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14400 | https://www.sigmamagic.com/blogs/what-are-quartiles/ | What are quartiles and where do we use them? | Blogs | Sigma Magic
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What are quartiles and where do we use them?
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What are the quartiles and what is the Interquartile Range (IQR). How do we calculate these values? Where are they used and why do I need to know about them? In this article, we will understand what they are and how we can calculate them and describe where they are typically used.
Introduction
In order to understand IQR, we first need to understand Quartiles (Q1, Q2, and Q3). Let’s say you got 89% marks in an exam. Looks like you did not get close to 100% so others may think your performance is not as good. But in fact, you actually came first in the class. The only reason you did not get 100% is because the exam had tough questions that no one in the class was able to answer. So, if you mention that you were 1st in class, then people understand that you did pretty well in the exam. So, even though you did not get 100% (percent), you are in fact the 100th percentile. That is 100% of the people in the class (excluding yourself of course) are behind you! So, when we say 75th percentile, that means that 25% of the people in the class did better than you and 75% of the people in the class did worse than you. Similarly, 50th percentile means that half the class did better than you and half the class did worse than you. It does not really matter how much you score but only the relative score you achieve compared to the rest of the class.
Theory
So, when we talk about quartiles, we are dividing the data set into 4 quarters. Each quarter is 25% of the total number of data points. The first quartile or Q1 is the value in the data set such that 25% of the data points are less than this value and 75% of the data set is greater than this value. The second quartile or Q2 is the value in the data set such that 50% of the data points are less than this value and 50% of the data set are greater than this value. The third quartile or Q3 is the value such that 75% of the values are less than this value and 25% of the values are greater than this value. The term Interquartile Range (IQR) refers to the difference between Q3 and Q1 (IQR = Q3 – Q1).
Let’s now calculate the IQR value for an example data set. Let’s say we have 11 data points as shown below:
Data Set: 5, 4, 2, 1, 7, 9, 8, 10, 12, 0, 15
The first step is to put the data in increasing order, we get the following…
Sorted Data Set: 0, 1, 2, 4, 5, 7, 8, 9, 10, 12, 15
If there are N data points, each quarter will contain (N+1)/4 data points. Since there are 11 data points, in our example we have 3 data points in each quarter. So, the first quartile is the value that is located at the 3rd data point (Q1 = 2 in this example). The second quartile is the value that is located at the 6th data point (Q2 = 7 in this example). The third quartile is the value that is located at the 9th data point (Q3 = 10 in this example). Hence, the IQR = 10-2 = 8.
How do we handle the case when the number of data points is not divisible by 4! In this case, we have to use interpolation to calculate the value of the quartile. Let’s say we have N = 5 data points, where the data set is as follows:
Data Set: 23, 45, 12, 18, 50
In this example, the first quartile is at the location (N+1)/4 = 1.5th data point. Hence, the first quartile is the value between the first and second data set. Hence, in this example, Q1 = (23 + 45)/2 = 34.
In summary, in order to calculate the quartiles, we first put the data set in increasing order and then calculate the quartiles as follows:
Q1 is the value of the data set located at the (N+1)/4th location, Q2 is the value of the data set located at the (N+1)/2nd location, and Q3 is the value of the data set that is located at the 3(N+1)/4th location.
Application
Some examples of the use of range in the real world are: temperature ranges for the day as reported on a weather report, min/max levels of water in a reservoir. However, in the presence of outliers, the range values can be significantly influenced by them. Hence, we prefer to use the IQR instead as we “ignore” the bottom 25% of the data points and the top 25% of the data points. IQR statistic is more robust with respect to outliers. In the presence of outliers, IQR is a better representation of the amount of spread in the data rather than the range. Some companies use the quartiles to benchmark other companies. For example, the median company pay for a given position is set at the first quartile of the top 20 companies in that region. The quartiles and IQR information is typically used when you create a box-plot of your data set.
Software
Using the Sigma Magic software, calculating the quartiles and IQR is relatively straightforward. Just add a new Basic Statistics template to Excel by clicking on Stat > Basic Statistics. Copy and paste the data for which you want to calculate the quartiles into the input area and then click on Compute Outputs. You could also calculate these values in Excel by using the formula =QUARTILE(C9:C28,1) for the first quartile, =QUARTILE(C9:C28,2) for second quartile, and =QUARTILE(C9:C28,3) for the third quartile. Note that Excel values may not match Sigma Magic and Minitab values as it uses a slightly different algorithm to calculate the quartiles. If you use the Minitab software, you can copy and paste the data into Minitab and then click on Stat > Basic Statistics > Display Descriptive Statistics. Then select the data column and then click on OK. This will print out the quartiles for the sample values. If you want the IQR value, you have to go back to the menu and click on Statistics and select the checkbox next to IQR in the statistics options.
Exercise
Calculate the quartiles and IQR for the data set given in the following Excel file: Basic Stats 1.
Using the Sigma Magic software, the analysis results will include the quartiles and IQR value. For this example, the first quartile Q1 = 29.25, the second quartile Q2 = 31 (also called the median), and the third quartile Q3 = 32. The Interquartile range IQR = 2.75.
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14401 | https://www.youtube.com/watch?v=e8XtUP0CliA | What is a standard basis?
David Friday
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Posted: 15 Jun 2021
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Transcript:
in this video we're going to go over some examples of what would be considered a standard basis for a set of vectors so we'll try to do a little variety in here so generally the way that we create a standard basis is we allow one what would normally be a variable to be one and then everything else in there would be a zero so just a few examples of this when we're dealing with vector space that is r three for example a uh sort of generic looking vector within uh r3 would probably take the form x1 x2 x3 where x1 x2 and x3 can each represent any real number so if we allow one of them to be one and everything else to be zero we would wind up with one vector being one zero zero a second vector being zero one zero and a third vector being zero zero one in fact with this in mind these are so frequently used as standard basis vectors that they uh they get special representations that are known as i hat j hat and k hat now this also lets us know that the dimension of this vector space would be equal to 3 because that's the number of vectors that we see in our basis well that's true for any basis not just a standard basis for another example if we wanted to consider the set of all polynomials of degree no more than three so uh generic looking thing in here would be something like a constant term a linear term a quadratic term and a cubic term so a standard basis would be taking any of these coefficients setting one of them to be equal to one and then have all of the rest of them be equal to zero so if we were to do so and i should probably do this using you know actual notation i could set a zero equal to one and all of the others equal to zero we would get one for that basis element i could set a one equal to one and all of the others equal to zero and get x as that element i could set a two equal to one set all of the others equal to zero and i could set a three equal to one and the others equal to zero so i see in this standard basis for p3 we see a total of four vectors so we can conclude that the dimension of the set of polynomials of degree no more than three would be equal to four now it's generally true that whatever number you see down here is going to be one less than whatever the dimension of that basis is so again now let's bump it up to the vector space being the set of uh let's do two by two matrices so generic 2x2 matrix would probably be something along the lines of a is equal to a11 a12 a21 and a22 i could set any of those equal to one and set all of the others equal to zero we would wind up with the following four standard basis elements so setting the one one entry to one and all others equal to zero we would get this setting a12 equal to one and all others equal to zero we would get zero one zero zero we could set a two one equal to one and set all of the other entries equal to zero and we could set a two two equal to one and set all of the others equal to zero through this it becomes rather easy to see that the dimension of the set of two by two matrices would be equal to four using these standard operations now that does take us to what about subspaces or things of that nature so with the two by two matrices in mind something that we have discussed in class a couple times is the idea of the set of symmetric two by two matrices so we'll define s to be the set of symmetric two by two matrices so a nice generic looking element for something like that might be something like x y y z basically for a symmetric 2 by 2 matrix we need to make sure that the 1 2 and 2 1 entries are equal to one another now if i were to set one of those variables equal to one while setting the other two equal to zero then the basis that i would get the standard basis for the set of two by two symmetric matrices would only contain three elements it would be necessary that the second element has two ones in it so it is as though we took the standard basis for the set of two by two matrices and combined these into one single matrix here so the dimension of the set of symmetric two by two matrices would be equal to three since there are only three elements in the basis here |
14402 | https://flexbooks.ck12.org/cbook/ck-12-algebra-ii-with-trigonometry-concepts/section/11.3/primary/lesson/using-and-writing-nth-term-rules-for-sequences-alg-ii/ | Skip to content
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11.3
Using and Writing nth Term Rules for Sequences
Written by:Lori Jordan | Kate Dirga
Fact-checked by:The CK-12 Editorial Team
Last Modified: Sep 01, 2025
You buy new furniture at zero percent interest on a monthly installment plan. The total of your furniture is $4800. The following sequence shows the balance you still owe on the furniture at the beginning of each month. How would you write a general rule for the sequence?
4800, 4600, 4400, 4200,...
nth Term
Recursive rules can help us generate multiple sequential terms in a sequence but are not helpful in determining a particular single term. Consider the sequence: @$\begin{align} 3, 5, 7, \ldots, a_n \end{align}@$. The recursive rule for this sequence is @$\begin{align}a_n=a_{n-1}+2\end{align}@$. What if we want to find the @$\begin{align}100^{th}\end{align}@$ term? The recursive rule only allows us to find a term in the sequence if we know the previous term. An @$\begin{align}n^{th}\end{align}@$term or general rule, however, will allow us to find the @$\begin{align}100^{th}\end{align}@$ term by replacing @$\begin{align}n\end{align}@$ in the formula with 100.
Let's solve the following problems.
Write the first three terms, the @$\begin{align}15^{th}\end{align}@$ term and the @$\begin{align}40^{th}\end{align}@$ term of the sequence with the general rule: @$\begin{align}a_n=n^2-1\end{align}@$.
We can find each of these terms by replacing @$\begin{align}n\end{align}@$ with the appropriate term number:
@$$\begin{align}a_1 &=(1)^2-1=0 \
a_2 &=(2)^2-1=3 \
a_3 &=(3)^2-1=8 \
a_{15} &=(15)^2-1=224 \
a_{40} &=(40)^2-1=1599\end{align}@$$
These terms can also be found using a graphing calculator. First press @$\begin{align}2^{nd}\end{align}@$ STAT (to get to the List menu). Arrow over to OPS, select option 5: seq( and type in (expression, variable, begin, end). For this particular problem, the calculator yields the following:
@$\begin{align}seq\left(x^2-1,x,1,3 \right) ={0 \ 3 \ 8}\end{align}@$ for the first three terms
@$\begin{align}seq\left(x^2-1,x,15,15 \right)={224}\end{align}@$ for the @$\begin{align}15^{th}\end{align}@$ term
@$\begin{align}seq\left(x^2-1,x,40,40 \right)={1599}\end{align}@$ for the @$\begin{align}40^{th}\end{align}@$ term
Write a general rule for the sequence: @$\begin{align} 5, 10, 15, 20,\ldots\end{align}@$
The previous problem illustrates how a general rule maps a term number directly to the term value. Another way to say this is that the general rule expresses the @$\begin{align}n^{th}\end{align}@$ term as a function of @$\begin{align}n\end{align}@$. Let’s put the terms in the above sequence in a table with their term numbers to help identify the rule.
Looking at the terms and term numbers together helps us to see that each term is the result of multiplying the term number by 5. The general rule is @$\begin{align}a_n=5n\end{align}@$
| @$\begin{align}n\end{align}@$ | 1 | 2 | 3 | 4 |
--- ---
| @$\begin{align}a\end{align}@$ | 5 | 10 | 15 | 20 |
Find the @$\begin{align}n^{th}\end{align}@$ term rule for the sequence: @$\begin{align}0, 2, 6, 12,\ldots\end{align}@$
Let’s make the table again to begin to analyze the relationship between the term number and the term value.
| @$\begin{align}n\end{align}@$ | 1 | 2 | 3 | 4 |
--- ---
| @$\begin{align}a_n\end{align}@$ | 0 | 2 | 6 | 12 |
| @$\begin{align}n(?)\end{align}@$ | @$\begin{align}(1)(0)\end{align}@$ | @$\begin{align}(2)(1)\end{align}@$ | @$\begin{align}(3)(2)\end{align}@$ | @$\begin{align}(4)(3)\end{align}@$ |
This time the pattern is not so obvious. To start, write each term as a product of the term number and a second factor. Then it can be observed that the second factor is always one less that the term number and the general rule can be written as @$\begin{align}a_n=n(n-1)\end{align}@$
Examples
Example 1
Earlier, you were asked how would you write a general rule for the sequence 4800, 4600, 4400, 4200, ...
Let’s put the terms in the sequence in a table with their term numbers to help identify the rule.
| @$\begin{align}n\end{align}@$ | 1 | 2 | 3 | 4 |
--- ---
| @$\begin{align}a\end{align}@$ | 4800 | 4600 | 4400 | 4200 |
Looking at the terms and term numbers together helps us to see that each term is the result of subtracting 200 times one less than the term from the first term. The general rule is @$\begin{align}a_n=a_{n-1} - 200(n - 1)\end{align}@$.
Example 2
Given the general rule: @$\begin{align}a_n=3n-13\end{align}@$, write the first five terms, @$\begin{align}25^{th}\end{align}@$ term and the @$\begin{align}200^{th}\end{align}@$ term of the sequence.
Plug in the term numbers as shown:
@$$\begin{align}a_1 &=3(1)-13=-10 \
a_2 &=3(2)-13=-7 \
a_3 &=3(3)-13=-4 \
a_4 &=3(4)-13=-1 \
a_5 &=3(5)-13=2 \
a_{25} &=3(25)-13=62 \
a_{200} &=3(200)-13=587 \end{align}@$$
Example 3
Write the general rule for the sequence: @$\begin{align}4, 5, 6, 7,\ldots\end{align}@$
Put the values in a table with the term numbers and see if there is a way to write the term as a function of the term number.
| @$\begin{align}n\end{align}@$ | 1 | 2 | 3 | 4 |
--- ---
| @$\begin{align}a_n\end{align}@$ | 4 | 5 | 6 | 7 |
| @$\begin{align}n\pm(?)\end{align}@$ | @$\begin{align}(1)+3\end{align}@$ | @$\begin{align}(2)+3\end{align}@$ | @$\begin{align}(3)+3\end{align}@$ | @$\begin{align}(4)+3\end{align}@$ |
Each term appears to be the result of adding three to the term number. Thus, the general rule is @$\begin{align}a_n=n+3\end{align}@$
Example 4
Write the general rule and find the @$\begin{align}35^{th}\end{align}@$ term of the sequence: @$\begin{align}-1, 0, 3, 8, 15,\ldots\end{align}@$
Put the values in a table with the term numbers and see if there is a way to write the term as a function of the term number.
| @$\begin{align}n\end{align}@$ | 1 | 2 | 3 | 4 | 5 |
--- --- --- |
| @$\begin{align}a_n\end{align}@$ | -1 | 10 | 3 | 8 | 15 |
| @$\begin{align}n(?)\end{align}@$ | @$\begin{align}(1)(-1)\end{align}@$ | @$\begin{align}(2)(0)\end{align}@$ | @$\begin{align}(3)(1)\end{align}@$ | @$\begin{align}(4)(2)\end{align}@$ | @$\begin{align}(5)(3)\end{align}@$ |
Each term appears to be the result of multiplying the term number by two less than the term number. Thus, the general rule is @$\begin{align}a_n=n(n-2)\end{align}@$.
Review
Use the @$\begin{align}n^{th}\end{align}@$ term rule to generate the indicated terms in each sequence.
@$\begin{align}2n+7\end{align}@$, terms @$\begin{align}1-5\end{align}@$ and the @$\begin{align}10^{th}\end{align}@$ term.
@$\begin{align}-5n-1\end{align}@$, terms @$\begin{align}1-3\end{align}@$ and the @$\begin{align}50^{th}\end{align}@$ term.
@$\begin{align}2^n-1\end{align}@$, terms @$\begin{align}1-3\end{align}@$ and the @$\begin{align}10^{th}\end{align}@$ term.
@$\begin{align}\left(\frac{1}{2}\right)^n\end{align}@$, terms @$\begin{align}1-3\end{align}@$ and the @$\begin{align}8^{th}\end{align}@$ term.
@$\begin{align}\frac{n(n+1)}{2}\end{align}@$, terms @$\begin{align}1-4\end{align}@$ and the @$\begin{align}20^{th}\end{align}@$ term.
Use your calculator to generate the first 5 terms in each sequence. Use MATH > FRAC, on your calculator to convert decimals to fractions.
@$\begin{align}4n-3\end{align}@$
@$\begin{align} -\frac{1}{2}n+5\end{align}@$
@$\begin{align}\left(\frac{2}{3}\right)^n+1\end{align}@$
@$\begin{align}2n(n-1)\end{align}@$
@$\begin{align}\frac{n(n+1)(2n+1)}{6}\end{align}@$
Write the @$\begin{align}n^{th}\end{align}@$ term rule for the following sequences.
@$\begin{align}3,5,7,9,\ldots\end{align}@$
@$\begin{align}1,7,25,79,\ldots\end{align}@$
@$\begin{align}6,14,24,36,\ldots\end{align}@$
@$\begin{align}6,5,4,3,\ldots\end{align}@$
@$\begin{align}2,5,9,14,\ldots\end{align}@$
Review (Answers)
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14403 | https://www.jacionline.org/article/0091-6749(86)90404-5/pdf | Postgraduate course
Neutrophil granules in health and disease
Judith Falloon, M.D., and John 1. Gallin, M.D. Bethesda, Md.
Once staining and microscopy techniques made the granules of the neutrophil visible, the pursuit of their function challienged those interested in the neutrophil and its role j.n health and in myriad disease states. Both a secretory function of the granules postulated by Paul Ehrlich’ in the late 1800s whereby neutrophils modify their environment and an intracellular diges-tive or lysosoimal function postulated by Ehrlich’s con-temporary Elie Metchnikoff’ appear to be of impor-tance. Current understanding of neutrophil granules has separated these functions according to granule type with the specific granules (also called secondary granules) serving the environmental modification or external secretory function and the azurophilic gran-ules (also called primary granules) acting as lysosomes by fusing with phagosomes to form phagolysosomes with a predominantly intracellular site of action.3 Al-though this is, of course, a conceptual oversimplifi-cation, since specific granule contents are found in phagolysosornes and azurophilic granules are clearly released into the extracellular milieu through degran-ulation and cell death,4 the concept of the separation of functions lhas proved useful. GRANULE CONTENTS AND FUNCTIONS The two main neutrophil granules, specific and azurophilic, are separated on the basis of differences in content, morphology, density, and appearance at different stages of neutrophil maturation (Fig. 1). During myelopoiesis, the first granules become visible at about the promyelocyte stage and stain azurophilic in the bone marrow; these are the primary or azuro-philic granules. Later on during maturation, synthesis of these granules ceases, and they are distributed to daughter cells during cell division, thus decreasing
From the Bacterial Diseases Section, Laboratory of Clinical In-vestigation, National Institute of Allergy and Infectious Diseases, National Institutes of Health, Bethesda. Received for publication Nov. 21, 1985. Accepted for pu,blication Jan. 9, 1986. Reprint requests: Judith Falloon, M.D., Bacterial Diseases Section, Laboratory of Clinical Investigation, National Institute of Allergy and Infectious Diseases, National Institutes of Health, Bldg. 10, Rm. llN112, Bethesda, MD 20892. FMLP:
N-formylmethionylleucylphenylanine
~1 CGD: Chronic granulomatous disease
the azurophilic granule content of the cells.” The granules produced second during neutrophil matura-tion, called secondary or specific granules, then come to predominate, comprising about two thirds of the granules of the mature ce11.5The differences between these granules have permitted the development of sub-cellular fractionation techniques leading to the sepa-ration of disrupted cells into granule-enriched frac-tions that have been used to probe further into granule content and function. The azurophilic granules sedi-ment heterogeneously on sucrose gradient9 and con-tain microbicidal enzymes, proteases, acid hydro-lases, and bactericidal cationic proteins (Table I).6-” These contents correlate with the postulated lysosomal function of azurophilic granules. The specific granule, a less dense granule that stains heavily for glycopro-tein, contains lysozyme, collagenase, lactoferrin, vi-tamin B,,-binding protein, and possibly histaminase and sialidase.6-‘3 Recently, the specific granule-en-riched fraction of neutrophils has been demonstrated to contain a unique b-type cytochrome’4. ” and pos-sibly a flavoprotein’6 believed to be of importance in the formation of the complete NADPH oxidase en-zyme responsible for the respiratory burst that pro-duces the toxic oxygen products important in neu-trophil function, including killing.17, ‘ Organelles that contain a preformed intracellular pool of re-ceptors important in the response of the phago-cyte to chemoattractants and interaction with serum complement proteins also cosediment with specific granules in gradients.“, ‘O A third granule type, agelatinase containing tertiary granule, has been de-scribed as a separate entity with a density slightly less than that of specific granules on sucrose gradients.” Further knowledge of this granule is scanty; it most likely sediments with the specific granule-enriched fraction in most gradients and is potentially the site
653 654 Falloon and Gallin J. ALLERGY CLIN. IMMUNOL. MAY 1986 FIG. 1. A human neutrophil stained for peroxidase. Five nuclear lobes (1 to 5/ are visible in the cytoplasm. Small, unstained specific granules (SJ are present throughout the cytoplasm. The larger, darkly stained structures are azurophilic granules that stain positively for peroxidase. irregularly shaped channels (at arrow] represent invaginations of the cell surface and are con-tinuous with the external environment. Fixation in aldehydes and osmium. Scale bar = 0.5 pm. (Micrograph courtesy of Dr. Marc Friedman, Georgetown University.) of certain contents currently attributed to specific granules. Neutrophil degranulation occurs in the midst of many other events during the complex response of the cell to inflammatory stimuli. As a neutrophil is ex-posed to chemoattractants such as N-formylated pep-tides released by bacteria,22 complement proteins, or leukotriene products, it develops a ruffled membrane and changes from a round to a polarized cell with anterior broad lammellipodium and posterior-narrow uropod. Cytoskeletal microtubules and microfila-ments reorganize, and there is anterior movement of granules while the nucleus moves toward the rear of the ce11.23Cells aggregate and adhere to endothelial surfaces. There is subsequent disaggregation, and cells become increasingly motile (chemokinetic) with directed movement (chemotaxis) toward the more concentrated source of chemoattractant. These events are accompanied by ion fluxes, changes in membrane potential and surface charge, increased glycolysis, al-terations in phospholipid and arachidonic acid metab-olism, protein phosphorylation, and production of toxic oxygen radicals.24. 25 In this way these cells are directed to tissue sites where they can alter and destroy bacteria and other target cells. At sites of inflam-mation, both phagocytosis and internal digestion of microorganisms or other particles occur, and the finely tuned exocytosis of granules, prompted also by chemoattractants, occurs. In order to understand further this eomplex neutrophil response, the role of granule contents in inflammatory and plasma membrane events has been the focus of recent inves-tigations . In keeping with their different functions, the extra-cellular release of specific and azurophilic granules is under separate control. In vitro, for example, che-moattractants and other substances can be used to cause specific granule secretion under conditions whereby azurophilic granule enzymes will not be re-leased.3. 4. lo. 26 Migration through filters in response to chemotactic stimuli results in preferential specific granule discharge,27 as does migration through tissues in vivo, as demonstrated in exudate cells harvested from human dermal suction blisters. The strongest stimuli for azurophilic granule degranulation also stimulate concomitant specific granule exocytosis.3 Specific granules, then, and probably the tertiary granules2’ appear readily released during cell stimu-lation so that their contents are available to modify the response of the cell to stimuli as well as to effect function. Investigations into the functions of granule contents have proceeded through isolation and characterization of granule proteins and through observations made on the neutrophils of patients with abnormal granules. In the specific granule, lactoferrin is an iron-chelating glycoprotein that probably has an antimicrobial func-tion via iron chelation or by direct bactericidal effect2 and may function in neutrophil adhesiveness,’ hy-droxyl radical formation,30 and in control of myelo-VOLUME 77 NUMBER !j
Neutrophil granules in health and disease 655
TABLE I. Neutrophil granule contents
Class 0f constituent
Microbicidal enzymes Neutral serine proteases Metalloproteinases Acid hydrolases Other
Azurophilic granules
Myeloperoxidase Lysozyme Elastase Cathepsin G Proteinase3 N-acetyl-P-glucosaminidase Cathepsin B Cathepsin D P-Glucuronidase P-Glycerophosphatase a-Mannosidase Bactericidal cationic proteins$
Specific granules
Lysozyme
Collagenaset
Lactoferrin Vitamin B,? binding proteins Cytochrome b Histaminase FMLP receptors5 C3bi receptors (CR3)S
From references 3and 10 with modifications.
TReleased asa latentenzyme.
$Note added in proof: defensins, antimicrobial peptides in azurophilic granules, have recently been described (Gang T, Selsted ME, Szklorek D, et al: JClin Invest 76:1427, 1985). 9Present in asubcellular fraction cosedimenting with specific granules in sucrose gradients.
poiesis by inhibition of the production of granulocyte-macrophage colony-stimulating factor.” Lysozyme,
present in both granules, digests glycopeptide debris in the cell,32 and hydrolyzes cell wall components of some bacteria.33 It may also modulate inflammation by suppressi:ng neutrophil chemotaxis and oxidative metabolism.‘4 The vitamin B,, binding protein of the specific granule remains enigmatic; its function is un-known. Cytochrome b, probably a specific granule membrane component, is believed to be a constituent of the membrane-associated electron transport chain comprising the NADPH oxidase responsible for the enhanced oxygen consumption and hydrogen per-oxide generation of the respiratory burst of the neu-trophil. I73I8 Unstimulated cells contain an inactive oxidase, which may consist of several subunits re-quiring assembly and then activation for activity. One subunit appears to be a cytochrome b, and one appears to be a flavoprotein, both believed to be partially located in the specific granule.14-16.35 Fu-sion of specific granule membrane containing cyto-chrome b w:ith the plasma membrane may be a key step in the formation of a complete oxidase enzyme,36 since the complete enzyme complex appears to be plasma membrane associated. Other specific granule functions include amplification of the inflammatory response by activation of the complement cascade, generation of the chemoattractant C5a,37 and release of a monocyte chemoattractant.3, ‘ Specific granule contents also promote neutrophil adherence to en-dothelium. It has recently been postulated that the specific gran-ule membrane contains receptors important in neutro-
phi1 function and that fusion of granule membrane
with plasma membrane during exocytosis may provide membrane and receptors to the cell surface.3, 4o Since membrane turnover may be important in chemotaxis4’ and new membrane synthesis appears unnecessary, the specific granule membrane pool may be a requirement for membrane replacement during chemotaxis. By correlating specific granule degranulation with in-creased surface expression of receptors and through binding assays to subcellular fractions enriched for specific granules, some plasma membrane receptors have been demonstrated to exist in a preformed pool that may be on the specific granule membrane. These receptors include the chemotactic peptide receptor, often called the FMLP receptor for its identification by the binding of this synthetic peptide, and a com-plement receptor called CR3 that binds the C3b hy-drolysis product called C3bi. 19, O, 42.43Activated neu-trophils have an asymmetric distribution of FMLP receptors on their cell surfaceti that could in part be caused by the discharging of specific granules at the leading edge of the cell during chemotaxis.3 This asymmetry, along with receptor turnover, may be cru-666 Falloon and Gallin J. ALLERGY CLIN. IMMUNOL. MAY 1996
TABLE II. Models of neutrophil specific granule deficiency
Neutrophil function Congenital-specific granule deficiency Neutrophil cytoplasts
Aggregation Disaggregation C3bi receptors Increased expression Chemotaxis FMLP receptors Increased expression Chemoattractant generation from serum by secretedproducts Phagocytosis Bactericidal activity
NL
Absent NL Absent 1NL Absent Absent NL NL Absent NL
J
NL
Absent NT NL
NL =normal; NT =not tested; &=decreased. Reproduced from reference 3with modifications.
cial for optimal sensing of a gradient of chemoattrac-tant. The specific granule membrane thus may play a central role in the responsiveness of the human neu-trophil to inflammatory stimuli. Azurophilic granule proteins have predominantly digestive and bactericidal functions. Elastase along with specific granule collagenase may permit pene-tration through tissues,45 and other enzymes serve to degrade phagocytosed material. Lysozyme and cat-ionic proteins have bactericidal activity.4647 MPO, aheme protein, has cytotoxic and antimicrobial activity and can inactivate humoral agents.48 It forms a part of the MPO-halide-hydrogen peroxide bactericidal system whereby the MPO-catalyzed oxidation of ahalide by hydrogen peroxide results in the production of a number of potent oxidants. These provide the ability to injure or kill via oxidation or halogenation of important mo1ecules.48 MPO also appears to con-tribute to regulation of the respiratory burst enzyme activity.49 In addition, it and other azurophilic granule proteins may modulate the inflammatory response by inactivating or digesting chemoattractants and lyso-somal enzymes.50 For example, an azurophilic granule product is able to destroy C5a activity,27 and the MPO system has been demonstrated to oxidize chemoat-tractant peptides and neutrophil secretory products.5’ The ability of the cell to digest chemoattractants may be important in chemotaxis.52 Azurophilic granules, then, serve lysosomal functions and are important in target cell death and in modulation of the inflammatory response. A novel role of neutrophil granules in the regulation of intracellular calcium has recently been proposed.53 Stimulation of the neutrophil by chemoattract&t stim-uli results in rapid rises in cytosolic free calcium be-lieved to represent release from nonmitochondrial in-tracellular stores.54 These calcium rises are important in regulating functional responses to stimuli. Since both specific and azurophilic granule-enriched sub-cellular fractions contain a magnesium-adenosine tri-phosphate-dependent calcium pump promoting cal-cium uptake, these granules may be an intracellular calcium storage site, thus linking granules with cell activation in yet another way.
MODELS OF GRANULE DEFICIENCY Cytoplasts
Our understanding of the role of neutrophil granules in the inflammatory response has been strengthened by experimental models of granule-deficient cells and by patients whose cells are lacking in granule contents (Table II). The model cell, the enucleated cytoplast, is a plasma membrane-bound bag of cytoplasm pinched off from a neutrophil and contains less than 10% of the original granules of the ce11.55, 56 It has various functional abnormalities including poor che-motaxis, slower bactericidal activity, abnormal shape change in response to chemoattractants, and abnormal disaggregation. The cytoplast preparation, unlike the neutrophil, is unable to increase the plasma membrane expression of FMLP or C3bi receptors in response to degranulating stimuli. O 56 Some of these abnormali-ties may relate to the absence of specific granule mem-brane to provide for the membrane turnover and recep-tor replenishment needed for cellular responses such as chemotaxis, shape change, and disaggregation.
Congenital specific granule deficiency
The model of specific granule deficiency, patients with recurrent bacterial infections (Table III) and a severe deficiency in the specific granule markers vi-tamin B,2 binding protein and lactoferrin, has been VOLUME 77 NUMBER 5 Neutrophil granules in health and disease 657 TABLE III. Clinical presentations of patients with neutrophil granule abnormalities
Granule type Disorder Clinical manifestations Screening test
Specific granules Specific granule de-Recurrent severe bacterial infections Peripheral blood smear (misshapen ficiency nucleus, no granules on Wright’s stain) and enzyme contents of neutrophils CGD (cytochrome bEarly onset recurrent infections with Nitroblue tetrazoleum reduction test deficient) catalase-positive bacteria and fungi, granulomatous lesions C3bi receptor Delayed separation of umbilical stump, Deficient binding of monoclonal an-(CR3) deficiency leukocytosis, recurrent bacteria1 in-tibodies against the C3bi receptor fections, periodontal disease to neutrophils Thermal injury Bacterial infections, infectious mortal-ity (relates to extent of burn) Azurophilic MPG deficiency Generally none; some patients with MPG stain of neutrophils granules other host defense defects such as di-abetes have bacterial or fungal infec-tion (particularly Candida albicans) Azurophilic and Chediak-Higashi Recurrent bacterial infections, partial Peripheral blood smear (character-specific syndrome oculocutaneous albinism, progression istic large lysosomal granules in granules to peripheral neuropathy, acceler-white cells) ated phase (organomegaly, coagulop-athy, neutropenia, lymphoma-like syndrome)
described.57 The cells of these patients are morpholog-ically abnonnal with misshapen nuclei. No granules are visible on Wright’s stain, but myeloperoxidase staining reveals azurophilic granules. The presence of empty specific granules has been suggested by electron microscopy studies5 Other abnormalities such as a reduction in the plasma membrane marker alkaline phosphatase, altered azurophilic granule com-plex carbohydrate staining by electron microscopy, and abnormal granule sedimentation in subcellular fractionation studies57-59 are present. In vitro, these neutrophils demonstrate abnormal chemotaxis and bacterial killing. Like cytoplasts, these cells do not increase receptors after stimulation.“. 6o In vivo, one patient has demonstrated deficient neutrophil and monocyte recruitment to skin windows, although monocyte chemotaxis was not deficient in vitro. In addition, secretory products from the cells of this pa-tient were unable to generate chemoattractants from serum.@’ Disaggregation was also abnormal.@’ These neutrophils were demonstrated to be lacking in ge-latinase activity, suggesting absence of tertiary gran-ules as well, Although the underlying defect in this disorder is complex and not well understood, the ab-
normalities documented point to a role of the specific
Dewald B, Baggiolini M, Gallin J: Unpublished observation.
granule and possibly the tertiary granule in chemo-taxis, receptor upregulation, chemoattractant gener-ation, disaggregation, and control of bacterial in-fections.
Specific granule deficiency secondary to thermal injury
Another model for specific granule-deficient cells is that of thermal injury. In bum patients a decrease in neutrophil lysozyme with increase in blood lacto-ferrin and lysozyme has been reported with neutrophil P-glucuronidase being normal, suggesting preferential specific granule degranulation.6’.62 Davis et a1.62 corre-lated this decrease in neutrophil lysozyme with abnor-malities of chemotaxis in bum patients. These abnor-malities have, in turn, been correlated with increased mortality.63 After a bum, preferential specific granule exocytosis may interfere with chemotaxis, perhaps by lack of membrane and preformed receptors for tum-over and, along with defects in oxidative metabo-lism,bl may contribute to compromised host defenses.
Neonatal neutrophils
Neonatal neutrophils comprise another system with decreased neutrophil lactoferrin but normal neutrophil MPO content.65 This suggests a relative specific gran-ule deficiency in comparison with adult cells. In ad-dition, neonatal neutrophils manifest a host of com-668 Falloon and Gallin J. ALLERGY CLIN. IMMUNOL. MAY 1996
plicated abnormalities when they are compared to adult cells. These include abnormalities of chemo-taxis,@ adherence, 66. 67 disaggregation,68 membrane potential,69 phagocytosis and microbicidal activity,” and oxidative metabolism.@ In particular, hydroxyl radical production appears diminished, which has been attributed to lactoferrin deficiency.65 The specific granule deficiency in these cells may be in part re-sponsible for the increased susceptibility of neonates to infection.
C3bi receptor (CR3) deficiency
Other patients have been described who appear to be deficient in a group of leukocyte membrane gly-coproteins including the C3bi receptor,” a receptor normally found in neutrophil plasma membrane and specific granule-enriched cell fractions. Patients with this abnormality present with delayed separation of the umbilical stump, recurrent and severe bacterial and fungal infections, impaired wound healing, dimin-ished pus formation, severe periodontal disease, and persistent leukocytosis. The severity of the clinical manifestations appears to be related to the degree of deficiency of the membrane glycoproteins.” In vitro, slightly differing abnormalities have been described, generally involving abnormalities of neutrophil ag-gregation, adherence, spreading, chemotaxis, and de-fective phagocytosis of some opsonized particles. Bactericidal activity is generally normal.” This defect involves deficiencies in a family of multimeric high-molecular-weight glycoprotein surface antigens with a common beta subunit and different alpha chains. The C3bi receptor is one of these molecules.74 Work in which the in vitro abnormalities of this disease are mimicked by adding a monoclonal antibody against the antigen to normal cells75,76suggests that the func-tional abnormalities are related to the missing gly-coprotein. Studies of these patients suggest that the C3bi receptor functions in neutrophil adherence, ag-gregation, and chemotaxis and that its absence from the granules and plasma membrane of the neutrophil is responsible for severe human disease.
CGD
Another patient group with abnormal granules is a subset of patients with CGD. CGD is a group of re-lated disorders characterized by recurrent infections and granulomatous lesions.77-79 The infections are due to an inability of the neutrophil to kill most bacteria and fungi because of an abnormality in oxidative me-tabolism with inability to produce toxic oxygen metab-olites, including hydrogen peroxide. This abnormality is believed to result from various genetically absent or defective constituents of the membrane-associated oxidase that normally triggers the respiratory burst or from defective activation of this oxidase. In one subset with X-linked inheritance, the patients’ specific gran-ule-enriched fractions appear deficient in cytochrome b, ‘O,” and another subset may be deficient in a fla-voprotein. 3s In a subset of patients with autosomal recessive inheritance, there is an absence of stimulus-induced phosphorylation of a protein distributed in plasma membrane and granule fractions that may in-teract with cytochrome b.82 In this way, abnormal specific granules may lead to an inability to mount anormal respiratory burst so that one of the important antibacterial mechanisms of the neutrophil is crippled.
Abnormalities of azurophilic granules: Myeloperoxidase deficiency
No patients lacking azurophilic granules have been described, but the inherited or acquired absence of azurophilic granule MPO is a relatively common neu-trophil abnormality occurring in about 1 in 2000 per-sons.83 It is interesting that the lack of this protein, believed to play a key role in bacterial killing by normal neutrophils, does not usually result in illness, although some patients have been described with
Candida albicans
infections. The patients with infec-tions generally have another illness such as diabetes mellitus contributing to compromised host defenses. In vitro, MPO-deficient cells undergo an augmented respiratory burst,49 and although bacterial killing is delayed, it is eventually normal, perhaps the result of the compensatory role of unusually vigorous produc-tion of toxic oxygen metabolites. The benign nature of MPO deficiency points out the complexity of neu-trophil bacterial killing and the importance of killing mechanisms other than the MPO-halide-hydrogen per-oxide system such as oxidative killing by other prod-ucts of oxygen and nonoxidative killing by exposure of bacteria to the interior of the phagolysosome and to granule contents.&, 47,84,”
The Chbdiak-Higashi syndrome
The Chediak-Higashi syndrome is an inherited ab-normality characterized by morphologically abnormal granules in neutrophils and other granule-containing cells.86 The giant granules of the cells are believed to be products of abnormal granule fusion,’ perhaps re-lated to abnormal microtubule metabolism88~89 or to altered membrane properties.” The neutrophils are functionally abnormal with decreased chemotaxis, ab-normal lysosomal degranulation, and delayed and de-ficient bactericidal capacity.‘lT 92The patients have in-creased susceptibility to bacterial infection as well as variable complicated systemic manifestations, includ-ing neutropenia, blood coagulation abnormalities, VOLUME 77 NUMBER 5
neuropathy, partial albinism, adenopathy, hepato-splenomegaly, and an increased incidence of lympho-proliferative disorders.86 Since the underlying defect in this syndrome is unknown, the role of the abnormal granules in the neutrophil defect is unclear. Granule deficiency secondary to leukocyte collection Granule exocytosis may be involved in the dimin-ished functional capacity of neutrophils intended for transfusion that were collected by filtration leuka-pheresis.“’ In this method cells are collected by re-versible adhesion to nylon fibers. Neutrophils become partially degranulated during this process with loss of specific granules being greater than that of azuro-philic granules. These cells are functionally abnormal with abnormal posttransfusion kinetics, chemotaxis, phagocytosis, and bactericidal activity. Because of these defects, as well as increased recipient side ef-fects with filtration leukapheresis cells, alternative collection methods are now prefemed.94 POSSIBLE ROLE OF NEUTROPHIL GRANULES IN SYSTEMIC DISEASE There is growing evidence that neutrophils and their granules contribute to a variety of systemic diseases through local release of damaging granule enzymes or toxic oxygen products. In particular, the neutrophil is believed important in the pathogenesis of both acute and chronic lung disease. In an acute lung disease, the adult respiratory distress syndrome, it is believed that activation of neutrophils by blood stream com-plement results in neutrophil trapping in the lung. The local release of toxic neutrophil products such as oxy-gen metabolites, proteolytic enzymes, and arachidonic acid products results in damaged alveolar endothelium and plasma exudation. 95-97 In some animal models lung damage is dependent on production of reactive oxi-dants and can be diminished by inactivators of these oxidants, particularly by scavengers of hydrogen per-oxide or hydroxyl radical.98, 99 In a chronic lung dis-ease, emphysema, it is believed that lung irritants induce alveolar macrophages to release neutrophil chemoattractants. Neutrophils arriving in the lung de-granulate, releasing enzymes such as elastase that con-tribute to the disruption of alveolar elastic fibers that results in emphysema. ‘O”Other systemic illnesses po-tentially related to neutrophil granule products include rheumatoid and crystalline arthritis, serum sickness, immune vasculitis and glomerulonephritis, myocar-dial infarction, and even malignancies at sites of chronic inflammation. ‘01-‘04The neutrophil and its granule components are thus potentially involved in many human disease states involving inflammation.
Neutrophil granules in health and disease 659
SUMMARY The granules of the neutrophil, in addition to con-tributing to its distinctive morphologic appearance, are critical to its unique functions. Specific granules ap-pear necessary for neutrophil recruitment to sites of inflammation, for upregulation of receptors important in the control of chemotaxis and the respiratory burst, for disaggregation, for bactericidal activity, and for chemoattractant generation. The azurophilic granules supply enzymes for digestive and bactericidal func-tions and supply MPO to the MPO-halide-hydrogen peroxide bactericidal system. Azurophilic granule contents also regulate inflammation by degrading in-flammatory products. Both granules may play a role in intracellular calcium regulation. In addition to these activities that protect the host from infection, granules also, under certain circumstances, contribute to dis-ease processes. For these reasons, greater knowledge about granule contents, control of degranulation, in-activation of toxic granule contents and products, and the role of granules in neutrophil membrane events and function has widespread implications for treat-ment of patients with neutrophil dysfunction syn-dromes and patients with multiple other systemic diseases.
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Anderson DC, Schmalsteig FC, Finegold MJ, Hughes BJ, Rothlein R, Miller LJ, Kohl S, Tosi MF, Jacobs RL, Waldrop TC, Goldman AS, Shearer WT, Springer TA: The severe and moderate phenotypes of heritable Mac-l, LFA-1 deficiency: their quantitative definition and relation to leukocyte dys-function and clinical features. J Infect Dis 152:668, 1985 73. Buescher ES, Gaither T, Nath J, Gallin JI: Abnormal adher-ence-related functions of neutrophils, monocytes, and Ep-stein-Barr virus-transformed B cells in a patient with C3bi receptor deficiency. Blood 65:1382, 1985 74. Sanchez-Madrid F, Nagy JA, Robbins E, Simon P, Springer TA: A human leukocyte differentiation antigen family with distinct o-subunits and a common B-subunit: the lymphocyte function-associated antigen (LFA-I), the C3bi complement receptor (OKMliMac-l), and the ~150.95 molecule. J Exp Med 158:1785, 1983 75. Harlan JM, Killen PD, Seneca1 FM, Schwartz BR, Yee EK, Taylor RF, Beatty PG, Price TH, Ochs HD: The role of the neutrophil membrane glycoprotein GP-150 in neutrophil ad-herence to endothelium in vitro. Blood 66: 167, 1985 76. Schwartz B, Ochs HD, Beatty PG, Harlan JM: A monoclonal antibody-defined membrane antigen complex is required for neutrophil-neutrophil aggregation. Blood 65: 1553, 1985 77. Gallin JI, Buescher ES, Seligmann BE, Nath J, Gaither T, Katz P: Recent advances in chronic granulomatous disease. Ann Intern Med 99:657, 1983 78. Tauber Al, Borregaard N, Simons E, Wright J: Chronic gran-ulomatous disease: a syndrome of phagocyte oxidase defi-ciencies. Medicine 62:286, 1983 79. Segal AW: Variations on the theme of chronic granulomatous disease. Lancet 1:1378, 1985 80. Ohno Y, Buescher ES, Roberts R, Metcalf JA, Gallin Jl: Areevaluation of cytochrome b and flavin adenine dinucleotide in neutrophils from patients with chronic granulomatous dis-ease and description of a family with probable autosomal recessive inheritance of cytochrome b deficiency. Blood (in press) 81. Segal AW, Cross AR, Garcia RC, Borregaard N, Valerius NH, Soothill JF, Jones OTG: Absence of cytochrome b-245 in chronic granulomatous disease: a multicenter European evaluation of its incidence and relevance. N Engl J Med 308:245, 1983 82. Segal A, Heyworth PG, Cockcraft S, Barrowman MM: Stim-ulated neutrophils from patients with autosomal recessive chronic granulomatous disease fail to phosphorylate a Mr-44,000 protein. Nature 316:547, 1985 83. Parry MF, Root RK, Metcalf JA, Delaney KK, Kaplow LS, Richar WJ: Myeloperoxidase deficiency: prevalence and clin-ical significance. Ann Intern Med 95:293, 1981 84. 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14404 | https://archive.lib.msu.edu/crcmath/math/math/f/f228.htm | Focus
Focus
A point related to the construction and properties of Conic Sections.
See alsoEllipse, Ellipsoid, Hyperbola, Hyperboloid, Parabola, Paraboloid, Reflection Property
References
Coxeter, H.S.M. and Greitzer, S.L. Geometry Revisited. Washington, DC: Math. Assoc. Amer., pp.141-144, 1967.
© 1996-9 _Eric W. Weisstein
1999-05-26_ |
14405 | https://www.youtube.com/watch?v=6Z8aQhV_aD4 | What is Chromatin?
The Explorer's Guide to Biology
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Description
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Posted: 20 Oct 2021
explorebiology.org/bio-dictionary
Chromatin is a complex of DNA and protein, found in eukaryotic cells, which packages DNA molecules into compact structures to fit within cells. Chromatin also regulates gene expression, controlling which genes are expressed when and where in the body, helps chromosomes correctly segregate into daughter cells after cell division, and prevents DNA from getting tangled, thus helping to stop DNA damage.
27 comments
Transcript:
The human genome contains over 3
billion base pairs of genetic information. If it were stretched out
end to end, it would span approximately 6 feet. But each human cell that houses
all this information is about 0.0002 inches across, approximately
360 thousand times smaller than the length of all of our genetic
information laid out side by side. So how in the world are we able to fit
that massive amount of information into a tiny cell? As always, nature came up
with a clever solution. Instead of all that DNA being stuffed into cells as
long, sinuous strands which wouldn't leave room for much else and would be
hard to untangle when the information is needed, the information is coiled up
around a series of proteins into a highly organized condensed structure
called chromatin. As you'll see, it's similar to the
process of turning wool into a knitted sweater, where each piece provides
structure and organization to the next. Let's explore this concept a bit more.
First, a small segment of the DNA double helix is wrapped around a
collection of proteins called histones, which forms a structure called a
nucleosome. Nucleusomes then fold on top of each
other into a single chromatin fiber. From here, the chromatin fibers are
further coiled and intertwined into a chromosome, which organizes our genetic
information in the form of genes, units of DNA that provide instructions to
make proteins. Each human cell has 23 pairs of chromosomes, or 46 total,
which contain approximately 20,000 to 25,000 genes, all of which live inside
the cell's nucleus. From here, all that neatly packaged information is
responsible for making sure each cell, and in turn our whole body, functions
properly. It turns out that this packaging trick doesn't just save space
inside a cell, but also makes sure DNA is able to do its job properly. Think
of it like a coiled up elastic band; when it is compacted together, no
individual strands are accessible, but you can pull away particular pieces of
the coil band as needed. Similarly, when DNA is tightly wound around
histones, also called heterochromatin, it isn't accessible to the proteins
required to read DNA, so those genes remain turned off. In response to
environmental changes, however, histones are altered through a variety
of molecular processes which triggers the DNA to uncoil into a more relaxed
state, known as euchromatin. Euchromatin is accessible to the DNA
reading proteins, thus turning the necessary gene on. These dynamic
changes in chromatin density, a process called chromatin remodeling, are
important for the process of cell division— when cells copy all of their
genetic material and then segregate and repackage it into the newly formed
daughter cell. In non-dividing cells, chromatin remodeling allows for tighter
control over when and where certain genes are activated, which ensures
complex organisms, like humans, run like a well-oiled machine. Beyond
ensuring the right DNA is accessible at the right time, this packing structure
also prevents DNA from getting tangled, which reduces DNA damage. The structure
of chromatin is one of biology's many ingenious solutions to a mathematical
challenge and simultaneously solves numerous cellular complexities. Number
one, it ensures vast amounts of genetic information can fit into each cell.
Number two, it enables complex organisms to diversify their
functionality. And number three, it helps prevent DNA damage. More levels
of biological intricacy, however, also means more opportunities for mistakes
to happen. Researchers are finding that genetic mutations affecting chromatin
structure are common in many types of diseases, especially cancer. While
scientists are still trying to uncover the precise roles these mutations play
in diseases, efforts to develop drugs that target chromatin machinery have
shown promise in early clinical studies, revealing their potential to provide a
new type of medicine for challenging health problems. |
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StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
StatPearls [Internet].
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Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
Major Neurocognitive Disorder (Dementia)
Prabhu D. Emmady; Caroline Schoo; Prasanna Tadi.
Author Information and Affiliations
Authors
Prabhu D. Emmady1; Caroline Schoo2; Prasanna Tadi3.
Affiliations
1 UNC school of Medicine, Atrium Health
2 VOLUNTEERS OF AMERICA PACE
3 Asram Medical College, Eluru, India
Last Update: November 19, 2022.
Continuing Education Activity
Dementia describes an overall decline in memory and other cognitive skills severe enough to reduce a person's ability to perform everyday activities. The progressive and persistent deterioration of cognitive function characterizes it. Affected patients often have memory loss and a partial or significant lack of insight into their deficits. This activity reviews the evaluation and management of dementia and highlights the role of the interprofessional team in caring for patients with this condition. It also reviews risk factors, pathophysiology, and challenges in diagnosing and managing this spectrum of cognitive disorders.
Objectives:
Assess risk factors associated with dementia and related conditions.
Assess the evaluation required for patients presenting with cognitive decline.
Evaluate the commonly available management options for patients with various cognitive decline syndromes.
Communicate the importance of improving care coordination among the interprofessional team to enhance care delivery and improve outcomes for patients affected by dementia.
Access free multiple choice questions on this topic.
Introduction
The definition of dementia has been updated in the DSM-5 criteria. It is no longer termed Dementia but is now called Major Neurocognitive Disorder (MND). However, due to the common use of the term dementia in society and medical literature, it is referred to as both Dementia and MND in this topic. It is worth noting the limitations of using the term dementia, including its common association exclusively with older patients, and that it is often used synonymously with Alzheimer disease. MND can affect younger individuals and does not always imply Alzheimer disease as the etiology of cognitive decline. MND is characterized by a significant decline in at least 1 of the cognition domains, including executive function, complex attention, language, learning, memory, perceptual-motor, or social cognition. The decline represents a change from a patient's prior level of cognitive ability, is persistent and progressive over time, and is not associated exclusively with an episode of delirium. In addition to the cognitive decline, there must also be a decline in the patient's ability to function and perform everyday tasks. The everyday function of a patient is often evaluated in terms of the ability to perform instrumental activities of daily living, such as managing finances or medications, or, if more severe, activities of daily living, such as grooming or feeding oneself. It is often a progressive disorder, and individuals often do not have insight into their deficits. Currently, no cure exists for any of the causes of dementia. The prevalence of dementia is expected to continue to increase along with the increasing numbers of the aging population. Currently, 47 million people in the world have dementia, and the number is expected to increase to 131 million by 2050. Alzheimer disease is the 5th leading cause of death for people over the age of 65 in the United States. Dementia is a significant public health burden and significantly increases the costs of care, both to the individual and society. The individual lifetime cost to care for an individual with dementia was nearly $200,000 more than an individual without dementia. In 2010, the costs of treating dementia in the United States were projected to be about $200 billion per year in the United States and $600 billion worldwide.
Etiology
Several conditions can cause MNDs, with Alzheimer dementia being the most common cause, accounting for about 70% of cases. The DSM-5 criteria for MND further delineate 13 etiological subtypes that indicate the possible etiology of the disorder. These subtypes include Alzheimer disease, vascular disease, frontotemporal lobar degeneration, Lewy body disease, Parkinson disease, HIV infection, Huntington disease, prion disease, substance and or medication use, traumatic brain injury, another medical condition, multiple etiologies, and unspecified. A patient may have more than 1 etiology contributing to MND. For example, there may be a mixed picture of Alzheimer disease with vascular disease in the same patient. Other medical conditions that can lead to dementia include progressive supranuclear palsy, corticobasal syndrome, and, less commonly, multiple system atrophy. The etiology is further characterized by "possibly" vs. "probably," assigning the degree of certainty as to the cause of the MND. It often takes time to distinguish the etiology. It can be aided by many factors, including the results of imaging studies, lab studies, genetic markers, patient comorbidities, medical and family history, and clinical findings.
Epidemiology
Alzheimer disease is the most common cause of dementia, as it is responsible for 70 to 80% of all cases of dementia. It can occur sporadically or be familial. Vascular dementia accounts for approximately 15% of all dementia cases. Its incidence increases with age and doubles every 5.3 years. Risk factors for vascular dementia include hypercholesteremia, diabetes mellitus, hypertension, and smoking. Lewy body dementia accounts for approximately 5% of dementia cases. The epidemiological data may not be completely accurate because the diagnosis of Lewy body dementia is often missed. Parkinson disease dementia accounts for approximately 10% of cases of dementia. Frontotemporal dementia is attributed to 25% of dementia cases in patients older than 65 years of age. It is, however, the second most common cause of dementia in patients younger than 65 years. There are, however, many limitations in the epidemiological studies of frontotemporal dementia, in part due to the inherent difficulty in identifying frontotemporal dementia. Creutzfeldt-Jakob disease is rare and occurs in about 1 to 2 cases per million per year globally. Mixed dementia is a condition in which patients have more than 1 type of dementia. In this condition, Alzheimer disease with vascular dementia is the most common coexistent dementia.
Pathophysiology
The pathophysiology of MND, or dementia, varies depending on the subtype. Most types, except vascular dementia, are caused by the accumulation of native proteins in the brain. Alzheimer disease is characterized by widespread atrophy of the cortex and deposition of amyloid plaques, and neurofibrillary tangles of hyperphosphorylated tau protein in neurons, which contribute to their degeneration. Lewy body dementia and Parkinson disease dementia are characterized by the intracellular accumulation of Lewy bodies, which are insoluble aggregates of alpha-synuclein protein in the brain. Frontotemporal dementia is characterized by various mutations leading to the deposition of ubiquitinated TDP-43 and hyperphosphorylated tau proteins in the frontal and temporal lobes, leading to dementia, early personality, behavioral changes, and aphasia depending on the subtype. Huntington disease is caused by an autosomal dominant inherited gene mutation.
Prion-related dementias are caused by misfolded prions, which are proteinaceous particles that are infectious in nature and self-spreading. Prion dementias include Creutzfeldt-Jakob disease and kuru, among other syndromes. HIV infection is associated with the development of neurocognitive disorders, in part due to the activation of macrophages and toxic inflammation leading to neurodegeneration in the brain. Alcohol consumption, particularly high doses, and prolonged use are associated with multiple cytotoxic processes within the brain. Vascular dementia is caused by ischemic injury to the brain (eg, stroke), leading to permanent neuronal death.
Histopathology
The pathological changes in the brain of patients with different types of dementia can be varied. However, there is often overlap and mixed presentations and findings. Neurodegeneration and vascular changes are seen in the brains of patients with vascular dementia. The findings can vary and are related to the underlying etiology of vascular compromise, including lacunar infarcts, hemorrhagic lesions, and microvascular disease. There is considerable overlap of neuropathological findings between dementia with Lewy body and Parkinson disease, and some overlap with Alzheimer disease. Lewy body dementia and Parkinson disease are characterized by the presence of Lewy bodies throughout the neocortex, brainstem, and limbic regions of the brain. Lewy bodies are intracellular aggregates of proteins, predominantly composed of alpha-synuclein proteins, which can be highlighted by various stains depending on location within the brain. There is also a loss of midbrain dopaminergic neurons and a loss of cholinergic neurons in ventral forebrain nuclei. Additionally, there are neuritic plaques made up of amyloid and neurofibrillary tangles, which can overlap findings in AD. Alzheimer disease is characterized by neuritic plaques composed of extracellular amyloid beta protein deposition and neurofibrillary tangles composed of hyperphosphorylated tau proteins. It is also common to see signs of vascular ischemic damage and hippocampal sclerosis. Frontotemporal dementia is characterized by atrophy in the frontal and or temporal lobes. There is neuronal loss, microvacuolation, and loss of myelin. Degeneration is found in the cortical and basal ganglia. Four pathological subtypes are named after the proteins that make up inclusions found in the brain tissue. FTLD-tau and FTLD-TDP are the most common. FTLD-FET is less common, and FTD-UPS is quite rare. Cruetzfield Jakob disease often does not need an autopsy for diagnosis; however, it shows loss of neurons, spongiform degeneration (vacuoles in the intraneuronal space), or plaques positive for PrPSc.
History and Physical
History must be obtained from the patient and their close friends, family members, or caregivers. Patients may present with symptoms of changes in behavior, getting lost in familiar neighborhoods, memory loss, mood changes, aggression, social withdrawal, self-neglect, cognitive difficulty, personality changes, difficulty performing tasks, forgetfulness, difficulty in communication, loss of independence, etc. A detailed history should include past medical, family, medication, and substance use history and defining observed symptoms of cognitive decline. Often, patients report different awareness of the deficits than caregivers or companions. In addition to further characterization of the cognitive changes, it is important to evaluate their current functional abilities and any changes in their ability to perform daily tasks. Evaluating any safety concerns arising from the cognitive changes is also vitally important. For example, is the patient still driving and doing so safely? Have there been any episodes of wandering or getting lost? Could they get out safely if there was a fire in the house? Can they still use the telephone? Are they vulnerable to financial or physical abuse? Are there firearms in the home to which they have access?
In addition to symptoms of dementia, the following atypical symptoms may be seen in the following conditions:
In patients with Lewy body dementia, symptoms of well-formed visual hallucinations, REM sleep behavior disorder, typical parkinsonian symptoms, and fluctuating cognition, attention, and alertness.
In patients with frontotemporal dementia, behavior changes, including disinhibition and apathy, and speech difficulties may be seen.
In patients with Creutzfeld-Jakob disease, myoclonus symptoms, visual changes, ataxia, and memory and behavior changes are seen.
In patients with Huntington disease, symptoms of chorea, irritability, and depression can be present.
In patients with vascular dementia, deficits can occur in stepwise declines.
In patients with Parkinson disease dementia, symptoms of parkinsonism characterized by bradykinesia, resting tremor, and muscle rigidity are found. In addition, visual hallucinations and delusions may also be seen, especially in the late stages.
Patients with atypical parkinsonian syndromes also bear mentioning. Multiple system atrophy, progressive supranuclear palsy, and corticobasal syndrome have symptoms of parkinsonism in addition to other characteristic findings. Multiple system atrophy has symptoms of autonomic failure and cerebellar ataxia. Progressive supranuclear palsy has symptoms of frequent falls (often backward) and vertical supranuclear gaze palsy. Corticobasal syndrome has progressive asymmetric muscle rigidity and alien limb phenomenon. The physical exam should be comprehensive, including a complete neurological exam including gait analysis.
Evaluation
The definitive diagnosis of the type of dementia can only be made at autopsy. A probable diagnosis can often be made using clinical history predominantly, sometimes aided by brain imaging and additional laboratory evaluation. Excluding treatable causes of cognitive impairment is also important in the initial evaluation. All domains of cognition must be assessed. There are multiple cognitive evaluation tools available for use in a clinical setting, including the Mini-mental status examination, Montreal Cognitive Assessment, Saint Louis University Mental Status, Addenbrooke's Cognitive Examination–Revised, the modified mini-mental state examination, Mini-Cog, and Rowland Universal Dementia Assessment Scale. See Table. Pain Assessment in Advanced Dementia Scale.
Each tool has different advantages; for example, the Montreal cognitive assessment takes approximately 10 minutes and is better suited for detecting mild cognitive impairment besides MND. In contrast, the Mini-Cog takes approximately 3 minutes to administer and is predominantly used to screen for MND. The Rowland Universal Dementia Assessment Scale is often used for cross-cultural evaluations and can be administered with an interpreter. None of these cognitive evaluations alone can diagnose MND, as a decline in function of daily tasks is also needed to meet the diagnostic criteria. These studies can be repeated over time to document the progression of decline. They can give an idea of the severity of the deficit along with specific cognitive domains that are affected. Specialized, more in-depth neuropsychological testing can provide further diagnostic information and help differentiate subtle differences or hard-to-diagnose cases. Laboratory tests to check in all patients during the evaluation of dementia include complete blood count, urinalysis, metabolic panel, Vitamin B12, folic acid, thyroid function tests, and serological tests for syphilis and HIV. Under certain circumstances, checking erythrocyte sedimentation rate, lumbar puncture, heavy metal screen, ceruloplasmin levels, Lyme disease titer, or serum protein electrophoresis may be appropriate.
Brain imaging is sometimes ordered, particularly if the age of onset is relatively early, atypical or rapidly progressing symptoms are present, or there is diagnostic uncertainty. A brain MRI without contrast is often the initial test ordered. It is valuable for evaluating signs of vascular or ischemic disease and localized regions or global atrophy that may be seen. A DaTscan uses a radiotracer to highlight dopamine transporter proteins in a SPECT scan on the presynaptic dopaminergic neurons. This scan can aid in differentiating pathologies that involve loss of the striatal dopaminergic pathway, including Parkinson disease, multiple system atrophy, progressive nuclear palsy, cortical-basal degeneration, and Lewy body dementia from other syndromes. Often reserved for academic settings, functional brain imaging with PET, SPECT, and fMRI can help in the early diagnosis and monitoring of patients with dementia, especially AD. These can also help differentiate the etiology of dementia. These are expensive, and routine use in clinical practice is not indicated. There are new CSF and blood tests under research to help identify Alzheimer disease and evaluate concentrations of amyloid and phosphorylated tau proteins as well as markers of neurodegeneration, including neurofilament light chain and glial fibrillar acidic proteins. These tests are not yet ready for regular clinical use.
Treatment / Management
FDA-approved medications to improve cognitive function include cholinesterase inhibitors and memantine. Cholinesterase inhibitors include donepezil, galantamine, and rivastigmine. Cholinesterase inhibitors prevent the breakdown of acetylcholine and aim to slow or delay the worsening of symptoms. Memantine is an NMDA antagonist and decreases the activity of glutamine. Donepezil is approved for all stages of Alzheimer disease, rivastigmine is approved for all stages of Alzheimer disease in its patch form, and mild to moderate stages with oral formulations. Galantamine is approved for mild to moderate stages and memantine for moderate to severe stages. Acetylcholinesterase inhibitors lead to a variable response among patients, with not all patients showing benefit. There are possible contraindications to their use and significant side effects, including the potential for cardiovascular complications, peptic ulcer disease, and weight loss. Memantine may have neuroprotective benefits as it serves as an uncompetitive NMDA receptor antagonist and can prevent neurotoxic and excessive calcium influx to the neuron. The benefits seen with the use of both acetylcholinesterase inhibitors and memantine are often modest, and many patients and providers choose to forgo pharmacologic treatment.
Aducanumab is a recombinant monoclonal antibody directed against amyloid beta that was recently approved by the FDA for the treatment of mild Alzheimer disease. Its approval remains highly controversial. The drug is costly and does not have clear, proven clinical benefits. It was approved based on positive clinical results seen in only 1 of the 2 phase III trials and aducanumab's effect on a surrogate endpoint (reducing amyloid beta plaques in the brain), which has not been proven clinically significant. Lifestyle modifications to optimize cognitive function include optimizing sleep, eating an anti-inflammatory diet, adequate exercise, treating hearing or vision loss, minimizing stress, and maintaining normal blood sugar, cholesterol, and blood pressure levels. Behavioral symptoms include irritability, anxiety, and depression. Antidepressants and sometimes antipsychotics can help with these symptoms. In addition, non-drug approaches like supportive care, memory training, physical exercise programs, and mental and social stimulation must be employed in symptom control. Patients and their families should be counseled about the disease and its consequences. They should be provided with all the necessary information about what to expect and how to react. Patients and their families should also be encouraged to seek social service consultations and to register with support groups and societies. Coaching caregivers on skills such as redirection and reassurance as opposed to repeated correction of patients confused due to dementia can avoid or de-escalate possible behavioral symptoms. Driving restrictions may have to be imposed.
Differential Diagnosis
The differential diagnosis for major neurocognitive disorder (dementia) include the following:
Delirium
Depression
Drug use
Normal age-associated memory changes
Mild cognitive impairment
Stress
Structural brain abnormalities like subdural hematoma, brain tumor, and normal pressure hydrocephalus
Infections like HIV and neurosyphilis
Thiamine deficiency
Vitamin B12 deficiency
Folic acid deficiency
Thyroid disorders
Metabolic abnormalities and derangements
Medication-induced
Vitamin E deficiency
Prognosis
The prognosis of dementia is poor. Dementia is often a progressive condition with no cure or treatment. The 1-year mortality rate was 30 to 40%, while the 5-year mortality rate was 60 to 65%. Men had a higher risk than women. Mortality rates among admitted patients with dementia were higher than those with cardiovascular diseases.
Complications
Dementia can affect many body systems and can lead to the following complications:
Inadequate nutrition
Pneumonia
Inability to perform self-care tasks
Personal safety challenges
Fractures due to falls
Hallucinations and delusions
Apathy
Agitation
Dysphagia
Death
Depression
Incontinence
Personality changes
Infections
Deterrence and Patient Education
The diagnosis of dementia can be stressful and overwhelming for patients and their families. Patient and caregiver education is vital to the clinical management of patients with dementia. Counseling must be given about regular clinic visits, medication compliance, a healthy diet, exercise, and sleep hygiene. Safety concerns become increasingly important as the disease progresses. Special attention should be paid to potential safety concerns, including when to retire from driving, the risk of wandering or getting lost, risks of fire or cooking mishaps, or lack of ability to prepare food for oneself. Patients with dementia often have a lack of insight into their limitations. Caregivers do better redirecting or reassuring patients rather than trying to correct them. Support groups can help with the reduction of issues like anxiety, frustration, anger, loneliness, and depression. The patients and caregivers should be counseled about the diagnosis and the prognosis. Creating an individualized care plan can empower the patient.
Enhancing Healthcare Team Outcomes
Dementia is a common condition, and its prevalence is expected to increase with time. Various underlying etiologies and disease states cause it, and each may present differently. An interprofessional approach is recommended when managing patients with dementia. Interventions, including care coordination and interprofessional communication, can help reduce hospitalization and decrease emergency department visits. Physicians must coordinate with other healthcare workers, including physicians, pharmacists, social workers, and nurses, when managing patients with dementia. Any medication change must be carefully coordinated with all physicians involved in the patient's care. Pharmacists can help with counseling about medication side effects and compliance, and attention should be paid to over-the-counter medications and supplements as much as prescribed medications. Safety in the current living situation must be reviewed during every clinic visit, and social workers may be consulted to assess the safety and adequacy of the living situation and caregiver support. Involving family members and caregivers is an important aspect of the care of patients with dementia. They must be encouraged to accompany the patient during clinic visits to provide an accurate clinical history and to reiterate and enact the plan at home.
Review Questions
Access free multiple choice questions on this topic.
Click here for a simplified version.
Comment on this article.
Figure
Pain Assessment in Advanced Dementia (PAINAD) Scale Contributed by D Schwerin, MD, FACEP, FAEMS, FAAEM
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: Disclosure: Prabhu Emmady declares no relevant financial relationships with ineligible companies.
: Disclosure: Caroline Schoo declares no relevant financial relationships with ineligible companies.
: Disclosure: Prasanna Tadi declares no relevant financial relationships with ineligible companies.
Copyright © 2025, StatPearls Publishing LLC.
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Emmady PD, Schoo C, Tadi P. Major Neurocognitive Disorder (Dementia) [Updated 2022 Nov 19]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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14407 | https://web.pdx.edu/~newsomj/uvclass/ho_planned%20contrasts.pdf | Newsom Psy 521/621 Univariate Quantitative Methods, Fall 2024 1 Planned Contrasts Following a significant one-way analysis of variance (ANOVA), the researcher may be interested in following up the analysis with some specific comparisons. In the case of the planned contrast or planned comparison, only a few predicted or a priori hypotheses are of interest, and familywise error is not likely to be a serious concern. Post hoc tests that adjust for familywise error typically follow a significant one-way ANOVA when many or all possible comparisons are of interest. Philosophically, the distinction between an a priori and post hoc test has to do with whether or not the group means compared were predicted to be different in advance or are decided after looking at the results. Statistically, the distinction also concerns whether there are a few or many contrasts conducted. Statisticians will cite either the philosophical or the statistical reason for deciding between the two approaches. We know that with many contrasts, familywise error becomes a problem, but, if not very many contrasts are performed (e.g., 2, 3, 4?) and there are a priori hypotheses about what groups may differ, then it is likely to be safe to use a planned contrast approach. Planned contrasts typically involve the comparison of just two means. More complicated tests can be conducted (e.g., in a three-group design, the average of two groups might be compared to the third group), but I will not get into demonstrating more complicated comparisons in this handout (see Keppel & Wickens, 2004, for more detail). The approach is to develop a set of weights that eliminate any group means that are not involved in the comparison by giving them a zero weight and to specify the group means to be compared by giving them opposite values, usually -1 and +1. Thus, the first step is to obtain the weighted sum, ˆ , that gives the appropriate difference between the two means that one wishes to compare. ˆ j j w Y In the formula, j Y represents the group mean for each cell and wi represents the contrast weights or “coefficients.” In a three-group design, a comparison between the first and second group means uses the weights of -1, +1, and 0. The third mean drops out because it is multiplied by 0. The next step is to compute the standard error: 1 2 ˆ / j s A j w s MS n The mean-square error, MSs/A, is then obtained from the full one-way (omnibus) ANOVA. Here, we use an estimate of error derived from using within-group variability of all cases in the study. This approach gives a more stable estimate of error and a more powerful statistical test than if we simply conducted a standard t test. The next step is simply to compute the t value using the familiar ratio of the difference to the standard error. ˆ ˆ contrast t s Although still fairly common practice, it is not advisable to conduct a standard t test after a significant ANOVA. Statistical power is lower with the standard t test compared than it is with the planned contrast version for two reasons: a) the sample size is smaller with the t test, because only the cases in the two groups are selected; and b) in the planned contrast the error term is smaller than it is with the standard t test because it is based on all the cases from the ANOVA. As with a standard t test, we can use a 1 I've followed the textbook (Myers, Well, & Lorch, 2010) by computing a planned t-test, but many books and some software packages us an F-test. In the F-test version, the contrast weights are used to compute the numerator sum of squares. The mean square for the contrast is then divided by the means square error in a familiar F ratio. The two tests are statistically equivalent, however, because 2 t F . Newsom Psy 521/621 Univariate Quantitative Methods, Fall 2024 2 Welch's robust approach if there are concerns about equal variances (see Myers, Well, & Lorch, p. 245, for details). I will use the study strategy example from the one-way ANOVA handout to illustrate the computation of the planned contrast: reading only (M = 6), retrieval practice (M = 9), and concept mapping (M = 6). The omnibus ANOVA was significant (see ANOVA Example handout), and it might be desirable to follow the test with a comparison of the means for reading only and retrieval practice groups for theoretical or policy reasons. Thus, the comparison involves the first two groups, and the contrast weights should be -1, +1, and 0. ˆ 1(6) 1(9) 0(6) 6 9 3 j j w Y 2 ˆ / 2 2 2 1 1 0 1.833 5 5 5 2 1.83 5 .732 .856 j s A j w s MS n ˆ ˆ / 3 .856 3.50 contrast t s The critical value for a t test with df = N – a = 15 -3 = 12 is 2.179. Because our calculated value of 3.50 is greater the critical value, the difference is significant. To obtain this contrast in SPSS, a contrast subcommand can be added to the one-way ANOVA using the following syntax and this produces a t test version of the planned comparison.2 For this example, which involves a very small sample size and very different variances, it would be wise to look at the Welch’s test (“Does not assume equal variances” row). ONEWAY recall BY groups /CONTRAST= -1 1 0. R The R function I use is an F-test version. The planned contrasts can also be stated in terms of an F test. In fact, the simple, old-fashioned way to conduct a planned comparison was to conduct an F test using only the two groups you want to compare, use the mean square from that analysis (or sum of squares 2 The MANOVA command can also be used, but the subcommand requires that you specify a -1 comparisons, where a is the number of levels (see the simple effects analysis handout for more details). Newsom Psy 521/621 Univariate Quantitative Methods, Fall 2024 3 because MSA(1vs2) = SS A(1vs2)/1), but divide it by the mean square error from the full omnibus ANOVA, with F = MS A(1vs2)/MSs/A. As it turns out, this method is equal to the t contrast method above, because t2 = F. I do not know of any R package that provides a Welch’s adjustment for unequal variances for the planned comparison, so you would have to rely on assumed equal variances (which is generally unproblematic for studies with larger sample sizes and less dramatic differences between variances). Note that if you have missing data, R may have problems running the analysis with the approach given below. And, if the conditions are not in ascending order in the data set, the analysis will not be correct. I had no missing data and my data were ordered, but for other data sets, you may need the first two statements I have commented out. #contrasts do not work with missing data, so use listwise deletion #d <- d[complete.cases(d), ] #need to sort data first #library(dplyr) #d = (d[order(d$groups, decreasing = FALSE), ] ) > #planned contrasts -- there must be a-1 contrasts for results to be correct > #I will ignore the second test, because it is not one that I wanted > #compare the first and second means > #(each command specifies number of cases in each group) > d$c1 <- rep(c(-1, 1, 0), each = 5) > #compare the first and third means > d$c3 <- rep(c(-1, 0, 1), each = 5) > > #request the first contrast, c1, and second contrast, c3 > anova(lm(recall ~ c1 + c3, d)) Analysis of Variance Table Response: recall Df Sum Sq Mean Sq F value Pr(>F) c1 1 22.5 22.5000 12.2727 0.004356 c3 1 7.5 7.5000 4.0909 0.065982 . Residuals 12 22.0 1.8333 --- Signif. codes: 0 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1 Notice that the square of the t-value from SPSS (assumed equal variances) equals the c1 F from R, (3.503)2 = 12.27). Write-Up (assumes omnibus ANOVA test already reported) A planned contrast, using the Welch’s adjustment for unequal variances, indicated that student recall in the retrieval practice group (M = 9.00, SD = 1.00) was significantly higher than in the reading only group (M = 6.00, SD = 2.00), t(5.882) = 3.00, p = .025. Note: In most instances in practice, the sample size will be sufficient (say > 15 per group) and the variances will not be dramatically different (e.g., not larger than a 4:1 ratio), so reporting the equal variances assumed value is unlikely lead to incorrect decisions (see Myers et al., 2010, p. 137). In this instance, it does not make a difference in the statistical decision even with the more extreme circumstances. |
14408 | https://www.expii.com/t/integrals-of-sums-differences-and-constant-multiples-235 | Integrals of Sums, Differences, and Constant Multiples - Expii
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Integrals of Sums, Differences, and Constant Multiples - Expii
Definite integrals are "additive and linear". In other words, integrals of sums, differences, and constant multiples are well-behaved. You can use this to break up an integral of a complicated combination of functions into simpler integrals of the individual functions.
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CalculusCommon Properties of Definite Integrals
Integrals of Sums, Differences, and Constant Multiples
Definite integrals are "additive and linear". In other words, integrals of sums, differences, and constant multiples are well-behaved. You can use this to break up an integral of a complicated combination of functions into simpler integrals of the individual functions.
Integrals of Sums, Differences, and Constant Multiples
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Cody Johnson
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We can break up the definite integral (over an interval [a,b]) of a complicated combination of functions by applying the following properties of integrals:
∫b a[f(x)+g(x)]d x=∫b a f(x)d x+∫b a g(x)d x(1)
∫b a[f(x)−g(x)]d x=∫b a f(x)d x−∫b a g(x)d x(2)
∫b a c f(x)d x=c∫b a f(x)d x(3)
For example,
∫1 0(x 2+4 x)d x=∫1 0 x 2 d x+∫1 0 4 x d x(1)=∫1 0 x 2 d x+4∫1 0 x d x(3)=[x 3 3]1 0+4[x 2 2]1 0=[1 3 3−0 3 3]+4[1 2 2−0 2 2]=1 3+4⋅1 2=1 3+2=7 3
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For two single variable functions f and g that are continuous except at a finite number of points on the interval [a,b], the following linearity principle holds: ∫b a[f(x)+g(x)]d x=∫b a f(x)d x+∫b a g(x)d x To prove this, suppose that F(x) is an antiderivative of f(x) and that G(x) is an antiderivative of g(x). Now, we may use the linearity property of derivatives, which follows from the corresponding property for limits: d d x[F(x)+G(x)]=d d x F(x)+d d x G(x)d d x[F(x)+G(x)]=f(x)+g(x) Thus, we see that F(x)+G(x) is an antiderivative of f(x)+g(x).
Using the Fundamental Theorem of Calculus, ∫b a[f(x)+g(x)]d x=[F(x)+G(x)]b a∫b a[f(x)+g(x)]d x=F(b)+G(b)−F(a)−G(a)∫b a[f(x)+g(x)]d x=[F(b)−F(a)]+[G(b)−G(a)] Using the Fundamental Theorem of Calculus one more time, we obtain our desired result: ∫b a[f(x)+g(x)]d x=∫b a f(x)d x+∫b a g(x)d x Another way to prove this is to work directly with the definition of an integral, and to add up the areas of the approximating rectangles. This leads us to exactly the same conclusion.
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14409 | https://www.drugsincontext.com/tinea-pedis-an-updated-review/ | Published Time: 2023-06-29T04:52:54+00:00
Tinea pedis: an updated review - Drugs in Context
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HomeArticlesTinea pedis: an updated review
Tinea pedis: an updated review
Alexander KC Leung, Benjamin Barankin, Joseph M Lam, Kin Fon Leong, Kam Lun Hon
Abstract
Background: Tinea pedis is one of the most common superficial fungal infections of the skin, with various clinical manifestations. This review aims to familiarize physicians with the clinical features, diagnosis and management of tinea pedis.
Methods: A search was conducted in April 2023 in PubMed Clinical Queries using the key terms ‘tinea pedis’ OR ‘athlete’s foot’. The search strategy included all clinical trials, observational studies and reviews published in English within the past 10 years.
Results: Tinea pedis is most often caused by Trichophyton rubrum and Trichophyton interdigitale. It is estimated that approximately 3% of the world population have tinea pedis. The prevalence is higher in adolescents and adults than in children. The peak age incidence is between 16 and 45 years of age. Tinea pedis is more common amongst males than females. Transmission amongst family members is the most common route, and transmission can also occur through indirect contact with contaminated belongings of the affected patient. Three main clinical forms of tinea pedis are recognized: interdigital, hyperkeratotic (moccasin-type) and vesiculobullous (inflammatory). The accuracy of clinical diagnosis of tinea pedis is low. A KOH wet-mount examination of skin scrapings of the active border of the lesion is recommended as a point-of-care testing. The diagnosis can be confirmed, if necessary, by fungal culture or culture-independent molecular tools of skin scrapings. Superficial or localized tinea pedis usually responds to topical antifungal therapy. Oral antifungal therapy should be reserved for severe disease, failed topical antifungal therapy, concomitant presence of onychomycosis or in immunocompromised patients.
Conclusion: Topical antifungal therapy (once to twice daily for 1–6 weeks) is the mainstay of treatment for superficial or localized tinea pedis. Examples of topical antifungal agents include allylamines (e.g. terbinafine), azoles (e.g. ketoconazole), benzylamine, ciclopirox, tolnaftate and amorolfine. Oral antifungal agents used for the treatment of tinea pedis include terbinafine, itraconazole and fluconazole. Combined therapy with topical and oral antifungals may increase the cure rate. The prognosis is good with appropriate antifungal treatment. Untreated, the lesions may persist and progress.
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Article Details
Article Type
Review
DOI
10.7573/dic.2023-5-1
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Keywords
athlete’s foot
dermatophytosis
interdigital
moccasin
Trichophyton interdigitale
Trichophyton rubrum
vesiculobullous
Categories
Dermatology
Infectious diseases
Publication Dates
Submitted: 06 May 2023;Accepted: 02 Jun 2023;Published: 29 Jun 2023.
Citation
Leung AKC, Barankin B, Lam JM, Leong KF, Hon KL. Tinea pedis: an updated review. Drugs Context. 2023;12:2023-5-1.
Copyright
Copyright © 2023 Leung AKC, Barankin B, Lam JM, Leong KF, Hon KL. Published by Drugs in Context under Creative Commons License Deed CC BY NC ND 4.0, which allows anyone to copy, distribute, and transmit the article provided it is properly attributed in the manner specified below. No commercial use without permission.
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14410 | https://math.stackexchange.com/questions/2010878/if-in-all-subsets-of-size-k-there-exist-at-least-one-pair-of-elements-in-a-relat | combinatorics - If in all subsets of size k there exist at least one pair of elements in a relation... - Mathematics Stack Exchange
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If in all subsets of size k there exist at least one pair of elements in a relation...
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I’ve recently got stuck on a theorem that seems to be true, but I am not sure if I am able to prove it:
∀x∈P k(N)∃{y 1,y 2}⊂x∧y 1≠y 2 φ(y 1,y 2)→∃x∈P k(N)∀{y 1,y 2}⊂x∧y 1≠y 2 φ(y 1,y 2)∀x∈P k(N)∃{y 1,y 2}⊂x∧y 1≠y 2 φ(y 1,y 2)→∃x∈P k(N)∀{y 1,y 2}⊂x∧y 1≠y 2 φ(y 1,y 2)
Where φ φ is any symmetric predicate, i.e. φ(y 1,y 2)⟺φ(y 2,y 1)φ(y 1,y 2)⟺φ(y 2,y 1), and P k(X)P k(X) is the set of all subsets of X X of size k k.
In simple words: for any symmetric relation on N N, if every subset of N N of size k contains at least one pair of related elements, then in some subset of size k all pairs of elements are related.
This statement is true for k=2 k=2 and for k=3 k=3 and while k=2 k=2 is easy, k=3 k=3 was proved by me using graphs (already a bit harder). I believe the proper method for such a theorem would be a proof by induction, however I can’t make the inductive step.
Any thoughts? I really appreciate any help or hint. I also believe this to be quite an interesting theorem.
EDIT:
My proof was like this:
I proved it by deducing without loss of generality how the graph of these relations would behave. I used for this 4 pairs of vertices (each pair having an edge between them, no other edges at the beginning). These edges are given, if by any deductive reasoning we obtain a triangle, then the theorem is proved. Each step of the proof consists of selecting 3 vertices that aren’t connected by any edge and choosing 2 of them and connecting them (in case you can’t do it without the loss of generality, you have to follow all possible routes). At each step you avoid at all cost creating a triangle. I was eventually forced to connect two vertices and by it created a triangle. Should I present the whole proof here? If so do you recommend any tool for drawing graphs and posting it here?
Proof: Ramsey’s theorem may be easily used to prove this statement. Solved.
combinatorics
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edited Nov 21, 2016 at 14:46
PJTraill
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Givng some details (if not the complete proof) about your proof for the k=3 k=3 case might give someone an idea.Henrik supports the community –Henrik supports the community 2016-11-12 18:50:47 +00:00 Commented Nov 12, 2016 at 18:50
2 For clarity, your "whichever" predicate is meant to model a symmetric relation? That is, an undirected graph on the nodes N N?hardmath –hardmath 2016-11-13 01:27:40 +00:00 Commented Nov 13, 2016 at 1:27
2 Isn't the relation φ(y 1,y 2)⟺y 1<y 2 φ(y 1,y 2)⟺y 1<y 2 a counterexample for the case k=2?k=2? Are you sure you don't want to assume that φ φ is a symmetric relation? In that case, what you're asking about is just a weak form of Ramsey's theorem.bof –bof 2016-11-13 02:40:49 +00:00 Commented Nov 13, 2016 at 2:40
1 @bof is the solution you're proposing to just say that R(k,k)R(k,k) is the size of the subgraph that imposes a requirement that there exist a clique, because there can't be any k k sized independent sets? Do I understand it correctly?user388646 –user388646 2016-11-13 10:03:18 +00:00 Commented Nov 13, 2016 at 10:03
1 @bof: I encourage you to write up your solution as an Answer, which I would upvote, unless you prefer the OP to try their hand at posting one in more detail than was given by the Comments. I suspect you would do a better job in this particular instance.hardmath –hardmath 2016-11-13 17:56:09 +00:00 Commented Nov 13, 2016 at 17:56
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The premise of your theorem is stronger than you realise, so that your conjecture can be proven as a corollary of a stronger theorem, which I state in graph-theoretical terms:
Any sufficiently large graph whose every subgraph of some given size is non-empty contains a complete subgraph of any desired size.
It would be helpful to flesh out your statement that it follows from Ramsey’s Theorem (of which the above is little more than a logical rearrangement) to explain how.
The proof below of the above reformulation is more or less in the spirit of the proof of Ramsey’s theorem (as given in Wikipedia), but somewhat clumsier, due perhaps to the less symmetrical formulation.
A little terminology
Let us write
G G for the graph,
G¯¯¯¯G¯ for its complement,
G.V G.V or simply V V for the vertices of G G,
x G x G for the neighbours of a vertex x x in G G,
an r r-clique, -subgraph for a clique or subgraph with r r vertices, where a clique is a complete subgraph,
C(G,k,r)C(G,k,r) for the assertion that every k k-subgraph of G G contains an r r-clique,
C(G,r)C(G,r) for C(G,|V|,r)C(G,|V|,r), i.e. the assertion that G G contains an r r-clique.
A stronger theorem
We assert that
∃f:N 2→N:|G.V|>f(k,r)∧C(G,k,2)⇒C(G,r)∃f:N 2→N:|G.V|>f(k,r)∧C(G,k,2)⇒C(G,r)
Proof from first principles
We proceed by induction on k k and, within k k, on r r. For k=1 k=1, the theorem is vacuously true, as |V|>2⇒¬C(G,1,2)|V|>2⇒¬C(G,1,2).
Now supposing the theorem proven for k k (in combination with all values of r r), we prove it for k+1 k+1; we therefore assume C(G,k+1,2)C(G,k+1,2) and that f(x,y)f(x,y) is defined for x⩽k x⩽k.
For r=2 r=2 trivially |V|>k∧C(G,k,2)⇒C(G,2)|V|>k∧C(G,k,2)⇒C(G,2).
Consider a vertex x x and its non-neighbours x G¯¯¯¯x G¯, among which we observe C(x G¯¯¯¯,k,2)C(x G¯,k,2), whence, by induction:
|x G¯¯¯¯|>f(k,r)⇒C(x G¯¯¯¯,r)|x G¯|>f(k,r)⇒C(x G¯,r)
Hence, in a counterexample to C(G,r+1)C(G,r+1), we have:
∀x∈V:|x G¯¯¯¯|<f(k,r+1)∀x∈V:|x G¯|<f(k,r+1)
By the induction on r r we have an r r-clique R R; if a vertex in G−R G−R is joined to all of R R we have a (k+1)(k+1)-clique. Otherwise every point in G−R G−R is not joined to at least one member of R R, which limits the size of the complement:
|G−R|=|G|−r<r f(k,r+1)|G−R|=|G|−r<r f(k,r+1)
i.e.
|G|<r(1+f(k,r+1))|G|<r(1+f(k,r+1))
Hence in this case we may take f(k,r)=r(1+f(k,r+1))f(k,r)=r(1+f(k,r+1)) and the theorem holds by induction for all k k and r r.
Q.E.D
Proof using Ramsey’s theorem
Ramsey’s Theorem tells us that for |G|>R(r,k)|G|>R(r,k), G G contains an r r-clique or independent set of size k k, or equivalently C(G,r)∨C(G¯¯¯¯,k)C(G,r)∨C(G¯,k) but the hypothesis C(G,k,2)C(G,k,2) is eqivalent to ¬C(G¯¯¯¯,k)¬C(G¯,k) so that the theorem I formulated above is a logical rearrangement of Ramsey’s theorem.
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edited Nov 21, 2016 at 15:27
answered Nov 21, 2016 at 12:18
PJTraillPJTraill
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I am wondering whether I ought to remove the proof from first principles, as it does indeed follow trivially from Ramsey’s theorem, which I had managed to overlook.PJTraill –PJTraill 2016-11-21 14:43:33 +00:00 Commented Nov 21, 2016 at 14:43
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14411 | https://math.colgate.edu/~integers/u25/u25.pdf | A25 INTEGERS 20 (2020) COMBINATORIAL IDENTITIES INVOLVING HARMONIC NUMBERS Necdet Batır Department of Mathematics, Nev¸ sehir Hacı Bekta¸ s Veli University, Nev¸ sehir, Turkey nbatir@hotmail.com Received: 12/13/18, Revised: 7/24/19, Accepted: 3/18/20, Published: 4/13/20 Abstract We present a new combinatorial identity with two parameters, and applying it, we establish several combinatorial identities involving harmonic numbers; some of them have already been considered previously, and the others are new. For example, we prove that n X k=1 (−1)k−1 k ✓n k ◆ Hn−k = H2 n + n X k=1 (−1)k k2$n k % and n X k=1 (−1)k−1 k2 ✓n k ◆ Hn−k = Hn[H2 n + H(2) n ] 2 − n−1 X k=0 (−1)k[Hn −Hk] (k + 1)(n −k) $n k %.
Introduction Let s 2 C. Then the generalized harmonic numbers H(s) n of order s are defined by H(s) n = n X k=1 1 ks , H(s) 0 = 0 and H(1) n = Hn; (1) see [1, 13]. These numbers have various applications in number theory, combina-torics, analysis, computer science and di↵erential equations. Recently, they have found applications in evaluating Feynman diagrams contributions of perturbed quantum field theory; see [7, 8]. Throughout this paper we let N0 = N [ {0}, Z−= {−1, −2, −3, · · · } and Z− 0 = Z−[ {0}. The polygamma functions (n)(s) (s 2 C\Z−) are defined by (n)(s) = dn+1 dsn+1 log Γ(s) = dn dsn (s), n = 0, 1, 2, · · · , (2) INTEGERS: 20 (2020) 2 where Γ(s) is the classical Euler gamma function, and (0)(s) = (s) is the digamma function. Let us recall some basic properties of these functions that will be used fre-quently in this work. A well-known relationship between the polygamma functions (n)(s) and the generalized harmonic numbers H(s) n is given by (m−1)(n + 1) − (m−1)(1) = (−1)m−1(m −1)!H(m) n , (3) for m 2 N and n 2 N0; see [6, pg. 22]. The digamma function and harmonic numbers Hn are related by (n + 1) = −γ + Hn (n 2 N); (4) see [23, pg.
31], where γ = 0.57721 · · · is the Euler-Mascheroni constant.
The digamma function possesses the following properties: ✓ s + 1 2 ◆ = 2 (2s) − (s) −2 log 2 (s 2 C\Z−), (5) and (s) − (1 −s) = −⇡cot(⇡s); (6) see [23, pg. 25]. The gamma function satisfies the reflection formula Γ(s)Γ(1 −s) = ⇡ sin(⇡s) (s 2 C\Z−), (7) and the duplication formula Γ ✓ s + 1 2 ◆ = Γ(2s)Γ(1/2) 22s−1Γ(s) ; (8) see [16, pgs. 346, 349]. The binomial coefficients $s t % (s, t 2 C\Z−) are defined by ✓s t ◆ = Γ(s + 1) Γ(t + 1)Γ(s −t + 1), (9) and they satisfy the following identities for n, k 2 N with k n.
✓n + 1 k ◆ = ✓n k ◆ + ✓ n k −1 ◆ and n k ✓n −1 k −1 ◆ = ✓n k ◆ .
(10) The beta function B(s, t) is defined by B(s, t) = 1 Z 0 us−1(1 −u)t−1du (<(s) > 0 , <(t) > 0).
The gamma and beta functions are related by B(s, t) = Γ(s)Γ(t) Γ(s + t) ; (11) INTEGERS: 20 (2020) 3 see [9, p.251].
In this paper, we shall frequently use the following form of the binomial coefficients: fn(s) := ✓s + n k ◆ = Γ(s + n + 1) k!Γ(s + n −k + 1).
Taking the logarithm of both sides of this equation, and then di↵erentiating with respect to s, gives f0 n(s) = ✓s + n k ◆ { (s + n + 1) − (s + n −k + 1)}.
(12) Let us also define gn(s) = ✓s + n n ◆ .
Then, di↵erentiating with respect to s yields g0 n(s) = ✓s + n n ◆ [ (s + n + 1) − (s + 1)].
(13) In the literature, there are many interesting papers dealing with finite sums involv-ing the binomial coefficients and harmonic numbers. For example, we have n X k=0 ✓n k ◆2 Hk = ✓2n n ◆ {2Hn −H2n} (see ), n X k=0 (−1)k ✓n k ◆ {Hk −2H2k} = 4n n ✓2n n ◆2 (see ), and n X k=0 (−1)k ✓n k ◆ H2 n+k = 1 n $2n n % ⇢ Hn −H2n −2 n ( (see ).
Over the decades, combinatorial identities involving harmonic numbers have at-tracted the interest of many mathematicians, and many remarkable identities have been discovered by using a variety of methods. In , the authors computed the following family of sums by using di↵erential operator and Zeilberger’s algorithm for definite hypergeometric sums.
n X k=0 ✓n k ◆m {1 + m(n −2k)Hk} m = 1, 2, 3, 4, 5.
Please see [2,3,4,5,12,14,15,17,18,21,22,24,25] and the references cited therein for more identities on this issue. The aim of this paper is to present further combi-natorial identities involving harmonic numbers. First, we establish a new general INTEGERS: 20 (2020) 4 combinatorial identity involving two parameters, and by di↵erentiating and inte-grating both sides of this identity with respect to these parameters, we obtain several harmonic number identities. Some of them have already been considered previously, and the others are new. Although many other combinatorial identities can be derived by using these identities, for briefness, we have selected here just some of them, and we intend to prepare a separate paper containing many other applications.
Now we are ready to present our main results.
Main Results Theorem 1. Let n 2 N, s 2 C\Z−and x 2 C. Then the following identity holds: n X k=0 ✓s + n k ◆ xk = (1 + x)n " 1 + s n−1 X k=0 1 k + 1 ✓s + k k ◆✓ x x + 1 ◆k+1# .
(14) Proof. We prove the theorem by mathematical induction. Clearly, (14) is valid for n = 1. We assume that it is valid for n, and then we show that it is also valid for n + 1. We have n+1 X k=0 ✓s + n + 1 k ◆ xk = n X k=0 ✓s + n + 1 k ◆ xk + ✓s + n + 1 n + 1 ◆ xn+1.
Using the first relation in (10), this becomes n+1 X k=0 ✓s + n + 1 k ◆ xk = n X k=0 ✓s + n k ◆ xk + n X k=1 ✓s + n k −1 ◆ xk + ✓s + n + 1 n + 1 ◆ xn+1.
Setting k −1 = k0 in the second sum on the right-hand side of this equation, and then dropping the prime on k, we get, after a simple computation, n+1 X k=0 ✓s + n + 1 k ◆ xk = n X k=0 ✓s + n k ◆ xk + x n X k=0 ✓s + n k ◆ xk + ✓s + n + 1 n + 1 ◆ − ✓s + n n ◆, xn+1.
By (10) we have $s+n+1 n+1 % − $s+n n % = $s+n 1+n % ; thus, we get n+1 X k=0 ✓s + n + 1 k ◆ xk = (1 + x) n X k=0 ✓s + n k ◆ xk + ✓s + n 1 + n ◆ x1+n.
(15) INTEGERS: 20 (2020) 5 Therefore, by induction hypothesis, we deduce that n+1 X k=0 ✓s + n + 1 k ◆ xk = (1 + x)n+1 + s(1 + x)n n−1 X k=0 ✓s + k k ◆ xk+1 (k + 1)(1 + x)k + ✓s + n n + 1 ◆ xn+1 = (1 + x)n+1 + s(1 + x)n n X k=0 ✓s + k k ◆ xk+1 (k + 1)(1 + x)k + ✓s + n 1 + n ◆ − s n + 1 ✓s + n n ◆, xn+1.
Since $s+n n+1 % − s n+1 $s+n n % = 0, this proves that (14) is also valid for n + 1. This completes the proof of Theorem 1.
Theorem 2. Let n 2 N, s 2 C\Z−and x 2 C. Then we have n X k=1 ✓s + n k ◆ { (s + n + 1) − (s + n −k + 1)}xk = (1 + x)n n−1 X k=0 $s+k k % k + 1 {1 + s[ (s + k + 1) − (s + 1)]} ✓ x x + 1 ◆k+1 .
(16) Proof. The proof follows from di↵erentiating both sides of (14) with respect to s, and then using formulas (12) and (13).
Corollary 1. For n 2 N and s 2 C\Z−we have n X k=0 (−1)k ✓s + n k ◆ [ (s + n + 1) − (s + n −k + 1)] = (−1)n n ✓s + n −1 n −1 ◆ [1 + s( (s + n) − (s + 1))].
(17) Proof. The proof immediately follows from (16) by taking x = −1.
Theorem 3. Let n 2 N, s 2 C\Z−and x 2 C. Then, we have n X k=0 ✓s + n k ◆⇢ ( (s + n + 1) − (s + n −k + 1))2 + 0(s + n + 1) − 0(s + n −k + 1) ( xk = (1 + x)n n−1 X k=0 1 k + 1 ✓s + k k ◆⇢ 2( (s + k + 1) − (s + 1)) + s ( (s + k + 1) − (s + 1))2 + 0(s + k + 1) − 0(s + 1) ,( ✓ x x + 1 ◆k+1 . (18) INTEGERS: 20 (2020) 6 Proof. Di↵erentiating both sides of (16) with respect to s, and then using (12) and (13), we conclude that (18) is valid. The next corollary follows from (18) by setting x = −1.
Corollary 2. For n 2 N and s 2 C\Z−we have n X k=0 (−1)k ✓s + n k ◆⇢ ( (s + n + 1) − (s + n −k + 1))2 + 0(s + n + 1) − 0(s + n −k + 1) ( = (−1)n n ✓s + n −1 n −1 ◆⇢ 2[ (s + n) − (s + 1)] + s h ( (s + n) − (s + 1))2 + 0(s + n) − 0(s + 1) i ( .
(19) Proof. If we write equation (18) at x = −1, the proof is completed.
Theorem 4. Let n 2 N and s 2 C\Z−. Then, we have n X k=1 ✓s + n k ◆(−1)k−1 k = Hn + s n−1 X k=0 (−1)k$s+k k % (k + 1)2$ n k+1 %.
(20) Proof. If we take the first term of the sum on the left-hand side of (14) to the right, and then divide both sides by x, and finally replace x by −x in the resulting equation, we obtain the following equality: n X k=1 (−1)k ✓s + n k ◆ xk−1 = −(1 −x)n −1 (1 −x) −1 + s n−1 X k=0 (−1)k$s+k k % k + 1 (1 −x)n−k−1xk. (21) Integrating both sides of (21) from x = 0 to x = 1 yields n X k=1 ✓s + n k ◆(−1)k k = − Z 1 0 (1 −x)n −1 (1 −x) −1 dx + s n−1 X k=0 (−1)k$s+k k % k + 1 Z 1 0 (1 −x)n−k−1xk dx.
(22) It is well-known that Z 1 0 (1 −x)n −1 (1 −x) −1 dx = Hn; see , and Z 1 0 (1 −x)n−k−1xk dx = (n −k −1)!k!
n!
= 1 (k + 1) $ n k+1 %.
Substituting these expressions in (22), we see that (20) is valid.
INTEGERS: 20 (2020) 7 Theorem 5. Let n 2 N and s 2 C\Z−. Then we have n X k=1 (−1)k−1 k ✓s + n k ◆ [ (s + n + 1) − (s + n −k + 1)] = n−1 X k=0 (−1)k$s+k k % (k + 1)2$ n k+1 % + s n−1 X k=0 (−1)k$s+k k % (k + 1)2$ n k+1 %[ (s + k + 1) − (s + 1)].
(23) Proof. If we di↵erentiate both sides of (20) with respect to s, and use (12) and (13), the desired conclusion follows.
Theorem 6. For n 2 N and s 2 C\Z−we have n X k=1 (−1)k−1 k ✓s + n k ◆⇢ ( (s + n + 1) − (s + n −k + 1))2 + 0(s + n + 1) − 0(s + n −k + 1) ( = 2 n−1 X k=0 (−1)k$s+k k % (k + 1)2$ n k+1 %[ (s + k + 1) − (s + 1)] + s n−1 X k=0 (−1)k$s+k k % (k + 1)2$ n k+1 % ⇢ ( (s + k + 1) − (s + 1))2 + 0(s + k + 1) − 0(s + 1) ( .
(24) Proof. If we di↵erentiate both sides of (23) with respect to s, the proof is follows.
Theorem 7. For n 2 N and s 2 C\Z−the following identity holds: n X k=1 ✓s + n k ◆(−1)k−1 k2 = H2 n + H(2) n 2 + s n−1 X k=0 (−1)k k + 1 ✓s + k k ◆Hn −Hk (n −k) $n k %.
(25) Proof. Integrating both sides of (21) from x = 0 to x = u, we get n X k=1 (−1)k ✓s + n k ◆uk k = Z u 0 (1 −x)n −1 x dx + s n−1 X k=0 (−1)k$s+k k % k + 1 Z u 0 (1 −x)n−k−1xkdx.
Divide both sides of this equation by u, and then integrate each side from u = 0 to u = 1 to obtain n X k=1 (−1)k ✓s + n k ◆1 k2 = Z 1 0 1 u Z u 0 (1 −x)n −1 x dxdu + s n−1 X k=0 (−1)k$s+k k % k + 1 Z 1 0 1 u Z u 0 (1 −x)n−k−1xkdxdu.
(26) INTEGERS: 20 (2020) 8 Integration by parts gives Z 1 0 1 u Z u 0 (1 −x)n −1 x dxdu = log u Z u 0 (1 −x)n −1 x dx / / / / u=1 u=0 − Z 1 0 log u u [(1 −u)n −1]du.
(27) The first term on the right-hand side of (27) is equal to zero. Therefore, applying integration by parts, we get Z 1 0 1 u Z u 0 (1 −x)n −1 x dxdu = −1 2 log2 u[(1 −u)n −1] / / / / u=1 u=0 −n 2 Z 1 0 log2 u(1 −u)n−1du.
(28) The first term on the right-hand side of (28) is also equal to zero, hence, we get by (11) and (3) Z 1 0 1 u Z u 0 (1 −x)n −1 x dxdu = −n 2 Z 1 0 log2 u(1 −u)n−1du = −n 2 d2 dt2 Z 1 0 ut(1 −u)n−1du / / / / t=0 = −n 2 d2 dt2 Γ(t + 1)Γ(n) Γ(t + n + 1) / / / / t=0 = −H2 n + H(2) n 2 .
(29) An application of integration by parts to the second integral on the right-hand side of (26) leads to Z 1 0 1 u Z u 0 (1 −x)n−k−1xkdxdu = log u Z u 0 (1 −x)n−k−1xkdx / / / / u=1 u=0 − Z 1 0 log u(1 −u)n−k−1ukdu.
The first term on the right-hand side of this equation is equal to zero. So, from (11) and (3) we get Z 1 0 1 u Z u 0 (1 −x)n−k−1xkdxdu = − Z 1 0 log u(1 −u)n−k−1ukdu = − Z 1 0 d dtut(1 −u)n−k−1du / / / / t=k = −d dt Z 1 0 ut(1 −u)n−k−1du / / / / t=k = −d dt Γ(t + 1)Γ(n −k) Γ(n + t −k + 1) / / / / t=k = Hn −Hk (n −k) $n k %.
(30) Inserting the values of the integrals given in (29) and (30) into (26), and using (3), we complete the proof of Theorem 7.
INTEGERS: 20 (2020) 9 Theorem 8. For n 2 N and s 2 C\Z−we have n X k=1 ✓s + n k ◆(−1)k−1 k2 [ (s + n + 1) − (s + n −k + 1)] = n−1 X k=0 (−1)k k + 1 ✓s + k k ◆Hn −Hk (n −k) $n k % + s n−1 X k=0 (−1)k k + 1 ✓s + k k ◆Hn −Hk (n −k) $n k %[ (s + k + 1) − (s + 1)].
(31) Proof. The proof follows from di↵erentiating both sides of (25) with respect to s, and then using (12) and (13).
Theorem 9. For all n 2 N and s 2 C\Z−we have n X k=1 ✓s + n k ◆(−1)k−1 k2 ⇢ ( (s + n + 1) − (s + n −k + 1))2 + 0(s + n + 1) − 0(s + n −k + 1) ( = 2 n−1 X k=0 (−1)k k + 1 ✓s + k k ◆Hn −Hk (n −k) $n k %[ (s + k + 1) − (s + 1)] + s n−1 X k=0 (−1)k k + 1 ✓s + k k ◆Hn −Hk (n −k) $n k % ⇢ ( (s + k + 1) − (s + 1))2 + 0(s + k + 1) − 0(s + 1) ( .
(32) Proof. The proof follows from di↵erentiating both sides of (31) with respect to s, and using (12) and (13).
Applications In this section, we present several applications of our main results, which are derived by taking particular values for the parameters s and x.
Identity 1. Let n 2 N. Then we have n X k=1 1 22k ✓2k k ◆ (2H2k −Hk) = 2n + 1 22n ✓2n n ◆ [2H2n+1 −Hn −2].
INTEGERS: 20 (2020) 10 Proof. Putting x = −1 and performing the replacement s ! s −n in (14), we get n X k=0 (−1)k ✓s k ◆ = (−1)n ✓s −1 n ◆ .
(33) If we di↵erentiate both sides of (33) with respect to s, and then let s = −1/2, the proof follows from (4)-(8).
Identity 2. For all n 2 N the following identity holds: n X k=1 (−1)k−1 k ✓n k ◆ = Hn.
(34) Proof. The proof follows from (20) by setting s = 0.
Remark 1. Identity 2 is well-known and due to Euler (see , and ).
Remark 2. Theorem 4 provides a generalization of (34).
Identity 3. Let n 2 N. Then it holds that n X k=1 (−1)k−1 k ✓n k ◆ Hn−k = H2 n + n X k=1 (−1)k k2$n k % .
(35) Proof. The proof follows from (23) by setting s = 0, and using (34) and (3).
Remark 3. Identity 3 can be compared with the following formula (see ): n X k=1 (−1)k−1 k ✓n k ◆ Hn+k = H2 n + n X k=1 1 k2$n+k k %.
Identity 4. Let n 2 N and x 2 C. Then we have n X k=0 ✓n k ◆ Hkxk = (1 + x)n " Hn − n X k=1 1 k(x + 1)k # .
(36) Proof. Taking s = 0 in (16), and using (4), we get n X k=0 ✓n k ◆ [Hn −Hn−k]xk = (1 + x)n n X k=1 1 k ✓ x 1 + x ◆k .
(37) If we replace x by 1/x and then multiply both sides of (37) by xn, we get n X k=0 ✓n k ◆ [Hn −Hn−k]xn−k = (1 + x)n n X k=1 1 k(x + 1)k , (38) which is equivalent with (36), since Pn k=0 $n k % Hn−kxn−k = Pn k=0 $n k % Hkxk.
INTEGERS: 20 (2020) 11 If we set x = 1 in (36), we get the following known result (see [6, 12, 21]).
Identity 5. Let n 2 N. Then, it holds that n X k=1 ✓n k ◆ Hk = 2n " Hn − n X k=1 1 k2k # .
Identity 6. For n 2 N and x 2 C we have n X k=1 ✓n k ◆h H2 k + H(2) k i xk = (1 + x)n " H2 n + H(2) n + 2 n X k=1 Hk−1 −Hn k(1 + x)k # .
(39) Proof. By setting s = 0 in (18), and using (3) and (4), we can readily deduce that n X k=0 ✓n k ◆h (Hn −Hn−k)2 −H(2) n + H(2) n−k i xk = 2(1 + x)n n X k=1 Hk−1 k ✓ x 1 + x ◆k .
(40) Expanding the quadratic term on the left-hand side of (40) and using (36), after some simplifications, we find that n X k=0 ✓n k ◆h H2 n−k + H(2) n−k i xk−(1 + x)n " H2 n + H(2) n −2Hn n X k=1 1 k ✓ x 1 + x ◆k# = 2(1 + x)n n X k=1 Hk−1 k ✓ x 1 + x ◆k .
(41) If we replace x by 1/x here, and then multiply both sides of (41) by xn, after some simplifications, we get the desired identity (39).
If we set x = −1 in (39), we get the following known result; see .
Identity 7. For n 2 N, we have n X k=1 (−1)k ✓n k ◆n H2 k + H(2) k o = −2 n2 .
The following identity is known and computer program Mathematica recognizes it.
Identity 8. For all n 2 N we have n X k=1 (−1)k k $n k % = (−1)n −1 n + 1 .
(42) INTEGERS: 20 (2020) 12 Proof. Setting s = 1 in (20), we get n X k=1 (−1)k−1 k ✓n + 1 k ◆ = Hn + n X k=1 (−1)k−1 k $n k % .
(43) Since n X k=1 (−1)k−1 k ✓n + 1 k ◆ = n+1 X k=1 (−1)k−1 k ✓n + 1 k ◆ −(−1)n n + 1 , if we use (34), it follows that n X k=1 (−1)k−1 k ✓n + 1 k ◆ = Hn+1 −(−1)n n + 1 .
This completes the proof of (42) by the help of (43).
Identity 9. Let n 2 N. Then we have n X k=1 (−1)k−1 k ✓n k ◆n H2 n−k + H(2) n−k o = H3 n + HnH(2) n + 2 n X k=1 (−1)k[Hn −Hk−1] k2$n k % .
Proof. Putting s = 0 in (24), we obtain that n X k=1 (−1)k−1 k ✓n k ◆n (Hn −Hn−k)2 −H(2) n + H(2) n−k o = 2 n−1 X k=1 (−1)kHk (k + 1)2$ n k+1 %.
Expanding the quadratic term here, it simplifies to h H2 n −H(2) n i n X k=1 (−1)k−1 k ✓n k ◆ −2Hn n X k=1 (−1)k−1 k ✓n k ◆ Hn−k + n X k=1 (−1)k−1 k ✓n k ◆n H2 n−k + H(2) n−k o = 2 n−1 X k=1 (−1)kHk (k + 1)2$ n k+1 %.
Using (34) and (35) here and simplifying the resulting equation, the desired con-clussion follows.
Identity 10. Let m, n 2 N. Then we have n X k=0 (−1)k ✓mn k ◆ Hmn−k = (−1)n m ✓mn n ◆ (m −1)H(m−1)n − 1 mn , .
(44) Proof. If we write equation (33) at s = mn, we get n X k=0 (−1)k ✓mn k ◆ = (−1)n ✓mn −1 n ◆ .
(45) INTEGERS: 20 (2020) 13 If we di↵erentiate both sides of (33) with respect to s, and then set s = mn (m 2 N), we, in view of (3), get n X k=0 (−1)k ✓mn k ◆ {Hmn −Hmn−k} = (−1)n ✓mn −1 n ◆ {Hmn−1 −Hmn−n−1}. (46) But if we use (45), this becomes n X k=0 (−1)k ✓mn k ◆ {Hmn −Hmn−k} = Hmn n X k=0 (−1)k ✓mn k ◆ − n X k=0 (−1)k ✓mn k ◆ Hmn−k = (−1)nHmn ✓mn −1 n ◆ − n X k=0 (−1)k ✓mn k ◆ Hmn−k.
Using this identity in (46) and simplifying the result, we complete the proof of (44).
The following identity is an immediate consequence of Identity 10 with m = 2 and m = 3.
Identity 11. For n 2 N, we have n X k=0 (−1)k ✓2n k ◆ H2n−k = (−1)n 2 ✓2n n ◆✓ Hn −1 2n ◆ and n X k=0 (−1)k ✓3n k ◆ H3n−k = (−1)n 3 ✓3n n ◆✓ 2Hn −1 3n ◆ .
Identity 12. For n 2 N we have n X k=1 (−1)kHk k $n k % = (−1)nHn+1 n + 1 + n+1 X k=1 (−1)k k2$n+1 k %.
Proof. Setting s = 1 in (23), and using (3) and (2) = 1 + (1), we obtain n X k=1 (−1)k−1 k ✓n + 1 k ◆ (Hn+1 −Hn−k+1) = n−1 X k=0 (−1)k (k + 1) $ n k+1 % + n−1 X k=0 (−1)k(Hk+1 −1) (k + 1) $ n k+1 % = n X k=1 (−1)k−1Hk k $n k % .
(47) INTEGERS: 20 (2020) 14 On the other hand, we have n X k=1 (−1)k−1 k ✓n + 1 k ◆ (Hn+1 −Hn−k+1) = Hn+1 "n+1 X k=1 (−1)k−1 k ✓n + 1 k ◆ −(−1)n n + 1 # − n+1 X k=1 (−1)k−1 k ✓n + 1 k ◆ Hn+1−k.
Using (34) and (35), we conclude from this equation that n X k=1 (−1)k−1 k ✓n + 1 k ◆ (Hn+1 −Hn−k+1) = (−1)n+1Hn+1 n + 1 + n+1 X k=1 (−1)k−1 k2$n+1 k % .
(48) Equating the right-hand sides of (47) and (48), we see that Identity 12 is valid.
Identity 13. For n 2 N we have n X k=0 (−1)k−14k$n k % $2k k % = 1 2n −1.
Proof. Setting s = −1 2 and x = −1 in (14), and using (8) and (9), the proof is completed.
Remark 4. Identity 13 is known and can be found in [22, Theorem 4.5] .
If we set s = 0 in (25), we get the following known result (see ).
Identity 14. Let n 2 N. Then we have n X k=1 ✓n k ◆(−1)k−1 k2 = H2 n + H(2) n 2 .
(49) Identity 15. For n 2 N we have n X k=1 (−1)kHn−k k $n k % = 1 −(−1)n (n + 1)2 − Hn n + 1.
Proof. If we take s = 1 in (25), we get n X k=1 ✓n + 1 k ◆(−1)k−1 k2 = H2 n + H(2) n 2 + n−1 X k=0 (−1)k(Hn −Hk) (n −k) $n k % .
(50) By (49), we arrive at n X k=1 ✓n + 1 k ◆(−1)k−1 k2 = n+1 X k=1 ✓n + 1 k ◆(−1)k−1 k2 − (−1)n (n + 1)2 = H2 n+1 + H(2) n+1 2 − (−1)n (n + 1)2 .
(51) INTEGERS: 20 (2020) 15 On the other hand, we have n−1 X k=0 (−1)k(Hn −Hk) (n −k) $n k % = Hn n−1 X k=0 (−1)k (n −k) $n k % − n−1 X k=0 (−1)kHk (n −k) $n k %.
If we substitute n −k = k0 in the sums on the right-hand side of this equation and then drop the prime on k, we get n−1 X k=0 (−1)k(Hn −Hk) (n −k) $n k % = Hn n X k=1 (−1)n−k k $n k % − n X k=1 (−1)n−kHn−k k $n k % .
Using (42) gives n−1 X k=0 (−1)k(Hn −Hk) (n −k) $n k % = Hn(1 −(−1)n) n + 1 −(−1)n n X k=1 (−1)kHn−k k $n k % .
(52) Using (51) and (52) in (50), we conculde that Identity 15 is valid.
Identity 16. For all n 2 N we have n X k=1 (−1)k−1 k2 ✓n k ◆ Hn−k = Hn ⇣ H2 n + H(2) n ⌘ 2 − n−1 X k=0 (−1)k(Hn −Hk) (k + 1)(n −k) $n k %.
(53) Proof. Setting s = 0 in (31) yields n X k=1 (−1)k−1 k2 ✓n k ◆ (Hn −Hn−k) = n−1 X k=0 (−1)k(Hn −Hk) (k + 1)(n −k) $n k %.
(54) By (49), we have n X k=1 (−1)k−1 k2 ✓n k ◆ (Hn −Hn−k) = Hn ⇣ (Hn)2 + H(2) n ⌘ 2 − n X k=1 (−1)k−1 k2 ✓n k ◆ Hn−k.
(55) Equating the right-hand sides of (54) and (55), The proof of (53) follows.
Identity 17. For all n 2 N the following identity holds: n X k=1 (−1)k−1 k2 ✓n k ◆n H2 n−k + H(2) n−k o = ⇣ H2 n + H(2) n ⌘2 2 −2 n−1 X k=0 (−1)k(Hn −Hk)2 (k + 1)(n −k) $n k % .
INTEGERS: 20 (2020) 16 Proof. Setting s = 0 in (32), we get n X k=1 (−1)k−1 k2 ✓n k ◆n (Hn −Hn−k)2 + H(2) n−k −H(2) n o = 2 n−1 X k=0 (−1)k(Hn −Hk)Hk (k + 1)(n −k) $n k % .
(56) Clearly, we have n X k=1 (−1)k−1 k2 ✓n k ◆n (Hn −Hn−k)2 + H(2) n−k −H(2) n o = h H2 n −H(2) n i n X k=1 (−1)k−1 k2 ✓n k ◆ −2Hn n X k=1 (−1)k−1 k2 ✓n k ◆ Hn−k + n X k=1 (−1)k−1 k2 ✓n k ◆n H2 n−k + H(2) n−k o .
Thus, by the help of (49) and (53), after some algebraic manipulations, we find that n X k=1 (−1)k−1 k2 ✓n k ◆n (Hn −Hn−k)2 + H(2) n−k −H(2) n o = − ⇣ H2 n + H(2) n ⌘2 2 −2Hn n−1 X k=0 (−1)k−1(Hn −Hk) (k + 1)(n −k) $n k % + n X k=1 (−1)k−1 k2 ✓n k ◆n H2 n−k + H(2) n−k o .
(57) If we equate the right-hand sides of (56) and (57), the conclusion follows.
Remark 5. We want to provide some comments about how we discovered the core identity (14) of this work. When we started writing this paper, our intention was to generalize the binomial theorem Pn k=0 $n k % xk = (1 + x)n. For this purpose, we replace n by n + s in the summand of this sum and define an = Pn k=0 $n+s k % xk.
First, we showed that an satisfies the following recurrence relation: a0 = 1 and an+1 = (1 + x)an + ✓n + s n + 1 ◆ xn+1 for n ≥1.
Using this relation, formula (14) was found.
References V. S. Adamchik and H. M. Srivastava, Some series of the zeta and related functions, Analysis 18 (1998), 131-144.
INTEGERS: 20 (2020) 17 N. Batir, On some combinatorial identities and harmonic sums, Int. J. Number Theory 13 (7) (2017), 1695–1709.
J. Choi and H. M. Srivastava, Some summation formulas involving harmonic numbers and generalized harmonic numbers, Math. Comput. Modelling 54 (2011), 2220-2234.
W. Chu and L. De Donno, Hypergeometric series and harmonic number identities, Adv. Appl.
Math. 34 (2005), 123-137.
W. Chu, Summation formulae involving harmonic numbers, Filomat 26(1), 2012, 143-152 M. W. Co↵ey, On some series repsentations of the Hurwitz zeta function, J. Comput. Appl.
Math. 216 (2008), 297-305.
M. W. Co↵ey, On a three dimentional symmetric Ising tetrahedron, and Clausen functions, J. Math. Phys. 4(49) (2008).
M. W. Co↵ey, A one dimentional digamma and polygamma series related to the evaluation of Feynman diagrams, J. Comp. Appl. Math. 183 (2005), 84-100.
P. Duren, Invitation to Classical Analysis, Amer. Math. Soc., 2012.
L. Euler, Demonstratio insignis theorematis numerici circa uncias potestatum binomialium, Nova Acta Acad. Sci. Petropol. 15 (1799/1802), 33-43; reprinted in Opera Omnia, Ser. I, Vol. 16(2), B. G. Teubner, Leipzig, 1935, pp. 104-116.
P. Flajolet and R. Sedgewick, Mellin transforms and asymptotics: Finite di↵erences and Rice’s integrals, Theoret. Comput. Sci 144 (1995), 101-124.
L. Gonzalez, A new approach for proving generating combinatorial identities, Int. J. Math.
Edu. Sci. Techology 41(3) (2010), 359-372.
G. R. Graham, D. E. Knuth and O. Patahnik, Concrete Mathematics, Addison-Wesley Pub-lishing Company, Reading, Massachusetts, 1959.
I -C. Huang, Applications of residues to combinatorial identities, Proc. Amer. Math. Soc.
125(4) (1997), 1011-1017.
T. Mansour, Combinatorial identities and inverse binomial coefficients, Adv. Appl. Math. 28 (2002), 196-202.
J. E. Marsden, Basic Complex Analysis, W. H. Freeman and Company, San Fransisco, 1973.
P. Paule and C. Schneider, Computer proofs of a new family of harmonic number identities, Adv. Appl. Math. 31 (2003), 359-378.
A. Sofo, Sums of derivatives of binomial coefficients, Adv. Appl. Math. 42 (2009), 123-134.
A. Sofo and N. Batir, On finite sums and integral representations, Integers 13 (2013), #A45.
A. Sofo, Some more identities involving rational sums, Appl. Anal. Discr. Math. 2 (2008), 56-66.
M. Z. Spivey, Combinatorial sums and finite di↵erences, Discrete Math. 307 (2007), 3130-3146.
R. Sprugnoli, Sums of reciprocals of the central binomial coefficients, Integers 6 (2006), #A27.
INTEGERS: 20 (2020) 18 H. M. Srivastava and J. Choi, Zeta and q-Zeta Functions and Associated Series and Integrals, Elsevier, Amsterdam, 2012.
C. Wei and X. Wang, Whipple-type 3F2- series and summations formulae involving general-ized harmonic numbers, Inter. J. Number Theory 14(9) (2018), 2385-2407.
C. Wei and X. Wang, Summation formulas involving generalized harmonic numbers, J. Dif-ference Equ. Appl. 22(10) (2016), 1554-1567. |
14412 | https://www.scribd.com/document/443436652/ModularArithmetic-Worksheets-Answers-pdf | Math Worksheets for Students | PDF | Number Theory | Abstract Algebra
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Math Worksheets for Students
The document discusses questions related to patterns that repeat over time, such as days of the week and seasons. It asks the reader to determine what day of the week or season it would be g…
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Worksheet 1 - Going round in circles 1
Worksheet 1 - Going round in circles
Most of these questions were taken f rom:
Charlie said:
"It's Sunday today, so it will be Sunday again in
7
days.. and in
770
days... and in
140
days... and in
35035
days... and in
14000000007
days!"
Alison said:
"and it will be Tuesday in
2
days... and in
72
days... and in
702
days... and in
779
days... and in
14777002
days!" 1)
Do you agree with all of Charlie's and Alison's statements? 2)
Charlie and Alison chose numbers that were easy to work with. Can you see why they were chosen? 3)
If today is Sunday, what day will it be in 15 days? 26 days? 234 days? 1000? 4)
If your birthday fell on a Sunday this year, what day will it fall on next year? 5)
If it is autumn now, what season will it be in 100 seasons? 6)
If it is 9 am now, what time will it be in 50 hours? 7)
If it is November, what month will it be in 1000 months? 8)
If it is midday now, will it be light or dark in 539 hours?
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Worksheet 1 - Going round in circles 2
You don't have to finish t hese questions.
9)
A railway l ine has 27 stat ions on a circu lar loop. If I fall asleep and travel through 312 stations, where will I end up in relation to where I started? 10)
If a running track is 400 metres around, where will I be in relation to the start after running 6 miles (approximately 9656 metres)? 11)
I was facing North and then spun around through 945° clockwise. In what direction was I facing at the end? 12)
If I get on at the bottom of a fairground wheel and the wheel turns through 5000°, whereabouts on the wheel will I be?
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Worksheet 2 - Remainders and congruences 1
Worksheet 2 Remainders and congruences
Instead of 13 = 1, in modular arithmetic we write 13 ≡ 1 (mod 12) and read it “13 is congruent to 1 modulo 12” or, to abbreviate, “13 is 1 modulo 12”.
Examples:
12 ≡ 0 (mod 12)
17 ≡ 5 (mod 12)
37 ≡ 1 (mod 12)
-
1 ≡ 11 (mod 12)
In general:
a ≡ b (mod n) if a
-b is a multiple of n. Equivalently:
a ≡ b (mod n) if a and b have the same remainder when divided by n (remainder
modulo n). When we work modulo n we replace all the numbers by their remainders modulo n, that is: 0, 1, 2,
…, n
-1. 1)
Find the remainders modulo 3 of: a)
31 b)
44 c)
75 d)
751 2)
Find the remainders modulo 2 of: a)
34
–
15 b)
141
–
78 c)
519
–
444 + 37
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Worksheet 2 - Remainders and congruences 2
3)
Find the remainders modulo 12 of: a)
31 + 28 + 31 + 30 b)
38 x 4 + 360 c)
66 + 5 + 26 4)
Which of the following congruences are true? a)
177 ≡ 17 (mod
2) b)
1322 ≡ 5294 (mod 12)
c)
16 + 30
≡
2 (mod 2) d)
16 + 30
≡
2 (mod 3)
You don't have to finish t hese questions.
5)
Which of the following congruences are true? a)
16 + 30
≡
2 (mod 12) b)
67 x 73 ≡
0 (mod 3) c)
14 x 15 x 16 ≡ 6
(mod 3)
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14413 | https://www.habloco.com/spanish-verbs/recibir-conjugation/ | Recibir Conjugation - Spanish Verbs - Habloco
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Home » Spanish Verb Conjugations » Basic Spanish Verbs » Recibir
Recibir conjugation
Table of Contents
Recibir is a Spanish verb which translates into English as “to receive”.
Basic Spanish Verbs, Spanish IR Verbs
Below are all of the conjugations for recibir in Spanish, in all three moods (indicative/indicativo, subjunctive/subjunctivo and imperative/imperativo) and all of the tenses, for each pronoun.
The vosotros pronoun is mainly used in mainland Spain, and is the informal second-person plural – it could be considered the Spanish version of “y’all”. It is rarely found in Latin America, where ustedes is used instead.
The vos form is used instead of tú in some Spanish speaking countries of South America, especially the Southern Cone (e.g. Argentina and Uruguay) and has a different conjugation.
Recibir Infinitive
| English Infinitive | to receive |
| Spanish Infinitive | recibir |
Recibir Gerund and Past Participle
The gerund (gerundio) is used with the continuous tenses, e.g. present continuous (está recibiendo) and past continuous (estaba recibiendo). The easiest way to think of it is the equivalent of english’s -ing form (e.g. receiving).
The past participle (participio) is used with perfect tense ‘haber’ verbs, e.g. he recibido and hubiera recibido. These are the equivalent of English’s ‘have’ (e.g. have received).
| Gerundio / Gerund | recibiendo |
| Participio / Past Participle | recibido |
Recibir Indicative Conjugations
The basic form of speech, el indicativo is used for making statements, talking about facts, events and things that are certain and objective.
Recibir Presente / Present
The present tense is as it sounds – it’s for talking about things that are currently going on, which are habitual, or which generally exist. In English, this would be “I receive” or “they receive”.
| Pronoun | Spanish |
--- |
| Yo | recibo |
| Tú | recibes |
| Él / Ella / Usted | recibe |
| Nosotros / as | recibimos |
| Vosotros / as | recibís |
| Ellos / Ellas / Ustedes | reciben |
| Vos | recibís |
Recibir Preterite / Pretérito Indefinido
Your simple past tense, e.g. “I received” or “she received” in English.
In Spanish, there are two past tenses where just one is used in English; the pretérite infefinido is typically used to refer to a concrete, specific moment in time.
| Pronoun | Spanish | English |
---
| Yo | recibí | I received |
| Tú | recibiste | You received |
| Él / Ella / Usted | recibió | He / she / you received |
| Nosotros / as | recibimos | We received |
| Vosotros / as | recibisteis | You received |
| Ellos / Ellas / Ustedes | recibieron | They / you received |
| Vos | recibiste | You received |
Recibir Imperfect / Pretérito Imperfecto
The pretérito imperfecto roughly translates as “I was receiving” or “she was receiving” in English, and is typically used to describe things and set a scene, talk about events without a specific timeframe, or talk about habitual events or states in the past.
| Pronoun | Spanish | English |
---
| Yo | recibía | I was receiving |
| Tú | recibías | You were receiving |
| Él / Ella / Usted | recibía | He was / she was / you were receiving |
| Nosotros / as | recibíamos | We were receiving |
| Vosotros / as | recibíais | You were receiving |
| Ellos / Ellas / Ustedes | recibían | They / you were receiving |
| Vos | recibías | You were receiving |
Recibir Perfect / Perfecto
The perfect tense is for talking about things which happened in the past but are still related to the present or continue into the present.
In English, these use the auxiliary verbs ‘have’ and ‘has’ – i.e. “I have received” and “she has received”.
| Pronoun | Spanish | English |
---
| Yo | he recibido | I have received |
| Tú | has recibido | You have received |
| Él / Ella / Usted | ha recibido | He has / she has / you have received |
| Nosotros / as | hemos recibido | We have received |
| Vosotros / as | habéis recibido | You have received |
| Ellos / Ellas / Ustedes | han recibido | They / you have received |
| Vos | has recibido | You have received |
Recibir Conditional / Condicional
The conditional is used in place of the English modal verb “would”, i.e. “I would receive” or “she would receive”. It can be used to talk about hypothetical situations.
| Pronoun | Spanish | Englush |
---
| Yo | recibiría | I would receive |
| Tú | recibirías | You would receive |
| Él / Ella / Usted | recibiría | He / she / you would receive |
| Nosotros / as | recibiríamos | We would receive |
| Vosotros / as | recibiríais | You would receive |
| Ellos / Ellas / Ustedes | recibirían | They / you would receive |
| Vos | recibirías | You would receive |
Recibir Future / Futuro
The future tense, simply put, replaces the English modal verb “will” – i.e. “I will receive” or “they will receive”.
It is more commonly used for making a hypothesis about the present. To talk about the future, Spanish speakers frequently use “ir + a + infinivo”, e.g. “van a recibir” means “They are going to receive”.
| Pronoun | Spanish | English |
---
| Yo | recibiré | I will receive |
| Tú | recibirás | You will receive |
| Él / Ella / Usted | recibirá | He / she / you will receive |
| Nosotros / as | recibiremos | We will receive |
| Vosotros / as | recibiréis | You will receive |
| Ellos / Ellas / Ustedes | recibirán | They / you will receive |
| Vos | recibirás | You will receive |
Recibir Subjunctive Conjugations
Recibir Present Subjunctive / Presente de Subjuntivo
| Pronoun | Spanish |
--- |
| Yo | reciba |
| Tú | recibas |
| Él / Ella / Usted | reciba |
| Nosotros / as | recibamos |
| Vosotros / as | recibáis |
| Ellos / Ellas / Ustedes | reciban |
| Vos | recibas |
Recibir Past Subjunctive / Imperfecto de Subjuntivo
There are two ways to form the imperfect subjunctive.
The first option sees verbs ending in -era (for -er and -ir verbs) and -ara (for -ar verbs), while the second sees verbs ending in -ese (for -er and -ir verbs) and -ase (for -ar verbs).
There is no difference between these two forms, and Spanish speakers use them interchangeably.
| Pronoun | Spanish era/ara | Spanish ese/ase |
---
| Yo | recibiera | recibiese |
| Tú | recibieras | recibiese |
| Él / Ella / Usted | recibiera | recibiese |
| Nosotros / as | recibiéramos | recibiésemos |
| Vosotros / as | recibierais | recibieseis |
| Ellos / Ellas / Ustedes | recibieran | recibiesen |
| Vos | recibieras | recibiese |
Recibir Future Subjunctive / Futuro de Subjuntivo
The future subjunctive is no longer used in modern-day Spanish, apart from in literary and legal contexts, and there is no need to learn it.
It is formed the same as the past/imperfect subjunctive, but with -e endings instead of -a endings.
| Pronoun | Spanish |
--- |
| Yo | recibiere |
| Tú | recibieres |
| Él / Ella / Usted | recibiera |
| Nosotros / as | recibiéremos |
| Vosotros / as | recibiereis |
| Ellos / Ellas / Ustedes | recibieren |
| Vos | recibieres |
Recibir Imperative Conjugations
Used for forming positive and negative commands, e.g. “receive!” and “don’t receive!”.
| Pronoun | Spanish Affirmative | Spanish Negative |
---
| Tú | recibe | no recibas |
| Él / Ella / Usted | reciba | no reciba |
| Nosotros / as | recibamos | no recibamos |
| Vosotros / as | recibid | no recibáis |
| Ellos / Ellas / Ustedes | reciban | no reciban |
| Vos | recibí | no recibas |
Recibir Compound Subjunctive Tenses
Recibir Subjunctive Perfect
| Pronoun | Spanish |
--- |
| Yo | haya recibido |
| Tú | hayas recibido |
| Él / Ella / Usted | haya recibido |
| Nosotros / as | hayamos recibido |
| Vosotros / as | hayáis recibido |
| Ellos / Ellas / Ustedes | hayan recibido |
| Vos | hayas recibido |
Recibir Subjunctive Past Perfect
| Pronoun | Spanish |
--- |
| Yo | hubiera recibido / hubiese recibido |
| Tú | hubieras recibido / hubieses recibido |
| Él / Ella / Usted | hubiera recibido / hubiese recibido |
| Nosotros / as | hubiéramos recibido / hubiésemos recibido |
| Vosotros / as | hubierais recibido / hubieseis recibido |
| Ellos / Ellas / Ustedes | hubieran recibido / hubiesen recibido |
| Vos | hubieras recibido / hubieses recibido |
Recibir Subjunctive Future Perfect
| Pronoun | Spanish |
--- |
| Yo | hubiere recibido |
| Tú | hubieres recibido |
| Él / Ella / Usted | hubiere recibido |
| Nosotros / as | hubiéremos recibido |
| Vosotros / as | hubiereis recibido |
| Ellos / Ellas / Ustedes | hubieren recibido |
| Vos | hubieres recibido |
Recibir Subjective Progressive Perfect
| Pronoun | Spanish |
--- |
| Yo | esté recibiendo |
| Tú | estés recibiendo |
| Él / Ella / Usted | esté recibiendo |
| Nosotros / as | estemos recibiendo |
| Vosotros / as | estéis recibiendo |
| Ellos / Ellas / Ustedes | estén recibiendo |
| Vos | estés recibiendo |
Recibir Subjunctive Past Progressive
| Pronoun | Spanish |
--- |
| Yo | estuviera recibiendo / estuviese recibiendo |
| Tú | estuvieras recibiendo / estuvieses recibiendo |
| Él / Ella / Usted | estuviera recibiendo / estuviese recibiendo |
| Nosotros / as | estuviéramos recibiendo / estuviésamos recibiendo |
| Vosotros / as | estuvierais recibiendo / estuvieseis recibiendo |
| Ellos / Ellas / Ustedes | estuviera recibiendo / estuviese recibiendo |
| Vos | estuvieras recibiendo / estuvieses recibiendo |
Recibir Subjunctive Future Progressive
| Pronoun | Spanish |
--- |
| Yo | estuviere recibiendo |
| Tú | estuvieres recibiendo |
| Él / Ella / Usted | estuviere recibiendo |
| Nosotros / as | estuviéremos recibiendo |
| Vosotros / as | estuviereis recibiendo |
| Ellos / Ellas / Ustedes | estuviere recibiendo |
| Vos | estuvieres recibiendo |
Recibir Vos Conjugation
Voseo is the practice of using ‘vos’ instead of ‘tú’ as the second-person singular pronoun, and is common throughout much of South America.
There are various versions of ‘voseo’ used throughout the Spanish-speaking world. The conjugations for the most common type – used throughout Argentina, parts of Bolivia, Colombia, Ecuador, Paraguya and Uruguay are below.
The present indicative (presente de indicativo) and affirmative imperative (imperativo) have different conjugations from the tú form, while all other tenses generally use the tú form.
| Tense | Vos Conjugation |
--- |
| Present IndicativePresente de Indicativo | Vos recibís |
| Simle Past / PreteritePreterite de Indicativo | Vos recibiste |
| Imperfect Past Preterite Imperfecto de Indicativo | Vos recibías |
| Conditional Condicional | Vos recibirías |
| Future Futuro de Indicativo | Vos recibirás |
| Present Subjunctive Presente de Subjunctivo | Vos recibas |
| Imperfect Subjunctive Imperfecto de Subjunctivo | Vos recibieras / Vos recibiese |
| Affirmative Imperative Imperativo | Vos recibí |
| Negative Imperative Imperativo Negativo | Vos no recibas |
Free Recibir Conjugation Printable
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14414 | https://crypto.stackexchange.com/questions/102165/fast-polynomial-multiplication-over-finite-field-gf2n | Fast polynomial multiplication over finite field GF(2^n) - Cryptography Stack Exchange
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Fast polynomial multiplication over finite field GF(2^n)
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I wonder if there is a more efficient polynomial multiplication than Karatsuba over the finite field GF(2 n)GF(2 n). Brief research on this topic gave me a few results on fast multiplication on GF(2 n)GF(2 n) using a polynomial over GF(2)GF(2), however, it was hard to find a fast 'polynomial' multiplication over GF(2^n). (i.e. Multiplication over GF(2 n)[X]GF(2 n)[X]).
A classic Karatsuba polynomial multiplication would give me a time complexity of O(N(log 3/log 2))O(N(log3/log2)) but I feel like there should be an algorithm with O(N log N)O(N logN) time complexity. Can anybody help me?
-edit
For my case, 10≤log N≤12 10≤logN≤12 and I can set n n to whatever I want. Also, the modulo arithmetic will be automatically done. (I don't need to care about the slow modulo arithmetic) I found an algorithm achieving O(N log N)O(N logN) time complexity by David Harvey, however it makes use of real number arithmetic with precision as far I understand.
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edited Oct 8, 2022 at 12:22
kodlu
25.5k 2 2 gold badges 30 30 silver badges 65 65 bronze badges
asked Oct 7, 2022 at 7:23
Lukie BoyLukie Boy
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Welcome to Cryptography. In Practice, the Karatsuba has applied a few levels. Then Montgomery or a similar reduction is applied. The performance and choice of algorithms really depend on the case. Are you asking for a specific case or a general case for the known complexity?kelalaka –kelalaka 2022-10-07 08:45:22 +00:00 Commented Oct 7, 2022 at 8:45
@kelalaka It is a specific case, with polynomial degree being 2^10, 2^11, 2^12. I can set the modulus according to the parameter which is power of two!Lukie Boy –Lukie Boy 2022-10-07 08:52:40 +00:00 Commented Oct 7, 2022 at 8:52
1 The O(n log n)O(n logn) for multiplication is recently achieved with a huge constant on the front. At least, we have seen that the complexity of multiplication is not bigger than O(n log n)O(n logn) with a big hidden constant. I've seen an improvement to Karatsuba-Ofman that is only applicable to hardware. Montgomery is a good choice, but it really depends on your application. Maybe you should edit your question with your aim and target and what you found...kelalaka –kelalaka 2022-10-07 09:24:04 +00:00 Commented Oct 7, 2022 at 9:24
1 While it's in a slightly different setting (integer multiplication vs polynomial multiplication), there have been some indications that you can beat Toom-Cook/Karatsuba using O(n log 2 n)O(n log 2n) techniques for general integer multiplication in a relevant parameter range (>1024>1024 bits), namely thisMark Schultz-Wu –Mark Schultz-Wu♦ 2022-10-08 17:17:05 +00:00 Commented Oct 8, 2022 at 17:17
1 As for multiplication over R:=G F(2 n)[x]R:=G F(2 n)[x], one can sometimes improve on Toom-Cook/Karatsuba via using NTT Friendly Rings. Roughly speaking, you embed R↪S R↪S into another ring S S that has fast multiplication. This embedding doesn't fully preserve algebraic structure, but provided you are computing some "small" formula in R R, often one can arrange such that this small formula is always evaluated correctly, i.e. one can use S S-arithmetic vs R R-arithmetic at no cost. S S arithmetic is often chosen to have O(n log 2 n)O(n log 2n) multiplication.Mark Schultz-Wu –Mark Schultz-Wu♦ 2022-10-08 17:19:37 +00:00 Commented Oct 8, 2022 at 17:19
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14415 | https://pmc.ncbi.nlm.nih.gov/articles/PMC12414446/ | Upadacitinib Therapy in Adolescent Severe Alopecia Areata: A Case Series and Narrative Review - PMC
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Clin Cosmet Investig Dermatol
. 2025 Sep 3;18:2141–2148. doi: 10.2147/CCID.S549823
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Upadacitinib Therapy in Adolescent Severe Alopecia Areata: A Case Series and Narrative Review
Yimeng Gao
Yimeng Gao
1 Department of Dermatology, State Key Laboratory of Complex Severe and Rare Diseases, Peking Union Medical College Hospital, Chinese Academy of Medical Sciences and Peking Union Medical College, National Clinical Research Center for Dermatologic and Immunologic Diseases, Beijing, 100730, People’s Republic of China
Find articles by Yimeng Gao
1, Chenyu Zhu
Chenyu Zhu
1 Department of Dermatology, State Key Laboratory of Complex Severe and Rare Diseases, Peking Union Medical College Hospital, Chinese Academy of Medical Sciences and Peking Union Medical College, National Clinical Research Center for Dermatologic and Immunologic Diseases, Beijing, 100730, People’s Republic of China
Find articles by Chenyu Zhu
1, Hongzhong Jin
Hongzhong Jin
1 Department of Dermatology, State Key Laboratory of Complex Severe and Rare Diseases, Peking Union Medical College Hospital, Chinese Academy of Medical Sciences and Peking Union Medical College, National Clinical Research Center for Dermatologic and Immunologic Diseases, Beijing, 100730, People’s Republic of China
Find articles by Hongzhong Jin
1,✉
Author information
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Copyright and License information
1 Department of Dermatology, State Key Laboratory of Complex Severe and Rare Diseases, Peking Union Medical College Hospital, Chinese Academy of Medical Sciences and Peking Union Medical College, National Clinical Research Center for Dermatologic and Immunologic Diseases, Beijing, 100730, People’s Republic of China
✉
Correspondence: Hongzhong Jin, Department of Dermatology, State Key Laboratory of Complex Severe and Rare Diseases, Peking Union Medical College Hospital, Chinese Academy of Medical Sciences and Peking Union Medical College, National Clinical Research Center for Dermatologic and Immunologic Diseases, Beijing, 100730, People’s Republic of China, Email jinhongzhong@263.net
Received 2025 Jul 5; Accepted 2025 Aug 21; Collection date 2025.
© 2025 Gao et al.
This work is published and licensed by Dove Medical Press Limited. The full terms of this license are available at and incorporate the Creative Commons Attribution – Non Commercial (unported, v4.0) License ( By accessing the work you hereby accept the Terms. Non-commercial uses of the work are permitted without any further permission from Dove Medical Press Limited, provided the work is properly attributed. For permission for commercial use of this work, please see paragraphs 4.2 and 5 of our Terms (
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PMCID: PMC12414446 PMID: 40922723
Abstract
Purpose
Alopecia areata (AA) is a common, immune-mediated, non-scarring form of hair loss. Janus kinase inhibitors provide considerable insight into the treatment of severe AA. However, the efficacy and safety of upadacitinib treatment of adolescents and pediatric patients with severe AA is unclear, especially in those without concomitant atopic diseases.
Patients and Methods
Adolescents with severe AA receiving upadacitinib were recruited from the Dermatology Outpatient department. This is a retrospective case series. Clinical characteristics, hair regeneration, and adverse events were analyzed to assess effectiveness and safety. We searched the PubMed, Web of Science Core Collection database and Cochrane Library to conduct a literature review on upadacitinib treatment of adolescents with severe AA.
Results
In the present study, three adolescents with severe AA without atopic comorbidity received upadacitinib 15 mg/day orally for 6 consecutive months. Regrowth of eyebrows and pubic hair began in the first and second months of treatment, respectively, and head hair regrowth was obvious after approximately 4 months of treatment. The median Severity of Alopecia Tool score dropped to 36.33 (representing 58.08% reduction) after 6 consecutive months of treatment. Our literature search identified seven papers covering a total of 26 adolescents with AA (including ours) aged from 9 to 17 years who had received upadacitinib treatment, 15 mg/day being the most commonly prescribed dosage. Upadacitinib contributed to improvement in comorbidities such as atopic dermatitis, vitiligo, and Crohn’s disease. No severe adverse events were detected.
Conclusion
Upadacitinib is an effective and safe treatment option for adolescents with severe AA. Our study provides more data on adolescents with severe AA without atopic comorbidities.
Keywords: adolescent, alopecia areata, Janus kinase inhibitor, upadacitinib
Introduction
Alopecia areata (AA), a common, immune-mediated, non-scarring form of hair loss, affects up to 2% of the general population in all ethnic-, sex-, and age-based groups.1 Besides patchy loss of head hair, AA may cause varying degrees of eyebrow, eyelash, pubic, and axillary hair loss. It can also cause nail dystrophy in severe cases. Baricitinib, an inhibitor of Janus kinase (JAK)1/2, has been approved for treatment of severe AA in adults by the Food and Drug Administration (FDA).2 However, there is still scanty clinical evidence for the use of JAK inhibitors in adolescents and pediatric patients with severe AA. The impairment of quality of life, urgent need for treatment, frequent relapses, and safety issues around medication make the treatment of adolescent and pediatric AA patients challenging.
Upadacitinib, a JAK 1 inhibitor, has been used to treat atopic dermatitis in patients aged 12 and older,3 despite its efficacy and safety in adolescents and pediatric patients with severe AA being unclear. Here, we report a case series of upadacitinib treatment of adolescents with severe AA and present the findings of a narrative review.
Materials and Methods
Adolescents with severe AA were recruited from the outpatient department of dermatology and completed 6 months of follow-up between November 2023 and December 2024. The inclusion criteria were as follows: (1) age from 12 to 17 years at the time of enrollment; (2) severe AA as evidenced by Severity of Alopecia Tool (SALT) scores over 50; (3) AA refractory to more than 3 consecutive months of at least one type of systemic therapy, including corticosteroids and immunosuppressive agents; and (4) patient strongly desirous of further treatment. The exclusion criteria were as follows: (1) contraindications to upadacitinib such as severe infection, tumor, thrombosis, and allergy; (2) aged less than 12 or over 18 years at the time of enrollment; (3) mild or moderate AA as evidenced by SALT scores less than 50; (4) no systemic therapy before initiation of a JAK inhibitor or treatment course less than 3 months; and (4) failure to attend for follow-up visits. All enrolled patients underwent full laboratory examinations, including antinuclear antibodies, thyroid function, plasma trace elements, immunoglobulin, ferritin, vitamin A, vitamin E, coagulation function, chest radiograph and screening for Hepatitis B virus and tuberculosis. All participants received upadacitinib 15 mg/day orally. SALT scores were used to assess AA severity and activity at monthly follow-ups. Adverse events were recorded during follow-up. The study was approved by the Ethics Committee of Peking Union Medical College Hospital, and complied with the Declaration of Helsinki.Written informed consent for publication was obtained from these patients’ legal guardians.
We conducted a literature review of the PubMed, Web of Science Core Collection database and Cochrane Library on 14 th August 11, 2025. The search terms were ((alopecia areata) OR (alopecia totalis) OR (alopecia universalis)) AND (upadacitinib) AND ((Adolescent) OR (children) OR (pediatric)). Articles that met the following inclusion criteria were analyzed: (1) pediatric and adolescent patients; (2) patients with AA received upadacitinib treatment; and (3) papers written in English. The exclusion criteria were as follows: (1) adult patients who did not meet the age criteria.; (2) incomplete data; and (3) patient with comorbidities that cannot receive upadacitinib treatment. Two authors independently screened and extracted information from the articles.
Results
Case Series
This is a retrospective case series. The cohort of this case series comprised three adolescents with severe AA (Table 1). Their mean age was 16.33 years (range 15–17 years) and the median duration of disease 6 years (range 1–10 years). None of the patients in this case series had a family history of AA or personal history of atopic dermatitis, allergic rhinitis, or asthma. The immunoglobulin E concentrations of these three patients were all within the normal range. All three had severe AA, with a median SALT score of 86.67 (range: 73–100, all > 50). Besides severe loss of head hair, two of the patients also had loss of eyebrow, eyelash, pubic, and axillary hair; additionally, one had nail dystrophy. The disease of all three patients was refractory to at least one type of systemic therapy, prior treatment having included systemic corticosteroids (n=3), cyclosporine (n=1), glycyrrhizin (n=1), traditional Chinese medicine (n=1), and topical minoxidil (n=3).
Table 1.
The Demographic Data and Clinical Response for Upadacitinib Treatment in This Case Series
| Patient | Age/Gender | Disease Duration | Location | Nail | Weight | Prior Treatment | Upadacitinib Treatment | Initial SALT Score | Response | Adverse Events |
:---: :---: :---: :---: :---:
| P1 | 15y/F | 1y | Hair loss | Nail dystrophy | 53kg | Systemic corticosteroids, cyclosporine, topical minoxidil | 15mg per day | SALT score 87 | Hair regrowth after 4 months’ treatment, SALT score 15 after 6 months | Acne |
| P2 | 17y/F | 7y | Hair, eyebrow, eyelash, pubic hair, and armpit hair loss | None | 48kg | Systemic corticosteroids, traditional Chinese medicine, topical minoxidil | 15mg per day | SALT score 73 | Eyebrow regeneration after 1month, pubic hair regeneration after 2 months, SALT score 37 after 6 months | Acne |
| P3 | 17y/M | 10y | Hair, eyebrow, eyelash, pubic hair, and armpit hair loss | None | 90kg | Systemic corticosteroids, glycyrrhizin, topical minoxidil | 15mg per day | SALT score 100 | Eyebrow regeneration after 1 month, SALT score 57 after 6 months | Acne |
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Abbreviations: y, year; F, female; M, male; kg, kilogram; SALT, severity of alopecia tool.
All three participants had received upadacitinib treatment at 15 mg/day for 6 consecutive months (Figure 1). Regrowth of eyebrow and pubic hair occurred earlier than regrowth of head hair, regrowth of eyebrows and pubic hair occurring in the first and second months after upadacitinib treatment, respectively, whereas obvious regeneration of head hair occurred approximately 4 months after initiating upadacitinib treatment. The median SALT score dropped to 36.33 (range: 15–57), representing a reduction of 58.08%, after 6 consecutive months of upadacitinib treatment (Figure 2). Additionally, our patient’s eyelashes regrew later than their head, eyebrow, and pubic hair. However, there was no improvement in nail dystrophy after upadacitinib treatment. All participants in this case series tolerated upadacitinib treatment well. Mild acne was the most common adverse event, developing in all three participants and resolving in response to topical fusidic acid. No moderate or severe adverse events were reported during follow-up. Upadacitinib treatment was continued after 6 months’ treatment.
Figure 1.
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The clinical manifestations for severe AA adolescent patients. (A) Patient 1 before treatment; (B) Patient 2 before treatment; (C) Patient 3 before treatment; (D) Patient 1 after 6 months of upadacitinib treatment; (E) Patient 2 after 4 months of upadacitinib treatment; (F) Patient 3 after 1 month of upadacitinib treatment.
Figure 2.
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SALT score improvement before and after upadacitinib treatment.
Literature Review
A total of seven papers were finally included after implementation of the inclusion and exclusion criteria.4–10 Reported details concerning upadacitinib for treatment of AA in adolescent patients are summarized in Table 2. Up to now, a total of 26 adolescents with AA, including the three in the present study, from five countries, comprising Canada, China, Korea, Italy, and Poland, have reportedly received upadacitinib treatment. The youngest reported patient with severe AA to receive upadacitinib treatment was 9 years old. This patient had significant regrowth of head hair after 3 months’ treatment and stopped upadacitinib treatment without relapse from the 5-month follow-up.9 The most commonly prescribed dosage of upadacitinib in adolescents with severe AA was 15 mg/day. Sporadic hair regeneration has been reported to occur as early as the second week after initiating upadacitinib treatment.9 However, the time of onset of hair regrowth in most reports is 4 to 6 weeks after initiating treatment4,6,8 and the longest period of treatment with upadacitinib is 12 months.5,10 Comorbidities in adolescents with severe AA receiving upadacitinib treatment have included atopic dermatitis (AD, n=10), vitiligo (n=1), thyroiditis (n=1), conjunctivitis (n=2), allergic rhinoconjunctivitis (n=1), Crohn’s disease (n=1) and eosinophilic gastroenteritis (n=1). Additionally, besides inducing significant regrowth of hair, upadacitinib treatment has contributed to improvements in AD, vitiligo, and Crohn’s disease. Our study provides more clinical data on response to upadacitinib treatment in adolescents with severe AA and without comorbidities. Overall, upadacitinib is a well-tolerated and safe treatment in adolescents with severe AA. Mild adverse events have been reported, including increased creatine phosphokinase (n=8), onset or exacerbation of acne (n=6), mild upper respiratory tract infection (n=1), and transient mild leukopenia (n=1). To our knowledge, severe adverse events have not been reported in adolescents with severe AA.
Table 2.
The Literature Review Concerning Upadacitinib for Treatment of AA In Adolescent Patients
| No. | Year | Authors | Country | Patient Number | Age/Gender | Disease Duration | Coexist Disease | Dosage | Response | Adverse Events |
--- --- --- --- ---
| 1 | 2022 | Bourkas AN et al4 | Canada | 1 | 14y/M | 13y | AD | NA | Improvement of AA and AD after 6 weeks. SALT score 0 after 5 months. | NA |
| 2 | 2023 | Kołcz K et al6 | Poland | 1 | 14y/F | 16m | AD | 15mg per day | Improvement of AA and AD after 4 weeks. SALT score 0 after 3 months. | Transient mild leukopenia |
| 3 | 2023 | Yu D et al9 | China | 1 | 9y/F | 7y | AD | 15mg per day | Sporadic hairs after 2 weeks. SALT score 9 after 5 months. AD complete remission. | None |
| 4 | 2024 | Ha GU et al5 | Korea | 1 | 15y/F | 11y | AD | 15mg per day | Improvement in the eyebrows after 2 months. SALT score 11.7 after 12 months. | None |
| 5 | 2024 | Mu Y et al7 | China | 1 | 9y/F | 2m | Vitiligo | 15mg per day | Recovered 70% of the white plaque lesion area and hair regrowth with white hair after 7 months. | Elevated level of creatine kinase |
| 6 | 2024 | Picone V et al8 | Italy | 15 | 14.6y/7M8F | 29.3m | AD (3), thyroiditis (1), conjunctivitis (2) | 15mg per day | pSALT score reduced as early as 4 weeks. 60% patients responded well. pSALT50, pSALT75 and pSALT90 responses of 100%, 67% and 44% patients in 10 months. | Temporarily increased creatine phosphokinase (5), acne onset or exacerbation (3), mild upper respiratory tract infection (1) |
| 7 | 2025 | Battilotti C et al10 | Italy | 3 | 13y/3M | 10.7m | AD (3), allergic rhinoconjunc- tivitis (1), CD (1), EGE (1) | 15mg per day | Improvement of AA, AD and CD. Partial regrowth of eyebrow and eyelashes after 1 month. Median SALT score dropped to 5.27 after 12 months. | Mild transient CPK elevation (2) |
| 8 | 2025 | Our study | China | 3 | 16.33y/1M2F | 6y | None | 15mg per day | Regrowth of eyebrows and pubic hair appearing in the first and second months. Median SALT score dropped to 36.33 after 6 months. | Acne (3) |
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Abbreviations: F, female; M, male; y, year; m, month; SALT, severity of alopecia tool; NA, not available; AD, atopic dermatitis; CD, Crohn’s disease; EGE, eosinophilic gastroenteritis.
Discussion
A common autoimmune, non-scarring form of hair loss, AA has a slightly higher prevalence in pediatric and adolescent patients than in adults. It has been estimated that pediatric and adolescent patients with AA are more susceptible to autoimmune and metabolic disorders than are pediatric and adolescent individuals in general.11 AD is one of the most common comorbidities in pediatric and adolescent patients with AA.11 A systematic review found that topical corticosteroids and contact immunotherapy are the first and second commonly preferred treatments for pediatric and adolescent patients with AA.12 However, the management of refractory, severe, or frequently relapsing AA in adolescent is sometimes complex.
The inhibitory effect of Janus kinase–signal transducers and activators of transcription signaling pathways is responsible for the favorable efficacy of JAK inhibitors in the treatment of both severe AA and AD. Up to now, there has been limited clinical evidence for the efficacy and safety of JAK inhibitors in adolescent and pediatric AA patients with and without comorbidities such as AD. Atopic diathesis, atopic comorbidities, and high serum concentrations of immunoglobulin E have been identified as risk factors for developing AA, especially coexisting AD. The genetic polymorphism of interleukins 13 and 4 and infiltration of mast cells and eosinophils in skin biopsies of patients with severe AA, and favorable responses to antihistamine and dupilumab therapy suggest that the mechanisms of Th1 and concomitant Th2 involved in AA.13 Our literature review showed that most reported adolescents with severe AA who had received upadacitinib treatment had concomitant AD. In one patient, upadacitinib treatment induced both hair regrowth and improvement in symptoms of AD, as well as relief of vitiligo. Another patient showed relief of symptoms of Crohn’s disease after upadacitinib treatment. The three patients in our study did not have a history of atopic diseases or atopic diathesis and had normal serum concentrations of immunoglobulin E, providing clinical evidence for the therapeutic efficacy of upadacitinib in adolescents with severe AA without AD. All three of our participants, none of whom had atopic comorbidity, responded well to upadacitinib treatment. In these patients, regrowth of eyebrow and pubic hair occurred earlier than did regrowth of head hair, the latter becoming obvious approximately 4 months after initiating treatment. Upadacitinib at 15 mg/day was the most commonly used treatment in reported adolescents with severe AA. Notably, the SALT score of our Patient 3 was still over 50 after 6 months of treatment. We gave further consideration to increasing this patient’s upadacitinib dosage because he weighed 90 kgs. Although one reported 9-year-old patient had significant regrowth of head hair after 3 months’ treatment and stopped upadacitinib treatment without relapse at the 5-month follow-up,9 our patients needed to continue on upadacitinib treatment for more than 6 months.
A JAK 1 inhibitor, upadacitinib downregulates gamma-interferon signaling by inhibiting JAK activity, thereby preventing breakdown of hair follicles via immune mechanisms. Upadacitinib reportedly induced rapid regrowth of hair and significant improvement in quality of life in a retrospective study of 25 adults with AA.14 The FDA has approved upadacitinib for patients with atopic dermatitis aged 12 and older and provided more clinical and safety evidence to support its application in adolescents with severe AA aged from 12 to 17 years. One study has reported adverse musculoskeletal and skin events in male patients receiving upadacitinib treatment, whereas musculoskeletal issues, infections, and abnormal laboratory tests were prevalent and severe in female patients receiving upadacitinib treatment.15 According to our research and literature review, adverse reactions are mild in adolescents with severe AA receiving upadacitinib treatment. Increased creatine phosphokinase was the most common adverse event recorded in adolescents with severe AA treated with Upadacitinib, followed by onset or exacerbation of acne. Additionally, the adverse reactions did not differ significantly between sexes.
Two Phase III trials over 52 weeks have found that baricitinib, an inhibitor of JAK 1/2, is an effective treatment for AA.16 The FDA has approved the use of baricitinib in adults with severe AA. There are also case reports supporting the off-label use of baricitinib for treating adolescents with severe AA. Compared with baricitinib, upadacitinib has the following advantages in treatment of adolescents with severe AA. First, upadacitinib more strongly inhibits the effect of JAK1/STAT on transcription than does baricitinib.17 In some studies, upadacitinib has been used as conversion therapy for insufficiently effective responses to baricitinib treatment, indicating the significance of JAK 1 inhibition in management of AA.18,19 Second, upadacitinib has been approved by the FDA for treatment of atopic dermatitis in children aged 12 and older, and it has been demonstrated that it is safer than baricitinib in adolescents aged 12 to 17 years. Third, upadacitinib may have better therapeutic effects on comorbidities of AA. AD has been identified as a risk factor for developing AA and is one of the most common comorbidities in pediatric and adolescent patients with AA. Although various JAK inhibitors have shown some efficacy in the management of AD, a network meta-analysis concluded that upadacitinib is the optimal option according to short-term studies.20 Further research is needed on the efficacy and safety of baricitinib and upadacitinib in the treatment of adolescents with severe AA.
This study has several limitations. First, as a retrospective case series, future randomized controlled trials with larger sample sizes are warranted to confirm the efficacy and safety of upadacitinib in adolescents with severe AA. Second, while the follow-up period exceeded 6 months, longer-term evaluation is required to comprehensively assess the treatment’s sustained efficacy. Moreover, systematic assessment of disease recurrence following treatment discontinuation should be conducted in subsequent studies.
Conclusion
In conclusion, in the present study, we found that upadacitinib at 15 mg/day is an effective and safe treatment option for adolescents with severe AA, both in those with concomitant AD and vitiligo, but also in those without comorbidities. However, further long-term follow-up is warranted.
Acknowledgments
The study was approved by the Ethics Committee of Peking Union Medical College Hospital, and complied with the Declaration of Helsinki. Written informed consent for publication was obtained from these patients’ legal guardians.
Funding Statement
Beijing Key Clinical Specialty Construction Project; National Key Clinical Specialty Project of China.
Disclosure
The authors report no conflicts of interest in this work.
References
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Home>Analytics>Performing Nonlinear Least Square and Nonlinear Regressions in R
Performing Nonlinear Least Square and Nonlinear Regressions in R
Analytics | August 18, 2018
Linear regression is a basic tool. It works on the assumption that there exists a linear relationship between the dependent and independent variable, also known as the explanatory variables and output. However, not all problems have such a linear relationship. In fact, many of the problems we see today are nonlinear in nature. A very basic example is our own decision making process which involves deciding an outcome based on various questions. For example, when we decide to have dinner, our thought process is not linear. It is based a combination of our tastes, our budget, our past experiences with a restaurant, alternatives available, weather conditions etc. There can be other simple nonlinear cases such as quadratic or exponential dependencies which are not too difficult to imagine. This is how non-linear regression came into practice – a powerful alternative to linear regression for nonlinear situations. Similar to linear regression, nonlinear regression draws a line through the set of available data points in such a way that the line fits to the data with the only difference that the line is not a straight line or in other words, not linear.
Non-linear Regression – An Illustration
InR, we have lm() function for linear regression while nonlinear regression is supported by nls() function which is an abbreviation for nonlinear least squares function. To apply nonlinear regression, it is very important to know the relationship between the variables. Looking at the data, one should be able to determine the generalized equation of the model which will fit the data. This model is then specified as the ‘formula’ parameter in nls() function. The function then determines the coefficients of the parameters in the model. Let’s try linear and nonlinear regression models on an exponential data. I will use the runif() function to generate an exponential set of values for y. Here I will use x as a sequence from 0 to 100.
I will also use a set.seed() function so that the values are regenerated for you.
#set a seed value
set.seed(23)
#Generate x as 100 integers using seq function
x<-seq(0,100,1)
#Generate y as ae^(bx)+c
y<-runif(1,0,20)exp(runif(1,0.005,0.075)x)+runif(101,0,5)
#How does our data look like? Lets plot it
plot(x,y)
This seems a fairly smooth non-linear plot. To illustrate the difference between linear and nonlinear models, let’s fit them both:
#Linear model
lin_mod=lm(y~x)
#Plotting the model
plot(x,y)
abline(lin_mod)
There is little overlap between the actual values and the fitted plot. Now let’s try the nonlinear model and specify the formula
nonlin_mod=nls(y~aexp(bx),start=list(a=13,b=0.1)) #a is the starting value and b is the exponential start
#This new plot can be made by using the lines() function
plot(x,y)
lines(x,predict(nonlin_mod),col=”red”)
This is a much better fit and clearly passes through most of the data. For more clarity, we will now calculate the errors for both the models
#Error calculation
error <- lin_mod$residuals
lm_error <- sqrt(mean(error^2)) #5.960544
error2=y-predict(nonlin_mod)
nlm_error <- sqrt(mean(error2^2)) #1.527064
The linear model has more than twice the error than that of nonlinear one. This shows that the nonlinear model fits better for nonlinear data.
Understanding the nls() function
There are a few parameters that the nls() function requires. I used two parameters to define the model in the above illustration – the formula and the start parameters. Nonlinear function requires us to look at the data first and estimate the model to fit in. This estimated model is specified as the formula parameter. We can also specify the coefficients as variables to be estimated. The next step involves specifying the start parameter. This parameter specifies the starting values of the coefficients we used in the formula. Here we have ‘a’ and ‘b’ as the coefficients. I took ‘a’ as the nearest integer to minimum value of y (which is approximately 13.19) and ‘b’ as the increment for the exponent. Using these values, the nls() function determines the optimal values of ‘a’ and ‘b’. It is very important to set the right starting parameter values otherwise the model may give us absurd results or even fail. Let’s see what are the estimated values of ‘a’ and ‘b’ for this dataset:
nonlin_mod
Nonlinear regression model
model: y ~ a exp(b x)
data: parent.frame()
a b
13.60391 0.01911
residual sum-of-squares: 235.5
Number of iterations to convergence: 15
Achieved convergence tolerance: 4.975e-07
The values for ‘a’ and ‘b’ estimated for this model are 13.60391 and 0.01911 respectively which are very close to those we provided as starting values. This shows that that the model estimated by the nls() function is y=13.60391e^(0.01911x). Further, the estimated values of ‘a’ and ‘b’ are very close to the starting values we provided. These results will remain the same if we keep ‘b’ as 0.01 or even 0.001 or keep ‘a’ as 10 or 100 or 1000. As long as the model is able to converge at the optimal estimation, some approximation is admissible. However, if the values of ‘a’ and ‘b’ are completely out of range, say 1 and 1, we get an error as the model fails. The right set of starting values need to be estimated by looking at the data before implementing the model.
Self-Starting Functions
The problem arises when one is beginning with nonlinear functions and does not know what value should be estimated for the parameters. To illustrate this problem, I will now use a non-linear dataset available in R. The Puromycin data shows the concentration and reaction rate for enzymatic reaction of Puromycin antibiotic. I will plot the data to understand the data and estimate the formula equation
attach(Puromycin)
plot(Puromycin$conc,Puromycin$rate)
This data is specific to biological reactions and can be estimated using the famous enzyme kinetics equation known as the Michaelis-Menten equation. For this, we will separate the dataset based on whether the state is “treated” or “untreated” and define a function for the equation
#Define a function to apply Michaelis-Menten equation
mm=function(conc,vmax,k) vmaxconc/(k+conc)
#Use the nls data over the first subset of treated data. I will set the starting values as 50 and 0.05
mm1=nls(rate~mm(conc,vmax,k),data=Puromycin,start=c(vmax=50,k=0.05),subset=state==”treated”)
#Use a similar function for the second subset of untreated data
mm2=nls(rate~mm(conc,vmax,k),data=Puromycin,start=c(vmax=50,k=0.05),subset=state==”untreated”)
Both the models, mm1 and mm2 make good estimations of the data and fit the model. However, it is hard to estimate the starting values looking at the plot of Puromycin conc. vs rate. The Puromycin concentration vs rate plot suggested that the minimum conc. on the x- axis is around 0.01 and the maximum rate (vmax) on the y-axis is around 200 yet I purposely used values which are very different from these estimations so that the model will fit while converging slowly. In this case, I am taking a risk on the estimation ability of the model. This is where “self-starting” functions come into the picture. As the name suggests, a self-starting function does not need a starting value for the parameters and do estimate themselves. We can rewrite the above two functions using the SSmicmen function which is a self starting function for Michaelis-Menten equation. The new models are:
mm3=nls(rate~SSmicmen(conc,vmax,k),data=Puromycin,subset=state==”treated”)
mm4=nls(rate~SSmicmen(conc,vmax,k),data=Puromycin,subset=state==”untreated”)
Let us compare the corresponding models by calling the model variables in R. We will first look at the models with state=”treated” which are mm1 and mm3 and compare the vmax and k values. We will then compare the models with state=”untreated” which are mm2 and mm4.:
#Print the model summary and estimated parameters for mm1
mm1
Nonlinear regression model
model: rate ~ mm(conc, vmax, k)
data: Puromycin
vmax k
212.68369 0.06412
residual sum-of-squares: 1195
Number of iterations to convergence: 7
Achieved convergence tolerance: 2.703e-06
#Print the model summary and estimated parameters for mm3
mm3
Nonlinear regression model
model: rate ~ SSmicmen(conc, vmax, k)
data: Puromycin
vmax k
212.68371 0.06412
residual sum-of-squares: 1195
Number of iterations to convergence: 0
Achieved convergence tolerance: 1.937e-06
#Print the model summary and estimated parameters for mm2
mm2
Nonlinear regression model
model: rate ~ mm(conc, vmax, k)
data: Puromycin
vmax k
160.28001 0.04771
residual sum-of-squares: 859.6
Number of iterations to convergence: 7
Achieved convergence tolerance: 2.039e-06
#Print the model summary and estimated parameters for mm4
mm4
Nonlinear regression model
model: rate ~ SSmicmen(conc, vmax, k)
data: Puromycin
vmax k
160.28012 0.04771
residual sum-of-squares: 859.6
Number of iterations to convergence: 5
Achieved convergence tolerance: 3.942e-06
The corresponding models have estimated the same coefficients up to the third decimal. This shows that self-starting functions fairly well in place of functions where I need to define the start parameters. The big limitation of estimating the starting parameters can be avoided using the self-starting functions. R has many self-starting functions available. A list of the same can be obtained by using the apropos function:
apropos(“^SS”)
“SSasymp” “SSasympOff” “SSasympOrig” “SSbiexp” “SSD”
“SSfol” “SSfpl” “SSgompertz” “SSlogis” “SSmicmen”
“SSweibull”
With the exception of the SSD function, there are 10 self-starting functions here in R. The final step in the model. I have given a brief description of what all these functions are defined for (in alphabetical order)
SSasymp asymptotic regression models
SSasympOff asymptotic regression models with an offset
SSasympOrig asymptotic regression models through the origin
SSbiexp biexponential models
SSfol first-order compartment models
SSfpl four-parameter logistic models
SSgompertz Gompertz growth models
SSlogis logistic models
SSmicmen Michaelis–Menten models
SSweibull Weibull growth curve models
Goodness of Fit
As an additional verification step, I will also check the goodness of fit of the model. This can be done by looking that the correlation between the values predicted by the model and the actual y values.
#Goodness of fit for first nonlinear function
cor(y,predict(nonlin_mod)) #0.9976462
#Goodness of fit for treated values of Puromycin function
cor(subset(Puromycin$rate,state==”treated”),predict(mm3)) #0.9817072
cor(subset(Puromycin$rate,state==”treated”),predict(mm1)) #0.9817072
#Goodness of fit for untreated values of Puromycin function
cor(subset(Puromycin$rate,state==”untreated”),predict(mm2)) #0.9699776
cor(subset(Puromycin$rate,state==”untreated”),predict(mm4)) #0.9699777
All our models have a high correlation value which indicates that the values are very close to each other and accurate. The corresponding model summary and estimation parameters also show the same observation.
