and then back into the core of the receiving optical fiber. Each change in the refractive index causes a certain amount of light to be reflected and therefore lost.
One way to overcome the effects of Fresno reflections in separated optical fibers is to use an index-matching gel, which is a transparent gel with a refractive index close to that of the core of the optical fibers being spliced. The gel fills the gap and reduces or eliminates the Fresno reflection. Matching gel is typically used for all mechanical splices.
Separation also causes loss because when light exists in the transmitting optical fiber, it spreads out like the light from a flashlight. Some of the light leaving the core of the transmitting optical fiber may enter the cladding of the receiving optical fiber, causing a loss. How much the light spreads out depends on several variables, including the distance between the optical fibers, the light source type, the launch type, the length and numerical aperture of the transmitting optical fiber, and the bends in the transmitting optical fiber.
Angular Misalignment
If the optical fibers in a splice meet each other at an angle, a loss from angular misalignment may occur. The amount of loss depends on the severity of the angular misalignment and the acceptance cones of the transmitting and receiving optical fibers. Because the NA of a multimode optical fiber is greater than the NA of a singlemode optical fiber, multimode splices tolerate angular misalignment better than singlemode splices.
The loss from angular misalignment occurs when light from the core of the transmitting optical fiber enters the cladding of the receiving optical fiber or enters the core of the receiving optical fiber at an angle exceeding the acceptance angle. Light entering the core of the receiving optical fiber at an angle exceeding the acceptance angle may not propagate the length of the receiving optical fiber.
Angular misalignment also prevents the end faces from contacting each other, resulting in Fresnel reflections exactly like those described in the end separation that we talked about earlier. This is why index-matching gel is used with every mechanical splice.
What is a mechanical splice?
Many manufacturers offer mechanical splices. Mechanical spices like the ones shown in the image above are typically permanent; they align the two cleaved optical fibers and hold them in place. Index-matching gel inside the mechanical splice reduces or eliminates any fresnel reflections or losses at the splice connection.
Mechanical splices can be used for both singlemode and multimode fiber cables. Mechanical splices do not outperform fusion splices. However, they can outperform mated connector pairs. The key advantage of a mechanical splice over a fusion splice is the low cost of the equipment required to perform the mechanical splice.
The assembly tool that holds the optical fibers and the mechanical splice is relatively inexpensive when compared to the price of a fusion splicer. However, the actual mechanical splices cost considerably more than the protective sleeve required for a fusion splice.
Steps to Perform Mechanical Splicing
Step 1: Fiber Preparation: Protective coatings, jackets, tubes, strength members, and other materials should all be removed, leaving only the naked fiber visible. The most important consideration is hygiene.
Step 2: Fiber Cleaning: It’s time to clean the raw fiber after it’s removed from its shell. The glass can be kept clean by wiping it with fiber optic cleaner and lint-free wipes.
Step 3: Cleaving the fiber: The process is comparable to fusion splicing cleaving; however, the accuracy of the cleave is less crucial.
Step 4: Joining fibers mechanically: This approach does not use heat. Simply place the fiber ends in the mechanical splice device and splice them together. Light coupling from one fiber end to the other will be aided by the index matching gel in the mechanical splice equipment.
Step 5: Fiber Protection: The final mechanical splice acts as its own splice protector.
Advantages of Mechanical Splicing
Mechanical splices do not require electricity.
Other than a fiber stripper and a fiber splitter, many mechanical fiber splice designs require no additional equipment.
Mechanical splicing is useful in cases where fusion splicing is not conceivable or practical. This makes them perfect for short-term connections.
Disadvantages of Mechanical Splicing
The insertion loss is much higher. The normal insertion loss of a mechanical splice is about 0.2 dB, which is much greater than the 0.02 dB loss of a standard fusion splice. Multimode fibers are usually spliced mechanically.
Mechanical splices struggle to meet the alignment tolerances of singlemode fibers.
Mechanical splices are only used under relatively safe conditions, such as in an office building.
What is Fusion Splicing?
Fusion splicing uses a high-voltage electric arc between two electrodes to heat and melt the fiber ends together. This creates a permanent splice that is very fragile and must be protected from the outside environment and bending. This is typically accomplished using heat-shrink tubing and a small metal rod commonly referred to as a protective sleeve. Prior to performing the splice, the optical fiber is passed through the center of the protective sleeve, and the sleeve is positioned so it will not interfere with the fusion splicing process. After the fusion splices are successfully completed, the optical fiber is removed from the fusion splicer, and the protective sleeve is positioned directly over the splice area. The protective sleeve is then placed in an electric oven that is typically built into the fusion splicer. The oven heats the tubing, shrinking it around the fusion splice.
Fusion splicers are more expensive than the assembly tools required for mechanical splicing. However, they provide the lowest-loss fiber splice possible. In addition, fusion splices do not produce fresnal reflections. Fusion splicing is the most accurate and durable method for joining two optical fibers. After the optical fibers are stripped and cleaved, they are placed into the fusion splicer, where the two optical fibers are aligned between two electrodes. Depending on the fusion splicer used, several alignment techniques are used to align the fiber optic cable.
Fusion splicers on the market today typically feature a display that allows you to see the optical fibers on two different axes. The cameras in the fusion splicer magnify the optical fibers so that the end faces can be elevated. Most fusion splicers available today also have the ability to approximate the loss of the splice after it is completed.
Steps to perform Fusion splicing
Step 1: Fiber stripping: You must first remove or peel off the protective polymer coating around the fiber optic cable before you can begin fusing it. A mechanical stripping device, similar to stripping pliers, is usually used for this purpose. Remember to clean the stripping equipment before starting the fusing process.
Step 2: Fiber Cleaning: It’s time to clean the raw fiber after it’s removed from its shell. The glass can be kept clean by wiping it with fiber optic cleaner and lint-free wipes.
Step 3: Fiber Cleaving: To make an effective fusion splice, you need a decent cleaver. Instead of cutting the fiber, the chopper knife cuts, pulls, or bends it to cause a neat break, with the end face remaining flat and perpendicular to the fiber axis.
Step 4: Fiber Fusion: After the fibers have been stripped and cleaved, use a fusion splicer to join them together. The ends of the fiber must first be aligned within the splicer. Melt the fibers with an electric arc after they’re correctly aligned, permanently fusing the ends together.
Step 5: Fiber Protection:The splice will not break during typical handling if the fiber is protected from bending and tensile loads. The splice is protected from the weather and breakage by a protective sleeve placed in an electric oven that is typically built into the fusion splicer. The oven heats the tubing, shrinking it around the fusion splice.
Advantages of Fusion Splicing
Fusion splicing is a compact process and has the lowest insertion loss and back reflection.
Fusion splicing is permanent and has the highest mechanical strength.
Fusion splicing can withstand a wide range of temperatures.
Fusion splicing prevents dust and other pollutants from entering the optical path.
Disadvantages of Fusion Splicing
If too much heat is applied to melt the fiber optic cable for termination, the connection will become brittle and cannot be used for a very long time.
Fusion splicing causes significant up-front costs for the splicer and additional instruments.
Fusion splicing cannot be used for temporary connections as it is a permanent connection.
Fusion Splice Troubleshooting
Now let's look at some variables of a fusion splice after it has been spliced, and let’s see what we can determine.
Here are some examples of what can happen when fusion splicing:
As you have now learned, how well a splice performs depends on many variables. As with mechanical splicing, these variables can be broken into two groups: intrinsic factors and extrinsic factors.
Intrinsic factors are the result of slight variations from one optical fiber to another. These variations can affect the performance of the splice even though the optical fibers are perfectly aligned when mated. Most optical fibers today have very little variation, which means there is a very small percentage of splice loss from intrinsic factors.
The most common factors in today's splice losses come from extrinsic factors related to the condition of the splice itself, external to the optical fiber. Oftentimes, they are caused by dirt and contamination. Microscopic particles of dirt can cause the misalignment of one or both optical fibers, creating a high-loss splice.
Let's consider five ways that can affect a fusion splice and why it is important to ensure these steps are followed in order to ensure a high-performance fusion splice.
We are going to look briefly at the following 5 ways:
Cleave Angle
Cleave Length
Bubbles
Necking
Contamination
Cleave Angle
The cleave angle is an important factor when it comes to fusion splicing. The endface of a properly cleaved optical fiber should be perpendicular to the optical fiber without any surface defects, and the cleave angle should not exceed 1.0 degrees. Here are some steps to consider in order to ensure a proper cleave angle.
Ensure the optical fiber and cleaver are clean.
Verify the cleave dimensions as stated in the manufacturer instructions.
Carefully align the endface of the optical fibers between the electrodes of the fusion splicer.
Perform the fusion splice.
Cleave Length
Cleave length is just as important as cleave angle. Each fusion splice manufacturer defines how much bare optical fiber should be exposed after the cleaving process is completed. Exposing too much fiber or not enough fiber can create a high-loss fusion splice. In order to ensure a proper cleave length, it is important that the steps below are carefully followed.
Ensure the optical fiber and cleaver are clean
Verify the dimensions as stated in the manufacturers’s specifications
Re-cleave both optical fibers if the cleave length is incorrect
Carefully align the endface of the optical fibers between the electrodes of the fusion splicer.
Perform the fusion splice.
Bubbles
Dirt or entrapped air may cause a bubble or bubbles, resulting in a possible high-loss fusion splice. In order to prevent bubbles in your fusion splice, consider the following steps:
Ensure you are using the cleaning products and cleaning processes required by the fusion splicer and cleaver manufacturer.
Verify that you have selected the appropriate splicing program or profile for the optical fiber you are splicing
Necking
This is the term used to describe a fusion splice where the diameter of the fused optical fiber is smaller near the electrodes than it was prior to being fused. Necking is typically the result of to much heat during the prefuse, causing glass to be transported away from the splice area and thus producing a high-loss splice. In order to prevent necking, consider the following steps:
Ensure you are using proper cleaning products and cleaning processes required by the fusion splicer and cleaver manufacturer.
Verify that the appropriate splicing program has been selected for the optical fibers you are splicing.
Ensure the optical fibers are properly cleaved, ensuring cleave angle and length.
Contamination
Contamination on the optical fiber or cleaver that is invisible to the human eye can still cause a fusion splice to exceed attenuation requirements, and this is especially true when it comes to singlemode fiber. Because you cannot see the contamination, make sure that you are using the types of cleaning products and the cleaning process required by the fusion splicer and cleaver manufacturer. It is also important to never let the cleaved fiber touch any surface after it has been cleaned; this will ensure the fiber does not attract dirt or dust on the surface.
Ensure you are using proper cleaning products and cleaning processes required by the fusion splicer and cleaver manufacturer.
Clean the fiber cable before it is cleaved.
Do not clean the fiber after cleaving; this can introduce contamination and affect the splice.
Fusion and mechanical splicing are two very important aspects of fiber optic cabling. Both options have there use cases, and depending on the application, one may be better to use than the other. Fusion splicing does remain the better-performing option, and trueCABLE offers four different fusion splicing pigtail options when fusion splicing is needed in your fiber optic installation.
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14883 | https://books.google.com/books/about/The_Developing_Human.html?id=-7dqAAAAMAAJ | The Developing Human: Clinically Oriented Embryology - Keith L. Moore, T. V. N. Persaud - Google Books
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The Developing Human: Clinically Oriented Embryology
Keith L. Moore, T. V. N. Persaud
Saunders, 2003 - Medical - 560 pages
This bestselling resource comprehensively covers human embryology and teratology, presenting all of the complex clinical and scientific concepts in an engaging, lucid, and practical way. Completely revised and updated, the New Edition emphasises the clinical aspects throughout by using a wealth of case studies, clinical correlations, and hundreds of outstanding illustrations.
Features a wealth of clinical case studies-and hundreds of color photgraphs-enabling readers to relate what they are learning to clinical practice.
Contains a chapter on birth defects that can be used as a "mini textbook" on the subject.
Integrates the molecular aspects of embryonic development, including information on stem cells
homeobox genes
gamete formation
regulation control and the molecules/receptors involved
gene activity and expression
and more.
Includes illustrations of new diagnostic procedures, including sonographs, MRIs, electron micrographs, 3D images, and clinical photographs.
Includes the new terminology developed for embryology-the Terminoligica Embryologica.
Presents completely revised and updated Clinically Oriented Questions and Answers.
Has been reviewed by leading geneticists and paediatricians to ensure that all of the information reflects the realities of clinical practice.
Features a wealth of clinical case studies-and hundreds of colour photgraphs-enabling readers to relate what they are learning to clinical practice.
Contains a chapter on birth defects that can be used as a "mini textbook" on the subject.
More »
From inside the book
Contents
Introduction to the Developing Human 1
First 15
Uterus Uterine Tubes and Ovaries 22
Copyright
24 other sections not shown
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Common terms and phrases
abdominalabnormalamniotic fluidaortic archarteriosusarteryatresiaatriumbirthbladderblastocystbloodbonebraincanalcardinal veincartilagecaudalcavitychorionic sacchromosomecleftclinicalcoelomcongenital anomaliesCourtesy of DrcranialdefectsderiveddifferentiatediverticulumectodermembryoEmbryologyembryonic discendodermepitheliumexternalextraembryonicfemalefetusFigurefoldsfollicleformationfourth weekfusegenesgeneticglandgroovegrowthhearthuman embryoinfantsintestinekidneylaterallayerLevel of sectionlimb budslungsManitobamaternalmaxillary prominencesmedianmesenchymemesenterymesodermMesonephric ductmusclenasalnerveneural foldsneural tubenewbornnormalnotochordoccursoocyteopticoval foramenpalatepharyngeal archplacentaplateposteriorpregnancyprenatalprimordialprimordiumpulmonaryseptumseptum primumsomitesspermspinal cordstagesyncytiotrophoblastsyndrometeratogenicthyroidTransverseUltrasoundumbilical cordurogenitalusuallyuterine tubeuterusveinventralventriclevertebralvesiclevesselsvilliwallWB Saundersyolk saczygote
References to this book
When Did I Begin?: Conception of the Human Individual in History, Philosophy ...
Norman M. Ford
Limited preview - 1991
The Human Embryo Research Debates:Bioethics in the Vortex of Controversy ...
Ronald M. Green
No preview available - 2001
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Bibliographic information
Title The Developing Human: Clinically Oriented Embryology
Developing Human: Clinically Oriented Embryology
AuthorsKeith L. Moore, T. V. N. Persaud
Edition 7, illustrated
Publisher Saunders, 2003
Original from the University of Michigan
Digitized Jul 24, 2008
ISBN 0721694128, 9780721694122
Length 560 pages
SubjectsScience
›
Life Sciences
›
Developmental Biology
Medical / Embryology
Medical / Pathophysiology
Science / Life Sciences / Developmental Biology
Export CitationBiBTeXEndNoteRefMan
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14884 | https://www.whattoexpect.com/first-year/milestones/babinski-reflex-in-babies | First Year > Milestones
What to Know About the Babinski Reflex
by Linda Rodgers
Medically Reviewed by Jesil Pazhayampallil, M.D., F.A.A.P. | November 1, 2021
Latest update: See more
Stocksy
The Babinski reflex, also known as the plantar reflex, helps doctors test a baby’s neurological responses. Here’s everything you need to know.
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In This Article
What is the Babinski reflex?
How do you test for the Babinski reflex in babies?
Is the Babinski reflex normal in babies?
What does an abnormal Babinski reflex mean?
As adorable as they are, just-born babies do many odd things — at least from a new parent’s perspective. But all those strange twitches are signs that your baby’s nervous system is developing just as it should be.
These newborn reflexes, sometimes spontaneous, sometimes in response to a touch or a loud noise, are your baby’s survival mechanisms. And while some may be obvious as to why they help babies thrive, others, like the Babinski reflex, are more subtle.
What is the Babinski reflex?
The Babinksi reflex, also known as the plantar reflex, is an automatic response in your baby’s foot after it’s been stroked from heel to toe. The reflex is named after a French neurologist, who was the first to describe this response at the end of the 19th century.
So what is the Babinski reflex used for? Doctors test all the newborn reflexes at your baby’s first checkup to see how well his nervous system works. The Babinski reflex lets doctors know that the brain is sending signals to the spinal cord, which causes the big toe to flex upward as the other toes fan out when the bottom of the foot is touched.
How long does the Babinski reflex last? Longer than the other newborn reflexes. It appears at birth and can disappear as early as 6 months or as late as 1 to 2 years old, after which the toes curl downward when the pediatrician strokes your toddler’s foot.
The Babinski reflex isn’t the same as the plantar grasp reflex (although Babinski is also known as the plantar reflex, which can be confusing). The plantar grasp reflex is just like the palmar grasp reflex, only with feet: Put your finger under your baby’s toes and they’ll curl over it. With the palmar grasp, those teeny-tiny fingers will grip yours when you touch your baby’s palm.
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How do you test for the Babinski reflex in babies?
Here’s how to check the Babinski reflex: Take your finger and stroke your baby’s foot from heel to toe. As you do so, the big toe will extend upward while the other toes spread out. The foot will probably flex upward too.
Where is the Babinski reflex tested? Well, the first check will happen at the hospital, though your baby’s pediatrician will look for it again no doubt during your baby's first well visits. Your health care provider will use a finger or a thin stick like a tongue depressor.
Is the Babinski reflex normal in babies?
It is super normal in babies, and a sign that the brain and spinal cord (aka the central nervous system) are working together well. Again, in babies and toddlers, a normal Babinski reflex means the big toe flexes upwards while the other toes flare out once the foot is stroked from heel to toe.
But if older children and adults have the same reflex, it could be a sign of something else going on that needs further testing. After age 2, your child’s toes should curl down once the foot is stroked. If they don’t, then the reflex, sometimes also called Babinski’s sign, could mean that something is wrong with the central nervous system.
What does an abnormal Babinski reflex mean?
Most healthy, full-term babies will show signs of the Babinski reflex when you run a finger along the sole of the foot. Even the majority of high-risk infants in the NICU — including premature babies and those with health conditions — have the Babinski reflex, according to researchers.
In babies, an abnormal Babinski reflex would mean that your baby’s foot doesn’t respond at all to being stroked or responds weakly, or one foot responds differently than the other. An abnormal Babinski reflex could indicate that there’s something wrong with the signals the brain is sending to the spinal cord.
Children older than age 2 and adults who still have the Babinski reflex or have an abnormal one may have a problem with the brain and/or spinal cord, including:
A brain tumor or injury
A stroke
Meningitis, an infection in the tissue surrounding the brain and spinal cord
Multiple sclerosis
A spinal cord injury
If your child's Babinski reflex isn't normal, a doctor will order more tests to uncover whatever may be wrong.
Even while your baby is lying still, he’s moving his whole body in small ways, including those newborn reflexes he's born with, that teach him about the world and his place in it. So while you're checking for the Babinski reflex, go ahead and savor your baby’s adorable little feet and tiny toes. Nuzzle them (they’ll never smell this good again!) or try a round of Little Piggy and other classic baby games. All that cuddling and attention will help your baby thrive — and bring you both closer.
From the What to Expect editorial team and Heidi Murkoff, author of What to Expect When You're Expecting. What to Expect follows strict reporting guidelines and uses only credible sources, such as peer-reviewed studies, academic research institutions and highly respected health organizations. Learn how we keep our content accurate and up-to-date by reading our medical review and editorial policy.
View Sources
What to Expect the First Year, 3rd edition, Heidi Murkoff.
WhatToExpect.com, Your Baby's First Checkup, March 2020.
WhatToExpect.com, What to Know if Your Baby Is in the NICU, September 2021.
WhatToExpect.com, What to Know About the Moro Reflex, October 2021.
American Academy of Family Physicians, Newborn Reflexes and Behavior, April 2020.
National Library of Medicine, Medline Plus, Infant Reflexes, October 2019.
KidsHealth From Nemours, Movement, Coordination, and Your Newborn, June 2019.
National Library of Medicine, National Center for Biotechnology Information, Babinski Reflex, August 2021.
National Institutes of Health, National Library of Medicine, Babinksi Reflex, January 2021.
Journal of Clinical Medical Research, Assessment of Primitive Reflexes in High-Risk Infants, December 2011.
American Psychological Association, APA Dictionary of Psychology, 2020.
RegisteredNurseRN.com, How to Assess Newborn Reflexes, July 2020.
American Academy of Pediatrics, Newborn Reflexes, March 2021
Children’s Hospital of Philadelphia, Newborn-Reflexes, 2021
International Journal of Pediatrics, The Grasp Reflex and Moro Reflex in Infants: Hierarchy of Primitive Reflex Responses, June 2012.
Journal of Neurology, Neurosurgery & Psychiatry, The Babinski Sign, October 2002.
National Institutes of Health, National Library of Medicine, Primitive Reflexes, March 2021.
Stanford Children’s Health, Neurological Exam for Children, 2021.
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14885 | https://askfilo.com/user-question-answers-trigonometry/how-do-you-convert-into-radians-35353837373234 | Question asked by Filo student
How do you convert 67.5∘ into π radians?
Views: 5,633 students
Updated on: Sep 24, 2023
Text SolutionText solutionverified iconVerified
Step 1. Recall the conversion factor between degrees and radians:
360∘=2π radians.
Step 2. Convert degrees into radians using the above conversion factor:
67.5∘=3602π⋅67.5 radians.
Step 3. Simplify the expression:
67.5∘=83π radians.
Step 4. The final answer is:
83π radians.
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| Question Text | How do you convert 67.5∘ into π radians? |
| Updated On | Sep 24, 2023 |
| Topic | All Topics |
| Subject | Trigonometry |
| Class | High School |
| Answer Type | Text solution:1 |
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14886 | https://www.youtube.com/watch?v=pwJyNZlEeLE | Bending of Cyclist ‖ Angle with which a cyclist has to bend with the vertical while taking a turn
MAGNUS JEE PHYSICS
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Posted: 25 Sep 2023
⦿ Chapter : Circular Motion ⦿ Topic : Bending of Cyclist ....
A cyclist provides himself the necessary centripetal force by leaning inward on a horizontal track, while going round a curve.
The cyclist leans in the direction of the center of curvature of the curve in bending. He avoids skidding and falling by bending. He does this in order to provide centripetal acceleration.
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Transcript:
bending of a cyclist this is a carb Road this is a carb Road R is the radius of this carb Road and O is the center and a cyclist will be taking a turn around these rough cardboard this is the carb Road along which a cyclist will be taking Ethan and we can draw the will we can draw the will of the cyclist or by cycle this is the wheel of the bicycle we can denote horizontal and vertical Direction and this is the will of the bicycle this is the will of the bicycle the weight of the bicycle that is mg is acting vertically downwards weight of the bicycle that is mg is acting vertically downwards R is the normal reaction acted on the wheel of the bicycle applied by the carb load now we can consider the two rectangular components of normal reaction are R makes an angle Theta with the vertical Direction and vertical component of normal election R is R cos Theta and the rectangular component of normal reaction are acting in the horizontal direction is R sine Theta now this is the direction along which the centripetal force FC is acting centripetal force FC Acts radially inwards now we can denote the imaginary curved line along which the bicycle will be taking a turn with a tangential velocity or translational velocity V is the linear velocity with weeds the bicycle will be trying to take a turn around the curved Road around the curb Road the reaction force under bracket r on the wheels of the bicycle the reaction force under bracket r on the wheel of the bicycle is resolved into two rectangular components is resolved into two rectangular components which are given by R cos Theta and R sine Theta which are given by R cos Theta and R sine Theta in equilibrium R cos Theta in equilibrium R cos Theta will be equal to what and here R sine Theta provides the necessary centripetal force R cos Theta will be equal to mg and R sine Theta provides the necessary centripetal force under bracket if C to the car to the bicycle R sine Theta provides the necessary centripetal force FC to the cyclist for taking a therefore R sine Theta is equal to FC therefore FC divided by m g is equal to R sine Theta divided by R cos Theta if we consider the forces acting in the vertical Direction then R cos Theta will be equal to mg and mg is the weight of the Y cyclist and R sine Theta Supply is the necessary in centripetal force fce for taking it on round the curved Road so therefore 1 by m g into MV Square by R will be equal to sine Theta divided by cos Theta or comma V Square by RG is equal to tan Theta therefore tan Theta is equal to V Square divided by r g tan Theta is equal to V Square divided by r g hence the cyclist has to bend the cyclase has to bin at an angle of at an angle of theta equal 10 inverse V Square divided by RG Theta is equal tan inverse V Square divided by r g inwards inwards from his vertical position inwards from his vertical position comma while turning the Curve Road while turning the carb road so we get tan Theta as equal V Square divided by RG therefore Theta is equal to tan inverse V Square by RG the cyclist has to lean or has to bend at an angle the cyclist has to bend or lean at an angle of theta equal tan inverse V Square by r g inverse from its vertical position while taking a turn |
14887 | https://www.wired.com/story/how-to-solve-a-rubiks-cube-step-by-step/ | How to Solve a Rubik's Cube, Step by Step
Earlier this year, while putting together a video about the world’s fastest solvers of the Rubik’s Cube, I decided to devote some time to learning to solve the classic puzzle myself. Tyson Mao, a cofounder of the World Cube Association, came to WIRED’s offices and spent about an hour teaching me his go-to beginner’s method. Afterwards he told me that, with practice, I could probably get my average solve time down to under a minute and a half. Ninety seconds is not fast by speedcubing standards (the world’s fastest cubers average well below 10 seconds per solve), but Mao said it would be a respectable time for a dabbler such as myself.
I began practicing the next day. My first time solving the cube on my own took me more than 20 minutes. Brutal. But I kept at it: For two weeks I spent at least 20 minutes a day scrambling my cube and solving it the way Mao had taught me. First I memorized a handful of algorithms (cuber lingo for defined sequences of moves known to advance a cube closer to its solved state). Then I practiced performing them faster and more precisely.
By day three I was solving the cube in under four minutes. I broke the two-minute barrier a couple days later, on a cross-country flight to Florida. (Planes are an ideal place to practice cubing.) The improvements came more slowly after that, but within a fortnight I’d lowered my average solve time to a little under 60 seconds.
In the time since we published the video about speedcubing, several viewers requested that WIRED create another video demonstrating the method I used when learning to solve the cube. So we made one! Above you’ll find a visual guide in which I walk you through the same solving method that Mao taught me. Below is a written tutorial that summarizes the points in the video, including the eight steps you’ll follow to solve the cube, an overview of cube notation, and descriptions of the algorithms you’ll need to memorize.
The tutorial below was originally created by Mao, so all credit goes to him. I’ve merely tweaked it for the sake of clarity.
One last thing: While the tutorial can function as a stand-alone document, it’s really intended as a supplement to the video. In time, you might come to rely solely on the written instructions, but don’t be discouraged if you find yourself referencing the video for help—especially when you’re starting out.
Before You Begin
Here are some things you should know about the Rubik’s Cube. Some of these points might strike you as trivial at first, but each affords some insight that will become clearer the more time you spend with the cube.
Cube Notation
Solving the cube will require you to turn its faces. Each face is represented by a letter. The direction of a given rotation is denoted by the presence or absence of a prime (’) symbol.
Right face: RLeft face: LUpward-pointing face: UDownward-pointing face: DFront face: FBack face: B
R, L, U, D, F, or B means to turn the corresponding face 90 degrees clockwise
R’, L’, U’, D’, F’, or B’ means to turn the corresponding face 90 degrees counterclockwise.
R2, L2, U2, D2, F2, or B2 means to turn the corresponding face 180 degrees.
Step One: Make the Daisy
The goal of this step is to place four white edge stickers around the yellow center. When you are finished with this step, the top of your cube should look like this:
Note: It doesn’t matter what color the grey squares are. Two things to keep in mind:
Step Two: Create the White Cross
For each petal on the “daisy,” match the non-white sticker to the center piece of the same color. Once matched, turn the face with the matching center two times. Repeat this process three more times. When you are finished, the bottom face of the cube will have a white cross.
Note: For the rest of the solve, the white cross will be on the bottom. If you ever find the white cross somewhere else, something has gone wrong.
Step Three: Solve the First Layer
Time to learn your first algorithms. The following “trigger moves” are the most basic of the bunch:
Look for white stickers on the top layer that face the sides. (If you find a white sticker on the top face of the cube, or on the bottom layer of the cube pointing outward, we’ll deal with it later.) Each white sticker should be on a corner piece with three stickers. Rotate the top face of the cube so that the sticker beside the white sticker that is also outward facing (i.e., not the sticker on the top) diagonally matches the center of the same color.
Once you’ve paired them, face the color-matched stickers toward you. If the matched sticker in the top layer is right of the center, perform the Right Trigger. If the matched sticker is left of center, perform the Left Trigger.
If you have a white sticker facing the top, position the white sticker over something that is not white (because it will disrupt whatever is underneath), and, depending on if the piece is on the right or on the left, perform the following algorithm:
R, U, R’, R, U, R’
Or
L’, U, L, L’, U, L
If you have an outward facing white sticker in the bottom layer, face it toward you and position the cube so that it is either in the bottom left or bottom right corner of the side facing you, and perform either the left or right trigger, respectively, to relocate it to the top face of the cube.
Step Four: Solve the Middle Layer
Identify edge pieces on the top layer that do not have yellow stickers. (If it has a yellow sticker, it belongs on the top and not in the middle.) Once you find an edge without a yellow sticker, rotate the top face of the cube until the outward facing sticker on that edge piece is directly over the center piece of the same color.
Once it matches, look at the upward-facing sticker on that edge piece. That sticker will match the center on either the left or the right.
If it matches on the right, perform the following algorithm:
U + Right Trigger
Doing so will disturb the first layer. Fix the displaced white corner sticker as you did in step three.
If it matches on the left, perform the following algorithm:
U’ + Left Trigger
Doing so will disturb the first layer. Fix the displaced white corner sticker as you did in step three.
Occasionally you will find no edge pieces in the top layer without yellow stickers but the middle layer is not solved. In such cases, displace them is-matched middle-layer edge piece by performing the left or right trigger. There should now be an edge piece in the top layer without a yellow sticker. Solve for it as described above.
Step Five: Create the Yellow Cross
The goal of this step is to create a yellow cross on the upward-pointing face of the cube. This entire step hinges on the following algorithm:
F U R U’ R’ F’
If your top face has no yellow edge pieces, perform F U R U’ R’ F’. If your top face has two yellow edge pieces such that they form a line with the center yellow piece, orient the cube such that the three yellow stickers form a vertical line and perform F U R U’ R’ F’. If your top face has two yellow edge pieces such that they form a backwards L, rotate the top face of the cube until the edge pieces are at the 12 and 9 positions of a clock and perform F U R U’ R’ F’. At this point, the top face of your cube should resemble a yellow cross.
Step Six: Solve the Yellow FaceThe goal of this step is to completely solve the top face of your cube. When you’re finished, that face should be entirely yellow. For this step, you will use the following algorithm:
R U R’ U R U2 R’
Begin by inspecting the top face of your cube. How many corners have yellow stickers on top?
If you have zero or two, hold the cube so a yellow sticker is in the upper right hand corner of the face in your left hand, i.e. here:
...and perform the algorithm R U R’ U R U2 R’.
If you have one corner with yellow on top, it will look like there’s a fish on the top face of your cube. Rotate that face until the fish is pointing down and to the left, like so:
...and perform the algorithm R U R’ U R U2 R’.
You might have to orient the fish and perform the algorithm one last time. Once you have, the yellow face will be entirely solved.
Step Seven: Position the Corners of the Cube
Time for a new algorithm:
L’ U R U’ L U R’ R U R’ U R U2 R’
The above algorithm swaps corners A and B. Note that the eighth step of the algorithm undoes the seventh. That’s intentional, because it will make memorizing the algorithm easier: Notice that R U R’ U R U2 R’ is the same algorithm you used in step six.
Use this new algorithm to position all four corners in the correct place. If you have to switch two corners diagonally, perform the algorithm once, then reposition and perform it a second time.
Step Eight: Position EdgesThe goal of this step is to cycle the position of the cube’s edge pieces. The following algorithms will cycle the positions of the edge pieces labeled X, Y, and Z in a clockwise or counter-clockwise fashion, respectively:
F2 U R’ L F2 L’ R U F2 (clockwise)
F2 U’ R’ L F2 L’ R U’ F2 (counter-clockwise)
If one face’s edge pieces are already correctly positioned, orient that face away from you and perform whichever algorithm will cycle the remaining edge pieces in the appropriate direction.
If all four edge pieces are misplaced, perform the counterclockwise algorithm once, position the side with the solved corners away from you, and perform it a second time.
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14888 | http://bgillette.com/wp-content/uploads/2014/08/Logic-6.pdf | I TRANsLATING ORD]NARY LANGUAGE INTO CATECORICAL PROPOSITIONS 2A'I I. TRANSLATING ORDINARY LANGUAGE INTO CATEGORICAL PROPOSITIONS - ategorical propositions can be found in everyday life Here is one example: Some footbatt coaches are persons of character \ryho always put their ptayers' hea[th first.
Gregg Easterbrook, "Concussion Hazards N4Lrst Be Addressed" \re have already seenhowlogic canhelp us make sense ofthe claims all around us---ut firstwe need to be able to paraphrase statements in ordinary langu age As we are .11 aware, ordinary language statements canbesubjectto differing interPretations lf :e$'ofthe types that you are most likely to encounter' Missing Plural Nouns -onsider the following statement: Some alcohotics are convicts.
.alegorical ProPosition: Some convicts are alcoholics.
Now consider a second statement: Sorne poljtical parties are disorganized.
Most people would have little difficultyunderstanding this example Onthe surface' ri appea;s t;be a standard-form categorical proposition' But this is deceivins Let's see '., hat happens if we switch the Position of the two terms: Some disorganized are potiticaI parties.
lVe no longer have an acceptable statement The problem is that the word "disor-:anized" is in adjectit e, not a noun. Adiectives are used to modify nouns' and they ,A2 C HAPTER 5 CATEGORlCAL PROPOSlTIONS oun, so that the resulting term will denote a class. For examPle: Sone politicaL parties are disorganized groups The term "disorganized groups" denotes a class of objects lf we now switch the terms, we get this result: Sotre rlisorqanizecl groups are p0iitical Farties' nouns.
