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15300 | https://math.stackexchange.com/questions/1942402/differentiability-of-piecewise-function-at-breakpoint | Stack Exchange Network
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Differentiability of piecewise function at breakpoint
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Problem
Show whether or not the function
$$ f(x) = \begin{cases} \frac{1-\cos(x)}{x} & x > 0 \ x^2 & x \leq 0 \ \end{cases} $$
is differentiable at $x=0$.
My progress
I'm going by the definition of point-differentiability I found on Wikipedia, thus I need to show that it is continuous at $x=0$, and that $f'(0)$ exists.
I've been able to show that it is continuous, because the left and right limits exist, and are equal to $f(0)$.
But how can I show that $f'(0)$ exists (or not)? My idea is that this is pretty trivial because $\frac{\mathrm d}{\mathrm dx}x^2 = 2x$ which evaluates just fine at $x=0$. So I conclude that the function is differentiable at $x=0$.
Question
Does this hold? Or am I possibly overlooking something?
real-analysis
derivatives
continuity
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asked Sep 26, 2016 at 16:33
AlecAlec
4,17422 gold badges3636 silver badges6868 bronze badges
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$\begingroup$ Show that both one-sided derivatives exist and are equal at $x = 0$. $\endgroup$
user307169
– user307169
2016-09-26 16:38:09 +00:00
Commented Sep 26, 2016 at 16:38
$\begingroup$ you need to make sure the other branch gets you a similar result... one ofthe ways to think about derivatives that $$f'(x) = \lim_{\delta \to 0} \frac{f(x+\delta) - f(x-\delta)}{2\delta}...$$ $\endgroup$
gt6989b
– gt6989b
2016-09-26 16:39:36 +00:00
Commented Sep 26, 2016 at 16:39
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3 Answers 3
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EDIT: Wrong Answer
A function. $f$ is differentiable at $x=0$ if
$$ \lim_{h \rightarrow 0} \dfrac{f(0+h) - f(0)}{h} $$
exists. Now, for given $f(x)$, $$\lim_{h \rightarrow 0} \dfrac{\frac{1-cos(h)}{h} - 0^2}{h}=\lim_{h \rightarrow 0} \dfrac{1-cos(h)}{h^2}= \lim_{h \rightarrow 0} \dfrac{1-cos(h)}{h}\lim_{h \rightarrow 0}\frac{1}{h}=\lim_{h \rightarrow 0}\frac{1}{h}$$ This limit does not exists so $f'(0)$ does not exist.
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edited Apr 6, 2022 at 17:23
answered Sep 26, 2016 at 16:47
Sahil KumarSahil Kumar
75144 silver badges99 bronze badges
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$\begingroup$ $\lim_{h \rightarrow 0} \dfrac{1-cos(h)}{h^2}=\dfrac{1}{2}$ $\endgroup$
abhishek
– abhishek
2022-04-06 13:37:42 +00:00
Commented Apr 6, 2022 at 13:37
1
$\begingroup$ @abhishek Thanks.. that was wrong answer $\endgroup$
Sahil Kumar
– Sahil Kumar
2022-04-06 17:22:51 +00:00
Commented Apr 6, 2022 at 17:22
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0
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You might be overthinking this.
A function. $f$ is differentiable at $x=0$ if
$$ \lim_{h \rightarrow 0} \dfrac{f(0+h) - f(0)}{h} $$
exists. Perhaps consider the left and right hand limits. Are they equal? What can you conclude if they are not?
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answered Sep 26, 2016 at 16:37
Demetri PananosDemetri Pananos
3,30433 gold badges3030 silver badges4141 bronze badges
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As I mentioned in my comment, show that both one-sided derivatives exist and are equal to each other at $x = 0$.
A limit exists if and only if both one-sided limits exist.
A derivative is a limit.
Therefore, the derivative exists if and only if both one-sided derivatives exist.
You already found that the derivative of $x^2$ at $x = 0$ is $0$. Now apply the fact in step 3 to (immediately) get that the left-hand derivative of $x^2$ at $x = 0$ is $0$.
Evaluate the derivative of $\dfrac{1-\cos x}x$ at $x = 0$. Let's call it $L$ (if it even exists at all - if it doesn't, then $f(x)$ is not differentiable at $x = 0$).
Apply the fact in step 3 to (immediately) get that the right-hand derivative of $\dfrac{1-\cos x}x$ is $L$.