Summary
Regression is a fundamental technique to estimate the relationships among variables and nonlinear regression is a handy technique if that relationship is nonlinear. It is similar to linear regression and provides a powerful method to fit a nonlinear curve based on the estimated formula while minimizing the error using nonlinear least squares method. There are a variety of other nonlinear models available such as SVM and Decision trees. Nonlinear regression is a robust technique over such models because it provides a parametric equation to explain the data. As the models becomes complex, nonlinear regression becomes less accurate over the data. This article gives an overview of the basics of nonlinear regression and understand the concepts by application of the concepts in R. Here is the complete R code used in the article.
set a seed value
set.seed(23)
Generate x as 100 integers using seq function
x<-seq(0,100,1)
Generate y as ae^(bx)+c
y<-runif(1,0,20)exp(runif(1,0.005,0.075)x)+runif(101,0,5)
How does our data look like? Lets plot it
plot(x,y)
Linear model
lin_mod=lm(y~x)
Plotting the model
plot(x,y)
abline(lin_mod)
nonlin_mod=nls(y~aexp(bx),start=list(a=13,b=0.1)) #a is the starting value and b is the exponential start
This new plot can be made by using the lines() function
plot(x,y)
lines(x,predict(nonlin_mod),col=”red”)
Error calculation
error <- lin_mod$residuals
lm_error <- sqrt(mean(error^2)) #5.960544
error2=y-predict(nonlin_mod)
nlm_error <- sqrt(mean(error2^2)) #1.527064
nonlin_mod
attach(Puromycin)
plot(Puromycin$conc,Puromycin$rate)
Define a function to apply Michaelis-Menten equation
mm=function(conc,vmax,k) vmaxconc/(k+conc)
Use the nls data over the first subset of treated data. I will set the starting values as 50 and 0.05
mm1=nls(rate~mm(conc,vmax,k),data=Puromycin,start=c(vmax=50,k=0.05),subset=state==”treated”)
Use a similar function for the second subset of untreated data
mm2=nls(rate~mm(conc,vmax,k),data=Puromycin,start=c(vmax=50,k=0.05),subset=state==”untreated”)
mm3=nls(rate~SSmicmen(conc,vmax,k),data=Puromycin,subset=state==”treated”)
mm4=nls(rate~SSmicmen(conc,vmax,k),data=Puromycin,subset=state==”untreated”)
Print the model summary and estimated parameters for mm1
mm1
Print the model summary and estimated parameters for mm3
mm3
Print the model summary and estimated parameters for mm2
mm2
Print the model summary and estimated parameters for mm4
mm4
Print the names of all functions in R which start with SS
apropos(“^SS”)
Goodness of fit for first nonlinear function
cor(y,predict(nonlin_mod)) #0.9976462
Goodness of fit for treated values of Puromycin function
cor(subset(Puromycin$rate,state==”treated”),predict(mm3)) #0.9817072
cor(subset(Puromycin$rate,state==”treated”),predict(mm1)) #0.9817072
Goodness of fit for untreated values of Puromycin function
cor(subset(Puromycin$rate,state==”untreated”),predict(mm2)) #0.9699776
cor(subset(Puromycin$rate,state==”untreated”),predict(mm4)) #0.9699777
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14417 | https://precalculusstudyguide2014.weebly.com/82-arithmetic-sequences-and-partial-sums.html | | |
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| Precalculus study guide | Home Unit 1 + 4.7 Inverse Trigonometric Functions + 5.3 Solving Trigonometric Equations + 5.4 Sum and Difference Formulas + 5.5 Multiple-Angle and Product-to-Sum Formulas + 6.1 Law of Sines + 6.2 Law of Consines Unit 2 + 8.1 Sequences and Series + 8.2 Arithmetic Sequences and Partial Sums + 8.3 Geometric Sequences and Series + 8.4 The Binomial Theorem Unit 3 + 9.1 Circles and Parabolas + 9.2 Ellipses + 9.3 Hyperbolas and Rotation of Conics + 9.4 Parametric Equations + 9.5 Polar Coordinates Unit 4 + 6.3 Vectors in the Plane + 6.4 Vectors and Dot Products + 6.5 Trigonometric Form of a Complex Number + 7.5 Operations with Matrices + 7.6 The Inverse of a Square Matrix + 7.7 The Determinant of a Square Matrix + 10.1 The Three- Dimensional Coordinate System + 10.2 Vectors in Space + 10.3 The Cross Product of Two Vectors Unit 5 + 11.1 Introduction to Limits + 11.2 Techniques for Evaluating Limits + 11.3 The Tangent Line Problem |
8.2 Arithmetic Sequences and Partial Sums
A sequence whose consecutive terms have a common difference is called an arithmetic sequence.a1, a2, a3, a4, ..., an, ...is arithmetic when there is a number d such thata2 - a1 = a3 - a2 = a4 - a3 = ⋅⋅⋅ = d.The number d is the common difference of the sequence.
Examples of Arithmetic Sequences
| | |
--- |
| Write the first five terms of the sequence. Determine whether or not the sequence is arithmetic. If it is, find the common difference. | |
The sequence 1, 4, 9, 16 ..., whose nth term is n², is not arithmetic. The difference between the first two terms is 3, but the difference between the second and third terms is 5.The nth term of an arithmetic sequence can be derived from the pattern.a1 = a1a2 = a1 + da3 = a1 + 2da4 = a1 + 3da5 = a1 + 4dan = a1 + (n -1)dThe nth term of an arithmetic sequence has the forman = a1 + (n - 1)dd is the common difference between consecutive terms of the sequence and a1 is the first term of the sequence.
Finding the nth Term of an Arithmetic Sequence
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| Find a formula for an for the arithmetic sequence.a1 = 5, a4 = 15 | |
Writing the Terms of an Arithmetic Sequence
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| Write the first five terms of the arithmetic sequence.a3 = 19, a15 = -1.7 | |
The Sum of a Finite Arithmetic SequenceThere is a simple formula for the sum of a finite arithmetic sequence.The sum of a finite arithmetic sequence with n terms is given bySn = (n/2)(a1 + an)This formula works only for arithmetic sequences.
Finding the Sum of a Finite Arithmetic Sequence
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| Find the sum of the finite arithmetic sequence.Sum of the integers from -100 to 30. | |
The sum of the first n terms of an infinite sequence is called the nth partial sum. The nth partial sum of an arithmetic sequence can be found by using the sum of a finite arithmetic sequence.
Finding a Partial Sum of an Arithmetic Sequence
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| Find the indicated nth partial sum of the arithmetic sequence.a1 = 100, a25 = 220, n = 25 | |
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| A hardware store makes a profit of $30,000 during its first year. The store owner sets a goal of increasing profits by $5000 each year for 4 years. Assuming that this goal is met, find the total profit during the first 5 years. | |
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14418 | https://www.freemathhelp.com/forum/threads/finding-the-apothem-area-and-perimeter-of-regular-polygon.51704/ | finding the apothem, area, and perimeter of regular polygon | Free Math Help Forum
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finding the apothem, area, and perimeter of regular polygon
Thread starter472flj
Start dateJun 2, 2007
4
472flj
New member
Joined Jun 2, 2007 Messages 2
Jun 2, 2007
#1
okay so the problem gives me a regular hexagon with the radii measuring 8 units.
i know that the formula to find the area of a regular polygon is area=1/2(apothme)(perimeter)
my question is how do i find the apothem so i acn solve for the area then for the perimeter and also how do i find the perimeter of it once i know the area.
S
soroban
Elite Member
Joined Jan 28, 2005 Messages 5,586
Jun 2, 2007
#2
Re: finding the apothem, area, and perimeter of regular poly
Hello, 472flj!
Given: regular hexagon with radius 8 units.
Find the area of the hexagon.
Click to expand...
Make a sketch . . .
You'll find that the regular hexgon is comprised of six equilateral triangles,
. . each with side 8.
The apothem is the altitude of the triangle.
So we have:
Code:
/|\
/ | \
8 / |a \
/ | \
/ | \
- - + - -
4
Use Pythagorus and solve for (\displaystyle a).
Got it?
T
TchrWill
Full Member
Joined Jul 7, 2005 Messages 856
Jun 2, 2007
#3
Re: finding the apothem, area, and perimeter of regular poly
472flj said:
okay so the problem gives me a regular hexagon with the radii measuring 8 units.
i know that the formula to find the area of a regular polygon is area=1/2(apothme)(perimeter)
my question is how do i find the apothem so i acn solve for the area then for the perimeter and also how do i find the perimeter of it once i know the area.
Click to expand...
Polygons
A polygon is a plane figure with three or more line segments and angles that are joined end to end so as to completely enclose an area without any of the line segments intersecting.
A convex polygon is one where the line segments joining any two points of the polygon remain totally inside the polygon, each interior angle being less than 180º.
A concave polygon is one where one or more line segments joining any two points of the polygon are outside of the polygon and one or more of the interior angles is greater than 180º. The inward pointing angle of a concave polygon is referred to as a reentrant angle. The angles less than 180º are called salient angles.
A regular polygon is one where all the sides have the same length and all the interior angles are equal.
A diagonal is a straight line connecting any two opposite vertices of the polygon.
Polygons are classified by the number of sides they have.
No. of sides.........Polygon Name
......3.....................Triangle
......4..................Quadrilateral
......5....................Pentogon
......6....................Hexagon
......7....................Heptagon
.....8......................Octagon
.....9......................Nonagon
....10.....................Decagon
....11....................Undecagon
....12....................Dodecagon
....13....................Tridecagon
....14....................Tetradecagon
....15....................Pentadecagon
......n........................n-gon
Regular Polygon Terminology
n = the number of sides
v = angle subtended at the center by one side = 360/n
s = the length of one side = R[2sin(v/2)] = r[2tan(v/2)]
R = the radius of the circumscribed circle = s[csc(v/2]/2 = r[sec(v/2)]
r = the radius of the inscribed circle = R[cos(v/2)] = s[cot(v/2)]/2
a = apothem = the perpendicular distance from the center to a side (the radius of the inscribed circle)
p = the perimeter = ns
Area = s^2[ncot(v/2)]/4 = R^2[nsin(v)]/2 = r^2[ntan(v/2)]
The formula for the area of a regular polygon is also A = (1/2 )ap = (1/2)ans, where a is the apothem, p is the perimeter, s is the side length and n is the number of sides..
The sum of all the interior angles in a polygon is 180(n - 2)
The sum of the exterior angles in a polygon is 360º.
The internal angle between two adjacent sides of a regular polygon is given by 180(n - 2)/n
The external angle between any side and the extended adjacent side of a regular polygon is given by 360/n.
You might be interested in why the sum of all the interior angles of a polygon is 180(n - 2).
Consider first the square, rectangle and trapazoid. Draw one ofthe diagonals in each of these figures.
What is created is two triangles within each figure.
The sum of the interior angles of any triangle is 180 deg.
Therefore, the sum of the interior angles of each of these 4 sided figues is 360 Deg.
Now consider a pentagon with 5 sides that can be divided up into 3 triangles.
Therefore, the sum of the interior angles of a pentagon is 540 Deg.
What about a hexagon. I tink you will soonsee that the sum of the interior angles is 720 Deg.
Do you notice anything?
n = number of sides........3........4........5........6
Sum of Int. Angles.........180....360....540....720
The sum of the interior angles is representable by 180(n - 2).
Consider also the sum of the exterior angles.
Each exterior angle is 180 - 180(n - 2)/n = (180 - 180n + 360)/n = 360/n.
Therefore, the sum of the exterior angles is 360n/n or 360 Deg.
4
472flj
New member
Joined Jun 2, 2007 Messages 2
Jun 3, 2007
#4
is there any possibility you can make this more of a step by step help please im still a bit confused
stapel
Super Moderator
Staff member
Joined Feb 4, 2004 Messages 16,582
Jun 3, 2007
#5
472flj said:
is there any possibility you can make this more of a step by step help please im still a bit confused
Click to expand...
You've been given a description (that is, an explanation) of the reasoning and a picture of the appropriate triangle, and have been told the formula to use and the values to plug into that formula. Please clarify which part of the Pythagorean Theorem you are needing broken down into further steps.
When you reply, please show all of your work, so we can see where you are bogging down. Thank you!
Eliz.
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14419 | https://blogformathematics.blogspot.com/2010/12/calculation-of-permutations-part-6-c.html | Loading [MathJax]/jax/output/HTML-CSS/fonts/TeX/fontdata.js
Calculation of permutations in a circular seating arrangement
In this post, we will study how the number of circular permutations from a different perspective. In the previous post, we studied circular permutations as the number of different arrangements that can be made of people sitting on a round table. In this post, we will learn how to calculate the number of different seating arrangements possible on a round table when a person does not have the same two neighbors.
There are five people A, B, C, D and E. They are seated on a table in two ways:
Both of the above arrangements are different, but in both the arrangements, each person has the same two neighbors. In the second arrangement, each person's neighbors on the left become their neighbors on the right and the neighbors at the right become the neighbors at the left. All the arrangements in which a person has the same two neighbors were included in the calculations shown in the previous post.
Here we will learn to calculate the number of circular permutations on the condition that each person should not have the same two neighbors.
Since for every circular permutation, there are two circular permutations in which a person will have the same two neighbors (as explained in the above diagram), therefore to obtain the total number of circular permutations in which no person has the same two neighbors, we will divide the total number of circular permutations by 2. Therefore,
Number of arrangement (permutations) in which no person has the same two neighbors
= (n - 1)! / 2
The above formula is also used to calculate the total number of different necklaces that can be formed from all different beads. This is because if we turn a necklace over, the arrangements of the beads becomes anticlockwise, and therefore there is no distinction between a clockwise and an anti clock wise direction in a necklace.
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14420 | https://pmc.ncbi.nlm.nih.gov/articles/PMC10167155/ | Renal Disease and Systemic Sclerosis: an Update on Scleroderma Renal Crisis - PMC
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. 2022 Jun 1;64(3):378–391. doi: 10.1007/s12016-022-08945-x
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Renal Disease and Systemic Sclerosis: an Update on Scleroderma Renal Crisis
Alice Cole
Alice Cole
1 UCL Centre for Rheumatology and Connective Tissue Diseases, Royal Free Campus, Rowland Hill Street, London, NW3 2PF UK
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1, Voon H Ong
Voon H Ong
1 UCL Centre for Rheumatology and Connective Tissue Diseases, Royal Free Campus, Rowland Hill Street, London, NW3 2PF UK
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1, Christopher P Denton
Christopher P Denton
1 UCL Centre for Rheumatology and Connective Tissue Diseases, Royal Free Campus, Rowland Hill Street, London, NW3 2PF UK
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1 UCL Centre for Rheumatology and Connective Tissue Diseases, Royal Free Campus, Rowland Hill Street, London, NW3 2PF UK
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Corresponding author.
Accepted 2022 May 11; Issue date 2023.
© The Author(s) 2022
Open AccessThis article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit
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PMCID: PMC10167155 PMID: 35648373
Abstract
Scleroderma renal crisis (SRC) is a life-threatening complication of systemic sclerosis (SSc) with a mortality of 20% at 6 months. Once the leading cause of mortality in scleroderma (SSc), it remains a serious complication, often necessitating level three care for patients affected. Whilst renal outcomes have significantly improved following the advent of angiotensin-converting enzyme inhibitor (ACEi) therapy, SRC remains a precarious challenge for clinicians, due to lack of preventative measures and the fact that patients can rapidly decline despite best medical management. Large cohort studies spanning decades have allowed clear identification of phenotypes particularly at risk of developing SRC thus allowing enhanced monitoring and early identification in those individuals. Novel urinary biomarkers for renal disease in SSc may offer a new window for early identification of SRC patients and response to treatment. Multiple studies have demonstrated increased activity of complement pathways in SRC with some anecdotal cases exhibiting serological response to treatment with eculizumab where ACEi and therapeutic plasma exchange (TPE) were not successful. Endothelin-1 blockade, a therapeutic strategy in other SSc vasculopathies, has shown potential as a target but clinical trials are yet to show a clear treatment benefit.Clear guidelines for the management of SRC are in place to standardise care and facilitate early collaboration between rheumatology and renal physicians. Outcomes following renal transplant have improved but the mortality of SRC remains high, indicating the need for continued exploration of the mechanisms precipitating and exacerbating SRC in order to develop novel therapies.
Keywords: Systemic sclerosis, Scleroderma renal crisis, Thrombotic microangiopathy, Complement, Acute kidney injury
Introduction
Scleroderma renal crisis (SRC) is a life-threatening complication of systemic sclerosis (SSc) with a mortality of 20% at 6 months. Once the leading cause of mortality in Scleroderma (SSc), it remains a serious complication, often necessitating level three care for patients affected. Whilst renal outcomes have significantly improved following the advent of angiotensin-converting enzyme inhibitor (ACEi) therapy, SRC remains a precarious challenge for clinicians, due to lack of preventative measures and the fact that patients can rapidly decline despite best medical management. Large cohort studies spanning decades have allowed clear identification of phenotypes particularly at risk of developing SRC thus allowing enhanced monitoring and early identification in those individuals. Emerging data surrounding the pathophysiology of SRC has suggested encouraging targets such as endothelin-1 and upregulated complement pathways which may lead to novel changes in our management of this patient group.
Overview
The earliest description considered to represent SRC originated from Auspitz in 1863 who described the rapid death of a patient with thickened skin and uraemia . In the late 1930s, some of the histological hallmarks of SRC such as intimal hyperplasia of the renal vessels and fibrinoid degeneration in interlobular arteries were described. In one of these cases, the patient had been diagnosed with SSc and obliterative endarteritis of the kidney . The histological abnormalities observed in the kidneys were also described in SSc patients who did not suffer from SRC . The term ‘renal-crisis’ was coined by Moore and Sheehan in 1952 . Treatment of SRC has historically consisted of aggressive anti-hypertensive therapy using methyldopa or propranolol with dialysis and in some cases bilateral nephrectomy. Mortality significantly reduced with the introduction of ACEi in the 1980s; however, European League Against Rheumatism Scleroderma Trials and Research (EUSTAR) data has shown no other significant impact on mortality in the post-ACE era. The frequency of SRC does seem to be reducing which may be due to the more widespread use of vasodilator therapy to treat complications of SSc such as Raynaud’s phenomenon and pulmonary arterial hypertension (PAH) . It may also be related to the more judicious use of glucocorticoids in patient with SSc due to the recognition that steroids may provoke SRC . Recent studies have also demonstrated that outcome of renal transplant in SRC have improved and are now comparable to other causes of end-stage renal failure (ESRF) .
Classifying Renal Disease in Scleroderma
Separate disease entities exist within the kidney in SSc. SRC classically presents with accelerated hypertension and acute kidney injury (AKI) defined as an increase in serum creatinine > 1.5 × baseline. Whilst SRC should certainly be included in the differential for such a presentation, other causes to consider include anti-neutrophil cytoplasmic antibody (ANCA)–associated vasculitis, membranous nephritis, other primary causes of thrombotic microangiopathies (TMA) such as thrombotic thrombocytopenic purpura (TTP) or disseminated intravascular coagulopathy (DIC). Other non-immune-mediated causes such as renal artery stenosis may mimic the presentation of SRC. There are a proportion of SSc patients who have unexplained renal abnormalities such as proteinuria and the significance of this is not fully recognised.
In 2015, the Scleroderma Clinical Trials Consortium (SCTC) working group conducted a scoping review and a consensus study to produce classification criteria for SRC which could be widely used in research. The group have identified a core set of variables which define SRC, and these are currently being used on real-world patients as part of the International Scleroderma Renal Crisis Survey II (ISRCS II) to validate the set and provide data on specificity. The main parameters outlined are AKI, hypertension, microangiopathic haemolytic anaemia (MAHA) and thrombocytopenia, target organ dysfunction, and renal histopathology (Table 1 ).
Table 1.
Classification criteria for SRC as defined by the Scleroderma Clinical Trials Consortium (SCTC) working group
| Domain |
| Blood pressure |
| Acute increase in blood pressure defined as any of the following: |
| - Systolic blood pressure ≥ 140 mm Hg |
| - Diastolic blood pressure ≥ 90 mm Hg |
| - An increase in systolic blood pressure of ≥ 30 mm Hg above normal |
| - An increase in diastolic blood pressure of ≥ 20 mm Hg above normal |
| Blood pressure measurement should be taken twice, separated by at least 5 min; if blood pressure readings are discordant, repeat readings should be taken until 2 consistent readings are obtained |
| Kidney injury |
| AKI defined as any of the following: |
| - Increase in serum creatinine of ≥ 26.5 μmoles/l (≥ 0.3 mg/dl) within 48 h |
| - Increase in serum creatinine to ≥ 1.5 times baseline, which is known or presumed to have occurred within the prior 7 days |
| - Urine volume < 0.5 ml/kg/h for 6 h |
| MAHA and thrombocytopenia |
| New or worsening anaemia not due to other causes |
| Schistocytes or other red blood cell fragments on blood smear |
| Thrombocytopenia ≤ 100,000 platelets/mm 3, confirmed by manual smear |
| Laboratory evidence of haemolysis, including elevated lactate dehydrogenase, reticulocytosis, and/or low or absent haptoglobin |
| A negative direct antiglobulin test |
| Target organ dysfunction |
| Hypertensive retinopathy (haemorrhages, hard and soft [cottonwool] exudates, and/or disc oedema, not attributable to other causes), confirmed by an ophthalmologist |
| Hypertensive encephalopathy, characterized by headache, altered mental status, seizures, visual disturbances, and/or other focal or diffuse neurologic signs not attributable to other causes |
| Acute heart failure, characterized by typical symptoms (e.g., breathlessness, ankle swelling, and fatigue) that may be accompanied by signs (e.g., elevated jugular venous pressure, pulmonary crackles, and peripheral oedema) |
| Acute pericarditis, diagnosed with at least 2 of the 4 following criteria: (1) pericarditis chest pain, (2) pericardial rub, (3) new widespread ST segment elevation or PR segment depression on electrocardiography, and (4) pericardial effusion (new or worsening) on cardiac echocardiography |
| Renal histopathology |
| Histopathologic findings on kidney biopsy consistent with SRC, which may include the following: |
| - Small vessel (arcuate and interlobular arteries) changes that predominate over glomerular alterations |
| - Glomerular changes of thrombotic microangiopathy may be present, with acute changes including fibrin thrombi and endothelial swelling, red blood cell fragments, and mesangiolysis, and chronic changes including double contours of the glomerular basement membrane |
| - Nonspecific ischemic changes with corrugation of the glomerular basement membrane, and even segmental or global sclerosis of glomeruli may occur |
| - Early vascular abnormalities include intimal accumulation of myxoid material, thrombosis, fibrinoid necrosis, and fragmented red blood cells, sometimes resulting in cortical necrosis |
| - Narrowing and obliteration of the vascular lumen lead to glomerular ischemia. Juxtaglomerular apparatus hyperplasia, while relatively rare (10%), can be observed |
| - Late changes are manifested by intimal thickening and proliferation (which lead to characteristic vascular ‘onion-skin’ lesions), glomerulosclerosis, and interstitial fibrosis |
| - Nonspecific tubular changes may also occur, including acute tubular injury in the early stage of injury, and later interstitial fibrosis and tubular atrophy. Since none of these findings is specific for SRC, the pathologic diagnosis must be supported by appropriate clinical and serologic data |
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Epidemiology
Initial studies reported the prevalence of SRC in early diffuse disease be as high as 25%; however, a 2016 meta-analysis demonstrated that the frequency is now as low as 5% , with the US Prospective Registry in Systemic Sclerosis (PRESS) cohort, reporting a 10% frequency . Studies in the United Kingdom (UK) have shown frequency of 14% in diffuse patients, whereas the frequency of SRC amongst limited patient remained low at 3% . Studies carried out in Japan have found the lowest reported frequencies of SRC between 1 and 3% [12, 13]. The highest rates of SRC have been observed in the USA, UK, and Australia . The distribution of SRC between patient populations is likely to link to the heterogeneity between groups and the varying prevalence of autoantibodies associated with SRC such as anti RNA polymerase III, as seen in Fig.1 . Studies have shown that the highest rates of anti RNA polymerase III have been observed in North America (14%) , correlating with SRC prevalence.
Fig. 1.
Open in a new tab
Worldwide prevalence of anti-RNA polymerase III antibody according to French systematic review and meta-analysis.
Reproduced with permission from Sobanski et al.
The 2016 meta-analysis by Turk et al. reported no significant change in SRC prevalence temporally; however, there was a non-significant (p = 0.16) reduction in SRC frequency observed in the diffuse group when analysing cohort by year. This non-significant finding would match with the general expert consensus that cases of SRC do appear to have fallen over time. A possible explanation for this could include the wider awareness of the patients at risk of SRC and the subsequent reduction in use of glucocorticoids and cyclosporine in these patients.
Renal crisis is classically reported to occur in the ‘early’ years of disease (less than 5 years from first non-Raynaud’s symptom), with 75% of cases occurring in years 1 to 4. A large German retrospective analysis found that the distribution of SRC between males and females was representative of the SSc cohort as whole, with a female predominance of 3:1. After univariable analysis, there was no significant difference between sex and risk of SRC (p = 0.063) . Clinical risk factors for the development of SRC, including diffuse disease, are discussed in a later section. One published case series found that 22% of SRC represented the patient’s first clinical presentation of the disease, even if other features of disease had preceded the crisis .
Late presentation of SRC should not be overlooked. Cases up to 20 years after diagnosis have been reported. It is possible that the presentation of SRC in these cases is affected by immunosuppression received for other complications of SSc, such as skin disease.
An analysis of the Genome Research in African American Scleroderma Patients (GRASP) cohort highlighted the severe disease burden amongst African Americans, demonstrating that the prevalence of SRC is 7%, 3.5 times higher than the 2% prevalence reported after analysis of the EUSTAR cohort . These figures are not adjusted for prevalence of anti-RNA polymerase III antibody as this serological test was only available after 2007 and thus missing for 40% of the GRASP cohort.
Pathogenesis
SRC arises from reduced blood supply to the kidney which is proposed to occur in susceptible individuals for a variety of reasons. Susceptible individuals with SSc have the unifying abnormal intra-renal features of vasculopathy, fibrosis, and autoimmunity which allow injury to the vessel wall to initiate an amplification loop of local damage and activation of the renin-aldosterone-angiotensin (RAA) axis and hence SRC, as demonstrated by Fig.2 . Autopsy specimens have shown that interstitial fibrosis, lymphocytosis, and chronic vasculopathy are often present in SSc without renal crisis but a yet-to-be-defined event, possibly vascular, triggers the endothelial activation and release of growth factors and cytokines which then leads to smooth muscle proliferative vasculopathy. Proliferative vasculopathy leads to glomerular ischaemia and sustained activation of the RAA axis with hyperplasia of the juxtaglomerular axis . The improvement observed after inhibition of the RAA axis with ACEi suggests that hypereninaemia plays a clear role in the pathogenesis of SRC; however, a prospective study showed that increased renin levels are not predictive of subsequent SRC , suggesting that other factors are involved. Other factors believe to play a role in reduction of glomerular blood flow include decreased cardiac output due to cardiac scleroderma or heart failure, direct effects of angiotensin II, glucocorticoids, and renal vasospasm, i.e. ‘renal Raynaud’s’ .
Fig. 2.
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Proposed pathogenesis of SRC.
Modified from Denton et al. . Created with BioRender.com
In 2019, a Japanese group proposed that there is further definition to be made in the pathophysiology of SRC . This group explained that there is clear pathological difference between two groups in SSc, with narrowly defined SSc (nd-SSc) vasculopathy causing intimal thickening and subsequent hypertension whereas SSc-TMA is associated with a lesion in the vessel wall leading to microvascular thrombosis. These pathologies overlap in presentation but do appear to have different disease trajectories; nd-SRC being associated with initial elevated blood pressure and serum creatinine associated with a milder thrombocytopenia later in the disease course. Conversely, SSc-TMA was associated with early and severe thrombocytopenia, followed by elevated blood pressure and creatine. In this study SSc-TMA was associated with steroids. The working group suggested we differentiate between the subtypes clinically by the sequence of thrombocytopenia, elevated BP, and elevated creatinine where possible. This is potentially more helpful than the current grouping of ‘hypertensive’ and ‘normotensive’ as it relates to mechanism of injury. However, if differentiation relies on histology via renal biopsy in the acutely unwell patient, this may produce a practical barrier to the uptake of this classification system.
Renal Biopsy Abnormalities
The overall pathological picture can be characterized by endothelial damage and thrombus formation. Unlike atypical haemolytic uraemic syndrome (aHUS), small vessel thrombus are more prevalent than glomerular thrombus . SRC demonstrates predominant small vessel involvement with early changes such as mucoid intimal oedema, thrombosis, and fibrinoid necrosis with later intimal thickening leading to obliteration of the lumen which gives an ‘onion skin’ appearance under microscopy. Work is currently ongoing to define the characteristic renal biopsy abnormalities observed in SRC as part of the ISRCC II study . Adventitial and peri-adventitial fibrosis is also observed which indicates a chronic vasculopathy process. Interestingly the extent of fibrosis does not reflect long-term renal outcome . In extensive histological studies of SRC, it has been found that the extent of acute vascular injury, glomerular ischaemic collapse, and C4d deposits are linked to delayed recovery or failure to recover renal function.
Despite biopsy findings providing good prognostic information, they are not routinely used in the acute setting as they are often not required to confirm diagnosis and the invasive procedure carries significant risk in the setting of hypertension and thrombocytopenia, so is reserved for cases where other diagnoses are being considered.
Complement
There is evidence that cases of SSc-TMA with normal ADAMTS13 levels (i.e. not TTP) have responded well to TPE, albeit alongside treatment with ACEi. This response may suggest that SSc-TMA is mediated by an unrestricted complement cascade, which could occur due to genetic mutations or autoantibodies against complement regulator proteins as seen in aHUS . A 2012 analysis of the EUSTAR database found 5.2% prevalence of hyocomplementaemia (defined as low C3 or C4 levels) in SSc patients but using multivariate analysis, this was not associated with any specific disease parameter such as SRC .
Further evidence of complement activity was found in a Spanish case series of 29 TMA patients, where immunofluorescence showed increased deposition of C5b-9 in the endothelium of renal arterioles and in glomeruli . Serum samples from the patients induced C5b-9 ex vivo and demonstrated increased soluble C5b-9 activity. Serum ratios of complement factor 3-d (C3d): complement factor 3 (C3) and Factor B: Factor Bb (FB:FBb) were also increased, which would agree with the hypothesis of increased alternate pathway activity. It is important to note, whilst revealing some important insights with regard to the role of complement in TMA, the 2 cases of SRC in this series did not respond to treatment with eculizumab.
There have however been multiple anecdotal cases of successful response to eculizumab, the monoclonal antibody directed against complement factor 5 (C5) which blocks production of C5b-9 and therefore formation of the membrane attack complex (MAC) complex [27, 28]. Cases treated with eculizumab had evidence of MAHA and showed a dramatic improvement in renal function following treatment but mortality was high due to the severity of SRC observed in these cases [27, 29, 30]. Genetic screening in these cases did not reveal any genetic mutations for the complement proteins or their regulators which has been observed in aHUS.
Whilst there is a growing body of evidence that aberrant complement activation is involved in SRC, we are yet to identify the specific mechanisms which result in the vascular structural abnormalities observed or trigger such changes, raising the question of cause or effect with regard to complement in SRC. However, based on the positive anecdotal evidence, it is reasonable to treat SRC patients with eculizumab in whom complement mediated TMA is suspected, who have not responded to ACEi or TPE.
Endothelin
The endothelin axis has a recognised role in SSc vasculopathy of digital ulceration and PAH. Studies directed at SRC showed increased levels of endothelin-1 and increased expression of endothelin A and B receptors in SRC [31–33]. Endothelin receptor antagonist have been explored in SRC; however, results with bosentan, a selective endothelin-1 receptor antagonist, did not improve renal outcome in SRC . The preliminary report of a phase II randomised controlled trial of zibotentan, an endothelin-A antagonist, in SSc-CKD did find a statistical change in urinary MCP-1 and stabilisation of estimated glomerular filtration rate (eGFR) .
Biomarkers
Adipose tissue–derived cytokines (adipokines) are thought to be important mediators of immunity. Liopcalin-2 levels were measured in treatment-naïve SSc patients and levels were significantly raised in cases of SRC (n = 2) and showed positive correlation with modified Rodnan skin score (mRSS) .
Endothelial damage is associated with expression of adhesion molecules such as soluble vascular cell adhesion molecule (sVCAM) which have previously been shown to correlate with disease severity and in the highest recorded case in one particular study, did precede a case of SRC .
Serum soluble CD147 has also been investigated to determine whether it has a role in SRC pathogenesis. One study found that despite there not being a difference in CD147 levels between limited and diffuse patients, higher levels of CD147 were associated with SRC (0.13 SSc, 0.0 control, p< 0.05) .
Animal Models of SRC
There are not yet any established animal models of SRC. The TβRIIΔk-fib transgenic mouse model replicates hypertension and large vessel fibrosis. The model demonstrates exaggerated fibrotic response to hypertensive injury and provides opportunity for further studies into the specific mechanism of injury in SRC .
Emerging
Studies have been carried out to determine whether SSc-specific autoantibodies not only stratify patient groups but also play an active role in the pathophysiology of SSc. There is evidence of autoimmunity towards AT(1)R and ET(A)R receptors on endothelial cells which increase TGF-beta expression .
Risk Factors
Scleroderma-Specific Antibodies
It has been widely described that the phenotype most at risk for development of SRC are those patients with early diffuse SSc with proximal skin thickening [16, 41]. Autoantibody profile certainly plays a predictive role in the development of SRC. Anti-RNA polymerase III has a higher prevalence in diffuse SSc and is strongly associated with SRC. Up to 50% of patients with anti-RNA pol III will go on to develop SRC [14, 42–44]. A study from the EUSTAR registry involving 2800 subjects demonstrated anti-RNA polymerase was independently associated with SRC (odds ratio 5.86, 95% confidence interval 2.6, 13.2) . Renal crisis occurs in 10% of patients with anti-topoisomerase (ATA) antibodies which is also associated with diffuse disease. In comparison, there a very few reported cases of SRC in limited anti-centromere antibody (ACA)–positive disease .
Outcomes in SRC depending on presence of anti-RNA polymerase III have been compared in a cohort from the Royal Free Hospital. Patients with anti-RNA polymerase III antibodies were more likely to require dialysis but were also more likely to discontinue dialysis (53% vs 26%, p = 0.01) and had better long-term survival (p = 0.003) .
Genetic Factors
HLA-DRB11407 and HLA-DRB11304 were identified as independent risk factors for SRC in a study examining over 1500 patients . There has also been suggestion of an association between anti-RNA polymerase III antibody and endothelin receptor A (EDNRA alleles H323H/C and E335E/A) polymorphism but the functional significance of this is yet to be determined .
A recent study exploring protein expression in SSc patients who were anti-RNA polymerase III positive found that there was increased expression of two candidate proteins, GPATCH2L and CTNND2, on biopsy staining in SRC patients compared to normal controls. This may help towards explaining why certain groups of anti-RNA polymerase patients are more susceptible to SRC that others .
Clinical Risk Factors
SRC is recognised in a subset of patients who are yet to evolve to diffuse cutaneous disease. These patients are likely to be in the early years of their disease and often display specific disease features suggestive of diffuse subtype such as tendon friction rub, polyarthritis, swollen hands, and/or carpel tunnel and will go on to develop skin thickening.
Historical studies have established that risk factors for development of SRC include diffuse disease, anaemia, pericardial effusion, and congestive heart failure . Rapid progression of skin thickening was also found to be an independent risk factor for SRC as is anti-RNA polymerase III antibody status, tendon friction rub, large joint contractures, heart enlargement , proteinuria, and corticosteroid use , as demonstrated in Table 2.
Table 2.
Odds ratio (OR) and hazard ratio (HR) in cohort studies analysing independent risk factors for development of SRC. CI, confidence interval 95%; DcSSc, diffuse cutaneous SSc; LcSSc, limited cutaneous SSc; DLCO, transfer factor as measured by spirometry
| | p value | OR | HR | CI | Study |
:--- :--- :--- |
| anti-RNA pol III | < 0.001 | 5.86 | | [2.6–13.2] | Moinzadeh et al. 2020 a |
| Chronic kidney disease | < 0.004 | 2.5 | | [1.34–4.6] |
| | < 0.001 | | 20.7 | [2.2–190.7] | Gordon et al. 2019 b |
| Proteinuria | < 0.001 | | 183 | [19.1–1750] |
| | < 0.001 | 5.55 | | [3.4–8.9] | Moinzadeh et al. 2020 a |
| DcSSc vs. LcSSc | 0.002 | 2.54 | | [1.42–4.5] |
| DLCO | < 0.001 | 4.41 | | [2.01–9.6] |
| Glucocorticoid use | 0.007 | 1.93 | | [1.20–3.1] |
| | 0.014 | 3.63 | | [1.30–10.05] | De Marco et al. 2002 c |
| | 0.49 | | 1.32 | [0.60–2.87] | Butikofer et al. 2020 d |
| Hypertension | 0.002 | | 2.22 | [1.34–3.6] |
| | < 0.001 | 13.1 | | [4.7–36.6] | Gordon et al. 2019 b |
| mRSS > 14 | | 3.08 | | [1.24–7.61] | Avouac et al. 2016 e |
| | 0.003 | 10 | | [2.21–45.9] | De Marco et al. 2002 c |
| ACE inhibitors | 0.003 | | 2.07 | [1.28–3.36] | Butikofer et al. 2020 d |
| Tendon friction rub | 0.15 | | 1.7 | [0.83–3.48] |
| | 0.0007 | | 2.33 | [1.03–6.19] | Avouac et al. 2016 e |
| Large joint contracture | 0.008 | 16.12 | | [2.07–125.2] | De Marco et al. 2002 c |
| Heart involvement | 0.048 | 2.93 | | [1.01–8.4] |
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a Moinzadeh et al. , b Gordon et al. , c De Marco et al. , d Butikofer et al. , e Avouac et al.
Glucocorticoids, particularly at high dose (> 15 mg/day), have long been associated with development of SRC . It has been suggested that glucocorticoids may directly contribute to SRC by inhibiting prostacyclin production and inducing activity of angiotensin-converting enzyme (ACE) . The patients most likely to be taking steroids are those with early and severe disease, who are also at increased risk of SRC which may confound data when discussing the link between steroids and SRC. Data from the ISRCS showed that every 1 mg of prednisolone a patient was taking prior to onset of SRC increased risk of death by 4% (hazard ratio 1.04, 95% CI 1.02, 1.07, p< 0.01) . High-dose steroid use is avoided, particularly in early diffuse disease.
By understanding the risk factors which predispose certain patients to development of SRC, the condition can be rapidly recognised and treatment with ACEi initiated promptly. Prompt treatment improves patient outcome .
Outcomes
Outcomes in SRC remain poor compared to other organ complications of SSc but have improved by 50% since the introduction of ACEi treatment in 1981 . Results from the ISRCS showed 36% mortality and 25% remain on dialysis at 1 year . Permanent dialysis is required in 19–40% of SRC cases . Interestingly in the post-ACE era, there has been no evidence of further improvement in morbidity and mortality, highlighting the need for novel treatments in SRC .
Recovery of renal function to achieve dialysis independence can occur up to 2 years after the initial event [7, 11] so decisions about renal transplant are delayed accordingly. Three to 17% of SRC cases will require renal transplant . Considerations prior to transplant include co-existing comorbidities, severity of SSc, and choice of immunosuppression following transplant as calcineurin inhibitors are vasoconstrictors and so can theoretically contribute towards further SRC . Survival for SRC patients is superior in the transplant population (54–91%) compared to those on dialysis (31–56%) with graft survival now similar to that of other ESRF .
Recurrence of SRC following renal transplant has been reported from 2 to 9% . Recurrence can also occur in the setting of treated SRC not requiring transplant. SRC has been incorporated into an internationally validated tool for predicting 5-year outcome in diffuse disease due to its impact on overall survival .
The Role of ACEi
Interestingly, it has been found that whilst ACEi reduce mortality as treatment for SRC, prior use of ACEi, or prophylactic use, has been associated with worse long-term outcomes and higher frequency of long-term dialysis after SRC [11, 55, 57]. The most widely accepted explanation for this is that small doses of ACEi are not sufficient to treat SRC but may mask the development of hypertension, an important clinical warning sign, leading to delay in treatment and a less reversible, more chronic process. Indeed, those with normotensive SRC have been shown to have worse long-term outcomes.
The ISRCS found a greater than twofold increased risk of mortality in SRC with prior exposure to ACEi ; however, many of the patients on ACEi were for indications other than pure prophylaxis of SRC (two case where prophylaxis was used due to glucocorticoid exposure in a high risk patient) so there is a possibility that the results were confounded by clinical severity . Further studies have not shown conclusive evidence to support the use of ACEi prophylactically.
In practice, cases considered to be particularly high risk such as those undergoing autologous haemopoietic stem cell transplant (AHSCT) due to the high dose of glucocorticoids and IV fluid used are given ACEi therapy a few weeks prior to initiating the AHSTC ; however, there is no conclusive evidence that ACEi prophylaxis in this context is beneficial.
Clinical Presentation
The two hallmark features of SRC are accelerated hypertension and AKI but patients can present with a variety of symptoms including headache, blurred vision, and nonspecific symptoms such as fatigue or dyspnoea. Severe SRC may be evidenced by seizures or symptomatic pericardial involvement at presentation.
It is important to point out that whilst an individual’s blood pressure may fall into the ‘normal’ range, the reading may represent a significant increase in average BP for that individual, e.g. if they normally have an average systolic of 80, an increase to 120 is significant and can represent SRC. It is recognised that around 10% of cases are ‘normotensive’ renal crisis, without a rise in systemic BP.
Cardiac complications of SRC are common and may be complicated by underlying cardiac scleroderma. Most symptoms are a response to the sudden increase in blood pressure caused by activation of the RAA axis and often improve in response to tight blood pressure control with therapeutic agents.
Diagnosis
The United Kingdom Scleroderma Study Group (UKSSG) has produced diagnostic criteria for SRC (Table 3).
Table 3.
UKSSG Diagnostic criteria for SRC 2016.
Reproduced with permission from Lynch et al.
| Diagnostic criteria (essential) |
| New onset BP > 150/85 mmHg or obtained at least twice over 24 h |
| Increase ≥ 20 mmHg from usual systolic BP |
| Acute kidney injury stage 1 or higher: |
| (> 50% increase in serum creatinine from stable baseline or an absolute increase of 26.5 µmol/L) |
| Supportive evidence (desirable) |
| MAHA on blood film, thrombocytopaenia and other biochemical findings consistent with haemolysis |
| Findings consistent with accelerated hypertension on retinal examination |
| Microscopic haematuria on urine dipstick and/or red blood cells on urine microscopy |
| Oliguria or anuria |
| Renal biopsy with typical features of SRC including onion skin proliferation within the walls of intrarenal arteries and arterioles, fibrinoid necrosis, glomerular shrinkage |
| Flash pulmonary oedema |
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Spectrum of Renal Disease
Some conditions can mimic SRC and indeed are difficult to distinguish both in presentation and response to therapy. There are several cases of TTP reported in SSc patients who were distinguishable by fever and haemorrhagic manifestations [59–61]. There is a possibility the two diseases are variations of the same process and if a diagnosis of TTP is made in SRC, an ACEi should be used regardless of whether TPE is also considered as a treatment for the TTP .
As discussed earlier, some groups have suggested that we define SRC cases by pathophysiology, rather than the presenting blood pressure. This can be done by observing the sequence and severity of thrombocytopenia, hypertension, and elevation in creatinine. This allows us to differentiate nd-SRC from SSc-TMA and indeed, other not strictly SSc-related causes of TMA, such as TTP (which would indicate TPE treatment), drug-induced TMA, and DIC. The clinical features, serological findings, and histological features or some important differential diseases, when assessing a patient with AKI, are highlighted in Table 4.
Table 4.
Summary of the clinical and serological features of SRC, TMA, ANCA-associated glomerular nephritis, and SLE nephritis
| | Serum markers | Urinalysis | Typical presentation | Patient cohort | Histopathology |
:--- :--- :--- |
| Scleroderma renal crisis (SRC) | Creatine increased AKI (150% typical) Anaemia (MAHA) Thrombocytopenia Haemolysis Negative DAT | Mild proteinuria (< 2 g/day)and haematuria May be urinary casts | Systolic blood pressure ≥ 140 mm Hg Diastolic blood pressure ≥ 90 mm Hg Acute | Early, diffuse SSc Anti-ARA antibody positive High dose glucocorticoid, tendon friction rub | Glomerular or extraglomerular TMA changes, rarely Juxtaglomerulus hyperplasia. Chronic ‘onion skin’ appearance |
| Thrombotic microangiopathies (TMA) (causes other than SSC) | AKI MAHA (thrombocytopenia and reticularcytosis, increased LDH and low haptoglobin) ADAMS-13 < 10% activity | Proteinuria haematuria | Fever Haemorrhagic manifestation, confusion, neurological deficit Acute | May occur as part of SRC In adults, often underlying comorbid cause. TTP less likely to see AKI | Intravascular fibrin thrombi with mucoid changes. Intimal proliferation of arterioles. Duplication of GBM with endocapillary hypercellularity |
| ANCA-associated glomerular nephritis | AKI ANCA (MPO/PR3) positive Eosinophilia Thrombophilia Elevated CRP | Proteinuria (often > 3 g/day) and Haematuria likely significant | Vasculitic rash, pulmonary lesions, peripheral neuropathy, fatigue, weight loss Acute/chronic | > 50 years old MPA, GPA, EGPA, RLV | Pauci immune necrotising glomerulonephritis; focal, crescentic, sclerotic or mixed |
| SLE-associated glomerular nephritis | AKI possible but not diagnostic dsDNA, anti-Smith Low complement (C3) | Proteinuria (> 4 g/day worst prognosis ) in 100%, microscopic haematuria in 80% | Known lupus or new features of disease. Possible nephrotic syndrome, 30% hypertension Acute/chronic | Early disease, increased risk in black patients, male patients | Well-defined grade I–V . Glomerular deposits with positive immunofluorescence. TMA in up to 25% |
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AKI acute kidney injury, DAT direct antiglobulin test, MAHA microangiopathic haemolytic anaemia, LDH lactate dehydrogenase, TTP thrombotic thrombocytopenic purpura, GBM glomerular basement membrane, ANCA anti-neutrophil cytoplasm antibodies, MPO myeloperoxidase antibody, PR3 anti-proetinase-3 antibodies, MPA microscopic polyangiitis, GPA granulomatosis with polyangiitis, EGPA eosinophilic granulomatosis with polyangiitis, RLV renal limited vasculitis, SLE systemic lupus erythematosus, dsDNA double-stranded DNA antibody
Laboratory Findings
Patients typically present with around 150% increase in creatinine from their baseline . The creatine value can continue to rise despite rapid correction of the blood pressure.
Urine dipstick can reveal haematuria and proteinuria which is normally mild (< 2 g/day). Casts may be present, and these findings are not specific to SRC and can be observed in other hypertensive disease.
MAHA is found in approximately 50% of SRC and is an indicator of TMA. Lab findings along with anaemia include thrombocytopenia and reticulocytosis. Thrombocytopenia is often marked, and the recovery of platelets is often the first sign of response to therapy and can occur even if creatinine continues to rise. Table two summarises the different characteristics of SRC and other observed causes of AKI.
Biomarkers as predictors for outcome have been explored. A retrospective study looking at 19 SRC patients found NTproBNP levels > 360 pg/l were highly correlated with patients requiring dialysis .
A recent novel finding is that of urinary proteins as candidate markers for renal disease in SSc. Proteins urinary intracellular adhesion molecule (ICAM-1) and urinary monocyte chemoattractant protein (MCP-1) appear to reflect renal involvement better than serum levels . MCP-1 in particular has previously been shown to correlate with SSc skin disease and lung fibrosis and may play a role locally in fibroblast differentiation. At present, the potential biomarkers have been explored in SSc-CKD but may provide insight into acute renal involvement in future prospective studies.
There is ongoing work assessing the role of doppler-measured ‘renal resistive index’ in SSc. The measurement not only appears to be correlated with systemic vasculopathy in SSc, specifically anti-centromere antibody and PAH development , but also provides useful information on renal vasculopathy and prediction of mortality .
Management
Monitoring
Whilst there are no specific preventative measures recognised for SRC, close observation, particularly in high-risk individuals, is key to ensure early detection and treatment. Observation should take the form of regular home recordings of blood pressure (at least twice weekly) with any sustained increase in BP of > 30 mmHg prompting medical attention.
Symptoms such as headache, visual changes, fatigue, or breathlessness should also prompt patients to check blood pressure. This strategy is reliant on patient autonomy so patient education early in the disease course will play a large role in its success. Initiatives such as a patient ‘warning card’ and access to specialist nurses are particularly helpful in this setting .
At regular routine clinical assessment, which would be at least 6 monthly as standard of care, urine dipstick and serum urea and creatinine should be monitored, along with blood pressure.
Pregnancy
As discussed, early stage SSc is the highest risk period for development of SRC and patients are advised to avoid pregnancy during this time . Pregnancy itself does not increase the risk of SRC . Other serious complications of pregnancy such as pre-eclampsia and HELLP (haemolysis, elevated liver enzymes, low platelet) syndrome can mimic SRC and renal biopsy may be required to differentiate the processes. Renal biopsy carries close to the same risk as the general population early in pregnancy but should be avoided in the third trimester unless it will determine management as the risk of complications increases with gestational age . An ACEi should be started immediately if SRC suspected [68, 70]. Captopril is the preferred choice due to lower risk of foetal renal complications. ACEi carries a teratogenic risk to the foetus but in this scenario, this is outweighed by the life-threatening complication of SRC to the mother. In patients with history of SRC planning a pregnancy, it is recommended that ACEi continue during the pregnancy and BP is optimised prior to conceiving. Again, this strategy is not without teratogenic risks to the foetus and this needs to be discussed clearly in pre-pregnancy counselling .
Current Therapies
If evidence of neurological or cardiac compromise is present at onset of SRC, rapid reduction of blood pressure is necessitated. However, if such symptoms are not present, a more gradual reduction in blood pressure can be afforded (10% reduction in systolic BP per day) and confers a better chance of renal recovery . Recommendations for SRC management developed by expert consensus of the UKSSG are shown in Fig.3.
Fig. 3.
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UKSSG guidelines on the diagnosis and management of scleroderma renal crisis Reproduced with permission from Lynch et al.
ACEi should be started as soon as SRC is diagnosed. If the patient is already taking an ACEi, then the dose should be increased. A long-acting ACEi such as ramipril is most used but short-acting agents such as captopril may be suitable in cases of haemodynamic compromise. A long-acting agent is preferable in most cases as it can be easily up-titrated to maximum dose, normally by doubling the dose at 24-h intervals.
Adequate control of blood pressure often takes 3–5 days. It is important to note that renal function can continue to deteriorate despite initiation of ACEi and correction of blood pressure. This should not prompt discontinuation of this important therapy. Additional agents can be added to achieve blood pressure control, including calcium channel blockers, alpha antagonists, and clonidine. Intravenous therapies such as GTN or Iloprost can be used with the latter having the added benefit of discouraging platelet and vascular endothelial activation .
ACEi should be continued life-long, even if the patient is dialysis-dependant as an ACEi will improve the chance of the patient subsequently managing to become dialysis-independent .
Angiotensin receptor blockers (ARB) can be used if ACEi is contraindicated or not tolerated; however, studies have suggested that ARBs are not clinically equivalent in treatment of SRC . Importantly, ARBs do not inhibit degradation of bradykinin, an agent which is needed in SRC due to its vasodilatory effects. Dual-agent therapy is associated with higher risk of adverse events ; therefore, ACEi alone is preferable.
Beta blockers should not be used in SRC due to their negative chronotropic effects on a circulatory system experiencing increased peripheral resistance and may lead to reduction in cardiac output. The use of beta blockers could also exacerbate renal vasospasm ‘renal Raynaud’s’.
It is important to consider differential diagnoses which may have specific treatments that differ to that of SRC. Factors that should prompt consideration of alternative pathologies include a normotensive presentation, significant urinary casts on microscopy, overlap disease phenotype such as SLE or vasculitis, and presentation with fever. In this case, renal biopsy is important and can be sought practically once clotting has normalised and the patient is in a stable condition.
Approximately 60% of SRC will require renal replacement therapy which is often delivered by haemofiltration in the acute setting, moving to haemodialysis or peritoneal dialysis as local resources allow.
TPE is used in settings where related pathologies are suspected such as TTP.
Immunosuppression
Mycophenolate mofetil (MMF) is an immunosuppressive agent commonly used in SSc and may be initiated following SRC if the patient it not already taking such medication. The rationale behind this being that SRC represents a degree of disease activity which then requires immunosuppression. Rat models of ischaemia and reperfusion to the kidney showed exaggerated production of reactive oxygen species (ROS) leading to cell necrosis . Immunostaining following administration of mycophenolate demonstrated a reduction in interleukin 6(IL-6) and inducible nitric oxygen synthase (iNOS) resulting in restored renal cortical oxygenation. Whether this translates to MMF treatment in SRC has not been proven, a recent retrospective cohort study database did not find any significant association between MMF use and SRC .
Conclusion
SRC is a well-documented complication of SSc but is rare and can present with a variety of symptoms so establishing a core classification criterion is going to be invaluable to future research. Whilst pathological features of SRC are recognised, they are not specific to SRC and there is still much to be understood about why certain predisposed individuals progress to SRC. There could be a role for endothelin and overactivation of the complement pathways, both treatable targets awaiting ongoing trial evidence. The development of sensitive biomarkers for renal disease in SSc may provide further insight into the pathogenesis of SRC whilst also providing tools for early detection and possibly prognosis in SRC. Early, specialist treatment and collaboration between rheumatology and renal physicians will enhance patient outcome. Advances in dialysis delivery and post-transplant management now mean that for those who do go on to require these treatments, the prognosis has improved.
Declarations
Conflict of Interest
The authors declare no competing interests.
Footnotes
Publisher's Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
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14421 | https://www.wikidoc.org/index.php/Absolute_value | Absolute value
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For the philosophical term, see value (philosophy).
In mathematics, the absolute value (or modulus which is Latin for a small measure) of a real number is its numerical value without regard to its sign. So, for example, 3 is the absolute value of both 3 and −3. In computer programming, the mathematical function used to perform this calculation is usually given the name abs().
Generalizations of the absolute value for real numbers occur in a wide variety of mathematical settings. For example an absolute value is also defined for the complex numbers, the quaternions, ordered rings, fields and vector spaces. The absolute value is closely related to the notions of magnitude, distance, and norm in various mathematical and physical contexts.
1 Real numbers
2 Complex numbers
3 Absolute value functions
4 Ordered rings
5 Distance
6 Derivatives
7 Fields
8 Vector spaces
9 Algorithms
10 Notes
11 References
12 See also
Real numbers
For any real number a the absolute value or modulus of a is denoted by | a | and is defined as
As can be seen from the above definition, the absolute value of a is always either positive or zero, but never negative.
From a geometric point of view, the absolute value of a real number is the distance along the real number line of that number from zero, and more generally the absolute value of the difference of two real numbers is the distance between them. Indeed the notion of an abstract distance function in mathematics can be seen to be a generalization of the absolute value of the difference (see "Distance" below).
The following proposition, gives an identity which is sometimes used as an alternative (and equivalent) definition of the absolute value:
Proposition 1:
The absolute value has the following four fundamental properties:
Proposition 2:
: | | |
--- |
| | Non-negativity |
| a| = 0 \iff a = 0 | Positive-definiteness |
| ab| = |a | \, | Multiplicativeness |
| a+b| \le |a| + |b| | Subadditivity |
Other important properties of the absolute value include:
Proposition 3:
: | | |
--- |
| | Symmetry |
| a - b| = 0 \iff a = b | Identity of indiscernibles (equivalent to positive-definiteness) |
| a - b| \le |a - c| +|c - b| | Triangle inequality (equivalent to subadditivity) |
| a/b| = |a| / |b| \mbox{ (if } b \ne 0) \, | Preservation of division (equivalent to multiplicativeness) |
| a-b| \ge b | | (equivalent to subadditivity) |
Two other useful inequalities are:
The above are often used in solving inequalities; for example:
: | | |
--- |
| x-3| \le 9 | |
| | |
Complex numbers
Since the complex numbers are not ordered, the definition given above for the real absolute value cannot be directly generalized for a complex number. However the identity given in Proposition 1:
can be seen as motivating the following definition.
For any complex number
where x and y are real numbers, the absolute value or modulus of z is denoted | z | and is defined as
It follows that the absolute value of a real number x is equal to its absolute value considered as a complex number since:
Similar to the geometric interpretation of the absolute value for real numbers, it follows from the Pythagorean theorem that the absolute value of a complex number is the distance in the complex plane of that complex number from the origin, and more generally, that the absolute value of the difference of two complex numbers is equal to the distance between those two complex numbers.
The complex absolute value shares all the properties of the real absolute value given in Propositions 2 and 3 above. In addition, If
and
is the complex conjugate of z, then it is easily seen that
and
with the last formula being the complex analogue of Proposition 1 mentioned above in the real case.
Since the positive reals form a subgroup of the complex numbers under multiplication, we may think of absolute value as an endomorphism of the multiplicative group of the complex numbers.
Absolute value functions
The real absolute value function is continuous everywhere. It is differentiable everywhere except for x = 0. It is monotonically decreasing on the interval (−∞, 0] and monotonically increasing on the interval [0, ∞). Since a real number and its negative have the same absolute value, it is an even function, and is hence not invertible.
The complex absolute value function is continuous everywhere but (complex) differentiable nowhere (One way to see this is to show that it does not obey the Cauchy-Riemann equations).
Both the real and complex functions are idempotent.
It is a nonlinear function.
Ordered rings
The definition of absolute value given for real numbers above can easily be extended to any ordered ring. That is, if a is an element of an ordered ring R, then the absolute value of a, denoted by | a |, is defined to be:
where −a is the additive inverse of a, and 0 is the additive identity element.
Distance
The absolute value is closely related to the idea of distance. As noted above, the absolute value of a real or complex number is the distance from that number to the origin, along the real number line, for real numbers, or in the complex plane, for complex numbers, and more generally, the absolute value of the difference of two real or complex numbers is the distance between them.
The standard Euclidean distance between two points
and
in Euclidean n-space is defined as:
This can be seen to be a generalization of | a − b |, since if a and b are real, then by Proposition 1,
while if
and
are complex numbers, then
: | | |
--- |
| a - b| \, | (a_1 + i a_2) - (b_1 + i b_2)|\, |
| | (a_1 - b_1) + i(a_2 - b_2)|\, |
| | |
The above shows that the "absolute value" distance for the real numbers or the complex numbers, agrees with the standard Euclidean distance they inherit as a result of considering them as the one and two-dimensional Euclidean spaces respectively.
The properties of the absolute value of the difference of two real or complex numbers: non-negativity, identity of indiscernibles, symmetry and the triangle inequality given in Propositions 2 and 3 above, can be seen to motivate the more general notion of a distance function as follows:
A real valued function d on a set X × X is called a distance function (or a metric) on X, if it satisfies the following four axioms:
: | | |
--- |
| | Non-negativity |
| | Identity of indiscernibles |
| | Symmetry |
| | Triangle inequality |
Derivatives
The derivative of the real absolute value function is the signum function, sgn(x), which is defined as
for x ≠ 0. The absolute value function is not differentiable at x = 0. Where the absolute value function of a real number returns a value without respect to its sign, the signum function returns a number's sign without respect to its value. Therefore x = sgn(x)abs(x). The signum function is a form of the Heaviside step function used in signal processing, defined as:
\begin{cases} 0, & x < 0 \\ \frac{1}{2}, & x = 0 \\ 1, & x > 0, \end{cases}
where the value of the Heaviside function at zero is conventional. So for all nonzero points on the real number line,
The absolute value function has no concavity at any point, the sign function is constant at all points. Therefore the second derivative of |x| with respect to x is zero everywhere except zero, where it is undefined.
The absolute value function is also integrable. Its antiderivative is
Fields
The fundamental properties of the absolute value for real numbers given in Proposition 2 above, can be used to generalize the notion of absolute value to an arbitrary field, as follows.
A real-valued function v on a field F is called an absolute value (also a modulus, magnitude, value, or valuation) if it satisfies the following four axioms:
: | | |
--- |
| | Non-negativity |
| | Positive-definiteness |
| | Multiplicativeness |
| | Subadditivity or the triangle inequality |
Where 0 denotes the additive identity element of F. It follows from positive-definiteness and multiplicativeness that v(1) = 1, where 1 denotes the multiplicative identity element of F. The real and complex absolute values defined above are examples of absolute values for an arbitrary field.
If v is an absolute value on F, then the function d on F × F, defined by d(a, b) = v(a − b), is a metric and the following are equivalent:
d satisfies the ultrametric inequality d(x, y) < max{d(x, z), d(y, z)}.
is bounded in R.
An absolute value which satisfies any (hence all) of the above conditions is said to be non-Archimedean, otherwise it is said to be Archimedean.
Vector spaces
Main article: Norm (mathematics)
Again the fundamental properties of the absolute value for real numbers can be used, with a slight modification, to generalize the notion to an arbitrary vector space.
A real valued function || · || on a vector space V over a field F, is called an absolute value (or more usually a norm) if it satisfies the following axioms:
For all a in F, and v, u in V,
: | | |
--- |
| \mathbf{v}\| \ge 0 | Non-negativity |
| \mathbf{v}\| = 0 \iff \mathbf{v} = 0 | Positive-definiteness |
| a \mathbf{v}\| = |a| \|\mathbf{v}\| | Positive homogeneity or positive scalability |
| \mathbf{v} + \mathbf{u}\| \le \|\mathbf{v}\| + \|\mathbf{u}\| | Subadditivity or triangle inequality |
The norm of a vector is also called its length or magnitude.
In the case of Euclidean space Rn, the function defined by
is a norm called the Euclidean norm. When the real numbers R are considered as the one-dimensional vector space R1, the absolute value is a norm, and is the p-norm for any p. In fact the absolute value is the "only" norm on R1, in the sense that, for every norm || · || on R1, || x || = || 1 || · | x |. The complex absolute value is a special case of the norm in an inner product space. It is identical to the Euclidean norm, if the complex plane is identified with the Euclidean plane R2.
Algorithms
In the C programming language, the abs(), labs(), llabs() (in C99), fabs(), fabsf(), and fabsl() functions compute the absolute value of an operand. Coding the integer version of the function is trivial, ignoring the boundary case where the largest negative integer is input:
int abs (int i) { if (i < 0) return -i; else return i; }
The floating-point versions are trickier, as they have to contend with special codes for infinities and not-a-number.
The function for absolute value in Fortran, Matlab, and GNU Octave is abs. It handles integer, real as well as complex numbers.
Using assembly language, it is possible to take the absolute value of a register in just three instructions (example shown for a 32-bit register on an x86 architecture, Intel syntax):
cdq xor eax, edx sub eax, edx
cdq extends the sign bit of eax into edx. If eax is nonnegative, then edx becomes zero, and the latter two instructions have no effect, leaving eax unchanged. If eax is negative, then edx becomes 0xFFFFFFFF, or -1. The next two instructions then become a two's complement inversion, giving the absolute value of the negative value in eax.
Notes
↑ Jean-Robert Argand, is credited with introducing the term "modulus" in 1806, see: Nahin, O'Connor and Robertson, and functions.Wolfram.com.
↑ functions.Wolfram.com credits Karl Weierstrass with introducing the notation in 1841.
↑ Schechter, p 260-261.
References
Nahin, Paul J.; An Imaginary Tale; Princeton University Press; (hardcover, 1998). ISBN 0-691-02795-1
O'Connor, J.J. and Robertson, E.F.; "Jean Robert Argand"
Schechter, Eric; Handbook of Analysis and Its Foundations, pp 259-263, "Absolute Values", Academic Press (1997) ISBN 0-12-622760-8
Template:MathWorld
Template:PlanetMath
See also
Absolute value (algebra)
Valuation (mathematics)
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What Is Aromatic Hydrocarbon? Definition and Examples
Aromatic Hydrocarbons are present in our everyday lives and often are even part of the air we breathe. But are they safe for humans and the environment? Let’s learn what is Aromatic Hydrocarbon and what substances are the best examples of these compounds. Aromatic Hydrocarbons – definition Source of Aromatic Hydrocarbons Properties of Aromatic Hydrocarbons […]
Aromatic Hydrocarbons are present in our everyday lives and often are even part of the air we breathe. But are they safe for humans and the environment? Let’s learn what is Aromatic Hydrocarbon and what substances are the best examples of these compounds.
Aromatic Hydrocarbons – definition
Source of Aromatic Hydrocarbons
Properties of Aromatic Hydrocarbons
Examples of Aromatic Hydrocarbons
Aromatic Hydrocarbons – definition
To put it simply, Aromatics Hydrocarbons are organic compounds with at least one aromatic ring and sometimes even more. They’re composed only of hydrogen and carbon atoms. It’s worth mentioning that the word “aromatic” is a little misleading, as nowadays the definition of aromatic compounds has nothing to do with their actual smell. However, years ago, scientists grouped molecules based on their scent and only later learned about their chemical properties.
Source of Aromatic Hydrocarbons
Where do Aromatic Hydrocarbons come from? Fossil fuels, such as petroleum and coal, are, without a doubt, the primary sources of these compounds. Whenever fossil fuels are burned, Aromatic Hydrocarbons are released into the air and just continue to float there. That causes significant issues as the vast majority of these chemicals are toxic to both humans and the environment and should be heavily limited or completely avoided.
Properties of Aromatic Hydrocarbons
Aromatic Hydrocarbons are quite distinctive because all of them have conjugated double bonds in their rings, which boosts their stability through resonance. They’re hydrocarbons and contain only carbon and hydrogen atoms. Carbon atoms are linked with each other in a circle and create so-called aromatic rings. Even though most Aromatic Hydrocarbons are known for their pleasant scent, they’re profoundly toxic and can cause cancer. On top of that, these compounds are highly flammable, and when burned, they release carbon dioxide, water, and heat. Check out what carbon sink is.
Examples of Aromatic Hydrocarbons
As we mentioned before, the majority of these compounds come from the burning of fossil fuels, so examples of Aromatic Hydrocarbons will include a lot of air pollutants. One of the main ones is benzene, also known as the simplest Aromatic Hydrocarbon because it has only one aromatic ring. It’s a colorless and sweetly-smelling liquid with highly flammable properties. Because of its simple molecular formula, it’s used mainly as a precursor in the manufacturing process of other, more complex chemicals. Other examples of well-known hydrocarbons include methane, propane, ethane, and butane.
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14423 | https://courses.lumenlearning.com/csn-precalculusv2/chapter/cr-14-solving-quadratic-equations-using-factoring-square-root-method-completing-the-square-and-quadratic-formula/ | CR.14: Solving Quadratic Equations Using Factoring, Square Root Method, Completing the Square, and Quadratic Formula
Learning Outcomes
Factor a quadratic equation to solve it.
Use the square root property to solve a quadratic equation.
Complete the square to solve a quadratic equation.
Use the quadratic formula to solve a quadratic equation.
Recall: identifying polynomials
Recall that a polynomial is an expression containing variable terms joined together by addition or subtraction. We can identify the degree of the polynomial by finding the term having the highest power (largest exponent on its variable). That term will be the leading term of the polynomial and its degree is the degree of the polynomial.
An equation containing a second-degree polynomial is called a quadratic equation. For example, equations such as 2x2+3x−1=0 and x2−4=0 are quadratic equations. They are used in countless ways in the fields of engineering, architecture, finance, biological science, and, of course, mathematics.
Often the easiest method of solving a quadratic equation is factoring. Factoring means finding expressions that can be multiplied together to give the expression on one side of the equation.
If a quadratic equation can be factored, it is written as a product of linear terms. Solving by factoring depends on the zero-product property which states that if a⋅b=0, then a=0 or b=0, where a and b are real numbers or algebraic expressions. In other words, if the product of two numbers or two expressions equals zero, then one of the numbers or one of the expressions must equal zero because zero multiplied by anything equals zero.
Multiplying the factors expands the equation to a string of terms separated by plus or minus signs. So, in that sense, the operation of multiplication undoes the operation of factoring. For example, expand the factored expression (x−2)(x+3) by multiplying the two factors together.
(x−2)(x+3)=x2+3x−2x−6=x2+x−6
The product is a quadratic expression. Set equal to zero, x2+x−6=0 is a quadratic equation. If we were to factor the equation, we would get back the factors we multiplied.
The process of factoring a quadratic equation depends on the leading coefficient, whether it is 1 or another integer. We will look at both situations; but first, we want to confirm that the equation is written in standard form, ax2+bx+c=0, where a, b, and c are real numbers and a≠0. The equation x2+x−6=0 is in standard form.
recall the greatest common factor
The greatest common factor (GCF) is the largest number or variable expression that can divided evenly into two or more numbers or expressions.
To find the GCF of an expression containing variable terms, first find the GCF of the coefficients, then find the GCF of the variables. The GCF of the variables will be the smallest degree of each variable that appears in each term.
For example, 4x is the GCF of 16x and 20x2 because it is the largest expression that divides evenly into both 16x and 20x2.
When factoring, remember to look first for a greatest common factor. If you can find one to factor out, it will usually make your factoring task less challenging.
We can use the zero-product property to solve quadratic equations in which we first have to factor out the greatest common factor (GCF), and for equations that have special factoring formulas as well, such as the difference of squares, both of which we will see later in this section.
A General Note: The Zero-Product Property and Quadratic Equations
The zero-product property states
If a⋅b=0, then a=0 or b=0,
where a and b are real numbers or algebraic expressions.
A quadratic equation is an equation containing a second-degree polynomial; for example
ax2+bx+c=0
where a, b, and c are real numbers, and a≠0. It is in standard form.
Solving Quadratics with a Leading Coefficient of 1
In the quadratic equation x2+x−6=0, the leading coefficient, or the coefficient of x2, is 1. We have one method of factoring quadratic equations in this form.
How To: Given a quadratic equation with the leading coefficient of 1, factor it
Find two numbers whose product equals c and whose sum equals b.
Use those numbers to write two factors of the form (x+k) or (x−k), where k is one of the numbers found in step 1. Use the numbers exactly as they are. In other words, if the two numbers are 1 and −2, the factors are (x+1)(x−2).
Solve using the zero-product property by setting each factor equal to zero and solving for the variable.
Example: Factoring and Solving a Quadratic with Leading Coefficient of 1
Factor and solve the equation: x2+x−6=0.
Show Solution
To factor x2+x−6=0, we look for two numbers whose product equals −6 and whose sum equals 1. Begin by looking at the possible factors of −6.
1⋅(−6)(−6)⋅12⋅(−3)3⋅(−2)
The last pair, 3⋅(−2) sums to 1, so these are the numbers. Note that only one pair of numbers will work. Then, write the factors.
(x−2)(x+3)=0
To solve this equation, we use the zero-product property. Set each factor equal to zero and solve.
(x−2)(x+3)=0(x−2)=0x=2(x+3)=0x=−3
The two solutions are x=2 and x=−3. We can see how the solutions relate to the graph below. The solutions are the x-intercepts of the graph of x2+x−6.
Try It
Factor and solve the quadratic equation: x2−5x−6=0.
Show Solution
(x−6)(x+1)=0;x=6,x=−1
Using Other Factoring Methods
Example: Setting it Equal to Zero and Factoring
Factor and solve the equation: x3=100x2.
Show Solution
First we must get everything equal to zero. If we divide both sides by x instead, we may miss one of our answers. That is why we must first set the equation equal to zero:
x3−100x2=0
Next we will factor out the GCF: x2(x−100)=0.
To solve this equation, we use the zero-product property. Set each factor equal to zero and solve.
x2(x−100)=0x2=0x=0(x−100)=0x=100
The two solutions are x=0 and x=100.
Try It
Factor and solve the quadratic equation: x20=2x19.
Show Solution
x19(x−2)=0;x=0,x=2
Example: Using the Method of Factor by Grouping
Factor and solve the equation: x3+11x2−121x−1331=0.
Show Solution
We will solve this we will use the the method of factoring by grouping. First group the terms.
x3+11x2−121x−1331=0
(x3+11x2)−(121x+1331)=0
The greatest common factor in the first grouping is x2.
The greatest common factor in the second grouping is 121.
Factor out x2 from the first grouping and 121 from the second grouping.
(x3+11x2)−(121x+1331)=0
x2(x+11)−121(x+11)=0
The common factor in the x2(x+11)−121(x+11)=0 is (x+11). We will now factor this out:
(x+11)(x2−121)=0
We can factor one more time:
(x+11)(x+11)(x−11)=0
To solve this equation, we use the zero-product property. Set each factor equal to zero and solve.
(x+11)(x+11)(x−11)=0(x+11)=0x=−11(x+11)=0x=−11(x−11)=0x=11
One solution is repeated twice, but we only need to write it once.
Therefore, the two solutions are x=−11 and x=11.
Try It
Factor and solve the quadratic equation: x3−12x2−144x+1728=0.
Show Solution
x3−12x2−144x+1728=0;x=−12,x=12
Using the Square Root Property
When there is no linear term in the equation, another method of solving a quadratic equation is by using the square root property, in which we isolate the x2 term and take the square root of the number on the other side of the equal sign. Keep in mind that sometimes we may have to manipulate the equation to isolate the x2 term so that the square root property can be used.
A General Note: The Square Root Property
With the x2 term isolated, the square root propty states that:
if x2=k,then x=±√k
where k is a nonzero real number.
Recall: principal square root
Recall that the princicpal square root of a number such as √9 is the non-negative root, 3.
And note that there is a difference between √9 and x2=9. In the case of x2=9, we seek all numberswhose square is 9, that is x=±3.
How To: Given a quadratic equation with an x2 term but no x term, use the square root property to solve it
Isolate the x2 term on one side of the equal sign.
Take the square root of both sides of the equation, putting a ± sign before the expression on the side opposite the squared term.
Simplify the numbers on the side with the ± sign.
Example: Solving a Simple Quadratic Equation Using the Square Root Property
Solve the quadratic using the square root property: x2=8.
Show Solution
Take the square root of both sides, and then simplify the radical. Remember to use a ± sign before the radical symbol.
x2=8x=±√8=±2√2
The solutions are x=2√2, x=−2√2.
Example: Solving a Quadratic Equation Using the Square Root Property
Solve the quadratic equation: 4x2+1=7
Show Solution
First, isolate the x2 term. Then take the square root of both sides.
4x2+1=74x2=6x2=64x=±√62
The solutions are x=√62, x=−√62.
Try It
Solve the quadratic equation using the square root property: 3(x−4)2=15.
Show Solution
x=4±√5
x=4√5, −4√5
Completing the Square
One method is known as completing the square. Using this process, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, a, must equal 1. If it does not, then divide the entire equation by a. Then, we can use the following procedures to solve a quadratic equation by completing the square.
We will use the example x2+4x+1=0 to illustrate each step.
Given a quadratic equation that cannot be factored and with a=1, first add or subtract the constant term to the right sign of the equal sign.
x2+4x=−1
2. Multiply the b term by 12 and square it.
12(4)=222=4
3. Add (12b)2 to both sides of the equal sign and simplify the right side. We have
x2+4x+4=−1+4x2+4x+4=3
4. The left side of the equation can now be factored as a perfect square.
x2+4x+4=3(x+2)2=3
5. Use the square root property and solve.
√(x+2)2=±√3x+2=±√3x=−2±√3
6. The solutions are x=−2+√3, x=−2−√3.
Properties of Equality and taking the square root of both sides
Remember that we are permitted, by the properties of equality, to add, subtract, multiply, or divide the same amount to both sides of an equation. Doing so won’t change the value of the equation but it will enable us to isolate the variable on one side (that is, to solve the equation for the variable).
The square root property gives us another operation we can do to both sides of an equation, taking the square root. We just have to remember when taking the square root (or any even root, as we’ll see later), to consider both the positive and negative possibilities of the constant.
The square root property
If √x=k
Then x=±k
Example: Solving a Quadratic by Completing the Square
Solve the quadratic equation by completing the square: x2−3x−5=0.
Show Solution
First, move the constant term to the right side of the equal sign by adding 5 to both sides of the equation.
x2−3x=5
Then, take 12 of the b term and square it.
12(−3)=−32(−32)2=94
Add the result to both sides of the equal sign.
x2−3x+(−32)2=5+(−32)2x2−3x+94=5+94
Factor the left side as a perfect square and simplify the right side.
(x−32)2=294
Use the square root property and solve.
√(x−32)2=±√294(x−32)=±√292x=32±√292
The solutions are x=32+√292, x=32−√292.
Try It
Solve by completing the square: x2−6x=13.
Show Solution
x=3±√22
Note that when solving a quadratic by completing the square, a negative value will sometimes arise under the square root symbol. Later, we’ll see that this value can be represented by a complex number (as shown in the video help for the problem below). We may also treat this type of solution as unreal, stating that no real solutions exist for this equation, by writing DNE. We will study complex numbers more thoroughly in a later module.
Using the Quadratic Formula
The fourth method of solving a quadratic equation is by using the quadratic formula, a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses when inserting a negative number.
We can derive the quadratic formula by completing the square. We will assume that the leading coefficient is positive; if it is negative, we can multiply the equation by −1 and obtain a positive a. Given ax2+bx+c=0, a≠0, we will complete the square as follows:
First, move the constant term to the right side of the equal sign:
ax2+bx=−c
2. As we want the leading coefficient to equal 1, divide through by a:
x2+bax=−ca
3. Then, find 12 of the middle term, and add (12ba)2=b24a2 to both sides of the equal sign:
x2+bax+b24a2=b24a2−ca
4. Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:
(x+b2a)2=b2−4ac4a2
5. Now, use the square root property, which gives
x+b2a=±√b2−4ac4a2x+b2a=±√b2−4ac2a
6. Finally, add −b2a to both sides of the equation and combine the terms on the right side. Thus,
x=−b±√b2−4ac2a
A General Note: The Quadratic Formula
Written in standard form, ax2+bx+c=0, any quadratic equation can be solved using the quadratic formula:
x=−b±√b2−4ac2a
where a, b, and c are real numbers and a≠0.
How To: Given a quadratic equation, solve it using the quadratic formula
Make sure the equation is in standard form: ax2+bx+c=0.
Make note of the values of the coefficients and constant term, a,b, and c.
Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula.
Calculate and solve.
Example : Solve A Quadratic Equation Using the Quadratic Formula
Solve the quadratic equation: x2+5x+1=0.
Show Solution
Identify the coefficients: a=1,b=5,c=1. Then use the quadratic formula.
x=−(5)±√(5)2−4(1)(1)2(1)=−5±√25−42=−5±√212
Recall adding and subtracting fractions
The form we have used to recall adding and subtracting fractions can help us write the solutions to quadratic equations.
ab±cd=ad±bcbd
Ex. The solutions
x=−b±√b2−4ac2a
can also be written as two separate fractions
x=−b2a±√b2−4ac2a.
Example: Solving a Quadratic Equation with the Quadratic Formula
Use the quadratic formula to solve x2+x+2=0.
Show Solution
First, we identify the coefficients: a=1,b=1, and c=2.
Substitute these values into the quadratic formula.
x=−b±√b2−4ac2a=−(1)±√(1)2−(4)⋅(1)⋅(2)2⋅1=−1±√1−82=−1±√−72
The solutions to the equation are unreal because the square root of a negative number does not exist in the real numbers. If asked to find all real solutions, we would indicate that they do not exist by writing DNE.
We may also express unreal solutions as complex numbers by writing
x=−1±i√72
or
x=−12+√72i and x=−12−√72i , where √−1 is expressed as i.
Try It
Solve the quadratic equation using the quadratic formula: 9x2+3x−2=0.
Show Solution
x=−23, x=13
Key Concepts
Many quadratic equations can be solved by factoring when the equation has a leading coefficient of 1 or if the equation is a difference of squares. The zero-factor property is then used to find solutions.
Many quadratic equations with a leading coefficient other than 1 can be solved by factoring using the grouping method.
Another method for solving quadratics is the square root property. The variable is squared. We isolate the squared term and take the square root of both sides of the equation. The solution will yield a positive and negative solution.
Completing the square is a method of solving quadratic equations when the equation cannot be factored.
A highly dependable method for solving quadratic equations is the quadratic formula based on the coefficients and the constant term in the equation.
The discriminant is used to indicate the nature of the solutions that the quadratic equation will yield: real or complex, rational or irrational, and how many of each.
The Pythagorean Theorem, among the most famous theorems in history, is used to solve right-triangle problems and has applications in numerous fields. Solving for the length of one side of a right triangle requires solving a quadratic equation.
Glossary
completing the square
: a process for solving quadratic equations in which terms are added to or subtracted from both sides of the equation in order to make one side a perfect square
discriminant
: the expression under the radical in the quadratic formula that indicates the nature of the solutions, real or complex, rational or irrational, single or double roots.
Pythagorean Theorem
: a theorem that states the relationship among the lengths of the sides of a right triangle, used to solve right triangle problems
quadratic equation
: an equation containing a second-degree polynomial; can be solved using multiple methods
quadratic formula
: a formula that will solve all quadratic equations
square root property
: one of the methods used to solve a quadratic equation in which the x2 term is isolated so that the square root of both sides of the equation can be taken to solve for x
zero-product property
: the property that formally states that multiplication by zero is zero so that each factor of a quadratic equation can be set equal to zero to solve equations
Section CR 14 Homework Exercises
How do we recognize when an equation is quadratic?
When we solve a quadratic equation, how many solutions should we always start seeking out? Explain why when solving a quadratic equation in the form ax2+bx+c=0 we may graph the equation y=ax2+bx+c and have no zeroes (x-intercepts).
When we solve a quadratic equation by factoring, why do we move all terms to one side, having zero on the other side?
In the quadratic formula, what is the name of the expression under the radical sign b2−4ac, and how does it determine the number of and nature of our solutions?
Describe two scenarios where using the square root property to solve a quadratic equation would be the most efficient method.
For the following exercises, solve the quadratic equation by factoring.
x2−4x−21=0
x2−9x+18=0
2x2+9x−5=0
6x2+17x+5=0
4x2−12x+8=0
3x2−75=0
8x2+6x−9=0
4x2=9
2x2+14x=36
5x2=5x+30
4x2=5x
7x2+3x=0
x3−9x=2
For the following exercises, solve the quadratic equation by using the square root property.
x2=36
x2=49
(x−1)2=25
(x−3)2=7
(2x+1)2=9
(x−5)2=4
For the following exercises, solve the quadratic equation by completing the square. Show each step.
x2−9x−22=0
2x2−8x−5=0
x2−6x=13
x2+23x−13=0
2+z=6z2
6p2+7p−20=0
2x2−3x−1=0
For the following exercises, determine the discriminant, and then state how many solutions there are and the nature of the solutions. Do not solve.
2x2−6x+7=0
x2+4x+7=0
3x2+5x−8=0
9x2−30x+25=0
2x2−3x−7=0
6x2−x−2=0
For the following exercises, solve the quadratic equation by using the quadratic formula. If the solutions are not real, state No Real Solution.
2x2+5x+3=0
x2+x=4
2x2−8x−5=0
3x2−5x+1=0
x2+4x+2=0
4+1x−1x2=0
x+4x+6=0
Beginning with the general form of the quadratic equation ax2+bx+c=0, solve for x by using the completing the square method, thus deriving the quadratic formula.
Show that the sum of the two solutions to the quadratic formula is −ba.
A person has a garden that has a length 10 feet longer than the width. Set up a quadratic equation to find the dimensions of the garden if its area is 119ft2. Solve the quadratic equation to find the length and width.
Abercrombie and Fitch stock had a price given as P=0.2t2−5.6t+50.2, where t is the time in months from 1999 to 2001. (t=1 is January 1999). Find the two months in which the price of the stock was $30.
Suppose that an equation is given p=−2x2+280x−1000, where x represents the number of items sold at an auction and p is the profit made by the business that ran the auction. How many items solve would make this profit a maximum? Solve this by graphing the expression in your graphing utility and finding the maximum using 2nd CALC maximum. To obtain a good window for the curve, set x [0,200] and y [0,10000].
A formula for the normal systolic blood pressure for a man age A, measured in mmHg, is given as P=0.006A2−0.02A+120. Find the age to the nearest year of a man whose normal blood pressure measures 125 mmHg.
The cost function for a certain company is C=60x+300 and the revenue is given by R=100x−0.5x2. Recall the the profit is revenue minus cost. Set up a quadratic equation and find two values of x (production level) that will create a profit of $300.
A falling object travels a distance given by the formula d=5t+16t2 ft, where t is measured in seconds. How long will it take for the object to travel 74 ft?
A vacanot lot is being converted into a community garden. The garden and walkway around its perimeter have an area of 378 ft2. Find the width of the walkway if the garden is 12ft wide by 15ft long.
An epidemiological study of the spread of a certain influenza strain that hit a small school population found that the total number of students, P, who contracted the flu t days after it broke out is given by the model P=−t2+13t+130, where 1<=t<=6. Find the day that 160 students had the flu. Recall that the restriction on t is at most 6. |
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6.262 | Spring 2011 | Graduate
Discrete Stochastic Processes
Course Description
Discrete stochastic processes are essentially probabilistic systems that evolve in time via random changes occurring at discrete fixed or random intervals. This course aims to help students acquire both the mathematical principles and the intuition necessary to create, analyze, and understand insightful models for a …
Discrete stochastic processes are essentially probabilistic systems that evolve in time via random changes occurring at discrete fixed or random intervals. This course aims to help students acquire both the mathematical principles and the intuition necessary to create, analyze, and understand insightful models for a broad range of these processes. The range of areas for which discrete stochastic-process models are useful is constantly expanding, and includes many applications in engineering, physics, biology, operations research and finance.
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14425 | https://blog.csdn.net/weixin_35903223/article/details/117239683 | C语言中心对称图形定义,中心对称图形的定义-CSDN博客
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C语言中心对称图形定义,中心对称图形的定义
最新推荐文章于 2022-03-24 10:37:32 发布
转载 于 2021-05-24 07:06:41 发布·1k 阅读
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#C语言中心对称图形定义
本文详细介绍了中心对称图形的概念及性质，列举了常见的中心对称图形，并对比了正偶边形与正奇数边形的区别。此外，还特别提到了反比例函数的图像双曲线为中心对称图形的特点。
中心对称图形∶
在平面内，把一个图形绕着某个点旋转180°，如果旋转后的图形与另一个图形重合，那么就说明这两个图形是中心对称图形，它们的形状关于这个点成中心对称，这个点叫做它的对称中心，旋转180°后重合的两个点叫做对称点。
对称中心图形的性质∶
1、对称中心平分中心对称图形内通过该点的任意线段且使中心对称图形的面积被平分。
2、成中心对称的两个图形全等。
3、成中心对称的两个图形上每一对对称点所连成的线段都被对称中心平分。
区分∶中心对称是两个图形间的位置关系，而中心对称图形是一种具有独特特征的图形。
常见的中心对称图形：
线段，矩形，菱形，正方形，平行四边形，圆,边数为偶数的正多边形，某些不规则图形等。
正偶边形是中心对称图形，正奇数边形不是中心对称图形。
正六角形是中心对称图形，等腰梯形不是中心对称图形，等边三角形(正三角形)，至少需旋转120度，而不是180度，所以它不是中心对称图形。反比例函数的图像双曲线是以原点为对称中心的中心对称图形。
相关资源：奇数阶 中心对称 幻方生成器资源
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14426 | https://www.youtube.com/watch?v=slFqL86q3EA | What is a Weighted Average?
Tyler DeWitt
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Posted: 23 Oct 2012
To see all my Chemistry videos, check out
What is a weighted average? How do you calculate a weighed average? It is a type of average based on the abundance of what you are measuring. Weighted averages are used for atomic mass and atomic weight, based on the mass numbers of various isotopes.
288 comments
Transcript:
Intro In this video, we're going to talk about weighted
averages. We'll talk about what a weighted average is,
what a weighted average means and how to calculate a weighted average. Here's an example where we calculate a weighted
average. Let's say that we have a group of kids here,
there are five kids, some boys and some girls and for simplicity sake let's just say that
each one weighs a hundred pounds. Now one of their father shows up and the father
is a bodybuilder. Check out these muscles, check out these eight
pack abs, I guess this means the guy has his shirt off which is kind of creepy but anyway
he's a big body builder and he weighs 300 pounds. So now we have a group of six people here. The kids 100 pounds each and the father 300
pounds. So now you can ask a question about this group
of people, you could say what is that average weight? We got these two different weights here, 100
pounds and 300 pounds, so you could calculate what I call a regular average. You take 100 pounds and 300 hundred pounds
because those are the two choices and you add them together and divide by two because
they're two different weights. You get 200 pounds and 200 pounds is right
smack dab in between 100 and 300. So that's one way that you could get the average
weight, 200 pounds. But here's the thing, is it really fair to
say that the average weight of this group is 200 pounds because there are five kids
that each weigh 100 pounds and there's only one father that weighs 300 pounds. So to say that the average weight is right
in the middle of those two weights kind of doesn't make a lot of sense. It feels like we should be able to take into
account the fact that there are many more kids who weigh a lot less than the one bodybuilder
father. This is where the idea of a weighted average
comes into play. A weighted average takes into account how
many things in each group you have. Here's how I would calculate the weighted
average in this group of people, okay? I would take the fact that there are five
kids here that weigh 100 pounds each so I do 100 + 100 + 100 + 100 + 100, that's five
hundreds for each of the five kids and then I add 300 for the weight of the bodybuilder
and I divide by six because there are six things total. Sometimes it's easier to express this with
multiplication, I've got five times a hundred plus one times three hundred divided by six
because there are six things and when we crank through these equations we end up with a different
answer than this, we end up with the 133 pounds. This is the answer to the weighted average. Now you'll notice this number 133 is much
less than 200 and 133 is a lot closer to the weight of the kids. That's because of the weighted average and
how it works. The weighted average will pull down the average
closest to whatever we have the most of, okay? So if we're just half and half it would be
200 pounds, we would have the same number of each, but since we have more of the kids
we take that 200 hundred and we pull the number down closer to the weight of what we have
more of. So that's why the weighted average for here
is a 133, very close to the weight of the kids. Weighted Average Example Now the weighted average could work the other
way as well. Let's say that we had a group where four people
were 300 pound bodybuilders and there's only one kid that weighed 100 pounds. In that case, we calculate the weighted average. Four 300 pound bodybuilders plus one 100 pound
kid divided by five total or we could do this with multiplication (4x300) plus (1x100) divided
by five and in that case we get 260 pounds, alright? This number is a lot higher than the number
200 that's right in between and it's much closer to the weight of the bodybuilders. It's much closer to 300 because they're more
of the body builders and there's only this one kid. So the weighted average should be closer to
the weight of what we have more of. Calculating a Weighted Average Now for weighted average, we don't just have
to have two things like the kids and the bodybuilders, we can have a number of different things more
than two and we can have a whole bunch of them. In this case I'm talking about a parking lot
that has three different kind of cars. They're all called Lemonas because they look
like lemons. We got the Lemona G that weighs 3,000 pounds,
the Lemona GX that weighs 4,000 pounds and the Lemona GXL which weighs 5,000 pounds. I have a different amounts of each of them. So how would I calculate the weighted average
here? What I do is I take the weight of Lemona G,
that's 3000 pounds, and I multiply it by how many Lemona G's there are. They're 32 times 3000 pounds each, I take
that and then I add the number of GX's. So I've got 5 of them, 5 times 4,000. And then the GXL's, I've got 7 of them, 7
times 5,000 pounds. Now I do this, I've accounted for the weights
of each one of these cars and now I've got to divide by the total number of cars I have
in this group and that's 44. So these are the G's, these are the GX's,
and these are the GXL's. Do the multiplication of each, add it up,
and I'll end up with 3,432 pounds. Now look at how this number compares to weights
of the individual cars. This number is closest to the weight of the
G and that makes sense because I have a lot more of the Lemona G's than I have of the
heavier GX's and GXL's. So this is the weighted average where as the
regular average for these cars would have been 3,000 plus 4,000 plus 5,000 divided by
3 which would have given us a number that was smack dab in the middle. But once again the weighted average here gives
us a number that takes into account how many of each type of these cars I have and gives
me a number that is much closer to the type of car that I have the most of. Using Percents Now a lot of times when you're talking weighted
average, you'll have to work with percent so instead of getting the number of things
you'll get the percentages of them like right here. Here's how you calculate a weighted average
using percent. Okay, what I'm going to do is I'm going to
take the weight of the Lemona G here and I'm going to multiply it by 73 percent expressed
as a decimal. So the decimal place in 73 percent would be
here. To turn this percentage into a decimal I'll
have to move the decimal place two spots to the left so I'm going to move it to here to
here, okay? The decimal place is going to end up right
here. So I'm going to do 0.73 and then multiply
that times the weight, 3,000 pounds, and that's the Lemona G. Now for the GX, I'll take 11 percent and express
that as a decimal. Move the decimal place two spots the left
so it's going to end up here, 0.11 and multiply that times the weight of the GX, 4,000 pounds
for that model the car. And finally I'll take 16 percent and express
it as a decimal, 0.16 and multiply it by the weight of the GXL model. Now when you're working with percentages,
you don't divide by anything. All you do is multiply the weights by the
percentages expressed in decimals and that's all you have to do. This is the same problem as the one that I
did before, we're just expressing these abundances or the amount that we have in percent so the
answer 3,432 pounds is going to be the same as what we got earlier. Anyway, that's how you do weighted average
using percent. Now I have to say this, weighted average doesn't
mean that you have to use weights of things, alright? You can calculate a weighted average with
the amount of money people make or like how much volume you have and different types of
containers, anything you can think of, any measurement you can think of you can do a
weighted average of. So I've been using examples of weights here,
weights of people, weights of cars but you don't just have to do weights for a weighted
average. The only reason you call it a weighted average
is because it takes into account the number of each thing you have and they kind of pull
the average closest to their value so that's why you call it a weighted average because
you give different weight to each type of thing that you have. Finally, if you're interested to see how this
equation here is the same as the one that I did here, you can finish this video up and
I'll show you at the very end how these are equivalent expressions. So these two expressions are equivalent ways
of writing the calculations for the weighted average for this information here. Let me show you how they're equivalent. We're going to take this and we want to end
up with something that looks like this. The first thing we can do is we can see these
three different things that are added together on the top of a fraction. We can separate them out, okay? So, I'm going to do 32 times 3,000 divided
by 44 plus 5 times 4,000 divided by 44 plus 7 times 5,000 divided by 44, okay? So I can split this into three pieces, each
of them have the same denominator. Now take a look at this. I've got a number over 44 in each one of these
cases and this is kind of a percentage because 44 is my total, and each one of these numbers
in the numerator are fractions of that total. So what I can do since these are multiplied
together is I can divide this and then just keep it multiplied by that. So 32 divided by 44 gives me 0.73, that's
this part, times 3,000 (you can put that in parentheses). Now I can do 5 divided by 44 and that gives
me 0.11 and I'm going to multiply this now times 4,000, put that in parentheses, and
finally 7 divided by 44 gives me 0.16 times 5,000 and check it out! What I end up with when I split these up and
I do the division is exactly the same as the number that I get when I take these values
and use the straight percentages here. So the point is whether you choose to use
the individual numbers of each thing you have or whether you choose to use the percentages,
the way you go about calculating a weighted average has the same math in it. Anyway, now that you know how to calculate
a weighted average using the numbers of the things or the percentages of the things, you
can now go on and learn how to calculate atomic mass. |
14427 | https://brainly.com/question/49841543 | [FREE] During testing for S. pneumoniae and S. pyogenes, why is optochin more effective on S. pneumoniae than S. - brainly.com
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During testing for S. pneumoniae and S. pyogenes, why is optochin more effective on S. pneumoniae than S. pyogenes?
Why is bacitracin more effective on S. pyogenes than S. pneumoniae?
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Optochin is more effective on S. pneumoniae because of its susceptibility to optochin's effects, while bacitracin is more effective on S. pyogenes due to its sensitivity to low concentrations, allowing for differentiation between these streptococci.
During testing for S. pneumoniae and S. pyogenes, optochin is more effective on S. pneumoniae because this bacterium is more susceptible to optochin's inhibitory effects, which allows for differentiation between S. pneumoniae (optochin sensitive) and other alpha-hemolytic streptococci (optochin resistant). Bacitracin, on the other hand, is more effective on S. pyogenes than S. pneumoniae, because S. pyogenes is susceptible to very low concentrations of bacitracin and can be identified by this susceptibility. This characteristic helps to distinguish S. pyogenes (Group A strep) from other beta-hemolytic streptococci. The use of these antibiotics is critical for rapid identification and effective treatment of infections caused by these bacteria.
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Optochin is more effective on S. pneumoniae due to its sensitivity, allowing differentiation from other alpha-hemolytic streptococci. Bacitracin is more effective on S. pyogenes because this bacterium is susceptible to low concentrations of bacitracin, enabling its identification from other beta-hemolytic streptococci. These antibiotic sensitivities are important for accurate diagnosis and treatment of infections.
Explanation
During testing for Streptococcus pneumoniae and Streptococcus pyogenes, optochin is used as a differentiating agent. Optochin is particularly effective against S. pneumoniae because this species is sensitive to it, meaning the presence of optochin will inhibit the growth of S. pneumoniae. This sensitivity helps in differentiating S. pneumoniae from other alpha-hemolytic streptococci, which are usually resistant to optochin.
On the other hand, bacitracin is more effective on S. pyogenes, which is also known as Group A Streptococcus. S. pyogenes is sensitive to low concentrations of bacitracin, making it susceptible to its effects. When testing, the zone of inhibition around a bacitracin disk helps distinguish S. pyogenes from other beta-hemolytic streptococci, which are generally resistant to bacitracin.
The effectiveness of these antibiotics in laboratory tests is crucial for rapid identification and treatment of infections caused by these bacteria. Understanding the differences in susceptibility to antibiotics like optochin and bacitracin can assist healthcare professionals in diagnosing and managing bacterial infections effectively.
Examples & Evidence
An example of this would be conducting a culture on a blood agar plate. If a sample shows a zone of inhibition around an optochin disk, it indicates the presence of S. pneumoniae. Conversely, a clear inhibition zone around a bacitracin disk suggests S. pyogenes in the sample.
The information regarding the susceptibility of these bacteria to optochin and bacitracin is supported by clinical microbiology guidelines and studies that detail the characteristics and antibiotic sensitivity patterns of S. pneumoniae and S. pyogenes.
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14428 | https://arxiv.org/html/2407.09662v1 | Analytical Expression for Continuum-continuum Transition Amplitude of Hydrogen-like Atoms with Angular-momentum Dependence
J B Ji1,∗, K Ueda1,2,†, M Han1,3 and H J Wörner1,§ 1 Laboratorium für Physikalische Chemie, ETH Zürich, 8093 Zürich, Switzerland 2 Department of Chemistry, Tohoku University, Sendai, 980-8578, Japan 3 J. R. Macdonald Laboratory, Department of Physics, Kansas State University, Manhattan, KS 66506, USA $^$ jiabao.ji@phys.chem.ethz.ch $^†$ kiyoshi.ueda@tohoku.ac.jp $^§$ hwoerner@ethz.ch , ,
Abstract
Attosecond chronoscopy typically utilises interfering two-photon transitions to access the phase information. Simulating these two-photon transitions is challenging due to the continuum-continuum transition term. The hydrogenic approximation within second-order perturbation theory has been widely used due to the existence of analytical expressions of the wave functions. So far, only (partially) asymptotic results have been derived, which fail to correctly describe the low-kinetic-energy behaviour, especially for high angular-momentum states. Here, we report an analytical expression that overcome these limitations. They are based on the Appell’s function and use the confluent hypergeometric function of the second kind as the intermediate states. We show that the derived formula can be applied to arbitrarily high angular momenta and quantitatively agrees with the numerical simulations using the time-dependent Schrödinger equation. Furthermore, we give an angular-momentum-dependent asymptotic form of the outgoing wavefunction and their continuum-continuum dipole transition amplitudes.
1 Introduction
The question “How long does the photoionsation take?” has intrigued researchers for decades ever since the early years of quantum theories. Its natural timescale of attosecond (as, s) has introduced a new field of science, known as the attosecond science, which is highlighted by the 2023 Nobel Prize in Physics . The last 30 years have witnessed the rapid development of the toolbox for investigating the photoionisation processes with phase information embedded. This includes, but is not limited to, high-harmonic generation (HHG) [2, 3] as a table-top XUV source thanks to the availability of the high-power lasers, the schemes of performing the phase measurement such as attosecond streaking [4, 5, 6] and the reconstruction of attosecond beating by interference of two-photon transitions (RABBIT) [7, 8], and three-dimensional momentum-resolved particle detectors such as the cold target recoil ion momentum spectrometer (COLTRIMS) [9, 10, 11], and so on. As a result, it has become possible to measure the Wigner time delay defined as [12, 13, 14]:
| | |
--- |
| | (1) |
where is the transition amplitude for photoionisation, and in the following discussions in this article, unless otherwise stated, we use the atomic units, where , and the atomic unit of time . Although this definition is rather straightforward, and the transition amplitudes (phases and moduli) of one-photon ionisation for atoms and molecules can be routinely calculated by e.g. random-phase approximation with exchange (RPAE) [15, 16], time-dependent local-density approximation (TDLDA) [17, 18] and Schwinger variational methods such as ePolyScat [19, 20], in reality, the schemes of the phase measurements involves two-photon transitions, which differs from one-photon ionisation. An illustration of the RABBIT measurement scheme is illustrated in figure 1. The attosecond pulse train (APT) created by HHG contains typically odd harmonics of the fundamental driving field (), which leads to the main bands (dashed lines in figure 1). If a copy of the driving field is spatially and temporally overlapped with the APT as the dressing field, an additional sideband (solid line in figure 1) lying between the two main bands appears, which corresponds to either absorbing one photon from the lower main band (the absorption pathway), or emitting one photon from the higher main band (the emission pathway). The transition amplitudes of the absorption and the emission pathways can be written as:
| | |
--- |
| | (2) |
where and in the subscript are for the absorption and the emission pathways, respectively. The relative time delay of the dressing field is denoted by (the dressing field has the light path longer than the attosecond pulse, where is the speed of light), meaning that at the time of the attosecond pulse burst the dressing field has the phase of . The transition amplitudes at are denoted as . The phase of the dressing field contributes to the absorption and the emission pathways oppositely, and the side-band intensity is expressed by:
| | |
--- |
| | (3) |
where is the phase difference between the emission and the absorption pathways. We can further write
| | |
--- |
| | (4) |
where corresponds to the relative phase between the harmonic frequencies in the attosecond pulse, also known as the attochirp, and is known as the two-photon ionisation time, since it corresponds to the phase difference of the two-photon transition amplitudes between the emission and the absorption pathways. More discussions and examples can be found in the book chapter .
From the theoretical point of view, the ab initio computation such as numerically solving the time-dependent Schrödinger equation (TDSE) is limited to the modelled potential with one active electron, e.g. the hydrogen-like atoms, while modelling the potential for molecules can be very challenging. A common approach for handling this problem is to separate the two-photon transition into two parts, i.e. the bound-continuum transition and the continuum-continuum transition, so that the two parts can be treated individually. This essentially stems from the lowest-order perturbation theory. The two-photon transition matrix element from an bound initial state to a final state in the continuum with momentum can be represented as :
| | |
--- |
| | (5) |
and the two-photon transition amplitude reads
| | |
--- |
| | (6) |
where and are the complex amplitudes (field strengths with phases included) of the XUV and dressing fields, respectively, and is the vector with modulus of and pointing towards the emission angle , either in the lab frame or in the molecular frame for molecules. Within the framework of dipole approximation, the dipole terms for the XUV and dressing fields are and , respectively. The sign refers to summing over all discrete states and integrating over all continuous states. If one assumes that the two-photon process is dominated by the pathway that the electron absorbs one XUV photon from the initial state to an intermediate state in the continuum (assuming that the XUV photon energy is higher than the photoionisation energy) with the corresponding kinetic energy, after which a continuum-continuum transition occurs due to the presence of the dressing IR field, where the electron either absorbs or emits one IR photon (multi-IR processes are ignored here, assuming the dressing field is relatively weak), as sketched in figure 1, then the two-photon transition amplitude can be expressed as:
| | |
--- |
| | (7) |
where the terms and refer to the bound-continuum and continuum-continuum transitions, respectively. Here we have decomposed the emission-angle-dependent transition amplitude into the spherical harmonics, i.e.
| | |
--- |
| | (8) |
where is the spherical harmonic (in the Condon–Shortley phase convention) with angular momentum and magnetic quantum number of and , respectively. For the continuum-continuum part, the transition amplitude can be expressed as the product of the angular part and the radial part:
| | |
--- |
| | (9) |
where is the polarisation of the dressing photon represented in the spherical harmonics. The -factor originates from the contour integral around the pole in equation (5), which is explained in detail in . The angular part can be calculated by the Clebsch–Gordan coefficients:
| | |
--- |
| | (10) |
where is the Clebsch–Gordan coefficient and is real in the Condon–Shortley phase convention. The latter integral between the radial wavefunctions, with the -term included, is denoted as in the following discussion, where and are the momenta of the intermediate and final states, respectively.
The radial wavefunctions in principle depend on the shape of the radial potential, which can be complicated for molecular cases. If we assume that the continuum-continuum transition is dominated by the long-range effect and is not sensitive to the local potential, then we can use the transition amplitudes derived from the hydrogen-like atoms. Therefore, instead of expression (1), the two-photon Wigner-like time delay is defined using the finite-difference method:
| | |
--- |
| | (11) |
which can be represented either in the spherical-coordinate basis or in the spherical-harmonic basis. If we further assume that the continuum-continuum transition amplitude in equation (7) does not depend on or , then the two-photon time delay can be separated into two parts:
| |
| |
| |
| |
| | (12) |
where we have approximated the derivative in expression (1) by its finite difference, and the continuum-continuum time delay is defined as
| | |
--- |
| | (13) |
By separating the two-photon time delay into the one-photon part and the continuum-continuum part, the problem is thus simplified as how to calculate the term , or equivalently the integral for the hydrogen-like atoms. Dahlström et al. reported two models based on the Wentzel–Kramers–Brillouin (WKB) approximation with asymptotic expansion, i.e. considering the phase modulation (P) and considering the phase and amplitude (modulus) modulation (P+A), both expressed by the analytical formulae involving the gamma function :
| | |
--- |
| | (14) |
and
| | |
--- |
| | (15) |
where and are the normalisation factors for the intermediate state and the final state, respectively. We note that the normalisation factor depends on normalising according to the cross section or normalising according to the flux. Since for the scattered wave the total flux is
| | |
--- |
| | (16) |
where is the scattering flux density, is the scattering amplitude, and is the differential cross section. The well-known solutions to the Coulomb wave equation are conventionally normalised according to the cross section, while the WKB approximation yields the flux-based normalisation factor, which contains an additional factor , which is clear from comparing their asymptotic oscillation moduli . The TDSE results lie typically between these two models, and they converge towards the same values for high kinetic energies . A number of works have explicitly [11, 25, 26, 27, 28, 29, 30] or implicitly (e.g. to assume the two-photon time-delay difference between two species with the same or similar electron kinetic energies is equal or almost equal to their one-photon time-delay difference) [31, 32] used this assumption and have shown good agreement with the experimental results on molecules, indicating that this approximation is reasonable.
Despite the great success of this elegant model in the past decade, some features of experimental observations cannot be explained by these models. For example, the phases of these models do not depend on the angular momentum of the intermediate or the final state, which causes that the angle-dependent time delay of two-photon ionisation is simply the angle-dependent time delay of one-photon ionisation plus an isotropic continuum-continuum time delay. This can be proven as follows. Using equations (7,9), we find
| | |
--- |
| | (17) |
Writing this formula into the spherical coordinates using equation (8), we have
| |
| |
| |
| |
| |
| | (18) |
where the term depends on the polarisation of the dressing field and does not contribute to the energy-dependent phase variation, i.e. the time delay. Hence, for helium one would expect an isotropic two-photon time delay, since its one-photon ionisation time delay has no angular dependence. This contradicts the experimental observation , and the mechanism was demonstrated in that the moduli of the transition amplitudes for and have preference depending on the increase or decrease of the electron kinetic energy, which is known as the Fano’s propensity rule in the continuum-continuum transition. The same mechanism was also confirmed by a more recent experimental work from the authors using the circular-XUV-circular-IR RABBIT, where the Fano’s propensity rule is embodied by the angle-dependent RABBIT phase difference between the co- and counter-rotating XUV and IR.
The missing piece was partially filled by Boll et al. who replaced the final-state wavefunction from the asymptotic formula by the regular solution of the Coulomb wave equation, which is expressed by the Kummer’s function, or the confluent hypergeometric function of the first kind . The result can also be written as an analytical expression using Gauss’ hypergeometric function . This model quantitatively reproduces Fano’s propensity rule at high electron kinetic energies and the angle-resolved time delay of helium [36, 37]. However, the agreement deteriorates at lower energies (typically below 10 eV), particularly for larger s, and in the same energy range the isotropic formulae proposed by Dahlström et al. also show large discrepancy compared to the TDSE. On the other hand, a lot of interesting phenomena in time delays, including shape resonance and two-center interference have been experimentally studied in this energy region, where the level of accuracy of the theory is partially limited by the reduced capability of the continuum-continuum delay models describing the relatively slow electrons. The general method for obtaining the -photon transition amplitude from the -photon transition amplitude has been known for a long time [38, 39], which involves expressing the Green’s function by the Sturmian expansion and extending into the continuum. We shall show its application as a modification of expression (5). We give its analytical expression using the Appell’s function, which quantitatively agrees with the TDSE and the experimental results in the whole energy range including the low-kinetic-energy region. We also note that the isotropic model can be extended to the -dependent model by expanding the WKB approximation to the next term, and the Fano’s propensity rule and the -dependent continuum-continuum transition phases can be nicely reproduced.
This article is structured as following. In section 2, we first define the functions for the intermediate and final states, then we derive the analytical formula for the transition amplitudes between the Tricomi’s and Kummer’s functions. In section 3, we show multiple results including the continuum-continuum transition phases and time delays, and the Fano’s propensity rule. These results are compared to TDSE and other theoretical approaches. We also compare our method to the experiment of circular-XUV-circular-IR RABBIT on helium . The conclusion and outlook can be found in section 4.
2 Method
2.1 Regular and irregular solutions to the Coulomb wave equation
The radial Coulomb electronic wave equation for the hydrogen atom (central charge ) with asymptotic kinetic energy and angular momentum reads (see , section (33.14) and , whereby equation (2) needs to be multiplied with for describing an attractive potential):
| | |
--- |
| | (19) |
where is the radial wavefunction. Equation (19) can be adapted for arbitrary by transforming and , where and are the asymptotic kinetic energy and the radius, respectively. For simplicity, in the discussion below we let , while the treatment is general for any . The regular solution of the outgoing radial wavefunction that satisfies the Coulomb wave equation can be written as (note that we have absorbed the factor into , compared to the definition in ):
| | |
--- |
| | (20) |
where is the momentum, and is the confluent hypergeometric function (Kummer’s function), also known as the function. Note that is a real function under such definition. The normalisation factor for the regular solution of the Coulomb wave equation with the Coulomb phase included is expressed as:
| | |
--- |
| | (21) |
such that oscillates with amplitude of 1 at . Since equation (19) is a second-order differential equation, it has two linearly-independent solutions, and the other one can be expressed using the confluent hypergeometric function of the second kind , also known as the Tricomi’s function:
| | |
--- |
| | (22) |
where we define the prefactor as:
| | |
--- |
| | (23) |
such that
| | |
--- |
| | (24) |
Its real part is another real function that satisfies the Coulomb wave equation, also known as the irregular solution:
| | |
--- |
| | (25) |
We note that and are linear combinations of and its complex conjugate , which are two linearly independent solutions. Asymptotically, approaches the spherical wave with a position-dependent phase (the Coulomb phase) :
| |
| |
| | (26) |
Near the origin, the regular solution approaches zero, while the irregular solution diverges.
Although the irregular solution has infinitely large probability density near the origin, the product does not diverge near the origin, as plotted in figure 2 (note that the normalisation for Coulomb waves here ensures that oscillates with modulus of 1 asymptotically), so the integral is finite. In the framework of the Green’s function method, the two photon ionisation amplitude is proportional to the following integral :
| | |
--- |
| | (27) |
where and are the radial wavefunctions of the initial (bound) and final (continuum) states, respectively, and is the Green’s function of the intermediate state that couples the initial and the final states. For the Coulomb potential, the initial and final states are the regular solutions to the Coulomb equation. The radial Green’s function in the three-dimensional space is [43, 44, 45]:
| | |
--- |
| | (28) |
where and () is the smaller (greater) variable between and . The Whittaker’s functions and are related to the confluent hypergeometric functions by , section (13.14)
| | |
--- |
| | (29) |
and
| | |
--- |
| | (30) |
If we assume that the bound-continuum transition occurs mainly in the short range due to the finite spatial extension of the initial wavefunction, and the continuum-continuum transition has major contribution from long range, we can replace and in the integral (27) by and , respectively. Thus, the two-photon transition amplitude is proportional to the product of
| | |
--- |
| | (31) |
and
| | |
--- |
| | (32) |
which corresponds to the bound-continuum and continuum-continuum transitions, respectively. Note that the additional term () comes from the transform from the Whittaker’s functions to the confluent hypergeometric functions. Therefore, equation (5) is slightly modified as:
| | |
--- |
| | (33) |
where for bound states and for continuum states is the outgoing-wave solution corresponding to the standing-wave solution . The former relation is also indicated by the fact that the confluent hypergeometric functions of the first kind () and second kind () truncate into the same associated Laguerre polynomials for the bound states. Compared to the approach of Boll et al. , integral (32) uses the intermediate of the outgoing-wave solution to the Coulomb wave equation instead of its asymptotic form, which is also used in Dahlström et al.’s approach . Therefore, we propose the following calculating procedure that if the final state from one-photon ionisation has partial wave with angular momentum and asymptotically approaches , we choose the intermediate state as
| |
| |
| | (34) |
and the final state as
| |
| |
| | (35) |
where and are the angular and linear momenta for the intermediate state and the final state, respectively. For the RABBIT experiment, where a photon from the dressing field is either absorbed (absorption pathway) or emitted (emission pathway), the final kinetic energy is , while can have both signs for each pathway, unless restricted by the magnetic quantum number in circular fields. For example, for the electron in hydrogen, if a linearly polarised XUV photon ionises it to the continuum state (-axis defined as the polarisation direction), and the dressing field is co-linearly polarised, both absorption and emission pathways lead to the superposition of and states as the final state; on the other hand, if it is ionised by a circularly polarised XUV photon and has as the intermediate state (-axis defined as the light propagation direction), and the dressing field is also circular with co-rotating polarisation, then for the absorption pathway only is allowed, while the emission pathway leads to the superposition of the and states. The dipole transition amplitude between the intermediate state and the final state is therefore , and reads:
| | |
--- |
| | (36) |
where and are the flux-based normalisation factors. In the following sections, we derive an analytical expression of equation (36) and related integrals. We will also give a simpler approximation that describes the asymptotic behaviour with consideration of the centrifugal energy following the WKB approximation.
2.2 General expression of integral containing the product of the confluent hypergeometric functions of the first and the second kind
Using the definition of and , we can write equation (36) into:
| |
| |
| |
| |
| | (37) |
where is a small positive number that ensures the integral converges according to the Abel-Dirichlet test . Note that we have used the fact that is real and equal to its complex conjugate. Although the similar integral involving two Kummer’s functions has already been discussed by Gordon almost a century ago and the more general case is known to have the analytical formulation using the Appell’s functions [48, 49, 50], to the best of our knowledge, the formula for the integral involving the product of the confluent hypergeometric functions of the first and second kind has yet to be explicitly demonstrated. Here we derive the analytical expression following the spirit of Gordon . Let us consider the integral with a more general form:
| | |
--- |
| | (38) |
The confluent hypergeometric function of the second kind can be represented as:
| | |
--- |
| | (39) |
with the following recurrence relations , equation (13.3.10):
| | | |
---
| | | (40) |
Using equation (40), we have the relation:
| |
| |
| | (41) |
For , which is the case for the transition between two continuum states (from equation (37) we have ), the integral finally comes to terms:
| | |
--- |
| | (42) |
In particular, we can write the two expressions for 1 and 3 explicitly:
| |
| |
| | (43) |
and
| |
| |
| |
| |
| | (44) |
In order to obtain , we use equation (39), thus
| | |
--- |
| | (45) |
By using the following formula for the function , equation (07.23.07.0003.01):
| | |
--- |
| | (46) |
and swapping the order of integration, we have:
| | |
--- |
| | (47) |
Using the definition of the generalised hypergeometric functions , equation (07.19.02.0002.01):
| | |
--- |
| | (48) |
we have:
| |
| |
| | (49) |
Let us use the integral representation of the Appell’s function , equation (07.36.07.0001.01):
| |
| |
| | (50) |
where we used . Compared to equation (49), we have , , , , , and . Therefore, we finally get
| |
| |
| |
| |
| | (51) |
where in the last step we used the relation , equation (5.11.1):
| | |
--- |
| | (52) |
With help of equations (41, 43, 44), we can calculate the multipole transition amplitudes between states expressed by the confluent hypergeometric functions of the first and the second kind. Therefore, equation (37) can be expressed as: 111Here we make a special note that although in equation (53) can be exactly zero, where the last two arguments in equation (51) are real and can be evaluated by analytic continuation when one of them has modulus greater than 1. However, due to the ambiguous phase of that occurs upon analytic continuation, one of the branches of the parameters may become incorrect, which means that taking the complex conjugates of all the first four arguments in does not return the complex conjugate of the result, although from its definition one would expect so. Therefore, it is recommended to numerically confirm that the case of is indeed the limit of .
| | |
--- |
| | (53) |
| |
One can find the similarity between our equations (41, 43, 44, 51) and the equation (3.1) in reference . In fact, we can write
| |
| |
| |
| |
| | (54) |
and one immediately sees that for the purpose of calculating the transition amplitude between two regular solutions to the Coulomb wave equation, also known as the Gordon’s integral, the result is given by the imaginary part of equation (54), which is also given in [48, 49, 50], while its real part corresponds to the transition amplitude between the regular and irregular solutions, where the regular and irregular solutions are defined to be real functions. We note that our formula has similarity to the general expression for the double- or multi-photon ionisation amplitude from bound states of hydrogen atoms given in references [53, 54], which use the Sturmaian expansion with modifications in the continuum.
2.3 Asymptotic formula for the CC transition amplitude with the centrifugal potential included
The purpose of this section is to demonstrate the mechanism with simpler mathematical expressions. Such approaches are usually available for describing the asymptotic behaviour at high kinetic energies, while we try to make the compromise that the formula applied at low kinetic energies should at least give a qualitative description, such that the main features are still captured. Let us consider the phase contribution of the centrifugal potential under the WKB approximation. The phase factor of the outgoing wave can be written as:
| |
| |
| |
| | (55) |
where comes from the centrifugal potential, and the expansion is made at . Thus, the outgoing radial wave function can be estimated as:
| |
| |
| |
| | (56) |
where scales with for the () wave, but with for waves with higher angular momenta. This explains why the phase contribution of the centrifugal potential should not be neglected, as is a rather slow convergence under the experimental context (typically , ). The dipole transition between two outgoing waves can thus be approximated as:
| |
| |
| | (57) |
where and . The factor of comes from the fact that the final state represented by the Kummer’s function contains the - and -parts, while the -part practically vanishes due to the large imaginary part in the gamma function . Here we have used the incomplete gamma function
| | |
--- |
| | (58) |
to include the case where the integral starts from instead of 0. This can be useful for the more complex scattering problem, where the (time-dependent) Schrödinger equation is numerically solved inside the radius , while the outgoing electron wavefunction is extrapolated beyond using the -dependent asymptotic formula, assuming that the potential is converged except the Coulomb () and centrifugal-like () parts. When , is the gamma function, and . The first term in equation (57) corresponds to equation (14) reported by Dahlström et al. [22, 23, 56], and the second term depends on the angular momenta of the initial and the final states.
We note that Dahlström et al. also reported another formula (equation (15)) that has the correction for the amplitude factor according to the WKB approximation:
| | |
--- |
| | (59) |
where the amplitude factor is
| | |
--- |
| | (60) |
Following the similar treatment, equation (57) is modified as
| | |
--- |
| | (61) |
where . The superscripts and in equations (57, 61) refer to including the phase variation and the phase and amplitude variations, respectively. For the hydrogen atom at low kinetic energies, the phase computed by TDSE typically lies between equation (57) and equation (61) with . Although the agreement may improve by choosing a suitable in a similar fashion as in the (P+A′) model of reference , it is better to evaluate according to equation (53).
2.4 Numerical evaluation
For numerical evaluation in this work, we used Python 3.8.10 with numpy 1.20.2 and mpmath 1.1.0 . The latter package is capable of calculating many special functions with complex arguments.
3 Results and discussion
3.1 Continuum-continuum transition phase for the hydrogen atom
In order to test the accuracy of the analytical formula and its asymptotic approximations, we compare to the time-dependent Schrödinger equation (TDSE) results reported in [23, 36], which are currently considered to be the benchmark for the CC transition amplitudes. As shown in figure 3, our equation (53) perfectly agrees with the TDSE values, which manifests the validity of our approach. Regarding the asymptotic approximations, the (P) and (P+A) formulae given by Dahlström et al. are independent of the angular momentum (isotropic), and the CC phases of and lie between the two approximations. The modified asymptotic approximation (P) given by equation (57) reproduces the phase difference between the and channels, indicating that the effect of the centrifugal potential is indeed captured by the expansion, although they both overestimate the absolute value of the phase at low kinetic energies, as in the case of the isotropic formula. On the other hand, the modified asymptotic approximation (P+A) given by equation (61) show very little difference between and , and both of them are well aligned with the corresponding isotropic formula. This is because at low kinetic energies the amplitude-variation effect overrides the phase-variation effect, while the former is independent of under the far-field expansion up to .
3.2 Fano’s propensity rule in continuum-continuum transition
Fano’s propensity rule in CC transition reported in references [34, 36] is tested using our equation (53) and its asymptotic approximations, as shown in figure 4. The analytical formula again shows excellent agreement with TDSE [23, 36], SOPT , and RPAE for intermediate states with various angular momenta. We note that the modified asymptotic approximation shows very good agreement from about 15 eV, where the absorption pathway converges seemingly faster than the emission pathway. The (P) and (P+A) approximations () give very similar moduli ratios, as the amplitude variation effect is mostly cancelled in the ratios. The Fano’s propensity rule is entirely absent from the isotropic asymptotic formulae, since they do not have -dependence. This indicates that the main effect from the centrifugal potential is well described by the expansion of the phase to the next order, which is a good approximation for .
3.3 Continuum-continuum transition time delay
The continuum-continuum transition time delay defined by the finite difference is
| | |
--- |
| | (62) |
where is the electron kinetic energy of the intermediate state, and is the phase of the transition amplitude. The results given by the analytical formula (53) are plotted in figure 5. The values for the and channels agree perfectly with the TDSE for hydrogen atoms. The (P) and (P+A) formulae reported by Dahlström et al. and with the modification introduced in this work lie below and above the curves, respectively. In particular, the (P+A) formula bends to the opposite direction at lower kinetic energies, although it matches the TDSE better at higher kinetic energies. This again manifests the validity of our analytical formula. Interestingly, although the phases of different -channels vary, as shown in figure 3, the phase differences between the emission and absorption pathways () are very close, which has recently been reported by Busto et al. . Hence, the key problem is to correctly determine the relative phase offset between channels and their moduli ratios, which can be accurately achieved by the analytical formula, as demonstrated in the previous sections.
3.4 Angle-dependent time delay for helium with circular-XUV-circular-IR RABBIT
An experimental test for the angular-momentum-dependent continuum-continuum transition amplitude are angle-resolved time-delay measurements on atoms [61, 33, 62]. For example, using the (isotropic) asymptotic wavefunction for the intermediate state and the Kummer’s function for the final state , Boll et al. showed that this analytical formula with partial approximation matches well with the linear-XUV-linear-IR RABBIT experiments regarding the angle-dependent phase . We note that due to the relatively large experimental uncertainties at emission angles close to , where different theoretical models differ most, it is not obvious to decide which model is the most accurate, especially at lower electron kinetic energies. Here we compare the theories with a circular-XUV-circular-IR RABBIT experiment reported recently by the present authors . The scheme for the generation and characterisation of the circular XUV attosecond pulse trains and the experimental setup for angle-resolved RABBIT have been described in our recent publications [35, 63, 64]. The benefit of the circular-XUV-circular-IR RABBIT scheme for helium is that the difference in the co- and counter-rotating XUV and IR provides additional information of the partial waves. For example, if the circular XUV excites one of the helium electrons from to the intermediate state (the -axis is defined as the light-propagation direction), in the continuum-continuum transition step, if the dressing field is co-rotating with the XUV, then the absorption pathway leads only to the final state due to the restriction of the magnetic quantum number, while the emission pathway leads to the superposition of the and states; on the contrary, if the dressing field is counter-rotating, then the emission pathway yields the final state, and the absorption pathway gives the superposition of the and states. Since there are only three partial waves contributing to a given sideband, their individual moduli and phases can be extracted by the global fitting of the two-dimensional interference pattern with only six parameters. By comparing the co-rotating and counter-rotating cases, one can separate the Wigner part and the continuum-continuum part of the photoionsation time delays for each final state . Theoretically, the angular dependence of the RABBIT phase is entirely absent from the isotropic asymptotic approximations, and the relative phase and modulus ratio between and play the key role, as explained in . The theoretical and experimental results are compared in figure 6. We choose SB18 with relatively low electron kinetic energy (3.3 eV), where the asymptotic formulae have poor accuracy and are not shown in the figure. Our proposed analytical formula (53) shows remarkably good agreement with the TDSE and experimental values for both the co- and counter-rotating cases, which is expected from the level of accuracy that it has shown in describing the moduli and phases of the transition amplitudes between different -states. The formula proposed by Boll et al. is known to have less accuracy at lower electron kinetic energies and agrees with the TDSE and experimental values semi-quantitatively. Considering the precision of the experimental values, we conclude that our analytical formula quantitatively reproduces the experimental results and has comparable accuracy with the TDSE computation, which outperforms the formula proposed by Boll et al. at low electron kinetic energies.
Table 1: Partial-wave fractions of helium one-photon ionisation with expansion center shifted by 1 Å, calculated using ePolyScat [19, 20].
| (eV) | | | | | |
--- --- --- |
| 3.3 | 0.0808 | 0.7484 | 0.1617 | 0.0088 | 0.0002 |
| 18.8 | 0.1891 | 0.2266 | 0.3970 | 0.1599 | 0.0274 |
In order to further check the sensitivity of our method regarding the shape of the potential, we performed calculations with helium one-photon photoionisation amplitudes with an expansion center displaced by 1 Å using ePolyScat [19, 20]. Since ePolyScat uses single-center expansions, when the atomic orbitals are displaced from the center of expansion, a series of (in principle infinite number of) partial waves are involved, in contrast to the centered helium atom, where the one-photon ionisation solely leads to the -wave. The fractions (defined as the modulus square ratio) of partial waves of the shifted helium atom with electron kinetic energies of 3.3 eV and 18.8 eV are summarised in table 1. At 3.3 eV, (-wave) is still dominating, with some contributions from and partial waves. As the electron kinetic energy increases to 18.8 eV, the distribution becomes more diffuse, with as the leading partial wave and comparable contributions from . Although all the of observables at asymptotic distance from the origin should be insensitive to the shift of the atom, depending on the model, this may introduce artefacts. When the atom is displaced from the origin in the spherical coordinate, the potential is no longer central, and the unshifted Coulomb waves are no longer its eigenfunctions. On the other hand, one may argue that since the continuum-continuum transition involves the outgoing electron wavepackets that escape from the atom, the transition amplitude should be less sensitive to the short-range potential, and the asymptotic behaviour is the same. The results using the analytical formula for the lower kinetic energy (3.3 eV) are shown as the dashed lines in figure 6, where the phase difference compared to the unshifted case is only marginal. For the higher kinetic energy (18.8 eV), we compare the results using the analytical formula (53), the formula proposed by Boll et al. , and the modified asymptotic formula ((P), equation (57), ) in figure 7. Since the electron kinetic energy is relatively high, all three models show reasonable agreement with the experimental values. For the shifted helium, the time delay from the modified asymptotic formula is practically unchanged, since it is by its nature only sensitive to the asymptotic behaviour, whereas the formula proposed by Boll et al. shows a non-negligible phase difference, indicating that the time delay may be less accurate when the potential deviates from the perfect central potential. The analytical formula shows much smaller difference, which implies that although taking the confluent hypergeometric function of the second kind may deviate from the reality in a non-central potential field, e.g. the potential of a molecular cation, the calculated time delay is largely converged.
4 Conclusion
In conclusion, we have proposed an analytical formula for calculating the continuum-continuum transition amplitudes for the hydrogen-like atoms which uses the confluent hypergeometric functions of the first and second kinds as the final and intermediate states, respectively. We first derived the integral between the confluent hypergeometric functions of the first and second kinds, which can be expressed using the Appell’s function. We note that this expression is an extension to the Gordon’s integral, as the regular solution of the Coulomb wave equation can be expressed by the imaginary part of the irregular solution, or the linear combination of the two independent irregular solutions, and indeed one can find similarities between this expression and the reported expressions for the Gordon’s integral. Then we showed that this formula excellently reproduces the phase and the Fano’s propensity rule in the continuum-continuum transition, compared to the TDSE and other theoretical approaches. In addition, we modified the WKB asymptotic approximation by expanding the terms to , to which the centrifugal potential contributes. The Fano’s propensity rule and the main feature of the phases in continuum-continuum transitions can be nicely captured with this relatively simple expression that involves only the gamma function. Finally, we used our method to calculate the relative RABBIT phases between the co- and counter-rotating XUV and IR at different electron emission angles. Even at kinetic energy as low as 3.3 eV, our proposed analytical formula quantitatively agrees with both the TDSE calculation and the experimental values. We also checked the case with the shifted expansion center for calculating the spherical harmonics, and the results indicates that our method is not very sensitive to the shape of the short-range potential field, which opens the possibility of adapting our method to molecules.
Acknowledgements
We thank David Busto and Mathieu Gisselbrecht for fruitful discussions. This work was funded by ETH Zürich. J.-B. Ji acknowledges the funding from the ETH grant 41-20-2. M. H.’s work was funded by the European Union’s Horizon 2020 research and innovation program under Marie Skłodowska-Curie agreement grant No. 801459, FP-RESOMUS.
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14429 | https://people.richland.edu/james/lecture/m116/functions/functions.html | 1.3 - Functions
1.3 - Functions
Definitions
RelationA rule that associates a value in the domain with a value in the range.FunctionA function is a relation (rule) that assigns each element in the domain to exactly one element in the range.DomainThe set of all the values which may be input into a function. That is, the set of all the values the independent variable may assume. Graphically, the domain is the set of all the x-coordinates.RangeThe set of all the values which are output when the function is evaluated at all the input values from the domain. That is, the set of all the values the dependent variable may assume. Graphically, the range is the set of all the y-coordinates.Independent VariableTypically, the independent variable is x. However, the independent variable is the variable which is free to assume different values independently of the other variable. Most of what we're going to do in this class will only involve one independent variable, but realize that it is possible to have more than one independent variable.Dependent VariableTypically, the dependent variable is y. However, the dependent variable is the variable which is determined based on the value of the independent variable(s). If a function is written as y=3x+2, then y depends on x, but x doesn't depend on y (in the form it's written in). So, x is independent of y, but y is dependent of x.Implied DomainThe set of all real numbers for which the expression is defined.
How is a function different from a relation?
Here are some guidelines for determining whether a relation is a function or not.
Each and every element in the domain (x) must be matched with an element in the range (y).
Every element in the domain (x) can only be matched with one element in the range (y).
Different elements in the domain can go to the same element in the range. A y-coordinate may be repeated, but an x-coordinate may not.
Some values in the range don't have to be used at all
Function Notation
Function notation is used to name functions for easy reference. Imagine if every function in the world had to start off with y=. Pretty soon, you would become confused about which y= you were talking about. This is the way antecedents work in English. If you just say "it" is red, you really don't have any clue which "it" you're talking about. You need some other way of naming things. Enter function notation.
Function Definition
f(x) = 3x + 2
g(x,y) = x 2 + 3y
In this example, the f is a function of x. That is, x is the independent variable, and the value of f depends on what x is. Also, g is a function of both x and y.
The notation f(x) does not mean f times x. It means the "value of f evaluated at x" or "value of f at x" or simply "f of x"
Function Evaluation
f(3) = 3(3) + 2 = 9 + 2 = 11
f(3) does not mean f times 3. It means the "value of f evaluated when x is 3".
f(t) = 3(t) + 2 = 3t + 2
Whatever is in parentheses on the left side of the function (t in this case) is substituted for the value of the independent variable on the right side.
f(x+h) = 3(x+h) + 2 = 3x + 3h + 2
Every occurrence of the independent variable is replaced by the quantity in parentheses. A common mistake is to take a quantity and apply linear or affine transformations to it.
f(x+h)
f(x+h) does not equal f(x) + h = 3x+2 + h
f(x+h) does not equal f(x) + f(h) = 3x+2 + 3h+2 = 3x + 3h + 4
f(x+h) does equal 3(x+h) + 2 = 3x + 3h + 2
f(3x)
f(3x) does not equal 3 f(x) = 3( 3x+2) = 9x + 6
It does equal 3(3x) + 2 = 9x + 2
You also specify which function you want to use when you use function notation. Consider
g(2,1) = 2 2 + 3(1) = 4 + 3 = 7
Since the order of the independent variables in the original definition was x and then y, the function g is evaluated when x=2 and y=1.
g(1,2) is something completely different. In this case, x=1 and y=2.
g(1,2) = 1 2 + 3(2) = 1 + 6 = 7
Okay, so it's not completely different, but it is found in a different manner.
Piecewise Definitions
Sometimes, functions are a little more complex than the simple functions we've described so far. If different rules are used for different values of the independent variable, then we can use a piecewise definition.
Consider the function shown and the following evaluations.
f(2.1) = 2(2.1) = 4.2
Since 2.1 is in the interval [1,3), we use the second piece of the definition, f(x) = 2x.
f(-2) = 3 - (-2)2 =3 - 4 = -1
Since -2 is in the interval (-∞,-1), we use the first piece of the definition, f(x) = 3-x 2
f(3) = 5 - (3) = 5 - 3 = 2
Since 3 is not included in the second piece, but it is included in the third piece on the interval [3,+∞) we use the third piece of the definition, f(x) = 5 - x.
f(0) is undefined
Since 0 doesn't fall into any of the domains, the function is undefined there.
When all of the domains are combined, the domain of the function f, is the set of all real numbers except [-1,1). You could also write it as (-∞,-1)U[1,+∞). The symbol U stands for the union of two sets.
Piecewise Functions and the Calculator
You can put piecewise functions into the graphing calculator. The TI82 and TI83 calculators have a [Test] key obtained by pressing [2 nd] [Math]. Under this key, you will find the different test operators (equal to, not equal to, greater than, greater than or equal to, less than, and less than or equal to).
When the calculator evaluates a test expression (x<1), it will return the value of 1 if the statement is true and the value of 0 if the statement is false. This works really well with multiplication, because multiplying by 1 won't change the expression, and multiplying by 0 will make the expression 0.
y = (1 - x 2)(x<-1)
Consider just the first part of the piecewise definition from above. If x<-1, then the calculator will return (1-x^2). If x is not less than negative one, then the calculator will return 0. That's not the proper value to return (it should be undefined), but it will graph on the x-axis and won't show up because the axis is already there. So, it has the semblance on graphing correctly.
The complete piecewise function can be defined for the calculator as:
y = (1-x 2)(x<-1) + 5x(1≤x and x<3) + (5-3x)(x≥3)
The "and" keyword can be found under [Test] Logic menu. Similarly, the ≤ and ≥ symbols can also be found under the [Test] menu. If you try to put in 1≤x<3 like we right, the calculator will interpret that as 1≤3 and always return true.
Be sure to use some kind of decimal setting or dot mode when you're viewing a piecewise function. Otherwise, you may get some weird results.
Finding the Domain
Implied Domain
We defined implied domain earlier. It is the set of all real numbers where the expression is defined.
Start off with all real numbers
Exclude any values of the independent variable which cause division by zero
Exclude any values of the independent variable which result in taking the square root of a negative number. Square root can be replaced with any even (4 th, 6 th, etc) root.
You do not need to state restrictions which result from the implied domain. In other words, if there is an (x-2) in the denominator, you do not need to state that x cannot be 2.
Other Exclusions
Sometimes, you will need to exclude other values.
Problem Constraints - these are used in applications. Here are some examples
If x is the length of the side of a triangle, then x can't be negative
If x is the position on a 12' teeter totter at which the fulcrum must be positioned, then x must be between 0 and 12.
If x is the number of people in a room, then x can't be negative.
Stated Restrictions - sometimes the problem will just come out and say you can't use a specific value, or that you can only use specific values. These aren't in the implied domain, so they need to be stated.
If you ever simplify a function, and a value which was in the implied domain is no longer in the implied domain, then it needs to become a stated restriction. Example, divide (x 2-4) by (x-2). The implied domain is x cannot equal 2 because that would cause division by zero. However, if you factor the numerator as (x-2)(x+2), then (x-2) in the numerator divides out with the (x-2) in the denominator and you're left with just (x+2). The fact that x cannot be 2 is no longer implied by a simple x+2 in the numerator, so you must now state that x cannot be 2.
Combining Domains
When you have a function which is a composition of several pieces, the values in the domain must be able to be used in all parts of the function. Let's say the numerator has a square root of x in it (so that x must be non-negative) and the denominator has an x-2 in it (so that x cannot be 2). When you combine those domains, you get all values of x that are non-negative except 2.
In other words, the domain for the function is the intersection of all the domains of the individual parts.
Now, let's say the numerator has a square root of x in it (so that x must be non-negative) and the denominator has an x+2 in it (so that x cannot be -2). When you combine those domains, you get all values of x that are non-negative except -2. Well, you don't need to state that x cannot be -2 since you've already said that x is non-negative. So, in this case, the domain would be all non-negative values of x. |
14430 | https://www.omnicalculator.com/math/sum-of-digits | Last updated:
Digit Sum Calculator
The digit sum calculator enables you to find the total sum of digits in any given set of numbers. You may view the results for each number separately or the entire set of numbers together. The outcome will be displayed as a table of the sum of digits. You also have the option to find the count of digits for the numbers and an example of the sum of digits to see how the calculator works!
What is digit sum?
We can obtain the sum of digits by adding the digits of a number by ignoring the place values. So, for example, if we have the number 567, we can calculate the digit sum as 5 + 6 + 7, which will give us 18.
In this manner, we can find the total sum of the digits of any positive integer, using our digit sum calculator. The calculator also works for numbers of different orders of magnitude.
The sum of digits calculator considers the number of occurrences of each digit in the given numbers. It calculates the partial sum of each digit by multiplying the number of occurrences of the digit and the digit itself. It then adds these partial sums to find the total sum of digits.
You may also input a set of consecutive integers to find the sum of digits in all those numbers! It's also interesting to note that numbers occurring in everyday life tend to follow the Benford's law. We have a dedicated tool if you are curious, check out Benford's law calculator!
How do I calculate the sum of digits of all numbers?
In order to find the sum of digits of the entire set of numbers, here's what you need to do:
Key in all the numbers into the digit sum calculator (up to 10 numbers).
Choose the Sum of digits option.
For the treatment of numbers, choose As a single group.
Voila! The sum of digits calculator will give you the total sum of all digits across all the numbers that you entered!
But if you're looking to calculate the total sum of the numbers and not just the digits, then you may want to use the addition calculator! Alternatively, you may also try finding the digital root of the numbers!
Sum of digits example
Let's say you want to find the sum of digits in the following numbers:
1111177770999
5555555555
0
After selecting the Sum of digits option, if you choose to treat each number separately, then you'll get the sum of digits as a table, as follows:
For the number 1111177770999:
| Digit | No. of occurrences | Sum |
---
| 1 | 5 | 5 |
| 7 | 4 | 28 |
| 0 | 1 | 0 |
| 9 | 3 | 27 |
| | Total | 60 |
For the number 5555555555:
| Digit | No. of occurrences | Sum |
---
| 5 | 10 | 50 |
| | Total | 50 |
For the number 0:
| Digit | No. of occurrences | Sum |
---
| 0 | 1 | 0 |
| | Total | 0 |
The No. of occurrences column also gives you the count of the number of digits. If you don't have any numbers handy, but you still want to try out this calculator, you may use our random number generator to get some numbers and input them here!
What is digit sum used for?
The most popular applications for the sum of digits lie in finding the divisibility of the given number. Here are a few rules that are based on the sum of digits:
Divisibility by 3 — If the sum of digits of a number is divisible by 3, then that number is also divisible by 3. For example, if we consider the number 123,453, the sum of its digits is 18, and we can further reduce the sum of digits of 18 to 9. Since 9 is divisible by 3, this means that 123,453 is also divisible by 3;
Divisibility by 9 — If the sum of digits of a number is divisible by 9, then that number is also divisible by 9. Looking at our earlier example, we can see that 123,453 is divisible by 9 too, since its sum of digits (18 and then subsequently 9) is divisible by 9!
Digit sums were also used in early computers as a means to check arithmetic or binary calculators.
FAQs
What is the formula for sum of consecutive numbers?
To find the sum of N consecutive numbers, we'll use the formula N × (first number + last number) / 2. So, for example, if we need to find the sum of numbers from 1 to 10, we will have 10 × (1 + 10) / 2, which will give us 55.
What is the sum of digits from 1 to 100?
The sum of all digits of all numbers from 1 to 100 is 901. This is obtained by adding digits 1 through 9 ten times for the units place, followed by adding digits 1 through 9 ten times again for the tens place, and then finally adding the digit 1 that's in the hundreds place of 100.
What is the sum of digits from 1 to 10?
The sum of all digits of all numbers from 1 to 10 is 46. This result is obtained by adding digits 1 through 9 once for the single-digit numbers, followed by adding the digit 1 in the tens place of 10.
What is the sum of digits from 1 to 1000?
The sum of all digits of all numbers from 1 to 1000 is 13,501. This result is obtained by doing the following:
Add digits 1 through 9 hundred times for the units place.
Add digits 1 through 9 hundred times again for the tens place.
Add digits 1 through 9 hundred times again for the hundreds place.
Finally, add the digit 1 that's in the thousands place of 1000.
Tada! We'll get the sum of digits of all integers from 1 to 1000 as 13,501!
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14431 | https://pmc.ncbi.nlm.nih.gov/articles/PMC5787910/ | Comparing the Efficacy of Low Dose and Conventional Dose of Oral Isotretinoin in Treatment of Moderate and Severe Acne Vulgaris - PMC
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J Res Pharm Pract
. 2017 Oct-Dec;6(4):233–238. doi: 10.4103/jrpp.JRPP_17_30
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Comparing the Efficacy of Low Dose and Conventional Dose of Oral Isotretinoin in Treatment of Moderate and Severe Acne Vulgaris
Gita Faghihi
Gita Faghihi
1 Skin Diseases and Leishmaniasis Research Center, Department of Dermatology, Isfahan University of Medical Sciences, Isfahan, Iran
Find articles by Gita Faghihi
1, Fatemeh Mokhtari
Fatemeh Mokhtari
1 Skin Diseases and Leishmaniasis Research Center, Department of Dermatology, Isfahan University of Medical Sciences, Isfahan, Iran
Find articles by Fatemeh Mokhtari
1, Nasrin Motamedi Fard
Nasrin Motamedi Fard
1 Skin Diseases and Leishmaniasis Research Center, Department of Dermatology, Isfahan University of Medical Sciences, Isfahan, Iran
Find articles by Nasrin Motamedi Fard
1,✉, Narges Motamedi
Narges Motamedi
2 Department of Community Medicine, Isfahan University of Medical Sciences, Isfahan, Iran
Find articles by Narges Motamedi
2, Sayed Mohsen Hosseini
Sayed Mohsen Hosseini
3 Department of Epidemiology and Biostatistics, Isfahan University of Medical Sciences, Isfahan, Iran
Find articles by Sayed Mohsen Hosseini
3
Author information
Article notes
Copyright and License information
1 Skin Diseases and Leishmaniasis Research Center, Department of Dermatology, Isfahan University of Medical Sciences, Isfahan, Iran
2 Department of Community Medicine, Isfahan University of Medical Sciences, Isfahan, Iran
3 Department of Epidemiology and Biostatistics, Isfahan University of Medical Sciences, Isfahan, Iran
✉
Address for correspondence: Dr. Nasrin Motamedi Fard, E-mail: motamedinasrin@yahoo.com
Received 2017 Apr; Accepted 2017 Jun.
Copyright: © 2018 Journal of Research in Pharmacy Practice
This is an open access article distributed under the terms of the Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License, which allows others to remix, tweak, and build upon the work non-commercially, as long as the author is credited and the new creations are licensed under the identical terms.
PMC Copyright notice
PMCID: PMC5787910 PMID: 29417084
Abstract
Objective:
This study was conducted to compare the effect of low-dose isotretinoin with its conventional dose in patients with moderate and severe acne.
Methods:
This was a clinical trial conducted on 60 male and female patients with moderate and severe acne vulgaris. The patients were divided into two treatment groups: 0.5 mg/kg/day isotretinoin capsule and low-dose isotretinoin capsule (0.25 mg/kg/day). Patients in both groups received 6-month treatment. At the end of the 6 th month and 12 th month (6 months after the end of the treatment), they were examined again, and their improvement was determined and compared.
Findings:
The average severity of acne in the two treatment groups did not differ significantly within any of the study periods. The most common side effects were nose dryness in the low-dose group (17%) and hair thinning and loss in the conventional-dose group (33.2%), although all the patients had dry lips.
Conclusion:
According to the same severity of the acne in two groups in different study periods, as well as fewer side effects and more patients' satisfaction, the low-dose isotretinoin can be considered in the treatment of acne.
K EYWORDS:Acne vulgaris, efficacy, Isotretinoin
I NTRODUCTION
Acne vulgaris is a common disease during adolescence. It is the most common inflammatory diseases of the sebaceous unit and a chronic inflammatory disease of the sebaceous glands.
There are different common treatments for acne vulgaris based on the severity of lesions. In that, topical solutions (e.g., clindamycin or erythromycin), topical ointment (e.g., tretinoin, benzoyl peroxide, and adapalene), and other keratolytic drugs (e.g., alpha-hydroxy acid), salicylic acid-containing medicine, and/or sulfur or azelaic acid-containing drugs are used to treat mild cases. In more severe cases, systemic treatments, such as tetracycline, doxycycline, azithromycin, minocycline, azithromycin, and cotrimoxazole are used. In cases with the risk of scar, oral isotretinoin is administered. In cases with associated hormonal abnormalities, androgen, estrogen, spironolactone, and dexamethasone are used.[3,4]
Despite all traditional and modern treatments, retinoid compounds are key components in the treatment of acne. In early 1980, the administration of oral isotretinoin was limited in patients with nodulocystic acne. Nevertheless, the use of oral isotretinoin was made wider with broadening relevant experience. It was also prescribed for patients with milder acne, who did not respond satisfactorily to common treatments, such as topical retinoids plus oral antibiotics. Patients with moderate acne, who exhibit the scar symptoms, are also a candidate for treatment with oral isotretinoin. Acne is mainly associated with physical and cosmetic morbidity. The psychological consequences of acne vary from depression and anxiety to work and interpersonal relationship disorders. Studies, used quality of life instruments, have shown that the treatment with isotretinoin results in a significant improvement of socialization and self-confidence. There may be a delay of 1–3 months before the initiation of treatment effects. In many cases, improvement continues even after discontinuation of the treatment. Therefore, the continuation of the treatment until the complete healing of lesions is not necessary.
Although isotretinoin is a very effective medication for the treatment of acne, its association with several complications necessitates the precise selection of patients. The recommended dose is 0.5–1 mg/kg/day for 4–6 months. A study conducted in India (2014) on 50 patients with moderate to severe acne showed the effectiveness of low-dose isotretinoin with lesser complications and greater cost-effectiveness. In this study, patients received 0.3–0.4 mg/kg/day isotretinoin. In addition, a study (2012) showed the effective treatment of severe acne with low-dose isotretinoin (0.1–0.3 mg/kg/day). Since, side effects of this medication are dose-dependent, the administration of low-dose isotretinoin (0.25 mg/kg/day) for 6 months seems logical; however, there is no relevant, comprehensive study with follow-up period in Iran, and although the patient's compliance is dependent on side effect and cost of drug and its cost and economic status of patients has turned it into a significant problem. Thus, the present study has been conducted to compare the low- and common-dose oral isotretinoin in patients with moderate and severe acne vulgaris.
M ETHODS
This study was a prospective randomized clinical trial conducted during 2014–2015. The population comprised patients (both male and female) with moderate to severe acne vulgaris referred for treatment to Alzahra Medical and Training Center, several clinics affiliated to Isfahan University of Medical Sciences and a privately-owned doctor's office. The inclusion criteria are as following: patient consent to participate in the study, no sensitivity to retinoids, no pregnancy, not willing to become pregnant, and absence of hormonal disorders in patients. Patients with the following criteria excluded from the study: Failure of the patient to attend follow-up sessions for any reason, and adoption of other supplementary therapies during the study.
The required sample size was estimated to be 30 patients using the sample size formula, serving to compare the ratios given the confidence level of 95% (Z1-α/2 = 1.96) and test power of 80% (Z1-β = 0.84). The minimum significant difference between the two groups was considered to be 0.25.
After obtaining a permit from the Ethics Committee of Isfahan University of Medical Sciences that Ethical code is 393894 and IRCT code is 2015061722780N1, the simple sampling was applied to patients (male and female) with moderate acne vulgaris and referred for treatment to one of the largest referral centers in isfahan (Iran's third largest city, located in the center of Iran). The patients were randomly treated by 0.5 and 0.25 mg/kg/day of oral isotretinoin (Roaccutane, F. HOFFMAN-LA ROCHE Switzerland) for 6 months. The effort was made every day to arrange equal numbers of patients in Groups 1 and 2 so as to fulfill the random sampling. The patients were advised to take the medication with meals, avoid fat-free diet, and yet refrain from the excessive fat intake. This process continued until the number of patients per group amounted to 30. The method of follow-up in two groups was an intention to treat.
At the baseline and 6 months later, the patients were examined for the severity of acne by a dermatologist that was blinded to intervention group. The severity of acne was determined through the Global Acne Grading System. This system was used in studies conducted by the British Association of Dermatologists and the American Society for Dermatologic Surgery (2013 and 2014). In this system, the numbers of lesions including comedones, papules, pustules, and nodules were counted on the forehead, right cheek, left cheek, nose, chin, chest, and upper back torso. Each type of lesion is given a grade based on its severity as follows: 1 - comedones; 2 - papules; 3 - pustules; and 4 - nodules. Moreover, each zone is given a factor as follows: 1 - nose, chin; 2 - forehead, right cheek, left cheek; and 3 - chest, back and upper back torso. The severity of lesion for each zone (local score) is calculated as follows: local score = Factor × grade (0–4). Global score is obtained from the sum of local scores in different zones. The severity of acne is considered based on the global score as follows: mild: 1–18; moderate: 19–30; severe: 31–38; and very severe: Over 38.
To prevent the complications of treatment, the female subjects in reproductive age were initially tested for pregnancy. They were included in the study if the results were negative. Then, the test was repeated every month. At baseline and at the end of 1 st, 2 nd, 4 th, and 6 th months, the levels of liver enzymes, blood cholesterol, and triglyceride were checked in all patients. The treatment discontinued in case the triglyceride level elevated over 400 mg/dl, cholesterol over 300 mg/dl, alkaline phosphatase over 246, alanine aminotransferase over 62, and aspartate aminotransferase over 80.
All patients were simultaneously treated with daily 250 mg of oral azithromycin in the first 2 weeks. Moreover, 0.25 mg of prednisolone was prescribed to treat patients with isotretinoin over the first week. To prevent complications such as dry lips and skin, identical topical emollient was administered on all patients. Furthermore, they were advised to apply sunscreen regularly and avoid sunlight as much as possible. After the end of treatment with isotretinoin, the patients went through local treatment with 2% clindamycin during the follow-up phase.
The patients were visited on a monthly basis. The cases with no new lesions were recorded in a data collection form. At the end, it was found out which group recovered from the lesions sooner than the other. The side effects of isotretinoin were examined and recorded at each visit by a dermatologist that was blinded to intervention group. The patient satisfaction level was assessed and recorded at the end of treatment by visual analogue scale (VAS) for satisfaction. VAS was a horizontal line. The patient rated his satisfaction by making a vertical mark on the line. There were two descriptors representing extremes of satisfaction (i.e., no satisfaction [0 point] and extreme satisfaction [5 points]). We analyzed complications by the questionnaire, physical examination, and laboratory tests that were conducted by a dermatologist that was blinded to intervention group.
The collected data were described through several tables, graphs and measures of central tendency and dispersion. Then, they were analyzed by Chi-square test, independent t-test, and we used two-way repeated measure ANCOVA to determine the significance difference between these two sets of observations within the same group and then to compare the significant difference between the low dose versus high-dose therapy. The SPSS 22 (released 2013, SPSS Inc, Chicago, IL) was also used as statistical software.
R ESULTS
Results of the frequency distribution of demographic variables (qualitative and quantitative) in each treatment group indicated that the average age of participants in the first treatment group (low dose) and second group (conventional dose) was 22.94 ± 6.25 and 23.1 ± 4.66 years, respectively. Therefore, there was no statistically significant difference between age results (P = 0.911). Similarly, the average age and duration of acne did not differ significantly in the two groups (P> 0.05). Frequency distributions of qualitative research variables including gender differed significantly in the two groups, because 5 participants in the first group (13.9%) and 8 participants in the second group (26.7%), which added up to 13 (19.7%), were male, whereas 31 participants in the first group (86.1%) and 22 (73.3%) in the second group (which added up to 53 or 80.3%) were female (P = 0.011). However, there was no significant difference between the two groups in terms of family history of acne and frequency of previous acne treatments (P> 0.05).
Results of determining and comparing the average severity of acne vulgaris in the first group prior to treatment (mean = 54.6, standard deviation [SD] = 2.9), 6 months into treatment (mean = 2.5, SD = 0.8), and 6 months following the treatment (mean = 8.7, SD = 0.8), showed that there was a significant difference in the average severity of acne in the first group before treatment and 6 months into the treatment as well as before treatment and 6 months following the treatment in the 12 th month (P< 0.001). According to the classification of severity of acne based on the global score, it was found that average severity of acne was extremely high in the first group before treatment (with average acne severity of 58.6), while 6 months into treatment and 6 months following the treatment severity of acne was mild (with respective average acne severity of 2.5 and 8.7). Figure 1 depicts the linear diagram of changes of average acne severity over the three periods under study. Frequency distributions of acne severity in the first and second groups before treatment, 6 months into treatment, and 6 months after treatment based on the gender and age variables revealed that there was no significant difference between the two groups in terms of severity of acne based on gender and age (P> 0.05).
Figure 1.
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Diagram of changes of acne average severity in the two treatment groups over three study periods
Results of determining and comparing the average severity of acne vulgaris in the second group (which received 0.5 mg isotretinoin capsules) before treatment (mean = 58.8, SD = 3.1), 6 months into treatment (mean = 1.8, SD = 0.7), and 6 months following the treatment (mean = 9.7, SD = 1.1) showed there was a significant difference in the average severity of acne in the second group before treatment and 6 months into the treatment as well as before the treatment and 6 months following the treatment in the 12 th month (P< 0.001). According to the classification of severity of acne based on the global score, it was found that average severity of acne was extremely high in the second group before treatment (with average acne severity of 58.87), while 6 months into treatment and 6 months following the treatment severity of acne was mild (with respective average acne severity of 1.87 and 9.73).
Results of comparing the average severity of acne in the two study groups showed that the average severity of acne in the two treatment groups did not differ significantly within any of the study periods (P> 0.05). Figure 1 shows the linear diagram of changes of average acne severity in the two groups over different periods.
Results of comparing the average severity of acne in the two treatment groups before treatment, 6 months into treatment, and 6 months after treatment based on the gender and age variables showed that there was no significance difference between the severity of acne in the two groups based on gender and age (P> 0.05).
The frequency distribution of the severity of acne in the first and the second groups before treatment, 6 months after treatment, and finally 6 months after completion of treatment in terms of duration of acne showed that no significant difference existed between the severity and duration of acne vulgaris in the first and second groups separately in any of the studied times (P> 0.05). In addition, comparison of the mean severity of acne in both treatment groups before treatment, 6 months after treatment, and 6 months after completion of treatment in terms of duration of acne showed no significant difference between the mean severities of acne in terms of the duration of acne in both groups (P> 0.05).
The frequency distribution of the severity of acne in the first and second groups separately, before treatment, 6 months after treatment, and finally 6 months after completion of treatment in terms of family history of acne showed that no significant difference existed between the severity of acne and family history of acne in none of the groups in any of the studied times (P> 0.05). In addition, comparison of the mean severity of acne in both treatment groups at different times in terms of family history of acne showed no significant difference between the mean severities of acne in terms of family history of acne (P> 0.05).
Considering the significance level of 5%, a significant difference can be observed between patients' satisfaction in the two treatment groups (P< 0.05), so the average of patients' satisfaction score in the first group (mean = 4.78, SD = 0.4) was significantly higher than the second group (mean = 4.43, SD = 0.6) (P = 0.02), which shows that the satisfaction of patients treated with lower doses of medication (Group I) was higher than patients treated with higher doses of the drug (Group II).
As shown in Table 1, considering the significance level of 5%, a significant difference can be seen in the frequency of side effects dry mouth, dry nose, repeated rhinorrhea, and hair thinning and loss in both groups (P< 0.05), so that the frequency of side effects in the group receiving a higher dose of the drug (the second group) was more than the group receiving the low dose of the drug (the first group). Table 1 shows the most common side effects were nose dryness in the first group (17%) and hair thinning and loss in the second group (33.2%) also all the patients had dry lips.
Table 1.
The frequency of adverse reactions in patients with acne in the study groups
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D ISCUSSION
There are different studies about this study that we mentioned some of them here. Amichai et al. and Lee et al. believed that doses of isotretinoin lower than 0.5 mg/kg/day may be effective for the treatment of some patients with acne.[15,16] In another study, Rasi et al. investigated the efficacy of low daily dose isotretinoin in moderate to severe acne patients. They found that low-dose isotretinoin was found to be a safe and effective choice for patients with moderate to severe scar prone acne vulgaris so their results are consistent with our results.
Ghalamkarpour and Nasiri studied about isotretinoin in the treatment of acne and this study was performed on patients with acne to examine the therapeutic effects, recurrence rate, and adverse effects of this drug. They concluded that oral isotretinoin appears to have favorable results and the least adverse effects in the treatment of carefully-selected patients with acne.
In addition, no significant difference was observed between the severity of acne and patients' sex and age at any time point in our results.
Furthermore, in the study of Duman et al. there were no statistically significant differences between control and acne groups with respect to age, sex, and Hospital Anxiety and Depression Scale (HAD) score.
A significant difference was observed between the two groups in terms of the frequency of the side effects such as dry mouth, nasal dryness, frequent nosebleeds, thinning hair, and hair loss. The above-mentioned side effects were more frequent in the high-dose group (second group) than in the low-dose group ( first group). Therefore, prescribing lower doses of this drug is more appropriate and effective in such patients.
Our results are supported by the findings of a study from India, a low-dose isotretinoin treatment (0.15–0.28 mg/kg/day) lead to clinically significant results in 87.54% of the participants, including 68.20% very good and 19.34% of good results.
Another study that is about safety and efficacy of low-dose isotretinoin in the treatment of moderate to severe acne vulgaris and conducted by Rao et al. In this study, they evaluated 50 participants diagnosed as having moderate to severe acne vulgaris and the participants were recruited over a period of 2 years and were followed up for 3 months to know the safety and efficacy of low-dose isotretinoin in the treatment of moderate to severe acne vulgaris. Hence, they recommended that judicious use of low-dose isotretinoin in patients with moderate to severe acne because acne not only scars the face, but also the mind and the heart.
Lee et al. evaluated the clinical efficacy and tolerability of low-dose and intermittent isotretinoin regimens and to compare them directly with conventional isotretinoin treatment. They suggested that when considering tolerability, efficacy and patient satisfaction, low-dose treatment is most suitable for patients with moderate acne which is consistent with this study results. The results of this study are consistent with the results of the previous study. This study shows the results of determining and comparing the average severity of acne vulgaris in the group receiving isotretinoin (0.5 mg capsules). The results show that in this group, there was a significant difference between the average acne intensity before and 6 months after treatment. Moreover, there was a significant difference between the average severity of disease before and 6 months after treatment.
The satisfaction level of the low-dose patients was higher than the high-dose patients. In addition, the frequency of complications was greater in the high-dose patients (the second group) than in the low-dose patients (the first group). On average, there was no significant between-groups different in terms of the severity of the acne in the aforementioned times. According to the same recurrence of patients with acne after follow-up at 6 months, more satisfied patients and fewer side effects can be considered low-dose isotretinoin in the treatment of acne.
A UTHORS' C ONTRIBUTION
Gita Faghihi and Fatemeh Mokhtari contribute to Concept and design of study. Nasrin Motamedi Fard contribute to intervention and data gathering. Narges Motamedi contribute to Drafting the article and revising it. Sayed Mohsen Hosseini contribute to analysis and interpretation of data.
Financial support and sponsorship
This study was supported by Isfahan University of Medical Sciences.
Conflicts of interest
There are no conflicts of interest.
Acknowledgments
This study was approved by Isfahan University of Medical Sciences. Appreciated the authors for their cooperation in this research plan.
R EFERENCES
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14432 | https://www.youtube.com/watch?v=FY7I_VYvLSk | Input Output Tables Writing Rules
Mario's Math Tutoring
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Posted: 20 Sep 2016
Introduction to Input Output Tables Writing Rules for the function in this free math video tutorial by Mario's Math Tutoring.
Timestamps:
00:00 Intro
0:26 Looking at How We Get From the Input (x) to the Output (y)
1:13 Example 2 We Look at the Difference From One Y Value to the Next
3:00 Example 3
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Transcript:
Intro one when you do this you're usually doing this like in a pre-algebra or algebra class you're looking for a pattern and so you say okay the X values these are what I'm putting into the function and the yv values that's what's coming out so if you just look at it here you can see hm I'm going from 7 to four 8 outputs 5 9 outputs 6 and 10 outputs s so by looking at it you can see that we're actually subtracting 3 7 Looking at How We Get From the Input (x) to the Output (y) - 3 is 4 8 - 3 is 5 9 - 3 is 6 and 10 - 3 is 7 so we can see that our equation or our function is y = x which is our input - three and that gives us our yv value our output so X is what we're putting in and we're taking away three and we get uh y as our output okay this one's kind of easy but we'll get to some a little more challenging ones here so number two what you can see here is it's a little bit uh trickier to tell what's happening like six out comes 17 7's the input 20 comes up out 8 23 926 we might not see a pattern right away but what you can do with these uh problems is you can look at what is happening to go from one yval to the next yvalue okay so here you can see I'm adding three to get from Example 2 We Look at the Difference From One Y Value to the Next 17 to 20 I'm adding three again to get from 20 to 23 and I'm adding three again to get from 23 to 26 so you can see that's uh consistent right and we can see that with the X values I keep going up by one up by one and up by one so looks like for every increase in one on the X okay the three is going up the Y is going up by three right so what we can do is we can say hm it looks like what I'm doing is I'm tripling the input okay I'm multiplying the input times 3 but we have to make a little bit of an adjustment so let's look at this 6 3 is 18 but you can see our output is 17 6 um 7 3 is 21 but again this is 20 8 3 is 24 this is 20 23 you can see we have to make an adjustment we have to subtract one so this is going to be the equation that represents you know this input output table so again what you do is you look at the change in y over the change in X so you make a fraction and that's giving you what's called the slope or the rate of change but at this level when they usually present these input output tables you haven't gotten yet to this equation that you'll be learning in a little bit which is the y = mx plus b so at this stage they're just trying to help you to understand you know how to recognize these patterns okay the last one number three that I've written here is again you can see what's happening is we're going up by one up by one again up by one again and with the X values we're going up by two this time see 4 to 6 6 to 8 and so here we can see that the change in y is an increase of one for every change in X which is an increase in two so what we have here is we have y = 1/2 Example 3 X so 1/2 times our input now you can see here 1/2 2 is 1 but we're getting 7 1/2 4 is 2 but we're getting eight so it looks like we have to adjust that by adding six more okay so this is what's called our Y intercept or initial condition and this is what the value is when X is zero so zero is like the starting point but I think for this video and just to kind of help you to get started with these input output tables is look at the how the Y's are changing look how the X's are changing and write a fraction the change in y over the change in X put that number in front of the x value which is the input and then test it out and see what adjustment you have to make you know do you have to subtract once add six so that you know so that it represents the the input output table so again this is just an introduction about how to write rules for functions given an input output table I've been working with some students lately uh in like pre-algebra algebra and they haven't really gotten yet into the slope intercept form if you want to learn more about this I'll have a link for a video that I did on slope intercept form so you can learn more I'll see you in the next video |
14433 | https://math.stackexchange.com/questions/2012794/integrals-of-functions-with-factors-like-sqrta2-x2 | Skip to main content
Integrals of functions with factors like a2−x2−−−−−−√
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When solving integrals which contain factors of the form a2−x2−−−−−−√ it is typical to make the substitution x=asin(θ) and then use the Pythagorean identity to simplify the integrand.
My question refers to approaches of this type, but I will use the following specific example as an illustration
∫dx1−x2−−−−−√.
If I substitute x=sin(θ), then dx=cos(θ)dθ, and I obtain:
∫dx1−x2−−−−−√=∫cos(θ)dθ1−sin2(θ)−−−−−−−−−√=∫cos(θ)dθcos(θ)=∫dθ=θ+C=arcsin(x)+C.
If I consult a table of integrals this is the solution given, and the approach used is the one recommended in elementary books. But I find this deeply troubling for the following simple reason:
cos2(θ)−−−−−−√=|cos(θ)|≠cos(θ), because the cosine function can take negative values. This is completely glossed over in the elementary calculus texts and YouTube videos I have looked at so far.
My guess is that actually the very first step is wrong. Instead of "making the substitution" x=sin(θ), we actually substitute θ=arcsin(x) subject to the restriction −π2≤θ≤π2...this solves our problem because, over that domain, cos(θ)≥0.
Is this correct? I don't really understand inverse functions well enough to be confident. The reason I am not sure is because in the explanations I have seen elsewhere, they do not mention this. Thanks.
calculus
integration
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edited Nov 14, 2016 at 0:44
A.Γ.
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asked Nov 14, 2016 at 0:17
LachyLachy
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Dont forget the +C, this is a phenomena that happens alot. What you are saying is essentially that −arcsin(x) is also the integral. However
−arcsinx=arcsinx+π
and this is absorbed in the +C.
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answered Nov 14, 2016 at 0:21
Rene SchipperusRene Schipperus
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The very first step is not wrong. Your primary variable x varies in the interval [−1,1]. To cover those values by sin(θ) one can pick θ, for example, between [−π/2,π/2] as you did, and get arcsin(x)+C. If you pick another interval, for example, [π/2,3π/2] then cos(θ) will be indeed negative, you will get the antiderivative −θ+C, however, in that interval θ=π−arcsin(x), so at the very end you will get the same positive arcsin as −θ+C=arcsin(x)+C−π.
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answered Nov 14, 2016 at 0:37
A.Γ.A.Γ.
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You are quite right. In fact you substitute x=asinθ but you observe that −π2<θ<π2 covers your domain for x, which is (−a,a) (because of the square root at the numerator). This gives you cosθ>0
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answered Nov 14, 2016 at 0:28
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∫dxa2−x2−−−−−−√
x=asinθ⇒dx=acosθdθ
∫dxa2−x2−−−−−−√=∫acosθdθacosθ=θ+c=arcsinxa+c
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edited Sep 6, 2022 at 13:48
answered Sep 6, 2022 at 13:43
Bouzari AbdelkaderBouzari Abdelkader
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14434 | https://booksnbackpacks.com/sum-latin-verb-to-be/ | Skip to content
Books 'n' Backpacks
Your Guide to Latin & Classics
Books 'n' Backpacks
Your Guide to Latin & Classics
The Irregular Latin Verb Sum (To Be): Conjugation, Uses, & More
ByLivia Updated
The Latin verb sum (“to be”) is one of the most frequently used words in Latin, but it is also highly irregular. This post covers how to conjugate sum, how to use it, and much more.
I’ll start with a quick overview before diving into the particulars. First and foremost, sum is used to express an equivalence between two things or to describe a noun’s characteristics. This is just like in English, where the verb to be tells us who someone is or what they are like.
In the following examples, forms of the verb to be are in bold.
Līvia est magistra. = Livia is a teacher.
Avēs sunt animālia. = Birds are animals.
Estis altī. = You (all) are tall.
Sum is the second most frequently used word in the Latin language, and this makes sense. We need the verb to be to talk about professions, nationalities, familial relationships, classifications, and so many other aspects of identity.
It’s safe to say that Latin will not make much sense if you don’t understand the verb sum. It is everywhere, from the simplest Latin novellas to the most advanced Roman texts.
The concept is simple, but there is a catch: sum is a highly irregular verb. The forms of to be in Latin often look quite unusual, so it is important to be able to recognize them.
Keep reading to learn all you ever wanted to know about the Latin verb sum (and more)!
How To Conjugate the Latin Verb Sum
Just like all Latin verbs, sum must be conjugated. You have already seen a few conjugated forms in the sample sentences above. For instance, est means “he/she/it is”, while sunt means “they are”.
👉 Need a refresher on conjugation? This post will explain everything.
As with any Latin verb, we will start by looking at the principal parts. The principal parts of sum are sum, esse, fuī, futūrus.
The first two principal parts are different from those of other verbs you may have learned. Normally a verb’s second principal part tells you its conjugation, but esse does not have any of the standard infinitive endings.
This is because sum, esse, fuī, futūrus is an irregular verb. In other words, it does not belong to any of the four Latin verb conjugations.
Sum is highly irregular in the present system (the present, future, and imperfect tenses). But luckily, it is completely regular in the perfect system (the perfect, pluperfect, and future perfect tenses).
In the following sections, I will list all the forms of the verb sum. If you are a beginning student, you may want to stop reading after the present, future, and imperfect tenses and jump down to the section on how to use the verb to be in Latin.
Present System (Indicative)
The present system includes the present, future, and imperfect tenses of the indicative mood. These are usually the first three tenses of sum that students learn.
These are also the most irregular tenses of sum, so you will have to do some memorization. But I do have some good news on that front.
Even in the present system, sum follows the expected patterns for active personal endings. That is, sum ends in m, es ends in s, and so forth.
👉 Not sure what this means? Read all about Latin verb personal endings here!
Now let’s look at the present tense of sum and its translation.
| PERSON | SINGULAR | PLURAL |
---
| 1 | sumI am | sumuswe are |
| 2 | esyou are | estisyou (y’all) are |
| 3 | esthe/she/it is | suntthey are |
Present indicative of sum
The present infinitive of sum is esse, which means “to be.”
And here is the future tense of sum and its translation:
| PERSON | SINGULAR | PLURAL |
---
| 1 | erōI will be | erimuswe will be |
| 2 | erisyou will be | eritisyou (y’all) will be |
| 3 | erithe/she/it will be | eruntthey will be |
Future indicative of sum
Finally, here is the imperfect tense of sum and its translation:
| PERSON | SINGULAR | PLURAL |
---
| 1 | eramI was | erāmuswe were |
| 2 | erāsyou were | erātisyou (y’all) were |
| 3 | erathe/she/it was | erantthey were |
Imperfect indicative of sum
And there you go. You have now seen the verb to be in Latin in three crucial tenses.
If you are a beginning student, I recommend that you now skip down to the uses of sum.
In a moment we will move on to the perfect tenses, but first we need to look at the present and imperfect subjunctive.
👉 Wondering what the subjunctive is? Read all about this Latin mood!
Present System (Subjunctive)
The present subjunctive is irregular and consists of sī– plus the active personal endings. The imperfect subjunctive, on the other hand, follows the standard rule: add the active personal endings to the present infinitive.
Here are the forms of the present and imperfect subjunctive. Subjunctives are hard to translate in isolation, but the present tense is something like I may be, you may be, etc. The imperfect tense can be translated as I might be, you might be, etc.
| Person | Present | Imperfect |
---
| 1 sing. | sim | essem |
| 2 sing. | sīs | essēs |
| 3 sing. | sit | esset |
| 1 plur. | sīmus | essēmus |
| 2 plur. | sītis | essētis |
| 3 plur. | sint | essent |
Present and imperfect subjunctive of sum
WATCH OUT: There is also an alternative form of the imperfect subjunctive that appears in some ancient authors. It is based on the future infinitive fore (discussed below).
You simply add the active personal endings to fore. Thus we find forem for essem, forēs for essēs, and so forth.
Perfect System (Indicative)
The perfect system includes the perfect, pluperfect, and future perfect tenses of the indicative mood, as well as the perfect and pluperfect tenses of the subjunctive.
Fortunately, the perfect system of the Latin verb to be is completely regular. You simply add the appropriate tense and mood endings to the perfect stem.
The perfect stem, as is standard, can be found by removing ī from the third principal part. This means that the perfect stem of sum is fu-.
Here is a conjugation chart showing the perfect indicative of sum:
| PERSON | SINGULAR | PLURAL |
---
| 1 | fuīI was / have been | fuimuswe were / have been |
| 2 | fuistīyou were / have been | fuistisyou (y’all) were / have been |
| 3 | fuithe/she/it was / has been | fuērunt / fuērethey were / have been |
Perfect indicative of sum
And here is the pluperfect indicative of sum:
| PERSON | SINGULAR | PLURAL |
---
| 1 | fueramI had been | fuerāmuswe had been |
| 2 | fuerāsyou had been | fuerātisyou (y’all) had been |
| 3 | fuerathe/she/it had been | fuerantthey had been |
Pluperfect indicative of sum
And then the future perfect indicative of sum:
| PERSON | SINGULAR | PLURAL |
---
| 1 | fuerōI will have been | fuerimuswe will have been |
| 2 | fuerisyou will have been | fueritisyou (y’all) will have been |
| 3 | fuerithe/she/it will have been | fuerintthey will have been |
Future perfect indicative of sum
Perfect System (Subjunctive)
Finally, let’s look at the perfect and pluperfect subjunctive. Just like with the perfect tenses of the indicative, you add the regular perfect and pluperfect subjunctive endings to the perfect stem (fu-).