Nonstandard Verbs everyday examples ofordr ry languaee statements contain verbs that must be lated into either "are" or "a not " Here are some examPles' 0rdinarY Language Statement: eiitt.,. pro..ttttt1t lhe c0fventio r !'r'ere arrested Standard-Form Trans[ation: aiitLn prot"'tt" at the conYention are peopie who were ar e ted' OrdinarY Language Statement: S0me studerlts vrould nrefer to cheat rathel'than learn the natetraL' Standard-Form Translation: Sone stLldents ''" 1lt"iLt t'ft" wouLd prefer to cheat rather than Learn the naterial OrdinarY Language Statement: Tres0asser5 t'till be prosecuied' Standard-Form Translation: All trespassers are oeople vro lrill face prosecution As these examples illustrate' you must be careful to translate the verb into either "r:' or "al'e not," and you rnu't rrr"k" 'o'" thut translation contaius terms that denote clas s ' ' /L-I TRANSLATING ORDINARY LANGUAGE ]NTO CATEGORICAL PROPOSITIONS \Iany ordinary language statements do not use any form ofthe verb "to be." In these ::.es you have to look closely to grasp the meaning ofthe statement. Here are some '-::.mples: 0rdinary Language Statement: So me assembty required.
Standard-Form Translation: Some parts of this jtem are parts that need assembling.
0rdinary Language Statement: \o pain, no gain.
Standard-Form Translation: \o exercjse routines without physicaI pain are exercise routines offerjng physicat gajn.
\'en short sentences in ordinary language can be misunderstood. The tradeoff of :::rting translations that are lengthyand repetitive is that they offer clarity, aswe shall .:: again in the next chapter.
Singular Propositions -- : examples so farhave contained pluralnouns denoting classes, but it is possible that . :.a ss has only one object. These cases occur in ordinary language when the proposi--::r is singular in naturej that is, something is asserted about a specific persoq place, :: :hing. A singular proposition can normallybe translated into a universal proposition.
i-i:re is one example: 0rdinary Language Statement: Al Gore is a Nobel Prize n inner.
Standard-Form Translation: Al[ persons identicaI to A[ Gore are persons who have won a NobeI Prjze.
The phrase "persons identical to Al Gore" may seem odd, but there is a reason for it.
!-:ce the subject is a single individual (el Gore), the subject term of the translation :st designate a class of objects which happens to have exactly one member. There - -.nly one person identical to Al Gore, and that is Al Gore himself So, the phrase -:.::sons identical to Al Gore" refers to a class ofobjects that has exactly one member, The phrase "persons identical to" is called a parameter. Aparameter must accurately ::rresent the intended meaning ofan ordinarylanguage statement, while at the same -.rre transforming it into a standard-form categorical proposition. Here are some pa-.::leters that you can use to translate singular propositions: ::rsons identicalto :.ings jdentjcal to -'nes id-"nticaL to -{hvays remember that a singular proposition refers to aspecfcperson (place, thing, ::r.Giventhis,thephrase"identicalto"istobetakenliterally.ThereisonlyoneEiflel 203 Singular propositionA proposition that asserts something about a spe crfc pers on, place, or thing places identicalto events identicaI to cases identicaI to 2O4 CHAPTER 5 CATEGOR]CAL PROPOSITIONS Tower, and it is in Paris. Ifyou go to Las Vegas, you will see a structure that looks ?er;y much like theEifrelTower (at one-third the size), but there is only one tower ideniiccl to the Eiffel Tower.
Here are some more singular propositions in ordinarylanguage and their translations: 0rdinary Language Statement: Shane is good aIDDR (DonceDanceRevalutio n) .
Standard-Form Translation: Af[ persons identicat to 5hane are persons good at DDR (Donce DanceRevoLution) .
0rdinary Language Statement: Hugo did not go to Hawaii [ast spring break.
Standard-Form Translation: No persons identicaI to Hugo are persons who went to Hawaij last spring break.
0rdinary Language Statement: tr.ry car i, r loe'5 ga dqe lor rcpo' ..
Standard-Form Transtation: Att thifgs identical to my car are things in Joe's garage for repairs.
0rdinary Language Statement: Leo was itl last night.
Standard-Form Transtation: All tjmes identrcalt0 [ast njght are tin]es that Leo was ill Parameters are used when translating singular proPositions They are not needed when the ordinary language statement has plural nouns.
Adverbs and Pronouns Some ordinary language statements contain adverbs that describe places ot times. For example, in the statement "Wherever there is smokethere is 6re," the word "wherever" is a spafialadverb. Spatial adverbs describewhere something happens Here are some spatial adverss: wherever, et erywhere' anywhere, somewhere' nowhere, upstairs, and underground' In the statement "Whenever you are audited by the IRS, you had better get legal help," the word "whenevet" is a temporal adverb. Temporal adverbs describe when something happens. Here ate some temporal adverbs: wh enever' never' always, anytime, yesterday, ar'd tomorrow.
Translating ordiuary language statements into standard-form categorical proposi-tions using these kinds ofadverbs is relatively straightforward: 0rdinary Language Statement: Wherever there is smoke, there is fire.
Standard-Form Translation: Att pLaces that have smoke are places that have fire.
0rdinary Language Statement: Whenever you are audjted by the IRS, you shoutd get legaI hetp.
Sta s:: 0-.1> I, TRANSLAT]NG ORDINARY LANGUAGE INTO CATEGORiCAL PROPOSITIONS 205 Standard-Form Transtation: Aii,l.., yo, are audited by the IRS are times that you should get legal he[p :rerything Here are translations ofthe last two examples: OrdinarY Language Statement: Whoever took my laptop is in big troubte' Standard-Form Transtation: Ati persons who took my [aptop are persons in big trouble Ordinary Language Statement: What goes around comes around' Standard-torm Translation: All thinqs that go around are things that come around' ''lt Is False That .
Here are some useful negation phrases: It is folse that . . .
It is not the case that .
It is nat true thot - . .
206 CHAPTER 5 CATEGORICAL PROPOSiTIONS Implied Quantifrers As we saw in Chapter 3, some statements in ordinarylanguage imply something with-out actually saying it. Important terms are either left out on purpose or simply over-Iooked. In these cases we have to supply the missing terms. Ifthe missing term is a quantifier word (all, no, some), then our translation into a standard-form categorical proposition must rely on a close reading ofthe intended meaning. Here is one example: Sharks are predators.
The statement connects a species of animals (s/rarks) with a specific characteristic (being a predator). Ls such, it refers to the entire subiect class and can be translated as follows: A[ sharks are predators.
Nowlet's look at another example that uses the same subject (sharks): I l'ere a'e slar(s ir rhe local aquariun.
It is unlikelythat the person makingthe assertion is claimingthat the entire class of sharks is in the local aquarium. Therefore, our translation will haye to use the quanti-fier "some": Some sharks are anjmals in the locaI aquanum.
We had to add the word "animals" because the phrase "in the local aquarium"would not by itselfdesignate a class ofobjects.
Howwould you translate the next statement?
A professor is a human being.
Although the statement contains the phrase "a professor," it appears likely that the assertion is about every professor. It can therefore be translated as follows: A[ professors are human beings.
What about this example?
A professor is not a machine.
This statement also refers to every professor, but it contains the word "not." It is temptinS to translate the statement as follows: Alt professors are not machines.
Tli to rei i In(orrect The correct form ofa universal affirmative categorical proposition is All S rire P, so we cannot add the word "not" using this form, The universal negative form solves our problem, No professors are machjnes. Correct Here is one more example to consider: A professor won the Nobel Prize.
Ea: class.
rioul, lrans \o \o be ti; neqat lhe s; lnLen nie nt s Thr Nor Ora:r ::1;e i ::: :; \c :. --::l I TRANSLATlNG ORDINARY LANGUAGE iNTO CATEGORlCAL PROPOSITIONS 2O7 This statement also contains the Phrase "a professor" but it is unlikely that it is meant to refer to everyProfessor, It can be translated as follows: Some professors are winners of the Nobel Prize' Earlier we had to make the subject term a plural noun in order for it to designate a class. Ofcourse, ifa specific professor had been named (e g , Professor Blake), then we rvould have used the information regarding singular proPositions to get the correct translation, Now try a more comPlex examPle: We wilL not be abie to finish atl the costumes by 5:00' A quick reading might suggest that the quantifier word "all"means fhat rhis should be trinslated as a-onilre.sal affirmatire ProPosition However, theword "not" indicates negation. Combining these two words gives us th phrase 'not all " It is unlikely that thJspeaker is claimi.rg that no costumes will be nished by 5:00 (lfthis had been inten,Ced, then we would expect the statement to be "We will not be able to Enish any costume by 5:00.") Therefore, the correct quantifier is "some," and the translated state-ment must include the word "not": Some costumes are not costumes that wilt be tinished by 5:00' This example illustrates why ordinary language statements often require a careful :eading in order to understand the meaning and to arrive at a correct translation' Nonstandard Quantifiers Ordinarylanguage statements might contain quantifiers that are nonstandard, because :hey are not one ofthe following: all, ro, or soze. Here is an example: Not every jnvestment banker is a crook.
ln this statementthe nonstandard quantifier "not every" probablymeans atleast one :nvestment banker is not a crook, Given this interPretation, the translation would be :he following: S0me investment bankers are rlot crooks Notice that we once againhad to change the subject and predicate terms into plural :'louns.
Here are some nonstandard quantifie rsi any' many, most, aJew, one, several, and not :r'er;I,. Let's take one from the list and look at another example: Not every novel about romance is interesting' In this statement the nonstandard quantifier "not every" means that there are some :ovels about romance that are not interesting Given this interPretation, the transla-:ron would be the following: Some novels about romance are not interestjng novels' 2O8 CHAPTER 5 CATEGORICAL PROPOS]T]ONS Here is another statement in ordinary language that uses a nonstandard quantifier: A few movies at the natl are worth watching.
Here the quantifier "a few" is likely to mean that at least one movie at the mall is worth watching. The translation would be the following: Some movies at the mall are movies v,/orth watching Since the phrase "worth watching" does not by itselfdesignate a class, we had to add the term "movies" to it.
Conditional Statements We have already encountered conditional statements when we looked at existential import. The A-proposition 'All scientists are PeoPle trained in m matics" can be translated as "I/a person is a scientist, fhe, that person is trained in hem atics." The E-proposition "No slackers are reliable workers" can be translated as "I/a person is a slacker, fhen that person is not a reliable worker." These translations are a result ofthe Boolean interpretation of universal cateSorical ProPositions.
As you might know from Chapter 3, the part ofthe conditional statement that fol-Iows the word "if" is called the anfecedenl and the part that follows the wold "then" is called lhe consequerr. Here are some simple examples: 0rdinary Language Statement: If a persor has $10 jn her crecking account, thef she is not rich Standard-Form Transtation: No persons having $t0 ln therr chec(inq accourlt are rich persons 0rdinary Language Statement: If a salesperson calts on the phone, ther I Just hang up.
Standard-torm Transtation: All calls from salespersons are calls where I hang up.
Sometimes ordinary Ianguage statemenIs do not have the word "if" at the begin-ning. When this occurs, we simply reposition the aPPropriale Part so the antecedent comes first: 0rdinary Language Statement: Pizza is a heatthy meal if jt has vegetable toppings Standard-Form Translation: All pizzas with veqetable toppings are healtry meals 0rdinary Language Statement: A dog is rot dargerous jf it has been ',r'e[[ trained.
Standard-Form Transtation: No vielltrained dogs are dangerous anima[s.
The conditional statement "If your cup of coffee is not Perfect, then you are not drinking a cup ofBigbucks coffee" poses a new kind ofproblem for translation To :iiiter: rall is .. add tn!ial r-, Fe 'lle :isa : rhe ,: iO--ats .!1: arl ] TRANSLATING ORDINARY LANGUAGE INTO CATEGORICAL PROPOSlTIONS 209 rsist us, we need to introduce transposition.lhis rule is a two-steP Procedure, First, ';: srvitch the positions ofthe antecedent and the consequent, and second, we negate :.::h of them. Let's work through it step by step and make any additional changes in - ..rding as we go to caPture the meaning ofthe statement: First Step: If you are not drjnking a cup of Bigbucks coffee. then your cup of coffee is not perfect.
Second Step: If you are drinking a cup of Bigbucks coffee, then your cup of coffee is perfect.
FinaI Translation: Atl cups of Bigbucks coffee are perfect cups of coffee.
\ow let's look at an example that is a little more challenging The conditional state-:ent "Ifmurderers do not get punished, then theydo not stop theirbehavior" requires : -.it ofrewritingto capture the meaningin standard-form cateSorical ProPosition. As --:lbre, we will take it step by step and apply the rule oftransPosition: First Step: If nurderers do not stop their behavjor, then murderers do not get punished Second Step: If n'rurderers stop their behavior, then murderers get punished.
tinaI Translation: Al[ murderers who have stopped their behavior are murderers who have been punished.
In orderto translate the statement'A citizen cannot be President unless the citizen is :: least 35 years old," we need to understand how the word "unless" gets translated.In rrost statements, the word "unless" means "ifnot." Substituting this into the original ::atement gives us this result, 'A citizen cannotbe president ifthe citizen is not at least 1-i years old." Next, we can place the antecedent at the beginning ofthe statement: "If ::re citizen is not at least 35 years old, then a citizen cannot be president." We are now ::1 a position to apply the two-step rule oftransposition: If a citizen can be presideft, then the citizen is at least 35 years old.
The Iast step completes the translationinto a standard-form categorical proposition: Alt cjtizens that can be president are citizens at least 35 years otd.
Exclusive Propositions Suppose you hear this announcement over a loudspeaker: 0n[y persons with tickets can enter the arena-The announcement means that admission into the arena is limited to those holding tickets. Therefore, anyone who does not have a ticket is excluded fuom efiering the arena, and we call th is an exclusiveproposifior. Anotherway ofsaying this is "Ifa Person 210 CHAPTTR 5 CAT E GO R]CA L PROPOSITIONS does not have a ticket, then that person cannot enter the arena." Applying transposi-tion to this statement, we get: If a person cdn enter the arena, then that person has a ticket.
This statement can nowbe translated into a standard-form categorical proposition: At[ persons who car] enter the arena are persons that have tickets.
Here are some other words lhat indicate an exclusive proposition: n0re but, solely, alone, and none excepf, Let's take the first one from the list and analyze a statement that contains the words "none but": None but studerts can see the movie for free.
Accordingto the statement, anyone who is not a student is excluded from seeing the movie for free. This can be rewritten as "Ifa person is not a student, then that person cannot see the moyie for free." Applying transposition to this statement we get: If a person can see the movie for free, thef that person is a student.
This statement can now be translated into a standard-form categorical proposition: A[[ persons who can see the movie for free are students.
Some ordinary language statements do not have the exclusive term at the beginning.
For example, "Lottery winners get lucky only once in their lives." In these cases, we have to rewrite the telms in order to designate the correct classes: All lottery r,vinners are persons who get [uc<y once jn their ljves.
"The Only" Although the words "only" and "the only" seem very much alike, they sometimes re-quire different kinds oftranslations. For example, the statement "The only true friends are people who want nothing from you" can be directly translated as 'A11 true friends are people who want nothing from you." However, if the words "the only" occurs in a diferent part of a statement/ then you rewrite the statement by placing it and the phrase following it at the beginning. Here is an example: Andrord phones are the onty phones jmported by her company The frrst step is to put "the only" phrase at the beginning: "The oniyphones imported by her company are Android phones." The hnal step is the translation into a standard-form categorical proposition: AIL p rones inrported by her company are Android phones.
Propositions Requiring Two Translations The examples so far could be translated as single statements. However, some state-ments need to be translated into compo&rd statementsr containing the word "and." For example, propositions that take the form 'Al1 except S are P" and 'All but S are P" are called exceptiue proposiflors, Here is one exceptive propositionr 'All except those under I, TRANSLATING ORD]NARY LANGUAGE INTO CATEGORICAL PROPOSITIONS 21 are allowed to gamble in Las Vegas." The meaning ofthe statement is quite clear: If you are under 21 you cannot gamble, and ifyou are 21 or older you can, In other words, the statement relates the predicate to both the class designated by subiect term afld to its complement. Hence the comPlete translationwill result in a comPound statement: No under-21 pers0ns are persons allowed to gamble in Las Vegas, and alt non under-21 persons are persons atlowed to gambte in Las Vegas Here is another example: Everyone but gamblers steep wet[ at night TransIation: No gamblers are people who steep wetl at night, and a[[ non-gamblers are people wl-o :leep we.[ at nighL.
Knowing the contextinwhich ordinary language statements occur canhelp in mak-ing correct translations. Whenwe have a conversation, we can ask questions to clear up any ambiguity. This option is obviously not available when we are reading something and the author is not present. When in doubt, it is befter to do more than less. In other rvords, ifthere are two reasonable interpretations ofthe meaning ofa statement, then you had best work out the details ofboth. For example, suppose you read the follow-ing: "The heavy snowfall affected the turnout. Few registered voters went to the polls today." Clearly, some registered voters went to the polls and some didnt. This can be translated as a comPound statement: Some regjstered voters are persons who went t0 the p0lts today, and some registered voters are not persons who went to the polls today.
Earlier, the nonstandard quantifier "a few" was translated as a single I-ProPosition ('A few movies at the mall are worth watching" was translated as "Some movies at the mall are movies worth watching.") However, sometimes "a few" should be translated as a compound statement. Again, the context is your best guide to which translation is appropriate.
Sometimes we should translate an exclusive ProPosition containing "only" as a com-pound statement. For examPle, the statement"Only Carlydesigned the weddingSown" makes two assertions. First, Carlydesigned theweddinggowry and second, no one else did. Also, since the statement asserts something about a sPecific person (an individual), our translation has to take that into account: At[ persons jdenticat to Carly are persons who designed the wedding gown.
and a[t persons who designed the wedding gown are persons identicaI to Carty.
We get the same results for the statement "The only person who designed the wed-ding gown is Carly." In this case, the statement is equivalent to "Only Carly designed the wedding gown," and therefore, it gets the same translation.
Here is one more example: Barack 0bama alone is the forty-fourth president of the United States.
21,1 ,12 CHAPTTR 5 CAT E GORlCAL PROPOSIT]ONS Thrs example contains two references The frrst is to an individual (Barack Obama)' and the second is to an elected office We can translate the statement as follows: president of the United States ate pers0ns idenlical lo Barack 0bama 'Iranslations lnto oPositions often require close and carelul reading, but the chance of misunderstanding lt makes us aware of t in ordrnary language, and it makes our spoken and written communication more Preclse' Translate the following ordinary Ianguage statements into standard-form categorical propositlons.
1. An apple is in the refrigerator' Answer: \ome aPPles are items in tlte refrigerator' ;i;l;;sh ;" ',i :ment is referring to a irticular appl: :l: ':: -'1, '"-e" is appro-prr"-,-" ,"t ,frt, ,r""rlation because it"has b n stipulated that it means "At least one " 2. Any medrcal doctor is well-educated' 3. No insects sing' 4. A flower is a Plant' 5. All haPPY PeoPle dance 6. Some bears hibernate 7. Some cars don't Pollute' 8. Amango is not avegetable' 9. It is not the case that every novel is a satire' 10. Every office worker is under pressure to perform' 11. A tsunami is dangerous' 12. Some PeoPle don t iaYwalk' 13. Not everyfinal exam in calculus is a challenging test 14. Every oPera is easy to understand' 15. Not everY dog is friendlY' 16. Any company that introduces green technologywill succeed' 17. Young children are not Protected from the dangers ofwar' 18. Ocean levels rise whenever glaciers melt I 214 CHAPTER 5 CATEGORICAL PROPOSITIONS 5l . Not all so.rp oper.r' are boring 52. Magicians are the only people capable ofkeeping a secret' 53. Whatever in.rprovement is made to the gas engir.re decreases our need for oil 54. Beautyis not skin deeP.
55. A pr actical joke rs not funny ifrt harms someone' 56. All sharks hunt 5 /. 5ome PeoPte oon t Do\vr' S8. Nol er ery computer is erPcnsive' 59. Most smokers ivish they could quit' 60. A11 good things nust come to an end 61. Beliefs worth having must withstand doubt 62. lfsomething is worth having, then it's B'orlh struggling for' 63. Fair-weather friends are not trustworthy' 64. Not all that glitters is gold.
65. l-r er1 end ing is a new begin rt ing 66. Whoever saves even one life saves the entire world' 67. The enemy ofmy enemy is my friend.
68. Evelythir.rg old is new again 69. It is false thatpeople over 30years ofage are not to be trusted' 70. Two snowflakes are neverthe same.
71. Whoever controls the media, controls the mind' Jin \'lorrison, quoted in Tdlli,g ItLte lt Is bv PrdlBo dcLl 72. Every unhappy family is unhappy in its own way' Leo Tolsioy, Atrtr, Ld,.,i,d 73. Whoever is wrnning at the moment will always seenl to be invincible' iieorge orwell, 7re or nellRca'ier.
74. Ifyou tell the truth, you don't have to remember anything' Nlark Tf iilNol"oo& 75. Whoever undertakes to set himself up as a judge in Truth and Knowledge is shipwrecked by the laughter ofthe gods' Alber! Einstern, quoted i nTte Ptu|"tat) Conpr i'11 to Mdth'nnti's |
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Factor Theorem: All You Need to Know
Updated: February 1, 2023
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Category: Advance Math
Update: This article was last updated on 3rd February 2025 to reflect the accuracy and up-to-date information on the page.
Moonpreneur
February 1, 2023
, Advance Math
Factoring equations can be tricky and challenging, especially when your child needs to learn all the methods to solve them – and let’s face it, memorizing long formulas isn’t often their priority.
But don’t worry! With the Factor Theorem, you’ll easily understand how best to approach these complex problems. In this blog post, we’ll discuss what exactly the Factor Theorem has to offer and why it is necessary for simplifying equations so that you and your child can gain a better understanding of this crucial mathematical concept.
What is a Factor Theorem?
The factor theorem states that if a polynomial p(x) can be divided by (x-a) with no remainder, then p(a) = 0. Conversely, if p(a) = 0, then (x-a) is a factor of p(x). Here, the Factor theorem provides a way to find the roots, or x-values, that make p(x) equal to zero.
Proof of Factor Theorem:
The proof of the factor theorem involves showing that if p(x) is divisible by (x-a), then p(a) = 0, and conversely, if p(a) = 0, then (x-a) is a factor of p(x).
To show that if p(x) is divisible by (x-a), then p(a) = 0, consider the polynomial division:
p(x) = (x-a) q(x) + r, where q(x) and r are polynomials, and the degree of r is less than the degree of (x-a), which is 1.
Setting x = a in the equation above, we get:
p(a) = (a-a) q(a) + r = r.
Since the degree of r is less than the degree of (x-a), we have r = 0. Hence, p(a) = 0.
To show that if p(a) = 0, then (x-a) is a factor of p(x), consider the polynomial:
p(x) – p(a) = (x-a) q(x) for some polynomial q(x).
Since p(a) = 0, we have:
p(x) = (x-a) q(x).
This shows that (x-a) is a factor of p(x).
Therefore, the factor theorem is proven.
Example of Factor Theorem: Factorize y2 – 5y + 6
Let p(y) = y2 – 5y + 6.
p(y) = (y – a)(y – b)
= y2 – by – ay + ab
On comparing the constants, we get ab = 6.
Next, the factors of 6 are 1, 2, and 3.
Now, p(2) = 22– (5 × 2) + 6 = 4 – 10 + 6 = 0.
So, (y – 2) is a factor of p(y).
Also, p(3) = 32 – (5 × 3) + 6 = 9 – 15 + 6 = 0.
So, (y – 3) is also a factor of y2 – 5y + 6.
Therefore, y2 – 5y + 6 = (y – 2)(y – 3)
Recommended Reading: Developing a Deeper Understanding of Math Concepts
Proof by Remainder Theorem:
We can also prove the factor theorem using the remainder theorem. The remainder theorem states that for a polynomial p(x) and a constant a, the remainder when p(x) is divided by (x-a) is equal to p(a).
Using this result, we can prove the factor theorem as follows:
If p(x) is divisible by (x-a), then the remainder when p(x) is divided by (x-a) is 0. Therefore, p(a) = 0.
Conversely, if p(a) = 0, then the remainder when p(x) is divided by (x-a) is 0. This means that (x-a) is a factor of p(x).
Therefore, the factor theorem is proven using the remainder theorem.
Recommended Reading: Vedic Math for Kids: Why It Is Important
How to Use Factor Theorem
Write the polynomial: Write the polynomial that you want to find the roots of or the polynomial that you want to factor.
Choose a value of “a”: Choose a constant “a” to test as a possible root of the polynomial.
Evaluate p(a): Substitute the value of “a” into the polynomial and evaluate the expression to get p(a).
Check if p(a) = 0: If p(a) = 0, then (x-a) is a factor of the polynomial.
Factor the polynomial: If (x-a) is a factor of the polynomial, you can use polynomial division to factor the polynomial. Divide the polynomial by (x-a) and write the result in the form p(x) = (x-a) q(x) + r, where q(x) is the quotient and r is the remainder. If r = 0, then (x-a) is a factor of the polynomial with no remainder.
Repeat steps 2-5: Repeat steps 2-5 with different values of “a” until you have found all the roots of the polynomial.
Construct the factors: Use the roots that you found in step 6 to construct the factors of the polynomial. The polynomial can be written as the product of the factors.
In summary, the factor theorem provides a way to find the roots of a polynomial and to factor the polynomial. The process involves evaluating the polynomial at a constant “a,” checking if the result equals zero, and if so, using polynomial division to factor the polynomial.
Recommended Reading: Top 5 Math Project Ideas for Kids
Why Do We Use Factor Theorem
The factor theorem is used for several reasons:
Finding roots of polynomials: The factor theorem provides a method for finding the roots, or x-values, that make a polynomial equal to zero.
Factoring polynomials: The factor theorem can be used to factor polynomials by finding the roots of the polynomial and then constructing the corresponding factors.
Simplifying polynomials: Through factoring polynomials by the factor theorem, one can reduce the expression and make it easier to manipulate.
Solving polynomial equations: The factor theorem can be employed to solve polynomial equations by factoring the polynomial from the equation on one side and finding the roots for the solution.
In other words, a factor theorem is one of the algebra tools which may be applied for finding the roots of polynomials, factoring polynomials, simplifying expressions, and solving polynomial equations.
Recommended Reading: Is Advanced Calculus The Hardest Math Class In High School? A Student’s Perspective
What are the other methods to find the factors of polynomials?
Synthetic Division: It is a division method of a polynomial by a linear factor which is used in order to find the factors of the polynomial.
Long Division: This is one way of doing division of polynomials by polynomials. These are used in finding the factors of a polynomial.
Rational Root Theorem: This is a theorem which says that when a polynomial has a rational root, then such a root has to divide both the leading coefficient and the constant term of the polynomial. A good theorem which limits the list of possible candidates for roots.
Descartes‘ Rule of Signs: According to this rule, the number of positive roots of a polynomial is less than or equal to the number of sign changes in the coefficients of the polynomial.
Factoring by Grouping: This is factoring a polynomial by grouping its terms together, then using the distributive property to factor terms.
Special Factoring Techniques There are several special factoring techniques. These include substitution, grouping, and sum and difference of cubes.
In a nutshell, the Factor Theorem is a very important mathematical concept that will help find the roots of polynomials and factor them.
The Factor Theorem can be proven by polynomial division or the remainder theorem. But, due to its effectiveness in solving complex polynomial problems, the Factor Theorem is very commonly applied
Want to get your child excited about math and sharpen his math skills?
Moonpreneur‘s online math curriculum is unique since it lets children understand their math skills through hands-on lessons, assists them in building real-life applications and excites them toward learning math. You can choose our Advanced Math or Vedic Math+Mental Math courses. Our Math Quiz for grades 3rd, 4th, 5th, and 6th help in further exciting and engaging in mathematics with hands-on lessons.
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Aditya Prakhar Singh
Aditya Prakhar Singh has been in the EdTech industry and has knowledge in coding, SEO, marketing, gaming, and more. He loves to write about technology trends.
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Leon
2 years ago
What distinguishes the Factor Theorem from other methods of factoring polynomials, such as synthetic division or long division?
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Marie
2 years ago
Reply to Leon
The Factor Theorem uniquely identifies factors by determining if a specific value is a root, directly leading to polynomial factorization, unlike synthetic or long division, which systematically break down polynomials without leveraging root relationships.
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Lukas
1 year ago
Can anyone demonstrate how the Factor Theorem applies in fields beyond mathematics, such as computer science, economics, or physics?
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Cora
1 year ago
Reply to Lukas
In computer science, the Factor Theorem aids in error detection and correction codes. In economics, it assists in modeling market behaviors. In physics, it underlies understanding complex waveforms and signal analysis.
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Unit 2: Roots and Radicals
MTH 65: Algebra 2
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Fri, 29 Aug 2025 03:15:02 GMT
Section 2.2: Simplify Square Roots
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Joshua Halpern
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6. Section 2.2: Simplify Square Roots
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Section 2.2: Simplify Square Roots
Last updated Aug 29, 2025
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Section 2.1P: Practice
Section 2.2P: Practice
Page ID 192805
Math Department
Clackamas Community College
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Be Prepared
2. Learning Objectives
1. Motivating Problem
2. Fun Fact
3. The Goal
Definition: SIMPLIFIED SQUARE ROOT
Use the Product Property to Simplify Square Roots
Definition: PRODUCT PROPERTY OF SQUARE ROOTS
Example 1
Solution
Try It 1
Definition: SIMPLIFY A SQUARE ROOT USING THE PRODUCT PROPERTY.
Example 2
Solution
Try It 2
Example 3
Solution
Try It 3
Example 4
Solution
Try It 4
Example 5
Solution
Try It 5
Example 6
Solution
Try It 6
Example 7
Solution
Try It 7
Example 8
Solution
Try It 8
Use the Quotient Property to Simplify Square Roots
Example 9
Solution
Try It 9
Example 10
Solution
Try It 10
Example 11
Solution
Try It 11
Example 12
Solution
Try It 12
Definition: QUOTIENT PROPERTY OF SQUARE ROOTS
Example 13
Solution
Try It 13
Example 14
Solution
Try It 14
Definition: SIMPLIFY A SQUARE ROOT USING THE QUOTIENT PROPERTY.
Example 15
Solution
Try It 15
Example 16
Solution
Try It 16
Example 17
Solution
Try It 17
Be Prepared
We will rely heavily on these skills throughout this section.
Find the prime factorization of 36
Identify: Is 49 a perfect square? What about 48?
Simplify −25
Multiply: 3⋅2⋅5
Learning Objectives
Motivating Problem
Melissa accidentally drops a pair of sunglasses from the top of a roller coaster, 64 feet above the ground. We will need to simplify 64 16 to determine the number of seconds it takes for the sunglasses to reach the ground.
Fun Fact
The ancient Babylonians developed a method for estimating square roots more than 4,000 years ago. Clay tablets show they could estimate roots like 2 accurately to several decimal places—without calculators!
The Goal
In this section, we’ll learn how to simplify square roots that are not perfect squares by factoring out perfect square factors. We’ll break down radicals using prime factorization and simplify expressions to a cleaner, more usable form.
In the last section, we estimated the square root of a number between two consecutive whole numbers. We can say that 50 is between 7 and 8. This is fairly easy to do when the numbers are small enough to use our table of square root values.
But what if we want to estimate 500? If we simplify the square root first, we’ll be able to estimate it easily. There are other reasons, too, to simplify square roots as you’ll see later in this chapter.
A square root is considered simplified if its radicand contains no perfect square factors.
Definition: SIMPLIFIED SQUARE ROOT
a is considered simplified if a has no perfect square factors.
So 31 is simplified. But 32 is not simplified, because 16 is a perfect square factor of 32.
Use the Product Property to Simplify Square Roots
The properties we will use to simplify expressions with square roots are similar to the properties of exponents. We know that (ab)m=a mb m. The corresponding property of square roots says that ab=a·b.
Definition: PRODUCT PROPERTY OF SQUARE ROOTS
If a, b are non-negative real numbers, then ab=a·b.
We use the Product Property of Square Roots to remove all perfect square factors from a radical.
Example 1
Simplify: 50.
Solution
Try It 1
Simplify: 48.
Answer
43
Notice in the previous example that the simplified form of 50 is 52, which is the product of an integer and a square root. We always write the integer in front of the square root.
Definition: SIMPLIFY A SQUARE ROOT USING THE PRODUCT PROPERTY.
Find the largest perfect square factor of the radicand. Rewrite the radicand as a product using the perfect-square factor.
Use the product rule to rewrite the radical as the product of two radicals.
Simplify the square root of the perfect square.
Example 2
Simplify: 500.
Solution
500 Rewrite the radicand as a product using the largest perfect square factor 100·5 Rewrite the radical as the product of two radicals 100·5 Simplify 105
Try It 2
Simplify:432.
Answer
123
We could use the simplified form 105 to estimate 500. We know 5 is between 2 and 3, and 500 is 105. So 500 is between 20 and 30.
The next example is similar to the previous ones, but with variables.
Example 3
Simplify: x 3.
Solution
x 3 Rewrite the radicand as a product using the largest perfect square factor x 2·x Rewrite the radical as the product of two radicals x 2·x Simplify xx
Try It 3
Simplify:b 5.
Answer
b 2b
We follow the same procedure when there is a coefficient in the radical, too.
Example 4
Simplify: 25y 5.
Solution
25y 5 Rewrite the radicand as a product using the largest perfect square factor.25y 4·y Rewrite the radical as the product of two radicals.25y 4·y Simplify.5y 2y
Try It 4
Simplify: 16x 7.
Answer
4x 3x
In the next example, both the constant and the variable have perfect square factors.
Example 5
Simplify: 72n 7.
Solution
72n 7 Rewrite the radicand as a product using the largest perfect square factor.36n 6·2n Rewrite the radical as the product of two radicals.36n 6·2n Simplify.6n 32n
Try It 5
Simplify: 32y 5.
Answer
4y 22y
Example 6
Simplify: 63u 3v 5.
Solution
63u 3v 5 Rewrite the radicand as a product using the largest perfect square factor.9u 2v 4·7uv Rewrite the radical as the product of two radicals.9u 2v 4·7uv Simplify.3uv 27uv
Try It 6
Simplify: 98a 7b 5.