Finally, $f(x)$ is differentiable at $x = 0$ if and only if $L = 0$.
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answered Sep 26, 2016 at 16:48
user307169user307169
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15303 | https://www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-linear-equations-functions/write-slope-intercept-equations/v/linear-equations-in-slope-intercept-form | Slope-intercept form problems (video) | Khan Academy
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8th grade math
Course: 8th grade math>Unit 3
Lesson 7: Writing slope-intercept equations
Slope-intercept equation from graph
Writing slope-intercept equations
Slope-intercept equation from graph
Slope-intercept equation from slope & point
Slope-intercept equation from two points
Slope-intercept from two points
Slope-intercept form problems
Slope-intercept form from a table
Slope-intercept form review
Math>
8th grade math>
Linear equations and functions>
Writing slope-intercept equations
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Slope-intercept form problems
CCSS.Math: HSF.LE.A.2
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About About this video Transcript
Learn how to solve problems involving writing an equation in slope-intercept form.Created by Sal Khan.
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ShevrillHD 8 years ago Posted 8 years ago. Direct link to ShevrillHD's post “Isn't double minus sign i...” more Isn't double minus sign incorrect? Shouldn't be there parentheses? Like 5 - (-2) instead of 5 - -2
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Kim Seidel 8 years ago Posted 8 years ago. Direct link to Kim Seidel's post “Yes, there should be pare...” more Yes, there should be parentheses separating the minus signs.
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Rebecca Flood 12 years ago Posted 12 years ago. Direct link to Rebecca Flood's post “How do I write the slope-...” more How do I write the slope-intercept form of the equation of the line through: (3,-2), if the slope is undefined?
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jmascaro 12 years ago Posted 12 years ago. Direct link to jmascaro's post “Hi, If a slope is undefin...” more Hi,
If a slope is undefined, it means that the x value does not change. Remember that slope is calculated as the change in the y value divided by the change in the x value. If x does not change, then we are dividing by zero, which is undefined. Therefore, an undefined slope is a vertical line through the point you were given.
You cannot really write it in a correct slope/intercept form. Usually this is written as:
x=whatever the x value of your point is.
In your case, the equation would be
x=3
And the graph would be a vertical line running through the 3 on the x axis.
Hope that helps :-)
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ajrulzabc123 a year ago Posted a year ago. Direct link to ajrulzabc123's post “on the third question sho...” more on the third question shouldn't you minis y2-y1 and x2-x1 instead of y1-y2 and x1-x2 like 0-6/5-2?
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Kim Seidel a year ago Posted a year ago. Direct link to Kim Seidel's post “It doesn't matter which p...” more It doesn't matter which point you label as (x1,y1) vs (x2,y2). Assuming you have no math errors, you will get the same slope. Finish the math that your started.
(0-6)/(5-2) = -6/3 = -2
That's the same slope calculated in the video.
Hope this helps.
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Golden Apples a year ago Posted a year ago. Direct link to Golden Apples's post “Wait, can you substitute ...” more Wait, can you substitute x for y and y for x and then have the y intercept be swapped out for the x intercept? so like x = my + b?
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Kim Seidel a year ago Posted a year ago. Direct link to Kim Seidel's post “If you solve the equation...” more If you solve the equation for "x" instead of "y", the the "b" value is the x-intercept. However, "m" is not the slope of the line. It is the reciprocal of the slope. For example, if the slope is 3/4, then solving for "x" will give you 4/3.
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Ryker Hatch 2 years ago Posted 2 years ago. Direct link to Ryker Hatch's post “At 7:54, Sal divides -3 b...” more At 7:54 , Sal divides -3 by 3 and then -3 by 6. Why does he divide -3 by 3 and -3 by 6, but not -3 by 5?
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Kim Seidel 2 years ago Posted 2 years ago. Direct link to Kim Seidel's post “Sal is multiplying: (5/6)...” more Sal is multiplying: (5/6)(-3) = -15/6
When we reduce fractions, we remove any common factors from the numerator and denominator. The common factor is 3. The simplifies the fraction into -5/2.
Sal chose to cross cancel (cancel out the common factor, then complete the multiplication.
Hope this helps.
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poodoo 2 years ago Posted 2 years ago. Direct link to poodoo's post “at 7:18 why did the negat...” more at 7:18 why did the negatives dissappear?