It is hard to translate subjunctive forms in isolation, but the perfect means something like I may have been, you may have been, etc., while the pluperfect is something like I might have been, you might have been, etc.
| Person | Perfect | Pluperfect |
---
| 1 sing. | fuerim | fuissem |
| 2 sing. | fueris | fuissēs |
| 3 sing. | fuerit | fuisset |
| 1 plur. | fuerimus | fuissēmus |
| 2 plur. | fueritis | fuissētis |
| 3 plur. | fuerint | fuissent |
Perfect and pluperfect subjunctive of sum
Infinitives, Participles, and Imperatives
The verb to be in Latin is highly irregular, as we have seen, but it is also defective. This means that it does not have all of the possible verb forms.
For instance, sum does not have a gerund or a supine. It is also missing some participles and infinitives.
The verb sum has three infinitives:
| Tense | Latin Form | Translation |
---
| Present | esse | to be |
| Perfect | fuisse | to have been |
| Future | futūrum esse / fore | to be about to be |
Infinitives of the Latin verb sum
Sum only has one surviving participle: futūrus, a, um “about to be, going to be”. This is a future participle. You can tell this from -ūrus, which is the standard future active participial ending.
Sum has both present imperatives and future imperatives (that is, command forms).
| Person | Present | Future |
---
| 2 sing. | esbe! | estōbe! (you shall be) |
| 3 sing. | none | estōhe/she/it shall be |
| 2 plur. | este(y’all) be! | estōtebe! (y’all shall be) |
| 3 plur. | none | suntōthey shall be |
Imperatives of the Latin verb sum
Uses of the Latin Verb Sum (To Be)
Now it’s time to look at concrete examples of the Latin verb to be in action. You will notice that the forms of sum appear at the end, middle, and even beginning of the sentence.
In general, verbs in Latin tend to go at the end of the sentence. But sum is a bit irregular in this regard, too. The word order is highly flexible, so don’t be surprised if you spot the verb to be popping up in unlikely places.
To Be (Linking Verb)
The primary use of the verb sum is as a copula or linking verb. In this context, sum links together and equates two nouns or a noun and an adjective.
Amīca mea est fīlia agricolae. = My friend is the daughter of a farmer.
Hic puer altus erit. = This boy will be tall.
Amāre est bonum. = To love is good.
Poēta esse volō. = I want to be a poet.
We can see sum as an equals sign (=). My friend equals the daughter of a farmer; the boy will equal a tall person.
In such circumstances, the noun or adjective linked by sum (the predicate nominative) needs to be in the same case as the subject. This is typically the nominative case.
Just like the English verb to be, sum can also be followed by a prepositional phrase.
Equus erat in agrō. = The horse was in the field.
Vir rogāvit cūr librī in mēnsā nōn essent. = The man asked why the books weren’t on the table.
Existential Use of Sum
We often think of sum as simply meaning “to be,” but it can also mean “to exist.” Our English verb to be can also have this existential sense, as we shall soon see.
A classic example is René Descartes’ famous phrase:
Cogitō ergō sum. = I think therefore I am / exist.
But this usage shows up in more mundane situations, too. Forms of sum can be the equivalent of English phrases like “there is” and “there are.”
“There” is not translated into Latin. Just the verb is enough.
est = there is
sunt = there are
erit / erunt = there will be
erat = there was
erant = there were
Take a look at the following example sentences and note the different ways that sum can be translated into English.
Sunt equī in agrō. = There are horses in the field.
Cūr est canis in sellā? = Why is there a dog on the chair?
Erant quī rēgem nōn amābant. = There were men who did not like the king. = Men existed who did not like the king.
Erit plūs pecūniae post bellum. = There will be more money after the war.
Sum as a helping verb
Sum can act as a helping verb, just like the verb to be can in English.
In English, for instance, you can say “the woman is working.” In this sentence, is isn’t acting as a linking verb or expressing existence. Instead, it is helping to convey the progressive nature of the woman’s action.
Is working is a verbal unit, which we would call the English “present progressive” tense.
Latin does not have a present progressive, but we can find the verb sum used to help create the passive voice of other Latin verbs. The perfect, pluperfect, and future perfect passive tenses are formed with sum and the perfect participle passive.
Here is a quick example with the 1st conjugation verb laudō “praise.” The 1st person singular perfect indicative passive form of laudō is laudātus sum.
On its own, laudātus means “having been praised.” When we combine it with sum, a super literal translation would be “I am having been praised,” i.e. I was praised at some point in the past.
This compound form should be translated as “I have been praised” or “I was praised.” As you can see, sum helps the participle to create the proper meaning.
If you are a beginning student, then you don’t need to worry about this at all yet. If you are more advanced, then be on the lookout for sum in passive forms!
Compound Verbs Built on Sum
The Latin verb to be can also form compound verbs. A prefix (usually a preposition) is added to the beginning of sum in order to change the meaning slightly.
One extremely common compound verb is possum, posse, potuī “be able, can.” It consists of the adjective potis “capable, able” plus sum.
Here are some other compound verbs formed from sum:
| Verb | Meaning |
--- |
| absum, abesse, āfuī, āfutūrus | be away, absent (used with ab + ablative) |
| adsum, adesse, adfuī, adfutūrus | be near, present |
| dēsum, dēesse, dēfuī, dēfutūrus | be absent, lacking |
| intersum, interesse, interfuī | be between, present; differ; be of interest |
| praesum, praeesse, praefuī | preside over (+ dative) |
| prōsum, prōdesse, prōfuī | be useful, benefit (+ dative) |
| supersum, superesse, superfuī | remain; abound |
Compound verbs formed from sum
The good news is that all of these verbs have the exact same forms as sum, just with the relevant prefix added.
For prōsum, you insert a D between prō and the form of sum if the form begins with E (so, prōdes, prōdest, etc.).
Turris duo mīlia ab urbe abest. = The tower is two miles away from the city.
Mihi pecūnia dēerat. = Money was lacking for me (i.e. I was lacking money).
Hic liber tibi nōn prōderit. = This book will not be useful to you.
Why is the Latin verb sum so weird?
As we have seen, the conjugation of the verb to be in Latin is very irregular. My students often ask me why, and here’s what I tell them.
One cause of the irregularity is the fact that sum is a suppletive verb. Basically, this means that its forms go back to two entirely different roots.
The present system of sum derives from the Proto-Indo-European root h1es- or es- meaning “be”. We see es or just s surviving in the present tense (sum, es, etc.), while in the future (erō, eris, etc.) and the imperfect (eram, erās, etc.) theS has changed to an R between vowels.
The perfect system and the future participle, on the other hand, come from the Proto-Indo-European root bhuhx– or bheu- meaning “become” or “grow.” Eventually, the bh turns into f, and we end up with fuī, futūrus, and all the forms with fu.
The two roots es- and bheu- account for a lot of the confusion in the conjugation of sum.
Fun fact: the exact same thing has happened with the English verb to be. Am, are, and is come from es-, while be, been, and being come from bheu-. Was and were derive from a third stem, wes- “remain, dwell.”
You may still be wondering: Why did this bizarre state of affairs occur? Historical linguistics has a lot to say on this topic, but here’s a quick answer. Basically, common words are more likely to be irregular.
Languages change over time, and irregularities in sound and form develop naturally. Over the generations speakers, just as naturally, tend to eliminate the irregularities to make words behave “normally.”
But this standardization does not happen as much with extremely common words. The reason is that speakers hear and use them constantly, so the irregular forms stick around.
The verb sum is the most frequently used verb in the Latin language, so it makes sense that its peculiarities have been preserved.
Frequently Asked Questions about the Latin Verb To Be
What conjugation is sum, esse?
Sum, esse is an irregular verb. This means that it does not belong to any of the four Latin verb conjugations. Instead, sum has its own unique forms in the present, imperfect, and future tenses.
Is sum active or passive?
Sum is neither active nor passive, because it is a linking verb. But it does use active personal endings, so it has more of an active flavor.
If you are parsing verbs for an assignment, ask your teacher for their preference. I tell my students they can either omit voice in the parsing of sum OR mark it as active.
Does sum have participles?
The Latin verb sum has one surviving participle: futūrus, a, um. Futūrus, which literally means “about to be” or “going to be”, is a future active participle.
Notice the familiar -ūrus ending, which signals the future active participle. The neuter futūrum “a thing about to be” is the source of the English word “future.”
What are the principal parts of the Latin verb to be?
The principal parts of sum “to be” are sum, esse, fuī, futūrus. Sum is the 1st person singular of the present indicative (“I am”). Esse is the present infinitive (“to be”). Fuī is the 1st person singular of the perfect indicative (“I was”). Futūrus is the future active participle (“about to be”).
What is the Latin verb for to be?
Sum, esse, fuī, futūrus is the Latin verb to be. If you want to translate the specific phrase “to be”, you should use the form esse. Esse is the present infinitive and thus means “to be” or “to exist.”
What is the difference between sum and est?
Sum is the first person singular form of the Latin verb to be and is translated as “I am.” Est is the third person singular and is translated as “he/she/it is.” Both forms are present tense and are extremely common, so you may see them even if you are a brand-new Latin learner.
Final Thoughts on the Verb To Be in Latin
Do you feel a bit more comfortable with the verb sum and its uses now? I hope so! This irregular verb shows up everywhere in Latin, so you will have lots of opportunities to test your new knowledge.
We have covered a lot of information in this post, so don’t worry if you are feeling overwhelmed. The most important thing to remember is that the verb sum is used in two main ways: first, as a linking verb (“to be”) and second, to express existence (“to exist”).
While you’re here, you may want to take a look at the following posts on learning Latin:
Awesome Resources for Latin Learners
Best Latin Dictionaries for Students
How To Say Happy Birthday in Latin
The Ultimate Guide to Latin Personal Pronouns
Livia
Rebecca Deitsch (aka Livia) is an assistant professor of Classical Languages and Literatures at Smith College. She has a PhD in Classical Philology from Harvard University and has taught Latin, Greek, and Classics courses to students of all ages at Harvard, Kenyon, Wellesley, and beyond. After 20 years of learning and 10 years of teaching Latin, she made this website to share all her language expertise with you!
6 Comments
Long, long ago, I used “y’all” in my Latin classes. My instructor – an immigrant from a weird part of northern Italy that grew up speaking German – had never heard that word before and was absolutely delighted by it.
Reply
Hi Steve, I’m glad you were able to share the delight of this word with your instructor. I grew up in Georgia, so “y’all” has always been part of my vocabulary. My students love using it in class, no matter where they are from!
Reply
2. Thank you for this exposition. I wonder why we have “eram” and “fui” meaning “I was”. Interpreting “fui” as “I have been” is easier because that gives us two different words for two different meanings. How does “fui” come be “I was” as well as “I have been”?
Reply
Hi Fred, this is a great question. There are two parts to the answer, one which involves the peculiarities of Latin and the other the peculiarities of English.
First, the Latin perfect tense is hybrid in the sense that it combines two Proto-Indo-European tenses: the aorist (simple past) and the present perfect. So, a perfect like vīdī can mean both I saw (simple past) and I have seen (present perfect), and laudāvī can mean both I praised (simple past) and I have praised (present perfect). Latin uses one tense where in English we have two separate ones, and this is why fuī can mean both I was (simple past) and I have been (present perfect).
But why is eram (imperfect) ALSO translated as I was? This is because English doesn’t tend to distinguish between the simple past (e.g. I praised) and the past progressive (e.g. I was praising) in the verb to be. You could technically translate eram as I was being to get the imperfect aspect, but this sounds weird in English, so we just say I was.
I hope this makes sense. To summarize, fuī, when translated as I was, emphasizes the completion of the action/state of being. Eram, also translated as I was, emphasizes the ongoing nature of the action/state of being.
Reply
3. Thank You, Livia! I’m now wondering about the latin word “incolunt”, used famously by Caesar. It is translated everywhere as “inhabit” or “live in” (I think). I wonder whether there might be some additional nuance to this word, as it resembles colony or colonize, Might the word carry in addition to its simplistic translations the connotation of carrying a developed culture into a new place, or pursuing the practices of a developed culture (“living”) in a particular place?
I promise not to inundate you with further questions of this sort. After reading this I think my real question has to do with whether Latin writers/speakers employ nuance as we do in English. I think the answer must be yes. I guess it will take a great deal of reading followed by writing in Latin for me to truly understand the language. Getting ahead of myself…
Reply
Hello again, Fred! This is a great question – ‘incolō’ comes from the verb ‘colō’, which originally meant “cultivate” or “till.” So the basic meaning is agricultural, something like “pursue farming practices in a given place.” This is where the relationship with ‘colony’ and ‘colonize’ comes in.
And to answer your second question: yes, Latin writers/speakers definitely employ nuance. In my opinion, one of the most fun parts of learning a new language is discovering all the little distinctions and shades of meaning.
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14435 | https://www.doubtnut.com/qna/643743591 | Prove that the tangent line to the parabola x2=2py at an arbitrary point makes an angle α with the x-axis whose tangent is equal to the x-coordinate of the point divided by the parameter p (i.e. tanα=x/p).
The direction of the speed of a practicle is just the direction of the tangent to its path. It may be seen from Fig. that tanα=gt/v0. But since t=x/v0, it follows that tanα=gx/v20=x/p When you have learned to differentiate, you will be able to solve the problem using the derivative in the following way: tanα=dydx=ddx(x22p)= =2x2p=xp
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14436 | https://ohsrehak.weebly.com/uploads/5/4/6/9/54699399/7-4_parallel_lines_and_proportional_parts.pdf | 1. If XM = 4, XN = 6, and NZ = 9, find XY. SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. Use the Triangle Proportionality Theorem. Substitute. Solve for MY. Find XY. ANSWER: 10 2. If XN = 6, XM = 2, and XY = 10, find NZ. SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. XY =10, So, MY = 10 – 8 =2. Use the Triangle Proportionality Theorem. eSolutions Manual - Powered by Cognero Page 1 7-4 Parallel Lines and Proportional Parts ANSWER: 10 2. If XN = 6, XM = 2, and XY = 10, find NZ. SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. XY =10, So, MY = 10 – 8 =2. Use the Triangle Proportionality Theorem. Solve for NZ. ANSWER: 24 3. In BC = 15, BE = 6, DC = 12, and AD = 8. Determine whether Justify your answer. SOLUTION: If BC = 15, then EC = 15 – 6 = 9. Use the Converse of the Triangle Proportionality Theorem. Therefore, . ANSWER: yes; 4. In JK = 15, JM = 5, LK = 13, and PK = 9. Determine whether Justify your answer.
eSolutions Manual - Powered by Cognero Page 2 7-4 Parallel Lines and Proportional Parts ANSWER: 24 3. In BC = 15, BE = 6, DC = 12, and AD = 8. Determine whether Justify your answer. SOLUTION: If BC = 15, then EC = 15 – 6 = 9. Use the Converse of the Triangle Proportionality Theorem. Therefore, . ANSWER: yes; 4. In JK = 15, JM = 5, LK = 13, and PK = 9. Determine whether Justify your answer. SOLUTION: JK = 15 and LK = 13. Therefore, MK = 15 – 5 = 10 and LP = 13 – 9 = 4. Use the Converse of the Triangle Proportionality Theorem. If BC = 15, then EC = 15 – 6 = 9. Use the Converse of the Triangle Proportionality Theorem. So, and are not parallel. ANSWER: no; is a midsegment of Find the value of x.
eSolutions Manual - Powered by Cognero Page 3 7-4 Parallel Lines and Proportional Parts So, and are not parallel. ANSWER: no; is a midsegment of Find the value of x.
5. SOLUTION: By the Triangle Midsegment Theorem, Substitute. ANSWER: 11 6. SOLUTION: By the Triangle Midsegment Theorem, Substitute. ANSWER: 10 7. MAPS Refer to the map. 3rd Avenue and 5th Avenue are parallel. If the distance from 3rd Avenue to City Mall along State Street is 3201 feet, find the distance between 5th Avenue and City Mall along Union Street. Round to the nearest tenth. SOLUTION: The distance between 5th Avenue and City Mall along State Street is 3201 − 1056 or 2145 feet.
Let x be the distance between 5th Avenue and City Mall along Union Street. Use the Triangle Proportionality Theorem. eSolutions Manual - Powered by Cognero Page 4 7-4 Parallel Lines and Proportional Parts ANSWER: 10 7. MAPS Refer to the map. 3rd Avenue and 5th Avenue are parallel. If the distance from 3rd Avenue to City Mall along State Street is 3201 feet, find the distance between 5th Avenue and City Mall along Union Street. Round to the nearest tenth. SOLUTION: The distance between 5th Avenue and City Mall along State Street is 3201 − 1056 or 2145 feet.
Let x be the distance between 5th Avenue and City Mall along Union Street. Use the Triangle Proportionality Theorem. The distance between 5th Avenue and City Mall along Union Street is 2360.3 ft. ANSWER: 2360.3 ft ALGEBRA Find x and y.
8. SOLUTION: We are given that and Solve for x. Solve for y. eSolutions Manual - Powered by Cognero Page 5 7-4 Parallel Lines and Proportional Parts The distance between 5th Avenue and City Mall along Union Street is 2360.3 ft. ANSWER: 2360.3 ft ALGEBRA Find x and y.
8. SOLUTION: We are given that and Solve for x. Solve for y. ANSWER: x = 5; y = 8 9. SOLUTION: We are given that . Solve for y. By Corollary 7.2, .
eSolutions Manual - Powered by Cognero Page 6 7-4 Parallel Lines and Proportional Parts ANSWER: x = 5; y = 8 9. SOLUTION: We are given that . Solve for y. By Corollary 7.2, . Solve for x. ANSWER: x = 20; y = 2 10. If AB = 6, BC = 4, and AE = 9, find ED. SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. Use the Triangle Proportionality Theorem. eSolutions Manual - Powered by Cognero Page 7 7-4 Parallel Lines and Proportional Parts ANSWER: x = 20; y = 2 10. If AB = 6, BC = 4, and AE = 9, find ED. SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. Use the Triangle Proportionality Theorem. Substitute. Solve for ED. ANSWER: 6 11. If AB = 12, AC = 16, and ED = 5, find AE. SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. AB =12. So, BC = 16 – 12 = 4. Use the Triangle Proportionality Theorem. Substitute. eSolutions Manual - Powered by Cognero Page 8 7-4 Parallel Lines and Proportional Parts ANSWER: 6 11. If AB = 12, AC = 16, and ED = 5, find AE. SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. AB =12. So, BC = 16 – 12 = 4. Use the Triangle Proportionality Theorem. Substitute. Solve for AE. ANSWER: 15 12. If AC = 14, BC = 8, and AD = 21, find ED. SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. Here, BC = 8. So, AB = 14 – 8 = 6. Let x be the length of the segment AE. So, ED = 21 – x. Use the Triangle Proportionality Theorem. S b tit t eSolutions Manual - Powered by Cognero Page 9 7-4 Parallel Lines and Proportional Parts ANSWER: 15 12. If AC = 14, BC = 8, and AD = 21, find ED. SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. Here, BC = 8. So, AB = 14 – 8 = 6. Let x be the length of the segment AE. So, ED = 21 – x. Use the Triangle Proportionality Theorem. Substitute. Solve for x. So, AE = 9 and ED = 21 – 9 = 12. ANSWER: 12 13. If AD = 27, AB = 8, and AE = 12, find BC. SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. Here, AE = 12. So, ED = 27 – 12 = 15. Use the Triangle Proportionality Theorem. eSolutions Manual - Powered by Cognero Page 10 7-4 Parallel Lines and Proportional Parts So, AE = 9 and ED = 21 – 9 = 12. ANSWER: 12 13. If AD = 27, AB = 8, and AE = 12, find BC. SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. Here, AE = 12. So, ED = 27 – 12 = 15. Use the Triangle Proportionality Theorem. Substitute in values and solve for BC. ANSWER: 10 Determine whether Justify your answer. 14. ZX = 18, ZV = 6, WX = 24, and YX = 16 SOLUTION: ZV = 6 and YX = 16. Therefore, VX = 18 – 6 = 12 and WY = 24 – 16 = 8. Use the Converse of the Triangle Proportionality Theorem. Since , then . eSolutions Manual - Powered by Cognero Page 11 7-4 Parallel Lines and Proportional Parts ANSWER: 10 Determine whether Justify your answer. 14. ZX = 18, ZV = 6, WX = 24, and YX = 16 SOLUTION: ZV = 6 and YX = 16. Therefore, VX = 18 – 6 = 12 and WY = 24 – 16 = 8. Use the Converse of the Triangle Proportionality Theorem. Since , then . ANSWER: yes; 15. VX = 7.5, ZX = 24, WY = 27.5, and WX = 40 SOLUTION: VX = 7.5 and WY = 27.5. So, ZV = 24 – 7.5 = 16.5 and YX = 40 – 27.5 = 12.5. Use the Converse of the Triangle Proportionality Theorem. Since ANSWER: yes; 16. ZV = 8, VX = 2, and YX = WY SOLUTION: Use the Converse of the Triangle Proportionality Theorem. eSolutions Manual - Powered by Cognero Page 12 7-4 Parallel Lines and Proportional Parts ANSWER: yes; 16. ZV = 8, VX = 2, and YX = WY SOLUTION: Use the Converse of the Triangle Proportionality Theorem. Because , and are not parallel. ANSWER: no; 17. WX = 31, YX = 21, and ZX = 4ZV SOLUTION: YX = 21, so WY = 31 – 21 = 10 and since ZX = 4ZV, then VX = 3ZV.
Use the Converse of the Triangle Proportionality Theorem. Because , we can say that and are not parallel.
ANSWER: no; are midsegments of Find the value of x.
18. SOLUTION: By the Triangle Midsegment Theorem, . By the Alternate Interior Angles Theorem, x = 57.
ANSWER: 57 eSolutions Manual - Powered by Cognero Page 13 7-4 Parallel Lines and Proportional Parts ANSWER: no; are midsegments of Find the value of x.
18. SOLUTION: By the Triangle Midsegment Theorem, . By the Alternate Interior Angles Theorem, x = 57.
ANSWER: 57 19. SOLUTION: By the Triangle Midsegment Theorem, . By the Alternate Interior Angles Theorem, .
ANSWER: 60 20. SOLUTION: By the Triangle Midsegment Theorem, Substitute. ANSWER: 50 eSolutions Manual - Powered by Cognero Page 14 7-4 Parallel Lines and Proportional Parts ANSWER: 50 21. SOLUTION: By the Triangle Midsegment Theorem, Substitute. ANSWER: 1.35 22. CCSS MODELING In Charleston, South Carolina, Logan Street is parallel to both King Street and Smith Street between Beaufain Street and Queen Street. What is the distance from Smith to Logan along Beaufain? Round to the nearest foot. SOLUTION: Let x be the distance from Smith to Logon along Beaufain. Use the Triangle Proportionality Theorem. Solve for x. So, the distance from Smith to Logan is 891 ft. ANSWER: about 891 ft 23. ART Tonisha drew the line of dancers shown below for her perspective project in art class. Each of the dancers is parallel. Find the lower distance between the first two dancers. SOLUTION: eSolutions Manual - Powered by Cognero Page 15 7-4 Parallel Lines and Proportional Parts So, the distance from Smith to Logan is 891 ft. ANSWER: about 891 ft 23. ART Tonisha drew the line of dancers shown below for her perspective project in art class. Each of the dancers is parallel. Find the lower distance between the first two dancers. SOLUTION: Distance between second dancer and third dancer = Let x be the lower distance between the first two dancers. Use the Triangle Proportionality Theorem. So, the lower distance between the first two dancers is or 1.2 inches.
ANSWER: 1.2 in. ALGEBRA Find x and y.
24. SOLUTION: eSolutions Manual - Powered by Cognero Page 16 7-4 Parallel Lines and Proportional Parts So, the lower distance between the first two dancers is or 1.2 inches.
ANSWER: 1.2 in. ALGEBRA Find x and y.
24. SOLUTION: We are given that and Solve for x. Solve for y. ANSWER: x = 2; y = 5 25. SOLUTION: We are given that and Solve for x. eSolutions Manual - Powered by Cognero Page 17 7-4 Parallel Lines and Proportional Parts ANSWER: x = 2; y = 5 25. SOLUTION: We are given that and Solve for x. Solve for y. ANSWER: x = 18; y = 3 ALGEBRA Find x and y.
26. SOLUTION: It is given that and Solve for x. eSolutions Manual - Powered by Cognero Page 18 7-4 Parallel Lines and Proportional Parts ANSWER: x = 18; y = 3 ALGEBRA Find x and y.
26. SOLUTION: It is given that and Solve for x. x = 10 Solve for y. ANSWER: x = 10; y = 3 27. SOLUTION: We are given that . Solve for y. eSolutions Manual - Powered by Cognero Page 19 7-4 Parallel Lines and Proportional Parts ANSWER: x = 10; y = 3 27. SOLUTION: We are given that . Solve for y. By Corollary 7.2, . Solve for x. ANSWER: x = 48; y = 72 CCSS ARGUMENTS Write a paragraph proof.
28. Corollary 7.1 SOLUTION: In Corollary 7.1, it is stated that, if three or more parallel lines intersect two transversals, then they cut off the transversals proportionally. A good approach to this proof it is apply the Triangle Proportionality theorem, one triangle at a time. Given: Prove: eSolutions Manual - Powered by Cognero Page 20 7-4 Parallel Lines and Proportional Parts ANSWER: x = 48; y = 72 CCSS ARGUMENTS Write a paragraph proof.
28. Corollary 7.1 SOLUTION: In Corollary 7.1, it is stated that, if three or more parallel lines intersect two transversals, then they cut off the transversals proportionally. A good approach to this proof it is apply the Triangle Proportionality theorem, one triangle at a time. Given: Prove: Proof: In By the Triangle Proportionality Theorem, AB and DE are proportional . In By the Triangle Proportionality Theorem, BC and EF are proportional. Therefore, ANSWER: Given: Prove: Proof: In By the Triangle Proportionality Theorem, AB and DE are proportional. In By the Triangle Proportionality Theorem, BC and EF are proportional. Therefore, 29. Corollary 7.2 SOLUTION: Corollary 7.2 states, if three or more parallel lines cut off congruent segments on one transversal, then they cut off congruent segments on every transversal. This proof can be approached by using Corollary 7.1 to establish that, since we have three parallel lines, then we know they cut off the transversals proportionally. If the ratio of one side of this proportion is equal to 1, since both parts are equal, then the other side of the proportion must also equal 1. Therefore, they are also equal, or congruent, parts. Given: eSolutions Manual - Powered by Cognero Page 21 7-4 Parallel Lines and Proportional Parts Proof: In By the Triangle Proportionality Theorem, AB and DE are proportional. In By the Triangle Proportionality Theorem, BC and EF are proportional. Therefore, 29. Corollary 7.2 SOLUTION: Corollary 7.2 states, if three or more parallel lines cut off congruent segments on one transversal, then they cut off congruent segments on every transversal. This proof can be approached by using Corollary 7.1 to establish that, since we have three parallel lines, then we know they cut off the transversals proportionally. If the ratio of one side of this proportion is equal to 1, since both parts are equal, then the other side of the proportion must also equal 1. Therefore, they are also equal, or congruent, parts. Given: Prove: Proof: From Corollary 7.1, Since AB = BC by definition of congruence. Therefore, = 1. By substitution, 1 = Thus, DE = EF. By definition of congruence, ANSWER: Given: Prove: Proof: From Corollary 7.1, Since AB = BC by definition of congruence. Therefore, = 1. By substitution, 1 = Thus, DE = EF. By definition of congruence, 30. Theorem 7.5 SOLUTION: Theorem 7.5 states, if a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. In order to prove that , we first need to establish that , which can be accomplished by showing that Then, by Segment Addition Postulate, we can state that CA = BA + CB and CE = DE + CD. Substitute these values in for CA and CE in the previous proportion and the simplify. Given: eSolutions Manual - Powered by Cognero Page 22 7-4 Parallel Lines and Proportional Parts SOLUTION: Theorem 7.5 states, if a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. In order to prove that , we first need to establish that , which can be accomplished by showing that Then, by Segment Addition Postulate, we can state that CA = BA + CB and CE = DE + CD. Substitute these values in for CA and CE in the previous proportion and the simplify. Given: Prove: Proof: , and because they are corresponding angles. By AA Similarity, From the definition of similar polygons, By the Segment Addition Postulate, CA = BA + CB and CE = DE + CD. By substitution, Rewriting as a sum, From simplifying, Thus, by subtracting one from each side. ANSWER: Given: Prove: Proof: , and because they are corresponding angles. By AA Similarity, From the definition of similar polygons, By the Segment Addition Postulate, CA = BA + CB and CE = DE + CD. By substitution, Rewriting as a sum, From simplifying, Thus, by subtracting one from each side. CCSS ARGUMENTS Write a two-column proof.
31. Theorem 7.6 SOLUTION: Theorem 7.6 states, if a line intersects two sides of a triangle and separates the sides into proportional corresponding segments, then the line is parallel to the third side of the triangle. Thinking backwards, how can we prove that two lines are parallel to each other? We can prove that by proving that a pair of corresponding angles, formed by these parallel lines, are congruent to each other. Using SAS Similarity theorem, prove that . Then, you can use congruent corresponding angles as a result of similar triangles. eSolutions Manual - Powered by Cognero Page 23 7-4 Parallel Lines and Proportional Parts CCSS ARGUMENTS Write a two-column proof.
31. Theorem 7.6 SOLUTION: Theorem 7.6 states, if a line intersects two sides of a triangle and separates the sides into proportional corresponding segments, then the line is parallel to the third side of the triangle. Thinking backwards, how can we prove that two lines are parallel to each other? We can prove that by proving that a pair of corresponding angles, formed by these parallel lines, are congruent to each other. Using SAS Similarity theorem, prove that . Then, you can use congruent corresponding angles as a result of similar triangles. Given: Prove: Proof: Statements (Reasons) 1. (Given) 2. (Add. Prop.) 3. (Subst.) 4. AB = AD + DB, AC = AE + EC (Seg. Add. Post.) 5. (Subst.) 6. (Refl. Prop.) 7. (SAS Similarity) 8. (Def. of polygons) 9. (If corr. angles are , then the lines are || .) ANSWER: Given: Prove: Proof: Statements (Reasons) 1. (Given) 2. (Add. Prop.) 3. (Subst.) 4. AB = AD + DB, AC = AE + EC (Seg. Add. Post.) 5. (Subst.) 6. (Refl. Prop.) 7. (SAS Similarity) 8 (Def of polygons) eSolutions Manual - Powered by Cognero Page 24 7-4 Parallel Lines and Proportional Parts 2. (Add. Prop.) 3. (Subst.) 4. AB = AD + DB, AC = AE + EC (Seg. Add. Post.) 5. (Subst.) 6. (Refl. Prop.) 7. (SAS Similarity) 8. (Def. of polygons) 9. (If corr. angles are , then the lines are || .) 32. Theorem 7.7 SOLUTION: Theorem 7.7 states that a midsegment of a triangle is parallel to one side of the triangle, and its length is half the length of that side. For this proof, use the given information that to prove that by AA Similarity. Then, since you know that D and E are both midpoints, then you can prove eventually prove that , using midpoint relationships and substitution. Then, using as a result of proving , then you can substitute into into and prove that , using algebra. Given: D is the midpoint of E is the midpoint of Prove: Proof: Statements (Reasons) 1. D is the midpoint of E is the midpoint of (Given) 2. (Midpoint Thm.) 3. AD = DB, AE = EC (Def. of segs.) 4. AB = AD + DB, AC = AE + EC (Seg. Add. Post.) 5. AB = AD + AD, AC = AE + AE (Subst.) 6. AB = 2AD, AC = 2AE (Subst.) 7. (Div. Prop.) 8. (Trans. Prop.) 9. (Refl. Prop.) 10. (SAS Similarity) 11. (Def. of polygons) 12. (If corr. angles are , the lines are parallel.) 13. (Def. of polygons) 14. (Substitution Prop.) 15. 2DE = BC (Mult. Prop.) 16. (Division Prop.) eSolutions Manual - Powered by Cognero Page 25 7-4 Parallel Lines and Proportional Parts 10. (SAS Similarity) 11. (Def. of polygons) 12. (If corr. angles are , the lines are parallel.) 13. (Def. of polygons) 14. (Substitution Prop.) 15. 2DE = BC (Mult. Prop.) 16. (Division Prop.) ANSWER: Given: D is the midpoint of E is the midpoint of Prove: Proof: Statements (Reasons) 1. D is the midpoint of E is the midpoint of (Given) 2. (Midpoint Thm.) 3. AD = DB, AE = EC (Def. of segs.) 4. AB = AD + DB, AC = AE + EC (Seg. Add. Post.) 5. AB = AD + AD, AC = AE + AE (Subst.) 6. AB = 2AD, AC = 2AE (Subst.) 7. (Div. Prop.) 8. (Trans. Prop.) 9. (Refl. Prop.) 10. (SAS Similarity) 11. (Def. of polygons) 12. (If corr. angles are , the lines are parallel.) 13. (Def. of polygons) 14. (Substitution Prop.) 15. 2DE = BC (Mult. Prop.) 16. (Division Prop.) Refer to 33. If ST = 8, TR = 4, and PT = 6, find QR. SOLUTION: Since , we know that and . Therefore, by AA Similarity, . eSolutions Manual - Powered by Cognero Page 26 7-4 Parallel Lines and Proportional Parts 14. (Substitution Prop.) 15. 2DE = BC (Mult. Prop.) 16. (Division Prop.) Refer to 33. If ST = 8, TR = 4, and PT = 6, find QR. SOLUTION: Since , we know that and . Therefore, by AA Similarity, . Use the definition of similar polygons to create a proportion: We know that SR = 8 + 4 =12. Substitute values and solve for QR. ANSWER: 9 34. If SP = 4, PT = 6, and QR = 12, find SQ. SOLUTION: Since , we know that and . Therefore, by AA Similarity, . Use the definition of similar polygons to set up a proportion: Substitute and solve for SQ: ANSWER: 8 35. If CE = t – 2, EB = t + 1, CD = 2, and CA = 10, find t and CE. eSolutions Manual - Powered by Cognero Page 27 7-4 Parallel Lines and Proportional Parts ANSWER: 8 35. If CE = t – 2, EB = t + 1, CD = 2, and CA = 10, find t and CE. SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. Use the Triangle Proportionality Theorem. Since CA = 10 and CD = 2, then DA =10-2= 8. Substitute and solve for t. Find CE. ANSWER: 3, 1 36. If WX = 7, WY = a, WV = 6, and VZ = a – 9, find WY. SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. Use the Triangle Proportionality Theorem. Since WY = a and WX = 7, XY = a – 7. eSolutions Manual - Powered by Cognero Page 28 7-4 Parallel Lines and Proportional Parts ANSWER: 3, 1 36. If WX = 7, WY = a, WV = 6, and VZ = a – 9, find WY. SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. Use the Triangle Proportionality Theorem. Since WY = a and WX = 7, XY = a – 7. Substitute and solve for a. So, a= WY = 21. ANSWER: 21 37. If QR = 2, XW = 12, QW = 15, and ST = 5, find RS and WV. SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. Use the Triangle Proportionality Theorem. Since QW = 15 and WX = 12, then QX = 3. Substitute and solve for RS. eSolutions Manual - Powered by Cognero Page 29 7-4 Parallel Lines and Proportional Parts So, a= WY = 21. ANSWER: 21 37. If QR = 2, XW = 12, QW = 15, and ST = 5, find RS and WV. SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. Use the Triangle Proportionality Theorem. Since QW = 15 and WX = 12, then QX = 3. Substitute and solve for RS. Additionally, we know that . Substitute and solve for WV. ANSWER: 8, 7.5 38. If LK = 4, MP = 3, PQ = 6, KJ = 2, RS = 6, and LP = 2, find ML, QR, QK, and JH. SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. Use the Triangle Proportionality Theorem. eSolutions Manual - Powered by Cognero Page 30 7-4 Parallel Lines and Proportional Parts SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. Use the Triangle Proportionality Theorem. Substitute and solve for ML. Also, we know that . Substitute and solve for Because , by AA Similarity, we know that . Substitute and solve for QK. Finally, by Triangle Proportionality Theorem, . Substitute and solve for JH. ANSWER: 2, 3, 6, 4 39. MATH HISTORY The sector compass was a tool perfected by Galileo in the sixteenth century for measurement. To draw a segment two-fifths the length of a given segment, align the ends of the arms with the given segment. Then draw a segment at the 40 mark. Write a justification that explains why the sector compass works for proportional measurement. eSolutions Manual - Powered by Cognero Page 31 7-4 Parallel Lines and Proportional Parts ANSWER: 2, 3, 6, 4 39. MATH HISTORY The sector compass was a tool perfected by Galileo in the sixteenth century for measurement. To draw a segment two-fifths the length of a given segment, align the ends of the arms with the given segment. Then draw a segment at the 40 mark. Write a justification that explains why the sector compass works for proportional measurement. SOLUTION: To prove that two corresponding sides of two triangles are the same ratio as another pair of corresponding sides, you need to first establish that the triangles are similar. Once this is completed, a proportion statement can be written, relating the proportional sides. Substitute in given values from the diagram to prove that . ANSWER: Determine the value of x so that 40. AB = x + 5, BD = 12, AC = 3x + 1, and CF = 15 eSolutions Manual - Powered by Cognero Page 32 7-4 Parallel Lines and Proportional Parts Determine the value of x so that 40. AB = x + 5, BD = 12, AC = 3x + 1, and CF = 15 SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. Use the Triangle Proportionality Theorem. Substitute. ANSWER: 3 41. AC = 15, BD = 3x – 2, CF = 3x + 2, and AB = 12 SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. Use the Triangle Proportionality Theorem. Substitute. ANSWER: 6 42. COORDINATE GEOMETRY has vertices A(–8, 7), B(0, 1), and C(7, 5). Draw Determine the coordinates of the midsegment of that is parallel to Justify your answer. eSolutions Manual - Powered by Cognero Page 33 7-4 Parallel Lines and Proportional Parts ANSWER: 3 41. AC = 15, BD = 3x – 2, CF = 3x + 2, and AB = 12 SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. Use the Triangle Proportionality Theorem. Substitute. ANSWER: 6 42. COORDINATE GEOMETRY has vertices A(–8, 7), B(0, 1), and C(7, 5). Draw Determine the coordinates of the midsegment of that is parallel to Justify your answer. SOLUTION: Use the midpoint formula to determine the midpoints of and . The midpoint of = . The midpoint of = . Use the distance formula. eSolutions Manual - Powered by Cognero Page 34 7-4 Parallel Lines and Proportional Parts The midpoint of = . Use the distance formula. The segment is parallel to because the slopes are both and the segment length is half of Thus, the segment is the midsegment of ANSWER: The endpoints of the midsegment are (–4, 4) and (–0.5, 6). Sample answer: The segment is parallel to because the slopes are both and the segment length is half of Thus, the segment is the midsegment of 43. HOUSES Refer to the diagram of the gable. Each piece of siding is a uniform width. Find the lengths of and SOLUTION: All the triangles are isosceles. Segment EH is the midsegment of triangle ABC . Therefore, segment EH is the half of the length of AC, which is 35 ÷ 2 or 17.5 feet. Similarly, FG is the midsegment of triangle BEH, so FG = 17.5 ÷ 2 or 8.75 feet. To find DJ, use the vertical altitude which is 12 feet. Let the altitude from B to the segment AC meet the segment DJ at K. Find BC using the Pythagorean Theorem. BC2 = BK2 + KC2 BC2 = 122 + 17.52 BC = eSolutions Manual - Powered by Cognero Page 35 7-4 Parallel Lines and Proportional Parts The endpoints of the midsegment are (–4, 4) and (–0.5, 6). Sample answer: The segment is parallel to because the slopes are both and the segment length is half of Thus, the segment is the midsegment of 43. HOUSES Refer to the diagram of the gable. Each piece of siding is a uniform width. Find the lengths of and SOLUTION: All the triangles are isosceles. Segment EH is the midsegment of triangle ABC . Therefore, segment EH is the half of the length of AC, which is 35 ÷ 2 or 17.5 feet. Similarly, FG is the midsegment of triangle BEH, so FG = 17.5 ÷ 2 or 8.75 feet. To find DJ, use the vertical altitude which is 12 feet. Let the altitude from B to the segment AC meet the segment DJ at K. Find BC using the Pythagorean Theorem. BC2 = BK2 + KC2 BC2 = 122 + 17.52 BC = Since the width of each piece of siding is the same, BJ = BC, which is about or 15.92 in. Now, use the Triangle Proportionality Theorem. ANSWER: 8.75 in., 17.5 in., 26.25 in. CONSTRUCTIONS Construct each segment as directed.
44. a segment separated into five congruent segments SOLUTION: Step 1: Construct an angle with vertex A, as shown below: Step 2: With your compass on vertex A, choose a radius and make an arc on the diagonal, as shown below: eSolutions Manual - Powered by Cognero Page 36 7-4 Parallel Lines and Proportional Parts Step 2: With your compass on vertex A, choose a radius and make an arc on the diagonal, as shown below: Step 3: With your compass on the new point formed on the diagonal, keep the same radius and make another arc further down the diagonal side of the angle. Continue this process until you have five arcs, like below: Step 4: Using a straight edge, draw a segment that connects each new point back to the horizontal side of the angle, perpendicular to that side, as shown below: Step 5. Label the points formed on the horizontal side of the angle and erase any extra length beyond the last point. ANSWER: Sample answer: eSolutions Manual - Powered by Cognero Page 37 7-4 Parallel Lines and Proportional Parts ANSWER: Sample answer: 45. a segment separated into two segments in which their lengths have a ratio of 1 to 3 SOLUTION: Step 1: Make an angle, with vertex A, as shown below: Step 2: With your compass on vertex A, make an arc that passes through the diagonal side of the angle. Connect this new point back to the horizontal side of the angle. Label B as the new point made on the horizontal side of the angle, as shown below. Step 3: Continue this process until you have four arcs. When you connect the points on the diagonal back to the horizontal, make sure the connecting lines are all parallel to each other. ( Since you want segment lengths at a ratio of 1 to 3, this can be created by 4 equal smaller segments, where three can be pieced together to make one that is 3/4 the original length.) Step 4: Label the first point B and the last point C. . eSolutions Manual - Powered by Cognero Page 38 7-4 Parallel Lines and Proportional Parts Step 4: Label the first point B and the last point C. . ANSWER: Sample answer: 46. a segment 3 inches long, separated into four congruent segments SOLUTION: Step 1: Copy a 3 inch segment. horizontally. Then, make an angle, with vertex A, as shown below: Step 2: With your compass on vertex A, make an arc that passes through the diagonal side of the angle. Connect this new point back to the horizontal side of the angle. Label B as the new point made on the horizontal side of the angle, as shown below. Step 3: Continue this process until you have four arcs. When you connect the points on the diagonal back to the horizontal, make sure the connecting lines are all parallel to each other. eSolutions Manual - Powered by Cognero Page 39 7-4 Parallel Lines and Proportional Parts Step 3: Continue this process until you have four arcs. When you connect the points on the diagonal back to the horizontal, make sure the connecting lines are all parallel to each other. Step 4: Label the points as shown. AB = BC = CD = DE ANSWER: Sample answer: 47. MULTIPLE REPRESENTATIONS In this problem, you will explore angle bisectors and proportions. a. GEOMETRIC Draw three triangles, one acute, one right, and one obtuse. Label one triangle ABC and draw angle bisector Label the second MNP with angle bisector and the third WXY with angle bisector b. TABULAR Complete the table at the right with the appropriate values. c. VERBAL Make a conjecture about the segments of a triangle created by an angle bisector. SOLUTION: a. When drawing the triangles, pay close attention to the directions and labeling instructions. Use a protractor, or construction tool, when making the angle bisectors, to ensure accurate measurement values for the table. S l eSolutions Manual - Powered by Cognero Page 40 7-4 Parallel Lines and Proportional Parts SOLUTION: a. When drawing the triangles, pay close attention to the directions and labeling instructions. Use a protractor, or construction tool, when making the angle bisectors, to ensure accurate measurement values for the table. Sample answer: b. Carefully measure the indicated lengths in centimeters. c. Look for a pattern in the table, specifically comparing the lengths of the ratios of sides for each triangle. Sample answer: The proportion of the segments created by the angle bisector of a triangle is equal to the proportion of their respective consecutive sides. ANSWER: a. Sample answer: eSolutions Manual - Powered by Cognero Page 41 7-4 Parallel Lines and Proportional Parts Sample answer: The proportion of the segments created by the angle bisector of a triangle is equal to the proportion of their respective consecutive sides. ANSWER: a. Sample answer: b. c. Sample answer: The proportion of the segments created by the angle bisector of a triangle is equal to the proportion of their respective consecutive sides. 48. CCSS CRITIQUE Jacob and Sebastian are finding the value of x in Jacob says that MP is one half of JL, so x is 4.5. Sebastian says that JL is one half of MP, so x is 18. Is either of them correct? Explain. SOLUTION: Jacob; sample answer: Since M is the midpoint of and P is the midpoint of , then is the midsegment of . Therefore, eSolutions Manual - Powered by Cognero Page 42 7-4 Parallel Lines and Proportional Parts c. Sample answer: The proportion of the segments created by the angle bisector of a triangle is equal to the proportion of their respective consecutive sides. 48. CCSS CRITIQUE Jacob and Sebastian are finding the value of x in Jacob says that MP is one half of JL, so x is 4.5. Sebastian says that JL is one half of MP, so x is 18. Is either of them correct? Explain. SOLUTION: Jacob; sample answer: Since M is the midpoint of and P is the midpoint of , then is the midsegment of . Therefore, ANSWER: Jacob; sample answer: is the midsegment, so 49. REASONING In AF = FB and AH = HC. If D is of the way from A to B and E is of the way from A to C, is DE sometimes, always, or never of BC? Explain. SOLUTION: Always; sample answer: Since FA=FB, then F is a midpoint of . Similarly, since AH=HC and H is the midpoint of . Therefore, FH is a midsegment of so and . Let BC = x, then Because , we know that FHCB is a trapezoid, so eSolutions Manual - Powered by Cognero Page 43 7-4 Parallel Lines and Proportional Parts ANSWER: Jacob; sample answer: is the midsegment, so 49. REASONING In AF = FB and AH = HC. If D is of the way from A to B and E is of the way from A to C, is DE sometimes, always, or never of BC? Explain. SOLUTION: Always; sample answer: Since FA=FB, then F is a midpoint of . Similarly, since AH=HC and H is the midpoint of . Therefore, FH is a midsegment of so and . Let BC = x, then Because , we know that FHCB is a trapezoid, so ANSWER: Always; sample answer: FH is a midsegment. Let BC = x, then FHCB is a trapezoid, so Therefore, CHALLENGE Write a two-column proof.
50. Given: AB = 4, BC = 4, and CD = DE Prove: SOLUTION: An effective strategy for this proof is to think of a way to get , by SAS Similarity. We already know that , so we need to establish that . You can show that 2BC = AC and 2DC = EC, through the given information and substitution into Segment Addition Postulate statements. Once this is done, you can prove that eSolutions Manual - Powered by Cognero Page 44 7-4 Parallel Lines and Proportional Parts SOLUTION: An effective strategy for this proof is to think of a way to get , by SAS Similarity. We already know that , so we need to establish that . You can show that 2BC = AC and 2DC = EC, through the given information and substitution into Segment Addition Postulate statements. Once this is done, you can prove that by transitive property. Once the triangles are proven similar, then the lines can be proven parallel by choosing a pair of congruent corresponding angles from the similar triangles. , Proof: Statements (Reasons) 1. AB = 4, BC = 4 (Given) 2. AB = BC (Subst.) 3. AB + BC = AC (Seg. Add. Post.) 4. BC + BC = AC (Subst.) 5. 2BC = AC (Subtraction property.) 6. AC = 2BC (Symm. Prop). 7. (Div. Prop.) 8. ED = DC (Given) 9. ED + DC = EC (Seg. Add. Post.) 10. DC + DC = EC (Subst.) 11. 2DC = EC (Subst.) 12. (Div. Prop.) 13. (Trans. Prop.) 14. (Reflexive Prop.) 15. (SAS Similarity) 16. (Def. of polygons) 17. (If corr. angles are , lines are ||.) ANSWER: Proof: Statements (Reasons) 1. AB = 4, BC = 4 (Given) 2. AB = BC (Sub.) 3. AB + BC = AC (Seg. Add. Post.) 4. BC + BC = AC (Subst.) 5. 2BC = AC (Sub.) 6. AC = 2BC (Symm. Prop). 7. (Div. Prop.) 8. ED = DC (Given) 9. ED + DC = EC (Seg. Add. Post.) 10. DC + DC = EC (Sub.) 11. 2DC = EC (Sub.) 12. (Div. Prop.) 13. (Trans. Prop.) 14. (Reflexive Prop.) 15. (SAS Similarity) eSolutions Manual - Powered by Cognero Page 45 7-4 Parallel Lines and Proportional Parts 8. ED = DC (Given) 9. ED + DC = EC (Seg. Add. Post.) 10. DC + DC = EC (Sub.) 11. 2DC = EC (Sub.) 12. (Div. Prop.) 13. (Trans. Prop.) 14. (Reflexive Prop.) 15. (SAS Similarity) 16. (Def. of polygons) 17. (If corr. angles are , lines are ||.) 51. OPEN ENDED Draw three segments, a, b, and c, of all different lengths. Draw a fourth segment, d, such that SOLUTION: By Corollary 7.1, we know that if we draw three parallel lines intersected by two transversals, then they will cut the transversals proportionally or , as seen in the diagram below. ANSWER: By Corollary 7.1, 52. WRITING IN MATH Compare the Triangle Proportionality Theorem and the Triangle Midsegment Theorem. SOLUTION: Both theorems deal with a parallel line inside the triangle. The Midsegment Theorem is a special case of the Converse of the Proportionality Theorem. ANSWER: Both theorems deal with a parallel line inside the triangle. The Midsegment Theorem is a special case of the Converse of the Proportionality Theorem. 53. SHORT RESPONSE What is the value of x? SOLUTION: By Corollary 7.2, .
eSolutions Manual - Powered by Cognero Page 46 7-4 Parallel Lines and Proportional Parts Converse of the Proportionality Theorem. ANSWER: Both theorems deal with a parallel line inside the triangle. The Midsegment Theorem is a special case of the Converse of the Proportionality Theorem. 53. SHORT RESPONSE What is the value of x? SOLUTION: By Corollary 7.2, . Solve for x. ANSWER: 8 54. If the vertices of triangle JKL are (0, 0), (0, 10) and (10, 10) then the area of triangle JKL is A 20 units2 B 30 units2 C 40 units2 D 50 units2 SOLUTION: So, the correct choice is D. ANSWER: D 55. ALGEBRA A breakfast cereal contains wheat, rice, and oats in the ratio 2 : 4: 1. If the manufacturer makes a mixture using 110 pounds of wheat, how many pounds of rice will be used? F 120 lb G 220 lb H 240 lb J 440 lb SOLUTION: Since the ratio of rice to wheat is 4: 2, we can set up a proportion to find the amount of rice needed for 110 pounds of wheat. eSolutions Manual - Powered by Cognero Page 47 7-4 Parallel Lines and Proportional Parts So, the correct choice is D. ANSWER: D 55. ALGEBRA A breakfast cereal contains wheat, rice, and oats in the ratio 2 : 4: 1. If the manufacturer makes a mixture using 110 pounds of wheat, how many pounds of rice will be used? F 120 lb G 220 lb H 240 lb J 440 lb SOLUTION: Since the ratio of rice to wheat is 4: 2, we can set up a proportion to find the amount of rice needed for 110 pounds of wheat. The correct answer is G, 220 lb. ANSWER: G 56. SAT/ACT If the area of a circle is 16 square meters, what is its radius in meters? A B C D 12π E 16π SOLUTION: Since the area of a circle can be found with , we can substitute in 16 for the area (A) and solve for r. Therefore, the answer is A. eSolutions Manual - Powered by Cognero Page 48 7-4 Parallel Lines and Proportional Parts The correct answer is G, 220 lb. ANSWER: G 56. SAT/ACT If the area of a circle is 16 square meters, what is its radius in meters? A B C D 12π E 16π SOLUTION: Since the area of a circle can be found with , we can substitute in 16 for the area (A) and solve for r. Therefore, the answer is A. ANSWER: A ALGEBRA Identify the similar triangles. Then find the measure(s) of the indicated segment(s).
57. SOLUTION: by the Vertical Angles Theorem. Since , by the Alternate Interior Angles Theorem. Therefore, by AA Similarity, . To find AB or x, write a proportion using the definition of similar polygons. eSolutions Manual - Powered by Cognero Page 49 7-4 Parallel Lines and Proportional Parts ANSWER: A ALGEBRA Identify the similar triangles. Then find the measure(s) of the indicated segment(s).
57. SOLUTION: by the Vertical Angles Theorem. Since , by the Alternate Interior Angles Theorem. Therefore, by AA Similarity, . To find AB or x, write a proportion using the definition of similar polygons. ANSWER: by AA Similarity; 6.25 58. SOLUTION: , since right angles are congruent.
, since . Therefore, by SAS Similarity, . Write a proportion using the definition of similar polygons to find the value of x. eSolutions Manual - Powered by Cognero Page 50 7-4 Parallel Lines and Proportional Parts ANSWER: by AA Similarity; 6.25 58. SOLUTION: , since right angles are congruent.
, since . Therefore, by SAS Similarity, . Write a proportion using the definition of similar polygons to find the value of x. Substitute this value for x to find RT and RS. ANSWER: by SAS Similarity; 15, 20 59. SOLUTION: by the Reflexive Property of Congruence. Since , by the Corresponding Angles Theorem. Therefore, by AA Similarity, . Write a proportion using the definition of similar polygons to find the value of x. eSolutions Manual - Powered by Cognero Page 51 7-4 Parallel Lines and Proportional Parts ANSWER: by SAS Similarity; 15, 20 59. SOLUTION: by the Reflexive Property of Congruence. Since , by the Corresponding Angles Theorem. Therefore, by AA Similarity, . Write a proportion using the definition of similar polygons to find the value of x. So, WT = 12.5. WT + TY = WY by the Segment Addition Postulate. Since WY = 20, you can solve for TY. ANSWER: by AA Similarity; 7.5 60. SURVEYING Mr. Turner uses a carpenter’s square to find the distance across a stream. The carpenter’s square models right angle NOL. He puts the square on top of a pole that is high enough to sight along to point P across the river. Then he sights along to point M. If MK is 1.5 feet and OK is 4.5 feet, find the distance KP across the stream. SOLUTION: By AA Similarity, . Use the Pythagorean Theorem to find MO. eSolutions Manual - Powered by Cognero Page 52 7-4 Parallel Lines and Proportional Parts ANSWER: by AA Similarity; 7.5 60. SURVEYING Mr. Turner uses a carpenter’s square to find the distance across a stream. The carpenter’s square models right angle NOL. He puts the square on top of a pole that is high enough to sight along to point P across the river. Then he sights along to point M. If MK is 1.5 feet and OK is 4.5 feet, find the distance KP across the stream. SOLUTION: By AA Similarity, . Use the Pythagorean Theorem to find MO. Write a proportion using corresponding sides of the two triangles: Therefore, the distance KP is about 13.5 feet. ANSWER: 13.5 ft COORDINATE GEOMETRY For each quadrilateral with the given vertices, verify that the quadrilateral is a trapezoid and determine whether the figure is an isosceles trapezoid. 61. Q(–12, 1), R(–9, 4), S(–4, 3), T(–11, –4) SOLUTION: eSolutions Manual - Powered by Cognero Page 53 7-4 Parallel Lines and Proportional Parts Therefore, the distance KP is about 13.5 feet. ANSWER: 13.5 ft COORDINATE GEOMETRY For each quadrilateral with the given vertices, verify that the quadrilateral is a trapezoid and determine whether the figure is an isosceles trapezoid. 61. Q(–12, 1), R(–9, 4), S(–4, 3), T(–11, –4) SOLUTION: Use the slope formula to find the slope of the sides of the quadrilateral. The slopes of exactly one pair of opposite sides are equal. So, this quadrilateral has only one pair of parallel sides. Therefore, the quadrilateral QRST is a trapezoid. Use the Distance Formula to find the lengths of the legs of the trapezoid. The lengths of the legs are equal. Therefore, QRST is an isosceles trapezoid. ANSWER: QRST is an isosceles trapezoid since RS = = QT. 62. A(–3, 3), B(–4, –1), C(5, –1), D(2, 3) SOLUTION: eSolutions Manual - Powered by Cognero Page 54 7-4 Parallel Lines and Proportional Parts The lengths of the legs are equal. Therefore, QRST is an isosceles trapezoid. ANSWER: QRST is an isosceles trapezoid since RS = = QT. 62. A(–3, 3), B(–4, –1), C(5, –1), D(2, 3) SOLUTION: Use the slope formula to find the slope of the sides of the quadrilateral. The slopes of exactly one pair of opposite sides are equal. So, they are parallel. Therefore, the quadrilateral ABCD is a trapezoid. Use the Distance Formula to find the lengths of the legs of the trapezoid. The lengths of the legs are not equal. Therefore, ABCD is not an isosceles trapezoid. ANSWER: ABCD is a trapezoid, but not isosceles since AB = and CD = 5. Point S is the incenter of Find each measure. 63. SQ SOLUTION: Since S is the incenter of then it is formed by the angle bisectors of each vertex is the angle bisector of eSolutions Manual - Powered by Cognero Page 55 7-4 Parallel Lines and Proportional Parts The lengths of the legs are not equal. Therefore, ABCD is not an isosceles trapezoid. ANSWER: ABCD is a trapezoid, but not isosceles since AB = and CD = 5. Point S is the incenter of Find each measure. 63. SQ SOLUTION: Since S is the incenter of , then it is formed by the angle bisectors of each vertex. is the angle bisector of therefore, by the Angle Bisector Theorem, SK= SQ. Use Pythagorean Theorem in the right triangle JSK. SK= SQ = 6 ANSWER: 6 64. QJ SOLUTION: Since S is the incenter of , then it is formed by the angle bisectors of each vertex. is the angle bisector of therefore, by the Angle Bisector Theorem, SK= SQ. Use Pythagorean Theorem in the right triangle JSK. Use Pythagorean Theorem in the right triangle JSQ to find QJ. Th f QJ 8 eSolutions Manual - Powered by Cognero Page 56 7-4 Parallel Lines and Proportional Parts SK= SQ = 6 ANSWER: 6 64. QJ SOLUTION: Since S is the incenter of , then it is formed by the angle bisectors of each vertex. is the angle bisector of therefore, by the Angle Bisector Theorem, SK= SQ. Use Pythagorean Theorem in the right triangle JSK. Use Pythagorean Theorem in the right triangle JSQ to find QJ. Therefore QJ=8. ANSWER: 8 65. m∠MPQ SOLUTION: Since S is the incenter of , then it is formed by the angle bisectors of each vertex. is the angle bisector of . Therefore, ANSWER: 56 66. m∠SJP SOLUTION: Since S is the incenter of , then it is formed by the angle bisectors of each vertex. is the angle bisector of . Therefore, eSolutions Manual - Powered by Cognero Page 57 7-4 Parallel Lines and Proportional Parts ANSWER: 56 66. m∠SJP SOLUTION: Since S is the incenter of , then it is formed by the angle bisectors of each vertex. is the angle bisector of . Therefore, and similarly, We know that sum of the measures of a triangle is 180. We know that because is an angle bisector of . Therefore, ANSWER: 37.5 Solve each proportion.
67. SOLUTION: Solve for x. ANSWER: 68.
eSolutions Manual - Powered by Cognero Page 58 7-4 Parallel Lines and Proportional Parts ANSWER: 37.5 Solve each proportion.
67. SOLUTION: Solve for x. ANSWER: 68. SOLUTION: Solve for x. ANSWER: 6.7 69. SOLUTION: Solve for x. ANSWER: 2.1 70. SOLUTION: Solve for x. eSolutions Manual - Powered by Cognero Page 59 7-4 Parallel Lines and Proportional Parts ANSWER: 2.1 70. SOLUTION: Solve for x. ANSWER: 3.6 71. SOLUTION: Solve for x. ANSWER: 8.7 eSolutions Manual - Powered by Cognero Page 60 7-4 Parallel Lines and Proportional Parts |
14437 | https://www.geeksforgeeks.org/digital-logic/basics-of-boolean-algebra-in-digital-electronics/ | Basics of Boolean Algebra in Digital Electronics
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Boolean algebra is a special mathematical way to express relations (logic) between variables. Its binary framework simplifies complex operations into true(1)/false(0) logic, making it indispensable for engineers and programmers.
Boolean algebra uses operators like AND, OR and NOT to model logical decisions, forming the basis of all digital systems.
It is essential for creating efficient logic gates, combinational circuits and memory units in hardware.
Techniques like Karnaugh maps and Boolean laws reduce circuit complexity, saving cost and power.
It carries conditional logic, search algorithms and database queries in software development.
Basic Boolean Operators and Logic Gates
Boolean operators are used to perform logical operations on Boolean values. Logic gates are physical devices or circuits used to implement the basic Boolean operators. Each logic gate performs a specific operation based on the Boolean logic.
1. Boolean Operators
AND ( ∧ ): The AND operator returns True (1) only if both operands are True (1).
OR ( ∨ ): The OR operator returns True (1) if at least one of the operands is True (1).
NOT ( ¬ ): The NOT operator reverses the value of the operand. It returns True (1) if the operand is False (0) and False (0) if the operand is True (1).
XOR (Exclusive OR) (⊕): The XOR operator returns True (1) if the operands are different and False (0) if they are the same.
2. Logic Gates
1. AND Gate: The AND gate implements the AND operation. It outputs 1 only if both inputs are 1.
| Truth Table | | |
---
| Input A | Input B | Output (A AND B) |
| | 0 | 0 |
| 0 | | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
2. OR Gate: The OR gate implements the OR operation. It outputs 1 if at least one input is 1.
| Truth Table | | |
---
| Input A | Input B | Output (A OR B) |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
3. NOT Gate: The NOT gate implements the NOT operation. It inverts the input, outputting 1 if the input is 0 and 0 if the input is 1.
| Truth Table | |
--- |
| Input A | Output (NOT A) |
| 0 | 1 |
| 1 | 0 |
4. NAND Gate: The NAND gate is the inverse of the AND gate. It outputs 0 only if both inputs are 1; otherwise, it outputs 1.
| Truth Table | | |
---
| Input A | Input B | Output (A NAND B) |
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
5. NOR Gate: The NOR gate is the inverse of the OR gate. It outputs 0 if at least one input is 1; otherwise, it outputs 1.
| Truth Table | | |
---
| Input A | Input B | Output (A NOR B) |
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 0 |
6. XOR Gate: The XOR gate implements the XOR operation. It outputs 1 if the inputs are different and 0 if they are the same.
| Truth Table | | |
---
| Input A | Input B | Output (A XOR B) |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
7. XNOR Gate: The XNOR gate is the inverse of the XOR operation. It outputs 1 if the inputs are same and 0 if they are the different.
| Truth Table | | |
---
| Input A | Input B | Output (A XOR B) |
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
Fundamental Laws & Theorems of Boolean Algebra
The fundamental laws and theorems of Boolean algebra simplify logical expressions and help in designing digital circuits. Some of the core laws and theorems in Boolean algebra are:
1. Commutative Law
The Commutative Law states that the order in which two variables are combined using the AND or OR operators does not affect the result. This law allows us to reorder terms without changing the output.
Commutative Law for AND:
A⋅B = B⋅A
Example:
1⋅0 = 0⋅1 = 0
Commutative Law for OR:
A + B = B + A
Example:
1 + 0 = 0 + 1 = 1
2. Associative Law
The Associative Law states that when three or more variables are combined using the AND or OR operators, the grouping of the variables does not affect the result. This law allows us to regroup terms without changing the output.
Associative Law for AND:
A⋅(B⋅C) = (A⋅B)⋅C
Example:
1⋅(0⋅1) = (1⋅0)⋅1 = 0
Associative Law for OR:
A + (B + C) = (A + B) + C
Example:
1 + (0 + 1) = (1 + 0) + 1 = 1
3. Distributive Law
The Distributive Law describes how the AND and OR operations distribute over each other. It is similar to how multiplication distributes over addition in arithmetic. This law allows the factoring of Boolean expressions, similar to factoring algebraic expressions.
Distributive Law for AND over OR:
A⋅(B + C) = (A⋅B) + (A⋅C)
Example:
1⋅(0 + 1) = (1⋅0) + (1⋅1) = 0 + 1 = 1
Distributive Law for OR over AND:
A + (B⋅C) = (A + B)⋅(A + C)
Example:
1 + (0⋅1) = (1 + 0)⋅(1 + 1) = 1⋅1 = 1
4. Identity Law
The Identity Law states that any variable ANDed with 1 or ORed with 0 will result in the original variable itself. This law shows that the identity elements for AND and OR operations are 1 and 0, respectively.
Identity Law for AND:
A⋅1 = A
Example:
1⋅1 = 1
Identity Law for OR:
A + 0 = A
Example:
1 + 0 = 1
5. Complement Law
The Complement Law involves the negation of a variable and provides the result when a variable is combined with its complement (opposite). This law shows that a variable ANDed with its complement will always be 0 and a variable ORed with its complement will always be 1.
Complement Law for AND:
A⋅A' = 0
Example:
1⋅1' = 1⋅0 = 0
Complement Law for OR:
A+A' = 1
Example:
1 + 1' = 1 + 0 = 1
6. Inversion Law
The Inversion Law is a unique principle in Boolean algebra, stating that the complement of the complement of any variable is equal to the variable itself.
(A') = A
Example:
(1')' = (0)' = 1
7. De Morgan's Theorems
De Morgan's Theorems provide a way to simplify expressions involving negations and are very useful in digital circuit design.
De Morgan’s First Theorem: The negation of an AND operation is equal to the OR operation of the negations of the operands.
(A⋅B)' = A' + B'
Example:
Expression: (1⋅0)'
Simplified: 1' + 0' = 0 + 1 = 1
De Morgan’s Second Theorem: The negation of an OR operation is equal to the AND operation of the negations of the operands.
(A+B)' = A'⋅B'
Example:
Expression: (1 + 0)'
Simplified: 1'⋅0' = 0⋅1 = 0
Example of Boolean Algebra
Let's solve a Boolean expression:
(A ∧ B) ∨ (C∧ (¬D))
Given A = B = 1, C = 0, D = 0.
Solution: Let's replace the expression with given values part by part.
(A ∧ B) = (1∧ 1) = 1
(C∧ (¬D)) = (0 ∧ (¬0)) = (0 ∧ 1) = 0
(A ∧ B) ∨ (C∧ (¬D)) = (1 ∨ 0) = 1
So, the final answer for the given Boolean expression is 1.
Let's understand it using truth table, assume:
Y= (A ∧ B) ∨ (C∧ (¬D))
| Input | | | | Output | | |
--- --- ---
| A | B | C | D | (A ∧ B) | (C ∧ (¬D)) | Y= (A ∧ B) ∨ (C ∧ (¬D)) |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 1 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 | 1 | 1 |
| 0 | 0 | 1 | 1 | 0 | 0 | 0 |
| 0 | 1 | 0 | 0 | 0 | 0 | 0 |
| 0 | 1 | 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 | 0 | 1 | 1 |
| 0 | 1 | 1 | 1 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 0 | 1 | 1 |
| 1 | 0 | 1 | 1 | 0 | 0 | 0 |
| 1 | 1 | 0 | 0 | 1 | 0 | 1 |
| 1 | 1 | 0 | 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 0 | 1 |
Logical circuit diagram of above example:
Applications of Boolean Algebra
Digital Circuit Design: Boolean algebra simplifies the design of combinational and sequential circuits like adders, multiplexers, flip-flops and registers, optimizing gate usage and performance.
Logical Expression Simplification: Boolean algebra reduces complex logical expressions, minimizing the number of gates required and improving circuit efficiency and power consumption.
Arithmetic Circuit Design: It is used to design binary adders, subtractors, multipliers and dividers, essential for binary arithmetic operations in digital systems.
Memory Elements Design: Boolean algebra helps in designing memory units like flip-flops, latches and registers, which store data and manage state transitions in sequential circuits.
Error Detection and Correction: It is used in error detection (e.g., parity checks) and correction techniques (e.g., Hamming code) to ensure data integrity in communication systems.
Control Systems and Logic Controllers: Boolean algebra is crucial in designing control logic for systems like finite state machines (FSMs) and programmable logic controllers (PLCs), which manage automated processes.
Optimizing Circuit Design: It helps reduce the complexity of digital circuits, lowering gate count, minimizing space and improving power efficiency in ICs, FPGAs and ASICs.
Cryptography and Security Systems: Boolean algebra is applied in designing encryption and decryption circuits, supporting secure data transmission and authentication in digital security systems.
Digital Signal Processing (DSP): It is used in processing digital signals for tasks like filtering, encoding and transforming audio, video and other signals in digital format.
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Error Detection Codes: The binary information is transferred from one location to another location through some communication medium. The external noise can change bits from 1 to 0 or 0 to 1. This change in values changes the meaning of the actual message and is called an error. For efficient data t
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Boolean Algebra and Logic Gates
Logic Gates - Definition, Types, Uses
Logic Gates are the fundamental building blocks in digital electronics. Used to perform logical operations based on the inputs provided to it and gives a logical output that can be either high(1) or low(0). The operation of logic gates is based on Boolean algebra or mathematics.There are basically s
10 min readBasic Conversion of Logic Gates
In the Digital System, logic gates are the basic building blocks. In these logic gates, we can find the gates having more than one input, but will have only one output. The connection between the input and the output of a gate is based on some logic. Based on this logic, different gates are develop
6 min readRealization of Logic Gate Using Universal gates
In Boolean Algebra, the NAND and NOR gates are called universal gates because any digital circuit can be implemented by using any one of these two i.e. any logic gate can be created using NAND or NOR gates only.Implementation of AND Gate using Universal GatesImplementation using NAND GatesThe AND ga
6 min readCanonical and Standard Form
Canonical Form - In Boolean algebra, the Boolean function can be expressed as Canonical Disjunctive Normal Form known as minterm and some are expressed as Canonical Conjunctive Normal Form known as maxterm. In Minterm, we look for the functions where the output results in "1" while in Maxterm we loo
6 min readTypes of Integrated Circuits
In this article, we will go through the Types of Integrated Circuits, we will start our article with the introductions of the ICs, then we will go through different types of ICs one by one, At last, we will conclude our article will their applications, advantages, disadvantages and some FAQs. Table
7 min read
Minimization Techniques
Minimization of Boolean Functions
Boolean functions are used to represent logical expressions in terms of sum of minterms or product of maxterms. Number of these literals (minterms or maxterms) increases as the complexity of the digital circuit increases. This can lead to large and inefficient circuits. By minimizing Boolean functio
4 min readIntroduction of K-Map (Karnaugh Map)
In many digital circuits and practical problems, we need to find expressions with minimum variables. We can minimize Boolean expressions of 3, 4 variables very easily using K-map without using any Boolean algebra theorems. It is a tool which is used in digital logic to simplify boolean expression. I
5 min read5 variable K-Map in Digital Logic
Prerequisite - Implicant in K-Map Karnaugh Map or K-Map is an alternative way to write a truth table and is used for the simplification of Boolean Expressions. So far we are familiar with 3 variable K-Map & 4 variable K-Map. Now, let us discuss the 5-variable K-Map in detail. Any Boolean Express
5 min readVarious Implicants in K-Map
An implicant can be defined as a product/minterm term in Sum of Products (SOP) or sum/maxterm term in Product of Sums (POS) of a Boolean function. For example, consider a Boolean function, F = AB + ABC + BC. Implicants are AB, ABC, and BC. There are various implicant in K-Map listed below :Prime Imp
5 min readDon't Care (X) Conditions in K-Maps
One of the most important concepts in simplifying output expressions using Karnaugh Maps (K-Maps) is the 'Don't Care' condition. The 'Don't Care' conditions allow us to treat certain cells in a K-Map as either 0, 1, or to ignore them altogether, which can help in forming larger and more efficient gr
4 min readQuine McCluskey Method
The Quine McCluskey method also called the tabulation method is a very useful and convenient method for simplification of the Boolean functions for a large number of variables (greater than 4). This method is useful over K-map when the number of variables is larger for which K-map formation is diffi
8 min readTwo Level Implementation of Logic Gates
The term "two-level logic" refers to a logic design that uses no more than two logic gates between input and output. This does not mean that the entire design will only have two logic gates, but it does mean that the single path from input to output will only have two logic gates.In two-level logic,
9 min read
Combinational Circuits
Half Adder in Digital Logic
A half adder is a combinational logic circuit that performs binary addition of two single-bit inputs, A and B, producing two outputs: SUM and CARRY. The SUM output which is the least significant bit (LSB) is obtained using an XOR gate while the CARRY output which is the most significant bit (MSB) is
3 min readFull Adder in Digital Logic
Full Adder is a combinational circuit that adds three inputs and produces two outputs. The first two inputs are A and B and the third input is an input carry as C-IN. The output carry is designated as C-OUT and the normal output is designated as S which is SUM. The C-OUT is also known as the majorit
5 min readHalf Subtractor in Digital Logic
A half subtractor is a digital logic circuit that performs the binary subtraction of two single-bit binary numbers. It has two inputs, A and B, and two outputs, Difference and Borrow. The Difference output represents the result of subtracting B from A, while the Borrow output indicates whether a bor
4 min readFull Subtractor in Digital Logic
A Full Subtractor is a combinational circuit used to perform binary subtraction. It has three inputs:A (Minuend)B (Subtrahend)B-IN (Borrow-in from the previous stage)It produces two outputs:Difference (D): The result of the subtraction.Borrow-out (B-OUT): Indicates if a borrow is needed for the next
3 min readParallel Adder and Parallel Subtractor
An adder adds two binary numbers one bit at a time using carry from each step. A subtractor subtracts one binary number from another using borrow when needed. A parallel adder adds all bits at once, making addition faster. Similarly, a parallel subtractor subtracts all bits at the same time for quic
5 min readSequential Binary Multiplier
In this article, we are going to learn how a sequential binary multiplier works with examples. So for that, we also need to learn a few concepts related to the sequential circuit, binary multipliers, etc. Finally solving the examples using a sequential binary multiplier method.Sequential CircuitA se
12 min readMultiplexers in Digital Logic
In this article we will go through the multiplexer, we will first define what is a multiplexer then we will go through its types which are 2x1 and 4x1, then we will go through the Implementation of the 2x1 mux and higher mux with lower order mux, at last we will conclude our article with some applic
10 min readEvent Demultiplexer in Node.js
Node.js is designed to handle multiple tasks efficiently using asynchronous, non-blocking I/O operations. But how does it manage multiple operations without slowing down or blocking execution? The answer lies in the Event Demultiplexer.The Event Demultiplexer is a key component of Node.js's event-dr
3 min readBinary Decoder in Digital Logic
A binary decoder is a digital circuit used to convert binary-coded inputs into a unique set of outputs. It does the opposite of what an encoder does. A decoder takes a binary value (such as 0010) and activates exactly one output line corresponding to that value while all other output lines remain in
5 min readEncoder in Digital Logic
An encoder is a digital circuit that converts a set of binary inputs into a unique binary code. The binary code represents the position of the input and is used to identify the specific input that is active. Encoders are commonly used in digital systems to convert a parallel set of inputs into a ser
7 min readCode Converters - Binary to/from Gray Code
In this article, we will go through Code Converters - Binary to/from Gray Code, we will start our article by defining Code converters, Binary code and Gray code, and then we will go through the conversion of binary code to gray code and vice versa.Table Of ContentCode ConvertersBinary CodeGray CodeC
5 min readMagnitude Comparator in Digital Logic
A magnitude digital Comparator is a combinational circuit that compares two digital or binary numbers in order to find out whether one binary number is equal, less than, or greater than the other binary number. We logically design a circuit for which we will have two inputs one for A and the other f
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Sequential Circuits
Introduction of Sequential Circuits
Sequential circuits are digital circuits that store and use the previous state information to determine their next state. Unlike combinational circuits, which only depend on the current input values to produce outputs, sequential circuits depend on both the current inputs and the previous state stor
7 min readDifference between Combinational and Sequential Circuit
In digital electronics, circuits are classified into two primary categories: The combinational circuits and the sequential circuits. Where the outputs depend on the current inputs are called combination circuit, combinational circuits are simple and effective for functions like addition, subtraction
4 min readLatches in Digital Logic
Latch is a digital circuit which converts its output according to its inputs instantly. To implement latches, we use different logic gates. In this article, we will see the definition of latches, latch types like SR, gated SR, D, gated D, JK and T with its truth table and diagrams and advantages and
7 min readFlip-Flop types, their Conversion and Applications
In this article, we will go through the Flip-Flop types, their Conversion and their Applications, First, we will go through the definition of the flip-flop with its types in brief, and then we will go through the conversion of the flip-flop with its applications, At last, we will conclude our articl
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Conversion of Flip-Flop
Conversion of S-R Flip-Flop into D Flip-Flop
Prerequisite - Flip-flop1. S-R Flip-Flop : S-R flip-flop is similar to S-R latch expect clock signal and two AND gates. The circuit responds to the positive edge of clock pulse to the inputs S and R. 2. D Flip-Flop : D Flip-Flop is a modified SR flip-flop which has an additional inverter. It prevent
1 min readConversion of S-R Flip-Flop into T Flip-Flop
Prerequisite - Flip-flop Here, we will discuss the process of conversion of S-R Flip-Flop into a T Flip-Flop using an example. Rules for conversion: Step-1: Find the characteristics table of required flip-flop and the excitation table of the existing (given) flip-flop. Step-2: Find the expression of
1 min readConversion of J-K Flip-Flop into T Flip-Flop
Prerequisite - Flip-flop 1. J-K Flip-Flop: JK flip-flop shares the initials of Jack Kilby, who won a Nobel prize for his fabrication of the world's first integrated circuit, some people speculate that this type of flip flop was named after him because a flip-flop was the first device that Kilby buil
1 min readConversion of J-K Flip-Flop into D Flip-Flop
A flip-flop is a basic component of digital electronics. This kind of circuit has two stable states and is frequently used in storing one bit of information. Various flip-flops such as SR (Set-Reset), D (Data or Delay), JK and T belong to this category. Each category has distinct features and functi
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Register, Counter, and Memory Unit
Counters in Digital Logic
A Counter is a device which stores (and sometimes displays) the number of times a particular event or process has occurred, often in relationship to a clock signal. Counters are used in digital electronics for counting purpose, they can count specific event happening in the circuit. For example, in
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Counters play a crucial role in digital logic circuits, enabling tasks such as clock frequency division and sequencing. This article explores the concept of ripple counters, a type of asynchronous counter, their operation, advantages, and disadvantages in digital logic design. What is a Counter?Coun
5 min readRing Counter in Digital Logic
A ring counter is a typical application of the Shift register. The ring counter is almost the same as the shift counter. The only change is that the output of the last flip-flop is connected to the input of the first flip-flop in the case of the ring counter but in the case of the shift register it
7 min readGeneral Purpose Registers
A register is a collection of flip-flops. Single bit digital data is stored using flip-flops. By combining many flip-flops, the storage capacity can be extended to accommodate a huge number of bits. We must utilize an n-bit register with n flip flops if we wish to store an n-bit word.General Purpose
8 min readShift Registers in Digital Logic
Pre-Requisite: Flip-FlopsFlip flops can be used to store a single bit of binary data (1 or 0). However, in order to store multiple bits of data, we need multiple flip-flops. N flip flops are to be connected in order to store n bits of data. A Register is a device that is used to store such informati
8 min readComputer Memory
Memory is the electronic storage space where a computer keeps the instructions and data it needs to access quickly. It's the place where information is stored for immediate use. Memory is an important component of a computer, as without it, the system wouldn€™t operate correctly. The computer€™s opera
9 min readRandom Access Memory (RAM)
Random Access Memory (RAM) is a type of computer memory that stores data temporarily. When you turn off your computer, the data in RAM disappears, unlike the data on your hard drive, which stays saved. RAM helps your computer run programs and process information faster. This is similar to how the br
11 min readRead Only Memory (ROM)
Memory plays a crucial role in how devices operate, and one of the most important types is Read-Only Memory (ROM). Unlike RAM (Random Access Memory), which loses its data when the power is turned off, ROM is designed to store essential information permanently.Here, we€™ll explore what ROM is, how it
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LMNs and GATE PYQs
LMN - Digital Electronics
Digital electronics deals with systems that use digital signals, represented as 0s and 1s, to process information. It is the backbone of modern devices like computers, smartphones, and calculators. Unlike analog electronics, which works with continuous signals, digital electronics offers higher accu
14 min readDigital Logic and Design - GATE CSE Previous Year Questions
The Digital Logic and Design(DLD) subject has high importance in GATE CSE exam because:moderate number of questions nearly 6-7% of the total papersignificant weightage (6-7 marks) across multiple years This can be seen in the table given below:YearApprox. Marks from Digital LogicNumber of QuestionsD
2 min read
Practice Questions - Digital Logic & Design
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14438 | https://www.chegg.com/homework-help/questions-and-answers/1-point-four-vertices-parallelepiped-2-9-8-b-1-5-6-c-5-12-1-d-4-11-10--three-edges-paralle-q123579763 | Solved (1 point) Four of the vertices of a parallelepiped | Chegg.com
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(1 point) Four of the vertices of a parallelepiped are A=(2,-9,8),B=(1,-5,6),C=(-5,-12,-1), and D=(4,-11,10).Three of the edges of the parallelepiped are AB,AC, and AD.Find the volume of the parallelepiped.Volume isFind the other four vertices of the parallelepiped.Vertices are(Enter your answer as a comma-separated list of points.)Point A is contained
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Question: (1 point) Four of the vertices of a parallelepiped are A=(2,-9,8),B=(1,-5,6),C=(-5,-12,-1), and D=(4,-11,10).Three of the edges of the parallelepiped are AB,AC, and AD.Find the volume of the parallelepiped.Volume isFind the other four vertices of the parallelepiped.Vertices are(Enter your answer as a comma-separated list of points.)Point A is contained
(1 point) Four of the vertices of a parallelepiped are A=(2,-9,8),B=(1,-5,6),C=(-5,-1 2,-1), and D=(4,-1 1,1 0). Three of the edges of the parallelepiped are A B,A C, and A D. Find the volume of the parallelepiped. Volume is Find the other four vertices of the parallelepiped. Vertices are (Enter your answer as a comma-separated list of points.) Point A is contained in three planar faces of the parallelepiped. Each of the other three faces of the parallelepiped is contained in a plane. Find the distances between A and these three planes. Distances are (Enter your answer as a comma-separated list of numbers.) Three planar faces of the parallelepiped meet at point A. Each of these faces is contained in a plane. Find the three acute pairwise angles between these three planes. Angles are (Enter your answer as a comma-separated list of numbers.)
There are 4 steps to solve this one.Solution Share Share Share done loading Copy link Step 1 The four vertices of parallelopiped are given as A=(2,−9,8) ; B=(1,−5,6) ; C=(−5,−12,−1) ; D=(4,−11,10).
Three of the edges are A B;A C;A D.
To find the Volum... View the full answer Step 2 UnlockStep 3 UnlockStep 4 UnlockAnswer Unlock Previous questionNext question
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14439 | https://www.thoughtco.com/history-of-the-telephone-alexander-graham-bell-1991380 | The History of the Telephone and How It Was Invented
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How the Telephone Was Invented
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The Most Important Inventions of the Industrial Revolution
Introduction
The American Industrial Revolution
Key Elements of the American Industrial Revolution
Top Inventors
Transportation
The Steam Engine
The Railroad
The Diesel Engine
The Airplane
The Automobile
Communication
The Telegraph
The Transatlantic Cable
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The Cotton Gin
The Sewing Machine
Electric Lights
The Electric Motor
This model of Bell's first telephone is a duplicate of the instrument through which speech sounds were first transmitted electrically (1875).Bettmann / Contributor / Getty Images
By
Mary Bellis
Mary Bellis
Inventions Expert
Mary Bellis covered inventions and inventors for ThoughtCo for 18 years. She is known for her independent films and documentaries, including one about Alexander Graham Bell.
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Updated on September 20, 2024
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Key Takeaways
Alexander Graham Bell beat Elisha Gray in patenting the telephone, dominating its history.
Bell's knowledge of sound helped him improve the telegraph, leading to the telephone's invention.
The invention of the telephone led to rapid industry growth, eventually forming AT&T's monopoly.
In the 1870s, Elisha Gray and Alexander Graham Bell independently designed devices that could transmit speech electrically. Both men rushed their respective designs for these prototype telephones to the patent office within hours of each other. Bell patented his telephone first and later emerged the victor in a legal dispute with Gray.
Today, Bell's name is synonymous with the history of the telephone, while Gray is largely forgotten. However, the story of who invented the telephone goes beyond these two men.
Bell's Biography
Alexander Graham Bell was born on March 3, 1847, in Edinburgh, Scotland. He was immersed in the study of sound from the beginning. His father, uncle, and grandfather were authorities on elocution and speech therapy for the deaf. It was understood that Bell would follow in the family footsteps after finishing college. But after Bell's two other brothers died of tuberculosis, Bell and his parents decided to immigrate to Canada in 1870.
After a brief period of living in Ontario, the Bells moved to Boston where they established speech-therapy practices specializing in teaching deaf children to speak. One of Alexander Graham Bell's pupils was a young Helen Keller, who when they met was not only blind and deaf but also unable to speak.
Although working with the deaf would remain Bell's principal source of income, he continued to pursue his own studies of sound on the side. Bell's unceasing scientific curiosity led to the invention of the photophone, significant commercial improvements in Thomas Edison's phonograph, and to the development of his own flying machine just six years after the Wright Brothers launched their plane at Kitty Hawk. As President James Garfield lay dying of an assassin's bullet in 1881, Bell hurriedly invented a metal detector in an unsuccessful attempt to locate the fatal slug.
The History of the Telephone
The telegraph and telephone are both wire-based electrical systems. Alexander Graham Bell's success with the telephone came as a direct result of his attempts to improve the telegraph. When he began experimenting with electrical signals, the telegraph had been an established means of communication for some 30 years. Although a highly successful system, the telegraph was basically limited to receiving and sending one message at a time.
Bell's extensive knowledge of the nature of sound and his understanding of music enabled him to consider the possibility of transmitting multiple messages over the same wire at the same time. Although the idea of a "multiple telegraph" had been in existence for some time, it was purely conjecture as no one had been able to fabricate one—until Bell. His "harmonic telegraph" was based on the principle that several notes could be sent simultaneously along the same wire if the notes or signals differed in pitch.
Talk With Electricity
By October 1874, Bell's research had progressed to the extent that he could inform his future father-in-law, Boston attorney Gardiner Greene Hubbard, about the possibility of a multiple telegraph. Hubbard, who resented the absolute control then exerted by the Western Union Telegraph Company, instantly saw the potential for breaking such a monopoly and gave Bell the financial backing he needed.
Bell proceeded with his work on the multiple telegraph but did not tell Hubbard that he and Thomas Watson, a young electrician whose services he had enlisted, were also developing a device that would transmit speech electrically. While Watson worked on the harmonic telegraph at the insistent urging of Hubbard and other backers, Bell secretly met in March 1875 with Joseph Henry, the respected director of the Smithsonian Institution, who listened to Bell's ideas for a telephone and offered encouraging words. Spurred on by Henry's positive opinion, Bell and Watson continued their work.
By June 1875, the goal of creating a device that would transmit speech electrically was about to be realized. They had proven that different tones would vary the strength of an electric current in a wire. To achieve success, therefore, they needed only to build a working transmitter with a membrane capable of varying electronic currents and a receiver that would reproduce these variations in audible frequencies.
"Mr. Watson, Come Here"
On June 2, 1875, while experimenting with the harmonic telegraph, the men discovered that sound could be transmitted over a wire completely by accident. Watson was trying to loosen a reed that had been wound around a transmitter when he plucked it by accident. The vibration produced by that gesture traveled along the wire into a second device in the other room where Bell was working.
The "twang" Bell heard was all the inspiration that he and Watson needed to accelerate their work. They continued to work into the next year. Bell recounted the critical moment in his journal:"I then shouted into M [the mouthpiece] the following sentence: 'Mr. Watson, come here—I want to see you.' To my delight, he came and declared that he had heard and understood what I said."
The first call had just been made, a pivotal moment in the history of the telephone.
The Telephone Network Is Born
Bell patented his device on March 7, 1876, and it quickly began to spread. By 1877, construction of the first regular telephone line from Boston to Somerville, Massachusetts, had been completed. By the end of 1880, there were over 49,000 telephones in the United States.1 The following year, telephone service between Boston and Providence,Rhode Island, had been established. Service between New York and Chicago started in 1892 and between New York and Boston in 1894. Transcontinental service began in 1915.
Bell founded his Bell Telephone Company in 1877. As the industry rapidly expanded, Bell quickly bought out competitors. After a series of mergers, the American Telephone and Telegraph Co.—the forerunner of today's AT&T—was incorporated in 1880. Because Bell controlled the intellectual property and patents behind the telephone system, AT&T had a de facto monopoly over the young industry. It would maintain its control over the U.S. telephone market until 1984 when a settlement with the U.S. Department of Justice forced AT&T to end its control over state markets.
Exchanges and Rotary Dialing
The first regular telephone exchange was established in New Haven, Connecticut, in 1878. Early telephones were leased in pairs to subscribers. The subscriber was required to put up his own line to connect with another. In 1889, Kansas City undertaker Almon B. Strowger invented a switch that could connect one line to any of 100 lines by using relays and sliders. The Strowger switch, as it came to be known, was still in use in some telephone offices well over 100 years later.
Strowger was issued a patent on March 11, 1891, for the first automatic telephone exchange. The first exchange using the Strowger switch was opened in La Porte, Indiana, in 1892. Initially, subscribers had a button on their telephone to produce the required number of pulses by tapping. Then an associate of Strowgers' invented the rotary dial in 1896, replacing the button. In 1943, Philadelphia was the last major area to give up dual service (rotary and button).
Pay Phones
In 1889, the coin-operated telephone was patented by William Gray of Hartford, Connecticut. Gray's payphone was first installed and used in the Hartford Bank. Unlike pay phones today, users of Gray's phone paid after they had finished their call.
Payphones proliferated along with the Bell System. By the time the first phone booths were installed in 1905, there were about 2.2 million phones; by 1980, there were more than 175 million.1 But with the advent of mobile technology, the public demand for payphones rapidly declined, and today there are fewer than 500,000 still operating in the United States.2
Touch-Tone Phones
Researchers at Western Electric, AT&T's manufacturing subsidiary, had experimented with using tones rather than pulses to trigger telephone connections since the early 1940s, but it wasn't until 1963 that dual-tone multifrequency signaling, which uses the same frequency as speech, was commercially viable. AT&T introduced it as Touch-Tone dialing and it quickly became the next standard in telephone technology. By 1990, push-button phones were more common than rotary-dial models in American homes.
Cordless Phones
In the 1970s, the very first cordless phones were introduced, another watershed moment in the history of telephones. In 1986, the Federal Communications Commission granted the frequency range of 47 to 49 MHz for cordless phones. Granting a greater frequency range allowed cordless phones to have less interference and need less power to run. In 1990, the FCC granted the frequency range of 900 MHz for cordless phones.
In 1994, digital cordless phones were introduced, followed by digital spread spectrum (DSS) in 1995. Both developments were intended to increase the security of cordless phones and decrease unwanted eavesdropping by enabling the phone conversation to be digitally spread out. In 1998, the FCC granted the frequency range of 2.4 GHz for cordless phones; the upward range is now 5.8 GHz.
Cell Phones
The earliest mobile phones were radio-controlled units designed for vehicles. They were expensive and cumbersome, and had extremely limited range. First launched by AT&T in 1946, the network would slowly expand and become more sophisticated, but it never was widely adopted. By 1980, it had been replaced by the first cellular networks.
Research on what would become the cellular phone network used today began in 1947 at Bell Labs, the research wing of AT&T. Although the radio frequencies needed were not yet commercially available, the concept of connecting phones wirelessly through a network of "cells" or transmitters was a viable one. Motorola introduced the first hand-held cellular phone in 1973.
Telephone Books
The first telephone book was published in New Haven, Connecticut, by the New Haven District Telephone Company in February 1878. It was one page long and held 50 names; no numbers were listed, as an operator would connect you. The page was divided into four sections: residential, professional, essential services, and miscellaneous.
In 1886, Reuben H. Donnelly produced the first Yellow Pages-branded directory featuring business names and phone numbers, categorized by the types of products and services provided. By the 1980s, telephone books, whether issued by the Bell System or private publishers, were in nearly every home and business. But with the advent of the Internet and of cell phones, telephone books have been rendered largely obsolete, relegated to the history of the telephone.
9-1-1
Prior to 1968, there was no dedicated phone number for reaching first responders in the event of an emergency. That changed after a congressional investigation led to calls for the establishment of such a system nationwide. The Federal Communications Commission and AT&T soon announced they would launch their emergency network in Indiana, using the digits 9-1-1 (chosen for its simplicity and for being easy to remember).
But a small independent phone company in rural Alabama decided to beat AT&T at its own game. On Feb. 16, 1968, the first 9-1-1 call was placed in Hayleyville, Alabama, at the office of the Alabama Telephone Company. The 9-1-1 network would be introduced to other cities and towns slowly; it wasn't until 1987 that at least half of all American homes had access to a 9-1-1 emergency network.
Caller ID
Several researchers created devices for identifying the number of incoming calls, including scientists in Brazil, Japan, and Greece, starting in the late 1960s. In the U.S., AT&T first made its trademarked TouchStar caller ID service available in Orlando, Florida, in 1984. Over the next several years, the regional Bell Systems would introduce caller ID services in the Northeast and Southeast. Although the service was initially sold as a pricey added service, caller ID today is a standard function found on every cell phone and available on almost any landline.
Additional Resources
Casson, Herbert N. The History of the Telephone. Chicago: A.C. McClurg & Co., 1910.
View Article Sources
"1870s to 1940s - Telephone."Imagining the Internet: a History and Forecast. Elon University School of Communications.
Kieler, Ashlee. “5 Things We Learned About Pay Phones & Why They Continue To Exist.”Consumerist, 26 Apr. 2016.
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Bellis, Mary. "How the Telephone Was Invented." ThoughtCo, Apr. 11, 2025, thoughtco.com/history-of-the-telephone-alexander-graham-bell-1991380.Bellis, Mary. (2025, April 11). How the Telephone Was Invented. Retrieved from Bellis, Mary. "How the Telephone Was Invented." ThoughtCo. (accessed September 16, 2025).
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14440 | https://www.vyssotski.ch/BasicsOfInstrumentation/LaplaceTransform.pdf | Chapter 7 Laplace Transform The Laplace transform can be used to solve differential equations. Be-sides being a different and efficient alternative to variation of parame-ters and undetermined coefficients, the Laplace method is particularly advantageous for input terms that are piecewise-defined, periodic or im-pulsive.
The direct Laplace transform or the Laplace integral of a function f(t) defined for 0 ≤t < ∞is the ordinary calculus integration problem Z ∞ 0 f(t)e−stdt, succinctly denoted L(f(t)) in science and engineering literature.
The L–notation recognizes that integration always proceeds over t = 0 to t = ∞and that the integral involves an integrator e−stdt instead of the usual dt. These minor differences distinguish Laplace integrals from the ordinary integrals found on the inside covers of calculus texts.
7.1 Introduction to the Laplace Method The foundation of Laplace theory is Lerch’s cancellation law R ∞ 0 y(t)e−stdt = R ∞ 0 f(t)e−stdt implies y(t) = f(t), or L(y(t) = L(f(t)) implies y(t) = f(t).
(1) In differential equation applications, y(t) is the sought-after unknown while f(t) is an explicit expression taken from integral tables.
Below, we illustrate Laplace’s method by solving the initial value prob-lem y′ = −1, y(0) = 0.
The method obtains a relation L(y(t)) = L(−t), whence Lerch’s cancel-lation law implies the solution is y(t) = −t.
The Laplace method is advertised as a table lookup method, in which the solution y(t) to a differential equation is found by looking up the answer in a special integral table.
7.1 Introduction to the Laplace Method 247 Laplace Integral.
The integral R ∞ 0 g(t)e−stdt is called the Laplace integral of the function g(t). It is defined by limN→∞ R N 0 g(t)e−stdt and depends on variable s. The ideas will be illustrated for g(t) = 1, g(t) = t and g(t) = t2, producing the integral formulas in Table 1.
R ∞ 0 (1)e−stdt = −(1/s)e−st t=∞ t=0 Laplace integral of g(t) = 1.
= 1/s Assumed s > 0.
R ∞ 0 (t)e−stdt = R ∞ 0 −d ds(e−st)dt Laplace integral of g(t) = t.
= −d ds R ∞ 0 (1)e−stdt Use R d dsF(t, s)dt = d ds R F(t, s)dt.
= −d ds(1/s) Use L(1) = 1/s.
= 1/s2 Differentiate.
R ∞ 0 (t2)e−stdt = R ∞ 0 −d ds(te−st)dt Laplace integral of g(t) = t2.
= −d ds R ∞ 0 (t)e−stdt = −d ds(1/s2) Use L(t) = 1/s2.
= 2/s3 Table 1.
The Laplace integral R ∞ 0 g(t)e−stdt for g(t) = 1, t and t2.
R ∞ 0 (1)e−st dt = 1 s R ∞ 0 (t)e−st dt = 1 s2 R ∞ 0 (t2)e−st dt = 2 s3 In summary, L(tn) = n!
s1+n An Illustration.
The ideas of the Laplace method will be illus-trated for the solution y(t) = −t of the problem y′ = −1, y(0) = 0. The method, entirely different from variation of parameters or undetermined coefficients, uses basic calculus and college algebra; see Table 2.
Table 2.
Laplace method details for the illustration y′ = −1, y(0) = 0.
y′(t)e−st = −e−st Multiply y′ = −1 by e−st.
R ∞ 0 y′(t)e−stdt = R ∞ 0 −e−stdt Integrate t = 0 to t = ∞.
R ∞ 0 y′(t)e−stdt = −1/s Use Table 1.
s R ∞ 0 y(t)e−stdt −y(0) = −1/s Integrate by parts on the left.
R ∞ 0 y(t)e−stdt = −1/s2 Use y(0) = 0 and divide.
R ∞ 0 y(t)e−stdt = R ∞ 0 (−t)e−stdt Use Table 1.
y(t) = −t Apply Lerch’s cancellation law.
248 Laplace Transform In Lerch’s law, the formal rule of erasing the integral signs is valid pro-vided the integrals are equal for large s and certain conditions hold on y and f – see Theorem 2. The illustration in Table 2 shows that Laplace theory requires an in-depth study of a special integral table, a table which is a true extension of the usual table found on the inside covers of calculus books. Some entries for the special integral table appear in Table 1 and also in section 7.2, Table 4.
The L-notation for the direct Laplace transform produces briefer details, as witnessed by the translation of Table 2 into Table 3 below. The reader is advised to move from Laplace integral notation to the L–notation as soon as possible, in order to clarify the ideas of the transform method.
Table 3.
Laplace method L-notation details for y′ = −1, y(0) = 0 translated from Table 2.
L(y′(t)) = L(−1) Apply L across y′ = −1, or multiply y′ = −1 by e−st, integrate t = 0 to t = ∞.
L(y′(t)) = −1/s Use Table 1.
sL(y(t)) −y(0) = −1/s Integrate by parts on the left.
L(y(t)) = −1/s2 Use y(0) = 0 and divide.
L(y(t)) = L(−t) Apply Table 1.
y(t) = −t Invoke Lerch’s cancellation law.
Some Transform Rules.
The formal properties of calculus integrals plus the integration by parts formula used in Tables 2 and 3 leads to these rules for the Laplace transform: L(f(t) + g(t)) = L(f(t)) + L(g(t)) The integral of a sum is the sum of the integrals.
L(cf(t)) = cL(f(t)) Constants c pass through the integral sign.
L(y′(t)) = sL(y(t)) −y(0) The t-derivative rule, or inte-gration by parts. See Theo-rem 3.
L(y(t)) = L(f(t)) implies y(t) = f(t) Lerch’s cancellation law. See Theorem 2.
1 Example (Laplace method) Solve by Laplace’s method the initial value problem y′ = 5 −2t, y(0) = 1.
Solution: Laplace’s method is outlined in Tables 2 and 3. The L-notation of Table 3 will be used to find the solution y(t) = 1 + 5t −t2.
7.1 Introduction to the Laplace Method 249 L(y′(t)) = L(5 −2t) Apply L across y′ = 5 −2t.
L(y′(t)) = 5 s −2 s2 Use Table 1.
sL(y(t)) −y(0) = 5 s −2 s2 Apply the t-derivative rule, page 248.
L(y(t)) = 1 s + 5 s2 −2 s3 Use y(0) = 1 and divide.
L(y(t)) = L(1) + 5L(t) −L(t2) Apply Table 1, backwards.
= L(1 + 5t −t2) Linearity, page 248.
y(t) = 1 + 5t −t2 Invoke Lerch’s cancellation law.
2 Example (Laplace method) Solve by Laplace’s method the initial value problem y′′ = 10, y(0) = y′(0) = 0.
Solution: The L-notation of Table 3 will be used to find the solution y(t) = 5t2.
L(y′′(t)) = L(10) Apply L across y′′ = 10.
sL(y′(t)) −y′(0) = L(10) Apply the t-derivative rule to y′, that is, replace y by y′ on page 248.
s[sL(y(t)) −y(0)] −y′(0) = L(10) Repeat the t-derivative rule, on y.
s2L(y(t)) = L(10) Use y(0) = y′(0) = 0.
L(y(t)) = 10 s3 Use Table 1. Then divide.
L(y(t)) = L(5t2) Apply Table 1, backwards.
y(t) = 5t2 Invoke Lerch’s cancellation law.
Existence of the Transform.
The Laplace integral R ∞ 0 e−stf(t) dt is known to exist in the sense of the improper integral definition1 Z ∞ 0 g(t)dt = lim N→∞ Z N 0 g(t)dt provided f(t) belongs to a class of functions known in the literature as functions of exponential order. For this class of functions the relation lim t→∞ f(t) eat = 0 (2) is required to hold for some real number a, or equivalently, for some constants M and α, |f(t)| ≤Meαt.
(3) In addition, f(t) is required to be piecewise continuous on each finite subinterval of 0 ≤t < ∞, a term defined as follows.
1An advanced calculus background is assumed for the Laplace transform existence proof. Applications of Laplace theory require only a calculus background.
250 Laplace Transform Definition 1 (piecewise continuous) A function f(t) is piecewise continuous on a finite interval [a, b] pro-vided there exists a partition a = t0 < · · · < tn = b of the interval [a, b] and functions f1, f2, . . . , fn continuous on (−∞, ∞) such that for t not a partition point f(t) = f1(t) t0 < t < t1, .
.
.
.
.
.
fn(t) tn−1 < t < tn.
(4) The values of f at partition points are undecided by equation (4). In particular, equation (4) implies that f(t) has one-sided limits at each point of a < t < b and appropriate one-sided limits at the endpoints.
Therefore, f has at worst a jump discontinuity at each partition point.
3 Example (Exponential order) Show that f(t) = et cos t + t is of expo-nential order, that is, show that f(t) is piecewise continuous and find α > 0 such that limt→∞f(t)/eαt = 0.
Solution: Already, f(t) is continuous, hence piecewise continuous.
From L’Hospital’s rule in calculus, limt→∞p(t)/eαt = 0 for any polynomial p and any α > 0. Choose α = 2, then lim t→∞ f(t) e2t = lim t→∞ cos t et + lim t→∞ t e2t = 0.
Theorem 1 (Existence of L(f)) Let f(t) be piecewise continuous on every finite interval in t ≥0 and satisfy |f(t)| ≤Meαt for some constants M and α. Then L(f(t)) exists for s > α and lims→∞L(f(t)) = 0.
Proof: It has to be shown that the Laplace integral of f is finite for s > α.
Advanced calculus implies that it is sufficient to show that the integrand is ab-solutely bounded above by an integrable function g(t). Take g(t) = Me−(s−α)t.
Then g(t) ≥0. Furthermore, g is integrable, because Z ∞ 0 g(t)dt = M s −α.
Inequality |f(t)| ≤Meαt implies the absolute value of the Laplace transform integrand f(t)e−st is estimated by f(t)e−st ≤Meαte−st = g(t).
The limit statement follows from |L(f(t))| ≤ R ∞ 0 g(t)dt = M s −α, because the right side of this inequality has limit zero at s = ∞. The proof is complete.
7.1 Introduction to the Laplace Method 251 Theorem 2 (Lerch) If f1(t) and f2(t) are continuous, of exponential order and R ∞ 0 f1(t)e−stdt = R ∞ 0 f2(t)e−stdt for all s > s0, then f1(t) = f2(t) for t ≥0.
Proof: See Widder [?].
Theorem 3 (t-Derivative Rule) If f(t) is continuous, lim t→∞f(t)e−st = 0 for all large values of s and f ′(t) is piecewise continuous, then L(f ′(t)) exists for all large s and L(f ′(t)) = sL(f(t)) −f(0).
Proof: See page 276.
Exercises 7.1 Laplace method.
Solve the given initial value problem using Laplace’s method.
1. y′ = −2, y(0) = 0.
2. y′ = 1, y(0) = 0.
3. y′ = −t, y(0) = 0.
4. y′ = t, y(0) = 0.
5. y′ = 1 −t, y(0) = 0.
6. y′ = 1 + t, y(0) = 0.
7. y′ = 3 −2t, y(0) = 0.
8. y′ = 3 + 2t, y(0) = 0.
9. y′′ = −2, y(0) = y′(0) = 0.
10. y′′ = 1, y(0) = y′(0) = 0.
11. y′′ = 1 −t, y(0) = y′(0) = 0.
12. y′′ = 1 + t, y(0) = y′(0) = 0.
13. y′′ = 3 −2t, y(0) = y′(0) = 0.
14. y′′ = 3 + 2t, y(0) = y′(0) = 0.
Exponential order.
Show that f(t) is of exponential order, by finding a constant α ≥0 in each case such that lim t→∞ f(t) eαt = 0.
15. f(t) = 1 + t 16. f(t) = et sin(t) 17. f(t) = PN n=0 cnxn, for any choice of the constants c0, . . . , cN.
18. f(t) = PN n=1 cn sin(nt), for any choice of the constants c1, . . . , cN.
Existence of transforms. Let f(t) = tet2 sin(et2). Establish these results.
19. The function f(t) is not of expo-nential order.
20. The Laplace integral of f(t), R ∞ 0 f(t)e−stdt, converges for all s > 0.
Jump Magnitude.
For f piecewise continuous, define the jump at t by J(t) = lim h→0+ f(t + h) −lim h→0+ f(t −h).
Compute J(t) for the following f.
21. f(t) = 1 for t ≥0, else f(t) = 0 22. f(t) = 1 for t ≥1/2, else f(t) = 0 23. f(t) = t/|t| for t ̸= 0, f(0) = 0 24. f(t) = sin t/| sin t| for t ̸= nπ, f(nπ) = (−1)n Taylor series.
The series relation L(P∞ n=0 cntn) = P∞ n=0 cnL(tn) often holds, in which case the result L(tn) = n!s−1−n can be employed to find a series representation of the Laplace transform.
Use this idea on the fol-lowing to find a series formula for L(f(t)).
25. f(t) = e2t = P∞ n=0(2t)n/n!
26. f(t) = e−t = P∞ n=0(−t)n/n!
252 Laplace Transform 7.2 Laplace Integral Table The objective in developing a table of Laplace integrals, e.g., Tables 4 and 5, is to keep the table size small. Table manipulation rules appear-ing in Table 6, page 257, effectively increase the table size manyfold, making it possible to solve typical differential equations from electrical and mechanical problems. The combination of Laplace tables plus the table manipulation rules is called the Laplace transform calculus.
Table 4 is considered to be a table of minimum size to be memorized.
Table 5 adds a number of special-use entries. For instance, the Heaviside entry in Table 5 is memorized, but usually not the others.
Derivations are postponed to page 270. The theory of the gamma func-tion Γ(x) appears below on page 255.
Table 4.
A minimal Laplace integral table with L-notation R ∞ 0 (tn)e−st dt = n!
s1+n L(tn) = n!
s1+n R ∞ 0 (eat)e−st dt = 1 s −a L(eat) = 1 s −a R ∞ 0 (cos bt)e−st dt = s s2 + b2 L(cos bt) = s s2 + b2 R ∞ 0 (sin bt)e−st dt = b s2 + b2 L(sin bt) = b s2 + b2 Table 5.
Laplace integral table extension L(H(t −a)) = e−as s (a ≥0) Heaviside unit step, defined by H(t) = 1 for t ≥0, 0 otherwise.
L(δ(t −a)) = e−as Dirac delta, δ(t) = dH(t).
Special usage rules apply.
L(floor(t/a)) = e−as s(1 −e−as) Staircase function, floor(x) = greatest integer ≤x.
L(sqw(t/a)) = 1 s tanh(as/2) Square wave, sqw(x) = (−1)floor(x).
L(a trw(t/a)) = 1 s2 tanh(as/2) Triangular wave, trw(x) = R x 0 sqw(r)dr.
L(tα) = Γ(1 + α) s1+α Generalized power function, Γ(1 + α) = R ∞ 0 e−xxαdx.
L(t−1/2) = rπ s Because Γ(1/2) = √π.
7.2 Laplace Integral Table 253 4 Example (Laplace transform) Let f(t) = t(t−1)−sin2t+e3t. Compute L(f(t)) using the basic Laplace table and transform linearity properties.
Solution: L(f(t)) = L(t2 −5t −sin 2t + e3t) Expand t(t −5).
= L(t2) −5L(t) −L(sin 2t) + L(e3t) Linearity applied.
= 2 s3 −5 s2 − 2 s2 + 4 + 1 s −3 Table lookup.
5 Example (Inverse Laplace transform) Use the basic Laplace table back-wards plus transform linearity properties to solve for f(t) in the equation L(f(t)) = s s2 + 16 + 2 s −3 + s + 1 s3 .
Solution: L(f(t)) = s s2 + 16 + 2 1 s −3 + 1 s2 + 1 2 2 s3 Convert to table entries.
= L(cos 4t) + 2L(e3t) + L(t) + 1 2L(t2) Laplace table (backwards).
= L(cos 4t + 2e3t + t + 1 2t2) Linearity applied.
f(t) = cos 4t + 2e3t + t + 1 2t2 Lerch’s cancellation law.
6 Example (Heaviside) Find the Laplace transform of f(t) in Figure 1.
1 3 1 5 5 Figure 1.
A piecewise defined function f(t) on 0 ≤t < ∞: f(t) = 0 except for 1 ≤t < 2 and 3 ≤t < 4.
Solution: The details require the use of the Heaviside function formula H(t −a) −H(t −b) = 1 a ≤t < b, 0 otherwise.
The formula for f(t): f(t) = 1 1 ≤t < 2, 5 3 ≤t < 4, 0 otherwise = 1 1 ≤t < 2, 0 otherwise + 5 1 3 ≤t < 4, 0 otherwise Then f(t) = f1(t) + 5f2(t) where f1(t) = H(t −1) −H(t −2) and f2(t) = H(t −3) −H(t −4). The extended table gives L(f(t)) = L(f1(t)) + 5L(f2(t)) Linearity.
= L(H(t −1)) −L(H(t −2)) + 5L(f2(t)) Substitute for f1.
254 Laplace Transform = e−s −e−2s s + 5L(f2(t)) Extended table used.
= e−s −e−2s + 5e−3s −5e−4s s Similarly for f2.
7 Example (Dirac delta) A machine shop tool that repeatedly hammers a die is modeled by the Dirac impulse model f(t) = PN n=1 δ(t −n). Show that L(f(t)) = PN n=1 e−ns.
Solution: L(f(t)) = L PN n=1 δ(t −n) = PN n=1 L(δ(t −n)) Linearity.
= PN n=1 e−ns Extended Laplace table.
8 Example (Square wave) A periodic camshaft force f(t) applied to a me-chanical system has the idealized graph shown in Figure 2.
Show that f(t) = 1 + sqw(t) and L(f(t)) = 1 s(1 + tanh(s/2)).
0 2 1 3 Figure 2.
A periodic force f(t) applied to a mechanical system.
Solution: 1 + sqw(t) = 1 + 1 2n ≤t < 2n + 1, n = 0, 1, . . ., 1 −1 2n + 1 ≤t < 2n + 2, n = 0, 1, . . ., = 2 2n ≤t < 2n + 1, n = 0, 1, . . ., 0 otherwise, = f(t).
By the extended Laplace table, L(f(t)) = L(1) + L(sqw(t)) = 1 s + tanh(s/2) s .
9 Example (Sawtooth wave) Express the P-periodic sawtooth wave repre-sented in Figure 3 as f(t) = ct/P −c floor(t/P) and obtain the formula L(f(t)) = c Ps2 − ce−P s s −se−P s.
0 c P 4P Figure 3.
A P-periodic sawtooth wave f(t) of height c > 0.
7.2 Laplace Integral Table 255 Solution: The representation originates from geometry, because the periodic function f can be viewed as derived from ct/P by subtracting the correct con-stant from each of intervals [P, 2P], [2P, 3P], etc.
The technique used to verify the identity is to define g(t) = ct/P −c floor(t/P) and then show that g is P-periodic and f(t) = g(t) on 0 ≤t < P. Two P-periodic functions equal on the base interval 0 ≤t < P have to be identical, hence the representation follows.
The fine details: for 0 ≤t < P, floor(t/P) = 0 and floor(t/P + k) = k. Hence g(t + kP) = ct/P + ck −c floor(k) = ct/P = g(t), which implies that g is P-periodic and g(t) = f(t) for 0 ≤t < P.
L(f(t)) = c P L(t) −cL(floor(t/P)) Linearity.
= c Ps2 − ce−Ps s −se−Ps Basic and extended table applied.
10 Example (Triangular wave) Express the triangular wave f of Figure 4 in terms of the square wave sqw and obtain L(f(t)) = 5 πs2 tanh(πs/2).
0 5 2π Figure 4.
A 2π-periodic triangular wave f(t) of height 5.
Solution: The representation of f in terms of sqw is f(t) = 5 R t/π 0 sqw(x)dx.
Details: A 2-periodic triangular wave of height 1 is obtained by integrating the square wave of period 2.
A wave of height c and period 2 is given by c trw(t) = c R t 0 sqw(x)dx. Then f(t) = c trw(2t/P) = c R 2t/P 0 sqw(x)dx where c = 5 and P = 2π.
Laplace transform details: Use the extended Laplace table as follows.
L(f(t)) = 5 π L(π trw(t/π)) = 5 πs2 tanh(πs/2).
Gamma Function.
In mathematical physics, the Gamma func-tion or the generalized factorial function is given by the identity Γ(x) = Z ∞ 0 e−ttx−1 dt, x > 0.
(1) This function is tabulated and available in computer languages like For-tran, C and C++. It is also available in computer algebra systems and numerical laboratories. Some useful properties of Γ(x): Γ(1 + x) = xΓ(x) (2) Γ(1 + n) = n!
for integers n ≥1.
(3) 256 Laplace Transform Details for relations (2) and (3): Start with R ∞ 0 e−tdt = 1, which gives Γ(1) = 1. Use this identity and successively relation (2) to obtain relation (3).
To prove identity (2), integration by parts is applied, as follows: Γ(1 + x) = R ∞ 0 e−ttxdt Definition.
= −txe−t|t=∞ t=0 + R ∞ 0 e−txtx−1dt Use u = tx, dv = e−tdt.
= x R ∞ 0 e−ttx−1dt Boundary terms are zero for x > 0.
= xΓ(x).
Exercises 7.2 Laplace transform.
Compute L(f(t)) using the basic Laplace table and the linearity properties of the transform.
Do not use the direct Laplace transform!
1. L(2t) 2. L(4t) 3. L(1 + 2t + t2) 4. L(t2 −3t + 10) 5. L(sin 2t) 6. L(cos 2t) 7. L(e2t) 8. L(e−2t) 9. L(t + sin 2t) 10. L(t −cos 2t) 11. L(t + e2t) 12. L(t −3e−2t) 13. L((t + 1)2) 14. L((t + 2)2) 15. L(t(t + 1)) 16. L((t + 1)(t + 2)) 17. L(P10 n=0 tn/n!) 18. L(P10 n=0 tn+1/n!) 19. L(P10 n=1 sin nt) 20. L(P10 n=0 cos nt) Inverse Laplace transform.
Solve the given equation for the function f(t). Use the basic table and linearity properties of the Laplace transform.
21. L(f(t)) = s−2 22. L(f(t)) = 4s−2 23. L(f(t)) = 1/s + 2/s2 + 3/s3 24. L(f(t)) = 1/s3 + 1/s 25. L(f(t)) = 2/(s2 + 4) 26. L(f(t)) = s/(s2 + 4) 27. L(f(t)) = 1/(s −3) 28. L(f(t)) = 1/(s + 3) 29. L(f(t)) = 1/s + s/(s2 + 4) 30. L(f(t)) = 2/s −2/(s2 + 4) 31. L(f(t)) = 1/s + 1/(s −3) 32. L(f(t)) = 1/s −3/(s −2) 33. L(f(t)) = (2 + s)2/s3 34. L(f(t)) = (s + 1)/s2 35. L(f(t)) = s(1/s2 + 2/s3) 36. L(f(t)) = (s + 1)(s −1)/s3 37. L(f(t)) = P10 n=0 n!/s1+n 38. L(f(t)) = P10 n=0 n!/s2+n 39. L(f(t)) = P10 n=1 n s2 + n2 40. L(f(t)) = P10 n=0 s s2 + n2 7.3 Laplace Transform Rules 257 7.3 Laplace Transform Rules In Table 6, the basic table manipulation rules are summarized.
Full statements and proofs of the rules appear in section 7.7, page 275.
The rules are applied here to several key examples.
Partial fraction expansions do not appear here, but in section 7.4, in connection with Heaviside’s coverup method.
Table 6.
Laplace transform rules L(f(t) + g(t)) = L(f(t)) + L(g(t)) Linearity.
The Laplace of a sum is the sum of the Laplaces.
L(cf(t)) = cL(f(t)) Linearity.
Constants move through the L-symbol.
L(y′(t)) = sL(y(t)) −y(0) The t-derivative rule.
Derivatives L(y′) are replaced in transformed equations.
L R t 0 g(x)dx = 1 sL(g(t)) The t-integral rule.
L(tf(t)) = −d dsL(f(t)) The s-differentiation rule.
Multiplying f by t applies −d/ds to the transform of f.
L(eatf(t)) = L(f(t))|s→(s−a) First shifting rule.
Multiplying f by eat replaces s by s −a.
L(f(t −a)H(t −a)) = e−asL(f(t)), L(g(t)H(t −a)) = e−asL(g(t + a)) Second shifting rule.
First and second forms.
L(f(t)) = R P 0 f(t)e−stdt 1 −e−Ps Rule for P-periodic functions.
Assumed here is f(t + P ) = f(t).
L(f(t))L(g(t)) = L((f ∗g)(t)) Convolution rule.
Define (f ∗g)(t) = R t 0 f(x)g(t −x)dx.
11 Example (Harmonic oscillator) Solve by Laplace’s method the initial value problem x′′ + x = 0, x(0) = 0, x′(0) = 1.
Solution: The solution is x(t) = sin t. The details: L(x′′) + L(x) = L(0) Apply L across the equation.
sL(x′) −x′(0) + L(x) = 0 Use the t-derivative rule.
s[sL(x) −x(0)] −x′(0) + L(x) = 0 Use again the t-derivative rule.
(s2 + 1)L(x) = 1 Use x(0) = 0, x′(0) = 1.
L(x) = 1 s2 + 1 Divide.
= L(sin t) Basic Laplace table.
x(t) = sin t Invoke Lerch’s cancellation law.
258 Laplace Transform 12 Example (s-differentiation rule) Show the steps for L(t2 e5t) = 2 (s −5)3 .
Solution: L(t2e5t) = −d ds −d ds L(e5t) Apply s-differentiation.
= (−1)2 d ds d ds 1 s −5 Basic Laplace table.
= d ds −1 (s −5)2 Calculus power rule.
= 2 (s −5)3 Identity verified.
13 Example (First shifting rule) Show the steps for L(t2 e−3t) = 2 (s + 3)3 .
Solution: L(t2e−3t) = L(t2) s→s−(−3) First shifting rule.
= 2 s2+1 s→s−(−3) Basic Laplace table.
= 2 (s + 3)3 Identity verified.
14 Example (Second shifting rule) Show the steps for L(sin t H(t −π)) = e−πs s2 + 1.
Solution: The second shifting rule is applied as follows.
L(sin t H(t −π)) = L(g(t)H(t −a) Choose g(t) = sin t, a = π.
= e−asL(g(t + a) Second form, second shifting theorem.
= e−πsL(sin(t + π)) Substitute a = π.
= e−πsL(−sin t) Sum rule sin(a + b) = sin a cosb + sin b cos a plus sin π = 0, cos π = −1.
= e−πs −1 s2 + 1 Basic Laplace table. Identity verified.
15 Example (Trigonometric formulas) Show the steps used to obtain these Laplace identities: (a) L(t cos at) = s2 −a2 (s2 + a2)2 (c) L(t2 cos at) = 2(s3 −3sa2) (s2 + a2)3 (b) L(t sin at) = 2sa (s2 + a2)2 (d) L(t2 sin at) = 6s2a −a3 (s2 + a2)3 7.3 Laplace Transform Rules 259 Solution: The details for (a): L(t cos at) = −(d/ds)L(cos at) Use s-differentiation.
= −d ds s s2 + a2 Basic Laplace table.
= s2 −a2 (s2 + a2)2 Calculus quotient rule.
The details for (c): L(t2 cos at) = −(d/ds)L((−t) cos at) Use s-differentiation.
= d ds −s2 −a2 (s2 + a2)2 Result of (a).
= 2s3 −6sa2) (s2 + a2)3 Calculus quotient rule.
The similar details for (b) and (d) are left as exercises.
16 Example (Exponentials) Show the steps used to obtain these Laplace identities: (a) L(eat cos bt) = s −a (s −a)2 + b2 (c) L(teat cos bt) = (s −a)2 −b2 ((s −a)2 + b2)2 (b) L(eat sin bt) = b (s −a)2 + b2 (d) L(teat sin bt) = 2b(s −a) ((s −a)2 + b2)2 Solution: Details for (a): L(eat cos bt) = L(cos bt)|s→s−a First shifting rule.
= s s2 + b2 s→s−a Basic Laplace table.
= s −a (s −a)2 + b2 Verified (a).
Details for (c): L(teat cos bt) = L(t cos bt)|s→s−a First shifting rule.
= −d dsL(cos bt) s→s−a Apply s-differentiation.
= −d ds s s2 + b2 s→s−a Basic Laplace table.
= s2 −b2 (s2 + b2)2 s→s−a Calculus quotient rule.
= (s −a)2 −b2 ((s −a)2 + b2)2 Verified (c).
Left as exercises are (b) and (d).
260 Laplace Transform 17 Example (Hyperbolic functions) Establish these Laplace transform facts about cosh u = (eu + e−u)/2 and sinh u = (eu −e−u)/2.
(a) L(cosh at) = s s2 −a2 (c) L(t cosh at) = s2 + a2 (s2 −a2)2 (b) L(sinh at) = a s2 −a2 (d) L(t sinh at) = 2as (s2 −a2)2 Solution: The details for (a): L(cosh at) = 1 2(L(eat) + L(e−at)) Definition plus linearity of L.
= 1 2 1 s −a + 1 s + a Basic Laplace table.
= s s2 −a2 Identity (a) verified.
The details for (d): L(t sinh at) = −d ds a s2 −a2 Apply the s-differentiation rule.
= a(2s) (s2 −a2)2 Calculus power rule; (d) verified.
Left as exercises are (b) and (c).
18 Example (s-differentiation) Solve L(f(t)) = 2s (s2 + 1)2 for f(t).
Solution: The solution is f(t) = t sin t. The details: L(f(t)) = 2s (s2 + 1)2 = −d ds 1 s2 + 1 Calculus power rule (un)′ = nun−1u′.
= −d ds (L(sin t)) Basic Laplace table.
= L(t sin t) Apply the s-differentiation rule.
f(t) = t sin t Lerch’s cancellation law.
19 Example (First shift rule) Solve L(f(t)) = s + 2 22 + 2s + 2 for f(t).
Solution: The answer is f(t) = e−t cos t + e−t sin t. The details: L(f(t)) = s + 2 s2 + 2s + 2 Signal for this method: the denom-inator has complex roots.
= s + 2 (s + 1)2 + 1 Complete the square, denominator.
7.3 Laplace Transform Rules 261 = S + 1 S2 + 1 Substitute S for s + 1.
= S S2 + 1 + 1 S2 + 1 Split into Laplace table entries.
= L(cos t) + L(sin t)|s→S=s+1 Basic Laplace table.
= L(e−t cos t) + L(e−t sin t) First shift rule.
f(t) = e−t cos t + e−t sin t Invoke Lerch’s cancellation law.
20 Example (Damped oscillator) Solve by Laplace’s method the initial value problem x′′ + 2x′ + 2x = 0, x(0) = 1, x′(0) = −1.
Solution: The solution is x(t) = e−t cos t. The details: L(x′′) + 2L(x′) + 2L(x) = L(0) Apply L across the equation.
sL(x′) −x′(0) + 2L(x′) + 2L(x) = 0 The t-derivative rule on x′.
s[sL(x) −x(0)] −x′(0) +2[L(x) −x(0)] + 2L(x) = 0 The t-derivative rule on x.
(s2 + 2s + 2)L(x) = 1 + s Use x(0) = 1, x′(0) = −1.
L(x) = s + 1 s2 + 2s + 2 Divide.
= s + 1 (s + 1)2 + 1 Complete the square in the de-nominator.
= L(cos t)|s→s+1 Basic Laplace table.
= L(e−t cos t) First shifting rule.
x(t) = e−t cos t Invoke Lerch’s cancellation law.
21 Example (Rectified sine wave) Compute the Laplace transform of the rectified sine wave f(t) = | sin ωt| and show it can be expressed in the form L(| sin ωt|) = ω coth πs 2ω s2 + ω2 .
Solution: The periodic function formula will be applied with period P = 2π/ω. The calculation reduces to the evaluation of J = R P 0 f(t)e−stdt. Because sin ωt ≤0 on π/ω ≤t ≤2π/ω, integral J can be written as J = J1 + J2, where J1 = Z π/ω 0 sin ωt e−stdt, J2 = Z 2π/ω π/ω −sin ωt e−stdt.
Integral tables give the result Z sin ωt e−st dt = −ωe−st cos(ωt) s2 + ω2 −se−st sin(ωt) s2 + ω2 .
Then J1 = ω(e−π∗s/ω + 1) s2 + ω2 , J2 = ω(e−2πs/ω + e−πs/ω) s2 + ω2 , 262 Laplace Transform J = ω(e−πs/ω + 1)2 s2 + ω2 .
The remaining challenge is to write the answer for L(f(t)) in terms of coth.
The details: L(f(t)) = J 1 −e−Ps Periodic function formula.
= J (1 −e−Ps/2)(1 + e−Ps/2) Apply 1 −x2 = (1 −x)(1 + x), x = e−Ps/2.
= ω(1 + e−Ps/2) (1 −e−Ps/2)(s2 + ω2) Cancel factor 1 + e−Ps/2.
= ePs/4 + e−Ps/4 ePs/4 −e−Ps/4 ω s2 + ω2 Factor out e−Ps/4, then cancel.
= 2 cosh(Ps/4) 2 sinh(Ps/4) ω s2 + ω2 Apply cosh, sinh identities.
= ω coth(Ps/4) s2 + ω2 Use coth u = cosh u/ sinh u.
= ω coth πs 2ω s2 + ω2 Identity verified.
22 Example (Half–wave rectification) Compute the Laplace transform of the half–wave rectification of sin ωt, denoted g(t), in which the negative cycles of sin ωt have been canceled to create g(t). Show in particular that L(g(t)) = 1 2 ω s2 + ω2 1 + coth πs 2ω Solution: The half–wave rectification of sin ωt is g(t) = (sin ωt + | sin ωt|)/2.
Therefore, the basic Laplace table plus the result of Example 21 give L(2g(t)) = L(sin ωt) + L(| sin ωt|) = ω s2 + ω2 + ω cosh(πs/(2ω)) s2 + ω2 = ω s2 + ω2 (1 + cosh(πs/(2ω)) Dividing by 2 produces the identity.
23 Example (Shifting rules) Solve L(f(t)) = e−3s s + 1 s2 + 2s + 2 for f(t).
Solution: The answer is f(t) = e3−t cos(t −3)H(t −3). The details: L(f(t)) = e−3s s + 1 (s + 1)2 + 1 Complete the square.
= e−3s S S2 + 1 Replace s + 1 by S.
= e−3S+3 (L(cos t))|s→S=s+1 Basic Laplace table.
7.3 Laplace Transform Rules 263 = e3 e−3sL(cos t) s→S=s+1 Regroup factor e−3S.
= e3 (L(cos(t −3)H(t −3)))|s→S=s+1 Second shifting rule.
= e3L(e−t cos(t −3)H(t −3)) First shifting rule.
f(t) = e3−t cos(t −3)H(t −3) Lerch’s cancellation law.
24 Example () Solve L(f(t) = s + 7 s2 + 4s + 8 for f(t).
Solution: The answer is f(t) = e−2t(cos 2t + 5 2 sin 2t). The details: L(f(t)) = s + 7 (s + 2)2 + 4 Complete the square.
= S + 5 S2 + 4 Replace s + 2 by S.
= S S2 + 4 + 5 2 2 S2 + 4 Split into table entries.
= s s2 + 4 + 5 2 2 s2 + 4 s→S=s+2 Prepare for shifting rule.
= L(cos 2t) + 5 2L(sin 2t) s→S=s+2 Basic Laplace table.
= L(e−2t(cos 2t + 5 2 sin 2t)) First shifting rule.
f(t) = e−2t(cos 2t + 5 2 sin 2t) Lerch’s cancellation law.
264 Laplace Transform 7.4 Heaviside’s Method This practical method was popularized by the English electrical engineer Oliver Heaviside (1850–1925). A typical application of the method is to solve 2s (s + 1)(s2 + 1) = L(f(t)) for the t-expression f(t) = −e−t +cos t+sin t. The details in Heaviside’s method involve a sequence of easy-to-learn college algebra steps.
More precisely, Heaviside’s method systematically converts a polyno-mial quotient a0 + a1s + · · · + ansn b0 + b1s + · · · + bmsm (1) into the form L(f(t)) for some expression f(t).
It is assumed that a0, .., an, b0, . . . , bm are constants and the polynomial quotient (1) has limit zero at s = ∞.
Partial Fraction Theory In college algebra, it is shown that a rational function (1) can be ex-pressed as the sum of terms of the form A (s −s0)k (2) where A is a real or complex constant and (s −s0)k divides the denomi-nator in (1). In particular, s0 is a root of the denominator in (1).
Assume fraction (1) has real coefficients. If s0 in (2) is real, then A is real. If s0 = α + iβ in (2) is complex, then (s −s0)k also appears, where s0 = α −iβ is the complex conjugate of s0. The corresponding terms in (2) turn out to be complex conjugates of one another, which can be combined in terms of real numbers B and C as A (s −s0)k + A (s −s0)k = B + C s ((s −α)2 + β2)k .
(3) Simple Roots.
Assume that (1) has real coefficients and the denomi-nator of the fraction (1) has distinct real roots s1, . . . , sN and distinct complex roots α1 +iβ1, . . . , αM +iβM. The partial fraction expansion of (1) is a sum given in terms of real constants Ap, Bq, Cq by a0 + a1s + · · · + ansn b0 + b1s + · · · + bmsm = N X p=1 Ap s −sp + M X q=1 Bq + Cq(s −αq) (s −αq)2 + β2 q .
(4) 7.4 Heaviside’s Method 265 Multiple Roots.
Assume (1) has real coefficients and the denomi-nator of the fraction (1) has possibly multiple roots. Let Np be the multiplicity of real root sp and let Mq be the multiplicity of complex root αq + iβq, 1 ≤p ≤N, 1 ≤q ≤M. The partial fraction expansion of (1) is given in terms of real constants Ap,k, Bq,k, Cq,k by N X p=1 X 1≤k≤Np Ap,k (s −sp)k + M X q=1 X 1≤k≤Mq Bq,k + Cq,k(s −αq) ((s −αq)2 + β2 q)k .
(5) Heaviside’s Coverup Method The method applies only to the case of distinct roots of the denominator in (1). Extensions to multiple-root cases can be made; see page 266.
To illustrate Oliver Heaviside’s ideas, consider the problem details 2s + 1 s(s −1)(s + 1) = A s + B s −1 + C s + 1 (6) = L(A) + L(Bet) + L(Ce−t) = L(A + Bet + Ce−t) The first line (6) uses college algebra partial fractions. The second and third lines use the Laplace integral table and properties of L.
Heaviside’s mysterious method.
Oliver Heaviside proposed to find in (6) the constant C = 1 2 by a cover–up method: 2s + 1 s(s −1) s+1 =0 = C .
The instructions are to cover–up the matching factors (s + 1) on the left and right with box , then evaluate on the left at the root s which makes the contents of the box zero. The other terms on the right are replaced by zero.
To justify Heaviside’s cover–up method, multiply (6) by the denominator s + 1 of partial fraction C/(s + 1): (2s + 1) (s + 1) s(s −1) (s + 1) = A (s + 1) s + B (s + 1) s −1 + C (s + 1) (s + 1) .
Set (s + 1) = 0 in the display. Cancellations left and right plus annihi-lation of two terms on the right gives Heaviside’s prescription 2s + 1 s(s −1) s+1=0 = C.
266 Laplace Transform The factor (s + 1) in (6) is by no means special: the same procedure applies to find A and B.
The method works for denominators with simple roots, that is, no repeated roots are allowed.
Extension to Multiple Roots.
An extension of Heaviside’s method is possible for the case of repeated roots. The basic idea is to factor–out the repeats. To illustrate, consider the partial fraction expansion details R = 1 (s + 1)2(s + 2) A sample rational function having repeated roots.
= 1 s + 1 1 (s + 1)(s + 2) Factor–out the repeats.
= 1 s + 1 1 s + 1 + −1 s + 2 Apply the cover–up method to the simple root fraction.
= 1 (s + 1)2 + −1 (s + 1)(s + 2) Multiply.
= 1 (s + 1)2 + −1 s + 1 + 1 s + 2 Apply the cover–up method to the last fraction on the right.
Terms with only one root in the denominator are already partial frac-tions. Thus the work centers on expansion of quotients in which the denominator has two or more roots.
Special Methods.
Heaviside’s method has a useful extension for the case of roots of multiplicity two. To illustrate, consider these details: R = 1 (s + 1)2(s + 2) A fraction with multiple roots.
= A s + 1 + B (s + 1)2 + C s + 2 See equation (5).
= A s + 1 + 1 (s + 1)2 + 1 s + 2 Find B and C by Heaviside’s cover– up method.
= −1 s + 1 + 1 (s + 1)2 + 1 s + 2 Multiply by s+1. Set s = ∞. Then 0 = A + 1.
The illustration works for one root of multiplicity two, because s = ∞ will resolve the coefficient not found by the cover–up method.
In general, if the denominator in (1) has a root s0 of multiplicity k, then the partial fraction expansion contains terms A1 s −s0 + A2 (s −s0)2 + · · · + Ak (s −s0)k .
Heaviside’s cover–up method directly finds Ak, but not A1 to Ak−1.
7.5 Heaviside Step and Dirac Delta 267 7.5 Heaviside Step and Dirac Delta Heaviside Function.
The unit step function or Heaviside func-tion is defined by H(x) = ( 1 for x ≥0, 0 for x < 0.
The most often–used formula involving the Heaviside function is the characteristic function of the interval a ≤t < b, given by H(t −a) −H(t −b) = ( 1 a ≤t < b, 0 t < a, t ≥b.
(1) To illustrate, a square wave sqw(t) = (−1)floor(t) can be written in the series form ∞ X n=0 (−1)n(H(t −n) −H(t −n −1)).
Dirac Delta.
A precise mathematical definition of the Dirac delta, denoted δ, is not possible to give here. Following its inventor P. Dirac, the definition should be δ(t) = dH(t).
The latter is nonsensical, because the unit step does not have a cal-culus derivative at t = 0. However, dH(t) could have the meaning of a Riemann-Stieltjes integrator, which restrains dH(t) to have meaning only under an integral sign. It is in this sense that the Dirac delta δ is defined.
What do we mean by the differential equation x′′ + 16x = 5δ(t −t0)?
The equation x′′ + 16x = f(t) represents a spring-mass system without damping having Hooke’s constant 16, subject to external force f(t). In a mechanical context, the Dirac delta term 5δ(t −t0) is an idealization of a hammer-hit at time t = t0 > 0 with impulse 5.
More precisely, the forcing term f(t) can be formally written as a Riemann-Stieltjes integrator 5dH(t−t0) where H is Heaviside’s unit step function.
The Dirac delta or “derivative of the Heaviside unit step,” nonsensical as it may appear, is realized in applications via the two-sided or central difference quotient H(t + h) −H(t −h) 2h ≈dH(t).
268 Laplace Transform Therefore, the force f(t) in the idealization 5δ(t −t0) is given for h > 0 very small by the approximation f(t) ≈5H(t −t0 + h) −H(t −t0 −h) 2h .
The impulse2 of the approximated force over a large interval [a, b] is computed from Z b a f(t)dt ≈5 Z h −h H(t −t0 + h) −H(t −t0 −h) 2h dt = 5, due to the integrand being 1/(2h) on |t −t0| < h and otherwise 0.
Modeling Impulses.
One argument for the Dirac delta idealization is that an infinity of choices exist for modeling an impulse. There are in addition to the central difference quotient two other popular difference quotients, the forward quotient (H(t + h) −H(t))/h and the backward quotient (H(t) −H(t −h))/h (h > 0 assumed). In reality, h is unknown in any application, and the impulsive force of a hammer hit is hardly constant, as is supposed by this naive modeling.
The modeling logic often applied for the Dirac delta is that the external force f(t) is used in the model in a limited manner, in which only the momentum p = mv is important. More precisely, only the change in momentum or impulse is important, R b a f(t)dt = ∆p = mv(b) −mv(a).
The precise force f(t) is replaced during the modeling by a simplistic piecewise-defined force that has exactly the same impulse ∆p. The re-placement is justified by arguing that if only the impulse is important, and not the actual details of the force, then both models should give similar results.
Function or Operator? The work of physics Nobel prize winner P.
Dirac (1902–1984) proceeded for about 20 years before the mathematical community developed a sound mathematical theory for his impulsive force representations.
A systematic theory was developed in 1936 by the soviet mathematician S. Sobolev.
The French mathematician L.
Schwartz further developed the theory in 1945. He observed that the idealization is not a function but an operator or linear functional, in particular, δ maps or associates to each function φ(t) its value at t = 0, in short, δ(φ) = φ(0). This fact was observed early on by Dirac and others, during the replacement of simplistic forces by δ. In Laplace theory, there is a natural encounter with the ideas, because L(f(t)) routinely appears on the right of the equation after transformation. This term, in the case 2Momentum is defined to be mass times velocity. If the force f is given by Newton’s law as f(t) = d dt(mv(t)) and v(t) is velocity, then R b a f(t)dt = mv(b) −mv(a) is the net momentum or impulse.
7.5 Heaviside Step and Dirac Delta 269 of an impulsive force f(t) = c(H(t−t0−h)−H(t−t0+h))/(2h), evaluates for t0 > 0 and t0 −h > 0 as follows: L(f(t)) = Z ∞ 0 c 2h(H(t −t0 −h) −H(t −t0 + h))e−stdt = Z t0+h t0−h c 2he−stdt = ce−st0 esh −e−sh 2sh !
The factor esh −e−sh 2sh is approximately 1 for h > 0 small, because of L’Hospital’s rule. The immediate conclusion is that we should replace the impulsive force f by an equivalent one f ∗such that L(f ∗(t)) = ce−st0.
Well, there is no such function f ∗!
The apparent mathematical flaw in this idea was resolved by the work of L. Schwartz on distributions. In short, there is a solid foundation for introducing f ∗, but unfortunately the mathematics involved is not elementary nor especially accessible to those readers whose background is just calculus.
Practising engineers and scientists might be able to ignore the vast lit-erature on distributions, citing the example of physicist P. Dirac, who succeeded in applying impulsive force ideas without the distribution the-ory developed by S. Sobolev and L. Schwartz. This will not be the case for those who wish to read current literature on partial differential equa-tions, because the work on distributions has forever changed the required background for that topic.
270 Laplace Transform 7.6 Laplace Table Derivations Verified here are two Laplace tables, the minimal Laplace Table 7.2-4 and its extension Table 7.2-5. Largely, this section is for reading, as it is designed to enrich lectures and to aid readers who study alone.
Derivation of Laplace integral formulas in Table 7.2-4, page 252.
• Proof of L(tn) = n!/s1+n: The first step is to evaluate L(tn) for n = 0.
L(1) = R ∞ 0 (1)e−stdt Laplace integral of f(t) = 1.
= −(1/s)e−st|t=∞ t=0 Evaluate the integral.
= 1/s Assumed s > 0 to evaluate limt→∞e−st.
The value of L(tn) for n = 1 can be obtained by s-differentiation of the relation L(1) = 1/s, as follows.
d dsL(1) = d ds R ∞ 0 (1)e−stdt Laplace integral for f(t) = 1.
= R ∞ 0 d ds (e−st) dt Used d ds R b a Fdt = R b a dF ds dt.
= R ∞ 0 (−t)e−stdt Calculus rule (eu)′ = u′eu.
= −L(t) Definition of L(t).
Then L(t) = −d dsL(1) Rewrite last display.
= −d ds(1/s) Use L(1) = 1/s.
= 1/s2 Differentiate.
This idea can be repeated to give L(t2) = −d dsL(t) and hence L(t2) = 2/s3.
The pattern is L(tn) = −d dsL(tn−1) which gives L(tn) = n!/s1+n.
• Proof of L(eat) = 1/(s −a): The result follows from L(1) = 1/s, as follows.
L(eat) = R ∞ 0 eate−stdt Direct Laplace transform.
= R ∞ 0 e−(s−a)tdt Use eAeB = eA+B.
= R ∞ 0 e−Stdt Substitute S = s −a.
= 1/S Apply L(1) = 1/s.
= 1/(s −a) Back-substitute S = s −a.
• Proof of L(cos bt) = s/(ss + b2) and L(sin bt) = b/(ss + b2): Use will be made of Euler’s formula eiθ = cos θ +i sin θ, usually first introduced in trigonometry. In this formula, θ is a real number (in radians) and i = √−1 is the complex unit.
7.6 Laplace Table Derivations 271 eibte−st = (cos bt)e−st + i(sin bt)e−st Substitute θ = bt into Euler’s formula and multiply by e−st.
R ∞ 0 e−ibte−stdt = R ∞ 0 (cos bt)e−stdt + i R ∞ 0 (sin bt)e−stdt Integrate t = 0 to t = ∞. Use properties of integrals.
1 s −ib = R ∞ 0 (cos bt)e−stdt + i R ∞ 0 (sin bt)e−stdt Evaluate the left side using L(eat) = 1/(s −a), a = ib.
1 s −ib = L(cos bt) + iL(sin bt) Direct Laplace transform defini-tion.
s + ib s2 + b2 = L(cos bt) + iL(sin bt) Use complex rule 1/z = z/|z|2, z = A + iB, z = A −iB, |z| = √ A2 + B2.
s s2 + b2 = L(cos bt) Extract the real part.
b s2 + b2 = L(sin bt) Extract the imaginary part.
Derivation of Laplace integral formulas in Table 7.2-5, page 252.
• Proof of the Heaviside formula L(H(t −a)) = e−as/s.
L(H(t −a)) = R ∞ 0 H(t −a)e−stdt Direct Laplace transform. Assume a ≥0.
= R ∞ a (1)e−stdt Because H(t −a) = 0 for 0 ≤t < a.
= R ∞ 0 (1)e−s(x+a)dx Change variables t = x + a.
= e−as R ∞ 0 (1)e−sxdx Constant e−as moves outside integral.
= e−as(1/s) Apply L(1) = 1/s.
• Proof of the Dirac delta formula L(δ(t −a)) = e−as.
The definition of the delta function is a formal one, in which every occurrence of δ(t −a)dt under an integrand is replaced by dH(t −a). The differential symbol dH(t −a) is taken in the sense of the Riemann-Stieltjes integral. This integral is defined in [?] for monotonic integrators α(x) as the limit Z b a f(x)dα(x) = lim N→∞ N X n=1 f(xn)(α(xn) −α(xn−1)) where x0 = a, xN = b and x0 < x1 < · · · < xN forms a partition of [a, b] whose mesh approaches zero as N →∞.
The steps in computing the Laplace integral of the delta function appear below.
Admittedly, the proof requires advanced calculus skills and a certain level of mathematical maturity.
The reward is a fuller understanding of the Dirac symbol δ(x).
L(δ(t −a)) = R ∞ 0 e−stδ(t −a)dt Laplace integral, a > 0 assumed.
= R ∞ 0 e−stdH(t −a) Replace δ(t −a)dt by dH(t −a).
= limM→∞ R M 0 e−stdH(t −a) Definition of improper integral.
272 Laplace Transform = e−sa Explained below.
To explain the last step, apply the definition of the Riemann-Stieltjes integral: Z M 0 e−stdH(t −a) = lim N→∞ N−1 X n=0 e−stn(H(tn −a) −H(tn−1 −a)) where 0 = t0 < t1 < · · · < tN = M is a partition of [0, M] whose mesh max1≤n≤N(tn −tn−1) approaches zero as N →∞. Given a partition, if tn−1 < a ≤tn, then H(tn−a)−H(tn−1−a) = 1, otherwise this factor is zero. Therefore, the sum reduces to a single term e−stn. This term approaches e−sa as N →∞, because tn must approach a.
• Proof of L(floor(t/a)) = e−as s(1 −e−as): The library function floor present in computer languages C and Fortran is defined by floor(x) = greatest whole integer ≤x, e.g., floor(5.2) = 5 and floor(−1.9) = −2. The computation of the Laplace integral of floor(t) requires ideas from infinite series, as follows.
F(s) = R ∞ 0 floor(t)e−stdt Laplace integral definition.
= P∞ n=0 R n+1 n (n)e−stdt On n ≤t < n + 1, floor(t) = n.
= P∞ n=0 n s (e−ns −e−ns−s) Evaluate each integral.
= 1 −e−s s P∞ n=0 ne−sn Common factor removed.
= x(1 −x) s P∞ n=0 nxn−1 Define x = e−s.
= x(1 −x) s d dx P∞ n=0 xn Term-by-term differentiation.
= x(1 −x) s d dx 1 1 −x Geometric series sum.
= x s(1 −x) Compute the derivative, simplify.
= e−s s(1 −e−s) Substitute x = e−s.
To evaluate the Laplace integral of floor(t/a), a change of variables is made.
L(floor(t/a)) = R ∞ 0 floor(t/a)e−stdt Laplace integral definition.
= a R ∞ 0 floor(r)e−asrdr Change variables t = ar.
= aF(as) Apply the formula for F(s).
= e−as s(1 −e−as) Simplify.
• Proof of L(sqw(t/a)) = 1 s tanh(as/2): The square wave defined by sqw(x) = (−1)floor(x) is periodic of period 2 and piecewise-defined. Let P = R 2 0 sqw(t)e−stdt.
7.6 Laplace Table Derivations 273 P = R 1 0 sqw(t)e−stdt + R 2 1 sqw(t)e−stdt Apply R b a = R c a + R b c .
= R 1 0 e−stdt − R 2 1 e−stdt Use sqw(x) = 1 on 0 ≤x < 1 and sqw(x) = −1 on 1 ≤x < 2.
= 1 s(1 −e−s) + 1 s(e−2s −e−s) Evaluate each integral.
= 1 s(1 −e−s)2 Collect terms.
An intermediate step is to compute the Laplace integral of sqw(t): L(sqw(t)) = R 2 0 sqw(t)e−stdt 1 −e−2s Periodic function formula, page 275.
= 1 s(1 −e−s)2 1 1 −e−2s .
Use the computation of P above.
= 1 s 1 −e−s 1 + e−s .
Factor 1 −e−2s = (1 −e−s)(1 + e−s).
= 1 s es/2 −e−s/2 es/2 + e−s/2 .
Multiply the fraction by es/2/es/2.
= 1 s sinh(s/2) cosh(s/2).
Use sinh u = (eu −e−u)/2, cosh u = (eu + e−u)/2.
= 1 s tanh(s/2).
Use tanh u = sinh u/ coshu.
To complete the computation of L(sqw(t/a)), a change of variables is made: L(sqw(t/a)) = R ∞ 0 sqw(t/a)e−stdt Direct transform.
= R ∞ 0 sqw(r)e−asr(a)dr Change variables r = t/a.
= a as tanh(as/2) See L(sqw(t)) above.
= 1 s tanh(as/2) • Proof of L(a trw(t/a)) = 1 s2 tanh(as/2): The triangular wave is defined by trw(t) = R t 0 sqw(x)dx.
L(a trw(t/a)) = 1 s(f(0) + L(f ′(t)) Let f(t) = a trw(t/a). Use L(f ′(t)) = sL(f(t)) −f(0), page 251.
= 1 sL(sqw(t/a)) Use f(0) = 0, (a R t/a 0 sqw(x)dx)′ = sqw(t/a).
= 1 s2 tanh(as/2) Table entry for sqw.
• Proof of L(tα) = Γ(1 + α) s1+α : L(tα) = R ∞ 0 tαe−stdt Direct Laplace transform.
= R ∞ 0 (u/s)αe−udu/s Change variables u = st, du = sdt.
274 Laplace Transform = 1 s1+α R ∞ 0 uαe−udu = 1 s1+α Γ(1 + α).
Where Γ(x) = R ∞ 0 ux−1e−udu, by definition.
The generalized factorial function Γ(x) is defined for x > 0 and it agrees with the classical factorial n! = (1)(2) · · · (n) in case x = n + 1 is an integer. In literature, α! means Γ(1 + α). For more details about the Gamma function, see Abramowitz and Stegun [?], or maple documentation.
• Proof of L(t−1/2) = rπ s : L(t−1/2) = Γ(1 + (−1/2)) s1−1/2 Apply the previous formula.
= √π √s Use Γ(1/2) = √π.
7.7 Transform Properties 275 7.7 Transform Properties Collected here are the major theorems and their proofs for the manipu-lation of Laplace transform tables.
Theorem 4 (Linearity) The Laplace transform has these inherited integral properties: (a) L(f(t) + g(t)) = L(f(t)) + L(g(t)), (b) L(cf(t)) = cL(f(t)).
Theorem 5 (The t-Derivative Rule) Let y(t) be continuous, of exponential order and let f ′(t) be piecewise continuous on t ≥0. Then L(y′(t)) exists and L(y′(t)) = sL(y(t)) −y(0).
Theorem 6 (The t-Integral Rule) Let g(t) be of exponential order and continuous for t ≥0. Then L R t 0 g(x) dx = 1 sL(g(t)).
Theorem 7 (The s-Differentiation Rule) Let f(t) be of exponential order. Then L(tf(t)) = −d dsL(f(t)).
Theorem 8 (First Shifting Rule) Let f(t) be of exponential order and −∞< a < ∞. Then L(eatf(t)) = L(f(t))|s→(s−a) .
Theorem 9 (Second Shifting Rule) Let f(t) and g(t) be of exponential order and assume a ≥0. Then (a) L(f(t −a)H(t −a)) = e−asL(f(t)), (b) L(g(t)H(t −a)) = e−asL(g(t + a)).
Theorem 10 (Periodic Function Rule) Let f(t) be of exponential order and satisfy f(t + P) = f(t). Then L(f(t)) = R P 0 f(t)e−stdt 1 −e−P s .
Theorem 11 (Convolution Rule) Let f(t) and g(t) be of exponential order. Then L(f(t))L(g(t)) = L Z t 0 f(x)g(t −x)dx .
276 Laplace Transform Proof of Theorem 4 (linearity): LHS = L(f(t) + g(t)) Left side of the identity in (a).
= R ∞ 0 (f(t) + g(t))e−stdt Direct transform.
= R ∞ 0 f(t)e−stdt + R ∞ 0 g(t)e−stdt Calculus integral rule.
= L(f(t)) + L(g(t)) Equals RHS; identity (a) verified.
LHS = L(cf(t)) Left side of the identity in (b).
= R ∞ 0 cf(t)e−stdt Direct transform.
= c R ∞ 0 f(t)e−stdt Calculus integral rule.
= cL(f(t)) Equals RHS; identity (b) verified.
Proof of Theorem 5 (t-derivative rule): Already L(f(t)) exists, because f is of exponential order and continuous.
On an interval [a, b] where f ′ is continuous, integration by parts using u = e−st, dv = f ′(t)dt gives R b a f ′(t)e−stdt = f(t)e−st|t=b t=a − R b a f(t)(−s)e−stdt = −f(a)e−sa + f(b)e−sb + s R b a f(t)e−stdt.
On any interval [0, N], there are finitely many intervals [a, b] on each of which f ′ is continuous. Add the above equality across these finitely many intervals [a, b]. The boundary values on adjacent intervals match and the integrals add to give Z N 0 f ′(t)e−stdt = −f(0)e0 + f(N)e−sN + s Z N 0 f(t)e−stdt.
Take the limit across this equality as N →∞. Then the right side has limit −f(0) + sL(f(t)), because of the existence of L(f(t)) and limt→∞f(t)e−st = 0 for large s. Therefore, the left side has a limit, and by definition L(f ′(t)) exists and L(f ′(t)) = −f(0) + sL(f(t)).
Proof of Theorem 6 (t-Integral rule): Let f(t) = R t 0 g(x)dx. Then f is of exponential order and continuous. The details: L( R t 0 g(x)dx) = L(f(t)) By definition.
= 1 sL(f ′(t)) Because f(0) = 0 implies L(f ′(t)) = sL(f(t)).
= 1 sL(g(t)) Because f ′ = g by the Fundamental theorem of calculus.
Proof of Theorem 7 (s-differentiation): We prove the equivalent relation L((−t)f(t)) = (d/ds)L(f(t)). If f is of exponential order, then so is (−t)f(t), therefore L((−t)f(t)) exists.
It remains to show the s-derivative exists and satisfies the given equality.
The proof below is based in part upon the calculus inequality e−x + x −1 ≤x2, x ≥0.
(1) 7.7 Transform Properties 277 The inequality is obtained from two applications of the mean value theorem g(b)−g(a) = g′(x)(b−a), which gives e−x+x−1 = xxe−x1 with 0 ≤x1 ≤x ≤x.
In addition, the existence of L(t2|f(t)|) is used to define s0 > 0 such that L(t2|f(t)|) ≤1 for s > s0. This follows from the transform existence theorem for functions of exponential order, where it is shown that the transform has limit zero at s = ∞.
Consider h ̸= 0 and the Newton quotient Q(s, h) = (F(s + h) −F(s))/h for the s-derivative of the Laplace integral. We have to show that lim h→0 |Q(s, h) −L((−t)f(t))| = 0.
This will be accomplished by proving for s > s0 and s + h > s0 the inequality |Q(s, h) −L((−t)f(t))| ≤|h|.
For h ̸= 0, Q(s, h) −L((−t)f(t)) = Z ∞ 0 f(t)e−st−ht −e−st + the−st h dt.
Assume h > 0. Due to the exponential rule eA+B = eAeB, the quotient in the integrand simplifies to give Q(s, h) −L((−t)f(t)) = Z ∞ 0 f(t)e−st e−ht + th −1 h dt.
Inequality (1) applies with x = ht ≥0, giving |Q(s, h) −L((−t)f(t))| ≤|h| Z ∞ 0 t2|f(t)|e−stdt.
The right side is |h|L(t2|f(t)|), which for s > s0 is bounded by |h|, completing the proof for h > 0. If h < 0, then a similar calculation is made to obtain |Q(s, h) −L((−t)f(t))| ≤|h| Z ∞ 0 t2|f(t)e−st−htdt.
The right side is |h|L(t2|f(t)|) evaluated at s + h instead of s. If s + h > s0, then the right side is bounded by |h|, completing the proof for h < 0.
Proof of Theorem 8 (first shifting rule): The left side LHS of the equality can be written because of the exponential rule eAeB = eA+B as LHS = Z ∞ 0 f(t)e−(s−a)tdt.
This integral is L(f(t)) with s replaced by s−a, which is precisely the meaning of the right side RHS of the equality. Therefore, LHS = RHS.
Proof of Theorem 9 (second shifting rule): The details for (a) are LHS = L(H(t −a)f(t −a)) = R ∞ 0 H(t −a)f(t −a)e−stdt Direct transform.
278 Laplace Transform = R ∞ a H(t −a)f(t −a)e−stdt Because a ≥0 and H(x) = 0 for x < 0.
= R ∞ 0 H(x)f(x)e−s(x+a)dx Change variables x = t −a, dx = dt.
= e−sa R ∞ 0 f(x)e−sxdx Use H(x) = 1 for x ≥0.
= e−saL(f(t)) Direct transform.
= RHS Identity (a) verified.
In the details for (b), let f(t) = g(t + a), then LHS = L(H(t −a)g(t)) = L(H(t −a)f(t −a)) Use f(t −a) = g(t −a + a) = g(t).
= e−saL(f(t)) Apply (a).
= e−saL(g(t + a)) Because f(t) = g(t + a).
= RHS Identity (b) verified.
Proof of Theorem 10 (periodic function rule): LHS = L(f(t)) = R ∞ 0 f(t)e−stdt Direct transform.
= P∞ n=0 R nP+P nP f(t)e−stdt Additivity of the integral.
= P∞ n=0 R P 0 f(x + nP)e−sx−nPsdx Change variables t = x + nP.
= P∞ n=0 e−nPs R P 0 f(x)e−sxdx Because f is P-periodic and eAeB = eA+B.
= R P 0 f(x)e−sxdx P∞ n=0 rn Common factor in summation.
Define r = e−Ps.
= R P 0 f(x)e−sxdx 1 1 −r Sum the geometric series.
= R P 0 f(x)e−sxdx 1 −e−Ps Substitute r = e−Ps.
= RHS Periodic function identity verified.
Left unmentioned here is the convergence of the infinite series on line 3 of the proof, which follows from f of exponential order.
Proof of Theorem 11 (convolution rule): The details use Fubini’s in-tegration interchange theorem for a planar unbounded region, and therefore this proof involves advanced calculus methods that may be outside the back-ground of the reader. Modern calculus texts contain a less general version of Fubini’s theorem for finite regions, usually referenced as iterated integrals. The unbounded planar region is written in two ways: D = {(r, t) : t ≤r < ∞, 0 ≤t < ∞}, D = {(r, t) : 0 ≤r < ∞, 0 ≤r ≤t}.
Readers should pause here and verify that D = D.
7.7 Transform Properties 279 The change of variable r = x + t, dr = dx is applied for fixed t ≥0 to obtain the identity e−st R ∞ 0 g(x)e−sxdx = R ∞ 0 g(x)e−sx−stdx = R ∞ t g(r −t)e−rsdr.
(2) The left side of the convolution identity is expanded as follows: LHS = L(f(t))L(g(t)) = R ∞ 0 f(t)e−stdt R ∞ 0 g(x)e−sxdx Direct transform.
= R ∞ 0 f(t) R ∞ t g(r −t)e−rsdrdt Apply identity (2).
= R D f(t)g(r −t)e−rsdrdt Fubini’s theorem applied.
= R D f(t)g(r −t)e−rsdrdt Descriptions D and D are the same.
= R ∞ 0 R r 0 f(t)g(r −t)dte−rsdr Fubini’s theorem applied.
Then RHS = L R t 0 f(u)g(t −u)du = R ∞ 0 R t 0 f(u)g(t −u)due−stdt Direct transform.
= R ∞ 0 R r 0 f(u)g(r −u)due−srdr Change variable names r ↔t.
= R ∞ 0 R r 0 f(t)g(r −t)dt e−srdr Change variable names u ↔t.
= LHS Convolution identity verified.
280 Laplace Transform 7.8 More on the Laplace Transform Model conversion and engineering.
A differential equation model for a physical system can be subjected to the Laplace transform in order to produce an algebraic model in the transform variable s. Lerch’s the-orem says that both models are equivalent, that is, the solution of one model gives the solution to the other model.
In electrical and computer engineering it is commonplace to deal only with the Laplace algebraic model. Engineers are in fact capable of hav-ing hour-long modeling conversations, during which differential equations are never referenced! Terminology for such modeling is necessarily spe-cialized, which gives rise to new contextual meanings to the terms input and output. For example, an RLC-circuit would be discussed with input F(s) = ω s2 + ω2 , and the listener must know that this expression is the Laplace transform of the t-expression sin ωt. Hence the RLC-circuit is driven by a sinu-soindal input of natural frequency ω. During the modeling discourse, it could be that the output is X(s) = 1 s + 1 + 10ω s2 + ω2 .
Lerch’s equivalence says that X(s) is the Laplace transform of e−t + 10 sin ωt, but that is extra work, if all that is needed from the model is a statement about the transient and steady-state responses to the input. |
14441 | https://www.learnmode.net/flip/video/4350 | 國小_數學_3-16-2 面積公式
影片描述
【講師】劉巧如
【講師簡介】
劉巧如，目前任教於彰化市大成國小
【影片簡介】
認識面積的普遍單位(1平方公分、1平方公尺)
認識正方形、長方形的面積公式。
正方形面積=邊長x邊長
長方形面積=長x寬
邊長1公分的正方形，面積是1平方公分。平方公分在國際上通用符號是「cm2」。認識面積的普遍單位(1平方公分、1平方公尺)
認識正方形、長方形的面積公式。
正方形面積=邊長x邊長
長方形面積=長x寬
邊長1公分的正方形，面積是1平方公分。平方公分在國際上通用符號是「cm2」。
相關推薦影片 |
14442 | https://vitrek.com/mti-instruments/knowledge-center/international-system-units-si/?srsltid=AfmBOopy7AqlglXc77nSPlfnrgTeDiFjZvPqqHoBbjHcgTTYQnEE_oxk | International System of Units (SI) - Vitrek
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Search for:
International System of Units (SI)
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International System of Units (SI)MTI Sales2025-05-27T00:53:50-07:00
International System of Units (SI)
The international system of units is the most widely used system of measurement, and is the modern form of the metric system. At its core, this system is built on seven primary units. This system specifies twenty prefixes to the unit symbols and name to state multiples and fractions of each unit.
The international system of units was introduced to the public in 1960 as a result to earlier research that began in 1948. SI is based on the meter-kilogram second system of units (MKS). SI was created to evolve as time progressed forward. It was intended that units and prefixes were to be created and unit definitions to be modified, on an international level, as the technology of the measuring tools we use on a daily basis continues to become more advanced and more precise. An excellent example of this evolving system would be the 24 th and 25 th General Conferences on Weights and Measures (CGPM) that took place in 2011 and 2014 where a proposal was introduced to alter the definition of “kilogram”. This was brought up because some began to believe that a kilogram was an invariant of nature rather than a measurement of mass.
As mentioned in the previous section, there was an incredible lack of coordination amongst the various professions in a number of areas across the world. To create unity, SI was born. The CGPM, established by the Meter Convention of 1875, took on the mission of bringing together numerous international organizations to agree upon the definitions and standards of the proposed new system. They were also brought together to agree upon the rules about how this new system would be presented and accepted around the world.
In theory, SI can be used for any type of physical measurement. Despite this, the fact could not be ignored that non-SI units of measurement would still appear in the technical, scientific and commercial literature for the foreseeable future. Furthermore, certain units of measurement have been engrained so deeply in history and certain cultures that they will continue to be used for a long time to come. The CIPM has categorized these units and incorporated them into the SI Brochure.
Below is the international system of units.
Base Units of the International System of Units (SI)
The SI system is built on seven fundamental units, each representing a different physical quantity. These units are the foundation for all other measurements in the system.
Quantity Unit name Unit symbol Definition
Length Meter m The distance traveled by light in a vacuum in 1/299792458 second.
Weight Kilogram kg This is the unit for weight. The mass of the international prototype kilogram.
Time Second s The duration of 9192631770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium 133 atom.
Current Ampere A The constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 m apart in vacuum, would produce between these conductors a force equal to 2×10-7 newtons per meter of length.
Thermodynamic temperature Kelvin K 1/273.16 of the thermodynamic temperature of the triple point of water.
Substance amount Mole mol The amount of substance of a system that contains as many elementary entities as there are atoms in 0.012 kilogram of carbon 12. (Limited to objects with clarified composition.) Elementary entities are subatomic particles that compose matter and energy.
Luminosity Candela cd The luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540×1012 hertz and that has a radiant intensity in that direction of 1/683 watt per steradian.
Supplementary Units
In addition to the base units, there are two supplementary units used in SI:
Quantity Unit name Unit symbol Definition
Plane angle Radian rad Radian describes the plane angle subtended by an arc of a circle with the same length as the radius of that circle corresponds to an angle of 1 radian.
Solid angle Steradian sr A steradian is a solid angle at the center of a sphere subtending a section on the surface equal in area to the square of the radius of the sphere.
Derived Units
Derived units are a combination of base units and supplementary units and the mathematical symbols of multiplication and division.Here are some examples of SI derived units:
Quantity Unit name Unit symbol
Area Square meter m 2
Volume Cubic meter m 3
Speed Meter per second m/s
Acceleration Meter per second squared m/s 2
Wavenumber Reciprocal meter m-1
Density Kilogram per cubic meter kg/m 3
Current density Ampere per square meter A/m 2
Magnetic field strength Ampere per meter A/m
Concentration (of amount of substance)Mole per cubic meter mol/m 3
Specific volume Cubic meter per kilogram m 3/kg
Luminance Candela per square meter cd/m 2
Common Derived Units with Special Names
Some derived units in the SI system have been given unique names for simplicity:
Quantity Unit name Unit symbol Composition
Frequency Hertz Hz 1Hz=1s-1
Force Newton N 1N=1kg･m/s 2
Pressure, stress Pascal Pa 1Pa=1N/m 2
Energy, work, amount of heat Joule J 1J=1N･m
Power, radiant flux Watt W W=1J/s
Electric charge, amount of electricity Coulomb C 1C=1A･s
Electric potential/electric potential difference, voltage, electromotive force Volt V 1V=1J/C
Resistance (electrical)Ohm Ω 1Ω=1V/A
Conductance (electrical)Siemens S 1S=1Ω-1
Magnetic Weber Wb 1Wb=1V･s
Magnetic flux density, magnetic induction Tesla T 1T=1Wb/m 2
Inductance Henry H 1H=1Wb/A
Celsius temperature Degree Celsius℃1t=T-To
Luminous flux Lumen lm 1lm=1cd･sr
Illuminance Lux lx 1lx=1lm/m 2
SI Unit Prefixes: A System for Scale
SI unit prefixes indicating integer powers of ten.The SI system uses prefixes to represent multiples and fractions of units. These prefixes help in expressing measurements that are either too large or too small to be conveniently written with base units alone.
Factor Prefix Symbol Factor Prefix Symbol
10 18 exa E 10-1 deci d
10 15 peta P 10-2 centi c
10 12 tera T 10-3 milli m
10 9 giga G 10-6 micro µ
10 6 mega M 10-9 nano n
10 3 kilo k 10-12 pico p
10 2 hecto h 10-15 femto f
10 deka da 10-18 atto a
Non-SI units
While the SI system is used internationally, there are still some non-SI units that are commonly used for specific measurements, like time, angle, and volume. These units are often used in everyday situations and scientific research.
Quantity Unit name Unit symbolDefinition
Time Minute min 1min=60s
Hour h 1h=60min
Day d 1d=24h
Plane angle Degree°1°= (π/180) rad
Minute′1′= (1/60) °
Second″1″= (1/60) ′
Volume Liter l, L 1l=1dm 3
Weight Metric ton t 1t=10 3 kg
Why the SI System is Crucial
The SI measurement system provides a universal language of measurement. It enables global consistency across science, industry, commerce, and education. From defining the kilogram using physical constants to standardizing everyday measurements, the International System of Units (SI) ensures precision and clarity in a connected world.
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14443 | https://www.chemteam.info/Mole/CombustionAnalysis2.html | Combustion Analysis: Problems 1 - 10
Go to Ten Examples
Go to a discussion of empirical and molecular formulas.
Return to Mole Table of Contents
Problem #1: 0.487 grams of quinine (molar mass = 324 g/mol) is combusted and found to produce 1.321 g CO2, 0.325 g H2O and 0.0421 g nitrogen. Determine the empirical and molecular formulas.
Problem #2: 95.6 mg of menthol (molar mass = 156 g/mol) are burned in oxygen gas to give 269 mg CO2 and 110 mg H2O. What is menthol's empirical formula?
Problem #3: 0.1005 g of menthol is combusted, producing 0.2829 g of CO2 and 0.1159 g of H2O. What is menthol's empirical formula? (Yes, the answer will be the same as #12.)
Go to answers for 1, 2, and 3
Problem #4: The combustion of 40.10 g of a compound which contains only C, H, Cl and O yields 58.57 g of CO2 and 14.98 g of H2O. Another sample of the compound with a mass of 75.00 g is found to contain 22.06 g of Cl. What is the empirical formula of the compound?
Solution #1:
1) Get grams of each element:
Carbon: (58.57 g) (12.011 / 44.009) = 15.985 g of C in 40.10 g sample
Hydrogen: (14.98 g) (2.016 / 18.015) = 1.6764 g of H in 40.10 g sample
Oxygen: we leave this to later (see below for an interesting solution path that involves determining the mass of oxygen.)
Chlorine: problem gives 22.06 g in 75.00 g sample
2) Let us determine the percent composition:
Carbon: 15.985 g / 40.10 g = 39.86%
Hydrogen: 1.6764 g / 40.10 = 4.18%
Chlorine: 22.06 / 75.00 = 29.41%
Oxygen: 100% − (39.86 + 4.18 +2 9.41) = 26.55%
3) Let us assume 100 g of the compound. In which case, the percentages above become grams. Now, let us determine the moles of each (I'll skip typing the calcs):
C: 3.32 mol
H: 4.147 mol
O: 1.66 mol
Cl: 0.83 mol
4) Divide each by 0.83
C: 4
H: 5
O: 2
Cl: 1
The empirical formula is C4H5ClO2
Solution #2:
1) Determine mass of all four compounds in the 40.10 g sample:
Carbon: 15.985 g
Hydrogen: 1.676 g
Here comes the interesting way that is different from solution #1:
Chlorine: (40.10 g) (22.06 g Cl / 75.00 g sample) = 11.795 g
Oxygen: 40.10 g − 15.985 g − 1.676 g − 11.795 g = 10.644 g
Pretty slick, heh? Notice how the oxygen is determined by subtraction after everything else is calculated. This is a common pattern in combustion analysis.
2) Determine the moles of each (I'll skip typing the calcs):
C = 1.331 mol
H = 1.663 mol
Cl = 0.3327 mol
O = 0.6653 mol
Note that there is no need to assume 100 g of the compound and work from the percent composition.
3) Divide all by the smallest to simplify:
C = 4
H = 5
Cl = 1
O = 2
The empirical formula is C4H5ClO2
Problem #5: The combustion of 1.38 grams of a compound which contains C, H, O and N yields 1.72 grams of CO2 and 1.18 grams of H2O. Another sample of the compound with a mass of 22.34 grams is found to contain 6.75 grams of O. What is the empirical formula of the compound?
1) Calculate grams of C and H:
carbon: (1.72 g) (12.011 / 44.009) = 0.4694 g of C
hydrogen: (1.18 g) (2.016 / 18.015) = 0.13205 g of H
oxygen: see next step
nitrogen: see next step
2) Calculate mass percent of each element:
carbon: 0.4694 g / 1.38 g = 34.01%
hydrogen: 0.13205 g / 1.38 g = 9.57%
oxygen: 6.75 g / 22.34 g = 30.215%
nitrogen: 100 − (34.01 + 9.57 + 30.215) = 26.205%
3) Assume 100 g of compound present. Therefore:
carbon: 34.01 g
hydrogen: 9.57 g
oxygen: 30.215 g
nitrogen: 26.205 g
4) Calculate moles:
carbon: 34.01 g / 12.011 g/mol = 2.832
hydrogen: 9.57 g / 1.008 g/mol = 9.494
oxygen: 30.215 g / 16.00 g/mol = 1.888
nitrogen: 20.205 g / 14.007 g/mol = 1.871
5) Look for smallest, whole-number ratio:
carbon: 2.832 / 1.871 = 1.5 (x 2 = 3)
hydrogen: 9.494 / 1.871 = 5 (x 2 = 10)
oxygen: 1.888 / 1.871 = 1 (x 2 = 1)
nitrogen: 1.871 / 1.871 = 1 (x 2 = 2)
6) Empirical formula:
C3H10N2O2
Problem #6: The combustion of 3.42 g of a compound is known to contain only nitrogen and hydrogen gave 9.82 g of NO2 and 3.85 g of water. Determine the empirical formula of this compound.