Answer
7a 3b 22ab
We have seen how to use the Order of Operations to simplify some expressions with radicals. To simplify 25+144, we must simplify each square root separately first, then add to get the sum of 17.
The expression 17+7 cannot be simplified—to begin, we’d need to simplify each square root, but neither 17 nor 7 contains a perfect square factor.
In the next example, we have the sum of an integer and a square root. We simplify the square root, but cannot add the resulting expression to the integer.
Example 7
Simplify: 3+32.
Solution
3+32 Rewrite the radicand as a product using the largest perfect square factor.3+16·2 Rewrite the radical as the product of two radicals.3+16·2 Simplify.3+42
The terms are not like, and so we cannot add them. Trying to add an integer and a radical is like trying to add an integer and a variable—they are not like terms!
Try It 7
Simplify: 5+75.
Answer
5+53
The next example includes a fraction with a radical in the numerator. Remember that in order to simplify a fraction you need a common factor in the numerator and denominator.
Example 8
Simplify: 4−48 2.
Solution
4−48 2 Rewrite the radicand as a product using thelargest perfect square factor.4−16·3 2 Rewrite the radical as the product of two radicals.4−16·3 2 Simplify.4−43 2 Factor the common factor from thenumerator.4(1−3)2 Remove the common factor, 2, from thenumerator and denominator.2(1−3)
Try It 8
Simplify: 10−75 5.
Answer
2−3
Use the Quotient Property to Simplify Square Roots
Whenever you have to simplify a square root, the first step you should take is to determine whether the radicand is a perfect square. A perfect square fraction is a fraction in which both the numerator and the denominator are perfect squares.
Example 9
Simplify: 9 64.
Solution
9 64 Since(3 8)2 3 8
Try It 9
Simplify: 25 16.
Answer
5 4
If the numerator and denominator have any common factors, remove them. You may find a perfect square fraction!
Example 10
Simplify: 45 80.
Solution
45 80 Simplify inside the radical first. Rewrite the expression to show the common factors of the numerator and denominator.5·9 5·16 Simplify the fraction by removing common factors.9 16 Simplify.(3 4)2=9 16 3 4
Try It 10
Simplify: 98 162.
Answer
7 9
In the last example, our first step was to simplify the fraction under the radical by removing common factors. In the next example, we will use the Quotient Property to simplify expressions under radicals. We divide the like bases by subtracting their exponents, a m a n=a m−n, a≠0.
Example 11
Simplify: m 6 m 4.
Solution
m 6 m 4 Simplify the fraction inside the radical first m 2 Divide the like bases by subtracting the exponents.Simplify.m
Try It 11
Simplify: x 14 x 10.
Answer
x 2
Example 12
Simplify: 48p 7 3p 3.
Solution
48p 7 3p 3 Simplify the fraction inside the radical first.16p 4 Simplify.4p 2
Try It 12
Simplify: 75x 5 3x.
Answer
5x 2
Remember the Quotient to a Power Property? It stated that we can raise a fraction to a power by raising the numerator and denominator to the power separately.
(a b)m=a m b m, b≠0
We can use a similar property to simplify the square root of a fraction. After removing all common factors from the numerator and denominator, if the fraction is not a perfect square, we simplify the numerator and denominator separately.
Definition: QUOTIENT PROPERTY OF SQUARE ROOTS
If a, b are non-negative real numbers and b≠0, then
a b=a b
Example 13
Simplify: 21 64.
Solution
21 64 We cannot simplify the fraction inside the radical. Rewrite using the quotient property.21 64 Simplify the square root of 64. The numerator cannot be simplified.21 8
Try It 13
Simplify: 19 49.
Answer
19 7
How to Use the Quotient Property to Simplify a Square Root.
Example 14
Simplify: 27m 3 196.
Solution
Try It 14
Simplify: 48x 5 100
Answer
2x 23x 5
Definition: SIMPLIFY A SQUARE ROOT USING THE QUOTIENT PROPERTY.
Simplify the fraction in the radicand, if possible.
Use the Quotient Property to rewrite the radical as the quotient of two radicals.
Simplify the radicals in the numerator and the denominator.
Example 15
Simplify: 45x 5 y 4.
Solution
45x 5 y 4 We cannot simplify the fraction inside the radical. Rewrite using the quotient property.45x 5 y 4 Simplify the radicals in the numerator and the denominator.9x 45x y 2 Simplify.3x 25x y 2
Try It 15
Simplify: 80m 3 n 6
Answer
4m5m n 3
Be sure to simplify the fraction in the radicand first, if possible.
Example 16
Simplify: 81d 9 25d 4.
Solution
81d 9 25d 4 Simplify the fraction in the radicand.81d 5 25 Rewrite using the quotient property.81d 5 25 Simplify the radicals in the numerator and the denominator.81d 4d 5 Simplify.9d 2d 5
Try It 16
Simplify: 64x 7 9x 3.
Answer
8x 2 3
Example 17
Simplify: 18p 5q 7 32pq 2.
Solution
18p 5q 7 32pq 2 Simplify the fraction in the radicand.9p 4q 5 16 Rewrite using the quotient property.9p 4q 5 16 Simplify the radicals in the numerator and the denominator.9p 4q 4q 4 Simplify.3p 2q 2q 4
Try It 17
Simplify: 50x 5y 3 72x 4y.
Answer
5yx 6
This page titled Section 2.2: Simplify Square Roots is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Math Department via source content that was edited to the style and standards of the LibreTexts platform.
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Problem in proper divisor algo
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I wrote two algos to get the sum of the proper divisors of a given number,to find perfect number or abundant number.
```c
long sum_divisors_1(int a)
{
int i, t;
long sum = 1;
for (i = 2, t = sqrt(a); i < t + 1; i++) {
if (a % i == 0) {
sum += i;
sum += a / i;
}
}
if (a % t == 0)
sum -= t;
return sum;
}
long sum_divisors_2(int a)
{
int i, sum;
sum = 0;
for (i = 1; i < (int) (a / 2 + 1); i++) {
if (a % i == 0)
sum += i;
}
return sum;
}
```
And I think they are both correct and the first one is faster. But I can only get the correct result from the second algo. Other parts of the code are the same.
Any suggestions? And how the proper divisors are found in real industrial programming?
Thanks in advance.
c
algorithm
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edited Jan 22, 2015 at 20:10
Zero Piraeus
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asked Sep 2, 2011 at 19:07
darkjhdarkjh
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2
Do you consider a itself as a divisor ?Yochai Timmer –Yochai Timmer 2011-09-02 19:20:17 +00:00 Commented Sep 2, 2011 at 19:20
You think both are correct, and meanwhile, that only the second gives the correct result? Either- or.user unknown –user unknown 2011-09-03 11:57:34 +00:00 Commented Sep 3, 2011 at 11:57
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Your problem lies here:
c
if (a % t == 0)
sum -= t;
Since you're casting t to an int from a floating point, it will round down to an integer value. This also assumes that t is the actual square root when it isn't. This will evaluate to true when a number has factors x&x+1 (the unit test I posted as well fails when i = 6 because it's square root is 2.45 and 2 is a factor).
The check really should be:
c
if (tt == a)
sum -= t;
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answered Sep 2, 2011 at 19:42
Austin SalonenAustin Salonen
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This is an old question but I was browsing.
There's a much faster algorithm to find the sum of proper divisors.
Find the prime factors of a number using Sieve of Eratosthenes (or Atkin). With wheel factorisation the first 1m prime numbers will take maybe 30ms.
Then the sum of all divisors is
```c
For each n
sum += (n ^ (m+1) - 1) / (n-1)
```
where n is the factor, m is the power of that factor.
Eg for 220 2^2 5 11 are the factors
So it's sum of
c
2 ^ (2+1) - 1 / 1
5 ^ (1+1) - 1 / 4
11 ^ (1+1) - 1 / 10
= 7 6 12
= 504
This is the sum of ALL divisors, so just subtract N
c
504-220 = 284
This should be a lot faster than trying all the numbers, especially if you precalculate the sieve and reuse it.
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answered Apr 19, 2013 at 10:22
Delta_ForeDelta_Fore
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Here's a simple unit test I wrote in C# that will quickly invalidate #1 given #2 is correct:
c
for(int i = 4; i < 28124; i++)
{
Assert.AreEqual(sum_divisors_2(i), sum_divisors_1(i), "Failed when i = {0}", i);
}
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edited Sep 2, 2011 at 19:47
community wiki
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Templatetypedef solved your problem; however the fastest possible way to compute the prime factors is to precompute all the prime factors up to sqrt(MAX_INT) with Eratostene sieve, store it into an array and then use it to factorize the number a. This is really really really much faster.
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answered Sep 2, 2011 at 19:48
SimoneSimone
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Complete Guide to Integers on ACT Math (Advanced)
Posted by Courtney Montgomery
ACT Math
Integers, integers, integers (oh, my)! You've already read up on your basic ACT integers and now you're hankering to tackle the heavy hitters of the integer world. Want to know how to (quickly) find a list of prime numbers? Want to know how to manipulate and solve exponent problems? Root problems? Well look no further!
This will be your complete guide to advanced ACT integers, including prime numbers, exponents, absolute values, consecutive numbers, and roots—what they mean, as well as how to solve the more difficult integer questions that may show up on the ACT.
Typical Integer Questions on the ACT
First thing's first—there is, unfortunately, no “typical” integer question on the ACT. Integers cover such a wide variety of topics that the questions will be numerous and varied. And as such, there can be no clear template for a standard integer question.
However, this guide will walk you through several real ACT math examples on each integer topic in order to show you some of the many different kinds of integer questions the ACT may throw at you.
As a rule of thumb, you can tell when an ACT question requires you to use your integer techniques and skills when:
#1: The question specifically mentions integers (or consecutive integers)
It could be a word problem or even a geometry problem, but you will know that your answer must be in whole numbers (integers) when the question asks for one or more integers.
(We will go through the process of solving this question later in the guide)
#2: The question involves prime numbers
A prime number is a specific kind of integer, which we will discuss later in the guide. For now, know that any mention of prime numbers means it is an integer question.
A prime number a is squared and then added to a different prime number, b. Which of the following could be the final result?
An even number
An odd number
A positive number
I only
II only
III only
I and III only
I, II, and III
(We'll go through the process of solving this question later in the guide)
#3: The question involves multiplying or dividing bases and exponents
Exponents will always be a number that is positioned higher than the main (base) number:
$4^3$, $(y^5)^2$
You may be asked to find the values of exponents or find the new expression once you have multiplied or divided terms with exponents.
(We will go through the process of solving this question later in the guide)
#4: The question uses perfect squares or asks you to reduce a root value
A root question will always involve the root sign: √
$√36$, $^3√8$
The ACT may ask you to reduce a root, or to find the square root of a perfect square (a number that is equal to an integer squared). You may also need to multiply two or more roots together.
We will go through these definitions as well as how all of these processes are done in the section on roots.
(We will go through the process of solving this question later in the guide)
(Note: A root question with perfect squares may involve fractions. For more information on this concept, look to our guide on fractions and ratios.)
#5: The question involves an absolute value equation (with integers)
Anything that is an absolute value will be bracketed with absolute value signs which look like this: | |
For example: $|-43|$ or $|z + 4|$
(We will go through how to solve this problem later in the guide)
Note: there are generally two different kinds of absolute value problems on the ACT—equations and inequalities.
About a quarter of the absolute value questions you come across will involve the use of inequalities (represented by > or <). If you are unfamiliar with inequalities, check out our guide to ACT inequalities.
The majority of absolute value questions on the ACT will involve a written equation, either using integers or variables. These should be fairly straightforward to solve once you learn the ins and outs of absolute values (and keep track of your negative signs!), all of which we will cover below.
We will, however, only be covering written absolute value equations in this guide. Absolute value questions with inequalities are covered in our guide to ACT inequalities.
We will go through all of these questions and topics throughout this guide in the order of greatest prevalence on the ACT.
We promise that your path to advanced integers will not take you a decade or more to get through (looking at you, Odysseus).
Exponents
Exponent questions will appear on every single ACT, and you'll likely see an exponent question at least twice per test. Whether you're being asked to multiply exponents, divide them, or take one exponent to another, you'll need to know your exponent rules and definitions.
An exponent indicates how many times a number (called a “base”) must be multiplied by itself.
So $3^2$ is the same thing as saying 33. And $3^4$ is the same thing as saying 3333. Here, 3 is the base and 2 and 4 are the exponents.
You may also have a base to a negative exponent. This is the same thing as saying: 1 divided by the base to the positive exponent.
For example, 4-3 becomes $1/{4^3}$ => $1/64$
But how do you multiply or divide bases and exponents? Never fear! Below are the main exponent rules that will be helpful for you to know for the ACT.
Exponent Formulas:
Multiplying Numbers with Exponents:
$x^a x^b = x^[a + b]$
(Note: the bases must be the same for this rule to apply)
Why is this true? Think about it using real numbers.
If you have $3^2 3^4$, you have:
If you count them, this give you 3 multiplied by itself 6 times, or $3^6$. So $3^2 3^4$ => $3^[2 + 4]$ => $3^6$.
(Note: the exponents must be the same for this rule to apply)
Why is this true? Think about it using real numbers.
If you have $3^52^5$, you have:
(33333)(22222) => (32)(32)(32)(32)(32)
So you have $(32)^5$, or $6^5$
If $3^x4^y=12^x$, what is y in terms of x?
${1/2}x$
x
2x
x+2
4x
We can see here that the base of the final answer is 12 and $3 4= 12$. We can also see that the final result, $12^x$, is taken to one of the original exponent values in the equation (x).
This means that the exponents must be equal, as only then can you multiply the bases and keep the exponent intact.
So our final answer is B, $y = x$
If you were uncertain about your answer, then plug in your own numbers for the variables.
Let's say that $x = 2$
$32 4y = 122$
$9 4y = 144$
$4y = 16$
$y = 2$
Since we said that $x = 2$ and we discovered that $y = 2$, then $x = y$.
So again, our answer is B, y = x
Dividing Exponents:
${x^a}/{x^b} = x^[a - b]$
(Note: the bases must be the same for this rule to apply)
Why is this true? Think about it using real numbers.
${3^6}/{3^4}$ can also be written as:
${(3 3 3 3 3 3)}/{(3 3 3 3)}$
If you cancel out your bottom 3s, you’re left with (3 3), or $3^2$
So ${3^6}/{3^4}$ => $3^[6 - 4]$ => $3^2$
The above $(x 10^y)$ is called "scientific notation" and is a method of writing either very large numbers or very small ones. You don't need to understand how it works in order to solve this problem, however. Just think of these as any other bases with exponents.
We have a certain number of hydrogen molecules and the dimensions of a box. We are looking for the number of molecules per one cubic centimeter, which means we must divide our hydrogen molecules by our volume. So:
$${810^12}/{410^4}$$
Take each component separately.
$8/4=2$, so we know our answer is either G or H. Now to complete it, we would say:
$10^12/10^4=10^[12−4]=10^8$
Now put the pieces together:
$2x10^8$
So our full and final answer is H, there are $2x10^8$ hydrogen molecules per cubic centimeter in the box.
Taking Exponents to Exponents:
$(x^a)^b=x^[ab]$
Why is this true? Think about it using real numbers.
$(3^2)^4$ can also be written as:
(33)(33)(33)(33)
If you count them, 3 is being multiplied by itself 8 times. So $(3^2)^4$=>$3^[24]$=>$3^8$
$(x^y)3=x^9$, what is the value of y?
2
3
6
10
12
Because exponents taken to exponents are multiplied together, our problem would look like:
$y3=9$
$y=3$
So our final answer is B, 3.
Distributing Exponents:
$(x/y)^a = x^a/y^a$
Why is this true? Think about it using real numbers. $(3/4)^3$ can be written as $(3/4)(3/4)(3/4)=9/64$
You could also say
$3^3/4^3= 9/64$
$(xy)^z=x^zy^z$
If you are taking a modified base to the power of an exponent, you must distribute that exponent across both the modifier and the base.
$(2x)^3$=>$2^3x^3$
In this case, we are distributing our outer exponent across both pieces of the inner term. So:
$3^3=27$
And we can see that this is an exponent taken to an exponent problem, so we must multiply our exponents together.
$x^[33]=x^9$
This means our final answer is E, $27x^9$
And if you're uncertain whether you have found the right answer, you can always test it out using real numbers. Instead of using a variable, x, let us replace it with 2.
$(3x^3)^3$
$(32^3)^3$
$(38)^3$
$24^3$
13,824
Now test which answer matches 13,824. We'll save ourselves some time by testing E first.
$27x^9$
$272^9$
$27512$
13,824
We have found the same answer, so we know for certain that E must be correct.
(Note: when distributing exponents, you may do so with multiplication or division—exponents do not distribute over addition or subtraction. $(x+y)^a$ is not $x^a+y^a$, for example)
Special Exponents:
It is common for the ACT to ask you what happens when you have an exponent of 0:
$x^0=1$ where x is any number except 0
(Why any number but 0? Well 0 to any power other than 0 equals 0, because $0^x=0$. And any other number to the power of 0 = 1. This makes $0^0$ undefined, as it could be both 0 and 1 according to these guidelines.)
Solving an Exponent Question:
Always remember that you can test out exponent rules with real numbers in the same way that we did in our examples above. If you are presented with $(x^3)^2$ and don’t know whether you are supposed to add or multiply your exponents, replace your x with a real number!
$(2^3)^2=(8)^2=64$
Now check if you are supposed to add or multiply your exponents.
$2^[2+3]=2^5=32$
$2^[32]=2^6=64$
So you know you’re supposed to multiply when exponents are taken to another exponent.
This also works if you are given something enormous, like $(x^19)^3$.
You don’t have to test it out with $2^19$! Just use smaller numbers like we did above to figure out the rules of exponents. Then, apply your newfound knowledge to the larger problem.
And exponents are down for the count. Instant KO!
Roots
Root questions are fairly common on the ACT, and they go hand-in-hand with exponents.
Why are roots related to exponents? Well, technically, roots are fractional exponents. You are likely most familiar with square roots, however, so you may have never heard a root expressed in terms of exponents before.
A square root asks the question: "What number needs to be multiplied by itself one time in order to equal the number under the root sign?"
So $√81=9$ because 9 must be multiplied by itself one time to equal 81.
In other words, $9^2=81$
Another way to write $√{81}$ is to say $^2√{81}$. The 2 at the top of the root sign indicates how many numbers (two numbers, both the same) are being multiplied together to become 81.
(Special note: you do not need the 2 on the root sign to indicate that the root is a square root. But you DO need the indicator for anything that is NOT a square root, like cube roots, etc.)
This means that $^3√27=3$ because three numbers, all of which are the same (333), are multiplied together to equal 27. Or $3^3=27$.
Fractional Exponents
If you have a number to a fractional exponent, it is just another way of asking you for a root.
So $4^{1/2}= √4$
To turn a fractional exponent into a root, the denominator becomes the value to which you take the root.
But what if you have a number other than 1 in the numerator?
$4^{2/3}$=>$^3√{4^2}$
The denominator becomes the value to which you take the root, and the numerator becomes the exponent to which you take the number under the root sign.
Distributing Roots
$√xy=√x√y$
Just like with exponents, roots can be separated out.
So $√30$ => $√2√15$, $√3√10$, or $√5√6$
$√x2√13=2√39$. What is the value of x?
1
3
9
13
26
We know that we must multiply the numbers under the root signs when root expressions are multiplied together. So:
$x13=39$
$x=3$
This means that our final answer is B, $x=3$ to get our final expression $2√39$
$√x√y=√xy$
Because they can be separated, roots can also come together.
So $√5√6$ => $√30$
Reducing Roots
It is common to encounter a problem with a mixed root, where you have an integer multiplied by a root (for example, $4√3$).
Here, $4√3$ is reduced to its simplest form because the number under the root sign, 3, is prime (and therefore has no perfect squares). But let's say you had something like $3√18$ instead.
Now, $3√18$ is NOT as reduced as it can be. In order to reduce it, we must find out if there are any perfect squares that factor into 18. If there are, then we can take them out from under the root sign.
(Note: if there is more than one perfect square that can factor into your number under the root sign, use the largest one.)
18 has several factor pairs. These are:
$118$
$29$
$36$
Well, 9 is a perfect square because $33=9$. That means that $√9=3$.
This means that we can take 9 out from under the root sign. Why? Because we know that $√{xy}=√x√y$.
So $√{18}=√2√9$. And $√9=3$. So 9 can come out from under the root sign and be replaced by 3 instead. $√2$ is as reduced as we can make it, since it is a prime number.
We are left with $3√2$ as the most reduced form of $√18$
(Note: you can test to see if this is true on most calculators. $√18=4.2426$ and $3√2=31.4142=4.2426$. The two expressions are identical.)
We are still not done, however. We wanted to originally change $3√18$ to its most reduced form. So far we have found the most reduced expression of $√18$, so now we must multiply them together.
$3√18=33√2$
$9√2$
So our final answer is $9√2$, this is the most reduced form of $3√{18}$.
You've rooted out your answers, you've gotten to the root of the problem, you've touched up those roots....
Absolute Values
Absolute values are quite common on the ACT. You should expect to see at least one question on absolute values per test.
An absolute value is a representation of distance along a number line, forward or backwards. This means that an absolute value equation will always have two solutions.
It also means that whatever is in the absolute value sign will be positive, as it represents distance along a number line and there is no such thing as a negative distance.
An equation $|x+4|=12$, has two solutions:
$x=8$
$x=−16$
Why -16? Well $−16+4=−12$ and, because it is an absolute value (and therefore a distance), the final answer becomes positive. So $|−12|=12$
When you are presented with an absolute value, instead of doing the math in your head to find the negative and positive solution, you can instead rewrite the equation into two different equations.
When presented with the above equation $|x+4|=12$, take away the absolute value sign and transform it into two equations—one with a positive solution and one with a negative solution.
So $|x+4|=12$ becomes:
$x+4=12$ AND $x+4=−12$
Solve for x
$x=8$ and $x=−16$
Now let's look at our absolute value problem from earlier:
As you can see, this absolute value problem is fairly straightforward. Its only potential pitfalls are its parentheses and negatives, so we need to be sure to be careful with them.
Solve the problem inside the absolute value sign first and then use the absolute value signs to make our final answer positive. (By process of elimination, we can already get rid of answer choices A and B, as we know that an absolute value cannot be negative.)
$|7(−3)+2(4)|$
$|−21+8|$
$|−13|$
We have solved our problem. But we know that −13 is inside an absolute value sign, which means it must be positive.
So our final answer is C, 13.
Absolutely fabulous absolute values are absolutely solvable. I promise this absolutely.
Consecutive Numbers
Questions about consecutive numbers may or may not show up on your ACT. If they appear, it will be for a maximum of one question. Regardless, they are still an important concept for you to understand.
Consecutive numbers are numbers that go continuously along the number line with a set distance between each number.
So an example of positive, consecutive numbers would be: 5, 6, 7, 8, 9
An example of negative, consecutive numbers would be: -9, -8, -7, -6, -5
(Notice how the negative integers go from greatest to least—if you remember the basic guide to ACT integers, this is because of how they lie on the number line in relation to 0)
You can write unknown consecutive numbers out algebraically by assigning the first in the series a variable, x, and then continuing the sequence of adding 1 to each additional number.
The sum of five positive, consecutive integers is 115. What is the first of these integers?
21
22
23
24
25
If x is our first, unknown, integer in the sequence, so you can write all four numbers as:
$x+(x+1)+(x+2)+(x+3)+(x+4)=115$
$5x+10=115$
$5x=105$
$x=21$
So x is our first number in the sequence and $x=21$:
This means our final answer is A, the first number in our sequence is 21.
(Note: always pay attention to what number they want you to find! If they had asked for the median number in the sequence, you would have had to continue the problem with $x=21$, $x+2=$median, $23=$median.)
You may also be asked to find consecutive even or consecutive odd integers. This is the same as consecutive integers, only they are going up every other number instead of every number. This means there is a difference of two units between each number in the sequence instead of 1.
An example of positive, consecutive even integers: 10, 12, 14, 16, 18
An example of positive, consecutive odd integers: 17, 19, 21, 23, 25
Both consecutive even or consecutive odd integers can be written out in sequence as:
$x,x+2,x+4,x+6$, etc. No matter if the beginning number is even or odd, the numbers in the sequence will always be two units apart.
What is the largest number in the sequence of four positive, consecutive odd integers whose sum is 160?
37
39
41
43
45
$x+(x+2)+(x+4)+(x+6)=160$
$4x+12=160$
$4x=148$
$x=37$
So the first number in the sequence is 37. This means the full sequence is:
37, 39, 41, 43
Our final answer is D, the largest number in the sequence is 43 (x+6).
When consecutive numbers make all the difference.
Remainders
Questions involving remainders are rare on the ACT, but they still show up often enough that you should be aware of them.
A remainder is the amount left over when two numbers do not divide evenly. If you divide 18 by 6, you will not have any remainder (your remainder will be zero). But if you divide 19 by 6, you will have a remainder of 1, because there is 1 left over.
You can think of the division as $19/6 = 3{1/6}$. That extra 1 is left over.
Most of you probably haven’t worked with integer remainders since elementary school, as most higher level math classes and questions use decimals to express the remaining amount after a division (for the above example, $19/6 = 3$ remainder 1 or 3.167). But you may still come across the occasional remainder question on the ACT.
How many integers between 10 and 40, inclusive, can be divided by 3 with a remainder of zero?
9
10
12
15
18
Now, we know that when a division problem results in a remainder of zero, that means the numbers divide evenly. $9/3 =3$ remainder 0, for example. So we are looking for all the numbers between 10 and 40 that are evenly divisible by 3.
There are two ways we can do this—by listing the numbers out by hand or by taking the difference of 40 and 10 and dividing that difference by 3. That quotient (answer to a division problem) rounded to the nearest integer will be the number of integers divisible by 3.
Let's try the first technique first and list out all the numbers divisible by 3 between 10 and 40, inclusive.
The first integer after 10 to be evenly divisible by 3 is 12. After that, we can just add 3 to every number until we either hit 40 or go beyond 40.
12, 15, 18, 21, 24, 27, 30, 33, 36, 39
If we count all the numbers more than 10 and less than 40 in our list, we wind up with 10 integers that can be divided by 3 with a remainder of zero.
This means our final answer is B, 10.
Alternatively, we could use our second technique.
$40−10=30$
$30/3$
$=10$
Again, our answer is B, 10.
(Note: if the difference of the two numbers had NOT be divisible by 3, we would have taken the nearest rounded integer. For example, if we had been asked to find all the numbers between 10 and 50 that were evenly divisible by 3, we would have said:
$50−10=40$
$40/3$
=13.333
$13.333$, rounded = 13
So our final answer would have been 13. And you can always test this by hand if you do not feel confident with your answer.)
Prime Numbers
Prime numbers are relatively rare on the ACT, but that is not to say that they never show up at all. So be sure to understand what they are and how to find them.
A prime number is a number that is only divisible by two numbers—itself and 1.
For example, 13 is a prime number because $113$ is its only factor. (13 is not evenly divisible by 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, or 12).
12 is NOT a prime number, because its factors are 1, 2, 3, 4, 6, and 12. It has more factors than just itself and 1.
1 is NOT a prime number, because its only factor is 1.
The only even prime number is 2. Standardized tests love to include the fact that 2 is a prime number as a way to subtly trick students who go too quickly through the test. If you assume that all prime numbers must be odd, then you may get a question on primes wrong.
A prime number x is squared and then added to a different prime number, y. Which of the following could be the final result?
An even number
An odd number
A positive number
I only
II only
III only
I and III only
I, II, and III
Now, this question relies on your knowledge of both number relationships and primes. You know that any number squared (the number times itself) will be an even number if the original number was even, and an odd number if the original number was odd. Why? Because an even an even = an even, and an odd an odd = an odd ($22=4$ & $33=9$).
Next, we are adding that square to another prime number. You’ll also remember that an even number + an odd number is odd, an odd number + an odd number is even, and an even number + an even number is even.
Knowing that 2 is a prime number, let’s replace x with 2. $2^2=4$. Now if y is a different prime number (as stipulated in the question), it must be odd, because the only even prime number is 2. So let’s say $y=5$.
$4+5=11$. So the end result is odd.
This means II is correct.
But what if both x and y were odd prime numbers? So let’s say that $x=3$ and $y=5$. So $3^2=9$ and 9+5=14$. So the end result is even.
This means I is correct.
Now, for option number III, our results show that it is possible to get a positive number result, since both our results were positive.
This means the final answer is E, I, II, and III
If you forgot that 2 was a prime number, you would have picked D, I and III only, because there would have been no possible way to get an odd number. Remembering that 2 is a prime number is the key to solving this question.
Another prime number question you may see on the ACT will ask you to identify how many prime numbers fall in a certain range of numbers.
How many prime numbers are between 20 and 40, inclusive?
Three
Four
Five
Six
Seven
This might seem intimidating or time-consuming, but I promise you do NOT need to memorize a list of prime numbers.
First, eliminate all even numbers from the list, as you know the only even prime number is 2.
Next, eliminate all numbers that end in 5. Any number that ends is 5 or 0 is divisible by 5.
Now your list looks like this:
21, 23, 27, 29, 31, 33, 37, 39
This is much easier to work with, but we need to narrow it down further.
(You could start using your calculator here, or you can do this by hand.) A way to see if a number is divisible by 3 is to add the digits together. If that number is 3 or divisible by 3, then the final result is divisible by 3.
For example, the number 23 is NOT divisible by 3 because $2+3=5$, which is not divisible by 3. However 21 is divisible by 3 because $2+1=3$, which is divisible by 3.
So we can now eliminate 21 $(2+1=3)$, 27 $(2+7=9)$, 33 $(3+3=6)$, and 39 $(3+9=12)$ from the list.
We are left with 23, 29, 31, 37.
Now, to make sure you try every necessary potential factor, take the square root of the number you are trying to determine is prime. Any integer equal to or less than a number's square root could be a potential factor, but you do not have to try any numbers higher.
Why? Well let’s take 36 as an example. Its factors are:
1, 2, 3, 4, 6, 9, 12, 18, and 36. But now look at the factor pairings.
1 & 36
2 & 18
3 & 12
4 & 9
6 & 6
(9 & 4)
(12 & 3)
(18 & 2)
(36 & 1)
After you get past 6, the numbers repeat. If you test out 4, you will know that 9 goes evenly into your larger number—no need to actually test 9 just to get 4 again!
So all numbers less than or equal to a potential prime’s square root are the only potential factors you need to test. And, since we are dealing with potential primes, we only need to test odd integers equal to or less than the square root. Why? Because all multiples of even numbers will be even, and 2 is the only even prime number.
Going back to our list, we have 23, 29, 31, 37.
Well the closest square root to 23 and 29 is 5. We already know that neither 2 nor 3 nor 5 factor evenly into 23 or 29. You’re done. Both 23 and 29 must be prime. (Why didn't we test 4? Because all multiples of 4 are even, as an even an even = an even.)
As for 31 and 37, the closest square root of these is 6. But because 6 is even, we don't need to test it. So we need only to test odd numbers less than six. And we already know that neither 2 nor 3 nor 5 factor evenly into 31 or 37.
So we are done. We have found all of our prime numbers.
So your final answer is B, there are four prime numbers (23, 29, 31, 37) between 20 and 40.
A different kind of Prime.
Steps to Solving an ACT Integer Question
Because ACT integer questions are so numerous and varied, there is no set way to approach them that is entirely separate from approaching other kinds of ACT math questions. But there are a few techniques that will help you approach your ACT integer questions (and by extension, most questions on ACT math).
#1. Make sure the question requires an integer.
If the question does NOT specify that you are looking for an integer, then any number—including decimals and fractions—are fair game. Always read the question carefully to make sure you are on the right track.
#2. Use real numbers if you forget your integer rules.
Forget whether positive, even consecutive integers should be written as x+(x+1) or x+(x+2)? Test it out with real numbers!
6, 8, 10 are consecutive even integers. If x=6, 8=x+2, and 10=x+4.
This works for most all of your integer rules. Forget your exponent rules? Plug in real numbers! Forget whether an even an even makes an even or an odd? Plug in real numbers!
#3. Keep your work organized.
Like with most ACT math questions, integer questions can seem more complex than they are, or will be presented to you in strange ways. Keep your work well organized and keep track of your values to make sure your answer is exactly what the question is asking for.
Got your list in order? Than let's get cracking!
Test Your Knowledge
1.
2.
3.
4.
5.
Answers: C, D, B, F, H
Answer Explanations:
1. We are tasked here with finding the smallest integer greater than $√58$. There are two ways to approach this—using a calculator or using our knowledge of perfect squares. Each will take about the same amount of time, so it's a matter of preference (and calculator ability).
If you plug $√58$ into your calculator, you'll wind up with 7.615. This means that 8 is the smallest integer greater than this (because 7.616 is not an integer).
Thus your final answer is C, 8.
Alternatively, you could use your knowledge of perfect squares.
$7^2=49$ and $8^2=64$
$√58$ is between these and larger than $√49$, so your closest integer larger than $√58$ would be 8.
Again, our answer is C, 8.
2. Here, we must find possible values for a and b such that $|a+b|=|a−b|$. It'll be fastest for us to look to the answers in order to test which ones are true. (For more information on how to plug in answers, check out our article on plugging in answers)
Answer choice A says this equation is "always" true, but we can see this is incorrect by plugging in some values for a and b. If $a=2$ and $b=4$, then $|a+b|=6$ and $|a−b|=|−2|=2$
6≠2, so answer choice A is wrong.
We can also see that answer choice B is wrong. Why? Because when a and b are equal, $|a−b|$ will equal 0, but $|a+b|$ will not.
If $a=2$ and $b=2$ then $|a+b|=4$ and $|a−b|=0$
$4≠0$
Now let's look at answer choice C. It's true that when $a=0$ and $b=0$ that $|a+b|=|a−b|$ because $0=0$. But is this the only time that the equation works? We're not sure yet, so let's not eliminate this answer for now.
So now let's try D. If $a=0$, but b=any other integer, does the equation work?