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Robin Heyer 2 years ago Posted 2 years ago. Direct link to Robin Heyer's post “When he divided -5 by -6 ...” more When he divided -5 by -6 and the result is 5/6?
The same rule applies for division as it does for multiplication. When you multiply or divide 2 negative values, the result will be positive. And because of that we can write the whole thing positive because the result will be the same and it is usually easier to read for us.
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Kartikeye 12 years ago Posted 12 years ago. Direct link to Kartikeye's post “I didn't understand the e...” more I didn't understand the equation f(1.5)=-3,f(-1)=2?
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Petrie (Peter S. Asiain III) 12 years ago Posted 12 years ago. Direct link to Petrie (Peter S. Asiain III)'s post “This is because our funct...” more This is because our function here is
f(x)=(x)-2
So each time we put in a value of x we multiply it by -2
Ex.
f(1.5)=(1.5)-2=-3
f(-1)=(-1)-2=2
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Khadiza Begum 2 years ago Posted 2 years ago. Direct link to Khadiza Begum's post “at 2:42 he did "Y1-Y2" bu...” more at 2:42 he did "Y1-Y2" but isn't y2 meant to come first?
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Bani Hundal 2 years ago Posted 2 years ago. Direct link to Bani Hundal's post “It doesn't matter whether...” more It doesn't matter whether its y2,y1/x2-x1 or y1-y2/x1-x2 because you'll get the same answer only if you do the calculation right.
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Rebecca Renner a year ago Posted a year ago. Direct link to Rebecca Renner's post “What is slope form” more What is slope form
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szhang045 a year ago Posted a year ago. Direct link to szhang045's post “slope intercept form is y...” more slope intercept form is y=mx+b, where m is the slope and b is the y-intercept
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Luke 3 years ago Posted 3 years ago. Direct link to Luke's post “7:30 0= 5/3(-3) + b why d...” more 7:30 0= 5/3(-3) + b why did sal divide -3 by -3 to get a +ve 1?
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WisteriaGrove 3 years ago Posted 3 years ago. Direct link to WisteriaGrove's post “Sal did not divide -3 by ...” more Sal did not divide -3 by -3. They were dividing the 0-5.
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Video transcript
In this video I'm going to do a bunch of examples of finding the equations of lines in slope-intercept form. Just as a bit of a review, that means equations of lines in the form of y is equal to mx plus b where m is the slope and b is the y-intercept. So let's just do a bunch of these problems. So here they tell us that a line has a slope of negative 5, so m is equal to negative 5. And it has a y-intercept of 6. So b is equal to 6. So this is pretty straightforward. The equation of this line is y is equal to negative 5x plus 6. That wasn't too bad. Let's do this next one over here. The line has a slope of negative 1 and contains the point 4/5 comma 0. So they're telling us the slope, slope of negative 1. So we know that m is equal to negative 1, but we're not 100% sure about where the y-intercept is just yet. So we know that this equation is going to be of the form y is equal to the slope negative 1x plus b, where b is the y-intercept. Now, we can use this coordinate information, the fact that it contains this point, we can use that information to solve for b. The fact that the line contains this point means that the value x is equal to 4/5, y is equal to 0 must satisfy this equation. So let's substitute those in. y is equal to 0 when x is equal to 4/5. So 0 is equal to negative 1 times 4/5 plus b. I'll scroll down a little bit. So let's see, we get a 0 is equal to negative 4/5 plus b. We can add 4/5 to both sides of this equation. So we get add a 4/5 there. We could add a 4/5 to that side as well. The whole reason I did that is so that cancels out with that. You get b is equal to 4/5. So we now have the equation of the line. y is equal to negative 1 times x, which we write as negative x, plus b, which is 4/5, just like that. Now we have this one. The line contains the point 2 comma 6 and 5 comma 0. So they haven't given us the slope or the y-intercept explicitly. But we could figure out both of them from these coordinates. So the first thing we can do is figure out the slope. So we know that the slope m is equal to change in y over change in x, which is equal to-- What is the change in y? Let's start with this one right here. So we do 6 minus 0. Let me do it this way. So that's a 6-- I want to make it color-coded-- minus 0. So 6 minus 0, that's our change in y. Our change in x is 2 minus 2 minus 5. The reason why I color-coded it is I wanted to show you when I used this y term first, I used the 6 up here, that I have to use this x term first as well. So I wanted to show you, this is the coordinate 2 comma 6. This is the coordinate 5 comma 0. I couldn't have swapped the 2 and the 5 then. Then I would have gotten the negative of the answer. But what do we get here? This is equal to 6 minus 0 is 6. 