Solution:
1) Calculate moles of N and moles of H in the combustion products:
Moles N
9.82 g NO2 / 46.0 g/mol = 0.213 mol NO2
(0.213 mol NO2) (1 mol N / 1 mol NO2) = 0.213 mol N
Moles H
3.85 g H2O / 18.0 g/mol = 0.213 mol H2O
(0.213 mol H2O) (2 mol H / 1 mol H2O ) = 0.428 mol H
2) Calculate the ratio of moles by dividing both by the smaller:
N ---> 0.213 / 0.213 = 1
H ---> 0.428 / 0.213 = 2
The empirical formula is NH2
Problem #7: A compound with a known molecular weight (146.99 g/mol) that contains only C, H, and Cl was studied by combustion analysis. When a 0.367 g sample was combusted, 0.659 g of CO2 and 0.0892 g of H2O formed. What are the empirical and molecular formulas?
Solution:
1) Carbon:
0.659 g of CO2 has 0.659 / 44 = 0.0150 moles of CO2
there is 1 mole of C in CO2 and all the C from the compound becomes CO2, so moles of C in the compound = 0.0150 moles
mass of C = 0.0150 x 12 = 0.1797 g
2) Hydrogen:
0.089 g of H2O has 0.0892 / 18 = 0.0050 moles of H2O
there are 2 moles of H in H2O, so moles of H in the compound = 0.0099 moles
mass of H = 0.0099 x 1.0079 = 0.0100 g
3) Chlorine:
mass of H + C = 0.1897 g
mass of sample = 0.3670 g
mass of Cl by difference = 0.1773 g
moles of Cl = 0.0050 moles
4) Smallest whole-number ratio:
molar ratio of C : H : Cl = 0.0150 : 0.0099 : 0.0050
divide the ratio by the smallest number
molar ratio of C : H : Cl = 3.00 : 1.98 : 1
5) Formulas:
empirical formula is C3H2Cl
this has an "empirical formula weight" of (36+2+35.5) = 73.5 g
which is 1/2 the molecular mass
so the molecular formula is twice the empirical formula
C6H4Cl2
Problem #8: A 2.52 g sample of a compound containing carbon, hydrogen, nitrogen, oxygen, and sulfur was burned in excess oxygen gas to yield 4.36 grams of CO2 and 0.892 grams of H2O as the only carbon and hydrogen products respectively. Another sample of the same compound of mass 4.14 g yielded 2.60 g of SO3 as the only sulfur containing product. A third sample of mass 5.66 g was burned under different conditions to yield 2.80 g of HNO3 as the only nitrogen containing product. Determine the empirical formula of the compound.
Solution #1:
1) Carbon:
(4.36 g) (12.011 g / 44.0 g) = 1.1902 g of C
1.1902 g / 2.52 g = 47.23%
2) Hydrogen:
(0.892 g) (2.016 g / 18.015 g) = 0.09982 g of H
0.09982 g / 2.52 g = 3.96%
3) Sulfur:
2.60 g x (32.065 g / 80.062 g) = 1.0413 g of S
1.0413 g / 4.14 g = 25.15%
4) Nitrogen:
2.80 x (14.007 / 63.012) = 0.6224 g of N
0.6224 g / 5.66 g = 11.00%
5) Oxygen:
100% − (47.23% + 3.96% + 25.15% + 11.00%) = 12.66%
6) Assume 100 g of compound present:
C = 47.23 g; H = 3.96 g; S = 25.15 g; N = 11.00 g; O = 12.66 g
7) Convert to moles:
C = 3.93; H = 3.93; S = 0.7843; N = 0.7853; O = 0.79125
8) Convert to lowest whole-number ratio by dividing each mole amount by 0.7843:
C = 5; H = 5; S = 1; N = 1; O = 1
Empirical formula is C5H5NOS
Repeat Problem #8: A 2.52 g sample of a compound containing carbon, hydrogen, nitrogen, oxygen, and sulfur was burned in excess oxygen gas to yield 4.36 grams of CO2 and 0.892 grams of H2O as the only carbon and hydrogen products respectively. Another sample of the same compound of mass 4.14 g yielded 2.60 g of SO3 as the only sulfur containing product. A third sample of mass 5.66 g was burned under different conditions to yield 2.80 g of HNO3 as the only nitrogen containing product. Determine the empirical formula of the compound.
Solution #2:
1) Carbon:
moles = mass / molar mass
molar mass of CO2 = 44.0 g/ mole
4.36 g of CO2 has 4.36 /44.0 = 0.09909 moles of CO2
there is 1 mole of C in CO2 and all the C from the compound becomes CO2
moles of C in the compound = 0.09909 moles
mass of C ---> 0.09909 x 12 = 1.1891 g
2) Hydrogen:
molar mass of H2O = 18 g/ mole
0.892 g of H2O has 0.892 / 18 = 0.04956 moles of H2O
there are 2 moles of H in H2O so moles of H in the compound = 0.09911 moles
mass of H ---> 0.09911 x 1.0079 = 0.100 g
3) Sulfur:
moles of SO3 ---> 2.60 / 80 = 0.0325
moles of S in 4.14 g of compound = 0.0325 mol
moles in 2.52 g of compound ---> (0.0325) (2.52 / 4.14) = 0.01978 moles
mass of S ---> 0.01978 x 32 = 0.6330 g
All the sulfur in the SO3 came from the 4.14 g sample.
Notice the scaling from 4.14 g of compound to 2.52 g.
4) Nitrogen:
moles of HNO3 ---> 2.80 / 63 = 0.04444 mol
moles of N in 5.66 g of sample = 0.04444
moles in 2.52 g of compound ---> (0.04444) (2.52 / 5.66) = 0.01979 moles
mass of N ---> 0.01979 x 14 = 0.2770 g
All the nitrogen in the HNO3 came from the 5.66 g sample.
Notice the scaling from 5.66 g of compound to 2.52 g.
5) Oxygen:
mass C + H + S + N = 1.1891 + 0.100 + 0.6330 + 0.2770 = 2.1991
mass of O by difference = 2.52 − 2.1991 = 0.3209 g
moles of O in 2.52 g = 0.3209 / 16 = 0.0201 moles
6) We now have all five mole amounts, so do the empirical formula:
molar ratio of C : H : S : N : O = 0.09909 : 0.09911 : 0.01978 : 0.01979 : 0.0201
divide by the smallest number to get whole-number ratio
C : H : S : N : O = 5 : 5 : 1 : 1 : 1
empirical formula is C5H5NOS
Problem #9: Burning 11.2 mL (measured at STP) of a gas known to contain only carbon and hydrogen, we obtained 44.0 mg CO2 and 0.0270 g H2O. Find the molecular formula of the gas.
Solution:
1) Determine mass of carbon and hydrogen:
C: (0.0440 g) (12.011 / 44.01) = 0.0120 g
H: (0.0270 g) (2.016 / 18.015) = 0.0030 g
2) Determine moles of carbon and hydrogen:
C: 0.0120 g / 12.0 g/mol = 0.00100 mol
H: 0.0030 g / 1.008 g/mol = 0.00300 mol
3) Determine lowest whole-number ratio:
C: 0.00100 mol / 0.00100 mol = 1
H: 0.00300 mol / 0.00100 mol = 3
empirical formula = CH3
4) Determine how many moles are in our 11.2 mL of gas:
PV = nRT
(1.00 atm) (0.0112 L) = (n) (0.08206) (273 K)
n = 0.00050 mol
5) The gas sample weighed this:
0.012 g + 0.003 g = 0.015 g
6) Get molecular weight of gas:
0.015 g / 0.00050 mol = 30 g/mol
7) The "empirical formula weight" of CH3 = 15
30 / 15 = 2
The molecular formula is C2H6
Problem #10: The osmotic pressure of a solution containing 2.04 g of an unknown molecular compound dissolved in 175.0 mL of solution at 25.0 °C is 2.13 atm. The combustion of 22.08 g of the unknown compound produced 36.26 g CO2 and 14.85 g H2O
Solution:
1) The osmotic pressure will allow us to calculate the molar mass of the substance:
π = iMRT
2.13 atm = (1) (x / 0.175 L) (0.08206 L atm / mol K) (298 K)
x = 0.015243 mole
2.04 g / 0.015243 mole = 133.83 g/mol
2) Let us calculate the amount of carbon and hydrogen. Then, by subtraction, we will check for oxygen:
carbon ---> 36.26 g x (12.011 / 44.01 ) = 9.8959 g
hydrogen ---> 14.85 g x (2.016 / 18.015 ) = 1.6618 g
oxygen ---> 22.08 minus (9.8959 + 1.6618) = 10.5223 g
3) Calculate moles of each element:
carbon ---> 9.8959 g / 12.011 g/mol = 0.8239 mol
hydrogen ---> 1.6618 g / 1.008 g/mol = 1.6486 mol
oxygen ---> 10.5223 g / 15.999 g/mol = 0.657685 mol
4) Determine a whole number ratio:
carbon ---> 0.8239 / 0.657685 = 1.25
hydrogen ---> 1.6486 / 0.657685 = 2.5
oxygen ---> 0.657685 / 0.657685 = 1
Look at it like this:
carbon ---> 1.25 = 5/4 (times 4 = 5)
hydrogen ---> 2.5 = 10/4 (times 4 = 10)
oxygen ---> 1 = 4/4 (times 4 = 4)
5) The empirical formula is:
C5H10O4
6) The "empirical formula weight" is 134. We calculated a molecular weight of 133.83. The molecular formula is:
C5H10O4
Problem #11: 1.5 L of a gaseous compound consisting of carbon and hydrogen is combusted with oxygen. When the resulting gaseous products are measured at the initial temperature and pressure, it was found that 3.0 L of CO2 and 1.5 L of H2O were formed. What is the formula of the compound?
Solution:
Because everything was measured at equal temperatures and pressures, Avogadro's Hypothesis can be used to solve this problem. The hypothesis is: "Equal volumes at equal temperatures and pressures contain equal number of molecules."
Just to make it a bit simpler to explain, let us assume a value of T and P such that the 3.0 L of each gas contains 3.0 moles. That means 1.5 moles of the compound was initially present.
3.0 moles of CO2 contains 3.0 moles of C and 1.5 moles of H2O contains 3.0 moles of H.
3.0 moles of C and 3.0 moles of H are contained in 1.5 moles of the compound. Therefore, 1.0 mol of the compound contains 2.0 mol of C and 2.0 mol of H.
The formula of the compound is C2H2.
Bonus Problem #1: A 6.20-g sample of an unknown compound containing only C, H, and O combusts in an oxygen rich environment. When the products have cooled to 20.0 °C at 1 bar, there are 8.09 L of CO2 and 3.99 mL of H2O. The density of water at 20.0 °C is 0.998 g/mL.
a) What is the empirical formula of the unknown compound?
b) If the molar mass is 168.2 g/mol, what is the molecular formula of the compound?
Solution:
1) Mass of H2O:
(3.99 mL) (0.998 g/mL) = 3.982 g
2) Mass of CO2:
PV = nRT
(1 bar) (8.09 L) = (n) (0.0831447 L bar / mol K) (293 K)
n = 0.33208276 mol
(0.33208276 mol) (44.009 g/mol) = 14.61463 g
3) We need grams of C and H before getting grams of O:
(14.61463 g) (12.011 g / 44.009 g) = 3.988646 g
(3.982 g) (2.016 g / 18.015 g) = 0.445613 g
4) Mass of O:
6.20 − (3.988646 + 0.445613) = 1.765741 g
5) Determine moles of all three components:
3.988646 g / 12.011 g/mol = 0.332083 mol
0.445613 g / 1.008 g/mol = 0.4420764 mol
1.765741 g / 16.00 g/mol = 0.110359 mol
6) Divide through by smallest:
0.332083 mol / 0.110359 mol = 3
0.4420764 mol / 0.110359 mol = 4
0.110359 mol / 0.110359 mol = 1
Empirical formula is C3H4O
7) The weight of C3H4O is 56.0636
168.2 / 56.0636 = 3
Molecular formula is C9H12O3
Bonus Problem #2: Given the following reaction:
BxHy(s) + O2(g) ---> (x/2)B2O3(s) + (y/2)H2O(g)
If 0.148 g BxHy yields 0.422 g B2O3 when burned in excess O2, what is the empirical formula of BxHy?
Solution:
1) Grams of B in 0.422 g of B2O3:
0.422 g times (21.622 / 69.619) = 0.131 g <--- also grams of B in BxHy
2) Grams of H in 0.148 g of BxHy:
0.148 − 0.131 = 0.017 g
3) Determine moles of each:
moles B in 0.131 g ---> 0.01212963
moles H in 0.017 g ---> 0.01686508
4) Look for smallest whole-number ratio:
B ---> 0.01212963 / 0.01212963 = 1
H ---> 0.01686508 / 0.01212963 = 1.4
Multiply by 5 gives 5 and 7
B5H7
Go to Ten Examples
Go to a discussion of empirical and molecular formulas.
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Coordinates: 32°42′54″N 117°09′45″W / 32.71500°N 117.16250°W / 32.71500; -117.16250
From Simple English Wikipedia, the free encyclopedia
| San Diego | |
--- |
| City | |
| Images from top, left to right: San Diego Skyline, Coronado Bridge, museum in Balboa Park, Serra Museum in Presidio Park and the Old Point Loma lighthouse | |
| Flag Seal | |
| Nickname(s): "America's Finest City", "Birthplace of California", "City in Motion" | |
| Motto: Semper Vigilans (Latin for "Ever Vigilant") | |
| Location of San Diego in San Diego County, California | |
| San Diego Location within California San Diego Location within the United States | |
| Coordinates: 32°42′54″N 117°09′45″W / 32.71500°N 117.16250°W / 32.71500; -117.16250 | |
| Country | United States of America |
| State | California |
| County | San Diego |
| Established | July 16, 1769 |
| Incorporated | March 27, 1850 |
| Named for | Saint Didacus of Alcalá |
| Government | |
| • Type | Strong Mayor |
| • Body | San Diego City Council |
| • Mayor | Todd Gloria (D) |
| • City Attorney | Heather Ferbert (D) |
| • City Council | List • Joe LaCava D-District 1 • Jennifer Campbell D-District 2 • Stephen Whitburn D-District 3 • Henry L. Foster D-District 4 • Marni von Wilpert D-District 5 • Kent Lee D-District 6 • Raul Campillo D-District 7 • Vivian Moreno D-District 8 • Sean Elo-Rivera D-District 9 |
| • State Assembly Members | List • Carl DeMaio R-75th District • Darshana Patel D-76th District • Tasha Boerner D-77th District • Chris Ward D-78th District • LaShae Sharp-Collins D-79th District • David Alvarez D-80th District |
| • State Senators | List • Steve Padilla D-18th District • Catherine Blakespear D-38th District • Akilah Weber D-39th District • Brian Jones R-40th District |
| Area | |
| • Total | 372.42 sq mi (964.57 km2) |
| • Land | 325.88 sq mi (844.04 km2) |
| • Water | 46.54 sq mi (120.53 km2) 12.50% |
| Elevation | 62 ft (19 m) |
| Highest elevation | 1,591 ft (485 m) |
| Lowest elevation | 0 ft (0 m) |
| Population (2020) | |
| • Total | 1,386,932 |
| • Rank | 8th in the United States 2nd in California |
| • Density | 4,255.96/sq mi (1,643.25/km2) |
| • Urban | 3,070,300 (US: 15th) |
| • Urban density | 4,550.5/sq mi (1,756.9/km2) |
| • Metro | 3,298,634 (US: 17th) |
| Demonym | San Diegan |
| Time zone | UTC−08:00 (PST) |
| • Summer (DST) | UTC−07:00 (PDT) |
| ZIP Codes | 92101–92124, 92126–92132, 92134–92140, 92142, 92143, 92145, 92147, 92149–92155, 92158–92161, 92163, 92165–92179, 92182, 92186, 92187, 92190–92199 |
| Area codes | 619/858 |
| FIPS code | 06-66000 |
| GNIS feature IDs | 1661377, 2411782 |
| Website | www.sandiego.gov |
San Diego (/ˌsæn diːˈeɪɡoʊ/) is the second largest city in the U.S. state of California and eighth largest in the United States. It is at the southwest corner of California, as well as the southwest corner of the continental United States. It was founded in 1769 and it is the oldest city in California. The mayor of San Diego as of 2020 is Todd Gloria.
Background
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It has comfortable weather most of the year. There are several military bases in and near San Diego. It has many beaches along the Pacific Ocean. The San Diego Zoo is very famous. The city has more small farms than any other city in the U.S., with about 7,000 farms.
San Diego is home to San Diego State University, the University of California, San Diego, and University of San Diego.
San Diego is on the international Mexico–United States Border.
San Diego is home to the San Diego Padres baseball team.
History
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The San Dieguito complex was started in San Diego 9,000 years ago. The Kumeyaay moved to the area around 1000 C.E. They created villages in what is now Old Town, San Diego.
The first European to visit San Diego was Juan Rodríguez Cabrillo. He claimed the land for the Spanish Empire and called it San Miguel. In November 1602, Sebastián Vizcaíno visited the harbor (Mission Bay and Point Loma). He named the area after Saint Didacus, who was called San Diego de Alcalá.
European colonization in San Diego started in 1769 when four groups of Spaniards came. In May 1769, Gaspar de Portolá started the Presidio of San Diego on a hill near the San Diego River above the Kumeyaay village of Cosoy. Later, Cosoy became part of the European land. This was the first time Europeans came to California. In July 1769, Mission San Diego de Alcalá was started by Franciscan friars under Saint Junípero Serra. Mission San Diego was the southern end of the trail El Camino Real. Both the Presidio and the Mission are National Historic Landmarks.
In 1821, Mexico won its independence from Spain. San Diego became part of the Mexican land of Alta California. In 1846, the United States started a war against Mexico. They wanted to gain the land of Alta California. The Battle of San Pasqual was fought in the San Pasqual Valley, which is now part of the city of San Diego. The Americans lost many people. After the Mexican–American War, Mexico and the United States signed the Treaty of Guadalupe Hidalgo in 1848. The treaty said Mexico had to give the United States the land of Alta California, including San Diego.
California became a U.S. state in 1850. In the same year, San Diego officially became a city. Joshua H. Bean, became the first mayor. In the late 1860s, Alonzo Horton moved to "New Town" (downtown San Diego). People moved there because its location on San Diego Bay was good for shipping products. New Town grew more popular than Old Town. It is important to the government and economy.
In 1916, the neighborhood of Stingaree, which had San Diego's first Chinatown and "Soapbox Row", was destroyed so people could build the Gaslamp Quarter.
In the early 20th century, San Diego hosted the World's Fair twice: the Panama–California Exposition in 1915 and the California Pacific International Exposition in 1935. Both fairs were held in Balboa Park. Many of the Spanish and Baroque-style buildings that were built for those fairs are still in the park There were many foreign animals at the 195 fair. This inspired the creation of San Diego Zoo. During the 1950s, there was a citywide festival called Fiesta del Pacifico. It showed the area's Spanish and Mexican past.
The southern part of the Point Loma peninsula was set aside for the military in 1852. Over the next several decades the Army set up Fort Rosecrans. The city was also an early center for aviation. As early as World War I, San Diego was calling itself "The Air Capital of the West". The city was home to important airplane developers.
During World War II, San Diego became a major hub of military and defense activity, There were many military bases and defense companies. The city's population grew quickly during and after World War II, more than doubling between 1930 (147,995) and 1950 (333,865). During the final months of the war, the Japanese planned to attack San Diego. They were going to attack on September 22, 1945, but it did not happen because Japan surrendered five weeks earlier.
After World War II, the military continued to play a major role in the local economy, but post–Cold War cutbacks negatively affected the defense industry. So, San Diego leaders wanted to focus on science and tourism.
In the early 1960s, Tom Hom became the first Asian American member of the San Diego City Council. Later, Leon Williams became the the first Black member of the city council.
In the early 1980s, Downtown San Diego experienced the opening of Horton Plaza, the revival of the Gaslamp Quarter, and the construction of the San Diego Convention Center. Petco Park, which is home to the San Diego Padres baseball team, opened in 2004. Outside of downtown, San Diego gained northern land and control of the San Ysidro Port of Entry.
As the Cold War ended, the military shrank and so did defense spending. San Diego is important to biotech and is home to telecommunications company Qualcomm. San Diego has many popular tourist attractions, such as the San Diego Zoo, SeaWorld San Diego, and Legoland California in Carlsbad.
Demographics
[change | change source]
2020 census
[change | change source]
In the 2020 census, there were 1,386,932 people, 515,676 households, and 310,098 families living in San Diego. The population density was 4,255.9 people per square mile (1,643.2/km²). There were 548,934 housing units. The breakdown by race was 46.4% White, 17.9% Asian, 5.9% Black, 0.9% Native American, 0.4% Pacific Islander, 14.1% from one other race, and 14.4% from two or more races. Hispanics and Latinos made up 29.7% of the people.
The median (middle) age was 35.8 years. The age breakdown was 18.9% under age 18, 67.1% from 18 to 65, and 14.0% over 65. The gender breakdown was 50.0% male and 50.0% female.
Of the households, 27.9% had children under age 18, 43.4% had a married couple, 7.8% had an unmarried couple, 27.4% had a woman with no partner, 21.5% had a man with no partner, and 27.5% had one person living alone. The average household size was 2.57 people.
As of January 2019[update], San Diego fifth-largest homeless population among major cities in the United States. In both the city and county, 8,102 people experienced homelessness. In the city alone, 4,887 individuals were experiencing homelessness. According to the San Diego Union-Tribune, homelessness is lessening.
Substance abuse statistics in San Diego
[change | change source]
Heroin, fentanyl, alcohol, and prescription drug addiction are common addiction types in San Diego. In 2017, there were 273 unintentional prescription-related deaths in San Diego County and 7,505 opioid-related visits to emergency rooms. From 2016 to 2017 there was a 115% increase of fentanyl-related deaths in San Diego. In 2017, there were 357 alcohol-related deaths in San Diego County. Of those deaths, 127 also involved prescription drugs, 15 involved a combination of heroin and prescription drugs, and involved heroin. 15,952 adults were admitted into drug treatment centers in San Diego in 2017.
References
[change | change source]
↑ "California City Nicknames List". www.seecalifornia.com. Retrieved 2020-12-29.
↑ "California Cities by Incorporation Date". California Association of Local Agency Formation Commissions. Archived from the original (Word) on November 3, 2014. Retrieved August 25, 2014.
↑ "City of San Diego City Charter, Article XV" (PDF). City of San Diego. Retrieved November 5, 2014.
↑ City News Service (November 8, 2024). "Ferbert secures victory over Maienschein in San Diego City Attorney race". ABC 10 News San Diego. Retrieved December 10, 2024.
↑ "City Council Offices". City of San Diego. Retrieved December 10, 2014.
↑ "2020 U.S. Gazetteer Files". United States Census Bureau. Retrieved October 30, 2021.
↑ "City of San Diego". Geographic Names Information System. United States Geological Survey. Retrieved October 16, 2014.
↑ "QuickFacts: San Diego city, California". census.gov. United States Census Bureau. Retrieved January 22, 2023.
↑ "List of 2020 Census Urban Areas". census.gov. United States Census Bureau. Retrieved January 8, 2023.
↑ "2020 Population and Housing State Data". United States Census Bureau. Retrieved 22 August 2021.
↑ "ZIP code(tm) Lookup". United States Postal Service. Retrieved November 19, 2014.
↑ "Discovering San Diego: The Oldest City In California". Cadanews. Retrieved July 14, 2025.
↑ "City of San Diego Official Website". www.sandiego.gov.
↑ "The Surprising Fun Facts About San Diego You Need To Know". California.com. Retrieved May 29, 2025.
↑ "Colleges in San Diego". Appily.com. Retrieved July 14, 2025.
↑ "Trend at San Diego-Tijuana border crossing sees ambulances involved in human smuggling". San Diego Tribune. Retrieved July 14, 2025.
↑ "Inside Pulse | How Did Baseball teams They Get Their Name – Part Six of Six". Archived from the original on 2010-01-09. Retrieved 2010-03-17.
↑ Catalysts to complexity: late Holocene societies of the California coast. Los Angeles: Cotsen Institute of Archaeology at UCLA. 2002. p. 30. ISBN 978-1-938770-67-8. OCLC 745176510.
↑ High, Gary and Jerri-Ann Jacobs High Tech (2007). San Diego Bay: A Story of Exploitation and Restoration. California Sea Grant College Program. ISBN 978-1-888691-17-7. The Kumeyaay could have derived from the San Dieguito or they may have arrived from the desert around 1000 C.E.
↑ Mogilner, Geoffrey. "Cosoy: Birthplace of New California". San Diego History Center | San Diego, CA | Our City, Our Story. Retrieved August 27, 2020.
↑ "Kosa'aay (Cosoy) History". www.cosoy.org. Archived from the original on March 5, 2016. Retrieved August 27, 2020.
↑ "San Diego Historical Society". Sandiegohistory.org. Archived from the original on May 5, 2009. Retrieved March 12, 2011.
↑ Mills, James (October 1967). "San Diego...Where California Began". Journal of San Diego History. 13 (4). Archived from the original on June 14, 2011. Retrieved February 17, 2017.
↑ Pourade, Richard F. 1960. The History of San Diego: The Explorers. Union-Tribune Publishing Company, San Diego.
↑ Ide, Arthur Frederick (Fall 1976). "San Diego: The Saint and the City". Journal of San Diego History. 22 (4).
↑ "Mission San Diego". Mission San Diego. Archived from the original on May 17, 2010. Retrieved July 1, 2010.
↑ "National Park Service, National Historical Landmarks Program: San Diego Presidio". Tps.cr.nps.gov. October 10, 1960. Archived from the original on July 21, 2011. Retrieved May 4, 2011.
↑ Griswold del Castillo, Richard (Winter 2003). "The U.S.-Mexican War in San Diego, 1846–1847". San Diego Historical Society Quarterly.
↑ Engstrand 2005, p. 80 harvnb error: no target: CITEREFEngstrand2005 (help)
↑ "Shady Ladies in the "Stingaree District" When The Red Lights Went Out in San Diego". San Diego History Center. Archived from the original on October 24, 2005. Retrieved March 8, 2011.
↑ "Balboa Park future is full of repair jobs". The San Diego Union-Tribune. March 18, 2015. Archived from the original on March 18, 2015. Retrieved August 7, 2018.
↑ Marjorie Betts Shaw. "The San Diego Zoological Garden: A Foundation to Build on". Journal of San Diego History. 24 (3, Summer 1978). Retrieved May 4, 2011.
↑ "CHAPTER 5: A Fiesta – Re-living the Days of the Dons | San Diego History Center". March 4, 2016. Archived from the original on March 4, 2016. Retrieved August 7, 2018.
↑ "Historic California Posts: Fort Rosecrans". California State Military Museum. Archived from the original on July 14, 2007. Retrieved November 26, 2012.
↑ Gerald A. Shepherd. "When the Lone Eagle returned to San Diego". Journal of San Diego History. 40 (s. 1 and 2, Winter 1992). Retrieved May 4, 2011.
↑ Moffatt, Riley. Population History of Western U.S. Cities & Towns, 1850–1990. Lanham: Scarecrow, 1996, 54.
↑ Naomi Baumslag, Murderous Medicine: Nazi Doctors, Human Experimentation, and Typhus, 2005, p.207
↑ Amy Stewart (April 25, 2011). "Where To Find The World's Most 'Wicked Bugs': Fleas". National Public Radio.
↑ Russell Working (June 5, 2001). "The trial of Unit 731". The Japan Times.
↑ "Milken Institute". Milken Institute. Retrieved July 1, 2010.
↑ "Renowned Visionary Tom Hom Honored At The Gaslamp Quarter's 150th Anniversary Gala!". gaslamp.org. Gaslamp Quarter Association. November 2017. Archived from the original on 2017-11-17. Retrieved 16 May 2018.Chinese America, History and Perspectives. Chinese Historical Society of America. 1998. p. 70. ISBN 9781885864055.
↑ Fox, Maura (March 5, 2025). "Leon Williams, who shaped San Diego and paved way for Black leaders, dies at 102". The San Diego Union-Tribune. Retrieved April 7, 2025.
↑ Erie, Steven P.; Kogan, Vladimir; MacKenzi, Scott A. (May 2010). "Redevelopment, San Diego Style: The Limits of Public–Private Partnerships". Urban Affairs Review. 45 (5): 644–678. doi:10.1177/1078087409359760. ISSN 1078-0874. S2CID 154024558.
↑ "About San Diego, California". www.sandiego.org. Retrieved 2024-12-02.
↑ Jump up to: 45.0 45.1 45.2 "DP1: PROFILE OF GENERAL POPULATION AND HOUSING CHARACTERISTICS". United States Census Bureau. Retrieved July 16, 2024.
↑ "P16: HOUSEHOLD TYPE". United States Census Bureau. Retrieved July 16, 2024.
↑ "2019 AHAR: Part 1 - PIT Estimates of Homelessness in the U.S. - HUD Exchange". www.hudexchange.info. United States Department of Housing and Urban Development. January 2020.
↑ "Homelessness on City of San Diego Streets Drops by 12 Percent in Annual Count". City of San Diego Official Website. April 28, 2020. Archived from the original on November 2, 2020. Retrieved November 3, 2020.
↑ Nelson, Blake (December 12, 2023). "The homeless population downtown just hit a two-year low. The result is a mixed bag". San Diego Union-Tribune. Retrieved December 16, 2023.
↑ "Drug Treatment Plan". Greenhouse Treatment Center. Retrieved 2020-11-19.
↑ "Insurance for Rehab & Payment Options". Greenhouse Treatment Center. Retrieved 2020-11-21.
| v t e City of San Diego | | |
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| By topic | Beaches Climate Companies Culture Economy History Media Parks People Sports | |
| Government | Mayor City Attorney City Council Libraries Port |
| Emergency services | Hospitals Fire Police + Crime |
| Transportation | Airport Metropolitan Transit System + Bus + Trolley |
| Education | List of primary and secondary schools San Diego USD Coronado USD Poway USD Escondido UHSD Grossmont UHSD San Dieguito UHSD Sweetwater UHSD Escondido Union ESD Del Mar Union ESD Lemon Grove ESD National ESD San Pasqual Union ESD San Ysidro ESD Santee SD Solana Beach ESD South Bay Union SD Valley Center-Pauma USD Universities and colleges Minato School (Weekend Japanese School) |
| San Diego County San Diego–Tijuana metropolitan area | | |
| v t e The 100 most populous cities of the United States | |
--- |
| | | | | | | --- --- | | 1. New York, New York 2. Los Angeles, California 3. Chicago, Illinois 4. Houston, Texas 5. Phoenix, Arizona 6. Philadelphia, Pennsylvania 7. San Antonio, Texas 8. San Diego, California 9. Dallas, Texas 10. Jacksonville, Florida 11. Fort Worth, Texas 12. San Jose, California 13. Austin, Texas 14. Charlotte, North Carolina 15. Columbus, Ohio 16. Indianapolis, Indiana 17. San Francisco, California 18. Seattle, Washington 19. Denver, Colorado 20. Oklahoma City, Oklahoma 21. Nashville, Tennessee 22. Washington, D.C. 23. El Paso, Texas 24. Las Vegas, Nevada 25. Boston, Massachusetts | 26. Detroit, Michigan 27. Louisville, Kentucky 28. Portland, Oregon 29. Memphis, Tennessee 30. Baltimore, Maryland 31. Milwaukee, Wisconsin 32. Albuquerque, New Mexico 33. Tucson, Arizona 34. Fresno, California 35. Sacramento, California 36. Atlanta, Georgia 37. Mesa, Arizona 38. Kansas City, Missouri 39. Raleigh, North Carolina 40. Colorado Springs, Colorado 41. Omaha, Nebraska 42. Miami, Florida 43. Virginia Beach, Virginia 44. Long Beach, California 45. Oakland, California 46. Minneapolis, Minnesota 47. Bakersfield, California 48. Tulsa, Oklahoma 49. Tampa, Florida 50. Arlington, Texas | 51. Aurora, Colorado 52. Wichita, Kansas 53. Cleveland, Ohio 54. New Orleans, Louisiana 55. Henderson, Nevada 56. Honolulu, Hawaii 57. Anaheim, California 58. Orlando, Florida 59. Lexington, Kentucky 60. Stockton, California 61. Riverside, California 62. Irvine, California 63. Corpus Christi, Texas 64. Newark, New Jersey 65. Santa Ana, California 66. Cincinnati, Ohio 67. Pittsburgh, Pennsylvania 68. Saint Paul, Minnesota 69. Greensboro, North Carolina 70. Jersey City, New Jersey 71. Durham, North Carolina 72. Lincoln, Nebraska 73. North Las Vegas, Nevada 74. Plano, Texas 75. Anchorage, Alaska | 76. Gilbert, Arizona 77. Madison, Wisconsin 78. Reno, Nevada 79. Chandler, Arizona 80. St. Louis, Missouri 81. Chula Vista, California 82. Buffalo, New York 83. Fort Wayne, Indiana 84. Lubbock, Texas 85. St. Petersburg, Florida 86. Toledo, Ohio 87. Laredo, Texas 88. Port St. Lucie, Florida 89. Glendale, Arizona 90. Irving, Texas 91. Winston-Salem, North Carolina 92. Chesapeake, Virginia 93. Garland, Texas 94. Scottsdale, Arizona 95. Boise, Idaho 96. Hialeah, Florida 97. Frisco, Texas 98. Richmond, Virginia 99. Cape Coral, Florida 100. Norfolk, Virginia | | |
| Cities ranked by United States Census Bureau population estimates for July 1, 2024. | |
| v t e Municipalities and communities of San Diego County, California, United States | | |
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| County seat: San Diego | | |
| Cities | Carlsbad Chula Vista Coronado Del Mar El Cajon Encinitas Escondido Imperial Beach La Mesa Lemon Grove National City Oceanside Poway San Diego San Marcos Santee Solana Beach Vista | |
| CDPs | Alpine Bonita Bonsall Borrego Springs Bostonia Boulevard Campo Camp Pendleton Mainside Camp Pendleton South Casa de Oro-Mount Helix Crest Del Dios Descanso Elfin Forest Eucalyptus Hills Fairbanks Ranch Fallbrook Granite Hills Harbison Canyon Harmony Grove Hidden Meadows Jacumba Hot Springs Jamul Julian Lake San Marcos Lakeside La Presa Mount Laguna Pala Pine Valley Potrero Rainbow Ramona Rancho San Diego Rancho Santa Fe San Diego Country Estates Spring Valley Valley Center Winter Gardens |
| Unincorporated communities | 4S Ranch Ballena Banner Casa de Oro De Luz De Luz Heights Dehesa Dulzura East Otay Mesa Flinn Springs Foster Guatay Jesmond Dene Kentwood-In-The-Pines Lincoln Acres Manzanita Oak Grove Ocotillo Wells Pala Mesa Palomar Mountain Pauma Valley Pine Hills Ranchita Rincon San Luis Rey Santa Ysabel Shelter Valley Tecate Warner Springs Wynola |
| Indian reservations | Barona Reservation Campo Indian Reservation Capitan Grande Reservation Ewiiaapaayp Indian Reservation Inaja and Cosmit Reservation Jamul Indian Village La Jolla Indian Reservation La Posta Reservation Los Coyotes Indian Reservation Manzanita Reservation Mesa Grande Reservation Pala Indian Reservation Pauma and Yuima Indian Reservation Rincon Indian Reservation San Pasqual Reservation Santa Ysabel Reservation Sycuan Band Reservation Viejas Reservation |
| Ghost towns | Banner City Branson City Coleman City Cuyamaca City Eastwood Jofegan Palm Spring Station Panhe San Felipe Station San Pasqual Stonewall Stratton Vallecito |
| v t e California county seats | |
--- |
| Consolidated city-county | San Francisco |
| Municipalities | Alturas Auburn Bakersfield Colusa Crescent City El Centro Eureka Fairfield Fresno Hanford Hollister Jackson Lakeport Los Angeles Madera Martinez Marysville Merced Modesto Napa Nevada City Oakland Oroville Placerville Red Bluff Redding Redwood City Riverside Sacramento Salinas San Bernardino San Diego San Jose San Luis Obispo San Rafael Santa Ana Santa Barbara Santa Cruz Santa Rosa Sonora Stockton Susanville Ukiah Ventura Visalia Willows Woodland Yreka Yuba City |
| CDPs | Bridgeport Downieville Independence Mariposa Markleeville Quincy San Andreas Weaverville |
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Hexagons in a Square
In the interior of a square of side-length 3 there are several regular hexagons whose sum of perimeters is equal to 42 (the hexagons may overlap). Prove that there are two perpendicular lines such that each one of them intersects at least five of the hexagons.
(Math Magazine, VOL. 84, NO. 2, APRIL 2011, problem 1842. Proposed by Bianca-Teodora Iordache, student, National College "Carol I," Craiova, Romania. Solution by CMC 328, Carleton College, Northfield, MN.)
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Copyright © 1996-2018 Alexander Bogomolny
In the interior of a square of side-length 3 there are several regular hexagons whose sum of perimeters is equal to 42 (the hexagons may overlap). Prove that there are two perpendicular lines such that each one of them intersects at least five of the hexagons.
Solution
Firrst observe that when we project a regular hexagon of side length a onto a line its shortest possible projection is a√3. To see that, note that the diameter of a circle inscribed into a regular hexagon of side a is a√3. In addition, the projection of a hexagon is not shorter than that of its inscribed circle.
Project all the hexagons onto one of the sides of the square. Since the sum of all the hexagons' perimeters is 42, the sum of all of their side-lengths is 7. Hence, their projection length on one edge of the square is at least 7√3 ≈ 12.124. Since all of these projections are onto a segment of length 3, and 7√3/4 > 3, there must be some region in the segment covered by at least five of the projections. Pick a point in this region and draw a line through this point perpendicular to the edge; this line must intersect at least five hexagons. By carrying out this construction for two perpendicular edges of the square, we get the desired two perpendicular lines.
Related material
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200 points have been cosen on a circle, all with integer number of degrees. Prove that the points there are at least one pair of antipodes, i.e., the points 180° apart.
If each point of the plane is colored red or blue then there are two points of the same color at distance 1 from each other.
The integers 1, 2, ..., 10 are written on a circle, in any order. Show that there are 3 adjacent numbers whose sum is 17 or greater.
Given a planar set of 25 points such that among any three of them there exists a pair at the distance less than 1. Prove that there exists a circle of radius 1 that contains at least 13 of the given points.
Prove that among any five points selected inside an equilateral triangle with side equal to 1, there always exists a pair at the distance not greater than .5.
Let A be any set of 19 distinct integers chosen from the arithmetic progression 1, 4, 7,..., 100. Prove that there must be two distinct integers in A whose sum is 104.
Prove that in any set of 51 points inside a unit square, there are always three points that can be covered by a circle of radius 1/7.
Five points are chosen at the nodes of a square lattice (grid). Why is it certain that at least one mid-point of a line joining a pair of chosen points, is also a lattice point?
Prove that there exist two powers of 3 whose difference is divisible by 1997.
If 9 people are seated in a row of 12 chairs, then some consecutive set of 3 chairs are filled with people.
Given any sequence of n integers, positive or negative, not necessarily all different, some consecutive subsequence has the property that the sum of the members of the subsequence is a multiple of n.
In every polyhedron there is at least one pair of faces with the same number of sides.
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Base 3 to Base 10 Conversion Table
Home
Tables
Base Conversion
Base 3 to Base 10
Base 3 to Base 10 Conversion Table
| Base 10 | Base 3 | Base 10 |
---
| 1 | 1 | 1 |
| 2 | 2 | 2 |
| 3 | 10 | 3 |
| 4 | 11 | 4 |
| 5 | 12 | 5 |
| 6 | 20 | 6 |
| 7 | 21 | 7 |
| 8 | 22 | 8 |
| 9 | 100 | 9 |
| 10 | 101 | 10 |
| 11 | 102 | 11 |
| 12 | 110 | 12 |
| 13 | 111 | 13 |
| 14 | 112 | 14 |
| 15 | 120 | 15 |
| 16 | 121 | 16 |
| 17 | 122 | 17 |
| 18 | 200 | 18 |
| 19 | 201 | 19 |
| 20 | 202 | 20 |
| 21 | 210 | 21 |
| 22 | 211 | 22 |
| 23 | 212 | 23 |
| 24 | 220 | 24 |
| 25 | 221 | 25 |
| 26 | 222 | 26 |
| 27 | 1000 | 27 |
| 28 | 1001 | 28 |
| 29 | 1002 | 29 |
| 30 | 1010 | 30 |
| 31 | 1011 | 31 |
| 32 | 1012 | 32 |
| 33 | 1020 | 33 |
| 34 | 1021 | 34 |
| 35 | 1022 | 35 |
| 36 | 1100 | 36 |
| 37 | 1101 | 37 |
| 38 | 1102 | 38 |
| 39 | 1110 | 39 |
| 40 | 1111 | 40 |
| 41 | 1112 | 41 |
| 42 | 1120 | 42 |
| 43 | 1121 | 43 |
| 44 | 1122 | 44 |
| 45 | 1200 | 45 |
| 46 | 1201 | 46 |
| 47 | 1202 | 47 |
| 48 | 1210 | 48 |
| 49 | 1211 | 49 |
| 50 | 1212 | 50 |
| 51 | 1220 | 51 |
| 52 | 1221 | 52 |
| 53 | 1222 | 53 |
| 54 | 2000 | 54 |
| 55 | 2001 | 55 |
| 56 | 2002 | 56 |
| 57 | 2010 | 57 |
| 58 | 2011 | 58 |
| 59 | 2012 | 59 |
| 60 | 2020 | 60 |
| 61 | 2021 | 61 |
| 62 | 2022 | 62 |
| 63 | 2100 | 63 |
| 64 | 2101 | 64 |
| 10 | 101 | 10 |
| 20 | 202 | 20 |
| 30 | 1010 | 30 |
| 40 | 1111 | 40 |
| 50 | 1212 | 50 |
| 60 | 2020 | 60 |
| 70 | 2121 | 70 |
| 80 | 2222 | 80 |
| 90 | 10100 | 90 |
| 100 | 10201 | 100 |
| 100 | 10201 | 100 |
| 200 | 21102 | 200 |
| 300 | 102010 | 300 |
| 400 | 112211 | 400 |
| 500 | 200112 | 500 |
| 600 | 211020 | 600 |
| 700 | 221221 | 700 |
| 800 | 1002122 | 800 |
| 900 | 1020100 | 900 |
| 1000 | 1101001 | 1000 |
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14448 | https://www.teacherspayteachers.com/Product/Probability-Compound-Events-Sample-Space-Tree-Diagrams-Guided-Notes-w-Doodles-11027600 | Probability Compound Events Sample Space & Tree Diagrams Guided Notes w/ Doodles
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Teach 7th-grade students how to find probabilities of compound events with sample space, lists, tree diagrams, and fundamental counting principle using this Probability Guided Notes & Doodles resource. Students create list of possible outcomes for events and then use it to solve probability problems. It contains 10 pages total, including guided notes, a practice coloring worksheet, a maze, and a real-life math application. It works well as graphic organizers, scaffolded notes, and interactive notebooks. And it's artsy—if your students love color by number, color by code, or sketch notes, they'll love these lessons.
CCSS Standards: 7.SP.C.8, 7.SP.C.8a, 7.SP.C.8b, 7.SP.C.8c
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Looking for a digital resource instead?
Grab the Probability of Simple and Compound Events Pixel Art Activities instead! Includes fun projects, digital pixel art, and more!
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Thank you so much for your support, Joy! I'm so happy that your students enjoyed this probability of compound events guided notes. I really appreciate you taking the time to write this feedback!
Thanks so much for sharing this! I'm really glad to hear this probability of compound events guided notes were helpful during your probability unit. :)
So glad to hear this probability of compound events guided notes was a success with your students, Amy! Thank you for your thoughtful review. Have a wonderful school year!
I’m overjoyed to know your students had fun and were engaged with the probability of compound events, Tamberly. Thanks for taking the time to leave feedback. Wishing you a fantastic school year ahead!
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14450 | https://www.reddit.com/r/askmath/comments/1ei126s/prove_that_a_composition_of_two_injective/ | Prove that a composition of two injective functions is also injective. : r/askmath
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•1 yr. ago
Clorxo
Prove that a composition of two injective functions is also injective.
Functions
Hi all, I am wondering how to best approach this proof. It intuitively makes sense in my head. I have provided my partial solution that deals with f o g as g o f should be a very similar proof. I was perhaps wondering if it would be more effective to use proof by contradiction instead?
Thanks in advance for your help!
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Understanding limits using intuitive examples
Limits are a fundamental concept in calculus that can be challenging to grasp initially. However, understanding them through intuitive examples and analogies can make the concept more accessible. Here are some explanations and examples from Redditors that can help you understand limits better:
Basic Intuition
Approaching a Value: Limits describe the behavior of a function as it approaches a specific value. For example, if you have a function f(x) = x^2, the limit as x approaches 2 would be 4, because as x gets closer to 2, f(x) gets closer to 4. "Limits are a tool to understand what happens when a function approaches any arbitrary point."
Real-Life Analogies
Cake Division: Imagine dividing a cake among three people. You cut it into four slices, give three to each person, and then cut the remaining slice into four again. This process can be repeated infinitely, and the amount of cake each person receives approaches but never exactly reaches one-third. "For example: lim(x->2) x 2 = 4. As x approaches x=x (in this case, 2), you're limit is 4."
Speedometer: A car's speedometer is a practical example of limits. It shows the instantaneous speed, which is calculated by taking the limit of the average speed over an infinitesimally small time interval. "Your car speedometer is a calculus machine. A derivative machine to be exact."
Formal Definitions
Epsilon-Delta Definition: The formal definition of a limit uses epsilon (ε) to define how close the function must be to the limit value and delta (δ) to define how close the input must be to the value it is approaching. "The definition means that f(x) can be made arbitrarily close to L as x is made close to (but not equal to) a."
Visual Explanation: Imagine drawing a box around a point on a graph. The limit exists if you can make the box as small as you want and still find a region around the x-value such that all the function values within that region are inside the box. "If you graph the function f, and someone draws a box centred at the point (a, L) of height epsilon, then you must be able to change the width of the box (delta) in such a way that every point that could possibly be in the box, is in the box."
Practical Uses
Derivatives and Integrals: Limits are used to define derivatives, which describe the instantaneous rate of change of a function, and integrals, which calculate the area under a curve. "Derivatives come from limits. Measuring anything’s rate change involves taking the limit of the difference quotient as the difference goes to zero."
Statistics: Limits are used to define probabilities and sampling distributions. "One definition of the probability that result A will happen: the limit of the number of times A happens, divided by the number of trials, as the number of trials approaches infinity."
Subreddits for Further Help
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14451 | https://www.youtube.com/watch?v=usIjbQqp-NI | Design Lec30: Fatigue Failure Criterions (Goodman, Gerber, Soderberg, ASME-Elliptic) (Fall 2021)
Pranav Bhounsule
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Posted: 13 Nov 2021
ME 370 UIC
Transcript:
a model we had and that's the goodman line which we said is a predictor of failure so as long as the points are outside the line the so the line says that all failures will be on that line but you can see that some points are above the line which is okay because you know this part is the failure right so oh sorry infinite life my bad infinite life and this part is the finite life now if you look at these points then you'll see that for this points which are away from the red line it's just predicting that these points are actually the finite life so these are actually the finite life but they are in the infinite life region which means that model doesn't predict does not do a good job of predicting these points okay and same is also true for these points so clearly goodman line is not the best way of explaining the data these points also there's a problem that it is not on the line so why do we use the goodman line we use the good one liner or equation because it's simple it is simple it is uh conservative for tension what that means is that for these points this line could have been drawn here right but then it's drawn on the left side which means that it sort of predicts that all these points above the line are going to have finite life and so that basically leads to a conservation in terms of design it is unfortunately non-conservative for compression and that is because you can see there are a lot of points which are all over the place so a better way to draw this line to make it conservative would be to put the line right here so if you made it put it not here but just below this so then it would be very conservative that it will predict uh the data for all the points and not just the points which are above okay so so there are issues with the goodman line and so what we're going to do is in this section we're going to see other this section is other [Music] that predict infinite slash finite life but only in the tension region okay so only in the tension region the reason we are not going to the compression region is most of the things in machine design and mechanical engineering tend to fall in the tension region and not so much in compression it's more of a civil engineering thing so i believe the book has more things details on compression but we only look at the explanations other explanations of these points only in the tension region okay so i'm going to put forth another figure from the book which start off x gives the different models for different models to predict failure okay let me just get it missing okay so you see a series of lines there a lot of lines okay these lines are different models which are used to explain the infinite fatigue line in tension region okay so we'll discuss some of these and what is shown here is first of all our reference lines we've already seen the language line we've seen the goodman line which connects sc to sut the the lines which we'll consider today in this section are the gerber line the asme elliptic line the solderberg line and that's it yeah just this three okay and then you know you can look at the model line and the walker line we'll ignore this because it's based on stress factor which is a different way of modeling the fatigue which we did not do it is the straight the stress the strain method which we did not quite discuss in this course okay so we'll not do this we did the empirical questions which is stress based uh and then swt walker is just i just ignore that because it doesn't look like it predicts it okay so let's uh review all these criteria for predicting infinite life and i'll just review goodman again i'm just going to put the equations down so it's all handy in one place so the goodman line as you see connects se to sut we said it's a simple and a conservative line and so for tension and it's not so good for compression okay so the equation of line which is the failure criterion is that sigma a which is now on the y-axis divided by the intercept which is sc plus sigma m and the intercept on the x axis which is s u t equals 1 so that's how we can model the line itself okay however what we want to do is we want to use this as a design criterion so what we in case of design criterion we want to find a factor of safety so that the line actually is over here and so then there's this margin which allows us to keep away from the actual fatigue line so that's so we need to replace essentially [Music] so factor of safety is greater than one so we replace sigma a with n times sigma a and this we'll call this nf for the fatigue failure so ns and then we'll replace sigma m as nf sigma m that just serves to bring the line down and that way we are on this line that's the design line and it's away from the actual failure line so replace that in that equation above and find a equation for nf which is our design factor for fatigue so nf comes out to be sigma a divided by sc plus sigma m divided by s u t the whole inverse and by the way that's not new we did this last time so nf is the safety factor for fatigue loads next we'll get to the gerber line and if you look at the gerber line it it joins the se to sut the same two points but it actually replaces with the parabola okay it doesn't look like a parabola there right it's difficult to see that but it's actually a parabolic equation so let's look at the gerber line next so what we have on the figure was first let's draw the two lines which you've seen in this course so one is the goodman line and the other one is the langer line so kerber okay so that's our lang langer line and then this is the goodman line s y s y s e s u okay so now we just need to draw a parabola from starting at s y and connect sorry starting at s we're starting at fc and joining scud you should dip like this so that's the curve per line so what you need to do is write the equation of a parabola and then put the conditions that it passes through s e and s u t and the equations you get is the equation for the gerber line now before i write the equation there are a few things here it's a parabolic equation as i said it is it's non conservative near the y-axis so it's a good predictor near the y-axis and it's only good for tensile stresses so when you say non-conservative it really is a is something bad right it's non-conservative means that it doesn't really fit the data well however it's very good as you go further away from the y-axis so it gets better on the closer to the away from the y-axis yes okay and sort of it's it's it's nice to have a parabolic question because if you go back to the plot i showed you earlier you see that these points are all stream over the place so if you are able to draw a parabola like this that's not right so it should be something like this then it becomes less conservative than the goodman and that's why there's a there's a new line okay so let's write the equation for it now so the failure criterion is the equation of the line it's sigma a divided by se plus sigma m divided by s u t square equals one now how would you use it for design well you'll find a factor of safety such that the actual line the curve is something like this okay so that factor of safety will ensure that the line is not along the purple line but it's parallel to it so that constitutes our design equation which is the same as before we take sigma a replace it with nf sigma a and take sigma m replace it with nf sigma m and then solve for nf and the expression is so when you put that in the equation you'll see that nf becomes a quadratic equation and to solve for the roots of the qatari equation you have to use that formula for solution of a quadratic machine you get two roots one of the root is negative and the other root is positive factor of safety cannot be negative so we stick with the positive root so the positive root is nf equals off you know some space for this sud square sigma a divided by sc and then this is the part in the the discriminant so it's 1 plus 2 sigma m se divided by sigma a s u t this should be square that's the formula okay sometimes it's easier to not use this complicated formula but we just put this values in the equation and just solve numerically for nf okay next next right here called the solar book line it's kind of interesting idea here if you look at this curve which i showed you uh there is the goodman line and let me just go a step back into our previous chapter so we had this goodman line and then we had the yield line right and so what happened was in this region there was some problems that we had to we saw that the yield line was below the goodman line okay and so we need to make this check with two criteria we need to check ny and nf and so what the shutterbug line does is actually connects se to sy and when it does so it uses that so that is the fatigue life criterion and which means that you no longer need to consider the line turning s y to s y that is the line deadline you only base it on this line okay and you can see that it's it's kind of very conservative in the sense that it misses on this region okay to say that this region is in the finite life region even though it's actually corresponds to infinite equation but it makes equations even more simpler than the using the langer and the goodman line so that's the idea behind the solderberg one so next one third one is solder book okay so again we have the anger goodman so the solderback line joins se with sy so it is simple clearly no need to check yielding okay because of the reason i explained just a while back and it is uh highly conservative so ultra conservative is the most conservative line in this whole set of lines we're going to talk today if you want to find the failure criterion it's just a matter of writing the equation of the line so as you see it passes through the two points sc and sy so the equation for line which passes through sc and sy is sc for the sigma well this i should write s sigma a sigma m so it passes through sc so that's the intercept and sigma m as the intercept s y equals one to make the to use this as a design equation what we do is we substitute sigma a as nf sigma a sigma m equals nf sigma m and solve for and if you do that nf comes out to be sigma a divided by se plus sigma m divided by s y raised to minus one okay so our last one and then we are done so as i said these uh if you look at the data again you see that it seems like it's advantageous to have like a curve right these points look like they live on a startup for like a curve but not a straight line right so it might be best to use a curve to describe those points but the same time if you look at the fact that you need to consider the yield line as well as the the goodman line that is this ambiguity in this region if you want to escape that what you can do is you can connect sc to sy not using a straight line but using a curve so if you if you do a curve it does both things which is it accounts for the fact that that points lie on a curve and also the fact that you do not really need to work with the yield line so the asme elliptic line is just that it is a curve connecting sc sy so the last curve is we're gonna talk is the asme elliptic line let's do all the lines on this plot now s y s y c s u d so that's the sort of bug line then we have the curved line again the final line which is the one joining this point to sew so those are all the lines this is of course the liquid one anchor so the advantage of this line is that for this curve is that it is considered yield as well as fatigue okay it is not as conservative as starter box it does the job of sort of work but it does so without being too conservative okay so then how do you find the equation of this line well you know it passes through sc and sy and we're going to use an ellipse elliptic equation so equation of an ellipse and then the equation is of this form so the failure criteria is sigma a divided by s e squared plus sigma m divided by s y square equals one if you want to use it for design the same thing we had before sigma a is replaced with nf sigma a sigma m is replaced with nf sigma m and then we solve for solve for nf you do that you get nf equals sigma a divided by sc plus sigma m divided by s phi square and the whole thing has to be raised to negative half so that's it so we have four criteria for design and the use of this will depend on how conservative you want to be with respect to your design so when usually the way this works is the bigger the factor of safety uh the heavier is going your design is going to be because you need more material and it's probably won't fail right and so you want to and sometimes having a heavier heavier design is not good especially when you want to design things which move if it's a stationary machinery it might just be okay but also the cost increases as you make things heavier so there are decisions to be made anyway so let's see how we can apply all these things into one problem so here is a problem which sort of helps you to understand how these equations are used so steel bar undergoes cyclic loading uh can we just use the line that is this conservative all the time yeah so the the problem with using the least conservative line is that the more conservative you get the higher the design the factor of safety and higher factor of safety means that your design is going to be expensive it's going to be heavier and you don't want those things especially for certain applications like i don't know for aerospace applications you want to design factor to be as low as possible so it just depends on what are your constraints as a designer okay a steel bar undergoes cyclic loading with nominal stress at the notch given as sigma max equals 60 k psi and sigma min equals minus 20 k psi so sigma max is 60 sigma min is minus uh stress concentration factor kf equals 1.2 the ultimate strength is 100 k psi yield strength is 85 k psi and fully corrected so this is not a c prime but sc it accounts for all the imperfections and then estimate the infinite life factor of safety which is essentially saying find nf using the goodman line gerber line solderable line in asme electric line okay so if you look at those equations all those equations involve so we need s sigma a sigma m s u t s y s e and these are the things we need in order to find the factor of uh factor of safety for in fatigue life right and so we for we know all these well the last three things we do not know sigma a sigma m and there's another thing which you need to account for which is the fact that there is a stress concentration the geometry of the stress concentration due to which there is a factor of the stress concentration due to fatigue and we need to account for that when we compute sigma a and sigma n so let's first find sigma a sigma m when there is no stress concentration right so the nominal value so i'm going to denote that as sigma a nominal and that is by a formula sigma max minus sigma min divided by two that's just the definition of sigma i so 60 minus sorry minus 28 divided by 2 so that's 40 then sigma terminal is sigma max plus sigma min divided by 2 that's 60 minus 60 plus minus 20 divided by 2 and that's 20 okay and so if that's the nominal stress if you want to find the maximum stress it's simply the factor of fatigue failure times sigma a nominal equals 1.2 times 40 which is 48. sigma m equals k f sigma m nominal which is 1.2 times 20 which is 24 and this is k psi aps okay so we have sigma sigma m now just a matter of using the equations to find the factor of safety so let's use the goodman line goodman equation so that says that that nf equals sigma a divided by fc plus sigma m divided by s u t negative 1 equals sigma a is 48 s e is see s se is 40 40 sigma m is 24 s ut i think is 100 comes out to be 0.69 the kerber equation just started on a new page uh okay okay so let's submit s ut is 100 sigma m was 24 sigma was 48 sc was 40 plus one plus 2 times 24 times 20 divided by 48 times 100 okay so doing all the math get zero points yes yeah so uh it's right in my notes it's point eight believe is the answer okay the next kind of evc let's do the solder book and that's simply nf equals sigma a divided by sc sigma m divided by s y and this one equals 48 40 plus 24 84. 85 inverse 0.69 okay the last one asme elliptic okay so let's summarize what we have good one solder work 0.69 0.8 point six seven point eight one this for this is the most and this is the least expensive |
14452 | https://www.mathdoubts.com/tan-45-degrees/ | $\tan{45^\circ}$ value
Exact value
$\tan{(45^\circ)} \,=\, 1$
Introduction
The value of tangent in a forty five degrees right triangle is called the tan of angle forty five degrees. The angle $45$ degrees is written $45^\circ$ in Sexagesimal system. Therefore, the tan of angle forty five degrees is written as $\tan{45^\circ}$ in trigonometry, and it is time to learn what the tan $45$ degrees value is.
$\tan{(45^\circ)} \,=\, 1$
The exact value of tan of $45$ degrees is equal to $1$, which is an integer. So, no need to write it in either fraction form or decimal form.
Other forms
The tan of $45$ degrees is alternatively written in different forms in trigonometry.
Circular system
According to the circular system, the tan of $45$ degrees is written as tan of pi divided by four radians. Hence, the tan of $\pi$ by $4$ radians value is exactly equal to one.
$\tan{\Big(\dfrac{\pi}{4}\Big)} \,=\, 1$
Centesimal system
In the same way, the tan of angle $45$ degrees is also written as tan of fifty grades in the centesimal system. So, the exact value of tan of fifty gradians is equal to one.
$\tan{(50^g)} \,=\, 1$
Proof
Learn how to find that the tan of angle forty five degrees is exactly equal to one by constructing a right angled triangle with an angle of $45$ degrees.
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14453 | https://www.khanacademy.org/computing/computer-science/cryptography/modarithmetic/a/the-euclidean-algorithm | The Euclidean Algorithm (article) | Khan Academy
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Computer science theory
Course: Computer science theory>Unit 2
Lesson 5: Modular arithmetic
What is modular arithmetic?
Modulo operator
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The Euclidean Algorithm
Google Classroom
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Recall that the Greatest Common Divisor (GCD) of two integers A and B is the largest integer that divides both A and B.
The Euclidean Algorithm is a technique for quickly finding the GCD of two integers.
The Algorithm
The Euclidean Algorithm for finding GCD(A,B) is as follows:
If A = 0 then GCD(A,B)=B, since the GCD(0,B)=B, and we can stop.
If B = 0 then GCD(A,B)=A, since the GCD(A,0)=A, and we can stop.
Write A in quotient remainder form (A = B⋅Q + R)
Find GCD(B,R) using the Euclidean Algorithm since GCD(A,B) = GCD(B,R)
Example:
Find the GCD of 270 and 192
A=270, B=192
A ≠0
B ≠0
Use long division to find that 270/192 = 1 with a remainder of 78. We can write this as: 270 = 192 1 +78
Find GCD(192,78), since GCD(270,192)=GCD(192,78)
A=192, B=78
A ≠0
B ≠0
Use long division to find that 192/78 = 2 with a remainder of 36. We can write this as:
192 = 78 2 + 36
Find GCD(78,36), since GCD(192,78)=GCD(78,36)
A=78, B=36
A ≠0
B ≠0
Use long division to find that 78/36 = 2 with a remainder of 6. We can write this as:
78 = 36 2 + 6
Find GCD(36,6), since GCD(78,36)=GCD(36,6)
A=36, B=6
A ≠0
B ≠0
Use long division to find that 36/6 = 6 with a remainder of 0. We can write this as:
36 = 6 6 + 0
Find GCD(6,0), since GCD(36,6)=GCD(6,0)
A=6, B=0
A ≠0
B =0, GCD(6,0)=6
So we have shown:
GCD(270,192) = GCD(192,78) = GCD(78,36) = GCD(36,6) = GCD(6,0) = 6
GCD(270,192) = 6
Understanding the Euclidean Algorithm
If we examine the Euclidean Algorithm we can see that it makes use of the following properties:
GCD(A,0) = A
GCD(0,B) = B
If A = B⋅Q + R and B≠0 then GCD(A,B) = GCD(B,R) where Q is an integer, R is an integer between 0 and B-1
The first two properties let us find the GCD if either number is 0. The third property lets us take a larger, more difficult to solve problem, and reduce it to a smaller, easier to solve problem.
The Euclidean Algorithm makes use of these properties by rapidly reducing the problem into easier and easier problems, using the third property, until it is easily solved by using one of the first two properties.
We can understand why these properties work by proving them.
We can prove that GCD(A,0)=A is as follows:
The largest integer that can evenly divide A is A.
All integers evenly divide 0, since for any integer, C, we can write C ⋅ 0 = 0. So we can conclude that A must evenly divide 0.
The greatest number that divides both A and 0 is A.
The proof for GCD(0,B)=B is similar. (Same proof, but we replace A with B).
To prove that GCD(A,B)=GCD(B,R) we first need to show that GCD(A,B)=GCD(B,A-B).
Suppose we have three integers A,B and C such that A-B=C.
Proof that the GCD(A,B) evenly divides C
The GCD(A,B), by definition, evenly divides A. As a result, A must be some multiple of GCD(A,B). i.e. X⋅GCD(A,B)=A where X is some integer
The GCD(A,B), by definition, evenly divides B. As a result, B must be some multiple of GCD(A,B). i.e. Y⋅GCD(A,B)=B where Y is some integer
A-B=C gives us:
X⋅GCD(A,B) - Y⋅GCD(A,B) = C
(X - Y)⋅GCD(A,B) = C
So we can see that GCD(A,B) evenly divides C.
An illustration of this proof is shown in the left portion of the figure below:
Proof that the GCD(B,C) evenly divides A
The GCD(B,C), by definition, evenly divides B. As a result, B must be some multiple of GCD(B,C). i.e. M⋅GCD(B,C)=B where M is some integer
The GCD(B,C), by definition, evenly divides C. As a result, C must be some multiple of GCD(B,C). i.e. N⋅GCD(B,C)=C where N is some integer
A-B=C gives us:
B+C=A
M⋅GCD(B,C) + N⋅GCD(B,C) = A
(M + N)⋅GCD(B,C) = A
So we can see that GCD(B,C) evenly divides A.
An illustration of this proof is shown in the figure below
Proof that GCD(A,B)=GCD(A,A-B)
GCD(A,B) by definition, evenly divides B.
We proved that GCD(A,B) evenly divides C.
Since the GCD(A,B) divides both B and C evenly it is a common divisor of B and C.
GCD(A,B) must be less than or equal to, GCD(B,C), because GCD(B,C) is the “greatest” common divisor of B and C.
GCD(B,C) by definition, evenly divides B.
We proved that GCD(B,C) evenly divides A.
Since the GCD(B,C) divides both A and B evenly it is a common divisor of A and B.
GCD(B,C) must be less than or equal to, GCD(A,B), because GCD(A,B) is the “greatest” common divisor of A and B.
Given that GCD(A,B)≤GCD(B,C) and GCD(B,C)≤GCD(A,B) we can conclude that:
GCD(A,B)=GCD(B,C)
Which is equivalent to:
GCD(A,B)=GCD(B,A-B)
An illustration of this proof is shown in the right portion of the figure below.
Proof that GCD(A,B) = GCD(B,R)
We proved that GCD(A,B)=GCD(B,A-B)
The order of the terms does not matter so we can say GCD(A,B)=GCD(A-B,B)
We can repeatedly apply GCD(A,B)=GCD(A-B,B) to obtain:
GCD(A,B)=GCD(A-B,B)=GCD(A-2B,B)=GCD(A-3B,B)=...=GCD(A-Q⋅B,B)
But A= B⋅Q + R so A-Q⋅B=R
Thus GCD(A,B)=GCD(R,B)
The order of terms does not matter, thus:
GCD(A,B)=GCD(B,R)
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Chris K 7 years ago Posted 7 years ago. Direct link to Chris K's post “I'm trying to wrap my hea...” more I'm trying to wrap my head around:
Find GCD(B,R) using the Euclidean Algorithm since GCD(A,B) = GCD(B,R)
How is it that GCD(A,B) = GCD(B,R)?
Maybe someone can help make this a little more intuitive for me?
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Cameron 7 years ago Posted 7 years ago. Direct link to Cameron's post “This may be an intuitive ...” more This may be an intuitive way to think about it.
Keep in mind that this is hand-wavy, and not a proof.
Suppose A and B are made up of chunks of size G
If we subtract B from A the difference will also be some number of chunks of size G
In fact, we can subtract several Bs from A and the difference will still be some number of chunks of size G
If we subtract as many Bs as we can from A, then the remaining amount is the remainder when we divide A by B.
(Remember from the quotient remainder theorem A - B Q = R )
And since it it just a difference of several Bs from A, that remainder is some number of chunks of size G.
So:
gcd(A,B) = G (A and B are made up of chunks of size G)
gcd(B,A-B) = G (A-B is also made up of chunks of size G)
gcd(B,R) = G (The remainder R which is just A-BQ is also made up of chunks of size G)
So:
gcd(A,B) = gcd(B,R)
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Rathony 11 years ago Posted 11 years ago. Direct link to Rathony's post “What are some practical a...” more What are some practical applications the Euclidean Algorithm might be used in?
I cannot imagine a situation where we might need to use it.
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ccmiint 9 years ago Posted 9 years ago. Direct link to ccmiint's post “Look up Quicksort using D...” more Look up Quicksort using Divide & conquer. I'm reading Grokking Algorithms, it has a simple example of it in chapter four.
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Haversine 5 years ago Posted 5 years ago. Direct link to Haversine's post “Would it not be faster to...” more Would it not be faster to just prime factorize both numbers and find the greatest common factor from there?
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Cameron 5 years ago Posted 5 years ago. Direct link to Cameron's post “Factorization large numbe...” more Factorization large numbers is generally a much harder problem than finding the gcd. In fact, factorizing large numbers is a problem that is considered so hard, that it forms a key part behind RSA encryption.
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Azimjon 6 years ago Posted 6 years ago. Direct link to Azimjon's post “How we can find modular i...” more How we can find modular inverse of A(mod C)?
I can't find method in this article.
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Venkata Sucheth Gerendla 7 years ago Posted 7 years ago. Direct link to Venkata Sucheth Gerendla's post “can u please elaborate ho...” more can u please elaborate how GCD(A,B) is equal to GCD(B,R)
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rashapredator 3 years ago Posted 3 years ago. Direct link to rashapredator's post “GCD(A,B) must be less tha...” more GCD(A,B) must be less than or equal to, GCD(B,C), because GCD(B,C) is the “greatest” common divisor of B and C.
didn't got this
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Cameron 3 years ago Posted 3 years ago. Direct link to Cameron's post “Let's review some termino...” more Let's review some terminology:
Suppose X divides Y evenly.
Then X is a "divisor" of Y.
Suppose X divides Z evenly.
Then it is a "divisor" of Z.
Since X divides both Y and Z, it is a "common divisor" of Y and Z.
Y and Z can have many "common divisors". The largest of these "common divisors" is the "greatest common divisor".
e.g.
1,2,3,6 are "common divisors" of 30 and 42
6 is the largest of the "common divisors" of 30 and 42, so 6 is the "greatest common divisor" of 30 and 42
So how does this apply to this situation ?
"Since the GCD(A,B) divides both B and C evenly it is a common divisor of B and C."
So, we know that GCD(A,B) is one of the "common divisors" of B and C. The biggest one of those "common divisors", is the "greatest common divisor" of B and C, namely GCD(B,C). You can be smaller than the biggest, or you can be the biggest, but you can't be bigger than the biggest. Which is why:
"GCD(A,B) must be less than or equal to, GCD(B,C), because GCD(B,C) is the “greatest” common divisor of B and C."
Hope this makes sense
1 comment Comment on Cameron's post “Let's review some termino...”
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Haversine 5 years ago Posted 5 years ago. Direct link to Haversine's post “How would using the Eucli...” more How would using the Euclidean Algorithm on variables work? i.e. Given that a is a multiple of 4, find the greatest common divisor of a^2+9a+24 and a+4?
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Tarun 4 years ago Posted 4 years ago. Direct link to Tarun's post “does the same thing apply...” more does the same thing apply to 3 terms?
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serena.pramos 2 years ago Posted 2 years ago. Direct link to serena.pramos's post “Can we use it to find out...” more Can we use it to find out how many modular inverses there is?
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Cameron 2 years ago Posted 2 years ago. Direct link to Cameron's post “Yes. If the gcd(A,B) != 1...” more Yes.
If the gcd(A,B) != 1 then A does not have a modular inverse mod B i.e. there are 0 modular inverses.
If the gcd(A,B) = 1 then A has a modular inverse mod B. There will be one A^-1 between 0 and B (this is typically the one we are concerned with). You can create infinite modular inverses by adding multiples of B to A^-1. If you take them mod B they will reduce down to the original A^-1.
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JigglyCuber200 5 years ago Posted 5 years ago. Direct link to JigglyCuber200's post “It says that GCF(270-192)...” more It says that GCF(270-192) = GCF(192-78) but 270-192=78.
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Cameron 5 years ago Posted 5 years ago. Direct link to Cameron's post “It's a "," not a "-" i.e....” more It's a "," not a "-"
i.e. gcd(270,192) = gcd(197,78)
The gcf or gcd is applied to two numbers, not one.
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14454 | https://byjus.com/maths/factors-of-18/ | Factors of 18 are the numbers that give the original number 18 when multiplied together in pairs. Basically, the factors are the numbers that divide the original number completely. The pair factor of 18 is a pair of numbers on multiplication, resulting in the original number. The pair factor of 18 can be positive as well as negative, but it cannot be a decimal or fraction. For example, the pair factor of 18 is written as (1, 18) or (-1, -18). If we multiply a pair of negative numbers, such as multiplying -1 and -18, results in an original number 18. In this article, we are going to learn what are the factors of 18, pair factors and the prime factors of 18 using the prime factorization method with many solved examples.
Table of Contents:
What are the Factors of 18?
Pair Factors of 18
Factors of 18 by Division Method
Prime Factorization of 18
Examples
FAQs
What are the Factors of 18?
The factors of 18 are the numbers that divide 18 completely without leaving a remainder value. In other words, the numbers which are multiplied together resulting in the number 18 are the factors of 18. As 18 is an even composite number, it has more factors other than 1 and 18. Hence, the factors of 18 are 1, 2, 3, 6, 9 and 18, and its negative factors are -1, -2, -3, -6, -9 and -18.
| |
| Factors of 18: 1, 2, 3, 6, 9 and 18. Prime Factorization of 18: 2 × 3 × 3 or 2 × 32. |
Pair Factors of 18
The numbers which are multiplied together resulting in 18 is called the pair factor of 18. As discussed above, the pair factors of 18 can be positive or negative. Thus, the positive and negative pair factors of 18 are given below:
Positive Pair Factors of 18:
| | |
--- |
| Positive Factors of 18 | Positive Pair Factors of 18 |
| 1 × 18 | (1, 18) |
| 2 × 9 | (2, 9) |
| 3 × 6 | (3, 6) |
Negative Pair Factor of 18:
| | |
--- |
| Negative Factors of 18 | Negative Pair Factors of 18 |
| -1 × -18 | (-1, -18) |
| -2 × -9 | (-2, -9) |
| -3 × -6 | (-3, -6) |
Hence, the positive pair factors of 18 are (1, 18), (2, 9) and (3, 6). Similarly, the negative pair factors of 18 are (-1, -18), (-2, -9), (-3, -6).
Factors of 18 by Division Method
In the division method, the factors of 18 are found by dividing 18 by different integers. If the integers divide 18 exactly and leave the remainder zero, then those integers are the factors of 18. Now, let us start dividing 18 by 1 and proceed with the different integers.
18/1 = 18 (Factor is 1 and Remainder is 0)
18/2 = 9 (Factor is 2 and Remainder is 0)
18/3 = 6 (Factor is 3 and Remainder is 0)
18/6 = 3 (Factor is 6 and Remainder is 0)
18/9 = 2 (Factor is 9 and Remainder is 0)
18/18 = 1 (Factor is 18 and Remainder is 0)
If we divide 18 by any numbers other than 1, 2, 3, 6, 9 and 18, it leaves a remainder. Thus, the factors of 18 are 1, 2, 3, 6, 9 and 18.
Prime Factorization of 18
The number 18 is a composite number. Now let us find the prime factors of it.
The first step is to divide the number 18 with the smallest prime factor,i.e. 2.
18 ÷ 2 = 9
Now, check whether 9 can be further divided by 2 or not.
9 ÷ 2 = 4.5
It gives a fractional value. But the factors should be a whole number.
Therefore, we will move to the next prime number, i.e. 3.
Now, divide 9 by 3.
9 ÷ 3 = 3
Again divide 3 by 3.
3 ÷ 3 = 1
We have received 1 at the end and further, we cannot proceed with the division method. So, the prime factors of 18 are 2 × 3 × 3 or we can also write them as 2 × 32, where 2 and 3 both are prime numbers.
Video Lesson on Prime Factors
| |
| Links Related to Factors |
| Factors And Multiples | Factor of 36 |
| Factor Theorem | Prime Factors |
Examples
Example 1:
Find the common factors of 18 and 17.
Solution:
The factors of 18 are 1, 2, 3, 6, 9 and 18.
The factors of 17 are 1 and 17.
As 17 is a prime number, the common factor of 18 and 17 is 1.
Example 2:
Find the common factors of 18 and 19.
Solution:
Factors of 18 = 1, 2, 3, 6, 9 and 18.
Factors of 19 = 1 and 19.
The common factor of 18 and 19 is 1, as 19 is a prime number.
Example 3:
Find the common factors of 18 and 9.
Solution:
The factors of 18 are 1, 2, 3, 6, 9 and 18.
The factors of 9 are 1, 3 and 9.
Therefore, the common factors of 18 and 9 are 1, 3 and 9.
Learn the factorization of more numbers here with us in BYJU’S and also download BYJU’S – The Learning App for a better experience.
Frequently Asked Questions on Factors of 18
Q1
What are the Factors of 18?
The factors of 18 are 1, 2, 3, 6, 9 and 18.
Q2
What is the prime factorization of 18?
The prime factorization of 18 is 2 × 3 × 3 or 2 × 32.
Q3
What are the positive pair factors of 18?
The positive pair factors of 18 are (1, 18), (2, 9) and (3, 6).
Q4
What are the negative pair factors of 18?
The negative pair factors of 18 are (-1, -18), (-2, -9) and (-3, -6).
Q5
Is 6 a factor of 18?
Yes, 6 is a factor of 18. As 6 divides 18 completely without leaving a remainder, and hence 6 is a factor of 18.
Quiz on Factors of 18
Q5
Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!
Select the correct answer and click on the “Finish” buttonCheck your score and answers at the end of the quiz
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14455 | https://www.ecb.torontomu.ca/~courses/ss1/Topic11.pdf | Signals and Systems I Topic 11 Laplace Transform Soosan Beheshti, Ryerson University Fourier Transform and LTIDE LTIDE ℎ(𝑡) FT of ℎ(𝑡) 𝐻(𝑗𝜔) Laplace Transform Soosan Beheshti, Ryerson University Laplace Transform Soosan Beheshti, Ryerson University 𝑅𝑒 −𝑎 𝑅𝑂𝐶 𝐼𝑚 Region of Convergence (ROC) of Laplace Transform Soosan Beheshti, Ryerson University 𝑥ଵ𝑡= 𝑒ିଶ௧𝑢(𝑡) 𝑥ଷ𝑡= 𝑒ଶ௧𝑢(−𝑡) 𝑥ଶ𝑡= −𝑒ିଶ௧𝑢(−𝑡) Soosan Beheshti, Ryerson University Laplace Transform & ROC 𝑅𝑒 −2 𝑅𝑂𝐶 𝐼𝑚 Soosan Beheshti, Ryerson University 𝑥ଵ𝑡= 𝑒ିଶ௧𝑢(𝑡) Laplace Transform & ROC −2 𝑅𝑒 𝑥ଶ𝑡= −𝑒ିଶ௧𝑢(−𝑡) Laplace Transform & ROC Soosan Beheshti, Ryerson University 𝑅𝑒 2 𝑅𝑂𝐶 Laplace Transform & ROC 𝑥ଷ𝑡= 𝑒ଶ௧𝑢(−𝑡) Soosan Beheshti, Ryerson University 𝑥ଵ𝑡= 𝑒ିଶ௧𝑢(𝑡) 𝑥ଷ𝑡= 𝑒ଶ௧𝑢(−𝑡) 𝑥ଶ𝑡= −𝑒ିଶ௧𝑢(−𝑡) Soosan Beheshti, Ryerson University Laplace Transform & ROC 𝑅𝑒 2 𝑅𝑂𝐶 −2 𝑅𝑒 𝑅𝑒 −2 𝑅𝑂𝐶 𝐼𝑚 𝑡 𝑒ି௧ 𝑒௧ 𝑅𝑒 𝑎 𝑅𝑂𝐶 −𝑎 𝐼𝑚 Laplace Transform & ROC Soosan Beheshti, Ryerson University 𝑅𝑒 −𝑎 𝑅𝑂𝐶 𝑅𝑒 𝑎 𝑅𝑂𝐶 𝑒ି𝑢(𝑡) 𝑡 Right sided 𝑒௧𝑢(−𝑡) 𝑡 Left sided Soosan Beheshti, Ryerson University Laplace Transform & ROC 𝑅𝑒 𝑎 𝑅𝑂𝐶 −𝑎 𝑒ି௧𝑢(|𝑡|) 𝑡 Two sided 𝑡 Bounded signal in time with finite values 𝑅𝑒 𝑅𝑂𝐶 Converge for all s 𝐼𝑚 Laplace Transform & ROC Soosan Beheshti, Ryerson University 𝑅𝑒 −2 𝑅𝑒 −3 𝑒ିଶ௧𝑢(𝑡) −𝑒ିଷ௧𝑢(𝑡) 𝑅𝑒 −2 𝑒ିଶ௧𝑢(𝑡) Laplace Transform & ROC Soosan Beheshti, Ryerson University Laplace Transform & ROC Soosan Beheshti, Ryerson University −2 𝑅𝑒 −3 𝑅𝑒 −𝑒ିଶ௧𝑢(−𝑡) 𝑒ିଷ௧𝑢(𝑡) −2 𝑅𝑒 −3 Laplace Transform & ROC Soosan Beheshti, Ryerson University 𝑅𝑒 −2 −2 𝑅𝑒 𝑅𝑒 −3 −3 𝑅𝑒 𝑒ିଶ௧𝑢(𝑡) −𝑒ିଶ௧𝑢(−𝑡) −𝑒ିଷ௧𝑢(𝑡) 𝑒ିଷ௧𝑢(−𝑡) (𝑎) (𝑏) (𝑐) (𝑑) Possible ROCs for ଵ ௦ାଶ Possible ROCs for ିଵ ௦ାଷ Laplace Transform & ROC Soosan Beheshti, Ryerson University 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 1 2 3 4 5 6 𝑥𝑡= 𝑒ିଶ௧𝑢𝑡−𝑒ିଷ௧𝑢(𝑡) 𝑅𝑒 −2 −3 Intersection: 𝑅𝑒{𝑠} > −2 𝑅𝑒 −2 −3 Laplace Transform & ROC Soosan Beheshti, Ryerson University -3 -2.5 -2 -1.5 -1 -0.5 0 -1 -0.5 0 0.5 1 1.5 2 2.5 𝑥𝑡= −𝑒ିଶ௧𝑢−𝑡−𝑒ିଷ௧𝑢(𝑡) 𝑅𝑒 −2 −3 Intersection: −3 < 𝑅𝑒𝑠< −2 0 20 40 60 80 100 120 -2 -1.5 -1 -0.5 0 𝑥𝑡= −𝑒ିଶ௧𝑢−𝑡+ 𝑒ିଷ௧𝑢(−𝑡) 𝑅𝑒 −2 −3 Intersection: 𝑅𝑒𝑠< −3 Laplace Transform & ROC Soosan Beheshti, Ryerson University 𝑅𝑒 𝐼𝑚 s= 𝑗𝜔 Laplace Transform & Fourier Transform Soosan Beheshti, Ryerson University Soosan Beheshti, Ryerson University Laplace Transform & ROC Soosan Beheshti, Ryerson University Laplace Transform & ROC Soosan Beheshti, Ryerson University Laplace Transform Properties Soosan Beheshti, Ryerson University Laplace Transform Properties Soosan Beheshti, Ryerson University Laplace Transform Properties Laplace Transform Properties Soosan Beheshti, Ryerson University 3 𝑦(𝑡) 1 2 𝑡 0 1 1 𝑥(𝑡) 𝑡 Laplace Transform Properties Soosan Beheshti, Ryerson University 3 𝑦(𝑡) 1 2 𝑡 0 1 1 𝑥(𝑡) 𝑡 0 2 1 𝑡 𝑧(𝑡) Laplace Transform Properties Soosan Beheshti, Ryerson University 𝑡 𝑥(𝑡) 1 2 1 𝑥ଶ(𝑡) 1 2 1 3 4 3.5 2.5 𝑡 −1 2 Laplace Transform Properties Soosan Beheshti, Ryerson University 𝑡 𝑥(𝑡) 1 2 1 𝑥ଶ(𝑡) 1 2 1 3 4 3.5 2.5 𝑡 −1 2 𝑑𝑥𝑡 𝑑𝑡 1 2 1 3 4 𝑡 −1 𝑡 𝑧𝑡= 𝑥(2𝑡) 1 0.5 1 𝑧𝑡−2.5 = 𝑥2 𝑡−2.5 = 𝑥(2𝑡−5) 1 2 1 3 4 3.5 2.5 𝑡 Laplace Transform Properties Soosan Beheshti, Ryerson University Laplace Transform Soosan Beheshti, Ryerson University 𝑅𝑒 −1 𝑅𝑂𝐶 𝐼𝑚 Laplace Transform Soosan Beheshti, Ryerson University Laplace Transform Soosan Beheshti, Ryerson University Laplace Transform Soosan Beheshti, Ryerson University 𝐻ଵ(𝑠) 𝐻ଶ(𝑠) 𝐻ଵ(𝑠)𝐻ଶ(𝑠) 1 𝑠+ 1 𝑒ିହ௦ 𝑒ିହ௦ 𝑠+ 1 Laplace Transform Soosan Beheshti, Ryerson University 𝐻ଵ(𝑠) 𝐻ଶ(𝑠) 𝐻ଵ(𝑠)+𝐻ଶ(𝑠) Laplace Transform Soosan Beheshti, Ryerson University 1 𝑠+ 3 s+2 𝑥(𝑡) 𝑦(𝑡) All pole system ℎଵ All zero system ℎଶ 1 𝑠+ 3 ℎଵ𝑡= 𝑒ି௧𝑢(𝑡) : s+2 ℎଶ𝑡= 𝛿′(𝑡)+ 2𝛿(𝑡) : Soosan Beheshti, Ryerson University Laplace Transform 𝑥(𝑡) 𝑦(𝑡) ℎ(𝑡) 𝑥(𝑡) 𝑦(𝑡) Soosan Beheshti, Ryerson University Laplace Transform ℎ(𝑡) 𝑥(𝑡) 𝑦(𝑡) Soosan Beheshti, Ryerson University Laplace Transform 𝑅𝑒 𝑅𝑂𝐶 𝐼𝑚 |
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Biochemical Bases of Endocrinology (II) Hormones and Other Chemical Intermediates
Estrogen Actions
Effects on genital organs. Estrogens stimulate the development of the ovary, Fallopian tubes, vagina, and uterus. Estrogens stimulate uterine growth and prepare the uterine lining for the action of progestational hormones. They induce endometrial development and increase its vascularization. They produce characteristic changes in the Fallopian tubes and the vaginal epithelium, and are responsible for the development and maintenance of the female secondary sexual characteristics.
Metabolic effects. Estrogens have anabolic action in female genital organs. They increase the uptake of water, sodium, amino acids, and glucose by the myometrium. These actions are secondary to stimulation of protein synthesis. The anabolic action on protein metabolism is weak in muscle, liver, and kidney.
Estrogens antagonize the effects of insulin in peripheral tissues, decreasing glucose tolerance, reducing plasma cholesterol concentration, and increasing HDL levels. They stimulate the liver synthesis of transport proteins for thyroid and sex hormones, and transcortin. They promote bone growth and epiphyseal closure at puberty, and inhibit osteoclast activity. In adult women, estrogens participate in bone remodeling, and play a crucial role in the maintenance of bone mass. The rapid decrease in estrogen secretion, common in postmenopausal women, is an important factor that contributes to osteoporosis.
Nongenomic actions. Estradiol also exerts effects that are not mediated by the nuclear receptor. For example, estradiol causes rapid increase in intracellular calcium concentration [Ca2+]i in endometrium, maturing oocytes, and granulosa cells. It also has direct action on the vascular system, and activates nitric oxide synthesis and produces vasodilation.
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Book2017, Medical BiochemistryAntonio Blanco, Gustavo Blanco
Review article
Three lactation-related hormones: Regulation of hypothalamus-pituitary axis and function on lactation
2021, Molecular and Cellular EndocrinologyYifan Ni, ... Jinzhi Zhang
4 Estrogen
Estrogen (E) is a steroid hormone with 18 carbon atoms. In the rodent, E is secreted by the ovary throughout pregnancy, and the level rises in late pregnancy. In women, E is secreted by the ovary during the first three months of pregnancy, after which the fetal adrenal, liver, and placenta coordinate to produce estrone (E1), 17β-estradiol (E2), and estriol (E3), of which E2 is the most common and potent form of estrogen in mammals (Neville et al., 2002; Barakat et al., 2016). E2 is also produced in many extragonadal sites, including the mesenchymal cells of adipose tissue such as breast, osteoblasts and chondrocytes of bone, as well as numerous sites in the brain. E2 acts as a paracrine or even intracrine factor (Simpson, 2003). E2 plays a beneficial role in fat formation, reduces osteoporosis, and prolongs hair growth (Barakat et al., 2016; Emoto et al., 1991). In addition, Es have been implicated in the pathogenesis of breast cancer (Nelson and Bulun, 2001). Besides, E promotes epithelial hyperplasia of the breast ducts, making the breast ducts extend and branch, leading to lactation.
4.1 Regulation of estrogen by the hypothalamus-pituitary axis
Estrogen synthesis and secretion is regulated by the hypothalamus-pituitary-gonadal axis. The ten-peptide gonadotropin-releasing hormone (GnRH) secreted by neuronal axons in the hypothalamic arcuate nucleus reaches the anterior pituitary through the pituitary portal vein system. The GnRH stimulates the pituitary to release luteinizing hormone (LH) and folliculotropin stimulating hormone (FSH) (Plewka et al., 2014). From menarche to menopause, estrogen is synthesized and secreted in a cyclical manner by the ovaries under the control of the pituitary gonadotrophins (LH, FSH) (Anderson and Clarke, 2004) (Fig. 3). The pituitary gonadotrophins are mainly synthesized by endometrial and granulosa cells. Under the action of LH, the endometrial cells convert cholesterol obtained from the blood into androstenedione which enters the granulosa cells by diffusion. FSH enhances the aromatase activity in granulosa cells, thus converting androstenedione into E.
The synthesis and secretion of E is mainly affected by the related hormones on the hypothalamus-pituitary axis, including GnRH and gonadotropin (FSH, LH). The secretion of pituitary LH is regulated by gonads, and E is involved in the classic feedback loop of the hypothalamus-pituitary-gonadal axis (Tsai et al., 1994). For example, in vivo and in vitro assays demonstrated that E secretion by follicles was downregulated by LH corresponding to that released at estrus, especially when FSH was substantially above LH level (Moor, 1974). It has also been shown that melatonin stimulates the secretion of gonadotropic hormones from the pituitary gland in sheep (Blasiak and Molik, 2015). This effect occurs within the central nervous system and is manifested by increased secretion of GnRH, LH, and FSH, which would increase E secretion.
4.2 The effect of E on lactation
Estrogen is an ovarian steroid hormone, which is known to regulate ductal morphogenesis and MECs proliferation. In pre-pubertal mice, increase in E is accompanied by swelling of the terminal end buds (TEB, distal ends of the mammary ducts) due to the proliferation of its cuboidal epithelial cells (Russo et al., 1982). By contrast, E promotes the formation of clusters by distal end of the ductal structures form into clusters, which are collectively referred to as the terminal ductal lobular unit (TDLU) (Cardiff and Wellings, 1999). In adult mammary glands of mice, rats and humans, E can cause bifurcation, ductal elongation and lateral duct ranching of TEBs and TDLUs forming an epithelial duct network (Need et al., 2014). E promotes rapid growth and expansion of ducts into the mammary fat pad (Arendt and Kuperwasser, 2015), via its receptor α (ERα) which functions as a transcription factor regulating the development of mammary glands (Feng et al., 2007). During successive gestation cycles, the deletion of whey acidic protein (WAP)-Cre-mediated ERα in mice mammary epithelium resulted in the loss of ductal collateral and lobular alveolar structures, ductal dilation, and suppressed the proliferation of alveolar cells (Feng et al., 2007; Park et al., 2011). The effects of PRL and E on the proliferation of MECs is mediated by Wnt-signaling pathway. This pathway has been implicated in the regulation of mammary gland development (Tong et al., 2016). The data reviewed here show that E regulates MECs via ER which is essential for the maintenance of normal duct morphology during puberty and the alveolar formation during pregnancy and lactation. In addition, E may directly stimulate the pituitary lactotrophs (Rutlin et al., 1977). E indirectly stimulates the pituitary to release PRL and increase the number of PRL receptors in the mammary gland (Blasiak and Molik, 2015). The presence of E is essential for lactogenesis. However, in many species, except in women, the secretion of E and P from the ovaries continues during pregnancy and which prevents both induction of labor and lactogenesis 2 (Neville et al., 2002). When the ratio of E to P is high, milk production is decreased. High levels of P during pregnancy suppress the stimulatory effect of E on PRL function (Blasiak and Molik, 2015).
In the ‘classical’ mechanism of estrogen action, E diffuses into the cell and binds to the nuclear ER (Deroo and Korach, 2006). The nuclear E and ER complex interacts with the activated protein 1 (AP1) or SP1 protein found in the promoter region of the E-responsive gene. The complex also binds to the sequence of the E response element directly or indirectly, leading to the recruitment of co-regulation proteins (coactivator or corepressor) to the promoters, mRNA levels increase or decrease and related protein production, as well as physiological responses.
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Estrogen in the Male: Nature, Sources, and Biological Effects
2019, Reference Module in Neuroscience and Biobehavioral PsychologyHans-Udo Schweikert
Abstract
Estrogens are also an important part of the male endocrine system. Opposed to the central role of the ovarian estrogen secretion in the female is the dominant conversion of circulating steroids of adrenal and testicular origin in peripheral tissues in the male. Rather than experimental approaches, mutations of estrogen receptors or steroid metabolizing enzymes permit an insight into the gonadotropin-regulatory and metabolic functioning of estrogens in the healthy human male: among them the maturation and homeostasis of the skeleton and the regulation of the gonadotropin secretion. Corresponding functions are the subject of research with genomically manipulated rodents (ER knockout mice) and receptor selective synthetic estrogenic steroids. It is obvious from animal studies and studies in men and women that peripherally generated estrogens and androgens interact in an array of sex hormone regulated organs and tissues, as the genital tract organs themselves, the pituitary, the sexually dimorphic breast, the prostate, the skeleton, the brain and sexually dimorphic growth of hair. Sexually dimorphic diseases as those of the cardiovascular system may deserve further evaluation under the aspect of the peripheral estrogen metabolism.
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Reference work2019, Reference Module in Neuroscience and Biobehavioral PsychologyHans-Udo Schweikert
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New insights into the molecular mechanisms underlying effects of estrogen on cholesterol gallstone formation
2009, Biochimica et Biophysica Acta (BBA) - Molecular and Cell Biology of LipidsHelen H. Wang, ... David Q.-H. Wang
Estrogens circulate in blood loosely bound to albumin in a nonsaturable and nonstoichiometric manner and tightly bound to the testosterone–estrogen-binding globulin, which is also called the sex hormone-binding globulin, a β-globulin . Plasma concentrations of estrogens are considerably lower than those of other gonadal steroids and vary over an almost 20-fold range during the cycle. Estradiol levels vary through the menstrual cycle and almost all of this type of estrogen comes from the ovary. In general, there are two peaks of estrogen secretion: one just before ovulation and the another during the midluteal phase [52–55]. The estradiol secretion rate is 36 μg/day (i.e., 133 nmol/day) in the early follicular phase, 380 μg/day just before ovulation, and 250 μg/day during the midluteal phase. After menopause, estrogen secretion declines to low levels. In addition, the estradiol production rate in men is about 50 μg/day (184 nmol/day). While estrogens are present in both men and women, they are usually present at significantly higher levels in women of reproductive age. Furthermore, estrogens promote the development of female secondary sex characteristics, such as breasts, and are also involved in the thickening of the endometrium and other aspects of regulating the menstrual cycle. In men, estrogens regulate certain functions of the reproductive system important to the maturation of sperm and may be necessary for a healthy libido. Hyperestrogenemia (elevated levels of estrogens in blood) may be a result of exogenous administration of estrogens or estrogen-like substances, or may be a result of physiologic conditions such as pregnancy.
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Journal2009, Biochimica et Biophysica Acta (BBA) - Molecular and Cell Biology of LipidsHelen H. Wang, ... David Q.-H. Wang
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Studies with Antibodies to Steroids
1971, Research on SteroidsR.L. VANDE WIELE, ... G. MIKHAIL
2 The role of estrogens and progesterone as triggers of the ovulatory LH release
In humans as well as in many laboratory animals, estrogens will trigger LH release and these observations have led to the hypothesis that the rising titer of estrogens is the factor that initiates the LH surge preceding ovulation. Progesterone on the other hand, has also been shown to produce LH release and a number of investigators have maintained the theory that progesterone, and not estrogens, sets off the preovulatory LH release. On the basis of available evidence, it is not possible to decide between these two theories, nor in fact, can the possibility be ruled out that factors other than estrogens and progesterone are the controlling factors in the timing of LH release. The pros and cons of these theories have recently been reviewed (13).
Studies carried out by us (5, 6) in which antibodies to steroids were used have added considerable new information to this controversy and have produced evidence that strongly favors the estrogen theory of LH release. It is in this type of situation that the immunologic approach has unique advantages over more classical approaches. Indeed the demonstration that either estrogen or progesterone elicits LH release, only proves that the steroids are capable of doing so, but does not necessarily have any bearing on the role of either estrogens or progesterone in the LH surge that precedes spontaneous ovulation. Antibodies, on the other hand, selectively inhibit the biological activity of one of the steroids and thereby present direct proof of the necessity of a specific hormone during the ovulatory process.
We have studied two experimental models to evaluate the relative importance of estrogens and progesterone in the pre-ovulatory LH release.
a) LH release in the immature rat treated with PMS (5)
The experimental design of this experiment is illustrated in Figure 6. When PMS is administered to 27-day-old immature rats, LH is released from the animal's own pituitary 50 hours after the administration of PMS and ovulation occurs 12 hours after the LH surge. Although the exact interval between LH release and the administration of PMS depends on specific laboratory conditions (such as the day/night schedule in the colony), under controlled conditions this interval is very predictable and the PMS treated immature animal offers, therefore, an excellent model to study the factors controlling LH release.
The effect of inactivation of the circulating estrogens by the administration of anti-E2 is illustrated in Figure 7. Treatment with 3.2 mg of anti-E2, prevented ovulation in virtually all animals even when the dose of PMS was increased to 45 International Units. Anti-testosterone, in sufficient amounts to block the biological activity of circulating testosterone, was inactive in this situation. The effect of anti-E2 was shown to be dose dependent, 1mg of antibody being sufficient to block ovulation in essentially all animals.
If anti-E2 exerts its effect solely through the inactivation of the estrogenic activity of 17β-estradiol, it should be possible to restore ovulation in animals treated with anti-E2, by the simultaneous administration of diethylstilbestrol (DES), since the biological activity of this estrogen is not blocked by anti-E2. Indeed, this proved to be the case; treatment with amounts of DES ranging from 0.5 to 10.0 μg, restored ovulation in approximately one-half of the animals. The fact that not all of the “blocked“ animals ovulated was not surprising since it would certainly be difficult to reproduce the individual estrogen pattern of secretion in the pre-ovulatory rat, by a single injection of DES in oil.
This experimental model also gave us an opportunity to study the time relationship between the triggering effect of estrogens and the actual release of LH. In the above mentioned experiments, the antibody was administered at the same time (but at a different site) as the PMS, i.e., 50 hours before the expected time of LH release. In subsequent experiments, in which the antibody was given closer and closer to the expected time of LH release, it was found that to block ovulation, the antibody had to be administered not later than 12 hours prior to the expected time of LH release. This long interval between the actual LH release and the triggering effect of estrogens is fascinating. A similarly long delay between the administration of estrogens and LH release has been found in sheep (14) as well as in humans (13).
b) LH release in the cyclic, adult female rat (6)
In another set of experiments we have studied the role of estrogens and progesterone as triggers of LH release in the adult cyclic rat. The sequence of events in the 4-day cycle rat, and the effects of treatment with anti-E2 and anti-P upon these, are illustrated in Figure 8. In the rat, the LH surge occurs in the afternoon of proestrus. The release of gonadotropin is preceded by follicular growth and a concomitant rise in the secretion of estrogens by the ovary (15). As a result of the rise of the estrogens, there is an increase in the weight of the uterus which displays a very typical ballooning due to accumulation of fluid in its lumen. The timing of this ballooning is indicated in the figure by the symbol BAL. Following LH release, progesterone levels rise, producing a relaxation of the cervical sphincter (16). The fluid is released from the uterus and the ballooning disappears. The amount of progesterone necessary to relax the cervical sphincter is extremely small, and the disappearance of the uterine ballooning is a very sensitive indicator of the presence of progesterone. Ovulation occurs in the night between proestrus and estrus and ova can be found in the tubes on the morning of estrus, at which time cornification of the vagina is also observed. The postovulatory period lasts 2 days (D1 and D2) but full activation of the corpus luteum and secretion of progesterone occurs only in fertile cycles.
Twenty-eight animals served as control animals and were submitted to repeated laparotomies to insure that these surgical manipulations did not interfere with the normal sequence of physiological events. When antibodies to 17β-estradiol were administered, the effect of the rise of estrogens was nullified as was evident from the absence of uterine ballooning and the disappearance of cornified cells from the vagina. In agreement with the estrogen theory of LH release, there was no LH surge and no ovulation. The absence of LH release was determined directly by radioimmunoassay. When the action of the antibodies faded, there was restoration of estrogen activity, ballooning reappeared and a new cycle was reinstituted. As in the PMS treated animal, to block ovulation it was necessary to administer the antibodies not later than 12 hours before the expected time of LH release. For this reason, the antibodies were administered in the afternoon of D2, 24 hours before the expected time of LH release.
A local effect of the antiserum upon the ovary was ruled out by a series of experiments in which HCG was administered at the time the LH surge would normally have occurred. More importantly, substituting the effect of the endogenous estrogens by the administration of DES, restored ovulation in 7 of the 9 animals. In contrast when progesterone was administered, LH discharge and ovulation occurred normally. Yet there was clear evidence that the biological activity of progesterone was nullified in the animals treated with anti-P, since the ballooning which normally disappears following LH release, persisted into the day of estrus. In a recent publication (13) we have presented evidence that in the human also, estrogens are the trigger that initiates the pre-ovulatory LH surge.
Our studies do not rule out the possibility that other steroids may act synergistically with estrogens; they indicate however, that estrogens are necessary and sufficient. It is possible for instance, that the small amounts of progesterone secreted during the LH surge, and prior to ovulation, potentiate the estrogen effect and conceivably prolong the surge secretion of LH.
Similar studies leading to the same conclusions were carried out in the sheep by Caldwell et al (17). These authors however, used active rather than passive immunization to inactivate the circulating estrogens.
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Book1971, Research on SteroidsR.L. VANDE WIELE, ... G. MIKHAIL
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Estrogen Actions in the Brain
2015, Estrogen Effects on Traumatic Brain InjuryLaura L. Carruth, Mahin Shahbazi
Additional Physiological Effects of Estrogens on the Body
In addition to their robust actions on the female reproductive system and sexual behavior, estrogens also alter several other physiological systems. They play a role in calcium metabolism, resulting in increased bone production. The increase in estrogen secretion during sexual development and puberty results in an increase in bone mineral density and the closing of the epiphyseal plate in long bones due to an increase in apoptosis of chondrocytes (Zhong et al., 2011). In women, more bone is made in the presence of high concentration of estrogens (Frenkel et al., 2010), and during menopause, when estrogen concentrations decrease, osteoporosis can result. Another aspect of E2 action that can affect overall body physiology is with body fluid regulation. E2 lowers the osmotic threshold for vasopressin (AVP) release resulting in an increased plasma volume (Stachenfeld, 2008). A recent finding suggests that in female rats brown adipose tissue thermogenesis is regulated via estradiol acting through ERα. Estradiol does this by inhibiting AMP-activated protein kinase in the ventromedial nucleus of the hypothalamus (Martínez de Morentin et al., 2014). This demonstrates the variety of actions that estrogens can exhibit in the body and brain.
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Book2015, Estrogen Effects on Traumatic Brain InjuryLaura L. Carruth, Mahin Shahbazi
Review article
Research progress of stem cell therapy for endometrial injury
2022, Materials Today BioJuan Cen, ... Yuqi Guo
4.2.3 Different functional drugs
For the treatment of endometrial injury, in addition to stem cell therapy, it has been found that it has a good synergistic effect with drugs with various functions such as antibacterial, anti-inflammatory, promoting angiogenesis, promoting cell proliferation, promoting stem cell homing, and promoting damage repair, so as to achieve better therapeutic effects.
Among them, estrogen is one of the most important female sex hormones, which plays a key role in the regulation and development of the female reproductive system and the formation of secondary sexual characteristics. For example, postoperative estrogen therapy can be used to prevent adhesion recurrence and promote endometrial regeneration [52–54]. In addition, estrogen promotes the levels of ESR1, MMP-9, EGF and IGF-1 by inhibiting the increase in the levels of serum TGF-β1, epidermal growth factor, and PDGF-BB mediated by IUA- . Therefore, these advantages provide a new option for the treatments of IUA with drug-loaded bioactive scaffolds. Studies found that compared with embryonic stem cell transplantation alone, its combination with estrogen therapy can improve the therapeutic effect . For instance, 17β-estradiol (E2) can be sustainably released for 21 days by a human amniotic extracellular matrix (HAECM) scaffold integrated with E2-loaded PLGA microspheres . The sustained release of E2 was achieved by encapsulating E2 in heparin-poloxamer (HP) micelles to construct a temperature-sensitive hydrogel. Prolonged estrogen release in the target area of IUA rats, accompanied by increased levels of VEGF, PI3K/Akt, and ERK1/2, significantly inhibited the endoplasmic reticulum stress . Antibiotics, proliferative drugs, antidiabetic drugs, drugs that can increase endometrial blood flow are listed in Table 3.
Table 3. Multifunctional platforms loaded with chemical drugs.
| Loaded drug | Mechanism | Diseases and Curative Effects | References |
--- --- |
| 17β-estradiol | Estrogen. Significantly inhibits IUA-increased TGF-β1, epidermal growth factor and PDGF-BB levels; promotes ESR1 levels; promotes PI3K/Akt and ERK1/2 signaling activation, and inhibits endoplasmic reticulum stress. | Postoperative estrogen therapy can be used to prevent recurrent adhesions; estrogen can increase the number of cells in endometrial damage and promote endometrial regeneration; inhibit endoplasmic reticulum stress-related apoptosis | |
| Sitagliptin | Dipeptidyl peptidase IV (DPP4) inhibitor. Promotes stem cell homing and enrichment to the site of tissue damage | Inhibits the expression of DIO2, a marker gene of senescent decidual cells, increases endometrial-media embryonic stem cells, and reduces decidual senescence | |
| Pentoxyphene | Medications that increase endometrial blood flow | Treats endometrial damage | |
| Tocopherol |
| Sildenafil |
| Vc | Regulatory factor. Promotes stem cell survival, promotes endometrial recovery, promotes keratin, vWF expression recovery, reduces IL-1β | Attenuates the cytotoxic effect of PF-127, promotes cell survival and growth during encapsulation of rat bone marrow mesenchymal stem cells, and promotes intimal regeneration. | |
| Mitomycin C | Antibiotics. | Inhibits the cell viability of endometrial stromal cells, promotes G1 cell cycle arrest and apoptosis, and inhibits the synthesis and secretion of type I collagen. | |
| Metformin | Anti-diabetic drugs. | Inhibits ER stress-induced apoptosis through PI3K/Akt and ERK1/2 pathways. | |
| Silver ions | Fungicide. Exhibits anti-infective effect. | Works synergistically with other ingredients facilitating endometrial regeneration, fertility restoration, and live birth of offspring | |
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Journal2022, Materials Today BioJuan Cen, ... Yuqi Guo
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THE ACTIONS OF FOLLICULAR FLUID FACTORS ON STEROIDOGENESIS BY CULTURED OVARIAN GRANULOSA CELLS
1983, Hormonal Steroids: Proceedings of the Sixth International Congress on Hormonal SteroidsF. LEDWITZ-RIGBY, B.W. RIGBY
(2) Estrogen secretion.
Incubation of granulosa cells from all size antral follicles with fluid from large follicles enhanced basal estrogen secretion several fold over that secreted in the presence of serum . 5 × 10− 7 M testosterone was added to all cultures as substrate. The androgen content of charcoal treated LFF1 (5 × 10− 10 M) had no influence on estrogen secretion when added to serum containing media. Fluid from small follicles only rarely exhibited minor stimulatory action. The addition of FSH to any of the media did not alter estrogen secretion to a statistically significant extent. The time course of stimulation for estrogen secretion was dramatically different from that of progesterone secretion. A 24 h incubation was required with cells from large follicles, while 72 h incubations were necessary before cells from small follicles exhibited an increase ,
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Dynamics of Primate Follicular Growth: A Physiological Perspective
2004, The Ovary (Second Edition)ANTHONY J. ZELEZNIK
MECHANISM BY WHICH A DEVELOPING FOLLICLE GAINS DOMINANCE OVER OTHERS
As follicles develop beyond the early antral stage under the influence of FSH, GC undergo striking developmental changes that transform the follicle from being steroidogenically quiescent to being capable of producing large quantities of estrogen. As is well known, estrogen is responsible for the endometrial proliferation that prepares the uterus for implantation and for the coordination of the timing of the ovulatory discharge of LH such that the stimulus for the follicle to ovulate occurs only when preovulatory follicle development is completed . As will now be discussed, not only does estrogen serve a coordinating role between follicle maturity and the timing of the onset of the LH surge but it also serves as the principal factor responsible for the establishment of dominance of a maturing follicle. Figure 3.2 summarizes the cellular control of follicular estrogen secretion.
The limiting step in follicular estrogen biosynthesis is the acquisition of aromatase by the GC that enables these cells to convert the thecally produced androgens into estrogen . FSH mediates the induction of aromatase . Thus an early manifestation of preovulatory follicular development is the acquisition of its ability to secrete estrogen. By collecting blood from the ovarian veins of monkeys and humans, it has been shown that asymmetrical production of estrogen (hence the presence of a selected follicle) is evident approximately 7 to 8 days before ovulation [18, 19]. When asymmetrical ovarian estrogen secretion becomes evident, systemic concentrations of estrogen in blood begin to rise indicating that the increasing estrogen concentrations during the mid- through late follicular phase are directly related to the presence and continued maturation of the selected follicle.
As discussed previously, estrogen is a highly efficient negative feedback modulator of gonadotropin secretion and it is this action of estrogen that is responsible for the acquisition of follicular dominance. Figure 3.3 illustrates a typical profile of plasma concentrations of FSH and estradiol during the follicular phase of the menstrual cycle.
It can be seen that as serum estrogen concentrations begin to rise approximately 6 to 7 days before midcycle, there is a corresponding decrease in plasma FSH concentrations such that immediately before the midcycle gonadotropin surge, FSH concentrations are about 50% of values seen during the early follicular phase, before the emergence of a maturing follicle. It may be hypothesized that this feedback system functions in the following manner. Since preantral folliculogenesis is inherently asynchronous because of the continuous exit of follicles from the primordial pool, there will always be a maturationally distinct distribution of early antral follicles within the ovaries that are available for continued maturation under the influence of FSH. As FSH concentrations rise after regression of the corpus luteum, follicles will be stimulated and when aromatase is induced sufficiently to cause peripheral estrogen concentrations to increase, FSH secretion will be inhibited thus depriving other follicles of their gonadotropic support. The preantral follicle that ultimately achieves dominance and ovulates is likely to be the one at the most advanced stage of maturation and/or the follicle with the greatest access to FSH at the time when FSH concentrations rise at the end of the previous menstrual cycle.
This model for follicle selection has been tested experimentally by manipulating the pattern of FSH secretion during the follicular phase of the menstrual cycle. If the model is correct, one would predict that administration of estrogen during the early follicular phase, before a leading follicle emerges, would cause a premature reduction in FSH concentrations that, in turn, would inhibit spontaneous follicular development. Studies in rhesus monkeys have shown that a premature elevation in systemic estrogen concentrations from 50 to 80 pg/ml on days 3 to 6 of the follicular phase resulted in a slight—but significant—decrease in plasma FSH concentrations and an interruption of spontaneous follicular development. Preovulatory follicular growth was arrested until the exogenous source estrogen was removed whereupon FSH concentrations once again became elevated, the follicular phase began anew, and ovulation occurred 12 to 14 days later . Similarly, in humans, oral administration of ethinyl estradiol on days 1 to 8 of the follicular phase led to a suppression of plasma gonadotropin concentrations and a lengthening of the follicular phase .
This model for follicular selection would also predict that interfering with the gonadotropin suppressing effects of estrogen should prevent the decrease in FSH concentrations during the mid- through late follicular phase of the menstrual cycle and result in the maturation of more than one preovulatory follicle. When rhesus monkeys were passively immunized with ovine antiestradiol gamma globulins to neutralize endogenously produced estrogen, serum FSH concentrations did not fall during the midfollicular phase of the menstrual cycle and multiple preovulatory follicles were present within the ovaries at the expected time of ovulation . In humans, it is well known that blockage of the biological actions of estrogen with the antie-strogen clomiphene citrate results in an augmentation of gonadotropin levels and the maturation of more than a single preovulatory follicle .
Thus there is solid experimental evidence that supports the hypothesis that the maturing follicle establishes its dominance over other follicles by feedback inhibition of FSH secretion. Although the aforementioned studies are consistent with the notion that this feedback inhibition is mediated by estrogen, it must be noted that the maturing follicle also produces the glycoprotein inhibin, which has been shown to suppress FSH secretion in primate and subprimate model systems [12, 24, 25]. Since inhibin and estrogen are produced because of preovulatory follicular growth, it has been difficult to establish the relative contributions of estradiol and inhibin with respect to the suppression of FSH secretion . However, it should be noted in the passive immunization studies described previously that neutralization of estrogen resulted in the maturation of more than a single follicle, presumably in the presence of elevated inhibin concentrations that would result from the ensuing ovarian hyperstimulation [22, 27]. In addition, the phenotype of a woman with an inactivating mutation of the aromatase gene is consistent with a primary role for estrogen in the regulation of FSH secretion in humans. This individual exhibited elevated FSH concentrations and the presence of multiple bilateral ovarian cysts (follicles), both of which were resolved by the administration of exogenous estradiol . Although these aforementioned studies indicate that estrogen is the principal regulator of FSH secretion during the follicular phase of the menstrual cycle, until passive immunization studies are conducted in primates with anti-inhibin antibodies and/or a genetic mutation in the ability to produce inhibin is identified, a conclusion regarding the contributions of inhibin and estrogen in the feedback regulation of FSH secretion in primates cannot be ascertained. It should be reinforced that there may be significant interspecies differences regarding the roles of inhibin and estrogen in the feedback control of FSH secretion and that inhibin may have a more pronounced role in the control of FSH secretion in some species [11, 24, 25]. Regardless whether inhibin or estrogen is the primary controller of FSH secretion during the follicular phase, because the maturing follicle produces both, the important fact is that the maturing follicle establishes its dominance over other follicles by feedback inhibition of FSH secretion.
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Book2004, The Ovary (Second Edition)ANTHONY J. ZELEZNIK
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Oestrogens
1970, Hormones and Human Breast CancerJohn Hayward
Publisher Summary
This chapter discusses the role of estrogens in breast cancer. Estrogen administration can stimulate the development of breast carcinoma in male mice to the rate found normally in females of the same strain. Therefore, the prolonged administration of estrogen to men might be associated with an increase in the incidence of breast cancer to the rate found in women. Human male breast cancer is rare, but when it occurs, it has very similar histological and behavioral characteristics to female breast cancer. Estrogens are commonly administered to treat carcinoma of the prostate. Certain factors such as marriage, parity, breast feeding, and early menopause seem to have some influence on the development of breast cancer. Each of these will be accompanied by an alteration in endogenous estrogen secretion and may exert their protective effect by this means. On the other hand, certain feminizing tumors of the ovary secrete abnormally high levels of estrogen.
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Book1970, Hormones and Human Breast CancerJohn Hayward
Related terms:
Gonadotropin-Releasing Hormone
Follicle-Stimulating Hormone
Progesterone
Estradiol
Luteinizing Hormone
Gonadotropin
Secretion (Process)
Progesterone Release
Estrus
Ovulation
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14457 | https://files.eric.ed.gov/fulltext/EJ1358674.pdf | LUMAT Special Issue 2022: Mathematical Thinking and Understanding in Learning of Mathematics LUMAT: International Journal on Math, Science and Technology Education Published by the University of Helsinki, Finland / LUMA Centre Finland | CC BY 4.0 Developing mathematical problem-solving skills in primary school by using visual representations on heuristics Susanna Kaitera and Sari Harmoinen Faculty of Education, University of Oulu, Finland Developing students’ skills in solving mathematical problems and supporting creative mathematical thinking have been important topics of Finnish National Core Curricula 2004 and 2014. To foster these skills, students should be provided with rich, meaningful problem-solving tasks already in primary school. Teachers have a crucial role in equipping students with a variety of tools for solving diverse mathematical problems. This can be challenging if the instruction is based solely on tasks presented in mathematics textbooks. The aim of this study was to map whether a teaching approach, which focuses on teaching general heuristics for mathematical problem-solving by providing visual tools called Problem-solving Keys, would improve students’ performance in tasks and skills in justifying their reasoning. To map students' problem-solving skills and strategies, data from 25 fifth graders’ pre-tests and post-tests with non-routine mathematical tasks were analysed. The results indicate that the teaching approach, which emphasized finding different approaches to solve mathematical problems had the potential for improving students’ performance in a problem-solving test and skills, but also in explaining their thinking in tasks. The findings of this research suggest that teachers could support the development of problem-solving strategies by fostering classroom discussions and using for example a visual heuristics tool called Problem-solving Keys. Keywords: mathematical problem-solving, heuristics, proportional reasoning 1 Introduction During the primary school years, students develop their understanding of concept of numbers and fluency in arithmetic skills (FNBE, 2016, p. 307). Learning mathematical procedures is important, but it is also crucial to equip students with strong problem-solving, reasoning, and thinking skills (e.g. Lester, 2003; Pehkonen et al., 2013) to give tools for functioning in a complex, unpredictable future. Mathematical problem-solving requires skills to apply variety of different solution strategies and models (Leppäaho, 2018, p. 374). It is not uncommon that while students may excel on routine exercises (those that they have already seen and practiced), they fail to solve problems that differ from those they have previously encountered (OECD, 2014). ARTICLE DETAILS LUMAT Special Issue Vol 10 No 2 (2022), 111–146 Pages: 36 References: 60 Correspondence: susanna.kaitera@oulu.fi LUMAT.10.2.1696 LUMAT 112 Traditional teaching approaches often focus on learning mathematical facts and procedures. Teachers could take advantage on creating learning environments, which engage students in investigating problems and seeking solutions in an active manner. (Pehkonen et al., 2013, 13.) Näveri et al. (2011, p. 169) point out that if teachers rely on using routine tasks in mathematics lessons, also the learning of students stays on the routine level. Mathematical thinking skills can be developed via problem-solving (e.g. Schoenfeld, 1985; Lester, 2003, Leppäaho, 2018), and on the other hand, problem-based teaching methods can be used to foster deeper understanding. The importance of developing mathematical reasoning and problem-solving skills is also recognised in international assessments, such as PISA and TIMSS. In PISA the problem-solving competence is defined as “an individuals’ capacity to engage in cognitive processing to understand and resolve problem situations where a method of solution is not immediately obvious” (OECD, 2014, p. 30). As Leppäaho (2018, p. 368) points out, mathematical problem-solving is learned only by practising it repeatedly. Mathematics can actually be taught through problem-solving (see for example Schoenfeld, 1985; Hiebert, 2003; Lester, 2013). This teaching method enables students themselves to engage with meaningful, rich problem tasks and instead of superficial procedure-learning, develop understanding of mathematical concepts and methods. Students should have possibilities to explore a variety of different and unfamiliar problems, even though they would not yet master certain methods or algorithms (Goldenberg et al., 2003, p. 28). Developing students’ mathematical thinking and problem-solving skills have been flagged as important goals in Finnish National Core Curricula for basic education (FNBE 2004; FNBE 2016). Students should be guided not only to solving problems, but also finding and modifying them (FNBE, 2004, p. 158). According to the mathematics curriculum in Finland, instruction should “support the development of the pupils’ skills in presenting their mathematical thinking and solutions to others in different ways and with the help of different tools” (FNBE, 2016, p. 307). Expressing mathematical ideas and justifying thinking can be challenging for primary-school aged students, but as Finnish Curriculum (FNBE, 2016, p. 306) underlines, it would be important to learn to communicate ideas and collaborate with peers. Collaborative problem-solving situations, identifying and discussing ideas and participating in explanation-building discourse can help learners in developing their thinking skills (Scardamalia & Bereiter, 2014, p. 3). Collaborative problem-solving situations are excellent opportunities to explore also complex problems, because KAITERA & HARMOINEN (2022) 113 different examples and explanations by group members enable better understanding (Sears & Reagin, 2013). In this research, fifth grade students were introduced general heuristics, which were understood to serve as stepping stones in solving non-routine mathematical problems. At the beginning of the school year, it appeared that many students seemed to struggle in mathematical tasks and especially in explaining their problem-solving processes in written form. Students were introduced to concrete tools called Problem-solving Keys, which were modified from Strategy Keys based on work by Herold-Blasius (2021). The aim was to provide students with a visual reminder of heuristics for mathematical problem-solving tasks. Similar heuristics were outlined also in the Singaporean Mathematics Syllabus 2013 and used as a reference when classifying and modifying the Keys for teaching purposes in Finland (Kaitera, 2021). The research aimed to map fifth graders’ skills and strategies before and after the intervention, which was designed to offer wide variety of mathematical problems and techniques to solve them. The interest was in finding out if the problem-oriented teaching approach influenced on how students solved mathematical problems, which required proportional reasoning. This research aimed to answer the following questions: 1. What kind of influence did teaching approach, which focused on mathematical problem-solving, have on students’ general performance in proportional reasoning tasks and abilities to explain thinking? 2. What kind of differences appeared in students’ use of erroneous and correct problem-solving strategies between pre- and post-tests? The study outlines possibilities to develop mathematics teaching towards a direction, in which students become more active participants in learning process and develop their mathematical problem-solving skills. Another aspect was to answer the 21st-century demands for analysing the teaching practises and creating knowledge as a practicing teacher (see for example Niemi & Nevgi, 2014). The study includes features of a teaching experiment and in this report is referred to as such. LUMAT 114 2 Theoretical framework 2.1 Mathematical problem-solving: focus on heuristics In many countries, mathematics curricula emphasize the importance of exploring versatile problem-solving activities. These have been a part of mathematics classrooms for a long time, but there is still confusion on what it means in practice. Teachers often understand it as solving word problems (e.g. Lester, 2003; Näveri et al., 2011), or even solving simple, routine arithmetic tasks presented in mathematics textbooks (Näveri et al., 2011). In this study, students were provided with non-routine tasks, which require skills to devise and implement a plan (Polya, 1945/1973) and combine previously learned solution strategies in a novel way (Lester, 2013; Leppäaho, 2018). An ability to solve mathematical problems in different contexts is an important skill, which can, and should be taught at schools. To be able to invent and test strategies, students need to have basic skills and understanding of problem-solving processes. As Leppäaho (2018, 374–375) points out, in addition to mathematical skills (e.g. how students can use different strategies), for example motivational aspects and reading and writing skills play important roles in an individual’s capacity in mathematical problem-solving situations. Mathematical problem-solving techniques are often called heuristics (Polya, 1945/1973; Schoenfeld, 1985; Goldenberg et al., 2003). Heuristics can be described as non-rigorous, general suggestions for strategies, which can be helpful when solving different types of problems. Learning these techniques and becoming familiar with different problem-solving methods helps students to tackle mathematical problems also in unfamiliar contexts. Heuristics were linked to everyday teaching by Polya in his book “How to solve it” (1945). Polya outlined a simple four-step problem-solving process, and the following phases are often referred to when defining heuristics: 1. Understanding the problem: what is being asked? What is known, what is unknown? 2. Creating a plan for solving the problem, considering whether the type of the problem is already familiar, choosing the most appropriate heuristic. 3. Solving the problem by carrying out the plan and assessing whether the steps are correct. KAITERA & HARMOINEN (2022) 115 4. Looking back and checking if the answer makes sense. (Polya, 1973, 5–6.) Important first steps of understanding a problem and choosing methods for solving the task are often forgotten when describing elements linked to mathematical problem-solving (Näveri et al., 2011, p. 169). School mathematics often emphasizes teaching certain algorithms to fit certain types of problems instead of providing a wider variety of general tools for problem-solving (Näveri et al., 2011; Leppäaho, 2018). Another important aspect linked to problem-solving can be derived from Polya’s views: he outlined the function of the last phase as not only reviewing the process but also discussing it (1973, p. 6). Heuristics are not the same as algorithms: they rarely prompt a solution, while carrying out an algorithm, which is suitable for a certain type of mathematical problem, leads to a rather unambiguous solution. According to Polya (1973, p. 113), heuristics cannot be used as a tool for rigorous proof. Instead, heuristics belong to a problem-solving process as a part of it. Whereas algorithms are usually constructed of certain predetermined steps, heuristics involve a decision-making process. Students make assumptions on whether a certain approach would work or not and try out different ways to implement the method: for example, in this study, making first a diagram or table provides numerous chances to proceed in solving the problem. Heuristics can be learned and practiced (Schoenfeld, 1985; Bruder & Collet, 2011) and are generally more applicable in different types of mathematical domains and problems than plain algorithms. Due to the nature of transferability, learning heuristics also supports the development of confidence in mathematical problem-solving (Goldenberg et al., 2003). The aim of teaching mathematics through problem-solving is to equip students with skills to apply previously learned techniques in non-routine and novel situations (Leppäaho, 2018, p. 379). Polya’s four-step model is still useful in today’s mathematics classroom and was referred to as a framework to underline different phases of problem-solving; mathematics is more than just filling in the textbook, it could be understood as an activity. Devising a plan and choosing the most appropriate heuristic were supported by visual tools called Problem-solving Keys, which are introduced in Chapter 3.2. LUMAT 116 2.2 Proportional reasoning as a problem-solving domain Fifth graders’ problem-solving skills were mapped by proportional reasoning tasks. It is an excellent domain to solve mathematical problems linked to everyday life. For example, adjusting the recipe, preparing juice from a concentrate, calculating the most beneficial buy or comparing discounts between two products, calculating the consumption of the petrol in a car trip, or using a map and its scale to calculate the distance between two targets require skills to reason proportionally. Traditional symbolic representations or algorithms linked to proportional reasoning are not familiar for Finnish fifth graders and was therefore chosen as a domain to assess students’ intuitive problem-solving skills and strategies in non-routine problems. Proportional reasoning is often described as a cornerstone to higher mathematical and scientific thinking and cognitive development (e.g. Lesh et al., 1988; Lamon, 2007; 2012). Understanding proportionality requires reasoning with ratios. In textbooks and mathematics dictionaries the word proportion is often defined as an equivalence of ratios or statement of equal ratios or fractions, written as follows: 𝑎𝑎 𝑏𝑏= 𝑐𝑐 𝑑𝑑 or a : b = c : d. Proportional reasoning requires skills to convey the same relationship for example in producing or comparing ratios or finding a missing value. Abilities to reason proportionally are a marker of a move towards more developed forms of reasoning and form a foundation for example for algebra. Previous research indicates that students are capable of solving proportional word problems already during their early years of primary school (e.g. Tourniaire, 1986; Van Dooren et al., 2005; Vanluydt et al., 2019). Understanding ratio and proportion requires the ability to reason with multiplicative relationships and distinguish them from relationships, which are additive in nature (Van Dooren et al., 2010; Son, 2013). In an additive approach, the student operates with an invariant difference between two values, whereas a multiplicative approach requires an understanding of an invariant ratio between two values (Van Dooren et al., 2010). Even if some proportional reasoning tasks can be solved by additive approaches, also in those situations students need to understand the co-varying situation of given values. Building-up or scaling-down by skip-counting until the anticipated value is reached represents one of the strategies, which KAITERA & HARMOINEN (2022) 117 often bases in additive reasoning. These types of solution methods can be supported as steps towards multiplicative and proportional strategies. Reasoning is an integral part of mathematical problem-solving and skills reach beyond solving routine problems. Reasoning requires logical and systematic thinking, being a process, which requires making conclusions on how to achieve certain goals; these conclusions guide problem-solving and decision-making behaviour (Leighton, 2004; Grønmo et al., 2013). Students make notions on patterns and regularities and use that information on making decisions on problem-solving approaches. Reasoning involves skills to make conjectures, logical deductions based on assumptions and rules, and abilities to justify results. (Grønmo et al., 2013, p. 27.) Teachers can help students to develop these skills by presenting mathematical problems linked to unfamiliar contexts and providing opportunities to solve open-ended or multi-step problems (e.g. Grønmo et al., 2013). This has not been typically encouraged in school culture (e.g. Pehkonen et al., 2013). Close-ended textbook examples do not necessarily support students’ skills to apply the learned procedures or algorithms outside the school context, and the applications to real-world situations can seem rare to them. 3 Teaching experiment: Heuristics for problem-solving Interest towards improving primary-aged students’ mathematical problem-solving skills was based on data, which was collected in Finland and Indonesia in 2014-2015 for Kaitera’s doctoral research. A preliminary analysis of the mentioned data indicated that Finnish students had severe difficulties in explaining their thinking in tasks. This led to wondering whether these skills could be developed by implementing a teaching approach, which provided tools for solving a wide variety of out-of-the-textbook problems. The teaching experiment was carried out during the following academic year in a class of fifth graders. The learning environment was designed to support the development of students’ mathematical problem-solving skills. The quasi-experimental design was conducted in real-world learning settings, attempting to discover aspects that could be useful for example for teachers aiming to develop mathematics teaching practices. Teaching heuristics for mathematical problem solving is often linked to working with students with challenges in learning mathematics (e.g. Gallagher Landi, 2001; Fuchs & Fuchs, 2003; Swanson et al., 2013). General heuristics are not associated directly to certain kinds of mathematical problems and therefore can facilitate integrating the given information with steps for action (Swanson et al., 2013, p. 170). LUMAT 118 This report suggests that any student would benefit from getting familiar with a range of generalisable problem-solving approaches instead of just learning a variety of algorithms fit for certain types of mathematical problems. 3.1 Participants and background for the research Research was carried out in a large urban school in Northern Finland with a class of 25 fifth graders (12 boys and 13 girls). In the beginning of academic year 2015-2016, students’ skills and strategies were mapped by a pre-test with proportional reasoning problems. At that time, students’ mean age was 11 years and 2 months (range from 10 years and 9 months to 11 years and 7 months). During the autumn semester, the class got familiar with a range of generalisable heuristics, which were used in solving a variety of mathematical problems. Participating class followed the guidelines of mathematics education outlined in the Finnish National Core Curriculum. Students had attended five years of elementary school, but not received any formal instructions in solving proportional reasoning tasks, which were the main domain for assessing the development of mathematical problem-solving skills in this research. The class-teacher had a degree as a Master of Education and had been teaching for 10 years in primary and secondary schools. She was working on her Doctoral research on mathematical problem-solving, and the study described in this report was carried out of an interest towards developing students’ problem-solving skills. At the beginning of the fifth school year, it appeared that many students seemed to struggle in mathematical tasks and especially in explaining their problem-solving processes in written form. Students were introduced to concrete tools called Problem-solving Keys, which were modified from Strategy Keys based on work by Herold-Blasius (2021). The aim was to provide students with a visual reminder of heuristics for mathematical problem-solving tasks. Similar heuristics were outlined also in the Singaporean Mathematics Syllabus 2013 and used as a reference when classifying and modifying the Keys for teaching purposes in Finland. Fifth graders had three mathematics lessons every week. Mathematics textbooks were used, but in addition to those, during the autumn semester the class spent on a weekly basis on average one mathematics lesson on working with mathematical tasks in a practical context and learning a variety of general heuristics for problem-solving. Out-of-the-textbook problems were solved during the spring semester, too, but learning heuristics was not the focus anymore. Post-test data was collected at the end of the fifth grade in 2016 by using the same test than in the beginning of the school KAITERA & HARMOINEN (2022) 119 year. At that time, the mean age of students was 12 years (range from 11 years and 7 months to 12 years and 5 months). 3.2 Framework for practicing mathematical problem-solving The central idea of study in a real-life context was to teach mathematics and general heuristics through solving a variety of out-of-the-textbook problems. Mathematical problems were sourced for example from everyday situations, children’s literacy, and local, national and international news. In addition to that, students created word problems for their peers and learned to solve them in various ways. Problems were often integrated into other subjects, such as Environmental Studies and other Science themes, Finnish as a mother tongue and Arts and Crafts. Exploration of mathematical problems followed a framework with different phases of problem-solving (Stein et al., 2008; OECD, 2014, p. 31): first, the task was presented by the teacher to the students (a launch phase), then students worked on problems either in small groups or individually (an exploration phase, planning and executing) and finally the outcomes were shared and discussed (a summarising and reflecting phase). In practice, the process was not a linear, step-by-step progressing path, but rather a flexible model for moving between different phases. Quite often discussing and sharing the ideas led to returning to the exploration phase and assessing the problem-solving approaches from new perspectives. Polya’s (1945/1973) four step model was followed especially during the exploration and summarising phases. Problems were solved in collaborative settings always when it was possible: this enabled discussion and made the importance of justifying thinking more visible. Heuristics or general techniques for solving mathematical problems were introduced to students by using a visual tool called Problem-solving Keys, which are based on for example Polya’s (1945/1973) and Bruder and Collet’s (2011) heuristics, and the same ideas were outlined in Singaporean Mathematics Curriculum 2013. These heuristics were modified into a concrete tool by Herold-Blasius and Rott (2016) and named as Strategy Keys. They describe these tools as “door openers” for a problem-solving process and reminders of general heuristics that students have learned (Herold-Blasius & Rott, 2016; Herold-Blasius, 2021). Keys were modified for teaching experiment purposes, translated in Finnish, and renamed as Problem-solving Keys. Keys that were used in this study were chosen based on their generalisability, transferability and fit for the mathematics curriculum LUMAT 120 for this age group. The guidance progressed by introducing one or two keys (heuristics) at the time, linking them in a variety of out-of-the-textbook problems. As the new heuristic was introduced and practised, the key linked to that particular heuristic was added to a student’s personal “Problem-solving key chain”. Each key was linked to a mathematical problem, which was often open-ended, or at least had multiple different solution paths to choose from. The problem was chosen so that the heuristic in that Key worked well in solving a particular problem: for example, Gravett’s The Rabbit Problem (2009) was based in Fibonacci’s approach and used when practising the problem solving by using a table. Literature often offers an excellent context to bring abstract and complicated concepts closer to real-world situations. The following Figure 1 shortly illustrates the keys which were chosen as a focus area in this study, and some prompts, which were presented in guiding the learning processes. Figure 1. Examples of Problem-solving Keys and prompts presented to students. In addition to the keys described in Figure 1, students had three additional keys, which were called “When I’m stuck” -keys: KAITERA & HARMOINEN (2022) 121 • Read the task again, • Guess and check, • Solve part of the problem. These ideas often enabled either using other heuristics or continuing with other steps in the process. Key called “Solve part of the problem” turned out to be well used. Breaking the problem down into more approachable steps and solving even a small part of the problem opened new insights on how to proceed in tasks. The notion is not new: Duncker (1945, p. 8) linked “reformulation of the original problem” as one of the important characteristics in the problem-solving process and referred to this reformulation as a step or phase on a path towards solution. As Kilpatrick (2016, p. 45) points out, it might be easier to solve the problem if it is broken into smaller pieces or modified into another form. It is important that teachers value attempts for intuitive problem-solving methods and will be able to guide the student forward. Children often use everyday logic and apply that also to mathematical problems. They can be invited to justify their thinking and invent proofs for their ideas. Later students should learn about mathematical proof and formalities. They need to recognise that there is a difference between a guess, a conjecture, and a proven assertion. It is important to encourage students to wonder why things are as they are and guide them in providing a logical chain of reasons as the explanation. (Goldenberg et al., 2003, p. 24.) An educated guess differs from a random guess by its metacognitive aspects. True mathematical problem solving is challenging, but at the same time rewarding for both students and the teacher, as Schoenfeld (1992, p. 354) points out. Students benefit from having opportunities to explain their thinking not only by using mathematical language, but also pictorial and natural language: possibilities to draw and write during the problem-solving process may strengthen the understanding of mathematical concepts and contribute to mathematical thinking skills (Joutsenlahti & Kulju, 2017). Open-ended problems or planted error tasks are excellent domains for developing students’ skills in negotiating and articulating their mathematical ideas to others. According to D’Ambrosio and Prevost (2008, p. 276) “all contributions should be valued and respected“. By assessing students’ solution methods, also the self-generated ones, teachers can correct the ones which are mathematically acceptable, or guide students forward in partially constructed explanations. Classroom discussions provide crucial information on students’ understanding on topic and problem-solving processes. Effective teaching includes LUMAT 122 listening to students’ ideas and explanations and using that information as a guide in making decisions on instruction (e.g. Lester, 2013; Ivars et al. 2020; Shaughnessy et al., 2021). These views were at the centre point of the study, because a variety of out-of-the-textbook problems enabled interesting mathematical discussions in the classroom. Conversations were emphasized as important steps in learning problem-solving. Students were advised and expected to show their thinking in tasks by writing down the calculations or drawing the stages in solving the problem in a mathematically understandable way. That can be surprisingly difficult even for the 10-12-year-old students, who have already attended several mathematics lessons per week for multiple years. 4 Mapping the problem-solving skills 4.1 Data collection instruments Students’ performance was assessed by individually completed paper-and-pencil tests, which were taken in the beginning and in the end of fifth grade. Tasks included different types of proportional reasoning problems and are presented in more detail in Table 1 and Table 2. Students had a 45-minute lesson to complete the pre- and post-tests, but most of them used 20-30 minutes for tasks. Multiple-choice questions 1-5 and 8 represented typical comparison problems, in which students needed to determine the relationship(s) of two or more ratios, for example by judging whether one ratio is greater or less than the other one(s) or are they equal. In task three with mixtures two of the given ratios were similar. The rationale for having two equivalent ratios in the task was to map whether students were favouring one of the choices over the other, in this case whether they took the first choice, 2:4 or rather chose 1:2, which is used in several everyday contexts. KAITERA & HARMOINEN (2022) 123 Table 1. Proportional reasoning tasks with ratios (multiple-choice questions) Item Description Context and source 1. Lemon tea Mum is making lemon tea. She mixes tea and sugar in a jug. Which one tastes the most sweet? Choose. 1 glass of tea and 1 spoon of sugar 4 glasses of tea and 4 spoons of sugar 1 glass of tea and 3 spoons of sugar Comparing ratios in mixtures (qualitative comparison), adapted from Noelting (1980) and Kaput and West (1994) 2. Lemon tea Which one tastes the most sweet? Choose. 1 glass of tea and 2 spoons of sugar 2 glasses of tea and 2 spoons of sugar 2 glasses of tea and 1 spoon of sugar 3. Lemon tea Which one tastes the most sweet? Choose. 2 glasses of tea and 3 spoons of sugar 1 glass of tea and 2 spoons of sugar 2 glasses of tea and 3 spoons of sugar 4. Lemon tea Which one tastes the most sweet? Choose. 2 glasses of tea and 3 spoons of sugar 1 glass of tea and 2 spoons of sugar 1 glass of tea and 3 spoons of sugar 5. Lemon tea Which one tastes the most sweet? Choose. 6 glasses of tea and 3 spoons of sugar 5 glass of tea and 2 spoons of sugar 5 glasses of tea and 3 spoons of sugar 8. Paint-mixture Green paint is made by mixing two buckets of blue paint and three buckets of yellow paint. The painter needs to get more paint. How many buckets of blue and yellow paint does he need to get the exactly same shade of green? Choose one option. 3 buckets of blue and 4 buckets of yellow paint 4 buckets of blue and 6 buckets of yellow paint 5 buckets of blue and 6 buckets of yellow paint 6 buckets of blue and 8 buckets of yellow paint Comparing ratios (quantitative comparison), similarity, adapted from Tourniaire (1986) Tasks 6A, 6B, 7 and 9 required proportional reasoning with ratios, inverse proportionality or similarity of mixtures. Students were explicitly asked to record their thinking in these tasks and explain their problem-solving processes by mathematical, pictorial and/or natural language in written form. LUMAT 124 Table 2. Tasks used in assessing strategies: Proportional reasoning tasks with ratios, inverse proportionality or similarity of mixtures Item Description Context and source 6A. Rectangles Students are building geometric shapes. They make two similar rectangles and triangles by using short and long sticks. How many sticks do they need in x? Determining a missing value with continuous ratio-preserving, geometric similarity, idea adapted from Mr. Tall and Mr. Small tasks by Karplus et al. (1974), Lamon (1993) and research by Son (2013) 6B. Triangles How many sticks do they need in x? 7. Painters Six painters paint a house in three days. If they all work at the same speed, how many painters would be needed to paint the same house in one day? Inverse proportionality, item was created for this research 9. Paint-mixture Orange paint is made by mixing four buckets of yellow paint and one bucket of red paint. To get exactly the same shade of orange, how many buckets of red would the painter need to mix to six buckets of yellow? Comparing ratios (quantitative comparison), similarity, adapted from Tourniaire (1986) Timing of pre-test was before students were introduced Problem-solving Keys, and therefore they were not used while solving the items. Post-test was in the end of the school year and students were allowed to use their “key chain”, if they wished, in a similar way they could during the ordinary mathematical tests as well. None of the students felt that they needed the Problem-solving Keys at the post-test in June. The aim of this study was not to map the role or usage of these tools for heuristics but would be another interesting viewpoint for the future research (see Herold-Blasius, 2021). Problem-solving Keys were on a very important role when teaching different KAITERA & HARMOINEN (2022) 125 ways to approach a wide variety of non-routine mathematical problems especially during the autumn semester. 4.2 Data analysis First, students’ overall test performance in tasks 1-9 was assessed by awarding points on correct and erroneous answers, but also on intermediate steps towards correct explanation. In multiple choice items 1-5 and 8, students received 1 point for a correct answer and 0 points for an erroneous one. In item 3, students were expected to choose both options A and B to gain 1 point, and 0,5 points were given, if they chose either A or B. Maximum points for multiple choice questions were 6. In items 6A, 6B, 7 and 9 students were expected to explain their thinking, and their answers were serving as a base for building a framework for correct and erroneous strategies from intuitive to more sophisticated ones. Maximum points for these items were 2 points for each. The in-between marks were the following: • 0 points: erroneous explanation and/or answer, or no answer provided • 0,5 points: some explanation towards correct answer provided, answer incorrect • 1 point: no explanation provided, answer correct • 1,5 points: some explanation provided, answer correct • 2 points: correct explanation provided, answer correct. With this grading, it was possible to gain a maximum of 8 points in items 6A, 6B, 7 and 9. This approach was close to rating used in school mathematics tests for this age group, and took also the partially correct answers into account. Numerical scores were used as indicators of overall performance and possible development between pre- and post-tests. Maximum points for the whole test were 14. Exploring and mapping the strategies that students used in task began by dismantling the data (students’ responses in items 6A, 6B, 7 and 9). This was done on a detailed level by creating codes based on how students justified their thinking and explained it by using numbers, drawings or written explanations. Coding was concluded with Grounded Theory methods, which provide systematic, yet flexible guidelines for collecting and analysing data (Charmaz, 2014; Birks & Mills, 2015; Chun Tie et al., 2019). Written explanations were worked through in three phases of analysis (Charmaz, 2014), and the framework for coding was created by classifying similar responses to sub-categories (focused coding phase) and core categories LUMAT 126 (theoretical coding phase). This scheme was used as the analytical tool to assess the strategies that students used in solving tasks and on the other hand, as an indicator on whether the teaching approach provoked a shift from intuitive to more sophisticated heuristics linked to proportional reasoning. Table 3 illustrates students’ correct answers in Task 6A, and how they were grouped as sub-categories during the focused coding phase. Table 3. Dismantling the data during the initial coding phase and sub-categories in focused coding phase: example from correct approaches in Task 6A Initial coding phase: observable behaviour Focused coding phase: sub-category Student understands that the long side on the second rectangle is “three times longer” than the corresponding side on the first rectangle. Demonstrates thinking by addition: 20+20+20=60 and 15+15+15=45, but cannot clearly explain where ”three times” comes from. Demonstration of relative thinking between given quantities but failing to provide mathematically understandable explanations. Student understands that 20 long sticks = 60 short sticks by comparing corresponding parts but cannot explain how he/she gets x=45. Student calculates the ratio between the sides of the first rectangle and applies the same logic to another picture. Demonstration of relative thinking between quantities e.g. by using ratio as a unit in calculations, but not necessarily able to create generalisable formulas. Student calculates that on the first rectangle the vertical side is ¾ of the horizontal side and applies the logic to the second rectangle to determine x (for example by deducting ¼ from 60). Student understands that long sticks are three times longer than short sticks, and is able to utilize the knowledge to solve missing value x. Student works with both rectangles simultaneously by using the ratio 3:4 to solve the missing value (ability to form generalisable calculations). Use of formal operations based on ratio or use of a certain algorithm, such as cross-multiplication or “rule of three”. Student uses a formula, such as a cross-multiplication algorithm, “rule of three” or equivalent to solve the task. Consistency for the coding scheme was ensured by comparing the original data in several phases of the coding by student to another student, student’s answer to anticipated strategy and strategy by strategy. This involved repeated visits to original answers to ensure that they were understood and interpreted correctly. Final scheme for coding can be found in Appendix 1. KAITERA & HARMOINEN (2022) 127 5 Results 5.1 Performance in tasks Tasks 1-5 and 8 were multiple choice mixture tasks, and even though the student’s choice of option could give some indication on solution strategy as well, this report focuses on analysis and classification of solution approaches in tasks 6A, 6B, 7 and 9. Students’ performance in tests provided insights on whether the teaching approach, which focused on mathematical problem-solving, improved students’ general skills in solving also proportional reasoning tasks. In the beginning of fifth grade, the mean for total score in the proportional reasoning test was 6,1 points (SD 2,5 p.). Boys (N=12) performed better than girls, their mean being 6,5 points (SD 2,1 p., minimum 3,5 p. and maximum 11 p.), whereas girls (N=12, one being absent) had a mean of 5,5 points (SD 2,9 p., minimum 0,5 p. and maximum 12 p.). After getting familiar with a variety of different heuristics (but not explicitly algorithms) for solving mathematical problems, the post-test in June indicated positive results: the mean score of students had risen to 8,9 points (SD 3,6 p.). It was interesting to notice that this time girls performed better than boys. Female students’ mean had risen from pre-tests’ 5,5 points to 9,3 points (SD 3,3 p., min. 5 p. and max. 13,5 p.). Male students also improved their performance: in the pre-test they had a mean of 6,5 points and in the post-test 8,4 points (SD 4 p., min. 0,5 p., max. 13,5 p.). Development of total points is illustrated in Figure 2. Figure 2. Total points in pre- and post-tests. LUMAT 128 The following Figure 3 visualises individual students’ performance. Blue marks indicate an individual’s total points in the beginning of the fifth grade, whereas orange marks are for post-test points in the end of the school year. Development of skills was visible especially among those students, who in the pre-test scored below the average points, but it seems that the intervention had a positive influence on skills of almost all students1. Figure 3. Development of total points by individual students in problem-solving pre-test in August and post-test in July. Difficulty of tasks is often linked to the number structure and numerical complexity. For example, mathematical problems with small, integer ratios are easier than tasks with non-integer ratios (e.g. Tourniaire, 1986). For assessing the difficulty of items, students’ answers were combined with a larger set of data from Finnish fifth graders’, which completed the same test. Difficulty of items was done by assessing frequencies of correct and erroneous answers by 95 students. Items in the test sheets 1 Student number 24 was absent during pre-test tasks 7-9 and the total points are not calculated. In the post-test, student 1 did not answer any of the questions 6-9, which affected the final score. Student 11 left several tasks unanswered, or it was not possible to determine the answer. KAITERA & HARMOINEN (2022) 129 were designed to get gradually more challenging, but it seems that the difficulty of items in this test did not match students’ skills and therefore the interpretations on students’ performance need to be addressed with reservations. Tasks 1-5 were too easy for fifth graders, and on the other hand the success rate in tasks 6A, 6B and 9 was 20-28%. With this setting, the difficulty of task 7 was fairly ideal (success rate 58%) and task 8 was almost too difficult. If the results of the post-test would be considered, too difficult items would appear to be more ideal also in tasks, which required skills to explain reasoning. Results indicate that there was no significant improvement in how students performed in multiple-choice mixture tasks in pre- and post-tests. High success rates suggest that tasks 1-5 were easy for fifth graders at the first place. Development of skills is visible in more difficult tasks 6, 7, 8 and 9, which are discussed in more detail in Chapter 5.2. These results indicate that the teaching approach with a focus on problem-solving may have had a positive influence on students’ abilities to solve tasks, which require proportional reasoning skills. Mathematical problems were usually not presented in a written form in a similar way as typical word problems in mathematics books. Students did not get any extra training in solving word problems and therefore the development of skills cannot be explained by them getting more fluent in solving mathematical problems presented in written form. A table describing students’ ability to solve tasks correctly can be accessed in Appendix 2 and will be discussed task by task in the next sub-chapter. 5.2 Strategies in tasks One of the aims of the study was to find out if teaching approach, which offered tools for heuristics, improved students’ skills in explaining their thinking in mathematical tasks. In addition to getting familiar with problem-solving techniques, the practical aim for the intervention was to build up students’ mathematical self-confidence so that they would become active in describing their problem-solving processes. Questions 6A, 6B, 7 and 9 were assessed as indicators, if students were able to give an understandable explanation on how they processed the task. Informal techniques and strategies provided an insight on how students understood problem-solving concepts and were able to progress even in an unfamiliar type of a problem. Explanations and heuristics were also assessed to see if there were differences in students’ use of correct and erroneous strategies between the pre-test and the post-test. LUMAT 130 In the beginning of fifth grade, students had major difficulties in describing their problem-solving path and often left the explanation completely out. Teaching approach, which encouraged students to describe their thinking even with partially complete explanations and solving problems one step after another, seemed to have a positive impact on their performance during the later phases of the academic year. 5.2.1 Task 6A: Rectangles Tasks 6A and 6B represented typical proportional reasoning problems with a missing value. According to Karplus et al. (1983, p. 21), these types of problems involve “reasoning in a system of two variables between which there exists a linear functional relationship”. To maintain proportional values, students carry out parallel transformations within or between variables (Son, 2013). The relation between quantities is invariant, whereas the quantities in the problem co-vary. In task 6A, a correct approach required the ability to compare corresponding parts between two rectangles. 17 students (68%) provided an answer to the question 6A, and six students (24%) were able to solve the task correctly. 10 students out of 17 were able to explain their solution process, whether the answer was erroneous or correct. Almost a quarter of all students (N=6) were skilled enough to explain their thinking with the correct approach. Seven students provided an answer but did not explain how they ended up in that. Eight students (32%) did not answer the question at all. It appears that the problem-based teaching approach had a positive impact on students’ skills: in the post-test 92% of students (N=23) answered the question and 68% (N=17) were able to provide a correct answer. 20 students out of 23, who answered the question, were able to explain their thinking in written form. Almost a half (N=12) of all students in the post-test approached the task with the correct strategy. Only three students answered the question but did not explain their thinking and two students (8%) did not answer the question at all. In pre-test 16% (N=4) were able to implement a correct ratio or unit factor approach in task 6A, demonstrating relative thinking between quantities in solving the unknown quantity. In post-test the number of students using a correct strategy had more than doubled, being 40% (N=10). Even though in many cases an explanation for the solution process did not include all the mathematically correct steps, students were demonstrating the understanding of long sticks being three times longer than the short sticks. KAITERA & HARMOINEN (2022) 131 During the pre-test, the most common erroneous strategy was additive reasoning (16%, N=4). It was typical to focus on dimensions within one rectangle, for example reasoning that because the difference between the sides of the first figure was five (20-15=5), the same difference applies for the second figure (60-5=55). In some cases, students calculated the perimeter of the first rectangle and tried to apply or modify the logic to find the missing value in the second rectangle. In both examples students failed to understand the relational nature of the task: if 20 long sticks equal the length of 60 short sticks, the same ratio should be maintained with 15 long sticks and x short sticks. According to the previous research, students often rely on additive strategies also in multiplicative situations (e.g. Tourniaire & Pulos, 1985; Nunes & Bryant, 1996; Van Dooren et al., 2010; Son, 2013). Still, it is not clear how students choose their preferations between additive and multiplicative relations (Vanluydt et al., 2019). Both approaches can be characterised as intuitive in nature, yet it is difficult to verbally describe reasoning; the given explanations are not necessarily in line with students’ actual solution processes (Degrande et al., 2020). Distinguishing multiplicative missing value problems from additive ones is challenging for students. Additive thinking is emphasized during the first years of school and the transition towards multiplicative ideas is not always straightforward. On the other hand, additive reasoning could support the development of multiplicative reasoning. Yet, the shift from additive to multiplicative thinking requires a qualitative change in thinking (e.g. Nunes & Bryant, 1996). In the beginning of the fifth grade, only one student approached the problem via multiplicative reasoning but ended up in an erroneous end-result. After the intervention, one fifth (N=5) of students turned into this approach. Even though these solution attempts were erroneous, they could be interpreted as a shift towards understanding the relative nature of the task. A more detailed description on the range of strategies that students used can be accessed in Appendix 3. Development of solution approaches and possible shifts between the strategies was visualised as individual students’ performance in tasks. In Figure 4, explanation categories are presented in an order, which suggests a hierarchy from erroneous and intuitive ones to more sophisticated and generalisable strategies. Light green area marks correct approaches. Opaque fill-ins in pre- and post-test markers indicate that the student was able to solve the task correctly. LUMAT 132 Figure 4. Individual students’ strategies in pre- and post-tests in task 6A. Correct solution approaches were rare in the pre-test, even though the task was relatively easy. In the end of the fifth school year the frequency for correct strategies had increased and students were able to approach the task by correct ratio or unit factor approach. 5.2.2 Task 6B: Triangles Task 6B was more difficult than 6A. It would have been possible to solve the task only by focusing on dimensions on one triangle and using Pythagorean theorem, but that is a topic for Finnish secondary school curriculum and therefore not expected that any of the students would use that algorithm. In the pre-test 16 students (64%) answered task 6B and only two (8%) of them solved the task correctly. Seven students of 16 explained their thinking process in writing, but only one of them was able to choose a correct strategy. Nine students gave an answer, but no explanation. Nine students (36%) did not answer the question 6B in pre-test. Before the intervention, students had difficulties in explaining their thinking, 72% of students (N=18) either leaving the explanation out (N=9) or not answering the question at all (N=9). By the end of the school year, the number of empty explanation spaces (36%, N=9) had decreased to half, even though the task was challenging. In the KAITERA & HARMOINEN (2022) 133 post-test 76% of students (N=19) answered the question and less than a quarter did not (N=6). 48% (N=12) of students were able to provide a correct answer. The majority, 16 students out of 19, tried to explain their thinking in a written form. Three students answered the question but did not provide any insights on the solution process. In the pre-test only one student was able to choose a correct strategy, but in the post-test the number increased to seven students (37% of 19 students answering this question). This was quite an interesting finding, because students had not encountered any similar mathematical problems during the academic year. In the pre-test, only one student was able to explain thinking by demonstrating mathematically correct reasoning. During the intervention the variety of correct solution strategies increased. Students came to conclusions by additive reasoning or more sophisticated multiplicative reasoning, and there were also some examples of abilities to create correct, generalisable formulas to solve these types of problems. None of the students solved the task by using ratio or unit factor. Students often relied on erroneous intuitive strategies, such as trying to solve the problem by random calculations on given numbers or basing the problem-solving process on visual observations on given pictures, and not mathematically valid concepts. The range of observable strategies in this task can be accessed in Appendix 3. Figure 5 illustrates the changes in used strategies that individual students had between from the pre-test and to the post-test. Figure 5. Individual students’ strategies in pre- and post-tests in task 6B. LUMAT 134 As one can see from Figure 5, task 6B was more difficult than 6A for this student group. After the teaching experiment, successful students chose usually correct additive or multiplicative reasoning, but many students left the explanation out still during the post-test. 5.2.3 Task 7: Painters Seventh task was based on inverse proportionality. Painting a house involved a situation, in which the time spent on painting was reduced from three days to one day, and students were calculating the number of people needed in painting work. In this task, it was crucial to understand that it would take three times as many painters to complete the work in 1/3 of the time. The analysis of students’ responses raised a question, whether many of them solved the task correctly without really understanding the concept. Due to the numerical structure in this task, it was possible to end up in a correct answer of 18 painters by simply multiplying the word problems’ given numbers, six and three. 67% (N=16) of students in the pre-test solved the problem correctly, and in the post-test the frequency had increased to 80% (N=20). Only one student in both tests did not answer the question at all. In the pre-test 70% of students (N=16) who answered the questions also explained their thinking, but only one of them was able to choose a correct strategy. In the post-test 24 out of 25 students gave an explanation on their solution process, and at that point 83% (N=20) of them used the correct approach. In the post-test none of them left the explanation slot empty. High success rates in all student groups are possibly linked also to the possible bias caused by the number structure. Majority of students based their explanation on this particular task simply stating 3x6=18 but did not provide any additional information on how they were thinking, or where the numbers came from. Only a few of the participants with correct answers were able to express that they understood the concept instead of performing a random calculation. They, for example, reasoned the number of painters by building up or scaling down with the figures (e.g., 6 painters=3 days, 12 painters=2 days, 18 painters=1 day) or used the addition or multiplication, but were rarely able to justify, why they chose certain procedures. Even though multiplicative reasoning was the most common correct strategy, it is difficult to assess whether the concept of inverse proportionality was really understood. Range of strategies in task 7 can be accessed in Appendix 4 and development of strategies between pre- and post-tests in Figure 6. KAITERA & HARMOINEN (2022) 135 Figure 6. Individual students’ strategies in pre- and post-tests in task 7. It is likely that the wording and the number structure of this task also guided the choice of erroneous approaches: students often relied on erroneous multiplicative reasoning, which was the most common erroneous strategy. To map the real understanding of inverse proportionality, the task could be worded for example by “Two painters paint the house in three days. If they all work at the same speed, how many painters would be needed to paint the same house in two days?” In this case, multiplying two by three would not result in a correct answer. 5.2.4 Task 9: Paint-mixtures Task 9 was a mixture task, in which students had to maintain the same ratio of paint buckets per mixture to determine the missing value (number of red paint buckets) for the similar mixture. This was a difficult test item for fifth graders, but on the other hand, provided interesting insights on students’ development of problem-solving skills. During the pre-test 29% of students (N=7) did not answer the question at all, whereas the percentage in the comparison group was lower, 14% (N=7). In the post-test, only two students did not answer the question and in both cases, they expressed their unwillingness to engage with the task at all. LUMAT 136 17% of students (N=4) had a correct answer in pre-test, but clearly struggled in providing explanations on their reasoning processes: 12 out of 17 students, who answered the question, left the explanation out. Only one student was able to choose the correct strategy in this task, the other four relied on erroneous approaches. Post-test results indicated significant improvement. 52% of students (N=13) ended up with a correct answer, and 10 out of 23 students answering this question also described their thinking with a correct strategy. When having a closer look on strategies that students used, in the pre-test only one student was able to provide an explanation while solving the problem correctly, turning into a building-up strategy. Development of skills was visible in the post-test: more students were able to not only explain their correct problem-solving process, but also use a more sophisticated strategy by working with the ratio. Even though students did not necessarily have skills to explain thinking with mathematically valid expressions, they became more confident in using different strategies. Figure 7 illustrates the ratio approach, in which student proceeds one step at the time. In this example, the student correctly reasons that because there is one red paint bucket in every four yellow paint buckets, you need to add 1,5 buckets of red to six buckets of yellow. Figure 7. Correct example in item 9 (student 42159 in post-test). During the pre-test, only one of the students was able to provide a correct explanation for the task, working with building-up strategy. In the end of the school year there were indications on improved skills of explaining thinking also visible: 32% (N=8) utilised either building-up or scaling-down strategy, ratio or unit factor approach (the most common) or even correct formal operations with generalisable formulas. In the post-test, five students (20%) were able to work through the task by expressing that for every two buckets of yellow you need 0,5 buckets of red paint. 54% (N=13) of students gave an erroneous answer in the pre-test. Three students relied on multiplicative reasoning but failed to understand the relative nature of the KAITERA & HARMOINEN (2022) 137 task. They multiplied the given amounts of yellow paint, 6x4=24 or only stated “Calculated by multiplication”, without providing a more detailed explanation. During the post-test the most common erroneous strategy was additive reasoning. 24% (N=6) of students chose that strategy. They often based their reasoning on the idea that “you need three more yellow than red”, focusing on the difference between the given numbers in the original paint mixture and ignoring the need to maintain the same relationship for the second paint. More detailed frequencies for the strategies visible in task 9 can be accessed in Appendix 4. Figure 8. Individual students’ strategies in pre- and post-tests in task 9. Assessing and classifying students’ strategies was not always straightforward. For example, student could state that multiplication was needed, but on the other hand, relied on additive reasoning when providing an answer: “Because in the beginning you needed three more yellow buckets than red buckets, so you just need to multiply it”, providing three as an answer. LUMAT 138 5.2.5 Students with the lowest and highest points To have a closer look on possible development of strategies of so-called low- and high-performing students, the performance of three students with lowest points and four students with the highest points in the pre-test were considered. Three low-performing students gained a maximum of 3,5 points in the pre-test and four high-performing students 8,5-12 points (for the overview of students’ performance, see Figure 3). Figure 9. Development of strategies of three students scoring the lowest points in pre-test. These three students tended to leave the answers completely out in the beginning of the fifth grade. By the end of the fifth grade, frequencies for solving the tasks correctly increased. With some individuals the difference was remarkable: for example, student 7 got correct answers in the post-test, but would still have needed a bit of support in explaining thinking (see Figure 9). After the problem-based teaching period, students were more willing to engage in attempts to solve mathematical problems, even though the strategies might not have been valid. With a correctly KAITERA & HARMOINEN (2022) 139 timed intervention the teacher has a change to support the shift from erroneous strategies to correct ones. If lacking the skills to explain thinking with mathematically correct processes, students often started to explore the dimensions between the given values by implementing intuitive methods. Consider the explanation in Figure 10 that student 22 gave in Task 6B: the answer was correct 40 sticks, and in this case, the student seemed to calculate the solution by exploring the given values and their relationships within the first triangle. This student calculated the difference between the hypotenuse and opposite side is multiplied by two to get the adjacent side. Figure 10. It was not uncommon that the answers were correct, but not necessarily based on generalisable ideas. In these kinds of examples, which are very common in primary school, students would benefit from opportunities to discuss their ideas with the teacher or with a peer; what is the purpose of short and long sticks in this task, and how should that information guide the solution process? If the strategy works with the given values, can that be generalised to all triangles with a 90-degree angle? How about all types of triangles? Findings of the study indicated that especially students with lower points benefited from exploring different problems and heuristics to approach them. Students with high points in the pre-test were able to develop their skills in explaining their ideas and on the other hand, to move towards more advanced strategies (see Figure 11). LUMAT 140 Figure 11. Development of strategies of four students scoring the highest points in pre-test. Problem-oriented teaching approach, which emphasized the importance of discussing and explaining ideas, no matter even if they are just partially constructed or immature, seemed to have a positive influence on how students communicated their thinking in written tasks. By the end of the school year there was a significant improvement on students’ reasoning skills, use of heuristics and abilities to explain their thinking. Some limitations on these observations needs to be addressed: with this research design, it is not possible to assess, whether the skills would have been improved by more traditional teaching approach as well. Another challenge is linked to the test items: several of them appeared to be too easy for Finnish students and tasks were completed in a shorter time than expected. A test with a wider variety of difficulty and more items would provide more reliable information on possible development of skills and strategies. 6 Discussion The study focused on exploring whether students benefitted from a problem-solving focused teaching approach, which introduced them to a general set of heuristics as a concrete tool called Problem-solving Keys. This tool worked as visual reminders of a variety of generalisable approaches for mathematical problems. The study aimed to KAITERA & HARMOINEN (2022) 141 explore whether this kind of an active, heuristics-based teaching approach would improve fifth graders’ performance and use of strategies and develop students' skills in explaining their thinking in mathematical tasks. The analysis indicated that skills to explain thinking improved. Before the intervention, students generally relied on intuitive strategies or opted to leave the justification completely out. After getting familiar with concrete tools for general heuristics, students became more confident in expressing their ideas and justifying their strategies and were also willing to help the others by explaining solution methods. Mathematical discourse helps students not only to develop their understanding of mathematical ideas, but also to build a personal relationship with mathematics on an emotional level (D’Ambrosio & Prevost, 2008). After the intervention students’ variety of heuristics increased and they were able to choose more sophisticated ones when solving different tasks. This can be interpreted as a positive development. This development was visible also in situations, in which the student still worked by implementing an erroneous strategy: in many cases, there was a shift from an intuitive approach towards a more sophisticated, yet erroneous approach. For example, an ability to base decisions on multiplication and demonstrate some understanding of the relative nature of the task can be interpreted as a step towards proportional reasoning, even though the student would not be able to expand the idea to cover the whole concept on a certain task. Findings of this study suggests that also the erroneous approaches can be viewed as hierarchical steps towards more sophisticated skills and correct reasoning. Teacher has a crucial role in recognising these small steps, for example student’s transition from erroneous intuitive approaches (for example drawing) towards additive and multiplicative reasoning and emerging skills in understanding relational nature of proportional reasoning tasks. In Polya’s model (1945/1973), the last phase of “looking back” provides opportunities to assess and discuss ideas that emerged during the problem-solving process. This research underlines the importance of discussing different approaches and heuristics already during the earlier phases of problem-solving. This increases students’ confidence in presenting also the partially correct ideas, which can be seen as steps forwards. Teacher’s role is to make sure that students are not left with the impression that any answer is mathematically valid, and to guide them towards correct methods and mathematically correct language. LUMAT 142 By teaching heuristics, students learn to solve complex word problems, reason mathematically in everyday situations and develop their thinking skills. Heuristics should be understood as general guidelines, methods, or possibilities to approach a diverse set of mathematical problems. Still, learning heuristics does not alone help students, and heuristics as such should not be reduced to learning certain techniques or sets of algorithms to choose from. Learning to describe and justify thought processes is equally important. It can be asked whether school mathematics in primary schools supports students’ development in explaining their thinking, or is the focus still on finding the correct answer? This is problematic when considering the transition to secondary school, where students are expected to be able to explain their thinking by using mathematical language. Teaching approach, which guides students in justifying their ideas by using various methods, develops mathematical problem-solving skills and creates an excellent foundation for learning more complex mathematical concepts. Classroom discussions enable the teacher to make decisions on which state students benefit from teacher’s guidance, and when it is more fruitful to let them find out the solution by themselves. A few limitations of this study need to be addressed. The data for this research was collected from one sub-urban, monolingual primary school in Northern Finland. With a larger sample from Finnish schools, it would have been possible to gain more generalisable results on whether the students’ performance and use of certain strategies would follow similar trends in schools in different areas. Another limitation is linked to the development of problem-solving skills and the possible effect that the teaching approach had on the results: it would have been beneficial to have the same pretest-posttest setting with a group of students without the intervention. At the point of implementing the teaching approach and collecting the data, this was not the main focus of the research, but the aim was to develop and assess the heuristics-based teaching approach, practicing teacher being also the researcher (e.g. Niemi & Nevgi, 2014). Further research is needed to understand the natural development of problem-solving skills and strategies, and whether and what kind of “out-of-the-textbook” approaches in mathematics classrooms could enhance these skills. Students should be provided rich mathematical problems and taught a variety of problem-solving heuristics to tackle the demands of the 21st century. Mathematics should work as a tool, which would help in facing everyday situations. Even though not everyone becomes a mathematician, students’ fluency as mathematical thinkers and problem solvers can be supported by paying attention in developing their skills KAITERA & HARMOINEN (2022) 143 already in primary school. Mathematics curriculum in Finland offers flexibility to shift from arithmetic “fill-in-the-book” exercises towards a meaningful problem-solving teaching approach. Teaching mathematics through problem-solving provides opportunities to develop a wide variety of problem-solving strategies and heuristics. Problem-solving Keys are one easily accessible tool to enhance these skills. Acknowledgements Strategy Keys as concrete tools were first introduced to the author by Raja Herold-Blasius in 2015 at the Joint Conference of ProMath and the GDM Working Group of Problem Solving in Halle, Germany. Thank you for the idea. References Baxter, G. P. & Junker, B. (2001). Designing Cognitive-Developmental Assessments: A Case Study in Proportional Reasoning. 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Helping Students with Learning Disabilities Make Sense of Word Problems. Intervention in School and Clinic, 37(1), 13-18. Goldenberg, E. P., Shteingold, N. & Feurzeig, N. (2003). Mathematical habits of mind of young children. In F. K. J. Lester (Eds.), Teaching mathematics through problem solving: Prekindergarten-Grade 6 (pp. 15-29). National Council of Teachers of Mathematics. Gravett, E. (2009). The Rabbit Problem. MacMillan Children's Books. Grønmo, L. S., Lindquist, M., Arora, A. & Mullis, I. V. S. (2013). TIMSS 2015 Mathematics Framework. In I. V. S. Mullis & M. O. Martin (Eds.), TIMSS 2015 assessment frameworks (pp. 11-27). TIMSS & PIRLS International Study Center, Lynch School of Education, Boston College and International Association for the Evaluation of Educational Achievement (IEA). Hart, K. (1984). Ratio and proportion. In K. Hart, M. Brown, D. Kerslake, D. Küchemann & G. Ruddock (Eds.), Chelsea Diagnostic Mathematics Test. Teacher's guide (pp. 93-100). NFER-Nelson. Herold-Blasius, R. (2021). Problemlösen mit Strategieschlüsseln. Eine explorative Studie zur Unterstützung von Problembearbeitungsprozessen bei Dritt- und Viertklässlern. Springer Spektrum. Herold-Blasius, R. & Rott, B. (2016). Using strategy keys as tool to influence strategy behaviour. A qualitative study. In T. Fritzlar, D. Assmus, K. Bräuning, A. Kuzle, & B. Rott (Eds.), Problem solving in mathematics education (Vol. 6, pp. 137–147). VTM. Hiebert, J. (2003). Signposts for teaching mathematics through problem solving. In F. K. J. Lester (Eds.), Teaching mathematics through problem solving: Prekindergarten-Grade 6 (pp. 53-61). National Council of Teachers of Mathematics. Ivars, P., Fernández, C. & Llinares, S. (2020). A Learning Trajectory as a Scaffold for Pre-service Teachers’ Noticing of Students’ Mathematical Understanding. International Journal of Science and Mathematics Education, 18(3), 529-548. Joutsenlahti, J. & Kulju, P. (2017). Multimodal Languaging as a Pedagogical Model—A Case Study of the Concept of Division in School Mathematics. Education Sciences, 7(1), 9. Kaitera, S. (2021). Mathematical problem-solving keys. Library of Open Educational Resources. Kaput, J. J. & West, M. M. (1994). Missing-value proportional reasoning problems: Factors affecting informal reasoning patterns. In G. Harel & J. Confrey (Eds.), The Development of Multiplicative Reasoning in the Learning of Mathematics (pp. 235–287). State University of New York Press. Karplus, E. F., Karplus, R. & Wollman, W. (1974). Intellectual Development Beyond Elementary School IV: Ratio, The Influence of Cognitive Style. School Science and Mathematics, 74(6), 476-482. Karplus, R., Pulos, S. & Stage, E. K. (1983). Early Adolescents' Proportional Reasoning on 'Rate' Problems. Educational Studies in Mathematics, 14(3), 219-233. Kilpatrick, J. (2016). Reformulating: Approaching Mathematical Problem Solving as Inquiry. In P. Felmer, E. Pehkonen & J. Kilpatrick (Eds.), Posing and Solving Mathematical Problems: Advances and New Perspectives (pp. 69-82). Springer. Lamon, S. (1993). Ratio and proportion: Children's cognitive and metacognitive processes. In T. P. Carpenter, E. Fennema & T. A. Romberg (Eds.), Rational numbers: An integration of research (pp. 131-156). Lawrence Erbaum Associates. KAITERA & HARMOINEN (2022) 145 Lamon, S. (2012). Teaching Fractions and Ratios for Understanding. Essential Content Knowledge and Instructional Strategies for Teachers (3rd Ed.). Routledge. Lamon, S. J. (2007). Rational Numbers and Proportional Reasoning. Toward a Theoretical Framework for Research. In F. Lester (Eds.), Second handbook of research on mathematics teaching and learning (pp. 629-668). Information Age Publishing. Langrall, C. W. & Swafford, J. (2000). Three balloons for two dollars: Developing proportional reasoning. Mathematics Teaching in the Middle School, 6, 254–261. Leighton, J. P. (2004). Defining and describing reason. In J. 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14458 | http://www2.cs.arizona.edu/patterns/weaving/webdocs/mo/G/Constraints.pdf | 181 Constrained Patterns Constraints limit what is possible. With respect to interlacement patterns, constraints impose both color and structural limitations. Constraints can take many forms. Expressed in terms of drawdowns, examples are: 1. The number of white cells and black cells must be equal.
2. No more than four consecutive cells in any row and column can be the same color.
3. Every cell must have at least one adjacent cell of the opposite color.
4. Constraints 1, 2, and 3 all must be satisfied.
Constraint 1 is a global constraint and is equivalent to requiring that a weave be balanced. This constraint cannot be satisfied by a drawdown with an odd number of cells. That is, of all drawdowns, only ones with even dimensions can possibly satisfy this constraint.
Constraint 2 is more local and in more familiar terms limits float length.
Constraint 3 is local; it specifies a property that all neighborhoods must have. [Cross-reference cellular automata.] Constraint 4 requires that three constraints be simultaneously satisfied. It is called a constraint set. In this sense, Constraints 1, 2, and 3 are constraint sets containing only one constraint.
Constraint Analysis Given a pattern, it generally is easy to determine if it satisfies a given constraint set. For example, whether or not a pattern satisfies Constraint Set 1 can be determined just by counting black and white cells. Similarly, whether or not a pattern satisfies Constraint Set 2 can be determined by examination or using the float-analysis feature of a weaving program.