Let's say that $b=2$, so $|a+b|=|0+2|=2$ and $|a−b|=|0−2|=|−2|=2$
$2=2$
We can also see that the same would work when $b=0$
$a=2$ and $b=0$, so $|a+b|=|2+0|=2$ and $|a−b|=|2−0|=2$
$2=2$
So our final answer is D, the equation is true when either $a=0$, $b=0$, or both a and b equal 0.
3. We are told that we have two, unknown, consecutive integers. And the smaller integer plus triple the larger integer equals 79. So let's find our two integers by writing the proper equation.
If we call our smaller integer x, then our larger integer will be $x+1$. So:
$x+3(x+1)=79$
$x+3x+3=79$
$4x=76$
$x=19$
Because we isolated the x, and the x stood in place of our smaller integer, this means our smaller integer is 19. Our larger integer must therefore be 20. (We can even test this by plugging these answers back into the original problem: $19+3(20)=19+60=79$)
This means our final answer is B, 19 and 20.
4. We are being asked to find the smallest value of a number from several options. All of these options rely on our knowledge of roots, so let's examine them.
Option F is $√x$. This will be the square root of x (in other words, a numberitself=x.)
Option G says $√2x$. Well this will always be more than $√x$. Why? Because, the greater the number under the root sign, the greater the square root. Think of it in terms of real numbers.
$√9=3$ and $√16=4$. The larger the number under the root sign, the larger the square root.
This means that G will be larger than F, so we can cross G off the list.
Similarly, we can cross off H. Why? Because $√xx$ will be even bigger than $2x$ and will thus have a larger number under the root sign and a larger square root than $√x$.
Option J will also be larger than option F because $√x$ will always be less than $√x$another number larger than 1 (and the question specifically said that x>1.)
Remember it using real numbers. $√16$ (answer=4) will be less than $16√16$ (answer=64).
And finally, K will be more than $√x$ as well. Why? Because K is the square of x (in other words, $xx=x^2$) and the square of a number will always be larger than that number's square root.
This means that our final answer is F, $√x$ is the least of all these terms.
5. Here, we are multiplying bases and exponents. We have ($2x^4y$) and we want to multiply it by ($3x^5y^8$). So let's multiply them piece by piece.
First, multiply your integers.
$23=6$
Next, multiply your x bases and their exponents. We know that we must add the exponents when multiplying two of the same base together.
$x^4x^5=x^[4+5]=x^9$
Next, multiply your y bases and their exponents.
$yy^8=y^[1+8]=y^9$
(Why is this $y^9$? Because y without an exponent is the same thing as saying $y^1$, so we needed to add that single exponent to the 8 from $y^8$.)
Put the pieces together and you have:
$6x^9y^9$
So our final answer is H, 6x9y9
Now celebrate because you rocked those integers!
The Take-Aways
Integers and integer questions can be tricky for some students, as they often involve concepts not tested in high school level math classes (have you had reason to use remainders much outside of elementary school?). But most integer questions are much simpler than they appear.
If you know your way around exponents and you remember your definitions—integers, consecutive integers, absolute values, etc.—you’ll be able to solve most any ACT integer question that comes your way.
What’s Next?
You've taken on integers, both basic and advanced, and emerged victorious. Now that you’ve mastered these foundational topics of the ACT math, make sure you’ve got a solid grasp of all the math topics covered by the ACT math section, so that you can take on the ACT with confidence.
Find yourself running out of time on ACT math? Check out our article on how to keep from running out of time on the ACT math section before it's pencil's down.
Feeling overwhelmed? Start by figuring out your ideal score and work to improve little by little from there.
Already have pretty good scores and looking to get a perfect 36? Check out our article on how to get a perfect ACT math score written by a 36 ACT-scorer.
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Engineering Electromagnetics 8th Edition Hayt Solutions Manual
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The document discusses the electrostatic interactions between multiple point charges, including their configurations, calculated electric fields, and forces acting on them in free space. It presents specific scenarios and provides equations for finding positions of additional charges to achieve desired electric field strengths at various points. The calculations incorporate principles of electrostatics, including superposition and vector forces related to charge distributions.
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![Image 22: We require |Ez| = |2Ey|, so 2(y + 2) = 5. Thus y = 1/2, and the field becomes:
Ez=0 =
⇢l
2⇡✏0
2.5ay 5az
(2.5)2 + 25
= 23ay 46az
2.18. a) Find E in the plane z = 0 that is produced by a uniform line charge, ⇢L, extending along the z
axis over the range L < z < L in a cylindrical coordinate system: We find E through
E =
Z L
L
⇢Ldz(r r0
)
4⇡✏0|r r0|3
where the observation point position vector is r = ⇢a⇢ (anywhere in the x-y plane), and where the
position vector that locates any di↵erential charge element on the z axis is r0
= zaz. So r r0
=
⇢a⇢ zaz, and |r r0
| = (⇢2
+ z2
)1/2
. These relations are substituted into the integral to yield:
E =
Z L
L
⇢Ldz(⇢a⇢ zaz)
4⇡✏0(⇢2 + z2)3/2
=
⇢L ⇢ a⇢
4⇡✏0
Z L
L
dz
(⇢2 + z2)3/2
= E⇢ a⇢
Note that the second term in the left-hand integral (involving zaz) has e↵ectively vanished because it
produces equal and opposite sign contributions when the integral is taken over symmetric limits (odd
parity). Evaluating the integral results in
E⇢ =
⇢L ⇢
4⇡✏0
z
⇢2
p
⇢2 + z2
L
L
=
⇢L
2⇡✏0⇢
L
p
⇢2 + L2
=
⇢L
2⇡✏0⇢
1
p
1 + (⇢/L)2
Note that as L ! 1, the expression reduces to the expected field of the infinite line charge in free
space, ⇢L/(2⇡✏0⇢).
b) if the finite line charge is approximated by an infinite line charge (L ! 1), by what percentage
is E⇢ in error if ⇢ = 0.5L? The percent error in this situation will be
% error =
"
1
1
p
1 + (⇢/L)2
⇥ 100
For ⇢ = 0.5L, this becomes % error = 10.6 %
c) repeat b with ⇢ = 0.1L. For this value, obtain % error = 0.496 %.
2.19. A uniform line charge of 2 µC/m is located on the z axis. Find E in rectangular coordinates at
P(1, 2, 3) if the charge extends from
a) 1 < z < 1: With the infinite line, we know that the field will have only a radial component
in cylindrical coordinates (or x and y components in cartesian). The field from an infinite line
on the z axis is generally E = [⇢l/(2⇡✏0⇢)]a⇢. Therefore, at point P:
EP =
⇢l
2⇡✏0
RzP
|RzP |2
=
(2 ⇥ 10 6
)
2⇡✏0
ax + 2ay
5
= 7.2ax + 14.4ay kV/m
where RzP is the vector that extends from the line charge to point P, and is perpendicular to
the z axis; i.e., RzP = (1, 2, 3) (0, 0, 3) = (1, 2, 0).
19
](
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![Image 24: This can be re-written and then evaluated using integral tables as
F =
⇢2
0 ax
4⇡✏0
Z 3/2
/2
dx
x
p
x2 + (/2)2
=
⇢2
0 ax
4⇡✏0
0
@ 1
(/2)
ln
"
/2 +
p
x2 + (/2)2
x
3`/2
/2
1
A
=
⇢2
0 ax
2⇡✏0
ln
"
(/2) 1 +
p
10
3(`/2) 1 +
p
2
=
⇢2
0 ax
2⇡✏0
ln
"
3(1 +
p
2)
1 +
p
10
=
0.55⇢2
0
2⇡✏0
ax N
2.21. Two identical uniform line charges with ⇢l = 75 nC/m are located in free space at x = 0, y = ±0.4
m. What force per unit length does each line charge exert on the other? The charges are parallel to
the z axis and are separated by 0.8 m. Thus the field from the charge at y = 0.4 evaluated at the
location of the charge at y = +0.4 will be E = [⇢l/(2⇡✏0(0.8))]ay. The force on a di↵erential length
of the line at the positive y location is dF = dqE = ⇢ldzE. Thus the force per unit length acting on
the line at postive y arising from the charge at negative y is
F =
Z 1
0
⇢2
l dz
2⇡✏0(0.8)
ay = 1.26 ⇥ 10 4
ay N/m = 126 ay µN/m
The force on the line at negative y is of course the same, but with ay.
2.22. Two identical uniform sheet charges with ⇢s = 100 nC/m2
are located in free space at z = ±2.0 cm.
What force per unit area does each sheet exert on the other?
The field from the top sheet is E = ⇢s/(2✏0) az V/m. The di↵erential force produced by this
field on the bottom sheet is the charge density on the bottom sheet times the di↵erential area
there, multiplied by the electric field from the top sheet: dF = ⇢sdaE. The force per unit area is
then just F = ⇢sE = (100 ⇥ 10 9
)( 100 ⇥ 10 9
)/(2✏0) az = 5.6 ⇥ 10 4
az N/m2
.
2.23. Given the surface charge density, ⇢s = 2 µC/m2
, in the region ⇢ < 0.2 m, z = 0. Find E at:
a) PA(⇢ = 0, z = 0.5): First, we recognize from symmetry that only a z component of E will be
present. Considering a general point z on the z axis, we have r = zaz. Then, with r0
= ⇢a⇢, we
obtain r r0
= zaz ⇢a⇢. The superposition integral for the z component of E will be:
Ez,PA =
⇢s
4⇡✏0
Z 2⇡
0
Z 0.2
0
z ⇢ d⇢ d
(⇢2 + z2)1.5
=
2⇡⇢s
4⇡✏0
z
"
1
p
z2 + ⇢2
0.2
0
⇢s
2✏0
z
1
p
z2
1
p
z2 + 0.04
With z = 0.5 m, the above evaluates as Ez,PA = 8.1 kV/m.
b) PB(⇢ = 0, z = 0.5). With z at 0.5 m, we evaluate the expression for Ez to obtain Ez,PB =
8.1 kV/m.
c) Show that the field along the z axis reduces to that of an infinite sheet charge at small values of
z: In general, the field can be expressed as
Ez =
⇢s
2✏0
1
z
p
z2 + 0.04
21
](
![Image 25: At small z, this reduces to Ez
.
= ⇢s/2✏0, which is the infinite sheet charge field.
d) Show that the z axis field reduces to that of a point charge at large values of z: The development
is as follows:
Ez =
⇢s
2✏0
1
z
p
z2 + 0.04
=
⇢s
2✏0
"
1
z
z
p
1 + 0.04/z2
.
⇢s
2✏0
1
1
1 + (1/2)(0.04)/z2
where the last approximation is valid if z >> .04. Continuing:
Ez
.
=
⇢s
2✏0
⇥
1 [1 (1/2)(0.04)/z2
]
⇤
=
0.04⇢s
4✏0z2
=
⇡(0.2)2
⇢s
4⇡✏0z2
This the point charge field, where we identify q = ⇡(0.2)2
⇢s as the total charge on the disk (which
now looks like a point).
22
](
Ad
Image 26: 2.24. a) Find the electric field on the z axis produced by an annular ring of uniform surface charge density
⇢s in free space. The ring occupies the region z = 0, a ⇢ b, 0 2⇡ in cylindrical coordinates:
We find the field through
E =
Z Z
⇢sda(r r0
)
4⇡✏0|r r0|3
where the integral is taken over the surface of the annular ring, and where r = zaz and r0
= ⇢a⇢. The
integral then becomes
E =
Z 2⇡
0
Z b
a
⇢s ⇢ d⇢ d (zaz ⇢a⇢)
4⇡✏0(z2 + ⇢2)3/2
In evaluating this integral, we first note that the term involving ⇢a⇢ integrates to zero over the
integration range of 0 to 2⇡. This is because we need to introduce the dependence in a⇢ by writing
it as a⇢ = cos ax + sin ay, where ax and ay are invariant in their orientation as varies. So the
integral now simplifies to
E =
2⇡⇢s z az
4⇡✏0
Z b
a
⇢ d⇢
(z2 + ⇢2)3/2
=
⇢s z az
2✏0
"
1
p
z2 + ⇢2
b
a
⇢s
2✏0
"
1
p
1 + (a/z)2
1
p
1 + (b/z)2
az
b) from your part a result, obtain the field of an infinite uniform sheet charge by taking appropriate
limits. The infinite sheet is obtained by letting a ! 0 and b ! 1, in which case E ! ⇢s/(2✏0) az
as expected.
2.25. Find E at the origin if the following charge distributions are present in free space: point charge, 12 nC
at P(2, 0, 6); uniform line charge density, 3nC/m at x = 2, y = 3; uniform surface charge density,
0.2 nC/m2
at x = 2. The sum of the fields at the origin from each charge in order is:
E =
(12 ⇥ 10 9
)
4⇡✏0
( 2ax 6az)
(4 + 36)1.5
+
(3 ⇥ 10 9
)
2⇡✏0
(2ax 3ay)
(4 + 9)
(0.2 ⇥ 10 9
)ax
2✏0
= 3.9ax 12.4ay 2.5az V/m
23
![Image 27: 2.26. Radially-dependent surface charge is distributed on an infinite flat sheet in the xy plane, and is char-
acterized in cylindrical coordinates by surface density ⇢s = ⇢0/⇢, where ⇢0 is a constant. Determine
the electric field strength, E, everywhere on the z axis.
We find the field through
E =
Z Z
⇢sda(r r0
)
4⇡✏0|r r0|3
where the integral is taken over the surface of the annular ring, and where r = zaz and r0
= ⇢a⇢.
The integral then becomes
E =
Z 2⇡
0
Z 1
0
(⇢0/⇢) ⇢ d⇢ d (zaz ⇢a⇢)
4⇡✏0(z2 + ⇢2)3/2
In evaluating this integral, we first note that the term involving ⇢a⇢ integrates to zero over the
integration range of 0 to 2⇡. This is because we need to introduce the dependence in a⇢
by writing it as a⇢ = cos ax + sin ay, where ax and ay are invariant in their orientation as
varies. So the integral now simplifies to
E =
2⇡⇢s z az
4⇡✏0
Z 1
0
d⇢
(z2 + ⇢2)3/2
=
⇢s z az
2✏0
"
⇢
z2
p
z2 + ⇢2
1
⇢=0
⇢s
2✏0z
az
2.27. Given the electric field E = (4x 2y)ax (2x + 4y)ay, find:
a) the equation of the streamline that passes through the point P(2, 3, 4): We write
dy
dx
=
Ey
Ex
=
(2x + 4y)
(4x 2y)
Thus
2(x dy + y dx) = y dy x dx
or
2 d(xy) =
1
2
d(y2
)
1
2
d(x2
)
So
C1 + 2xy =
1
2
y2 1
2
x2
or
y2
x2
= 4xy + C2
Evaluating at P(2, 3, 4), obtain:
9 4 = 24 + C2, or C2 = 19
Finally, at P, the requested equation is
y2
x2
= 4xy 19
b) a unit vector specifying the direction of E at Q(3, 2, 5): Have EQ = [4(3) + 2(2)]ax [2(3)
4(2)]ay = 16ax + 2ay. Then |E| =
p
162 + 4 = 16.12 So
aQ =
16ax + 2ay
16.12
= 0.99ax + 0.12ay
24
](
Ad
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Engineering Electromagnetics 8th Edition Hayt Solutions Manual
CHAPTER 2 2.1. Threepoint charges are positioned in the x-y plane as follows: 5nC at y = 5 cm, -10 nC at y = 5 cm, 15 nC at x = 5 cm. Find the required x-y coordinates of a 20-nC fourth charge that will produce a zero electric field at the origin. With the charges thus configured, the electric field at the origin will be the superposition of the individual charge fields: E0 = 1 4⇡✏0 15 (5)2 ax 5 (5)2 ay 10 (5)2 ay = 1 4⇡✏0 ✓ 3 5 ◆ [ax ay] nC/m The field, E20, associated with the 20-nC charge (evaluated at the origin) must exactly cancel this field, so we write: E20 = 1 4⇡✏0 ✓ 3 5 ◆ [ax ay] = 20 4⇡✏0⇢2 ✓ 1 p 2 ◆ [ax ay] From this, we identify the distance from the origin: ⇢ = q 100/(3 p 2) = 4.85. The x and y coordinates of the 20-nC charge will both be equal in magnitude to 4.85/ p 2 = 3.43. The coodinates of the 20-nC charge are then (3.43, 3.43). 2.2. Point charges of 1nC and -2nC are located at (0,0,0) and (1,1,1), respectively, in free space. Determine the vector force acting on each charge. First, the electric field intensity associated with the 1nC charge, evalutated at the -2nC charge location is: E12 = 1 4⇡✏0(3) ✓ 1 p 3 ◆ (ax + ay + az) nC/m in which the distance between charges is p 3 m. The force on the -2nC charge is then F12 = q2E12 = 2 12 p 3 ⇡✏0 (ax + ay + az) = 1 10.4 ⇡✏0 (ax + ay + az) nN The force on the 1nC charge at the origin is just the opposite of this result, or F21 = +1 10.4 ⇡✏0 (ax + ay + az) nN 12 Engineering Electromagnetics 8th Edition Hayt Solutions Manual Full Download: This sample only, Download all chapters at: alibabadownload.com
2.3. Point chargesof 50nC each are located at A(1, 0, 0), B( 1, 0, 0), C(0, 1, 0), and D(0, 1, 0) in free space. Find the total force on the charge at A. The force will be: F = (50 ⇥ 10 9 )2 4⇡✏0 RCA |RCA|3 + RDA |RDA|3 + RBA |RBA|3 where RCA = ax ay, RDA = ax + ay, and RBA = 2ax. The magnitudes are |RCA| = |RDA| = p 2, and |RBA| = 2. Substituting these leads to F = (50 ⇥ 10 9 )2 4⇡✏0 1 2 p 2 + 1 2 p 2 + 2 8 ax = 21.5ax µN where distances are in meters. 2.4. Eight identical point charges of Q C each are located at the corners of a cube of side length a, with one charge at the origin, and with the three nearest charges at (a, 0, 0), (0, a, 0), and (0, 0, a). Find an expression for the total vector force on the charge at P(a, a, a), assuming free space: The total electric field at P(a, a, a) that produces a force on the charge there will be the sum of the fields from the other seven charges. This is written below, where the charge locations associated with each term are indicated: Enet(a, a, a) = q 4⇡✏0a2 2 6 6 6 4 ax + ay + az 3 p 3 | {z } (0,0,0) + ay + az 2 p 2 | {z } (a,0,0) + ax + az 2 p 2 | {z } (0,a,0) + ax + ay 2 p 2 | {z } (0,0,a) + ax |{z} (0,a,a) + ay |{z} (a,0,a) + az |{z} (a,a,0) 3 7 7 7 5 The force is now the product of this field and the charge at (a, a, a). Simplifying, we obtain F(a, a, a) = qEnet(a, a, a) = q2 4⇡✏0a2 1 3 p 3 + 1 p 2 + 1 (ax + ay + az) = 1.90 q2 4⇡✏0a2 (ax + ay + az) in which the magnitude is |F| = 3.29 q2 /(4⇡✏0a2 ). 2.5. Let a point charge Q1 = 25 nC be located at P1(4, 2, 7) and a charge Q2 = 60 nC be at P2( 3, 4, 2). a) If ✏ = ✏0, find E at P3(1, 2, 3): This field will be E = 10 9 4⇡✏0 25R13 |R13|3 + 60R23 |R23|3 where R13 = 3ax +4ay 4az and R23 = 4ax 2ay +5az. Also, |R13| = p 41 and |R23| = p 45. So E = 10 9 4⇡✏0 25 ⇥ ( 3ax + 4ay 4az) (41)1.5 + 60 ⇥ (4ax 2ay + 5az) (45)1.5 = 4.58ax 0.15ay + 5.51az b) At what point on the y axis is Ex = 0? P3 is now at (0, y, 0), so R13 = 4ax + (y + 2)ay 7az and R23 = 3ax + (y 4)ay + 2az. Also, |R13| = p 65 + (y + 2)2 and |R23| = p 13 + (y 4)2. Now the x component of E at the new P3 will be: Ex = 10 9 4⇡✏0 25 ⇥ ( 4) [65 + (y + 2)2]1.5 + 60 ⇥ 3 [13 + (y 4)2]1.5 13
2.5b (continued) Toobtain Ex = 0, we require the expression in the large brackets to be zero. This expression simplifies to the following quadratic: 0.48y2 + 13.92y + 73.10 = 0 which yields the two values: y = 6.89, 22.11 2.6. Two point charges of equal magnitude q are positioned at z = ±d/2. a) find the electric field everywhere on the z axis: For a point charge at any location, we have E = q(r r0 ) 4⇡✏0|r r0|3 In the case of two charges, we would therefore have ET = q1(r r0 1) 4⇡✏0|r r0 1|3 + q2(r r0 2) 4⇡✏0|r r0 2|3 (1) In the present case, we assign q1 = q2 = q, the observation point position vector as r = zaz, and the charge position vectors as r0 1 = (d/2)az, and r0 2 = (d/2)az Therefore r r0 1 = [z (d/2)]az, r r0 2 = [z + (d/2)]az, then |r r1|3 = [z (d/2)]3 and |r r2|3 = [z + (d/2)]3 Substitute these results into (1) to obtain: ET (z) = q 4⇡✏0 1 [z (d/2)]2 + 1 [z + (d/2)]2 az V/m (2) b) find the electric field everywhere on the x axis: We proceed as in part a, except that now r = xax. Eq. (1) becomes ET (x) = q 4⇡✏0 xax (d/2)az |xax (d/2)az|3 + xax + (d/2)az |xax + (d/2)az|3 (3) where |xax (d/2)az| = |xax + (d/2)az| = ⇥ x2 + (d/2)2 ⇤1/2 Therefore (3) becomes ET (x) = 2qx ax 4⇡✏0 [x2 + (d/2)2] 3/2 c) repeat parts a and b if the charge at z = d/2 is q instead of +q: The field along the z axis is quickly found by changing the sign of the second term in (2): ET (z) = q 4⇡✏0 1 [z (d/2)]2 1 [z + (d/2)]2 az V/m In like manner, the field along the x axis is found from (3) by again changing the sign of the second term. The result is 2qd az 4⇡✏0 [x2 + (d/2)2] 3/2 14
2.7. A 2µC point charge is located at A(4, 3, 5) in free space. Find E⇢, E , and Ez at P(8, 12, 2). Have EP = 2 ⇥ 10 6 4⇡✏0 RAP |RAP |3 = 2 ⇥ 10 6 4⇡✏0 4ax + 9ay 3az (106)1.5 = 65.9ax + 148.3ay 49.4az Then, at point P, ⇢ = p 82 + 122 = 14.4, = tan 1 (12/8) = 56.3 , and z = z. Now, E⇢ = Ep · a⇢ = 65.9(ax · a⇢) + 148.3(ay · a⇢) = 65.9 cos(56.3 ) + 148.3 sin(56.3 ) = 159.7 and E = Ep · a = 65.9(ax · a ) + 148.3(ay · a ) = 65.9 sin(56.3 ) + 148.3 cos(56.3 ) = 27.4 Finally, Ez = 49.4 V/m 2.8. A crude device for measuring charge consists of two small insulating spheres of radius a, one of which is fixed in position. The other is movable along the x axis, and is subject to a restraining force kx, where k is a spring constant. The uncharged spheres are centered at x = 0 and x = d, the latter fixed. If the spheres are given equal and opposite charges of Q coulombs: a) Obtain the expression by which Q may be found as a function of x: The spheres will attract, and so the movable sphere at x = 0 will move toward the other until the spring and Coulomb forces balance. This will occur at location x for the movable sphere. With equal and opposite forces, we have Q2 4⇡✏0(d x)2 = kx from which Q = 2(d x) p ⇡✏0kx. b) Determine the maximum charge that can be measured in terms of ✏0, k, and d, and state the separation of the spheres then: With increasing charge, the spheres move toward each other until they just touch at xmax = d 2a. Using the part a result, we find the maximum measurable charge: Qmax = 4a p ⇡✏0k(d 2a). Presumably some form of stop mechanism is placed at x = xmax to prevent the spheres from actually touching. c) What happens if a larger charge is applied? No further motion is possible, so nothing happens. 2.9. A 100 nC point charge is located at A( 1, 1, 3) in free space. a) Find the locus of all points P(x, y, z) at which Ex = 500 V/m: The total field at P will be: EP = 100 ⇥ 10 9 4⇡✏0 RAP |RAP |3 where RAP = (x+1)ax +(y 1)ay +(z 3)az, and where |RAP | = [(x+1)2 +(y 1)2 +(z 3)2 ]1/2 . The x component of the field will be Ex = 100 ⇥ 10 9 4⇡✏0 (x + 1) [(x + 1)2 + (y 1)2 + (z 3)2]1.5 = 500 V/m And so our condition becomes: (x + 1) = 0.56 [(x + 1)2 + (y 1)2 + (z 3)2 ]1.5 15
2.9b) Find y1if P( 2, y1, 3) lies on that locus: At point P, the condition of part a becomes 3.19 = ⇥ 1 + (y1 1)2 ⇤3 from which (y1 1)2 = 0.47, or y1 = 1.69 or 0.31 2.10. A charge of -1 nC is located at the origin in free space. What charge must be located at (2,0,0) to cause Ex to be zero at (3,1,1)? The field from two point charges is given generally by ET = q1(r r0 1) 4⇡✏0|r r0 1|3 + q2(r r0 2) 4⇡✏0|r r0 2|3 (1) where we let q1 = 1nC and q2 is to be found. With q1 at the origin, r0 1 = 0. The position vector for q2 is then r0 2 = 2ax. The observation point at (3,1,1) gives r = 3ax + ay + az. Eq. (1) becomes 1 4⇡✏0 1(3ax + ay + az) (32 + 1 + 1)3/2 + q2[(3 2)ax + ay + az] (1 + 1 + 1)3/2 Requiring the x component to be zero leads to q2 = 35/2 113/2 = 0.43 nC 2.11. A charge Q0 located at the origin in free space produces a field for which Ez = 1 kV/m at point P( 2, 1, 1). a) Find Q0: The field at P will be EP = Q0 4⇡✏0 2ax + ay az 61.5 Since the z component is of value 1 kV/m, we find Q0 = 4⇡✏061.5 ⇥ 103 = 1.63 µC. b) Find E at M(1, 6, 5) in cartesian coordinates: This field will be: EM = 1.63 ⇥ 10 6 4⇡✏0 ax + 6ay + 5az [1 + 36 + 25]1.5 or EM = 30.11ax 180.63ay 150.53az. c) Find E at M(1, 6, 5) in cylindrical coordinates: At M, ⇢ = p 1 + 36 = 6.08, = tan 1 (6/1) = 80.54 , and z = 5. Now E⇢ = EM · a⇢ = 30.11 cos 180.63 sin = 183.12 E = EM · a = 30.11( sin ) 180.63 cos = 0 (as expected) so that EM = 183.12a⇢ 150.53az. d) Find E at M(1, 6, 5) in spherical coordinates: At M, r = p 1 + 36 + 25 = 7.87, = 80.54 (as before), and ✓ = cos 1 (5/7.87) = 50.58 . Now, since the charge is at the origin, we expect to obtain only a radial component of EM . This will be: Er = EM · ar = 30.11 sin ✓ cos 180.63 sin ✓ sin 150.53 cos ✓ = 237.1 16
2.12. Electrons arein random motion in a fixed region in space. During any 1µs interval, the probability of finding an electron in a subregion of volume 10 15 m2 is 0.27. What volume charge density, appropriate for such time durations, should be assigned to that subregion? The finite probabilty e↵ectively reduces the net charge quantity by the probability fraction. With e = 1.602 ⇥ 10 19 C, the density becomes ⇢v = 0.27 ⇥ 1.602 ⇥ 10 19 10 15 = 43.3 µC/m3 2.13. A uniform volume charge density of 0.2 µC/m3 is present throughout the spherical shell extending from r = 3 cm to r = 5 cm. If ⇢v = 0 elsewhere: a) find the total charge present throughout the shell: This will be Q = Z 2⇡ 0 Z ⇡ 0 Z .05 .03 0.2 r2 sin ✓ dr d✓ d = 4⇡(0.2) r3 3 .05 .03 = 8.21 ⇥ 10 5 µC = 82.1 pC b) find r1 if half the total charge is located in the region 3 cm < r < r1: If the integral over r in part a is taken to r1, we would obtain 4⇡(0.2) r3 3 r1 .03 = 4.105 ⇥ 10 5 Thus r1 = 3 ⇥ 4.105 ⇥ 10 5 0.2 ⇥ 4⇡ + (.03)3 1/3 = 4.24 cm 2.14. The electron beam in a certain cathode ray tube possesses cylindrical symmetry, and the charge density is represented by ⇢v = 0.1/(⇢2 + 10 8 ) pC/m3 for 0 < ⇢ < 3 ⇥ 10 4 m, and ⇢v = 0 for ⇢ > 3 ⇥ 10 4 m. a) Find the total charge per meter along the length of the beam: We integrate the charge density over the cylindrical volume having radius 3 ⇥ 10 4 m, and length 1m. q = Z 1 0 Z 2⇡ 0 Z 3⇥10 4 0 0.1 (⇢2 + 10 8) ⇢ d⇢ d dz From integral tables, this evaluates as q = 0.2⇡ ✓ 1 2 ◆ ln ⇢2 + 10 8 3⇥10 4 0 = 0.1⇡ ln(10) = 0.23⇡ pC/m b) if the electron velocity is 5 ⇥ 107 m/s, and with one ampere defined as 1C/s, find the beam current: Current = charge/m ⇥ v = 0.23⇡ [pC/m] ⇥ 5 ⇥ 107 [m/s] = 11.5⇡ ⇥ 106 [pC/s] = 11.5⇡ µA 17
2.15. A sphericalvolume having a 2 µm radius contains a uniform volume charge density of 105 C/m3 (not 1015 as stated in earlier printings). a) What total charge is enclosed in the spherical volume? This will be Q = (4/3)⇡(2 ⇥ 10 6 )3 ⇥ 105 = 3.35 ⇥ 10 12 C. b) Now assume that a large region contains one of these little spheres at every corner of a cubical grid 3mm on a side, and that there is no charge between spheres. What is the average volume charge density throughout this large region? Each cube will contain the equivalent of one little sphere. Neglecting the little sphere volume, the average density becomes ⇢v,avg = 3.35 ⇥ 10 12 (0.003)3 = 1.24 ⇥ 10 4 C/m3 2.16. Within a region of free space, charge density is given as ⇢v = (⇢0r/a) cos ✓ C/m3 , where ⇢0 and a are constants. Find the total charge lying within: a) the sphere, r a: This will be Qa = Z 2⇡ 0 Z ⇡ 0 Z a 0 ⇢0r a cos ✓ r2 sin ✓ dr d✓ d = 2⇡ Z a 0 ⇢0r3 a dr = 0 b) the cone, r a, 0 ✓ 0.1⇡: Qb = Z 2⇡ 0 Z 0.1⇡ 0 Z a 0 ⇢0r a cos ✓ r2 sin ✓ dr d✓ d = ⇡ ⇢0a3 4 ⇥ 1 cos2 (0.1⇡) ⇤ = 0.024⇡⇢0a3 c) the region, r a, 0 ✓ 0.1⇡, 0 0.2⇡. Qc = Z 0.2⇡ 0 Z 0.1⇡ 0 Z a 0 ⇢0r a cos ✓ r2 sin ✓ dr d✓ d = 0.024⇡⇢0a3 ✓ 0.2⇡ 2⇡ ◆ = 0.0024⇡⇢0a3 2.17. A uniform line charge of 16 nC/m is located along the line defined by y = 2, z = 5. If ✏ = ✏0: a) Find E at P(1, 2, 3): This will be EP = ⇢l 2⇡✏0 RP |RP |2 where RP = (1, 2, 3) (1, 2, 5) = (0, 4, 2), and |RP |2 = 20. So EP = 16 ⇥ 10 9 2⇡✏0 4ay 2az 20 = 57.5ay 28.8az V/m b) Find E at that point in the z = 0 plane where the direction of E is given by (1/3)ay (2/3)az: With z = 0, the general field will be Ez=0 = ⇢l 2⇡✏0 (y + 2)ay 5az (y + 2)2 + 25 18
We require |Ez|= |2Ey|, so 2(y + 2) = 5. Thus y = 1/2, and the field becomes: Ez=0 = ⇢l 2⇡✏0 2.5ay 5az (2.5)2 + 25 = 23ay 46az 2.18. a) Find E in the plane z = 0 that is produced by a uniform line charge, ⇢L, extending along the z axis over the range L < z < L in a cylindrical coordinate system: We find E through E = Z L L ⇢Ldz(r r0 ) 4⇡✏0|r r0|3 where the observation point position vector is r = ⇢a⇢ (anywhere in the x-y plane), and where the position vector that locates any di↵erential charge element on the z axis is r0 = zaz. So r r0 = ⇢a⇢ zaz, and |r r0 | = (⇢2 + z2 )1/2 . These relations are substituted into the integral to yield: E = Z L L ⇢Ldz(⇢a⇢ zaz) 4⇡✏0(⇢2 + z2)3/2 = ⇢L ⇢ a⇢ 4⇡✏0 Z L L dz (⇢2 + z2)3/2 = E⇢ a⇢ Note that the second term in the left-hand integral (involving zaz) has e↵ectively vanished because it produces equal and opposite sign contributions when the integral is taken over symmetric limits (odd parity). Evaluating the integral results in E⇢ = ⇢L ⇢ 4⇡✏0 z ⇢2 p ⇢2 + z2 L L = ⇢L 2⇡✏0⇢ L p ⇢2 + L2 = ⇢L 2⇡✏0⇢ 1 p 1 + (⇢/L)2 Note that as L ! 1, the expression reduces to the expected field of the infinite line charge in free space, ⇢L/(2⇡✏0⇢). b) if the finite line charge is approximated by an infinite line charge (L ! 1), by what percentage is E⇢ in error if ⇢ = 0.5L? The percent error in this situation will be % error = " 1 1 p 1 + (⇢/L)2 # ⇥ 100 For ⇢ = 0.5L, this becomes % error = 10.6 % c) repeat b with ⇢ = 0.1L. For this value, obtain % error = 0.496 %. 2.19. A uniform line charge of 2 µC/m is located on the z axis. Find E in rectangular coordinates at P(1, 2, 3) if the charge extends from a) 1 < z < 1: With the infinite line, we know that the field will have only a radial component in cylindrical coordinates (or x and y components in cartesian). The field from an infinite line on the z axis is generally E = [⇢l/(2⇡✏0⇢)]a⇢. Therefore, at point P: EP = ⇢l 2⇡✏0 RzP |RzP |2 = (2 ⇥ 10 6 ) 2⇡✏0 ax + 2ay 5 = 7.2ax + 14.4ay kV/m where RzP is the vector that extends from the line charge to point P, and is perpendicular to the z axis; i.e., RzP = (1, 2, 3) (0, 0, 3) = (1, 2, 0). 19
b) 4 z 4: Here we use the general relation EP = Z ⇢ldz 4⇡✏0 r r0 |r r0|3 2.19b (continued) where r = ax + 2ay + 3az and r0 = zaz. So the integral becomes EP = (2 ⇥ 10 6 ) 4⇡✏0 Z 4 4 ax + 2ay + (3 z)az [5 + (3 z)2]1.5 dz Using integral tables, we obtain: EP = 3597 (ax + 2ay)(z 3) + 5az (z2 6z + 14) 4 4 V/m = 4.9ax + 9.8ay + 4.9az kV/m The student is invited to verify that when evaluating the above expression over the limits 1 < z < 1, the z component vanishes and the x and y components become those found in part a. 2.20. A line charge of uniform charge density ⇢0 C/m and of length , is oriented along the z axis at/2 < z < /2. a) Find the electric field strength, E, in magnitude and direction at any position along the x axis: This follows the method in Problem 2.18. We find E through E = Z/2 /2 ⇢0 dz(r r0 ) 4⇡✏0|r r0|3 where the observation point position vector is r = xax (anywhere on the x axis), and where the position vector that locates any di↵erential charge element on the z axis is r0 = zaz. So r r0 = xax zaz, and |r r0 | = (x2 + z2 )1/2 . These relations are substituted into the integral to yield: E = Z/2 /2 ⇢0dz(xax zaz) 4⇡✏0(x2 + z2)3/2 = ⇢0 x ax 4⇡✏0 Z/2 /2 dz (x2 + z2)3/2 = Ex ax Note that the second term in the left-hand integral (involving zaz) has e↵ectively vanished because it produces equal and opposite sign contributions when the integral is taken over symmetric limits (odd parity). Evaluating the integral results in Ex = ⇢0 x 4⇡✏0 z x2 p x2 + z2/2 /2 = ⇢0 2⇡✏0x/2 p x2 + (/2)2 = ⇢0 2⇡✏0x 1 p 1 + (2x/)2 b) with the given line charge in position, find the force acting on an identical line charge that is oriented along the x axis at /2 < x < 3/2: The di↵erential force on an element of the x-directed line charge will be dF = dqE = (⇢0 dx)E, where E is the field as determined in part a. The net force is then the integral of the di↵erential force over the length of the horizontal line charge, or F = Z 3/2/2 ⇢2 0 2⇡✏0x 1 p 1 + (2x/`)2 dx ax 20
This can bere-written and then evaluated using integral tables as F = ⇢2 0 ax 4⇡✏0 Z 3/2 /2 dx x p x2 + (/2)2 = ⇢2 0 ax 4⇡✏0 0 @ 1 (/2) ln " /2 + p x2 + (/2)2 x #3/2/2 1 A = ⇢2 0 ax 2⇡✏0 ln " (/2) 1 + p 10 3(/2) 1 + p 2 # = ⇢2 0 ax 2⇡✏0 ln " 3(1 + p 2) 1 + p 10 # = 0.55⇢2 0 2⇡✏0 ax N 2.21. Two identical uniform line charges with ⇢l = 75 nC/m are located in free space at x = 0, y = ±0.4 m. What force per unit length does each line charge exert on the other? The charges are parallel to the z axis and are separated by 0.8 m. Thus the field from the charge at y = 0.4 evaluated at the location of the charge at y = +0.4 will be E = [⇢l/(2⇡✏0(0.8))]ay. The force on a di↵erential length of the line at the positive y location is dF = dqE = ⇢ldzE. Thus the force per unit length acting on the line at postive y arising from the charge at negative y is F = Z 1 0 ⇢2 l dz 2⇡✏0(0.8) ay = 1.26 ⇥ 10 4 ay N/m = 126 ay µN/m The force on the line at negative y is of course the same, but with ay. 2.22. Two identical uniform sheet charges with ⇢s = 100 nC/m2 are located in free space at z = ±2.0 cm. What force per unit area does each sheet exert on the other? The field from the top sheet is E = ⇢s/(2✏0) az V/m. The di↵erential force produced by this field on the bottom sheet is the charge density on the bottom sheet times the di↵erential area there, multiplied by the electric field from the top sheet: dF = ⇢sdaE. The force per unit area is then just F = ⇢sE = (100 ⇥ 10 9 )( 100 ⇥ 10 9 )/(2✏0) az = 5.6 ⇥ 10 4 az N/m2 . 2.23. Given the surface charge density, ⇢s = 2 µC/m2 , in the region ⇢ < 0.2 m, z = 0. Find E at: a) PA(⇢ = 0, z = 0.5): First, we recognize from symmetry that only a z component of E will be present. Considering a general point z on the z axis, we have r = zaz. Then, with r0 = ⇢a⇢, we obtain r r0 = zaz ⇢a⇢. The superposition integral for the z component of E will be: Ez,PA = ⇢s 4⇡✏0 Z 2⇡ 0 Z 0.2 0 z ⇢ d⇢ d (⇢2 + z2)1.5 = 2⇡⇢s 4⇡✏0 z " 1 p z2 + ⇢2 #0.2 0 = ⇢s 2✏0 z 1 p z2 1 p z2 + 0.04 With z = 0.5 m, the above evaluates as Ez,PA = 8.1 kV/m. b) PB(⇢ = 0, z = 0.5). With z at 0.5 m, we evaluate the expression for Ez to obtain Ez,PB = 8.1 kV/m. c) Show that the field along the z axis reduces to that of an infinite sheet charge at small values of z: In general, the field can be expressed as Ez = ⇢s 2✏0 1 z p z2 + 0.04 21
At small z,this reduces to Ez . = ⇢s/2✏0, which is the infinite sheet charge field. d) Show that the z axis field reduces to that of a point charge at large values of z: The development is as follows: Ez = ⇢s 2✏0 1 z p z2 + 0.04 = ⇢s 2✏0 " 1 z z p 1 + 0.04/z2 # . = ⇢s 2✏0 1 1 1 + (1/2)(0.04)/z2 where the last approximation is valid if z >> .04. Continuing: Ez . = ⇢s 2✏0 ⇥ 1 [1 (1/2)(0.04)/z2 ] ⇤ = 0.04⇢s 4✏0z2 = ⇡(0.2)2 ⇢s 4⇡✏0z2 This the point charge field, where we identify q = ⇡(0.2)2 ⇢s as the total charge on the disk (which now looks like a point). 22
2.24. a) Findthe electric field on the z axis produced by an annular ring of uniform surface charge density ⇢s in free space. The ring occupies the region z = 0, a ⇢ b, 0 2⇡ in cylindrical coordinates: We find the field through E = Z Z ⇢sda(r r0 ) 4⇡✏0|r r0|3 where the integral is taken over the surface of the annular ring, and where r = zaz and r0 = ⇢a⇢. The integral then becomes E = Z 2⇡ 0 Z b a ⇢s ⇢ d⇢ d (zaz ⇢a⇢) 4⇡✏0(z2 + ⇢2)3/2 In evaluating this integral, we first note that the term involving ⇢a⇢ integrates to zero over the integration range of 0 to 2⇡. This is because we need to introduce the dependence in a⇢ by writing it as a⇢ = cos ax + sin ay, where ax and ay are invariant in their orientation as varies. So the integral now simplifies to E = 2⇡⇢s z az 4⇡✏0 Z b a ⇢ d⇢ (z2 + ⇢2)3/2 = ⇢s z az 2✏0 " 1 p z2 + ⇢2 #b a = ⇢s 2✏0 " 1 p 1 + (a/z)2 1 p 1 + (b/z)2 # az b) from your part a result, obtain the field of an infinite uniform sheet charge by taking appropriate limits. The infinite sheet is obtained by letting a ! 0 and b ! 1, in which case E ! ⇢s/(2✏0) az as expected. 2.25. Find E at the origin if the following charge distributions are present in free space: point charge, 12 nC at P(2, 0, 6); uniform line charge density, 3nC/m at x = 2, y = 3; uniform surface charge density, 0.2 nC/m2 at x = 2. The sum of the fields at the origin from each charge in order is: E = (12 ⇥ 10 9 ) 4⇡✏0 ( 2ax 6az) (4 + 36)1.5 + (3 ⇥ 10 9 ) 2⇡✏0 (2ax 3ay) (4 + 9) (0.2 ⇥ 10 9 )ax 2✏0 = 3.9ax 12.4ay 2.5az V/m 23
2.26. Radially-dependent surfacecharge is distributed on an infinite flat sheet in the xy plane, and is char- acterized in cylindrical coordinates by surface density ⇢s = ⇢0/⇢, where ⇢0 is a constant. Determine the electric field strength, E, everywhere on the z axis. We find the field through E = Z Z ⇢sda(r r0 ) 4⇡✏0|r r0|3 where the integral is taken over the surface of the annular ring, and where r = zaz and r0 = ⇢a⇢. The integral then becomes E = Z 2⇡ 0 Z 1 0 (⇢0/⇢) ⇢ d⇢ d (zaz ⇢a⇢) 4⇡✏0(z2 + ⇢2)3/2 In evaluating this integral, we first note that the term involving ⇢a⇢ integrates to zero over the integration range of 0 to 2⇡. This is because we need to introduce the dependence in a⇢ by writing it as a⇢ = cos ax + sin ay, where ax and ay are invariant in their orientation as varies. So the integral now simplifies to E = 2⇡⇢s z az 4⇡✏0 Z 1 0 d⇢ (z2 + ⇢2)3/2 = ⇢s z az 2✏0 " ⇢ z2 p z2 + ⇢2 #1 ⇢=0 = ⇢s 2✏0z az 2.27. Given the electric field E = (4x 2y)ax (2x + 4y)ay, find: a) the equation of the streamline that passes through the point P(2, 3, 4): We write dy dx = Ey Ex = (2x + 4y) (4x 2y) Thus 2(x dy + y dx) = y dy x dx or 2 d(xy) = 1 2 d(y2 ) 1 2 d(x2 ) So C1 + 2xy = 1 2 y2 1 2 x2 or y2 x2 = 4xy + C2 Evaluating at P(2, 3, 4), obtain: 9 4 = 24 + C2, or C2 = 19 Finally, at P, the requested equation is y2 x2 = 4xy 19 b) a unit vector specifying the direction of E at Q(3, 2, 5): Have EQ = [4(3) + 2(2)]ax [2(3) 4(2)]ay = 16ax + 2ay. Then |E| = p 162 + 4 = 16.12 So aQ = 16ax + 2ay 16.12 = 0.99ax + 0.12ay 24
2.28 An electricdipole (discussed in detail in Sec. 4.7) consists of two point charges of equal and opposite magnitude ±Q spaced by distance d. With the charges along the z axis at positions z = ±d/2 (with the positive charge at the positive z location), the electric field in spherical coordinates is given by E(r, ✓) = ⇥ Qd/(4⇡✏0r3 ) ⇤ [2 cos ✓ar + sin ✓a✓], where r >> d. Using rectangular coordinates, determine expressions for the vector force on a point charge of magnitude q: a) at (0, 0, z): Here, ✓ = 0, ar = az, and r = z. Therefore F(0, 0, z) = qQd az 4⇡✏0z3 N b) at (0, y, 0): Here, ✓ = 90 , a✓ = az, and r = y. The force is F(0, y, 0) = qQd az 4⇡✏0y3 N 2.29. If E = 20e 5y (cos 5xax sin 5xay), find: a) |E| at P(⇡/6, 0.1, 2): Substituting this point, we obtain EP = 10.6ax 6.1ay, and so |EP | = 12.2. b) a unit vector in the direction of EP : The unit vector associated with E is (cos 5xax sin 5xay), which evaluated at P becomes aE = 0.87ax 0.50ay. c) the equation of the direction line passing through P: Use dy dx = sin 5x cos 5x = tan 5x ) dy = tan 5x dx Thus y = 1 5 ln cos 5x + C. Evaluating at P, we find C = 0.13, and so y = 1 5 ln cos 5x + 0.13 2.30. For fields that do not vary with z in cylindrical coordinates, the equations of the streamlines are obtained by solving the di↵erential equation E⇢/E = d⇢(⇢d ). Find the equation of the line passing through the point (2, 30 , 0) for the field E = ⇢ cos 2 a⇢ ⇢ sin 2 a : E⇢ E = d⇢ ⇢d = ⇢ cos 2 ⇢ sin 2 = cot 2 ) d⇢ ⇢ = cot 2 d Integrate to obtain 2 ln ⇢ = ln sin 2 + ln C = ln C sin 2 ) ⇢2 = C sin 2 At the given point, we have 4 = C/ sin(60 ) ) C = 4 sin 60 = 2 p 3. Finally, the equation for the streamline is ⇢2 = 2 p 3/ sin 2 . 25 Engineering Electromagnetics 8th Edition Hayt Solutions Manual Full Download: This sample only, Download all chapters at: alibabadownload.com
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14895 | https://books.google.com/books/about/Difference_Equations.html?id=tQORIGHnSf8C | Difference Equations: From Rabbits to Chaos - Paul Cull, Mary Flahive, Robert O. Robson - Google Books
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Difference Equations: From Rabbits to Chaos =========================================== Paul Cull, Mary Flahive, Robert O. Robson Taylor & Francis US, Apr 12, 2005 - Mathematics - 392 pages Difference equations are models of the world around us. From clocks to computers to chromosomes, processing discrete objects in discrete steps is a common theme. Difference equations arise naturally from such discrete descriptions and allow us to pose and answer such questions as: How much? How many? How long? Difference equations are a necessary part of the mathematical repertoire of all modern scientists and engineers. In this new text, designed for sophomores studying mathematics and computer science, the authors cover the basics of difference equations and some of their applications in computing and in population biology. Each chapter leads to techniques that can be applied by hand to small examples or programmed for larger problems. Along the way, the reader will use linear algebra and graph theory, develop formal power series, solve combinatorial problems, visit Perron—Frobenius theory, discuss pseudorandom number generation and integer factorization, and apply the Fast Fourier Transform to multiply polynomials quickly. The book contains many worked examples and over 250 exercises. While these exercises are accessible to students and have been class-tested, they also suggest further problems and possible research topics. Paul Cull is a professor of Computer Science at Oregon State University. Mary Flahive is a professor of Mathematics at Oregon State University. Robby Robson is president of Eduworks, an e-learning consulting firm. None has a rabbit. More » Preview this book »
Contents
Fibonacci Numbers 1
Homogeneous Linear Recurrence Relations 11
Finite Difference Equations 33
Generating Functions 67
Nonnegative Difference Equations 101
Leslies Population Matrix Model 137
Matrix Difference Equations 179
Modular Recurrences 217
Some Nonlinear Recurrences 297
A Worked Examples 337
B Complex Numbers 347
Highlights of Linear Algebra 353
Roots in the Unit Circle 361
References369
Index381
Copyright
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Computational Complexity 253
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Title Difference Equations: From Rabbits to Chaos
Undergraduate Texts in Mathematics, ISSN 0172-6056
AuthorsPaul Cull, Mary Flahive, Robert O. Robson
Edition illustrated
Publisher Taylor & Francis US, 2005
ISBN 0387232338, 9780387232331
Length 392 pages
SubjectsMathematics
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Algebra
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General
Mathematics / Algebra / General
Mathematics / Algebra / Linear
Mathematics / Applied
Mathematics / Calculus
Mathematics / Combinatorics
Mathematics / Differential Equations / General
Mathematics / Discrete Mathematics
Mathematics / General
Mathematics / Mathematical Analysis
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Covering subsets of the integers by congruences
Michael Filaseta, Wilson Harvey
To cite this version:
Michael Filaseta, Wilson Harvey. Covering subsets of the integers by congruences. Acta Arithmetica, 2018, 182 (1), pp.43-72. 10.4064/aa161214-4-10. hal-04167632 ACTA ARITHMETICA
Online First version
Covering subsets of the integers by congruences
by
Michael Filaseta (Columbia, SC) and Wilson Harvey (Monroe, LA)
Introduction. In 1950, P. Erd˝ os introduced the concept of cov-ering systems (or coverings for short) of the integers to show that a positive proportion of the positive odd integers cannot be expressed as a prime plus a power of 2. His argument was based on the observation that
n ≡ 0 (mod 2) ⇒ 2n ≡ 1 (mod 3) ,n ≡ 0 (mod 3) ⇒ 2n ≡ 1 (mod 7) ,n ≡ 1 (mod 4) ⇒ 2n ≡ 2 (mod 5) ,n ≡ 3 (mod 8) ⇒ 2n ≡ 8 (mod 17) ,n ≡ 7 (mod 12) ⇒ 2n ≡ 11 (mod 13) ,n ≡ 23 (mod 24) ⇒ 2n ≡ 121 (mod 241) ,
where the congruences on the left form a covering of the integers, that is, a finite set of congruences with distinct moduli > 1 having the property that every integer satisfies at least one congruence in the set. The rest of the Erd˝ os argument is fairly simple. One takes a positive odd integer N
satisfying the congruences on the right above, that is, the following: (1.1) x ≡ 1 (mod 3) , x ≡ 1 (mod 7) , x ≡ 2 (mod 5) ,x ≡ 8 (mod 17) , x ≡ 11 (mod 13) , x ≡ 121 (mod 241) .
If N = 2 n + p where n is an integer ≥ 0 and p is a prime, then p = N − 2n
and the implications above imply that p ∈ S := {3, 5, 7, 13 , 17 , 241 }. What Erd˝ os had in mind to handle the case that p ∈ S is unclear. However, with a little effort it can be shown that if N satisfies the congruences in (1.1) and n is an integer ≥ 0, then N − 2n cannot equal a prime in the set S.More precisely, N ≡ 1 (mod 3) implies N − 2n ≡ 0 or 2 (mod 3), and the
2010 Mathematics Subject Classification : Primary 11A07; Secondary 11B25.
Key words and phrases : covering system, subsets of integers. Received 14 December 2016; revised 23 July 2017. Published online .
DOI: 10.4064/aa161214-4-10 c©Instytut Matematyczny PAN, 2017 2M. Filaseta and W. Harvey
condition N ≡ 1 (mod 7) implies N − 2n ≡ 0, 4 or 6 (mod 7). As no prime p
in S satisfies both p ≡ 0 or 2 (mod 3) and p ≡ 0, 4 or 6 (mod 7), the result of Erd˝ os follows. Coverings of the integers have found a number of interesting applications; for example, see [2, 4, 5, 11, 15, 16, 17]. The literature on the subject is ex-tensive (cf. ). There has in particular been a great deal of interest in two old problems on the subject. The first, recently resolved by Bob Hough , is whether the minimum modulus of a covering system of the integers consist-ing of congruences can be arbitrarily large. Bob Hough has shown that this minimum modulus is ≤ 10 16 . In the other direction, Pace Nielsen has shown that the minimum modulus can be as large as 40, and later his ideas were extended by Tyler Owens who showed the minimum modulus can be as large as 42. The second problem is to determine whether or not there is a covering of the integers consisting only of odd moduli. This prob-lem remains open, though recent work of Bob Hough and Pace Nielsen shows that every covering of the integers has some modulus divisible by either 2 or 3. With this in mind, Ognian Trifonov (private communication) has raised the question as to what nice sets of integers can be covered by a finite collection of congruences where all moduli are distinct and arbitrarily large or where all moduli are distinct odd numbers > 1. To formalize this discussion better, we make the following definitions.
Definition 1.1 . A covering of a set S ⊆ Z is a finite set {x ≡ aj
(mod mj ) : 1 ≤ j ≤ r} of congruences such that the moduli mj are distinct integers > 1 and each s ∈ S satisfies at least one of them. An odd covering of a set S ⊆ Z is a covering C of S where each mj is odd. An exact covering of a set S ⊆ Z is a covering C of S for which each s ∈ S satisfies exactly one of the congruences in C. If S = Z, we speak of a covering of the integers , an odd covering of the integers , and an exact covering of the integers , respectively. We emphasize that, in this paper, as defined above, coverings will always refer to systems of congruences where the moduli are distinct and > 1. This will allow for our results, which include these conditions, to be stated more succinctly. We also freely use the terminology that a set S ⊆ Z is covered by a set of congruences (or a set of congruences covers S) to mean that the set of congruences form a covering of S as defined above. On the other hand, we will still have use of sets C of congruences with repeated moduli such that every element of some set S satisfies at least one congruence from C. To be consistent, we do not refer to such C as coverings of S (see, for example, Lemma 2.2 used in the proof of Proposition 2.10, and the statement of Theorem 2.5). The idea of covering subsets of the integers is not really new. For ex-ample, the motivating paper by Erd˝ os demonstrates that (1.1) is not Covering subsets of the integers by congruences 3
only an odd covering of the powers of 2 but a covering of the powers of 2 where the moduli are primes. Hence, the idea that one can cover a subset of
Z in some nice way is in fact part of the point of the original use of coverings given by Erd˝ os . Analogously to (1.1) providing a covering of the powers of 2, A. Granville has pointed out that (1.2) x ≡ 1 (mod 2) , x ≡ 0 (mod 3) , x ≡ 1 (mod 7) ,x ≡ 0 (mod 17) , x ≡ 2 (mod 19) , x ≡ 15 (mod 23) provides a covering of the Fibonacci numbers. As in (1.1), the moduli in (1.2) are primes. It is also of interest that the sum of the reciprocals of the moduli in (1.1) is 0 .816 . . . < 1 and the sum of the reciprocals of the moduli in (1.2) is 1 .131 . . . > 1. For coverings of the integers, it is well-known and not difficult to see that the sum of the reciprocals of the moduli is always > 1. We have not addressed the literature on exact coverings of the integers. A classic argument (cf. ) using a complex variable shows that there is no exact covering of the integers; more precisely, if a finite system of congru-ences x ≡ aj (mod mj ), where 1 ≤ j ≤ r and m1 ≤ · · · ≤ mr, is such that every integer satisfies exactly one congruence in the system, then mr = mr−1
so that the system is not a covering of the integers. On the other hand, there can be exact coverings for subsets of the integers. The concept of covering subsets of Z sparks a great deal of questions, and our hope here is to raise some interest in the topic. Some explicit examples of coverings of subsets of Z that we obtain include:
• There is an odd covering of the set of primes.
• There is an odd covering of the numbers that are sums of two squares.
• There is an exact covering of the powers of 2 consisting of moduli only divisible by primes larger than any prescribed value and with the sum of the reciprocals of the moduli smaller than any prescribed ε > 0.
• There is a covering of the Fibonacci numbers consisting of moduli only divisible by primes larger than any prescribed value and with the sum of the reciprocals of the moduli smaller than any prescribed ε > 0.
• There is an exact odd covering of the Fibonacci numbers. Our demonstration of the second example above involves a lengthy con-struction using 64731 congruences, though no real attempt has been made to minimize this number. In the third and fourth examples, the conditions stated imply that we can find such coverings where the moduli are odd and with the minimum modulus larger than any prescribed value. These two examples, in particular, might suggest that subsets N = {n1, n 2, . . . } of the integers for which nj increases sufficiently fast are more easily covered using large moduli, but we show in the next section that this is not in general true. More precisely, we show that there are sets N for which the nj increase as 4 M. Filaseta and W. Harvey
fast as one wants but which cannot be covered by congruences with moduli all > 10 16 . An analogous result is obtained for odd coverings of subsets of the integers provided no odd covering of the integers exists. Among the questions we have not addressed are the following. Does there exist a covering of the set of primes or of the set of numbers which are sums of two squares that uses only moduli that are greater than an arbitrary fixed bound M ? Does there exist an odd covering of the set of squarefree numbers? Does there exist an odd covering of the set of squarefree numbers using moduli that are all > 10 100 ? Although we were able to obtain some results for subsets of the form Sf = {f (n) : n ∈ Z+}, where f (x) ∈ Z[x], we are far from a general result in this direction. In particular, is it true that for every such Sf , there is an odd covering of Sf ? Although it will be clear that part of our approach for Fibonacci numbers generalizes to other recursive sequences, we have not obtained a result for arbitrary recursive sequences in the integers.
Preliminary results. In this section, we discuss some basic results in the general spirit of understanding coverings of subsets of Z, and then turn to some further results that follow from recent work from and .
Proposition 2.1 . If there is no odd covering of the integers, then there is no odd covering of the odd integers. Proof. Suppose there exists an odd covering C of the odd integers. Thus,
C = {x ≡ ai (mod mi) : 1 ≤ i ≤ r}, where each mi is odd and 1 < m 1 <
· · · < m r. To establish the proposition, it suffices to show that C covers all the integers. Define M = ∏rj=1 mj , and note that M is odd. Let n ∈ Z. Then either n
is odd and covered by a congruence in C, or n + M is odd and thus covered by a congruence in C. We deduce that n + εM ≡ ak (mod mk) for some
k ∈ { 1, . . . , r } and ε ∈ { 0, 1}. But mk | M , so n ≡ n + εM ≡ ak (mod mk). Thus, n necessarily satisfies a congruence from C.
Lemma 2.2 . Let t ∈ Z and S ⊆ Z. Given a set C = {x ≡ ak (mod mk) : 1 ≤ k ≤ r} of congruences such that every element of S satisfies at least one congruence in C, the set Ct = {x ≡ ak + t (mod mk) : 1 ≤ k ≤ r} has the property that every element of St = {s + t : s ∈ S} satisfies at least one congruence in Ct.Proof. Fix x0 ∈ St. We may write x0 = s0 + t for some s0 ∈ S. The conditions in the lemma imply there exists k0 ∈ { 1, . . . , r } with s0 ≡ ak0
(mod mk0 ). Further, x0 = s0 + t ≡ ak0 + t (mod mk0 ), so x0 satisfies a congruence in Ct.The following is a simple consequence of the previous two results. Covering subsets of the integers by congruences 5
Corollary 2.3 . If there is no odd covering of the integers, then there is no odd covering of the even integers.
Along the lines of basic results like Lemma 2.2, we note that if C is a set of congruences with the least common multiple of the moduli equal to L,then C is a covering of Z if and only if C is a covering of a set S containing
L consecutive integers. In particular, C is a covering of Z if and only if C is a covering of Z+.
Definition 2.4 . Let P be a property that can be satisfied by some subsets of Z+. We say that there exist arbitrarily thin sets S satisfying P if for all functions f with f (x) → ∞ as x → ∞ , there exists a set S, depending on f , satisfying P and an x0 ∈ R+ such that
|{ s ∈ S : s ≤ x}| ≤ f (x) for all x ≥ x0.
Theorem 2.5 . Let r ∈ Z+. Let N be a possibly infinite set of congru-ences x ≡ a (mod m), with 0 ≤ a < m , such that for any finite set C of congruences from N , with each modulus appearing in C at most r times, there exists some integer which fails to satisfy every congruence in C. Then there exist arbitrarily thin sets S ⊆ Z+ such that for any finite set C of congruences from N , with each modulus appearing in C at most r times, there exists some element of S which fails to satisfy every congruence in C.Proof. Let C(M ) denote the set of all finite subsets
C = {x ≡ ak (mod mk) : 1 ≤ k ≤ s} ⊆ N with m1 · · · ms ≤ M.
Given the definition of N , we have 0 ≤ aj < m j for each j. Note that for any M , there are finitely many sets C in C(M ). As every finite subset C
of N belongs to C(M ) for some positive integer M , we deduce that the finite subsets of N are countably many. We take the finite subsets C of N where each modulus appears in C
at most r times, and we order them C1, C 2, . . . . By the conditions in the theorem, for each Cj , there is some integer nj which fails to satisfy every congruence in Cj . Then for every subset C ⊆ N , with each modulus ap-pearing in C at most r times, there exists some element of S = {n1, n 2, . . . }
which fails to satisfy every congruence in C.Observe that if a number n fails to satisfy every congruence in Ck, then so does any number that is of the form n plus a multiple of the product of the moduli in Ck. Hence, the values nk may grow at any desired rate, implying that the set S can be constructed to be arbitrarily thin. The case r = 1 is of particular importance to us. If we take N equal to the set of congruences x ≡ a (mod m) where m > 10 16 and a ∈ [0 , m ) or if
N is the set of congruences x ≡ a (mod m) where m is odd and > 1 and
a ∈ [0 , m ), then we obtain respectively the following two results. 6 M. Filaseta and W. Harvey
Corollary 2.6 . There are arbitrarily thin sets S ⊆ Z for which no covering of S exists using only moduli greater than 10 16 .
Corollary 2.7 . If there is no odd covering of the integers, then there are arbitrarily thin sets S ⊆ Z for which no odd covering of S exists.
There is an alternative approach to establishing Corollaries 2.6 and 2.7: they both follow from the following.
Theorem 2.8 . There exist arbitrarily thin sets S ⊆ Z+ such that if C
is a set of congruences that covers S, then C is a covering of the integers. Proof. We give an explicit construction of S. Let x ≥ 3. For simplicity, we describe a set S ⊆ Z+ which has (log log x)2 elements up to x with the property in the theorem, and then we briefly indicate how to modify it to obtain thinner sets S. Define
S = {n + uu! − 1 : n ∈ Z+, u ∈ Z+, u ≥ n}.
One easily checks that u! ≥ 2u−1 for all positive integers u. If s = n + uu! − 1
∈ S and s ≤ x, then u2u−1
≤ x so that u log log x. On the other hand,
u ≥ n, so also n log log x. It follows that there are (log log x)2 elements of S up to x.Suppose C is a set of congruences that covers S. Let n′ ∈ Z+. Then
n′ + uu! − 1 ∈ S where we can choose the integer u ≥ n′ as we want. We choose a prime u that is larger than m for each modulus m appearing in a congruence in C. With φ(x) denoting the Euler φ-function, we deduce
uφ(m) ≡ 1 (mod m) for each such m. Also, 1 ≤ φ(m) ≤ m ≤ u implies φ(m)divides u! for each modulus m appearing in a congruence in C. We deduce then that uu! ≡ 1 (mod m) for each such m. Since C covers S, there is a congruence x ≡ a (mod m) in C for which
n′ + uu! − 1 ≡ a (mod m).
Hence, uu! ≡ 1 (mod m) implies that n′ satisfies the congruence x ≡ a
(mod m) in C. Recalling the remark after Corollary 2.3, we see that C is a covering of the integers. More generally, one can repeat this argument with
uu! in the definition of S replaced by uw(u)! with w(u) tending to infinity as quickly as one wants to obtain a set S as thin as one wants to complete the proof of the theorem. Corollary 2.6 makes use of the result by B. Hough mentioned in the introduction. Our next result similarly relies on this work. To clarify a distinction in these two results, we note that Corollary 2.6 is a statement about the existence of thin sets S, whereas our next result is a statement about all sets S which are sufficiently dense in the set of integers. Covering subsets of the integers by congruences 7
Proposition 2.9 . Let S ⊆ Z+ satisfy
(2.1) lim sup
X→∞
|{ s ∈ S : s ≤ X}|
X = 1 .
Then S cannot be covered by a set of congruences with minimum modulus
10 16 .Proof. Fix S ⊆ Z+, and suppose
C = {x ≡ aj (mod mj ) : 1 ≤ j ≤ r}, where 10 16 < m 1 < · · · < m r,
is a covering of S. From , we know that C is not a covering of the integers, so there is an integer x0 with x0 6 ≡ aj (mod mj ) for each 1 ≤ j ≤ r. Let L
be the least common multiple of m1, . . . , m r (note that the product of these moduli will also suffice here). Thus, L ≡ 0 (mod mj ) for each j. Hence, for every integer k, the number x0(k) = x0 + kL satisfies x0(k) 6 ≡ aj (mod mj )for each 1 ≤ j ≤ r. It follows that lim inf
X→∞
∣∣{x ∈ Z+ : x ≤ X, x 6 ≡ aj (mod mj ), ∀j ∈ { 1, . . . , r }}∣ ∣
X ≥ 1
L .
Thus, asymptotically at least 1 /L of the positive integers are not covered by C. Since C is a covering of S, we can deduce that the left-hand side of (2.1) is at most 1 − 1/L < 1. The proposition follows. In connection with the above result, it should be noted that there are sets with density arbitrarily close to 1 which can be covered using moduli that are larger than any prescribed value. For example, if we want to use only moduli > M , then we can take z ≥ M ,
S = {n ∈ Z+ : ∃p ∈ (z, e z ] such that p | n}
and C = {x ≡ 0 (mod p) : z < p ≤ ez }. The asymptotic density of the set of positive integers not in S is lim
X→∞
|{ s 6 ∈ S : s ∈ Z ∩ [1 , X ]}|
X = ∏
z<p ≤ez
(
1 − 1
p
)
∼ log zz
as z tends to infinity. Hence, by choosing z sufficiently large, the density of S can be made > 1 − ε for any fixed ε > 0. The following proposition, concerning an odd covering of the prime num-bers, makes use of recent work of J. Harrington .
Proposition 2.10 . There exists an odd covering of the prime numbers. Proof. In , J. Harrington established the existence of a system C of congruences where the moduli are all odd and > 1, the modulus 3 is used exactly twice, no other modulus is repeated, and every integer satisfies at least one of the congruences. By Lemma 2.2 with S = Z and an appropriate choice of t, we may suppose that x ≡ 0 (mod 3) is one of the congruences 8 M. Filaseta and W. Harvey
in C. The only prime p ≡ 0 (mod 3) is p = 3. Thus, since every integer satisfies a congruence in C, every prime p 6 = 3 must satisfy a congruence in C that is different from x ≡ 0 (mod 3). We construct an odd covering
C′ of the primes as in the proposition by removing x ≡ 0 (mod 3) from C
and including in C′ instead the congruence x ≡ 3 (mod m) where m is any odd integer > 1 that does not appear as a modulus in C. Therefore, this last congruence is satisfied by the prime 3, and the other congruences in C′
cover the remaining primes.