Constraint Set 3 requires a little more work, since it may be necessary, in general, to examine a large number of individual neighborhoods.
And, of course, determining whether or not a pattern satisfies Constraint Set 4 can be determined by checking each of the constraints in its constraint set.
It is important to realize that there are constraint sets that no patterns satisfy. For example, a constraint set that contains Constraint 1 and a constraint that patterns must have an odd number of cells cannot be satisfied — it is unsatisfiable. In this example, it is obvious that the two constraints are mutually exclusive. In general, it may be difficult to determine whether or not a constraint set can be satisfied any pattern.
182 Neighborhood Constraints As far as weave structure is concerned, neighborhood constraints are interesting, since they have a strong effect on appearance.
Neighborhood constraints can be characterized by neighborhood templates.
As with drawdown automata, there are many kinds of neighborhoods that can be used. The von Neumann 5-cell neighborhood is used in what follows. This neighborhood is small enough to be computationally tractable but large enough to characterize a wide range of structural characteristics.
Neighborhood constraints can be pictured like this: None of these constraints taken alone is satisfiable, simply because they all require every cell to be the same color (white in the first, black in the other two) while simultaneously requiring that them to be surrounded by cells of other colors.
Taken in combination in constraint sets, however, they may be satisfiable.
For example, the constraint set consisting of the templates is satisfied by plain weave (and only plain weave).
On the other hand, the constraint set consisting of the templates is unsatisfiable because it requires every cell to be black and at the time to have adjacent white cells.
Neighborhood constraints can be looked at in several ways: • Does a pattern satisfy a given neighborhood constraint set?
• What neighborhood constraint set does a given pattern satisfy?
• What patterns satisfy a specific neighborhood constraint set?
The first question is easy to answer: as mentioned above, it’s only necessary to compare the cell neighborhoods to the templates in the constraint set. The second question also is generally easy to answer by cataloging the 183 neighborhoods of all cells, although there are some issues to be addressed, such as how to handle cells at the borders that do not have complete neighborhoods.
The third question is, in general, much harder to answer. It is, nonetheless, interesting. For example, it would be interesting to know what patterns satisfy the same neighborhood constraint set that a 2/2 twill does. The problem is hard because there is no is known way to construct patterns from constraint sets that does not involve a large amount of computation.
Neighborhood Analysis In the first article on constraints, we introduced the concept of neighbor-hood constraints . In this article, we’ll look at the problem of determining the neighborhood constraint set of a pattern.
Consider the following pattern: All that is necessary to determine the constraint set for this pattern is to examine every cell and record the template for its neighborhood.
For example, the template for the cell outlined above is This process is straightforward except for cells at the edges, which have incomplete neighborhoods. There are several ways to handle such cells: 184 1. Don’t include the edge cells in the analysis.
2. Assume that the pattern repeats so that the edges wrap around.
3. Don’t assume the pattern repeats (for example, the Morse-Thue carpet does not ) but include partial neighborhoods of the edge cells.
Method 1 amounts to analyzing a sub-pattern, shown by the blue outline below: The problem with this approach is that the constraint set obtained may not be complete. For example, the unit motif for plain weave is a 2 × 2 pattern: This pattern only has edge cells. If they are ignored, there is no constraint set at all, which obviously is incorrect.
Method 2 can be handled by augmenting the pattern, adding cells around the edges that correspond to what would appear if the pattern were contained in a repeat: Now the analysis can proceed for the cells enclosed in the red rectangle above; 185 effectively there are no edge cells.
This method is fine for repeating patterns, but it produces erroneous results for aperiodic patterns such as the Morse-Thue carpet.
Method 3 tries to deal with this situation by adding edges with unknown cell colors: In this case, an edge cells such as the one outlined below has a neighborhood template with a cell of unspecified color: Here is that template: It can be added this partial constraint to the set. But note that there is other cells in the pattern with complete templates that have the same three cells as the partial constraint: 186 This neighborhood, “covers” the incomplete one, it is not necessary to keep the incomplete one.
If partial constraints remain after analyzing all cells, one possibility is to just “force them” by arbitrarily coloring the unspecified cells.
What to do about an aperiodic patterns is an open question. One can analyze a portion of it using Method 3. But how can one tell if the constraint set obtained is complete? Would analyzing a larger portion add to the constraint set?
In the case of the Morse-Thue carpet, analyzing a modest portion yields a constraint set with 18 templates. Analyzing larger portions do not increase the size of the constraint set. It seems reasonable, examining the method by which the Morse-Thue carpet is constructed, that this constraint set applies to the entire, unlimited pattern.
But for other patterns, such as random ones, there is no basis for such an assumption. In fact, the constraint set for a randomly generated pattern may include all 32 possible constraints.
On the other hand, what is the point of trying to determine the neighbor-hood constraint set for a pattern without structure?
Representing Constraint Sets Neighborhood constraint sets can be represented in several ways. For human understanding, graphical methods work best. For computer processing, textual representations or numerical codes are more appropriate.
187 Graphical Representations In previous articles, we showed templates as neighborhoods laid out according to their natural geometrical interpretation, as in Any constraint set then can be represented by a collection of such template images. For example, the constraint set for plain weave is If a constraint set is large, this kind of representation takes a fair amount of space for images of a size sufficient to be readily understood. For example, the constraint set containing all constraints is A less useful but more compact graphical representation is as a bar of 32 cells, each cell corresponding to one of the constraints. If a cell is in a constraint set, it is colored gray, otherwise white. Gray is used so the black dividing lines can be used as a guide to cell position. For example, the cell bar for the plain-weave constraint set is The problem with the cell-bar representation is that the templates are coded by position, so that to determine the constraints, it’s necessary to know where individual templates are in the bar and the order of the templates (which is as shown in the image for all templates, reading left to right and top to bottom.
Determining the actual templates in this way is tedious and error prone, so the cell-bar representation is not appropriate for that purpose. It is suitable, how-ever, for getting an idea of the number of constraints in a set and comparing 188 patterns of different constraint sets.
Textual Representations The graphic representation as a series of templates has a natural counter-part as a list of 5-bit binary strings in which a bit is 1 if the corresponding cell in the neighborhood is black and 0 otherwise.
A convention is needed to determine the order of the bits in the bit string.
The convention we’ll use here numbers the cells starting with the center cell and continuing clockwise around the outer cells: 5 4 1 2 3 Therefore the plain-weave constraint set has the textual representation 01111 10000 (Since this represents a set, the order of the constraints is not important, but a useful convention is to order the binary strings by magnitude, as we have done in this example.) A more compact textual representation of constraint sets is as 32-bit binary strings in which a bit is one if the corresponding constraint is in the set and 0 otherwise. For example, the plain-weave constraint set represented in this way is 00000000000000011000000000000000 Note that although the cell-bar representation is difficulty for a human being to interpret in its entirety, the 32-bit binary string representation presents no problem for a program: It’s just another decoding task of the kind that programs have to handle all the time.
Numerical Codes A variation on the textual representations is to think of bit strings as base-2 integers. These base-2 integers then can be converted to conventional base-10 integers. For example, in terms of 5-bit constraints, the plain-weave constraint set has the numerical codes 15 16 189 while the 32-bit representation has the numerical code 98305.
For computer programs, these are just other ways of encoding and present no more problems than the textual forms. Base-10 integers have advantages for programming in some situations because all commonly used programming languages support integer arithmetic. However, the equivalents of 32-bit bit strings can be very large: as large as 4,294,967,296, which is beyond the range of integer arithmetic in most programming languages.
Numerical codes are somewhere between graphical representations, suit-able for human beings, and textual representations, suitable for computers. For human beings, they do have value as labels, if arbitrary, and are only about one-third the length of the corresponding binary strings, as well as being easier to differentiate than bit strings. von Neumann Constraint Pattern Catalog [This section has not been published on the Web.] Wolfram has shown that only 171 repeating patterns are needed to characterize all the von Neumann neighborhood constraint sets . (Rotations, reflections, and color reversals are omitted.) The pages that follow show unit weaves for these patterns in the order given in Reference 1. Below each pattern are its dimensions and, if weavable from a drawup, the loom resources required. At the bottom is a cell bar showing the constraints involved .
Some of the patterns that are unweavable as drawups can be drafted using color-and-weave effects. Obvious examples are the stripes.
All of the patterns that are weavable from drawn-up drafts “hang together” . [Cross reference; consider changing “weavable”to “draftable” here and elsewhere.] 190 191 192 193 194 195 196 197 198 199 200 |
14459 | https://pmc.ncbi.nlm.nih.gov/articles/PMC9159128/ | Patients' Knowledge and Information Needs about Isotretinoin Therapy Use in Jordan - PMC
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. 2022 Feb 14;2022:9443884. doi: 10.1155/2022/9443884
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Patients' Knowledge and Information Needs about Isotretinoin Therapy Use in Jordan
Anan S Jarab
Anan S Jarab
1 Department of Clinical Pharmacy, Faculty of Pharmacy, Jordan University of Science and Technology, P.O. Box 3030, Irbid 22110, Jordan
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1,✉, Sayer Al-Azzam
Sayer Al-Azzam
1 Department of Clinical Pharmacy, Faculty of Pharmacy, Jordan University of Science and Technology, P.O. Box 3030, Irbid 22110, Jordan
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1, Shriefa Almutairi
Shriefa Almutairi
1 Department of Clinical Pharmacy, Faculty of Pharmacy, Jordan University of Science and Technology, P.O. Box 3030, Irbid 22110, Jordan
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1, Tareq L Mukattash
Tareq L Mukattash
1 Department of Clinical Pharmacy, Faculty of Pharmacy, Jordan University of Science and Technology, P.O. Box 3030, Irbid 22110, Jordan
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1 Department of Clinical Pharmacy, Faculty of Pharmacy, Jordan University of Science and Technology, P.O. Box 3030, Irbid 22110, Jordan
Academic Editor: Giuseppe Alessandro Scardina
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Corresponding author.
Received 2021 Nov 16; Revised 2022 Jan 19; Accepted 2022 Jan 25; Collection date 2022.
Copyright © 2022 Anan S. Jarab et al.
This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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PMCID: PMC9159128 PMID: 35685597
Abstract
Background
Despite being the first-line treatment for severe or moderate acne, isotretinoin has several serious side effects that necessitate the evaluation of patients' knowledge about isotretinoin side effects and its proper use.
Objective
The current study aim was to explore information needs about isotretinoin by evaluating patients' knowledge about the appropriate use of isotretinoin and its associated side effects.
Methods
In addition to the sociodemographic variables, a validated online questionnaire was adopted from the literature to evaluate patients' knowledge about isotretinoin use and its potential side effects. Independent t-test and one-way analysis of variance (ANOVA) test were implemented to find the correlation between the study variables and the knowledge score.
Results
The most recognized side effect of isotretinoin therapy was dryness (98.1%). The study patients showed good knowledge about isotretinoin use with a mean knowledge score of 8.1 (SD = 0.7). However, more than half of them (61.0%) mistakenly thought that isotretinoin therapy should be taken continuously for more than 6 months without stop, and some of them did not know that isotretinoin is recommended to be taken with fatty meal (24%) and sunblock (24.6%). Female gender (8.2 (SD = 0.8)) and using isotretinoin for more than 6 months (8.3 (SD = 1.2)) were significantly associated with a higher knowledge score of isotretinoin use (p=0.01), when compared with male patients (7.8 (SD = 0.7)) and less than 6-month use of isotretinoin (7.7 (SD = 0.7)).
Conclusions
The lack of patients' information about the potential side effects, duration of therapy, and some instructions on isotretinoin use, such as taking the medication with fatty meal and sunblock, shed the light on the necessity to prepare leaflets, educational brochures, and educational posts via social media in order to improve patients' knowledge about isotretinoin therapy and its optimal use.
1. Introduction
As stated by the Global Burden of Disease study, acne vulgaris (AV) affects about 85% of individuals between the ages of 12 and 25 years, and its effect extends to younger children [1, 2]. Isotretinoin is approved as first-line therapy for the treatment of severe or moderate acne that does not respond to other medications [2–4]. Nevertheless, there are many side effects of isotretinoin that range from mild to life-threatening [5, 6]. The clinical adverse effects of isotretinoin can be categorized into mucocutaneous such as cheilitis and facial dermatitis and systemic such as teratogenicity [2–4]. Other potential side effects include dryness, dyslipidemia, nosebleeds, dizziness, eye inflammation, joint and back pain, depression, and abnormal liver function . Congenital defects are estimated to occur in up to 35% of infants exposed to the drug in utero, and neurocognitive deficits have been reported to affect 30–60% of children exposed to isotretinoin prenatally . On the contrary, most patients do not have enough knowledge about the potential side effects of isotretinoin and how to deal with them, and there are no specific risk minimization tools other than the warnings in the product leaflet , which increase the chance of unwanted adverse effects [6, 10, 11]. A recent Saudi study reported that more than half of the patients did not recognize hyperlipidemia as a side effect of isotretinoin. Furthermore, more than one-third of them did not know that isotretinoin can elevate liver enzymes' level . Another study conducted in Saudi Arabia reported that patients who were prescribed isotretinoin were not sufficiently aware of its proper use . Nearly half of the patients in the later study did not know that they should not donate blood while using isotretinoin. In addition, the majority of the surveyed women did not know that they must stop taking isotretinoin at least one month before becoming pregnant . A recent study showed that patients' knowledge about the importance of using contraceptives in women of reproductive age while on isotretinoin is suboptimal , with only few proportions of women adhering to the use of contraception during isotretinoin use [14, 15].
The low safety profile of isotretinoin therapy sheds the light on the necessity to improve patients' awareness about the medication and its optimal use. The current study is the first one which evaluated patients' knowledge about the appropriate use of isotretinoin and its associated side effects in Jordan. Findings of the present study are expected to contribute to a deeper understanding of the current situation in the region, help developing more effective interventions that aim at raising patient's awareness and knowledge about isotretinoin use, and ultimately enhancing safety and health outcomes among isotretinoin users.
1.1. Aim of the Study
The aim of this study was to explore the gaps in patients' knowledge about isotretinoin use and its associated side effects.
2. Methods
2.1. Study Design and Subjects
The current cross-sectional study was conducted in Jordan in the period from January through May 2019. An online questionnaire was distributed using Facebook groups and Twitter. A cover letter that clarified the objectives of the study and the inclusion criteria was attached to the questionnaire. Patients were included in the study if they were 13 years or older, had acne, used isotretinoin as treatment and live in Jordan. The cover letter also informed the patients that their participation is voluntary and the data will be kept confidential. Based on the Raosoft program, a minimum sample of 267 patients was required to produce 90% confidence interval and less than 5% margin of error. To adjust for any dropout, 10% of the sample size was added, giving a total sample size of 294 patients.
2.2. Study Instrument
After extensive literature review [9, 16–18], the current study survey was designed. The questionnaire was designed in Arabic language and contained three parts. The first part included information about sociodemographic variables such as age, gender, marital status, education level, profession, daily dose of isotretinoin, and the duration since isotretinoin course initiation. The second part evaluated patients' knowledge about different isotretinoin adverse effects including skin dryness, teratogenicity, dyslipidemia, itching, rash, nosebleeds, joint and back pain, dizziness, eye inflammation, depression, and increased liver enzymes. The third part included nine yes/no questions that evaluated patients' knowledge about isotretinoin use including the need for an authorized prescription, the appropriate duration of isotretinoin use, isotretinoin use instructions such as the necessity to take isotretinoin with water and after fatty meal, the importance of using sunblock with isotretinoin, and the necessity to make regular laboratory checkout in addition to the awareness of isotretinoin use during pregnancy, lactation, and blood donation. A score of one was given for each correct answer, and the scores were summed out of nine. The survey was reviewed by experts in the field including a dermatologist and two professors of pharmacy, and changes were made when deemed appropriate. The study survey was also piloted on twenty patients to ensure for the clarity of the questionnaire. The piloted patients were excluded from the main study.
2.3. Ethical Approval
Ethical approval of this study was attained from the Institutional Review Board (IRB) of King Abdullah University Hospital and Jordan University of Science and Technology (JUST). The IRB reference number is 48/118/2018.
2.4. Data Analysis
Data were coded and entered into SPSS software program version 22 for statistical analysis. Descriptive statistics were made for quantitative variables using frequencies and percentages for categorical variables and mean and standard deviation for continuous variables. Independent t-test was conducted to find the association between the dichotomous predictors and the continuous outcome knowledge or awareness score, while one-way analysis of variance (ANOVA) test was used to find the correlation between the categorical predictors of three or more categories and the knowledge score.
3. Results
A total of 367 patients completed the online survey. Most of the participants were female (89.1%), in the age group of 16 to 25 years (87.4%), not married (94%), had high education level (93.8%), were students (63%), were receiving 10–20 mg isotretinoin daily (85.2%), and they were taking isotretinoin for less than 6 months (55.3%). Demographic characteristics of the study participants are presented in Table 1.
Table 1.
Demographic characteristics of the participating patients (n = 367).
| Characteristics | N (%) |
:---: |
| Gender | |
| Male | 40 (10.9%) |
| Female | 327 (89.1%) |
| Age (years) | |
| 16–25 | 321 (87.4%) |
| 26–35 | 37 (10.1%) |
| 36–45 | 5 (1.4%) |
| ≥46 | 4 (1.1%) |
| Marital status | |
| Married | 22 (6%) |
| Others† | 345 (94%) |
| Education level | |
| Low education‡ | 23 (6.2%) |
| High education | 344 (93.8%) |
| Occupation | |
| Unemployed | 17 (4.6%) |
| Student | 231 (63%) |
| Medical field | 98 (26.7%) |
| Nonmedical field | 21 (5.7%) |
| Daily dose of isotretinoin | |
| 10–20 mg | 313 (85.2%) |
| 20–40 mg | 50 (13.7%) |
| 40–60 mg | 4 (1.1%) |
| Period since starting the current isotretinoin course (months) | |
| Less than or equal to six | 203 (55.3%) |
| More than six | 164 (44.7%) |
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†Single, separated, divorced, and widowed. ‡Low education: less than the university level.
The majority of the patients (67.6%) knew about isotretinoin therapy from their physicians, while the rest of the patients knew about this therapy from other sources including pharmacists, friends, family members, and internet resources. Most of the patients (93.2%) advised others to use isotretinoin based on their old experience.
The most recognized side effect of isotretinoin therapy was dryness (98.1%), followed by teratogenicity (66.5%) and dyslipidemia (58.9%). Other side effects including itching, rash, nosebleeds, joint and back pain, dizziness, eye inflammation, depression, and increased liver enzymes were recognized by less than half of the patients. Among the seventeen married female patients who participated in this study, only one patient used isotretinoin without oral contraceptives, two patients were not informed by the physician or the pharmacist about the risk of teratogenicity of isotretinoin, and three of them did not make pregnancy test on a monthly basis.
The study patients showed good knowledge about isotretinoin use with a mean total knowledge score of 8.1 (SD = 0.7). As shown in Table 2, more than half of the patients (61.0%) mistakenly thought that isotretinoin therapy should be taken continuously for more than 6 months without stop, and nearly one-fourth of them did not know that isotretinoin is recommended to be taken with fatty meal (24%) and sunblock (24.6%).
Table 2.
Patients' knowledge about isotretinoin use.
| Knowledge item | Yes | No |
:---:
| Isotretinoin can be used without prescription† | 23 (6.3%) | 344 (93.7%) |
| Isotretinoin can be used for more than 6 months without stop† | 224 (61.0%) | 143 (39.0%) |
| No need to have laboratory monitoring during isotretinoin use† | 16 (4.4%) | 351 (95.6%) |
| Isotretinoin should be given with plenty of water to avoid dryness | 363 (98.9%) | 4 (1.1%) |
| Isotretinoin is recommended to be given with fatty meal | 279 (76.0%) | 88 (24.0%) |
| Isotretinoin is recommended to be used with sunblock | 277 (75.4%) | 90 (24.6%) |
| Isotretinoin can be used during pregnancy† | 8 (2.1%) | 359 (97.9%) |
| Blood donation is allowed during isotretinoin therapy† | 6 (1.6%) | 361 (98.4%) |
| Lactation is allowed during isotretinoin therapy† | 3 (0.8%) | 364 (99.2%) |
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†The correct answer is no.
As shown in Table 3, the t-test reported that female gender (8.2 (SD = 0.8)) and more than 6-month use of isotretinoin (8.3 (SD = 1.2)) were significantly associated with a higher knowledge score of isotretinoin use (p=0.01), when compared with male patients (7.8 (SD = 0.7)) and less than 6-month use of isotretinoin (7.7 (SD = 0.7)).
Table 3.
Association between sociodemographics and the total knowledge score.
| Characteristics | Mean (SD)§ | p value |
:---:
| Gender | | |
| Male | 7.8 (0.7) | 0.01 ∗ |
| Female | 8.2 (0.8) |
| Marital status | | |
| Married | 8.2 (0.6) | 0.92 |
| Others† | 8.1 (0.8) |
| Age | | |
| 16–25 | 8.2 (0.8) | 0.65 |
| 26–35 | 8.1 (0.8) |
| 36–45 | 7.9 (0.7) |
| ≥46 | 7.7 (0.5) |
| Education level | | |
| Low education‡ | 8.6 (0.7) | 0.41 |
| High education | 8.0 (0.7) |
| Occupation | | |
| Unemployed | 8.6 (0.5) | 0.85 |
| Student | 8.1 (0.8) |
| Medical profession | 8.2 (0.7) |
| Nonmedical profession | 8.2 (0.7) |
| Duration of usage | | |
| Less than or equal to six | 7.7 (0.7) | 0.01 ∗ |
| More than six | 8.3 (1.2) |
| Doses | | |
| 10–20 mg | 8.1 (0.8) | 0.43 |
| 20–40 mg | 8.2 (0.7) |
| 40–60 mg | 7.7 (0.9) |
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§One-way analysis of variance (ANOVA) and independent t-test were applied where applicable ∗ p< 0.05 showed a significant difference compared to other groups. †Single, separated, divorced, and widowed. ‡Low education: less than the university level.
Results showed that only 1.4% of the patients were asked by their physicians to sign a consent form before starting isotretinoin therapy. However, nearly 84.5% of the patients were informed by their physicians about the expected side effect of isotretinoin and how to manage it. Most of the patients (63.5%) made a monthly follow-up with their physicians, 17.2% made a follow-up every two months, and 12.0% made a follow-up every three months or more. Approximately one-third of the patients (34.4%) did not inform their physicians before using other drugs with isotretinoin.
4. Discussion
In 1971, clinical trials' results revealed that isotretinoin was ineffective for treating skin cancer, but could be useful for the management of AV. However, due to concerns about its side effects and its potential to cause birth defects in particular, the medication was not released to the market until 1982 . Isotretinoin is, by far, the most cost-effective drug when compared with other medications used for severe acne treatment. However, it has been associated with some serious and teratogenic adverse effects that patients must be aware of [2–4]. Therefore, the current study, which is the first one in Jordan, was conducted to evaluate patients' knowledge about isotretinoin use and its accompanying adverse effects.
The majority of this study patients (67.6%) heard about isotretinoin therapy from their physicians. Similar results were reported by Kara Polat (64.4%) and Al-Harbi (61.9%) , but higher percentages were reported in studies conducted by Younis and Al-Harbi (89.6%) , Bakheet et al. (93%) , and Imam et al. (81.9%) . On the contrary, an earlier Saudi study reported that the most identified source of information for isotretinoin therapy was friends, followed by the internet and social media. However, most of the current study patients advised others to use isotretinoin based on their old experience, which might increase the potential of medication side effects.
Consistent with earlier study findings [9, 12, 20–22], dryness was the most commonly recognized side effect of isotretinoin therapy. The majority of the current study participants were not able to recognize side effects of isotretinoin therapy such as itching, rash, nosebleeds, joint and back pain, dizziness, eye inflammation, depression, and increased liver enzymes. In comparison, only 28.6% and 33.5% of the patients participated in a Saudi study recognized depression and abnormal liver function as side effects of isotretinoin, respectively . In addition, around half of the isotretinoin users enrolled in another study were not able to recognize depression as a side effect of isotretinoin therapy . Comparable with earlier Saudi study finding , some patients were not able to recognize dyslipidemia as a side effect of isotretinoin therapy in the current study. Although the majority of the present study patients recognized teratogenicity as a side effect, higher percentages were reported in earlier studies [9, 12]. Lack of knowledge about isotretinoin side effects in this study could be attributed to the availability of isotretinoin without prescription, which could lead to the missing out of physician explanation about isotretinoin side effects and might make the patients feel that the drug is safe. Furthermore, inadequate community pharmacists' knowledge regarding isotretinoin toxicity may have an impact, given the vital role which pharmacists play as medication experts in providing the information needed about isotretinoin to the patients .
Given the teratogenicity of isotretinoin and the fact that 50% of isotretinoin users are women of reproductive age [3, 23], precautious interventions must be implemented while using isotretinoin among these patients. In other words, married female patients who are prescribed isotretinoin must perform pregnancy test before and during the treatment course in order to avoid birth defects . Consistent with earlier Saudi and Canadian studies [25, 26], some married female patients in the present study were not informed by the physician or the pharmacist about the teratogenicity of isotretinoin and did not perform pregnancy test on a monthly basis while using the drug. These findings shed the light on the necessity to implement patient counseling and educational programs which improve patients' information about the potential teratogenic effect and other potential side effects of isotretinoin therapy.
The current study patients' knowledge about isotretinoin use was similar to earlier studies' findings [21, 22] and better than other studies' reports [10, 11, 20]. In the present study, the patients demonstrated poor knowledge about the appropriate duration of isotretinoin therapy and its combination with fatty meal and sunblock. In order to achieve the optimal therapeutic response to isotretinoin therapy, it should be taken for four to six months depending on the daily dose given . More than half of the current study patients mistakenly thought that isotretinoin should be taken continuously for more than six months without stop. A similar finding was reported in two studies that were conducted in Saudi Arabia [9, 21]. In contrast, a study conducted in another region of Saudi Arabia reported that the majority of the patients used isotretinoin for less than six months . Moreover, patients who used isotretinoin for more than six months in the present study were found to have significantly better knowledge about isotretinoin therapy. It seems that the increased experience associated with longer duration of isotretinoin use could justify this finding. Lastly, female patients significantly reported a higher knowledge score than male patients in the present study, which could be justified by the fact that females are more caring for skin and more interested in cosmetics and beauty products when compared to males.
Results of the present study showed that only 1.4% of the patients signed up a consent form before starting isotretinoin therapy. Comparable findings were reported in a Canadian study, where only 13% of the patients were requested to sign a consent form . Emphasis should be placed on improving physicians' knowledge about the recent isotretinoin dispensing protocol to combine a consent form with isotretinoin therapy. Results also revealed that more than one-third of the patients did not perform follow-up with their physicians on a monthly basis and did not inform them before using other drugs concurrently with isotretinoin. Therefore, a regular monitoring and medication review should be considered during the course of isotretinoin therapy.
In summary, healthcare providers should provide patient counseling, along with regular monitoring, to improve patients' knowledge and awareness about the appropriate use of isotretinoin and its potential side effects, in order to enhance patients' safety, prevent isotretinoin-associated harms, and ultimately optimize treatment outcomes among isotretinoin users.
4.1. Limitations
A larger sample size would make the findings of this study more robust. The use of self-reported questionnaire could make the respondent misunderstand the question which may cause bias to the study findings.
5. Conclusion
The current study demonstrates a margin for improvement in patients' knowledge about isotretinoin and its use. Therefore, it is essential to use educational brochures, medication leaflets, and educational posts on social media which aim at improving patients' knowledge on isotretinoin therapy and its associated side effects, particularly for male patients and those with less duration of isotretinoin use.
Data Availability
The data are available at the corresponding author disk and could be provided upon request.
Additional Points
What's Known. (i) Despite being the first-line treatment of severe or moderate acne, isotretinoin has several serious side effects that necessitate the evaluation of patients' knowledge about isotretinoin side effects and its proper use. (ii) The low safety profile of isotretinoin therapy sheds the light on the necessity to improve patients' awareness about the medication and its optimal use. What This Study Adds. (i) The current study revealed lack of patients' information about the potential side effects, duration of isotretinoin use, and some instructions regarding isotretinoin use including taking the medication with fatty meal and sunblock. (ii) Findings of the present study should help exploring the information need about isotretinoin therapy and its use which should be targeted in the future disease management program which aim to enhance safety and health outcomes among isotretinoin users.
Conflicts of Interest
The authors declare that there are no conflicts of interest.
References
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Associated Data
This section collects any data citations, data availability statements, or supplementary materials included in this article.
Data Availability Statement
The data are available at the corresponding author disk and could be provided upon request.
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Prove that gcd(n,n+1)=1 for all positive integers n.
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The greatest common divisor (gcd) of n and n+1 is always 1 for all positive integers n.
Explanation
To prove that gcd(n, n+1) = 1 for all positive integers n, we can use the Euclidean algorithm. Let's assume that there exists a positive integer k such that k is the gcd of n and n+1. This means that k divides both n and n+1 without leaving a remainder.
Using the Euclidean algorithm, we can write:
gcd(n, n+1) = gcd(n, (n+1) - n)
gcd(n, n+1) = gcd(n, 1)
Since 1 is a positive integer, the gcd(n, 1) is equal to 1 for any positive integer n. Therefore, gcd(n, n+1) = 1 for all positive integers n.
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The greatest common divisor (gcd) of any positive integer n and its successor n+1 is always 1. This is because they are consecutive integers, and the only positive integer that divides both is 1. Thus, gcd(n,n+1)=1 for all positive integers n.
Explanation
To prove that gcd(n,n+1)=1 for all positive integers n, we can use the properties of the greatest common divisor (gcd) and a simple logical argument.
Understanding gcd: The greatest common divisor of two numbers a and b is the largest positive integer that divides both a and b without leaving a remainder.
Consecutive integers: The integers n and n+1 are consecutive. This means that they are two numbers that are one apart.
Divisibility argument: If a number k divides both n and n+1, then it must also divide the difference between them:
gcd(n,n+1)=gcd(n,(n+1)−n)=gcd(n,1)
This simplification shows that the only common divisor between any integer n and 1 is 1 itself.
Conclusion: Since gcd(n,1)=1, we conclude that gcd(n,n+1)=1 for all positive integers n. Thus, we have proved that consecutive integers are coprime (having no common divisor other than 1).
Examples & Evidence
For example, if n=5, then n+1=6. The factors of 5 are {1, 5} and the factors of 6 are {1, 2, 3, 6}. The only common factor is 1. Similarly, for n=10, n+1=11: the factors of 10 are {1, 2, 5, 10} and those of 11 are {1, 11}. Again, the only common factor is 1.
This statement is supported by the fundamental properties of integers and the definition of coprimality. For any two consecutive integers n and n+1, there are no common factors besides 1, which can be verified by the nature of divisibility.
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Community Answer Give the prime decomposition of 2¹ - 1 . Evaluate gcd(n, n+1) and LCM[n, n+1] where n is a positive integer. . PROVE: If a and b are positive integers such that [a, b] = (a, b), then a = b.
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Community Answer Prove or disprove. a) For all integers a, b, n, n is divisible by a and b iff n is divisible by a · b. b) For all integers a, b, m, n, gcd(ma, nb) = gcd(m, n). gcd(a, b). c) For all m, n,k € Z with k > d) For all m, n,k € Z with k > 1, if m = n, then gcd(m, k) = gcd(n, k). 1, if ged(m, k) = gcd(n, k), then m = =k n.
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Anatomy and Physiology 2e
22.4 Gas Exchange
Anatomy and Physiology 2e
22.4
Gas Exchange
Search for key terms or text.
Learning Objectives
By the end of this section, you will be able to:
Compare the composition of atmospheric air and alveolar air
Describe the mechanisms that drive gas exchange
Discuss the importance of sufficient ventilation and perfusion, and how the body adapts when they are insufficient
Discuss the process of external respiration
Describe the process of internal respiration
The purpose of the respiratory system is to perform gas exchange. Pulmonary ventilation provides air to the alveoli for this gas exchange process. At the respiratory membrane, where the alveolar and capillary walls meet, gases move across the membranes, with oxygen entering the bloodstream and carbon dioxide exiting. It is through this mechanism that blood is oxygenated and carbon dioxide, the waste product of cellular respiration, is removed from the body.
Gas Exchange
In order to understand the mechanisms of gas exchange in the lung, it is important to understand the underlying principles of gases and their behavior. In addition to Boyle’s law, several other gas laws help to describe the behavior of gases.
Gas Laws and Air Composition
Gas molecules exert force on the surfaces with which they are in contact; this force is called pressure. In natural systems, gases are normally present as a mixture of different types of molecules. For example, the atmosphere consists of oxygen, nitrogen, carbon dioxide, and other gaseous molecules, and this gaseous mixture exerts a certain pressure referred to as atmospheric pressure (Table 22.2). Partial pressure (Px) is the pressure of a single type of gas in a mixture of gases. For example, in the atmosphere, oxygen exerts a partial pressure, and nitrogen exerts another partial pressure, independent of the partial pressure of oxygen (Figure 22.21). Total pressure is the sum of all the partial pressures of a gaseous mixture. Dalton’s law describes the behavior of nonreactive gases in a gaseous mixture and states that a specific gas type in a mixture exerts its own pressure; thus, the total pressure exerted by a mixture of gases is the sum of the partial pressures of the gases in the mixture.
Partial Pressures of Atmospheric Gases
| Gas | Percent of total composition | Partial pressure (mm Hg) |
---
| Nitrogen (N2) | 78.6 | 597.4 |
| Oxygen (O2) | 20.9 | 158.8 |
| Water (H2O) | 0.4 | 3.0 |
| Carbon dioxide (CO2) | 0.04 | 0.3 |
| Others | 0.06 | 0.5 |
| Total composition/total atmospheric pressure | 100% | 760.0 |
Table
22.2
Figure
22.21
Partial and Total Pressures of a Gas
Partial pressure is the force exerted by a gas. The sum of the partial pressures of all the gases in a mixture equals the total pressure.
Partial pressure is extremely important in predicting the movement of gases. Recall that gases tend to equalize their pressure in two regions that are connected. A gas will move from an area where its partial pressure is higher to an area where its partial pressure is lower. In addition, the greater the partial pressure difference between the two areas, the more rapid is the movement of gases.
Solubility of Gases in Liquids
Henry’s law describes the behavior of gases when they come into contact with a liquid, such as blood. Henry’s law states that the concentration of gas in a liquid is directly proportional to the solubility and partial pressure of that gas. The greater the partial pressure of the gas, the greater the number of gas molecules that will dissolve in the liquid. The concentration of the gas in a liquid is also dependent on the solubility of the gas in the liquid. For example, although nitrogen is present in the atmosphere, very little nitrogen dissolves into the blood, because the solubility of nitrogen in blood is very low. The exception to this occurs in scuba divers; the composition of the compressed air that divers breathe causes nitrogen to have a higher partial pressure than normal, causing it to dissolve in the blood in greater amounts than normal. Too much nitrogen in the bloodstream results in a serious condition that can be fatal if not corrected. Gas molecules establish an equilibrium between those molecules dissolved in liquid and those in air.
The composition of air in the atmosphere and in the alveoli differs. In both cases, the relative concentration of gases is nitrogen > oxygen > water vapor > carbon dioxide. The amount of water vapor present in alveolar air is greater than that in atmospheric air (Table 22.3). Recall that the respiratory system works to humidify incoming air, thereby causing the air present in the alveoli to have a greater amount of water vapor than atmospheric air. In addition, alveolar air contains a greater amount of carbon dioxide and less oxygen than atmospheric air. This is no surprise, as gas exchange removes oxygen from and adds carbon dioxide to alveolar air. Both deep and forced breathing cause the alveolar air composition to be changed more rapidly than during quiet breathing. As a result, the partial pressures of oxygen and carbon dioxide change, affecting the diffusion process that moves these materials across the membrane. This will cause oxygen to enter and carbon dioxide to leave the blood more quickly.
Composition and Partial Pressures of Alveolar Air
| Gas | Percent of total composition | Partial pressure (mm Hg) |
---
| Nitrogen (N2) | 74.9 | 569 |
| Oxygen (O2) | 13.7 | 104 |
| Water (H2O) | 6.2 | 40 |
| Carbon dioxide (CO2) | 5.2 | 47 |
| Total composition/total alveolar pressure | 100% | 760.0 |
Table
22.3
Ventilation and Perfusion
Two important aspects of gas exchange in the lung are ventilation and perfusion. Ventilation is the movement of air into and out of the lungs, and perfusion is the flow of blood in the pulmonary capillaries. For gas exchange to be efficient, the volumes involved in ventilation and perfusion should be compatible. However, factors such as regional gravity effects on blood, blocked alveolar ducts, or disease can cause ventilation and perfusion to be imbalanced.
The partial pressure of oxygen in alveolar air is about 104 mm Hg, whereas the partial pressure of oxygenated blood in pulmonary veins is about 100 mm Hg. When ventilation is sufficient, oxygen enters the alveoli at a high rate, and the partial pressure of oxygen in the alveoli remains high. In contrast, when ventilation is insufficient, the partial pressure of oxygen in the alveoli drops. Without the large difference in partial pressure between the alveoli and the blood, oxygen does not diffuse efficiently across the respiratory membrane. The body has mechanisms that counteract this problem. In cases when ventilation is not sufficient for an alveolus, the body redirects blood flow to alveoli that are receiving sufficient ventilation. This is achieved by constricting the pulmonary arterioles that serves the dysfunctional alveolus, which redirects blood to other alveoli that have sufficient ventilation. At the same time, the pulmonary arterioles that serve alveoli receiving sufficient ventilation vasodilate, which brings in greater blood flow. Factors such as carbon dioxide, oxygen, and pH levels can all serve as stimuli for adjusting blood flow in the capillary networks associated with the alveoli.
Ventilation is regulated by the diameter of the airways, whereas perfusion is regulated by the diameter of the blood vessels. The diameter of the bronchioles is sensitive to the partial pressure of carbon dioxide in the alveoli. A greater partial pressure of carbon dioxide in the alveoli causes the bronchioles to increase their diameter, as does a decreased level of oxygen in the blood supply, allowing carbon dioxide to be exhaled from the body at a greater rate. As mentioned above, a greater partial pressure of oxygen in the alveoli causes the pulmonary arterioles to dilate, increasing blood flow.
Gas Exchange
Gas exchange occurs at two sites in the body: in the lungs, where oxygen is picked up and carbon dioxide is released at the respiratory membrane, and at the tissues, where oxygen is released and carbon dioxide is picked up. External respiration is the exchange of gases with the external environment, and occurs in the alveoli of the lungs. Internal respiration is the exchange of gases with the internal environment, and occurs in the tissues. The actual exchange of gases occurs due to simple diffusion. Energy is not required to move oxygen or carbon dioxide across membranes. Instead, these gases follow pressure gradients that allow them to diffuse. The anatomy of the lung maximizes the diffusion of gases: The respiratory membrane is highly permeable to gases; the respiratory and blood capillary membranes are very thin; and there is a large surface area throughout the lungs.
External Respiration
The pulmonary artery carries deoxygenated blood into the lungs from the heart, where it branches and eventually becomes the capillary network composed of pulmonary capillaries. These pulmonary capillaries create the respiratory membrane with the alveoli (Figure 22.22). As the blood is pumped through this capillary network, gas exchange occurs. Although a small amount of the oxygen is able to dissolve directly into plasma from the alveoli, most of the oxygen is picked up by erythrocytes (red blood cells) and binds to a protein called hemoglobin, a process described later in this chapter. Oxygenated hemoglobin is red, causing the overall appearance of bright red oxygenated blood, which returns to the heart through the pulmonary veins. Carbon dioxide is released in the opposite direction of oxygen, from the blood to the alveoli. Some of the carbon dioxide is returned on hemoglobin, but can also be dissolved in plasma or is present as a converted form, also explained in greater detail later in this chapter.
External respiration occurs as a function of partial pressure differences in oxygen and carbon dioxide between the alveoli and the blood in the pulmonary capillaries.
Figure
22.22
External Respiration
In external respiration, oxygen diffuses across the respiratory membrane from the alveolus to the capillary, whereas carbon dioxide diffuses out of the capillary into the alveolus.
Although the solubility of oxygen in blood is not high, there is a drastic difference in the partial pressure of oxygen in the alveoli versus in the blood of the pulmonary capillaries. This difference is about 64 mm Hg: The partial pressure of oxygen in the alveoli is about 104 mm Hg, whereas its partial pressure in the blood of the capillary is about 40 mm Hg. This large difference in partial pressure creates a very strong pressure gradient that causes oxygen to rapidly cross the respiratory membrane from the alveoli into the blood.
The partial pressure of carbon dioxide is also different between the alveolar air and the blood of the capillary. However, the partial pressure difference is less than that of oxygen, about 5 mm Hg. The partial pressure of carbon dioxide in the blood of the capillary is about 45 mm Hg, whereas its partial pressure in the alveoli is about 40 mm Hg. However, the solubility of carbon dioxide is much greater than that of oxygen—by a factor of about 20—in both blood and alveolar fluids. As a result, the relative concentrations of oxygen and carbon dioxide that diffuse across the respiratory membrane are similar.
Internal Respiration
Internal respiration is gas exchange that occurs at the level of body tissues (Figure 22.23). Similar to external respiration, internal respiration also occurs as simple diffusion due to a partial pressure gradient. However, the partial pressure gradients are opposite of those present at the respiratory membrane. The partial pressure of oxygen in tissues is low, about 40 mm Hg, because oxygen is continuously used for cellular respiration. In contrast, the partial pressure of oxygen in the blood is about 100 mm Hg. This creates a pressure gradient that causes oxygen to dissociate from hemoglobin, diffuse out of the blood, cross the interstitial space, and enter the tissue. Hemoglobin that has little oxygen bound to it loses much of its brightness, so that blood returning to the heart is more burgundy in color.
Considering that cellular respiration continuously produces carbon dioxide, the partial pressure of carbon dioxide is lower in the blood than it is in the tissue, causing carbon dioxide to diffuse out of the tissue, cross the interstitial fluid, and enter the blood. It is then carried back to the lungs either bound to hemoglobin, dissolved in plasma, or in a converted form. By the time blood returns to the heart, the partial pressure of oxygen has returned to about 40 mm Hg, and the partial pressure of carbon dioxide has returned to about 45 mm Hg. The blood is then pumped back to the lungs to be oxygenated once again during external respiration.
Figure
22.23
Internal Respiration
Oxygen diffuses out of the capillary and into cells, whereas carbon dioxide diffuses out of cells and into the capillary.
Everyday Connection
Hyperbaric Chamber Treatment
A type of device used in some areas of medicine that exploits the behavior of gases is hyperbaric chamber treatment. A hyperbaric chamber is a unit that can be sealed and expose a patient to either 100 percent oxygen with increased pressure or a mixture of gases that includes a higher concentration of oxygen than normal atmospheric air, also at a higher partial pressure than the atmosphere. There are two major types of chambers: monoplace and multiplace. Monoplace chambers are typically for one patient, and the staff tending to the patient observes the patient from outside of the chamber (Figure 22.24). Some facilities have special monoplace hyperbaric chambers that allow multiple patients to be treated at once, usually in a sitting or reclining position, to help ease feelings of isolation or claustrophobia. Multiplace chambers are large enough for multiple patients to be treated at one time, and the staff attending these patients is present inside the chamber. In a multiplace chamber, patients are often treated with air via a mask or hood, and the chamber is pressurized.
Figure
22.24
Hyperbaric Chamber
(credit: “komunews”/flickr.com)
Hyperbaric chamber treatment is based on the behavior of gases. As you recall, gases move from a region of higher partial pressure to a region of lower partial pressure. In a hyperbaric chamber, the atmospheric pressure is increased, causing a greater amount of oxygen than normal to diffuse into the bloodstream of the patient. Hyperbaric chamber therapy is used to treat a variety of medical problems, such as wound and graft healing, anaerobic bacterial infections, and carbon monoxide poisoning. Exposure to and poisoning by carbon monoxide is difficult to reverse, because hemoglobin’s affinity for carbon monoxide is much stronger than its affinity for oxygen, causing carbon monoxide to replace oxygen in the blood. Hyperbaric chamber therapy can treat carbon monoxide poisoning, because the increased atmospheric pressure causes more oxygen to diffuse into the bloodstream. At this increased pressure and increased concentration of oxygen, carbon monoxide is displaced from hemoglobin. Another example is the treatment of anaerobic bacterial infections, which are created by bacteria that cannot or prefer not to live in the presence of oxygen. An increase in blood and tissue levels of oxygen helps to kill the anaerobic bacteria that are responsible for the infection, as oxygen is toxic to anaerobic bacteria. For wounds and grafts, the chamber stimulates the healing process by increasing energy production needed for repair. Increasing oxygen transport allows cells to ramp up cellular respiration and thus ATP production, the energy needed to build new structures.
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Book title: Anatomy and Physiology 2e
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14462 | https://www.cuemath.com/numbers/factors-of-84/ | LearnPracticeDownload
Factors of 84
Factors of 84 are integers that can be divided evenly into 84. It has total 12 factors of which 84 is the biggest factor and the prime factors of 84 are 2, 3 and 7. The Prime Factorization of 84 is 22 × 3 × 7.
Factors of 84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 and 84
Negative Factors of 84: -1, -2, -3, -4, -6, -7, -12, -14, -21, -28, -42 and -84
Prime Factors of 84: 2, 3, 7
Prime Factorization of 84: 2 × 2 × 3 × 7 = 22 × 3 × 7
Sum of Factors of 84: 224
| | |
--- |
| 1. | What Are the Factors of 84? |
| 2. | How to Calculate Factors of 84? |
| 3. | Factors of 84 by Prime Factorization |
| 4. | Factors of 84 in Pairs |
| 5. | Important Notes |
| 6. | FAQs on Factors of 84 |
What are the Factors of 84?
Factors of a given number are the numbers which divide the given number exactly without any remainder.
Hence, the factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 and 84.
84 is strictly smaller than the sum of its factors (except 84).
1 + 2 + 3 + 4 + 6 + 7 + 12 + 14 + 21 + 28 + 42 = 140
So, 84 is known as an abundant number.
How to Calculate the Factors of 84?
Various methods such as prime factorization and the division method can be used to calculate the factors of 84. In prime factorization, we express 84 as a product of its prime factors and in the division method, we see what numbers divide 84 exactly without a remainder.
Hence, the factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 and 84.
Explore factors using illustrations and interactive examples.
Factors of 120 - The factors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120
Factors of 24 - The factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24
Factors of 96 - The factors of 96 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96
Factors of 72 - The factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
Factors of 42 - The factors of 42 are 1, 2, 3, 6, 7, 14, 21, 42
Factors of 60 - The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Factors of 84 by Prime Factorization
Prime factorization means expressing a number in terms of the product of its prime factors. We can do use the division method or factor tree to do this.
1. Prime Factorization by Division method
The number 84 is divided by the smallest prime number which divides 84 exactly, i.e. it leaves a remainder 0. The quotient is then divided by the smallest or second smallest prime number and the process continues till the quotient becomes indivisible.
Since 84 is even, it will be divisible by 2.
Let us divide 84 by the prime number 2
84 ÷ 2 = 42
42 is again even, let's again divide by 2
42 ÷ 2 = 21
Now, 21 is an odd number. So it won't be divisible by 2. Let's check the next prime number i.e. 3
21 ÷ 3 =7
7 is a prime number. So, it's not further divisible.
Prime factorization of 84 = 2 × 2 × 3 × 7
2. Prime Factorization by Factor Tree
The other way of prime factorization as taking 84 as the root, we create branches by dividing it by prime numbers. This method is similar to above division method. The difference lies in presenting the factorization.
The figure below shows the factor tree of 84. The composite numbers will have branches as they are further divisible. We continue making branches till we are left with only prime numbers.
The numbers inside the circles are the prime factors of 84
Now that we have done the prime factorization of our number, we can multiply them and get the other factors. Can you try and find out if all the factors are covered or not? And as you might have already guessed it, for prime numbers, there are no other factors.
Factors of 84 in Pairs
The factor pairs of a number are the two numbers which, when multiplied, give the required number.
Considering the number 84, we have
1 × 84 = 84
2 × 42 = 84
3 × 28 = 84
4 × 21 = 84
6 × 14 = 84
7 × 12 = 84
So, the factor pairs of 84 are : (1, 84), (2, 42), (3, 28), (4, 21), (6, 14), (7, 12)
The product of two negative numbers also results in a positive number i.e. (-ve) × (-ve) = (+ve).
So, (-1, -84), (-2, -42), (-3, -28), (-4, -21), (-6, -14), (-7, -12). are also factor pairs of 84.
Our focus in this article will be on the positive factors.
Important Notes:
Every number has a finite number of factors
The factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 and 84
As 84 is strictly smaller than the sum of its factors (leaving 84), it is called an abundant number.
Download FREE Study Materials
Worksheet on Factors
Factors of 84
Worksheet on Factors
Factors of 84 Solved Examples
Example 1:Rooney wants to find the number which is three more than the greatest factor of 84, other than 84 Can you find that number?
Solution:
The factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 and 84.
Out of these factors, the greatest factor of 84, other than 84 is 42
A number which is three more than 42 = 42 + 3 = 45
Hence, 45 is the required number
2. Example 2:It's time for a Christmas Party! Jennifer baked 40 brownies and 84 cookies for her guests. She wants to pack them in boxes such that each box has equal number of brownies and cookies. What is the least number of boxes she would require for this?
Solution:
To find the least number of boxes such that we can cookies and brownies equally in those boxes, we need to find the GCF (Gretest Common Factor) of 40 and 84
The factors of 40 are 1, 2, 4, 5, 8, 10, 20 and 40
The factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 and 84
Common factors of 40 and 84 = 1, 2, 4
So, GCF (40, 84) = 4
Hence, the least number of boxes required is 4
3. Example 3:Julia has two pieces of cloth. The first one is 90 inches wide and the second one is 84 inches wide. She wants to cut the two pieces into equal width so that they are as wide as possible. What width should she consider for this?
Solution:
As we need to cut both size clothes into maximum equal width, we are going to find GCF (90,84)
The factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, and 90
The factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 and 84
The common factors of 90, 84 are 1, 2, 3, and 6. Out of these, 6 is greatest.
Hence, she should consider to cut both the pieces into a width of 6 inches.
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FAQs on Factors of 84
What are the Factors of 84?
The factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84 and its negative factors are -1, -2, -3, -4, -6, -7, -12, -14, -21, -28, -42, -84.
What are the Prime Factors of 84?
The prime factors of 84 are 2, 3, 7.
How Many Factors of 84 are also common to the Factors of 44?
Since, the factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84 and the factors of 44 are 1, 2, 4, 11, 22, 44.
Hence, [1, 2, 4] are the common factors of 84 and 44.
What is the Sum of all the Factors of 84?
Sum of all factors of 84 = (22 + 1 - 1)/(2 - 1) × (31 + 1 - 1)/(3 - 1) × (71 + 1 - 1)/(7 - 1) = 224
What is the Greatest Common Factor of 84 and 72?
The factors of 84 and 72 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84 and 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72 respectively.
Common factors of 84 and 72 are [1, 2, 3, 4, 6, 12].
Hence, the Greatest Common Factor of 84 and 72 is 12.
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14463 | http://elearn.psgcas.ac.in/nptel/courses/video/105103096/lec14.pdf | Hydraulics Prof. Dr. Arup Kumar Sarma Department of Civil Engineering Indian Institute of Technology, Guwahati Module No. # 03 Energy and Momentum Principle Lecture No. # 02 Computation of Critical Depth Friends, today we shall be discussing about computation of critical depth. We have already discussed regarding how to compute normal depth and also we have discussed about the concept of specific energy. Today, we shall be going to see, how we can calculate or how we can compute the critical depth, and how these things are coming into use in our different hydraulics calculations. Well, before going to this, let me just recapitulate what we did in the last class. Well, we started with concept of specific energy. That means, just to recall, that specific energy is nothing but energy per unit weight of flowing fluid, rather per unit weight of flowing fluid with respect to channel bed. So, this is what we call as specific energy. Then, we did discuss about the specific energy diagram. Well, this specific energy is a function of the depth. So, if we plot the depth and the specific energy that diagram basically we call as specific energy diagram. Well, then we did discuss about another diagram that we call as a depth discharge diagram. Well, for a given specific energy, we can have a relationship between the discharge and the depth. So, that diagram, that is, for a constant specific energy, we get a single line for depth and discharge a curve, we get rather and then again for a different energy level, we get different lines, a set of curve we get that we call as a depth discharge diagram. (Refer Slide Time: 01:38) Well, then again we did discuss about critical depth, which is very important. Today, we are going to discuss about the computation of critical depth. So, critical depth is the depth which corresponds to the minimum specific energy. So, we got specific energy and then, we could see that for a particular depth of flow, for a given discharge, the specific energy in a channel become minimum. That particular depth is called a critical depth. Well, then we got another term that is called alternate depth. That also, we did discuss. What you mean by alternate depth that for a particular specific energy. That means, in a channel, we can get a specific energy for two different depths. We can have same specific energy and these two different depths are called alternate depth. I mean, one of these two depth is in the super-critical region and another, is in the sub-critical region. So, when the flow is in sub-critical condition, then for a particular depth, suppose we are getting a specific energy. Then, say for the same specific energy, we can get another depth when the flow is in a super-critical region. So, that way these depths are called alternative. So, that also we did discuss in our last class. Then, we talked about another term that is called critical slope. Well, when a channel is carrying a particular discharge Q, then for a particular slope, we can have uniform flow for a particular slope. We can have uniform flow at critical depth; we know that critical depth is already defined. So, we know that this is that critical depth and in a channel for a particular slope, we can adjust the slope. We can have a slope where we are having uniform flow for a particular discharge and that uniform flow depth, is equal to the critical depth. So, that particular slope we refer as critical slope. That means, critical slope is the slope where uniform flow occur with critical depth. Well, now when the slope of channel is flatter, then the critical slope, we call this as mild slope. When the slope of the channel is steeper, then the critical slope, we call that as a steep slope. So, that way we got critical slope, we got mild slope, then we got steep slope. Well, after that, we did discuss the condition of critical flow. That means how we will understand that flow is critical. Well, we can define the minimum specific energy, means it is critical flow, but how we will understand. So, we need some definite index to show that it is the critical flow condition. Basically, one condition is definitely that when specific energy is minimum for a given discharge. That is called critical flow condition. Then, another condition is that the when Froude number, suppose in a channel, we are observing the velocity, we are observing the depth and then we can very easily calculate the Froude number. That we all know what is Froude number. So, when that Froude number is unity, then we call, that is the critical flow condition, that we did derive in the last class. Similarly, we could see another point also for a given specific energy discharge is maximum at critical flow condition. So, that is also another condition. Then, we got another point related to that kinetic energy is half of the hydraulic depth. So, that sort of relation also, we did obtain and then we got another point, that is, of course, we did not discuss in detail that point, but still we just marked that point. That is specific force. Of course, specific force is minimum at critical flow condition. So, all those different conditions, we got for critical flow condition and which indicate that the flow is critical. Now, after that we came to computation of critical depth rather computation of critical flow, means computation of critical depth. So, we started with that. We rather end up our last class with that computation of critical flow. There we could see that for a trapezoidal channel, if our channel is of trapezoidal shape, then we could see that we get relations which do not permit us to calculate the critical depth directly or explicitly. We can calculate and in that case, we need to go for trial and error procedure. So, let us start from that particular point, that is, computation of critical depth. So, you can concentrate into the slide. (Refer Slide Time: 08:54) When it is trapezoidal channel, then we know that very basic relation for critical depth is Q square T by gA cube. This is equal to 1 that we did derive in the last class. If it is a trapezoidal channel like this, then we get that Q square T, that is, the top width T. If the side slope is z, then the top width T is B. That means the bottom with B plus twice of ZY c twice of ZY c. So, this part and that way, we are getting this Q square T by g and A cube. If we write on this side, then area is B plus ZY c into Y c cube, so A cube. So, that is what we are getting and from this equation, it is clear that we cannot have this in the form that Y c is equal to some other known term. So, we cannot calculate it directly. Well, what we can do then for this sort of trapezoidal channel? We need to go for trial and error procedure. We put some critical depth value and we try to see whether the left hand side and right side is matching or not. Then, if we keep on changing this one and we try to get the value of Y c by trial and error procedure, well that procedure is definitely tiresome and time consuming. Of course, now with the advent of computer, we can get it very easily, but otherwise, if we go for hand calculation, this is really problematic. That is why some graphical procedures were also developed. So, we have some graphical procedure and then in the graphical procedure, we use a term that is called section factor. In the computation of normal depth also, we got a term called section factor, which is basically coming from the sectional dimension of the flow section. Here, this section factor also we use, but this section factor for critical depth computation is not the same expression, that we got for normal depth. It is different expression. So, if we start from this relation Q square T by ZA cube equal to 1, then we can write it that Q square by Z equal to A square into A by T. So, that is given us. We know that A by T is nothing, but hydraulic depth D. So, as A by T is equal to hydraulic depth T. So, we can have, that is equal to A square, that is area square into hydraulic depth. That we can write as Q by root over g taking, I mean square root of both side, we can write as Q by root over g is equal to A root over D. This particular expression is called section factor Z for critical depth computation using this section factor. In fact, we can find this section factor, if we know the Q and g is, of course, known. So, once Q is given, we can find this value. Then, we have some graphical platform which we could find the critical depth value. However, nowadays these are not used. Utmost, the basic reason is that we have computer. We can go for trial and error method very easily and that is why, these methods are not having that popularity as it had earlier. So, then similarly from this expression, we can see, we can rewrite this expression. The Z is equal to a root over D. A is also a function of Y and D hydraulic depth is also a function of Y. So, that way, Z square we can write that A square into D. What we had? So, that we can write as C, I mean 1 co-efficiency, then Y to the power M. This is the depth. So, that way, from this relation, we are writing it in terms of one exponent M. That is called hydraulic exponent M for critical depth computation. Just as an information, I am just sharing these things that this sort of relation are using this hydraulic exponent M, then plotting it. This expression again in graphical form, we could solve for the critical depth, but as I am telling that nowadays, we have other easier procedure. So, we normally, do not go for this sort of graphical procedure. Well, then from this particular slide, we could see that computation of critical depth, of course, we are talking about the expression what we are getting for trapezoidal channel. For this sort of channel computation of critical depth, we cannot get directly and that is why we are talking about all these graphical procedure or trial and error procedure. Is it the same situation for other shape of channel? That is, if our channel is say rectangular, if our channel is say triangular, then what will happen? So, let us just see, what sort of expression we get for solving critical depth in these different types of channels. (Refer Slide Time: 14:47) First, let us talk about rectangular channel. So, this is a rectangular channel. Then, say this is B bed width. In D, if this depth is Y c, that is the critical depth. Now, starting from the very general relation, Q square T by gA cube is equal to say 1. This is a general expression for critical depth. So, we can start from that and here, our top width T is nothing, but equal to the B. So, what we can write that this implies Q square. Then, we can write it as B and g and area is nothing, but B into Y c. So, we can write this as B cube into Y c cube. This is equal to 1. So, this area cube and from that we can write it in a different form. So, this B cube and B is there. So, it can be written Q square by B square and g, let it remain like that 1 by Y c cube equal to 1. Now, this Q square by B square, that term can be written as in the term of unit discharge. So, discharge per unit width, that is Q is the discharge. So, per unit width means Q by B. That can be written as unit discharge and symbol used for, that is small q. So, this can be written as Q square by g and that can be written as equal to Y c cube. This implies that Y c is equal to Q square by g whole to the power 1 third. So, that way for a rectangular channel, we can see that we can calculate the critical depth Yc directly. If we know the Q value, small q or say capital Q and if we know the B value bed width B, then we can directly calculate this critical depth Y c is equal to Q square by g whole to the power 1 third. Well, one important point. Though, it is obvious that we need to mention or rather, if we mention this is better that for a channel, if someone ask that a channel is getting this much of discharge and it is having this much of slope, what is the critical depth. Normally, when someone asks to compute critical depth, many student I have seen that things about the slope here, basically critical depth is not depending on the slope. So, critical depth is a not a function of slope. Rather, we can calculate the critical depth as in terms of Q and this g value. So, that way critical depth can be calculated directly. Well, now let us see another relationship. (Refer Slide Time: 18:21) Another relationship at critical condition, suppose the channel is flowing in a critical, so water is flowing in a channel with critical condition or at a particular section. We have a critical depth. Now, what is the energy at that point? We know that energy or the specific energy is minimum, that is known, but I mean, can we express that energy in terms of that critical depth, because we know that energy is a function of Y, depth Y. So, at critical condition, can we have a definite relationship between this specific energy and the depth? Well, let us just explore that. So, in this slide, we will be discussing about, that is E versus Y relationship for critical condition and this we are discussing for, say rectangular channel as we are starting with rectangular channel. Well, we know that, that is why, say relationship of critical depth Y c and specific energy, we are meaning, but specific energy we can always had as E minimum because this particular condition, we are talking about minimum specific energy because it is the critical flow condition. Well, so, we can start like say E minimum is equal to Y c plus v square by twice gY c plus v square by twice g as we are writing E minimum. So, we are writing here Y c. We are writing here, critical depth. Well, then V square that we can write as, V is nothing, but equal to Q by area and for rectangular channel this is nothing, but Q by B into Y c. This is always critical depth. So, this is equal to, we can write as Q by B. That part, we can write as Q by Y c. So, this part, we can write this is equal to Y c plus V square. We can write as Q square by Y c square and then, it is twice g. From this expression, what we can write that Y c plus, we had one expression Q square by g. So, I am keeping it like Q square by g. If we go to our previous slide, we could see that q square by g. This expression Q square by g is nothing, but Y c cube. So, this Q square by g, we can write as Y c cube. This is plus. So, this is equal to Y c plus Y c cube by 2Y c square and that will lead us to, say Y c plus Y c by 2T Y c plus Y c by 2. All these we are writing about E minimum. So, this is nothing, but we can write that 3 by 2 into Y c. So, Yc minimum, we can write as E minimum is equal to 3 by 2 into Y c or this can be written in the form that Y c is equal to 2 by 3 E minimum. So, once we know the energy E minimum, then we can calculate the Y c directly from this or if we know the Y c, we can calculate E minimum directly from this. So, this relation shows that, there is a definite relation for critical depth computation or for computation of minimum specific energy in critical condition. (Refer Slide Time: 23:01) Now, let us see if our channel is not rectangular. Rather, say it is a triangular channel. So, let me draw a triangular channel. Suppose, we are having flow Y c and then, side slope is Z. So, this top width here, it will say this part will be ZY c. So, total will be twice Z into Y c and this Z is basically, we are writing about the side slope Z. So, in triangular channel, again if we start from the various basic relation, that is Q square T by gA cube and area we can write as half of base in altitude. So, base of this triangular is twice ZY c and altitude is equal to Y c. So, this is equal to ZY c square. So, area we can write like that. So, Q square T by gA cube equal to 1. That will lead us to Q square, then top width is equal to twice Z into Y c. Then, it is g and area we can write as ZY c square whole cube. Now, from this, we can express Y c in terms of the other value. I mean in terms of the other value. So, let us try doing, that is, will be Y c cube Y c to the power 6. This part will be and this is 1Y c. So, ultimately, we can write Y c to the power 5Y c to the power 5 is equal to square. Then, twice Z and here, it will be g and Z to the power cube. So, finally, we can have it in the form that Q square, twice Q square by gZ square. So, Y c to the power 5, we are getting in the form that twice Q square by gZ square. Using this relationship again, say Y c is equal to twice Q square by gZ square whole to the power 1 fifth. Now, this relation we can use for computing critical depth directly. So, in a triangular channel, if the side slope Z is known, Q is known, then we can directly calculate the Y c. Now, let us see, can we have some relationship again for computing minimum specific energy or specific energy at the critical condition that on the basis of this critical depth. (Refer Slide Time: 26:33) So, this slide will be discussing about E versus Y relationship for critical condition is triangular channel. Let me again start from, say E is equal to specific energy E minimum. We are talking about E minimum is equal to Y c plus v square by twice g. Starting from that we can write it as Y c plus v square is equal to Q square by a square and that a square relationship as we could get here, that we can see is ZY c square. So, we can write ZY c square whole square and then, twice g. We just need to remember what relationship we got for this Y c? So, that is Y c is equal to twice Q square by gZ square. Y c we got as twice Q square gZ square Y c to the power 5. Rather, Y c to the power 5 is equal to twice Q square by gZ square. Now, we need to express this particular expression. To have this expression, that is this expression twice Q square by gZ square, so that we can have it terms of Y c. So, let us just rewrite this and this from Y minimum is equal to Y c plus. If we write here twice Q square, just have it in this form. In this form, we can write here another two. So, it will become 4. Then, this ZY c square is already there. So, let me write it as Z square and then g is here and Y c, I am writing separately, Y c to the power 4. So, twice Q square by gZ square that expression we are getting here and that expression is nothing, but it is Y c to the power 5. So, we can rewrite this expression as Y c plus Y c to the power 5 divided by 4Y c to the power 4. This is nothing equal to Y c plus 1 by 4Y c or Y c by 4 and that we can write as 5 by Y c 5 by 4Y c. So, we can write as E minimum, we can write as 5 by 4Y c. So, that is another relation that we can have for triangular channel. Y c is equal to 4 by 5 E minimum. (Refer Slide Time: 30:27) Well, from these two analyses, we could see that for triangular channel and for rectangular channel, we can express minimum specific energy, that is, the energy at critical condition in terms of Y. Now, if it is a parabolic channel, let us see what sort of expression we can have or can we solve with directly, can we solve critical depth directly for parabolic channel. So, let us take a parabolic channel like this and the in parabolic channel, if this depth is Y c, then this top width can be expressed in terms of Y c as, say K some coefficient and Y c to the power half. This is our top width from the very basic expression of parabola. We can write it like, that top width is equal to KY c to the power half and the area, that we can write as area is equal to 2 by 3 KY c to the power half, that is what the top width and into Y c two third. So, this becomes equal to 2 by 3 KY c to the power 3 by 2. So, that is the expression for area and K is a constant of the parabola. So, using this expression again, if we start from the very basic relation, that is, Q square T by gA cube is equal to 1. If we start from that again, then what we can write Q square and T top width is nothing, but KY c to the power half. So, Q square T. Then, gA cube, we can write as 2 by 3 KY c to the power 3 by 2, but this entire expression will have to be power to cube. So, this we can again simplify, rather our basic objective is to find out Y c. Well, let me write one more step here, that is Q square K into Y c to the power half g. Then, this 2Q means 8 divided by 27 and then, we can write K cube and then, this will be Y c to the power half Y c to the power half. Sorry, this will be Y c to the power 9 by 2. Now, from this, if we write the Y c, directly this is equal to 1 and from that, let me go this side, that is Y c to the power 9 by 2. If it is coming here, then it will be 8 by 2. So, it will be Y c to the power 8 by 2 or we can write Y c to the power 4 directly. This expression, this Y c and this Y c combining, it will be Y c to the power 4. This is equal to Q square by 8 by 27g and this K, it is K square. So, that expression, we are getting, that is Y c to the power 4 is equal to Q square by 8 by 27 gK square. Well, now using this expression, can we again just find one relationship between the minimum specific energy and Y and the critical depth Y c again. We need to recall this relationship. This can be written in this form also, Y c is equal to 27 by Q square by gK square 4 to the power one fourth. This can be written like that also ok. Now, let me go to the next slide and see how we can calculate Y c or how we can express Y c, in terms of minimum specific energy. (Refer Slide Time: 34:49) Well, again minimum specific energy E minimum is equal to Y c plus v square by twice g. So, Y c plus v square is nothing, but Q square by A square and twice g. Now, let me write this A square term and let me have this term here. So, that readily, we can write it in area terms is equal to 2 by KY c to the power 3 by 2. So, area is equal to Z square whole to the power 1 fourth. Starting from that what we can write? This implies E minimum is equal to Y c and let me again rewrite this expression in the form. So, that we can write Y c for this particular expression, Y c plus say Q square. Let it remain like that and then, area we will be writing as 2 by 3 KY c to the power 3 by 2 square and twice g. So, this is equal to Y c plus. This we can write as 4 by 9 K square and Y c to the power 6 by 2, that is Y c cube and then, it is twice g. Then, Q square is remaining there. So, this expression, if we just try to write it in this form, then we can just write it like that Y c plus, it will be 9 by 4, but we need to get it as 27 by 8. Let me write it as 9 by 4. Then, it is Q square, then K square g is there. So, this 2 and this 4, this will become 8, 2 and 4. This will become 8 and then, this is Y c cube, but still to express this in the form, that is 27 by 8, we can write it in the form that Y c plus. This we will write at 27 Q square, then 8k square g. This part is there and then, we will multiply it by 3Y c cube. So, finally, this expression we can combine and we can write this as Yc plus. This part is again Y c to the power 4. So, untimely it will be one third of Y c and this can be written as 4 by 3Y c. So, the expression can be written as E minimum is equal to four third of Y4 by 3Y c 4 by 3Y c. So, this is one relationship that we can have for a parabolic channel as such. We can see now that and computation of critical depth is not always, I mean very difficult, rather for most of the cases like rectangular channel, like triangular channel and then, like parabolic channel. Even, we can get it very directly, that is if we know why, if we know the discharge, if we know the channel section, then we can get the critical depth expression directly. So, computation is rather easy as compared to the computation of normal depth. Only for trapezoidal channel, we need to go for trial and error procedure. (Refer Slide Time: 39:59) Well, let me go to another important relationship. We have seen that Y c can be calculated and thus, energy at that Y c can also be calculated and if we just recall our specific energy diagram, it was like that. Then, you can just concentrate on to the graph. Then for a specific energy E, we can have two depths, say this is Y 1 and this Y 2. These two depths, we call as alternate depth, other than the minimum specific energy. When it is minimum specific energy, then the depth we get is called critical depth. This is what our critical depth, but other than minimum specific energy, if we go to any other energy level, then we get two different depths for a specific energy, for a given discharge Q. These two depths, one is in super-critical condition and other is in sub-critical condition. Now, let us see, can we derive some relationship between the critical depth and these alternate depths. So, in this slide, we will be discussing about relationship between critical depth and alternate depth. Well, let me write that as in these two points, specific energy is equal. So, what we can write, that is Y 1 plus, suppose velocity corresponding to this point is this depth is V 1. So, Y 1 plus V 1 square by twice g and that is equal to Y 2 plus V 2 square by twice g. So, from this, we can say that Y 2 minus Y 1 is equal to V 2 square by twice g. Sorry, V 1 square by twice g minus V 2 square by twice g. Now, for rectangular channel, what we can write for rectangular channel? This relationship is very popular and for rectangular channel, we can have this in this form, that is if it is Y c, then this velocity V at any point can be written as Q. Basically, first we can write it as area into, sorry Q by area and this is velocity equal to Q by area. Area is nothing but B into Y c. So, this can be written as Q by Y c. So, writing this, whatever may be the velocity, see here the velocity is V 1 and in the second section, velocity is or when the depth is higher, velocity is V 2. Definitely, it is V 2 will be less than V 1, but the discharge, there is the unit discharge or the discharge of the channel is not changing. So, if we are talking about this particular line, we are talking about this specific energy line for a given discharge. So, when the discharge is same and if it is a prismatic channel, then our B is also same. So, that way, it is Q. This small q is not changing. So, whether it is V 1 square or V 2 square, that we can write as in terms of Q square by Y c Q square by Y c and twice g minus Q square by Y 1. Rather, at this point, it will be Y 1. At the other point, it will be Y 2 Q square by Y 1 minus Q square by twice g into Y 2. So, taking this advantage, what we can write that we can write as 1 by twice gQ square by twice g. Rather, we can write Q square by twice g and then, it is 1 by Y 1 minus 1 by Y 2 by Y 1 minus 1 by Y 2. Oh! Here, I did a mistake because this is V square. I am writing V is equal to Q by Y 1. So, V square is equal to Q square by Y 1 square and this is equal to Y 2 square. So, it will be Y 1 square minus Y 2 square. So, this can be written as, again we know that Q square by g, that particular expression is nothing, but Y c cube, if we recall our earlier expression. So, Q square by g can be written as Y c cube and then, it will be 1 by 2. This part is Y 1 square Y 2 square. This will be Y 2 square minus Y 1 square and that can be written as Y 2 minus Y 1 into Y 2 plus Y 1 divided by twice Y 1 square Y 2 square and this here on the left hand side, we had Y 2 minus Y 1. So, this Y 2 minus Y 1, if we cancel, then we can write that this Y c cube is equal to this part twice Y 1 square Y 2 square divided by Y 2 plus Y 1. So, this expression finally, can be written as Y c cube is equal to twice Y 1 square Y 2 square divided by Y 1 plus Y 2. So, this is one of the very popular expressions. That is used when we want to know that what will be the, suppose if we know that a particular depth, we have Y 1, then we want to know what is the alternate depth. Then, if we know the critical depth, we can calculate that. Suppose, we know Y 1 and Y 2 at a particular energy level, we know that this is Y 1 and that is Y 2, then knowing Y 1 and Y 2 also, we can calculate critical depth. That is why while we are discussing the computation of critical depth, we are discussing this relation as well because this can be used for computing the value of critical depth. (Refer Slide Time: 47:28) Well, then, let me discuss another topic very briefly. Of course, this topic is critical depth in compound channel and this need to be discussed very elaborately, but it is beyond the scope of this particular course. That is why, just to give you some idea or just to have some discussion on this topic, let me discuss it briefly. Well this, let us draw one compound channel and let us see how we can have some peculiar relationship for critical depth computation in compound channel. In fact, Chaudhry and Ballamudi in 1988 did this sort of analysis by symmetrical compound section. What we mean that these are identical. This left hand side, right sides are identical. This is a compound channel and then, say depth flow at any level, may be Y here, then this B we can write as B m. That is the main channel and always we know that in a compound channel, when the depth of flow will be less or discharge will be less, then flow will be concentrated, sorry the flow will be flowing through the main channel. Of course, concentrated flow goes through. The flow concentration is always in the central part and then, on the side, the flow will be less. Rather, the flow velocity will be less. Then, this level we call as depth up to the floodplain Y f and this portion is call width of the floodplain B f. As it is symmetrical, so this side also it will be floodplain width, the same bf. So far, Manning’s roughness is concerned; here we can write that this is nm, main channel roughness. This is, say floodplain roughness and then in the compound channel, using this term, some non-dimensional term can be derived say Y r, that is, equal to Y depth of flow at any point divided by Y f. This is say non-dimensional term regarding depth. Then, we can have breadth related non-dimensional term, that is, equal to floodplain width divided by say, sorry floodplain width divided by width of the main channel B m. Then, we can have another term B f. This is equal to B f divided by Y f. Then, we can have this n r, that is, equal to nm divided by n f. Now, this is just non-dimensional term which is used for obtaining or for analyzing this critical flow computation. Of course, as I have already mentioned that we do not have that much of scope to go in detail of these analyses, but still we can have some of the understanding. Well, now we know that Y c, as we have already discussed say Y c is equal to, we can have Q square by gY c cube is equal to Q square by g. So, if it is Q square by g, means small q is equal to q by B and when we talked about this rectangular portion, that is the lower portion of the compound channel, this particular portion. Then, it is just simple rectangular channel and we can have this expression, that is, Q square by g. That Q square we can write Q square by B square small q square. So, q square by g and Y c cube is equal to Q square by g. So, this is say, B m. Now, this we can write as at critical condition, it is like that Y c cube is equal to, so this can be written as gB m square. Then, Y c cube by Q square is equal to 1. I mean from that we can write this one. Now, let us see, if for a particular discharge, if our critical depth lies below this channel, below this floodplain depth, then we can use this expression directly because up to this point, we do not have anything, nothing like compound channel. Here, it is just simple rectangular channel. Now, we can calculate if in place of Y, say that means, upto Y c is equal to Y f. If your Y c remains Y f, then we can have this sort of relationship. So, we can write one expression that gB m square, then say, Y f cube by Q square. Let us define this as K. Now, from this expression, we can have some understanding that if our K is equal to 1. So, K is equal to 1 means Y c. This implies that Y c is equal to Y f. Now, if K is less than 1, then 1 means Y c, that is it is less than y means Y f is less than Y c. So, Y c is greater than Y f. Well, now what it indicate Y c greater than Y f means critical depth. If you just refer this diagram, critical depth will be greater than Y f. That means, greater than this particular depth. So, when critical depth is greater than this particular depth, that means we cannot have critical depth below this main channel, below the floodplain level. In that case, that will not be valid because up to this portion only, we can have rectangular. So, this we can consider, this is rectangular channel. So, this implies that critical flow cannot occur within the, let me write critical flow, cannot occur in the rectangular portion. Now, if we have another condition that K is greater than 1. So, K greater than 1 will indicate Y f is greater than Y c. Now, if our floodplain depth B f, sorry if our floodplain depth Y f is greater than critical depth, that means the critical depth will occur or critical flow will occur within the rectangular channel. So, this will lead us to that condition that critical flow occurs in the rectangular portion. So, when critical flow occurs in the rectangular portion, then this we can calculate directly. Now, when critical flow is occurring somewhere here, when the flow is over floodplain that means from this analysis, what we are getting that, if the flow is critical depth is occurring, what the flow is occurring above the floodplain. Then only, that is when the flow depth somewhere here, then we can have some complexities in computing critical flow, when it is below this. Yes, we can do it directly and for that another term was defined. (Refer Slide Time: 56:48) That term is called C, which basically relates, that is Y r b r. Those non-dimensional relationships, B f and N r and this C 1 mean, when this K expression is equal to C, then we can have critical flow condition. Well, that C, we can plot like this. That means, on this side, if we plot Y r, basically for a section for other things remaining the same. Say, our B r, as we could see, just to again go back, B r is equal to B f by B f by B m. It should be written B m; sorry B r is equal to B f by B m. So, as we can see that this B f B m, suppose for a given section, if these are constant B f Y f these are constant, but this Y r will be varying with that depth, that means when depth changes, Y r is varying N m and N f. This nr ratio is also remaining the same. So, ultimately from this expression, if for a given section, for a given roughness, everything remaining the same C becomes a function of Y r. So, we can plot Y r versus C and C and Y r. Y r will start from 1. If you just go back to this expression, Y r is nothing, but Y by Y f. So, when Y is equal to Y f, this Y is equal to Y f. The depth flow is this much. We are getting it below the rectangular portion. Now, our point or we are now interested when it is above this thing. So, when Y will be exceeding this Y f, then Y r become higher than 1. So, when it is higher than 1, then we are interested what will happen then. We are talking about this term C. So, this is say 1. This is 0123. If we get like that, depending on the section, we can have different sort of curve. This will be 1, say 1.1, 1.2. That way the ratio is gradually increasing and then, we can have a curve like this. We can have a curve like this and then, when we calculate the K value, that K value, what we got earlier, this K value. Then, if we see that in this curve, K can be greater than, there we are getting a value C max, maximum value of C for a given section. If our K is, suppose K value we are calculating and K is greater than C max, this can be one condition. Say, this is case 1. Case 1, we are getting K is greater than C max. Then, we can have another condition that now, here we can see when case 1, K is greater than C max means, this on the floodplain, when the flow is above floodplain, we are not getting any intercept of this K value with the C max. That means, it directly means that no critical flow will occur in the upper portion of the compound channel because we are talking about upper portion of the compound channel as Y r ratio is higher. So, no critical depth will occur in the upper portion of the compound channel. Well, critical flow will occur only in the lower portion of the compound channel, that is, the condition. When we have, say if our case 2, if we can have this K is less than C max and we can have K is greater than 1. That means, it is above 1 and is less than C max means somewhere here, this K value. Now, if it is like that, that means, we are finding there will be one critical depth where K is equal to C. Here, one point and this is above the floodplain Y r ratio 1.1 and 1.2 between. So, it is higher than that. So, at some point, it will be there and another critical depth we are getting here, where it is just likely higher than Y r. Y r is slightly higher than 1 means just near the floodplain, we are getting this thing. If I draw this compound section, that means, Y r is just greater than 1 means just here somewhere, we are getting one critical depth, say Y c. We are getting 1Y c and another Yc we will be getting here which correspond to this value. So, this condition indicates two critical depths. One will be there above the floodplain, 2Y c above the floodplain. So, that is what we are getting. Of course, as K is greater than 1, from this condition, we got when K is greater than 1, there will be another critical depth below the floodplain. So, 2Y c and 1Y c below floodplain, so that also we are getting. Then, we can have another case 3. When our K is less than C max and also K is less than equal to 1. That means, if we are coming below this 1, then it is going like this and we are getting one critical depth above the floodplain. So, this will lead to one critical depth above the floodplain and obviously as K is less than 1, we will not get any critical depth below the floodplain level. That is why from these conditions, we can see that there will be a, if draw the specific energy diagram, this is a floodplain level. (Refer Slide Time: 1:03:06) For case 1, there will be no critical depth here and there will be a change in the diagram. This is the critical depth. This is case 1. Then, for case 2 what will happen, there will be one critical depth here. Then, energy diagram is going like that and then, it is coming like that. So, there will be three critical depths. Two above the floodplain and one below the floodplain 1 2 3 and here, it is only one. For the third, this is case 2 and for the third case, what will happen, there will be a critical depth above the floodplain, but there will be no critical depth below the floodplain. This is what one critical depth. So, that is why in compound section, we can have three critical depth conditions, three critical flow conditions. For all these specific energy is of course minimum. Here, specific energy is minimum in the case 1. Specific energy is minimum for two cases. In the case two, but for case 1, specific energy is not even minimum. So, this sort of different conditions, we can have for computing critical flow in compound channel. Well, this much is, I think sufficient who have some understanding of critical depth computation in compound channel. In the next class, we will be going in more detail into the critical flow computation. We will be solving some numerical. That way, we will try to see how critical depth computation can be done for the different types of channel and how this critical flow computation helps us in studying other aspect of hydraulic engineering. Thank you very much. |
14464 | https://www.exodusbooks.com/samples/aops/51888toc.pdf | CONTENTS Contents How to Use This Book iii Acknowledgements vii 1 Counting Is Arithmetic 1 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1 1.2 Counting Lists of Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 1.3 Counting with Addition and Subtraction . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 1.4 Counting Multiple Events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13 1.5 Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19 1.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21 2 Basic Counting Techniques 27 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27 2.2 Casework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27 2.3 Complementary Counting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35 2.4 Constructive Counting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38 2.5 Counting with Restrictions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41 2.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45 3 Correcting for Overcounting 49 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49 3.2 Permutations with Repeated Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49 3.3 Counting Pairs of Items . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
53 ix Excerpt from "Introduction to Counting & Probability" ©2013 AoPS Inc. www.artofproblemsolving.com Copyrighted Material CONTENTS 3.4 Counting with Symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
57 3.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
62 4 Committees and Combinations 65 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
65 4.2 Committee Forming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
65 4.3 How to Compute Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
69 4.4 Our First Combinatorial Identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
72 4.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
76 5 More With Combinations 81 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
81 5.2 Paths on a Grid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
81 5.3 More Committee-type Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
84 5.4 Distinguishability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
88 5.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
92 6 Some Harder Counting Problems 97 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
97 6.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
97 6.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 7 Introduction to Probability 111 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 7.2 Basic Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 7.3 Equally Likely Outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 7.4 Counting Techniques in Probability Problems . . . . . . . . . . . . . . . . . . . . . . . . 118 7.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 8 Basic Probability Techniques 123 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 8.2 Probability and Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 8.3 Complementary Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 x Excerpt from "Introduction to Counting & Probability" ©2013 AoPS Inc. www.artofproblemsolving.com Copyrighted Material CONTENTS 8.4 Probability and Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 8.5 Probability with Dependent Events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 8.6?
Shooting Stars—a hard problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 8.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 9 Think About It!
145 9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 9.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 9.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150 10 Geometric Probability 153 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 10.2 Probability Using Lengths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 10.3 Probability Using Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 10.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 11 Expected Value 165 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 11.2 Definition of Expected Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166 11.3 Expected Value Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166 11.4? A Funky Game . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 11.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 12 Pascal’s Triangle 175 12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 12.2 Constructing Pascal’s Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 12.3 Those Numbers Look Familiar!
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 12.4 An Interesting Combinatorial Identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180 12.5 Another Interesting Combinatorial Identity . . . . . . . . . . . . . . . . . . . . . . . . . 184 12.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186 13 The Hockey Stick Identity 191 13.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 xi Excerpt from "Introduction to Counting & Probability" ©2013 AoPS Inc. www.artofproblemsolving.com Copyrighted Material CONTENTS 13.2 The Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192 13.3 A Step-by-Step Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 13.4 A Clever Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198 13.5 The Identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 13.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 14 The Binomial Theorem 209 14.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 14.2 A Little Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 14.3 The Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 14.4 Applications of the Binomial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 14.5 Using the Binomial Theorem in Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . 214 14.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216 15 More Challenging Problems 219 15.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 15.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 Hints to Selected Problems 231 Index 241 xii Excerpt from "Introduction to Counting & Probability" ©2013 AoPS Inc. www.artofproblemsolving.com Copyrighted Material |
14465 | https://learninglab.si.edu/standards/CCSS.Math.Content.HSF-IF.A/531 | Standards::Understand the concept of a function and use function notation
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Interpreting Functions
High School — Functions
Common Core State Standards (CCSS) Initiative: Mathematics Standards
CCSS.Math.Content.HSF-IF.AUnderstand the concept of a function and use function notation
CCSS.Math.Content.HSF-IF.A.1Understand that a function from one set (called the domain) to another set (called the range) assigns to each element of the domain exactly one element of the range. If f is a function and x is an element of its domain, then f(x) denotes the output of f corresponding to the input x. The graph of f is the graph of the equation y = f(x).
CCSS.Math.Content.HSF-IF.A.2Use function notation, evaluate functions for inputs in their domains, and interpret statements that use function notation in terms of a context.
CCSS.Math.Content.HSF-IF.A.3Recognize that sequences are functions, sometimes defined recursively, whose domain is a subset of the integers.
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14466 | https://mathwo.github.io/assets/files/maccool.pdf | At Right Angles | Vol. 6, No. 3, November 2017 28 Features Complex …… NAPOLEON’S THEOREM …… Made Simple SHAILESH SHIRALI N apoleon’s Theorem states the following. Let ABC be an arbitrary triangle. With the three sides of the triangle as bases, construct three equilateral triangles, each one outside △ABC. Next, mark the centres P, Q, R of these three equilateral triangles. Napoleon’s theorem asserts that △PQR is equilateral, irrespective of the shape of △ABC. (See Figure 1.) A C B D E F P Q R Figure 1 Keywords: Napoleon, equilateral triangle, rotation, parallelogram, basic proportionality theorem In an earlier issue of At Right Angles, we had studied a gem of Euclidean geometry called Napoleon's Theorem, a result discovered in post-revolution France. We had offered proofs of the theorem that were computational in nature, based on trigonometry and complex numbers. We continue our study of the theorem in this article, and offer proofs that are more geometric in nature; they make extremely effective use of the geometry of rotations.
Part 2 29 At Right Angles | Vol. 6, No. 3, November 2017 In Part 1 of this article, we had considered computational proofs of Napoleon’s theorem. In the trigonometric proof, we derived an expression for the length of one side of △PQR in terms of the sides and the angles of the △ABC (i.e., in terms of a, b, c, A, B, C). After going through the computations, we discovered that the resulting expression is symmetric in the parameters of the parent triangle. This fact suffices to prove that triangle PQR is equilateral.
Now we study an extremely elegant pure geometry proof of Napoleon’s theorem; it makes very effective use of rotational geometry. In the literature, it is ascribed to an Irish mathematician, MacCool .
Before proceeding, we make a comment about rotations. Figure 2 shows a segment AB being subjected to two different rotations, both centred at a point O. The first one is through an angle of +30◦(the positive sign tells us that the rotation is in a counterclockwise direction); it takes segment AB to segment A1B1. The second one is through an angle of −30◦(the negative sign tells us that the rotation is in a clockwise direction); it takes segment AB to segment A2B2. Note that segments AB, A1B1 and A2B2 have equal length.
Now we get back to the proof of Napoleon’s theorem. Consider a rotation through an angle of −30◦, centred at B (see Figure 3; the rotation is in a clockwise direction). Our interest is in what this rotation ‘does’ to points R and P, i.e., where it takes these two points. Since ∡ABR = 30◦and ∡DBP = 30◦, it follows that the image R1 of R lies on side AB, and the image P1 of P lies on side BD.
We argue as follows. The steps of the reasoning are laid out in itemised form at the right side of the diagram.
To see why BR1/BA = 1/√ 3 = BP1/BD, you will first need to understand why BR/BA = 1/√ 3 = BP/BD. But this follows from the basic geometry of an equilateral triangle.
We leave the details for you to fill in.
From the fact that BR1/BA = BP1/BD, we deduce (using the basic proportionality theorem) that R1P1 ∥AD, R1P1 AD = 1 √ 3.
(1) Since RP = R1P1, it follows that: RP AD = 1 √ 3.
(2) In just the same way, we consider a rotation through an angle of +30◦, centred at C. Then, if the rotation takes Q and P to Q2 and P2, respectively, it follows that Q2 lies on side AC, and P2 lies on side CD; and arguing as earlier, we conclude that Q2P2 ∥AD, Q2P2 AD = 1 √ 3, (3) and QP AD = 1 √ 3.
(4) From (2) and (4), we conclude that RP = QP.
At this stage, we can proceed in two different ways. One way is to say that the same argument can be repeated for another pair of sides of △PQR and to conclude that equality therefore holds for the lengths of that pair of sides of △PQR, and to A B A1 B1 A2 B2 O • ∡AOA1 = +30◦= ∡BOB1 • ∡AOA2 = −30◦= ∡BOB2 • Segments AB, A1B1 and A2B2 have equal length Figure 2 30 At Right Angles | Vol. 6, No. 3, November 2017 A C B D E F P Q R R1 P1 P2 Q2 (a) R1P1 = RP (b) BR1 = BR (c) BR1/BA = 1/√3 (d) BP1 = BP (e) BP1/BD = 1/√3 (f) BR1/BA = BP1/BD (g) R1P1 ∥AD (h) R1P1/AD = 1/√3 Figure 3 conclude from this that all three sides of the triangle have the same length. From this it follows that △PQR is equilateral. (We do not actually have to repeat all the steps of the argument. All we need to say is that since the argument worked for this particular pair of sides, it will also work for another pair of sides. Note that this is an appeal to symmetry.) Another way is to say that RP = QP and ∡RPQ = 60◦; this is so because R1P1 is parallel to Q2P2, and we had obtained these two segments by rotations of segments RP and QP through 30◦and 30◦respectively, the first one through a rotation of −30◦(i.e., 30◦in a clockwise direction), and the second one through a rotation of +30◦(i.e., 30◦ in an anticlockwise direction). So the two rotations are in opposite directions. After the two rotations, the resulting segments are parallel to each other, which means that prior to the rotations they must have been inclined at an angle of (+30)◦−(−30)◦= 60◦to each other. This suffices to prove that △PQR is equilateral.
□ This proof is to be admired for its elegance and its compactness! It shows just how much can be accomplished using arguments belonging only to elementary geometry.
In Part 3 of this series, we will consider generalisations and further aspects of Napoleon’s theorem.
References 1. M. R. F. Smyth, “MacCool’s Proof of Napoleon’s Theorem”, Irish Math. Soc. Bulletin 59 (2007), 71–77, SHAILESH SHIRALI is Director of Sahyadri School (KFI), Pune, and Head of the Community Mathematics Centre in Rishi Valley School (AP). He has been closely involved with the Math Olympiad movement in India. He is the author of many mathematics books for high school students, and serves as Chief Editor for At Right Angles. He may be contacted at shailesh.shirali@gmail.com.
Irish Math. Soc. Bulletin 63 (2009), 63–66 63 MacCool’s Proof of Morley’s Miracle M. R. F. SMYTH One of the most beautiful results in plane geometry is known as Morley’s Miracle (1899). In essence it states that the triangle XY Z in the figure below is always equilateral. It features prominently on the front cover of the popular work but is “still not as well-known as it deserves to be” . The excellent web article continues to track its development and also hosts a wide variety of proofs. None of the early proofs was easy but since 1990 elementary ones have emerged which are backward in the sense that they start from the equilateral triangle and eventually reconstruct the original. Finding a direct proof that matches them in brevity and simplicity has always been an elusive goal .
A B C Z Y X α α α β β β γ γ γ So I was amazed to find just such a proof in MacCool’s notebooks and indeed it was so short that I nearly missed it. At first glance 64 M. R. F. Smyth he seemed to be merely doodling, but moments later he had finished the proof and was working on something completely different.
Those readers who haven’t heard of MacCool’s notebooks may be surprised to learn that I am still less than halfway through the first one. Translation from the Ogham script is proving a long slow process and I am deeply indebted to one correspondent who reviewed and improved upon my original efforts, often spotting intricacies that I had overlooked. Although the gist of his arguments is always clear MacCool delights in recording only a minimum of information, and this particular proof was little more than a sketch decorated with jottings of line segments and angles. Like all the rest so far it is based solely on straight line geometry and similar triangles, but anyone interested in more advanced concepts may be pleased to know that diagrams containing circles begin to appear early in book two.
In his doodle the unit of measure is the perpendicular DX, and the lengths of BX and CX are s and s′. E and F are points on BC where ∠BXE = ∠FXC = 60o. P and P ′ are where BP = s and CP ′ = s′ and S is constructed so that BS = s and ∠SBX = 120o.
This makes the four marked angles 60o (even if ∆ABC is obtuse).
The rest of his construction is self-explanatory.
Now by (vi) and (vii) 2ST = 2SU + 2UT = s + 2 ¡ s −2s−1¢ = 3s −4s−1 and ∆BQV ∼∆BDX yields V Q = 1 −4s−2 so PQ = PV + V Q = 3 −4s−2 thus 2ST = sPQ.
Then by (iv) and (v) AY = µAC s′ ¶ µPS ST ¶ = 2AC.PS ss′PQ .
If W is the midpoint of PS then since ∆BSP is isosceles ∆BWP and ∆EDX will have identical angles, hence ∆BWP ∼∆EDX giving XE = 2s/PS. Therefore XE.AY = 4AC s′PQ and, by symmetry, XE.AY XF.AZ = µ 4AC s′PQ ¶ µsP ′Q′ 4AB ¶ = sAC.P ′Q′ s′AB.PQ = 1 because PQ(AB/s) = P ′Q′(AC/s′) is the height of ∆ABC. How-ever this means AZ : AY = XE : XF and as ∠ZAY = α = ∠EXF MacCool’s Proof of Morley’s Miracle 65 α α α β β β γ γ γ α β β 1 1 1 1 1 2 s s s 2/s 2/s s/2 s −2/s s′ s′ P ′ Q′ 120o + β A B C Z Y X D E F L P Q R S T U V Notes (i) α + β + γ = 60o (so ̸ EXF = α) (ii) ̸ CY A = ̸ CLP ′ = 120o + β (iii) ̸ TSB = ̸ SPX = ̸ BXE = ̸ FXC = 60o (iv) AY = (AC/s′) LP ′ (as ∆CP ′L ∼∆CAY ) (v) LP ′ = PS/ST (as ̸ P ′LY = ̸ SPT = 60o −β) (vi) XR = RV = 2/s (as ∆BDX ∼∆PRX ∼∆PV R) (vii) SU = s/2 (as ∆BSU is half of an equilateral triangle) then ∆AZY ∼∆XEF. Hence ∠Y ZA = ∠FEX = 60o + β and ∠AY Z = ∠XFE = 60o + γ. Analogous arguments for ∆BXZ and ∆CY X show ∠ZXB = 60o + γ, ∠CXY = 60o + β and ∠BZX = ∠CY X = 60o +α. All the angles in the doodle may now be deduced in terms of α, β, γ and it transpires that every angle of ∆XY Z is 60o.
Here are some comments on the proof leading to a slight variation that may help to make it more intuitive. The underlying idea is to treat it as a series of left/right linkages. The results that 2ST = sPQ and XE = 2s/PS are clearly “internal” to the left hand side. On the other hand LP ′ has a foot in each camp since it can be expressed both in terms of objects from the left PS/ST and objects from the right s′AY/AC. Equating these expressions gives a “cross-linkage” XE.PQ = 4AC/(s′AY ) and its companion 4AB/(sAZ) = XF.P ′Q′ which may then be combined to form the complicated looking quo-tient above. Even MacCool seems to have been shocked by the final 66 M. R. F. Smyth devastating cross-linkage PQ.AB/s = P ′Q′.AC/s′ which reduces this quotient to unity. After that the rest is plain sailing.
References A. Bogomolny, Morley’s Miracle from Interactive Mathematics Miscellany and Puzzles H. S. M. Coxeter, and S. L. Greitzer, Geometry Revisited, Mathematical Association of America, 1967.
R. K. Guy, The Lighthouse Theorem, Morley and Malfatti — A budget of paradoxes, Amer. Math. Monthly, 114 (2007), 97–141.
M. R. F. Smyth, 15 Harberton Avenue, Belfast BT9 6PH malcolm.smyth@ntlworld.com Received on 20 March 2009 and in revised form on 11 May 2009.
Irish Math. Soc. Bulletin 59 (2007), 71–77 71 MacCool’s Proof of Napoleon’s Theorem A sequel to The MacCool/West Point 1 M. R. F. SMYTH I came across this incredibly short proof in one of MacCool’s note-books. Napoleon’s Theorem is one of the most often proved results in mathematics, but having scoured the World Wide Web at some length I have yet to find a proof that comes near to matching this particular one for either brevity or simplicity.
MacCool refers to equilateral triangles as e-triangles and he uses κ to denote the distance from a vertex of an e-triangle with unit side to its centroid. Naturally κ is a universal constant. He also treats anti-clockwise rotations as positive and clockwise rotations as negative.
Theorem 1. If exterior e-triangles are erected on the sides of any triangle then their centroids form a fourth e-triangle.
L M N R B C P N' L' M'' L'' A Q 1Irish Math. Soc. Bulletin 57 (2006), 93–97 72 M. R. F. Smyth Proof. Let ABC be any triangle and construct the three exterior e-triangles with centroids L, M, N as shown. Rotate LN by −30◦ about B to give L′N ′ and LM by +30◦about C giving L′′M ′′. Since all four marked angles are 30◦it follows that L′, N ′, L′′, M ′′ will lie on BP, BA, CP, CA respectively and κ = BL′: BP = BN ′: BA = CL′′: CP = CM ′′: CA. Then by similarity L′N ′ = κAP = L′′M ′′ and L′N ′ ∥AP ∥L′′M ′′ so LN = LM and the angle between them is 30◦+ 30◦= 60◦. Hence ∆LMN is an e-triangle.
□ Theorem 1 is the classical Napoleon theorem. MacCool refers to the resultant e-triangle as the outer triangle to distinguish it from the inner triangle whose vertices are the centroids of the internally erected e-triangles.
The proof shows that each side of the outer triangle is equal to κAP. Since it could equally well have used BQ or CR instead this means AP = BQ = CR. The common length of these three lines is central to the next result. Also required is the fact that the centroid lies one third of the way along any median. This important prop-erty is easily deduced by observing that the medians of any triangle dissect it into six pieces of equal area.
Theorem 2. The centroids of the outer triangle and the original triangle are coincident.
L M N R B C P A Q D O Proof. Let D be the mid point of BC, O be the centroid of ∆ABC, and L be the centroid of ∆BPC. Then DA = 3DO and DP = 3DL so ∆DLO and ∆DPA are similar, giving AP ∥OL and AP = 3 OL.
MacCool’s Proof of Napoleon’s Theorem 73 Likewise BQ = 3 OM and CR = 3 ON. Since AP = BQ = CR the distances from O to the vertices of ∆LMN are equal. As ∆LMN is equilateral O must be its centroid.
□ Next MacCool fixes ∆BPC and allows A to vary continuously throughout the plane. He notes that the proofs of these two theo-rems still apply whenever A drops below the level of BC, in effect making the angle at A reflexive and the angles at B and C negative.
Essentially this is because the three e-triangles always retain their original orientation. For the orientation of an e-triangle to change under continuous deformation its area must first become zero which means that it must shrink to a point, but for the e-triangles in ques-tion this can only happen at B or C. So long as A avoids those two points no orientational changes to the e-triangles can occur.
However one subtle change does take place as A drops below BC in that the orientation of ∆ABC itself changes. When that happens the e-triangles become internal rather than external. This has the following consequence.
Theorem 3. The inner triangle is an e-triangle whose centroid co-incides with the centroid of the original triangle.
The next result gives an alternative proof that AP = BQ = CR.
Only the “external” proof is given since the “internal” case is handled by exactly the same proof with the assumption that A lies below rather than above BC.
Theorem 4. Suppose external (internal) e-triangles are erected on the sides of a given triangle. Then the three lines joining each vertex of the given triangle to the remote vertex of the opposite e-triangle are equal in length, concurrent, and cut one another at angles of 60◦.
R A B C P Q X' X 74 M. R. F. Smyth Proof. Let ∆ABC be given and CBP, ACQ, BAR be the external e-triangles. Clearly ∆ABQ is a +60◦rotation of ∆ARC about A, ∆BCR is a +60◦rotation of ∆BPA about B and ∆CAP is a +60◦ rotation of ∆CQB about C. It follows that AP = BQ = CR and all angles of intersection are 60◦. To prove concurrency assume BQ and CR cut at X and construct BX′ by rotating BX through +60◦ about B as shown. Since ∠BXR = 60◦and BX = BX′ it follows that X′ must lie on CR.
However a rotation of the line CX′R through −60◦about B will map C 7→P, R 7→A, and X′ 7→X.
Therefore A, X, and P are collinear which means that AP, BQ, CR must be concurrent.
□ MacCool next studies the areas of the various triangles. He uses (UV W) to denote the algebraic area of ∆UV W.
In other words (UV W) is equal to the area of ∆UV W when the orientation of ∆UV W is positive, and minus that value whenever the orientation is negative.
Lemma 5. In the diagram below BPC, ACQ, and ARB are e-triangles whose mean area is Ω, and Z is constructed so that AZBQ is a parallelogram. Then AZP is also an e-triangle and 2(AZP) = 3Ω+ 3(ABC).
R Q A B C P 60 60 60 60 60 60 Z Proof. As AZBQ is a parallelogram ∠ZAP is alternate to an angle of 60◦so it too is 60◦. Also AP = BQ = AZ so AZP must be an e-triangle. Clearly (AZP) = (ABP) + (BZP) + (AZB) by tesselation = (ABP) + (APC) + (ABQ) as (APC) = (BZP) and (ABQ) = (AZB).
MacCool’s Proof of Napoleon’s Theorem 75 Now (BCR) = (ABP) and (BCQ) = (APC) and (ARC) = (ABQ) therefore 2(AZP) = (ABP)+(APC)+(ABQ)+(BCR)+(BCQ)+(ARC) = 3Ω+ 3(ABC).
□ The diagram below shows two e-triangles, one with unit side and the other with side κ. Although I have found no evidence that Mac-Cool was familiar with Pythagoras, he inferred from this diagram that 3κ2 = 1 and he deduced that the areas of the inner and outer triangles were one third the area of an e-triangle of side AP.
3κ2 = 1 1 1 κ κ κ The area of the smaller equilateral triangle is clearly κ2 that of the larger, from which it follows that κ must satisfy the equation : Theorem 6. The mean area of the three e-triangles plus (minus) the area of the original triangle equals twice the area of the outer (inner) triangle.
Proof. Let ∆be the area of the outer triangle.
As explained on the previous page (AZP) = 3∆.
Applying Lemma 5 now yields 2∆= Ω+(ABC). Alternatively, if ∆is the area of the inner triangle this equation still holds, but there is a caveat. The orientations of ∆AZP and the inner triangle don’t change as long as A avoids the point P where the latter shrinks to a point, but ∆ABC has changed its orientation and so the value of (ABC) is now negative. Hence rewriting the equation in positive terms, 2∆= Ω−(ACB).
□ Corollary 7. The area of the outer triangle is that of the inner triangle plus that of the original one.
Finally MacCool presents a generalisation of Theorem 1.
Lemma 8. Let A, B, C be non-collinear and X any point between A and C. Construct P and Q on BX such that ∠PAB = ∠XBC and ∠QCB = ∠XBA. Then the triangles PAB and QBC are directly 76 M. R. F. Smyth similar, moreover P and Q coincide if and only if AX : XC = AB2 : BC2.
X C B A Q P Proof. Clearly ∆PAB and ∆QBC are directly similar.
Suppose BC = λAB and XC = µAX. Then (QBC) = λ2(PAB) whereas (PBC) = µ(PAB). If P and Q coincide then clearly µ = λ2. Con-versely if µ = λ2 then (PBC) = (QBC) so (PQC) = 0 which implies P = Q.
□ Note that if AB and BC have equal length then ∆PAB and ∆PBC are similar (but not directly similar) for all points P on the bisector of ∠ABC. Also the lines AB and BC (extended) divide the plane into four zones, and if a point O exists such that ∆OAB and ∆OBC are directly similar then O must lie in the zone that includes the line segment AC. This leads to a key result.
Corollary 9. If the points A, B, C are non-collinear then there exists a unique point O such that the triangles OAB and OBC are directly similar.
Theorem 10 (Generalised Napoleon). Let ABC and A′B′C′ be di-rectly similar triangles with a common vertex C = B′. Suppose A′′, B′′, C′′ are chosen such that the triangles AA′A′′, BB′B′′, CC′C′′ are directly similar.
Then so too are the triangles A′′B′′C′′ and ABC.
Proof. There are 3 separate cases. First if B′ is midway between B and C′ then ABB′A′ is a parallelogram and the result follows easily. Otherwise if B, B′, C′ are collinear take O to be the point where AA′ cuts BB′. Then ∆A′B′C′ is a dilation of ∆ABC and it is clear that ∆A′′B′′C′′ may be obtained from ∆ABC by a rotation MacCool’s Proof of Napoleon’s Theorem 77 A'' A' O C' C=B' B A B'' C'' of ∠AOA′′(= ∠BOB′′ = ∠COC′′) about O followed by a dilation of size OA′′/OA. So once again the result holds. Finally if B, B′, C′ aren’t collinear apply Corollary 9 to ∆BB′C′ (aka BCC′) giving the point O such that OBB′ and OCC′ are directly similar. Let θ = ∠BOB′ = ∠COC′ and λ = OB′:OB = OC′:OC. Let τ be the transformation that first rotates through the angle θ about O and then dilates by the scaling factor λ. Clearly τ preserves directly similar figures and maps B 7→B′, C 7→C′ so as ABC and A′B′C′ are directly similar it must also map A 7→A′. Thus ∠AOA′ = θ and OA′ :OA = λ from which it follows that ∆OAA′ is directly similar to both ∆OBB′ and ∆OCC′. Then OAA′′A′, OBB′′B′, OCC′′C′ are directly similar quadrilaterals so OAA′′, OBB′′, OCC′′ are directly similar triangles. Thus OA′′ : OA = OB′′ : OB = OC′′ : OC = µ and ∠AOA′′ = ∠BOB′′ = ∠COC′′ = φ for some µ and φ. That means the quadrilateral OA′′B′′C′′ may be obtained from OABC by rotating it through φ about O and dilating the result by the scaling factor µ. Therefore ∆A′′B′′C′′ and ∆ABC are directly similar.
□ The wheel has come full circle. To derive Napoleon’s Theorem from this result take ∆ABC to be equilateral and choose A′′ so that ∆AA′A′′ is isosceles with base AA′ and base angles of 30◦.
M. R. F. Smyth, 15 Harberton Avenue, Belfast BT9 6PH malcolm.smyth@ntlworld.com Received on 5 January 2007.
Irish Math. Soc. Bulletin Number 72, Winter 2013, 75–77 ISSN 0791-5578 MACCOOL’S SECOND PROOF OF MORLEY’S MIRACLE M.R.F. SMYTH In memory of Kenneth Beales and Trevor West Abstract. Here is a traditional proof of Morley’s miracle that is unrivalled for brevity and simplicity. It stems from a sadly neglected mathematical gem published in 1914.
1. Introduction That the triangle XY Z in the figure below is always equilateral is formally known as Morley’s trisector theorem and informally as Morley’s miracle.
Its modern discovery dates back to 1899 and since then it has been proved many times by a wide variety of meth-ods. The website tracks developments and plays host to roughly twenty proofs including MacCool’s original effort.
A X Y However as explained, the proof there was based entirely on straight line geometry and similar triangles. It opined that Mac-Cool’s second notebook which was marked “Advanced” and con-tained diagrams of circles might hold an alternative proof. And so indeed it has proved, although it has taken me a very long time to decipher the Ogham. So whilst I have yet to find any evidence that MacCool was familiar with Pythagoras, the result that we know to-day as the inscribed angle theorem [Euclid: Book 3, Prop 22] does 2010 Mathematics Subject Classification. 51M04.
Key words and phrases. Morley, trisector.
Received on 17-8-2013; revised 9-12-2013.
c ⃝2013 Irish Mathematical Society 75 76 ROGER SMYTH indeed appear in his “Advanced” notebook, and soon afterwards comes the second proof of Morley’s theorem. This is even shorter and easier than his “Basic” one, and completely debunks the urban myth that all purely geometric proofs must necessarily be longer and more complex than the “backward” ones.
2. Proof His proof runs as follows. In any triangle ABC let X be the Morley vertex adjacent to BC. First construct the points P and Q on AB and AC respectively such that |BP| = |BX| and |CQ| = |CX|.
Then construct the right-angled triangle PRX with hypotenuse PX and ∠XPR = 30◦as shown below. The six marked segments will all have equal length. Produce PR and the trisector CS to meet in Y . Note that the three right-angled triangles, ∆RXY and ∆SXY and ∆SQY , (which MacCool calls wedges) have equal hypotenuses and an equal (marked) side therefore they are congruent. Evidently α + β + γ = 60◦.
β β β γ γ γ Q A B C Y X P R S 30o 60o 3α Z b b b b b b b b b Now ∠QXP = 360◦−2(90◦−β) −2(90◦−γ) = 120◦−2α. So ∠YXR = ∠SXY = ∠YQS = 1 2(∠QXP −60◦) = 30◦−α.
As ∆PQX is isosceles its base angles ∠XPQ and ∠PQX are both 30◦+ α so ∠YPQ = α and ∠PQY = (30◦+ α) −(30◦−α) = 2α.
MACCOOL’S SECOND PROOF OF MORLEY’S MIRACLE 77 Therefore ∠QYP = 180◦−3α. And now for the advanced bit.
Finn spots this is supplementary to ∠BAC making APYQ a cyclic quadrilateral. Consequently ∠YAQ = ∠YPQ = α which fixes Y as the Morley vertex adjacent to AC. Next he performs a similar construction (shown in outline) starting from a right-angled triangle on hypotenuse QX and angles 30◦and 60◦at Q and X respectively.
This generates three more wedges which are clearly congruent to the first three, plus the Morley vertex Z adjacent to AB. In particular |XY | = |XZ| and ∠YXZ = 60◦from which he deduces that ∆XYZ is equilateral.
3. Conclusion After scanning numerous proofs the only “modern” one I’ve seen that is remotely like this is given in and attributed to W. E. Philip.
William Edward Philip was Third Wrangler at Cambridge in 1894, but despite many references to in the literature the beauty of his proof seems to have been strangely overlooked. Indeed also contains a version of Leon Bankoff’s 1962 trigonometric proof, long regarded as the easiest non-backward approach to the theorem. As years passed without anyone finding a short, simple, non-backward geometric proof a mistaken belief has proliferated that no such proof exists. So, as the centenary of its publication approaches, the time seems ripe to call attention to and bring it back centre stage.
References A. Bogomolny: Morley’s Miracle from Interactive Mathematics Miscellany and Puzzles F. Glanville Taylor & W. L. Marr: The six trisectors of each of the angles of a triangle, Proc. Edin. Math. Soc. 32 (1914), 119–131 M. R. F. Smyth: MacCool’s proof of Morley’s Miracle, Irish Math. Soc. Bul-letin 63 (2009), 63–66 Roger Smyth was a research student of Trevor West and received his Ph.D from Dublin University in 1972. Thereafter he moved into Information Technology and worked in Queen’s University Belfast and the Northern Ireland Department of Health. His primary line is Fredholm theory in Banach algebras, but he maintains a recreational interest in simple Euclidean geometry.
(Roger Smyth) 15 Harberton Avenue, Belfast E-mail address: malcolm.smyth@ntlworld.com Irish Math. Soc. Bulletin Number 74, Winter 2014, 94–96 ISSN 0791-5578 S. Louridas and M. Rassias: Problem-Solving and Selected Topics in Euclidean Geometry: In the Spirit of the Mathematical Olympiads, Springer, 2013.
ISBN:978-1-4614-7272-8, EUR 42.79, 235 pp.
REVIEWED BY JIM LEAHY The success of the International Mathematical Olympiad (IMO) has helped to revive interest in Euclidean geometry and to halt somewhat its decline during the second half of the twentieth century.
Consequently there is a constant trickle of new publications on the subject of which the book under review is one. Both authors have connections with the IMO. Sotirios E. Louridas has been a coach of the Greek Mathematical Olympiad team while Michael Th. Rassias is a winner of a silver medal at the IMO 2003 in Tokyo and holds a Master of Advanced Study from the University of Cambridge.
The book has six chapters with a foreword by Fields Medalist Michael H. Freedman. Chapter 1, Introduction, is short with a little history of geometry and containing Euclid’s axioms and postulates.
Chapter 2 deals with the basic concepts of logic and covers methods of proof including proof by analysis, by synthesis, proof by contra-diction and proof by induction with examples. The one induction example is more a problem in number theory than geometry having the theorem of Pythagoras as a starting point. There is no other problem in the book that uses proof by induction. Chapter 3 covers geometrical transformations, viz. translations, symmetry, rotations, homothety and inversion. These are illustrated with examples and some theorems with proofs. The section on inversion will be found particularly useful to students and teachers as it gives several exam-ples of its power in solving certain types of problems. Some of the later IMO type problems in the book also use inversion, something not common in many publications on Euclidean geometry.
Chapter 4 is a collection of thirty-eight theorems some of which are proved. The selection of theorems is excellent. Knowledge of these theorems together with the theorems of Euclid would go a long way towards solving many a geometrical problem. The proof of Received on 15-9-2014.
c ⃝2014 Irish Mathematical Society 94 BOOK REVIEW 95 Feuerbach’s theorem, Theorem 4.21 in the book, contains an error and the proof of Morley’s theorem, Theorem 4.11, is not correct.
In the latter case if the word ‘isosceles’, used twice, is replaced by ‘equilateral’ the proof would be correct but incomplete. Interestingly the construction at the beginning of this proof is similar to the construction used in MacCool’s proof of Morley’s theorem .
Chapter 5 offers sixty-five problems divided into three categories, problems with basic theory, problems with more advanced theory and geometrical inequalities. There is little difference between the first two categories and many of the problems are of IMO standard.
The solutions follow in chapter 6 forming the main body of the work.
Reading through the solutions is not easy. In some cases parts of the solutions seem to have been omitted and a good deal has been left to the reader usually without any comment from the authors. The statement of problem 6.2.25 p.169 is false as the wrong angles are designated as being equal. The solution uses the correct two angles but if you were attempting to solve the problem without consulting the solution, which you would expect a reader to do, your work would be in vain. The problems are restated before each solution in chapter 6 and the error is repeated. The solution of problem 6.2.22 p.164 is also incorrect since it would require the side of an inscribed pentagon to also be a tangent to the circumscribing circle! Obviously there are serveral misprints in this solution. On the other hand some of the proofs are quite innovative and the solution of problem 6.2.15 p.151 is an excellent example of the use of inversion.
There is an appendix on the Golden Section which is a reprinting of an article by Dirk Jan Struik in . I fail to see the point of this as it is a popular article containing all the usual material of such articles which can be found in many publications and on the internet. Besides the Golden Section is not mentioned anywhere in the main text of the book. There is also a useful index of symbols used in the text, a subject index and a list of references, ninety-nine in all, including a reference to Wiles’ paper on the solution of Fermat’s Last Theorem! Since there are no references in the text, apart from acknowledging authors of problems, the references should rightly be called a bibliography.
To summarise, this is not a book showing how to solve problems in geometry except in the sense of learning from seeing problems solved.
This is not a criticism as much can be learned in this way particularly 96 JIM LEAHY if a solution has been attempted beforehand. The book is beautifully produced, the quotions at the head of each chapter adding to the publication. However the work is unfortunately marred by poor edit-ing and proofreading. I counted over sixty errors, omissions, typos or misprints, mostly the latter, which does not make for easy read-ing. In addition some solutions have no diagram. Woody Guthrie, the American folk singer, once said that he liked books with errors as it made them more human. I doubt he ever read a mathematics text.
References M.R.F. Smyth: MacCool’s second proof of Morley’s miracle, Irish Mathemat-ical Society Bulletin, Number 72, Winter 2013.
Th. M. Rassias (ed): Mathematics in education , University of LaVerne Press, California, 1992, pp. 123-131.
Jim Leahy lectured in the Department of Mathematics and Statistics , Uni-versity of Limerick, until his retirement. He has been involved with the Irish Mathematical Olympiad since its inception and continues to actively participate.
School of Mathematical Sciences, University of Limerick E-mail address: leahyn@indigo.ie |
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Tangent to the parabola y2=4ax is of the form y=mx+am where m is the slope of the tangent.
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To show that the tangent to the parabola y2=4ax is of the form y=mx+am, we will follow these steps: Step 1: Find the derivative of the parabolaThe equation of the parabola is given by:y2=4axTo find the slope of the tangent line, we differentiate both sides with respect to x.ddx(y2)=ddx(4ax)Using the chain rule on the left side, we have:2ydydx=4aNow, solving for dydx:dydx=4a2y=2ay Hint: Remember that the derivative gives the slope of the tangent line at any point on the curve. Step 2: Identify the point of tangencyLet the point of tangency on the parabola be (at2,2at). Here, t is a parameter that represents the point on the parabola. Hint: The coordinates (at2,2at) are derived from the parametric equations of the parabola. Step 3: Substitute the point into the slope formulaUsing the point (at2,2at), we can find the slope m of the tangent line at this point:m=2a2at=1t Hint: The slope m is derived from substituting the coordinates of the point of tangency into the derivative. Step 4: Write the equation of the tangent lineUsing the point-slope form of the equation of a line, we have:y−y1=m(x−x1)Substituting (x1,y1)=(at2,2at) and m=1t:y−2at=1t(x−at2) Hint: The point-slope form is a useful way to express the equation of a line given a point and a slope. Step 5: Rearranging the equationRearranging the equation gives:y−2at=xt−aty=xt+2at−aty=xt+a(1t) Hint: Pay attention to how terms are combined and rearranged to isolate y. Step 6: Final form of the tangent equationNow, we can rewrite the equation as:y=mx+amwhere m=1t. Hint: This final form matches the required format, confirming that the tangent line to the parabola is indeed of the specified form. ConclusionThus, we have shown that the tangent to the parabola y2=4ax is of the form y=mx+am, where m is the slope of the tangent.
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Tangent to the parabola y2=4ax is of the form y=mx+am where m is the slope of the tangent.
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From a point P, two tangents are drawn to the parabola y2=4ax. If the slope of one tagents is twice the slope of other, the locus of P is
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14468 | https://www.diva-portal.org/smash/get/diva2:1665796/FULLTEXT01.pdf | Discrete Applied Mathematics 316 (2022) 75–86 Contents lists available at ScienceDirect Discrete Applied Mathematics journal homepage: www.elsevier.com/locate/dam On Hamiltonicity of regular graphs with bounded second neighborhoods Armen S. Asratian, Jonas B. Granholm ∗ Department of Mathematics, Linköping University, Sweden a r t i c l e i n f o Article history: Received 14 July 2021 Received in revised form 9 February 2022 Accepted 22 February 2022 Available online 27 April 2022 Keywords: Regular graphs Hamilton cycles Locally connected graphs a b s t r a c t Let G(k) denote the set of connected k-regular graphs G, k ≥2, where the number of vertices at distance 2 from any vertex in G does not exceed k. Asratian (2006) showed (using other terminology) that a graph G ∈G(k) is Hamiltonian if for each vertex u of G the subgraph induced by the set of vertices at distance at most 2 from u is 2-connected.
We prove here that in fact all graphs in the sets G(3), G(4) and G(5) are Hamiltonian.
We also prove that the problem of determining whether there exists a Hamilton cycle in a graph from G(6) is NP-complete. Nevertheless we show that every locally connected graph G ∈G(k), k ≥6, is Hamiltonian and that for each non-Hamiltonian cycle C in G there exists a cycle C′ of length |V(C)|+ℓin G, ℓ∈{1, 2}, such that V(C) ⊂V(C′). Finally, we note that all our conditions for Hamiltonicity apply to infinitely many graphs with large diameters.
© 2022 The Authors. Published by Elsevier B.V. This is an open access article under the CC BY license (
1. Introduction In this paper we continue our investigation of interconnections between local properties of a graph and its Hamiltonic-ity (see, for example, [4–11]).
We use for terminology and notation not defined here, and consider finite undirected graphs without loops and multiple edges only. A Hamilton cycle of a graph G = (V(G), E(G)) is a cycle containing every vertex of G. A graph that has a Hamilton cycle is called Hamiltonian. There is a vast literature in graph theory devoted to obtaining sufficient conditions for Hamiltonicity (see, for example, the surveys [19,20,29]). Our paper is devoted to the investigation of Hamiltonicity of k-regular graphs, that is, graphs where all vertices have the same degree k. It is known that for any fixed k ≥3, determining whether a k-regular graph has a Hamilton cycle is an NP-complete problem.
Dirac proved that a graph G on at least three vertices is Hamiltonian if the degree of every vertex of G is at least |V(G)|/2. Nash-Williams showed that the result of Dirac can be relaxed for regular graphs as follows: A k-regular graph G with k ≥2 is Hamiltonian if k ≥|V(G)|−1 2 . Jackson showed that if G is 2-connected then the bound |V(G)|−1 2 can be replaced by 1 3|V(G)|. Better bounds were obtained for 3-connected k-regular graphs (see e.g. [12,26,28]).
All the above mentioned conditions for a regular graph G contain a global parameter of G, namely the number of vertices, and only apply to graphs with large vertex degrees ( ≥constant · |V(G)| ) and small diameters ( o(|V(G)|) ).
Another type of sufficient conditions for Hamiltonicity of a graph G, which contain no global parameter of G, was obtained by Chartrand and Pippert , using the concept of local connectedness. A graph G is called locally connected if for every vertex u of G the subgraph induced by the set of neighbors of u is connected. The result of Chartrand and Pippert is devoted to graphs with maximum degree at most four and for regular graphs implies the following: ∗Corresponding author.
E-mail addresses: armen.asratian@liu.se (A.S. Asratian), jonas.granholm@liu.se (J.B. Granholm).
0166-218X/© 2022 The Authors. Published by Elsevier B.V. This is an open access article under the CC BY license ( licenses/by/4.0/).
A.S. Asratian and J.B. Granholm Discrete Applied Mathematics 316 (2022) 75–86 Proposition 1.1.
All connected, locally connected, k-regular graphs are Hamiltonian if k ≤4.
Locally connected 5-regular graphs were considered in : Theorem 1.2 (Kikust ). Every connected, locally connected 5-regular graph G is Hamiltonian.
The situation is different for locally connected k-regular graphs with k ≥6.
Theorem 1.3 (Irzhavski ). For any fixed k ≥6, determining whether a locally connected k-regular graph is Hamiltonian is an NP-complete problem.
Some other results on locally connected graphs can be found in [1,2,18,21].
Oberly and Sumner showed that the concept of local connectedness is fruitful for Hamiltonicity of claw-free graphs, that is, graphs that have no induced subgraph isomorphic to K1,3.
Theorem 1.4 (Oberly and Sumner ).
Every connected, locally connected, claw-free graph on at least 3 vertices is Hamiltonian.
Other properties of claw-free graphs were found in [4,11,14,17,24,30,35,36]. The results on locally connected graphs can be formulated in terms of balls. For a vertex u of a graph G, the ball of radius r centered at u is the subgraph of G induced by the set Mr(u) of vertices at distance at most r from u. Clearly, G is locally connected if and only if every ball of radius 1 in G is 2-connected.
Some Hamiltonian properties of a graph have been obtained using the structure of balls of radius 2 (see, for example, [5–11,27,34]). In particular, Asratian obtained the following result: Theorem 1.5 (Asratian ). Let G be a connected k-regular graph where k ≥|M2(u)|−1 2 for every vertex u ∈V(G) and every ball of radius two in G is 2-connected. Then G is Hamiltonian.
Theorem 1.5 is a generalization of the result of Nash-Williams since a connected k-regular graph G with k ≥|V(G)|−1 2 is 2-connected and any ball of radius 2 in G is the graph G itself.
Let G(k) denote the set of connected k-regular graphs G where the number of vertices at distance 2 from any vertex in G does not exceed k. Then one can show (see Lemma 2.1) that Theorem 1.5 can be reformulated as follows1: Theorem 1.6.
Let G be a graph in G(k) such that every ball of radius two in G is 2-connected. Then G is Hamiltonian.
The starting point of our present research was the following question: Is it possible to omit the condition on 2-connectedness of balls of radius 2 in Theorem 1.6 without losing the Hamiltonicity of G? This question was motivated by the fact that the set G(k), k ≥3, contains infinitely many Hamiltonian graphs where no ball of radius 2 is 2-connected (see Examples 2.1 and 2.2 in Section 2).
In the present paper we find new classes of Hamiltonian graphs not satisfying known conditions for Hamiltonicity, e.g. the conditions of Theorems 1.2 and 1.4. We prove that all graphs in the sets G(3), G(4) and G(5) are Hamiltonian, that is, the condition of 2-connectedness of balls of radius 2 in Theorem 1.6 can be omitted if k ≤5. Moreover we characterize all graphs in G(3), G(4) and G(5) where not all balls of radius 2 are 2-connected. Our results on 3- and 4-regular graphs imply Proposition 1.1, and our result on 5-regular graphs (Theorem 3.4) and Kikust’s theorem (Theorem 1.2) are incomparable to each other in the sense that neither theorem implies the other.
We also show that in contrast with the sets G(3), G(4) and G(5), the set G(k) contains non-Hamiltonian graphs for any k ≥6. Furthermore, we prove that the problem of determining whether there exists a Hamilton cycle in a graph from G(6) is NP-complete. Despite this and Theorem 1.3, we show that every locally connected graph G ∈G(k), k ≥6, is Hamiltonian and for any k ≥30 the set G(k) contains infinitely many locally connected graphs not satisfying the conditions of Theorem 1.4. We also prove that if C is a non-Hamiltonian cycle in a locally connected graph G ∈G(k) then there exists a cycle C′ of length |V(G)| + ℓin G, ℓ∈{1, 2}, such that V(C) ⊂V(C′). The paper is concluded with a conjecture.
2. Definitions and preliminary results The distance between vertices u and v in G is denoted by dG(u, v) or simply d(u, v). The greatest distance between any two vertices in a connected graph G is the diameter of G, denoted by diam(G). For each vertex u ∈V(G) and integer r ≥1 we denote by Nr(u) and Mr(u) the set of all vertices v ∈V(G) with d(u, v) = r and d(u, v) ≤r, respectively. The set N1(u) is called the neighborhood of u and usually is denoted by N(u). The number of vertices in N(u) is the degree of u, denoted by d(u). The set N2(u) is called the second neighborhood of u. The ball of radius r centered at u, denoted by Gr(u), is the subgraph of G induced by the set Mr(u).
1 This formulation will be more convenient for our further descriptions and proofs.
76 A.S. Asratian and J.B. Granholm Discrete Applied Mathematics 316 (2022) 75–86 Let C be a cycle of a graph G. We denote by − → C the cycle C with a given orientation, and by ← − C the cycle C with the reverse orientation. If u, v ∈V(C) then u− → C v denotes the consecutive vertices of C from u to v in the direction specified by − → C . The same vertices in reverse order are given by v← − C u. We use u+ to denote the successor of u on − → C and u−to denote its predecessor. Analogous notation is used with respect to paths instead of cycles.
A graph is called complete r-partite (r ≥2) if its vertices can be partitioned into r nonempty independent sets V1, . . . , Vr such that two vertices are adjacent if and only if they do not belong to the same set Vi for any i ∈{1, . . . , r}. A complete r-partite graph with independent sets V1, . . . , Vr of sizes n1, . . . , nr is denoted by Kn1,...,nr .
Lemma 2.1.
The following properties are equivalent for a connected k-regular graph G, k ≥3: (i) G ∈G(k).
(ii) k ≥|M2(u)|−1 2 for every vertex u of G.
(iii) d(u) + d(v) ≥|M2(w)| −1 for any induced path uwv in G.
(iv) |N(u) ∩N(v)| ≥|M2(w) \ (N(u) ∪N(v))| −1 for any induced path uwv.
Proof.
Let G be a connected k-regular graph. For any u ∈V(G), |M2(u)| = 1 + k + |N2(u)|. Therefore the condition k ≥ |M2(u)|−1 2 is equivalent to the condition |N2(u)| ≤k which means that G ∈G(k). Thus (i) is equivalent to (ii).
Evidently, (ii) is equivalent to (iii) because d(u) = k = d(v). Finally, (iii) is equivalent to (iv), since d(u) + d(v) = |N(u) ∩N(v)| + |N(u) ∪N(v)|.
□ Lemma 2.2.
Let G be a graph in G(k). If v is a cut vertex of a ball of radius 2 in G, then v is a cut vertex of the ball G2(v), the subgraph G2(v) −v has exactly two components with k vertices each, and every neighbor of v is adjacent to all other vertices of the component of G2(v) −v it lies in.
Proof. If v is a cut vertex of the ball G2(v), then the subgraph G2(v)−v has at least two components, at most 2k vertices, and each neighbor of v has k−1 neighbors in this subgraph. This is only possible if G2(v)−v has exactly two components with k vertices each and each neighbor of v is adjacent to all other vertices of the component of G2(v) −v it lies in.
Now suppose that v is a cut vertex of the ball G2(u) for some vertex u ̸= v. Then uv ∈E(G). Let F denote the component of G2(u) −v where u lies, and let N(v) ∩V(F) = {u1, . . . , up}, where u = u1 and p < k. Since G is k-regular, v has k −p neighbors that do not lie in F, we will call these vertices x1, . . . , xk−p. Furthermore, since G is k-regular, u has at least k −p neighbors in G2(u) −M1(v), and x1 has at least p neighbors outside G2(u); all of these vertices are at distance 2 from v. Since there are at most k vertices at distance 2 from v, this means that u is adjacent to the vertices v, u2, . . . , up and exactly k −p other vertices which we will call w1, . . . , wk−p, and the vertex x1 is adjacent to exactly p vertices outside G2(u) which we will call y1, . . . , yp. Moreover, it is clear that the vertices x1, . . . , xk−p are adjacent to each other and all of them are adjacent to the vertices y1, . . . , yp. Also, since we now know that all neighbors of u2, . . . , up lie in G2(u), it is easy to see that the vertices u2, . . . , up are all adjacent to each other and to w1, . . . , wk−p. It remains to show that G2(v)−v is not connected, that is, that wiyj / ∈E(G) for all i, j. But this follows from the fact that N(wi) is contained in G2(u) for all i, and we know that yj does not lie in G2(u) for all j. The lemma follows.
□ Lemma 2.3.
Let G be a connected, locally connected graph and C a cycle in G. Then for every vertex w ∈V(C) that has a neighbor outside C there is a pair of vertices v, z (depending on w) such that v ∈N(w) \ V(C), z ∈V(C), and wz, vz ∈E(G).
Proof. Let u be a vertex in N(w)\ V(C). Since G is locally connected, the ball G1(w) is 2-connected. Then there is a path P in G1(w) −w with origin w+ and terminus u. Let z be the vertex on V(P) ∩V(C) such that all other vertices on the path z− → P u do not belong to C, and let v denote the successor of z on − → P . Then v ∈N(w)\V(C), z ∈V(C) and wz, vz ∈E(G).
□ We also need the following result obtained in .
Proposition 2.4 (). If every ball of radius r in a graph G is t-connected, t ≥2, then all balls of any radius r′ > r in G are t-connected, too.
Now we will show that for any k ≥3 the set G(k) contains infinitely many Hamiltonian graphs. First we describe in G(k) two infinite classes of Hamiltonian claw-free graphs where no ball of radius 2 is 2-connected.
Example 2.1.
We define a graph G(k, n), where k ≥3, n ≥2, as follows. Its vertex set is V0 ∪V1 ∪· · · ∪Vn, where V0, V1, . . . , Vn are pairwise disjoint sets, V0 = {v1, . . . , vn} and Vi = {vi 1, . . . , vi k}, i = 1, . . . , n. The set Vi induces in G(k, n) a complete graph, i = 1, . . . , n. For each i = 2, 3, . . . , n, the vertex vi is adjacent to the first ⌊k 2 ⌋ vertices in Vi and the last ⌈k 2 ⌉ vertices in Vi−1. Finally, v1 is adjacent to the first ⌊k 2 ⌋ vertices in V1 and the last ⌈k 2 ⌉ vertices in Vn.
Example 2.2.
Now we define a graph H(k, n), where k ≥3, n ≥2, as follows. Its vertex set is W0 ∪W1 ∪· · · ∪Wn, where W0, W1, . . . , Wn are pairwise disjoint sets, W0 = {u1, . . . , un, v1, . . . , vn} and |Wi| = k −1, for i = 1, . . . , n. The set Wi 77 A.S. Asratian and J.B. Granholm Discrete Applied Mathematics 316 (2022) 75–86 induces in H(k, n) a complete graph, i = 1, . . . , n. For each i = 1, . . . , n, the vertices ui and vi are adjacent to all vertices in Wi. Finally, unv1 ∈E(H(k, n)) and uivi+1 ∈E(H(k, n)), for i = 1, . . . , n −1.
Clearly, all graphs in { G(k, n) : n ≥2 } and { H(k, n) : n ≥2 } are Hamiltonian and have n(k+1) vertices, connectivity 2 and diameter ⌊3n 2 ⌋, k = 3, 4, . . . . Furthermore, they are not locally connected, since the subgraph induced by the neighborhood N(v1) in each of these graphs is not connected. Finally, the graph G(3, n) is isomorphic to H(3, n), for n ≥2.
Let us also note that for every k ≥3 the set G(k) contains Hamiltonian graphs that are not claw-free. First we mention graphs obtained from a complete bipartite graph Kk+1,k+1 by removing a perfect matching of Kk+1,k+1. Next note that for every divisor d ≥3 of the integer k a complete r-partite graph Kd,...,d, where r = 1 + k d, is Hamiltonian and is not claw-free. For example, the graphs K3,3,3,3,3, K4,4,4,4, K6,6,6 and K12,12 belong to the set G(12). Also note that the required graphs can be obtained from a complete bipartite graph Kk,k, k ≥4, with bipartition (V1, V2) as follows: choose an integer t < k/2 and a matching E(t) = {x1y1, . . . , x2ty2t} consisting of 2t edges, where x1, . . . , x2t ∈V1, and y1, . . . , y2t ∈V2, and construct a graph (G −E(t)) + {x1x2, y1y2, . . . , x2t−1x2t, y2t−1y2t}.
It is not difficult to verify that such a graph is not claw-free and all its balls of radius 2 are 2-connected. We have a stronger result in the case k = 4n, n ≥1 (see Corollary 3.3 and Proposition 4.5).
3. Hamiltonicity of graphs in G(k), k ≤5 It is evident that all graphs in the set G(2) are Hamiltonian.
Theorem 3.1.
Every graph G in the set G(3) is Hamiltonian. Moreover, either G satisfies the conditions of Theorem 1.6, or G is isomorphic to H(3, n), for some n ≥2.
Proof. Let G be a graph in G(3). If every ball of radius two in G is 2-connected then G satisfies the conditions of Theorem 1.6 and is thus Hamiltonian.
Suppose that not every ball of radius two in G is 2-connected, and let v1 be a cut vertex of a ball of radius 2 in G. Then by Lemma 2.2, v1 is a cut vertex of the ball G2(v1), the subgraph G2(v1) −v1 has exactly two components with 3 vertices each, and each neighbor of v1 is adjacent to all other vertices of the component of G2(v1) −v1 it lies in.
Since G is 3-regular, one of the components of G2(v1) −v1 will contain two neighbors of v1, denoted w1 and w2, and one other vertex which we call u1, and the vertices u1, w1 and w2 will be adjacent to each other. In G, the vertex u1 will have one additional neighbor, which we call v2. Clearly d(v1, v2) > 2, since v1 is a cut vertex of G2(v1)−v1. But then v2 is a cut vertex of the ball G2(v2), and one of the components of G2(v2) −v2 contains only one neighbor of v2, namely u1. Let F1 denote the subgraph induced by the set {v1, u1, v2} ∪W1, where W1 = {w1, w2}.
Assume that we have already found in G t ≥1 isomorphic subgraphs F1, . . . , Ft and disjoint sets W0, W1, . . . , Wt such that • W0 = {u1, . . . , ut, v1, . . . , vt, vt+1} and |W1| = · · · = |Wt|, • V(Fi) = {vi, ui, vi+1} ∪Wi, for i = 1, . . . , t, • vi is a cut vertex of the ball G2(vi), for i = 1, . . . , t + 1.
If vt+1 ̸= v1, then using for vt+1 the same argument as for v1 we will find a new subgraph Ft+1 isomorphic to F1, where V(Ft+1) = {vt+1, ut+1, vt+2} ∪Wt+1 and the vertex vt+2 is adjacent to ut+1, (Wt+1 ∪{ut+1}) ∩(⋃t i=1 V(Fi)) = ∅and vt+2 is a cut vertex of the ball G2(vt+2). Put W0 := W0 ∪{ut+1, vt+2}.
Since G is finite, after finite number of steps we will find a vertex vn+1 such that n ≥2 and vn+1 = v1. Then ⋃n i=1 Fi = G and G is isomorphic to H(3, n) and thus Hamiltonian. The proof of the theorem is complete.
□ Now we will show that all graphs in the set G(4) are Hamiltonian. Moreover we will characterize all graphs in G(4) that do not satisfy the conditions of Theorem 1.6.
Let F1, . . . , Fn, where n ≥2, be a sequence of disjoint graphs where each Fi is isomorphic to one of the graphs in Fig. 1, and let u1 i and u2 i be the vertices in Fi corresponding to the vertices a and b in this isomorphism. Consider a graph obtained from F1, . . . , Fn by identifying the vertices u2 i and u1 i+1 and denoting the obtained new vertex by ui+1, i = 1, . . . , n, where u1 n+1 = un+1 = u1. The set of all graphs obtained in such a way we denote by F(4, n). Clearly, G(4, n) ∈F(4, n) and the diameter of any G ∈F(4, n) satisfies ⌊3n 2 ⌋≤diam(G) ≤⌊5n 2 ⌋, for each n ≥2. All graphs from the set F(4, 2) can be seen in Fig. 2.
Theorem 3.2.
Every graph G in the set G(4) is Hamiltonian. Moreover, either G satisfies the conditions of Theorem 1.6, or G is isomorphic to H(4, n) or one of graphs in F(4, n), for some n ≥2.
Proof. Let G be a graph in G(4). If every ball of radius two in G is 2-connected then G satisfies the conditions of Theorem 1.6 and is thus Hamiltonian.
78 A.S. Asratian and J.B. Granholm Discrete Applied Mathematics 316 (2022) 75–86 Fig. 1. The graphs T1, T2, T3, and T4.
Fig. 2. The set F(4, 2).
Suppose that not every ball of radius two in G is 2-connected, and let v1 be a cut vertex of a ball of radius 2 in G. Then by Lemma 2.2, v1 is a cut vertex of the ball G2(v1), the subgraph G2(v1) −v1 has exactly two components with 4 vertices each, and each neighbor of v1 is adjacent to all other vertices of the component of G2(v1) −v1 it lies in.
Since G is 4-regular, v1 will have either one neighbor in one of the components of G2(v1) −v1 and three in the other, or two neighbors in each component.
Case 1. v1 has one neighbor in one of the components of G2(v1) −v1 and three in the other.
Consider the component of G2(v1) −v1 that contains three neighbors of v1, denoted w1, w2, w3, and one other vertex which we will call u1. By Lemma 2.2, the vertices u1, w1, w2, w3 will be adjacent to each other. In G, the vertex u1 will have one additional neighbor, which we will call v2. Clearly d(v1, v2) > 2, since v1 is a cut vertex of G2(v1). But then v2 is a cut vertex of the ball G2(v2), and one of the components of G2(v2) −v2 contains only one neighbor of v2, namely u1.
Denote by F1 the subgraph of G induced by the set {v1, u1, v2} ∪W1 where W1 = {w1, w2, w3}.
By repeating the argument that we used in the proof of Theorem 3.1 we can show that the graph G is a union of n ≥2 subgraphs F1, . . . , Fn such that F2, . . . , Fn are isomorphic to F1, |V(Fi) ∩V(Fi+1)| = 1 for i = 1, . . . , n (with Fn+1 = F1) and |V(Fi) ∩V(Fj)| = ∅, if |i −j| ̸= 1 mod n.
It is easy to see that the resulting graph is isomorphic to H(4, n).
Case 2. v1 has two neighbors in each component of G2(v1) −v1.
Let w1 and w2 be the neighbors of v1 in one of the components of G2(v1)−v1, and let u1 and u2 be the other two vertices in that component. As noted above, w1 and w2 are adjacent to each other and to u1 and u2. Let S = {w1, w2, u1, u2}.
Case 2.1 u1u2 ∈E(G).
In this case u1 and u2 each have one neighbor outside S.
Case 2.1.1 u1 and u2 are adjacent to the same vertex v2 / ∈S.
In this case the graph F1 induced by the set S ∪{v1, v2} is isomorphic to T1. Clearly d(v1, v2) > 2, since v1 is a cut vertex of G2(v1).
Case 2.1.2 u1 and u2 are adjacent to different vertices outside S.
Let x1 be the neighbor of u1 outside S, and let x2 be the neighbor of u2 outside S. Clearly x1x2 ∈E(G), since otherwise there are five vertices at distance two from u1, namely x2, v1 and three neighbors of x1 outside S, contradicting the conditions of the theorem. Let y1 and y2 be the two neighbors of x1 outside S ∪{x2}. Then w1, w2, u2 and the vertices in the set N(y1) \ {x2, y2} are at distance 2 from x1. This means that the latter set can contain at most one vertex, so y1 is adjacent to x2 and y2. Similarly we see that y2 is also adjacent to x2. Now y1 and y2 have one neighbor each outside {x1, x2, y1, y2}. Let v2 and z be the neighbors of y1 and y2, respectively, outside this set. If v2 ̸= z then there are five vertices, namely w1, w2, u2, v2 and z, at distance 2 from x1, a contradiction. Thus v2 = z, which means that the subgraph F1 induced by S ∪{x1, x2, y1, y2, v1, v2} is isomorphic to T2. Note that v2 ̸= v1, since otherwise G = G2(v1), contradicting the fact that v1 is a cut vertex of G2(v1). This means that u1 and u2 are at distance 3 from v2, so v2 is a cut vertex of G2(v2).
79 A.S. Asratian and J.B. Granholm Discrete Applied Mathematics 316 (2022) 75–86 Case 2.2 u1u2 / ∈E(G).
In this case u1 and u2 each has two neighbors outside S, and these neighbors are at distance 2 from w1. Clearly N(u1) = N(u2), since otherwise there are more than four vertices at distance 2 from w1. Let x1 and x2 be the neighbors of u1 and u2 outside S.
Case 2.2.1 x1x2 ∈E(G).
In this case x1 and x2 both have one more neighbor. Let v2 be the neighbor of x1 outside S ∪{x2}. If x2v2 / ∈E(G), then there are five vertices at distance 2 from x1, namely w1, w2, and the three other neighbors of v2. Thus x2v2 ∈E(G), which means that the subgraph induced by S ∪{x1, x2, v1, v2} is isomorphic to T3. Note that v2 / ∈M1(v1), since otherwise G = G2(v1), contradicting the fact that v1 is a cut vertex of G2(v1). This means that w1 and w2 are at distance 3 from v2, so v2 is a cut vertex of G2(v2).
Case 2.2.2 x1x2 / ∈E(G).
In this case x1 and x2 each has two neighbors outside S, and these neighbors are at distance 2 from u1. Clearly N(x1) = N(x2), since otherwise there are more than four vertices at distance 2 from u1. Let y1 and y2 be the neighbors of x1 and x2 outside S. Then the vertices w1, w2, x2, and any neighbors of y1 and y2 that are not adjacent to x1 are at distance 2 from x1. Thus y1 and y2 are adjacent to each other and to one common additional vertex v2, which means that the subgraph induced by S ∪{x1, x2, y1, y2, v1, v2} is isomorphic to T4. Note that v2 ̸= v1, since otherwise G = G2(v1), contradicting the fact that v1 is a cut vertex of G2(v1). This means that u1 and u2 are at distance 3 from v2, so v2 is a cut vertex of G2(v2).
To conclude Case 2, we see that in all subcases we found a subgraph F1 that is isomorphic to one of the graphs T1, T2, T3, T4 and has only two vertices of degree 2, namely v1 and v2.
Assume that we have already found t ≥1 subgraphs F1, . . . , Ft each of which is isomorphic to one of the graphs T1, T2, T3, T4 such that • Fi has only two vertices of degree 2 in Fi, namely vi and vi+1, i = 1, . . . , t, • vi is a cut vertex of the ball G2(vi), for i = 1, . . . , t + 1, • if t ≥2 then V(Fi) ∩V(Fi+1) = {vi+1}, for i = 1, . . . , t −1.
If vt+1 ̸= v1, then using for vt+1 the same argument as for v1 we will find in G a subgraph Ft+1 isomorphic to one of the graphs T1, T2, T3, T4 such that V(Ft+1) ∩V(Ft) = {vt+1} and vt+1 is a cut vertex of the ball G2(vt+1).
Since G is finite, after finite number of steps we will find subgraphs F1, . . . , Fn and vertices v1, . . . , vn+1 such that vn+1 = v1 and ⋃n i=1 Fi = G. This means that G ∈F(4, n) and therefore G is Hamiltonian. The proof of the theorem is complete.
□ Note that Theorem 3.1 and Theorem 3.2 imply Proposition 1.1. This holds because every connected, locally connected, k-regular graph with k ∈{3, 4} is the square of a cycle Cn of length n, n ≥4, (see ), and the square of Cn belongs to G(3) if n = 4 and to G(4) if n > 4.
Corollary 3.3.
The set G(4) contains an infinite set of Hamiltonian graphs that are not claw-free or locally connected.
Proof. Consider the set of all graphs in { F(4, n) : n ≥2 } containing an induced subgraph T3 or T4. Clearly this set is infinite and no graph in this set is claw-free or locally connected.
□ Theorem 3.4.
Every graph G in the set G(5) is Hamiltonian. Moreover, either G satisfies the conditions of Theorem 1.6, or G is isomorphic to one of the graphs G(5, n) and H(5, n), for some n ≥2.
Proof. Let G be a graph in G(5). If every ball of radius two in G is 2-connected then G satisfies the conditions of Theorem 1.6 and is thus Hamiltonian.
Suppose that not every ball of radius two in G is 2-connected, and let v1 be a cut vertex of a ball of radius 2 in G. Then by Lemma 2.2, v1 is a cut vertex of the ball G2(v1), the subgraph G2(v1) −v1 has exactly two components with 5 vertices each, and each neighbor of v1 is adjacent to all other vertices of the component of G2(v1)−v1 it lies in. Since G is 5-regular, v1 will have either one neighbor in one of the components of G2(v1) −v1 and four in the other, or two neighbors in one component and three in the other.
Case 1. v1 has one neighbor in one of the components of G2(v1) −v1 and four in the other.
Consider the component of G2(v1) −v1 which contains four neighbors of v1. Let W1 = {w1, w2, w3, w4} denote the set of four neighbors of v1 in this component, and let u1 denote the fifth vertex in the same component. By Lemma 2.2, all of these five vertices are adjacent to each other. In G, the vertex u1 will have one additional neighbor, which we will call v2. Clearly d(v1, v2) > 2, since v1 is a cut vertex of G2(v1). But then v2 is a cut vertex of the ball G2(v2), and one of the components of G2(v2) −v2 contains only one neighbor of v2, namely u1. Denote by F1 the subgraph of G induced by the set {v1, u1, v2} ∪W1.
By repeating the argument that we used in the proof of Theorem 3.1 we can show that in G there are n ≥2 subgraphs F1, . . . , Fn and n vertices v1, . . . , vn such that ⋃n i=1 Fi = G, each Fi is isomorphic to F1, 2 ≤i ≤n, V(Fi) ∩V(Fi+1) = {vi+1}, for i = 1, . . . , n, and V(Fi) ∩V(Fj) = ∅, for all |i −j| ̸= 1 mod n, where we consider vn+1 = v1.
80 A.S. Asratian and J.B. Granholm Discrete Applied Mathematics 316 (2022) 75–86 It is easy to see that the resulting graph is isomorphic to H(5, n) and therefore is Hamiltonian.
Case 2. v1 has two neighbors in one of the components of G2(v1) −v1 and three in the other.
Consider the component of G2(v1)−v1 which has two neighbors of v1. Let w1 and w2 be the two neighbors of v1 in this component, and let the other three vertices in the component be z1, z2, and z3. As noted above, w1 and w2 are adjacent to each other and to z1, z2, and z3. At distance 2 from w1 are the three other neighbors of v1, and all neighbors of z1, z2, and z3 outside the set S = {w1, w2, z1, z2, z3}. Thus z1, z2, and z3 can together have at most two neighbors outside S. Since G is 5-regular, this means that each of the vertices z1, z2, and z3 must be adjacent to at least one of the other two, which means that one of them, say z1, must be adjacent to both z2 and z3.
Suppose that z2z3 / ∈E(G). Then z2 and z3 are both adjacent to two vertices x1 and x2 outside S, and z1 will be adjacent to one of them, say x1. Since G is 5-regular, x2 will have at least one neighbor y outside S ∪{x1} that is not adjacent to x1.
If x1x2 ∈E(G), then y will have at least three neighbors outside M1(x2), and these vertices together with w1, w2, and z1 are all at distance 2 from x2, contradicting the fact that G ∈G(5). If instead x1x2 / ∈E(G), then y will have at least two neighbors outside M1(x2), and these vertices together with w1, w2, z1 and x1 are all at distance 2 from x2, contradicting the fact that G ∈G(5).
Thus we can conclude that z2z3 ∈E(G). Since z1, z2, and z3 are all adjacent to each other, each of them has exactly one neighbor outside S, and as mentioned before they can together have at most two neighbors outside S.
Suppose that z1, z2, and z3 are not all adjacent to the same vertex outside S. Then w.l.o.g. we can assume that z1 is adjacent to a vertex x1 / ∈S and z2 and z3 are adjacent to another vertex x2 / ∈S. We can see that x1x2 ∈E(G), since otherwise z1 would have six vertices at distance 2: v1, x2, and the four remaining neighbors of x1. Thus x1 has three neighbors apart from z1 and x2, we will call them y1, y2, and y3. Since G is 5-regular, y1 must have a neighbor z4 outside {x1, x2, y1, y2, y3}. Then x1 has five vertices at distance 2, namely w1, w2, z2, z3, and z4. Thus all remaining neighbors of y1, y2, and y3 must be in {x2, y1, y2, y3, z4}, which means that y1, y2, and y3 are all adjacent to x2. But then d(x2) = 6, a contradiction.
Thus, we can conclude that z1, z2, and z3 are all adjacent to the same vertex outside S, denoted v2. Set V1 = {w1, w2, z1, z2, z3}. Clearly, the subgraph induced by V1 is a complete graph.
Assume that we have already found in G a sequence of disjoint sets V0, V1, . . . , Vt such that V0 = {v1, . . . , vt+1}, and for i = 1, . . . , t, the subgraph induced by Vi is complete, vi is adjacent to two vertices in Vi, and the other vertices in Vi are adjacent to vi+1.
If vt+1 ̸= v1 then using the same argument as we used for v1 we will find a set Vt+1 and a vertex vt+2 / ∈Vt+1 such that the subgraph induced by Vt+1 is complete, Vt+1 ∩(⋃t i=0 Vi ) = ∅and vt+2 is adjacent to those vertices in Vt+1 that are not neighbors of vt+1. Then put V0 := V0 ∪{vt+2}.
Since G is finite, after finite number of steps we will find n + 1 disjoint sets V0, V1, . . . , Vn and a vertex vn+1 such that V0 = {v1, . . . , vn}, vn+1 = v1 and V0 ∪V1 ∪· · · ∪Vn = V(G). Thus G is isomorphic to G(5, n) and therefore Hamiltonian.
The proof of the theorem is complete.
□ Remark 3.1. The graphs G(5, n) and H(5, n) are not locally connected for n ≥2 and therefore do not satisfy the conditions of Kikust’s theorem (Theorem 1.2). On the other hand there exist locally connected, 5-regular graphs which do not satisfy Theorem 3.4. Therefore Theorem 3.4 and Kikust’s theorem are incomparable to each other in the sense that neither theorem implies the other.
4. Hamiltonicity of graphs in G(k), k ≥6 The next result shows that the condition of 2-connectedness of balls of radius 2 in Theorem 1.6 cannot be omitted in the case k ≥6.
Proposition 4.1.
For any k ≥6, the set G(k) contains non-Hamiltonian graphs.
Proof. Pick three integers k1, k2, k3 ≥2 such that k1 + k2 + k3 = k. We define three disjoint graphs H1, H2, H3 as follows: For each i = 1, 2, 3, the vertex set of Hi is Ui ∪Vi ∪V ′ i ∪U′ i ∪{wi, w′ i} where • the sets Ui, Vi, V ′ i , U′ i , and {wi, w′ i} are pairwise disjoint, • |Vi| = |V ′ i | = ki and |Ui| = |U′ i | = k −ki, • each of the sets Ui, Vi, V ′ i and U′ i induces a complete subgraph, • every vertex in Ui is adjacent to wi and to all vertices in Vi, • every vertex in U′ i is adjacent to w′ i and to all vertices in V ′ i .
• every vertex in Vi is adjacent to exactly one vertex in V ′ i , and vice versa.
We also let ˜ H and ˜ H′ be two copies of Kk, disjoint from each other and from H1, H2, and H3.
Now we will use the parts defined above to construct a non-Hamiltonian graph in G(k). We will start with the graph ˜ H ∪H1 ∪H2 ∪H3 ∪˜ H′. To this graph we will add edges such that for each i = 1, 2, 3 the vertex wi is adjacent to ki vertices of ˜ H and each vertex of ˜ H is adjacent to exactly one of w1, w2, and w3, and similarly for ˜ H′ and w′ 1, w′ 2, and w′ 3. The 81 A.S. Asratian and J.B. Granholm Discrete Applied Mathematics 316 (2022) 75–86 Fig. 3. A non-Hamiltonian graph in G(6).
Fig. 4. The transformation from G to G′.
resulting graph lies in G(k) but is not Hamiltonian. The unique graph for k = 6 (with k1 = k2 = k3 = 2) can be seen in Fig. 3.
□ For k = 6 we furthermore have the following: Theorem 4.2.
The problem of determining whether there exists a Hamilton cycle in a graph from G(6) is NP-complete.
Proof. Akiyama, Nishizeki, and Saito proved that the problem of determining whether there exists a Hamilton cycle in a 2-connected, 3-regular, bipartite, planar graph is NP-complete. Thus, to prove the theorem we only need to provide a polynomial-time reduction of this problem to our problem.
Let G be any 2-connected, 3-regular, bipartite, planar graph. We will construct a graph G′ ∈G(6) that is Hamiltonian if and only if G is. Let (X, Y) be a bipartition of G. We set V(G′) = ⋃ x∈X Ux ∪ ⋃ y∈Y Vy ∪{ we : e ∈E(G) }, where each Ux is a set of four vertices and each Vy is a set of six vertices such that Ux ∩Vy = ∅= Ux ∩Uz = Vy ∩Vs for all x, z ∈X and y, s ∈Y. The edges of G′ are defined as follows: • each of the sets Ux and Vy induces a complete graph in G′, for x ∈X, y ∈Y, and • if e = xy is an edge in G with x ∈X and y ∈Y, then the vertex we in G′ is adjacent to all vertices in Ux and exactly two vertices in Vy with the additional requirement that if e′ and e′′ are two other edges in G incident to y, then every vertex in Vy is adjacent to exactly one of the vertices we, we′, and we′′, see Fig. 4.
It is easy to verify that G′ ∈G(6). It is also clear that G′ can be constructed from G in polynomial time.
Before we prove that G′ is Hamiltonian if and only if G is, we will introduce some practical notation. Let e = xy ∈E(G).
In any Hamilton cycle of G′, the vertex we is incident to exactly two edges of the cycle. If one of these edges joins we to a vertex of Ux and the other edge joins we to a vertex of Vy, we say that we is crossed; otherwise it is not crossed. If e1, e2, e3 are three edges incident to a vertex in G, then {we1, we2, we3} is a cut set of G′. This means that in any Hamilton cycle of G′, exactly two of we1, we2, we3 are crossed.
Suppose there is a Hamilton cycle C′ in G′. Then we will construct in G a subgraph C by including an edge e ∈E(G) in C if we is crossed by C′. By the previous discussion, C will be a 2-regular spanning subgraph of G. Also, for any two 82 A.S. Asratian and J.B. Granholm Discrete Applied Mathematics 316 (2022) 75–86 Fig. 5. The transformation from a Hamilton cycle in G to one in G′.
vertices z1 and z2 in G we can find a path with edges e1, . . . , ep such that we1, . . . , wep are crossed by C′, since C′ is a Hamilton cycle in G′. Thus C is connected, so it will form a Hamilton cycle of G.
Suppose now that G is Hamiltonian, and let − → C denote a Hamilton cycle in G with a given orientation. Consider an edge e = xy in − → C and assume that the edge preceding x in the cycle is d, that the edge succeeding y is f , and that the other two edges incident to x and y are a and b, respectively. It is not hard to find a path from wd to wf in G′ that covers all vertices of Ux ∪Vy ∪{wd, we, wf , wb}, see Fig. 5. By representing the Hamilton cycle − → C as − → C = x1y1x2y2 · · · xnynx1 and using this method for every pair xkyk, we can construct a Hamilton cycle of G′.
We can conclude that this construction gives us a polynomial time reduction of the Hamilton cycle problem for 2-connected cubic bipartite planar graphs to the Hamilton cycle problem for graphs in G(6), and thus that the latter problem is NP-complete.
□ By Proposition 1.1 and Theorem 1.2, all connected, locally connected, k-regular graphs are Hamiltonian if k ≤5. We now formulate a result describing a new class of Hamiltonian, locally connected, k-regular graphs with k ≥6.
Theorem 4.3.
Every locally connected graph G in G(k), k ≥6, is Hamiltonian. Moreover, for every non-Hamiltonian cycle C in G there exists a cycle of length |V(C)| + ℓin G, ℓ∈{1, 2}, containing the vertices of C.
Proof. Since G is locally connected, every vertex of G lies on a triangle. Suppose that for some n < |V(G)|, G has a cycle C of length n but G has no cycle of length n + 1 or n + 2 containing the vertices of C.
(1) Consider a vertex w1 on C with N(w1)\V(C) ̸= ∅. Such a vertex exists because G is connected. By Lemma 2.3, there exists a pair of vertices v, z such that v ∈N(w1) \ V(C), z ∈V(C) and w1z, vz ∈E(G).
Set W = N(v) ∩V(C) = {w1, . . . , wp}, p ≥2, where the vertices w1, . . . , wp occur on − → C in the order of their indices.
(We consider wp+1 = w1.) Claim 1.
|N(v) ∩N(w+ 1 )| ≥2.
Proof.
Suppose that N(v) ∩N(w+ 1 ) = {w1}. By (1), vw+ 1 / ∈E(G) and, by Lemma 2.1, |N(v) ∩N(w+ 1 )| ≥|M2(w1) \ (N(v) ∪N(w+ 1 ))| −1. Then N(v) ∩N(w+ 1 ) = {w1} implies M2(w1) \ (N(v) ∪N(w+ 1 )) = {v, w+ 1 }. This means that w+ 1 is adjacent to every vertex in M2(w1) \ (M1(v) ∪{w+ 1 }). By (1), the vertex v is not adjacent to z+. Then z+ belongs to the set M2(w1) \ (M1(v) ∪{w+ 1 }) and, therefore, w+ 1 z+ ∈E(G). But then G has the cycle w1vz← − C w+ 1 z+− → C w1 of length n + 1 containing V(C), a contradiction.
□ Claim 2. w+ i = w− i+1 for each i = 1, . . . , p, that is, n = 2p and v is adjacent to every second vertex of C.
Proof. Set W + = {w+ 1 , . . . , w+ p }. We will count the number of edges between W + and W which we denote by e(W +, W).
By the assumption (1), the set W + ∪{v} is independent and N(w+ i )∩N(v)∩(V(G)\V(C)) = ∅, for 1 ≤i ≤p. Moreover, for each i = 1, . . . , p, we have d(v, w+ i ) = 2 and wi ∈N(v)∩N(w+ i ), so by the hypothesis of this theorem and by Lemma 2.1, |N(v) ∩N(w+ i )| ≥|M2(wi) \ (N(v) ∪N(w+ i ))| −1.
(2) Obviously, N(wi) ∩W + ⊆M2(wi) \ (N(v) ∪N(w+ i ) ∪{v}).
(3) 83 A.S. Asratian and J.B. Granholm Discrete Applied Mathematics 316 (2022) 75–86 Thus, |N(wi) ∩W +| ≤|M2(wi) \ (N(v) ∪N(w+ i ))| −1. This and (2) imply that |N(wi) ∩W +| ≤|N(v) ∩N(w+ i )|. Hence, e(W +, W) = p ∑ i=1 |N(wi) ∩W +| ≤ p ∑ i=1 |N(v) ∩N(w+ i )| = e(W +, W).
(4) It follows, for each i = 1, . . . , p, that M2(wi) \ (N(v) ∪N(w+ i ) ∪{v}) = N(wi) ∩W + ⊆W +.
(5) Noting that p ≥2 and, by Claim 1, |N(w+ 1 ) ∩N(v)| ≥2, we now prove by contradiction that w+ i = w− i+1 for each i = 1, . . . , p.
Assume without loss of generality that w+ 1 ̸= w− 2 , whence w− 2 / ∈W +. Observe that w− 2 ∈N(w+ 2 ), because otherwise, by (5), w− 2 ∈W +. Clearly, the assumption (1) implies w− 2 w− 3 / ∈E(G). Hence w+ 2 ̸= w− 3 . Repetition of this argument shows that w+ i ̸= w− i+1 and w+ i w− i ∈E(G) for all i ∈{ 1, . . . , p }. By Claim 1, N(w+ 1 ) ∩N(v) contains a vertex x ̸= w1. Clearly, x ∈V(C), because otherwise G contains a cycle of length n+2 containing the vertices of C. Then x = wi for some i ≥2 and G contains the cycle w1vwiw+ 1 − → C w− i w+ i − → C w1 of length n + 1 containing V(C). This contradiction proves that w+ i = w− i+1 for each i = 1, . . . , p, and n = 2p.
□ We continue to prove the theorem. First we will show that the vertex w+ i has no neighbor outside the cycle C, for each i = 1, . . . , p. Suppose that the condition N(w+ t ) \ V(C) ̸= ∅holds for some t. Since the vertex w1 was picked arbitrarily in V(C), we can conclude, using similar arguments, that w+ t has a neighbor u ∈N(w+ t ) \ V(C) that is adjacent to every second vertex of C as well. More precisely, N(u)∩V(C) = W +. But then G has the cycle w1vw2w+ 1 uw+ 2 − → C w1 of length n+2 containing V(C). This contradiction implies that N(w+ i ) ⊆W and d(w+ i ) ≤p, for each i = 1, . . . , p. On the other hand, W ⊆N(v), that is, d(v) ≥p. These condition and k-regularity of G imply that N(v) = W, d(v) = p = k and N(w+ i ) = W, for each i = 1, . . . , p. Then w1 is adjacent to each w+ i , i = 1, . . . , p, as well as to v. But this implies that d(w1) > p, a contradiction.
Thus G has a cycle of length n + 1 or n + 2 containing V(C). This implies that G will also have a Hamilton cycle, which completes the proof of the theorem.
□ We continue with two results concerning the graphs in G(k).
Proposition 4.4.
For every k ≥30, the set G(k) contains an infinite class of locally connected graphs that are not claw-free.
Proof. Let k be an integer, k ≥30. Then k = 5p + t for some p ≥6 and 0 ≤t ≤4. We will show that for any n ≥2, the set G(k) contains a locally connected k-regular graph of diameter n which is not claw-free.
Consider first a graph D(n, p) which is defined as follows: its vertex set is V1 ∪· · ·∪V2n, where V1, . . . , V2n are pairwise disjoint sets of cardinality 2p and two vertices in V1 ∪· · · ∪V2n are adjacent if and only if they both belong to V1 ∪V2n or to Vi ∪Vi+1 for some i ∈{1, . . . , 2n −1}. Denote by Gi the subgraph of D(n, p) induced by the set Vi. Clearly, Gi is a complete graph on 2p vertices and therefore admits a proper edge coloring with 2p −1 colors, i = 1, . . . , 2n.
Now from the graph D(n, p) where the edges of the subgraphs G1, . . . , G2n are properly colored with 2p −1 colors, delete all edges having the first p −1 −t colors (note that the conditions p ≥6 and 0 ≤t ≤4 imply that p −t −1 ≥1).
It is not difficult to verify that the resulting graph is (5p + t)-regular, locally connected and is not claw-free. Finally, the resulting graph belongs to the set G(k) because every vertex of this graph has exactly 5p −t −1 vertices at distance 2 and this number does not exceed k = 5p + t.
□ Remark 4.1.
Using the same argument as above, one can additionally show that Proposition 4.4 holds for any k ∈{10, 15, 16, 20, 21, 22, 25, 26, 27, 28}.
Proposition 4.5.
For every integer k = 4n, n ≥2, the set G(k) contains an infinite class of graphs that satisfy the conditions of Theorem 1.6 but are not locally connected or claw-free.
Proof. Let k = 4n, for some n ≥2. In order to prove the proposition, we will show that for any integer p ≥2 there is a graph of diameter ⌊3p 2 ⌋which satisfies the conditions of Theorem 1.6 but is not locally connected or claw-free.
First consider a graph Q (4n, p) which is defined as follows: (i) its vertex set is V1 ∪· · · ∪V3p, where V1, . . . , V3p are pairwise disjoint sets of cardinality |V1| = |V2| = 2n, |V3| = 2 and |V3i+1| = 2n, |V3i+2| = 2n −1, |V3i+3| = 2, for i = 1, . . . , p −1, (ii) two vertices in V1 ∪· · · ∪V3p are adjacent if and only if they both belong to V1 ∪V3p or to Vi ∪Vi+1 for some i ∈{1, . . . , 3p −1}.
Let Mj be a perfect matching in the subgraph induced by the set Vj, j = 1, 2, 3. Then it is not difficult to verify that the graph Q (4n, p) −(M1 ∪M2 ∪M3) is a 4n-regular graph of diameter ⌊3p 2 ⌋which satisfies the conditions of Theorem 1.6 but is not locally connected or claw-free. The resulting graph for n = 2 and p = 2 can be seen in Fig. 6.
□ 84 A.S. Asratian and J.B. Granholm Discrete Applied Mathematics 316 (2022) 75–86 Fig. 6. A Hamiltonian graph from G(8) with 2-connected balls of radius 2 that is not locally connected and not claw-free.
Proposition 4.5 shows that Theorem 1.6 does not follow from any known result on Hamiltonicity of claw-free or locally connected graphs.
The following property was proved in : Proposition 4.6 (Gordon, Orlovich, Potts, and Strusevich ). Let G be a connected k-regular graph with k ≥9 where every edge belongs to at least k −4 triangles. Then G is fully cycle extendable, that is, every vertex of G belongs to a triangle and for every non-Hamiltonian cycle C in G there exists a cycle C′ of G such that V(C) ⊂V(C′) and |V(C′)| = |V(C)| + 1.
We will show now that all graphs satisfying the conditions of Proposition 4.6 belong to the set G(k).
Proposition 4.7.
Let G be a connected k-regular graph with k ≥9 where every edge belongs to at least k −4 triangles. Then G ∈G(k).
Proof. Let G be k-regular graph with k ≥9 where every edge belongs to at least k −4 triangles. Consider an arbitrary vertex v in G. We will show that |N2(v)| ≤k. Let e(N1(v), N2(v)) denote the number of edges between N1(v) and N2(v).
Clearly, every vertex u ∈N1(v) has at most 3 neighbors in N2(v), since the edge uv belongs to at least k −4 triangles.
This implies that e(N1(v), N2(v)) ≤3k. Furthermore, for every neighbor w of u in N2(v) the edge uw belongs to at most 2 triangles with the third vertex in N2(v). Therefore the vertex w must have at least k −6 neighbors in N1(v) different from u, since the edge uw belongs to at least k −4 triangles. This implies that e(N1(v), N2(v)) ≥(k −5)|N2(v)|. Then 3k ≥e(N1(v), N2(v)) ≥(k −5)|N2(v)| which implies that |N2(v)| ≤k if k ≥9. Thus G ∈G(k).
□ Taking into consideration Theorem 4.3, Proposition 4.6 and Proposition 4.7, we believe that the following conjecture is true: Conjecture 1. Let G be a locally connected graph in G(k), k ≥9. Then G is fully cycle extendable.
Acknowledgment The authors thank Carl Johan Casselgren for helpful suggestions on this manuscript.
References S.A. van Aardt, A.P. Burger, M. Frick, C. Thomassen, J.P. de Wet, Hamilton cycles in sparse locally connected graphs, Discrete Appl. Math. 257 (2019) 276–288.
S.A. van Aardt, M. Frick, O.R. Oellermann, J.P. de Wet, Global cycle properties in locally connected, locally traceable and locally hamiltonian graphs, Discrete Appl. Math. 205 (2016) 171–179.
T. Akiyama, T. Nishizeki, N. Saito, NP-Completeness of the Hamiltonian cycle problem for bipartite graphs, J. Info. Process. 3 (2) (1980) 73–76.
A.S. Asratian, Every 3-connected, locally connected, claw-free graph is Hamilton-connected, J. Graph Theory 23 (2) (1996) 191–201.
A.S. Asratian, New local conditions for a graph to be Hamiltonian, Graphs Combin. 22 (2) (2006) 153–160.
A.S. Asratian, H.J. Broersma, J. van den Heuvel, H.J. Veldman, On graphs satisfying a local Ore-type condition, J. Graph Theory 21 (1) (1996) 1–10.
85 A.S. Asratian and J.B. Granholm Discrete Applied Mathematics 316 (2022) 75–86 A.S. Asratian, J.B. Granholm, N.K. Khachatryan, A localization method in Hamiltonian graph theory, J. Combin. Theory Ser. B 148 (2021a) 209–238.
A.S. Asratian, J.B. Granholm, N.K. Khachatryan, Some local–global phenomena in locally finite graphs, Discrete Appl. Math. 293 (2021b) 166–176.
A.S. Asratian, N.K. Khachatryan, Investigation of the Hamiltonian property of a graph using neighborhoods of vertices (Russian), Akademiya Nauk Armenian SSR. Doklady 81 (3) (1985) 103–106.
A.S. Asratian, N.K. Khachatryan, Some localization theorems on Hamiltonian circuits, J. Combin. Theory Ser. B 49 (2) (1990) 287–294.
A.S. Asratian, N. Oksimets, Graphs with Hamiltonian balls, Australas. J. Combin. 17 (1998) 185–198.
H.J. Broersma, J. van den Heuvel, B. Jackson, H.J. Veldman, Hamiltonicity of regular 2-connected graphs, J. Graph Theory 22 (2) (1996) 105–124.
G. Chartrand, R.E. Pippert, Locally connected graphs, Asopis pro Pstování Matematiky 99 (2) (1974) 158–163.
L. Clark, Hamiltonian Properties of connected locally connected graphs, Congr. Numer. 32 (1981) 199–204.
R. Diestel, Graph Theory. 5th edition, Springer, 2016.
G.A. Dirac, Some theorems on abstract graphs, Proc. Lond. Math. Soc. s3-2 (1) (1952) 69–81.
R. Faudree, E. Flandrin, Z. Ryjáček, Claw-free graphs – a survey, Discrete Math. 164 (1–3) (1997) 87–147.
V.S. Gordon, Y.L. Orlovich, C.N. Potts, V.A. Strusevich, Hamiltonian properties of locally connected graphs with bounded vertex degree, Discrete Appl. Math. 159 (16) (2011) 1759–1774.
R.J. Gould, Advances on the Hamiltonian problem – A survey, Graphs Combin. 19 (1) (2003) 7–52.
R.J. Gould, Recent advances on the Hamiltonian problem: Survey III, Graphs Combin. 30 (1) (2014) 1–46.
G.R.T. Hendry, A strengthening of Kikust’s theorem, J. Graph Theory 13 (2) (1989) 257–260.