Powers of 2 and the Fibonacci numbers. In this section, we obtain a theorem for coverings of the powers of 2 and a similar result for coverings of the Fibonacci numbers. The main distinction in the covering results obtained is that, for the powers of 2, we are able to construct an exact covering. At the end of this section, however, we demonstrate an example of an exact odd covering of the Fibonacci numbers.
Theorem 3.1 . Let P ≥ 2, M ≥ 2, and ε > 0. There exists a finite set of congruences
(3.1) x ≡ aj (mod mj ) for 1 ≤ j ≤ r,
with distinct moduli mj > 1, that satisfies each of the following:
(i) For each n ≥ 0, the number 2n satisfies exactly one of the congruences in (3.1) . (Thus, the congruences form an exact covering of the powers of 2.)(ii) Each prime divisor of each mj is > P . (In particular, the congruences form an odd covering of the powers of 2.)(iii) Each mj is > M . (Hence, the minimum modulus is arbitrarily large. )(iv) The sum of the reciprocals of the moduli mj is < ε . (Therefore, the sum of the reciprocals of the moduli is arbitrarily small. )
Comment. The main reason for stating (iii) above is to emphasize that the minimum modulus can be arbitrarily large; this is a consequence of (ii) since each mj is > 1.
Proof of Theorem 3.1. Define A(n) = 2 2n
− 1 and B(n) = 2 2n
By K. Zsigmondy’s Theorem , for each n ≥ 1, there is a prime p dividing
A(n) which fails to divide A(k) for every integer k ∈ [0 , n ). Alternatively, we can use that (3.2) A(n) = B(n − 1) A(n − 1) = B(n − 1) B(n − 2) · · · B(0) ,B(n − 1) − A(n − 1) = 2 to see that, for n ≥ 1, the number B(n − 1) > 1 is a divisor of A(n) that is relatively prime to A(k) for each k ∈ [0 , n ). For n ≥ 1, we let dn be any Covering subsets of the integers by congruences 9
such divisor of A(n), that is, a divisor > 1 that is relatively prime to each
A(k) with 0 ≤ k < n . In particular, note that the values of dn are pairwise relatively prime. For n a positive integer, dn has the property that the powers of 2 repeat modulo dn with period 2 n. To clarify, first, if r = 2 n, then 2+r ≡ 2 (mod dn) for every integer ` ≥ 0,
which follows since the same congruence with the modulus replaced by A(n)is easily seen to hold and dn | A(n). Second, the minimum positive integer r
for which 2 r ≡ 1 (mod dn) must divide 2 n, which can be seen by observing that if two different values of r satisfy 2 r ≡ 1 (mod dn), then so does their greatest common divisor. Thus, the minimum r satisfying 2 r ≡ 1 (mod dn)is a power of 2 that is ≤ 2n. Finally, 2 r ≡ 1 (mod dn) cannot hold for r apower of 2 that is < 2n since dn does not divide A(k) with 0 ≤ k < n . Thus, indeed the powers of 2 repeat modulo dn with period 2 n.In a moment, we will consider a product D of distinct dj . Observe that if n is the largest subscript of dj in the product D, then the period of the sequence of powers of 2 modulo D is 2 n. Furthermore, in this case, the numbers 2 i for 0 ≤ i < 2n are distinct modulo D.The coprimality of dn and dk for 0 ≤ k < n is enough to ensure that dn
tends to infinity with n and that in fact the minimum prime divisor of dn
tends to infinity with n. We now fix k ∈ Z+ such that every dn with n ≥ k has all prime divisors > P and dn > M for each n ≥ k. In particular, from (3.2), the n − k + 1 numbers (3.3) dk, d k+1 , . . . , d n−1, d n
are pairwise relatively prime divisors > 1 of A(n). We consider the set Sn of 2 n−k integers formed by taking arbitrary prod-ucts of distinct numbers from the first n − k divisors of A(n) listed in (3.3). Since these divisors are pairwise relatively prime, these 2 n−k integers are distinct. Thus, the size of Sn is 2 n−k. For each D = dns where s ∈ Sn, we consider a congruence of the form x ≡ 2a (mod D). We take a ∈ [0 , 2n) so that distinct s ∈ Sn are assigned distinct a. Observe that the congruence
x ≡ 2a (mod D) covers the integers 2 m satisfying m ≡ a (mod 2 n) and no other powers of 2. We use these 2 n−k congruences formed from the 2 n−k
elements of Sn as described to cover 2 m for m from 2 n−k residue classes modulo 2 n.Given the above, we begin with n = k. Thus, initially, we have one congruence formed from the unique element of Sk, and this congruence cov-ers 2 m for m in this unique residue class modulo 2 k. This residue class corresponds to two residue classes modulo 2 k+1 . We then take n = k + 1 and cover two more residue classes modulo 2 k+1 . Thus, at this point four 10 M. Filaseta and W. Harvey
different residue classes are covered modulo 2 k+1 . A straightforward induc-tion argument shows that if we continue in this manner, taking n = k + j
for increasing values of j, we cover ( j + 1)2 j different residue classes mod-ulo 2 k+j . We stop this process when j = J = 2 k − 1 since at that point we will be covering (J + 1)2 J = 2 k22k −1 = 2 2k +k−1
different residue classes modulo 2 k+J = 2 2k +k−1. In other words, after con-sidering all j ∈ { 0, 1, . . . , J }, we cover 2 m for m from every residue class modulo 2 k+J , and thus the congruences formed cover every power of 2. Fur-thermore, each power of 2 satisfies at most one of the congruences in this construction. Thus, we obtain a covering system of the powers of 2 satisfying (i)–(iii). Finally, we address (iv). Recall that we can take dn = B(n−1) = 2 2n−1
+1 for every positive integer n. Our set of congruences constructed from Sn
above, with n = k + j where 0 ≤ j ≤ 2k − 1, consist of |Sn| = 2 j moduli divisible by dn. If we take dn = B(n − 1) = 2 2n−1
1, the sum of the reciprocals of all moduli in our covering of the powers of 2 is (3.4) ≤
2k−1
∑
j=0
2j
22k+j−1 + 1 ≤
∞
∑
j=k
2j
22j−1 + 1 .
The series ∞∑
m=0
2m
22m−1 + 1 converges, for example, by the inequality 2m
22m−1 + 1 ≤ 122m−1−m ≤ 12m−2 for every integer m ≥ 0.
It follows that the last series in (3.4) tends to 0 as k → ∞ . Thus, by choosing
k sufficiently large, we deduce that (iv) also holds. We turn next to a covering of the Fibonacci numbers Fn. Thus, F0 = 0,
F1 = 1 and Fn+1 = Fn + Fn−1 for n ≥ 1. Our argument will make use of the Lucas numbers Ln defined by L0 = 2, L1 = 1, and Ln+1 = Ln + Ln−1 for
n ≥ 1. Take α = (1 + √5) /2 and β = (1 − √5) /2. Then we have the classical formulas
Fn = αn − βn
α − β and Ln = αn + βn
α + β = αn + βn
for all n ≥ 0. Our next lemma easily follows from these identities, and we omit the proof.
Lemma 3.2 . Let u and v denote integers with u ≥ v ≥ 0. The Fibonacci and Lucas numbers satisfy the following properties: Covering subsets of the integers by congruences 11
(a) Fu+v = FuLv − (−1) vFu−v.
(b) L2u = L2
u
− 2( −1) u.
We will also make use of the following lemmas.
Lemma 3.3 . Let j be a positive integer. The Fibonacci numbers modulo the Lucas number L2j are periodic with period dividing 2j+2 .Proof. Let n be a positive integer. Taking u = n + 2 j+1 + 2 j and v = 2 j
in Lemma 3.2(a), we have
Fn+2 j+2 = Fu+v = FuLv − (−1) vFu−v = FuLv − (−1) vFn+2 j+1 .
With u′ = n + 2 j and v = 2 j , we deduce from Lemma 3.2(a) that
Fn+2 j+1 = Fu′+v = Fu′ Lv − (−1) vFu′−v = Fu′ Lv − (−1) vFn.
Thus,
Fn+2 j+2 = FuLv − (−1) v(Fu′ Lv − (−1) vFn) = ( Fu − (−1) vFu′ )Lv + Fn,
so that Fn+2 j+2 − Fn is divisible by Lv = L2j .
Lemma 3.4 . For j and k non-negative integers with j 6 = k, we have
gcd( L2j , L 2k ) = 1 .Proof. Since L1 = 1, it follows from Lemma 3.2(b) and induction that
L2j and L2k are odd. We suppose as we may that k > j ≥ 1. Let p be a prime divisor of L2j . In particular, p > 2. Lemma 3.2(b) with u = 2 j implies
L2j+1 ≡ − 2 (mod p) and by induction L2j+t ≡ 2 (mod p) for every integer
t ≥ 2. Hence, L2k ≡ ± 2 (mod p). Since p > 2, we deduce p - L2k .The following are consequences of the previous two lemmas.
Corollary 3.5 . Let P = L2a1 · · · L2ak for some integer k > 0 and some aj satisfying 1 ≤ a1 < · · · < a k. The Fibonacci numbers modulo P are periodic with period dividing 2ak +2 .
Corollary 3.6 . The minimum prime divisor of L2j goes to infinity with j.
Our next lemma is an easy consequence of Lemma 3.2 (b) and induction, and we leave out further details of its proof.
Lemma 3.7 . For every positive integer j, we have L2j ≥ 10 2j−3
1 .
Theorem 3.8 . Let P ≥ 2, M ≥ 2, and ε > 0. There exists a finite set of congruences
(3.5) x ≡ aj (mod mj ) for 1 ≤ j ≤ r,
with distinct moduli mj > 1, that satisfies each of the following: 12 M. Filaseta and W. Harvey
(i) For each n ≥ 0, the Fibonacci number Fn satisfies at least one of the congruences in (3.5) . (Thus, the congruences form a covering of the Fibonacci numbers. )(ii) Each prime divisor of each mj is > P . (In particular, the congruences form an odd covering of the Fibonacci numbers. )(iii) Each mj is > M . (Hence, the minimum modulus is arbitrarily large. )(iv) The sum of the reciprocals of the moduli mj is < ε . (Therefore, the sum of the reciprocals of the moduli is arbitrarily small. )
Proof. The argument is similar to the proof of Theorem 3.1. From Corol-lary 3.6, there is a k ∈ Z+ such that for every j ≥ k, L2j has each of its prime divisors > P . Momentarily, we fix such a k ≥ 3, and consider n ≥ k.In the end, we will want to choose k large. Lemma 3.2(a) with u = v implies
F2n+1 = F2n L2n = F2n−1 L2n−1 L2n = · · · = L21 · · · L2n .
Combining the above with Lemma 3.4, we see that the Lucas numbers (3.6) L2k , L 2k+1 , . . . , L 2n
are relatively prime divisors of F2n+1 . Note that k ≥ 3 easily implies L2j > 1for every j ≥ k.We consider the set Sn of the 2 n−k distinct integers formed by taking arbitrary products of distinct numbers from the first n − k divisors of F2n+1
listed in (3.6). For each D = D(s, n ) = s · L2n where s ∈ Sn, we consider a congruence of the form x ≡ Fa (mod D) where 0 ≤ a < 2n+2 . We take
a so that distinct s ∈ Sn are assigned distinct a ∈ [0 , 2n+2 ). Let Cn denote the set of these 2 n−k congruences corresponding to the 2 n−k elements of Sn.By Corollary 3.5, the congruence x ≡ Fa (mod D) covers the integers Fm
satisfying m ≡ a (mod 2 n+2 ). Thus, Cn covers the set of Fm for m from at least 2 n−k residue classes modulo 2 n+2 .With the above set-up, we begin with n = k, so that the set Ck just described contains one congruence which covers the set of Fm for m from at least one residue class modulo 2 k+2 . This one residue class modulo 2 k+2
corresponds to two residue classes, say r1 and r2, modulo 2 k+3 . We then select a ∈ [0 , 2k+3 ) in creating the two congruences x ≡ Fa (mod D) for
Ck+1 so that each of these a’s belongs to neither r1 nor r2. Thus, Ck ∪ Ck+1
covers the integers belonging to a total of four residue classes modulo 2 k+3 .Similarly, we create four congruences for Ck+2 that cover Fm for m from at least four residue classes modulo 2 k+4 so that the congruences in Ck ∪
Ck+1 ∪ Ck+2 cover Fm for m from at least 12 residue classes modulo 2 k+4 .Inductively, for 0 ≤ j ≤ J with J = 2 k+2 − 1, we create 2 j congruences for
Ck+j so that the congruences in Ck ∪ Ck+1 ∪ · · · ∪ Ck+j cover Fm for m from at least ( j + 1)2 j residue classes modulo 2 k+j+2 . Observe that when j = J,the congruences in Ck ∪ · · · ∪ Ck+J cover Fm for m from at least 2 k+2 k+2 +1 Covering subsets of the integers by congruences 13
residue classes modulo 2 k+2 k+2 +1 . Thus, we obtain a collection of distinct congruences from C = Ck ∪ · · · ∪ Ck+J that satisfy (i) and (ii). If we take k sufficiently large, for example so that L2k > M , then (iii) is also satisfied. For (iv), observe that Ck+i consists of 2 i congruences with each modulus ≥ L2k+i . For any integer k, we note that 2 k−3 ≥ k − 3. Since
k ≥ 3, we deduce from Lemma 3.7 that the sum of the reciprocals of the moduli in C is bounded above by
J
∑
i=0
2i
10 2k+i−3 + 1 ≤
J
∑
i=0
2k+i−3
10 2k+i−3 + 1 <
∞
∑
t=k−3
( 210
)t
= 54 ·
( 15
)k−3
.
Thus, taking k sufficiently large, we can also ensure that (iv) holds. Observe that Theorem 3.1 establishes the existence of an exact cover-ing of the powers of 2, whereas Theorem 3.8 establishes a covering of the Fibonacci numbers which is not necessarily exact. In the general context of the other conditions in these results, we were not able to strengthen Theo-rem 3.8 to give an exact covering of the Fibonacci numbers. On the other hand, it is of some interest to note that an exact odd covering of the Fi-bonacci numbers does exist. In the first column of Table 1 below, we give such a covering and leave the details of the verification to the reader. The second column lists the Fibonacci numbers covered by each congruence in the first column.
Table 1. An exact odd covering of the Fibonacci numbers Congruence nfor which Fn
satisfies the congruence
x≡2 (mod 3) n≡3,5,6 (mod 8)
x≡0 (mod 7) n≡0 (mod 8)
x≡13 (mod 21) n≡7,9,10 (mod 16)
x≡3 (mod 141) n≡4,12 (mod 32)
x≡1 (mod 329) n≡1,2,31 (mod 32)
x≡843 (mod 987) n≡20 (mod 32)
x≡610 (mod 2207) n≡15 ,49 (mod 64)
x≡1597 (mod 103729) n≡17 ,47 (mod 64)
x≡311184 (mod 311187) n≡60 (mod 64)
x≡317811 (mod 726103) n≡28 (mod 64)
x≡2584 (mod 3261) n≡18 (mod 128)
x≡1464 (mod 7609) n≡50 (mod 128)
x≡6112 (mod 22827) n≡82 (mod 128)
x≡50712 (mod 51089) n≡114 (mod 128) 14 M. Filaseta and W. Harvey
Sums of two squares. In this section, we prove
Theorem 4.1 . There is an odd covering of the set of integers which are sums of two squares.
We will make use of the following classical result. We omit the proof. Recall that pe ‖ n means that pe | n and pe+1 - n, where we allow for e = 0.
Lemma 4.2 . An n ∈ Z+ is a sum of two squares if and only if each prime p ≡ 3 (mod 4) satisfies pe ‖ n for some even non-negative integer e.
For vectors ~a = 〈a1, . . . , a t〉 and ~m = 〈m1, . . . , m t〉, with a1, . . . , a t arbi-trary integers and with m1, . . . , m t positive pairwise relatively prime inte-gers, we denote by J~a, ~ mK the unique congruence x ≡ A (mod M ), given by the Chinese Remainder Theorem, where M = m1 · · · mt and where
A ∈ [0 , M ) ∩ Z, corresponding to simultaneously satisfying the congruences
x ≡ a1 (mod m1), x ≡ a2 (mod m2), . . . , x ≡ at (mod mt).
We stress that we are interested only in the case where the mj are positive pairwise relatively prime integers. For convenience later, we do not require that aj ∈ [0 , m j ). Our next two lemmas are similar in nature. The basic idea behind these lemmas has been used by a number of authors and can be traced back to at least C. E. Krukenberg’s 1971 dissertation .
Lemma 4.3 . Let p be a prime. Let S ⊆ Z+ have the property that given any integers m ≥ 1 and u with p - m and any s1 ∈ S, there is an s2 ∈ S such that s2 ≡ s1 (mod m) and s2 ≡ u (mod p). Let C1, C2 and C = C1 ∪ C2 be sets of congruences given by
C1 = {J〈aj 〉, 〈mj 〉K : 1 ≤ j ≤ r} and C2 = {J〈bj , b j 〉, 〈p, m ′
j
〉K : 1 ≤ j ≤ s},
where p is a prime, each modulus mj appearing in C1 is at least 2 and relatively prime to p, and each modulus pm ′
j
appearing in C2 is non-divisible by p2 (and hence exactly divisible by p). Suppose that every integer in
{s′ ∈ S : s′ ≡ a (mod p) for some a ∈ { 0, 1, . . . , p − 2}}
satisfies at least one congruence in C. Then every integer in S satisfies at least one of the r + ts + q congruences given below:
(i) J〈aj 〉, 〈mj 〉K for 1 ≤ j ≤ r,
(ii) J〈pi−1 − 1 + bj pi−1, b j 〉, 〈pi, m ′
j
〉K for 1 ≤ i ≤ t and 1 ≤ j ≤ s,
(iii) J〈j, −1〉, 〈q, p t−j 〉K for 0 ≤ j ≤ q − 1,where q is an arbitrary fixed prime such that q - (pm ′
1
· · · m′
s
) and where
t ∈ Z satisfies t ≥ q.
We will turn to the proof shortly. We note first that S = Z+ has the property required of S in the second sentence of the lemma. Perhaps less Covering subsets of the integers by congruences 15
clear is that the set S of integers which are sums of two squares has this property, and this will be the S of interest to us (and we will only require
p = 5). We explain next why the integers that are sums of two squares can be used for S in Lemma 4.3. As a preliminary result, we let a and b be integers with gcd( a, 2b) = 1. We show first that there are infinitely many n ∈ Z such that an +b is a prime that is 1 modulo 4. Since a is odd, there is an n0 ∈ Z such that an 0 + b ≡ 1(mod 4). Let n = n0 + 4 k where k is an integer to be determined. Thus,
an + b = an 0 + b + 4 ak . Since an 0 + b ≡ 1 (mod 4) and gcd( a, b ) = 1, we deduce gcd( an 0 + b, 4a) = 1. Dirichlet’s Theorem on primes in arithmetic progressions implies now that there are infinitely many integers k such that, with n chosen as above, an + b is a prime congruent to 1 modulo 4. To see that the integers that are sums of two squares can be used for S
in Lemma 4.3 with p an arbitrary prime, let m and u be as stated and let s1
be a sum of two squares. We suppose as we may that u ∈ { 0, 1, . . . , p − 1}. If
s1 = 0, then we define v = 1. If s1 6 = 0, we define v as a positive integer such that if qe | s1 where q is a prime and e is a positive integer, then e + 2 ≤ 2v.Set d = gcd( s1, m 2v), s′ = s1/d and m′ = m2v/d . Note that gcd( s′, m ′) = 1. Also, the definition of v implies that if q is a prime and e is a positive integer satisfying qe ‖ d, then qe ‖ s1. From Lemma 4.2, we see that since s1 and m2v
are each a sum of two squares, so are d, s′ and m′. Since p - m, we have
p - d. Also, since p - m, there is a k0 ∈ Z such that s1 + k0m2v ≡ u (mod p2). Let k = k0 + p2, where is an integer to be chosen. We take
s2 = s1 + km 2v = s1 + k0m2v + p2m 2v = d(s′ + k0m′ + p2m′).
Observe that s2 ≡ s1 (mod m) and s2 ≡ s1 + k0m2v ≡ u (mod p). We are left with showing s2 is a sum of two squares. We consider a couple cases. If u = 0, then
s2 = dp 2
( s′ + k0m′
p2 + m′`
)
.
The expression ( s′ +k0m′)/p 2 is an integer relatively prime to m′. If m′ is odd, then our preliminary result above with a = m′ and b = ( s′ +k0m′)/p 2 implies that we can take so that a + b is a prime congruent to 1 modulo 4. As s2 is then a product of the 3 numbers d, p2 and a` + b each of which is a sum of two squares, so is s2. If m′ is even, then m is even and our definition of v implies 4 | m′. Also, p - m, so p is odd, and p2 ≡ 1 (mod 4). Since gcd( s′, m ′) = 1,
s′ is odd. Since s′ is a sum of two squares, s′ ≡ 1 (mod 4). Thus, Dirichlet’s Theorem implies there is an ∈ Z such that ( s′ + k0m′)/p 2 + m′ is a prime that is 1 modulo 4. Again, we see that s2 is a sum of two squares. If u 6 = 0, then s′ + k0m′ is not divisible by p. Further, it is relatively prime to m′. If m′ is even, then, as in the last case, 4 | m′, s′ ≡ 1 (mod 4) and there is an such that s′ + k0m′ + p2m′ is a prime that is 1 modulo 4. 16 M. Filaseta and W. Harvey
Thus, s2 is a sum of two squares. So suppose m′ is odd. If p is odd, then we use our preliminary result above with a = p2m′ and b = s′ + k0m′ to see that we can take with a + b a prime congruent to 1 modulo 4. Again, we deduce that s2 is then a sum of two squares. Suppose now that p = 2. In this case, u ∈ { 0, 1} and u 6 = 0, so u = 1. Hence, s1 + k0m2v ≡ u (mod p2)implies s1 + k0m2v ≡ 1 (mod 4). Since p - m, we see that m is odd. Thus, d
is odd. Since d is a sum of two squares, d ≡ 1 (mod 4). We deduce that
s′ + k0m′ ≡ d(s′ + k0m′) ≡ s1 + k0m2v ≡ 1 (mod 4) .
Since s′ +k0m′ is relatively prime to 4 m′, we can choose ` in such a way that
s′ + k0m′ + p2m′= s′ + k0m′ + 4 m′ is a prime congruent to 1 modulo 4. Again, s2 is a sum of two squares. Before proceeding to the proof of Lemma 4.3, we note that if m1, . . . , m r
are distinct and m′
1
, . . . , m ′
s
are distinct, then the various moduli appearing in (i)–(iii) are all distinct. The condition q - (pm ′
1
· · · m′
s
) makes this easier to see; however, it is more than one needs. We note also that if one is interested in not introducing new prime divisors into the product of the moduli, this generally can be done by instead taking q to be an integer not divisible by
p and different from each modulus m′
j
.
Proof of Lemma 4.3. For each u ∈ { 0, 1, . . . , p − 1}, we define
C(u)2 = {J〈bj 〉, 〈m′
j
〉K : 1 ≤ j ≤ s, b j ≡ u (mod p)}.
We begin by showing that for each u 6 = p − 1, each element of S satisfies a congruence in C1 ∪ C(u)2 .Fix u ∈ { 0, 1, . . . , p − 2}, and let Tu denote the set of j ∈ { 1, . . . , s } for which bj ≡ u (mod p). Let n ∈ S, and set
m =
( r∏
j=1
mj
)( ∏
j∈Tu
m′
j
)
.
By the condition on S in the lemma, there is an n0 ∈ S such that n0 ≡ u
(mod p) and for some integer n′ we have
n0 = n + mn ′ = n +
( r∏
j=1
mj
)( ∏
j∈Tu
m′
j
)
n′.
Since each integer in S that is u modulo p satisfies a congruence in C, either
n0 ≡ aj (mod mj ) for some j ∈ { 1, . . . , r } or n0 ≡ bj (mod pm ′
j
) for some
j ∈ { 1, . . . , s }. In the former case, n, being congruent to n0 modulo each mj ,satisfies a congruence in C1. In the latter case, n ≡ n0 ≡ bj (mod m′
j
) for some j ∈ { 1, . . . , s }. Also, bj ≡ n0 ≡ u (mod p), so j ∈ Tu. Thus, in this case, n satisfies a congruence in C(u)2 . Hence, each integer in S satisfies a congruence in C1 ∪ C(u)2 .Covering subsets of the integers by congruences 17
Note that (ii) with i = 1 corresponds to the congruences in C2. Thus, every integer congruent to 0 , 1, . . . , p − 3, or p − 2 modulo p satisfies a con-gruence from either (i) or (ii). We restrict ourselves now to integers n ∈ S
that are congruent to p − 1 modulo p. Observe that either n = −1 or there is some ≥ 1 such that (4.1) n ≡ − 1 (mod p) and n 6 ≡ − 1 (mod p`+1 ).
First, we consider n 6 = −1 and the case in (4.1) where ` ≤ t − 1. Then, with
as above, n ≡ p − 1 + pu (mod p+1 ) where u 6 ≡ p − 1 (mod p). Suppose
n does not satisfy a congruence in C1. Then since each integer in S satisfies a congruence in C1 ∪ C(u)2 , there is a j ∈ Tu such that n ≡ bj (mod m′
j
). Since j ∈ Tu, we have bj ≡ u (mod p), which implies up ≡ bj p (mod p+1 ). Hence, n ≡ p − 1 + pbj (mod p+1 ). Thus, n satisfies (ii) with i = ` + 1. So far, we know that n satisfies a congruence in (i) or (ii) provided
n 6 = −1 and the value of in (4.1) satisfies ≤ t − 1. Now, suppose n = −1, or ` in (4.1) is ≥ t. Either of these implies that n ≡ − 1 (mod pi) for every integer i ∈ [1 , t ]. Also, n ≡ k (mod q) for some k ∈ { 0, 1, . . . , q − 1}. Since 1 ≤ t − k ≤ t, we obtain n ≡ − 1 (mod pt−k), so n satisfies the congruence in (iii) corresponding to j = k.
Lemma 4.4 . Let , r and s be integers with ≥ 1, 0 ≤ r ≤ ` and 0 ≤ s
≤ . Let b1, . . . , b and m1, . . . , m ` be integers, with each mj > 1. Let
b′
1
, . . . , b ′
r
, b′′
1
, . . . , b ′′
s
, m′
1
, . . . , m ′
r
and m′′
1
, . . . , m ′′
s
be such that {(b′
j
, m ′
j
) : 1 ≤ j ≤ r} and {(b′′
j
, m ′′
j
) : 1 ≤ j ≤ s} are both subsets of {(bj , m j ) : 1 ≤
j ≤ }. Let p be an odd prime such that p - (3 m1 · · · m), and let a, w and
t be integers with w ≥ 1 and t ≥ p − 1. Suppose n is an integer which sat-isfies the congruence J〈a, b 1, . . . , b 〉, 〈3w, m 1, . . . , m〉K. Then n satisfies at least one of the following congruences:
(i) J〈a + 2(3 w + 3 w+1 + · · · + 3 i−2), b ′
1
, . . . , b ′
r
〉, 〈3i, m ′
1
, . . . , m ′
r
〉K for w + 1 ≤
i ≤ t,
(ii) J〈a + 2(3 w + 3 w+1 + · · · + 3 i−2) + 3 i−1, b ′′
1
, . . . , b ′′
s
〉, 〈3i, m ′′
1
, . . . , m ′′
s
〉K for
w + 1 ≤ i ≤ t,
(iii) J〈a + 2(3 w + 3 w+1 + · · · + 3 t−k−1), k 〉, 〈3t−k, p 〉K for 0 ≤ k ≤ p − 1,where empty sums are interpreted as equal to 0.Proof. Let n be an integer satisfying the congruence
J〈a, b 1, . . . , b 〉, 〈3w, m 1, . . . , m〉K.
If n ≡ a (mod 3 w+1 ), then n satisfies (i) with i = w + 1. If n ≡ a + 3 w
(mod 3 w+1 ), then n satisfies (ii) with i = w + 1. Since n ≡ a (mod 3 w), we are left with the case n ≡ a + 2 · 3w (mod 3 w+1 ). Again, there are three possibilities. If n ≡ a + 2 · 3w (mod 3 w+2 ), then n satisfies (i) with i = w + 2. If n ≡ a + 2 · 3w + 3 w+1 (mod 3 w+2 ), then n satisfies (ii) with i = w + 2. 18 M. Filaseta and W. Harvey
We are left then with n ≡ a + 2(3 w + 3 w+1 ) (mod 3 w+2 ). Continuing in this manner, using a congruence from (i) and a congruence from (ii) with successive values of i, we are left with n satisfying either (i) or (ii) unless n ≡
a+2(3 w+3 w+1 +· · · +3 t−1) (mod 3 t). For such n, there is a k ∈ { 0, 1, . . . , p −1}
such that both n ≡ k (mod p) and n ≡ a + 2(3 w + 3 w+1 + · · · + 3 t−k−1)(mod 3 t−k); the first of these congruences determines k and the second follows from n ≡ a + 2(3 w + 3 w+1 + · · · + 3 t−1) (mod 3 t). Thus, in this case,
n satisfies a congruence in (iii).
Proof of Theorem 4.1. Let S = {0, 1, 2, 4, 5, 8, 9, . . . } be the set of integers that are sums of two squares. We begin by using a single congruence to cover every element n of S that is divisible by 3. Lemma 4.2 implies that each such
n satisfies n ≡ 0 (mod 9). Hence, we cover these n by using the congruence (C1) J〈0〉, 〈9〉K;here, and throughout the proof, we indicate congruences in our final covering of S by labelling them with (C ∗) with ∗ replaced by a number. Now, we use (C2) J〈1〉, 〈3〉K
to cover those n in S that are 1 modulo 3. Thus, it remains to cover the elements of S that are 2 modulo 3. We separate the remaining elements of S into three groups, depending on whether they are 2, 5 or 8 modulo 9. We work first with those n ∈ S
for which n ≡ 2 (mod 9). These we break up into five groups depending on their residues modulo 5. The n congruent to 0, 1 or 2 modulo 5 (that also satisfy n ≡ 2 (mod 9)) are covered by the three congruences (C3) J〈0〉, 〈5〉K, J〈2, 1〉, 〈3, 5〉K, J〈2, 2〉, 〈9, 5〉K.
We separate the n ∈ S for which n ≡ 2 (mod 9) and n ≡ 3 (mod 5), into seven groups depending on their residue classes modulo 7. We take advantage of Lemma 4.2 again to see that the n ∈ S divisible by 7 satisfy n ≡ 0 (mod 49). We can therefore cover the n in all seven groups by using the congruences (C4) J〈0〉, 〈49 〉K, J〈1〉, 〈7〉K, J〈2, 2〉, 〈3, 7〉K, J〈3, 3〉, 〈5, 7〉K,
J〈2, 3, 4〉, 〈3, 5, 7〉K, J〈2, 5〉, 〈9, 7〉K, J〈2, 3, 6〉, 〈9, 5, 7〉K.
As of now, we are left with covering the n ∈ S which satisfy J〈2, 4〉, 〈9, 5〉K,
J〈5〉, 〈9〉K or J〈8〉, 〈9〉K. We return to the first of these later in the argument. Next, we cover the elements of S satisfying J〈5〉, 〈9〉K. We break these up into their five residue classes modulo 5. The first two congruences in (C3) will already cover the n ∈ S for which n ≡ 5 (mod 9) and n is either 0 or 1 modulo 5. We cover the n ∈ S that satisfy J〈5, 2〉, 〈9, 5〉K next. For this, we use Lemma 4.4 with a = 5, w = 2, r = 0, s = 1, b′′
1
= 2 and m′′
1
= 5. The values of t and p do not play a significant role, though they will need to be Covering subsets of the integers by congruences 19
sufficiently large. We take t = p = 53. Thus, our congruences here are (C5)
J〈5 + 2(3 2 + 3 3 + · · · + 3 i−2)〉, 〈3i〉K for 3 ≤ i ≤ 53 ,
J〈5 + 2(3 2 + 3 3 + · · · + 3 i−2) + 3 i−1, 2〉, 〈3i, 5〉K for 3 ≤ i ≤ 53 ,
J〈5 + 2(3 2 + 3 3 + · · · + 3 52 −k), k 〉, 〈353 −k, 53 〉K for 0 ≤ k ≤ 52 .
Next, we cover the integers in S satisfying J〈5, 3〉, 〈9, 5〉K by grouping them into seven residue classes modulo 7. All integers in S divisible by 7 have been covered by the first congruence in (C4). For the integers satis-fying J〈5, 3〉, 〈9, 5〉K and that are either 1, 2, 3 or 4 modulo 7, we can use
J〈1〉, 〈7〉K, J〈2, 2〉, 〈3, 7〉K, J〈3, 3〉, 〈5, 7〉K and J〈2, 3, 4〉, 〈3, 5, 7〉K from (C4). For the integers in J〈5, 3〉, 〈9, 5〉K that are 5 modulo 7, we apply Lemma 4.4 with
a = 5, w = 2, r = 0, s = 2, b′′
1
= 3, b′′
2
= 5, m′′
1
= 5 and m′′
2
= 7. We note that we can take t = p = 53 as before. For Lemma 4.4 here, we are reusing the congruences
J〈5 + 2(3 2 + 3 3 + · · · + 3 i−2)〉, 〈3i〉K for 3 ≤ i ≤ 53 ,
J〈5 + 2(3 2 + 3 3 + · · · + 3 52 −k), k 〉, 〈353 −k, 53 〉K for 0 ≤ k ≤ 52 ,
which appear in (C5), and making use of the additional congruences (C6) J〈5 + 2(3 2 + 3 3 + · · · + 3 i−2) + 3 i−1, 3, 5〉, 〈3i, 5, 7〉K for 3 ≤ i ≤ 53 .
For the remaining integers in S satisfying J〈5, 3〉, 〈9, 5〉K which are 6 mod-ulo 7, we apply Lemma 4.4 again, this time with a = 5, w = 2, r = 0, s = 1,
b′′
1
= 6, m′′
1
= 7, and t = p = 53. For this application of Lemma 4.4, after reusing congruences from (C5) as above, we additionally use (C7) J〈5 + 2(3 2 + 3 3 + · · · + 3 i−2) + 3 i−1, 6〉, 〈3i, 7〉K for 3 ≤ i ≤ 53 .
Thus far, we have covered all the integers in S satisfying J〈5〉, 〈9〉K except for those that are 4 modulo 5. Combining what we now know, we are left with covering the n ∈ S which satisfy J〈2, 4〉, 〈9, 5〉K, J〈5, 4〉, 〈9, 5〉K or J〈8〉, 〈9〉K.We deal with the latter next. We break up the integers in S satisfying J〈8〉, 〈9〉K into groups depending on the residue class modulo 5. We use J〈0〉, 〈5〉K and J〈2, 1〉, 〈3, 5〉K from (C3) to cover those integers in S satisfying J〈8〉, 〈9〉K that are 0 or 1 modulo 5. The remaining integers in S satisfying J〈8〉, 〈9〉K must satisfy J〈8, 2〉, 〈9, 5〉K,
J〈8, 3〉, 〈9, 5〉K or J〈8, 4〉, 〈9, 5〉K.We group those satisfying J〈8, 2〉, 〈9, 5〉K into the seven residue classes modulo 7. We reuse the first three congruences in (C4) so that we are left with elements in S satisfying J〈8, 2〉, 〈9, 5〉K that also satisfy one of
J〈8, 2, 3〉, 〈9, 5, 7〉K, J〈8, 2, 4〉, 〈9, 5, 7〉K,
J〈8, 2, 5〉, 〈9, 5, 7〉K, J〈8, 2, 6〉, 〈9, 5, 7〉K.
We examine these in reverse order. 20 M. Filaseta and W. Harvey
We group those integers in S satisfying J〈8, 2, 6〉, 〈9, 5, 7〉K into their eleven residue classes modulo 11. Since 11 ≡ 3 (mod 4), Lemma 4.2 im-plies that (C8) J〈0〉, 〈121 〉K
covers every integer in S divisible by 11. We cover nine further residue classes modulo 11, of the integers in S satisfying J〈8, 2, 6〉, 〈9, 5, 7〉K, using (C9)
J〈1〉, 〈11 〉K, J〈2, 2〉, 〈3, 11 〉K, J〈8, 3〉, 〈9, 11 〉K,
J〈2, 4〉, 〈5, 11 〉K, J〈2, 2, 5〉, 〈3, 5, 11 〉K, J〈8, 2, 6〉, 〈9, 5, 11 〉K,
J〈6, 7〉, 〈7, 11 〉K, J〈2, 6, 8〉, 〈3, 7, 11 〉K, J〈8, 6, 9〉, 〈9, 7, 11 〉K.
To cover the residue class of integers in S that are 10 modulo 11 and satisfy
J〈8, 2, 6〉, 〈9, 5, 7〉K, we apply Lemma 4.4 with a = 8, w = 2, r = 1, s = 2,
b′
1
= 10, b′′
1
= 6, b′′
2
= 10, m′
1
= 11, m′′
1
= 7, m′′
2
= 11 and t = p = 59. This leads to (C10)
J〈8 + 2(3 2 + · · · + 3 i−2), 10 〉, 〈3i, 11 〉K for 3 ≤ i ≤ 59 ,
J〈8 + 2(3 2 + · · · + 3 i−2) + 3 i−1, 6, 10 〉, 〈3i, 7, 11 〉K for 3 ≤ i ≤ 59 ,
J〈8 + 2(3 2 + · · · + 3 58 −k), k 〉, 〈359 −k, 59 〉K for 0 ≤ k ≤ 58 .
Thus, the integers in S satisfying J〈8, 2, 6〉, 〈9, 5, 7〉K will be covered by the congruences in (C8)–(C10). Next, we turn to the integers in S satisfying J〈8, 2, 5〉, 〈9, 5, 7〉K. We split these up into congruence classes modulo 11 as well. Those congruent to 0, . . . , 6 (mod 11) are covered by (C8) and the first six congruences in (C9). Those congruent to 7, 8 or 9 modulo 11 are covered by one of (C11) J〈2, 5, 7〉, 〈5, 7, 11 〉K, J〈2, 2, 5, 8〉, 〈3, 5, 7, 11 〉K,
J〈8, 2, 5, 9〉, 〈9, 5, 7, 11 〉K.
To cover those that are 10 modulo 11, we apply Lemma 4.4 with a = 8,
w = 2, r = 1, s = 2, b′
1
= 10, b′′
1
= 2, b′′
2
= 10, m′
1
= 11, m′′
1
= 5, m′′
2
= 11 and t = p = 59. For this, we reuse congruences from (C10). We are then left with the additional congruences (C12) J〈8 + 2(3 2 + · · · + 3 i−2) + 3 i−1, 2, 10 〉, 〈3i, 5, 11 〉K for 3 ≤ i ≤ 59 .
Our congruences then cover the integers in S satisfying J〈8, 2, 5〉, 〈9, 5, 7〉K.Next, we turn to the integers in S satisfying J〈8, 2, 4〉, 〈9, 5, 7〉K. Again, consider their residue classes modulo 11 and use (C8) and the first six con-gruences in (C9) to cover those that are in the residue classes 0 , . . . , 6modulo 11. Furthermore, our use of Lemma 4.4 leading to (C12) cov-ers the integers in S satisfying J〈8, 2, 4〉, 〈9, 5, 7〉K that are 10 modulo 11. To finish covering the integers in S satisfying J〈8, 2, 4〉, 〈9, 5, 7〉K, we are left now with covering those which satisfy one of J〈8, 2, 4, 7〉, 〈9, 5, 7, 11 〉K,Covering subsets of the integers by congruences 21
J〈8, 2, 4, 8〉, 〈9, 5, 7, 11 〉K and J〈8, 2, 4, 9〉, 〈9, 5, 7, 11 〉K. To cover those satisfy-ing J〈8, 2, 4, 7〉, 〈9, 5, 7, 11 〉K, we break them up into the seven residue classes 4, 11 , 18 , . . . , 46 modulo 49. We cover these by using (C13)
J〈2, 4〉, 〈3, 49 〉K, J〈2, 11 〉, 〈5, 49 〉K, J〈8, 18 〉, 〈9, 49 〉K,
J〈2, 2, 25 〉, 〈3, 5, 49 〉K, J〈8, 2, 32 〉, 〈9, 5, 49 〉K,
J〈2, 39 , 7〉, 〈3, 49 , 11 〉K, J〈2, 46 , 7〉, 〈5, 49 , 11 〉K.
The first five of these congruences also cover the integers in S satisfying
J〈8, 2, 4, 8〉, 〈9, 5, 7, 11 〉K which are 4, 11, 18, 25 or 32 modulo 49. To cover the integers in S satisfying J〈8, 2, 4, 8〉, 〈9, 5, 7, 11 〉K which are 39 or 46 modulo 49, we use (C14) J〈8, 39 , 8〉, 〈9, 49 , 11 〉K, J〈2, 2, 46 , 8〉, 〈3, 5, 49 , 11 〉K.
The first five of the congruences in (C13) also cover the integers in S satis-fying J〈8, 2, 4, 9〉, 〈9, 5, 7, 11 〉K which are 4, 11, 18, 25 or 32 modulo 49. Those in the residue classes 39 and 46 modulo 49 satisfy one of (C15) J〈8, 2, 39 , 9〉, 〈9, 5, 49 , 11 〉K, J〈8, 46 〉, 〈27 , 49 〉K,
J〈17 , 2, 46 〉, 〈27 , 5, 49 〉K, J〈26 , 2, 46 , 9〉, 〈27 , 5, 49 , 11 〉K.
We have now finished covering the integers in S satisfying J〈8, 2, 4〉, 〈9, 5, 7〉K.We turn to the integers in S satisfying J〈8, 2, 3〉, 〈9, 5, 7〉K. We break up these integers into 13 residue classes modulo 13 and cover the integers in 12 of these residue classes using (C16)
J〈0〉, 〈13 〉K, J〈2, 1〉, 〈3, 13 〉K, J〈8, 2〉, 〈9, 13 〉K, J〈2, 3〉, 〈5, 13 〉K,
J〈2, 2, 4〉, 〈3, 5, 13 〉K, J〈8, 2, 5〉, 〈9, 5, 13 〉K, J〈3, 6〉, 〈7, 13 〉K,
J〈2, 3, 7〉, 〈3, 7, 13 〉K, J〈8, 3, 8〉, 〈9, 7, 13 〉K, J〈2, 3, 9〉, 〈5, 7, 13 〉K,
J〈2, 2, 3, 10 〉, 〈3, 5, 7, 13 〉K, J〈8, 2, 3, 11 〉, 〈9, 5, 7, 13 〉K.
For those in the residue class 12 modulo 13, we apply Lemma 4.4 with a = 8,
w = 2, r = 1, s = 2, b′
1
= 12, b′′
1
= 2, b′′
2
= 12, m′
1
= 13, m′′
1
= 5, m′′
2
= 13 and t = p = 59. We reuse the last set of congruences in (C10) and make use of the additional congruences (C17) J〈8 + 2(3 2 + · · · + 3 i−2), 12 〉, 〈3i, 13 〉K for 3 ≤ i ≤ 59 ,
J〈8 + 2(3 2 + · · · + 3 i−2) + 3 i−1, 2, 12 〉, 〈3i, 5, 13 〉K for 3 ≤ i ≤ 59 .
The last set of congruences in (C10) and the congruences in (C16) and (C17) cover then the integers in S satisfying J〈8, 2, 3〉, 〈9, 5, 7〉K. Therefore we have now covered all the integers in S satisfying J〈8, 2〉, 〈9, 5〉K.Next, we will cover the integers in S satisfying J〈8, 3〉, 〈9, 5〉K. Any of the previous congruences used can be applied here as long as they intersect the class 8 modulo 9 and 3 modulo 5. In particular, the first five congruences 22 M. Filaseta and W. Harvey
in (C4) imply that we need only concern ourselves with integers in S satisfy-ing J〈8, 3, 5〉, 〈9, 5, 7〉K and J〈8, 3, 6〉, 〈9, 5, 7〉K. By combining this information with the congruences (C8)–(C10), what remains to be covered of the integers in S satisfying J〈8, 3〉, 〈9, 5〉K are those satisfying one of
J〈8, 3, 5, 4〉, 〈9, 5, 7, 11 〉K, J〈8, 3, 5, 5〉, 〈9, 5, 7, 11 〉K,
J〈8, 3, 5, 6〉, 〈9, 5, 7, 11 〉K, J〈8, 3, 5, 7〉, 〈9, 5, 7, 11 〉K,
J〈8, 3, 5, 8〉, 〈9, 5, 7, 11 〉K, J〈8, 3, 5, 9〉, 〈9, 5, 7, 11 〉K,
J〈8, 3, 5, 10 〉, 〈9, 5, 7, 11 〉K, J〈8, 3, 6, 4〉, 〈9, 5, 7, 11 〉K,
J〈8, 3, 6, 5〉, 〈9, 5, 7, 11 〉K, J〈8, 3, 6, 6〉, 〈9, 5, 7, 11 〉K.
Of these, the integers in S satisfying
J〈8, 3, 5, 4〉, 〈9, 5, 7, 11 〉K and J〈8, 3, 6, 4〉, 〈9, 5, 7, 11 〉K
will be covered last. To cover the integers in S satisfying J〈8, 3, 5, 5〉, 〈9, 5, 7, 11 〉K, we break them up into residue classes modulo 19 and make use of the 24 divisors of 32 · 5 · 7 · 11 to form the congruences we want. To specify, we use (C18)
J〈8, j 〉, 〈3j , 19 〉K, J〈8, 3, j + 3 〉, 〈3j , 5, 19 〉K,
J〈8, 5, j + 6 〉, 〈3j , 11 , 19 〉K, J〈8, 3, 5, j + 9 〉, 〈3j , 5, 11 , 19 〉K,
J〈8, 5, j + 12 〉, 〈3j , 7, 19 〉K, J〈8, 3, 5, j + 15 〉, 〈3j , 5, 7, 19 〉K,
J〈5, 5, 18 〉, 〈7, 11 , 19 〉K, for j ∈ { 0, 1, 2},
to cover the integers in S satisfying J〈8, 3, 5, 5〉, 〈9, 5, 7, 11 〉K.For the integers in S satisfying J〈8, 3, 5, 6〉, 〈9, 5, 7, 11 〉K, we make use of (C19)
J〈0〉, 〈17 〉K, J〈2, 1〉, 〈3, 17 〉K, J〈3, 2〉, 〈5, 17 〉K,
J〈8, 3〉, 〈9, 17 〉K, J〈2, 3, 4〉, 〈3, 5, 17 〉K,
J〈8, 3, 5〉, 〈9, 5, 17 〉K, J〈6, 6〉, 〈11 , 17 〉K,
J〈2, 6, 7〉, 〈3, 11 , 17 〉K, J〈3, 6, 8〉, 〈5, 11 , 17 〉K,
J〈8, 6, 9〉, 〈9, 11 , 17 〉K, J〈2, 3, 6, 10 〉, 〈3, 5, 11 , 17 〉K,
J〈8, 3, 6, 11 〉, 〈9, 5, 11 , 17 〉K, J〈5, 12 〉, 〈7, 17 〉K,
J〈2, 5, 13 〉, 〈3, 7, 17 〉K, J〈3, 5, 14 〉, 〈5, 7, 17 〉K,
J〈8, 5, 15 〉, 〈9, 7, 17 〉K, J〈2, 3, 5, 16 〉, 〈3, 5, 7, 17 〉K.
Thus, these integers have been covered by breaking them up into their residue classes modulo 17. We break up the integers in S satisfying J〈8, 3, 5, 7〉, 〈9, 5, 7, 11 〉K into their residue classes modulo 13. Observe that moduli with largest prime divisor 13 have been used in (C16) and (C17). Here, the moduli will be different in that the largest two prime divisors will be 11 and 13. We cover the residue Covering subsets of the integers by congruences 23
classes 0 , 1, . . . , 11 modulo 13 by using (C20)
J〈7, 0〉, 〈11 , 13 〉K, J〈2, 7, 1〉, 〈3, 11 , 13 〉K, J〈8, 7, 2〉, 〈9, 11 , 13 〉K,
J〈3, 7, 3〉, 〈5, 11 , 13 〉K, J〈2, 3, 7, 4〉, 〈3, 5, 11 , 13 〉K,
J〈8, 3, 7, 5〉, 〈9, 5, 11 , 13 〉K, J〈5, 7, 6〉, 〈7, 11 , 13 〉K,
J〈2, 5, 7, 7〉, 〈3, 7, 11 , 13 〉K, J〈8, 5, 7, 8〉, 〈9, 7, 11 , 13 〉K,
J〈3, 5, 7, 9〉, 〈5, 7, 11 , 13 〉K, J〈2, 3, 5, 7, 10 〉, 〈3, 5, 7, 11 , 13 〉K,
J〈8, 3, 5, 7, 11 〉, 〈9, 5, 7, 11 , 13 〉K.
For the integers in S satisfying J〈8, 3, 5, 7〉, 〈9, 5, 7, 11 〉K and lying in the residue class 12 modulo 13, we apply Lemma 4.4 with a = 8, w = 2, r = 1,
s = 3, b′
1
= 12, b′′
1
= 3, b′′
2
= 7, b′′
3
= 12, m′
1
= 13, m′′
1
= 5, m′′
2
= 11, m′′
3
= 13 and t = p = 59. Thus, we reuse congruences from (C10) and (C17) and add (C21) J〈8 + 2(3 2 + 3 3 + · · · + 3 i−2) + 3 i−1, 3, 7, 12 〉, 〈3i, 5, 11 , 13 〉K
for 3 ≤ i ≤ 59 to our collection of congruences. To cover the integers in S satisfying J〈8, 3, 5, 8〉, 〈9, 5, 7, 11 〉K, we can reuse the congruences in (C19) to cover those that fall into one of the residue classes 0 , 1, . . . , 5 and 12 , 13 , . . . , 16 modulo 17. To cover those in the re-maining residue classes modulo 17, we use (C22)
J〈8, 3, 5, 6〉, 〈9, 5, 7, 17 〉K, J〈5, 8, 7〉, 〈7, 11 , 17 〉K,
J〈2, 5, 8, 8〉, 〈3, 7, 11 , 17 〉K, J〈3, 5, 8, 9〉, 〈5, 7, 11 , 17 〉K,
J〈8, 5, 8, 10 〉, 〈9, 7, 11 , 17 〉K, J〈2, 3, 5, 8, 11 〉, 〈3, 5, 7, 11 , 17 〉K.
Note that (C8) is the only congruence thus far involving a modulus divisible by 121. To cover the integers in S that satisfy the congruence
J〈8, 3, 5, 9〉, 〈9, 5, 7, 11 〉K, we break them up into 11 residue classes mod-ulo 121. These integers are thus covered by (C23)
J〈2, 9〉, 〈3, 121 〉K, J〈3, 20 〉, 〈5, 121 〉K, J〈8, 31 〉, 〈9, 121 〉K,
J〈2, 3, 42 〉, 〈3, 5, 121 〉K, J〈8, 3, 53 〉, 〈9, 5, 121 〉K,
J〈5, 64 〉, 〈7, 121 〉K, J〈2, 5, 75 〉, 〈3, 7, 121 〉K,
J〈3, 5, 86 〉, 〈5, 7, 121 〉K, J〈8, 5, 97 〉, 〈9, 7, 121 〉K,
J〈2, 3, 5, 108 〉, 〈3, 5, 7, 121 〉K, J〈8, 3, 5, 119 〉, 〈9, 5, 7, 121 〉K.
We cover the integers in S satisfying J〈8, 3, 5, 10 〉, 〈9, 5, 7, 11 〉K by breaking them up into residue classes modulo 23. These are then covered by using 24 M. Filaseta and W. Harvey
(C24)
J〈8, j 〉, 〈3j , 23 〉K, J〈8, 3, j + 3 〉, 〈3j , 5, 23 〉K,
J〈8, 5, j + 6 〉, 〈3j , 7, 23 〉K, J〈8, 10 , j + 9 〉, 〈3j , 11 , 23 〉K,
J〈8, 3, 5, j + 12 〉, 〈3j , 5, 7, 23 〉K, J〈8, 3, 10 , j + 15 〉, 〈3j , 5, 11 , 23 〉K,
J〈8, 5, 10 , j + 18 〉, 〈3j , 7, 11 , 23 〉K, J〈3, 5, 10 , 21 〉, 〈5, 7, 11 , 23 〉K,
J〈2, 3, 5, 10 , 22 〉, 〈3, 5, 7, 11 , 23 〉K, for j ∈ { 0, 1, 2}.
Next, we turn to covering integers in S satisfying J〈8, 3, 6, 6〉, 〈9, 5, 7, 11 〉K.We break these up into residue classes modulo 17. The first 12 congruences from (C19) cover those integers in the residue classes 0 , 1, . . . , 11 modulo 17. To cover those that are 12, 13 or 14 modulo 17, we use (C25)
J〈8, 3, 6, 6, 12 〉, 〈9, 5, 7, 11 , 17 〉K, J〈8, 13 〉, 〈27 , 17 〉K,
J〈17 , 3, 13 〉, 〈27 , 5, 17 〉K, J〈26 , 6, 13 〉, 〈27 , 7, 17 〉K,
J〈8, 6, 14 〉, 〈27 , 11 , 17 〉K, J〈17 , 3, 6, 14 〉, 〈27 , 5, 7, 17 〉K,
J〈26 , 3, 6, 14 〉, 〈27 , 5, 11 , 17 〉K.
For those integers in S satisfying J〈8, 3, 6, 6〉, 〈9, 5, 7, 11 〉K and lying in the residue class 15 modulo 17, we cover separately the three residue classes 8, 17 and 26 modulo 27 that they belong to, covering the first two directly and using Lemma 4.4 to cover the third. For Lemma 4.4, we want a = 26,
w = 3, r = 1, s = 2, b′
1
= 15, b′′
1
= 3, b′′
2
= 15, m′
1
= 17, m′′
1
= 5, m′′
2
= 17, and t = p = 59. We reuse the last 59 congruences in (C10). This leads to the additional congruences (C26)
J〈8, 6, 6, 15 〉, 〈27 , 7, 11 , 17 〉K, J〈17 , 3, 6, 6, 15 〉, 〈27 , 5, 7, 11 , 17 〉K,
J〈26 + 2(3 3 + · · · + 3 i−2), 15 〉, 〈3i, 17 〉K for 4 ≤ i ≤ 59 ,
J〈26 + 2(3 3 + · · · + 3 i−2) + 3 i−1, 3, 15 〉, 〈3i, 5, 17 〉K for 4 ≤ i ≤ 59 .
For the remaining integers in S satisfying J〈8, 3, 6, 6〉, 〈9, 5, 7, 11 〉K, which are 16 modulo 17, we break them up into residue classes modulo 13 and note that we have not used any moduli yet where the largest two prime divisors are 13 and 17. We are able then to cover these integers by using (C27)
J〈0, 16 〉, 〈13 , 17 〉K, J〈2, 1, 16 〉, 〈3, 13 , 17 〉K, J〈8, 2, 16 〉, 〈9, 13 , 17 〉K,
J〈3, 3, 16 〉, 〈5, 13 , 17 〉K, J〈2, 3, 4, 16 〉, 〈3, 5, 13 , 17 〉K,
J〈8, 3, 5, 16 〉, 〈9, 5, 13 , 17 〉K, J〈6, 6, 16 〉, 〈7, 13 , 17 〉K,
J〈2, 6, 7, 16 〉, 〈3, 7, 13 , 17 〉K, J〈8, 6, 8, 16 〉, 〈9, 7, 13 , 17 〉K,
J〈6, 9, 16 〉, 〈11 , 13 , 17 〉K, J〈2, 6, 10 , 16 〉, 〈3, 11 , 13 , 17 〉K,
J〈8, 6, 11 , 16 〉, 〈9, 11 , 13 , 17 〉K, J〈3, 6, 12 , 16 〉, 〈5, 7, 13 , 17 〉K.
Now, we cover the integers in S satisfying J〈8, 3, 6, 5〉, 〈9, 5, 7, 11 〉K by breaking them up into their residue classes modulo 19 and beginning with congruences similar to the last case but with 17 replaced by 19. In particular, Covering subsets of the integers by congruences 25
(C18) covers those integers in S satisfying J〈8, 3, 6, 5〉, 〈9, 5, 7, 11 〉K that are 0, 1, . . . , 10 or 11 modulo 19. For those congruent to 12, 13 or 14 modulo 19, we use (C28)
J〈8, 3, 6, 5, 12 〉, 〈9, 5, 7, 11 , 19 〉K, J〈8, 13 〉, 〈27 , 19 〉K,
J〈17 , 3, 13 〉, 〈27 , 5, 19 〉K, J〈26 , 6, 13 〉, 〈27 , 7, 19 〉K,
J〈8, 5, 14 〉, 〈27 , 11 , 19 〉K, J〈17 , 3, 6, 14 〉, 〈27 , 5, 7, 19 〉K,
J〈26 , 3, 5, 14 〉, 〈27 , 5, 11 , 19 〉K.
For those congruent to 15 modulo 19, we consider their three residue classes modulo 27, covering those in the first two residue classes directly and cov-ering those in the third residue class using Lemma 4.4 with a = 26, w = 3,
r = 1, s = 2, b′
1
= 15, b′′
1
= 3, b′′
2
= 15, m′
1
= 19, m′′
1
= 5, m′′
2
= 19, and
t = p = 59. We again reuse the last collection of congruences in (C10). Thus, we make use of (C29)
J〈8, 6, 5, 15 〉, 〈27 , 7, 11 , 19 〉K, J〈17 , 3, 6, 5, 15 〉, 〈27 , 5, 7, 11 , 19 〉K,
J〈26 + 2(3 3 + · · · + 3 i−2), 15 〉, 〈3i, 19 〉K for 4 ≤ i ≤ 59 ,
J〈26 + 2(3 3 + · · · + 3 i−2) + 3 i−1, 3, 15 〉, 〈3i, 5, 19 〉K for 4 ≤ i ≤ 59 .
For those that are 16 modulo 19, we consider their seven residue classes modulo 49 and use (C30)
J〈6, 16 〉, 〈49 , 19 〉K, J〈2, 13 , 16 〉, 〈3, 49 , 19 〉K,
J〈8, 20 , 16 〉, 〈9, 49 , 19 〉K, J〈3, 27 , 16 〉, 〈5, 49 , 19 〉K,
J〈34 , 5, 16 〉, 〈49 , 11 , 19 〉K, J〈2, 3, 41 , 16 〉, 〈3, 5, 49 , 19 〉K,
J〈8, 3, 48 , 16 〉, 〈9, 5, 49 , 19 〉K.
For those that are 17 modulo 19, we consider their eleven residue classes modulo 121 and use (C31)
J〈5, 17 〉, 〈121 , 19 〉K, J〈2, 16 , 17 〉, 〈3, 121 , 19 〉K,
J〈8, 27 , 17 〉, 〈9, 121 , 19 〉K, J〈3, 38 , 17 〉, 〈5, 121 , 19 〉K,
J〈6, 49 , 17 〉, 〈7, 121 , 19 〉K, J〈2, 3, 60 , 17 〉, 〈3, 5, 121 , 19 〉K,
J〈8, 3, 71 , 17 〉, 〈9, 5, 121 , 19 〉K, J〈2, 6, 82 , 17 〉, 〈3, 7, 121 , 19 〉K,
J〈8, 6, 93 , 17 〉, 〈9, 7, 121 , 19 〉K, J〈2, 3, 6, 104 , 17 〉, 〈3, 5, 7, 121 , 19 〉K,
J〈8, 3, 6, 115 , 17 〉, 〈9, 5, 7, 121 , 19 〉K.
We break up the integers in S satisfying J〈8, 3, 6, 5〉, 〈9, 5, 7, 11 〉K which are 18 modulo 19 into residue classes modulo 13. Observe that we have not used moduli which have their two largest prime divisors 13 and 19. We can cover these integers with the congruences (C32) J〈0, 18 〉, 〈13 , 19 〉K, J〈2, 1, 18 〉, 〈3, 13 , 19 〉K,
J〈8, 2, 18 〉, 〈9, 13 , 19 〉K, J〈3, 3, 18 〉, 〈5, 13 , 19 〉K,26 M. Filaseta and W. Harvey
(
C32)
[cont .]
J〈2, 3, 4, 18 〉, 〈3, 5, 13 , 19 〉K, J〈8, 3, 5, 18 〉, 〈9, 5, 13 , 19 〉K,
J〈6, 6, 18 〉, 〈7, 13 , 19 〉K, J〈2, 6, 7, 18 〉, 〈3, 7, 13 , 19 〉K,
J〈8, 6, 8, 18 〉, 〈9, 7, 13 , 19 〉K, J〈3, 6, 9, 18 〉, 〈5, 7, 13 , 19 〉K,
J〈2, 3, 6, 10 , 18 〉, 〈3, 5, 7, 13 , 19 〉K, J〈8, 3, 6, 11 , 18 〉, 〈9, 5, 7, 13 , 19 〉K,
J〈3, 6, 5, 12 , 18 〉, 〈5, 7, 11 , 13 , 19 〉K.
Of the integers in S satisfying J〈8, 3〉, 〈9, 5〉K, we are now left with those which satisfy one of
J〈8, 3, 5, 4〉, 〈9, 5, 7, 11 〉K, J〈8, 3, 6, 4〉, 〈9, 5, 7, 11 〉K.
We handle those satisfying J〈8, 3, 5, 4〉, 〈9, 5, 7, 11 〉K next by considering the different residue classes modulo 29. There are 24 moduli that we can use here of the form 29 d where d | (9 · 5 · 7 · 11), and we use all of them to cover those integers in 24 of the 29 residue classes. Since J〈8, 3, 5, 4〉, 〈9, 5, 7, 11 〉K
is equivalent to
x ≡ 2798 (mod 9 · 5 · 7 · 11) ,
we can express these congruences as (C33) J〈2798 , j 〉, 〈dj , 29 〉K, where 0 ≤ j ≤ 23 and where
{di : 0 ≤ i ≤ 23 } = {d ∈ Z+ : d | (9 · 5 · 7 · 11) }.
Note that the order of the divisors di in (C33) does not matter. We cover the integers in S satisfying J〈8, 3, 5, 4〉, 〈9, 5, 7, 11 〉K in four more residue classes modulo 29 by applying Lemma 4.4 four times. For those congruent to 24 modulo 29, we apply Lemma 4.4 with a = 8, w = 2, r = 1, s = 2, b′
1
= 24,
b′′
1
= 3, b′′
2
= 24, m′
1
= 29, m′′
1
= 5, m′′
2
= 29, and t = p = 59. By the last line of congruences in (C10), this gives the additional congruences (C34) J〈8 + 2(3 2 + · · · + 3 i−2), 24 〉, 〈3i, 29 〉K for 3 ≤ i ≤ 59 ,
J〈8 + 2(3 2 + · · · + 3 i−2) + 3 i−1, 3, 24 〉, 〈3i, 5, 29 〉K for 3 ≤ i ≤ 59 .
For those congruent to 25 modulo 29, we apply Lemma 4.4 with a = 8,
w = 2, r = 2, s = 2, b′
1
= 5, b′
2
= 25, b′′
1
= 4, b′′
2
= 25, m′
1
= 7, m′
2
= 29,
m′′
1
= 11, m′′
2
= 29, and t = p = 59. In this case, with the congruences from (C10), we need only make use of the additional congruences (C35) J〈8 + 2(3 2 + · · · + 3 i−2), 5, 25 〉, 〈3i, 7, 29 〉K for 3 ≤ i ≤ 59 ,
J〈8 + 2(3 2 + · · · + 3 i−2) + 3 i−1, 4, 25 〉, 〈3i, 11 , 29 〉K for 3 ≤ i ≤ 59 .
For the integers in S satisfying J〈8, 3, 5, 4〉, 〈9, 5, 7, 11 〉K that are 26 mod-ulo 29, we apply Lemma 4.4 with a = 8, w = 2, r = 3, s = 3, b′
1
= 3, b′
2
= 5,
b′
3
= 26, b′′
1
= 3, b′′
2
= 4, b′′
3
= 26, m′
1
= 5, m′
2
= 7, m′
3
= 29, m′′
1
= 5,
m′′
2
= 11, m′′
3
= 29, and t = p = 59. In this case, we additionally get Covering subsets of the integers by congruences 27
(C36) J〈8 + 2(3 2 + · · · + 3 i−2), 3, 5, 26 〉, 〈3i, 5, 7, 29 〉K for 3 ≤ i ≤ 59 ,
J〈8+2(3 2+ · · · +3 i−2)+3 i−1, 3, 4, 26 〉, 〈3i, 5, 11 , 29 〉K for 3 ≤ i ≤ 59 .
For the integers in S satisfying J〈8, 3, 5, 4〉, 〈9, 5, 7, 11 〉K that are 27 mod-ulo 29, we apply Lemma 4.4 with a = 8, w = 2, r = 3, s = 4, b′
1
= 5, b′
2
= 4,
b′
3
= 27, b′′
1
= 3, b′′
2
= 5, b′′
3
= 4, b′′
4
= 27, m′
1
= 7, m′
2
= 11, m′
3
= 29, m′′
1
= 5,
m′′
2
= 7, m′′
3
= 11, m′′
4
= 29, and t = p = 59. In this case, we additionally get (C37) J〈8 + 2(3 2 + · · · + 3 i−2), 5, 4, 27 〉, 〈3i, 7, 11 , 29 〉K for 3 ≤ i ≤ 59 ,
J〈8 + 2(3 2 + · · · + 3 i−2) + 3 i−1, 3, 5, 4, 27 〉, 〈3i, 5, 7, 11 , 29 〉K
for 3 ≤ i ≤ 59 .
For the integers in S satisfying J〈8, 3, 5, 4〉, 〈9, 5, 7, 11 〉K that are 28 mod-ulo 29, we consider their residue classes modulo 37. There are 48 divisors of 9 · 5 · 7 · 11 · 29. If these divisors are ordered from least to greatest, the 37th divisor is 2233. The congruence
J〈8, 3, 5, 4, 28 〉, 〈9, 5, 7, 11 , 29 〉K
is equivalent to
x ≡ 6263 (mod 9 · 5 · 7 · 11 · 29) .
Therefore we can obtain a covering of these integers in the residue class 28 modulo 29 with the congruences (C38) J〈6263 , j 〉, 〈d′
j
, 37 〉K, where 0 ≤ j ≤ 36 and where
{d′
i
: 0 ≤ i ≤ 36 } = {d ∈ Z+ : d | (9 · 5 · 7 · 11 · 29) , d ≤ 2233 }.
Analogously to the situation in (C33), the order of the d′
i
in (C38) can be arbitrary. Next, we describe a covering of the integers in S satisfying
J〈8, 3, 6, 4〉, 〈9, 5, 7, 11 〉K.
We can replace the congruences which we just listed for integers in S satis-fying J〈8, 3, 5, 4〉, 〈9, 5, 7, 11 〉K by congruences which cover 29 residue classes modulo 31 instead of residue classes modulo 29. The congruence J〈8, 3, 6, 4〉,
〈9, 5, 7, 11 〉K is equivalent to
x ≡ 818 (mod 9 · 5 · 7 · 11) ,
so this will lead to some changes needed in the congruences. For the first 24 residue classes modulo 31, we use (C39) J〈818 , j 〉, 〈dj , 31 〉K, where 0 ≤ j ≤ 23 and the dj are as in (C33). We make use of Lemma 4.4 and the last line of congruences from (C10) again to cover the integers that are 24 modulo 31. This leads to the additional congruences 28 M. Filaseta and W. Harvey
(C40) J〈8 + 2(3 2 + · · · + 3 i−2), 24 〉, 〈3i, 31 〉K for 3 ≤ i ≤ 59 ,
J〈8 + 2(3 2 + · · · + 3 i−2) + 3 i−1, 3, 24 〉, 〈3i, 5, 31 〉K for 3 ≤ i ≤ 59 .