P.A. Irzhavski, Hamiltonicity of locally connected graphs: complexity results (Russian), Vestsi Natsyyanal’na Akadèmii Navuk Belarusi. Seryya Fizika-Matèmatychnykh Navuk 4 (2014) 37–43.
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T. Kaiser, P. Vrána, Hamilton cycles in 5-connected line graphs, European J. Combin. 33 (5) (2012) 924–947.
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H.-J. Lai, Y. Shao, M. Zhan, Hamiltonian N2-locally connected claw-free graphs, J. Graph Theory 48 (2) (2005) 142–146.
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M.M. Matthews, D.P. Sumner, Hamiltonian Results in K1, 3-free graphs, J. Graph Theory 8 (1) (1984) 139–146.
C.St.J.A. Nash-Williams, Hamiltonian arcs and circuits, in: M. Capobianco, J.B. Frechen, M. Krolik (Eds.), Recent Trends in Graph Theory, Volume 186 of Lecture Notes in Mathematics, Springer, Berlin, Heidelberg, 1971, pp. 197–210.
D.J. Oberly, D.P. Sumner, Every connected, locally connected nontrivial graph with no induced claw is Hamiltonian, J. Graph Theory 3 (4) (1979) 351–356.
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86 |
14469 | https://medlineplus.gov/vitaminddeficiency.html | Published Time: Tue, 16 Sep 2025 04:15:18 GMT
Vitamin D Deficiency: MedlinePlus
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Vitamin D Deficiency
Also called: Hypovitaminosis D, Low Vitamin D
Summary
What is vitamin D deficiency?
Vitamin D deficiency means that your body is not getting enough vitamin D to stay healthy.
Why do I need vitamin D and how do I get it?
Vitamin D helps your body absorb calcium. Calcium is one of the main building blocks of bone. Vitamin D also has a role in your nervous, muscle, and immune systems.
You can get vitamin D in three ways: through your skin, from your diet, and from supplements. Your body forms vitamin D naturally after exposure to sunlight. But too much sun exposure can lead to skin aging and skin cancer, so many people try to get their vitamin D from other sources.
How much vitamin D do I need?
The amount of vitamin D you need each day depends on your age. The recommended amounts, in international units (IU), are:
Birth to 12 months: 400 IU
Children 1-13 years: 600 IU
Teens 14-18 years: 600 IU
Adults 19-70 years: 600 IU
Adults 71 years and older: 800 IU
Pregnant and breastfeeding women: 600 IU
People at high risk of vitamin D deficiency may need more. Check with your health care provider about how much you need.
What causes vitamin D deficiency?
You can become deficient in vitamin D for different reasons:
You don't get enough vitamin D in your diet
You don't absorb enough vitamin D from food (a malabsorption problem)
You don't get enough exposure to sunlight
Your liver or kidneys cannot convert vitamin D to its active form in the body
You take medicines that interfere with your body's ability to convert or absorb vitamin D
Who is at risk of vitamin D deficiency?
Some people are at higher risk of vitamin D deficiency:
Breastfed infants, because human milk is a poor source of vitamin D. If you are breastfeeding, give your infant a supplement of 400 IU of vitamin D every day.
Older adults, because your skin doesn't make vitamin D when exposed to sunlight as efficiently as when you were young, and your kidneys are less able to convert vitamin D to its active form.
People with dark skin, which has less ability to produce vitamin D from the sun.
People with conditions that make it difficult to absorb nutrients from food, such as Crohn's disease, ulcerative colitis, and celiac disease.
People who have obesity, because their body fat binds to some vitamin D and prevents it from getting into the blood.
People who have had gastric bypass surgery, a type of weight loss surgery which creates a bypass of part of the small intestine. Since vitamin D is absorbed there, bypassing part of it makes it harder to absorb enough vitamin D.
People with chronic kidney or liver disease, which can affect your ability to change vitamin D into a form your body can use.
People who take medicines that affect vitamin D levels, including certain cholesterol, anti-seizure, steroid, and weight-loss medicines.
Talk with your provider if you are at risk for vitamin D deficiency. There is a blood test that can measure how much vitamin D is in your body.
What problems does vitamin D deficiency cause?
Vitamin D deficiency can lead to a loss of bone density, which can contribute to osteoporosis and fractures (broken bones).
Severe vitamin D deficiency can also lead to other diseases:
In children, it can cause rickets. Rickets is a rare disease that causes the bones to become soft and bend. African American infants and children are at higher risk of getting rickets.
In adults, severe vitamin D deficiency leads to osteomalacia. Osteomalacia causes weak bones, bone pain, and muscle weakness.
How can I get more vitamin D?
There are a few foods that naturally have some vitamin D:
Fatty fish such as salmon, tuna, and mackerel
Beef liver
Cheese
Mushrooms
Egg yolks
You can also get vitamin D from fortified foods. You can check the food labels to find out whether a food has vitamin D. Foods that often have added vitamin D include:
Milk
Breakfast cereals
Orange juice
Other dairy products, such as yogurt
Soy drinks
Vitamin D is in many multivitamins. There are also vitamin D supplements, both in pills and in a liquid for babies.
If you have vitamin D deficiency, the treatment is with supplements. Check with your provider about how much you need to take, how often you need to take it, and how long you need to take it.
Can too much vitamin D be harmful?
Getting too much vitamin D (known as vitamin D toxicity) can be harmful. Signs of toxicity include nausea and vomiting, poor appetite, constipation, weakness, and weight loss.
Very high levels of vitamin D can damage the kidneys. It also raises the level of calcium in your blood. High levels of blood calcium (hypercalcemia) can cause confusion, kidney failure, and irregular heartbeat (arrhythmia).
Most cases of vitamin D toxicity happen when someone overuses vitamin D supplements. You cannot get too much vitamin D from sun exposure because the skin limits the amount of vitamin D it makes.
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25-hydroxy vitamin D test (Medical Encyclopedia) Also in Spanish
Osteomalacia (Medical Encyclopedia) Also in Spanish
Rickets: MedlinePlus Health Topic (National Library of Medicine) Also in Spanish
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Vitamin D (Harvard School of Public Health)
Vitamin D Test (National Library of Medicine) Also in Spanish
Clinical Trials
ClinicalTrials.gov: Vitamin D Deficiency (National Institutes of Health)
Journal Articles References and abstracts from MEDLINE/PubMed (National Library of Medicine)
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14470 | https://www.geogebra.org/m/DV6Ehjnx | Outline
Visualizing Fractions
Author:EDC Interactive Math
Topic:Addition, Area, Diagrams, Division, Fractions, Mathematics, Multiplication, Numbers, Rational Numbers, Subtraction
A collection of resources that include visual representations of fractions.
Contact us at ptierneyfife@edc.org if this resource should be updated, or with questions or suggestions. We appreciate your help maintaining the quality of this resource.
Table of Contents
Proper Fractions
Multiple Models for Proper Fractions
Multiple Models for Proper Fractions – 2 Fractions
Visualize Equivalent Proper Fractions
Improper Fractions
Area Models for Improper Fractions
Area Models for Improper Fractions – 2 Fractions
Fractions Set Model
Fractions on a Number Line
Where is the Fraction? (0 to 1)
Where is the Fraction? (0 to 2)
Where is the Number?
Locating Fractions on a Number Line (1)
Locating Fractions on a Number Line (2)
Comparing Fractions
Comparing Rational Numbers – Fractions
Fraction Relationships with Visual Models
Adding Fractions
Fraction Strip Addition
Multiplying Fractions
Multiplying Unit Fractions - Area Model
Multiplying Fractions - Area Model
Multiplying Proper Fractions – Area Model Problems (1)
Multiplying Fractions - Area Model
Field of Fractions
Dividing Fractions
Fraction Division Problems with Bar Models
Next
Multiple Models for Proper Fractions
New Resources
Midpoint Coordinates
bewijs stelling van Pythagoras
apec
רישום חופשי
Fraction Subtraction
Discover Resources
Inscribe a Square
Van_Schooten1
lesson 12-2
Untitled
Cyclic 2
Conics: Exploring the Discriminant
Discover Topics
Definite Integral
Scalene Triangles
Hypergeometric Distribution
Intersection
Means |
14471 | https://sciencenotes.org/what-are-noble-gases-definition-and-properties/ | What Are Noble Gases? Definition and Properties
The noble gases are the elements in group 18 on the periodic table. Atoms of these elements have filled valence electron shells, making them relatively inert, colorless, odorless, monatomic gases at room temperature and pressure.
Why Are Noble Gases Called Noble?
The term “noble gas” comes from a translation of the German word Edelgas, which means noble gas. German chemist Hugo Erdmann coined the phrase in 1898. Like a nobleman might consider it undignified to associate with commoners, noble gases tend not to react with other elements.
Other names for noble gases include rare gases, inert gases, and aerogens. When referencing the periodic table, the noble gases are IUPAC group 18 (group 0 under the old method), CAS group VIIIA, the helium group, or the neon group.
List of Noble Gases
There are either six or seven noble gas elements, depending on whether or not you include element 118, oganesson.
The first six elements occur naturally. Radon and oganesson are radioactive elements. Oganesson is a man-made (synthetic) element that doesn’t entirely fit into the group. While it may have a filled valence shell (7p6), it is predicted to be a metallic solid at room temperature.
Noble Gas Properties
Elements in the noble gas group share common chemical and physical properties:
Common Misconceptions
The most common misconception about the noble gases is that they cannot form chemical bonds and compounds. While their atoms normally have filled valence shells, it’s possible to remove one or more electrons or (less commonly) add electrons. Under certain conditions, the noble gases can form diatomic gases, clathrates, fluorides, chlorides, metal complexes, and other compounds. Usually, compounds form under extremely high pressures. Examples of noble gas compounds include argon fluorohydride (HArF) and xenon hexafluoride (XeF6).
Another misconception is that the noble gases are rare. As with the rare earths, the rare gases aren’t particularly uncommon. Argon is the third or fourth most abundant gas in the atmosphere (depending on the amount of water vapor). It accounts for 1.3% of the atmospheric mass or 0.94% of its volume. Neon, krypton, helium, and xenon are trace elements in air. The gases may be more abundant deeper within the earth. Helium is found in natural gas, while xenon occurs in vapors from some mineral springs and may bind with iron and nickel in the Earth’s core.
Noble Gas Uses
The noble gases have several important uses. They are used as an inert atmosphere to protect specimens and minimize chemical reactions. Their low melting and boiling points make them useful as refrigerants. The noble gases are important in lighting applications, such as high-intensity lamps, neon lights, car headlamps, and excimer lasers. Helium is used in balloons, in breathing gas mixtures for deep-sea diving, and to cool superconducting magnets. The gases, especially xenon, are used in ion drives. At present, oganesson has no practical uses, but it might help scientists make even heavier elements someday.
Noble Gas Sources
Neon, argon, krypton, and xenon come from fractional distillation of liquefied air. The primary source of helium is cryogenic separation of natural gas. Radon comes from radioactive decay of radium, thorium, uranium, and other heavy radioactive elements. Oganesson is a man-made element synthesized by striking a target with accelerated particles. In the future, noble gases may be sourced from other planets. For example, helium and xenon are much more abundant on Jupiter and other gas planets than on Earth.
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14472 | https://ecampusontario.pressbooks.pub/math3080prep/chapter/2-2-finding-the-domain-of-a-function-defined-by-an-equation/ | 2.2 Finding the Domain of a Function Defined by an Equation – Math 3080 Preparation
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Contents
Acknowledgements
About this Book
Introduction
Chapter 1: Functions and Function Notation
1.1 Introduction to Functions and Function Notation
1.2 Determining Whether a Relation Represents a Function
1.3 Using Function Notation
1.4 Representing Functions Using Tables
1.5 Finding Input and Output Values of a Function
1.6 Determining Whether a Function is One-to-One
1.7 Identifying Basic Toolkit Functions
1.8 Review and Summary
1.9 Practice Questions
1.10 Chapter 1 Example Solutions
1.11 Practice Question Solutions
Chapter 2: Domain and Range
2.1 Introduction to Domain and Range
2.2 Finding the Domain of a Function Defined by an Equation
2.3 Using Notations to Specify Domain and Range
2.4 Finding Domain and Range from Graphs
2.5 Finding Domains and Ranges of the Toolkit Functions
2.6 Graphing Piecewise-Defined Functions
2.7 Review and Summary
2.8 Practice Questions
2.9 Chapter 2 Example Solutions
2.10 Practice Question Solutions
Chapter 3: Transformation of Functions
3.1 Transformation of Functions
3.2 Graphing Functions Using Vertical and Horizontal Shifts
3.3 Graphing Functions Using Reflections about the Axes
3.4 Determining Even and Odd Functions
3.5 Graphing Functions Using Stretches and Compressions
3.6 Review and Summary
3.7 Practice Questions
3.8 Chapter 3 Example Solutions
3.9 Practice Question Solutions
Chapter 4: Factoring Polynomials
4.1 Factoring Polynomials
4.2 Factoring the Greatest Common Factor of a Polynomial
4.3 Factoring a Trinomial with Leading Coefficient 1
4.4 Factoring by Grouping
4.5 Factoring a Perfect Square Trinomial
4.6 Factoring a Difference of Squares
4.7 Factoring the Sum and Difference of Cubes
4.8 Factoring Expressions with Fractional or Negative Exponents
4.9 Review and Summary
4.10 Chapter Exercises
4.11 Chapter 4 Example Solutions
4.12 Practice Question Solutions
Math 3069 Review
Foundations of Mathematics Review
Versioning History
Math 3080 Preparation
2.2 Finding the Domain of a Function Defined by an Equation
In 1.3 Using Function Notation, we were introduced to the concepts of domain and range. In this section, we will practice determining domains and ranges for specific functions. Keep in mind that, in determining domains and ranges, we need to consider what is physically possible or meaningful in real-world examples, such as tickets sales and year in the horror movie example above. We also need to consider what is mathematically permitted. For example, we cannot include any input value that leads us to take an even root of a negative number if the domain and range consist of real numbers. Or in a function expressed as a formula, we cannot include any input value in the domain that would lead us to divide by 0.
We can visualize the domain as a “holding area” that contains “raw materials” for a “function machine” and the range as another “holding area” for the machine’s products. See Figure 2-2.
Figure 2-2
We can write the domain and range in interval notation, which uses values within brackets to describe a set of numbers. In interval notation, we use a square bracket [ when the set includes the endpoint and a parenthesis ( to indicate that the endpoint is either not included or the interval is unbounded. For example, if a person has $100 to spend, he or she would need to express the interval that is more than 0 and less than or equal to 100 and write (0,100] We will discuss interval notation in greater detail later.
Let’s turn our attention to finding the domain of a function whose equation is provided. Oftentimes, finding the domain of such functions involves remembering three different forms. First, if the function has no denominator or an even root, consider whether the domain could be all real numbers. Second, if there is a denominator in the function’s equation, exclude values in the domain that force the denominator to be zero. Third, if there is an even root, consider excluding values that would make the radicand negative.
Before we begin, let us review the conventions of interval notation:
The smallest term from the interval is written first.
The largest term in the interval is written second, following a comma.
Parentheses, ( or ), are used to signify that an endpoint is not included, called exclusive.
Brackets, [ or ], are used to indicate that an endpoint is included, called inclusive.
See Figure 2-3 for a summary of interval notation.
Figure 2-3
Example 1: Finding the Domain of a Function as a Set of Ordered Pairs
Find the domain of the following function: {(2,10),(3,10),(4,20),(5,30),(6,40)}
Solution
Example 2: Find the Domain of a Function
Find the domain of the function: {(−5,4),(0,0),(5,−4),(10,−8),(15,−12)}
Solution
How To
Given a function written in equation form, find the domain.
Identify the input values.
Identify any restrictions on the input and exclude those values from the domain.
Write the domain in interval form, if possible.
Example 3: Finding the Domain of a Function
Find the domain of the function f(x)\=x2−1
Find the domain of the function: f(x)\=5−x+x3
Solution
How To
Given a function written in an equation form that includes a fraction, find the domain.
Identify the input values.
Identify any restrictions on the input. If there is a denominator in the function’s formula, set the denominator equal to zero and solve for x. If the function’s formula contains an even root, set the radicand greater than or equal to 0, and then solve.
Write the domain in interval form, making sure to exclude any restricted values from the domain.
Example 4: Finding the Domain of a Function Involving a Denominator
Find the domain of the function f(x)\=x+12−x
Find the domain of the function:f(x)\=1+4x2x−1
Solution
How To
Given a function written in equation form including an even root, find the domain.
Identify the input values.
Since there is an even root, exclude any real numbers that result in a negative number in the radicand. Set the radicand greater than or equal to zero and solve for x.
The solution(s) are the domain of the function. If possible, write the answer in interval form.
Example 5: Finding the Domain of a Function with an Even Root
Find the domain of the function f(x)\=7−x
Find the domain of the function f(x)\=5+2x
Solution
Question & Answer
Can there be functions in which the domain and range do not intersect at all?
Yes. For example, the function f(x)\=−1x has the set of all positive real numbers as its domain but the set of all negative real numbers as its range. As a more extreme example, a function’s inputs and outputs can be completely different categories (for example, names of weekdays as inputs and numbers as outputs, as on an attendance chart), in such cases the domain and range have no elements in common.
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14473 | https://fa.ewi.tudelft.nl/~hart/onderwijs/set_theory/Jech/05-AC_and_cardinal_arithmetic.pdf | The Axiom of Choice and Cardinal Arithmetic The Axiom of Choice Axiom of Choice (AC). Every family of nonempty sets has a choice func-tion.
If S is a family of sets and ∅/ ∈S, then a choice function for S is a func-tion f on S such that (5.1) f(X) ∈X for every X ∈S.
The Axiom of Choice postulates that for every S such that ∅/ ∈S there exists a function f on S that satisfies (5.1).
The Axiom of Choice differs from other axioms of ZF by postulating the existence of a set (i.e., a choice function) without defining it (unlike, for instance, the Axiom of Pairing or the Axiom of Power Set). Thus it is often interesting to know whether a mathematical statement can be proved without using the Axiom of Choice. It turns out that the Axiom of Choice is independent of the other axioms of set theory and that many mathematical theorems are unprovable in ZF without AC.
In some trivial cases, the existence of a choice function can be proved outright in ZF: (i) when every X ∈S is a singleton X = {x}; (ii) when S is finite; the existence of a choice function for S is proved by induction on the size of S; (iii) when every X ∈S is a finite set of real numbers; let f(X) = the least element of X.
On the other hand, one cannot prove existence of a choice function (in ZF) just from the assumption that the sets in S are finite; even when every X ∈S has just two elements (e.g., sets of reals), we cannot necessarily prove that S has a choice function.
Using the Axiom of Choice, one proves that every set can be well-ordered, and therefore every infinite set has cardinality equal to some ℵα. In particular, 48 Part I. Basic Set Theory any two sets have comparable cardinals, and the ordering |X| ≤|Y | is a well-ordering of the class of all cardinals.
Theorem 5.1 (Zermelo’s Well-Ordering Theorem). Every set can be well-ordered.
Proof. Let A be a set. To well-order A, it suffices to construct a transfinite one-to-one sequence ⟨aα : α < θ⟩that enumerates A. That we can do by induction, using a choice function f for the family S of all nonempty subsets of A. We let for every α aα = f(A −{aξ : ξ < α}) if A −{aξ : ξ < α} is nonempty. Let θ be the least ordinal such that A = {aξ : ξ < θ}. Clearly, ⟨aα : α < θ⟩enumerates A.
⊓ ⊔ In fact, Zermelo’s Theorem 5.1 is equivalent to the Axiom of Choice: If every set can be well-ordered, then every family S of nonempty sets has a choice function. To see this, well-order S and let f(X) be the least element of X for every X ∈S.
Of particular importance is the fact that the set of all real numbers can be well-ordered. It follows that 2ℵ0 is an aleph and so 2ℵ0 ≥ℵ1.
The existence of a well-ordering of R yields some interesting counterex-amples. Well known is Vitali’s construction of a nonmeasurable set (Exer-cise 10.1); another example is an uncountable set of reals without a perfect subset (Exercise 5.1).
If every set can be well-ordered, then every infinite set has a countable subset: Well-order the set and take the first ω elements. Thus every infinite set is Dedekind-infinite, and so finiteness and Dedekind finiteness coincide.
Dealing with cardinalities of sets is much easier when we have the Axiom of Choice. In the first place, any two sets have comparable cardinals. Another consequence is: (5.2) if f maps A onto B then |B| ≤|A|.
To show (5.2), we have to find a one-to-one function from B to A. This is done by choosing one element from f−1({b}) for each b ∈B.
Another consequence of the Axiom of Choice is: (5.3) The union of a countable family of countable sets is countable.
(By the way, this often used fact cannot be proved in ZF alone.) To prove (5.3) let An be a countable set for each n ∈N. For each n, let us choose an 5. The Axiom of Choice and Cardinal Arithmetic 49 enumeration ⟨an,k : k ∈N⟩of An. That gives us a projection of N × N onto ∞ n=0 An: (n, k) →an,k.
Thus ∞ n=0 An is countable.
In a similar fashion, one can prove a more general statement.
Lemma 5.2. | S| ≤|S| · sup{|X| : X ∈S}.
Proof. Let κ = |S| and λ = sup{|X| : X ∈S}. We have S = {Xα : α < κ} and for each α < κ, we choose an enumeration Xα = {aα,β : β < λα}, where λα ≤λ. Again we have a projection (α, β) →aα,β of κ × λ onto S, and so | S| ≤κ · λ.
⊓ ⊔ In particular, the union of ℵα sets, each of cardinality ℵα, has cardinal-ity ℵα.
Corollary 5.3. Every ℵα+1 is a regular cardinal.
Proof. This is because otherwise ωα+1 would be the union of at most ℵα sets of cardinality at most ℵα.
⊓ ⊔ Using the Axiom of Choice in Mathematics In algebra and point set topology, one often uses the following version of the Axiom of Choice. We recall that if (P, <) is a partially ordered set, then a ∈P is called maximal in P if there is no x ∈P such that a < x. If X is a nonempty subset of P, then c ∈P is an upper bound of X if x ≤c for every x ∈X.
We say that a nonempty C ⊂P is a chain in P if C is linearly ordered by <.
Theorem 5.4 (Zorn’s Lemma). If (P, <) is a nonempty partially ordered set such that every chain in P has an upper bound, then P has a maximal element.
Proof. We construct (using a choice function for nonempty subsets of P), a chain in P that leads to a maximal element of P. We let, by induction, aα = an element of P such that aα > aξ for every ξ < α if there is one.
Clearly, if α > 0 is a limit ordinal, then Cα = {aξ : ξ < α} is a chain in P and aα exists by the assumption. Eventually, there is θ such that there is no aθ+1 ∈P, aθ+1 > aθ. Thus aθ is a maximal element of P.
⊓ ⊔ 50 Part I. Basic Set Theory Like Zermelo’s Theorem 5.1, Zorn’s Lemma 5.4 is equivalent to the Axiom of Choice (in ZF); see Exercise 5.5.
There are numerous examples of proofs using Zorn’s Lemma. To mention only of few: Every vector space has a basis.
Every field has a unique algebraic closure.
The Hahn-Banach Extension Theorem.
Tikhonov’s Product Theorem for compact spaces.
The Countable Axiom of Choice Many important consequences of the Axiom of Choice, particularly many concerning the real numbers, can be proved from a weaker version of the Axiom of Choice.
The Countable Axiom of Choice. Every countable family of nonempty sets has a choice function.
For instance, the countable AC implies that the union of countably many countable sets is countable. In particular, the real line is not a countable union of countable sets. Similarly, it follows that ℵ1 is a regular cardinal. On the other hand, the countable AC does not imply that the set of all reals can be well-ordered.
Several basic theorems about Borel sets and Lebesgue measure use the countable AC; for instance, one needs it to show that the union of count-ably many Fσ sets is Fσ. In modern descriptive set theory one often works without the Axiom of Choice and uses the countable AC instead. In some instances, descriptive set theorists use a somewhat stronger principle (that follows from AC): The Principle of Dependent Choices (DC). If E is a binary relation on a nonempty set A, and if for every a ∈A there exists b ∈A such that b E a, then there is a sequence a0, a1, . . . , an, . . . in A such that (5.4) an+1 E an for all n ∈N.
The Principle of Dependent Choices is stronger than the Countable Axiom of Choice; see Exercise 5.7.
As an application of DC we have the following characterization of well-founded relations and well-orderings: Lemma 5.5.
(i) A linear ordering < of a set P is a well-ordering of P if and only if there is no infinite descending sequence a0 > a1 > . . . > an > . . .
in A.
The Axiom of Choice and Cardinal Arithmetic 51 (ii) A relation E on P is well-founded if and only if there is no infinite sequence ⟨an : n ∈N⟩in P such that (5.5) an+1 E an for all n ∈N.
Proof. Note that (i) is a special case of (ii) since a well-ordering is a well-founded linear ordering.
If a0, a1, . . . , an, . . . is a sequence that satisfies (5.5), then the set {an : n ∈N} has no E-minimal element and hence E is not well-founded.
Conversely, if E is not well-founded, then there is a nonempty set A ⊂P with no E-minimal element. Using the Principle of Dependent Choices we construct a sequence a0, a1, . . . , an, . . . that satisfies (5.5).
⊓ ⊔ Cardinal Arithmetic In the presence of the Axiom of Choice, every set can be well-ordered and so every infinite set has the cardinality of some ℵα. Thus addition and multipli-cation of infinite cardinal numbers is simple: If κ and λ are infinite cardinals then κ + λ = κ · λ = max{κ, λ}.
The exponentiation of cardinals is more interesting. The rest of Chapter 5 is devoted to the operations 2κ and κλ, for infinite cardinals κ and λ.
Lemma 5.6. If 2 ≤κ ≤λ and λ is infinite, then κλ = 2λ.
Proof.
2λ ≤κλ ≤(2κ)λ = 2κ·λ = 2λ.
⊓ ⊔ (5.6) If κ and λ are infinite cardinals and λ < κ then the evaluation of κλ is more complicated. First, if 2λ ≥κ then we have κλ = 2λ (because κλ ≤ (2λ)λ = 2λ), but if 2λ < κ then (because κλ ≤κκ = 2κ) we can only conclude (5.7) κ ≤κλ ≤2κ.
Not much more can be claimed at this point, except that by Theorem 3.11 in Chapter 3 (κcf κ > κ) we have (5.8) κ < κλ if λ ≥cf κ.
If λ is a cardinal and |A| ≥λ, let (5.9) [A]λ = {X ⊂A : |X| = λ}.
Lemma 5.7. If |A| = κ ≥λ, then the set [A]λ has cardinality κλ.
52 Part I. Basic Set Theory Proof. On the one hand, every f : λ →A is a subset of λ × A, and |f| = λ.
Thus κλ ≤|[λ × A]|λ = |[A]λ|. On the other hand, we construct a one-to-one function F : [A]λ →Aλ as follows: If X ⊂A and |X| = λ, let F(X) be some function f on λ whose range is X. Clearly, F is one-to-one.
⊓ ⊔ If λ is a limit cardinal, let (5.10) κ<λ = sup{κµ : µ is a cardinal and µ < λ}.
For the sake of completeness, we also define κ<λ+ = κλ for infinite successor cardinals λ+.
If κ is an infinite cardinal and |A| ≥κ, let (5.11) [A]<κ = Pκ(A) = {X ⊂A : |X| < κ}.
It follows from Lemma 5.7 and Lemma 5.8 below that the cardinality of Pκ(A) is |A|<κ.
Infinite Sums and Products Let {κi : i ∈I} be an indexed set of cardinal numbers. We define (5.12) i∈I κi = i∈I Xi , where {Xi : i ∈I} is a disjoint family of sets such that |Xi| = κi for each i ∈I.
This definition does not depend on the choice of {Xi}i; this follows from the Axiom of Choice (see Exercise 5.9).
Note that if κ and λ are cardinals and κi = κ for each i < λ, then i<λ κi = λ · κ.
In general, we have the following Lemma 5.8. If λ is an infinite cardinal and κi > 0 for each i < λ, then (5.13) i<λ κi = λ · supi<λ κi.
Proof. Let κ = supi<λ κi and σ = i<λ κi. On the one hand, since κi ≤κ for all i, we have i<λ κ ≤λ·κ. On the other hand, since κi ≥1 for all i, we have λ = i<λ 1 ≤σ, and since σ ≥κi for all i, we have σ ≥supi<λ κi = κ.
Therefore σ ≥λ · κ.
⊓ ⊔ 5. The Axiom of Choice and Cardinal Arithmetic 53 In particular, if λ ≤supi<λ κi, we have i<λ κi = supi<λ κi.
Thus we can characterize singular cardinals as follows: An infinite cardinal κ is singular just in case κ = i<λ κi where λ < κ and for each i, κi < κ.
An infinite product of cardinals is defined using infinite products of sets.
If {Xi : i ∈I} is a family of sets, then the product is defined as follows: (5.14) i∈I Xi = {f : f is a function on I and f(i) ∈Xi for each i ∈I}.
Note that if some Xi is empty, then the product is empty. If all the Xi are nonempty, then AC implies that the product is nonempty.
If {κi : i ∈I} is a family of cardinal numbers, we define (5.15) i∈I κi = i∈I Xi , where {Xi : i ∈I} is a family of sets such that |Xi| = κi for each i ∈I.
(We abuse the notation by using both for the product of sets and for the product of cardinals.) Again, it follows from AC that the definition does not depend on the choice of the sets Xi (Exercise 5.10).
If κi = κ for each i ∈I, and |I| = λ, then i∈I κi = κλ. Also, infinite sums and products satisfy some of the rules satisfied by finite sums and products.
For instance, i κλ i = ( i κi)λ, or i κλi = κ P i λi. Or if I is a disjoint union I = j∈J Aj, then (5.16) i∈I κi = j∈J i∈Aj κi .
If κi ≥2 for each i ∈I, then (5.17) i∈I κi ≤ i∈I κi.
(The assumption κi ≥2 is necessary: 1+1 > 1·1.) If I is finite, then (5.17) is certainly true; thus assume that I is infinite. Since i∈I κi ≥ i∈I 2 = 2|I| > |I|, it suffices to show that i κi ≤|I| · i κi. If {Xi : i ∈I} is a disjoint family, we assign to each x ∈ i Xi a pair (i, f) such that x ∈Xi, f ∈ i Xi and f(i) = x. Thus we have (5.17).
Infinite product of cardinals can be evaluated using the following lemma: 54 Part I. Basic Set Theory Lemma 5.9. If λ is an infinite cardinal and ⟨κi : i < λ⟩is a nondecreasing sequence of nonzero cardinals, then i<λ κi = (supi κi)λ.
Proof. Let κ = supi κi. Since κi ≤κ for each i < λ, we have i<λ κi ≤ i<λ κ = κλ.
To prove that κλ ≤ i<λ κi, we consider a partition of λ into λ disjoint sets Aj, each of cardinality λ: (5.18) λ = j<λ Aj.
(To get a partition (5.18), we can, e.g., use the canonical pairing function Γ : λ × λ →λ and let Aj = Γ(λ × {j}).) Since a product of nonzero cardinals is greater than or equal to each factor, we have i∈Aj κi ≥supi∈Aj κi = κ, for each j < λ. Thus, by (5.16), i<λ κi = j<λ i∈Aj κi ≥ j<λ κ = κλ.
⊓ ⊔ The strict inequalities in cardinal arithmetic that we proved in Chapter 3 can be obtained as special cases of the following general theorem.
Theorem 5.10 (K¨ onig). If κi < λi for every i ∈I, then i∈I κi < i∈I λi.
Proof. We shall show that i κi ≱ i λi. Let Ti, i ∈I, be such that |Ti| = λi for each i ∈I. It suffices to show that if Zi, i ∈I, are subsets of T = i∈I Ti, and |Zi| ≤κi for each i ∈I, then i∈I Zi ̸= T .
For every i ∈I, let Si be the projection of Zi into the ith coordinate: Si = {f(i) : f ∈Zi}.
Since |Zi| < |Ti|, we have Si ⊂Ti and Si ̸= Ti. Now let f ∈T be a function such that f(i) / ∈Si for every i ∈I. Obviously, f does not belong to any Zi, i ∈I, and so i∈I Zi ̸= T .
⊓ ⊔ Corollary 5.11. κ < 2κ for every κ.
Proof. 1 + 1 + . . .
κ times < 2 · 2 · . . .
κ times .
⊓ ⊔ Corollary 5.12. cf(2ℵα) > ℵα.
The Axiom of Choice and Cardinal Arithmetic 55 Proof. It suffices to show that if κi < 2ℵα for i < ωα, then i<ωα κi < 2ℵα.
Let λi = 2ℵα.
i<ωα κi < i<ωα λi = (2ℵα)ℵα = 2ℵα.
⊓ ⊔ Corollary 5.13. cf(ℵℵβ α ) > ℵβ.
Proof. We show that if κi < ℵℵβ α for i < ωβ, then i<ωβ κi < ℵℵβ α . Let λi = ℵℵβ α .
i<ωβ κi < i<ωβ λi = (ℵℵβ α )ℵβ = ℵℵβ α .
⊓ ⊔ Corollary 5.14. κcf κ > κ for every infinite cardinal κ.
Proof. Let κi < κ, i < cf κ, be such that κ = i<cf κ κi. Then κ = i<cf κ κi < i ℵα, and therefore 2ℵα ≥ℵα+1, for all α. The Generalized Continuum Hypothesis (GCH) is the statement 2ℵα = ℵα+1 for all α. GCH is independent of the axioms of ZFC. Under the assumption of GCH, cardinal exponentiation is evaluated as follows: Theorem 5.15. If GCH holds and κ and λ are infinite cardinals then: (i) If κ ≤λ, then κλ = λ+.
(ii) If cf κ ≤λ < κ, then κλ = κ+.
(iii) If λ < cf κ, then κλ = κ.
Proof. (i) Lemma 5.6.
(ii) This follows from (5.7) and (5.8).
(iii) By Lemma 3.9(ii), the set κλ is the union of the sets αλ, α < κ, and |αλ| ≤2|α|·λ = (|α| · λ)+ ≤κ.
⊓ ⊔ The beth function is defined by induction: ℶ0 = ℵ0, ℶα+1 = 2ℶα, ℶα = sup{ℶβ : β < α} if α is a limit ordinal.
Thus GCH is equivalent to the statement ℶα = ℵα for all α.
We shall now investigate the general behavior of the continuum func-tion 2κ, without assuming GCH.
56 Part I. Basic Set Theory Theorem 5.16.
(i) If κ < λ then 2κ ≤2λ.
(ii) cf 2κ > κ.
(iii) If κ is a limit cardinal then 2κ = (2<κ)cf κ.
Proof. (ii) By Corollary 5.12, (iii) Let κ = i<cf κ κi, where κi < κ for each i. We have 2κ = 2 P i κi = i 2κi ≤ i 2<κ = (2<κ)cf κ ≤(2κ)cf κ ≤2κ.
⊓ ⊔ For regular cardinals, the only conditions Theorem 5.16 places on the continuum function are 2κ > κ and 2κ ≤2λ if κ < λ. We shall see that these are the only restrictions on 2κ for regular κ that are provable in ZFC.
Corollary 5.17. If κ is a singular cardinal and if the continuum function is eventually constant below κ, with value λ, then 2κ = λ.
Proof. If κ is a singular cardinal that satisfies the assumption of the theorem, then there is µ such that cf κ ≤µ < κ and that 2<κ = λ = 2µ. Thus 2κ = (2<κ)cf κ = (2µ)cf κ = 2µ.
⊓ ⊔ The gimel function is the function (5.19)(גκ) = κcf κ.
If κ is a limit cardinal and if the continuum function below κ is not eventually constant, then the cardinal λ = 2<κ is a limit of a nondecreasing sequence λ = 2<κ = limα→κ 2|α| of length κ. By Lemma 3.7(ii), we have cf λ = cf κ.
Using Theorem 5.16(iii), we get (5.20) 2κ = (2<κ)cf κ = λcf λ.
If κ is a regular cardinal, then κ = cf κ; and since 2κ = κκ, we have (5.21) 2κ = κcf κ.
Thus (5.20) and (5.21) show that the continuum function can be defined in terms of the gimel function: Corollary 5.18.
(i) If κ is a successor cardinal, then 2κ =(גκ).
(ii) If κ is a limit cardinal and if the continuum function below κ is even-tually constant, then 2κ = 2<κ ·(גκ).
(iii) If κ is a limit cardinal and if the continuum function below κ is not eventually constant, then 2κ =(ג2<κ).
⊓ ⊔ 5. The Axiom of Choice and Cardinal Arithmetic 57 Cardinal Exponentiation We shall now investigate the function κλ for infinite cardinal numbers κ and λ.
We start with the following observation: If κ is a regular cardinal and λ < κ, then every function f : λ →κ is bounded (i.e., sup{f(ξ) : ξ < λ} < κ).
Thus κλ = α<κ αλ.
and so κλ = α<κ |α|λ.
In particular, if κ is a successor cardinal, we obtain the Hausdorffformula (5.22) ℵℵβ α+1 = ℵℵβ α · ℵα+1.
(Note that (5.22) holds for all α and β.) In general, we can compute κλ using the following lemma. If κ is a limit cardinal, we use the notation limα→κ αλ to abbreviate sup{µλ : µ is a cardinal and µ < κ}.
Lemma 5.19. If κ is a limit cardinal, and λ ≥cf κ, then κλ = (limα→κ αλ)cf κ.
Proof. Let κ = i<cf κ κi, where κi < κ for each i. We have κλ ≤ ( i<cf κ κi)λ = i κλ i ≤ i(limα→κ αλ) = (limα→κ αλ)cf κ ≤(κλ)cf κ = κλ.
⊓ ⊔ Theorem 5.20. Let λ be an infinite cardinal. Then for all infinite cardi-nals κ, the value of κλ is computed as follows, by induction on κ: (i) If κ ≤λ then κλ = 2λ.
(ii) If there exists some µ < κ such that µλ ≥κ, then κλ = µλ.
(iii) If κ > λ and if µλ < κ for all µ < κ, then: (a) if cf κ > λ then κλ = κ, (b) if cf κ ≤λ then κλ = κcf κ.
Proof. (i) Lemma 5.6 (ii) µλ ≤κλ ≤(µλ)λ = µλ.
(iii) If κ is a successor cardinal, we use the Hausdorffformula. If κ is a limit cardinal, we have limα→κ αλ = κ. If cf κ > λ then every f : λ →κ is bounded and we have κλ = limα→κ αλ = κ. If cf κ ≤λ then by Lemma 5.19, κλ = (limα→κ αλ)cf κ = κcf κ.
⊓ ⊔ Theorem 5.20 shows that all cardinal exponentiation can be defined in terms of the gimel function: 58 Part I. Basic Set Theory Corollary 5.21. For every κ and λ, the value of κλ is either 2λ, or κ, or(גµ) for some µ such that cf µ ≤λ < µ.
Proof. If κλ > 2λ · κ, let µ be the least cardinal such that µλ = κλ, and by Theorem 5.20 (for µ and λ), µλ = µcf µ.
⊓ ⊔ In the Exercises, we list some properties of the gimel function.
A cardinal κ is a strong limit cardinal if 2λ < κ for every λ < κ.
Obviously, every strong limit cardinal is a limit cardinal. If the GCH holds, then every limit cardinal is a strong limit.
It is easy to see that if κ is a strong limit cardinal, then λν < κ for all λ, ν < κ.
An example of a strong limit cardinal is ℵ0. Actually, the strong limit cardi-nals form a proper class: If α is an arbitrary cardinal, then the cardinal κ = sup{α, 2α, 22α, . . . } (of cofinality ω) is a strong limit cardinal.
Another fact worth mentioning is: (5.23) If κ is a strong limit cardinal, then 2κ = κcf κ.
We recall that κ is weakly inaccessible if it is uncountable, regular, and limit. We say that a cardinal κ is inaccessible (strongly) if κ > ℵ0, κ is regular, and κ is strong limit.
Every inaccessible cardinal is weakly inaccessible. If the GCH holds, then every weakly inaccessible cardinal κ is inaccessible.
The inaccessible cardinals owe their name to the fact that they cannot be obtained from smaller cardinals by the usual set-theoretical operations.
If κ is inaccessible and |X| < κ, then |P(X)| < κ. If |S| < κ and if |X| < κ for every X ∈S, then | S| < κ.
In fact, ℵ0 has this property too. Thus we can say that in a sense an inaccessible cardinal is to smaller cardinals what ℵ0 is to finite cardinals.
This is one of the main themes of the theory of large cardinals.
The Singular Cardinal Hypothesis The Singular Cardinal Hypothesis (SCH) is the statement: For every singular cardinal κ, if 2cf κ < κ, then κcf κ = κ+.
Obviously, the Singular Cardinals Hypothesis follows from GCH. If 2cf κ ≥ κ then κcf κ = 2cf κ. If 2cf κ < κ, then κ+ is the least possible value of κcf κ.
The Axiom of Choice and Cardinal Arithmetic 59 We shall prove later in the book that if SCH fails then a large cardinal axiom holds. In fact, the failure of SCH is equiconsistent with the existence of a certain large cardinal.
Under the assumption of SCH, cardinal exponentiation is determined by the continuum function on regular cardinals: Theorem 5.22. Assume that SCH holds.
(i) If κ is a singular cardinal then (a) 2κ = 2<κ if the continuum function is eventually constant below κ, (b) 2κ = (2<κ)+ otherwise.
(ii) If κ and λ are infinite cardinals, then: (a) If κ ≤2λ then κλ = 2λ.
(b) If 2λ < κ and λ < cf κ then κλ = κ.
(c) If 2λ < κ and cf κ ≤λ then κλ = κ+.
Proof. (i) If κ is a singular cardinal, then by Theorem 5.16, 2κ is either λ or λcf κ where λ = 2<κ. The latter occurs if 2α is not eventually constant below κ. Then cf λ = cf κ, and since 2cf κ < 2<κ = λ, we have λcf λ = λ+ by the Singular Cardinals Hypothesis.
(ii) We proceed by induction on κ, for a fixed λ. Let κ > 2λ. If κ is a successor cardinal, κ = ν+, then νλ ≤κ (by the induction hypothesis), and κλ = (ν+)λ = ν+ · νλ = κ, by the Hausdorffformula.
If κ is a limit cardinal, then νλ < κ for all ν < κ. By Theorem 5.20, κλ = κ if λ < cf κ, and κλ = κcf κ if λ ≥cf κ, In the latter case, 2cf κ ≤2λ < κ, and by the Singular Cardinals Hypothesis, κcf κ = κ+.
⊓ ⊔ Exercises 5.1. There exists a set of reals of cardinality 2ℵ0 without a perfect subset.
[Let ⟨Pα : α < 2ℵ0⟩be an enumeration of all perfect sets of reals. Construct disjoint A = {aα : α < 2ℵ0} and B = {bα : α < 2ℵ0} as follows: Pick aα such that aα / ∈{aξ : ξ < α} ∪{bξ : ξ < α}, and bα such that bα ∈Pα −{aξ : ξ ≤α}. Then A is the set.] 5.2. If X is an infinite set and S is the set of all finite subsets of X, then |S| = |X|.
[Use |X| = ℵα.] 5.3. Let (P, <) be a linear ordering and let κ be a cardinal. If every initial segment of P has cardinality < κ, then |P| ≤κ.
5.4. If A can be well-ordered then P(A) can be linearly ordered.
[Let X < Y if the least element of X △Y belongs to X.] 5.5. Prove the Axiom of Choice from Zorn’s Lemma.
[Let S be a family of nonempty sets. To find a choice function on S, let P = {f : f is a choice function on some Z ⊂S}, and apply Zorn’s Lemma to the partially ordered set (P, ⊂).] 60 Part I. Basic Set Theory 5.6. The countable AC implies that every infinite set has a countable subset.
[If A is infinite, let An = {s : s is a one-to-one sequence in A of length n} for each n. Use a choice function for S = {An : n ∈N} to obtain a countable subset of A.] 5.7. Use DC to prove the countable AC.
[Given S = {An : n ∈N}, consider the set A of all choice functions on some Sn = {Ai : i ≤n}, with the binary relation ⊃.] 5.8 (The Milner-Rado Paradox). For every ordinal α < κ+ there are sets Xn ⊂α (n ∈N) such that α = S n Xn, and for each n the order-type of Xn is ≤κn.
[By induction on α, choosing a sequence cofinal in α.] 5.9. If {Xi : i ∈I} and {Yi : i ∈I} are two disjoint families such that |Xi| = |Yi| for each i ∈I, then |S i∈I Xi| = |S i∈I Yi|.
[Use AC.] 5.10. If {Xi : i ∈I} and {Yi : i ∈I} are such that |Xi| = |Yi| for each i ∈I, then | Q i∈I Xi| = | Q i∈I Yi|.
[Use AC.] 5.11. Q 0<n<ω n = 2ℵ0.
5.12. Q n<ω ℵn = ℵℵ0 ω .
5.13. Q α<ω+ω ℵα = ℵℵ0 ω+ω.
5.14. If GCH holds then (i) 2<κ = κ for all κ, and (ii) κ<κ = κ for all regular κ.
5.15. If β is such that 2ℵα = ℵα+β for every α, then β < ω.
[Let β ≥ω. Let α be least such that α + β > β. We have 0 < α ≤β, and α is limit. Let κ = ℵα+α; since cf κ = cf α ≤α < κ, κ is singular. For each ξ < α, ξ + β = β, and so 2ℵα+ξ = ℵα+ξ+β = ℵα+β. By Corollary 5.17, 2κ = ℵα+β, a contradiction, since ℵα+β < ℵα+α+β.] 5.16. Q α<ω1+ω ℵα = ℵℵ1 ω1+ω.
[ℵℵ1 ω1+ω ≤(Q∞ n=0 ℵω1+n)ℵ1 = Q n ℵℵ1 ω1+n = Q n(ℵℵ1 ω1 ·ℵω1+n) = ℵℵ1 ω1 ·Q n ℵω1+n = Q α<ω1+ω ℵα.] 5.17. If κ is a limit cardinal and λ < cf κ, then κλ = P α<κ |α|λ.
5.18. ℵℵ1 ω = ℵℵ0 ω · 2ℵ1.
5.19. If α < ω1, then ℵℵ1 α = ℵℵ0 α · 2ℵ1.
5.20. If α < ω2, then ℵℵ2 α = ℵℵ1 α · 2ℵ2.
5.21. If κ is regular and limit, then κ<κ = 2<κ. If κ is regular and strong limit then κ<κ = κ.
5.22. If κ is singular and is not strong limit, then κ<κ = 2<κ > κ.
The Axiom of Choice and Cardinal Arithmetic 61 5.23. If κ is singular and strong limit, then 2<κ = κ and κ<κ = κcf κ.
5.24. If 2ℵ0 > ℵω, then ℵℵ0 ω = 2ℵ0.
5.25. If 2ℵ1 = ℵ2 and ℵℵ0 ω > ℵω1, then ℵℵ1 ω1 = ℵℵ0 ω .
5.26. If 2ℵ0 ≥ℵω1, then(גℵω) = 2ℵ0 and(גℵω1) = 2ℵ1.
5.27. If 2ℵ1 = ℵ2, then ℵℵ0 ω ̸= ℵω1.
5.28. If κ is a singular cardinal and if κ <(גλ) for some λ < κ such that cf κ ≤cf λ then(גκ) ≤(גλ).
5.29. If κ is a singular cardinal such that 2cf κ < κ ≤λcf κ for some λ < κ, then (גκ) =(גλ) where λ is the least λ such that κ ≤λcf κ.
Historical Notes The Axiom of Choice was formulated by Zermelo, who used it to prove the Well-Ordering Theorem in . Zorn’s Lemma is as in Zorn ; for a related prin-ciple, see Kuratowski . (Hausdorffin , pp. 140–141, proved that every partially ordered set has a maximal linearly ordered subset.) The Principle of De-pendent Choices was formulated by Bernays in .
K¨ onig’s Theorem 5.10 appeared in J. K¨ onig . Corollary 5.17 was found independently by Bukovsk´ y and Hechler. The discovery that cardinal expo-nentiation is determined by the gimel function was made by Bukovsk´ y; cf. .
The inductive computation of κλ in Theorem 5.20 is as in Jech [1973a].
The Hausdorffformula (5.22): Hausdorff.
Inaccessible cardinals were introduced in the paper by Sierpi´ nski and Tar-ski ; see Tarski for more details.
Exercise 5.1: Felix Bernstein.
Exercise 5.8: Milner and Rado .
Exercise 5.15: L. Patai.
Exercise 5.17: Tarski [1925b].
Exercises 5.28–5.29: Jech [1973a]. |
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This is a good question, given the way calculus is currently taught, which for me says more about the sad state of math education, rather than the material itself. All calculus textbooks and teachers claim that they are trying to teach what calculus is and how to use it. However, in the end most exams test mostly for the students' ability to turn a word problem into a formula and find the symbolic derivative for that formula. So it is not surprising that virtually all students and not a few teachers believe that calculus means symbolic differentiation and integration.
My view is almost exactly the opposite. I would like to see symbolic manipulation banished from, say, the first semester of calculus. Instead, I would like to see the first semester focused purely on what the derivative and definite integral (not the indefinite integral) are and what they are useful for. If you're not sure how this is possible without all the rules of differentiation and antidifferentiation, I suggest you take a look at the infamous "Harvard Calculus" textbook by Hughes-Hallett et al. This for me and despite all the furor it created is by far the best modern calculus textbook out there, because it actually tries to teach students calculus as a useful tool rather than a set of mysterious rules that miraculously solve a canned set of problems.
I also dislike introducing the definition of a derivative using standard mathematical terminology such as "limit" and notation such as h→0h→0. Another achievement of the Harvard Calculus book was to write a math textbook in plain English. Of course, this led to severe criticism that it was too "warm and fuzzy", but I totally disagree.
Perhaps the most important insight that the Harvard Calculus team had was that the key reason students don't understand calculus is because they don't really know what a function is. Most students believe a function is a formula and nothing more. I now tell my students to forget everything they were ever told about functions and tell them just to remember that a function is a box, where if you feed it an input (in calculus it will be a single number), it will spit out an output (in calculus it will be a single number).
Finally, (I could write on this topic for a long time. If for some reason you want to read me, just google my name with "calculus") I dislike the word "derivative", which provides no hint of what a derivative is. My suggested replacement name is "sensitivity". The derivative measures the sensitivity of a function. In particular, it measures how sensitive the output is to small changes in the input. It is given by the ratio, where the denominator is the change in the input and the numerator is the induced change in the output. With this definition, it is not hard to show students why knowing the derivative can be very useful in many different contexts.
Defining the definite integral is even easier. With these definitions, explaining what the Fundamental Theorem of Calculus is and why you need it is also easy.
Only after I have made sure that students really understand what functions, derivatives, and definite integrals are would I broach the subject of symbolic computation. What everybody should try to remember is that symbolic computation is only one and not necessarily the most important tool in the discipline of calculus, which itself is also merely a useful mathematical tool.
ADDED: What I think most mathematicians overlook is how large a conceptual leap it is to start studying functions (which is really a process) as mathematical objects, rather than just numbers. Until you give this its due respect and take the time to guide your students carefully through this conceptual leap, your students will never really appreciate how powerful calculus really is.
ADDED: I see that the function θ↦sinθθ↦sinθ is being mentioned. I would like to point out a simple question that very few calculus students and even teachers can answer correctly: Is the derivative of the sine function, where the angle is measured in degrees, the same as the derivative of the sine function, where the angle is measured in radians. In my department we audition all candidates for teaching calculus and often ask this question. So many people, including some with Ph.D.'s from good schools, couldn't answer this properly that I even tried it on a few really famous mathematicians. Again, the difficulty we all have with this question is for me a sign of how badly we ourselves learn calculus. Note, however, that if you use the definitions of function and derivative I give above, the answer is rather easy.
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edited Sep 27, 2010 at 13:41
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Deane Yang
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I emphatically agree that students don't know what a function is. But then again, it is a deceptively deep concept. As for modern treatments that emphasize other things than the standard, did you ever look at "Calculus in Context", the five colleges calculus? Available at math.smith.edu/Local/cicintro/cicintro.html My problem with this approach though is that even if you can convince me easily that it's the right thing to do mathematically, how will it mesh with the courses in other disciplines that students take, which will expect much more traditional material.
– Thierry Zell
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Sep 27, 2010 at 12:55
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Harry, that is exactly how any pure mathematician, including me, would do it. But that's the hard way. For an engineer or physicists, who thinks in units and dimensional analysis and views the derivative as a "sensitivity" as I've described above, the answer is dead obvious.
– Deane Yang
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Sep 27, 2010 at 17:51
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Alexander and Pietro, unfortunately I said "I would like to see...", which means I don't really get to banish symbolic methods for a whole semester. In fact, I advise being pragmatic and teaching in a fashion that will not alienate you from your department or school administration. That said, if you want to slip in more understanding (which I claim actually helps students learn the symbolic methods better), I recommend taking problems from the Harvard calculus textbook, as well as their precalculus text ("Functions Modeling Change"). Especially those where no formula is given for the function.
– Deane Yang
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Sep 28, 2010 at 2:07
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Your description of a function as a box seems to miss the most important part: that whenever you put a given number in, you always get the same output. That is, the box behavior should be single-valued. (Otherwise, we might imagine a black-box that accepts a given input and outputs a random number, perhaps different every time, and although this accords with your description, it is not a function.)
– Joel David Hamkins
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Jan 24, 2011 at 11:36
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I do consulting in the financial sector, and first and second derivatives are widely used but never symbolic differentiation. The derivatives are always computed numerically and under the hood. What your finance students need to know is how to interpret and use these numbers. This is presented rather well in my view in the first few chapters of the Harvard Calculus text.
– Deane Yang
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Jan 30, 2011 at 4:23
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I'm teaching Calc 1 this semester, and I've stumbled onto something that I like very much.
First of all, I start (always) by having my students draw bunches of tangent lines to graphs, compute slopes and draw the "slope graphs" (they also do "area graphs", but that's not relevant to this answer). They build up a bit of intuition about slope and slope graphs.
Then (after a few days of this) I ask them to give me unambiguous instructions about how to draw a tangent line. They find, of course, that they are stumped.
In the past, I went from this to saying "we can't get a tangent line, but maybe we can get an approximately tangent line" and develop the limit formula.
This semester, I said, "we have an intuitive notion of tangency; suppose someone offered a definition of tangency -- what properties would it satisfy?" We had a discussion with the following result: tangency at point x=ax=a should satisfy:
tangency (of one function with another) should be an equivalence relation
if two linear functions are tangent at x=ax=a, they are equal.
a quadratic has a horizontal tangent line at its vertex.
if ff and gg are tangent at x=ax=a, then f(a)=g(a)f(a)=g(a).
if f1f1 is tangent to f2f2 at x=ax=a and g1g1 is tangent to g2g2 at x=ax=a
then f1+g1f1+g1 is tangent to f2+g2f2+g2 at x=ax=a and similarly for the products.
the evident rule for composition.
Using these rules, we showed that if ff has a tangent line at x=ax=a, it has only one.
So we can define f′(a)f′(a) to be the slope of the tangent line at x=ax=a, if it exists!
The axioms are enough to prove the product rule, the sum rule and the chain rule. So we get derivatives of all polynomials, etc., assuming only that tangency can be defined.
Then (limits having presented themselves in the computation of area) I defined
ff to be tangent to gg if limx→af(x)−g(x)x−a=0limx→af(x)−g(x)x−a=0. We derive the limit formula for the derivative, and check the axioms.
EDIT: Here's some more detail, in case you're wondering about implementing this yourself. I had the initial discussion about tangency in class, writing on the board. A day or so later, I handed out group projects in which the axioms were clearly stated and numbered, and the basic properties (as outlined above) given as problems.
The students' initial impulse is to argue from common sense, but I insisted on argument directly from the axioms. There was one day that was kind of uncomfortable, because that is very unfamiliar thinking. I had them work in class several days, and eventually they really took to it.
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edited Oct 9, 2010 at 1:27
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Jeff Strom
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This is very nice.
– Deane Yang
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Sep 27, 2010 at 14:20
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I second Deane's comment.
– Mark Meckes
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Sep 27, 2010 at 17:12
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Is there really a unique equivalence relation satisfying these rules? I do not see how these rules could ever access a function which is not a polynomial. If not, saying that you can define f'(a) to be the slope of the tangent line at x=a presupposes that you have chosen one of the many equivalence relations which satisfy these properties.
– Steven Gubkin
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Nov 3, 2010 at 15:41
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See this MO question: mathoverflow.net/questions/44774/…
– Steven Gubkin
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Nov 4, 2010 at 14:23
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@Jeff: The MO question I linked to shows that for C^\infty functions, these axioms do characterize differentiation.
– Steven Gubkin
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Nov 11, 2010 at 15:21
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I'm going to answer this part:
does anyone out there actually use this definition to calculate a derivative that couldn't be obtained by a known symbolic rule?
Yes. sin(x)sin(x).
My point is that of course we can just learn the derivative of this function, but then we could just learn the derivative of any function. So looking for a "complicated function" that needs the limit definition is pointless: we could just extend our list of examples to include this function. It's a bit like the complaint that there's no closed form for a generic elliptic integral: all we really mean is that we haven't given it a name yet.
In fact, one could do x2x2 like this, or even xx, but I think that sin(x)sin(x) has a good pedagogical value. If you can get them first to ponder the question, "What is sin(x)sin(x)?" then it might work. I'm teaching a course at the moment where I'm trying to get my students out of the "black box" mentality and start thinking about how one builds those black boxes in the first place. One of my starting points was "What is sin(x)sin(x)?". Or more precisely, "What is sin(1)sin(1)?". If you take that question, it can lead you to all sorts of interesting places: polynomial approximation of continuous functions, for example, and thence to Weierstrass' approximation theorem.
Many students will just want the rules. But if the students refuse to learn, that's their problem. My job is to provide them with an environment in which they can learn. Of course, I should ensure that what they are trying to learn is within their grasp, but they have to choose to grasp it. So I'm not going to give them a full exposition on the deep issues involving the ZF axioms if all I want is for them to have a vague idea of a "set" and a "function", but I am going to ensure that what I say is true (or at the least is clearly flagged as a convenient lie).
Here's a quote from Picasso (of all people) on teaching:
So how do you go about teaching them something new? By mixing what they know with what they don't know. Then, when they see vaguely in their fog something they recognise, they think, "Ah, I know that." And it's just one more step to, "Ah, I know the whole thing.". And their mind thrusts forward into the unknown and they begin to recognise what they didn't know before and they increase their powers of understanding.
We all remember professors who forgot to mix the new in with the old and presented the new as completely new. We must also avoid the other extreme: that of not mixing in any new things and simply presenting the old with a new gloss of paint.
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edited Mar 16 at 8:34
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2 revs, 2 users 94%Andrew Stacey
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+1 for "if the students refuse to learn, that's their problem. My job is to provide them with an environment in which they can learn."
– Mark Meckes
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Sep 27, 2010 at 14:44
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@Mark and Sean, I have to admit that I'm a little put off by the cheerleading for this particular phrase -- stripped from the context Andrew provided it, it comes across rather as, "I don't do a bad job of teaching, my students do a bad job of learning." I think this is an attitude those of us who teach should be careful to avoid, in general. (Of course, everyone who has ever taught has come across specific cases where it might be applied.)
– JBL
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Sep 28, 2010 at 13:12
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@JBL: point taken, although I disagree with your restatement. "My job is to provide them with an environment in which they can learn" is the sentiment of someone who takes doing that job well seriously. I really liked the line because of several recent conversations about students who don't take notes, skip class, rush through homework, and don't ask questions. Such students are the exception rather than the rule, but they can get under one's skin. For times like that, I thought Andrew's line would be a good substitute for the glib old saw about a horse and water.
– Mark Meckes
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Sep 28, 2010 at 13:57
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@Mark: Yes, I didn't think that either you or Sean agreed with my rephrasing, just that it seems to me that this statement (in isolation) has a little of that ring to it. Students who behave as you describe are extremely irritating, but I think teachers would do well to avoid sounding like we think this is the norm :)
– JBL
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Sep 28, 2010 at 15:06
@JBL: I agree with your underlying point, but disagree with the point you actually made. I think that Mark quoted enough context by quoting the second sentence. I also disagree with your rephrasing because my original statement said, "If the students ..." whereas your rephrasing seems more like "When the students ...". There's loads more to say on this, but MO is (thankfully) a lousy place to say it. So I will content myself with saying that I know Mark and I know that he takes teaching very seriously so I see his "cheerleading" of that sentence coming from the best possible motives.
– Andrew Stacey
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Sep 28, 2010 at 16:07
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While I think that ideally, even in a freshman course of calculus, students should receive some historical notions about the development of the ideas of infinitesimal calculus, I think that, even in a freshman course of calculus, the true definition of derivative of a function should be given, that is, via the first order approximation. A function f:(a,b)→Rf:(a,b)→R is differentiable at xx if there exists mm such that
f(x+h)=f(x)+mh+o(h)ash→0.
f(x+h)=f(x)+mh+o(h)ash→0.
The fact that the coefficient mm (the derivative) can be characterized, and sometimes efficiently computed, as a limit of a quotient, has certainly to be observed, and should be applied immediately to treat some elementary functions like x2x2, 1/x1/x or exex, as usual. But I would never give it as a definition.
I think there is a philosophical issue here. It may seem simpler to define something as the result of a procedure for getting it, compared with defining it via a characteristic property. But the latter way is superior, and on a long distance, simpler. And in the case of students who will stop there their mathematical education, then, I prefer they at least see the true idea behind, rather that being able to compute the derivative of cos(ex)cos(ex) : when will that be of use for them?
The definition via first order expansion is very natural, and more understandable to the freshman student, as it has a more direct geometrical meaning. It reflects the physical idea of linearity of small increments (like in Hooke's law of elasticity, etc). It is much closer to the practical use of derivatives in approximations. It makes easier all the elementary theorems of calculus (consider how needlessly complicated becomes the proof of the theorem for the derivative of a composition by introducing a useless quotient). Finally, it is closer to the generalization to Fréchet differential, which is a good thing for those students that will continue their study in maths.
A funny remark, from my experience. Ask students that received the definition of derivative as limit of incremental quotient, to compute limx→0sin(x)/xlimx→0sin(x)/x. Will anybody say, it's the derivative of sin(x)sin(x) at 00, that is cos(0)=1cos(0)=1? No, they will try and use the "rule of de L'Hopital"!
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edited Mar 16 at 9:12
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From my prof back in the days: "Some of you might have heard of a thing called L'Hopital's rule. It has a lot of hypotheses that no-one ever checks, and students always apply it when the quotient is in the wrong form, so I won't teach it and you'd better not use it." And now I do the same... (I didn't mind when he said that because I was one of the ones who'd never heard of it.)
– Thierry Zell
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Sep 27, 2010 at 12:42
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@Thierry If I was chairman of the mathematics department when your professor was teaching and I'd heard that,he'd have been fired on the spot unless he had tenure-and if he had tenure,I'd have made sure he never taught it again. He should have been ashamed of himself.Really. If you think this material is usually presented so lousily,do something about it!
– The Mathemagician
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Sep 27, 2010 at 16:53
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@Andrew: and if I was the chairman in another maths department, I'd immediately engage him with double salary. De'LHopital himself would be embarassed to know somebody's still wasting time with such an awkward theorem like that thing that brings his name. Theorems, like cakes, don't always come out well; that thing came out very badly, and left a mess in the oven. Today, it may be at most of some historical interest. Teach the Landau notation instead! (Btw, as you probably know, Edmund Landau was fired from Göttingen in 1933, with the pretext of his way of teaching a calculus course.)
– Pietro Majer
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Sep 27, 2010 at 19:13
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@Pietro: what you give is essentially Caratheodory's definition, as alluded to in my answer. It's so close to the usual definition that I don't really believe that students have a significantly easier time with it. However, I believe that when you teach calculus, this definition inspires you and you do a very good job teaching it, more so than you would with the standard definition. I suspect that most "the students find it easier when..." statements are like this, but that's fine -- finding the version that you can get behind enthusiastically and explain well is part of good teaching.
– Pete L. Clark
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Sep 27, 2010 at 19:49
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About that funny remark: even saying limx→0sin(x)/xlimx→0sin(x)/x is the derivative of sin(x)sin(x) at 0 may be viewed as cheating, since the typical textbook approach is to use a geometric argument to prove limx→0sin(x)/x=1limx→0sin(x)/x=1 and then use that limit to prove that ddxsin(x)=cos(x)ddxsin(x)=cos(x).
– Mark Meckes
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Sep 28, 2010 at 14:27
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I agree with the above comments.
The point of my comment-question "What competing definition do you have in mind?" was to emphasize something that seems to be under-emphasized in the question itself: the reason we speak of derivatives as limits is because that's the definition of the derivative, and we want to give a definition of the concept that is going to be discussed for much of the semester.
[It is possible to give other definitions of a derivative, but they are all variations on the same theme and, in particular, all use either the concept of limit or the (equivalent!) concept of continuity. For instance, Caratheodory has a nice definition of the derivative in terms of functions vanishing to first order, but this is not going to be any more palatable to the freshman calculus student.]
[Added: I admit that I forgot about nonstandard analysis when I wrote the above paragraph. That indeed has a somewhat different feel from the usual limits and continuity. One the one hand, although I have never taught calculus this way, I rather doubt that doing so would suddenly make the difficult concepts of continuity and differentiability go over easily. On the other hand, I certainly couldn't decide to teach a nonstandard approach to calculus because it would be...nonstandard. The curriculum among different sections, different classes and different departments has to have a certain minimal level of coherence, and at the moment the majority of the grad students and faculty in every math department I have ever seen are not familiar enough with nonstandard analysis to field questions from students who have learned calculus by this approach.]
If we don't give a definition of the most important concept in the course, then we lose all pretense of developing things in a logical sequence. In particular, it's hard to see how to discuss the derivations of any of the basic rules the students will actually be using to compute derivatives, and thus we would be forced to reduce calculus to a (long!) list of algorithms based on certain unexplained rules.
Nevertheless I take your question seriously, since I have taught a fair amount of freshman calculus in recent years. It is absolutely correct that a lot of students get impatient, angry and/or confused at the limit definition of the derivative (or really, at anything having to do with limits and/or continuity). I do derivations of things like the product rule and the power rule rather quickly in class, because I know that something like half the class isn't following and doesn't care to follow. And yet I do them anyway (not all of them, but more than half) because, to me, not to do them makes the course something I could not bring myself to teach (and, by the way, would put it well below the level of the AP calculus class I had in high school: I feel honorbound to give to my calculus students not too much less than was given to me). Thus there is a real disconnect between the calculus class that I want to teach and the calculus class that something like half of the students want to take. It's discouraging.
I would be happy to hear that I am making a false dichotomy between giving the limit definition of the derivative and just giving algorithms to solve problems. I definitely experiment with different kinds of explanation beyond (and instead of!) just a formal proof. Here are some things I have tried:
1) Take the definition of continuity as primary, and define the limit of a function at a point as the value at which one can (re)define the function to make it continuous. I think this should be helpful, since I think most people have an intuitive idea of a "continuous, unbroken curve" and much less of the limit of a function at a point.
2) Emphasize physical reasoning. The last time I taught freshman calculus, I spent the entire first day talking about velocities: first average velocity, then instantaneous velocity. If a differentiation rule has a plausible physical interpretation -- e.g. the chain rule says that rates of change should multiply -- then I often give it.
3) Emphasize "chemical reasoning", i.e., dimensional analysis:. I often give the independent variable and the dependent variable units and emphasize that the units of the derivative are
different from the units of the original function. In this way one can see that the conjectured product rule (fg)′=f′g′(fg)′=f′g′ is dimensionally wrong and thus nonsense. (And again, the chain rule is "obvious" from a unit conversion perspective.) Similarly dimensional analysis should stop you from saying that the volume of a cylinder is πrhπrh.
Unfortunately none of these things have worked with the portion of the class that doesn't want to hear anything but how to solve the problems.
Added: To more directly address your specific question: yes, there are problems one can ask of freshman calculus students which require them to use the limit definition of the derivative rather than (just) the differentiation rules, but I do not recommend asking many of these questions, since the students find them very difficult. A personal example: when I was teaching Math 1A (first semester calculus) as a graduate student at Harvard, we had communal exams but the course head (who was a tenured professor of mathematics, hence a very brilliant person) had the final say. On the first exam, we decided that one of the questions was too hard, so at the last minute the course head replaced it with the following one (which he did not show to us):
Consider the function f(x)f(x) defined as xasin(1x2)xasin(1x2) for x≠0x≠0 and f(0)=0f(0)=0. What is the smallest integer value of aa such that ff is (i) continuous, (ii) differentiable, (iii) twice differentiable?
I had the good fortune to grade this problem. Out of 200200 or so exams, the median score was 0.50.5 out of 1212. About three students wrote down the right numerical answer for part (iii), but this was not supported by any work or reasoning whatsoever.
Added: by the way, it's not as though the above question is "bad" in the sense that it's not testing mathematical competence and depth of understanding of calculus. I think it absolutely is, just at a level way above that which one should be testing in a freshman class for non math majors. For the next few years, when the story came up in a social setting involving mathematical hotshots, after telling it I would press them for an answer to part c) on the spot. Most people I asked did not get it. (Note that I would not of course give them pen and paper and a quiet spot to think about the problem for some period of time. I generally required an answer after a minute or so. Let's hold PhD mathematicians to higher standards than freshman non-majors after all!) For instance, I watched a cloud pass over one Fields Medalist's face as he got very confused. After a while though I stopped using this as a pop quiz in addition to a story: I can't explicitly remember why, but I'd like to think it dawned me how obnoxious it was to put people on the spot like that...
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edited Jan 29, 2011 at 22:12
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+1. Differentiation can be done without limits too, en.wikipedia.org/wiki/Formal_derivative I interpreted the question as distinguishing between derivatives in analysis and "generating functions". Even most trigonometric functions have combinatorial meaning and so their derivatives can be computed formally. But as you say that misses the point of calculus (continuity, physical reasoning etc.).
– Gjergji Zaimi
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Sep 27, 2010 at 8:37
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The freshman-level example is especially biased because students tend to believe that any problem with two letters in it is very hard. But I've had a reasonable degree of success with a problem of this type, using a value for a (disclaimer: at a good school, though no Harvard). But it's also because I'd spent some time on this in class; you can't spring this on students out of the blue like that professor did and expect they'll do well.
– Thierry Zell
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Sep 27, 2010 at 11:41
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Teaching undergraduate calculus using nonstandard analysis is not out of the question. I haven't done it but I know others who have, using for example Henle and Kleinberg's Infinitesimal Calculus.
– Timothy Chow
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Sep 27, 2010 at 14:50
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@Mariano: no, I was dead serious. The point was that this person was far too bright to realize that this was a ridiculously hard question for freshman calculus.
– Pete L. Clark
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Sep 27, 2010 at 19:33
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A much simpler version of your problem is the following (I actually used this problem in the past): Consider f(x)=x2sin(1x) for x≠0 and f(0)=0. Find f′(0). This function is differentiable at zero but f′ is not eve continuous at 0 (so no power series representation), so I doubt that any other approach than the limit would work. I think this is one of the simplest examples which explains why the limit is needed. Also, what if one discovers a completelly new great function f, how does one find f′?
– Nick S
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Sep 27, 2010 at 22:45
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I wanted to add one further point to the many good answers already given here: "black box" symbolic computation, in the absence of understanding the formal definitions, can work when everything goes right, but is very unstable with respect to student errors (which are sadly all too common). Knowledge of definitions provides a crucial extra layer of defence against such errors. (Of course, it is not the only such layer; for instance, good mathematical or physical intuition and conceptual understanding are also very important layers of defence, as is knowledge of key examples. But it is a key layer in situations which are too foreign, complicated, or subtle for intuition or experience to be a good guide.)
For instance, without knowing the formal definition of the derivative, a student could very easily start with a true formula such as
(x2)′=2x
and do something like "substitute x=3" to obtain the false formula
(9)′=6.
(An example that I have actually seen: someone attempted to prove Fermat's last theorem by starting with the equation
an+bn=cn
and then differentiating with respect to n. Ironically, a variant of this type of trick actually works when solving FLT over polynomial rings, but that's another story...)
Now, without bringing in the definition of a derivative (and of a function), how could you explain to the student what went wrong here in a way that the student will actually remember? Saying that one can use the law of substitution or the trick of differentiating both sides in some situations, but not in others, is likely to be recalled inaccurately, if at all (and may have the side effect that the student may view such basic moves as substitution as somehow being "suspect", thus avoiding it in the future).
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edited Nov 2, 2010 at 16:37
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People say I'm mean for asking for the derivative of π2, but I think it's a memorable example for the students. Another place the blind symbol manipulation goes wrong is on sin−1x=arcsinx. Many students are willing to assume that if f(x)=g(x) then f′(x)=g′(x), which is only a consequence for some meanings of the first equation.
– Douglas Zare
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Nov 6, 2010 at 16:43
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@Terry: I agree that the black-box use is an issue, but some might argue that it's possible to fix without necessarily going all the way to a limit definition, because the root problem is conceptual understanding of functions, and more precisely of the derivative as a function. A simple aphorism would suffice: "chug then plug, don't plug and chug!" More seriously, one could discuss the difference between the derivative as a function and the derived number at a point (slope) purely graphically, with no references to limits.
– Thierry Zell
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Nov 7, 2010 at 0:01
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@Douglas: I didn't think the derivative of π2 was a mean thing to ask! Who thinks so? students? colleagues? Never mind, I'll be sure to borrow it for next time. (Even meaner would be to ask for the derivative of e2, btw.) As for the blind symbol manipulation, I cannot believe that it took me all these years before seeing for the first time (in an exam) that the derivative of arctanx was arcsec2x. In retrospect, I should have been expecting this for a long time!
– Thierry Zell
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Nov 7, 2010 at 0:09
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I think this type of error is avoidable if you introduce x not as a variable, but as a special symbol for identity function. Even more, you can use bold x for the identity function and normal x for a value.
– Anixx
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Jan 2, 2011 at 18:35
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When my father was a judge at a high school math fair, a student gave a presentation on calculus. During the question period after the presentation, he asked the student "If f(x) = 3^2, what is the derivative of f(x)?" The student said "6". My father then asked "If f(x) = 9, what is the derivative of f(x)?" The student said "0". He asked the first question again and the student still said "6". Of course this student did not go on to the next round of the math fair.
– KConrad
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Jan 23, 2011 at 20:31
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The derivative of x|x| is best computed via the "limit" definition. A more general example is xf(x) where f is any continuous function, and we are computing the derivative at x=0.
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answered Sep 27, 2010 at 15:33
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Igor Belegradek
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If you want example with nonzero derivative consider x|x|+x.
– Igor Belegradek
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Sep 27, 2010 at 17:12
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For the more general example, one should also ask that f is not differentiable at 0. (By varying such f, we can get any derivative at all, or none. Taking f(x)=xsin(1/x), extended continuously, gives a very interesting example that's come up elsewhere on this page.)
– Toby Bartels
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Apr 3, 2011 at 23:08
This example can be done straightforwardly without limits, using infinitesimals (either informally or in NSA).
– user21349
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Oct 6, 2012 at 2:10
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I am surprised that no answer has explicitly mentioned the fundamental theorem calculus yet: that is a classic, and important, instance of calculating the derivative using the limit definition. So, for example, the integral sine function
∫x0sinttdt
has important applications in signal processing and the cumulative distribution function of the normal distribution N(a,σ2)
1√2πσ2∫x−∞e−(t−a)22σ2dt
is the bread and butter of probability and statistics. Both functions are not elementary and their derivatives, while significant, would be impossible to calculate by other means.
I also disagree with the comment that piecewise defined functions "are not good at all" for illustrating the definition of the derivative based on limits. In fact, piecewise polynomial functions, in the form of splines, are used in mechanical engineering (e.g. to design the shape of the car body), and provide a neat opportunity to relate conceptual and computational aspects of derivatives.
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answered Sep 28, 2010 at 1:26
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Victor Protsak
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+1, the guys who deal in splines always make a big deal out of left and right continuity for making the approximant cn (for whatever value of n is needed by the application). I too feel that the error function and the sine integral are too important not to at least be given a passing mention in the context of the Fundamental Theorem.
– J. M. isn't a mathematician
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Sep 28, 2010 at 1:55
I agree with both points here.
– Deane Yang
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Sep 28, 2010 at 2:08
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The way that Calculus is traditionally taught gives a false impression that every function worth looking at can be differentiated using the rules of differentiation. This comes from a misconception that any function worth looking at can be described by an algebraic formula, or using trigonometric or logarithmic functions.
That's just not the case: the most common everyday functions don't have any formulas. Some examples:
Price of a company stock over several decades.
Volume of water in a water tower over the course of a week
Median price of a house in your area (adjusted for inflation), over the course of 100 years.
US National Debt over the last two hundred years.
US Deficit
For such functions, rate of change has a very real meaning. I find that students who had Calculus in high-school are stumped if I give them an example like that and ask them to graph the rate at which, say, the US national debt has changed throughout US history, and how that relates to the deficit.
Understanding the derivative as both rate of change and the slope of the tangent line helps, and the only good way to tie those concepts is with using limits.
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answered Jan 29, 2011 at 19:25
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Anna Varvak
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@Anna: I agree heartily. To take it a step further: I have often thought that the traditional calculus sequence lacks an applied component which severely limits its usefulness to those who are not going on to physics and math. As you say, when given a real world function of interest, you are generally not given an algebraic expression for it. Rather, in order to apply the methods of calculus in a quantitative way, there needs to be a step where you create a mathematical model of the function. I was never taught how to do this step myself, and it seems not to be at all trivial...
– Pete L. Clark
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Jan 29, 2011 at 22:38
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Once or twice I tried to remedy this by asking exam questions like: "Write down an explicit function which has local minima at ±2 and approaches infinity as x approaches ±infinity." I was hoping that a student would see that a simple function satisfying these conditions is a fourth degree polynomial with double roots at ±2, thus f(x)=(x−2)2(x+2)2. But they had a lot of trouble with this, and the answer to my question "Does teaching the standard curriculum give them the tools to answer this question?" was "No." But wouldn't it be great if students could do this?
– Pete L. Clark
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Jan 29, 2011 at 22:44
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@Pete: I have been thinking a lot about how the calculus we teach is too neat and not applied enough these days, and I've been thinking about it not because of my calculus classes, but because of my higher-level courses where I've had to use "messy" or "exotic" stuff like Taylor expansions pretty often to motivate some avenues of investigation. I wish my students were more comfortable with this, and expected less of the neat closed-form formula results. Though of course, we should teach them to appreciate when closed-form stuff happens too!
– Thierry Zell
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Mar 12, 2011 at 1:25
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I appreciate your point, but your examples are all discontinuous functions, so in fact the notion of a limit doesn't quite work. For example, the US deficit can only change in steps of one cent, the amount of water only in steps of one molecule. These could all be used as examples to show that the Cauchy-Weierstrass limit does not actually connect to reality. Perhaps a better point to make with these examples would be that students could benefit by understanding discrete/numerical calculus as well as the classical calculus of continuous real functions.
– user21349
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Oct 6, 2012 at 2:47
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The definition of derivatives is useful in exercises about functional equations. Ever solve f(x+y)=f(x)f(y) ? A more elaborate one is [f(x):f(y):f(z):f(t)]=[x:y:z:t], functions preserving the cross-ratio (= anharmonic ratio).
However, we should not neglect the interest of the black-box side of mathematics. We should remember that it is this aspect which has made mathematics so much unavoidable in Science. Somehow, it participates to the ``unreasonable effectiveness of Mathematics in the Natural Sciences'' (E. Wigner's famous statement). After all, the definition of derivatives has the same status as the constructions of Z,Q,R,C. One can spend a year without thinking about them, while using these fundamental objects every hour, by applying rules. Do you remember the construction of the polynomial algebra k[X] ? How would you define π ? In a more advanced situation, chemists have efficient rules to deal with characters of representations of finite groups, and they do not need to read a justification, or to remember it, even though the first Chapter of J.-P. Serre's book was intended to be read by his chemist wife. Mathematics is the tool box of Science. It is even a tool box for itself, in the sense that new topics use the older ones. To go further, we must accept older truths. Of course, it is way better to accept them for good reasons, that is, because we have completely understood the definitions. But if the half of a classroom, who does not intend to do mathematical research, neglects the definition and prefer focussing on the rules, there is no problem at all, provided they apply the rules correctly. There are many ways to learn rules, one of them being solving a lot of exercises.
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answered Sep 27, 2010 at 7:15
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Denis Serre
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+1. But, I think a point some other people have been making is that students should at least see the formal definition. But I agree that you shouldn't beat it over the head of students who aren't happy...
– Matthew Daws
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Sep 27, 2010 at 12:07
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If scientists want their students to learn a sequence of formal rules, I think they should be the ones to teach them. Personally I want to teach mathematics, and mindless manipulation of symbols is not math.
– Steven Gubkin
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Nov 11, 2010 at 15:31
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An example I like is exp(−1x2) and the "bump functions" one can construct with it.
First of all, this example is important in differential geometry (e.g. Whitney's embedding theorem) and complex analysis (as an example of a real C∞ function which isn't holomorphic).
In second place, even in first year calculus it's an important illustration of the concept of derivative and of Taylor's theorem. It's important in my opinion to understand why all derivatives at zero are zero (i.e. because it goes to zero faster then any polinomial) but even so the function is changing values.
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answered Sep 27, 2010 at 10:03
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Pablo Lessa
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The example is well-chosen, but your parenthesis sounds misleading to me :there exist functions that go to zero faster than any polynomial at zero, while they are not even twice derivable.
– Benoît Kloeckner
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Sep 27, 2010 at 15:53
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You're right! Thank you. To be honest, I didn't consider this possibility at the time of writing. I'll leave the parenthesis as is, since it still has some content and you're comment is right below.
– Pablo Lessa
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Sep 27, 2010 at 16:57
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@ Benoi: Are there examples of such functions which are continuous in a neighborhood of 0? I ask because the only examples I can find are along the lines of the characteristic function of the rationals time exp(-1/x^2).
– Steven Gubkin
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Nov 2, 2010 at 22:59
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Take f(x)=exp(−1/x2) and g a continuous nowhere differentiable function. The function h(x)=f(x)g(x) is continuous, goes to zero faster then any polynomial when x→0 but isn't differentiable at any point other than 0. Hence h isn't twice differentiable.
– Pablo Lessa
Commented
Nov 9, 2010 at 12:10
exp(−1/x2) is at the heart of all distribution theory.
– Jochen Wengenroth
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Sep 9, 2020 at 9:41
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Since this is community wiki, I'll feel free to share a possibly relevant anecdote; feel free to delete if you don't think this is an answer.
I once had a freshman calculus student ask me if they'd be required to learn the "Greek method" for calculating derivatives. When I looked puzzled, he explained to me that the "Greek method" involved taking a limit as h goes to zero, and that this was the method the ancient Greeks had used back before anyone realized that all you have to do is bring down the exponent and then lower it by one.
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answered Dec 14, 2014 at 23:25
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Steven Landsburg
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Of course the Greek method uses a Δ rather than an h.
– user9072
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Dec 15, 2014 at 6:49
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@quid: Well, of course. The Greeks were very primitive and had not yet discovered the letter h.
– Steven Landsburg
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Dec 15, 2014 at 6:56
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We always learn by students. Not really in the topic, but I need to tell this recent one. A student showed me a couple of exercises on limits he had done (a bit worried). "Is it 1?" "Correct!" I said. "And this one? I got -2..." "Correct too, very good" said I. And he... "But isn't this in contradiction with the principle of uniqueness of the limit?".
– Pietro Majer
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Dec 15, 2014 at 13:36
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This is a question that I also struggle with sometimes. On the one hand, I understand the value of sweeping things under the carpet when students are not ready for them yet. When I learned Calculus in High School, we talked about -- but never properly defined -- limits (I'm can't recall if we did the limit derivatives). Yet, we managed to go pretty far into the material, e.g. establishing recurrence relations for integrals of the type ∫bae−xsin(nx). This lack of definition was a very frustrating point for me, and when I finally learned about (ϵ,δ) two years later, a wave of relief washed over me. Yet, I'm pretty sure that my cohorts did not feel the same way, hence my sympathy for teachers who want to keep things simple by hiding the definition.
At the same time, I don't want my Calc course to be a series of magic tricks, so I always insist on the logical construction of the course: we want to investigate slopes of tangents. We want to work exactly, not approximately. This is why we'll get into limits in the first place (not very historical, but a logical development). So what do I do?
I briefly cover (ϵ,δ) without really applying it. Just to show the difference between a "wordy" definition and a mathematical one.
I insist on the fact that the limit laws are derived fro this rigorous definition. (You can sketch the proof for the sum of limits for instance).
This sets up for students how the mathematical edifice is built: abstract definitions to formalize intuition, big gun theorems proved rigorously from these definition (limit laws, derivative laws...). Add a few examples to the mix and then you're set up for practical, mechanical computations (the stuff that computers do).
I am upfront about the fact that I don't expect my students to use the (ϵ,δ) definition, though I like them to memorize it. The payoff will be later.
I also stress that, historically, calculus was done without this definition for a long time: so it can be done, they will be able to do it, but it also has its limitations when dealing with more abstract material.
In a course that is set up in this way, it is quite natural to cover the limit definition of derivatives. There are a lot of good reasons why one should do that anyway, some of which have already been addressed. Functions which are defined piecewise do require this, and that includes important examples like exp(−1/x2) and fun ones like x2sin(1/x). The rigorous derivation of the derivative of sinx is another good example.
There are also wrong ways of doing this. In the comments, Holger pointed to the case study in the Notices article Teaching mathematics graduate students how to teach. Here, the problem asked to use the definition of derivative to compute the slope of a certain cubic at a point. By the time the exam rolls around, you have easier ways of doing this, so of course the students would feel that this is an arbitrary and confusing question.
[Actually, I took so long to write this that I've been ninja'd by Pietro on this example.]
One example that I have yet to see though is Taylor series: defining the derivative in this way makes it obvious that
f(a+h)≈f(a)+f′(a)h+o(h)
and sets you up for the higher order ones. Yes, you can see that from the graph too, but at that level most of my students have a terrible time reasoning from an abstract graph.
Given how fundamental these ideas are, especially in Physics, I can never stress enough these kinds of relationship in my course.
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edited Sep 27, 2010 at 12:48
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Thierry Zell
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I hope my answer is read as a response to the question asked, rather than as either a defense of or disagreement with the choices the pedagogists (is that a word?) make.
I think one of the main reasons to teach derivatives in terms of the h→0 limit is that it captures the dual notions of "instantaneous velocity" and "slope", which are respectively physical and geometric.
(Ok, now I will mention some personal opinions about teaching calculus. I love physics, and sometimes pretend to be a physicist, so for me the geometric/physical meanings of calculus are very important. So I would love if they were emphasized more. Unfortunately, we do not do enough in introductory calculus classes in that direction, and it is very hard to present functions and ask students to find the slopes of their graphs without essentially teaching them these black-box techniques. So I don't know whether it's worth it: maybe we should just do the algebraic part of calculus --- it's the only thing we tend to test anyway. I also don't really think that MO is the best place to get into that discussion, though, and I don't think that OP intended as such.)
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answered Sep 27, 2010 at 6:29
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Theo Johnson-Freyd
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Interesting answer. I must stress the connections with Physics than I realized, since a student asked me just last week if I taught Physics also. You mention that we tend to only test the algebraic part of calculus; clearly, we should put our money where our mouth is. Because I emphasize the theory, I make sure there are at least some non-algebraic questions on my tests (you can make up some easy ones). If a prof thinks that the audience can only handle the algebraic stuff, I'm ok with it, but then you should be teaching only that.
– Thierry Zell
Commented
Sep 27, 2010 at 12:00
I agree that there should be at least somewhat more physics in calculus courses than their currently is. To do it otherwise seems like another example of the decortication of calculus: the subject takes place in a certain intellectual context and as a response to certain scientific problems. A lot of students get really worried when you talk physics though, since they think you're expecting them to have some outside knowledge that they in fact don't have...
– Pete L. Clark
Commented
Jan 29, 2011 at 22:18
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...The irony of this for me personally is that I haven't taken a physics class since high school, so all the physics I now know is that which comes up when explaining closely related topics in mathematics. For instance, if I am supposed to talk about first order differential equations, I feel compelled to talk at least a little about second order differential equations, especially those of the form F(x) = c x'', because this is the best motivation I know as to what solving differential equations means and why it's important:...
– Pete L. Clark
Commented
Jan 29, 2011 at 22:21
...the equation gives a law governing the behavior of some object over time, and the solution to the equation tells you what the consequences of that law are for the behavior of the object. In particular, my favorite example is m x'' = - kx. To solve this equation without writing down Hooke's Law, a picture of a spring, carefully explaining the physical significance of the minus sign...what a disservice that would be! By making it physical, you give the student a chance to use her intuition: "Well, if I were forced to satisfy this differential equation, what would I do?"
– Pete L. Clark
Commented
Jan 29, 2011 at 22:27
Look at Morse and Feshbach's Methods for Theoretical Physics for an intuitive explanation of the form of the Schrodinger equation for the wave function for a free particle.
– Tom Copeland
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Mar 22, 2016 at 6:13
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So far no one's mentioned (or did I miss it?) that if you make students compute
limw→5w6−56w−5,
then some of them will use L'Hopital's rule to do that, if you don't tell them not to.
Here's an example of something I have students do with the limit definition of the derivative:
They find all sorts of creative ways of getting things wrong when doing this. Here's another:
I think after they've done several like this, they actually do learn what this is for, and that it's not being used as a way to avoid quick and efficient ways of computing derivatives.
But I have them thinking about instantaneous rates of change without using limits on the first day of the course:
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answered Nov 6, 2010 at 23:22
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Michael Hardy
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Fantastic handouts! I think you would get a lot more upvotes on this answer if people clicked on the links.
– Steven Gubkin
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Jan 28, 2011 at 15:20
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Sorry about the broken links. Those were for the duration of the semester. I've reinstated the first one. To be continued.......
– Michael Hardy
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Mar 12, 2011 at 21:36
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OK, for now the one that ends with "13th.pdf" can be considered superseded by this: wnk.hamline.edu/~mjhardy/1170/homework/5th.pdf
– Michael Hardy
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Mar 12, 2011 at 21:38
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THe "handout" referred to there is here: wnk.hamline.edu/~mjhardy/1170/handouts/March.2nd.2011.pdf
– Michael Hardy
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Mar 12, 2011 at 21:38
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All links give 404 errors.
– Julien Puydt
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May 19, 2012 at 13:06
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Well, the definition of derivative is probably one of the best application of the notion of limit, from a didactical point of view. If you define the derivative as a limit process then students who understand it will not miss the geometric flavour: the slope of the tangent line is the limit as h→0 of the slope of the line through (x,f(x)) and (x+h,f(x+h)). I think this is beautiful and relatively simple, once you get the students to think about it for a minute. Plus, it answers the question "When do we agree that the graph of f admit the existence of a tangent line at (x,f(x))"?
Of course one has to keep in mind that for most students the useful thing to learn is how to compute practically a derivative without using the definition but rather applying a collection of rules. Nevertheless I think it is important to give them an idea of where all these rules come from. Think about those students who want to get a a math major? No?
In Italy in the so called "scientific high school", the schools that provide you with the widest and most basic education (you learn a bit of everything) with a focus in math, physics, chemistry perhaps, ecc.. we are taught the limit using the ϵ−δ definition, and the derivative from its definition. This is to say that I think it is possible to have students learn this theoretical aspects of calculus, if high school kids do.
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edited Sep 27, 2010 at 12:29
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Tommaso Centeleghe
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As you probably know, the Italian government is now planning to gradually change the high school teaching programs into three main topics: "Religion", "Use of guns", "Commercials". So the content of the maths programs of the Italian high school is becoming soon a historical topic. If this is the trend, I guess (I hope) Italy itself will become soon a historical topic too :-(
– Pietro Majer
Commented
Sep 27, 2010 at 12:22
I also learnt calculus in high school using the ϵ-δ definition. This was an ordinary public (state-run) high school in the United States.
– Toby Bartels
Commented
Apr 3, 2011 at 21:11
It can create issues for students (due to their typical lack of logic skills in first year) to think about derivatives giving a criterion for the existence of tangent lines. This is because the first thing they think of after a line is a circle, and the tangent to a circle is hard to describe in this way.
– Glen Wheeler
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Oct 21, 2014 at 4:03
@GlenWheeler, as far as the concept of the slope of the tangent line via a limit goes, it works just as well to consider lim(x,y)∈C(x,y)→(x0,y0)x≠x0y−y0x−x0; this makes sense for any curve, though it fails for points on vertical segments.
– LSpice
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Mar 7, 2020 at 22:27
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A problem I like to give students to solve shortly after introducing the derivative is to evaluate f′(2) for f(x)=xx. Of course, this function can be rewritten as f(x)=exlnx but in my experience students don't think of this. In fact, students who have seen Calculus before almost universally reach the solution f′(2)=4 which they get from the mistaken idea that f′(x)=x⋅xx−1=xx. The only students that usually get this problem correct are those that haven't yet learned any of the computational methods and only know the definition.
I teach the limit definition and emphasize the physical and geometric interpretations, and then move from that to the concept of the tangent line and linear approximation. I think these concepts encapsulate most of what is significant (intuitively) about the definition. I dislike exam questions that require students to compute derivatives using the limit definition when they know a "better" way to do it. It isn't too hard to write a problem where no formula for the function is given and ask students questions about the sign or approximate magnitude of the derivative or whether or not the function should even have a derivative. For students who to do not intend to pursue mathematics, this seems appropriate to me. Even those who become mathematicians will almost surely see these ideas again in complete detail in an elementary analysis course.
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answered Nov 7, 2010 at 6:31
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Jeremy West
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How do you differentiate x^x using only the limit definition? This seems like a tall order to me.
– Steven Gubkin
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Nov 11, 2010 at 15:35
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I only meant that they would numerically estimate it at x = 2 using the limit definition. It is easy to estimate using the definition, but if they try to differentiate and plug in 2 they will probably get the wrong answer.
– Jeremy West
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The standard definitions of limit, continuity and derivative are things of
beauty mathematically - flexible and well-honed like fine woodworking
tools. But to get calculus students to care, and appreciate their
meaning and significance, takes some motivation.
A pretty good way to motivate ϵ-δ is that it has to do
with determining what control on input error (δ) is needed to
guarantee meeting a given tolerance for output error (ϵ).
How accurately do you have to aim a spacecraft to ensure it enters
Martian orbit without burning up the way Beagle 2 did, costing hundreds
of millions?
Students can appreciate this is a serious question, and that it is
fair to insist they be able to handle simple examples like f(x)=−100x+50,
ϵ=10−2. (In large lectures for freshmen, I wouldn't do
much more than Lipschitz examples or something carefully designed so δ is easy to find without cases. Many calculus students are
adults but, ahem, need practice with inequalities.)
One can tell engineering students who just want the formulas
that they'll be surprised to find that in a couple of years they'll be
estimating ``sensitivity coefficients'' numerically from
black-box software or experiment. Gee, sensitivity coefficients are just derivatives, and they'll be estimated from the definition, not symbol-pushing.
Speaking of which, it's nice to express the error in the definition
explicitly:
f(x)−f(a)x−a=f′(a)+Ea(x),
and do the algebra that occurs to few to write
f(x)=f(a)+f′(a)(x−a)+Ea(x)(x−a)This makes the nature of linear approximation a bit more apparent.
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answered Nov 6, 2010 at 22:30
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Bob Pego
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One way to avoid limits without losing too much is to teach the calculus of finite differences. Conceptually, the move from numbers to lists-of-numbers as first-class mathematical objects is easier than the move from numbers to real-valued-functions-of-a-real-variable, and the easier move also forms a good stepping stone to the harder one. One can develop the calculus of finite differences mutatis mutandis and thereby make the transition to infinitesimal calculus essentially painless. (So, for example, one should work not with polynomials per se, but with linear combinations involving rising or falling powers).
All the black box rules have their analogues, and all are reasonably easy to see and/or prove. Passing the limit, when it happens, comes as a welcome simplification.
Aside from the conceptual challenge of functions themselves, students find limits difficult because of their quantifier complexity. I have never understood why standard algebra pedagogy suppresses quantifiers, thus, for example, leaving many students unable to distinguish between unknowns (literals bound by existential quantifiers), variables (literals bound by universal quantifiers) and constants (literals that belong to the language itself). Students who miscalculate the derivative of π2, mentioned elsewhere, don't get this distinction. People who become mathematicians usually "got it" without anyone spelling all this out, and then they learned about quantifiers studying logic in college, so they regard quantifiers as sophisticated and advanced. But most students don't "get it," and I think this accounts for the huge attitude downturn when they get to algebra.
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answered May 19, 2012 at 6:02
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David Feldman
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It's funny that actually many students believe that the symbiosis is always the other way around, i.e. derivatives are used to compute limits (l'Hopital etc.). My favorite example of an elegant calculation of a derivative using the limit definition comes from basic physics. Ask the students why the acceleration of an object performing uniform circular motion is always perpendicular to the velocity. One could come up with a non-elegant solution by writing the equations of motion and using a derivatives table, or one could observe the nice geometric proof of considering an infinitesimal isosceles triangle formed by the two velocity vectors that are a few seconds apart and notice that Δ→v is the base of this triangle and points toward the center of the circle.
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answered Sep 27, 2010 at 5:53
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Gjergji Zaimi
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If you use the derivative table for a scalar product (applied to the square product), from v.v=cte, you get 2v.a=0 -- that doesn't look that non-elegant. In fact, I use this very example to show my students (which are more into physics) that this rule gets results they're already well-acquainted with. That convinces them more than the proof...
– Julien Puydt
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May 19, 2012 at 13:16
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sinxx at x=0 should be a good example.
P.S.: Talking about esoteric definitions, if you can introduce stationary point without derivatives, you can then introduce derivatives using sheaves, like you would introduce vectors on a smooth manifold. It would broaden the consciousness of your freshmen, he-he ^_^
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edited Sep 27, 2010 at 7:29
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Aleksei Averchenko
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I believe it gives a good conceptual or practical reasoning to why we would want to study calculus in the first place. As it was first introduced to me, we can always talk about the average speed a car has traveled over a certain distance. Even very small distances, but apart from a speedometer, how can we say 'I'm travelling XX km/h right now'. There enters the limit definition, where we want the instantaneous rate of change!
If we only presented the formal rules for differentiation, we run in to the same problem as high school students who dislike math present "But my calculator can just do it! Why do I need to learn this?!". If the fundamentals are not taught, one day they will be forgotten.
There are certainly other rigorous approaches to the derivative out there. The delta-epsilon method, which most students in their first year simply struggle to grasp as easily as the h→0 definition. This approach is typically reserved for the math majors who go on to take a course in analysis, not the general first calculus course for all science majors.
While I do not use this definition in practice, I am primarily not calculating derivatives, so take that for what it's worth I suppose.
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answered Sep 27, 2010 at 8:07
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Alex
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The answer I give my students is that mathematicians want to know what a word (in this case 'derivative') means in all cases, and the definition of the derivative is a communal agreement about what to say in strange cases such as the absolute value function. (Well, since I banish symbolic stuff from the first two weeks, I say 'function whose graph has a sharp corner like this one (draws on board)'.)
If students press further, I point out that in a literature class they are expected to learn the communal agreement on the difference between a 'simile' and a 'metaphor'. It helps that I am at a liberal arts institution and not a technical one.
Let me also use this opportunity to share a pedagogical trick:
I find it helpful (third time I've tried it) to break up the definition of f′(2) into two parts:
1) Define a new function E2 by the formula E2(x)=f(x)−f(2)x−2.
2) Take the limit of E2 at 2.
To pull this off, you do need to take the function E2 somewhat seriously; graph it, write formulas for it, et c.
Rationale:
1) It always helps to break up complicated definitions into smaller pieces.
2) It emphasizes that you take limits of functions (in the sense of machines that accept a single number as input and gives a single number of output) rather than of symbolic expressions.
3) Students get to really understand why a discontinuous function or something like the absolute value function is not differentiable (at the relevant point).
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answered Sep 28, 2010 at 21:54
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Alexander Woo
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This semester I am teaching Calculus I following Rogawski's textbook. The chapter on limits spends some serious effort on getting comfortable with average rates of change, going as far as creating a table of values (easy to do with a graphing calculator) and numerically estimating the limit, before going on to derivatives in the next chapter devoted to derivatives. By the way, I don't see how this approach could accomplish a worthy goal stated in Rationale 3.
– Victor Protsak
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Sep 29, 2010 at 7:12
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It is worth noting that there is a lot of historical precedent for teaching it as a limit, which occurs already in Euclid. I.e. Euclid characterizes the tangent to a circle as the unique line such that between it and any other line through the same point, one can interpose a secant (Prop. 16, Book III). (Strictly, he says equivalently that one cannot interpose another line between the tangent and the circle itself, i.e. every other line through the point is a secant.) Thus the tangent is the limit of those secants. Thus I believe one can easily say that the limiting point of view is the original one of Euclid. From this point of view, the idea of limit is the one used so fruitfully by the Greeks, and the contribution of the mathematicians of later times is to make that notion more precise.
On the other hand, if you want to avoid the conceptual difficulty students have with limits, you can follow Descartes instead, at least for derivatives of polynomials, and characterize the tangent line as the unique line such that subtracting its equation from the original function gives a polynomial with a double root at the given point. This leads to motivating the Zariski cotangent space, as M/M^2.
Both points of view also have a nice dynamic interpretation as realizing the tangent as the unique line intersecting the curve doubly at the point, understood as the limit of the two secant intersections,and measured by the presence of a double root.
But if you want a defense of limits, I suggest Euclid Prop. 16, Book III as ample precedent.
If you want a defense of making students practice using the limit definition, I propose that as noted above, this is the only way to get them to appreciate the fundamental theorem of calculus. That theorem cannot be appreciated by memorizing rules for derivatives, One must understand the definition and apply it to an abstractly defined area function. I suggest that one reason many students do not understand why the fundamental theorem of calculus is true, is that (again as noted above) they have not grasped either what an abstractly defined function is, nor what a derivative truly means.
So if you want them to understand the relation between the derivative and the integral, then I agree with others that they need to know what a function is and derivative is. The reasoning here is that once someone understands something, he can use it in more settings than could possibly be covered by any set of rules.
Another practical benefit of testing the use of the h-->0 definition to obtain derivatives of simple functions, is that it forces practice in algebra, trig identities, and exponentials, skills which most of my students are almost completely lacking.
However, I recommend you teach it any way that makes sense to you. after all you understand it, so whatever you say based on that understanding will be useful. Make up your mind what seems important to you, and go for it!
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Just to add to this answer, the idea to use limits and functions to teach calculus comes from Felix Klein, at the end of the 1800s. The idea was also suggested by Poincare and Borel when they were revising the school math curriculum of France. Interesting that it goes back to Euclid!
– Manya
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If it is just a question of definition but not a question of computation, I have heard when I was a student the following definition:
Let f be a real continuous function, class C0. Let
Δf(x,x′)=f(x′)−f(x)x′−xdefined onR2−{x=x′}If Δf admits a continuous extension on the diagonal {x=x′} then it is unique, and f is said to be of class C1. The function f′x)=Δf(x,x) is then called the derivative of f
Of course this is the standard definition, nothing new under the sun, but the ϵ−δ calculus if hidden, and of course, not for long :-)
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answered Jan 26, 2011 at 6:28
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Patrick I-Z
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It's an interesting alternative to the usual, but ultimately I'm afraid it's even more challenging than the usual one. Of course, here I'm not talking about using the definition to perform computations, but even to grasp the intuitive meaning, resorting to functions of two variables strikes me as more than a student can comfortably handle.
– Thierry Zell
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Jan 27, 2011 at 16:14
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This also has the technical problem of not covering the case when a function is differentiable but not continuously so. (Of course, this might appear in a context where one is uninterested in such functions.) We can fix this by making it slightly more complicated: ask whether Δf(a,−), the restriction of Δf to a given vertical line {x=a}, admits a continuous extension; if so, then this is unique, and we define f′(a) to be the new value Δf(a,a).
– Toby Bartels
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Apr 3, 2011 at 21:22
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Another alternative way of teaching calculus is via infinitesimals (for example the book Elementary Calculus, An Infintesimal Approach by Keisler). The way of thinking about calculus via infinitesimals is obviously very natural, and mathematicians (e.g. the pioneers of calculus, Euler etc...) have used arguments using infinitesimals long before they should have been really allowed to do so. Keisler's book (and in general the area of `Non-standard analysis') makes rigorous our intuition regarding infinitesimals, and is a set of rules that teach us how to formally reason with them. In my opinion this system is intuitive, but the student can never really have a proper understanding of what they are doing "from the ground up" with out some basic knowledge of model theory. The limit approach is less intuitive, but at least a student doesn't have to just accept some rules without truly understanding what's behind them. Possibly this infinitesimal approach is a half way house between teaching it properly with limits and just teaching rules of differentiation to people who aren't interested.
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Edward Nelson's "IST" version of non-standard analysis is vastly more user-friendly. Alain Robert's book on non-standard analysis takes that viewpoint, and is a marvel of lucidity.
– paul garrett
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14475 | https://historyoftheworlds.com/2024/07/13/zero-sum-game-in-derivatives-markets-the-competition-where-one-investors-gain-is-anothers-loss/ | Zero-Sum Game in Derivatives Markets: The Competition Where One Investor’s Gain is Another’s Loss – The History of the World
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Purpose
Zero-Sum Game in Derivatives Markets: The Competition Where One Investor’s Gain is Another’s Loss
In the realm of financial investments, derivatives markets stand out as a clear illustration of zero-sum games. A zero-sum game is a situation in game theory where one participant’s gain or loss is exactly balanced by the losses or gains of another participant. Essentially, the total wealth in the system remains constant; it is merely redistributed among the players.
Understanding Derivatives
Derivatives are financial instruments whose value is derived from the value of an underlying asset, which can be stocks, bonds, commodities, currencies, interest rates, or market indexes. Common types of derivatives include futures, options, and swaps. These instruments are used for various purposes such as hedging, speculation, and arbitrage.
Zero-Sum Nature of Derivatives
In derivatives markets, every contract has two sides: the buyer (or holder) and the seller (or writer). The payoff structure of derivatives ensures that whatever one party gains, the other party loses. This creates a perfect example of a zero-sum game. Here’s a closer look at how this dynamic works:
Futures Contracts: In a futures contract, one party agrees to buy an asset at a future date for a price agreed upon today, while the other party agrees to sell the asset at that future date for the same price. If the market price of the asset at the future date is higher than the agreed-upon price, the buyer profits and the seller incurs an equivalent loss. Conversely, if the market price is lower, the seller profits and the buyer loses.
Options Contracts: Options give the buyer the right, but not the obligation, to buy or sell an asset at a predetermined price before or at expiration. If a call option (the right to buy) increases in value because the underlying asset’s price rises, the holder of the option gains while the writer of the option, who must sell at the predetermined price, incurs a loss. Similarly, for put options (the right to sell), if the asset’s price falls, the holder of the put option profits while the writer loses.
Swaps: In a swap agreement, two parties exchange cash flows or other financial instruments. For instance, in an interest rate swap, one party might agree to pay a fixed interest rate while receiving a floating interest rate from the other party. The net payoff to one party is exactly the loss of the other, depending on the movement of interest rates.
Strategic Implications
The zero-sum nature of derivatives markets has significant strategic implications for investors. Since one participant’s gain is another’s loss, the markets are highly competitive, and success requires skill, information, and sometimes a bit of luck. Here are some strategic considerations:
Information Advantage: Investors with better information or more accurate market forecasts have a significant advantage in zero-sum games. Sophisticated models, timely data, and market insights can tilt the balance in favor of one party.
Risk Management: While the zero-sum nature suggests a stark win-lose outcome, derivatives are often used for risk management. Hedgers use derivatives to offset potential losses in their portfolios, transferring risk to speculators who are willing to take on that risk for potential profit.
Speculation: Speculators thrive in zero-sum environments, betting on price movements to make profits. Their strategies can range from technical analysis and market trends to more complex quantitative models.
Regulation and Fair Play: Given the zero-sum nature, the derivatives markets are heavily regulated to ensure fair play and to prevent market manipulation, which could unfairly tip the balance in favor of one party.
Conclusion
The competition in derivatives markets epitomizes the concept of a zero-sum game, where the gains of one investor are precisely offset by the losses of another. Understanding this dynamic is crucial for investors as they navigate these markets, employing strategies that leverage information, manage risk, and capitalize on market movements. While the zero-sum nature creates intense competition, it also fosters a rich environment for strategic thinking and financial innovation.
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14476 | https://www.popularmechanics.com/science/math/a30706968/viral-triangle-brain-teaser-solved/ | We use technologies that provide information about your interactions with this site to others for functionality, analytics, targeted advertising, and other uses. Learn more in our Privacy Notice.
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We Spent All Day Arguing About This Triangle Brain Teaser. Can You Solve It?
We Spent All Day Arguing About This Triangle Brain Teaser. Can You Solve It?
Mathematicians reveal the real answer. See if you’re right.
By Andrew Daniels
Gear-obsessed editors choose every product we review. We may earn commission if you buy from a link. Why Trust Us?
There’s nothing quite like a maddening math problem, mind-bending optical illusion, or twisty logic puzzle to halt all productivity in the Popular Mechanics office. We’re curious people by nature, but we also collectively share a stubborn insistence that we’re right, dammit, and so we tend to throw work by the wayside whenever we come upon a problem with several seemingly possible solutions.
This triangle brain teaser isn’t new—shoutout to Popsugar for unearthing it a couple years ago—but based on some shady Internet magic, the tweet below reappeared in my feed today and kick-started a new debate on our staff-wide Slack channel, a place traditionally reserved for workshopping ideas, but instead mostly used for yelling about other stuff that we occasionally turn into content.
View full post on X
Because I’m a masochist, I drew the triangle again and asked everyone on staff to promptly drop what they were doing and attempt to solve the simple question: How many triangles can you find?
I’ll spare you the full conversation—trust me, nobody wants to see that—but the team’s responses ranged all over the place. Some editors saw four triangles. Others saw 12. A few saw 6, 16, 22. Even more saw 18. One wiseguy counted the triangles in the A’s in the question itself, while another seemed to be having an existential crisis: “None of these lines are truly straight, just curves—thus you cannot define any of them as a triangle,” he said. “There are no triangles in this photo. Life has no meaning.”
We then posed the problem to our Instagram followers, whose replies also ran the gamut, from 5 to 14 to 37. While we acknowledge the high probability of trolling here, it’s clear that people respond to the problem many different ways.
View full post on Instagram
I could’ve listened to my colleagues explain their questionable processes all day, but instead, I reached out to several geometry experts to see if we could arrive at a consensus answer. Turns out virtually all of the mathematicians I contacted found the same solution—but not all of them figured it out in the same way.
If you don’t want to know the answer just yet, stop reading and try to solve the problem first. I’ll meet you back here when you’re done.
Hey, that was quick. Ready for the answer? Unlike some viral math problems that are purposely vague and open for interpretation, this one actually does have a slam-dunk, no-doubt-about-it solution, and it’s 18. Let’s hear from some of the geometry experts as to why.
“I would approach this just like one approaches any mathematical problem: reduce it and find structure,” says Sylvester Eriksson-Bique, Ph.D., a postdoctoral fellow with the University of California Los Angeles’s math department.
The only way to form triangles in the figure I drew, Erikkson-Bisque says, is if the top vertex (corner) is part of the triangle. The base of the triangle will then have to be one of the three levels below. “There are three levels, and on each you can choose a base among six different ways. This gives 18, or 3 times 6 triangles.”
Let’s look at the master triangle again.
Andrew Daniels
“It’s convenient to generalize to the case where there are n lines passing through the top vertex, and p horizontal lines,” says Francis Bonahon, Ph.D., a professor of mathematics at the University of Southern California.
In our case, n = 4, and p = 3. Any triangle we find in the drawing should have one top vertex and two others on the same horizontal line, so for each horizontal line, the number of triangles with two vertices on that line is equal to the number of ways we can choose these vertices, Bonahon says—namely the number of ways we can choose two distinct points out of n, or “n choose 2.”
Remember high school math? That’s n(n-1)/2. And since there are p horizontal lines, says Bonahan, this gives p n(n-1)/2 possible triangles. In our case, that’s 3x4(4-1)/ 2=18.
Here’s a handy breakdown of how to find each possible triangle:
Kory Kennedy
Johanna Mangahas, Ph.D., an assistant math professor at the University at Buffalo, also came to 18—first through simple brute-force counting, then through the same crafty combinatorics as above—but admits our triangle brain teaser isn’t quite as cool as this one from Po-Shen Loh, Ph.D., a math professor at Carnegie Mellon University in Pittsburgh, as featured in the New York Times last year:
Po-Shen Loh
This one has a slicker mathematical answer, she says, because here, counting triangles is the same thing as counting combinations of three lines chosen out of six [6-choose-3 = (654)/(321)].
“In that case, every pair of lines intersects and there are no triple-or-more intersections, so any choice of three always gives a triangle,” says Mangahas. In the picture I sent her, some lines are parallel, so they can’t be part of the same triangle. “If you took the same seven lines and shook them up a bit, probabilistically they’d most likely land like [Loh’s] problem and you’d have more triangles and a similar cute answer.” (For the record: 35.)
Whew. I haven’t shared this new triangle problem with my coworkers yet. But it’s only a matter of time before they discover it—and argue some more.
🚨IMPORTANT UPDATE 1/30/20🚨: Since publishing this story, many, many readers have reached out to let me know that while 18 is indeed an acceptable answer to this problem, it isn’t the only one, due to some unintentional oversight on my part. I could have made this much easier on readers—and, crucially, much easier on my inbox—had I just sketched the triangle on plain, white computer paper. But no.
I unfortunately drew this triangle on lined paper, and lots of smart people have correctly pointed out that, well, actually, if you count the light blue parallel lines in the image in addition to the dark blue lines written in marker, there are actually more than 18 total triangles here—considerably more. I never specified to only use those dark blue lines, and thus, I am wrong. You are right.
One reader, Ralph Linsangan, totally owned me by sending this image, in which he marks every additional triangle found under the technicality, flagging 17 additional triangles for a total of 35. Behold:
Ralph Linsangan
That kind of dedication is just one of many reasons I love Popular Mechanics readers. We can’t get anything past you guys. Until the next teaser!
🚨YET ANOTHER TRIANGLE UPDATE 1/31/20🚨: Since posting the last update, I’ve heard from even more of you, continuing to chide me—and your fellow readers—for not considering additional possible triangles. Let’s hear from reader Derek Schneider, who sent in another graphic suggesting there are 45 triangles.
If we follow the original rules however, I count and additional 9 that are definite (in green) and one that could be open to interpretation depending on how you visually place the top vertex (in purple)…I would personally count it.
Derek Schneider
Reader Poingly, meanwhile, wrote in to say we’ve been making a “grave error” in counting the triangles all along:
Take the bottom right corner, for instance, it shows one arrow for one triangle. However, these light blue lines could conceivably form as many as THREE triangles in this one corner alone:
Poingly
While some of these MAY be somewhat debatable (ie, where EXACTLY do the light blue lines intersect the dark ones and do they technically form a triangle or a quadrilateral), I have counted SEVEN ADDITIONAL triangles that may be made in this way. This bring the total number of triangles up to 42. The bad news is that we missed some triangles. The good news is that this confirms that life clearly DOES have meaning, as evidenced by the exact number: 42.
Outstanding point, Poingly. Reader James Goodrich took it another step further, suggesting we open our minds to consider what a triangle could be:
Well, according to your reader, who pointed out 17 additional triangles (using the "Andrew didn't specify what lines can comprise the 3 edges of a triangle" clause), failed to clearly find quite a lot more. Take, for example, the bottom-left mini-triangle in the 30 Jan 2020 "Important Update" addendum. Would not the areas of the mini-triangle and the area of the rhombus adjacent to it, combined, make for another triangle?
Another idea for consideration: Triangles have 3 angles (who would have guessed?); however, I would postulate that how you describe a triangle, by way of said angles, would generate different triangles. Given a triangle T, with vertices A, B, and C, t-one might indeed be described by ABC, with B being the central angle. I suggest that t-two, being described by BAC, is different. Similarly for BCA.
If we then take a particular case, right-angle triangles, we can derive sine, cosine, and tangent functions (SOH, CAH, TOA). If we were to apply that to the triangle (and relax the right-angle requirement, it might mean that BAC is different than CAB. Of course, exceptions are made for isoscolese and equilateral triangles (the latter would only have 3 distinct triangle definitions).
I haven't quite thought of how to quantify each suggestion (and applying the latter after the former would increase the count still), so I don't have an easy number for you to use in an updated important update (if you found my ideas worthwhile to update).
I did, James. And I’ll be waiting. Begrudgingly, I decided to take one last stab at figuring out just how many additional triangles there could be given our new chaotic rules, and arrived at 43, for a total of 61:
Andrew Daniels
I’m quite sure, however, that someone reading this will very quickly tell me I’m wrong yet again and deliver proof of even more hidden triangles, sending me down another rabbit hole on the long and winding path to eventual insanity. (Side note: I haven’t seen my wife in three days. Please tell her I love her.) So I’m issuing one last challenge: If you can find the most possible triangles in the original image, show me your work, and definitively prove your supremacy, I will update this story one final time and crown you the Triangle King or Queen, now and forever. Godspeed.
4 More Puzzles to Try to Solve
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${a_1, a_2, a_3, ...}$ is a non-constant arithmetic sequence, while both $a_1, a_2, a_6$ and $a_1, a_4$, an are finite geometric sequences. Find $n$
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$\begingroup$
Hello fellow math lovers,
This question is a fairly trivial one from the New Jersey Undergraduate Mathematics competition. I'm having trouble understanding the solution and I'm fairly frustrated because it seems, on the surface, so simple.
Problem: The sequence $a_1, a_2, a_3, \dotsc$ is a non-constant arithmetic sequence, while both $a_1, a_2, a_6$ and $a_1, a_4, a_n$ are finite geometric series. Find $n$.
Solution: Let $a_k = a + (k-1)d$ for all $k$. The fact that $a_1, a_2, a_6$ is a geometric sequence tells us that $$ (a + d)^2 = a_2^2 = a_1 a_6 = a(a + 5d). $$ From this it follows easily that $d = 3a$. Hence, $a_4 = a + 3d = 10a$ and $a_n = a + (n-1)(3a) = (3n-2)a$. Since $a_1, a_4, a_n$ is also a geometric sequence, we must have $$ (10a)^2 = a \cdot (3n - 2)a \longrightarrow 100 = 3n - 2 \longrightarrow n = 34. $$
(Original image here.)
My question is in regards to the squaring the second term of the geometric sequence step. How is that equal to $a_1 a_6$ and why was it important to do this to being with? The rest is very clear and easy to follow but the crux move is leaving me at a loss. If there is an easier way to understand and solve this problem please let me know.
-Ernie
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edited Aug 26, 2018 at 11:07
Noa Even
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$\begingroup$ It is the Geometric Mean of $a_1$ and $a_6$. So $a_2^2=a_1 \cdot a_6$ $\endgroup$
prog_SAHIL
– prog_SAHIL
2018-08-26 08:12:34 +00:00
Commented Aug 26, 2018 at 8:12
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$$a_2 = a_1 r$$ $$a_6 = a_2 r$$
$$r = \frac{a_2}{a_1}=\frac{a_6}{a_2}$$
Hence we have $$a_2^2=a_1a_6$$
It enables us to find a relationship between $a$ and $d$.
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answered Aug 26, 2018 at 7:52
Siong Thye GohSiong Thye Goh
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We know $a_1=a, $ and $a_2=a+d$. Hence $a^2=(a+d)^2$
If you imagine an 'extract' from a geometric sequence $$\dots,ar^{n-1},ar^n,ar^{n+1},\dots$$ you can see that for any three terms in a geometric sequence, the square of the middle term is equal to the product of the two adjacent terms.
This is important in the context of your solution because it establishes the value of $d$ in relation to $a$, since $(a+d)^2=a(a+5d)\implies d=3a$. Thus we know that $(a_1,a_4,a_n)=(a,10a,a_n)$ and thus $a_n=100a$. From here, we can work out the value of $n$.
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answered Aug 26, 2018 at 7:56
user574848user574848
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$\begingroup$ Thank you friend. This is detailed. $\endgroup$
Maths2020
– Maths2020
2018-08-27 00:58:56 +00:00
Commented Aug 27, 2018 at 0:58
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Given three consecutive members of a geometric sequence, the square of the middle is equal to the product of the other two (Prove this by writing all three values in terms of the first member).
This fact was used to find a connection between a1 and d.
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answered Aug 26, 2018 at 8:07
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14478 | https://www.gastrojournal.org/article/S0016-5085(25)00901-1/fulltext | Canadian Association of Gastroenterology Clinical Practice Guideline for the Endoscopic Management of Nonvariceal Nonpeptic Ulcer Upper Gastrointestinal Bleeding - Gastroenterology
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GuidelinesVolume 169, Issue 5p863-891 October 2025 Open access
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Canadian Association of Gastroenterology Clinical Practice Guideline for the Endoscopic Management of Nonvariceal Nonpeptic Ulcer Upper Gastrointestinal Bleeding
Alan N.Barkun
Alan N.Barkun
Correspondence
Alan N. Barkun, MD, CM, FRCPC, FACP, FACG, AGAF, MSc (Clinical Epidemiology), McGill University Health Centre, McGill University, 1650 Cedar Avenue, D7.346, Montreal, Quebec, Canada, H3G 1A4.
alan.barkun@muhc.mcgill.ca
Footnotes
∗
Authors share co-first authorship.
Affiliations
Division of Gastroenterology and Hepatology, McGill University Health Centre, McGill University, Montreal, Quebec, Canada
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1,∗alan.barkun@muhc.mcgill.ca ∙ Loren Laine
Loren Laine
Footnotes
∗
Authors share co-first authorship.
Affiliations
Section of Digestive Diseases, Yale School of Medicine, New Haven, Connecticut; Veterans Affairs Connecticut Healthcare System, West Haven, Connecticut
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2,∗ ∙ Grigorios I.Leontiadis
Grigorios I.Leontiadis
Affiliations
Department of Medicine and Farncombe Family Digestive Health Research Institute, McMaster University, Hamilton, Ontario, Canada
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3 ∙ … ∙ Ian M.Gralnek
Ian M.Gralnek
Affiliations
Institute of Gastroenterology and Hepatology, Emek Medical Center, Afula, Israel
Rappaport Family Faculty of Medicine, Technion Israel Institute of Technology, Haifa, Israel
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4,5 ∙ Nicholas Carman
Nicholas Carman
Affiliations
SickKids Inflammatory Bowel Disease Centre, Division of Gastroenterology, Hepatology and Nutrition, The Hospital for Sick Children (SickKids), Toronto, Ontario, Canada
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6 ∙ Mostafa Ibrahim
Mostafa Ibrahim
Affiliations
Department of Gastroenterology and Hepatology, Theodor Bilharz Research Institute, Cairo, Egypt
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7 ∙ Michael Sey
Michael Sey
Affiliations
Lawson Health Research Institute, London Health Sciences Centre, London, Ontario, Canada
Division of Gastroenterology, Western University, London, Ontario, Canada
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8,9 ∙ Ali A.Alali
Ali A.Alali
Affiliations
Department of Medicine, Faculty of Medicine, Kuwait University, Jabriya, Kuwait
Thunayan Alghanim Gastroenterology Center, Amiri Hospital, Sharq, Kuwait
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10,11 ∙ Matthew W.Carroll
Matthew W.Carroll
Affiliations
Division of Pediatric Gastroenterology, Hepatology and Nutrition, Department of Pediatrics, University of Alberta, Edmonton, Alberta, Canada
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12 ∙ Lawrence Hookey
Lawrence Hookey
Affiliations
Division of Gastroenterology, Department of Medicine, Queen's University, Kingston, Ontario, Canada
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13 ∙ Mark Borgaonkar
Mark Borgaonkar
Affiliations
Department of Medicine, Memorial University, St Johns, Newfoundland, Canada
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14 ∙ David Armstrong
David Armstrong
Affiliations
Department of Medicine and Farncombe Family Digestive Health Research Institute, McMaster University, Hamilton, Ontario, Canada
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3 ∙ James Y.W.Lau
James Y.W.Lau
Affiliations
Department of Surgery, The Chinese University of Hong Kong, New Territories, Hong Kong
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15 ∙ Nauzer Forbes
Nauzer Forbes
Affiliations
Department of Medicine, University of Calgary, Calgary, Canada
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16 ∙ Rapat Pittayanon
Rapat Pittayanon
Affiliations
Department of Medicine, Faculty of Medicine, Chulalongkorn University, Bangkok, Thailand
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17 ∙ Frances Tse
Frances Tse
Correspondence
Correspondence Address correspondence to: Frances Tse, MD, MPH, FRCPC, AGAF, CAGF, Division of Gastroenterology, Department of Medicine, Farncombe Family Digestive Health Research Institute, Hamilton Health Sciences Centre, McMaster University, 1280 Main Street West, Suite 2F53, Hamilton, Ontario, Canada, L8S4K1.
tsef@mcmaster.ca
Affiliations
Department of Medicine and Farncombe Family Digestive Health Research Institute, McMaster University, Hamilton, Ontario, Canada
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3tsef@mcmaster.ca … Show more Show less
Affiliations & Notes Article Info
1 Division of Gastroenterology and Hepatology, McGill University Health Centre, McGill University, Montreal, Quebec, Canada
2 Section of Digestive Diseases, Yale School of Medicine, New Haven, Connecticut; Veterans Affairs Connecticut Healthcare System, West Haven, Connecticut
3 Department of Medicine and Farncombe Family Digestive Health Research Institute, McMaster University, Hamilton, Ontario, Canada
4 Institute of Gastroenterology and Hepatology, Emek Medical Center, Afula, Israel
5 Rappaport Family Faculty of Medicine, Technion Israel Institute of Technology, Haifa, Israel
6 SickKids Inflammatory Bowel Disease Centre, Division of Gastroenterology, Hepatology and Nutrition, The Hospital for Sick Children (SickKids), Toronto, Ontario, Canada
7 Department of Gastroenterology and Hepatology, Theodor Bilharz Research Institute, Cairo, Egypt
8 Lawson Health Research Institute, London Health Sciences Centre, London, Ontario, Canada
9 Division of Gastroenterology, Western University, London, Ontario, Canada
10 Department of Medicine, Faculty of Medicine, Kuwait University, Jabriya, Kuwait
11 Thunayan Alghanim Gastroenterology Center, Amiri Hospital, Sharq, Kuwait
12 Division of Pediatric Gastroenterology, Hepatology and Nutrition, Department of Pediatrics, University of Alberta, Edmonton, Alberta, Canada
13 Division of Gastroenterology, Department of Medicine, Queen's University, Kingston, Ontario, Canada
14 Department of Medicine, Memorial University, St Johns, Newfoundland, Canada
15 Department of Surgery, The Chinese University of Hong Kong, New Territories, Hong Kong
16 Department of Medicine, University of Calgary, Calgary, Canada
17 Department of Medicine, Faculty of Medicine, Chulalongkorn University, Bangkok, Thailand
∗
Authors share co-first authorship.
Publication History:
Received February 28, 2025; Accepted April 29, 2025; Published online August 8, 2025
Footnotes:
Conflicts of interest The conflicts of interest of all participants were managed according to the CAG policies ( which are based on the recommendations of the Institute of Medicine (now the National Academy of Medicine) and the Guidelines International Network. Detailed conflict of interest management methods are in Appendix 1.
Funding The Canadian Association of Gastroenterology (CAG), a nonprofit medical specialty society representing gastroenterologists, entirely funded the development of these guidelines. CAG staff members facilitated panel appointments and coordinated meetings, but had no role in selecting guideline questions or formulating recommendations. The panellists volunteered their expertise without receiving any payments and the methodologists received funding support from the CAG.
DOI: 10.1053/j.gastro.2025.04.041 External LinkAlso available on ScienceDirect External Link
Copyright: © 2025 The Author(s). Published by Elsevier Inc. on behalf of the AGA Institute.
User License: Creative Commons Attribution (CC BY 4.0) | Elsevier's open access license policy
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Abstract
Keywords
Abbreviations used in this paper
Interpretation of Strong and Conditional Recommendations
Recommendations
Values and Preferences
Explanations and Other Considerations
Aim of This Guideline and Specific Objectives
Description of the Health Problems
Methods
Recommendations
What Are Others Saying and What Is New in This Guideline?
Revision or Adaptation of This Guideline
Acknowledgments
Supplementary Material (6)
References
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Abstract
Keywords
Abbreviations used in this paper
Interpretation of Strong and Conditional Recommendations
Recommendations
Values and Preferences
Explanations and Other Considerations
Aim of This Guideline and Specific Objectives
Description of the Health Problems
Methods
Recommendations
What Are Others Saying and What Is New in This Guideline?
Revision or Adaptation of This Guideline
Acknowledgments
Supplementary Material (6)
References
Article metrics
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Abstract
Background & Aims
Nonvariceal, nonpeptic ulcer bleeding, arising from etiologies such as malignant tumors, Mallory-Weiss tears (MWTs), Dieulafoy’s lesions, and gastric antral vascular ectasia, constitutes a significant and increasing proportion of upper gastrointestinal bleeding cases. These evidence-based guidelines, developed by the Canadian Association of Gastroenterology with international collaboration, are the first to specifically address the endoscopic management of these conditions, aiming to support patients, clinicians, and others in making informed decisions.
Methods
The Canadian Association of Gastroenterology formed a guideline panel with a balanced representation to minimize potential bias from conflicts of interest. The Cochrane Gut Group supported the guideline-development process, including conducting literature searches and performing systematic reviews. The panel prioritized clinical questions and outcomes according to their importance for clinicians and adult patients. The Grading of Recommendations Assessment, Development and Evaluation approach was used, including developing the Grading of Recommendations Assessment, Development and Evaluation Evidence-to-Decision frameworks, which underwent public comment.
Results
The panel formulated 19 conditional recommendations for adult patients with nonvariceal, nonpeptic ulcer bleeding due to malignant tumors, MWTs, Dieulafoy’s lesions, and gastric antral vascular ectasia.
Conclusions
For patients with active bleeding from malignant tumors, the panel suggested topical hemostatic agents over conventional endoscopic hemostatic therapy; it also suggested the administration of oncologic therapy after the endoscopic intervention. In patients with active bleeding from MWTs (oozing and spurting), the panel suggested endoscopic band ligation or endoscopic through-the-scope clip over epinephrine injection alone. For nonbleeding MWTs with visible vessels, adherent clots, flat pigmented spots, or clean-based ulcers, the panel suggested against endoscopic hemostatic therapy. For Dieulafoy’s lesions, the panel suggested mechanical modalities with endoscopic band ligation or through-the-scope clip, contact thermocoagulation, or injection of sclerosants over epinephrine injection alone. For patients with gastric antral vascular ectasia, the panel suggested endoscopic band ligation over argon plasma coagulation.
Guideline Endorsement
This guideline has been formally endorsed by leading international endoscopy societies: the American Society for Gastrointestinal Endoscopy, the European Society of Gastrointestinal Endoscopy, the Sociedad Interamericana de Endoscopía Digestiva, and the World Endoscopy Organization, as well as by the American Gastroenterological Association.
Keywords
Gastrointestinal Neoplasms
Mallory-Weiss Syndrome
Dieulafoy’s Lesions
Gastric Antral Vascular Ectasia
Gastrointestinal Hemorrhage
Practice Guidelines
GRADE
Equity
Abbreviations used in this paper
APC (argon plasma coagulation)
ARR (absolute risk reduction)
CAG (Canadian Association of Gastroenterology)
DEP (Doppler endoscopic probe)
DL (Dieulafoy’s lesion)
EBL (endoscopic band ligation)
EtD (evidence to decision)
GAVE (gastric antral vascular ectasia)
GI (gastrointestinal)
GRADE (Grading of Recommendations Assessment, Development and Evaluation)
Hgb (hemoglobin)
HR (hazard ratio)
MD (mean difference)
MWT (Mallory-Weiss tear)
NVNPUB (nonvariceal nonpeptic ulcer bleeding)
PICO (patient population, intervention, comparator, outcome)
PUB (peptic ulcer bleeding)
RCT (randomized controlled trial)
RD (risk difference)
RFA (radiofrequency ablation)
RR (risk ratio)
UGI (upper gastrointestinal)
UGIB (upper gastrointestinal bleeding)
THA (topical hemostatic agent)
TTSC (through-the-scope clip)
Upper gastrointestinal bleeding (UGIB) accounts for more than 300,000 hospital admissions annually in the United States.1
1.
Peery, A.F. ∙ Crockett, S.D. ∙ Murphy, C.C. ...
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Although peptic ulcer disease causes 30%–50% of UGIB cases, there has been a notable rise in nonvariceal nonpeptic ulcer bleeding (NVNPUB), accounting for more than one-third to two-thirds of UGIB cases and, sometimes surpassing peptic ulcer bleeding (PUB).2–7
2.
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NVNPUB can result from various causes, including malignant tumors, Mallory-Weiss tears (MWTs), Dieulafoy’s lesions (DLs), gastric antral vascular ectasia (GAVE), esophagitis, gastritis, duodenitis, and other vascular lesions. Recent US data indicate a 30% decrease in hospitalizations for PUB, while hospitalizations for UGIB due to malignancy, DLs, and angiodysplasia increased by 50%, 33%, and 32%, respectively.7
7.
Wuerth, B.A. ∙ Rockey, D.C.
Changing epidemiology of upper gastrointestinal hemorrhage in the last decade: a nationwide analysis
Dig Dis Sci. 2018; 63:1286-1293
Crossref
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PubMed
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Endoscopic hemostatic interventions play a crucial role in managing NVNPUB. Most previous guidelines on nonvariceal UGIB primarily focused on PUB, often overlooking or grouping other causes with peptic ulcers.8–12
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This guideline is the first to specifically address endoscopic management of NVNPUB, providing health care professionals with evidence-based strategies that address the distinct challenges posed by NVNPUB. This guideline was based on original systematic reviews of evidence and was conducted under the auspices of the Canadian Association of Gastroenterology (CAG) with international collaborators. The panel followed best practices for guideline development recommended by the Institute of Medicine and the Guidelines International Network.13–16
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The panel used the Grading of Recommendations Assessment, Development and Evaluation (GRADE) approach to assess the certainty of the evidence and formulate recommendations.17–23
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Interpretation of Strong and Conditional Recommendations
The strength of a recommendation is categorized as either strong, indicated by the phrase “the guideline panel recommends ... ” or conditional, indicated by the phrase “the guideline panel suggests . . .” Table 1 provides GRADE’s interpretation of strong and conditional recommendations by patients, clinicians, health care policy makers, and researchers.
| Implications for | Strong recommendation | Conditional recommendation |
---
| Patients | Most individuals in this situation would want the recommended course of action, and only a small proportion would not. | The majority of individuals in this situation would want the suggested course of action, but many would not. |
| Clinicians | Most individuals should receive the intervention. Formal decision aids are not likely to be needed to help individual patients make decisions consistent with their values and preferences. | Recognize that different choices will be appropriate for individual patients and that you must help each patient arrive at a management decision consistent with his or her values and preferences. Decision aids may be useful in helping individuals to make decisions consistent with their values and preferences. |
| Policy makers | The recommendation can be adopted as policy in most situations. Adherence to this recommendation, according to the guideline, could be used as a quality criterion or performance indicator. | Policy making will require substantial debate and involvement of various stakeholders. Performance measures should assess whether decision making is appropriate. |
| Researchers | The recommendation is supported by credible research or other convincing judgments that make additional research unlikely to alter the recommendation. On occasion, a strong recommendation is based on low or very low certainty in the evidence. In such instances, further research may provide important information that alters the recommendations. | This recommendation will likely be strengthened (for future updates or adaptation) by additional research. An evaluation of the conditions and criteria (and the related judgments, research evidence, and additional considerations) that determined the conditional recommendation will help identify possible research gaps. |
Table 1
Interpretation of Strong and Conditional Recommendations
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Recommendations
The list of recommendations is provided in Table 2.
| Recommendation | Strength of recommendation and certainty of evidence |
--- |
| Malignant UGIB | |
| 1A | In patients with active bleeding from malignant UGI tumors, we suggest conventional endoscopic hemostatic therapy (eg, injection, thermal devices, mechanical devices, or a combination thereof) over no endoscopic hemostatic therapy (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝). |
| 1B | In patients with active bleeding from malignant UGI tumors, we suggest THAs over no endoscopic hemostatic therapy (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝). |
| 2 | In patients with active bleeding from malignant UGI tumors, we were unable to reach a recommendation for or against any specific type of conventional endoscopic hemostatic therapy over another. |
| 3 | In patients with active bleeding from malignant UGI tumors, we suggest THAs over conventional endoscopic hemostatic therapy (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝). |
| 4 | In patients with active bleeding from malignant UGI tumors, we suggest administering oncologic therapy after endoscopic hemostatic therapy rather than not providing oncologic therapy after endoscopic hemostatic therapy (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝). |
| MWTs | |
| 5 | In patients with active bleeding from MWTs (spurting or oozing), we suggest endoscopic hemostatic therapy over no endoscopic hemostatic therapy (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝). |
| 6A | In patients with active bleeding from MWTs (spurting or oozing), we suggest EBL or endoscopic TTSC placement over epinephrine injection alone (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝). |
| 6B | In patients with active bleeding from MWTs (spurting or oozing), we suggest EBL or endoscopic TTSC placement (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝). |
| 7A | In patients with MWTs with nonbleeding visible vessels, we suggest against endoscopic hemostatic therapy (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝). |
| 7B | In patients with MWTs with nonbleeding adherent clots, we suggest against endoscopic hemostatic therapy (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝). |
| 7C | In patients with MWTs with nonbleeding clean-based ulcers or flat pigmented spots, we suggest against endoscopic hemostatic therapy (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝). |
| DLs | |
| 8A | In patients with UGIB from DL, we suggest either EBL with or without epinephrine injection or endoscopic TTSC placement with or without epinephrine injection (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝). |
| 8B | In patients with UGIB from DL, we suggest either mechanical devices (EBL or endoscopic TTSC placement) with or without epinephrine injection or contact thermal devices (heater probe and bipolar electrocoagulation) with or without epinephrine injection (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝). |
| 8C | In patients with UGIB from DL, we cannot make a recommendation for or against the use of cap-mounted clips over conventional endoscopic hemostatic therapy. |
| 8D | In patients with UGIB from DL, we suggest either mechanical devices (EBL or endoscopic TTSC placement) with or without epinephrine injection or injection of sclerosants with or without epinephrine injection (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝). |
| 9A | In patients with upper GI bleeding from DL, we suggest against epinephrine injection alone over mechanical devices (EBL or endoscopic TTSC placement) (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝). |
| 9B | In patients with UGIB from DL, we suggest against epinephrine injection alone over thermal devices (heater probe, bipolar or multipolar electrocoagulation) (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝). |
| GAVE | |
| 10 | In patients with GAVE, we suggest against RFA over APC (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝). |
| 11 | In patients with GAVE, we suggest EBL over APC (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝). |
Table 2
List of Recommendations With Strength of Recommendation and Certainty of Evidence
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Values and Preferences
The guideline panel rated further bleeding as critical for decision making regarding malignant UGIB, MWTs, and DLs. This composite outcome includes the failure to achieve immediate hemostasis and any rebleeding. For GAVE, the panel rated the number of units of blood transfusions needed and changes in hemoglobin (Hgb) levels as critical for decision making, as this condition typically results in anemia related to chronic blood loss rather than acute bleeding. These outcomes are highly valued, and a strong emphasis is placed on interventions that can effectively address them.
Explanations and Other Considerations
These recommendations also consider cost and cost-effectiveness, impact on health equity, acceptability, and feasibility.
Aim of This Guideline and Specific Objectives
This guideline aims to provide evidence-based recommendations for the endoscopic management of adults with NVNPUB, with a focus on malignant UGIB, MWTs, DLs, and GAVE. Other causes of NVNPUB have been excluded from this iteration. The target audience includes health care providers managing UGIB, patients, and decision makers. Policy makers involved in developing local, national, or international programs for optimizing the management of NVNPUB will also find these guidelines valuable. This document can also serve as a basis for adaptation by local, regional, or national guideline panels.
Description of the Health Problems
Malignant UGIB is associated with high morbidity and mortality.3
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Methods
The guideline panel developed and graded the recommendations and assessed the certainty of the evidence following the GRADE approach.21
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The overall guideline-development process, including funding, panel formation, management of conflicts of interest, internal and external review, and organizational approval, was guided by CAG policies and procedures derived from the Guidelines International Network–McMaster Guideline Development Checklist and was intended to meet recommendations for trustworthy guidelines by Institute of Medicine and the Guidelines International Network.13–16
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Organization, Panel Composition, Planning, and Coordination
The CAG coordinated the work of this panel, with project oversight managed by a steering committee (A.B., L.L., G.L., F.T.) reporting to the CAG Guideline Committee. The CAG was also responsible for selecting and appointing members to the guideline panel and vetting and retaining methodologists (F.T., N.F., M.C., N.C.) to conduct systematic reviews and coordinate the guideline development process using the GRADE approach. Details on panel membership are provided in Appendix 1.
The international panel included gastroenterologists and a surgeon with clinical and research expertise on the guideline topic from Canada, the United States, Hong Kong, Thailand, Israel, Egypt, and Kuwait. It also included methodologists with expertise in evidence appraisal and guideline development. One of the panel co-chairs (A.B.) was a content expert and the other (F.T.) was an expert in guideline-development methodology. The panel also incorporated input from 2 patient representatives who reviewed and provided feedback on the patient population, intervention, comparator, outcome (PICO) questions and the recommendations. The panel’s activities were facilitated through a combination of web-based tools ( and www.gradepro.org) and online meetings.
Guideline Funding and Management of Conflicts of Interest
The development of these guidelines was funded by the CAG, a nonprofit medical specialty society representing gastroenterologists. CAG staff members facilitated panel appointments and meetings, but did not participate in selecting guideline questions or formulating recommendations. The panelists volunteered their expertise and the methodologists received funding support from the CAG. Conflicts of interests for all participants were managed according to CAG policies, following the recommendations of the Institute of Medicine and the Guidelines International Network.50
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Before appointment to the panel, individuals disclosed financial and nonfinancial interests, which were reviewed by the CAG Guideline Committee to identify and manage potential conflicts, as detailed in Appendix 1.
Recusal was used to manage conflicts of interest. Panel members with a current direct financial interest in commercial entities that could be impacted by the guidelines were recused from making judgments about relevant recommendations. The evidence-to-decision (EtD) framework for each recommendation specifies which individuals were recused. Methodologists with material interest in commercial entities potentially impacted by the guidelines were recused from performing systematic reviews or rating the certainty of evidence related to those products.
Selection of Questions and Outcomes of Interest
The steering committee brainstormed the questions, which the panel prioritized, as outlined in Table 3. The questions address all types of endoscopic therapies with published studies relevant to the conditions in question, including randomized controlled trials (RCTs) and observational studies. Some questions have been divided into sub-questions to compare different modalities.
| Question no. | Question |
--- |
| Malignant UGIB | |
| 1 | Should patients with active bleeding from malignant UGI tumors receive endoscopic hemostatic therapy vs no endoscopic hemostatic therapy? |
| 2 | Should patients with active bleeding from malignant UGI tumors receive one conventional endoscopic hemostatic therapy vs another conventional endoscopic hemostatic therapy? |
| 3 | Should patients with active bleeding from malignant UGI tumors receive THAs vs conventional endoscopic hemostatic therapies? |
| 4 | Should patients with active bleeding from malignant UGI tumors receive oncologic therapy after endoscopic hemostatic therapy vs no oncologic therapy after endoscopic hemostatic therapy? |
| MWTs | |
| 5 | Should patients with active bleeding from MWTs (spurting or oozing) receive endoscopic hemostatic therapy vs no endoscopic hemostatic therapy? |
| 6 | Should patients with active bleeding from MWTs (spurting or oozing) receive one endoscopic hemostatic therapy vs another endoscopic hemostatic therapy? |
| 7 | Should patients with no active bleeding from MWTs (nonbleeding visible vessels, adherent clots, flat pigmented spots, clean-based ulcers) receive endoscopic hemostatic therapy vs no endoscopic hemostatic therapy? |
| DLs | |
| 8 | Should patients with UGIB from DL receive one endoscopic hemostatic therapy vs another endoscopic hemostatic therapy? |
| 9 | Should patients with UGIB from DL receive injection of epinephrine alone vs other endoscopic hemostatic therapies? |
| GAVE | |
| 10 | Should patients with GAVE receive one endoscopic hemostatic therapy vs another endoscopic hemostatic therapy? |
| 11 | Should patients with GAVE receive EBL vs other endoscopic hemostatic therapies? |
Table 3
Clinical Questions Formulated and Prioritized
Open table in a new tab
The panel selected outcomes of interest for each question a priori, following the GRADE approach.52
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the panel rated the number of units of blood transfusion and changes in Hgb levels as critical for decision making. Table 4 outlines the critical and important outcomes for all questions.
| Variable | Outcomes |
--- |
| Malignant UGIB | |
| 1–3 | Further bleeding (30-d): critical |
| | Overall mortality (6-mo): important |
| | Failure to achieve immediate hemostasis: important |
| | Rebleeding (30-d): important |
| | No. of units of blood transfusions needed: important |
| | Length of hospitalization: important |
| | Readmissions: important |
| | Adverse effects – important |
| 4 | Further bleeding (30-d, 6-mo): critical (only 6-month) |
| | Overall mortality (6-mo): critical |
| | Failure to achieve immediate hemostasis: important |
| | Rebleeding (30-d): important |
| | No. of units of blood transfusions needed: important |
| | Length of hospitalization: important |
| | Readmissions: important |
| | Adverse effects: important |
| MWTs | |
| 5–7 | Further bleeding (7-d, 30-d): critical (only 7-d) |
| | Overall mortality: important |
| | Failure to achieve immediate hemostasis: important |
| | Rebleeding (7-d, 30-d): important |
| | No. of units of blood transfusions needed: important |
| | Length of hospitalization: important |
| | Adverse effects: important |
| DL | |
| 8–9 | Further bleeding (7-d, 30-d): critical (only 7-d) |
| | Failure to achieve immediate hemostasis: important |
| | Overall mortality: important |
| | Rebleeding (7-d, 30-d): important |
| | Adverse effects: important |
| | Additional hemostatic therapy (eg repeat endoscopic hemostatic therapy, radiologic embolization, surgery): important |
| GAVE | |
| 10–11 | Change in units of blood transfusions needed: critical |
| | Change in Hgb level: critical |
| | Overall mortality: important |
| | Number of endoscopic sessions required for obliteration of lesions: important |
| | Adverse effects: important |
Table 4
Critical or Important Outcomes for Decision Making
Open table in a new tab
Evidence Review and Development of Recommendations
The Cochrane Gut Group at McMaster University conducted systematic searches of the published English-language literature, including MEDLINE, Embase, Cochrane Database of Systematic Reviews, and Cochrane Central Register of Controlled Trials from inception through March 6, 2024. Search strategies and the Preferred Reporting Items for Systematic Reviews and Meta-Analyses57
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flow diagram are included in Appendix 2. The search aimed to identify studies assessing any endoscopic hemostatic therapy for NVNPUB, including malignant UGIB, MWTs, DLs, and GAVE.
Methodologists performed duplicate screening, data extraction, and risk of bias assessment. RCTs, meta-analyses, and observational studies were sought to address the guideline questions. RCTs that included patients with nonvariceal UGIB were considered, provided they offered subgroup data specifically for the conditions. Observational studies with fewer than 10 patients for any specific type of endoscopic therapy were excluded to ensure the reliability and robustness of the evidence being synthesized. Evidence on values, preferences, and costs was also sought.
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The EtD table addressed the magnitude of the effects of interventions, patients’ values and preferences, the balance between desirable and undesirable effects, resource utilization, health equity, acceptability, and feasibility. The guideline panel reviewed draft EtD tables for accuracy and completeness.
During a 2-day online meeting, the panel developed recommendations based on the evidence summarized in the EtD frameworks. For each recommendation, the panel took an individual patient perspective and reached a consensus on every domain of the EtD table.
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Interpretation of Strong and Conditional Recommendations
According to the GRADE approach, recommendations are categorized as “strong” or “conditional.” For strong recommendations, the words “the guideline panel recommends. . . ” are used, and for conditional recommendations, “the guideline panel suggests . . .” Table 1 provides GRADE’s interpretation of strong and conditional recommendations.
Document Review
All panel members reviewed and revised the draft recommendations before making them available online on February 3, 2025, for external review by stakeholders, including allied organizations, medical professionals, patients, and the public. Comments were received, and although relevant feedback was addressed, the recommendations remain unchanged. On February 20, 2025, the CAG Guideline Committee confirmed adherence to the guideline-development process, and on February 25, 2025, the CAG Executive Board approved the submission of the guidelines for publication under the imprimatur of CAG. The guidelines were then subjected to the peer review process of Gastroenterology.
How to Use This Guideline
This guideline is intended to aid clinicians in selecting endoscopic treatment options, while also supporting policy making, education, advocacy, and identifying future research needs. Patients may find them useful as well. It is not intended to define a standard of care. Clinicians should make decisions based on each patient’s clinical presentation, ideally through a shared decision-making process that considers the patient’s values and preferences. Decision making may be influenced by specific clinical settings and available resources. This guideline may not cover all suitable care methods, and recommendations may become outdated as new evidence emerges. Each recommendation includes statements about its underlying values and preferences, along with qualifying remarks that are essential for accurate interpretation and should be included when the guideline is quoted or translated. The guideline is further supported by EtD frameworks and summary-of-findings tables, enhancing their usability.
Recommendations
Malignant Upper Gastrointestinal Bleeding
Question 1: Should patients with active bleeding from malignant UGI tumors receive endoscopic hemostatic therapy vs no endoscopic hemostatic therapy?
Recommendation 1A: In patients with active bleeding from malignant UGI tumors, we suggest conventional endoscopic hemostatic therapy (eg, injection, thermal devices, mechanical devices, or a combination thereof) over no endoscopic hemostatic therapy (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝).
Recommendation 1B: In patients with active bleeding from malignant UGI tumors, we suggest topical hemostatic agents (THAs) over no endoscopic hemostatic therapy (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝).
Recommendation 1A. Conventional Endoscopy Hemostatic Therapy vs No Endoscopic Hemostatic Therapy
Evidence summary
We found no RCTs or observational studies that directly addressed this question. Six comparative cohort studies were excluded due to seriously biased comparisons between patients treated for active bleeding and those untreated due to inactive bleeding.24–26
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Twelve cohort studies provided data on the outcomes of treating malignant UGIB with various conventional endoscopic therapies, including injection therapy (ie, epinephrine, saline, sclerosant, ethanol, and fibrin glue), thermocoagulation (ie, argon plasma coagulation [APC], heater probe, bipolar devices, coagulation forceps, and Nd-YAG laser), and mechanical devices (through-the-scope clips [TTSCs]), either alone or in combination.24–26
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The majority of patients (62%) received APC in these studies.24–26
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Yet, most studies did not provide subgroup data for each intervention. The validity of subgroup comparisons was further compromised by the small sample sizes and selection bias present in these nonrandomized studies. Due to significant heterogeneity in populations, interventions, outcomes, duration of follow-up, and study designs, a proportional meta-analysis of these studies was deemed inappropriate and potentially misleading. Instead, we summarized the results as ranges and presented them in forest plots without pooled estimates following the recommendations of the Cochrane Non-Randomized Studies Methods Group.58
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These details, along with the EtD framework, are provided in Appendix 3.
Benefits, harms, and burden
Based on data from 12 single-arm cohort studies, outcomes for patients treated with conventional endoscopic therapies were variable: 30-day further bleeding occurred in 22%–87% of patients (approximately 50% for most studies), failure to achieve immediate hemostasis in 0%–69% (approximately 20% for most studies), 30-day rebleeding in 17%–53% (approximately 30% for most studies), and mortality in 13%–93%, with most studies reporting 30-day mortality.24–26
24.
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Abu-Sbeih, H. ∙ Szafron, D. ∙ Elkafrawy, A.A. ...
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Kim, Y.I. ∙ Choi, I.J. ∙ Cho, S.J. ...
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Koh, K.H. ∙ Kim, K. ∙ Kwon, D.H. ...
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Google Scholar
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Full Text (PDF)
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PubMed
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Two studies reported 0% adverse effects in 81 patients treated with APC.25
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Without comparative data, estimating the effects of conventional endoscopic therapies compared with no therapy was impossible. To address this, an expert evidence survey was conducted using the GRADE expert evidence approach.69
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The panel estimated that patients with active bleeding from malignant UGI tumors who do not receive endoscopic therapy have a risk of further bleeding of at least 85%, failure to achieve immediate hemostasis of at least 50%, and 30-day rebleeding of at least 35% for those who stopped bleeding spontaneously. The panel judged that conventional endoscopic hemostatic therapies likely have variable magnitudes of desirable and undesirable effects.
Certainty in the evidence of effects
Very low certainty of evidence due to serious risk of bias, serious inconsistency, and serious imprecision.
Other evidence-to-decision criteria and considerations
There was no research evidence on patients’ values and preferences in the context of malignant UGIB. The panel agreed, based on their experience, that patients place a high value on reducing further bleeding (critical outcome) to avoid prolonged or repeated hospitalizations, additional interventions, and decreased quality of life. However, the panel recognized possible variability and uncertainty in patients’ values and preferences regarding blood transfusions and 6-month mortality, influenced by their overall prognosis and care goals. Patients may prefer less invasive procedures if their goal is comfort care, and those aiming for extended survival might choose more aggressive treatments. Costs may vary depending on the intervention. There was no research evidence on cost-effectiveness. The panel judged conventional endoscopic therapies as probably acceptable and feasible with no adverse impact on equity-deserving groups.
Conclusions and research needs
The panel judged that the balance of effects likely favors conventional endoscopic hemostatic therapy in patients with active bleeding from malignant UGI tumors. As a result, the panel issued a conditional recommendation for conventional endoscopic hemostatic therapies over no endoscopic hemostatic therapy, recognizing that the certainty of the evidence was very low. This conditional recommendation would be more applicable to tumors with spurting bleeding rather than those with diffuse oozing. When tumors are diffusely oozing over a large area, conventional endoscopic therapies can be challenging to apply, and some (eg, thermocoagulation and sclerosants) may lead to further tissue injury. Therefore, some patients with diffusely oozing tumors may not be suitable for conventional endoscopic hemostatic intervention. Future research should focus on well-designed RCTs or observational studies comparing conventional endoscopic therapy with no endoscopic therapy, potentially in combination with other hemostatic interventions, such as radiation, embolization, or surgery. In addition, research on patient values and preferences would enhance shared decision making.
Recommendation 1B. Topical Hemostatic Agents vs No Endoscopic Hemostatic Therapy
Evidence summary
We found no RCTs that directly addressed this question.
Nineteen studies provided single-arm cohort-type data that addressed this question, including 4 RCTs and 15 observational studies.70–83
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Despite the differences in composition, all of these agents function as mechanical barriers to control bleeding, justifying their inclusion in pooled proportional meta-analyses. These results, along with the EtD framework, are detailed in Appendix 3.
Benefits
A comparative cohort study suggested that THAs, compared with no endoscopic therapy, may reduce 30-day rebleeding (risk ratio [RR], 0.67; 95% CI, 0.14–3.17; absolute risk reduction [ARR], 99 fewer per 1000; 95% CI, 258 fewer to 651 more) and 6-month mortality (RR, 0.33; 95% CI, 0.04–2.69; ARR, 201 fewer per 1000; 95% CI, 288 fewer to 507 more), but these estimates were very imprecise.80
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Full Text (PDF)
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PubMed
Google Scholar
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Pittayanon, R. ∙ Prueksapanich, P. ∙ Rerknimitr, R.
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PubMed
Google Scholar
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Full Text (PDF)
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PubMed
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PubMed
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Hemostatic spray powder TC-325 for GI bleeding in a nationwide study: survival and predictors of failure via competing risks analysis
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Full Text (PDF)
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PubMed
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Full Text (PDF)
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PubMed
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Full Text (PDF)
Scopus (33)
PubMed
Google Scholar
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Cahyadi, O. ∙ Bauder, M. ∙ Meier, B. ...
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PubMed
Google Scholar
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Chen, Y.I. ∙ Barkun, A. ∙ Nolan, S.
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PubMed
Google Scholar
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Chen, Y.I. ∙ Wyse, J. ∙ Lu, Y. ...
TC-325 hemostatic powder versus current standard of care in managing malignant GI bleeding: a pilot randomized clinical trial
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Most deaths were not related to GIB. Estimating the effects of THA vs no endoscopic therapy was challenging due to limited comparative data. To address this, expert survey results from PICO question 1A were applied, and the panel concluded that THA likely has large desirable effects over no endoscopic therapy.
Harms and burden
A comparative cohort study suggested THA, compared with no endoscopic therapy, may increase blood transfusions (MD, 5.60 units; 95% CI, 4.59 fewer to 15.79 more units) and length of hospitalization (MD, 4.40 days; 95% CI, 10.85 fewer to 19.65 more days).80
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However, the data were skewed, and these outcomes can be subjective, influenced by factors such as the health system’s structure, institutional protocols, clinical judgment, and patient-specific factors.80
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Few studies reported adverse effects related to THA, but a systematic review of THAs in UGIB, including patients with malignant UGIB, found a pooled adverse event rate of 2% (95% CI, 1%–3%), with abdominal distention and bleeding being observed most frequently.89
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Alali, A.A. ∙ Moosavi, S. ∙ Martel, M. ...
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The panel judged that THA likely has moderate undesirable effects over no endoscopic therapy.
Certainty in the evidence of effects
Very low certainty of evidence due to serious risk of bias and serious imprecision.
Other evidence-to-decision criteria and considerations
No research evidence was available on patients’ values and preferences, but the panel assumed that avoiding further bleeding is critical to patients. The panel considered the cost of THA to be high. There was no research evidence on the cost-effectiveness of THA compared with no endoscopic therapy. The panel judged THA probably acceptable with no adverse impact on equity-deserving groups, but its feasibility may vary due to access challenges in some countries.
Conclusions and research needs
The panel judged that the balance of effects likely favors THA in patients with active bleeding from malignant UGI tumors (oozing or spurting). Thus, the panel issued a conditional recommendation for THA over no endoscopic hemostatic therapy, recognizing the very low certainty of the evidence, which was based predominantly on TC-325. If TC-325 is chosen as a THA, it is crucial that TC-325 be applied to actively bleeding lesions only. The panel highlighted the need for more comparative studies, potentially in combination with other hemostatic interventions, such as radiation, embolization, or surgery. Because most evidence was derived from TC-325, further research on other THAs is needed.
Question 2: Should patients with active bleeding from malignant UGI tumors receive one conventional endoscopic hemostatic therapy vs another conventional endoscopic hemostatic therapy?
Recommendation 2: In patients with active bleeding from malignant UGI tumors, we were unable to reach a recommendation for or against any specific type of conventional endoscopic hemostatic therapy over another.
Recommendation 2. Specific Type of Conventional Endoscopic Therapy vs Another Type of Conventional Endoscopic Therapy
Evidence summary
We did not find any studies that directly compared various conventional endoscopic hemostatic therapies with each other.
Conclusions and research needs
Due to the lack of evidence, the panel was unable to reach a recommendation for or against any specific type of conventional endoscopic therapy over another in patients with active bleeding from malignant UGI tumors. Because current evidence suggests that THA may be more effective than conventional hemostatic therapies for malignant UGIB, the panel did not consider clinical trials comparing conventional endoscopic therapies a high research priority.
Question 3: Should patients with active bleeding from malignant UGI tumors receive THAs vs conventional endoscopic hemostatic therapies?
Recommendation 3: In patients with active bleeding from malignant UGI tumors, we suggest THAs over conventional endoscopic hemostatic therapy (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝).
Recommendation 3. Topical Hemostatic Agents vs Conventional Endoscopic Hemostatic Therapy
Evidence summary
We identified 2 systematic reviews that indirectly addressed this question.89
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Pittayanon, R. ∙ Khongka, W. ∙ Linlawan, S. ...
Hemostatic powder vs standard endoscopic treatment for gastrointestinal tumor bleeding: a multicenter randomized trial
Gastroenterology. 2023; 165:762-772.e2
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The panel judged that THA likely has moderate undesirable effects compared with conventional endoscopic hemostatic therapy.
Certainty in the evidence of effects
Very low certainty of evidence due to serious risk of bias, serious inconsistency, and very serious imprecision. The evidence was not downgraded for indirectness because most patients had bleeding from UGI tumors, as opposed to lower GI tumors.
Other evidence-to-decision criteria and considerations
No research evidence existed on patients’ values and preferences, but the panel assumed that avoiding further bleeding was critical for decision making for patients. A cost minimization study published after our search period found that using THA compared with standard endoscopic therapy resulted in a cost-saving of US$1613 for malignant UGIB, considering device-related costs, incremental facility costs, and additional physician and staff member time.92
92.
Shah, E.D. ∙ Law, R.
Valuing innovative endoscopic techniques: hemostatic powder for the treatment of GI tumor bleeding
Gastrointest Endosc. 2024; 100:49-54
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In addition, a cost-utility analysis published later also found that using TC-325 as first-line treatment for malignant GIB in the United Kingdom resulted in a cost savings of £245.88 compared with standard endoscopic therapy.93
93.
Cooper, D.M. ∙ Norton, B. ∙ Hawkes, N.D. ...
Hemostatic powder TC-325 as first-line treatment option for malignant gastrointestinal bleeding: a cost-utility analysis in the United Kingdom
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The panel judged THA likely to be acceptable, with no adverse impact on equity-deserving groups, although its feasibility may vary due to access challenges in some countries. Conclusions and research needs. The panel judged that the balance of effects likely favors THA in patients with active bleeding (oozing or spurting) from malignant UGI tumors. The data on hospitalization length were skewed and may not be generalizable to all settings. As a result, the panel prioritized the potential benefits of reducing further bleeding over the potential burden and costs. Therefore, the panel issued a conditional recommendation for THA over conventional endoscopic hemostatic therapy, recognizing the very low certainty of the evidence, which was predominantly based on TC-325. The panel emphasized the need for more comparative studies to increase the certainty of the evidence. In addition, RCTs should compare the efficacy and safety of THAs with other nonendoscopic hemostatic therapies.
Question 4: Should patients with active bleeding from malignant UGI tumors receive oncologic therapy after endoscopic hemostatic therapy vs no oncologic therapy after endoscopic hemostatic therapy?
Recommendation 4: In patients with active bleeding from malignant UGI tumors, we suggest administering oncologic therapy after endoscopic hemostatic therapy rather than not providing oncologic therapy after endoscopic hemostatic therapy (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝).
Recommendation 4. Oncologic Therapy vs No Oncologic Therapy After Endoscopic Hemostatic Therapy
Evidence summary
We defined oncologic therapies as surgery, chemotherapy, and radiation therapy. The goals of these therapies can vary based on cancer type, stage, and the patient's overall health, but generally include curative intent, disease control, symptom management, palliative care, prolongation of survival, and prevention of complications, such as bleeding. Although radiologic embolization is not typically considered a primary oncologic therapy for UGI cancers, it can play a crucial role in a multimodal approach.
We did not find any RCTs addressing this question directly. However, there were observational data from an RCT and a multicenter retrospective cohort study.79
79.
Pittayanon, R. ∙ Khongka, W. ∙ Linlawan, S. ...
Hemostatic powder vs standard endoscopic treatment for gastrointestinal tumor bleeding: a multicenter randomized trial
Gastroenterology. 2023; 165:762-772.e2
Full Text
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81.
Pittayanon, R. ∙ Rerknimitr, R. ∙ Barkun, A.
Prognostic factors affecting outcomes in patients with malignant GI bleeding treated with a novel endoscopically delivered hemostatic powder
Gastrointest Endosc. 2018; 87:994-1002
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In the RCT, patients with malignant GI bleeding were randomized to receive TC-325 or standard endoscopic treatment.79
79.
Pittayanon, R. ∙ Khongka, W. ∙ Linlawan, S. ...
Hemostatic powder vs standard endoscopic treatment for gastrointestinal tumor bleeding: a multicenter randomized trial
Gastroenterology. 2023; 165:762-772.e2
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The proportion of patients undergoing additional nonendoscopic hemostatic or oncologic treatments, such as surgery, chemotherapy, radiation, embolization, or a combination thereof, within 1 month after the initial endoscopy was similar in the 2 arms (50.9% in the TC-325 arm vs 62.7% in the standard endoscopic treatment arm).79
79.
Pittayanon, R. ∙ Khongka, W. ∙ Linlawan, S. ...
Hemostatic powder vs standard endoscopic treatment for gastrointestinal tumor bleeding: a multicenter randomized trial
Gastroenterology. 2023; 165:762-772.e2
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The high rate of additional treatment was likely due to the good performance status of the included patients (Eastern Cooperative Oncology Group score 0–2).79
79.
Pittayanon, R. ∙ Khongka, W. ∙ Linlawan, S. ...
Hemostatic powder vs standard endoscopic treatment for gastrointestinal tumor bleeding: a multicenter randomized trial
Gastroenterology. 2023; 165:762-772.e2
Full Text
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Scopus (9)
PubMed
Google Scholar
Both studies reported on 6-month overall mortality (critical outcome) and 30-day rebleeding, but not on 6-month further bleeding (critical outcome).79
79.
Pittayanon, R. ∙ Khongka, W. ∙ Linlawan, S. ...
Hemostatic powder vs standard endoscopic treatment for gastrointestinal tumor bleeding: a multicenter randomized trial
Gastroenterology. 2023; 165:762-772.e2
Full Text
Full Text (PDF)
Scopus (9)
PubMed
Google Scholar
,81
81.
Pittayanon, R. ∙ Rerknimitr, R. ∙ Barkun, A.
Prognostic factors affecting outcomes in patients with malignant GI bleeding treated with a novel endoscopically delivered hemostatic powder
Gastrointest Endosc. 2018; 87:994-1002
Full Text
Full Text (PDF)
Scopus (30)
PubMed
Google Scholar
No studies provided direct evidence of adverse effects from oncologic therapy after endoscopic hemostatic therapy compared to without it. Three systematic reviews of noncomparative cohort studies provided indirect evidenfce on the toxicity of palliative radiotherapy for symptomatic locally advanced gastric cancer, chemotherapy in advanced gastric cancer, and complications after surgical resection in patients with gastric cancer.94–96
94.
Chen, G. ∙ Wang, J. ∙ Chen, K. ...
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Tey, J. ∙ Soon, Y.Y. ∙ Koh, W.Y. ...
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Wagner, A.D. ∙ Syn, N.L. ∙ Moehler, M. ...
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The EtD framework is available in Appendix 3.
Benefits
In a post-hoc observational analysis of an RCT, multivariable analysis identified the Charlson comorbidity index (hazard ratio [HR], 1.17; 95% CI, 1.05–1.32) and the receipt of additional nonendoscopic hemostatic treatments or oncologic treatments, such as surgery, chemotherapy, radiation, embolization, or a combination of these (HR, 0.16; 95% CI, 0.06–0.43), as independent predictors of 6-month survival, after adjustment for Eastern Cooperative Oncology Group score, Glasgow-Blatchford score, and upper GI lesions; TC-325 application was not an independent predictor.79
79.
Pittayanon, R. ∙ Khongka, W. ∙ Linlawan, S. ...
Hemostatic powder vs standard endoscopic treatment for gastrointestinal tumor bleeding: a multicenter randomized trial
Gastroenterology. 2023; 165:762-772.e2
Full Text
Full Text (PDF)
Scopus (9)
PubMed
Google Scholar
Receiving additional nonendoscopic hemostatic treatments or oncologic treatments did not predict 30-day rebleeding (HR, 1.1; 95% CI, 0.24–5.06).79
79.
Pittayanon, R. ∙ Khongka, W. ∙ Linlawan, S. ...
Hemostatic powder vs standard endoscopic treatment for gastrointestinal tumor bleeding: a multicenter randomized trial
Gastroenterology. 2023; 165:762-772.e2
Full Text
Full Text (PDF)
Scopus (9)
PubMed
Google Scholar
In the retrospective cohort study, factors associated with 6-month survival included low Eastern Cooperative Oncology Group scores 0–2 (HR, 0.14; 95% CI, 0.04–0.47); cancer stage I–III (HR, 0.31; 95% CI, 0.01–0.96); and receipt of definitive hemostatic treatments, including surgery, chemotherapy, radiotherapy, or radiologic embolization (HR, 0.24; 95% CI, 0.09–0.59), after adjusting for comorbidity, type of cancer bleeding, and coagulopathy.81
81.
Pittayanon, R. ∙ Rerknimitr, R. ∙ Barkun, A.
Prognostic factors affecting outcomes in patients with malignant GI bleeding treated with a novel endoscopically delivered hemostatic powder
Gastrointest Endosc. 2018; 87:994-1002
Full Text
Full Text (PDF)
Scopus (30)
PubMed
Google Scholar
However, these observational analyses are prone to residual confounding, as the comparison was not randomized. The association between oncologic treatments after endoscopic hemostatic therapy and improved 6-month survival may be confounded by baseline health and cancer stage. Nevertheless, the panel judged that oncologic therapies likely have large desirable effects over no oncologic therapies.
Harms and burden
A systematic review of 7 cohort studies involving 161 patients with symptomatic locally advanced gastric cancer found grade 3 to 4 toxicities in up to 15% of those treated with radiotherapy alone and up to 25% with chemoradiotherapy.95
95.
Tey, J. ∙ Soon, Y.Y. ∙ Koh, W.Y. ...
Palliative radiotherapy for gastric cancer: a systematic review and meta-analysis
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Another systematic review of 60 RCTs involving 11,698 patients receiving systemic chemotherapy for advanced gastric cancer reported treatment discontinuation due to toxicity ranging from 8% to 21%.96
96.
Wagner, A.D. ∙ Syn, N.L. ∙ Moehler, M. ...
Chemotherapy for advanced gastric cancer
Cochrane Database Syst Rev. 2017; 8:CD004064
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In addition, a systematic review of single-arm cohort studies including 32,067 patients who underwent gastrectomy for gastric cancer reported postoperative complication rates between 12.5% and 51.0%, with infectious complications, anastomotic leakage, and pneumonia being the most common.94
94.
Chen, G. ∙ Wang, J. ∙ Chen, K. ...
Relationship between postoperative complications and the prognosis of gastric carcinoma patients who underwent surgical resection: a systematic review and meta-analysis
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Google Scholar
The panel judged that oncologic therapies likely have moderate undesirable effects over no oncologic therapies.
Certainty in the evidence of effects
Very low certainty of evidence due to serious risk of bias, serious indirectness from different types and early stages of GI cancers, and the inclusion of both oncologic treatments and radiologic embolization as the intervention, as well as very serious imprecision.
Other evidence-to-decision criteria and considerations
There was no research evidence on patients’ values and preferences regarding oncologic therapies, but the panel assumed patients might have variable views on adverse effects from these therapies, particularly when considering improving quality of life vs increasing survival.97
97.
Meropol, N.J. ∙ Egleston, B.L. ∙ Buzaglo, J.S. ...
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98.
Weeks, J.C. ∙ Cook, E.F. ∙ O'Day, S.J. ...
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In the context of malignant GI bleeding, patients may choose interventions that align with their personal goals—opting for less invasive procedures if prioritizing quality of life or more aggressive treatments for prolonged survival. Therefore, engaging patients in shared decision making is crucial. The cost of oncologic therapies can vary widely due to factors such as tumor location, specific treatments, and individual patient circumstances (eg, staging and comorbidities). No data existed on the cost-effectiveness of oncologic therapies after endoscopic hemostatic therapies. Although the panel judged that oncologic therapies were probably acceptable and feasible, they could increase inequity due to cancer health disparities. Factors such as race, gender, age, geography, socioeconomic status, cultural beliefs, social support, and health literacy could impact equitable access to these treatments.99
99.
Patel, M.I. ∙ Lopez, A.M. ∙ Blackstock, W. ...
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To address cancer health disparities, it is crucial to ensure equitable access to treatment for all populations. This involves culturally sensitive community outreach, policy changes to reduce socioeconomic barriers, and investment in research focused on underserved groups.99
99.
Patel, M.I. ∙ Lopez, A.M. ∙ Blackstock, W. ...
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Conclusions and research needs
Although endoscopic therapies may reduce further bleeding from malignant UGI tumors in the short term, they appear to have little or no benefit by themselves on long-term mortality. For patients with active bleeding from malignant UGI tumors, the panel determined that oncologic therapies after endoscopic hemostatic therapy likely offer a net benefit, despite the very low certainty of the evidence. As a result, the panel issued a conditional recommendation in favor of administering oncologic therapy after endoscopic hemostatic therapy, as opposed to not providing it. This conditional recommendation placed a higher value on the potential reduction in 6-month mortality (critical outcome) over the potential burdens, harms, costs, and negative impact on equity. The panel emphasized a patient-centered, multidisciplinary approach to decision making, as optimal oncologic care requires collaboration among gastroenterologists, oncologists, surgeons, radiologists, and palliative care specialists. Patients should be involved in discussions about personalized treatment risks and benefits. Given the ethical and practical challenges of conducting an RCT on this topic, the panel emphasized the need for well-designed observational studies to determine optimal timing for oncologic therapy and called for more comparative studies to strengthen the evidence.