Recalling our application of Lemma 4.4 to obtain (C35), we want here to use the same information except b′
1
= 6 and m′
2
= m′′
2
= 31. Reusing congru-ences in (C10), we cover the integers in S satisfying J〈8, 3, 6, 4〉, 〈9, 5, 7, 11 〉K
that are in 25 modulo 31 with the additional congruences (C41) J〈8 + 2(3 2 + · · · + 3 i−2), 6, 25 〉, 〈3i, 7, 31 〉K for 3 ≤ i ≤ 59 ,
J〈8 + 2(3 2 + · · · + 3 i−2) + 3 i−1, 4, 25 〉, 〈3i, 11 , 31 〉K for 3 ≤ i ≤ 59 .
Similarly, for the integers in S satisfying J〈8, 3, 6, 4〉, 〈9, 5, 7, 11 〉K that are 26 modulo 31, we mirror what we did to obtain (C36) and use additionally (C42) J〈8 + 2(3 2 + · · · + 3 i−2), 3, 6, 26 〉, 〈3i, 5, 7, 31 〉K for 3 ≤ i ≤ 59 ,
J〈8 + 2(3 2 + · · · + 3 i−2) + 3 i−1, 3, 4, 26 〉, 〈3i, 5, 11 , 31 〉K
for 3 ≤ i ≤ 59 .
We cover the integers in S satisfying J〈8, 3, 6, 4〉, 〈9, 5, 7, 11 〉K that are 27 modulo 31 by applying Lemma 4.4 as in (C37) but with b′
1
= 6, b′′
2
= 6 and
m′
3
= m′′
4
= 31, by using congruences in (C10), and additionally (C43) J〈8 + 2(3 2 + · · · + 3 i−2), 6, 4, 27 〉, 〈3i, 7, 11 , 31 〉K for 3 ≤ i ≤ 59 ,
J〈8 + 2(3 2 + · · · + 3 i−2) + 3 i−1, 3, 6, 4, 27 〉, 〈3i, 5, 7, 11 , 31 〉K
for 3 ≤ i ≤ 59 .
The remaining integers in S satisfying J〈8, 3, 6, 4〉, 〈9, 5, 7, 11 〉K are 28, 29 or 30 modulo 31. The integers in each of these residue classes can be covered using a list of congruences similar to (C38) but with 29 replaced by 31, and 37 replaced by 41, 43 and 47. Observe that
J〈8, 3, 6, 4, 28 〉, 〈9, 5, 7, 11 , 31 〉K, J〈8, 3, 6, 4, 29 〉, 〈9, 5, 7, 11 , 31 〉K,
J〈8, 3, 6, 4, 30 〉, 〈9, 5, 7, 11 , 31 〉K
are equivalent to
x ≡ 38933 (mod 9 · 5 · 7 · 11 · 31) , x ≡ 7748 (mod 9 · 5 · 7 · 11 · 31) ,x ≡ 83978 (mod 9 · 5 · 7 · 11 · 31) ,
respectively. We cover the integers in S satisfying these congruences then by using (C44)
J〈38933 , j 〉, 〈d′′
j
, 41 〉K, where 0 ≤ j ≤ 40 ,
J〈7748 , j 〉, 〈d′′
j
, 43 〉K, where 0 ≤ j ≤ 42 ,
J〈83978 , j 〉, 〈d′′
j
, 47 〉K, where 0 ≤ j ≤ 46 and where
{d′′
0
, d ′′
1
, . . . , d ′′
47
} = {d ∈ Z+ : d | (9 · 5 · 7 · 11 · 31) }.Covering subsets of the integers by congruences 29
The order of the divisors d′′
j
does not matter, but we take d′′
0
< d ′′
1
< · · · < d ′′
47
to be explicit. We have now completely covered all the integers in S which satisfy
J〈8, 3〉, 〈9, 5〉K. We still need congruences that cover the n ∈ S satisfying one of J〈2, 4〉, 〈9, 5〉K, J〈5, 4〉, 〈9, 5〉K and J〈8, 4〉, 〈9, 5〉K, or equivalently the n
that satisfy J〈2, 4〉, 〈3, 5〉K. Of significance is that in every one of the 2210 congruences appearing in (C1)–(C44), no modulus is divisible by 5 2. We set
r = 1169, which is the number of moduli appearing in (C1)–(C44) which are not divisible by 5; and we set s = 1041, the number of moduli divisible by 5. We apply Lemma 4.3 with p = 5. Recall that after the statement of Lemma 4.3, we showed that the set S of integers that are sums of two squares satisfies the condition in the second sentence of the lemma. As noted above, every modulus appearing in (C1)–(C44) can be expressed in the form given by the elements of C1 in Lemma 4.3 or of the form given by the elements of C2. Also, the congruences in C = C1 ∪ C2 cover the set of integers in S
that belong to one of the congruence classes 0, 1, 2 and 3 modulo 5. We take
q = t = 61. Note that the congruences in C1 correspond to the congruences given in (i) of Lemma 4.3 and the congruences in C2 correspond to those in (ii) with i = 1. Hence, Lemma 4.3 now implies that we can cover every element of S by combining these congruences with (C45) J〈5i−1 − 1 + bj 5i−1, b j 〉, 〈5i, m ′
j
〉K for 2 ≤ i ≤ 61 and 1 ≤ j ≤ 1041 ,
J〈j, −1〉, 〈61 , 561 −j 〉K for 0 ≤ j ≤ 60 .
Thus, from (C1)–(C45), we obtain a set of 64731 congruences, with dis-tinct odd moduli > 1, that cover the set of integers that are sums of two squares, completing the proof of Theorem 4.1.
Concluding remarks. The recent works by B. Hough and by B. Hough and P. Nielsen establish that certain conditions on the mod-uli (that the moduli are all distinct and large or the moduli are > 1 and relatively prime to 6) ensure a system of congruences cannot cover the in-tegers. Motivated by this, we have provided here some initial insights into the notion of using a set of congruences to cover subsets of the integers. As noted at the end of the introduction, there are many questions in this direction that are still unanswered, which we hope will provide a source of future investigations.
Acknowledgements. The authors are grateful to the National Security Agency for funding during research for this paper. In addition, the authors express their gratitude to the referee for some helpful remarks on exposition in this paper. 30 M. Filaseta and W. Harvey
References
P. Erd˝ os, On integers of the form 2k + p and some related problems , Summa Brasil. Math. 2 (1950), 113–123. M. Filaseta, C. Finch and M. Kozek, On powers associated with Sierpi´ nski numbers, Riesel numbers and Polignac’s conjecture , J. Number Theory 128 (2008), 1916–1940. M. Filaseta, K. Ford, S. Konyagin, C. Pomerance and G. Yu, Sieving by large integers and covering systems of congruences , J. Amer. Math. Soc. 20 (2007), 495–517. M. Filaseta, M. Kozek, C. Nicol and J. Selfridge, Composites that remain composite after changing a digit , J. Combin. Number Theory 2 (2010), 25–36. R. L. Graham, A Fibonacci-like sequence of composite numbers , Math. Mag. 37 (1964), 322–324. A. Granville, private communication, December 13, 2016. R. K. Guy, Unsolved Problems in Number Theory , 3rd ed., Problem Books in Math., Springer, New York, 2004. J. Harrington, Two questions concerning covering systems , Int. J. Number Theory 11 (2015), 1739–1750. B. Hough, Solution of the minimum modulus problem for covering systems , Ann. of Math. 181 (2015), 361–382. B. Hough and P. Nielsen, Covering systems with restricted divisibility , arXiv:1703.02133 (2017). L. Jones, When does appending the same digit repeatedly on the right of a positive integer generate a sequence of composite integers? , Amer. Math. Monthly 118 (2011), 153–160. C. E. Krukenberg, Covering sets of the integers , doctoral dissertation, Univ. of Illinois at Urbana-Champaign, IL, 1971. P. P. Nielsen, A covering system whose smallest modulus is 40 , J. Number Theory 129 (2009), 640–666. T. Owens, A covering system with minimum modulus 42 , master’s thesis, Brigham Young Univ., Provo, UT, 2014 H. Riesel, N˚ agra stora primtal , Elementa 39 (1956), 258–260. A. Schinzel, Reducibility of polynomials and covering systems of congruences , Acta Arith. 13 (1967), 91–101. W. Sierpi´ nski, Sur un probl` eme concernant les nombres k · 2n + 1, Elem. Math. 15 (1960), 73–74. K. Zsigmondy, Zur Theorie der Potenzreste , Monatsh. Math. Phys. 3 (1892), 265–284. Michael Filaseta Department of Mathematics University of South Carolina Columbia, SC 29208, U.S.A. E-mail: filaseta@math.sc.edu Wilson Harvey Walker 3-33 University of Louisiana at Monroe Monroe, LA 71209, U.S.A. E-mail: harvey@ulm.edu Abstract (will appear on the journal’s web site only) A number of results are established showing that certain subsets of the integers can be covered by congruences with distinct moduli satisfying var-ious restrictions. For example, the primes, the powers of 2, the Fibonacci numbers, and the sums of two squares can each be covered by congruences with distinct odd moduli > 1. |
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Charles law | Volume and Temperature Relationship
Azubuike faith
14 min readDec 18, 2023
Introduction:
Charles’s Law, a fundamental principle in the field of thermodynamics, provides valuable insights into the behavior of gases. Named after the French scientist Jacques Charles, this law describes the relationship between the volume and temperature of an ideal gas when pressure remains constant. In this blog, we will delve into the intricacies of Charles’s Law, its historical context, and its practical applications.
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Historical Background:
Jacques Charles, born in 1746, made significant contributions to the understanding of gas behavior. In the late 18th century, he conducted experiments with gases, specifically focusing on the relationship between temperature and volume. Charles's observations led to the formulation of what we now know as Charles's Law.
The historical background of Charles's Law is rooted in the pioneering work of Jacques Charles, a French scientist, and inventor. Born in 1746, Charles made significant contributions to the field of physics, with one of his most notable achievements being the formulation of Charles's Law in the late 18th century.
In 1787, Charles conducted experiments on the thermal expansion of gases, a topic that had intrigued scientists for centuries. Charles observed that, at constant pressure, the volume of a gas is directly proportional to its absolute temperature. This groundbreaking discovery led to the formulation of Charles's Law, a fundamental principle in the study of gas behavior.
Charles's Law was later refined and popularized by Joseph Louis Gay-Lussac, another French chemist, in the early 19th century. The law played a crucial role in advancing the understanding of gas properties and laid the foundation for the development of the broader field of thermodynamics.
Today, Charles's Law is a fundamental concept in physics and chemistry, taught in classrooms around the world. Its application extends beyond theoretical understanding to practical uses in fields such as engineering, meteorology, and industrial processes. The historical journey of Charles's Law reflects the collaborative and evolving nature of scientific exploration, with each discovery building upon the foundations laid by those who came before.
Statement of Charles's Law:
Charles's Law can be succinctly stated as follows: the volume of a gas is directly proportional to its absolute temperature when pressure remains constant. Mathematically, it can be expressed as V ∝ T or V/T = k, where V is volume, T is temperature, and k is a constant. It can also be written as V1/T1 = V2/T2 where V1 and V2 are initial and final volume while T1 and T2 are initial and final temperature respectively.
Interpreting the Law:
This law implies that as the temperature of a gas increases, its volume will also increase proportionally, and vice versa. The constant k in the equation represents the situation when the gas is at absolute zero temperature (0 Kelvin), where the volume theoretically becomes zero.
Illustrative Example:
Consider a balloon filled with a certain amount of gas at room temperature. If the temperature is then increased, Charles's Law predicts that the volume of the gas inside the balloon will expand. Conversely, if the temperature is decreased, the volume will contract.
Graphical Representation:
Charles's Law is a fundamental principle in thermodynamics that describes how gases tend to expand when heated. Named after the French scientist Jacques Charles, this law is a crucial component of the ideal gas law, providing insight into the relationship between temperature and volume. Graphical representation plays a pivotal role in elucidating the nuances of Charles's Law, offering a visual understanding of the gas behavior under changing thermal conditions.
At its core, Charles's Law states that the volume of a given amount of gas is directly proportional to its absolute temperature, provided the pressure and quantity of gas remain constant. This relationship can be expressed mathematically as V/T = k, where V is the volume, T is the absolute temperature, and k is a constant. To illustrate this concept graphically, scientists often employ temperature-volume (T-V) graphs.
The T-V graph for Charles's Law typically produces a straight line when the data points are plotted. As the temperature of the gas increases, the volume also increases proportionally. This linear relationship signifies the direct proportionality between temperature and volume, a key characteristic of Charles's Law.
One common way to visualize Charles's Law is by using a Cartesian coordinate system, where the x-axis represents temperature (in Kelvin) and the y-axis represents volume. The resulting graph showcases a positive slope, indicating the direct correlation between temperature and volume. Observing the graph, one can discern that as the gas is heated, it expands, occupying a larger volume.
The graphical representation of Charles's Law becomes even more insightful when comparing different gases under similar conditions. While the proportional relationship holds true for all gases, the slopes of their T-V graphs may differ due to variations in their molecular properties. This aspect of Charles's Law is particularly valuable in understanding the behavior of different gases when subjected to temperature changes.
Moreover, the T-V graph allows scientists and researchers to extrapolate data and make predictions about the gas's behavior at various temperatures. By extending the line on the graph, one can estimate the volume of the gas at temperatures beyond the experimental data points, providing a comprehensive understanding of the gas's response to temperature variations.
Essentially, graphical representation is a powerful tool for comprehending Charles's Law. The T-V graph visually captures the direct proportionality between temperature and volume, offering a clear depiction of how gases behave when exposed to changing thermal conditions. This graphical approach not only aids in conceptualizing the law but also facilitates predictions and comparisons between different gases, contributing to a deeper understanding of the principles governing gas behavior.
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Applications in the Real World:
Charles's Law finds applications in various scientific and industrial fields. One practical application is in weather balloons, which ascend to different altitudes and experience changes in temperature. Understanding how the volume of the gas in the balloon responds to temperature fluctuations is crucial for accurate data collection.
Charles's Law has numerous real-world applications, ranging from everyday occurrences to crucial industrial processes.
One notable application is in the automotive industry, specifically in tire pressure. According to Charles's Law, as the temperature of a gas increases, its volume also increases proportionally. This principle is critical for understanding and predicting changes in tire pressure due to temperature variations. As the temperature rises, the air inside a tire expands, leading to an increase in volume. This expansion can result in higher tire pressure, influencing vehicle performance and safety. Therefore, engineers and motorists must consider Charles's Law when determining optimal tire pressure for different weather conditions.
In the realm of weather forecasting and meteorology, Charles's Law plays a vital role in understanding atmospheric phenomena. The Earth's atmosphere is composed of gases, and their behavior conforms to the principles of gas laws. As temperature changes occur in different atmospheric layers, Charles's Law helps meteorologists predict alterations in volume, pressure, and density. This knowledge is crucial for anticipating weather patterns, including the formation of storms and the dynamics of air masses.
Medical applications also benefit from Charles's Law, particularly in the field of respiratory therapy. Devices such as ventilators and respiratory humidifiers rely on the principles of gas laws to regulate the volume and temperature of inhaled and exhaled air. Healthcare professionals use these principles to optimize respiratory treatments for patients with conditions such as asthma or chronic obstructive pulmonary disease (COPD).
In the domain of cryogenics, where extremely low temperatures are employed, Charles's Law becomes instrumental. Industries that deal with liquefied gases, such as the production and storage of liquid nitrogen or oxygen, rely on this law to predict volume changes as temperatures plummet. Engineers must carefully consider these factors to design and maintain cryogenic systems effectively.
Additionally, Charles's Law finds applications in various scientific research endeavors, including studies related to space exploration. Understanding how gases behave at different temperatures is essential for designing spacecraft and life support systems that can function in the extreme conditions of outer space.
Summarily, Charles's Law transcends theoretical physics, finding practical applications in diverse fields. From ensuring road safety through tire pressure regulation to advancing medical treatments and enabling space exploration, the real-world implications of this gas law underscore its significance in enhancing our understanding of the physical world and driving technological advancements.
Moreover, Charles's Law is integral to the operation of devices such as air conditioners and refrigerators. These appliances manipulate gas volumes to control temperature, and an understanding of Charles's Law is vital for their efficient functioning.
Limitations and Considerations:
While this law provides valuable insights into gas behavior, it is essential to recognize its limitations and consider various factors that may impact its applicability.
Ideal Gas Assumption:
Charles' Law is based on the assumption that the gas behaves ideally, meaning there are no intermolecular forces or other deviations from the ideal gas behavior. In reality, gases deviate from this ideal behavior at high pressures and low temperatures. Hence, the law is most accurate under conditions where the ideal gas assumption holds true.
Pressure Must Remain Constant:
One of the limitations of Charles' Law is that it holds only when the pressure on the gas remains constant. In many practical scenarios, pressure changes may occur, affecting the accuracy of the law's predictions. Deviations from constant pressure conditions can lead to discrepancies between the expected and observed results.
Temperature Scale Consistency:
Charles' Law relies on a consistent temperature scale, typically the Kelvin scale. It is crucial to use an absolute temperature scale to avoid negative temperatures, which would result in undefined or unrealistic volume values. Celsius or Fahrenheit scales must be converted to Kelvin for accurate application of the law.
Molecular Interaction:
The law assumes negligible molecular volume and no intermolecular forces. In reality, gas molecules occupy space, and there are interactions between them. At high pressures, the volume occupied by the gas molecules themselves becomes significant, impacting the accuracy of the law.
Limited Applicability to Real Gases:
Real gases exhibit deviations from ideal behavior, especially at extreme conditions. Charles' Law may not accurately predict the behavior of real gases, particularly when approaching conditions of high pressure or low temperature. Corrections and adjustments are often needed to account for these deviations.
Considerations for Practical Application:
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Calibration and Instrumentation:
Accurate measurements of temperature and volume are crucial for applying Charles' Law. Calibration of instruments and ensuring they operate within their specified ranges contribute to reliable results.
Experiment Design:
Careful consideration of experimental conditions is essential. The selection of appropriate pressure levels, temperature ranges, and gas types impacts the reliability and relevance of the data obtained using Charles' Law.
In essence, while Charles' Law provides valuable insights into the relationship between temperature and volume of gases, its limitations must be acknowledged. Understanding the ideal gas assumptions, pressure conditions, and real gas behavior is crucial for its accurate application in various scientific and industrial contexts. Scientists and engineers must exercise caution and consider these factors when utilizing Charles' Law to make informed predictions and decisions.
Common Misconceptions:
Despite its significance, there are several common misconceptions surrounding this law that can lead to misunderstandings. Let's explore and debunk some of these misconceptions.
Misconception 1: Temperature Always Increases Volume
One common misconception is that if you increase the temperature of a gas, its volume will always increase. While Charles's Law states that there is a direct proportional relationship between the temperature and volume of a gas at constant pressure, this assumes that other factors, such as pressure and quantity of gas, remain constant. If pressure changes, the relationship between temperature and volume may not follow Charles's Law precisely.
Misconception 2: Charles's Law Applies to All Gases Equally
Another misconception is the idea that Charles's Law applies universally to all gases in the same way. In reality, different gases may behave differently due to variations in their molecular properties. The ideal gas law incorporates these differences by considering the specific gas constant, which adjusts for the molecular weight of the gas. So, while Charles's Law provides a useful framework, it's essential to consider the specific characteristics of the gas in question.
Misconception 3: Charles's Law Only Works at Extreme Temperatures
Some may mistakenly believe that Charles's Law is only applicable under extreme temperatures. However, this law applies across a broad range of temperatures, from low to high. It becomes particularly relevant in situations where temperature changes significantly impact gas behavior, such as in industrial processes, weather patterns, or everyday applications like inflating a balloon.
Misconception 4: Charles's Law Is Invalid at Low Temperatures
Conversely, some may think that Charles's Law becomes invalid at low temperatures. In reality, as long as the gas remains in a gaseous state, Charles's Law holds true. At lower temperatures, gases may approach their condensation point and behave differently, but within the gas phase, the law remains a valuable tool for understanding their behavior.
Misconception 5: Volume Can Decrease Indefinitely with Decreasing Temperature
A common misunderstanding is the belief that as temperature decreases, the volume of a gas can shrink indefinitely. While Charles's Law suggests a decrease in volume with decreasing temperature, it does not account for the point at which the gas undergoes a phase change (liquefaction). At certain temperatures and pressures, gases transition to liquids, altering the relationship between temperature and volume.
Ultimately, understanding Charles's Law is crucial for comprehending gas behavior, but it's equally important to recognize its limitations and the conditions under which it applies accurately. Dispelling these common misconceptions enables a more nuanced and accurate grasp of the principles governing gas properties.
FAQ:
Q1. Can Charles law be experimentally proven?
Answer: While Charles's Law is widely accepted, experimental verification is crucial in scientific inquiry.
To experimentally prove Charles's Law, one can conduct a simple laboratory experiment. Begin with a gas sample confined in a container with a movable piston, maintaining constant pressure. By varying the temperature while keeping pressure constant, measurements of volume and temperature can be recorded. The data should exhibit a linear relationship, demonstrating that as temperature increases, so does the volume, validating Charles's Law.
Moreover, real-world applications, such as weather balloons, rely on Charles's Law. As a balloon ascends in the atmosphere, the temperature decreases, causing the volume of the gas within the balloon to contract. This practical application further supports the experimental evidence of Charles's Law.
Charles's Law can be experimentally proven through controlled conditions in a laboratory setting and through the observation of natural phenomena. The law's validation not only enhances our understanding of gas behavior but also contributes to the development of technologies essential in various scientific fields.
Q2. Does Charles law apply to all gases?
Answer: Charles's Law, a fundamental principle in thermodynamics, describes the relationship between the volume and temperature of an ideal gas, assuming pressure remains constant. While this law is a valuable tool for understanding gas behavior, it is essential to recognize its limitations.
Charles's Law primarily applies to ideal gases, which adhere to certain assumptions, such as negligible molecular volume and attractive forces between particles. Real gases, however, deviate from these ideal conditions, especially at high pressures or low temperatures. Under such circumstances, the intermolecular forces become significant, impacting the gas's behavior and causing deviations from the predictions of Charles's Law.
Additionally, certain gases exhibit deviations even under normal conditions. For instance, diatomic gases like oxygen and nitrogen may deviate slightly due to their molecular structures and intermolecular forces.
While Charles's Law is a valuable tool for understanding the behavior of ideal gases, it is crucial to consider the nature of the specific gas in question. Real gases and those with unique molecular structures may deviate from the predictions of Charles's Law, emphasizing the importance of a nuanced understanding of gas behavior in different conditions.
Q3. Does Charles law use kelvin?
Answer: Charles's Law, a fundamental principle in thermodynamics, describes the relationship between the volume and temperature of a gas at constant pressure. While the law itself doesn't explicitly use Kelvin, the Kelvin scale is often employed when working with gas laws due to its absolute temperature measurement.
Charles's Law is expressed mathematically as V1/T1 = V2/T2, where V represents volume and T represents temperature. The temperatures involved can be in any scale, but using Kelvin is advantageous for scientific calculations. Unlike Celsius or Fahrenheit, Kelvin starts from absolute zero, the point where molecular motion theoretically ceases.
In Kelvin, the temperature is directly proportional to the average kinetic energy of gas particles. This makes it a preferred scale for gas law calculations as it simplifies mathematical relationships and avoids negative temperatures, ensuring consistency in calculations.
While Charles's Law itself doesn't specify the temperature scale to be used, Kelvin is commonly employed in gas law applications. Its absolute nature aligns seamlessly with the underlying principles of Charles's Law, providing a consistent and scientifically sound framework for understanding the behavior of gases at different temperatures and volumes.
Q4. Does Charles law depend on the identity of the gas?
Answer: According to this law, as long as pressure remains constant, the volume of a gas is directly proportional to its absolute temperature. However, Charles's Law does not depend on the identity of the gas.
The law's universality arises from the assumption that all gases behave similarly under identical conditions. In other words, it doesn't matter if the gas is oxygen, nitrogen, or any other element; Charles's Law holds true as long as pressure is constant. This generalization simplifies calculations and allows scientists to make predictions about the behavior of gases without delving into the specific properties of individual elements.
The law's foundation lies in the kinetic theory of gases, where gas particles are considered to be point masses with negligible volume. Consequently, the interaction between gas particles is minimal, making the gas behavior consistent across different elements.
Charles's Law is a universal principle applicable to all gases, irrespective of their identity. Its simplicity and broad applicability make it a cornerstone in understanding and predicting the behavior of gases in various scientific and industrial contexts.
Q5. Does Charles law apply on liquids?
Answer: Charles's Law primarily applies to ideal gases, stating that the volume of a gas is directly proportional to its absolute temperature when pressure is constant. However, this law doesn't directly extend to liquids.
Liquids, unlike gases, have a relatively fixed volume and are not easily compressible. Charles's Law relies on the assumption that gas molecules have negligible volume compared to the space between them, allowing for significant volume changes with temperature variations. In contrast, the molecules in liquids are closely packed, limiting their ability to undergo significant volume changes in response to temperature changes.
That said, the behavior of liquids with temperature changes is not entirely disregarded. While not a direct application of Charles's Law, liquids generally expand when heated and contract when cooled. This phenomenon is attributed to the increased kinetic energy of liquid molecules with higher temperatures, causing them to move more and occupy a larger space.
Charles's Law does not directly apply to liquids as it does to ideal gases, but the impact of temperature on liquid volume is acknowledged through the general observation of thermal expansion and contraction in liquids.
Disclaimer: I have explained to the best of my knowledge, the subject of discussion. Hence, this is for educational purposes only. Also, this article contains some affiliate links through which I earn a commission upon every successful purchase. This is to support this channel. You are not under compulsion to make any purchase.
Conclusion:
Charles's Law, a cornerstone of gas behavior principles, continues to play a pivotal role in scientific research and everyday applications. Its simplicity and applicability make it a key component of thermodynamics, providing a framework to comprehend how gases respond to temperature variations under constant pressure. As we navigate a world influenced by technological advancements, Charles's Law remains a timeless guide in understanding and harnessing the behavior of gases.
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二次函数性质
百度百科上传时间：2024-03-04
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图文简介： 2023年科普中国×百度百科科学100词合作建设
定义
一般地，自变量x和因变量y之间存在如下关系：
一般式：y=ax2+bx+c(a≠0，a 、b、c为常数)，则称y为x的二次函数。
顶点式：y=a(x-h)2+k(a≠0，a、h、k为常数)
交点式（与x轴）：y=a(x-x1)(x-x2)(a≠0，x1、x2为常数)
重要知识：a，b，c为常数，a≠0，且a决定函数的开口方向，a>0时，开口方向向上，a<0时，开口方向向下2。IaI还可以决定开口大小,IaI越大开口就越小,IaI越小开口就越大。
二次函数表达式的右边通常为二次。
x是自变量，y是x的二次函数
一元二次方程求根公式
当b2-4ac>0 时
当b2-4ac=0时
x1=x2=-b/2a
表达式
①一般式
y=ax 2+bx+c(a,b,c为常数,a≠0)
②顶点式
[抛物线的顶点 P(h，k) ]：y=a(x-h)2+k(a,h,k为常数,a≠0)
③交点式
[仅限于与x轴有交点 A(x1,0) 和 B(x2,0) 的抛物线]：y=(x-x 1)(x-x2)(x1,x2为常数,)3
转化
3种形式的转化∶
①一般式和顶点式
对于二次函数 y=ax2+bx+c，其顶点坐标为(-b/2a,(4ac-b2)/4a)，即
h=-b/2a=(x 1+x2)/2
k=(4ac-b 2 )/4a
②一般式和交点式
x1,x2=[-b±√(b2-4ac)]/2a(即一元二次方程求根公式)4
有关性质
抛物线的性质
1.抛物线是轴对称图形。对称轴为直线 x = -b/2a。
对称轴与抛物线唯一的交点为抛物线的顶点P。
特别地，当b=0时，抛物线的对称轴是y轴（即直线x=0）
2.抛物线有一个顶点P，坐标为P ( -b/2a ，(4ac-b2)/4a )
当-b/2a=0，〔即b=0〕时，P在y轴上；当Δ= b2-4ac=0时，P在x轴上。
3.二次项系数a决定抛物线的开口方向和大小。
当a>0时，抛物线开口向上；当a<0时，抛物线开口向下。
|a|越大，则抛物线的开口越小。
4.一次项系数b和二次项系数a共同决定对称轴的位置。
当a与b同号时（即ab>0），对称轴在y轴左；
当a与b异号时（即ab<0），对称轴在y轴右。
5.常数项c决定抛物线与y轴交点。
抛物线与y轴交于（0，c）
6.抛物线与x轴交点个数
Δ= b2-4ac>0时，抛物线与x轴有2个交点。
Δ= b2-4ac=0时，抛物线与x轴有1个交点。
Δ= b2-4ac<0时，抛物线与x轴没有交点。X的取值是虚数（x= -b±√b2－4ac 乘上虚数i，整个式子除以2a）
当a>0时，函数在x= -b/2a处取得最小值f(-b/2a)=〔4ac-b2〕/4a；在{x|x<-b/2a}上是减函数，在{x|x>-b/2a}上是增函数；抛物线的开口向上；函数的值域是{y|y≥4ac-b2/4a}相反不变
当b=0时，抛物线的对称轴是y轴，这时，函数是偶函数，解析式变形为y=ax2+c(a≠0)
7.定义域：R
值域：（对应解析式，且只讨论a大于0的情况，a小于0的情况请读者自行推断）①[(4ac-b2)/4a，正无穷）；②[k，正无穷）
奇偶性：非奇非偶 （当且仅当b=0时，函数解析式为f（x）=ax2+c, 此时为偶函数）
周期性：无
解析式：
①y=ax2+bx+c[一般式]
⑴a≠0，a、b、c为常数。
⑵a>0，则抛物线开口朝上；a<0，则抛物线开口朝下；
⑶极值点：（-b/2a，(4ac-b2)/4a）；
⑷Δ=b2-4ac,
Δ>0，图象与x轴交于两点：
（[-b+√Δ]/2a，0）和（[-b-√Δ]/2a，0）；
Δ=0，图象与x轴交于一点：
（-b/2a，0）；
Δ<0，图象与x轴无交点；
②y=a(x-h)2+k[配方式]
此时，对应极值点为（h，k），其中h=-b/2a，k=(4ac-b2)/4a。5
二次函数的性质
特别地，二次函数（以下称函数）y=ax2+bx+c(a≠0)，
当 y=0 时，二次函数为关于x的一元二次方程（以下称方程），
即 ax2+bx+c=0(a≠0)
此时，函数图像与x轴有无交点即方程有无实数根。
函数与x轴交点的横坐标即为方程的根。
1．二次函数 y=ax2，y=ax2+k,y=a(x-h)2，y=a(x-h)2+k，y=ax2+bx+c(各式中，a≠0)的图象形状相同，只是位置不同，它们的顶点坐标及对称轴如下表：
|| ||
当 h>0 时，y=a(x-h)2 的图象可由抛物线 y=ax^2 向右平行移动 h 个单位得到，
当 h<0 时，则向左平行移动|h|个单位得到．
当 h>0,k>0 时，将抛物线 y=ax^2 向右平行移动h个单位，再向上移动k个单位，就可以得到y=a(x-h)2+k的图象；
当 h>0,k<0 时，将抛物线 y=ax2_ 向右平行移动h个单位，再向下移动|k|个单位可得到 _y=a(x-h)2+k 的图象；
当 h<0,k>0 时，将抛物线向左平行移动|h|个单位，再向上移动k个单位可得到 y=a(x-h)2+k的图象；
当 h<0,k<0 时，将抛物线向左平行移动|h|个单位，再向下移动|k|个单位可得到 y=a(x-h)2+k的图象；
因此，研究抛物线 y=ax2+bx+c(a≠0)的图象，通过配方，将一般式化为y=a(x-h)2+k的形式，可确定其顶点坐标、对称轴，抛物线的大体位置就很清楚了．这给画图象提供了方便．
2．抛物线 y=ax2+bx+c(a≠0)_的图象：当a>0时，开口向上，当a<0时开口向下，对称轴是直线_ x=-b/2a，_顶点坐标是_(-b/2a，[4ac-b2]/4a)．
3．抛物线 y=ax2+bx+c(a≠0)，_若_ a>0，当 x ≤ -b/2a 时，y随x的增大而减小；当 x ≥ -b/2a 时，y随x的增大而增大．若a<0，当x ≤ -b/2a时，y随x的增大而增大；当x ≥ -b/2a时，y随x的增大而减小．
4．抛物线y=ax2+bx+c(a≠0)的图象与坐标轴的交点：
(1)图象与y轴一定相交，交点坐标为(0，c)；
(2)当△=b2-4ac>0，图象与x轴交于两点A(x1，0)和B(x2，0)，其中的x1,x2是一元二次方程ax2+bx+c=0
(a≠0)的两根．这两点间的距离AB=|x2-x1| 另外，抛物线上任何一对对称点的距离可以由2x|A+b/2a|（A为其中一点的横坐标）
当△=0．图象与x轴只有一个交点；
当△<0．图象与x轴没有交点．当a>0时，图象落在x轴的上方，x为任何实数时，都有y>0；当a<0时，图象落在x轴的下方，x为任何实数时，都有y<0．
5．抛物线y=ax2+bx+c的最值（也就是极值）：如果a>0(a<0)，则当x= -b/2a时，y最小(大)值=(4ac-b2)/4a．
顶点的横坐标，是取得极值时的自变量值，顶点的纵坐标，是极值的取值．
6．用待定系数法求二次函数的解析式
(1)当题给条件为已知图象经过三个已知点或已知x、y的三对对应值时，可设解析式为一般形式：
y=ax2+bx+c(a≠0)．
(2)当题给条件为已知图象的顶点坐标或对称轴时，可设解析式为顶点式：y=a(x-h)2+k(a≠0)．
(3)当题给条件为已知图象与x轴的两个交点坐标时，可设解析式为两根式：y=a(x-x1)(x-x2)(a≠0)．
7．二次函数知识很容易与其它知识综合应用，而形成较为复杂的综合题目。因此，以二次函数知识为主的综合性题目是中高考的热点考题，往往以大题形式出现。
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