Mallory Weiss Tears
Question 5: Should patients with active bleeding from MWTs (spurting or oozing) receive endoscopic hemostatic therapy vs no endoscopic hemostatic therapy?
Recommendation 5: In patients with active bleeding from MWTs (spurting or oozing), we suggest endoscopic hemostatic therapy over no endoscopic hemostatic therapy (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝).
Recommendation 5. Endoscopic Hemostatic Therapy vs No Endoscopic Hemostatic Therapy
Evidence summary
We identified 2 RCTs that addressed this question. One included patients with UGIB from MWTs with “active bleeding” or nonbleeding visible vessels, comparing epinephrine and polidocanol injections with no endoscopic hemostatic therapy.100
100.
Llach, J. ∙ Elizalde, J.I. ∙ Guevara, M.C. ...
Endoscopic injection therapy in bleeding Mallory-Weiss syndrome: a randomized controlled trial
Gastrointest Endosc. 2001; 54:679-681
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Only the “active bleeding” subgroup was included in our systematic review.100
100.
Llach, J. ∙ Elizalde, J.I. ∙ Guevara, M.C. ...
Endoscopic injection therapy in bleeding Mallory-Weiss syndrome: a randomized controlled trial
Gastrointest Endosc. 2001; 54:679-681
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Another RCT included patients with UGIB and provided data on a subgroup of patients with MWTs with spurting or oozing bleeding, comparing multipolar electrocoagulation with no endoscopic hemostatic therapy.101
101.
Laine, L.
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N Engl J Med. 1987; 316:1613-1617
Crossref
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No study provided data for further bleeding (critical outcome). One study reported failure to achieve immediate hemostasis,101
101.
Laine, L.
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N Engl J Med. 1987; 316:1613-1617
Crossref
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and the other provided data for rebleeding.100
100.
Llach, J. ∙ Elizalde, J.I. ∙ Guevara, M.C. ...
Endoscopic injection therapy in bleeding Mallory-Weiss syndrome: a randomized controlled trial
Gastrointest Endosc. 2001; 54:679-681
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All rebleeding occurred within 7 days. To avoid double-counting 7-day and 30-day outcomes, we included the rebleeding outcomes as 7-day results only. This approach was based on our a priori decision that 7-day further bleeding was the critical outcome. One study provided data for the length of hospitalization.100
100.
Llach, J. ∙ Elizalde, J.I. ∙ Guevara, M.C. ...
Endoscopic injection therapy in bleeding Mallory-Weiss syndrome: a randomized controlled trial
Gastrointest Endosc. 2001; 54:679-681
Full Text
Full Text (PDF)
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Google Scholar
Both studies provided data on adverse effects.100
100.
Llach, J. ∙ Elizalde, J.I. ∙ Guevara, M.C. ...
Endoscopic injection therapy in bleeding Mallory-Weiss syndrome: a randomized controlled trial
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101.
Laine, L.
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Only 2 outcomes—blood transfusions and 30-day mortality—could be pooled for meta-analyses.
Four comparative cohort studies evaluated endoscopic hemostatic therapy vs no endoscopic hemostatic therapy in patients with UGIB due to MWT.102–105
102.
Bataller, R. ∙ Llach, J. ∙ Salmeron, J.M. ...
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Meta-analyses of these studies were deemed inappropriate as they compared patients with active bleeding who received endoscopic therapy to those without active bleeding who did not receive endoscopic treatment.
The EtD framework is available in Appendix 4.
Benefits
Meta-analyses of 2 RCTs suggested that endoscopic hemostatic therapy, compared with no endoscopic hemostatic therapy, may reduce failure to achieve immediate hemostasis (RR, 0.06; 95% CI, 0.004 to 0.91; ARR, 823 fewer per 1000; 95% CI, 871 fewer to 79 fewer), 7-day rebleeding (RR, 0.20; 95% CI, 0.03 to 1.58; ARR, 174 fewer per 1000; 95% CI, 211 fewer to 126 more), blood transfusions (MD, –2.20 units; 95% CI, –5.23 to 0.84 units), and length of hospitalization (MD, –2.10 days; 95% CI, –2.20 to 2.00 days), but these effects were very imprecise.100
100.
Llach, J. ∙ Elizalde, J.I. ∙ Guevara, M.C. ...
Endoscopic injection therapy in bleeding Mallory-Weiss syndrome: a randomized controlled trial
Gastrointest Endosc. 2001; 54:679-681
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101.
Laine, L.
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Crossref
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Endoscopic hemostatic therapy may have little or no impact on 30-day mortality (RR, 0.24; 95% CI, 0.03 to 2.05; ARR, 58 fewer per 1000; 95% CI, 75 fewer to 81 more), but the effect was very imprecise.100
100.
Llach, J. ∙ Elizalde, J.I. ∙ Guevara, M.C. ...
Endoscopic injection therapy in bleeding Mallory-Weiss syndrome: a randomized controlled trial
Gastrointest Endosc. 2001; 54:679-681
Full Text
Full Text (PDF)
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PubMed
Google Scholar
,101
101.
Laine, L.
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N Engl J Med. 1987; 316:1613-1617
Crossref
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Google Scholar
Given the natural history of MWT with relatively low rebleeding rates, the panel prioritized failure to achieve immediate hemostasis over 7-day rebleeding and judged that endoscopic hemostatic therapy likely has large desirable effects compared with no endoscopic hemostatic therapy.
Harms and burden
Both RCTs reported no adverse effects, with or without endoscopic hemostatic therapy, making relative and absolute effects nonestimable.100
100.
Llach, J. ∙ Elizalde, J.I. ∙ Guevara, M.C. ...
Endoscopic injection therapy in bleeding Mallory-Weiss syndrome: a randomized controlled trial
Gastrointest Endosc. 2001; 54:679-681
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101.
Laine, L.
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The panel judged that endoscopic hemostatic therapy likely has trivial undesirable effects. This assessment was also supported by indirect evidence from PUB.8
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Certainty in the evidence of effects. Very low certainty of evidence due to serious risk of bias and very serious imprecision.
Other evidence-to-decision criteria and considerations
There was no research evidence on patients’ values and preferences in the context of MWTs. However, the panel assumed there is probably no important uncertainty or variability in how much patients value the main outcomes, such as avoiding failure to achieve immediate hemostasis or rebleeding. The costs of endoscopic hemostatic therapy were considered moderate. Although cost-effectiveness studies were lacking, the panel considered endoscopic hemostatic therapy for actively bleeding MWTs as likely cost-effective by achieving immediate hemostasis, reducing rebleeding rates, and decreasing blood transfusions and length of hospitalization. The panel judged endoscopic hemostatic therapy as likely acceptable and feasible with no adverse impact on equity-deserving groups.
Conclusions and research needs
For patients with active bleeding (oozing or spurting) from MWTs, the panel judged that endoscopic hemostatic therapy likely offers a net benefit over no endoscopic hemostatic therapy, despite very low certainty of evidence. Therefore, the panel issued a conditional recommendation for endoscopic hemostatic therapy over no endoscopic hemostatic therapy. The panel highlighted the need for more comparative studies to increase the certainty of the evidence. In addition, future RCTs should compare the efficacy and safety of different endoscopic hemostatic therapies and stratify randomization based on stigmata.
Question 6: Should patients with active bleeding from MWTs (spurting or oozing) receive one endoscopic hemostatic therapy vs another endoscopic hemostatic therapy?
Recommendation 6A: In patients with active bleeding from MWTs (spurting or oozing), we suggest EBL or endoscopic TTSC placement over epinephrine injection alone (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝).
Recommendation 6B: In patients with active bleeding from MWTs (spurting or oozing), we suggest EBl or endoscopic TTSC placement (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝).
Recommendation 6A. Mechanical Modalities (Endoscopic Band Ligation/Endoscopic Through-the-Scope Clip Placement) vs Epinephrine Injection Alone
Evidence summary
We identified 3 RCTs that addressed this question.103
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Chung, I.K. ∙ Kim, E.J. ∙ Hwang, K.Y. ...
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One study did not specify a follow-up period or the timing of rebleeding.103
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To avoid double-counting 7-day and 30-day outcomes, we considered all rebleeding and further bleeding outcomes as 7-day outcomes. One study reported failure to achieve immediate hemostasis and length of hospitalization.107
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A prospective, randomized trial of endoscopic band ligation vs. epinephrine injection for actively bleeding Mallory-Weiss syndrome
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Two studies reported units of blood transfused and adverse effects.106
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No comparative cohort studies addressing this question were identified.
We also conducted proportional meta-analyses of single-arm cohort-type data from 7 studies on EBL/TTSC and 4 studies on epinephrine injection alone.103
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The results of the proportional meta-analyses were provided as supplementary data in the Evidence Profile Table, but were not included in the EtD framework due to the absence of direct comparison. The EtD framework is in Appendix 4.
Benefits
Meta-analyses of 3 RCTs suggested that EBL/TTSC, compared with epinephrine injection alone, may reduce the risk of 7-day further bleeding (RR, 0.32; 95% CI, 0.06–1.64; ARR, 95 fewer per 1000; 95% CI, 131 fewer to 89 more) and 7-day rebleeding (RR, 0.31; 95% CI, 0.03–3.13; ARR, 80 fewer per 1000; 95% CI, 103 fewer to 248 more), with no significant subgroup differences among EBL, TTSC, or EBL/TTSC.103
103.
Chung, I.K. ∙ Kim, E.J. ∙ Hwang, K.Y. ...
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However, these estimates were very imprecise due to small sample sizes and very low event rates.103
103.
Chung, I.K. ∙ Kim, E.J. ∙ Hwang, K.Y. ...
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Park, C.H. ∙ Min, S.W. ∙ Sohn, Y.H. ...
A prospective, randomized trial of endoscopic band ligation vs. epinephrine injection for actively bleeding Mallory-Weiss syndrome
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Due to zero events in the EBL/TTSC group for failure to achieve immediate hemostasis, calculating the RR and CI with continuity corrections introduced instability and potential bias.107
107.
Park, C.H. ∙ Min, S.W. ∙ Sohn, Y.H. ...
A prospective, randomized trial of endoscopic band ligation vs. epinephrine injection for actively bleeding Mallory-Weiss syndrome
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However, when considering the opposite outcome (immediate hemostasis), EBL/TTSC appeared to have no impact compared with epinephrine injection alone (RR, 1.01; 95% CI, 0.94–1.09).107
107.
Park, C.H. ∙ Min, S.W. ∙ Sohn, Y.H. ...
A prospective, randomized trial of endoscopic band ligation vs. epinephrine injection for actively bleeding Mallory-Weiss syndrome
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Thirty-day mortality was not estimable due to double-zero events.103
103.
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107.
Park, C.H. ∙ Min, S.W. ∙ Sohn, Y.H. ...
A prospective, randomized trial of endoscopic band ligation vs. epinephrine injection for actively bleeding Mallory-Weiss syndrome
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Full Text
Full Text (PDF)
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Overall, the panel judged that EBL/TTSC likely has moderate desirable effects compared with epinephrine injection alone.
Harms and burden
Compared with epinephrine injection, EBL/TTSC may increase blood transfusions (MD, 0.85 units; 95% CI, –0.34 to 2.04 units) and length of hospitalization (MD, 0.80 days; 95% CI, 0.13 to 1.47 days).106
106.
Huang, S.P. ∙ Wang, H.P. ∙ Lee, Y.C. ...
Endoscopic hemoclip placement and epinephrine injection for Mallory-Weiss syndrome with active bleeding
Gastrointest Endosc. 2002; 55:842-846
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,107
107.
Park, C.H. ∙ Min, S.W. ∙ Sohn, Y.H. ...
A prospective, randomized trial of endoscopic band ligation vs. epinephrine injection for actively bleeding Mallory-Weiss syndrome
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Full Text
Full Text (PDF)
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These estimates were very imprecise and were not considered clinically meaningful by the panel. In addition, the outcomes related to blood transfusion requirements and length of hospitalization can be subjective and influenced by individual clinical practices. We were unable to assess the adverse effects of EBL/TTSC compared with epinephrine injection alone due to double-zero events.106
106.
Huang, S.P. ∙ Wang, H.P. ∙ Lee, Y.C. ...
Endoscopic hemoclip placement and epinephrine injection for Mallory-Weiss syndrome with active bleeding
Gastrointest Endosc. 2002; 55:842-846
Full Text
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Google Scholar
,107
107.
Park, C.H. ∙ Min, S.W. ∙ Sohn, Y.H. ...
A prospective, randomized trial of endoscopic band ligation vs. epinephrine injection for actively bleeding Mallory-Weiss syndrome
Gastrointest Endosc. 2004; 60:22-27
Full Text
Full Text (PDF)
Scopus (55)
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Google Scholar
The panel judged that EBL/TTSC likely has trivial undesirable effects compared with epinephrine injection alone.
Certainty in the evidence of effects. Very low certainty of evidence due to serious risk of bias and very serious imprecision.
Other evidence-to-decision criteria and considerations
The panel assumed patients likely do not have important uncertainty or variability in valuing the critical outcome of further bleeding. Costs of EBL/TTSC were considered moderate, but no studies have assessed its cost-effectiveness compared with epinephrine injection alone in MWT. The panel judged EBL/TTSC as likely acceptable, feasible, and unlikely to adversely impact equity-deserving groups.
Conclusions and research needs
For patients with active bleeding from MWT (oozing or spurting), the panel judged that EBL/TTSC likely offers a net benefit over epinephrine injection alone, despite the evidence being of very low certainty. This conditional recommendation placed a higher value on the potential benefits of reducing 7-day further bleeding than the potential burden and costs. The panel stressed the importance of conducting more comparative studies to increase the certainty of the evidence. In addition, future RCTs should investigate how different stigmata (oozing, spurting) and the size of MWT respond to each intervention. Cost-effectiveness analyses should also be performed to assess the economic impact of each treatment option.
Recommendation 6B. Endoscopic Band Ligation vs Endoscopic Through-the-Scope Clip Placement
Evidence summary
We found 1 RCT and 1 comparative cohort study that addressed this question.108
108.
Cho, Y.S. ∙ Chae, H.S. ∙ Kim, H.K. ...
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Lecleire, S. ∙ Antonietti, M. ∙ Iwanicki-Caron, I. ...
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Both studies reported further bleeding and all of the important outcomes.108
108.
Cho, Y.S. ∙ Chae, H.S. ∙ Kim, H.K. ...
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,112
112.
Lecleire, S. ∙ Antonietti, M. ∙ Iwanicki-Caron, I. ...
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Aliment Pharmacol Ther. 2009; 30:399-405
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Scopus (0)
PubMed
Google Scholar
Because all rebleeding occurred within 7 days, we included all rebleeding and further bleeding outcomes as 7-day outcomes.
We also conducted proportional meta-analyses using single-arm cohort-type data from 5 studies on EBL and 5 studies on TTSC.103
103.
Chung, I.K. ∙ Kim, E.J. ∙ Hwang, K.Y. ...
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106.
Huang, S.P. ∙ Wang, H.P. ∙ Lee, Y.C. ...
Endoscopic hemoclip placement and epinephrine injection for Mallory-Weiss syndrome with active bleeding
Gastrointest Endosc. 2002; 55:842-846
Full Text
Full Text (PDF)
Scopus (80)
PubMed
Google Scholar
107.
Park, C.H. ∙ Min, S.W. ∙ Sohn, Y.H. ...
A prospective, randomized trial of endoscopic band ligation vs. epinephrine injection for actively bleeding Mallory-Weiss syndrome
Gastrointest Endosc. 2004; 60:22-27
Full Text
Full Text (PDF)
Scopus (55)
PubMed
Google Scholar
108.
Cho, Y.S. ∙ Chae, H.S. ∙ Kim, H.K. ...
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World J Gastroenterol. 2008; 14:2080-2084
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109.
Higuchi, N. ∙ Akahoshi, K. ∙ Sumida, Y. ...
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Surg Endosc. 2006; 20:1431-1434
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110.
Shimoda, R. ∙ Iwakiri, R. ∙ Sakata, H. ...
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Dig Endosc. 2009; 21:20-23
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111.
Yamaguchi, Y. ∙ Yamato, T. ∙ Katsumi, N. ...
Endoscopic hemoclipping for upper GI bleeding due to Mallory-Weiss syndrome
Gastrointest Endosc. 2001; 53:427-430
Full Text
Full Text (PDF)
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PubMed
Google Scholar
112.
Lecleire, S. ∙ Antonietti, M. ∙ Iwanicki-Caron, I. ...
Endoscopic band ligation could decrease recurrent bleeding in Mallory-Weiss syndrome as compared to haemostasis by hemoclips plus epinephrine
Aliment Pharmacol Ther. 2009; 30:399-405
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Scopus (0)
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The results were provided as supplementary data in the Evidence Profile Table, but were not included in the EtD framework due to the lack of direct comparative data. The EtD framework is in Appendix 4.
Benefits, harms, and burden
Based on 1 RCT, the relative and absolute effects of EBL compared with TTSC on 7-day further bleeding and 7-day rebleeding were highly uncertain (RR, 2.10; 95% CI, 0.21 to 21.39; ARR, 52 more per 1000; 95% CI, 38 fewer to 971 more).108
108.
Cho, Y.S. ∙ Chae, H.S. ∙ Kim, H.K. ...
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The relative effects for failure to achieve immediate hemostasis, overall mortality, and adverse effects were not estimable due to double-zero events.107
107.
Park, C.H. ∙ Min, S.W. ∙ Sohn, Y.H. ...
A prospective, randomized trial of endoscopic band ligation vs. epinephrine injection for actively bleeding Mallory-Weiss syndrome
Gastrointest Endosc. 2004; 60:22-27
Full Text
Full Text (PDF)
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Google Scholar
For “immediate hemostasis,” EBL appeared to have no impact compared with TTSC (RR, 1.00; 95% CI, 0.91 to 1.10).108
108.
Cho, Y.S. ∙ Chae, H.S. ∙ Kim, H.K. ...
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The estimates for blood transfusions (MD, 1.30 units; 95% CI, –0.36 to 2.96 units) and length of hospitalization (MD, 0.60 days; 95% CI, –2.02 to 3.22 days) were very imprecise and were not deemed clinically meaningful by the panel.108
108.
Cho, Y.S. ∙ Chae, H.S. ∙ Kim, H.K. ...
Endoscopic band ligation and endoscopic hemoclip placement for patients with Mallory-Weiss syndrome and active bleeding
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A comparative cohort study also found highly uncertain effects of EBL compared with TTSC on 7-day further bleeding and 7-day rebleeding (risk difference [RD], –0.19; 95% CI, –0.34 to 0.03; ARR, 190 fewer per 1000; 95% CI, 340 fewer to 30 more).112
112.
Lecleire, S. ∙ Antonietti, M. ∙ Iwanicki-Caron, I. ...
Endoscopic band ligation could decrease recurrent bleeding in Mallory-Weiss syndrome as compared to haemostasis by hemoclips plus epinephrine
Aliment Pharmacol Ther. 2009; 30:399-405
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Scopus (0)
PubMed
Google Scholar
The relative effects for failure to achieve immediate hemostasis, overall mortality, and adverse effects were not estimable due to double-zero events.112
112.
Lecleire, S. ∙ Antonietti, M. ∙ Iwanicki-Caron, I. ...
Endoscopic band ligation could decrease recurrent bleeding in Mallory-Weiss syndrome as compared to haemostasis by hemoclips plus epinephrine
Aliment Pharmacol Ther. 2009; 30:399-405
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Scopus (0)
PubMed
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EBL appeared to have no impact on “immediate hemostasis” compared with TTSC (1.00; 95% CI, 0.95 to 1.06).112
112.
Lecleire, S. ∙ Antonietti, M. ∙ Iwanicki-Caron, I. ...
Endoscopic band ligation could decrease recurrent bleeding in Mallory-Weiss syndrome as compared to haemostasis by hemoclips plus epinephrine
Aliment Pharmacol Ther. 2009; 30:399-405
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Scopus (0)
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For units of blood transfused, the MD was 0.00 (95% CI, –1.31 to 1.31)112
112.
Lecleire, S. ∙ Antonietti, M. ∙ Iwanicki-Caron, I. ...
Endoscopic band ligation could decrease recurrent bleeding in Mallory-Weiss syndrome as compared to haemostasis by hemoclips plus epinephrine
Aliment Pharmacol Ther. 2009; 30:399-405
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and for the length of hospitalization, the MD was –0.10 (95% CI, –1.67 to 1.47).112
112.
Lecleire, S. ∙ Antonietti, M. ∙ Iwanicki-Caron, I. ...
Endoscopic band ligation could decrease recurrent bleeding in Mallory-Weiss syndrome as compared to haemostasis by hemoclips plus epinephrine
Aliment Pharmacol Ther. 2009; 30:399-405
Crossref
Scopus (0)
PubMed
Google Scholar
Due to the low event rates and the fragility of results in both the RCT and observational study, the panel judged that there was very serious imprecision, rendering the relative and absolute effects of EBL compared with TTSC for active bleeding MWT highly uncertain. Nevertheless, both EBL and TTSC appeared to be highly effective in achieving “immediate hemostasis” with low rates of failure to achieve immediate hemostasis, 7-day rebleeding and 7-day further bleeding (critical outcome).
Certainty in the evidence of effects
Very low certainty of evidence due to serious risk of bias and very serious imprecision.
Other evidence-to-decision criteria and considerations
The panel assumed there is no important uncertainty or variability in how patients value the critical outcome of avoiding further bleeding. The panel considered the relative costs negligible. No published cost-effectiveness analyses were available. The panel judged EBL compared with TTSC as probably acceptable and feasible and unlikely to negatively affect equity-deserving groups.
Conclusions and research needs
For patients with active bleeding from MWT (oozing or spurting), the panel could not determine whether the balance of effects favored EBL or TTSC due to the very low certainty of the evidence. Consequently, the panel made a conditional recommendation for either EBL or TTSC. The panel emphasized the need for more comparative studies to increase the certainty of the evidence.
Question 7: Should patients with no active bleeding from MWTs (nonbleeding visible vessels, adherent clots, flat pigmented spots, clean-based ulcers) receive endoscopic hemostatic therapy vs no endoscopic hemostatic therapy?
Recommendation 7A: In patients with MWTs with nonbleeding visible vessels, we suggest against endoscopic hemostatic therapy (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝).
Recommendation 7B: In patients with MWTs with nonbleeding adherent clots, we suggest against endoscopic hemostatic therapy (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝).
Recommendation 7C: In patients with MWTs with nonbleeding clean-based ulcers or flat pigmented spots, we suggest against endoscopic hemostatic therapy (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝).
Recommendation 7A. Endoscopic Hemostatic Therapy vs No Endoscopic Hemostatic Therapy in Non-bleeding Visible Vessels; and Recommendation 7B. Endoscopic Hemostatic Therapy vs No Endoscopic Hemostatic Therapy in Adherent Clots
Evidence summary
We found 1 RCT and 2 retrospective cohort studies that addressed the above 2 questions.100
100.
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The RCT included patients with MWT with nonbleeding visible vessels and compared epinephrine and polidocanol injections against no endoscopic hemostatic therapy.100
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Both cohort studies included patients with various stigmata of MWT and compared endoscopic hemostatic therapies (eg, epinephrine injection, EBL, TTSC, or fibrin glue injection) with medical treatment for UGIB due to MWT, providing subgroup data on nonbleeding visible vessels or adherent clots.103
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Patients who received endoscopic treatment had more severe bleeding, as indicated by higher rates of transfusion.104
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Because these were nonrandomized studies, there may be inherent selection bias that could influence the results.
None of the studies provided data on the critical outcome of further bleeding or the important outcomes of mortality, units of blood transfusions, and length of hospitalization. However, all studies provided data on 7-day rebleeding. One RCT reported adverse effects.100
100.
Llach, J. ∙ Elizalde, J.I. ∙ Guevara, M.C. ...
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We also conducted a proportional meta-analysis of single-arm cohort-type data from 3 studies for 7-day rebleeding.100
100.
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As indirect evidence of the adverse effects of endoscopic hemostatic therapy in patients with nonbleeding stigmata, we performed a proportional meta-analysis of 9 studies involving patients with bleeding MWT who received endoscopic hemostatic therapy.100–102
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Gastrointest Endosc. 2002; 55:842-846
Full Text
Full Text (PDF)
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PubMed
Google Scholar
107.
Park, C.H. ∙ Min, S.W. ∙ Sohn, Y.H. ...
A prospective, randomized trial of endoscopic band ligation vs. epinephrine injection for actively bleeding Mallory-Weiss syndrome
Gastrointest Endosc. 2004; 60:22-27
Full Text
Full Text (PDF)
Scopus (55)
PubMed
Google Scholar
108.
Cho, Y.S. ∙ Chae, H.S. ∙ Kim, H.K. ...
Endoscopic band ligation and endoscopic hemoclip placement for patients with Mallory-Weiss syndrome and active bleeding
World J Gastroenterol. 2008; 14:2080-2084
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Higuchi, N. ∙ Akahoshi, K. ∙ Sumida, Y. ...
Endoscopic band ligation therapy for upper gastrointestinal bleeding related to Mallory-Weiss syndrome
Surg Endosc. 2006; 20:1431-1434
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,111
111.
Yamaguchi, Y. ∙ Yamato, T. ∙ Katsumi, N. ...
Endoscopic hemoclipping for upper GI bleeding due to Mallory-Weiss syndrome
Gastrointest Endosc. 2001; 53:427-430
Full Text
Full Text (PDF)
Scopus (78)
PubMed
Google Scholar
,113
113.
Dwivedi, M. ∙ Misra, S.P.
Mallory-Weiss syndrome: clinical features and management
J Assoc Physicians India. 1999; 47:397-399
PubMed
Google Scholar
In addition, to assess the adverse effects of no endoscopic hemostatic therapy, we performed a proportional meta-analysis using single-arm cohort-type data from 3 studies involving patients with flat-pigmented spot or clean-based ulcers who did not receive this therapy.100–102
100.
Llach, J. ∙ Elizalde, J.I. ∙ Guevara, M.C. ...
Endoscopic injection therapy in bleeding Mallory-Weiss syndrome: a randomized controlled trial
Gastrointest Endosc. 2001; 54:679-681
Full Text
Full Text (PDF)
Scopus (56)
PubMed
Google Scholar
101.
Laine, L.
Multipolar electrocoagulation in the treatment of active upper gastrointestinal tract hemorrhage. A prospective controlled trial
N Engl J Med. 1987; 316:1613-1617
Crossref
PubMed
Google Scholar
102.
Bataller, R. ∙ Llach, J. ∙ Salmeron, J.M. ...
Endoscopic sclerotherapy in upper gastrointestinal bleeding due to the Mallory-Weiss syndrome
Am J Gastroenterol. 1994; 89:2147-2150
PubMed
Google Scholar
The results of the proportional meta-analyses were provided as supplementary data in the Evidence Profile Table.
The EtD framework is available in Appendix 4.
Benefits, harms, and burden
Based on 1 RCT, the relative and absolute effects of endoscopic hemostatic therapy compared with no endoscopic hemostatic therapy on 7-day rebleeding in patients with MWT and visible vessels were highly uncertain due to the small sample size and very low event rates (RR, 0.30; 95% CI, 0.04 to 2.31; ARR, 262 fewer per 1000; 95% CI, 360 fewer to 491 more).100
100.
Llach, J. ∙ Elizalde, J.I. ∙ Guevara, M.C. ...
Endoscopic injection therapy in bleeding Mallory-Weiss syndrome: a randomized controlled trial
Gastrointest Endosc. 2001; 54:679-681
Full Text
Full Text (PDF)
Scopus (56)
PubMed
Google Scholar
A systematic review of 2 comparative cohort studies also found highly uncertain effects of endoscopic hemostatic therapy compared with no endoscopic hemostatic therapy on 7-day rebleeding in patients with nonbleeding visible vessels or adherent clots due to very low event rates with zero event in the group receiving no endoscopic hemostatic therapy (RD, 0.03; 95% CI, –0.06 to 0.11; ARR, 30 more per 1000; 95% CI, 60 less to 110 more).103
103.
Chung, I.K. ∙ Kim, E.J. ∙ Hwang, K.Y. ...
Evaluation of endoscopic hemostasis in upper gastrointestinal bleeding related to Mallory-Weiss syndrome
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104.
Lee, S. ∙ Ahn, J.Y. ∙ Jung, H.Y. ...
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In 1 study, none of the patients experienced rebleeding, regardless of whether they received endoscopic hemostatic therapy or medical treatment.103
103.
Chung, I.K. ∙ Kim, E.J. ∙ Hwang, K.Y. ...
Evaluation of endoscopic hemostasis in upper gastrointestinal bleeding related to Mallory-Weiss syndrome
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In the other study, there was 1 case of rebleeding in a patient with a nonbleeding visible vessel who had undergone endoscopic hemostatic therapy.104
104.
Lee, S. ∙ Ahn, J.Y. ∙ Jung, H.Y. ...
Effective endoscopic treatment of Mallory-Weiss syndrome using Glasgow-Blatchford score and Forrest classification
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Google Scholar
In contrast, no rebleeding cases occurred in patients with adherent clots.103
103.
Chung, I.K. ∙ Kim, E.J. ∙ Hwang, K.Y. ...
Evaluation of endoscopic hemostasis in upper gastrointestinal bleeding related to Mallory-Weiss syndrome
Endoscopy. 2002; 34:474-479
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PubMed
Google Scholar
,104
104.
Lee, S. ∙ Ahn, J.Y. ∙ Jung, H.Y. ...
Effective endoscopic treatment of Mallory-Weiss syndrome using Glasgow-Blatchford score and Forrest classification
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Crossref
Scopus (9)
PubMed
Google Scholar
Proportional meta-analyses of single-arm cohort-type data suggested that no endoscopic hemostatic therapy was associated with very low 7-day rebleeding rates (pooled rebleeding rates of 6% for nonbleeding visible vessels and 0% for adherent clots).100
100.
Llach, J. ∙ Elizalde, J.I. ∙ Guevara, M.C. ...
Endoscopic injection therapy in bleeding Mallory-Weiss syndrome: a randomized controlled trial
Gastrointest Endosc. 2001; 54:679-681
Full Text
Full Text (PDF)
Scopus (56)
PubMed
Google Scholar
,103
103.
Chung, I.K. ∙ Kim, E.J. ∙ Hwang, K.Y. ...
Evaluation of endoscopic hemostasis in upper gastrointestinal bleeding related to Mallory-Weiss syndrome
Endoscopy. 2002; 34:474-479
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PubMed
Google Scholar
,104
104.
Lee, S. ∙ Ahn, J.Y. ∙ Jung, H.Y. ...
Effective endoscopic treatment of Mallory-Weiss syndrome using Glasgow-Blatchford score and Forrest classification
J Dig Dis. 2016; 17:676-684
Crossref
Scopus (9)
PubMed
Google Scholar
Based on 1 RCT, adverse effects related to endoscopic hemostatic therapy were not estimable due to double-zero events.100
100.
Llach, J. ∙ Elizalde, J.I. ∙ Guevara, M.C. ...
Endoscopic injection therapy in bleeding Mallory-Weiss syndrome: a randomized controlled trial
Gastrointest Endosc. 2001; 54:679-681
Full Text
Full Text (PDF)
Scopus (56)
PubMed
Google Scholar
Based on proportional meta-analyses of single-arm cohort-type data, the pooled rate of adverse effects with endoscopic hemostatic therapy was 5% (95% CI, 2%–12%),100–102
100.
Llach, J. ∙ Elizalde, J.I. ∙ Guevara, M.C. ...
Endoscopic injection therapy in bleeding Mallory-Weiss syndrome: a randomized controlled trial
Gastrointest Endosc. 2001; 54:679-681
Full Text
Full Text (PDF)
Scopus (56)
PubMed
Google Scholar
101.
Laine, L.
Multipolar electrocoagulation in the treatment of active upper gastrointestinal tract hemorrhage. A prospective controlled trial
N Engl J Med. 1987; 316:1613-1617
Crossref
PubMed
Google Scholar
102.
Bataller, R. ∙ Llach, J. ∙ Salmeron, J.M. ...
Endoscopic sclerotherapy in upper gastrointestinal bleeding due to the Mallory-Weiss syndrome
Am J Gastroenterol. 1994; 89:2147-2150
PubMed
Google Scholar
,106–109
106.
Huang, S.P. ∙ Wang, H.P. ∙ Lee, Y.C. ...
Endoscopic hemoclip placement and epinephrine injection for Mallory-Weiss syndrome with active bleeding
Gastrointest Endosc. 2002; 55:842-846
Full Text
Full Text (PDF)
Scopus (80)
PubMed
Google Scholar
107.
Park, C.H. ∙ Min, S.W. ∙ Sohn, Y.H. ...
A prospective, randomized trial of endoscopic band ligation vs. epinephrine injection for actively bleeding Mallory-Weiss syndrome
Gastrointest Endosc. 2004; 60:22-27
Full Text
Full Text (PDF)
Scopus (55)
PubMed
Google Scholar
108.
Cho, Y.S. ∙ Chae, H.S. ∙ Kim, H.K. ...
Endoscopic band ligation and endoscopic hemoclip placement for patients with Mallory-Weiss syndrome and active bleeding
World J Gastroenterol. 2008; 14:2080-2084
Crossref
Scopus (44)
PubMed
Google Scholar
109.
Higuchi, N. ∙ Akahoshi, K. ∙ Sumida, Y. ...
Endoscopic band ligation therapy for upper gastrointestinal bleeding related to Mallory-Weiss syndrome
Surg Endosc. 2006; 20:1431-1434
Crossref
Scopus (23)
PubMed
Google Scholar
,111
111.
Yamaguchi, Y. ∙ Yamato, T. ∙ Katsumi, N. ...
Endoscopic hemoclipping for upper GI bleeding due to Mallory-Weiss syndrome
Gastrointest Endosc. 2001; 53:427-430
Full Text
Full Text (PDF)
Scopus (78)
PubMed
Google Scholar
,113
113.
Dwivedi, M. ∙ Misra, S.P.
Mallory-Weiss syndrome: clinical features and management
J Assoc Physicians India. 1999; 47:397-399
PubMed
Google Scholar
and without endoscopic hemostatic therapy was 0% (95% CI, 0%–0.04%).100–102
100.
Llach, J. ∙ Elizalde, J.I. ∙ Guevara, M.C. ...
Endoscopic injection therapy in bleeding Mallory-Weiss syndrome: a randomized controlled trial
Gastrointest Endosc. 2001; 54:679-681
Full Text
Full Text (PDF)
Scopus (56)
PubMed
Google Scholar
101.
Laine, L.
Multipolar electrocoagulation in the treatment of active upper gastrointestinal tract hemorrhage. A prospective controlled trial
N Engl J Med. 1987; 316:1613-1617
Crossref
PubMed
Google Scholar
102.
Bataller, R. ∙ Llach, J. ∙ Salmeron, J.M. ...
Endoscopic sclerotherapy in upper gastrointestinal bleeding due to the Mallory-Weiss syndrome
Am J Gastroenterol. 1994; 89:2147-2150
PubMed
Google Scholar
The panel judged these effect estimates as highly uncertain due to low event rates and the fragility of the RCT and observational data. For nonbleeding visible vessels in MWT, the panel determined that the desirable and undesirable effects of endoscopic hemostatic therapy compared with no therapy were unclear. For adherent clots, the panel determined that both the desirable and undesirable effects of endoscopic hemostatic therapy were trivial.
Certainty in the evidence of effects
Very low certainty of evidence due to serious risk of bias and very serious imprecision.
Other evidence-to-decision criteria and considerations
The panel assumed patients likely do not have important uncertainty or variability in valuing the main outcome of rebleeding. The costs of endoscopic hemostatic therapy were deemed moderate, yet no studies have assessed its cost-effectiveness compared with no endoscopic hemostatic therapy. The panel judged endoscopic hemostatic therapy likely to be acceptable, feasible, and unlikely to adversely impact equity-deserving groups.
Conclusions and research needs
For patients with MWT with nonbleeding visible vessels, the panel could not determine whether the balance of effects favored endoscopic hemostatic therapy or no endoscopic hemostatic therapy due to very low certainty evidence and the fragility of results. For patients with MWT with adherent clots, the panel judged that the balance of effects does not favor either endoscopic hemostatic therapy or no endoscopic hemostatic therapy. Although the evidence comparing endoscopic and no endoscopic therapy for MWT with nonbleeding visible vessels was extremely limited, fragile, and of very low certainty, the rebleeding rates without endoscopic therapy were very low for both nonbleeding visible vessels and adherent clots—far lower than in ulcers with the same stigmata. Given these very low rebleeding rates, the panel would not anticipate meaningful ARRs to be demonstrated with endoscopic hemostatic therapy and expressed concerns about potentially inducing bleeding with such therapy.
Given the uncertainty surrounding the benefits and the potential harms of endoscopic hemostatic therapy and its moderate costs, the panel issued a conditional recommendation against endoscopic hemostatic therapy for MWT with nonbleeding vessels or adherent clots. This decision aligns with PUB guidelines concerning adherent clots, which advise against endoscopic intervention for lesions with similarly low rebleeding rates.8
8.
Barkun, A.N. ∙ Almadi, M. ∙ Kuipers, E.J. ...
Management of nonvariceal upper gastrointestinal bleeding: guideline recommendations from the International Consensus Group
Ann Intern Med. 2019; 171:805-822
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PubMed
Google Scholar
,11
11.
Laine, L. ∙ Barkun, A.N. ∙ Saltzman, J.R. ...
ACG clinical guideline: upper gastrointestinal and ulcer bleeding
Am J Gastroenterol. 2021; 116:899-917
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Google Scholar
However, for nonbleeding visible vessels, this recommendation diverges from PUB guidelines, as the rebleeding risk in MWT appears to be much lower, thereby justifying the decision against endoscopic therapy. The panel highlighted the need for more comparative studies to increase the certainty of the evidence. In addition, understanding the natural history of visible vessels or adherent clots in the context of MWT, especially in the era of proton pump inhibitor treatment, is crucial.
Recommendation 7C. Endoscopic Hemostatic Therapy vs No Endoscopic Hemostatic Therapy in Clean-Based Ulcers or Pigmented Spots
Evidence summary
We did not find any RCT that addressed this question. One comparative cohort study included 52 patients who did not receive endoscopic hemostatic therapy and only 2 who did, making comparative analysis unfeasible.104
104.
Lee, S. ∙ Ahn, J.Y. ∙ Jung, H.Y. ...
Effective endoscopic treatment of Mallory-Weiss syndrome using Glasgow-Blatchford score and Forrest classification
J Dig Dis. 2016; 17:676-684
Crossref
Scopus (9)
PubMed
Google Scholar
Proportional meta-analyses of single-arm cohort-type data were conducted on 2 studies involving patients with MWT who had flat pigmented spots or clean-based ulcers and received endoscopic hemostatic therapy and on 4 studies involving those who did not receive endoscopic hemostatic therapy.102
102.
Bataller, R. ∙ Llach, J. ∙ Salmeron, J.M. ...
Endoscopic sclerotherapy in upper gastrointestinal bleeding due to the Mallory-Weiss syndrome
Am J Gastroenterol. 1994; 89:2147-2150
PubMed
Google Scholar
,104
104.
Lee, S. ∙ Ahn, J.Y. ∙ Jung, H.Y. ...
Effective endoscopic treatment of Mallory-Weiss syndrome using Glasgow-Blatchford score and Forrest classification
J Dig Dis. 2016; 17:676-684
Crossref
Scopus (9)
PubMed
Google Scholar
,114
114.
Inaspettato, G. ∙ Prattico, F. ∙ Rodella, L. ...
Mallory-Weiss syndrome: retrospective analysis of 97 cases
Giornale Italiano di Endoscopia Digestiva. 1996; 19:113-118
Google Scholar
,115
115.
Bharucha, A.E. ∙ Gostout, C.J. ∙ Balm, R.K.
Clinical and endoscopic risk factors in the Mallory-Weiss syndrome
Am J Gastroenterol. 1997; 92:805-808
PubMed
Google Scholar
Overall, 6 patients received endoscopic hemostatic therapy, and 154 did not.102
102.
Bataller, R. ∙ Llach, J. ∙ Salmeron, J.M. ...
Endoscopic sclerotherapy in upper gastrointestinal bleeding due to the Mallory-Weiss syndrome
Am J Gastroenterol. 1994; 89:2147-2150
PubMed
Google Scholar
,104
104.
Lee, S. ∙ Ahn, J.Y. ∙ Jung, H.Y. ...
Effective endoscopic treatment of Mallory-Weiss syndrome using Glasgow-Blatchford score and Forrest classification
J Dig Dis. 2016; 17:676-684
Crossref
Scopus (9)
PubMed
Google Scholar
,114
114.
Inaspettato, G. ∙ Prattico, F. ∙ Rodella, L. ...
Mallory-Weiss syndrome: retrospective analysis of 97 cases
Giornale Italiano di Endoscopia Digestiva. 1996; 19:113-118
Google Scholar
,115
115.
Bharucha, A.E. ∙ Gostout, C.J. ∙ Balm, R.K.
Clinical and endoscopic risk factors in the Mallory-Weiss syndrome
Am J Gastroenterol. 1997; 92:805-808
PubMed
Google Scholar
None of the studies provided data on the critical outcome of further bleeding, or important outcomes of mortality, units of blood transfusions received and the length of hospitalization. However, all studies provided data on 7-day rebleeding.
There was no direct evidence of adverse effects from endoscopic hemostatic therapy in patients with MWT with flat-pigmented spots or clean-based ulcers. Therefore, indirect evidence was obtained through proportional meta-analyses of single-arm cohort-type data from 9 studies on patients with bleeding MWT who underwent endoscopic hemostatic therapy.100–102
100.
Llach, J. ∙ Elizalde, J.I. ∙ Guevara, M.C. ...
Endoscopic injection therapy in bleeding Mallory-Weiss syndrome: a randomized controlled trial
Gastrointest Endosc. 2001; 54:679-681
Full Text
Full Text (PDF)
Scopus (56)
PubMed
Google Scholar
101.
Laine, L.
Multipolar electrocoagulation in the treatment of active upper gastrointestinal tract hemorrhage. A prospective controlled trial
N Engl J Med. 1987; 316:1613-1617
Crossref
PubMed
Google Scholar
102.
Bataller, R. ∙ Llach, J. ∙ Salmeron, J.M. ...
Endoscopic sclerotherapy in upper gastrointestinal bleeding due to the Mallory-Weiss syndrome
Am J Gastroenterol. 1994; 89:2147-2150
PubMed
Google Scholar
,106–109
106.
Huang, S.P. ∙ Wang, H.P. ∙ Lee, Y.C. ...
Endoscopic hemoclip placement and epinephrine injection for Mallory-Weiss syndrome with active bleeding
Gastrointest Endosc. 2002; 55:842-846
Full Text
Full Text (PDF)
Scopus (80)
PubMed
Google Scholar
107.
Park, C.H. ∙ Min, S.W. ∙ Sohn, Y.H. ...
A prospective, randomized trial of endoscopic band ligation vs. epinephrine injection for actively bleeding Mallory-Weiss syndrome
Gastrointest Endosc. 2004; 60:22-27
Full Text
Full Text (PDF)
Scopus (55)
PubMed
Google Scholar
108.
Cho, Y.S. ∙ Chae, H.S. ∙ Kim, H.K. ...
Endoscopic band ligation and endoscopic hemoclip placement for patients with Mallory-Weiss syndrome and active bleeding
World J Gastroenterol. 2008; 14:2080-2084
Crossref
Scopus (44)
PubMed
Google Scholar
109.
Higuchi, N. ∙ Akahoshi, K. ∙ Sumida, Y. ...
Endoscopic band ligation therapy for upper gastrointestinal bleeding related to Mallory-Weiss syndrome
Surg Endosc. 2006; 20:1431-1434
Crossref
Scopus (23)
PubMed
Google Scholar
,111
111.
Yamaguchi, Y. ∙ Yamato, T. ∙ Katsumi, N. ...
Endoscopic hemoclipping for upper GI bleeding due to Mallory-Weiss syndrome
Gastrointest Endosc. 2001; 53:427-430
Full Text
Full Text (PDF)
Scopus (78)
PubMed
Google Scholar
,113
113.
Dwivedi, M. ∙ Misra, S.P.
Mallory-Weiss syndrome: clinical features and management
J Assoc Physicians India. 1999; 47:397-399
PubMed
Google Scholar
A proportional meta-analysis was also performed of single-arm cohort-type data from 3 studies on patients with flat-pigmented spot or clean-based ulcers who did not receive endoscopic hemostatic therapy.100–102
100.
Llach, J. ∙ Elizalde, J.I. ∙ Guevara, M.C. ...
Endoscopic injection therapy in bleeding Mallory-Weiss syndrome: a randomized controlled trial
Gastrointest Endosc. 2001; 54:679-681
Full Text
Full Text (PDF)
Scopus (56)
PubMed
Google Scholar
101.
Laine, L.
Multipolar electrocoagulation in the treatment of active upper gastrointestinal tract hemorrhage. A prospective controlled trial
N Engl J Med. 1987; 316:1613-1617
Crossref
PubMed
Google Scholar
102.
Bataller, R. ∙ Llach, J. ∙ Salmeron, J.M. ...
Endoscopic sclerotherapy in upper gastrointestinal bleeding due to the Mallory-Weiss syndrome
Am J Gastroenterol. 1994; 89:2147-2150
PubMed
Google Scholar
The EtD framework is in Appendix 4.
Benefits, harms, and burden
The pooled rebleeding rate for patients who received endoscopic hemostatic therapy was 0% (95% CI, 0%–0.5%; range, 0%–0%).104
104.
Lee, S. ∙ Ahn, J.Y. ∙ Jung, H.Y. ...
Effective endoscopic treatment of Mallory-Weiss syndrome using Glasgow-Blatchford score and Forrest classification
J Dig Dis. 2016; 17:676-684
Crossref
Scopus (9)
PubMed
Google Scholar
,115
115.
Bharucha, A.E. ∙ Gostout, C.J. ∙ Balm, R.K.
Clinical and endoscopic risk factors in the Mallory-Weiss syndrome
Am J Gastroenterol. 1997; 92:805-808
PubMed
Google Scholar
For those who did not receive endoscopic hemostatic therapy, the pooled rebleeding rate was also 0% (95% CI, 0%–0.02%; range, 0%–0%).102
102.
Bataller, R. ∙ Llach, J. ∙ Salmeron, J.M. ...
Endoscopic sclerotherapy in upper gastrointestinal bleeding due to the Mallory-Weiss syndrome
Am J Gastroenterol. 1994; 89:2147-2150
PubMed
Google Scholar
,104
104.
Lee, S. ∙ Ahn, J.Y. ∙ Jung, H.Y. ...
Effective endoscopic treatment of Mallory-Weiss syndrome using Glasgow-Blatchford score and Forrest classification
J Dig Dis. 2016; 17:676-684
Crossref
Scopus (9)
PubMed
Google Scholar
,114
114.
Inaspettato, G. ∙ Prattico, F. ∙ Rodella, L. ...
Mallory-Weiss syndrome: retrospective analysis of 97 cases
Giornale Italiano di Endoscopia Digestiva. 1996; 19:113-118
Google Scholar
,115
115.
Bharucha, A.E. ∙ Gostout, C.J. ∙ Balm, R.K.
Clinical and endoscopic risk factors in the Mallory-Weiss syndrome
Am J Gastroenterol. 1997; 92:805-808
PubMed
Google Scholar
The pooled rate of adverse effects with endoscopic hemostatic therapy was 5% (95% CI, 2%–12%),100–102
100.
Llach, J. ∙ Elizalde, J.I. ∙ Guevara, M.C. ...
Endoscopic injection therapy in bleeding Mallory-Weiss syndrome: a randomized controlled trial
Gastrointest Endosc. 2001; 54:679-681
Full Text
Full Text (PDF)
Scopus (56)
PubMed
Google Scholar
101.
Laine, L.
Multipolar electrocoagulation in the treatment of active upper gastrointestinal tract hemorrhage. A prospective controlled trial
N Engl J Med. 1987; 316:1613-1617
Crossref
PubMed
Google Scholar
102.
Bataller, R. ∙ Llach, J. ∙ Salmeron, J.M. ...
Endoscopic sclerotherapy in upper gastrointestinal bleeding due to the Mallory-Weiss syndrome
Am J Gastroenterol. 1994; 89:2147-2150
PubMed
Google Scholar
,106–109
106.
Huang, S.P. ∙ Wang, H.P. ∙ Lee, Y.C. ...
Endoscopic hemoclip placement and epinephrine injection for Mallory-Weiss syndrome with active bleeding
Gastrointest Endosc. 2002; 55:842-846
Full Text
Full Text (PDF)
Scopus (80)
PubMed
Google Scholar
107.
Park, C.H. ∙ Min, S.W. ∙ Sohn, Y.H. ...
A prospective, randomized trial of endoscopic band ligation vs. epinephrine injection for actively bleeding Mallory-Weiss syndrome
Gastrointest Endosc. 2004; 60:22-27
Full Text
Full Text (PDF)
Scopus (55)
PubMed
Google Scholar
108.
Cho, Y.S. ∙ Chae, H.S. ∙ Kim, H.K. ...
Endoscopic band ligation and endoscopic hemoclip placement for patients with Mallory-Weiss syndrome and active bleeding
World J Gastroenterol. 2008; 14:2080-2084
Crossref
Scopus (44)
PubMed
Google Scholar
109.
Higuchi, N. ∙ Akahoshi, K. ∙ Sumida, Y. ...
Endoscopic band ligation therapy for upper gastrointestinal bleeding related to Mallory-Weiss syndrome
Surg Endosc. 2006; 20:1431-1434
Crossref
Scopus (23)
PubMed
Google Scholar
,111
111.
Yamaguchi, Y. ∙ Yamato, T. ∙ Katsumi, N. ...
Endoscopic hemoclipping for upper GI bleeding due to Mallory-Weiss syndrome
Gastrointest Endosc. 2001; 53:427-430
Full Text
Full Text (PDF)
Scopus (78)
PubMed
Google Scholar
,113
113.
Dwivedi, M. ∙ Misra, S.P.
Mallory-Weiss syndrome: clinical features and management
J Assoc Physicians India. 1999; 47:397-399
PubMed
Google Scholar
and without endoscopic hemostatic therapy was 0% (95% CI, 0%–0.04%).100–102
100.
Llach, J. ∙ Elizalde, J.I. ∙ Guevara, M.C. ...
Endoscopic injection therapy in bleeding Mallory-Weiss syndrome: a randomized controlled trial
Gastrointest Endosc. 2001; 54:679-681
Full Text
Full Text (PDF)
Scopus (56)
PubMed
Google Scholar
101.
Laine, L.
Multipolar electrocoagulation in the treatment of active upper gastrointestinal tract hemorrhage. A prospective controlled trial
N Engl J Med. 1987; 316:1613-1617
Crossref
PubMed
Google Scholar
102.
Bataller, R. ∙ Llach, J. ∙ Salmeron, J.M. ...
Endoscopic sclerotherapy in upper gastrointestinal bleeding due to the Mallory-Weiss syndrome
Am J Gastroenterol. 1994; 89:2147-2150
PubMed
Google Scholar
The panel judged that the desirable effects of endoscopic hemostatic therapy were trivial, and the undesirable effects were small.
Certainty in the evidence of effects
Very low certainty of evidence due to serious risk of bias, very serious imprecision, and serious indirectness due to the lack of comparative data.
Other evidence-to-decision criteria and considerations
The panel assumed patients likely do not have important uncertainty or variability in valuing the main outcome of rebleeding. The costs of endoscopic hemostatic therapy were deemed moderate. There was no published cost-effectiveness analysis. However, when benefits are minimal, even modest procedure costs can lead to poor cost-effectiveness ratios. The panel judged endoscopic hemostatic therapy likely to be acceptable, feasible, and unlikely to adversely impact equity-deserving groups.
Conclusions and research needs
For patients with MWT with flat pigmented spots or clean-based ulcers, the panel determined that the balance of desirable and undesirable effects probably favors no endoscopic hemostatic therapy. These patients have a very low risk of rebleeding, which significantly reduces the potential benefits and increases the potential harms of endoscopic intervention. As a result, the panel issued a conditional recommendation against endoscopic hemostatic therapy based on very low certainty evidence. This aligns with PUB guidelines, which advise against endoscopic intervention for such lesions.8
8.
Barkun, A.N. ∙ Almadi, M. ∙ Kuipers, E.J. ...
Management of nonvariceal upper gastrointestinal bleeding: guideline recommendations from the International Consensus Group
Ann Intern Med. 2019; 171:805-822
Crossref
Scopus (396)
PubMed
Google Scholar
,11
11.
Laine, L. ∙ Barkun, A.N. ∙ Saltzman, J.R. ...
ACG clinical guideline: upper gastrointestinal and ulcer bleeding
Am J Gastroenterol. 2021; 116:899-917
Crossref
Scopus (338)
PubMed
Google Scholar
Dieulafoy’s Lesion
Question 8: Should patients with UGIB from DL receive one endoscopic hemostatic therapy vs another endoscopic hemostatic therapy?
Recommendation 8A: In patients with UGIB from DL, we suggest either EBL with or without epinephrine injection or endoscopic TTSC placement with or without epinephrine injection (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝).
Recommendation 8B: In patients with UGIB from DL, we suggest either mechanical devices (EBL or endoscopic TTSC placement) with or without epinephrine injection or contact thermal devices (heater probe and bipolar electrocoagulation) with or without epinephrine injection (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝).
Recommendation 8C: In patients with UGIB from DL, we cannot make a recommendation for or against the use of cap-mounted clips over conventional endoscopic hemostatic therapy.
Recommendation 8D: In patients with UGIB from DL, we suggest either mechanical devices (EBL or endoscopic TTSC placement) with or without epinephrine injection or injection of sclerosants with or without epinephrine injection (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝).
Recommendation 8A. Endoscopic Band Ligation With or Without Epinephrine Injection vs Endoscopic Through-the-Scope Clip Placement With or Without Epinephrine Injection
Evidence summary
We identified 2 systematic reviews that addressed this question.116
116.
Barakat, M. ∙ Hamed, A. ∙ Shady, A. ...
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Yuan, Y. ∙ Wang, C. ∙ Hunt, R.H.
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The first review combined observational data with RCT data in its meta-analyses, introducing potential bias due to the inherent differences in study designs.116
116.
Barakat, M. ∙ Hamed, A. ∙ Shady, A. ...
Endoscopic band ligation versus endoscopic hemoclip placement for Dieulafoy's lesion: a meta-analysis
Eur J Gastroenterol Hepatol. 2018; 30:995-996
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The second review included 1 RCT relevant to this question.117
117.
Yuan, Y. ∙ Wang, C. ∙ Hunt, R.H.
Endoscopic clipping for acute nonvariceal upper-GI bleeding: a meta-analysis and critical appraisal of randomized controlled trials
Gastrointest Endosc. 2008; 68:339-351
Full Text
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To provide a more robust comparison between EBL and TTSC in patients with DL, we conducted our own systematic reviews and meta-analyses that included 2 RCTs and 2 comparative cohort studies, analyzed separately.118–121
118.
Ahn, D.W. ∙ Lee, S.H. ∙ Park, Y.S. ...
Hemostatic efficacy and clinical outcome of endoscopic treatment of Dieulafoy's lesions: comparison of endoscopic hemoclip placement and endoscopic band ligation
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119.
Ji, J.S. ∙ Kim, H.K. ∙ Kim, S.S. ...
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Epinephrine injection was used in some patients in both groups to reduce active bleeding and improve visualization, but no subgroup data for patients treated with and without epinephrine were available.118–121
118.
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Hemostatic efficacy and clinical outcome of endoscopic treatment of Dieulafoy's lesions: comparison of endoscopic hemoclip placement and endoscopic band ligation
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Full Text
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Ji, J.S. ∙ Kim, H.K. ∙ Kim, S.S. ...
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One comparative cohort study included in the first review was excluded from our meta-analyses, as it only provided data for 3 patients treated with EBL and 9 patients treated with TTSC, falling below our predetermined threshold of 10 patients for each treatment group.122
122.
Chung, I.K. ∙ Kim, E.J. ∙ Lee, M.S. ...
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All studies reported further bleeding, failure to achieve immediate hemostasis, rebleeding, and mortality.118–121
118.
Ahn, D.W. ∙ Lee, S.H. ∙ Park, Y.S. ...
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Because all rebleeding episodes occurred within 7 days, we classified all rebleeding and further bleeding events as 7-day outcomes to avoid double counting 7-day and 30-day outcomes. One RCT did not report the need for additional hemostatic therapy.120
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One RCT reported adverse effects.121
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In addition, we conducted proportional meta-analyses of single-arm cohort-type data from 10 studies on EBL118–121
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The results of the proportional meta-analyses were presented as supplementary data in the Evidence Profile Table, but were not included in the EtD framework due to the lack of direct comparison. The EtD framework is in Appendix 5.
Benefits, harms, and burden
Based on meta-analyses of 2 RCTs, the relative and absolute effects of EBL compared with TTSC on 7-day further bleeding and rebleeding in patients with DL were highly uncertain due to the small sample size and very low event rates (RR, 0.66; 95% CI, 0.09–5.03; ARR, 22 fewer per 1000; 95% CI, 59 fewer to 260 more).120
120.
Liu, D. ∙ Lu, F. ∙ Ou, D. ...
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Thirty-day mortality and adverse effects were not estimable due to double-zero events. Failure to achieve immediate hemostasis was also not estimable due to double-zero events. When considering “immediate hemostasis,” EBL showed no difference compared with TTSC (RR, 1.00; 95% CI, 0.92–1.09; ARR, 1 fewer per 1000; 95% CI, 1 fewer to 1 fewer). EBL also appeared to have no impact on additional hemostatic therapy (RR, 1.00; 95% CI, 0.07–14.34; ARR, 0 fewer per 1000; 95% CI, 72 fewer to 1000 more), although this estimate was very imprecise.121
121.
Park, C.H. ∙ Joo, Y.E. ∙ Kim, H.S. ...
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A systematic review of 2 comparative cohort studies suggested that EBL may reduce 7-day further bleeding, rebleeding, and additional hemostatic therapy in patients with DL.118
118.
Ahn, D.W. ∙ Lee, S.H. ∙ Park, Y.S. ...
Hemostatic efficacy and clinical outcome of endoscopic treatment of Dieulafoy's lesions: comparison of endoscopic hemoclip placement and endoscopic band ligation
Gastrointest Endosc. 2012; 75:32-38
Full Text
Full Text (PDF)
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119.
Ji, J.S. ∙ Kim, H.K. ∙ Kim, S.S. ...
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The RR was the same for all 3 outcomes (RR, 0.22; 95% CI, 0.05–0.95, ARR, 166 fewer per 1000; 95% CI, 202 fewer to 11 fewer).118
118.
Ahn, D.W. ∙ Lee, S.H. ∙ Park, Y.S. ...
Hemostatic efficacy and clinical outcome of endoscopic treatment of Dieulafoy's lesions: comparison of endoscopic hemoclip placement and endoscopic band ligation
Gastrointest Endosc. 2012; 75:32-38
Full Text
Full Text (PDF)
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,119
119.
Ji, J.S. ∙ Kim, H.K. ∙ Kim, S.S. ...
Clinical outcome of endoscopic management of duodenal Dieulafoy's lesions: endoscopic band ligation versus endoscopic hemoclip placement
Surg Endosc. 2016; 30:3526-3531
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Failure to achieve immediate hemostasis and 30-day mortality were not estimable due to double-zero events. Compared with TTSC, EBL showed no difference in achieving immediate hemostasis (RR, 1.00; 95% CI, 0.95–1.06; ARR, 1 fewer per 1000; 95% CI, 1 fewer to 1 fewer).118
118.
Ahn, D.W. ∙ Lee, S.H. ∙ Park, Y.S. ...
Hemostatic efficacy and clinical outcome of endoscopic treatment of Dieulafoy's lesions: comparison of endoscopic hemoclip placement and endoscopic band ligation
Gastrointest Endosc. 2012; 75:32-38
Full Text
Full Text (PDF)
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,119
119.
Ji, J.S. ∙ Kim, H.K. ∙ Kim, S.S. ...
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The panel prioritized the systematic review of RCTs over cohort studies, given the limitations and inherent selection bias with the latter. They judged the effect estimates of 7-day further bleeding and rebleeding highly uncertain due to low event rates and the fragility of the RCT and observational data. Yet, both EBL and TTSC appeared highly effective in achieving immediate hemostasis. Therefore, the desirable effects of EBL compared with TTSC remain unclear, while the undesirable effects of EBL relative to TTSC were deemed trivial.
Certainty in the evidence of effects
Very low certainty of evidence due to serious risk of bias and very serious imprecision.
Other evidence-to-decision criteria and considerations
There was no research evidence on patients’ values and preferences in the context of DL. However, the panel assumed patients likely do not have important uncertainty or variability in valuing the critical outcome of further bleeding. The cost difference between EBL and TTSC was considered negligible, with no cost-effectiveness studies comparing the 2. The panel judged EBL probably acceptable, feasible, and unlikely to adversely impact equity-deserving groups.
Conclusions and research needs
For patients with UGIB from DL, the panel could not determine whether the balance of effects favored EBL or TTSC due to very low certainty evidence. Consequently, the panel issued a conditional recommendation for either EBL or TTSC, both of which can be performed with or without epinephrine injection. They highlighted the need for more comparative studies. However, conducting RCTs on DL is challenging due to their rarity, intermittent nature of bleeding, and interobserver variability in diagnosis, as they may be misdiagnosed as small ulcers. Until more evidence becomes available, clinicians should consider individual patient factors (eg, lesion location and accessibility), the availability of devices, and clinician expertise when choosing between these 2 interventions. The panel expressed concerns that TTSC might be less effective than EBL, as EBL can aspirate and ligate vessels originating from the submucosal layers.
Recommendation 8B. Mechanical Modalities (Endoscopic Band Ligation/Endoscopic Through-the-Scope Clip Placement) With or Without Epinephrine Injection vs Contact Thermocoagulation With or Without Epinephrine Injection
Evidence summary
We did not find any RCTs addressing this question. Two comparative cohort studies compared mechanical vs contact thermal therapies in patients with UGIB from DL.126
126.
Matsui, S. ∙ Kamisako, T. ∙ Kudo, M. ...
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One prospective cohort study compared EBL with bipolar electrocoagulation in patients with nonvariceal UGIB and provided subgroup data on DL.126
126.
Matsui, S. ∙ Kamisako, T. ∙ Kudo, M. ...
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Gastrointest Endosc. 2002; 55:214-218
Full Text
Full Text (PDF)
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Another study provided subgroup data from 2 RCTs and prospective cohort studies, comparing Doppler endoscopic probe (DEP)-guided treatments vs visually guided hemostasis.138
138.
Nulsen, B. ∙ Jensen, D.M. ∙ Kovacs, T.O.G. ...
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This study included 77 patients who were treated with either mechanical modalities, such as TTSC or cap-mounted clips or thermal modalities, like heater probe or multipolar electrocoagulation, regardless of DEP use.138
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Only 7 patients received cap-mounted clips, and no subgroup data were provided to differentiate between TTSC or cap-mounted clips.138
138.
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Patients in the DEP cohort were more likely to undergo mechanical treatments than the visually guided cohort (100% vs 49.2%; P< .001), which could introduce a confounding factor.138
138.
Nulsen, B. ∙ Jensen, D.M. ∙ Kovacs, T.O.G. ...
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Epinephrine injection was used in some patients in both groups, but no subgroup data were provided.126
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Matsui, S. ∙ Kamisako, T. ∙ Kudo, M. ...
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Gastrointest Endosc. 2002; 55:214-218
Full Text
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138.
Nulsen, B. ∙ Jensen, D.M. ∙ Kovacs, T.O.G. ...
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One study provided data on 7-day further bleeding, 7-day rebleeding, and failure to achieve immediate hemostasis,126
126.
Matsui, S. ∙ Kamisako, T. ∙ Kudo, M. ...
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Gastrointest Endosc. 2002; 55:214-218
Full Text
Full Text (PDF)
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while another reported 30-day rebleeding and adverse effects.138
138.
Nulsen, B. ∙ Jensen, D.M. ∙ Kovacs, T.O.G. ...
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Both studies reported additional hemostatic therapy, but neither reported mortality.126
126.
Matsui, S. ∙ Kamisako, T. ∙ Kudo, M. ...
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Gastrointest Endosc. 2002; 55:214-218
Full Text
Full Text (PDF)
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138.
Nulsen, B. ∙ Jensen, D.M. ∙ Kovacs, T.O.G. ...
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118.
Ahn, D.W. ∙ Lee, S.H. ∙ Park, Y.S. ...
Hemostatic efficacy and clinical outcome of endoscopic treatment of Dieulafoy's lesions: comparison of endoscopic hemoclip placement and endoscopic band ligation
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Full Text (PDF)
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Abstract
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The results were provided as supplementary data in the Evidence Profile Table. However, due to the lack of direct comparative data, they were not included in the EtD framework, which is available in Appendix 5.
Benefits, harms, and burden
Based on 1 cohort study, the relative and absolute effects of mechanical modalities (EBL) compared with thermal modalities on 7-day further bleeding (RD, –0.21; 95% CI, –0.45 to 0.02; ARR, 210 fewer per 1000; 95% CI, 450 fewer to 20 more), failure to achieve immediate hemostasis (RD, –0.14; 95% CI, –0.36 to 0.07; ARR, 140 fewer per 1000; 95% CI, 360 fewer to 70 more), and 7-day rebleeding (RD, –0.07; 95% CI, –0.25 to 0.11; ARR, 70 fewer per 1000; 95% CI, 250 fewer to 110 more) were highly uncertain due to the small sample size and very low event rates.126
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Matsui, S. ∙ Kamisako, T. ∙ Kudo, M. ...
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Based on 1 cohort study, the relative and absolute effects of mechanical modalities (TTSC) compared with thermal modalities on 30-day rebleeding were also highly uncertain (RR, 0.81; 95% CI, 0.40 to 1.64; ARR, 61 fewer per 1000; 95% CI, 194 fewer to 206 more).138
138.
Nulsen, B. ∙ Jensen, D.M. ∙ Kovacs, T.O.G. ...
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A meta-analysis of the 2 cohort studies suggested that mechanical modalities may reduce the need for additional hemostatic therapy, with surgery being the most common form (RR, 0.14; 95% CI, 0.03 to 0.77; ARR, 153 fewer per 1000; 95% CI, 172 fewer to 41 fewer), but the effect estimate was very imprecise.126
126.
Matsui, S. ∙ Kamisako, T. ∙ Kudo, M. ...
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,138
138.
Nulsen, B. ∙ Jensen, D.M. ∙ Kovacs, T.O.G. ...
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For adverse effects, mechanical modalities appeared to have little to no impact compared with thermal modalities (RR, 1.35; 95% CI, 0.13 to 14.23; ARR, 11 more per 1000; 95% CI, 28 fewer to 427 more), but the effect estimate was very imprecise.138
138.
Nulsen, B. ∙ Jensen, D.M. ∙ Kovacs, T.O.G. ...
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The panel judged the desirable and undesirable effects of mechanical modalities as uncertain compared with contact thermal modalities.
Certainty in the evidence of effects
Very low certainty of evidence due to serious risk of bias and very serious imprecision. The evidence was not further downgraded for indirectness related to the small number of patients who received cap-mounted clips.
Other evidence-to-decision criteria and considerations
The panel assumed patients likely do not have important uncertainty or variability in valuing the critical outcome of further bleeding. The cost difference between mechanical and thermal modalities was considered negligible, and no cost-effectiveness studies were found. The panel judged mechanical modalities as probably acceptable, feasible, and unlikely to adversely impact equity-deserving groups.
Conclusions and research needs
For patients with UGIB from DL, the panel was unable to ascertain whether the balance of effects favored mechanical (EBL/TTSC) or contact thermal modalities due to very low certainty evidence from cohort studies and highly imprecise results. Because both modalities appeared effective and safe, the panel issued a conditional recommendation for using either mechanical or contact thermal modalities, with or without epinephrine injection. The panel emphasized the need for more comparative studies, including the need for combination with epinephrine injection.
Recommendation 8C. Cap-Mounted Clip vs Conventional Endoscopic Hemostatic Therapy
Evidence summary
We found 1 RCT comparing cap-mounted clips with conventional endoscopic hemostatic therapy (TTSC or bipolar electrocoagulation) in patients with nonvariceal UGIB.142
142.
Jensen, D.M. ∙ Kovacs, T. ∙ Ghassemi, K.A. ...
Randomized controlled trial of over-the-scope clip as initial treatment of severe nonvariceal upper gastrointestinal bleeding
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Full Text
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A small subgroup of patients had DLs (2 in the cap-mounted clip group and 3 in the standard endoscopic treatment group), but the data were insufficient for meaningful analysis.142
142.
Jensen, D.M. ∙ Kovacs, T. ∙ Ghassemi, K.A. ...
Randomized controlled trial of over-the-scope clip as initial treatment of severe nonvariceal upper gastrointestinal bleeding
Clin Gastroenterol Hepatol. 2021; 19:2315-2323.e2
Full Text
Full Text (PDF)
Scopus (54)
PubMed
Google Scholar
Most evidence for cap-mounted clips in treating DLs came from case series and case reports, with those involving fewer than 10 patients excluded from this guideline’s evidence synthesis. A retrospective cohort study evaluated cap-mounted clips for nonvariceal UGIB, perforations, and fistulas.143
143.
Qiu, J. ∙ Xu, J. ∙ Zhang, Y. ...
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This study also used propensity score matching to compare cap-mounted clips with “standard endoscopic therapy” for DL, although it did not define the latter, aside from excluding epinephrine injection as a monotherapy.143
143.
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Further bleeding, failure to achieve immediate hemostasis, mortality, and the need for additional hemostatic therapy were not reported for the “standard endoscopic therapy” group. All rebleeding occurred within 5 days and was the only outcome available for comparative analysis.143
143.
Qiu, J. ∙ Xu, J. ∙ Zhang, Y. ...
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Adverse effects were not reported.143
143.
Qiu, J. ∙ Xu, J. ∙ Zhang, Y. ...
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Another cohort study included patients from 2 RCTs and prospective cohort studies, comparing DEP-guided treatments with visually guided hemostasis.138
138.
Nulsen, B. ∙ Jensen, D.M. ∙ Kovacs, T.O.G. ...
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It included 77 patients with DLs treated with either mechanical modalities like TTSC or cap-mounted clips or thermal modalities, such as heater probe or bipolar electrocoagulation, regardless of DEP use.138
138.
Nulsen, B. ∙ Jensen, D.M. ∙ Kovacs, T.O.G. ...
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However, only 7 patients received cap-mounted clips, and no subgroup data were provided.138
138.
Nulsen, B. ∙ Jensen, D.M. ∙ Kovacs, T.O.G. ...
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This study was excluded from our evidence synthesis as it did not meet our predefined threshold of 10 patients per treatment group.138
138.
Nulsen, B. ∙ Jensen, D.M. ∙ Kovacs, T.O.G. ...
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A large retrospective study provided indirect evidence regarding the safety of cap-mounted clips in 1517 patients with refractory bleeding, perforation, fistula, and anastomotic dehiscence.144
144.
Kobara, H. ∙ Mori, H. ∙ Fujihara, S. ...
Outcomes of gastrointestinal defect closure with an over-the-scope clip system in a multicenter experience: an analysis of a successful suction method
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The EtD framework is in Appendix 5.
Benefits, harms, and burden
Based on data from a comparative cohort study, the relative and absolute effects of cap-mounted clips vs conventional endoscopic therapy on 7-day rebleeding in patients with DL were highly uncertain due to the small sample size and very low event rates (RR, 0.50; 95% CI, 0.14–1.73; ARR, 150 fewer per 1000; 95% CI, 258 fewer to 219 more).143
143.
Qiu, J. ∙ Xu, J. ∙ Zhang, Y. ...
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Among the 20 patients treated with cap-mounted clips for DL, 15.0% had 7-day further bleeding, none failed to achieve immediate hemostasis, and 10.0% required additional hemostatic therapy.143
143.
Qiu, J. ∙ Xu, J. ∙ Zhang, Y. ...
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Based on 1 cohort study, the overall and severe adverse event rates of cap-mounted clips were 1.7% and 0.59%, respectively.144
144.
Kobara, H. ∙ Mori, H. ∙ Fujihara, S. ...
Outcomes of gastrointestinal defect closure with an over-the-scope clip system in a multicenter experience: an analysis of a successful suction method
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The panel judged the effect estimate of 7-day rebleeding as highly uncertain due to low event rates, small sample size, and the inherent bias associated with observational design. As a result, the desirable effects of cap-mounted clips over conventional endoscopic hemostatic therapy remained unclear. Due to a lack of comparative data, the undesirable effects of cap-mounted clips were also uncertain.
Certainty in the evidence of effects
Very low certainty of evidence due to serious risk of bias, very serious imprecision, and serious indirectness due to the lack of a clear definition for “standard endoscopic therapy” in the cohort study.
Other evidence-to-decision criteria and considerations
The panel assumed that patients likely do not have important uncertainty or variability in valuing the critical outcome of further bleeding. The costs of cap-mounted clips compared with conventional endoscopic therapy were considered moderate. There were no cost-effectiveness studies. The panel deemed cap-mounted clips to be probably acceptable and unlikely to adversely impact equity-deserving groups. Nevertheless, feasibility may vary depending on access and training to ensure the safe and effective application of these systems, as well as proper management of any complications or failures.
Conclusions and research needs
For patients with UGIB from DL, the panel was unable to ascertain whether the balance of effects favored cap-mounted clips or conventional endoscopic therapy due to very low certainty evidence from a cohort study and highly imprecise results. The panel also expressed concerns about the higher upfront costs of the cap-mounted clips and the challenges involved in retrieval if misapplication occurs. Consequently, the panel was unable to make a recommendation for or against the use of cap-mounted clips over conventional endoscopic hemostatic therapy for DL. They emphasized the need for well-designed RCTs.
Recommendation 8D. Mechanical Modalities (Endoscopic Band Ligation/Endoscopic Through-the-Scope Clip Placement) With or Without Epinephrine Injection vs Injection of Sclerosants With or Without Epinephrine
Evidence summary
We did not identify any RCTs that compared EBL with injection of sclerosants. One RCT included 107 patients with UGIB due to DL and compared aethoxysklerol injection with TTSC placement alone or combined with aethoxysklerol injection.145
145.
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Epinephrine injection was not used in this study. The study defined “successful endoscopic hemostasis” as the cessation of bleeding during the index endoscopy without subsequent rebleeding and “unsuccessful endoscopic hemostasis” as any rebleeding occurring within 48 hours post index endoscopy. The study did not provide outcome data for further bleeding, failure to achieve immediate hemostasis, need for additional hemostatic therapy, mortality, or adverse effects.145
145.
Cui, J. ∙ Huang, L.Y. ∙ Liu, Y.X. ...
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We performed proportional meta-analyses using single-arm cohort-type data from 13 studies on TTSC39
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Full Text
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Google Scholar
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Park, C.H. ∙ Joo, Y.E. ∙ Kim, H.S. ...
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PubMed
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129.
Ding, Y.J. ∙ Zhao, L. ∙ Liu, J. ...
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World J Gastroenterol. 2010; 16:631-635
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PubMed
Google Scholar
130.
Jiang, Y. ∙ Hu, J. ∙ Li, P. ...
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Google Scholar
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Lara, L.F. ∙ Sreenarasimhaiah, J. ∙ Tang, S.J. ...
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Ljubicic, N.
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Hepatogastroenterology. 2006; 53:224-227
PubMed
Google Scholar
133.
Massinha, P. ∙ Cunha, I. ∙ Tome, L.
Dieulafoy lesion: predictive factors of early relapse and long-term follow-up
GE Port J Gastroenterol. 2020; 27:237-243
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Scopus (10)
PubMed
Google Scholar
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Park, C.H. ∙ Sohn, Y.H. ∙ Lee, W.S. ...
The usefulness of endoscopic hemoclipping for bleeding Dieulafoy lesions
Endoscopy. 2003; 35:388-392
Crossref
Scopus (90)
PubMed
Google Scholar
135.
Park, S.H. ∙ Lee, D.H. ∙ Park, C.H. ...
Predictors of rebleeding in upper gastrointestinal dieulafoy lesions
Clin Endosc. 2015; 48:385-391
Crossref
Scopus (0)
PubMed
Google Scholar
136.
Sone, Y. ∙ Kumada, T. ∙ Toyoda, H. ...
Endoscopic management and follow up of Dieulafoy lesion in the upper gastrointestinal tract
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Yamaguchi, Y. ∙ Yamato, T. ∙ Katsumi, N. ...
Short-term and long-term benefits of endoscopic hemoclip application for Dieulafoy's lesion in the upper GI tract
Gastrointest Endosc. 2003; 57:653-656
Full Text
Full Text (PDF)
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and 3 studies on injection of sclerosants,133
133.
Massinha, P. ∙ Cunha, I. ∙ Tome, L.
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The EtD framework is in Appendix 5.
Benefits, harms, and burden
Based on data from 1 RCT, the relative and absolute effects of TTSC vs injection of sclerosants on 7-day rebleeding in patients with DLs were highly uncertain due to the small sample size and very low event rates (RR, 0.80; 95% CI, 0.36–1.77; ARR, 57 fewer per 1000; 95% CI, 181 fewer to 218 more).145
145.
Cui, J. ∙ Huang, L.Y. ∙ Liu, Y.X. ...
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Crossref
Scopus (0)
PubMed
Google Scholar
No comparative data were available for other outcomes.
The study also reported a higher risk of 7-day rebleeding with TTSC alone compared with when combined with sclerosant injection (RR, 6.77; 95% CI, 0.80–51.80; ARR, 192 more per 1000; 95% CI, 4 fewer to 1000 more).145
145.
Cui, J. ∙ Huang, L.Y. ∙ Liu, Y.X. ...
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Scopus (0)
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However, the effect estimate was very imprecise due to very low event rates and the small sample size.145
145.
Cui, J. ∙ Huang, L.Y. ∙ Liu, Y.X. ...
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Scopus (0)
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As this comparison was not part of our predefined question and the study had serious risk of bias—such as unclear allocation sequence generation and concealment—the panel deemed it inappropriate to draw conclusions regarding the effects of combining TTSC with sclerosant injection over either treatment alone.
Proportional meta-analyses of single-arm cohort-type data suggested that both TTSC and injection of sclerosants had similar 7-day further bleeding rates (14%; 95% CI, 11%–19% vs 14%; 95% CI, 8%–23%) and adverse effects (0%).39
39.
Parra-Blanco, A. ∙ Takahashi, H. ∙ Mendez Jerez, P.V. ...
Endoscopic management of Dieulafoy lesions of the stomach: a case study of 26 patients
Endoscopy. 1997; 29:834-839
Crossref
PubMed
Google Scholar
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118.
Ahn, D.W. ∙ Lee, S.H. ∙ Park, Y.S. ...
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The panel judged the effect estimates as highly uncertain due to low event rates, small sample size, and high risk of bias. As a result, the desirable effects of TTSC over injection of sclerosants remained unclear. Due to a lack of comparative data, the undesirable effects of TTSC compared with injection of sclerosants were also uncertain.
Certainty in the evidence of effects
Very low certainty of evidence due to serious risk of bias and very serious imprecision. Specifically, the panel expressed concerns about the lack of reporting of allocation sequence generation and baseline characteristics of the included patients in the RCT.145
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Other evidence-to-decision criteria and considerations
The panel assumed patients likely do not have important uncertainty or variability in valuing the main outcome of rebleeding. The costs of TTSC compared with injection of sclerosants were considered negligible. There were no cost-effectiveness studies. The panel judged TTSC to be probably acceptable, feasible, and unlikely to adversely impact equity-deserving groups.
Conclusions and research needs
For patients with UGIB from DL, the panel was unable to determine whether the balance of effects favored mechanical modalities (EBL/TTSC) or injection of sclerosants due to very low certainty evidence from an RCT and highly imprecise results. Because both interventions appeared to be effective and safe, the panel issued a conditional recommendation for using either mechanical modalities or injection of sclerosants, with or without epinephrine injection. The panel highlighted the need for more comparative studies, including the need for combination with epinephrine injection.
Question 9: Should patients with UGIB from DL receive injection of epinephrine alone vs other endoscopic hemostatic therapies?
Recommendations 9A: In patients with UGIB from DL, we suggest against epinephrine injection alone over mechanical devices (EBL or endoscopic TTSC placement) (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝).
Recommendation 9B: In patients with UGIB from DL, we suggest against epinephrine injection alone over thermal devices (eg, heater probe, bipolar or multipolar electrocoagulation) (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝).
Recommendation 9A. Epinephrine Injection Alone vs Mechanical Modalities (Endoscopic Band Ligation/Endoscopic Through-the-Csope Clip Placement)
Evidence summary
We identified 3 RCTs that addressed this question.122
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We conducted meta-analyses of these 3 RCTs.122
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Two of them reported further bleeding, failure to achieve immediate hemostasis, and mortality,122
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while all 3 reported rebleeding and the need for additional hemostatic therapy. Notably, only 1 detailed the timing of the rebleeding, noting that all rebleeding events occurred within 48 hours after the initial endoscopy.124
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However, both studies were excluded from evidence synthesis as they involved fewer than 10 patients for a specific type of endoscopic therapy.
The EtD framework is in Appendix 5.
Benefits, harms, and burden
Meta-analyses of 3 RCTs suggested that epinephrine injection alone may increase the risk of 7-day further bleeding (RR, 4.37; 95% CI, 1.43–13.33; ARR, 361 more per 1000; 95% CI, 46 more to 1000 more), failure to achieve immediate hemostasis (RR, 2.49; 95% CI, 0.52–11.82; ARR, 106 more per 1000; 95% CI, 34 fewer to 773 more), 7-day rebleeding (RR, 7.46; 95% CI, 1.81–30.76; ARR, 170 more per 1000; 95% CI, 21 more to 783 more), and additional hemostatic therapy (RR, 10.47; 95% CI, 2.62–41.86; ARR, 249 more per 1000; 95% CI, 43 more to 1000 more), with no significant subgroup differences between EBL, TTSC, or EBL/TTSC.122
122.
Chung, I.K. ∙ Kim, E.J. ∙ Lee, M.S. ...
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124.
Alis, H. ∙ Oner, O.Z. ∙ Kalayci, M.U. ...
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Park, C.H. ∙ Sohn, Y.H. ∙ Lee, W.S. ...
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Thirty-day mortality and adverse effects were not estimable due to double-zero events.
The panel judged the desirable effects of epinephrine injection alone compared with EBL or TTSC as trivial, while the undesirable effects were large.
Certainty in the evidence of effects
Very low certainty of evidence due to serious risk of bias and very serious imprecision.
Other evidence-to-decision criteria and considerations
The panel assumed patients likely do not have important uncertainty or variability in valuing the critical outcome of further bleeding. The costs of epinephrine injection alone compared with EBL or TTSC were considered negligible. No cost-effectiveness studies were found. The panel judged epinephrine injection alone to be probably acceptable and feasible and unlikely to adversely impact equity-deserving groups.
Conclusions and research needs
For patients with UGIB from DL, the panel judged that the balance of effects likely favored EBL or TTSC over epinephrine injection alone. As a result, the panel made a conditional recommendation against epinephrine injection alone over mechanical modalities (EBL or TTSC). Although this conditional recommendation serves as guidance for current clinical practice, there is a need for more well-designed comparative studies to improve the precision of available data for the formulation of a stronger recommendation.
Recommendation 9B. Epinephrine Injection Alone vs Thermocoagulation
Evidence summary
We did not identify any RCTs that addressed this question. We found 1 comparative cohort study involving 21 patients with DL, which compared epinephrine injection alone with epinephrine injection and heater probe coagulation.139
139.
Cheng, C.L. ∙ Liu, N.J. ∙ Lee, C.S. ...
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All rebleeding cases occurred during hospitalization, although the exact timing was not specified. For consistency and to prevent double counting between 7-day and 30-day outcomes, we classified these rebleeding events under the 7-day outcome.
We did not identify any study that assessed noncontact thermocoagulation in DL.
The EtD framework is available in Appendix 5.
Benefits, harms, and burden
Based on 1 cohort study, epinephrine injection alone may increase the risk of 7-day further bleeding (RD, 0.45; 95% CI, 0.15 to 0.76; ARR, 450 more per 1000; 95% CI, 150 more to 760 more), failure to achieve immediate hemostasis (RD, 0.27; 95% CI, –0.01 to 0.56; ARR, 270 more per 1000; 95% CI, 10 fewer to 560 more), 7-day rebleeding (RD, 0.18; 95% CI, –0.08 to 0.44; ARR, 180 more per 1000; 95% CI, 80 fewer to 440 more), and the need for additional hemostatic therapy (RD, 0.45; 95% CI, 0.15 to 0.76; ARR, 450 more per 1000; 95% CI, 150 more to 760 more).139
139.
Cheng, C.L. ∙ Liu, N.J. ∙ Lee, C.S. ...
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139.
Cheng, C.L. ∙ Liu, N.J. ∙ Lee, C.S. ...
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Epinephrine injection alone may have little to no impact on 30-day mortality (RD, 0.09; 95% CI, –0.13 to 0.31; ARR, 90 more per 1000; 95% CI, 130 fewer to 310 more), but the estimate was also very imprecise.
The panel judged the desirable effects of epinephrine injection alone compared with contact thermocoagulation as trivial, while the undesirable effects were large.
Certainty in the evidence of effects
Very low certainty of evidence due to serious risk of bias and very serious imprecision.
Other evidence-to-decision criteria and considerations
The panel assumed patients likely do not have important uncertainty or variability in valuing the critical outcome of further bleeding. The costs of epinephrine injection alone compared with contact thermocoagulation were considered negligible. No cost-effectiveness studies were found. The panel judged epinephrine injection alone to be probably acceptable and feasible and unlikely to adversely impact equity-deserving groups.
Conclusions and research needs
For patients with UGIB from DL, the panel judged that the balance of effects likely favored contact thermocoagulation over epinephrine injection alone. As a result, the panel made a conditional recommendation against epinephrine injection alone over contact thermocoagulation. Although this conditional recommendation serves as guidance for current clinical practice, there is a need for well-designed comparative studies to improve the precision of available data, thereby facilitating the formulation of a stronger recommendation. In addition, RCTs comparing thermocoagulation with mechanical modalities are needed to inform clinical decision making, as both modalities appear superior to epinephrine injection alone.
Gastric Antral Vascular Ectasia
Question 10: Should patients with GAVE receive one endoscopic hemostatic therapy vs another endoscopic hemostatic therapy?
Recommendation 10: In patients with GAVE, we suggest against radiofrequency ablation (RFA) over APC (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝).
Evidence summary
We identified 5 RCTs that compared EBL with APC in patients with GAVE (presented in PICO question 11).150–154
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Benefits, harms, and burden
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St Romain, P. ∙ Boyd, A. ∙ Zheng, J. ...
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St Romain, P. ∙ Boyd, A. ∙ Zheng, J. ...
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The minimally important difference in Hgb level in patients with GAVE is unknown. However, based on indirect evidence from 2 studies that assessed health-related quality of life in postmenopausal women and patients with arthritis, a 2-g/dL decrease was found to be associated with a statistically significant and clinically meaningful decline in health utility.178
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Consequently, the panel used a minimally important difference of 2 g/dL as our decision threshold.
The panel judged that the desirable effects of RFA, in terms of fewer treatment sessions, and its undesirable effects, in terms of change in Hgb level (a critical outcome) compared to APC in GAVE, were trivial.
Certainty in the evidence of effects
Very low certainty of evidence due to serious risk of bias, very serious imprecision, and serious indirectness, particularly concerning the surrogate outcome of change in hemoglobin level in relation to the patient-important outcome of transfusion requirements.
Other evidence-to-decision criteria and considerations
There is no research evidence on patients’ values and preferences in the context of GAVE. However, the panel assumed patients likely do not have important uncertainty or variability in valuing the critical outcomes of change in hemoglobin levels and the units of blood transfused, as well as important outcomes like the number of endoscopy sessions and hospitalizations. The costs of RFA compared with APC were considered large. There were no cost-effectiveness studies in patients with GAVE who had not failed prior APC treatments. The panel judged RFA probably acceptable and unlikely to adversely impact equity-deserving groups. However, feasibility may vary based on accessibility and expertise.
Conclusions and research needs
The panel judged that the balance of effects does not favor either RFA or APC in patients with GAVE. However, considering the large costs associated with RFA and its variable feasibility due to expertise and accessibility, the panel made a conditional recommendation against RFA over APC, particularly in patients who have not failed other treatments. The panel emphasized the need for RCTs that compare the effectiveness and safety of RFA vs APC or EBL in GAVE patients, focusing on long-term outcomes, such as recurrence rates and transfusion requirements. These studies should also examine health care resource utilization, including hospitalization, transfusion requirements, and follow-up procedures associated with each treatment. In addition, research should assess quality-adjusted life-years and patient-important outcomes to better understand the impact of endoscopic interventions on quality of life.
Question 11: Should patients with GAVE receive EBL vs other endoscopic hemostatic therapies?
Recommendation 11: In patients with GAVE, we suggest EBL over APC (conditional recommendation, very low certainty of evidence ⊕⊝⊝⊝).
Evidence summary
We identified 5 RCTs that addressed this question, involving patients with cirrhosis or portal hypertension.150–154
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None of the studies conducted adjusted analyses to account for variations in prognostic factors between groups that could potentially affect the outcomes.
The details of our assessment of these studies and the EtD framework are in Appendix 6.
Benefits
Meta-analyses of 2 RCTs suggested that, compared with APC, EBL may increase Hgb level by 0.7 g/dL (95% CI, 0.08 lower to 1.49 higher) over 6 months.150
150.
Abd Al-Wahab, N. ∙ Amer, K. ∙ Ibrahim, A.
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EBL may also reduce the units of blood transfusions (MD, 3.30 lower; 95% CI, 4.45 lower to 2.15 lower), decrease the number of endoscopic sessions required for the obliteration of lesions (MD, 1.51 lower; 95% CI, 4.69 lower to 1.66 more), and lower the number of hospitalizations (MD, 1.30 lower; 95% CI, 2.24 lower to 0.36 lower).150
150.
Abd Al-Wahab, N. ∙ Amer, K. ∙ Ibrahim, A.
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However, these estimates were very imprecise.150
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Similarly, meta-analyses of 5 comparative cohort studies suggested that EBL may increase Hgb level by 0.54 g/dL (95% CI, 0.30 higher to 0.77 higher).181–185
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EBL may also reduce the units of blood transfusions (MD, 1.90 lower; 95% CI, 5.74 lower to 1.94 higher), decrease the number of endoscopic sessions required for the obliteration of lesions (MD, 1.53 lower; 95% CI, 3.94 lower to 0.89 more), and reduce the number of hospitalizations (MD, 2.10 lower; 95% CI, 3.82 lower to 0.38 lower), but these estimates were very imprecise.181–185
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EBL appeared to have little to no impact on mortality (RR, 0.69; 95% CI, 0.31–1.58; ARR, 93 fewer per 1000; 95% CI, 207 fewer to 174 more).182
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The panel judged that the desirable effects of EBL compared with APC were small.
Harms and burden
Meta-analyses of 2 RCTs suggested that EBL, compared with APC, may increase the risk of adverse effects (RD, 0.16; 95% CI, –0.29 to 0.61; ARR, 160 more per 1000; 95% CI, 290 fewer to 610 more).150
150.
Abd Al-Wahab, N. ∙ Amer, K. ∙ Ibrahim, A.
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Similarly, meta-analyses of 5 RCTs also suggested that EBL may increase the risk of adverse effects (RR, 1.93; 95% CI, 0.60–6.18; ARR, 32 more per 1000; 95% CI, 14 fewer to 179 more).181–185
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However, these estimates were very imprecise due to low event rates. The adverse effects associated with EBL included post-banding ulcers, hypertrophied polyps, abdominal pain, nausea, and vomiting. The panel considered these adverse effects to be minor. As a result, the panel judged that the undesirable effects of EBL compared with APC were small.
Certainty in the evidence of effects
Very low certainty of evidence due to serious risk of bias, very serious imprecision, and serious indirectness. The indirectness relates to using change in hemoglobin level as a surrogate outcome. Also, the RCTs included only patients with cirrhosis, who mostly presented with overt bleeding rather than occult bleeding with a diffuse punctate endoscopic pattern.
Other evidence-to-decision criteria and considerations
The panel assumed patients likely do not have important uncertainty or variability in valuing the critical outcomes and important outcomes. The costs of EBL compared with APC were considered negligible. No cost-effectiveness studies were found. The panel judged EBL to be probably acceptable and feasible and unlikely to adversely impact equity-deserving groups.
Conclusions and research needs
The panel judged that the balance of effects probably favored EBL over APC in patients with GAVE. Consequently, the panel issued a conditional recommendation for EBL over APC. The panel stressed the importance of conducting large, high-quality RCTs. Future research should also explore potential subgroup differences, focusing on different endoscopic patterns and patients with and without cirrhosis. These studies should prioritize long-term outcomes, such as recurrence rates and the durability of treatment effects in maintaining hemoglobin levels and reducing transfusion requirements. In addition, they should also examine health care resource utilization, including hospitalization, transfusion requirements, and follow-up procedures associated with each treatment.
What Are Others Saying and What Is New in This Guideline?
This guideline is the first to specifically address the endoscopic management of NVNPUB caused by conditions such as malignant UGIB, MWTs, DLs, and GAVE. In addition, this guideline, in contrast to many, is based entirely on high-quality systematic reviews and adopted the GRADE approach with evidence profiles and EtD framework for each recommendation, thereby enhancing the transparency and trustworthiness of the decision-making process.
Recent guidelines from the United States, Canada, Europe, and Asia on nonvariceal UGIB have focused primarily on PUB and have not specifically addressed these conditions.8
8.
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The European Society of Gastrointestinal Endoscopy’s 2015 guidelines suggested consideration of endoscopic hemostasis for malignant UGIB, actively bleeding MWT, DLs, and upper GI angioectasias, but no specific modalities were recommended due to insufficient evidence.10
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The older American Society of Gastrointestinal Endoscopy 2004 guidelines advised endoscopic therapy for vascular abnormalities, such as DLs and vascular malformations, as well as for ongoing or severely bleeding MWTs, but they did not specify particular modalities or provide recommendations for malignant UGIB.186
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Limitations of This Guideline
The limitations of this guideline are inherent in the very low certainty of the evidence available for the questions addressed. For 2 recommendations concerning malignant UGIB (PICO questions 1A and 1B), there was no published direct or relevant indirect evidence. Consequently, the panel was surveyed to gather unpublished collective data to inform the decision-making process. It is important to note that the interpretation of these survey data is limited by factors, such as recall bias and variations in individual provider practices. This process is explicitly noted for relevant recommendations. Furthermore, this guideline focuses on the endoscopic management of these conditions and does not cover pre- and postendoscopic management. The consideration of the impact of antithrombotic therapy on NVNPUB was beyond the scope of this guideline, and readers are encouraged to refer to the joint American College of Gastroenterology–CAG clinical practice guideline on management of anticoagulants and antiplatelets during acute GI bleeding and the peri-endoscopic period.187
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We acknowledge that not all aspects of managing these conditions are included in this guideline. The panel prioritized questions for which there was clinical uncertainty or new information might guide decision making. Future updates could include new recommendations as evidence becomes available. The panel suggested actions based on the best evidence available when developing these guidelines. Some recommendations may change as new evidence emerges. Finally, the recommendations are intended to help clinicians and patients make informed decisions, but they should not replace careful consideration of the individual clinical circumstances and patients’ values and preferences.
Revision or Adaptation of This Guideline
Plans for Updating This Guideline
After this guideline is published, the CAG will maintain it by monitoring new evidence, reviewing it with experts, and revising it regularly.
Updating or adapting recommendations locally
Although this guideline was developed for global application, adaptation will be necessary in many circumstances based on resource availability, feasibility, and acceptability of interventions. The EtD frameworks and the “adolopment” model should guide adaptations.188
188.
Schunemann, H.J. ∙ Wiercioch, W. ∙ Brozek, J. ...
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Local guideline groups may leverage the evidence already gathered and appraised in these frameworks. By adding relevant local information, they can efficiently create local recommendations, requiring fewer resources than developing a guideline de novo.
Acknowledgments
The authors acknowledge Yuhong Yuan for supporting the literature search but not qualifying for authorship, and Nosheen Maqsood from the Canadian Association of Gastroenterology for administrative support of these guidelines. The steering committee (Alan N. Barkun, Loren Laine, Grigorios I. Leontiadis, and Frances Tse) drafted the patient population, intervention, comparator, outcome (PICO) questions, which were reviewed and approved by all panel members. Methodologists performed evidence synthesis, assessed the certainty of evidence, developed evidence profiles and evidence-to-decision frameworks, and facilitated panel discussions. Panelists formulated the recommendations. The steering committee prepared the initial manuscript, which was later revised with input from all panel members. In addition, 2 patients with a history of gastrointestinal bleeding reviewed the PICO questions and provided feedback on the final manuscript.
CRediT Authorship Contributions
Alan N. Barkun, MD, MPH (Conceptualization: Equal; Methodology: Equal; Writing – review & editing: Supporting)
Loren Laine, MD (Methodology: Equal; Writing – review & editing: Supporting)
Grigorios I. Leontiadis, MD, PhD (Methodology: Supporting; Writing – review & editing: Supporting)
Ian M. Gralnek, MD, MSHS, FESGE, FASGE (Methodology: Supporting; Writing – review & editing: Supporting)
Nicholas Carman, MD (Methodology: Supporting)
Mostafa Ibrahim, MD (Methodology: Supporting; Writing – review & editing: Supporting)
Michael Sey, MD, MPH, FRCPC, AGAF, CAGF (Methodology: Supporting; Writing – review & editing: Supporting)
Ali A. Alali, MD (Methodology: Supporting; Writing – review & editing: Supporting)
Matthew W. Carroll, MD (Methodology: Supporting; Writing – review & editing: Supporting)
Lawrence Hookey, MD (Methodology: Supporting; Writing – review & editing: Supporting)
Mark Borgaonkar, MD, MPH (Methodology: Supporting; Writing – review & editing: Supporting)
David Armstrong, MD (Methodology: Supporting; Writing – review & editing: Supporting)
James Y.W. Lau, MD, MBBS, FRCS, FHKCS (Methodology: Supporting; Writing – review & editing: Supporting)
Nauzer Forbes, MD, MPH (Methodology: Supporting; Writing – review & editing: Supporting)
Rapat Pittayanon, MD, MPH (Methodology: Supporting; Writing – review & editing: Supporting)
Frances Tse, MD, MPH, FRCPC, AGAF, CAGF (Methodology: Lead; Project administration: Lead; Writing – original draft: Lead; Writing – review & editing: Lead)
Supplementary Material (6)
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Appendix 4
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Appendix 5
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Appendix 6
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最新推荐文章于 2023-11-25 23:00:46 发布
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于 2018-01-13 10:34:03 发布
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版权声明：本文为博主原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接和本声明。
文章标签：
#高等数学
#函数
#单调性
#极值
高等数学
专栏收录该内容
29 篇文章
一、单调性的判定法
如果函数y=f(x)y=f(x)在[a,b][a,b]上单调增加(单调减少)，那么它的图形是一条沿xx轴正在上升(下降)的曲线.可知，这时曲线上各点的切线斜率是非负的(非正的)，即y′=f′(x)≥0（y′=f(x)≤0）y′=f′(x)≥0（y′=f(x)≤0），反之亦然.
可见，函数的单调性与其导数的符号有密切的关系.
由此可得
定理：设函数y=f(x)y=f(x)在[a,b][a,b]上连续，在(a,b)(a,b)内可导.
(1)如果(a,b)(a,b)内f′(x)≥0f′(x)≥0，且等号仅在有限多个点处成立，那么函数y=f(x)y=f(x)在[a,b][a,b]上单调增加；
(2)如果(a,b)(a,b)内f′(x)≤0f′(x)≤0，且等号仅在有限多个点处成立，那么函数y=f(x)y=f(x)在[a,b][a,b]上单调减少；
二、函数极值点
函数极值的定义：设函数f(x)f(x)在点x0x0的某邻域U(x0)U(x0)内有定义。如果对于去心邻域˚U(x0)U˚(x0)内的任一x，有
f(x)f(x0)）
那么就称f(x0)是函数f(x)的一个极大值(极小值).
联系函数的单调性可知，函数的极值点一般出现在其导数为0的点(驻点，又称稳定点)，或者不可导点(导数不存在)
要使驻点是极值点，必须满足该点两侧的导函数正负符号相反.
需要明确的是，函数的极值点并不是一个点，而是极值所对应的x值.
在上图中，x1、x2、x4、x5、x6均为驻点，且是极值点，而x3是一个驻点，但不是极值点.
下面是可导函数取极值的充分必要条件说明
①第一充分条件：
设函数f(x)在x0处连续，且在x0的某去心邻域˚U(x0,δ)内可导
(1)若x∈(x0−δ,x0)时，f′(x)>0；而x∈(x0,x0+δ)时f′(x)<0，则f(x)在x0处取得极大值；
(2)若x∈(x0−δ,x0)时，f′(x)<0；而x∈(x0,x0+δ)时f′(x)>0，则f(x)在x0处取得极小值；
(3)若x∈˚U(x0,δ)时，f′(x)的符号保持不变，则f(x)在x0处没有极值.
②第二充分条件：
设函数f(x)在x0处具有二阶导数且f′(x0)=0,f″(x0)≠0，则
(1)当f″(x0)<0时，函数f(x)在x0处取得极大值；
(2)当f″(x0)>0时，函数f(x)在x0处取得极小值.
③必要条件
设函数f(x)在x0处可导，且在x0处取得极值，则f′(x0)=0
这个条件说明：可导函数f(x)的极值点必是它的驻点。但反过来，函数的驻点却不一定是极值点。所以，函数的驻点只是可能的极值点.此外，函数在它的导数不存在的点处也可能取得极值.
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f(x1)<f(x2)
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很多同学给老师留言说，函数的概念我懂，可是就是不会做题，高中数学必修一函数章节那个点，一定要把函数的基础学扎实了，然后利用技巧方法去解关于函数的单调性与最值这个点，不管是函数那个点都要学好基础。
大部分同学都能掌握必修一函数章节内容，重点还在于抽象函数的定义域，这就需要同学们仔细思考，回顾函数概念才能弄懂的；函数值域的求法，重点掌握换元法和方程组法、数形结合法，这些方法是最基础的。另外对于函数的...
【高等数学】单调性、极值
写一封信
11-03
479
第三章 part2 单调性与极值
weixin_45564209的博客
03-02
607
单调性与极值的概念题型型一型二 函数的零点或方程根的个数问题型三 不等式证明
单调性与极值的概念
这两个注解注意一下
极值点的判别步骤与方法
第二充分条件的证明
题型
型一
直接利用极限的保号性（去心邻域）
去心邻域左右区间分一下
拆分出来，最终目的是再次利用等价无穷小凑出一个f’’(x)的定义的样子
这个其实挺难想的
观察式子得出第一个结论
利用这个结论再次将已经利用过的式子拆分成两个有意义的样子，再次得出一个值
照样根据分母等于0推出分母也等于0
根据极限值大于0你就可以直接用.
c/c++中的内置数据类型极值
xiaoyong298的专栏
11-30
2916
c/c++中经常会涉及到获取内置数据类型（int，float，double）的极大值或极小值，结合STL的numeric_limits可以很轻松地实现，实际上c语言也自带了一些极值的宏。下面分别介绍通过STL的numeric_limits获取极值和通过c语言的头文件的宏获取极值的方法。
1.STL的numeric_limits方法
numeric_limits是STL中表示内置数据类型的算术性
函数单调性与凹凸性
笨笨的博客
04-13
3064
单调性定义:
函数的导数大于0就是增函数，小于0就是减函数
题型
结论:
求出驻点和导数不存在的点，然后把它的区间分成不同的区间，在这几个区间上讨论函数的单调性
由结论考查的题型:
1.这个题是求出驻点也就是导数等于0的点
2.这个题不仅要求出驻点，还要求出导数不存在的点
这几种题型考查的概率不太大，因为它要和凹凸性结合
凹凸性
凹凸函数的三个定义:
从它的曲线上任取两点，过这两点的直线在这个曲线的上头，就叫凹函数。
从它的曲线上任取两点，过这两点的直线在这个...
人工智能数学基础：利用导数判断函数单调性、凹凸性、极值、最值和描绘函数图形
老猿Python
07-17
9218
本文介绍了利用导数判断函数单调性、凹凸性、极值相关的概念和定理，通过本文的介绍，可以熟悉通过导数判断函数单调性、凹凸性、极值以及求最值的原理和方法。最后，通过一阶导数和二阶导数确定了函数的单调性、凹凸性、极值点之后，就可以描绘出函数的几何图形。
函数单调性与凹凸性，极值和最值
最新发布
whcwhc111111的博客
11-25
2262
9.函数在（x0-a,x0+a）内连续er且在邻域内可导（x0可以不可导，参考第八条），x0的导函数左增右减极大值，左减右增极小值，如果单调性不变则不是极值。7.可导函数极值点是驻点，但驻点不一定为极值点（驻点为导数为0的点，比如x的三次方在0处是驻点但不是极值点）2.函数在某个区间凹，两点函数值和的二分之一大于两点和的二分之一的函数值，如果相反，那么函数是凸的。10.函数在x0处有二阶导，一阶导等于0二阶导不为0，二阶导小于0极大值，二阶导大于0极小值。6.如果函数可导，那么在极值点导数为0。
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Tools & Reference>Emergency Medicine
Pediatric Dehydration Treatment & Management
Updated: Dec 18, 2023
Author: Alex Koyfman, MD; Chief Editor: Muhammad Waseem, FAAP, FACEP, FAHA, MBBS, MS more...;)
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Sections
Pediatric Dehydration
Sections Pediatric Dehydration
Overview
Practice Essentials
Pathophysiology
Etiology
Epidemiology
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Presentation
History
Physical Examination
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DDx
Workup
Approach Considerations
Obtaining Vascular Access
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Treatment
Approach Considerations
Mild Volume Depletion
Moderate Volume Depletion
Severe Volume Depletion
Pharmacologic Therapy
Consultations
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Medication
Tables)
References;)
Treatment
Approach Considerations
Address emergent airway, breathing, and circulatory problems first. Obtain intravenous access, and give a 20 mL/kg isotonic fluid bolus (Ringer lactate or normal saline) to children with severe volume depletion. This should not delay transport to the appropriate facility. Reassessment of perfusion, cardiac function, mentation should take place after each intervention. At times, cardiac failure can mimic volume depletion leading to further deterioration of clinical findings after fluid administration.
Failure to diagnose appendicitis, intussusception, or small bowel obstruction places patients at risk of serious complications (including death).
Antidiarrheal medications have adverse effects and are generally not recommended without medical supervision.
Next:
Mild Volume Depletion
Patients with minimal to mild volume depletion should be encouraged to continue an age-appropriate diet and adequate intake of oral fluids. Oral rehydration solution (ORS) should be used. Children should be given sips of ORS (5 mL or 1 teaspoon) every 2 minutes. As an estimate for the amount of fluid to replace, the goal should be to drink 10 mL/kg body weight for each watery stool and estimate volume of emesis for each episode of vomiting. [8, 15, 17]
If commercially prepared ORS is not available, the following recipe may be used:
In 1 L of water, add 2 level tablespoons of sugar or honey, a quarter teaspoon of table salt (NaCl), and a quarter teaspoon of baking soda (bicarbonate of soda)
If baking soda is not available, use another quarter teaspoon of salt instead
If available, add one-half cup of orange juice, coconut water, or a mashed ripe banana to the drink
Use a safe water source, boil water if source is questioned
Inpatient therapy generally is not indicated for mild volume depletion. However, it is prudent to arrange outpatient follow-up evaluation within 48 hours, with instructions to return sooner if symptoms worsen.
Guidelines from the American Academy of Pediatrics on fruit juice in infants, children and adolescents recommend against the use of fruit juice in the treatment of dehydration or the management of diarrhea.
A study by Freedman et al indicated that in children with mild gastroenteritis and minimal dehydration, better oral rehydration results can be achieved by substituting dilute apple juice (initially) and a preferred fluid (later) for electrolyte maintenance solution.
Previous
Next:
Moderate Volume Depletion
The literature supports use of oral rehydration for the moderately dehydrated child. Similar outcomes have been achieved in randomized studies comparing ORS with intravenous fluid therapy with fewer complications and higher parent satisfaction in the ORS groups. Moreover, ORS can typically be initiated sooner than IV fluid therapy. However, children must be cooperative and have caregivers available to instruct and administer the oral fluids.
With ORS, patients should receive approximately 50-100 mL/kg body weight over 2-4 hours, again starting with 5 mL every 5 minutes. If the child can tolerate this amount and asks for more fluids, the amount given can gradually be increased. Once the fluid deficit has been corrected, parents should be instructed on how to replace volume losses at home if the child continues to have vomiting or diarrhea.
Children in whom ORS fails should be given a bolus (20 mL/kg) of isotonic fluid intravenously. This may be followed by 1.5-2 times maintenance therapy. Over the next few hours, the patient may be transitioned to oral rehydration as tolerated, at which point the intravenous therapy may be discontinued.
Children with moderate volume depletion may require inpatient treatment if they are unable to tolerate oral fluids despite rehydration. Hospitalization may also be required for treatment of the underlying cause of the fluid deficit.
Previous
Next:
Severe Volume Depletion
Patients with severe volume depletion should receive intravenous isotonic fluid boluses (20-60 mL/kg). In children with difficult peripheral access, perform intraosseous or central access promptly. Fluid boluses should be repeated until vital signs, perfusion, and capillary refill have normalized.
If a patient reaches 60-80 mL/kg in isotonic crystalloid boluses and is not significantly improved, consider other causes of shock (eg, sepsis, hemorrhage, cardiac disease). In addition, consider administering vasopressors and instituting advanced monitoring, such as a bladder catheter, central venous pressure, and measuring mixed venous oxygen saturation.
Although physicians typically give normal saline for these initial boluses, it is important to remember to check a bedside glucose level for patients who appear lethargic or altered. Treat hypoglycemia promptly. The appropriate dose is 0.25 g/kg IV (2.5 mL/kg of 10% dextrose or 1 mL/kg of 25% dextrose) with reassessment of glucose level after administration of dextrose.
Once vital sign abnormalities are corrected, initiate maintenance fluid therapy plus additional fluid to make up for any continued losses. Daily requirements for maintenance fluids can be approximated as follows:
If the patient weighs less than 10 kg, give 100 mL/kg/d
If the patient weighs less than 20 kg, give 1000 mL/d plus 50 mL/kg/d for each kilogram between 10 and 20 kg
If the patient weighs more than 20 kg, give 1500 mL/d, plus 20 mL/kg/d for each kilogram over 20 kg
Divide the total by 24 to obtain the hourly rate
Daily fluid requirements may be met using dextrose 5% in half-normal saline solution. For patients with significant hyponatremia or hypernatremia, it is preferable to use dextrose 5% in normal saline. Dextrose is important to include because these patients generally have a notable ketosis.
The emergency physician also should consider daily sodium and potassium requirements as follows:
Sodium 2-3 mEq/kg/d
Potassium 2-3 mEq/kg/d
Isonatremic and hyponatremic volume depletion states may be treated with normal saline or other isotonic solutions. The goal for correction rates for either hyponatremic or hypernatremic patients should be no more than 0.5 mEq/L/h or no more than 8mEq/L per 24 hour period to prevent the devastating CNS complications of over-rapid correction (central pontine myelinolysis and cerebral edema). Full correction of severe sodium abnormalities usually should be staged over 24-48 hours or longer.
Although a potassium deficit is present in all cases of volume depletion, it is not usually clinically significant; few patients with moderate dehydration require supplemental potassium. However, failure to correct for hypokalemia during volume repletion may result in clinically significant hypokalemia.
Add potassium to fluids when the patient has documented hypokalemia. For all other patients, avoid adding potassium to fluids until the patient has received resuscitation and has demonstrated adequate urine output.
Children with severe volume depletion, especially those with hypernatremia or hyponatremia, require inpatient therapy. Children with severe hyperosmolar states, severe electrolyte derangements, or associated renal failure may require admission to a critical care unit.
Previous
Next:
Pharmacologic Therapy
The emergency medicine literature now supports the use of a single dose of oral ondansetron in combination with oral rehydration for patients with dehydration, nausea, and vomiting. [16, 20, 21] However, the use of an antiemetic should not shift the focus away from adequate fluid resuscitation.
Acute gastroenteritis is typically a self-limited condition that does not require antibiotics. Chronic infectious cases of diarrhea may require antimicrobial agents after appropriate stool studies have indicated the etiology. Antidiarrheal agents are not recommended. When dehydration is caused by other disease processes, such as diabetic ketoacidosis or sepsis, appropriate pharmacologic therapy should be initiated as soon as possible.
Previous
Next:
Consultations
Infants and children who present to the ED with mild to moderate dehydration may respond to fluid boluses and may be discharged home with close follow-up with their primary care provider. Patients who are severely volume depleted or who are unable to tolerate oral fluids must be admitted, with a pediatric consultation.
If the child is in shock, is unable to drink fluids, or does not respond to intravenous bolus therapy, significant abnormalities requiring correction may exist. In such patients, obtain pediatric consultation for admission and further therapy. If renal tubular acidosis or other primary renal or endocrine disorder is suspected, specialty consultation may be indicated.
Previous
Medication
References
Vega RM, Avva U. Pediatric Dehydration. StatPearls [Internet]. 2023 Jan. [QxMD MEDLINE Link].[Full Text].
Steiner MJ, DeWalt DA, Byerley JS. Is this child dehydrated?. JAMA. 2004 Jun 9. 291(22):2746-54. [QxMD MEDLINE Link].
Mange K, Matsuura D, Cizman B, et al. Language guiding therapy: the case of dehydration versus volume depletion. Ann Intern Med. 1997 Nov 1. 127(9):848-53. [QxMD MEDLINE Link].
Sterns RH, Cappuccio JD, Silver SM, Cohen EP. Neurologic sequelae after treatment of severe hyponatremia: a multicenter perspective. J Am Soc Nephrol. 1994 Feb. 4 (8):1522-30. [QxMD MEDLINE Link].
Singhi SC, Shah R, Bansal A, Jayashree M. Management of a child with vomiting. Indian J Pediatr. 2013 Apr. 80(4):318-25. [QxMD MEDLINE Link].
Glaser NS, Ghetti S, Casper TC, Dean JM, Kuppermann N. Pediatric diabetic ketoacidosis, fluid therapy, and cerebral injury: the design of a factorial randomized controlled trial. Pediatr Diabetes. 2013 Mar 13. [QxMD MEDLINE Link].[Full Text].
Trainor JL, Glaser NS, Tzimenatos L, et al. Clinical and Laboratory Predictors of Dehydration Severity in Children With Diabetic Ketoacidosis. Ann Emerg Med. 2023 Aug. 82 (2):167-78. [QxMD MEDLINE Link].
King CK, Glass R, Bresee JS, Duggan C. Managing acute gastroenteritis among children: oral rehydration, maintenance, and nutritional therapy. MMWR Recomm Rep. 2003 Nov 21. 52:1-16. [QxMD MEDLINE Link].
Wathen JE, MacKenzie T, Bothner JP. Usefulness of the serum electrolyte panel in the management of pediatric dehydration treated with intravenously administered fluids. Pediatrics. 2004 Nov. 114(5):1227-34. [QxMD MEDLINE Link].
Lozon MM. Pediatric vascular access and blood sampling techniques. Roberts JR, Hedges JR.Clinical Procedures in Emergency Medicine. 4th ed. Philadelphia: WB Saunders; 2004. 357-8.
Jauregui J, Nelson D, Choo E, Stearns B, Levine AC, Liebmann O, et al. The BUDDY (Bedside Ultrasound to Detect Dehydration in Youth) study. Crit Ultrasound J. 2014. 6 (1):15. [QxMD MEDLINE Link].
Jauregui J, Nelson D, Choo E, Stearns B, Levine AC, Liebmann O, et al. The BUDDY (Bedside Ultrasound to Detect Dehydration in Youth) study. Crit Ultrasound J. 2014. 6(1):15. [QxMD MEDLINE Link].[Full Text].
KoÅodziej M, Jalali A, Åukasik J. Point-of-care ultrasound to assess degree of dehydration in children: a systematic review with meta-analysis. Arch Dis Child. 2023 Jun 14. [QxMD MEDLINE Link].
Kaminecki I, Huang DM, Shipman PC, Gibson RW. Point-of-Care Ultrasonography for the Assessment of Dehydration in Children: A Systematic Review. Pediatr Emerg Care. 2023 Oct 1. 39 (10):786-96. [QxMD MEDLINE Link].
Spandorfer PR, Alessandrini EA, Joffe MD, Localio R, Shaw KN. Oral versus intravenous rehydration of moderately dehydrated children: a randomized, controlled trial. Pediatrics. 2005 Feb. 115(2):295-301. [QxMD MEDLINE Link].
Freedman SB, Adler M, Seshadri R, Powell EC. Oral ondansetron for gastroenteritis in a pediatric emergency department. N Engl J Med. 2006 Apr 20. 354(16):1698-705. [QxMD MEDLINE Link].
Santillanes G, Rose E. Evaluation and Management of Dehydration in Children. Emerg Med Clin North Am. 2018 May. 36 (2):259-273. [QxMD MEDLINE Link].
Heyman MB, Abrams SA, SECTION ON GASTROENTEROLOGY, HEPATOLOGY, AND NUTRITION., COMMITTEE ON NUTRITION. Fruit Juice in Infants, Children, and Adolescents: Current Recommendations. Pediatrics. 2017 Jun. 139 (6):[QxMD MEDLINE Link].
Freedman SB, Willan AR, Boutis K, Schuh S. Effect of Dilute Apple Juice and Preferred Fluids vs Electrolyte Maintenance Solution on Treatment Failure Among Children With Mild Gastroenteritis: A Randomized Clinical Trial. JAMA. 2016 May 10. 315 (18):1966-74. [QxMD MEDLINE Link].
Kersten H. Oral ondansetron decreases the need for intravenous fluids in children with gastroenteritis. J Pediatr. 2006 Nov. 149(5):726. [QxMD MEDLINE Link].
Alhashimi D, Alhashimi H, Fedorowicz Z. Antiemetics for reducing vomiting related to acute gastroenteritis in children and adolescents. Cochrane Database Syst Rev. 2006 Oct 18. CD005506. [QxMD MEDLINE Link].
American Academy of Pediatrics. Practice parameter: the management of acute gastroenteritis in young children. American Academy of Pediatrics, Provisional Committee on Quality Improvement, Subcommittee on Acute Gastroenteritis. Pediatrics. 1996 Mar. 97(3):424-35. [QxMD MEDLINE Link].
Barkin RM, Ward DG. Infectious diarrheal disease and dehydration. Marx JA.Rosen's Emergency Medicine: Concepts and Clinical Practice. 6th ed. Philadelphia, Pa: Mosby/Elsevier; 2006. Vol 3: 2623-34.
Danewa AS, Shah D, Batra P, Bhattacharya SK, Gupta P. Oral Ondansetron in Management of Dehydrating Diarrhea with Vomiting in Children Aged 3 Months to 5 Years: A Randomized Controlled Trial. J Pediatr. 2016 Feb. 169:105-9.e3. [QxMD MEDLINE Link].
Hom J, Sinert R. Evidence-based emergency medicine/systematic review abstract. Comparison between oral versus intravenous rehydration to treat dehydration in pediatric gastroenteritis. Ann Emerg Med. 2009 Jul. 54(1):117-9. [QxMD MEDLINE Link].
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Tables
Table. Physical Examination Findings in Pediatric Dehydration;)
Table. Physical Examination Findings in Pediatric Dehydration
| | |
--- |
| Symptom | Degree of Dehydration |
| Mild (< 3% body weight lost) | Moderate (3-9% body weight lost) | Severe (>9% body weight lost) |
| Mental status | Normal, alert | Restless or fatigued, irritable | Apathetic, lethargic, unconscious |
| Heart rate | Normal | Normal to increased | Tachycardia or bradycardia |
| Quality of pulse | Normal | Normal to decreased | Weak, thready, impalpable |
| Breathing | Normal | Normal to increased | Tachypnea and hyperpnea |
| Eyes | Normal | Slightly sunken | Deeply sunken |
| Fontanelles | Normal | Slightly sunken | Deeply sunken |
| Tears | Normal | Normal to decreased | Absent |
| Mucous membranes | Moist | Dry | Parched |
| Skin turgor | Instant recoil | Recoil < 2 seconds | Recoil >2 seconds |
| Capillary refill | < 2 seconds | Prolonged | Minimal |
| Extremities | Warm | Cool | Mottled, cyanotic |
| Adapted from King CK, Glass R, Bresee JS, et al. Managing acute gastroenteritis among children: oral rehydration, maintenance, and nutritional therapy. MMWR Recomm Rep. Nov 21 2003;52(RR-16):1-16. |
Back to List
Contributor Information and Disclosures
Author
Alex Koyfman, MD Assistant Professor, Department of Emergency Medicine, University of Texas Southwestern Medical Center, Parkland Memorial HospitalAlex Koyfman, MD is a member of the following medical societies: American Academy of Emergency Medicine, American College of Emergency Physicians, Society for Academic Emergency MedicineDisclosure: Nothing to disclose.
Coauthor(s)
Carrie Ng, MD Resident Physician, Department of Pediatrics, Bellevue Hospital Center, New York University School of MedicineCarrie Ng, MD is a member of the following medical societies: American Academy of Pediatrics, American College of Emergency Physicians, American Institute of Ultrasound in MedicineDisclosure: Nothing to disclose.
Mark P Foran, MD, MPH Assistant Professor of Emergency Medicine, New York University School of Medicine; Attending Emergency Physician, Bellevue Hospital Center and NYU Langone Medical CenterMark P Foran, MD, MPH is a member of the following medical societies: American Academy of Emergency Medicine, American College of Emergency Physicians, American Public Health Association, Society for Academic Emergency MedicineDisclosure: Nothing to disclose.
Specialty Editor Board
Kirsten A Bechtel, MD Associate Professor of Pediatrics, Section of Pediatric Emergency Medicine, Yale University School of Medicine; Co-Director, Injury Free Coalition for Kids, Yale-New Haven Children's HospitalKirsten A Bechtel, MD is a member of the following medical societies: American Academy of PediatricsDisclosure: Nothing to disclose.
Chief Editor
Muhammad Waseem, FAAP, FACEP, FAHA, MBBS, MS Professor of Emergency Medicine and Clinical Pediatrics, Weill Cornell Medical College; Attending Physician, Departments of Emergency Medicine and Pediatrics, Lincoln Medical and Mental Health Center; Adjunct Professor of Emergency Medicine, Adjunct Professor of Pediatrics, St George's University School of Medicine, GrenadaMuhammad Waseem, FAAP, FACEP, FAHA, MBBS, MS is a member of the following medical societies: American Academy of Pediatrics, American Academy of Urgent Care Medicine, American College of Emergency Physicians, American Heart Association, American Medical Association, Association of Clinical Research Professionals, Public Responsibility in Medicine and Research, Society for Academic Emergency Medicine, Society for Simulation in HealthcareDisclosure: Nothing to disclose.
Additional Contributors
Timothy E Corden, MD Associate Professor of Pediatrics, Co-Director, Policy Core, Injury Research Center, Medical College of Wisconsin; Associate Director, PICU, Children's Hospital of WisconsinTimothy E Corden, MD is a member of the following medical societies: American Academy of Pediatrics, Phi Beta Kappa, Society of Critical Care Medicine, Wisconsin Medical SocietyDisclosure: Nothing to disclose.
Acknowledgements
Richard G Bachur, MD Associate Professor of Pediatrics, Harvard Medical School; Associate Chief and Fellowship Director, Attending Physician, Division of Emergency Medicine, Children's Hospital of Boston
Richard G Bachur, MD is a member of the following medical societies: American Academy of Pediatrics, Society for Academic Emergency Medicine, and Society for Pediatric Research
Disclosure: Nothing to disclose.
Ann G Egland, MD Consulting Staff, Department of Operational and Emergency Medicine, Walter Reed Army Medical Center
Ann G Egland, MD is a member of the following medical societies: American College of Emergency Physicians, American Medical Association, Association of Military Surgeons of the US, Medical Society of Virginia, and Society for Academic Emergency Medicine
Disclosure: Nothing to disclose.
Terrance K Egland, MD Director, Business Planning and Development, Bureau of Medicine and Surgery
Terrance K Egland, MD is a member of the following medical societies: American Academy of Pediatrics
Disclosure: Nothing to disclose.
James Li, MD Former Assistant Professor, Division of Emergency Medicine, Harvard Medical School; Board of Directors, Remote Medicine
Disclosure: Nothing to disclose.
Alison Wiley Lozner, MD Resident Physician, Harvard Affiliated Emergency Medicine Residency, Brigham and Women's Hospital; Clinical Fellow in Emergency Medicine, Harvard Medical School
Alison Wiley Lozner, MD is a member of the following medical societies: American Academy of Emergency Medicine and American College of Emergency Physicians
Disclosure: Nothing to disclose.
James Kimo Takayesu, MD, MSc Assistant Professor in Surgery, Director of Undergraduate Medical Education, Consulting Staff, Massachusetts General Hospital; Associate Residency Director, Harvard Affiliated Emergency Medicine Residency Partners
James Kimo Takayesu, MD, MSc is a member of the following medical societies: Alpha Omega Alpha, American College of Emergency Physicians, Sigma Xi, and Society for Academic Emergency Medicine
Disclosure: Nothing to disclose.
Mary L Windle, PharmD Adjunct Associate Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug Reference
Disclosure: Nothing to disclose.
Wayne Wolfram, MD, MPH Associate Professor, Department of Emergency Medicine, Mercy St Vincent Medical Center
Wayne Wolfram, MD, MPH is a member of the following medical societies: American Academy of Emergency Medicine, American Academy of Pediatrics, and Society for Academic Emergency Medicine
Disclosure: Nothing to disclose.
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Sections Pediatric Dehydration
Overview
Practice Essentials
Pathophysiology
Etiology
Epidemiology
Show All
Presentation
History
Physical Examination
Show All
DDx
Workup
Approach Considerations
Obtaining Vascular Access
Show All
Treatment
Approach Considerations
Mild Volume Depletion
Moderate Volume Depletion
Severe Volume Depletion
Pharmacologic Therapy
Consultations
Show All
Medication
Tables)
References;)
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Types of Sentences
Types of Sentences
Essay Writing Guides
Written by
Daniel Parker
Written by
Adam Jason
Last updated:
April 3, 2020
October 25, 2023
8 min read
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On average, a person says about 7,000 words every day. However, what really forms our speech is not really about words, but rather about sentences and how we use them.
We use different types of sentences in both written and oral speech. Each type serves a specific designated purpose, and, though you may not realize it, all sentence types are different from one another.
How many types of sentences exist in the English language? What are their purposes and how is each formed? Knowing the answers to these questions is important. You will be able to bring your writing skills to a whole new level, and make your speech more thoughtful, effective, and purposeful — which is especially important for students. In this article, the EssayPro research paper writing service has made a comprehensive guide on the different types of sentences and their characteristics.
What Are the Four Types of Sentences?
In the English language, we can identify four types of sentences. They are classified based on their purposes:
The declarative sentence type is used for making a statement;
The interrogative type is used to ask a question;
Imperative sentences are used to tell somebody to do something (i.e. give a command or an order);
The exclamatory type is used to express (e.g. surprise).
Apart from their purposes, these sentence types require the use of different punctuation marks. Also, if we are talking about oral speech rather than writing, every type uses different intonations to highlight their purposes.
In the chart below, you can find a short overview of all these types with their key characteristics from our dissertation writing services:
| | | | |
--- --- |
| Sentence Type | Purpose | Final Punctuation Mark | Example |
| Declarative | Make a statement | . | I am a student. |
| Interrogative | Ask a question | ? | Are you a student? |
| Imperative | Give a command | . or ! | Welcome the new student. |
| Exclamatory | Form an expression | ! | There are so many students here! |
Now, when you have grasped the general idea, let’s take a close look at each sentence type separately.
Also, read our: PUNCTUATION GUIDE to bring your writing skills to a whole new level.
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Declarative Sentence
What is a declarative sentence? The main goal of this type of sentence is to make a statement. So, in a nutshell, any sentence that tells us something can be attributed to this category. It doesn’t matter what kind of information it delivers, whether it is a proven fact or a theoretical statement, the only thing that matters is if it declares something – if it does, it is a declarative sentence.
The standard order of words in such sentences is as follows:
Subject + verb + object… = where the subject is usually a noun or pronoun (a person, thing, place, etc.); the verb is the action or state of being; and the object is any word (or multiple words) that are influenced by the verb.
Example: The girl (subject) lost (verb) her favourite doll (object).
As a rule, a declarative sentence ends with a period. Sentences that fit the definition for being of declarative type, can be further categorized into two groups – positive and negative. Respectively, the difference between them is whether you intend to make a positive or negative statement:
| | |
--- |
| Positive | Negative |
| Jessica likes history lectures. | Jessica does not like history lectures. |
| He plays on a football team. | He doesn’t play on a football team. |
Where are declarative sentences used most often, you might ask? This type of sentence is the most common type of sentence. We use it in oral and written speech whenever we intend to share some information. Therefore, declarative sentences are widely used in all types of academic papers, written documents, dialogues, etc.
Interrogative Sentence
What is an interrogative sentence? A declarative sentence aims to share information, whereas an interrogative one strives to receive information. According to the interrogative sentence definition, any sentence that asks a question can be attributed to this category and will always end with a question mark.Unlike the other types of sentences, interrogative sentences have a different word order:(wh-word or how) + auxiliary verb + subject
Example: Where (wh-word) is (auxiliary verb) Kate (subject)?
Similar to other types of sentences, interrogative sentences can be either positive or negative. Here are a few interrogative sentence examples of both:
| | |
--- |
| Positive | Negative |
| Does Jessica like history lectures? | Doesn’t Jessica like history lectures? |
| Did he play on a football team? | Didn’t he play on a football team? |
Where can you use this type of sentence? Generally, interrogative sentences can have a variety of applications in your speech. However, when it comes to writing, especially academic papers, questions don’t always fit in context. For example, if you write a narrative essay, you will likely not be required to ask questions, as the purpose of a narrative essay paper is to provide information, not to collect it. Interrogative sentences are more commonly used in persuasive essays to encourage readers to reflect upon or reinforce the effect of the author’s arguments (e.g. “Did you know that…?”).
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Imperative Sentence
What is an imperative sentence? The main goal of these sentences is to tell others to do something, or, in other words, give a command. Imperative sentences can end with either a period or an exclamation mark.
The word order and form of such a sentence are different from other types. It often doesn’t have a subject, because an imperative sentence, by default, speaks to the recipient or reader (if it is a written text). Generally, such sentences consist of a base verb + any additional details.
These sentences can also be negative and positive, here are a few imperative sentence examples to help you grasp the idea:
| | |
--- |
| Positive | Negative |
| Attend history lectures! | Do not attend history lectures! |
| Join a football team. | Don’t join a football team. |
As for possible applications of imperative sentences, they are mostly used in oral speech, or, if we are talking about writing, can be used in dialogues between characters, or in the form of a “call to action” that encourages readers to do something.
Exclamatory Sentence
What is an exclamatory sentence? The last of the four sentence types is the exclamatory sentence. It is used to express a strong surprise of emotion and always ends with an exclamation mark.Here are a couple of examples of how the basic order of words in such sentences might look:
What (+ adjective) + noun + subject + verb
How (+ adjective/adverb) + subject + verb
For example: What wonderful (adjective) weather (subject)! or How generous (adjective) you (subject) are (verb)!
Unlike previous types, exclamatory sentences do not have a negative form.Look at these exclamatory sentence examples to see how they are formed:
What a beautiful painting!
I feel terrible!
What an excellent idea it was to throw him a surprise party!
How nice it was!
Exclamatory sentences express powerful emotions, and, respectively, strive to evoke the same emotions in readers. In many cases, using this type of sentence in academic papers is inappropriate. However, if you are writing a descriptive or narrative essay, exclamatory sentences are great tools for helping your story to become even more vivid by delivering the right emotions to the reader.
Extra Tips on Variety
What is the key to having the perfect writing style? Some may say it is a solid vocabulary, others may not and suggest that it is being able to include an abundance of details. Furthermore, people might suggest that it is the number of ideas, examples, and arguments you include in your writing. But, if you put it all together, it turns out that the true key to literary mastery is variety!
So, here is our best tip for empowering your writing – add more variety. It may seem a bit tricky at first. But, as soon as you fully understand the concept behind every sentence type and get a bit more comfortable with each, you can try experimenting with them.
Here are some of the best tips on how you can use different types of sentences to your benefit in writing:
Make a hook with a question. The introduction of a paper should be intriguing and engaging to make the reader want to continue reading. A good way to draw attention is to put a hook, in the form of a question, at the beginning of your introduction.
Example: “Have you ever thought how much benefit school uniforms can bring to students?”
Use imperative sentences to establish the right guidance. The right command, presented in the right tone, can have a powerful effect on readers and stimulate their interest.
Example: “The way people thought of uniforms in schools has often been controversial. Some like the idea of it, while others don’t. However, studies show that uniforms have more pros than cons. Get ready to change your opinion!”
Try different word orders. Although we have shared some basic formulas for shaping sentences with you in this article, they don’t necessarily always have to follow the “subject + verb” scenario. In fact, changing the order of words can occasionally add variety to your style and make your text look and sound better.
Example: Instead of writing - “I know what it means to be a part of a large team, so I know how hard it can be to find compromises.” Try using a different format - “Knowing what it means to be a part of a large team, I do know how hard it can be to find compromises.” See the difference?
Use a question to summarize key points. Here’s another way to use interrogative sentences – put them at the beginning of a paragraph to create a quick summary of your ideas.
Example: “What was the result of a study on the pros of uniforms? – that’s what we are going to talk about.”
This trick will help you to create a smoother transition between paragraphs.
Use different structures. As a rule, most writing tips say the same thing – “Keep it Simple!” While this advice makes sense, sometimes adding a bit of variety won’t hurt. To keep readers engaged, writers often use this trick. They alternate simple, compound, and complex constructions. This tip helps to significantly improve the readability of the text. The thing is that people find it hard and tiring to read the same sentence types one after another.
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What Are The Four Types Of Sentences?
The four types of sentences are:
Declarative: These sentences make statements and end with a period.
Interrogative: These sentences ask questions and always end with a question mark.
Imperative: These sentences give commands or make requests and end with a period or an exclamation mark depending on the intensity.
Exclamatory: These sentences express strong emotions or excitement and end with an exclamation mark.
Which One Of The Following Types Of Sentences Asks A Question And Always Ends With A Question Mark?
The type of sentence that asks a question and always ends with a question mark is an "Interrogative" sentence.
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Mia
November 1, 2024
Eye of the Tiger gonna help me pass calc? Sure, I’ll give it a shot but not holding my breath here 💀
Lucy
October 30, 2024
Nice choice of songs! I know almost all of them and the playlist for studying is epic! Florence and the Machine - Dog Days Are Over is a cray cray :)
Sofia
October 30, 2024
Absolutely loving this playlist! 🎧 Big thanks for putting this together – totally recommend this to anyone needing that extra motivation boost! 🙌
Katty
October 30, 2024
The song that motivates me the most is MÅNESKIN - Honey!
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Daniel Parker
is a seasoned educational writer focusing on scholarship guidance, research papers, and various forms of academic essays including reflective and narrative essays. His expertise also extends to detailed case studies. A scholar with a background in English Literature and Education, Daniel’s work on EssayPro blog aims to support students in achieving academic excellence and securing scholarships. His hobbies include reading classic literature and participating in academic forums.
Adam Jason
is an expert in nursing and healthcare, with a strong background in history, law, and literature. Holding advanced degrees in nursing and public health, his analytical approach and comprehensive knowledge help students navigate complex topics. On EssayPro blog, Adam provides insightful articles on everything from historical analysis to the intricacies of healthcare policies. In his downtime, he enjoys historical documentaries and volunteering at local clinics.
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14482 | https://www.youtube.com/watch?v=BAoRwWflYbc | Proof that Square Root of 2 is Irrational. Proof by Contradiction.
Simple Science and Maths
17100 subscribers
219 likes
Description
14052 views
Posted: 24 Oct 2019
This is the formal proof that the square-root of 2 is irrational. This is a proof by contradiction.
Credits:
Kevin MacLeod: Carefree
23 comments
Transcript:
[Music] okay in this video we're going to go through the formal proof that root two is irrational so the form of proof for that is by contradiction so we're gonna try and prove that it is rational and we'll end up with some sort of contradiction which you'll see at the end so to start off we assume that it is in fact a rational number well for root two to be rational rational means it can be written as a fraction P over Q where P and Q have no common factors okay so now into the proof if the square root of two is P over Q this implies that if we were to square both sides two would be equal to P squared over Q squared work in true just a bit of algebra isolating or getting rid of the fraction this implies that 2q squared is equal to P squared what this means is P squared is an even number because P squared can be written as 2 times some other number so therefore P squared is even now if P squared is even that means P is also even if you think about any even square number for example 36 the square root is 6 64 the square root is 8 so all the even numbers when you square them the answer is also even so if P is even P can be written as 2 times some number K any even number can be written as 2 times another number K is an element of Z so it's some integer that's the first thing that we need to first thing that we need to say the second thing then is using this here so 2q squared is equal to P squared well now instead of P I'm going to write in to K so 2q squared is equal to 2 K brackets squared ok 2k squared is 4 K squared so that means 2q squared is equal to 4 K squared divide across by 2 Q squared is equal to 2k squared well this is similar to this we've now just shown that Q squared is even therefore Q squared is even and that again same as a pair means that Q is even which again means that Q can be written as 2 times a number let's call it m so Q is can be written as 2 times M now if we take these two things here so Q is equal to 2m P is equal to 2 K and we put them back into this let's see what happens so square root of 2 is equal to P over Q now P is 2 K and Q is 2m that means the square root of 2 is equal to 2 K over 2m now we have a common factor so it's a common factor of 2 and that is a contradiction remember at the start we said if P if the square root of 2 is rational then it can be written as P over Q where P and Q have no common factors now we've just proven that if it was in fact rational it would have a common factor of 2 and that is the contradiction so therefore the square root of 2 is irrational ok if you have any questions just ask in the comments below and thanks for watching and I'll see you in the next video |
14483 | https://controlaltbackspace.org/percent/ | Understanding Percentages
Figures expressed in percentages are ubiquitous,
and interpreting and doing calculations with them
is an important part of the basic mathematical literacy useful in everyday life.
Yet few points of arithmetic are as confusing and full of traps for the unwary
as percentages.
I recently got fed up with being unable to think clearly about percentages myself,
so I figured I’d write this article
to force myself to understand them completely.
Hopefully it will help you, too!
This article is written for adults
and secondary-school students
who generally understand arithmetic and basic algebra,
but could use a refresher on percentages.
I’m going to explain why each fact about percentages is true
to make it more memorable,
so if you’ve never heard of percentages,
the order of operations,
or the distributive property,
this is probably not the right introduction for you.
What is a percentage?Permalink
This may seem like a silly thing to need to remind yourself of,
but before we think about anything else,
it really is worth taking a minute
to remind yourself of the definition of a percentage,
because thinking clearly about the definition
will often allow you to work out the answers
to otherwise difficult questions.
N% is simply the number N divided by 100,
i.e., a fraction with N as the numerator and 100 as the denominator.
Or put another way, it’s a decimal, times 100.
So 10% means 1010010100 or 0.1;
250% is 250100250100 or 2.5.
Percentages are commutativePermalink
Here’s my favorite unexpected fact about percentages:
8% of 50 is the same thing as 50% of 8.
While counterintuitive at first glance,
this is easy to understand if you keep the definition of a percentage in mind:
these simplify to (8×1100)×50(8×1100)×50
and (50×1100)×8(50×1100)×8,
and multiplication is commutative and associative
(the order and grouping don’t matter).
This often comes in handy when doing mental math:
8% of 50 feels awkward enough
that you might go looking for a calculator
if you didn’t notice the shortcut,
but 50% of 8 is trivial.
Percent of, percent increase, and percent decreasePermalink
These small words make a huge difference to the meaning of a percentage.
BasicsPermalink
Percent of is the simplest:
this simply means to multiply some number by a percentage.
So if we take 20% of 50, that’s 50×0.2=1050×0.2=10.
Generally, P% of N is N(P100)N(P100).
Percent increase means to multiply some number by a percentage
(or, take that percent of the number)
and then add the result to the original figure.
So a 20% increase from a baseline of 50 is 50+50(0.2)=6050+50(0.2)=60.
Generally, a P% increase to N is N+N(P100)N+N(P100).
Percent decrease means to multiply some number by a percentage
and then subtract the result from the original figure.
So a 20% decrease from a baseline of 50 is 50−50(0.2)=4050−50(0.2)=40.
Generally, a P% decrease to N is N−N(P100)N−N(P100).
Percent change can be used to refer to both percent increase and percent decrease.
A positive percent change is an increase;
a negative percent change is a decrease
of the absolute value of the percent change
(e.g., a −15% change is a 15% decrease).
Converting percent change to percent ofPermalink
Percents increase and decrease
can be converted to percents of by adding the percentage change to 100%.
You can see this equivalence clearly
by applying the distributive property
to the formula for percent increase given above:
N increased by P%=N+N(P100)=N(1+P100)=N(100%+P%)
So a 150% increase is 100%+150%=250% of the original.
To convert in the opposite direction,
switch the sign on the 100%:
if a figure is 120% of something,
it’s a 120%−100%=20% increase.
The same formula works for decreases
if you treat them as negative percent change:
a 30% decrease is 100%−30%=70% of the original,
and 60% of something is a 60%−100%=−40% change or 40% decrease.
Another mental math tip:
The conversion from percent change to percent of is almost always beneficial.
The most straightforward approach to percent change
requires one multiplication and then one addition,
and you have to hold onto the original number during the entire problem.
If you convert the change to a percent of,
the addition is guaranteed to be to or from 100 (easy),
and you don’t have to remember anything during the subsequent multiplication.
Percentages equal to or greater than 100%Permalink
All this likely feels straightforward so far.
The trick is what happens once we reach 100%.
The math still works the same way,
but the answers don’t make as much intuitive sense anymore:
100% of 50: This is just 50.
100% increase from 50: This is 100, 50+50(1.00).
100% decrease from 50: This is 0, 50−50(1.00).
In summary, 100% of a number is a no-op
(though depending on context,
there’s a good chance it means
whoever gave you this figure was confused
and intended a 100% increase).
A 100% increase means the number doubles.
A 100% decrease means there’s nothing left,
no matter what the original number was.
It does not mean the number halved,
as you might naïvely expect if you invert the increase case without checking the math;
people frequently mix this up.
To halve a number, you need a 50% decrease,
but to double it, you need a 100% increase.
Similarly with percentages greater than 100%:
150% of 50: This is 75, 50×1.5.
150% increase from 50: This is 125, 50+50×1.5.
150% decrease from 50: This is not well-defined.
You could argue it’s −25,
but that probably doesn’t make a whole lot of sense in context.
A 100% decrease is normally the largest meaningful decrease.
Other wordsPermalink
When you see something that isn’t clearly phrased as of, increase, or decrease,
you need to determine how to interpret it to avoid errors.
Examples:
20% off or a 20% reduction is a 20% decrease.
An 18% surcharge or tip is 18% of the amount it’s applied to, but represents an 18% increase to your bill.
A 30% adjustment or change is either an increase or decrease, depending on context.
5% as many means 5% of.
Percentage pointsPermalink
When talking about changes in figures already expressed as percentages,
things get dicey.
If you say 62% of people voted in the last election,
and this year there was a 20% decrease,
technically you’re saying that the number represented by the percentage decreased by 20%,
so that 0.62−0.62∗0.2=49.6% of people voted.
You probably meant to say that 42% of people voted;
to express this idea,
you should say that voting decreased by 20 percentage points
(that is, 0.2 should be subtracted from the previous percentage).
This only applies to increases and decreases.
If you say that 80% as many people voted in this election as last election,
and last election 62% of people voted,
there is no ambiguity.
(That said, this is a terrible way to express statistics for people to read,
as it’s hard to interpret these numbers;
nobody wants to have to multiply 80% by 62%
to figure out the all-important value of
what proportion of people voted this election.)
Sequentially applied percentagesPermalink
People are often tempted to take shortcuts with percentages
that result in incorrect answers.
Most of these involve incorrectly combining the effects of several percentages.
Percentage increases or decreases cannot be addedPermalink
Suppose you’re in a store and see that something’s on sale for 20% off
(i.e., its price at checkout will be 20% lower than quoted on the tag),
and you also have a 10% off coupon.
It’s very tempting to add these up and calculate a 30% decrease,
but this is not the same thing.
To see this, suppose the item is $50.
20% off and then 10% off:
$50−$50×0.2=$40;
then $40−$40×0.1=$36.
30% off:
$50−$50×0.3=$35.
(Fast way to do this in your head, applying our tricks from earlier:
a 30% decrease from X is 70% of X,
and 70% of 50 is the same as 50% of 70,
and half of 70 is 35.)
The problem here is that percentage changes are relative to some other number,
and after applying the first percentage decrease,
the base number has become lower,
so the second percentage decrease has a smaller effect.
A good way to understand this intuitively is to consider what happens
if you have a 50% off coupon and find a clearance rack at 50% off.
You probably (correctly) expect you can get a total of 75% off
any items from that rack;
you don’t expect you can walk out the door with the entire rack
for free (50% + 50% = 100% off)
and sell it on eBay for hundreds of dollars.
If you want to correctly combine two percentage decreases into a single percentage decrease,
you need to “pre-apply” the first decrease to the second decrease
prior to summing them up,
to account for the baseline being lower
after the application of the first decrease.
So to combine 20% and 10%,
we would decrease 10% by 20% for 8%,
and then add our reduced 8% to 20% for a 28% decrease.
Now $50−$50×0.28=$36,
as we got when we applied them separately.
Similarly, if we decrease 50% by 50% and add it to 50%, we get 75%.
In general, to combine two sequential decreases of A% and B%
into a combined C% decrease:
C%=A%+B%(100%−A%)
Note however that we do get the same result,
without doing anything special,
if we first take 10% off and then 20% off,
rather than 20% off and then 10% off,
because percentages are commutative;
we can apply the reductions in any order
as long as we apply them separately
or combine them as described above.
Tip:
As you might expect, stores occasionally do this wrong.
On the off chance that you notice, you generally should not complain
unless an increase is involved somewhere,
as the error comes out in your favor.
If you’re feeling honest,
let me just say this will be a loooong argument,
and you’re not likely to find a chalkboard at the cash register.
An N% increase and an N% percent decrease do not cancel outPermalink
As a corollary,
suppose you have a $50 item that’s not on sale,
you have a 10% off coupon,
and you know 10% tax will be added at the register.
That makes your cost $50, right?
Not so,
because the 10% change is applied to different baseline values:
the decrease is applied to a higher figure than the increase.
A 10% decrease followed by a a 10% increase comes out
as $50−$50×0.1=$45,
then $45+$45×0.1=$49.50.
Surprisingly, the order we apply the changes in still doesn’t matter.
If we take the 10% increase first, then the 10% decrease,
we end up with $50+$50×0.1=$55,
then $55−$55×0.1=$49.50 still.
This is easier to intuit if you realize that in either order,
the subtraction is applied to the larger number.
(It still feels wrong that it’s systematically biased downwards, though,
at least to me!
If you know of a clever proof that makes this intuitive, let me know.)
The “error” (percentage decrease) introduced by a successive N% increase and N% decrease, or vice versa, is exactly N%2.
Proof:
This surprisingly simple result can be derived as follows.
Suppose we have a baseline value B.
An N% decrease in B can be expressed as B(100%−N%),
since B decreased by N% is the same as 100%−N% of B.
A subsequent N% increase similarly multiplies this value by (100%+N%),
for a total of:
B(100%−N%)(100%+N%)
Now distribute (note that 100% times 100% is 100%, since 100%=1):
B(100%+100%⋅N%−100%⋅N%−(N%)2)=B(100%−(N%)2)
B(100%−X) is equivalent to an X% decrease in B,
so this is the same thing as an N%2 decrease. ∎
Percentages by another order of magnitudePermalink
Two other concepts are much less used than percentages,
but work very similarly and are worth familiarizing yourself with:
per mille (‰):
This is a number divided by 1000, rather than a number divided by 100.
25% = 250‰.
basis point (bp):
This is one-one hundredth of a percentage point.
Basis points are most commonly used to describe changes in rates in financial markets.
If an interest rate is 4.0% and it increases by 25 basis points,
the new interest rate is 4.25%.
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14484 | https://www.gov.uk/government/publications/methanol-properties-incident-management-and-toxicology/methanol-toxicological-overview | Skip to main content
UK Health
Security
Agency
Guidance
Methanol: toxicological overview
Updated 11 October 2024
© Crown copyright 2024
This publication is licensed under the terms of the Open Government Licence v3.0 except where otherwise stated. To view this licence, visit nationalarchives.gov.uk/doc/open-government-licence/version/3 or write to the Information Policy Team, The National Archives, Kew, London TW9 4DU, or email: psi@nationalarchives.gov.uk.
Where we have identified any third party copyright information you will need to obtain permission from the copyright holders concerned.
This publication is available at
Main points
Kinetics and metabolism
Methanol is readily absorbed by all routes and distributed in the body water.
Methanol undergoes extensive metabolism, but small quantities are excreted unchanged by the lungs and in the urine.
Health effects of acute exposure
Methanol is toxic following ingestion, inhalation, or dermal exposure.
Exposure may initially result in CNS depression, followed by an asymptomatic latent period.
Metabolic acidosis and ocular toxicity, which may result in blindness, are subsequent manifestations of toxicity.
Coma and death may occur following substantial exposures.
Long-term effects may include blindness and, following more substantial exposures, permanent damage to the CNS.
Health effects of chronic exposure
Long-term inhalation exposure to methanol has resulted in headaches and eye irritation in workers.
Methanol is considered not to be a mutagen or carcinogen in humans.
Methanol is considered not to be a reproductive toxicant in humans.
Summary of health effects
Methanol may be acutely toxic following inhalation, oral or dermal exposure.
Acute methanol toxicity often follows a characteristic series of features; initially central nervous system (CNS) depression and gastrointestinal tract (GI) irritation may be observed. This is typically followed by a latent period of varying duration from 12 to 24 hours and occasionally up to 48 hours. Subsequently, a severe metabolic acidosis develops with nausea, vomiting and headache.
Ocular toxicity ranges from photophobia and misty or blurred vision to markedly reduced visual acuity and complete blindness following high levels of exposure. Ingestion of as little as 4 to 10mL of methanol in adults may cause permanent damage.
Coma and death may occur after substantial exposures. The minimal lethal dose following ingestion is considered to be in the range of 300 to 1,000mg/kg. Severe intoxication may cause permanent damage to the CNS, manifesting as a Parkinsonian-like condition and permanent blindness.
Chronic inhalation exposure to low concentrations of methanol has resulted in headache and eye irritation in workers.
Methanol has no structural alerts for mutagenicity and in-vitro animal studies are negative. The limited data on the mutagenicity of methanol in humans, suggests that methanol is non-genotoxic.
Methanol is not classified as a mutagen or carcinogen in humans.
Methanol is not classified as a human reproductive toxicant. However, fetal toxicity may arise secondary to maternal toxicity. Findings from animal studies may indicate possible risks to the human fetus at early stages of development due to the similarity of early embryonic processes; however, non-primate metabolism of methanol is distinct from human metabolism, and this should be considered when determining risks to humans. It is unlikely that exposure to low concentrations of methanol would result in adverse effects in the fetus.
Kinetics and metabolism
Absorption
Methanol is readily absorbed by inhalation, ingestion, and dermal exposure (1, 2). Following ingestion, methanol is absorbed within 30 to 60 minutes depending upon the presence or absence of food in the GI tract (1). Around 60 to 80% of inhaled methanol is absorbed in the lung of humans (2).
Distribution
Distribution is rapid and occurs throughout body water as indicated by a volume of distribution of approximately 0.6L/kg (2). Individual tissue and organ concentrations are dependent on their water content (2). There is no protein binding and methanol is poorly distributed to fatty tissues (3).
Metabolism
The liver is the primary site of metabolism for methanol. Through a series of oxidative steps methanol is oxidised to methanal (HCHO, formaldehyde), methanoic acid (H•COOH, formic acid) and finally detoxified to carbon dioxide (CO2). The majority of an ingested dose of methanol (96.9%) is converted to carbon dioxide by this route (1). The main enzyme groups involved in each step are alcohol dehydrogenase, aldehyde dehydrogenase and folate-dependent mechanisms, respectively. Methanoate (formate) or methanoic acid (formic acid) may be formed, dependent on pH (2). The term “formic acid” – and not methanoic acid – persists in the literature and will therefore be used in this text for compatibility.
Mechanism of toxicity
In humans and primates, toxicity of methanol is mediated through metabolites and not the parent molecule. Formic acid is considered to be the key toxicant; and in animal species with a poor ability to metabolise this product (primates and humans) fatal toxicity may occur from a profound anion gap metabolic acidosis and neuronal toxicity (2, 4). Un-dissociated formic acid readily crosses the blood–brain barrier leading to CNS toxicity; aggressive alkaline therapy is required to maintain formic acid in the dissociated form (3).
As a moderate inhibitor of cytochrome-c oxidase, formate may cause tissue oxygen use to be impaired, leading to anaerobic respiration with subsequent increased lactate production, which may further contribute to the acidosis (3).
The relative affinity of alcohol dehydrogenase for ethanol is much greater than for methanol (20 : 1) (2). This difference has been exploited therapeutically in cases of poisoning, where alcohol is administered under medical supervision to reduce the formation of formic acid. A selective enzyme inhibitor such as Fomepizole may also be used to block the metabolism of methanol (5).
Excretion
The majority of metabolised methanol is ultimately excreted as carbon dioxide (1). A minor portion of methanol is excreted unchanged in the urine or in exhaled air. The half-life for the systemic clearance of methanol has been reported to be 2.5 to 3 hours for doses less than 100mg/kg bodyweight (bw) methanol, increasing to 24 hours or longer for doses greater than 1,000mg/kg bw methanol (1).
Sources and Route of Human Exposure
Methanol is produced endogenously, as a result of intermediary metabolism. It is also present in the diet, notably in fruit and vegetables and their juices. Exposure to methanol at levels found in the diet would not be expected to result in adverse effects (1). Low concentrations of methanol may be found in alcoholic beverages, although concentrations in products of illegal distillation may be much higher (1). Exposure may also occur by the intentional or accidental ingestion of consumer products containing methanol, such as antifreeze, brake fluid and window cleaning solutions (6).
Inhalant misuse of volatile methanol-containing products may be a significant source of exposure to methanol (7). In some countries methanol is used as a denaturant for ethanol; methanol intoxication has occurred in individuals intentionally consuming such mixtures (8).
Methanol is produced in large quantities worldwide and is used extensively as a solvent. Other notable uses are as a chemical intermediate and as a denaturant. Individuals who work in industries in which methanol is used may be chronically exposed to methanol vapours (3). Workplace exposure limits (WELs) are enforced to protect workers from the harmful effects of methanol; in the UK the long-term WEL is 266mg/m3 (200ppm) and the short-term WEL is 333mg/m3 (250ppm) (9).
Health effects of acute or single exposure
Human data
General toxicity
Humans (and primates) are particularly sensitive to methanol toxicity when compared to non-primates. The severity of toxicity following exposure has been correlated with the degree of metabolic acidosis rather than to the concentration of methanol (3). This is due to toxicity being determined primarily by the rate of formic acid formation and hepatic folate status which governs its detoxification. The key features include metabolic acidosis, ocular toxicity, CNS depression and coma (see table 1). Convulsions, coma, shock, persistent acidosis, bradycardia and renal failure are indicators of a poor prognosis (5). Morbidity and mortality are high in methanol intoxication (4).
Table 1: Key features of methanol toxicity
| Features | Comments |
--- |
| Initial CNS depression | Initial intoxication may resemble that arising from ethanol ingestion but of shorter duration and less pronounced. GI irritation may also occur |
| Asymptomatic latent period | May last 12–24 hours following ingestion, but occasionally up to 48 hours. Patients describe no overt signs or symptoms during this period |
| Severe metabolic acidosis | Nausea, vomiting and headaches may occur with acidosis |
| Ocular toxicity | Visual disturbances ranging from mild photophobia and “snowfield” vision to markedly reduced visual acuity and blindness may develop 12 to 48 hours after ingestion. Visual impairment usually takes the form of central scotoma or complete blindness secondary to optic atrophy |
| Delayed onset neuropathy | Symptoms occur 12 to 24 hours after exposure and include seizures, coma or cerebral oedema that may develop as a result of metabolic acidosis. Tremor, dementia, rigidity and bradykinesia have been observed |
References
(2-5, 10)
Inhalation
Inhalation of methanol vapour can cause acute toxicity, as described in the general toxicity section. Toxicity has been associated with the inhalation of methanol vapour at concentrations greater than 400mg/m3 (2). Deliberate inhalation of volatile preparations containing methanol may cause toxicity, in a series of four case reports, one patient was found on ophthalmic examination to have hyperaemic discs and decreased visual acuity (2, 11).
Ingestion
Ingestion of methanol can cause severe acute toxicity, as described in the general toxicity section.
There is significant variability within humans on the reported oral toxicity and lethality of methanol. The minimal lethal dose following ingestion is considered to be in the range of 300 too 1,000mg/kg (2). In one review, the minimum lethal dose following ingestion has been reported at 15mL of a 40% volume/volume (v/v) methanol solution (10). Another individual is reported to have survived ingestion of 500mL of the same solution. A significant confounding factor may be the concomitant ingestion of ethanol, which may have mitigated some of the methanol toxicity. Ingestion of as little as 4 tp 10mL methanol in adults may cause permanent blindness (3, 5). CNS effects may occur from doses as low as 3 to 20mL of methanol (12). Other effects reported include acute pancreatitis and renal failure (5).
In one clinical case, a pregnant woman (at 35 weeks’ gestation) was reported to have ingested 250 to 500mL of methanol (13). After 1 hour of uncomplicated labour on day 6 of admission and treatment, the patient delivered a child who had no signs of distress and with Apgar scores of 9/10 at 1 minute and 10/10 after 5 minutes. The clinical course was uneventful in both the child and mother and no visual disturbances developed in the child within a follow up of 10 years (13). This case highlights the potential for a positive outcome following acute maternal intoxication with methanol where interventions are initiated rapidly.
Dermal or ocular exposure
Methanol may be absorbed across the skin and can result in systemic toxicity. Methanol is also irritating to skin and may cause dry skin and redness (14).
Percutaneous absorption has been noted to cause toxicity in children. In a case series of 48 intoxicated patients, 30 had severe respiratory depression, 14 were comatose, 11 had seizures and 7 had anuria or severe oliguria; there were 12 deaths (2). In Egypt, a number of neonates died of severe metabolic acidosis following dermal exposure to methanol which was the main constituent of a compress used to relieve fever. The compresses were made using a local product termed “red-alcohol” which, on analysis, was found to have contained methanol (70 to 90% v/v) (15).
Contact of methanol with the eyes may result in irritation only; the ocular toxicity described previously is mediated by systemic and not local ocular exposure (3).
Delayed effects following acute exposure
Signs and symptoms of methanol poisoning may be delayed for up to 12 to 24 hours post-exposure. Ocular effects generally develop 12 to 48 hours following methanol ingestion (1). Visual impairment or blindness may be permanent. Damage to the CNS is often in the form of lesions in basal ganglia especially the putamen, which may result in long-term neurological deficits ranging from moderate polyneuropathy to tremors, rigidity, spasticity and hypokinesis, as well as Parkinsonian-like extrapyramidal syndrome with mild dementia (2, 16-18). Long-term effects of methanol exposure may be reversible; in one case of acute intoxication, a follow up at 1 month showed increased cognitive function and only a mild lower extremity tremor (17).
Ethanol co-ingestion may lead to a delay in the onset of the metabolic toxic features due to competitive co-inhibition of alcohol dehydrogenase (please refer to the kinetics and metabolism section for more detail) (5).
Animal data and in-vitro data
Due in part to metabolic differences, lower order animal species, such as the rat, exhibit different responses to methanol than humans. Methanol – and not its metabolites – is the key toxicant, with features of CNS depression a common finding. The key findings in humans of metabolic acidosis and ocular toxicity are normally not seen. Thus, extrapolation from animal studies to human findings must be performed with caution.
Non-human primates, such as rhesus monkeys (Macaca mulatta), are sensitive to methanol and acidosis, ocular findings have been reported. Consequently, primate data is the focus of much of the animal toxicology section.
Inhalation
Methanol has been demonstrated as toxic through inhalation exposure in a number of animal species. Acute inhalation exposure has been associated with degeneration and necrosis of parenchymal tissues and neurons, accompanied by capillary congestion and oedema in rats, rabbits and monkeys (2). In one early study using primates, death was reported following exposure to 1,310mg/m3 (1,000ppm); however, the duration of exposure was not cited (19). This is at odds with a more recent study which did not report any ocular toxicity by ophthalmic examination in monkeys exposed to 6,500mg/m3 for 6 hours a day, 5 days a week for a total of 4 weeks (2). Considering the chronic exposures (described below) the results from the former study need to be considered with some caution as exposures in excess of 1,310mg/m3 (1,000ppm) have been tolerated by monkeys in other studies.
Most animal and in-vitro data in the literature concerns chronic exposure to methanol.
Ingestion
A minimum lethal dose of methanol of 3g/kg bw has been reported for the rhesus monkey (20). The authors note, however, that the series of experiments was too small to give more than an approximate lethal dose, especially since there is likely to be considerable inter-individual variation in their response to methanol. The authors conclude that, although approximate, the primate data would suggest that the single oral lethal dose is of the same order of magnitude as that for humans. Clinical observations in the animals were considered to have been akin to those in humans. Inebriation was not observed below lethal doses, but CNS depression was apparent at higher dose levels. This was followed by a latent period and progressive weakness, coma and death from respiratory failure. Two out of four monkeys receiving lethal doses of methanol had ocular changes. In one animal, receiving a dose of methanol of 6g/kg bw, a small monocular retinal haemorrhage was noted prior to death and 29 hours after dosing. The other animal, receiving 3g/kg bw, had slight but definite blurring of the temporal disc margins which were blurred everywhere except nasally at 31.5 hours after dosing. At the time of assessment, both animals were apparently too weak to resist handling, suggesting vascular changes did not arise from neck stricture. Animals receiving lethal doses were noted as severely acidotic within 24 hours (20).
Dermal or ocular exposure
Methanol has been demonstrated to be toxic through dermal exposure in a number of animal species. In one early study using primates, following dosing with either 0.5 or 1.3mL/kg bw applied four times daily, toxicity was noted on the first day, with death occurring on the second day (19). The exposure model used minimised concomitant inhalation exposure (19).
Health effects following chronic or repeated exposure
Human data
General toxicology
In contrast to the widely reported toxicity of acute exposure to methanol, reports of effects following chronic exposure are infrequently reported (2).
Inhalation
Data on chronic exposure to methanol is limited. There have been reports of dizziness, headaches, GI, and visual disturbance in individuals exposed to methanol above the WEL of 260mg/m3 (200ppm) but not in those exposed to lower levels (1).
In one study, blurred vision, headache, nausea, dizziness, and eye irritation were experienced by workers using “spirit duplicators” (early document copying machines) at concentrations greater than the WEL (2). The duration of exposure and the number of individuals exposed were not reported (2).
Dermal or ocular exposure
Long-term or repeated dermal exposures to methanol may cause dermatitis (14).
There is insufficient data on chronic ocular exposure.
Genotoxicity
There are no studies in the literature that describe mutagenic or chromosomal effects of methanol in humans. There are no structural alerts for methanol and in-vitro studies are negative (see the next section on animal and in-vitro data).
The European Food Safety Authority (EFSA) recently evaluated the limited available evidence and concluded that methanol was not genotoxic (21).
Carcinogenicity
There is no data in the literature to indicate that methanol is carcinogenic in humans. Based on limited animal data, the lack of structural alerts and the lack of genotoxicity, methanol is not considered to be a carcinogen.
Reproductive and developmental toxicity
There is insufficient human data upon which to evaluate the developmental toxicity of methanol (22). Methanol is not classified on the basis of its reproductive toxicity and is not considered to be a reproductive or developmental toxicant in humans. However, fetal toxicity may arise secondary to maternal toxicity. Findings from animal studies may indicate possible risks to the fetus at early stages of development, but non-primate metabolism of methanol is distinct from that of humans as indicated previously. It is unlikely that exposure to low concentrations of methanol would result in adverse effects in the fetus.
Animal data
Inhalation
In a chronic inhalation study, monkeys were exposed to methanol concentrations of 13, 130 or 1,300mg/m3 (10, 100 and 1,000ppm, respectively) for 22 hours a day for up to 29 months. Bodyweight values and haematological and pathological examinations did not reveal any dose-dependent effects except for hyperplasia of reactive astroglias in the nervous system. This effect was not correlated with dose or exposure time and was found to be a reversible effect within a recovery test (2).
Ingestion
There is insufficient data on the toxicity of methanol in vivo following chronic ingestion of methanol.
Genotoxicity
Methanol is not classified as a mutagenic compound; the data indicates that it does not damage genetic material (2).
Methanol gave negative results for mutagenicity in a series of Ames tests in Salmonella typhimurium strains TA98, TA100, TA1535, TA1537 and TA1538 and in Escherichia coli strain WP2uvrA (23).
Mice exposed by inhalation to methanol at 1,050 or 5,200mg/m3 had no increase in the frequency of micronuclei in red blood cells or sister chromosome exchanges (SCEs), chromosomal aberrations or micronuclei in lung cells (2).
In a study designed to measure DNA adducts from exogenous sources, isotope-labelled methanol was administered orally to rats at doses of 500 or 2,000mg/kg bw/day for 5 days (21). An increasing number of adducts in all tissues was observed with an increase in administered dose (21). In a recent evaluation EFSA considered that no conclusions could be drawn from this and similar studies, noting that the method was not suitably robust, and that DNA adducts are biomarkers of exposure of organs and tissues to methanol and not direct indicators of mutagenic effect. EFSA concluded that the reliable in-vivo and in-vitro data did not suggest that methanol is genotoxic (21).
Carcinogenicity
There is no evidence from animal studies to suggest that methanol is a carcinogen, although because of major differences in the metabolism of methanol between rodents and humans the lack of an appropriate animal model is recognised (2). In a recent evaluation EFSA concluded that the available evidence is inadequate to determine the carcinogenic potential of methanol (21).
Reproductive and developmental toxicity
Developmental effects have been reported in rodents exposed to high concentrations (>2g/kg bw) of methanol through inhalation and ingestion (1, 2, 21, 22, 24). The differences in metabolism in rodents when compared with primates must be considered when relating these findings to possible human exposures, as must the high dose levels used in these studies.
There is little data available on the reproductive and developmental effects of exposure to methanol in non-human primates. In the only reported study macaque monkeys were exposed to 0, 260, 780 or 2,340mg/m3 (0, 200, 600 or 1,800ppm) methanol vapour for 2.5 hours a day for 7 days a week prior to breeding and throughout pregnancy (approximately 120 days). The mothers remained healthy throughout the study and the methanol exposure did not affect menstrual cycles, number of matings to conception or conception rate. There was a significant reduction in the length of pregnancy in treated animals (6 to 8 days less than in the controls); however, the reduction was not dose related. The maternal methanol exposure did not have an apparent effect on the birthweight or health of the offspring (1, 21).
References
Committee on Toxicity of Chemicals in Food Consumer Products and the Environment (COT). COT Statement on the Effects of Chronic Dietary Exposure to Methanol, 2011.
International Programme on Chemical Safety (IPCS). Methanol. Environmental Health Criteria 196, 1997. World Health Organization: Geneva.
International Programme on Chemical Safety (IPCS). Methanol. Poisons Information Monograph. PIM 335, 2001. World Health Organization: Geneva.
Fein DM and Y-J Sue. Methanol ingestion. Pediatr Rev, 2011; 32(12): 549–50.
National Poisons Information Service (NPIS). Methanol. TOXBASE® , 2014 (accessed 11/2014).
National Poisons Information Service (NPIS). Antifreeze and screenwash products. (accessed 11/2014).
Givens M, K Kalbfleisch and S Bryson. Comparison of methanol exposure routes reported to Texas poison control centers. West J Emerg Med, 2008; 9(3): 150–53.
Kruse JA. Methanol and ethylene glycol intoxication. Crit Care Clin, 2012; 28(4): 661–711.
Health and Safety Executive (HSE). Workplace exposure limits (second edition). EH40/2005, 2011.
Tephly TR. The toxicity of methanol. Life Sci, 1991; 48(11): 1031–41.
Frenia ML and JL Schauben. Methanol inhalation toxicity. Ann Emerg Med, 1993; 22(12): 1919–23.
Environmental Protection Agency (EPA). Toxicological review of methanol (noncancer). (CAS No. 67-56-1). In support of summary information on the Integrated Risk Information System (IERFS). Available at 2013.
Hantson P, JY Lambermont and P Mahieu. Methanol poisoning during late pregnancy. J Toxicol Clin Toxicol, 1997; 35(2): 187–91.
International Programme on Chemical Safety (IPCS). Methanol. International Chemical Safety Card (ICSC): 0057, 2004. World Health Organization: Geneva.
Darwish A, et al. Investigation into a cluster of infant deaths following immunization: evidence for methanol intoxication. Vaccine, 2002; 20(29–30): 3585–9.
Fujita M, et al. Methanol intoxication: differential diagnosis from anion gap-increased acidosis. Intern Med, 2004; 43(8): p. 750–54.
Fontenot AP and VS Pelak. Development of neurologic symptoms in a 26-year-old woman following recovery from methanol intoxication. Chest, 2002; 122(4): 1436–9.
Suit PF and ML Estes. Methanol intoxication: clinical features and differential diagnosis. Cleve Clin J Med, 1990; 57(5): 464–71.
McCord CP. Toxicity of methyl alcohol (methanol) following skin absorption and inhalation. Ind Eng Chem, 1931; 23(8): 931–6.
Gilger AP and AM Potts. Studies on the visual toxicity of methanol. V. The role of acidosis in experimental methanol poisoning. Am J Ophthalmol, 1955, 39(2, Part 2): 63–86.
European Food Safety Authority (EFSA) Panel on Food Additives and Nutrient Sources added to Food. Scientific opinion on the re-evaluation of aspartame (E 951) as a food additive. 2013. EFSA J, 2013; 11(12): 3496 (263 pp). doi:10.2903/j.efsa.2013.3496.
Shelby M, et al. NTP-CERHR expert panel report on the reproductive and developmental toxicity of methanol. Reprod Toxicol, 2004; 18(3): 303–90.
Shimizu H, et al. The results of microbial mutation test for forty-three industrial chemicals. Sangyo Igaku, 1985; 27(6): 400–419.
Nelson BK, et al. Teratological assessment of methanol and ethanol at high inhalation levels in rats. Fundam Appl Toxicol, 1985; 5(4): 727–36.
Email chemcompendium@ukhsa.gov.uk if you have any questions about this guidance or enquiries@ukhsa.gov.uk if you have any other questions.
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Interpreting a limit as a derivative
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Asked 12 years, 9 months ago
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I am being stumped by the following question:
Evaluate the limit by interpreting each as a derivative:
lim x→π 6 cos(2 x)−1 2 x−π 6 lim x→π 6 cos(2 x)−1 2 x−π 6
The only way I can think to solve this is using L'Hopital's rule. I have done that and got the correct answer −3–√−3. But, I can not figure out how to do it the was it is described.
Any help would be greatly appreciated.
calculus
limits
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edited Dec 24, 2012 at 18:29
Zev Chonoles
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Hint: Write down the definition of the derivative of f(x)=cos 2 x f(x)=cos2 x at π 6 π 6
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answered Dec 24, 2012 at 18:24
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I sure did not realize how simple the question actually was. Thank you Richard –Richard 2012-12-24 18:29:26 +00:00 Commented Dec 24, 2012 at 18:29
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ResourcesAops Wiki 2025 AIME I Problems/Problem 6
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2025 AIME I Problems/Problem 6
Contents
1 Problem
2 Diagram
3 Video solution by grogg007
4 Solution 1
5 Solution 2 (Trigonometry)
6 Solution 3 (Fastest formula)
7 Solution 4(Double-angle Formula)
8 Solution 5
9 Solution 6 (Brahmagupta's)
10 Video Solution(Fast and Easy)
11 See also
Problem
An isosceles trapezoid has an inscribed circle tangent to each of its four sides. The radius of the circle is , and the area of the trapezoid is . Let the parallel sides of the trapezoid have lengths and , with . Find
Diagram
Video solution by grogg007
Solution 1
To begin with, because of tangents from the circle to the bases, the height is The formula for the area of a trapezoid is Plugging in our known values we have Next, we use Pitot's Theorem which states for tangential quadrilaterals Since we are given is an isosceles trapezoid we have Using Pitot's we find, Finally we can use the Pythagorean Theorem by dropping an altitude from D, Noting that we find,
~mathkiddus
Solution 2 (Trigonometry)
Draw angle bisectors from the bottom left vertex to the center of the circle. Call the angle formed . Drawing a line from the center of the circle to the midway point of the bottom base of the trapezoid makes a right angle, and the other angle has to be . Then draw a line segment from the center of the circle to the top left vertex, then you have a right triangle. The smaller angle of this triangle is . This means . This also means . Note that . The area of the trapezoid is . .
-alwaysgonnagiveyouup
Solution 3 (Fastest formula)
Denote the radius of the inscribed circle as , and the parallel sides as and . By formula, we get , where . Also, by formula, , where . Therefore, r 2+s 2=(r+s)2−2 r s=24 2−2⋅36=504
Solution 4(Double-angle Formula)
Let ∠O A B=α, tan(α)=6 s. By the double - angle formula for tangent tan(2 α)=2 tanα 1−tan 2α=2×6 s 1−(6 s)2=12 s s 2−36 s 2=12 s s 2−36.
Since ∠D A B=2 α, tan(2 α)=12 s−r=12 s s(s−r)=12 s s 2−s r.
Set 12 s s 2−s r=12 s s 2−36. Since s≠0, we can cancel out 12 s from both sides of the equation, getting s 2−s r=s 2−36. Subtracting s 2 from both sides, we have −s r=−36, so s r=36.
Assume (r+s)2=576. Using the formula (r+s)2=r 2+2 r s+s 2, then r 2+s 2=(r+s)2−2 r s.
Substitute r s=36 and (r+s)2=576 into the formula: r 2+s 2=576−2×36=576−72=504.
So the final answer is 504. By Airbus 320-214.
Formula reference to here:
~Mitsuihisashi14
~ LaTeX by alwaysgonnagiveyouup
Solution 5
The height of the trapezoid is clearly the diameter of the circle or . We let the larger base be , and the smaller one be . We have by the area of a trapezoid: . Now, by Pitot's theorem, and letting the legs of the trapezoid be , we have that . Now, by pythag we have that . Now, by systems of equations, we find that , and . Now, we have that
-jb2015007
Solution 6 (Brahmagupta's)
The area of the trapezoid is Note that external tangents of a circle are equal. Let and be the two external tangents making up the bottom base and and be the two external tangents making up the top base. Since the trapezoid is isosceles, So the legs of the trapezoid have length From here, we can apply Brahmagupta's Formula to the trapezoid since isosceles trapezoids are cyclic: Substituting and solving, we get Since we know we can just solve this system to get and The answer is
~grogg007
Video Solution(Fast and Easy)
~MC
See also
2025 AIME I (Problems • Answer Key • Resources)
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HomeLipiduria – with special relevance to Fabry disease
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Lipiduria – with special relevance to Fabry disease
Gavin J. BeckerGavin J. Becker Department of Nephrology and University of Melbourne, Department of Medicine, The Royal Melbourne Hospital, Parkville, Melbourne, Victoria, Australia Email author Search for this author in: De Gruyter De Gruyter Brill| Google Scholar and Kathleen NichollsKathleen Nicholls Department of Nephrology and University of Melbourne, Department of Medicine, The Royal Melbourne Hospital, Parkville, Melbourne, Victoria, Australia Search for this author in: De Gruyter De Gruyter Brill| Google Scholar
Published/Copyright:June 30, 2015
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From the journalClinical Chemistry and Laboratory Medicine (CCLM)Volume 53 Issue s2
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Becker, Gavin J. and Nicholls, Kathleen. "Lipiduria – with special relevance to Fabry disease" Clinical Chemistry and Laboratory Medicine (CCLM), vol. 53, no. s2, 2015, pp. s1465-s1470.
Becker, G. & Nicholls, K. (2015). Lipiduria – with special relevance to Fabry disease. Clinical Chemistry and Laboratory Medicine (CCLM), 53(s2), s1465-s1470.
Becker, G. and Nicholls, K. (2015) Lipiduria – with special relevance to Fabry disease. Clinical Chemistry and Laboratory Medicine (CCLM), Vol. 53 (Issue s2), pp. s1465-s1470.
Becker, Gavin J. and Nicholls, Kathleen. "Lipiduria – with special relevance to Fabry disease" Clinical Chemistry and Laboratory Medicine (CCLM) 53, no. s2 (2015): s1465-s1470.
Becker G, Nicholls K. Lipiduria – with special relevance to Fabry disease. Clinical Chemistry and Laboratory Medicine (CCLM). 2015;53(s2): s1465-s1470.
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Abstract
Examination of the urine under the microscope using polarised light is invaluable for detecting and identifying lipid particles. Attention to the shape of these Maltese cross bearing bodies can distinguish conventional fat particles from Fabry bodies with great sensitivity and specificity across a wide phenotypic spectrum. This could be a cheap and rapid tool for screening subjects suspected of having Fabry disease for renal involvement. It remains to be seen whether there is value in integrating polarised light into automated urine microscopy machines, but potentially this could greatly help the pathologist or nephrologist in identifying unusual urinary particles, and broaden the capacity for larger scale screening.
Keywords: albuminuria; Fabry; lipiduria; microscopy; renal
Introduction
Though organic lipids were once defined as “naturally occurring fat-like substances – soluble in organic solvents but not in water” the inclusion of lipids with polar moieties, hence soluble in water, lead to the refined definition “actual or potential derivatives of free fatty acids and their metabolites” . However, it is the lipids that are insoluble in water that form particles in human urine; hence we will confine this discussion to these compounds. Such lipid particles are not normally seen in the urine, and this paper will attempt to provide answers to the following questions regarding lipiduria: Does lipiduria reflect tissue histology or pathology? How is lipiduria detected by urinary microscopy? Can lipiduria cause confusion in urinary microscopy? How is this relevant to Fabry disease?
Lipids in tissue histology
To process tissues for conventional microscopy the fact that the tissue undergoes a series of “insults” is often forgotten. Fixation in either an aqueous or solvent based fixative is usually the first step, to prevent deterioration by autolysis and to increase tissue rigidity for later sectioning. These fixatives alter the natural characteristics of the tissue and may alter the nature of contained lipids and lipoproteins. The main alternative is freezing, but frozen sections ultimately result in quite poor histological definition, with a finite useful lifespan for further study. Microwave fixation, as often used in immunohistochemistry, utilises heat, which is damaging to lipids. The chemically fixed tissue is then imbedded by immersion in a fluid which sets – usually paraffin. This involves heating the tissue and the use of agents, such as graded alcohols, xylene or similar lipid solvents, to allow penetration of the paraffin wax. The section is then cut, and the paraffin removed by de-waxing with lipid solvents, a process which once again removes lipids but is helpful to allow later aqueous stains to penetrate. Finally aqueous stains, such as haematoxylin and eosin, are used to demonstrate the histology of the dewaxed, lipid removed, sections . It should be no surprise that lipids in tissues are then represented by “vacuoles” or “cholesterol clefts”. The lipids are not seen – only the spaces they once occupied.
If lipids are to be sought histologically, e.g. to be sure the vacuoles seen in fatty liver cells actually contain fat, fresh frozen tissue is imbedded in an agent that requires little heat and is water soluble (such as Carbowax™, Dow Chemicals or Tissue Tek™, Fisher Scientific), and then stained with such agents as Sudan or Oil red which preferentially bind to lipids and fats . The resulting tissue definition is poor.
One of the odd characteristics of cholesterol esters in fresh frozen tissue (or fluids) is that when observed under polarised light, they display distinctive double refraction resulting in the appearance of a “Maltese” or “Amalfi” cross, so named because of the resemblance to the emblem of the mediaeval Knights Hospitaller of St John. This birefringence, scientifically termed “conic focal anisotropism” is of great importance to detection of lipids in urine .
There are renal diseases where lipid “seen” in tissue sections (or actually not seen in conventionally processed tissues – though the vacuoles or clefts remain) has been of significant diagnostic value. In Minimal Change disease where lipid vacuoles are present in tubular cells, associated with heavy proteinuria but minimal glomerular abnormality, this appearance once lead to the diagnostic term “Lipoid nephrosis”. With cholesterol embolic disease of the kidney, diagnostic clefts are left in vessels where the cholesterol was once lodged. Fabry disease is a genetically acquired condition in which the cytoplasm of glomerular podocytes and other cells is packed with glycosphingolipid particles, leading to a foamy appearance with conventional microscopy . First described simultaneously and independently by Johannes Fabry (1860–1930) and William Anderson (1842–1900) in 1898 – depending on where you are in the world it bears either Fabry alone or both as eponymous names, or the term “angiokeratosis corporis diffusum” because of the characteristic angiokeratotic skin “rash” seen in the peri-umbilical and lower torso region in many cases. It is due to an X-linked deficiency of the action of the enzyme α-galactose A, resulting in lysosomal accumulation of the glycosphingolipid ceramide trihexoside, now called globotriaosylceramide (GL3). The clinical picture varies widely with GL3 deposits particularly troublesome in skin (angiokeratosis), kidney (albuminuria, haematuria, and renal failure), nerve vessels (neuropathy, both peripheral and autonomic), heart (arrhythmia, ischaemia and cardiomyopathy) and cerebral vessels (transient ischaemic attacks, strokes and micro-infarcts). This variability in phenotype results in late diagnosis in many instances, which is particularly unfortunate now that enzyme replacement therapy, effective in delaying or preventing these complications, is now available [6, 7].
Renal biopsy is one of the methods by which the condition has been diagnosed, with lipid vacuoles in fixed section cells (foam cells), Maltese cross appearance under polarised light in frozen sections and electron microscopy demonstrating very characteristic whorled myelin bodies in the cytoplasm of cells especially glomerular podocytes, but renal biopsy is invasive, expensive and has risks for the patient.
Detection of lipid particles in urine
As no fixation/imbedding/sectioning/dewaxing is required, fresh urine specimens still retain any lipid particles, and these can easily be seen exhibiting the characteristic Maltese cross appearance in polarised light [4, 8]. Their presence has been well described in association with heavy proteinuria including Minimal Change disease but also other albuminuric conditions. Lipiduria may also be occasionally seen in polycystic kidney disease and can be seen associated with haematuria, where it has been suggested that the lipids arise from degraded red cell membranes . Lipiduria can also occur in disorders, such as chyluria [10, 11], with lipid emulsion therapy associated with intra-abdominal trauma as well as with prostatitis and oil contamination of urine specimens .
In nephrotic syndrome lipids in urine may be seen as free floating fat droplets (Figure 1), packed into renal tubular epithelial cells or macrophages (oval fat bodies) (Figure 2) and within tubular casts (fatty casts) (Figure 3). The uncommonly seen cholesterol crystals, flat plates (Figure 4), interestingly do not demonstrate birefringence hence no Maltese cross appearance. In Fabry disease, cells packed with GL3 vacuoles, “Mulberry cells”, have long been known to be characteristic [14–16], however, they are usually considered insufficiently common to use as a sensitive diagnostic tool.
Figure 1:
Fatty droplets seen in phase contrast (A) and with polarised light (B), hence Maltese cross appearances.
(From reference , reproduced with permission).
Figure 2:
Oval fat bodies in phase contrast (A) and polarised light (B).
(From reference , reproduced with permission).
Figure 3:
Fatty cast in phase contrast (A) and polarised light (B).
(From reference , reproduced with permission).
Figure 4:
Cholesterol crystal – phase contrast.
(From reference , reproduced with permission).
Confusion with lipid particles in urine
With conventional microscopy there are a variety of bodies which may be confused with lipid particles . Haemoglobin – free red cells (ghosts) may appear similar to large fat droplets but they do not exhibit birefringence with polarised light. Candida, free nuclei and other cell debris may look like lipid droplets but again do not form Maltese crosses under polarised light. However, starch particles as may fall from latex gloves do exhibit birefringence and a Maltese cross appearance, but their conventional light microscopy appearance is quite different as they are irregular in size and shape and often have a central dimple.
Urine microscopy in Fabry disease
As mentioned, Fabry disease varies tremendously, and as has been recently realised, can also cause disease in the heterozygous females who have the abnormal gene on one of their X chromosomes. It can thus be difficult to detect or diagnose. Enzyme and genetic testing is expensive, slow and not available in some circumstances. Mulberry cells in the urine are helpful but uncommon. We therefore asked the question “can urine microscopy help, across a wide phenotypic spectrum?” .
Over decades, a large cohort of Fabry patients had been collected, initially to document the manifestations of the disease in our environment [18, 19], and also in anticipation of the development of enzyme replacement therapy, which fortunately has eventuated . This gave us an opportunity to examine the urine of 29 patients, deliberately selected across a wide phenotype, all enzyme proven, most genotyped. We compared examination of their urine with 21 control subjects: 20 with other renal disease, one normal. Of the Fabry patients, nearly half (46%) were male, ages ranged from 18 to 67 years and 51% were already on enzyme replacement therapy. From a renal point of view they also varied widely. In total four females and two males had normal GFR and albumin excretion, 15 had a urinary albumin/creatinine ratio (A/Cr) >3–30 mg/mL, albuminuria A/Cr >30–100 mg/mmol and >100 mg/mmol was present in nine and five, respectively. Four had GFR <60 mL/min/1.73 m² and one was on maintenance dialysis.
The “control” group had a wide range of renal disease, mainly glomerular (lupus in 5, 2 each with diabetes, focal glomerulosclerosis, crescentic glomerulonephritis and vasculitis), and a variety of other diagnoses. As well there were three with acute tubular necrosis (ATN) and one normal individual.
Fresh midstream urine specimens from both groups were centrifuged and resuspended as 25 μL of the thus ×20 concentrated specimen. This 25 μL specimen was examined under polarised light in a counting chamber. The number of Maltese cross particles/μL was graded: None = 0, <100=1, >100 with no clumps =2 and >100 with clumps =3.
Three types of Maltese cross particles were found. Unfortunately, but perhaps educationally, subjects had been issued with starch-covered latex gloves to collect their urine specimens. Maltese cross particles easily identifiable as starch granules were seen in many specimens. Rounded, non-lamellated particles containing symmetrical Maltese crosses were seen in all but one albuminuric individual, both in Fabry and other renal diseases, but not in one patient with membranous nephropathy in remission (A/Cr 26 mg/mmol). They were not seen in the ATN patients or in the normal individual. These were thus typical urinary lipid particles (Figures 1–3).
Importantly a different type of Maltese cross particle, mostly spherical or non-spherical with protrusions of different size and shape, less frequently a structure with an internal spiral pattern always with an asymmetrical, truncated or atypical Maltese cross, was seen in 28/29 Fabry patients (Figure 5). No particles with this appearance were seen in the 21 “controls”. In the Fabry patients there was a direct relationship between the severity of albuminuria and the graded number of particles seen.
Figure 5:
Fabry disease: a lipid containing particle with protrusions (A) and a round particle with internal spiral pattern (C), both particles with asymmetrical Maltese crosses under polarised light (B, D).
(Courtesy of G.B. Fogazzi and G. Garigali).
To further explore the identity of the particles in the Fabry urine we used immuno- histochemistry to identify cells as podocytes using anti-human podocalyxin antibody (PHM5, Clone 18.29; Chemcon catalogue, Millipore, Billerica, MA, USA) and GL3 by an anti CD 77 antibody (Abcam, Cambridge, UK). By this technique we were able to identify GL3 in renal podocytes in patients Fabry urine . We suggest that using the examination of urinary sediment with polarised light microscopy may be a useful way to detect Fabry disease particularly in situations where enzyme or genetic analysis is unavailable.
Corresponding author: Gavin J. Becker, Department of Nephrology and University of Melbourne, Department of Medicine, The Royal Melbourne Hospital, Grattan Street, Parkville, Melbourne, Victoria, Australia, E-mail: Gavin.Becker@mh.org.au
Acknowledgments
The support of A. Menarini Diagnostics for G.B. to attend a meeting in Rome to discuss this work is gratefully acknowledged. Dr. G.B. Fogazzi provided the excellent photomicrographs.
Author contributions: All the authors have accepted responsibility for the entire content of this submitted manuscript and approved submission.
Financial support: None declared.
Employment or leadership: None declared.
Honorarium: None declared.
Competing interests: The funding organisation(s) played no role in the study design; in the collection, analysis, and interpretation of data; in the writing of the report; or in the decision to submit the report for publication.
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Received:2015-5-28
Published Online:2015-6-30
Published in Print:2015-11-1
©2015 by De Gruyter
Articles in the same Issue
Articles in the same Issue
Frontmatter
Editorial
Urinary sediment: still an important diagnostic tool
Introduction
A history of urine microscopy
The modern view on some urinary sediment particles
Lipiduria – with special relevance to Fabry disease
Cylindruria
Crystalluria
External quality control programs on urinary sediment
External quality assessment of urine particle identification: a Northern European experience
The Italian External Quality Assessment (EQA) program on urinary sediment: results of the period 2012–2015
Recent advances in the automation of urinary sediment analysis
Screening of presumptive urinary tract infections by the automated urine sediment analyser sediMAX
Automated urine screening devices make urine sediment microscopy in diagnostic laboratories economically viable
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Keywords for this article
albuminuria; Fabry; lipiduria; microscopy; renal
Articles in the same Issue
Frontmatter
Editorial
Urinary sediment: still an important diagnostic tool
Introduction
A history of urine microscopy
The modern view on some urinary sediment particles
Lipiduria – with special relevance to Fabry disease
Cylindruria
Crystalluria
External quality control programs on urinary sediment
External quality assessment of urine particle identification: a Northern European experience
The Italian External Quality Assessment (EQA) program on urinary sediment: results of the period 2012–2015
Recent advances in the automation of urinary sediment analysis
Screening of presumptive urinary tract infections by the automated urine sediment analyser sediMAX
Automated urine screening devices make urine sediment microscopy in diagnostic laboratories economically viable
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Given the parametric equations x=3cosθy=2sinθ 1. Sketch the curve indicated by the parametric equation. Make sure to include the direction. (2 pts) 2. Eliminate the parameter to obtain the standard form of the rectangular equation. ( 2 pts)
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14490 | https://www.effortlessmath.com/math-topics/radical-inequalities/?srsltid=AfmBOoo4bhKWmaWnwW5LMWdeRUABdrlszSLnWSZCHwJoQwR0DGIJpmXx | How to Solve Radical Inequalities? - Effortless Math: We Help Students Learn to LOVE Mathematics
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How to Solve Radical Inequalities?
An inequality that contains a variable within a radical is called a "radical inequality". To solve a radical inequality, we can use a graph or algebra.
A step-by-step guide tosolving radical inequalities
In mathematics, a radical is the opposite of an exponent, denoted by the symbol √√, also known as the root.
An inequality that has variables inside the radicand is called a radical inequality. In other words, a radical inequality is an inequality that has a variable or variables inside the radical symbol.
We can solve radical inequalities using algebra. Some radical inequalities also have variables outside the radical, and we can use algebra to calculate them as well. The following steps can be used to solve radical inequalities:
Step 1:Check the index of the radical.
If the index is even, the final calculated value of the radicand cannot be negative and must be positive. This is called domain restriction.
Step 2: If the index is even, consider the value of the radicand as positive. Solve for the variable x in radicands.
Therefore, we solve for the variable x 𝑥 for this radicand when it is greater than or equal to zero. That is, we consider the radicand as x≥0 𝑥≥0 from the radical inequality n√x<d 𝑥⎯⎯√𝑛<𝑑 and calculate the variable x 𝑥. If the index is odd, however, then consider the radicand as x<d 𝑥<𝑑.
Step 3: Solve the original inequality expression algebraically and also remove the radical symbol from the expression.
We eliminate the radical by taking the index and using it as the exponent in terms of both sides of the inequality. (i.e., n√x<d→(n√x)n<d n). Note here that when using the index as an exponent on the radical expression, it nullifies the radical symbol, thus removing it.
Step 4: Test the values to check the solution.
To test the values of x, we consider a random value that satisfies the inequality. And we will also consider values outside the equality so that we can confirm the correctness of our solution.
Solving Radical Inequalities– Example 1:
solve 3+√4 x−4≤7.
Solution:
To solve this radical inequality, first, we check the index of the given radical inequality. Since the index value is not given, the index value is 2. Since the index is even, the radicand of the square root will be greater than or equal to zero.
4 x−4≥0
4 x≥4
x≥1………….. (1)
We now solve the radical inequality algebraically and also remove the radical symbol to simplify it. First, we isolate the radical.
3+√4 x−4≤7→√4 x−4≤4
Now, we remove the radical symbol by taking the index as an exponent on both sides of the inequality.
(√4 x−4)2≤4 2
4 x−4≤16
4 x≤20
x≤5………….. (2)
Here, we got two inequalities for the value of x from equations 1 and 2. So we combine them both and write it as a compound inequality. Then our final answer is:
1≤x≤5
Exercises forSolving Radical Inequalities
Solve.
3√x+3≥2
−2√x+1≤−6
4 3√x+1≥12
x≥5
x≥8
x≥26
by: Effortless Math Team about 3 years ago (category: Articles)
Effortless Math Team
3 weeks ago
Effortless Math Team
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14491 | https://etda.libraries.psu.edu/files/final_submissions/21215 | The Pennsylvania State University The Graduate School
ON THE SQUAREFREE VALUES OF POLYNOMIALS
A Dissertation in Mathematics by James M. Kowalski
© 2020 James M. Kowalski Submitted in Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy May 2020 The dissertation of James M. Kowalski was reviewed and approved by the following: Robert C. Vaughan Professor of Mathematics Dissertation Advisor, Chair of Committee Mihran Papikian Associate Professor of Mathematics Yuriy Zarkhin Professor of Mathematics Sean Hallgren Professor of Computer Science and Engineering Alexei Novikov Professor of Mathematics Co-Chair of Graduate Program ii Abstract
When considering the density of squarefree numbers among the values of a polynomial, one would expect a “local-to-global” type result to hold, involving the proportion of values not divisible by p2 for each prime p. Such a result has been established in many cases in the existing literature, and in this dissertation we add several classes of degree three, multivariable polynomials. iii Table of Contents
List of Symbols vi Acknowledgments vii Chapter 1 Introduction 1
1.1 The Expected Result . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Previous Research . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
Chapter 2 Preliminary Results 6
2.1 The Möbius Function . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Hua’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.3 Other Necessary Results . . . . . . . . . . . . . . . . . . . . . . . . . 10
Chapter 3 The General Procedure 11
3.1 The Sum S(1 , D ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 3.2 The Sum S(D, B 3/2) . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
Chapter 4 Sums of Cubic Polynomials 15
4.1 Required Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 4.2 Proof of the Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 4.3 Proofs of the Corollaries . . . . . . . . . . . . . . . . . . . . . . . . . 21
Chapter 5 Sums of Two Cubes 23
5.1 Proof of the Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
Chapter 6 A Special Case for Cubic Forms 26
iv 6.1 Estimating ρ(d2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 6.2 A Bound on the Number of Solutions to F (x) = F (y) . . . . . . . . . 29 6.3 Estimating ∑ R(n)2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 6.4 Proof of the Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
Chapter 7 General Cubic Forms 46
7.1 Estimating ρ(d2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 7.2 Estimating ∑ R(n)2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 7.3 Proof of the Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
Chapter 8 Further Research 52 Bibliography 53
vList of Symbols
a|b a divides b; b = ma for some integer m
x, y x = ( x1, x 2, . . . , x s) and y = ( y1, y 2, . . . , y s)
e(x) e(x) = exp(2 πix ) = ∑∞
n=0 (2 πix )n
n!
ε a small positive real number which may be different each appearance
f (x) = O(g(x)) there are real numbers A > 0 and x0 > 0 where |f (x)| ≤ Ag (x) for all x > x 0
f (x) g(x) equivalent to f (x) = O(g(x)) #A the cardinality of the set A
B the box [a1, b 1] × · · · × [as, b s] in Zs
Vol (B) ∏si=1 (bi − ai)
BB the box [Ba 1, Bb 1] × · · · × [Ba s, Bb s] in Zs, B scaled by BNg(B) the number of squarefree g(x) where x ∈ BB
ρ(d) the number of x ∈ [1 , d ]s with g(x) ≡ 0 (mod d)
R(n) the number of x ∈ BB with g(x) = n
vi Acknowledgments
First, I’d like to thank my advisor, Professor Robert C. Vaughan, for his guidance and support throughout writing this dissertation. This dissertation would not have been possible without his expertise and assistance. I’d like to also thank the members of my committee for their feedback in reviewing this dissertation. I’d like to thank Sid Graham for encouraging me to pursue higher mathematics. I’d like to thank the math faculty at Central Michigan University for preparing me for my studies. I’d like to thank my parents for encouraging me to keep learning. I’d like to thank my wife, Theresa, for tolerating the many years being the spouse of a grad student. Finally, I’d like to thank my friends for providing some much-needed distractions. vii Dedication
To my son, J. viii Chapter 1 | Introduction
In this dissertation, we will explore the proportion of squarefree numbers among the values of certain multivariable polynomials, when the arguments are allowed to run over all integers. A squarefree number is one which has no square divisors (other than 1). For example, 14 = 2 · 7 is squarefree, while 12 = 2 2 · 3 is not. For any polynomial g(x) in s variables, with no square factors or fixed square divisors, it would be reasonable to guess that g(x) should be squarefree infinitely often. Furthermore, one could expect an asymptotic formula for how often g(x) is squarefree.
1.1 The Expected Result
Let ρ(n) be the number of x in (Z/n Z)s where g(x) ≡ 0 ( mod n), and let B be some box [a1, b 1] × · · · × [as, b s] in Zs. Then by BB we mean the box [Ba 1, Bb 1] × · · · ×
[Ba s, Bb s]. We will define
Ng(B) = # {x ∈ BB : g(x) is squarefree }.
For a given prime p, the proportion of x in Zs where p2 divides g(x) is ρ(p2)
p2s
, which follows from the definition of ρ(p2). If we assume that g(x) being divisible by p2 is an independent event for each prime, then the proportion of x in Zs where g(x) is not divisible by any p2 would be ∏
p
(
1 − ρ(p2)
p2s
)
.
1It would follow then that
lim
B→∞
Ng(B)
Vol (B)Bs = ∏
p
(
1 − ρ(p2)
p2s
)
.
While a priori we cannot assume this independence, we would expect to have
Ng(B) = Vol (B)Bs ∏
p
(
1 − ρ(p2)
p2s
)
o(Bs). (1.1)
1.2 Previous Research
Indeed, for many classes of polynomials, results similar to 1.1 have been established in the literature. Early work in this subject pertained to single-variable polynomials. The first such result was that of Gegenbauer in 1885 [13, 34], who gave an asymptotic formula for the number of squarefree integers in the interval [1 , B ]. This could be interpreted as Ng(B) when g(x) = x. The next result was in 1922 by Nagel , who showed that there are infinitely many k-free values when g(x) is a degree k
irreducible polynomial. Additionally, Nagel showed that if g(x) is a linear polynomial, there are infinitely many squarefree values. In 1931, Estermann showed that when
g(x) = x2 + 1 there is a positive proportion of squarefree values. In 1933, Ricci gave an asymptotic formula for the number of k-free values of
g(x) when k is at least the degree of g. Erdős in 1953 showed that when the degree
d of g is at least 3, with no (d − 1) power linear factors, and when d is a power of 2 if
g is not identically divisible by 2d−1, then g(x) is (d − 1) power free infinitely often. Erdős further asked if the expected results still held when the domain of g(x) was restricted to primes. In the same year, Cugilani gave a bound on size of the gaps between integers q1 < q 2 < . . . < g n < · · · where g(qi) is k-free. Hooley, gave an asymptotic formula similar to 1.1 for the number of (d − 1) -free values of an irreducible polynomial of degree d. In 1977 [24, 25], he made progress on Erdős’ conjecture, giving an asymptotic formula for the amount of (d − 1) -free values of degree d irreducible polynomials for prime inputs, when d ≥ 55 , and for smaller values of d with some additional conditions. Nair gave estimates on the k-free values of g(x) on intervals (x, x + h] in 1976 and 1979 . He and Huxley improved on these in 1980, and established the same type of results from prime inputs. Filaseta [9 –11] proceeded to improve these 2results between 1988 and 1993. In 1998, Granville gave a formula for the proportion of squarefree values for
g(x) having any degree, provided there are no repeated roots or fixed square divisors, assuming the abc conjecture. In 2007, Lee and Murty extend this result to k-free values, and giving an asymptotic formula with an error term. For k = 2 , they required an “abscissa conjecture.” In 2014, Murty and Pasten removed this dependence. In 2004, Helfgott gave an unconditional asymptotic formula for squarefree values of irreducible polynomials of degree 3, and in 2007 gave the probability that g(p) is (d − 1) -free when g has degree d. In 2014 , he gave an asymptotic formula for the number of primes where g(p) is squarefree for any cubic polynomial without repeated roots. Heath-Brown established the formula for k-free values of polynomials of degree
d ≥ 3 given that k ≥ (3 d + 2) /4. In 2011, Browning improved this restriction to k ≥ (3 d + 1) /4, and gave the corresponding formula for prime inputs. In 2013, Heath-Brown established the formula for g(x) = xd + c when g(x) is irreducible and k ≥ (5 d + 3) /9. Finally, in 2018 Reuss gave an asymptotic formula for the number of primes p where g(p) is (d − 1) -free if g has degree d and has no fixed divisor
pd−1.Recently, progress has been made toward establishing this asymptotic formula for multi-variable polynomials. The first such result was by Greaves in 1992, where he established the formula for binary forms with nonzero discriminants, nonzero diagonal terms, and irreducible factors of bounded degree. In 1994, Filaseta showed that if g(x, y ) is a binary form with degree d and no fixed k-th power factors, then the asymptotic formula holds for k-free values when k ≥ (2 √2 − 1) d/ 4.Granville, in 1998 gave the proportion of squarefree values for a binary form of any degree with no repeated linear factors and no fixed square divisor, assuming the
abc conjecture. In 2003 Poonen showed that if B = {x ∈ Zs : 0 ≤ xi ≤ Bi} then
lim
B1→∞
· · · lim
Bs→∞
Ng
Vol B = ∏
p
(
1 − ρ(p2)
p2s
)
,
a slightly different formula than 1.1, so long as g has no square factor in Q[x]
assuming the abc conjecture. He also notes that the abc conjecture is unnecessary if each irreducible factor of g has degree at most 3. In 2004, Helfgott established the asymptotic formula unconditionally for squarefree values of binary forms with degree 3at most 6. Later, in 2009, Hooley [26, 27] gave the asymptotic formula for the k-free values of a polynomial in two variables (not necessarily a form) provided that g(x, y ) is irreducible of degree d, not a polynomial in some linear combination of x and y, has no fixed k-power divisors, and is a product of linear factors in some extension field of
Q, and k ≥ d/ 2 − 1.Browning, in his 2011 paper , shows that for g(x, y ) there is a positive proportion of (x, y ) where g(x, y ) is k-free when
k >
7d/ 16 if g is homogeneous
39 d/ 64 otherwise. This is an improvement over previous results when d ≥ 18 in the homogeneous case and when d = 5 , 11 or d ≥ 13 in the non-homogeneous case. Recently Xiao has obtained several results of this type. In 2017 , he showed that if a form of degree d in s variables factors into a product of linear forms, and has nonzero discriminant, then the asymptotic formula holds for k-free values when
k ≥ (d − 2) /s . In the same year he established the asymptotic formula for k-free values of binary forms of degree at least 2 with nonzero discriminant, no fixed k-th power divisor, d the largest degree among the irreducible factors, and
k > min
{
7d
18 ,
⌈
d
2
⌉
− 2
}
.
He and Stewart gave a formula for the number of k-free h ≤ Z that were represented by a degree D form with largest irreducible factor degree d, under the same restrictions as his previous result. Note that this formula counts elements of the range as opposed to the domain, and so differs from the usual result by the number of representations of h. Lastly, he and Lapkova gave a general version of the formula for k-free values as long as g has degree d, no fixed square factors, and k ≥ (3 d + 1) /4.Other results have recently been established for specific classes of multi-variable polynomials. In 2014, Bhargava gave a formula for squarefree values of “invariant polynomials” arising from certain algebraic invariants. In 2016, Bhargava, Shankar, and Wang gave a formula for the squarefree values of discriminants of monic integer 4polynomials. Ordering by height, they obtained a slightly different density
∏
p
(
1 − 1
p + (p − 1) 2
p2(p + 1)
)
.
For “incomplete norm forms,” a type of multivariable polynomial determined by irreducible one variable polynomials, Maynard gave a formula for the number of
primes represented.
1.3 Main Results
In this text, we establish formulas in the form of 1.1 for various classes of polynomials. In Chapter 4, we will consider sums of single-variable polynomials of degree 3, that is
g(x) = f1(x1) + · · · + fs(xs)
where each fi is a cubic polynomial in xi. We are able to obtain the desired asymptotic formula when s is at least 3. In Chapter 5, we examine the specific case when s = 2
and f1(x) = x3, f2(y) = y3.In Chapters 6 and 7, we establish the formula for cubic forms. In Chapter 6, we consider the special case where one variable appears only as a cube, that is
g(x) = x31 + F (x2, . . . , x s)
where F is a cubic form in s − 1 variables. For this class of cubic forms, we are able to obtain the desired formula when s is at least three (although note that s = 2 is essentially equivalent to the case discussed in Chapter 5). In Chapter 7, we consider more general cubic forms, but still need to insist on some conditions, including nonsingularity, for our method to yield results. 5Chapter 2 | Preliminary Results
Throughout this dissertation, we will make use of several common results from analysis and number theory. We will list the most common ones here.
2.1 The Möbius Function
When studying squarefree numbers, the Möbius function, given by
μ(n) =
(−1) k n = p1p2 · · · pk (all distinct)
0 otherwise is quite useful. We can immediately see that |μ(n)| = 1 if and only if n is squarefree. Another identity that μ(n) satisfies is
∑
d|n
μ(d) =
1 n = 1 0 n > 1.
We give the proof for this identity from . When n = 1 , the identity is obviously true. If n > 1, then n can be written as a product of primes n = pa1
1
· · · pak
k
. In the sum ∑
d|n
μ(d), the only nonzero terms will come from d = 1 and when d is a product of distinct primes. Therefore,
∑
d|n
μ(d) = μ(1) + μ(p1) + · · · + μ(pk) + μ(p1p2) + · · · + μ(pk−1pk)+ · · · + μ(p1p2 · · · pk)
6= 1 +
(
k
1
)
(−1) +
(
k
2
)
(−1) 2 + · · · +
(
kk
)
(−1) k = (1 − 1) k = 0
From this identity, we obtain ∑
d2|n
μ(d) = |μ(n)|
which we make use of in our method. If n < 0, this identity also holds, as μ(−n) = μ(n).
2.2 Hua’s Lemma
There are two versions of Hua’s Lemma that we make use of in this text. The first version deals with sums of kth powers.
Lemma 2.1. Let f (α) = ∑Nm=1 e(αm k). Then for 1 ≤ j ≤ k
∫10
|f (α)|2j
dα N 2j −j+ε.
The proof is from Chapter 2 of and makes use of Weyl’s Lemma.
Lemma 2.2 (Weyl) . Let
T (φ) =
Q∑
x=1
e(φ(x))
where φ is an arbitrary arithmetical function. Then
|T (φ)|2j
≤ (2 Q)2j −j−1 ∑
|h1|<Q
· · · ∑
|hj|<Q
Tj
where
Tj = ∑
x∈Ij
e(∆ j (φ(x); h1, . . . , h j ))
and the intervals Ij = Ik(h1, . . . , h j ) (possibly empty) satisfy
I1(h1) ⊂ [1 , Q ], I j (h1, . . . , h j ) ⊂ Ij−1(h1, . . . , h j−1).
Here ∆j is the jth iterate of the forward difference operator, so that
∆1(φ(α); β) = φ(α + β) − φ(α)
7and
∆j+1 (φ(α); β1, . . . , β j+1 ) = ∆ 1(∆ j (φ(α); β1, . . . , β j ); βj+1 ).
It follows that
∆j (αk; β1, . . . , β j ) = β1 · · · βj pj (α; β1, . . . , β j )
where pj is a polynomial in α with degree k − j which has leading coefficient k!/(k − j)! .First we prove Weyl’s Lemma.
Proof. By induction on j. For brevity, write ∆j (x) for ∆j (φ(x); h1, . . . h j ). Obviously
|T (φ)|2 =
Q
∑
x=1
Q−x
∑
h1=1 −x
e(∆ 1(x)) =
Q−1
∑
h1=1 −Q
∑
x∈I1
e(∆ 1(x))
where I1 = [1 , Q ] ∩ [1 − h1, Q − h1].
Now if the conclusion of the lemma is assumed for a particular value of j, then by Cauchy’s inequality,
|T (φ)|2j+1
≤ (2 Q)2j+1 −2j−2(2 Q)j ∑
h1,...,h j
|Tj |2
and obviously
|Tj |2 = ∑
|h|<Q
∑
x∈Ij+1
e(∆ j (x + h) − ∆j (x))
with Ij+1 = Ij ∩ { x : x + h ∈ Ij }.
Now the proof of Hua’s Lemma follows.
Proof. By induction on j. The case j = 1 is immediate from Parseval’s identity. Now suppose the lemma holds for 1 ≤ j ≤ k−1. By Weyl’s lemma with φ(x) = αx k,
|f (α)|2j
≤ (2 N )2j −j−1 ∑
h1,...,h j
|hi|≤ N
∑
x∈Ij
e(αh 1 · · · hj pj (x; h1, . . . , h j ))
where pj (x; h1, . . . , h j ) is a polynomial in x of degree k − j with integer coefficients. Hence
|f (α)|2j
(2 N )2j −j−1 ∑
h
che(αh ) (2.1) 8where ch is the number of solutions of the equation
h1 · · · hj pj (x; h1, . . . , h j ) = h
with |hi| < N and x ∈ Ij . Clearly
c0 Nj . ch N ε(h 6 = 0) .
By writing
|f (α)|2j
= f (α)2j−1
f (−α)2j−1
one also obtains
|f (α)|2j
= ∑
h
bhe(−αh ) (2.2) where bn is the number of solutions of
xk
1
· · · + xk
2j−1
− yk
1
− · · · − yk
2j−1
= h
with xi, y i ≤ N . Thus ∑
h
bh = f (0) 2j
= N 2j
and, on the inductive hypothesis,
b0 =
∫ 10
|f (α)|2j
dα N 2j −j+ε.
By 2.1, Parseval’s identity, and 2.2,
∫ 10
|f (α)|2j+1
dα (2 N )2j −j−1 ∑
h
chbh.
Moreover,
∑
h
chbh c0b0 + N ε ∑
h6=0
bh
N j N 2j −j+ε + N εN 2j
,
which gives the desired conclusion. The more general version, which applies to sums of polynomials, is Theorem 4 in 9Hua’s book . All of Chapter 3 in that book is dedicated to proving this version.
Lemma 2.3. Let f (x) be a polynomial of degree k with integer coefficients. Set
T (α) =
P∑
x=1
e(f (x)α).
Then for 1 ≤ ν ≤ k,
∫10
|T (α)|2ν
dα P 2ν −ν log( P )A
where A and the implicit constant depend on k, ν, and the coefficients of f (x).
Note that when f (x) = xk, the two versions of the lemma agree.
2.3 Other Necessary Results
In Chapter 4, we need the following estimate for exponential sums, which is Corollary 2.1 in .
Lemma 2.4. Suppose p is a prime. Suppose g(X) = anXn + · · · + a0 is a polynomial with integer coefficients having 0 < n < p and p - an. Then
∣∣∣∣∣∣
p−1∑
x=0
e(g(x)/p )
∣∣∣∣∣∣
≤ (n − 1) p1/2.
10 Chapter 3 | The General Procedure
For each of the primary cases described in this dissertation, our procedure begins in the same manner. Given a polynomial g(x) in s variables and a box B = [ a1, b 1] ×· · · × [as, b s], we want to find an asymptotic formula for
Ng(B) = # {x ∈ BB : g(x) is squarefree }.
Setting R(n) = # {x ∈ BB : g(x) = n}, we can use the previously described properties of the Möbius function to write
Ng(B) = ∑
n B3
μ(n)2R(n) = ∑
n B3
n6=0
∑
d2|n
μ(d)
R(n)
where the sums in n are taken over all possible values of g(x) with x ∈ BB, with
n = 0 excluded as μ(0) = 0 . Note that there is a constant A where |g(x)| ≤ AB 3 for all x ∈ BB. Interchanging the order of summation gives us
Ng(B) = ∑
d B3/2
μ(d) ∑
n B3
d2|nn6=0
R(n).
We then introduce a parameter D to split the domain of d:
Ng(B) = ∑
d≤D
μ(d) ∑
n B3
d2|nn6=0
R(n) + ∑
D<d B3/2
μ(d) ∑
n B3
d2|nn6=0
R(n).
11 We will refer to these sums as
S(1 , D ) = ∑
d≤D
μ(d) ∑
n B3
d2|nn6=0
R(n)
and
S(D, B 3/2) = ∑
D<d B3/2
μ(d) ∑
n B3
d2|nn6=0
R(n)
and will analyze them individually. Note that there is an implicit constant involved in the domain of S(D, B 3/2), but in each application this will end up in the error term, so there is no need to write it explicitly.
3.1 The Sum S(1 , D )
For a given value of d, we can expand the inner sum according to residue classes mod
d2 as
∑
n B3
d2|n
R(n) = ∑
x∈BB
d2|g(x)
1= ∑
1≤y1,y 2,...,y s≤d2
d2|g(y)
∑
x∈BB
xi≡yi(mod d2)
1= ∑
1≤y1,y 2,...,y s≤d2
d2|g(y)
s
∏
i=1
( B(bi − ai)
d2 + O(1)
)
= ρ(d2)
s
∏
i=1
( B(bi − ai)
d2 + O(1)
)
= ρ(d2)BsVol (B)
d2s + O
(
ρ(d2)
( Bs−1
d2s−2 + 1
))
.
It follows, then, that
S(1 , D ) = ∑
d≤D
μ(d) ∑
n B3
d2|n
R(n) − ∑
d≤D
μ(d)R(0)
12 = BsVol (B) ∑
d≤D
μ(d)ρ(d2)
d2s + O
∑
d≤D
μ(d)2ρ(d2)
( Bs−1
d2s−2 + 1
)
DR (0)
= BsVol (B)
∞
∑
d=1
μ(d)ρ(d2)
d2s + O
∑
d≤D
μ(d)2ρ(d2)
( Bs−1
d2s−2 + 1
)
+Bs ∑
d>D
μ(d)2ρ(d2)
d2s + DR (0)
.
In later chapters, we will obtain estimates for ρ(d2) for each case that allow us to express the error term as a power of B. We will also need to justify the convergence of the infinite sum in the last line above. Additionally, we will show that R(0) Bs−1
so that DR (0) is absorbed into the rest of the error term.
3.2 The Sum S(D, B 3/2)
We apply the Cauchy-Schwarz inequality to S(D, B 3/2) to write
S(D, B 3/2) = ∑
D<d B3/2
μ(d) ∑
n B3
d2|nn6=0
R(n)
≤
∑
D<d B3/2
μ(d) ∑
n B3
d2|nn6=0
1
1/2
∑
D<d B3/2
μ(d) ∑
n B3
d2|nn6=0
R(n)2
1/2
.
For the first factor, we have the simple estimate
∑
D<d B3/2
∑
n B3
d2|nn6=0
1 B3
D ,
and for the second factor, we note that
∑
D<d B3/2
∑
n B3
d2|nn6=0
R(n)2 = ∑
n B3
n6=0
R(n)2 ∑
D<d B3/2
d2|n
1 Bε ∑
n B3
R(n)2
13 where the term with n = 0 is included in the final sum. Therefore, we have
S(D, B 3/2) B3/2+ ε
D1/2
∑
n B3
R(n)2
1/2
.
As with S(1 , D ) the sum ∑
n B3
R(n)2 will be approximated separately for each type of polynomial. Then these estimates will be combined and a suitable choice for D will be made. 14 Chapter 4 | Sums of Cubic Polynomials
In this chapter, we will establish the desired asymptotic formula when g(x) is the sum of at least three cubic polynomials.
Theorem 4.1. Suppose s ≥ 3, g(x) = f1(x1) + · · · + fs(xs) with each fi a polynomial of degree 3 with integer coefficients. Then for s ≥ 4 we have
Ng(B) = BsVol (B) ∏
p
(
1 − ρ(p2)
p2s
)
O(Bs− 2s−14s−1 +ε)
and
Ng(B) = B3Vol (B) ∏
p
(
1 − ρ(p2)
p6
)
O(B3− 122 +ε)
when s = 3 .
We also establish the following two corollaries.
Corollary 4.2. If ∏
p
(
1 − ρ(p2)
p2s
)
= 0 ,
it must be the case that there exists a prime p1 where fi(x) ≡ fi(0) ( mod p21) for each
i and for all x, and f1(0) + · · · + fs(0) is divisible by p21.
Corollary 4.3. If fi(xi) = x3
i
for each i, then
∏
p
(
1 − ρ(p2)
p2s
)
0
and so there are infinitely many integer s-tuples x for which g(x) = x31 + · · · + x3
s
is squarefree.
15 4.1 Required Estimates
In Chapter 3, we saw that our method requires estimates for ρ(d2) and ∑
n B3
R(n)2.We will obtain those estimates here.
Lemma 4.4. There is a prime p0 such that when p > p 0 and s ≥ 2, we have
ρ(p2) p2s−2.
Proof. Note that
ρ(p2) = 1
p2
p2
∑
a=1
p2
∑
y1=1
p2
∑
y2=1
· · ·
p2
∑
ys=1
e(( f1(y1) + f2(y2) + · · · + fs(ys)) a/p 2)= 1
p2
p2
∑
a=1
s
∏
i=1
p2
∑
y=1
e(fi(y)a/p 2)
.
Now, letting
Si(q, a ) =
q
∑
y=1
e(fi(y)a/q ),
the above is equivalent to
1
p2
p2
∑
a=1 (a,p )=1
s
∏
i=1
Si(p2, a ) +
p−1
∑
b=1
s
∏
i=1
Si(p2, pb ) +
s
∏
i=1
Si(p2, p 2)
.
Clearly Si(p2, p 2) = p2 and Si(p2, pb ) = pS i(p, b ). It follows from Lemma 2.4 that
Si(p, b ) p1/2. For the remaining term, there must be a prime p0 so that when p > p 0,
p does not divide any of the coefficients of any of the fi. In this case, we have
Si(p2, a ) =
p
∑
u=1
p−1
∑
v=0
e(fi(u + vp )a/p 2)=
p
∑
u=1
p−1
∑
v=0
e(fi(u)a/p 2 + f ′
i
(u)av/p )=
p
∑
u=1
e(fi(u)a/p 2)
p−1
∑
v=0
e(f ′
i
(u)av/p )
16 = p
p
∑
u=1
f′
i(u)≡0 (mod p)
e(fi(u)a/p 2)
and so
|Si(p2, a )| ≤ 2p.
This gives us
ρ(p2) 1
p2
(
p2+ s + p3s/ 2 + p2s)
ps + p3s/ 2−1 + p2s−2.
As long as s ≥ 2, we then have ρ(p2) p2s−2.
Lemma 4.5. The series ∞∑
d=1
μ(d)ρ(d2)
d2s
converges absolutely, since ρ(d2) d2s−2+ ε for squarefree d, and so is equal to the product ∏
p
(
1 − ρ(p2)
p2s
)
.
Proof. Since ρ(d) is a multiplicative function, for squarefree d we have
ρ(d2) = ∏
p|d
ρ(p2).
From the previous lemma, each ρ(p2) p2s−2, and so ρ(d2) d2s−2+ ε since the product of the implicit constants is O(dε). Then, when s ≥ 2,
N
∑
d=1
∣∣∣∣∣
μ(d)ρ(d2)
d2s
∣∣∣∣∣
N
∑
d=1
1
d2+ ε
and so the series converges absolutely. Next we estimate the sum of R(n)2. The result is different when s = 3 than when
s ≥ 4.
Lemma 4.6. For s = 3 , we have
∑
n
R(n)2 B7/2+ ε
17 and for s ≥ 4 ∑
n
R(n)2 B2s−3+ ε.
Proof. Since the sum represents the solutions for g(x) = g(y), that is,
∑
n
R(n)2 = # {x, y ∈ BB : f1(x1) + · · · + fs(xs) = f1(y1) + · · · + fs(ys)}
we can express it as
∫ 10
∑
x,y∈BB
e(α(f1(x1) + · · · + fs(xs) − f1(y1) − · · · − fs(ys))) dα
=
∫ 10
s
∏
i=1
∑
Ba i≤xi≤Bb i
e(αf i(xi))
s
∏
i=1
∑
Ba i≤xi≤Bb i
e(−αf i(yi))
dα
s
∏
i=1
∫ 10
∣∣∣∣∣∣∑
Ba i≤xi≤Bb i
e(αf i(xi))
∣∣∣∣∣∣
2s
1/s
by Hölder’s inequality. We will consider two cases on ai and bi.Case 1: If 0 ≤ ai ≤ bi or ai ≤ bi ≤ 0, then
∣∣∣∣∣∣∑
Ba i≤xi≤Bb i
e(αf i(xi))
∣∣∣∣∣∣
2s
∣∣∣∣∣∣∑
1≤xi≤Bb i
e(αf i(xi))
∣∣∣∣∣∣
2s
+
∣∣∣∣∣∣∑
1≤xi≤Ba i
e(αf i(xi))
∣∣∣∣∣∣
2s
or
∣∣∣∣∣∣∑
1≤xi≤− Bb i
e(αf i(xi))
∣∣∣∣∣∣
2s
+
∣∣∣∣∣∣∑
1≤xi≤− Ba i
e(αf i(xi))
∣∣∣∣∣∣
2s
.
For s = 3 , integrating gives us an expression of the form
∫ 10
∣∣∣∣∣∣∑
1≤xi≤Q
e(αf i(xi))
∣∣∣∣∣∣
6
dα
which is
1
∑
0
∣∣∣∣∣∣∑
1≤xi≤Q
e(αf i(xi))
∣∣∣∣∣∣
4
dα
1/2
1
∑
0
∣∣∣∣∣∣∑
1≤xi≤Q
e(αf i(xi))
∣∣∣∣∣∣
8
dα
1/2
by the Cauchy-Schwarz inequality. Applying Hua’s lemma (Lemma 2.3), this is 18 O(Q7/2+ ε). Therefore, in this case, for s = 3
∫ 10
∣∣∣∣∣∣∑
Ba i≤xi≤Bb i
e(αf i(xi))
∣∣∣∣∣∣
6
B7/2+ ε.
For larger s, we use the trivial estimate
∣∣∣∣∣∣∑
1≤xi≤Q
e(αf i(xi))
∣∣∣∣∣∣ ≤ Q
and write
∫ 10
∣∣∣∣∣∣∑
1≤xi≤Q
e(αf i(xi))
∣∣∣∣∣∣
2s
dα Q2s−8
∫ 10
∣∣∣∣∣∣∑
1≤xi≤Q
e(αf i(xi))
∣∣∣∣∣∣
8
dα Q2s−3+ ε
and so in this case ∫ 10
∣∣∣∣∣∣∑
Ba i≤xi≤Bb i
e(αf i(xi))
∣∣∣∣∣∣
2s
B2s−3+ ε.
Case 2: When ai < 0 < b i, we have
∣∣∣∣∣∣∑
Ba i≤xi≤Bb i
e(αf i(xi))
∣∣∣∣∣∣
2s
∣∣∣∣∣∣∑
1≤xi≤− Ba i
e(αf i(xi))
∣∣∣∣∣∣
2s
1 +
∣∣∣∣∣∣∑
1≤xi≤Bb i
e(αf i(xi))
∣∣∣∣∣∣
2s
and so the argument is the same as in the first case.
4.2 Proof of the Theorem
If x2, . . . , x s are chosen in advance, since f1 is a cubic polynomial, there are at most 3 choices for x1 so that f1(x1) + f2(x2) + · · · + fs(xs) = 0 , and so R(0) Bs−1.
Recall from Chapter 3 that
S(1 , D ) = BsVol (B)
∞
∑
d=1
μ(d)ρ(d2)
d2s
+O
∑
d≤D
μ(d)2ρ(d2)
( Bs−1
d2s−2 + 1
)
Bs ∑
d>D
μ(d)2ρ(d2)
d2s + DR (0)
.
19 From Lemma 4.5 and the beginning of this section, this is
BsVol (B) ∏
p
(
1 − ρ(p2)
p2s
)
+
∑
d≤D
(Bs−1dε + d2s−2+ ε) + Bs ∑
d>D
d−2+ ε + DB s−1
= BsVol (B) ∏
p
(
1 − ρ(p2)
p2s
)
O(Bs−1D1+ ε + D2s−1+ ε + BsD−1+ ε).
We also showed in Chapter 3 that
S(D, B 3/2) B3/2+ ε
D1/2
∑
n B3
R(n)2
1/2
.
When n = 3 , this is
B3+1 /4+ εD−1/2
and for n ≥ 4 this is
Bs+εD−1/2.
Putting both of these estimates together gives
Ng(B) = B3Vol (B) ∏
p
(
1 − ρ(p2)
p6
)
O(B2D1+ ε + D5+ ε + B3D−1+ ε + B3+1 /4+ εD−1/2)
for s = 3 . Setting D = Bγ , the error term becomes
B2+ γ+ε + B5γ+ε + B13 /4−γ/ 2+ ε
which is minimized when γ = 13 /22 resulting in
Ng(B) = B3Vol (B) ∏
p
(
1 − ρ(p2)
p2s
)
O(B3−1/22+ ε)
completing the proof. When s ≥ 4, we have
Ng(B) = BsVol (B) ∏
p
(
1 − ρ(p2)
p2s
)
+O(Bs−1D1+ ε +D2s−1+ ε +BsD−1+ ε +Bs+εD−1/2).
20 Again, setting D = Bγ the error term is
Bs−1+ γ+ε + B2sγ −γ+ε + Bs−γ+ε + Bs−γ/ 2+ ε.
Setting γ = 2s
4s−1
gives
Ng(B) = BsVol (B) ∏
p
(
1 − ρ(p2)
p2s
)
O(Bs− 2s−14s−1 +ε)
which was the desired result.
4.3 Proofs of the Corollaries
To prove the first corollary, we note that the infinite product cannot diverge to zero, as a consequence of Lemma 4.5, and so the only for it to be equal to zero is if one of the factors were zero. In this case, there would be a p1 so that ρ(p21) = p2s
1
. In the proof of the lemma, we saw that
ρ(p2) = 1
p2
p2
∑
a=1
p2
∑
y1=1
· · ·
p2
∑
ys=1
e(( f1(y1) + f2(y2) + · · · + fs(ys)) a/p 2).
To maximize this, we would need to have
p2
∑
a=1
e(( f1(y1) + f2(y2) + · · · + fs(ys)) a/p 2) = p2
for each choice of (y1, . . . , y s). This can only happen if p2 divides f1(y1) + · · · + fs(ys)
for all choices of (y1, . . . , y s).Setting y1 = y2 = · · · = ys = p21 shows that f1(0) + · · · + fs(0) must be divisible by
p21. Then, fixing a j ≤ s and setting xi = p21 for all i 6 = j, we see that we must have
fj (x) ≡ fj (0) (mod p2) for all x.To prove the second corollary, suppose g(x1, . . . , x s) = x31 + · · · + x3
s
. Similarly to the first corollary, since
{(x1, . . . , x s) : 1 ≤ xi ≤ p2} = p2s
so the only way for the infinite product to be zero is for there to be a prime p1 where 21 p21 divides x31 + · · · + x3
s
for all choices of xi. However, if x1 = 1 and x2 = · · · = xs = p21,we see that g(x1, . . . , x s) ≡ 1 ( mod p21). Therefore, ρ(p2) < p 2s and so the infinite product must be positive. 22 Chapter 5 | Sums of Two Cubes
In the previous chapter, we were able to prove our formula for sums of 3 or more cubic polynomials. Unfortunately, our method will not work for a sum of two polynomials, because of a conflict between two of our bounds. For any s, we must have
∑
n
R(n)2 Bs
since the sum represents the number of solutions to g(x) = g(y), and for any choice of x, setting x = y gives a solution. For the case when s = 2 , the best we could hope for, then, is B2. But this gives us the estimate
S(D, B 3/2) B5/2+ εD−1/2
which is only o(B2) if D is larger than B. But in this case the error term for S(1 , D ),which contains the term BD 1+ ε would not be o(B2).This case was already established by Greaves , as it is a binary cubic form, but in this chapter we show how our method can be altered to deal with cases with this kind of obstruction.
Theorem 5.1. If g(x, y ) = x3 + y3, then
Ng(B) = B2Vol (B) ∏
p
(
1 − ρ(p2)
p4
)
O(B2(log B)−1).
23 5.1 Proof of the Theorem
Let P = (log B)/2 and set D to be the product of all primes bounded by P ,
D = ∏
p≤P
p.
Then we can write the number of pairs (x, y ) ∈ BB where x3 + y3 is squarefree as
Ng(B) ≥ N1 − N2
= # {(x, y ) ∈ BB : p2|x3 + y3 =⇒ p > P }− #{(x, y ) ∈ BB : there exists a prime p > P where p2|x3 + y3}.
Note, then, that N1 − Ng(B) ≤ N2 and so Ng(B) = N1 + O(N2). The cardinality of the second set above, N1, can be expressed as
N2 = ∑
P <p B3/2
∑
(x,y )∈BB
p2|x3+y3
1.
Since x3 + y3 factors into (x + y)( x2 − xy + y2) we can consider the following cases where p2|x3 + y3.First, we could have p2|x + y. Such a p must be B1/2 as x and y are bounded. For a given p, there are ( Bp2 + 1 )
B choices for x and y. Then, we must have
N2 ∑
P <p B1/2
(
Bp2 + 1
)
B B2(log B)−1.
Next, we could have p|x + y but p2 - x + y and so p|x2 − xy + y2. These conditions together imply that either p = 3 , or p|x and p|y, and so there are at most ( Bp + 1 )2
choices for x and y. In this case we then have
N2 ∑
P <p B
(
Bp + 1
)2
B2(log B)−1.
Finally, in the case where p - x + y and p2|x2 − xy + y2, given p and x, there are at most two choices for y (mod p2). This is because we must have p - x or else we would have p|y and so p|x + y. Therefore, given p, there are ( Bp2 + 1 )
B choices for x and 24 y. Since p B2, we have
N2 ∑
P <p B2
( Bp2 + 1
)
B B2(log B)−1
and so N2 B2(log B)−1 in each case. For N1, we proceed essentially as in the usual method,
N1 = ∑
(x,y )∈BB
∑
d2|(x3+y3,D 2)
μ(d) = ∑
d|D
μ(d) ∑
u,v ≤d2
d2|u3+v3
∑
x≡u(mod d2)
y≡v(mod d2)
1,
which is equivalent to
∑
d|D
μ(d)ρ(d2)B2Vol (B)
d4 + O(BD 1+ ε + D3+ ε).
We can write the sum as a product, giving us
B2Vol (B) ∏
p≤P
(
1 − ρ(p2)
p4
)
O(BD 1+ ε + D3+ ε).
Using the result from Lemma 4.5, we can bound the product
∏
p>P
(
1 − ρ(p2)
p4
)−1
1 + 1
P log P ,
and so we can extend to an infinite product
N1 = B2Vol (B) ∏
p
(
1 − ρ(p2)
p4
)
O
( B2
P log P + BD 1+ ε + D3+ ε
)
.
Since we set P = (log B)/2 we have D = O(B1/2+ ε) giving the desired estimate
Ng(B) = B2Vol (B) ∏
p
(
1 − ρ(p2)
p4
)
O
( B2
log B
)
.
25 Chapter 6 | A Special Case for Cubic Forms
In this chapter, we will consider the special case where g(x) is a cubic form with an isolated variable. We follow the method described in Chapter 3 to prove the following theorem.
Theorem 6.1. Let g(x) be a cubic form with g(x) = x31 + F (x2, . . . , x s) where F is a cubic form in s − 1 variables, and F is not of the form AL(x)3 where L(x) is a linear form. Then
Ng(B) = BsVol (B) ∏
p
(
1 − ρ(p2)
p2s
)
O(Bs− 18s−2 +ε).
If F is a linear form cubed, one would expect a similar result to follow from the work of Greaves . As we saw before, we require estimates for ρ(d2) and ∑
n
R(n)2, which we will now establish.
6.1 Estimating ρ(d2)
We will use an elementary argument to bound ρ(p2), which will require the following two theorems from Hardy and Wright .
Lemma 6.2 (Theorem 107) . If f (x) is a polynomial of degree n and has more than
n roots (mod p) then f (x) ≡ 0 (mod p).
Here by f (x) ≡ 0 ( mod p) we mean each that each coefficient of f is divisible by
p.26 Lemma 6.3 (Theorem 123) . The number of solutions to f (x) ≡ 0 ( mod pa) corre-sponding to a solution ξ of f (x) ≡ 0 (mod pa−1) is
• zero, if f ′(ξ) ≡ 0 (mod p) and ξ is not a solution to f (x) ≡ 0 (mod pa),
• one, if f ′(ξ) 6 ≡ 0 (mod p),
• p, if f ′(ξ) ≡ 0 (mod p) and ξ is a solution to f (x) ≡ 0 (mod pa).
Lemma 6.3 is commonly referred to as a version of Hensel’s Lemma (see for a general statement). We will assume that p > 3 and is large enough that it does not divide any of the coefficients of F (y). For a fixed choice of (y1, . . . , y s−1), we have the polynomial congruence in xx3 + F (y1, . . . , y s−1) ≡ 0 (mod p2).
First, if we consider the above congruence modulo p, we see that there can be at most
3 solutions for x by Lemma 6.2 since the coefficient of x3, 1, is not divisible by p.If F (y1, . . . , y s−1) 6 ≡ 0 ( mod p) then ξ 6 ≡ 0 ( mod p) for any solution ξ and so
f ′(ξ) = 3 ξ2 6 ≡ 0 ( mod p) provided p > 3. Therefore, in this case each solution modulo
p lifts to at most one solution modulo p2 by Lemma 6.3 and hence ρ(p2) p2s−2 as there are O(( p2)s−1) choices for the yi.If F (y1, . . . , y s−1) ≡ 0 ( mod p), however, then we must have x ≡ 0 ( mod p) and so there are p solutions for x modulo p2. We then want to find a bound for the number of (y1, . . . , y s−1) where this occurs. Assume that y2, . . . , y s−1 have already been chosen, and write F (y1, . . . , y s−1) as a polynomial in y1, so
C0y31 + C1(y2, . . . , y s−1)y21 + C2(y2, . . . , y s−1)y1 + C2(y2, . . . , y s−1) ≡ 0 (mod p)
where C0 is constant, C1 is a linear form, C2 is a quadratic form, and C3 is a cubic form. Again appealing to Lemma 6.2, this congruence either has at most 3 solutions for y1 or each coefficient is divisible by p. In the first case, the solutions will lift to at most 3p choices for y1 modulo p2 and so ρ(p2) p · p · (p2)s−2 = p2s−2 where each factor represents the number of choices for x, y1, and (y2, . . . , y s−1) respectively. In the alternative case, we must have
C0 ≡ C1(y2, . . . , y s−1) ≡ C2(y2, . . . , y s−1) ≡ C3(y2, . . . , y s−1) ≡ 0 (mod p).
27 Since C0 is constant and p was chosen not to divide any coefficient of F , we must have
C0 = 0 . Write C1(y2, . . . , y s−1) = d2y2 + · · · + ds−1ys−1. If dj 6 = 0 for some j then we have the congruence
dj yj ≡ − ∑
i6=j
dj yi (mod p).
If (y2, . . . , y j−1, y j+1 , . . . , y s−1) are chosen in advance, there is only one yj that will satisfy this congruence, which corresponds to p choices for yj modulo p2. In this case,
ρ(p2) p · p2 · p · (p2)s−3 = p2s−2, where each factor represents the number of choices for x, y1, yj , and (y2, . . . , y j−1, y j+1 , . . . , y s−1) respectively. If each coefficient of C1 is zero, we must turn to C2. Note that C0, C1, and C2
cannot all be identically zero, as we insisted that y1 appear explicitly in F . We write
2C2(y2, . . . , y s−1) = ∑
i,j
dij yiyj
where dij = dji . There must be some indices m and n, not necessarily distinct, where
dmn 6 = 0 . Suppose all of the variables except x, y1, and ym are chosen in advance, and write
dmm y2
m
2 ym
∑
j6=m
dmj yj + ∑
i,j 6=m
dij yiyj ≡ 0 (mod p).
Once again, using Lemma 6.2, this congruence either has at most two solutions in ym
modulo p, which would lead to at most 2p solutions modulo p2, or we must have
dmm ≡ 0 (mod p)
and ∑
j6=m
fmj yj ≡ 0 (mod p).
In the first case, we see that ρ(p2) p2s−2, counting as in the prior cases. In the second case, it is necessary that m 6 = n, since dmm is the coefficient of the y1y2
m
term of F , and so could not be both nonzero and divisible by p. Then, we write the second congruence as
dmn yn ≡ − ∑
j6=mj6=n
dmj yj (mod p).
For a given choice of yj for each j 6 = n, there is only one solution to this congruence in yn module p, which lifts to p solutions modulo p2. In this case, y1 and ym can be 28 any value, and again we have ρ(p2) p2s−2. Since we’ve accounted for every possible case, we have ρ(p2) p2s−2, as desired As we showed in Chapter 4, if d is squarefree, it follows that ρ(d2) d2s−2+ ε.
6.2 A Bound on the Number of Solutions to F (x) =
F (y)
In order to find an upper bound for
∑
n
R(n)2 = # {x1, x 2, x, y : x31 + F (x) = x32 + F (y)}
we will first find an upper bound for
{x, y : F (x) = F (y)}.
We aim to prove the following.
Theorem 6.4. If F (x) is a cubic form in t ≥ 2 variables, satisfying the above conditions, and is not of the form AL(x)3 where L(x) is a linear form, then the number N (P ) of solutions of
F (x) = F (X)
with x, X ∈ P B∗ satisfies
N (P ) P 2t−2+ ε.
Here t is s − 1, and P B∗ is [P a 2, P b 2] × · · · × [P a s, P b s]. We will suppose, for simplicity, that P B∗ = [ −P, P ]t. In the end, the only difference will be in the implicit constants, so there is no harm in this. We will begin with two results from Estermann .
Lemma 6.5. Suppose that a, b, and n are positive integers and let Q(n; a, b ) denote the number of solutions of ax 2 + by 2 = n in positive integers x and y. Then
Q(n; a, b ) ≤ 2d(n).
Lemma 6.6. Suppose that a, b, m, and n are positive integers and let R(n; a, b )
denote the number of solutions of ax 2 − by 2 = n in positive integers x and y with
29 ax 2 ≤ m. Then
R(n; a, b ) ≤ 2d(n)(1 + log m).
Another minor lemma we will need is as follows.
Lemma 6.7. Suppose that n is a positive integer, b is any integer, and S(P ; n, b )
denotes the number of integers y ∈ [−P, P ] such that ny + b is a perfect square. Let
k2 be the largest square dividing (n, b ). Then
S(P ; n, b ) nε
1 + k
√
Pn
.
Note that setting n = 1 and b = 0 shows that this lemma is best possible.
Proof. Write (n, b ) = k2l, n1 = n/ (n, b ), and b1 = b/ (n, b ) so that l is squarefree and
(n1, b 1) = 1 . If x satisfies x2 = ny + b, then kl |x. We wish to bound the number of pairs z, y with z ≥ 0, y ∈ [−P, P ] and lz 2 = n1y + b1. Since (n1, b 1) = 1 , we have
(l, n 1) = 1 . Let R denote the set of residue classes r modulo n1 such that lr 2 ≡ b1
( mod n1). Since (lb 1, n 1) = 1 we have #R nε
1
. Let r ∈ R . Then it suffices to bound the number of solutions with z ≡ r ( mod n1). Let z0 be the least such solution and
y0 the corresponding value of y. Thus for any other solution z, we have z = z0 + n1v
where v ≥ 0. Hence
n1(y − y0) = ( lz 2 − b1) − (lz 20 − b1)= lz 2 − lz 20
= l(z0 + n1v)2 − lz 20
= 2 lz 0n1v + ln 21v2
= ln 1v(2 z0 + n1v)
and so ln 1v2 = ( y − y0) − 2lvz 0 ≤ 2P . Therefore, the number of possible v is
1 +
√
Pln 1
,
from which the desired result follows. Next is the main lemma necessary for the theorem. 30 Lemma 6.8. Let P ≥ 2 and Q ≥ 2, and suppose that a, b, c, d, e, and f are integers in [−Q, Q ] and let ∆ = 4 ac − b2 and
Θ = 4 acf + ebd − ae 2 − cd 2 − f b 2.
Let N (P, Q ) denote the number of solutions of
ax 2 + bxy + cy 2 + dx + ey + f = 0
in integers x and y with x, y ∈ [−P, P ]. Then we have the following cases: (i) if (a2 + c2)∆Θ 6 = 0 , then
N (P, Q ) (P Q )ε
(ii) if (a2 + c2)∆ 6 = 0 , Θ = 0 , and −∆ is not a perfect square, then
N (P, Q ) 1
(iii) if a = c = 0 and ∆Θ 6 = 0 , then
N (P, Q ) (P Q )ε
(iv) if (a2 + c2)∆ 6 = 0 , Θ = 0 , and −∆ is a perfect square, then
N (P, Q ) P
(v) if a 6 = 0 , 2ae − bd = ∆ = 0 , and d2 − 4af is not a perfect square, then
N (P, Q ) = 0
(vi) if a 6 = 0 , 2ae − bd 6 = 0 , ∆ = 0 , and k2 is the largest square dividing |2ae − bd |,then
N (P, Q ) Qε
(
1 + k
√
P
|2ae − bd |
)
(vii) if a 6 = 0 , 2ae − bd = ∆ = 0 and d2 − 4af is a perfect square, then
N (P, Q ) P
31 (viii) if a = c = Θ = 0 and ∆ 6 = 0 , then
N (P, Q ) P
and for completeness, we also state (ix) if a = c = ∆ = Θ = 0 , then
N (P, Q ) P 2.
Proof. If a = c = 0 , then ∆ = −b2 and Θ = b(ed − f b ). Hence if ∆ 6 = 0 , so that b 6 = 0 ,multiplication by b gives
(bx + e)( by + d) + bf − ed = 0
and this has (P Q )ε solutions when Θ 6 = 0 and P solutions when Θ = 0 . This deals with cases (iii) and (viii) and so, apart from the trivial case (xi), we can suppose that a2 + c2 6 = 0 , and thus, without loss of generality, that a 6 = 0 . Completing the square gives the equation
(2 ax + by + d)2 + ∆ y2 + 2(2 ae − bd )y + 4 af − d2 = 0 .
Suppose ∆ = 0 . Then we want 2( bd − 2ae )y + d2 − 4af to be a perfect square. If
2ae − bd = 0 and d2 − 4af is not a perfect square, then we have no solutions and are in case (v). If bd 6 = 2 ae , then we are in case (vi) and this follows from Lemma 6.7. If
bd = 2 ae and d2 − 4af is a perfect square, then x is determined by y and we get case (vii). Thus, we may assume that ∆ 6 = 0 . Completing the square once more gives
∆(2 ax + by + d)2 + (∆ y + 2 ae − bd )2 + 4 aΘ = 0 . (6.1) If Θ = 0 and −∆ is not a perfect square, the solutions are only possible with
2ax + by + d = 0 ∆y + 2 ae − bd = 0
and so y is determined by the second equation above and then x is determined by the 32 first. This gives case (ii). If Θ = 0 and −∆ is a nonzero square, say m2, then equation 6.1 becomes
(2 amx + mby + md )2 − (m2y + 2 ae − bd )2 = 0 .
Any choice of y gives O(1) solutions for x, which gives case (iv). For the final case, case (i), we write equation 6.1 in the form
∆X2 + Y 2 = −4aΘ.
If ∆ > 0 and −4aΘ ≤ 0, then there is at most one solution in integers X and Y . If
∆ > 0 and −4aΘ > 0, then we can appeal to Lemma 6.5, and we see that the number of solutions in X and Y is O(Qε).For each such pair X, Y , there are O(1) pairs x, y with ∆y + 2 ae − bd = Y and
2ax + by + d = X. If ∆ < 0 and −4aΘ > 0, then we write the equation as
Y 2 − (−∆) X2 = −4aΘ
and use Lemma 6.6 instead. On the other hand, if 4aΘ > 0, then we write
(−∆) X2 − Y 2 = 4 aΘ
and proceed in the same way. We will now give a proof of Theorem 6.4
Proof. Now suppose that t ≥ 2. We will write F (x) as
F (x) =
t∑
i=1
t∑
j=1
t∑
k=1
cijk xixj xk
where the coefficients are symmetric, that is cijk = cπ(ijk ) where π(ijk ) is any permu-tation of i, j, and k. We will first transform the form, if necessary, so that at least two of the variables, say x1 and x2, have nonzero x31 and x32 terms. If two variables already appear in this way, then we’re done. Otherwise, we have c111 = 0 (replace 1 with 2 if x31 has a nonzero coefficient). If there is a j where c11 j 6 = 0 , then replace xj with
xj + λx 1 where λ is an integer at our disposal. Then x31 will appear with coefficient 33 (3 c11 j λ + 3 c1jj λ2 + cjjj λ3), taking into account the different permutations of 11 j and
1jj . We can choose λ so that this is nonzero. If c11 j = 0 for every j, we can do the same with c1jj . If one of these is nonzero, then replacing xj with xj + λx 1 gives us x31(3 λ2c1jj + λ3cjjj ) which we can make nonzero as well. If we also have c1jj = 0 for all j, then since x1 appears explicitly there must be
j 6 = k where c1jk 6 = 0 . Now, replace xj with xj + λx 1 and xk with xk + μx 1, where λ
and μ are integers at our disposal. Now we have
(6 c1jk λμ + 3 cjkk λ2μ + 3 cjjk λμ 2 + cjjj λ3 + ckkk μ3)x31.
Again, this can be made nonzero with a suitable choice for λ and μ. This process can be repeated, if necessary, for x2. The transformations are invertible, and so this new form represents the same values as the old one, although our domain is now a parallelepiped. This new domain can be contained in [−cP, cP ]t for some constant c,however, so our results are unaffected. We now will distinguish two variables x and y where x3 and y3 appear explicitly and denote the remaining variables by z. Thus
F (x) = Ax 3 + Bx 2y + Cxy 2 + Dy 3
x2L1(z) + xy L2(z) + y2L3(x) + xQ1(z) + yQ2(z) + C1(z)
where Li, Qj , and C1 are linear, quadratic, and cubic forms, respectively, which are not all identically zero, as the variables in z must appear explicitly. Additionally,
AD 6 = 0 per our choice of x and y. In fact, we can assume that the Li and the Qj are not all identically zero, otherwise, we would have F (x) = C0(x, y ) + C1(z) where C0 is a binary cubic form, and the number of solutions to F (x) − F (X) = 0 is the number of solutions to C0(x, y ) − C 0(X, Y ) = m which is bounded by the number of solutions where m = 0 and so follows from the two variable case (which we establish first). Let R(n) here denote the number of solutions of F (x) = n with x ∈ [−P, P ]t.Then sorting by the choice of z, we have
R(n) = ∑
z∈[−P,P ]t−2
R(n, z)
34 where R(n, z) is the number of solutions of
Ax 3 + Bx 2y + Cxy 2 + Dy 3
x2L1(z) + xy L2(z) + y2L3(z) + xQ1(z) + yQ2(z) + C1(z) = n
with x, y, ∈ [−P, P ]. The object of our theorem, the number of solutions of F (x) =
F (X), can then be written as
N (P ) = ∑
n
R(n)2 = ∑
n
∑
z∈[−P,P ]t−2
R(n, z)
2
.
Applying the Cauchy-Schwarz inequality, we have
N (P ) ≤ (2 P + 1) t−2 ∑
z∈[−P,P ]t−2
∑
n
R(n, z)2.
The double sum represents the number of solutions to
Ax 3 + Bx 2y + Cxy 2 + Dy 3 + x2L1(z) + xy L2(z) + y2L3(z)+xQ1(z) + yQ2(z) = AX 3 + BX 2Y + CXY 2 + DY 3
+X2L1(z) + XY L2(z) + Y 2L3(z) + XQ1(z) + Y Q2(z).
This equation has t+2 variables, and we want to show that there are O(P t+ε) solutions. Setting g = X − x and h = Y − y, the above equation becomes
Ag (3 x2 + 3 gx + g2) + B(x2h + 2 gxy + 2 ghx + g2y + g2h)+ C(2 hxy + gy 2 + h2x + 2 ghy + gh 2) + Dh (3 y2 + 3 hy + h2) + L1(z)(2 xg + g2)+ L2(z)( xh + yg + gh ) + L3(z)(2 yh + h2) + gQ1(z) + hQ2(z) = 0 .
We will rewrite this as
ax 2 + bxy + cy 2 + dx + ey + f = 0
where
a = 3 Ag + Bh, b = 2 Bg + 2 Ch, c = Cg + 3 Dh,
35 d = 3 Ag 2 + 2 Bgh + Ch 2 + 2 L1(z)g + L2(z)h, e = Bg 2 + 2 Cgh + 3 Dh 2 + L2(z)g + 2 L3(z)h,
and
f = Ag 3 + Bg 2h + Cgh 2 + Dh 3 + L1(z)g2 + L2(z)gh + L3(z)h2 + Q1(z)g + Q2(z)h.
Note that these coefficients depend on g, h, and z, but not on x or y. We will apply Lemma 6.8 multiple times, for the cases (i) through (ix). We start by specifically addressing the case when t = 2 . In this case, each term involving z vanishes. If for a given choice of g and h any of cases (i), (ii), (iii), or (v) of Lemma 6.8 hold, we are done, as the number of solutions in x and y is O(P ε).In case (iv), the equation Θ = 0 is a polynomial in g where the coefficient of g5
is A(3 AC − B2). If 3AC − B2 6 = 0 , then g is determined by h so we are done. If
3AC − B2 = 0 , then the g4 term of Θ = 0 is (9 A2D + 2 ABC − B2)g4h which simplifies to A(9 AD − BC )g4h and so either h = 0 , g is determined by h or 9AD − BC = 0 . In the first two cases, we are done. In the third case, we look at the h5 term of Θ = 0 .Either h depends on g or we have
3AC − B2 = 9 AD − BC = C2 − 3BD = 0 ,
but this would give ∆ = 0 , which is excluded in case (iv). For cases (vi) and (vii), ∆ = 0 gives us
(B2 − 3AC )g2 + ( BC − 9AD )gh + ( C2 − 3BD )h2 = 0 .
If any of these coefficients are nonzero, then g or h must be zero, or g is determined by h, or vice versa. If all three are zero, then we have the case
F (x, y ) = 13BC (Bx + Cy )3
which we have excluded. For cases (vii) and (ix), a = c = 0 implies that
g = −Bh
3A , h = − Cg
3D .
36 In case (vii), B and C cannot both be zero as ∆ 6 = 0 , so we have g determined by
h or both are zero, so we are done. In case (ix), if either B or C were zero, then both g and h would be zero and we would be done, se we suppose BC 6 = 0 . Then g is determined by h, so we write g = λh
where
λ = − B
3A = −3DC = − CB .
In this case we then have
F (x, y ) = A(x − λy )3
which we’ve excluded. Now consider t > 2. In the notation of the lemma, if for a given choice of g, h,and z cases (i), (ii), (iii), or (v) hold, then we are done. Now consider case (iv). Given that Θ = 0 , the coefficient of g5 is A(3 AC − B2). If
3AC − B2 6 = 0 , then g is determined by z and h so we are done. If 3AC − B2 = 0 ,then Θ = 0 has a term (9 A2D + 2 ABC − B2)g4h = A(9 AD − BC )g4h and so h = 0
or g is determined by h. Finally, by considering the h5 term, either we are done or
3AC − B2 = BC − 9AD = C2 − 3BD = 0 .
In this case a = BC (Bg + Ch ), b = 2( Bg + Ch ), and c = CB (Bg + Ch ). But then ∆ = 0 ,which is contrary to the assumption in this case. For the remaining cases, if it should happen that a = 0 and c 6 = 0 , we will interchange x and y so that a 6 = 0 and c = 0 .For cases (vi) and (vii), ∆ = 0 gives us
(B2 − 3AC )g2 + ( BC − 9AD )gh + ( C2 − 3BD )h2 = 0 .
If any of the coefficients are nonzero, then g or h must be zero, or g is determined by
h, or vice versa. Otherwise, we have
B2 − 3AC = BC − 9AD = C2 − 3BD = 0 .
Since AD 6 = 0 , we also have BC 6 = 0 and so
A = B2
3C , D = C2
3B
37 and so
a = BC (Bg + Ch ), c = CB (Bg + Ch )
and hence
ax 2 + bxy + cy 2 = 1
BC (Bg + Ch )( Bx + Cy )2.
The above immediately implies that 3C divides B2 and 3B divides C2, and so all of the following expressions are integer valued. Since a 6 = 0 , we have Bg + Ch 6 = 0 . We also have
d = 1
C (Bg + Ch )2 + 2 L1(z)g + L2(z)h, e = 1
B (Bg + Ch )2 + L2(z)g + 2 L3(z)h,
and
f = 13BC (Bg + Ch )3 + L1(z)g2 + L2(z)gh + L3(z)h2 + Q1(z)g + Q2(z)h.
Recall that in case (vi) a 6 = 0 , 2ae − bd 6 = 0 , ∆ = 0 , and in case (vii) a 6 = 0 ,
2ae − bd = ∆ = 0 , d2 − 4af = m2 for some m. Moreover,
2ae − bd = 2Bc (Bg + Ch )
(
1
B (Bg + Ch )2 + L2(z)g + 2 L3(z)h
)
− 2( Bg + Ch )
(
1
C (Bg + Ch )2 + 2 L1(z)g + L2(z)h
)
so that
2ae − bd = ( Bg + Ch )
(
2BC L3(z)h +
(
Bg C − h
)
L2(z) − 2L1(z)g
)
.
In case (vi), let j = |2ae − bd | that 0 < j P 3. Then g, h, and z satisfy
(Bg + Ch )(2 BL3(z)h + ( Bg − Ch )L2(z) − 2CL1(z)g) = ±Cj.
Then in this case, given g, h, and z, the number of choices for x and y is
P ε
(
1 + k
√
Pj
)
where k2 divides j. The total contribution from the P ε term is O(P t + ε), which is 38 acceptable. We need to determine the contribution from the P εk√ Pj term. We can presume that there is a j ∈ N and a δ ∈ Z \ { 0} such that
Bg + Ch = δ
and
(BL2(z) − 2CL1(z)) g + (2 BL3(z) − CL2(z)) h = ±Cj/δ.
Given g, the first of these equations determines h. Since in the second equation we have Cj/δ 6 = 0 and the left side is a linear form in z, at least one of the zj appears explicitly and so is determined by the other variables. Thus, given j, the total number of possible δ, g, h and z is O(P t−2+ ε). Thus the total contribution is
P t−2+ ε ∑
j P3
∑
k2|j
k
√
Pj P t−2+ ε ∑
k P3/2
P 1/2 ∑
l P3/k 2
l−1/2 P t+2 ε
and so we are done in this case. In case (vii) we have
(Bg + Ch )
(
2BC L3(z)h +
(
Bg C − h
)
L2(z) − 2L1(z)g
)
= 0 .
Since a 6 = 0 we have Bg + Ch 6 = 0 . Thus
2Bc L3(z)h +
(
Bg C − h
)
L2(z) − 2L1(z)g = 0
so that
(BL2(z) − 2CL1(z)) g + (2 BL3(z) − CL2(z)) h = 0 .
If either BL2(z) − 2CL1(z) or 2BL3(z) − CL2(z) is not identically zero, then either g
is determined by h and z, or h is determined by g and z, or there are O(P t−3) choices of z for which BL2(z) − 2CL1(z) = 0 or 2BL3(z) − CL2(z) = 0 . Then we are done by the case (vii) of the lemma. If both BL2(z) − 2CL1(z) and 2BL3(z) − CL2(z) are identically zero, then we have
0 = ax 2 + bxy + cy 2 + dx + ey + f
= Bg + Ch
3BC (3( Bx + Cy )2 + 3( Bx + Cy )( Bg + Ch ) + ( Bg + Ch )2)
39 + L1(z)g2 + L2(z)gh + L3(z)h2 + Q1(z)g + Q2(z)h. (6.2) Since both BL2(z) − 2CL1(z) and 2BL3(z) − CL2(z) are identically zero, this becomes
G(3 X2 + 3 XG + G2) + G2
B2 L1(z) + GB Q1(z) + hB (BQ2(z) − CQ1(z)) = 0
where X = Bx + Cy and G = Bg + Ch . If G = 0 or h = 0 , then g is determined by h and we are done, since we are in case (vii) of the lemma. If G 6 = 0 but
BQ2(z) − CQ1(z) = 0 identically, then
3X2 + 3 XG + G2 + GB L1(z) + 1
B Q1(z) = 0 .
Applying the lemma to this polynomial, since 3, 1 and 4 · 3 · 1 − 32 are nonzero and
4 · 3 · 1 − 32 = 3 is not a square, we have P (P ε) choices for X and G given z, and so again we are done. If G 6 = 0 , h 6 = 0 and BQ2(z) − CQ1(z) is not identically zero, then G divides
h(BQ2(z) − CQ1(z)) and so given h and z there are O(P t−1+ ε) choices for g. Then, applying the result of the lemma from case (vii), we are done. For case (viii), note that a = c = 0 implies that
g = −Bh
3A and h = − Cg
3D .
Since B and C cannot both be zero, as ∆ 6 = 0 , we have g determined by h. Therefore, there are O(P t−1) choices for g, h, and z, and so appealing to the lemma gives O(P t)
solutions. Case (ix) is similar to case (vii), but now a = b = c = 0 . Then
g = −Bh
3A , h = − Cg
3D .
Thus B = 0 would imply that g = 0 and so h = 0 and so we are done. Likewise if
C = 0 , and so we can suppose that BC 6 = 0 . Then g is determined by h and vice versa. It follows that g = λh where
λ = − B
3A = −3DC = − CB
40 and
3Ag 2 + 2 Bgh + Ch 2 = 0
Bg 2 + 2 Cgh + 3 Dh 2 = 0
Ag 3 + Bg 2h + Cgh 2 + Dh 3 = 0
and so our equation reduces to
dx + ey + f = 0
where
d = 2 L1(z)g + L2(z)h, e = L2(z)g + 2 L3(z)h
and
f = L1(z)g2 + L2(z)gh + L3(z)h2 + Q1(z)g + Q2(z)h.
If d 6 = 0 or e 6 = 0 , the either x or y is fixed by the other variables and we are done. Thus we can suppose that d = e = 0 . Then f = 0 and so
2L1(z)λ + L2(z) = L2(z)λ + 2 L3(z) = 0 (6.3) and
L1(z)λ2 + L2(z)λ + L3(z) + Q1)( z)λ + Q2(z) = 0 . (6.4) Substituting equations 6.3 into equation 6.4 gives
Q1(z)λ + Q2(z) = 0 . (6.5) If any of the equations 6.3 or 6.5 does not hold identically, then they hold for at most
O(P t−3) choices for z and we are done. If they all hold identically, then substituting this into the original equation F (X, Y, z) = F (x, y, z) gives us F (x + λh, y + h, z) =
F (x, y, z) identically for all x, y, h, and z. Taking h = −y, this becomes F (x, y, z) =
F (x − λy, 0, z), and we can appeal to induction on t, unless F (x − λy, 0, z) can be expressed as a linear form cubed, but then in that case the original F (x, y, z) could have been as well. 41 6.3 Estimating ∑ R(n)2
The sum of interest in this case is the number of solutions to x3 + F (x) = y3 + F (y)
where (x, x) and (y, y) are elements of BB. We can write this as
∫10
∑
x,y, x,y
e(α(x3 + F (x) − y3 − F (y))) dα =
∫10
∣∣∣∣∣∑
x
e(αx 3)
∣∣∣∣∣
2∣∣∣∣∣∑
x
e(αF (x))
∣∣∣∣∣
2
dα.
Applying the Cauchy-Schwarz inequality, this is
∫10
∣∣∣∣∣∑
x
e(αF (x))
∣∣∣∣∣
2
dα
1/2∫10
∣∣∣∣∣∑
y
e(αy 3)
∣∣∣∣∣
4∣∣∣∣∣∑
y
e(αF (y))
∣∣∣∣∣
2
dα
1/2
.
The first integral above is the number of solutions of F (x) = F (y) and so by the result from the previous section it must be O(B2s−4+ ε). Therefore, the expression above is
Bs−2+ ε
∫10
∣∣∣∣∣∑
y
e(αy 3)
∣∣∣∣∣
4∣∣∣∣∣∑
y
e(αF (y))
∣∣∣∣∣
2
dα
1/2
.
For the remaining integral, we will modify the proof of Lemma 2.1. We consider
∣∣∣∣∣∑
y
e(αy 3)
∣∣∣∣∣
2
= ∑
h
∑
y
e(α(( y + h)3 − y3))
where the values of h are O(B) (since h = y1 − y2 for some y1, y 2 ∈ [Ba 1, Bb 1]) and each sum in y runs over some subset of [Ba 1, Bb 1] depending on h. Simplifying and isolating the term with h = 0 shows that the above is
B + ∑
h6=0
∑
y
e(α(3 y2h + 3 yh 2 + h3)) .
Squaring, we get
∣∣∣∣∣∑
y
e(αy 3)
∣∣∣∣∣
4
B2 +
∣∣∣∣∣∣∑
h6=0
∑
y
e(α(3 y2h + 3 yh 2 + h3))
∣∣∣∣∣∣
2
,
which, if we set j = y2 − y1 so that y2 = y1 + j, becomes, after applying the Cauchy-42 Schwarz inequality,
B2 + B ∑
h6=0
∑
j
∑
y
e(α(3(( y + j)2h + 3( y + j)h2 + h3) − 3( y2h + 3 yh 3 + h3))) = B2 + B ∑
h6=0
∑
j
∑
y
e(αjh (6 y + 3( j + h))) .
Again, separating the term where j = 0 , we have
B3 + B ∑
h6=0
∑
j6=0
∑
y
e(αjh (6 y + 3( j + h))) .
Setting u = 6 yjh + 3 jh (j + h) we have
B3 + B ∑
u
c(u)e(αu ),
where c(u) is the number of solutions to 6yjh + 3 jh (h + j) = u in j, h, and y. Since
h and j must divide u, and once h and j are chosen there at most one solution in y,we have c(u) | u|ε for u 6 = 0 and c(0) B2. Putting this back into the integral, we have
∫10
∣∣∣∣∣∑
y
e(αy 3)
∣∣∣∣∣
4∣∣∣∣∣∑
y
e(αF (y))
∣∣∣∣∣
2
dα
B3
∫10
∣∣∣∣∣∑
y
e(αF (y))
∣∣∣∣∣
2
dα + B
∫10
∑
u
c(u)e(αu )
∣∣∣∣∣∑
y
e(αF (y))
∣∣∣∣∣
2
dα.
In the second integral above, if we separate the term where u = 0 , we get another copy of the first term, so the above is
B3
∫10
∣∣∣∣∣∑
y
e(αF (y))
∣∣∣∣∣
2
dα + B ∑
u6=0
c(u)
∫10
e(αu )
∣∣∣∣∣∑
y
e(αF (y))
∣∣∣∣∣
2
dα.
The first integral above is again the number of solutions to F (x) = F (y) and so is
O(B2s−4+ ε) by Theorem 6.4. The second integral represents the number of solutions to the equation F (x) − F (y) = u. If x and y are chosen, there is only one choice for
u, and so this integral is O(B2s−2). Therefore, we have
∑
n B3
R(n)2 B2s−5/2+ ε.
43 6.4 Proof of the Theorem
As in Chapter 4, if x2, . . . , x s are chosen in advance, there is now at most one possible choice for x1 so that x31 + F (x2, . . . , x s) = 0 , and so again R(0) Bs−1. From the previous section, we have
S(D, B 3/2)
( B3/2+ ε
D1/2
) (
B2s−5/2+ ε)1/2
= Bs+1 /4+ εD−1/2.
In Section 6.1, we showed that, for squarefree d, we have ρ(d2) d2s−2+ ε. Therefore
S(1 , D ) = BsVol (B)
∞
∑
d=1
μ(d)ρ(d2)
d2s
O
∑
d≤D
μ(d)2ρ(d2)
( Bs−1
d2s−2 + 1
)
Bs ∑
d>D
μ(d)2ρ(d2)
d2s + DR (0)
= BsVol (B) ∏
p
(
1 − ρ(p2)
p2s
)
+O
∑
d≤D
d2s−2+ ε
( Bs−1
d2s−2 + 1
)
Bs ∑
d>D
d2s−2+ ε
d2s + DB s−1
= BsVol (B) ∏
p
(
1 − ρ(p2)
p2s
)
+O
∑
d≤D
Bs−1dε + ∑
d≤D
d2s−2+ ε + Bs ∑
d>D
d−2+ ε + DB s−1
= BsVol (B) ∏
p
(
1 − ρ(p2)
p2s
)
O (
Bs−1D1+ ε + D2s−1+ ε + BsD−1+ ε)
.
Combining these, we have
Ng(B) = BsVol (B) ∏
p
(
1 − ρ(p2)
p2s
)
O(Bs−1D1+ ε + D2s−1+ ε + BsD−1+ ε + Bs+1 /4+ εD−1/2).
Setting D = Bγ , this becomes
Ng(B) = BsVol (B) ∏
p
(
1 − ρ(p2)
p2s
)
+O(Bs−1+ γ+ε +B2sγ −γ+ε +Bs−γ+ε +Bs+1 /4−γ/ 2+ ε).
44 The optimal choice for γ here is 4s+1 8s−2 , and with this choice we have
Ng(B) = BsVol (B) ∏
p
(
1 − ρ(p2)
p2s
)
O(Bs− 18s−2 +ε).
45 Chapter 7 | General Cubic Forms
In this chapter, we will establish the desired result for cubic forms with seven or more variables, satisfying a bound on an associated exponential sum. For simplicity we will redefine BB to be the set of integer points in [0 , B ]s. We suppose g(x) is a cubic form in s variables. We define
S(g, α ) = ∑
x∈BB
e(αg (x)) ,
and we define a new cubic form F in 2s variables by
F (x1, . . . , x 2s) = g(x1, . . . , x s) − g(xs+1 , . . . , x 2s). (7.1) We will require the following two hypotheses.
Hypothesis 7.1. Let θ be independent of B and satisfy 0 < θ < 1. Let η > 0 be any fixed positive number. Then for every α, either
|S(F, α )| < B 2s− 12 sθ +η
or α has a rational approximation such that
(a, q ) = 1 , 1 ≤ q B2θ, |qα − a| < B −3+2 θ.
This is alternative (B) from Lemma 13.4 in Davenport .
Hypothesis 7.2. Let gj (x) = ∂g (x1,...,g s)
∂x j
. Then for all but finitely many primes p,
g(x), g1(x), . . . , g s(x) have no nontrivial factor in common modulo p.
46 We define a new cubic form F in 2s variables by
F (x1, . . . , x 2s) = g(x1, . . . , x s) − g(xs+1 , . . . , x 2s).
Then we have the following theorem.
Theorem 7.3. If g(x) satisfies Hypothesis 7.2 and F (x) satisfies Hypothesis 7.1, then for s ≥ 7,
Ng(B) = Bs ∏
p
(
1 − ρ(p2)
p2s
)
O (
Bs− 8s+3 16 s−4 +δ)
where δ > 0 is an arbitrary small real number.
7.1 Estimating ρ(d2)
We start with the following lemma.
Lemma 7.4. For any form f (x), we have ρ(p) ps−1 and ρ(p2) p2s−1.Proof. We will prove the first inequality by induction on the degree and the number of variables. For degree one, assume that p is large enough that none of the coefficients of f are divisible by p. Our form is then c1x1 + · · · + csxs, and so for each choice of
x1, . . . , x s−1, the value of xs is determined uniquely. Now suppose that the form f (x) has degree d, and that
f (x) = g0xds + g1(y)xd−1
s
· · · + gd−1(y)xs + gd(y)
where y = ( x1, . . . , x s−1) and gj is a form of degree j. Choose values for y. If there is a j ≤ d − 1 such that gj (y) 6 ≡ 0 ( mod p) then there are at most d choices for xs.Otherwise, gj (y) ≡ 0 ( mod p) for all j ≤ d − 1, and so f (x) ≡ 0 ( mod p) only if
gd(y) ≡ 0 ( mod p) as well. Then, by induction on d and s, the number of y for which
gd(y) ≡ 0 (mod p) is O(ps−2).The second inequality follows from the first as there at most ps solutions to f (x) ≡ 0(mod p2) for each solution modulo p.Lemma 3.4 in Chapter 4 of states
Lemma 7.5. Let u1(x), . . . , u t(x) be polynomials in n variables over Fq, each of total degree at most e, and without common factor. Then the number of their common zeros is at most qn−2e3.
47 From this and the argument used to prove Lemma 7.4, we can deduce that if g(x)
satisfies Hypothesis 7.2, then ρ(p) ps−2 and ρ(p2) p2s−2. It follows, then, that for squarefree d, we have ρ(d2) d2s−2+ ε.
7.2 Estimating ∑ R(n)2
Given that
S(F, α ) = ∑
x∈(BB)2
e(αF (x))
in this context, we have
∑
n B3
R(n)2 =
∫10
S(F, α )dα.
Let 0 < θ < 1 and call ξ(θ) the set of numbers α ∈ [0 , 1] having a rational approxima-tion satisfying
(a, q ) = 1 , 1 ≤ q B2θ, |qα − a| < B −3+2 θ.
Then, we call the set of α belonging to ξ(θ) the major arcs and label it M. We also call its complement relative to [0 , 1] the minor arcs , labeled m. Note that both M
and m depend on θ. The following is an adaptation of Lemma 15.1 in .
Lemma 7.6. Suppose F (x) is any cubic form in 2s variables that satisfies Hypothesis 7.1. Then given a number 0 < ∆ < 1, where 1 < s
18
∆3 , we have
∫
m
|S(F, α )|dα B2s−3+∆ .
Proof. Choose a set of numbers
θ = θ0 < θ 1 < · · · < θ h = 34 + δ (7.2) for some small δ > 0. Then every real α lies in ξ(θh), since one can always find a and
q where
q ≤ B3/2, |qα − a| < B −3/2.
Since ξ(θ0) is the major arcs, and the minor arcs are their complement, the set m can 48 be regarded as the union of
ξ(θh) \ ξ(θh−1), ξ (θh−1) \ ξ(θh−2), . . . , ξ (θ1) \ ξ(θ0).
In each set ξ(θg) \ ξ(θg−1) with 1 ≤ g ≤ h, Hypothesis 7.1 holds with θ = θg−1, so
|S(F, α )| B2s− 12 sθ g−1+η.
The measure of this set ξ(θg) \ ξ(θg−1) is bounded by the measure of ξ(θg) which is
∑
q≤B2θg
q
∑
a=1
q−1B−3+2 θg
B−3+4 θg ,
and so the contribution over this set to the integral is
B2s− 12 sθ g−1−3+4 θg +η.
Since we require this to be O(B2s−3+∆ ), we must have 7.2 and
2s − 12sθ g−1 − 3 + 4 θg + η < 2s − 3 + ∆ ,
or equivalently
θg−1 < θ g < s
8θg−1 + ∆ − η
4 .
If s < 8, then we need 8 − s
8 θg−1 < ∆ − η
4
to hold for all g ≤ h and then we require
θh < s
8 · 88 − s · ∆ − η
4 + ∆ − η
4 = 88 − s · ∆ − η
4 .
By 7.2, this is 8 − s
8
( 34 + δ
)
< ∆ − η
4 .
49 Since δ and η can be made arbitrarily small, we only need
8 − s
8 · 3 < ∆,
or
3 − 3s
8 < ∆
which is possible if
1 < s
8 + ∆3 .
Lemma 7.7. Under the same conditions as the previous lemma,
∫
M
|S(C, α )|dα Bs−3+4 θ.
Proof. For any α, we have the simple estimate |S(C, α )| Bs. Then, as before, the measure of M = ξ(θ) is O(B−3+4 θ), which gives the desired result. Combining the results of these two lemmas, when ∆ is taken to be 4θ, we have
∑
n B3
R(n)2 =
∫10
S(F, α )dα B2s−3+4 θ
since F (x) has 2s variables.
7.3 Proof of the Theorem
If g(x) is a cubic form in one variable, then R(0) = 1 and hence is O(Bs−1). In , Heath-Brown shows that R(0) Bs−1− 1
s−1
log (B)2s when s > 1, which is more than sufficient for our needs. Using the previously obtained estimates, we have
S(D, B 3/2) B3/2+ ε
D1/2
∑
n B3
R(n)2
1/2
Bs+ε+2 θD−1/2
50 and
S(1 , D ) = BsVol (B) ∏
p
(
1 − ρ(p2)
p2s
)
+O (
Bs−1D1+ δ + D2s−1+ δ + BsD−1+ δ + DR (0) )
.
Combining these, we have
Ng(B) = BsVol (B) ∏
p
(
1 − ρ(p2)
p2s
)
O (
Bs−1D1+ δ + D2s−1+ δ + BsD−1+ δ + Bs+ε+2 θD−1/2)
.
Setting D = Bγ , the error term becomes
O (
Bs−1+ γ+γδ + B2sγ −γ+γδ + Bs−γ+γδ + Bs+ε+2 θ−γ/ 2)
.
Setting θ = 18 , (and hence ∆ = 12 ) so that Lemma 7.6 holds with s ≥ 14 , then setting
γ = 4s+1 8s−2 we have
Ng(B) = BsVol (B) ∏
p
(
1 − ρ(p2)
p2s
)
O (
Bs− 8s+3 16 s−4 +δ)
.
51 Chapter 8 | Further Research
The results obtained in this dissertation are part of a quite large problem - determining the density of squarefree numbers amongst the values of an arbitrary polynomial of several variables - and as such there are numerous avenues for further work. In Chapter 4, we gave an asymptotic formula for the number of squarefree values among sums of 3 or more cubic polynomials. As such a formula was established by Hooley [23, 24] for a single cubic polynomial, the sum of two cubic polynomials presents an obvious gap. One would also suspect that sums of higher degree polynomials, or even sums of
k-th powers, would be a reasonable first place to start on the higher degree cases. In Chapter 6, we had to exclude the case where g(x) = x31 + AL(x2, . . . , x s)3. One would expect to be able to adapt the work of Greaves - or even the method applied in Chapter 5 - to yield results in this case. The result we obtained in Chapter 7 for general cubic forms is fairly weak. It should be possible to remove the dependence on the Hypotheses and to reduce the number of variables required. 52 Bibliography
Apostol, T. M. Introduction to analytic number theory . Springer-Verlag, New York-Heidelberg, 1976. Undergraduate Texts in Mathematics. Bhargava, M. The geometric sieve and the density of squarefree values of invariant polynomials, 2014. Bhargava, M., Shankar, A., and Wang, X. Squarefree values of polynomial discriminants i, 2016. Browning, T. D. Power-free values of polynomials. Arch. Math. (Basel) 96 , 2 (2011), 139–150. Cugiani, M. Sugli intervalli fra i valori dell’argomento pei quali un polinomio risulta libero da potenze. Riv. Mat. Univ. Parma 4 (1953), 95–103. Davenport, H. Analytic methods for Diophantine equations and Diophantine inequalities , second ed. Cambridge Mathematical Library. Cambridge University Press, Cambridge, 2005. With a foreword by R. C. Vaughan, D. R. Heath-Brown and D. E. Freeman, Edited and prepared for publication by T. D. Browning. Erdös, P. Arithmetical properties of polynomials. J. London Math. Soc. 28
(1953), 416–425. Estermann, T. Einige Sätze über quadratfreie Zahlen. Math. Ann. 105 , 1 (1931), 653–662. Filaseta, M. An elementary approach to short interval results for k-free numbers. J. Number Theory 30 , 2 (1988), 208–225. Filaseta, M. Short interval results for squarefree numbers. J. Number Theory 35 , 2 (1990), 128–149. Filaseta, M. Short interval results for k-free values of irreducible polynomials.
Acta Arith. 64 , 3 (1993), 249–270. Filaseta, M. Powerfree values of binary forms. J. Number Theory 49 , 2 (1994), 250–268. 53 Gegenbauer, L. Asymptotische gesetze der zahlentheorie. Denkschriften Österreich. Akad. Wiss. Math.-Natur. Cl. 49 (1885), 37–80. Granville, A. ABC allows us to count squarefrees. Internat. Math. Res. Notices , 19 (1998), 991–1009. Greaves, G. Power-free values of binary forms. Quart. J. Math. Oxford Ser. (2) 43 , 169 (1992), 45–65. Hardy, G. H., and Wright, E. M. An introduction to the theory of numbers ,sixth ed. Oxford University Press, Oxford, 2008. Revised by D. R. Heath-Brown and J. H. Silverman, With a foreword by Andrew Wiles. Heath-Brown, D. R. Cubic forms in ten variables. Proc. London Math. Soc. (3) 47 , 2 (1983), 225–257. Heath-Brown, D. R. Counting rational points on algebraic varieties. In
Analytic number theory , vol. 1891 of Lecture Notes in Math. Springer, Berlin, 2006, pp. 51–95. Heath-Brown, D. R. Power-free values of polynomials. Q. J. Math. 64 , 1 (2013), 177–188. Helfgott, H. A. On the square-free sieve. Acta Arithmetica 115 , 4 (2004), 349–402. Helfgott, H. A. Power-free values, large deviations and integer points on irrational curves. J. Théor. Nombres Bordeaux 19 , 2 (2007), 433–472. Helfgott, H. A. Square-free values of f (p), f cubic. Acta Math. 213 , 1 (2014), 107–135. Hooley, C. On the power free values of polynomials. Mathematika 14 (1967), 21–26. Hooley, C. On power-free numbers and polynomials. I. J. Reine Angew. Math. 293(294) (1977), 67–85. Hooley, C. On power-free numbers and polynomials. II. J. Reine Angew. Math. 295 (1977), 1–21. Hooley, C. On the power-free values of polynomials in two variables. In Analytic number theory . Cambridge Univ. Press, Cambridge, 2009, pp. 235–266. Hooley, C. On the power-free values of polynomials in two variables. II. J. Number Theory 129 , 6 (2009), 1443–1455. 54 Hua, L. K. Additive theory of prime numbers . Translations of Mathematical Monographs, Vol. 13. American Mathematical Society, Providence, R.I., 1965. Huxley, M. N., and Nair, M. Power free values of polynomials. III. Proc. London Math. Soc. (3) 41 , 1 (1980), 66–82. Lang, S. Algebraic number theory , second ed., vol. 110 of Graduate Texts in Mathematics . Springer-Verlag, New York, 1994. Lapkova, K., and Xiao, S. Y. Density of power-free values of polynomials.
Mathematika 65 , 4 (2019), 1038–1050. Lee, J.-J., and Murty, M. R. Dirichlet series and hyperelliptic curves. Forum Math. 19 , 4 (2007), 677–705. Maynard, J. Primes represented by incomplete norm forms, 2015. Montgomery, H. L., and Vaughan, R. C. Multiplicative number the-ory. I. Classical theory , vol. 97 of Cambridge Studies in Advanced Mathematics .Cambridge University Press, Cambridge, 2007. Murty, M. R., and Pasten, H. Counting squarefree values of polynomials with error term. Int. J. Number Theory 10 , 7 (2014), 1743–1760. Nagel, T. Zur Arithmetik der Polynome. Abh. Math. Sem. Univ. Hamburg 1 , 1 (1922), 178–193. Nair, M. Power free values of polynomials. Mathematika 23 , 2 (1976), 159–183. Nair, M. Power free values of polynomials. II. Proc. London Math. Soc. (3) 38 ,2 (1979), 353–368. Poonen, B. Squarefree values of multivariable polynomials. Duke Math. J. 118 ,2 (2003), 353–373. Reuss, T. Power-free values of polynomials. Bull. Lond. Math. Soc. 47 , 2 (2015), 270–284. Ricci, G. Ricerche aritmetiche sui polinomi. Rendiconti del Circolo Matematico di Palermo 57 (1933), 433–75. Schmidt, W. Equations over finite fields: an elementary approach , second ed. Kendrick Press, Heber City, UT, 2004. Stewart, C. L., and Xiao, S. Y. On the representation of k-free integers by binary forms, 2016. 55 Vaughan, R. C. The Hardy-Littlewood method , second ed., vol. 125 of Cambridge Tracts in Mathematics . Cambridge University Press, Cambridge, 1997. Xiao, S. Y. Power-free values of binary forms and the global determinant method. Int. Math. Res. Not. IMRN , 16 (2017), 5078–5135. Xiao, S. Y. Square-free values of decomposable forms. Canad. J. Math. 70 , 6 (2018), 1390–1415. 56 Vita
James M. Kowalski
Education
Doctor of Philosophy Mathematics, May 2020, Pennsylvania State University
Master of Arts Mathematics, May 2014, Central Michigan University
Bachelor of Science Mathematics, May 2011, Central Michigan University
Bachelor of Music Education Instrumental Music, May 2011, Central Michigan University
Publications
2020 - Accepted On the proportion of squarefree numbers among sums of cubic polynomials, The Ramanujan Journal , to appear. |
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Home > Blog > Data Analytics
Operating Income vs. Net Income : Analysis & Differences
Are you tired of scratching your head over operating income vs. net income?
Spreadsheet is here to your rescue. Its powerful formulas, functions, and visualization capabilities will help you analyze these financial figures effortlessly.
But before we dive into the spreadsheets, let me summarize it in simple terms.
Operating income signifies a company’s earnings from its fundamental operations. It’s like the star of the show, highlighting the company’s revenue generation and expense management prowess.
In contrast, net income represents what remains after deducting all expenses, including taxes, interest, and non-operating elements. It’s like the cherry on top, offering insight into the company’s overall profitability.
Now, here comes the fun part. Comparing operating vs. net income will uncover valuable insights about a company’s financial health. You can identify trends, assess profitability, and make informed decisions.
Table of Contents:
What is Operating Income?
Operating Income Example
Operating Income Formula
How to Calculate Operating Income?
What is Net Income?
Net Income Example
Net Income Formula
How to Calculate Net Income?
Operating Income vs. Net Income: Key Differences
Operating Income vs. Net Income: Comparison
How to Evaluate the Results of Operating Income vs. Net Income?
How To Manage Operating Income And Net Income?
Wrap Up
So, let’s crunch those numbers and unravel the mysteries of operating income vs. net income.
Video Tutorials:
Operating Income vs. Net Income in Excel
Operating Income vs. Net Income in Google Sheets
What is Operating Income?
Definition: Operating income, also known as operating profit, mirrors a company’s earnings from core business activities. It is calculated by deducting operating expenses like wages and rent from gross income.
It excludes interest and taxes, focusing solely on operational efficiency. This metric aids in assessing a company’s ability to generate profit through its fundamental operations. Thus, it is a pivotal indicator of financial health and efficiency.
Operating Income Example
Company XYZ reports the following figures for the fiscal year:
Total Sales Revenue: $850,000
Cost of Goods Sold (COGS): $500,000
Operating Expenses: $150,000 (includes costs like marketing, salaries, and utilities)
Operating Income Formula
Operating Income=Total Revenueˆ’Cost of Goods Sold (COGS)ˆ’Operating Expenses
Where:
Total Revenue is the total amount earned from sales.
Cost of Goods Sold (COGS) is the direct cost of producing goods or services sold.
Operating Expenses include costs related to running the business, such as salaries, rent, and utilities.
How to Calculate Operating Income?
Determine Gross Profit
Formula:
Gross Profit=Total Revenue – Cost of Goods Sold (COGS)
Example: If Total Revenue is $500,000 and COGS is $300,000:
Gross Profit=$500,000-$300,000=$200,000
Subtract Operating Expenses
Formula:
Operating Income=Gross Profit-Operating Expenses
Example: If Operating Expenses are $100,000:
Operating Income=$200,000-$100,000=$100,000
Summary
Operating Income: The resulting figure represents the profit earned from core business operations before interest, taxes, and non-operating items.
What is Net Income?
Definition: Net income, the bottom line of a company’s income statement, represents total profits after all expenses. Calculated by subtracting all costs, taxes, and interest from total revenue, net income reflects the company’s overall profitability.
It’s a crucial indicator for investors, indicating the actual earnings available to shareholders after all obligations. Net income serves as a key measure of a company’s financial success and sustainability in the long run.
Net Income Example
ABC Corp reports the following financial figures for the year:
Total Revenue: $1,200,000
Cost of Goods Sold (COGS): $700,000
Operating Expenses: $250,000
Interest Expense: $50,000
Taxes: $70,000
Net Income Formula
Net Income=Total Revenue-Cost of Goods Sold (COGS)-Operating Expenses-Interest Expense-Taxes
Where:
Total Revenue is the total income from sales.
Cost of Goods Sold (COGS) is the direct cost of producing goods or services sold.
Operating Expenses include costs like salaries, rent, and utilities.
Interest Expense is the cost of interest on borrowed funds.
Taxes are the income taxes owed.
How to Calculate Net Income?
Calculate Gross Profit:
Gross Profit=Total Revenue–COGS
Gross Profit=$1,200,000-$700,000=$500,000
Calculate Operating Income:
Operating Income=Gross Profit-Operating Expenses
Operating Income=$500,000-$250,000=$250,000
Subtract Interest Expense:
Income Before Taxes=Operating Income-Interest Expense
Income Before Taxes=$250,000-$50,000=$200,000
Subtract Taxes:
Net Income=Income Before Taxes-Taxes
Net Income=$200,000-$70,000=$130,000
Result:
Net Income: $130,000
Operating Income vs. Net Income: Key Differences
These metrics may sound similar, but they have some differences that can make or break your company’s success. So, what sets them apart?
Scope
Operating income delves deep into the company’s core operations, excluding non-operational factors. On the other hand, net income paints a broader financial canvas, encapsulating all revenues and expenses. As a result, it offers a comprehensive overview beyond mere operations.
Components
Operating income comprises the difference between operating revenue and operating expenses, focusing on the financial intricacies directly tied to a company’s day-to-day business operations. This concept is closely related to the analysis of marginal revenue vs marginal cost, where understanding the incremental benefits and costs is crucial for decision-making.
In contrast, net income encompasses a wider array of components, extending beyond operating realms to include taxes, interest, and other non-operational factors.
Inclusiveness
The key differentiator lies in inclusiveness. Operating income deliberately excludes non-operational elements, providing a purer assessment of a company’s operational prowess. Meanwhile, net income embraces all aspects, presenting a more holistic view of the company’s financial health.
Financial Assessment
While operating income zeros in on profitability from core operations, net income casts a wider net. Net income reflects the overall profitability, considering all revenues and expenses, whether from operations or non-operational facets.
Place on the Income Statement
Operating income takes a prominent position above net income on the income statement. Acting as a precursor, it showcases a company’s operational performance. Net income, situated at the bottom line, summarizes the overall profitability, encapsulating all financial elements.
Use in Financial Analysis
Operating income proves invaluable for analysts assessing a company’s operational efficiency and profitability, especially when using tools like Power BI for income statement analysis. It allows a deep dive into the core of a business. Conversely, net income is a pivotal metric for a broader financial analysis. It is handy for investors to evaluate a company’s overall financial performance analysis and attractiveness.
Operating Income vs. Net Income: Comparison
Understanding the distinction between these financial metrics is like deciphering the secret language of profitability. It enables us to grasp the true essence of a company’s financial performance.
Understand the definitions: Begin by grasping the operating income and net income definitions. Operating income focuses on core operations, while net income encompasses all revenues and expenses.
Review the income statement: Examine the company’s income statement. Operating income is a precursor to net income, providing insight into operational performance before broader financial considerations.
Assess operational efficiency: Evaluate operational efficiency using operating income. This metric isolates the profitability derived exclusively from day-to-day business activities, providing insights into the company’s operational health.
Consider non-operating items: Factor in non-operating items when looking at net income. Understand how interest, taxes, and other non-operational elements influence the company’s overall profitability.
Evaluate overall profitability: While operating income sheds light on operational success, net income gives a broader perspective on overall profitability. Assess both metrics to understand the company’s holistic financial standing.
Analyze trends over time: Study trends over time. Analyzing operating and net income trends unveils patterns and offers insights into a company’s financial trajectory.
Assess non-operating impact: Distinguish between the impact of operating and non-operating factors. Identify areas where non-operating items significantly contribute or detract from the overall financial performance.
Consider industry benchmarks: Benchmark your company’s operating and net income against industry standards. This comparative analysis provides context, indicating how the company fares within its sector.
Evaluate profit margins: Calculate profit margins using both metrics. Operating margin gauges operational efficiency, while net margin reflects overall profitability, aiding in a comprehensive assessment.
Understand the context: Lastly, understand the broader context. Consider the company’s strategic goals, market conditions, and industry nuances to interpret the significance of these metrics. This contextual understanding is crucial for a comprehensive comparison.
How to Evaluate the Results of Operating Income vs. Net Income?
Data visualization is the superhero of data analysis, saving us from drowning in a sea of numbers. Charts and graphs help to spot trends and outliers in data, making it accessible and understandable.
But alas, the trusty sidekick, Excel, falls short when visualizing complex data relationships.
So, how can we analyze net income vs operating income in Excel?
ChartExpo.
ChartExpo creates stunning, insightful data visualizations that make data come alive. Consequently, you can easily analyze operating income vs. net income in Excel.
Let’s learn how to install ChartExpo in Excel
Open your Excel application.
Open the worksheet and click the “Insert” menu.
You’ll see the “My Apps” option.
In the office Add-ins window, click “Store” and search for ChartExpo on my Apps Store.
Click the “Add” button to install ChartExpo in your Excel.
ChartExpo charts are available both in Google Sheets and Microsoft Excel. Please use the following CTA’s to install the tool of your choice and create beautiful visualizations in a few clicks in your favorite tool.
Assume you have the financial data table below.
| | | | |
--- --- |
| Business Unit | Operating Income | Net Income | Profit Margin |
| Unit A | 500000 | 400000 | 20 |
| Unit B | 600000 | 450000 | 25 |
| Unit C | 700000 | 500000 | 30 |
| Unit D | 800000 | 550000 | 28 |
| Unit E | 900000 | 600000 | 33 |
Follow these steps to visualize this data with ChartExpo and glean the operating income vs. net income insights.
To get started with ChartExpo, install ChartExpo in Excel.
Now Click on My Apps from the INSERT menu.
Choose ChartExpo from My Apps, then click Insert.
Once it loads, scroll through the charts list to locate and choose the “Multi-Axis Line Chart”.
Click the “Create Chart From Selection” button after selecting the data from the sheet, as shown.
ChartExpo will generate the visualization below for you.
Click on Settings and change the “Data Representation” of Operating Income into Bar as follows.
If you want to add anything to the chart, click the Edit Chart button:
Click the pencil icon next to the Chart Header to change the title.
It will open the properties dialog. Under the Text section, you can add a heading in Line 1 and enable Show.
Give the appropriate title of your chart and click the Apply button.
Let’s add the Prefix (e.g., $ sign) with the “Net Income” values.
Click the highlighted pencil icon. Expand the “Label” properties to add the Prefix value (e.g., $ sign).
Let’s add the $ sign with the “Operating Income” values.
Click the highlighted pencil icon. Expand the “Label” properties to add the Prefix value ($ sign).
You can change the precision value of the Profit Margin to zero:
Let’s add the % sign with the “Profit Margin” values.
Click the highlighted pencil icon. Expand the “Label” properties to add the Prefix value ($ sign).
Click on Save Changes and again click on Edit for more changes as follows:
Change the Legend shape of Operating Income into a column and click the “Apply” button.
Change the Legend shape of Net Income to Line and Circle and click the “Apply” button.
Click the “Save Changes” button to persist the changes made to the chart.
Your final chart will look like the one below.
Insights
Unit E is the most lucrative, boasting a 33% profit margin.
Unit C follows with a 30% margin, while Unit A, B, and D trail with 20%, 25%, and 28% margins, respectively.
Despite variations, all units contribute to the company’s financial success. Unit E is the top performer, closely followed by Unit C.
How To Manage Operating Income And Net Income?
Monitor Revenue and Expenses
Operating Income: Focus on increasing revenue while controlling costs of goods sold (COGS) and operating expenses.
Net Income: Track all expenses, including interest and taxes, to ensure they don’t disproportionately impact your profit.
Optimize Costs
Operating Income: Regularly review and optimize operational costs such as labor, materials, and overhead.
Net Income: Look for ways to reduce interest expenses and tax liabilities through effective financial management and tax planning.
Improve Efficiency
Operating Income: Enhance business processes and productivity to increase efficiency and reduce waste.
Net Income: Implement cost-control measures and streamline operations to improve overall profitability.
Forecast and Budget
Operating Income: Create accurate forecasts and budgets to plan for expected revenues and expenses.
Net Income: Include all financial aspects in your budget, such as debt service and tax obligations, to prevent unexpected financial shortfalls.
Regular Financial Review
Operating Income: Conduct periodic reviews of income statements to assess the performance of core operations.
Net Income: Analyze the impact of all financial activities, including interest and taxes, on the overall profitability.
Adjust Pricing Strategies
Operating Income: Reevaluate pricing strategies to ensure they cover costs and align with market conditions.
Net Income: Ensure pricing adjustments also account for changes in cost structure and tax implications.
Strategic Investments
Operating Income: Invest in growth opportunities that enhance core operations and generate additional revenue.
Net Income: Make strategic investments that provide long-term benefits and optimize returns after accounting for all financial obligations.
Operating Income vs. Net Income FAQ
Is net income and operating income the same?
No, they differ. Operating income focuses on the core financial operations of a business, excluding non-operational elements. Conversely, net income encompasses all aspects, offering a comprehensive financial overview.
Why is operating income better than net income?
Operating income hones in on core operations, providing a purer measure of operational efficiency. Net income includes non-operational elements, offering a broader but sometimes muddled financial picture.
What is operating income vs. net income?
Operating income reflects a company’s profitability before taxes and interest, isolating core operational efficiency. Conversely, net income encompasses all revenue and expenses, providing a comprehensive financial snapshot, including non-operational elements.
Wrap Up
Operating income vs. net income serves distinct roles in financial analysis. Operating income reflects a company’s core profitability, excluding interest and taxes. Net income, encompassing all costs, provides a comprehensive overview of overall financial performance.
Analyzing both metrics in a spreadsheet aids in dissecting a company’s financial health and facilitates strategic decision-making. Moreover, understanding the differences clarifies the impact of financial activities on the company’s bottom line.
The spreadsheet’s versatility allows for detailed analysis, providing insights into income components and profitability drivers. Accurate analysis facilitates better-informed decisions and strategic planning for sustainable financial growth. Incorporating visualizations enhances data interpretation, making financial insights more accessible.
While spreadsheet offers robust number-crunching capabilities, it falters in the visualization department. Enter ChartExpo, the visual maestro that transforms this numerical dance into a vibrant spectacle. ChartExpo gives life to your financial data, unlocking new dimensions of comprehension and empowering informed decision-making.
Do not hesitate.
Let the numbers groove and the insights flow with ChartExpo as your guide in the financial analysis arena.
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14493 | https://www.news-medical.net/life-sciences/Northern-Blot-RNA-Blot.aspx | Published Time: 2016-04-25T23:43:00-04:00
Northern Blot / RNA Blot
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3Currently rated 4.7 by 3 people
1
Northern Blot / RNA Blot
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By Susha Cheriyedath, M.Sc.
Introduction
The first blotting technique developed by Dr. Edwin Southern is called the Southern blot. This technique was subsequently modified so as to detect other target molecules. These techniques being variations of Southern blot were named northern blot, western blot, and eastern blot. Northern blot is used to detect specific RNA sequences in a sample. Therefore it is also called the RNA blot.
Pimp your brain: Northern Blot Play
Principle
In northern blot, the sample RNA is separated on the basis of size with the help of gel electrophoresis. The separated RNA fragments are transferred to a support membrane and then treated with a labeled DNA probe. If the sample contains the complementary RNA sequence, the probe will bind to the membrane and it can be visualized using various methods.
Procedure
The key steps in the northern blot technique are as follows:
The RNA sample is first separated by size using agarose gel electrophoresis. RNA molecules form streaks rather than bands on the gel as there are several small fragments of RNA.
The RNA molecules are then transferred to a special blotting paper usually made of nitrocellulose. Membranes made up of nylon can also be used. The separation pattern of the RNA molecules in the blotting paper remains the same as that in the gel.
The blot is then exposed to a labeled, single-stranded DNA probe. The bases in this probe will pair with their complementary RNA sequences in the blot producing a double-stranded DNA-RNA molecule. Although the probe cannot be seen at this stage, since it is labeled with an enzyme or radioactive tag, it can be seen after appropriate treatment in the next step.
Next, the probe is exposed to a colorless substrate which gets converted by the enzyme to a colored product and is visible on an X-ray film. In case the probe is radioactive, it can directly be seen on the X-ray film.
Applications
Some of the key applications of northern blotting are listed below:
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Gene expression studies – to observe overexpression of cancer-causing genes and gene expression in case of transplant rejection
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Limitations of northern blot
A few limitations of the northern blot technique are discussed below:
The standard northern blot method has lower sensitivity when compared to that of RT-PCR and nuclease protection assays.
The technique requires a large amount of target RNA sample sequence while newer techniques like real time RT-PCR need only a single copy of RNA for amplification.
The northern blot technique is very sensitive to even slight degradation of the RNA samples. Degradation drastically affects data quality and quantitation.
The technique becomes laborious in cases where multiple probes need to be added. A second probe can be added to the system only after removing the first probe completely. This can make the process highly complex. After treating the system with harsh chemicals to remove the first probe, the accuracy of the data from the subsequent rounds may also be compromised.
Northern blot can only measure steady state mRNA accumulation and not RNA stability or transcription rates.
Summary
The northern blot technique is a classical way of analyzing the presence of a specific RNA sequence in a sample. Northern blotting is relatively cheap and simple to perform in the lab. Recent advancements in buffers and hybridization membranes have enabled high sensitivity blotting. However, breakthroughs in PCR technology in recent times have enabled a much more simple, quick, and precise identification as well as quantitation of RNA.
References
Further Reading
All RNA Content
mRNA Medicine: what’s next after the COVID-19 vaccine?
What is RNA?
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Types of RNA: mRNA, rRNA and tRNA
More...
Last Updated: Aug 23, 2018
Currently rated 4.7 by 3 people
Written by
Susha Cheriyedath
Susha is a scientific communication professional holding a Master's degree in Biochemistry, with expertise in Microbiology, Physiology, Biotechnology, and Nutrition. After a two-year tenure as a lecturer from 2000 to 2002, where she mentored undergraduates studying Biochemistry, she transitioned into editorial roles within scientific publishing. She has accumulated nearly two decades of experience in medical communication, assuming diverse roles in research, writing, editing, and editorial management.
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Please use one of the following formats to cite this article in your essay, paper or report:
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Cheriyedath, Susha. (2018, August 23). Northern Blot / RNA Blot. News-Medical. Retrieved on August 31, 2025 from
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Alzheimer’s disease is a brain disorder that slowly destroys memory and thinking skills, and eventually, the ability to carry out the simplest tasks. In most people with Alzheimer’s, symptoms first appear later in life. Estimates vary, but experts suggest that more than 6 million Americans, most of them age 65 or older, may have Alzheimer’s.
Alzheimer’s is currently ranked as the seventh leading cause of death in the United States and is the most common cause of dementia among older adults.
Dementia is the loss of cognitive functioning — thinking, remembering, and reasoning — and behavioral abilities to such an extent that it interferes with a person’s daily life and activities. Dementia ranges in severity from the mildest stage, when it is just beginning to affect a person’s functioning, to the most severe stage, when the person must depend completely on others for help with basic activities of daily living.
Share this infographic and help spread the word about understanding different types of dementia.
The causes of dementia can vary depending on the types of brain changes that may be taking place. Other forms of dementia include Lewy body dementia, frontotemporal disorders, and vascular dementia. It is common for people to have mixed dementia — a combination of two or more types of dementia. For example, some people have both Alzheimer’s and vascular dementia.
Alzheimer’s disease is named after Dr. Alois Alzheimer. In 1906, Dr. Alzheimer noticed changes in the brain tissue of a woman who had died of an unusual mental illness. Her symptoms included memory loss, language problems, and unpredictable behavior. After she died, he examined her brain and found many abnormal clumps (now called amyloid plaques) and tangled bundles of fibers (now called neurofibrillary, or tau, tangles).
These plaques and tangles in the brain are still considered some of the main features of Alzheimer’s. Another feature is the loss of connections between neurons in the brain. Neurons transmit messages between different parts of the brain, and from the brain to muscles and organs in the body.
Participating in Alzheimer's disease clinical trials
Everybody — those with Alzheimer’s or mild cognitive impairment as well as healthy volunteers with or without a family history of Alzheimer’s — may be able to take part in clinical trials and studies. Participants in Alzheimer’s clinical research can help scientists learn how the brain changes in healthy aging and in Alzheimer’s.
Many volunteers are needed to participate in the hundreds of active clinical trials and studies that are testing ways to better understand, diagnose, treat, and prevent Alzheimer’s. Researchers need participants of different ages, sexes, races, and ethnicities to ensure that results are meaningful for many people. To learn more about clinical trials, watch this video from the NIH National Library of Medicine.
NIA leads the federal government’s research efforts on Alzheimer’s. NIA-funded Alzheimer’s Disease Research Centers throughout the United States conduct a wide range of research, including trials and studies of the causes, diagnosis, and management of the disease. NIA also sponsors the Alzheimer’s Clinical Trials Consortium, which is designed to accelerate and expand research and therapies in Alzheimer’s and related dementias.
To learn more about Alzheimer’s clinical trials and studies:
Talk to your health care provider about local studies that may be right for you.
Search the Alzheimers.gov Clinical Trials Finder for options near you or sign up for email alerts about new trials and studies.
Sign up for a registry or matching service to be invited to participate in trials and studies.
Contact an Alzheimer’s Disease Research Center or a memory or neurology clinic in your community.
Learn more about participating in clinical trials. Watch videos of participants in Alzheimer’s clinical trials talking about their experiences.
How does Alzheimer's disease affect the brain?
Scientists continue to unravel the complex brain changes involved in Alzheimer’s. Changes in the brain may begin a decade or more before symptoms appear. During this very early stage of Alzheimer’s, toxic changes are taking place in the brain, including abnormal buildups of proteins that form amyloid plaques and tau tangles. Previously healthy neurons stop functioning, lose connections with other neurons, and die. Many other complex brain changes are thought to play a role in Alzheimer’s as well.
The damage initially appears to take place in the hippocampus and the entorhinal cortex, which are parts of the brain that are essential in forming memories. As more neurons die, additional parts of the brain are affected and begin to shrink. By the final stage of Alzheimer’s, damage is widespread and brain tissue has shrunk significantly.
Signs and symptoms of Alzheimer's disease
Memory problems are typically one of the first signs of cognitive impairment related to Alzheimer’s. Some people with memory problems have a condition called mild cognitive impairment (MCI). With MCI, people have more memory problems than normal for their age, but their symptoms do not interfere with their everyday lives. Movement difficulties and problems with the sense of smell have also been linked to MCI. Older people with MCI are at greater risk for developing Alzheimer’s, but not all of them do so. Some may even revert to normal cognition.
The first symptoms of Alzheimer’s vary from person to person. For many, decline in nonmemory aspects of cognition, such as word finding, vision/spatial issues, and impaired reasoning or judgment may signal the very early stages of the disease. Researchers are studying biomarkers (biological signs of disease found in brain images, cerebrospinal fluid, and blood) to detect early changes in the brains of people with MCI and in cognitively normal people who may be at greater risk for Alzheimer’s. More research is needed before these techniques can be used broadly and routinely to diagnose Alzheimer’s in a health care provider’s office.
Stages of Alzheimer's disease
Mild Alzheimer’s disease
As Alzheimer’s worsens, people experience greater memory loss and other cognitive difficulties. Problems can include wandering and getting lost, trouble handling money and paying bills, repeating questions, taking longer to complete normal daily tasks, and personality and behavior changes. People are often diagnosed in this stage.
Moderate Alzheimer’s disease
In this stage, damage occurs in areas of the brain that control language, reasoning, conscious thought, and sensory processing, such as the ability to correctly detect sounds and smells. Memory loss and confusion grow worse, and people begin to have problems recognizing family and friends. They may be unable to learn new things, carry out multistep tasks such as getting dressed, or cope with new situations. In addition, people at this stage may have hallucinations, delusions, and paranoia and may behave impulsively.
Severe Alzheimer’s disease
Ultimately, plaques and tangles spread throughout the brain, and brain tissue shrinks significantly. People with severe Alzheimer’s cannot communicate and are completely dependent on others for their care. Near the end of life, the person may be in bed most or all of the time as the body shuts down.
What causes Alzheimer’s disease?
In recent years, scientists have made tremendous progress in better understanding Alzheimer’s and the momentum continues to grow. Still, scientists don’t yet fully understand what causes Alzheimer’s disease in most people. The causes probably include a combination of age-related changes in the brain, along with genetic, environmental, and lifestyle factors. The importance of any one of these factors in increasing or decreasing the risk of developing Alzheimer’s may differ from person to person.
The basics of Alzheimer’s disease
Scientists are conducting studies to learn more about plaques, tangles, and other biological features of Alzheimer’s. Advances in brain imaging techniques enable researchers to see the development and spread of abnormal amyloid and tau proteins in the living brain, as well as changes in brain structure and function. Scientists are also exploring the very earliest steps in the disease process by studying changes in the brain and body fluids that can be detected years before Alzheimer’s symptoms appear. Findings from these studies will help improve our understanding of the causes of Alzheimer’s and make diagnosis easier.
One of the great mysteries of Alzheimer’s is why it largely affects older adults. Research on normal brain aging is exploring this question. For example, scientists are learning how age-related changes in the brain may harm neurons and affect other types of brain cells to contribute to Alzheimer’s damage. These age-related changes include atrophy (shrinking) of certain parts of the brain, inflammation, blood vessel damage, production of unstable molecules called free radicals, and mitochondrial dysfunction (a breakdown of energy production within a cell).
Alzheimer's disease genetics
In most cases, Alzheimer’s does not have a single genetic cause. Instead, it is likely influenced by multiple genes in combination with lifestyle and environmental factors. Changes in genes, called genetic variations, may increase or decrease a person’s risk of developing the disease.
Scientists currently know of more than 80 genetic regions associated with Alzheimer’s. Of the genetic variants associated with Alzheimer’s so far, only three are known to cause the disease. Although it happens rarely, when someone inherits an altered version of one of these genes — APP, PSEN1, or PSEN2 — they will likely develop Alzheimer’s before age 65 and sometimes much earlier.
People with Down syndrome also have a higher risk of developing Alzheimer’s earlier in life. Down syndrome results from having an extra chromosome 21, which carries the APP gene that produces the amyloid precursor protein. Too much of this protein leads to build-up of beta-amyloid plaques in the brain. Estimates suggest that 50% or more of people living with Down syndrome will develop Alzheimer’s with symptoms appearing in their 50s and 60s.
Another genetic variation, in the APOE gene, which has several forms, is known to influence the risk of Alzheimer’s. Specifically, APOE ε4 increases a person’s risk of developing Alzheimer’s and is also associated with developing Alzheimer’s earlier in life for certain populations. APOE ε2 may provide some protection against Alzheimer’s.
Changes in different genes, along with other biomedical, lifestyle, and environmental factors, play a role in potentially developing Alzheimer’s. Still, it is never known for certain if any individual will or will not develop the disease.
For more about Alzheimer’s genetics research, see NIA’s Alzheimer’s Disease Genetics Fact Sheet.
Health, environmental, and lifestyle factors
Research suggests that a host of factors beyond genetics may play a role in the development and course of Alzheimer’s. There is a great deal of interest, for example, in the relationship between cognitive decline and vascular conditions, such as heart disease, stroke, and high blood pressure, as well as metabolic diseases, such as diabetes and obesity. Ongoing research will help us understand whether and how reducing risk factors for these conditions may also reduce the risk of Alzheimer’s.
A nutritious diet, physical activity, social engagement, and mentally stimulating pursuits have all been associated with helping people stay healthy as they age. These factors might also help reduce the risk of cognitive decline and Alzheimer’s. Researchers are testing some of these possibilities in clinical trials.
How is Alzheimer’s disease diagnosed?
Doctors use several methods and tools to help determine whether a person who is having memory problems has Alzheimer’s.
To diagnose Alzheimer’s, doctors may:
Ask the person and a family member or friend questions about overall health, use of prescription and over-the-counter medicines, diet, past medical problems, ability to carry out daily activities, and changes in behavior and personality.
Conduct tests of memory, problem solving, attention, counting, and language.
Order blood, urine, and other standard medical tests to help identify other possible causes of the problem.
Administer tests to determine if depression or another mental health condition is causing or contributing to a person’s symptoms.
Collect cerebrospinal fluid via a spinal tap or order blood tests to measure the levels of proteins associated with Alzheimer’s and related dementias.
Perform brain scans, such as CT, MRI, or PET (positron emission tomography), to support an Alzheimer’s diagnosis or to rule out other possible causes for symptoms.
These tests may be repeated to give doctors information about how the person’s memory and other cognitive functions are changing over time.
People with memory and thinking concerns should talk to their doctor to find out whether their symptoms are due to Alzheimer’s or to another cause, such as stroke, tumor, Parkinson’s disease, sleep disturbances, side effects of medication, an infection, or another type of dementia. Some of these conditions may be treatable and, possibly, reversible.
If the diagnosis is Alzheimer’s, beginning treatment as early as possible in the disease process may help preserve daily functioning for a while. An early diagnosis also helps families plan for the future. They can take care of financial and legal matters, address potential safety issues, learn about living arrangements, and develop support networks.
In addition, an early diagnosis provides people with more opportunities to participate in clinical trials or studies testing possible new treatments for Alzheimer’s.
For more information, visit How Is Alzheimer’s Disease Diagnosed?
How is Alzheimer’s disease treated?
Alzheimer’s is complex, and it is therefore unlikely that any one drug or other intervention will successfully treat it in all people living with the disease. In ongoing clinical trials, scientists are developing and testing several possible treatment interventions.
While there is currently no cure for Alzheimer’s, medications are emerging to treat the progression of the disease by targeting its underlying causes. There are also medications that may temporarily improve or stabilize memory and thinking skills in some people and may help manage certain symptoms and behavioral problems.
Additionally, people with Alzheimer’s also may experience sleeplessness, depression, anxiety, agitation, and other behavioral and psychological symptoms. Scientists continue to research why these symptoms occur and are exploring new medications and non-drug strategies to manage them. Research shows that treating these symptoms may make people with Alzheimer’s feel more comfortable and also help their caregivers. Antidepressants, antipsychotics, and anti-anxiety drugs may be helpful for some people with Alzheimer’s, but experts agree that these medicines should be used only after other strategies to promote physical and emotional comfort, such as avoiding stressful situations, have been tried. It’s important to talk with a doctor about what treatment will be most effective in your situation.
For more information, visit How Is Alzheimer's Disease Treated?
Clinical trials on Alzheimer’s disease treatments
Volunteers are needed for clinical trials that are testing ways to treat Alzheimer’s. By joining one of these studies, you may help scientists discover new Alzheimer’s treatments and contribute useful information to help people living with Alzheimer's disease. Find clinical trials near you
Support for families and Alzheimer's disease caregivers
Caring for a person with Alzheimer’s can have significant physical, emotional, and financial costs. The demands of day-to-day care, changes in family roles, and decisions about placement in a care facility can be difficult. NIA supports efforts to evaluate programs, strategies, approaches, and other research to improve the quality of care and life for those living with dementia and their caregivers.
Becoming well-informed about the disease is one important long-term strategy. Programs that teach families about the various stages of Alzheimer’s and about ways to deal with difficult behaviors and other caregiving challenges can help.
Good coping skills, a strong support network, and respite care are other things that may help caregivers handle the stress of caring for a loved one with Alzheimer’s. For example, staying physically active provides physical and emotional benefits.
Some caregivers have found that joining a support group is a critical lifeline. These support groups enable caregivers to find respite, express concerns, share experiences, get tips, and receive emotional comfort. Many organizations sponsor in-person and online support groups, including groups for people with early-stage Alzheimer’s and their families.
For more information, see Alzheimer’s Caregiving.
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Email Address
For more information about Alzheimer’s disease
NIA Alzheimer’s and related Dementias Education and Referral (ADEAR) Center
800-438-4380
adear@nia.nih.gov
www.nia.nih.gov/alzheimers
The NIA ADEAR Center offers information and publications for download (PDF) about Alzheimer’s and related dementias for families, caregivers, and health professionals. ADEAR Center staff answer telephone, email, and written requests and make referrals to local and national resources.
Alzheimers.gov
www.alzheimers.gov
Explore the Alzheimers.gov website for information and resources on Alzheimer’s and related dementias from across the federal government.
Eldercare Locator
800-677-1116
eldercarelocator@USAging.org
MedlinePlus
National Library of Medicine
www.medlineplus.gov
Alzheimer's Association
800-272-3900
866-403-3073 (TTY)
info@alz.org
www.alz.org
Alzheimer’s Foundation of America
866-232-8484
info@alzfdn.org
www.alzfdn.org
This content is provided by the NIH National Institute on Aging (NIA). NIA scientists and other experts review this content to ensure it is accurate and up to date.
Content reviewed: April 5, 2023
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14495 | https://www.socratica.com/pages/platonic-solids | Skip to main content
What is a Platonic solid?
And what is a Polyhedron?
A polyhedron is a 3D shape that consists of a bunch of polygons glued together at the edges to form a volume so there are no gaps, openings, or overlaps.
A regular polyhedron is a polyhedron where all of the faces are the same shape and size, and it has the same number of polygons meeting at each corner.
A Platonic solid is one of the 5 regular polyhedron that are "convex". That is, there are no inward dips on the outside.
Polyhedra Vocabulary
Three Key Words
V = Vertices
A point on a polyhedron is called a vertex. A point is 0-dimensional, so a vertex does not have a size. The number of vertices in a polyhedron is denoted by "V".
### E = Edges
A line segment connecting two vertices is called an edge. Edges are 1-dimensional, and they have a length. In math, people use "E" for the number of edges.
### F = Faces
The polygons that encase a polyhedron are called faces. In a Platonic solid, each face is a regular polygon and all the faces are identical. The number of faces is denoted by "F".
Remember
Vertices are the corner points where edges meet.
Edges are the straight lines connecting two vertices.
Faces are the flat surfaces enclosed by edges.
Platonic Solids
The 5 Regular Polyhedra
The ancient scholar Plato believed the universe was built from the 5 solids. Four of the solids were used for earth, air, water, and fire, while the remainder was "the fifth element". This is why they are called the Platonic solids.
Tetrahedron
4 vertices, 6 edges, 4 faces
The tetrahedron is the simplest of all Platonic solids. The ancient Greek philosopher Plato associated it with the element fire due to its sharp and piercing nature. The tetrahedron is not a pyramid - a pyramid has a square bottom and triangular sides, but ALL sides of a tetrahedron are triangles.
Cube
8 vertices, 12 edges, 6 faces
The cube, also known as a hexahedron if you're feeling fancy, is the most familiar Platonic solid. Be careful to not call it a box! Boxes can be short, thin, and a wide variety of shapes. A cube has 6 identical square faces.
Octahedron
6 vertices, 12 edges, 8 faces
The prefix "oct" means 8, as in octopus (8 legs), octagon (8 sides), and octave (8 notes). The Octahedron can be described as two pyramids joined at their bases. One place in nature where you will find octahedrons is in crystals! Pyrite, diamonds, and other crystals often form octahedrons.
Icosahedron
12 vertices, 30 edges, 20 faces
The icosahedron is composed of 20 equilateral triangles. The word is a combination of the Greek "eíkosi" which means twenty, and "hedra" meaning face. This solid has the highest number of faces among the 5 Platonic solids. It makes you wonder - if you were to roll all 5 Platonic solids, would the icosahedron roll the longest?
Dodecahedron
20 vertices, 30 edges, 12 faces
The dodecahedron is built using 12 equal sized pentagons. This solid is often associated with the cosmos or universe in Platonic philosophy, symbolizing a sense of mystery and wonder. The name comes from the Greek words for "twelve faces", or "dodeka" + "hedra".
Euler's Formula
A Connection between V, E, and F
The mathematician Leonhard Euler showed that for every convex polyhedron, the number of vertices (V), edges (E), and faces (F) have this relationship:
V−E+F=2
For example, a dodecahedron has V=20, E=30, and F=12. Plugging these in gives you:
20−30+12=2
If the polyhedron is not convex or has holes, the relationship may differ, but for simple convex polyhedra like the Platonic solids, Euler's formula holds true.
Why are there only 5 Platonic Solids?
Considering that a platonic solid is a convex polyhedron in which faces are congruent regular polygons and there's the same number of faces meeting at each vertex, the following limit the number of possible Platonic solids:
Regular Polygon Face: The interior angle of a regular polygon must be less than 360∘ because at least three faces must meet at each vertex to form a 3D shape.
Angle Sum Around a Vertex: For a convex polyhedron, the sum of the angles around a vertex must be less than 360∘. This condition limits the number of faces that can meet at a vertex depending on the type of polygon that forms the faces.
Let's go through the possible polygons:
Equilateral Triangle (interior angle = 60):
Three triangles: 3×60=180∘ (forms a tetrahedron)
Four triangles: 4×60∘=240∘ (forms an octahedron)
Five triangles: 5×60∘=300∘ (forms an icosahedron)
Square (interior angle = 90):
Three squares: 3×90∘=270∘ (forms a cube or regular hexahedron)
Pentagon (interior angle = 108∘):
Three pentagons: 3×108∘=324∘ (forms a dodecahedron)
Hexagon (interior angle = 120∘) or polygons with more sides:
Three hexagons: 3×120∘=360∘ (flat tiling, does not form a 3D shape)
As the number of sides of the polygon increases, the interior angle also increases, and the sum of the angles around a vertex exceeds 360∘, making it impossible to form a 3D shape.
Thus, there are only five combinations of regular polygons that meet the criteria for Platonic solids, resulting in the five unique Platonic solids:
Tetrahedron: 4 faces, each an equilateral triangle.
Cube (Hexahedron): 6 faces, each a square.
Octahedron: 8 faces, each an equilateral triangle.
Dodecahedron: 12 faces, each a regular pentagon.
Icosahedron: 20 faces, each an equilateral triangle.
Nets
A net of a three-dimensional object is a two-dimensional representation that, when folded along specific lines or edges, forms the 3D object. It is essentially a flat layout of all the faces of the object, connected in such a way that they can be folded into the shape without any overlaps or gaps. For example, the net of a cube consists of six squares arranged in a cross-like pattern, where each square represents one face of the cube. When folded correctly, the squares come together to form the cube.
Nets are a valuable tool in geometry and spatial reasoning because they help visualize and understand how 3D objects are constructed from their surfaces. They are also used to calculate surface area, as each face of the object is represented in the net, allowing for easy measurement and summation of the areas. Understanding nets can provide deeper insights into the properties of shapes and their dimensions, making them an essential concept in mathematics and engineering.
Here are thenets of the 5 platonic solids.
Review
| | | | | |
--- ---
| Platonic Solid | Number of Faces | Face Shape | Number of Vertices | Number of Edges |
| Tetrahedron | 4 | Equilateral Triangle | 4 | 6 |
| Cube (Hexahedron) | 6 | Square | 8 | 12 |
| Octahedron | 8 | Equilateral Triangle | 6 | 12 |
| Icosahedron | 20 | Equilateral Triangle | 12 | 30 |
| Dodecahedron | 12 | Regular Pentagon | 20 | 30 |
“In the grand blueprint of the cosmos, the gods employed geometry as their divine language. The tetrahedron sparked the flames of fire, the cube laid the earth's foundation, the octahedron whispered to the winds, and the icosahedron flowed through the waters. As for the dodecahedron, the gods used it to roll the dice on the universe's fate. Such is the playful wit of celestial architects.”
Plato's "Lost" Tweets
The Simplex
In mathematics, a simplex is a generalization of a triangle and tetrahedron. The first four simplexes are made from familiar shapes: a point, a line segment, a triangle, and a tetrahedron.
Learn More |
14496 | https://math.libretexts.org/Bookshelves/Differential_Equations/Introduction_to_Partial_Differential_Equations_(Herman)/06%3A_Problems_in_Higher_Dimensions/6.09%3A_Curvilinear_Coordinates | 6.9.1
6.9.2
6.9.3
6.9.4
6.9.5
6.9.6
6.9.1
6.9.2
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6.9: Curvilinear Coordinates
Last updated
: Sep 4, 2024
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6.8: Schrödinger Equation in Spherical Coordinates
6.10: Problems
Page ID
: 90958
Russell Herman
University of North Carolina Wilmington
( \newcommand{\kernel}{\mathrm{null}\,})
In order to study solutions of the wave equation, the heat equation, or even Schrödinger’s equation in different geometries, we need to see how differential operators, such as the Laplacian, appear in these geometries. The most common coordinate systems arising in physics are polar coordinates, cylindrical coordinates, and spherical coordinates. These reflect the common geometrical symmetries often encountered in physics.
In such systems it is easier to describe boundary conditions and to make use of these symmetries. For example, specifying that the electric potential is 10.0 V10.0 V on a spherical surface of radius one, we would say ϕ(x,y,z)=10ϕ(x,y,z)=10 for x2+y2+z2=1x2+y2+z2=1. However, if we use spherical coordinates, (r,θ,ϕ)(r,θ,ϕ), then we would say ϕ(r,θ,ϕ)=10ϕ(r,θ,ϕ)=10 for r=1r=1, or ϕ(1,θ,ϕ)=10ϕ(1,θ,ϕ)=10. This is a much simpler representation of the boundary condition.
However, this simplicity in boundary conditions leads to a more complicated looking partial differential equation in spherical coordinates. In this section we will consider general coordinate systems and how the differential operators are written in the new coordinate systems. This is a more general approach than that taken earlier in the chapter. For a more modern and elegant approach, one can use differential forms.
We begin by introducing the general coordinate transformations between Cartesian coordinates and the more general curvilinear coordinates. Let the Cartesian coordinates be designated by (x1,x2,x3)(x1,x2,x3) and the new coordinates by (u1,u2,u3)(u1,u2,u3). We will assume that these are related through the transformations
x1=x1(u1,u2,u3)x2=x2(u1,u2,u3)x3=x3(u1,u2,u3)
x1=x1(u1,u2,u3)x2=x2(u1,u2,u3)x3=x3(u1,u2,u3)(6.9.1)
Thus, given the curvilinear coordinates (u1,u2,u3)(u1,u2,u3) for a specific point in space, we can determine the Cartesian coordinates, (x1,x2,x3)(x1,x2,x3), of that point. We will assume that we can invert this transformation: Given the Cartesian coordinates, one can determine the corresponding curvilinear coordinates.
In the Cartesian system we can assign an orthogonal basis, {i,j,k}{i,j,k}. As a particle traces out a path in space, one locates its position by the coordinates (x1,x2,x3)(x1,x2,x3). Picking x2x2 and x3x3 constant, the particle lies on the curve x1=x1= value of the x1x1 coordinate. This line lies in the direction of the basis vector ii. We can do the same with the other coordinates and essentially map out a grid in three dimensional space as sown in Figure 6.9.16.9.1. All of the xi−xi− curves intersect at each point orthogonally and the basis vectors {i,j,k}{i,j,k} lie along the grid lines and are mutually orthogonal. We would like to mimic this construction for general curvilinear coordinates. Requiring the orthogonality of the resulting basis vectors leads to orthogonal curvilinear coordinates.
As for the Cartesian case, we consider u2u2 and u3u3 constant. This leads to a curve parametrized by u1:r=x1(u1)i+x2(u1)j+x3(u1)ku1:r=x1(u1)i+x2(u1)j+x3(u1)k. We call this the u1u1-curve. Similarly, when u1u1 and u3u3 are constant we obtain a u2u2-curve and for u1u1 and u2u2 constant we obtain a u3u3-curve. We will assume that these curves intersect such that each pair of curves intersect orthogonally as seen in Figure 6.9.26.9.2. Furthermore, we will assume that the unit tangent vectors to these curves form a right handed system similar to the {i,j,k}{i,j,k} systems for Cartesian coordinates. We will denote these as {ˆu1,ˆu2,ˆu3}{u^1,u^2,u^3}.
We can determine these tangent vectors from the coordinate transformations. Consider the position vector as a function of the new coordinates,
r(u1,u2,u3)=x1(u1,u2,u3)i+x2(u1,u2,u3)j+x3(u1,u2,u3)k.
r(u1,u2,u3)=x1(u1,u2,u3)i+x2(u1,u2,u3)j+x3(u1,u2,u3)k.
Then, the infinitesimal change in position is given by
dr=∂r∂u1du1+∂r∂u2du2+∂r∂u3du3=3∑i=1∂r∂uidui.
dr=∂r∂u1du1+∂r∂u2du2+∂r∂u3du3=∑i=13∂r∂uidui.
We note that the vectors ∂r∂ui∂r∂ui are tangent to the uiui-curves. Thus, we define the unit tangent vectors
ˆui=∂r∂ui|∂r∂ui|.
u^i=∂r∂ui∣∣∂r∂ui∣∣.
Solving for the original tangent vector, we have
∂r∂ui=hiˆui,
∂r∂ui=hiu^i,
where
hi≡|∂r∂ui|.
hi≡∣∣∣∂r∂ui∣∣∣.
The hihi ’s are called the scale factors for the transformation. The infinitesimal change in position in the new basis is then given by
dr=3∑i=1hiuiˆui.
dr=∑i=13hiuiu^i.
Note
The scale factors, hi≡|∂r∂ui|hi≡∣∣∂r∂ui∣∣.
Example 6.9.16.9.1
Determine the scale factors for the polar coordinate transformation.
Solution
The transformation for polar coordinates is
x=rcosθ,y=rsinθ.
x=rcosθ,y=rsinθ.
Here we note that x1=x,x2=y,u1=rx1=x,x2=y,u1=r, and u2=θu2=θ. The u1u1-curves are curves with θ=θ= const. Thus, these curves are radial lines. Similarly, the u2u2-curves have r=r= const. These curves are concentric circles about the origin as shown in Figure 6.9.36.9.3.
The unit vectors are easily found. We will denote them by ˆuru^r and ˆuθu^θ. We can determine these unit vectors by first computing ∂r∂ui∂r∂ui. Let
r=x(r,θ)i+y(r,θ)j=rcosθi+rsinθj.
r=x(r,θ)i+y(r,θ)j=rcosθi+rsinθj.
Then,
∂r∂r=cosθi+sinθj∂r∂θ=−rsinθi+rcosθj.
∂r∂r=cosθi+sinθj∂r∂θ=−rsinθi+rcosθj.(6.9.2)
The first vector already is a unit vector. So,
ˆur=cosθi+sinθj.
u^r=cosθi+sinθj.
The second vector has length r since |−rsinθi+rcosθj|=r. Dividing ∂r∂θ by r, we have
ˆuθ=−sinθi+cosθj.
We can see these vectors are orthogonal (ˆur⋅ˆuθ=0) and form a right hand system. That they form a right hand system can be seen by either drawing the vectors, or computing the cross product,
(cosθi+sinθj)×(−sinθi+cosθj)=cos2θi×j−sin2θj×i=k.
Since
∂r∂r=ˆur,∂r∂θ=rˆuθ,
The scale factors are hr=1 and hθ=r.
Once we know the scale factors, we have that
dr=3∑i=1hiduiˆui.
The infinitesimal arclength is then given by the Euclidean line element
ds2=dr⋅dr=3∑i=1h2idu2i
when the system is orthogonal. The h2i are referred to as the metric coefficients.
Example 6.9.2
Verify that dr=drˆur+rdθˆuθ directly from r=rcosθi+rsinθj and obtain the Euclidean line element for polar coordinates.
Solution
We begin by computing
dr=d(rcosθi+rsinθj)=(cosθi+sinθj)dr+r(−sinθi+cosθj)dθ=drˆur+rdθˆuθ.
This agrees with the form dr=∑3i=1hiduiˆui when the scale factors for polar coordinates are inserted.
The line element is found as
ds2=dr⋅dr=(drˆur+rdθˆuθ)⋅(drˆur+rdθˆuθ)=dr2+r2dθ2.
This is the Euclidean line element in polar coordinates.
Also, along the ui-curves,
dr=hiduiˆui, (no summation).
This can be seen in Figure 6.9.5 by focusing on the u1 curve. Along this curve, u2 and u3 are constant. So, du2=0 and du3=0. This leaves dr=h1du1ˆu1 along the u1-curve. Similar expressions hold along the other two curves.
We can use this result to investigate infinitesimal volume elements for general coordinate systems as shown in Figure 6.9.5. At a given point (u1,u2,u3) we can construct an infinitesimal parallelepiped of sides hidui, i=1,2,3. This infinitesimal parallelepiped has a volume of size
dV=|∂r∂u1⋅∂r∂u2×∂r∂u3|du1du2du3.
The triple scalar product can be computed using determinants and the resulting determinant is call the Jacobian, and is given by
J=|∂(x1,x2,x3)∂(u1,u2,u3)|=|∂r∂u1⋅∂r∂u2×∂r∂u3|=|∂x1∂u1∂x2∂u1∂x3∂u1∂x1∂u2∂x2∂u2∂x3∂u2∂x1∂u3∂x2∂u3∂x3∂u3|.
Therefore, the volume element can be written as
dV=Jdu1du2du3=|∂(x1,x2,x3)∂(u1,u2,u3)|du1du2du3.
Example 6.9.3
Determine the volume element for cylindrical coordinates (r,θ,z), given by
x=rcosθ,y=rsinθ,z=z.
Solution
Here, we have (u1,u2,u3)=(r,θ,z) as displayed in Figure 6.9.6. Then, the Jacobian is given by
J=∣∂(x,y,z)∂(r,θ,z)∣=|∂x∂r∂y∂r∂z∂r∂x∂θ∂y∂θ∂z∂θ∂x∂z∂y∂z∂z∂z|=|cosθsinθ0−rsinθrcosθ0001|=r
Thus, the volume element is given as
dV=rdrdθdz.
This result should be familiar from multivariate calculus.
Another approach is to consider the geometry of the infinitesimal volume element. The directed edge lengths are given by dsi=hiduiˆui as seen in Figure 6.9.2. The infinitesimal area element of for the face in direction ˆuk is found from a simple cross product,
dAk=dsi×dsj=hihjduidujˆui׈uj.
Since these are unit vectors, the areas of the faces of the infinitesimal volumes are dAk=hihjduiduj.
The infinitesimal volume is then obtained as
dV=|dsk⋅dAk|=hihjhkduidujduk|ˆui⋅(ˆuk׈uj)|.
Thus, dV=h1h2h3du1du1du3. Of course, this should not be a surprise since
J=|∂r∂u1⋅∂r∂u2×∂r∂u3|=|h1ˆu1⋅h2ˆu2×h3ˆu3|=h1h2h3.
Example 6.9.4
For polar coordinates, determine the infinitesimal area element.
Solution
In an earlier example, we found the scale factors for polar coordinates as hr=1 and hθ=r. Thus, dA=hrhθdrdθ=rdrdθ. Also, the last example for cylindrical coordinates will yield similar results if we already know the scales factors without having to compute the Jacobian directly. Furthermore, the area element perpendicular to the z-coordinate gives the polar coordinate system result.
Next we will derive the forms of the gradient, divergence, and curl in curvilinear coordinates using several of the identities in section ??. The results are given here for quick reference.
Note
Gradient, divergence and curl in orthogonal curvilinear coordinates.
∇ϕ=3∑i=1ˆuihi∂ϕ∂ui=ˆu1h1∂ϕ∂u1+ˆu2h2∂ϕ∂u2+ˆu3h3∂ϕ∂u3⋅∇⋅F=1h1h2h3(∂∂u1(h2h3F1)+∂∂u2(h1h3F2)+∂∂u3(h1h2F3))∇×F=1h1h2h3|h1ˆu1h2ˆu2h3ˆu3∂∂u1∂∂u2∂∂u3F1h1F2h2F3h3|.∇2ϕ=1h1h2h3(∂∂u1(h2h3h1∂ϕ∂u1)+∂∂u2(h1h3h2∂ϕ∂u2)+∂∂u3(h1h2h3∂ϕ∂u3))
Note
Derivation of the gradient form.
We begin the derivations of these formulae by looking at the gradient, ∇ϕ, of the scalar function ϕ(u1,u2,u3). We recall that the gradient operator appears in the differential change of a scalar function,
dϕ=∇ϕ⋅dr=3∑i=1∂ϕ∂uidui.
Since
dr=3∑i=1hiduiˆui
we also have that
dϕ=∇ϕ⋅dr=3∑i=1(∇ϕ)ihidui.
Comparing these two expressions for dϕ, we determine that the components of the del operator can be written as
(∇ϕ)i=1hi∂ϕ∂ui
and thus the gradient is given by
∇ϕ=ˆu1h1∂ϕ∂u1+ˆu2h2∂ϕ∂u2+ˆu3h3∂ϕ∂u3.
Note
Derivation of the divergence form.
Next we compute the divergence,
∇⋅F=3∑i=1∇⋅(Fiˆui).
We can do this by computing the individual terms in the sum. We will compute ∇⋅(F1ˆu1).
Using Equation (6.9.17), we have that
∇ui=ˆuihi.
Then
∇u2×∇u3=ˆu2׈u3h2h3=ˆu1h2h3
Solving for ˆu1, gives
ˆu1=h2h3∇u2×∇u3.
Inserting this result into ∇⋅(F1ˆu1) and using the vector identity 2c from section ??,
∇⋅(fA)=f∇⋅A+A⋅∇f,
we have
∇⋅(F1ˆu1)=∇⋅(F1h2h3∇u2×∇u3)=∇(F1h2h3)⋅∇u2×∇u3+F1h2h2∇⋅(∇u2×∇u3)
The second term of this result vanishes by vector identity 3c,
∇⋅(∇f×∇g)=0.
Since ∇u2×∇u3=ˆ11h2h3, the first term can be evaluated as
∇⋅(F1ˆu1)=∇(F1h2h3)⋅ˆu1h2h3=1h1h2h3∂∂u1(F1h2h3).
Similar computations can be carried out for the remaining components, leading to the sought expression for the divergence in curvilinear coordinates:
∇⋅F=1h1h2h3(∂∂u1(h2h3F1)+∂∂u2(h1h3F2)+∂∂u3(h1h2F3)).
Example 6.9.5
Write the divergence operator in cylindrical coordinates.
Solution
In this case we have
∇⋅F=1hrhθhz(∂∂r(hθhzFr)+∂∂θ(hrhzFθ)+∂∂θ(hrhθFz))=1r(∂∂r(rFr)+∂∂θ(Fθ)+∂∂θ(rFz))=1r∂∂r(rFr)+1r∂∂θ(Fθ)+∂∂θ(Fz)
We now turn to the curl operator. In this case, we need to evaluate
∇×F=3∑i=1∇×(Fiˆui)
Again we focus on one term, ∇×(F1ˆu1). Using the vector identity 2e,
∇×(fA)=f∇×A−A×∇f,
we have
∇×(F1ˆu1)=∇×(F1h1∇u1)=F1h1∇×∇u1−∇(F1h1)×∇u1
The curl of the gradient vanishes, leaving
∇×(F1ˆu1)=∇(F1h1)×∇u1.
Since ∇u1=ˆu1h1, we have
∇×(F1ˆu1)=∇(F1h1)׈u1h1=(3∑i=1ˆuihi∂(F1h1)∂ui)׈u1h1=ˆu2h3h1∂(F1h1)∂u3−ˆu3h1h2∂(F1h1)∂u2.
The other terms can be handled in a similar manner. The overall result is that
∇×F=ˆu1h2h3(∂(h3F3)∂u2−∂(h2F2)∂u3)+ˆu2h1h3(∂(h1F1)∂u3−∂(h3F3)∂u1)+ˆu3h1h2(∂(h2F2)∂u1−∂(h1F1)∂u2)
This can be written more compactly as
∇×F=1h1h2h3|h1ˆu1h2ˆu2h3ˆu3∂∂u1∂∂u2∂∂u3F1h1F2h2F3h3|
Example 6.9.6
Write the curl operator in cylindrical coordinates.
Solution
∇×F=1r|ˆerrˆeθˆez∂∂r∂∂θ∂∂zFrrFθFz|=(1r∂Fz∂θ−∂Fθ∂z)ˆer+(∂Fr∂z−∂Fz∂r)ˆeθ+1r(∂(rFθ)∂r−∂Fr∂θ)ˆez.
Finally, we turn to the Laplacian. In the next chapter we will solve higher dimensional problems in various geometric settings such as the wave equation, the heat equation, and Laplace’s equation. These all involve knowing how to write the Laplacian in different coordinate systems. Since ∇2ϕ=∇⋅∇ϕ, we need only combine the results from Equations (6.9.17) and (6.9.19) for the gradient and the divergence in curvilinear coordinates. This is straight forward and gives
∇2ϕ=1h1h2h3(∂∂u1(h2h3h1∂ϕ∂u1)+∂∂u2(h1h3h2∂ϕ∂u2)+∂∂u3(h1h2h3∂ϕ∂u3)).
The results of rewriting the standard differential operators in cylindrical and spherical coordinates are shown in Problems ?? and ??. In particular, the Laplacians are given as
Definition 6.9.1: Cylindrical Coordinates
∇2f=1r∂∂r(r∂f∂r)+1r2∂2f∂θ2+∂2f∂z2.
Definition 6.9.2: Spherical Coordinates
∇2f=1ρ2∂∂ρ(ρ2∂f∂ρ)+1ρ2sinθ∂∂θ(sinθ∂f∂θ)+1ρ2sin2θ∂2f∂ϕ2.
6.8: Schrödinger Equation in Spherical Coordinates
6.10: Problems |
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2024 AMC 12B Problems/Problem 13
Contents
1 Problem 13
2 Solution 1 (Easy and Fast)
3 Solution 2 (Coordinate Geometry and AM-GM Inequality)
4 Solution 3
5 Video Solution 1 by SpreadTheMathLove
6 See also
Problem 13
There are real numbers and that satisfy the system of equationsWhat is the minimum possible value of ?
Solution 1 (Easy and Fast)
Adding up the first and second equation, we get: All squared values must be greater than or equal to . As we are aiming for the minimum value, we set the two squared terms to be .
This leads to
~mitsuihisashi14
Solution 2 (Coordinate Geometry and AM-GM Inequality)
The distance between 2 circle centers is The 2 circles must intersect given there exists one or more pairs of (x,y), connecting and any pair of the 2 circle intersection points gives us a triangle with 3 sides, then Note that they will be equal if and only if the circles are tangent,
Applying the AM-GM inequality () in the steps below, we get
Therefore, .
~luckuso,~
Solution 3
~Kathan
Video Solution 1 by SpreadTheMathLove
See also
2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 12Followed by
Problem 14
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25
All AMC 12 Problems and Solutions
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
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14498 | https://aviation.stackexchange.com/questions/58232/when-calculating-lift-using-the-the-nasa-lift-equation-can-any-value-be-increa | aerodynamics - When calculating lift, using the the NASA lift equation, can any value be increased, ie. velocity or surface area, for increased lift? - Aviation Stack Exchange
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When calculating lift, using the the NASA lift equation, can any value be increased, ie. velocity or surface area, for increased lift?
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I am trying to understand the breakdown of the NASA lift formula, which is written as such
L=C L ρ v 2 2 A L=C L ρ v 2 2 A
If I take for example C L=0.87, v=20m/s, ρ=1.22kg/m 3, and A=9.02m 2, my equation looks like this so far: Lift = (1/2) 0.87 1.22 20^2 9.02. Can any one of those values be increased for greater lift? If I doubled my velocity would I double the lift that the velocity provides? But if I added 25% to the surface area of the wing, my drag also increases. How does that factor into lift? By decreasing velocity?
Does this equation scale in that way? I apologize if it doesn't make sense, I am not sure how to word the question well.
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edited Dec 19, 2018 at 17:55
fooot
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asked Dec 19, 2018 at 17:38
DerrickDerrick
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Can any one of those values be increased for greater lift?
Certainly. But increased lift will not be the only effect.
If I doubled my velocity would I double the lift that the velocity provides? But if I added 25% to the surface area of the wing, my drag also increases. How does that factor into lift? By decreasing velocity?
It doesn't directly factor. The formula gives you the approximate lift given certain conditions. It doesn't tell you if those conditions are reasonable or achievable.
You're correct that increasing the wing area or speed will increase drag. But this equation doesn't tell you how much. You'd need to look elsewhere to find out the specifics. The drag increase will not be linear with either one.
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answered Dec 19, 2018 at 17:46
BowlOfRedBowlOfRed
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Thanks for your reply. I am trying to make sense of building a model of a cargo plane and need to redesign the wing a bit. My best option to increase lift overall would probably be increase wing surface area and velocity, even if drag increases as well? Also, do you know what it is called when things like air and water don't scale?Derrick –Derrick 2018-12-19 19:05:56 +00:00 Commented Dec 19, 2018 at 19:05
@Derrick You're looking for the Reynolds number with regard to scaling. If you're interested in designing a model aircraft that will fly, you should get an introductory textbook for aerodynamics. You can change any property of the wing, but it will have other effects besides just increasing lift.zaen –zaen 2018-12-19 21:25:05 +00:00 Commented Dec 19, 2018 at 21:25
@zaen Thank you! I'll look into it.Derrick –Derrick 2018-12-19 22:51:52 +00:00 Commented Dec 19, 2018 at 22:51
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MS-PS1-4 Phase Changes & States of Matter Lesson Plans Middle School
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Description
The phase change lesson plans introduce the states of matter, energy changes, and molecule movement between phases. There are stations and a "choose your own story" to help gather background knowledge. Students practice their understanding in a station lab. The final assessments are critical thinking questions about the lesson phenomenon and a phase change model they draw.
Included are:
Teacher's guide (secured PDF)
Graphing Model Template and instructions (secured PDF)
Drawing Model Template and instructions (secured PDF)
Student packet (secured PDF)
Water Droplet Choose Your Own Story (secured PDF with clickable links)
The student packet includes additional handouts and may be shared with students digitally on a secured school site.
The Phase Change Model and Water Droplet Choose Your Own Story in these lesson plans can be purchased separately.
Standards/Objectives:
✅NGSS: MS-PS1-4
This product includes the following activities:
⚪Lesson Phenomenon
⚪Phase Change Station Notes
⚪Choose Your Own Story- Water Droplet (can be purchased separately)
⚪Phase Change Stations Lab
⚪State of Change Model (can be purchased separately)
⚪Critical Thinking Question
Looking for more related content?
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⚫MS-PS1-2 Physical and Chemical Changes Lesson Plans
⚫MS-PS1-3 Natural to Synthetic Materials Lesson Plans
⚫MS-PS1-5 The Law of Conservation of Mass Lesson Plans
⚫MS-PS1-6 Thermal Energy and Heat Design Lesson Plans
⚫Atoms Background Stations & Note Sheet Bundle
⚫Atoms & Periodic Table of Elements Escape Puzzle Room
⚫MS-PS1 Atoms and Matter Activity Bundle
⚫Who Am I? Atoms Vocabulary Review Game
⚫Who Am I? Water Cycle Vocabulary Review Game
⚫States of Matter Vocabulary Note Sheet
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©Tonya Schield @ The Inquisitive Tortoise
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MS-PS1-4 Phase Changes & States of Matter Lesson Plans Middle School
The Inquisitive Tortoise
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$4.00
Add to cart
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Specs
What's Included
Grade
6 th - 8 th
Subject
General Science, Physical Science, Science
Standards
NGSS MS-PS1-4
Tags
Activities, Homeschool, Lessons, Projects
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MS-PS1 Atoms and Matter Lesson Plan Bundle Middle School
This bundle is available while I create a unit plan on atoms, matter, and energy. Each objective includes lesson plans and student packets on: modeling molecules and atoms, identifying chemical reactions, natural to synthetic materials, phase changes (energy changes and motion of molecules), conserv
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NGSS MS-PS1-4
Develop a model that predicts and describes changes in particle motion, temperature, and state of a pure substance when thermal energy is added or removed. Emphasis is on qualitative molecular-level models of solids, liquids, and gases to show that adding or removing thermal energy increases or decreases kinetic energy of the particles until a change of state occurs. Examples of models could include drawings and diagrams. Examples of particles could include molecules or inert atoms. Examples of pure substances could include water, carbon dioxide, and helium.
Meet the Teacher-Author
### The Inquisitive Tortoise
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I am a licensed science teacher and have multiple years of experience in middle school science in public school, private school, charter, special education, and been a substitute teacher. Through these experiences, I discovered my passion for writing curriculum and was hired to help write a STEM course and develop 3D science curriculum. I have volunteered for EdHeads to help create their STEM Mysteries, like the "Decreasing Hedgehogs." Creating science content is incredibly rewarding.
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Apple Valley, Minnesota, United States
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