2 minus 5 is negative 3. So this becomes negative 6 over 3, which is the same thing as negative 2. So that's our slope. So, so far we know that the line must be, y is equal to the slope-- I'll do that in orange-- negative 2 times x plus our y-intercept. Now we can do exactly what we did in the last problem. We can use one of these points to solve for b. We can use either one. Both of these are on the line, so both of these must satisfy this equation. I'll use the 5 comma 0 because it's always nice when you have a 0 there. The math is a little bit easier. So let's put the 5 comma 0 there. So y is equal to 0 when x is equal to 5. So y is equal to 0 when you have negative 2 times 5, when x is equal to 5 plus b. So you get 0 is equal to -10 plus b. If you add 10 to both sides of this equation, let's add 10 to both sides, these two cancel out. You get b is equal to 10 plus 0 or 10. So you get b is equal to 10. Now we know the equation for the line. The equation is y-- let me do it in a new color-- y is equal to negative 2x plus b plus 10. We are done. Let's do another one of these. All right, the line contains the points 3 comma 5 and negative 3 comma 0. Just like the last problem, we start by figuring out the slope, which we will call m. It's the same thing as the rise over the run, which is the same thing as the change in y over the change in x. If you were doing this for your homework, you wouldn't have to write all this. I just want to make sure that you understand that these are all the same things. Then what is our change in y over our change in x? This is equal to, let's start with the side first. It's just to show you I could pick either of these points. So let's say it's 0 minus 5 just like that. So I'm using this coordinate first. I'm kind of viewing it as the endpoint. Remember when I first learned this, I would always be tempted to do the x in the numerator. No, you use the y's in the numerator. So that's the second of the coordinates. That is going to be over negative 3 minus 3. This is the coordinate negative 3, 0. This is the coordinate 3, 5. We're subtracting that. So what are we going to get? This is going to be equal to-- I'll do it in a neutral color-- this is going to be equal to the numerator is negative 5 over negative 3 minus 3 is negative 6. So the negatives cancel out. You get 5/6. So we know that the equation is going to be of the form y is equal to 5/6 x plus b. Now we can substitute one of these coordinates in for b. So let's do. I always like to use the one that has the 0 in it. So y is a zero when x is negative 3 plus b. So all I did is I substituted negative 3 for x, 0 for y. I know I can do that because this is on the line. This must satisfy the equation of the line. Let's solve for b. So we get zero is equal to, well if we divide negative 3 by 3, that becomes a 1. If you divide 6 by 3, that becomes a 2. So it becomes negative 5/2 plus b. We could add 5/2 to both sides of the equation, plus 5/2, plus 5/2. I like to change my notation just so you get familiar with both. So the equation becomes 5/2 is equal to-- that's a 0-- is equal to b. b is 5/2. So the equation of our line is y is equal to 5/6 x plus b, which we just figured out is 5/2, plus 5/2. We are done. Let's do another one. We have a graph here. Let's figure out the equation of this graph. This is actually, on some level, a little bit easier. What's the slope? Slope is change in y over change it x. So let's see what happens. When we move in x, when our change in x is 1, so that is our change in x. So change in x is 1. I'm just deciding to change my x by 1, increment by 1. What is the change in y? It looks like y changes exactly by 4. It looks like my delta y, my change in y, is equal to 4 when my delta x is equal to 1. So change in y over change in x, change in y is 4 when change in x is 1. So the slope is equal to 4. Now what's its y-intercept? Well here we can just look at the graph. It looks like it intersects y-axis at y is equal to negative 6, or at the point 0, negative 6. So we know that b is equal to negative 6. So we know the equation of the line. The equation of the line is y is equal to the slope times x plus the y-intercept. I should write that. So minus 6, that is plus negative 6 So that is the equation of our line. Let's do one more of these. So they tell us that f of 1.5 is negative 3, f of negative 1 is 2. What is that? Well, all this is just a fancy way of telling you that the point when x is 1.5, when you put 1.5 into the function, the function evaluates as negative 3. So this tells us that the coordinate 1.5, negative 3 is on the line. Then this tells us that the point when x is negative 1, f of x is equal to 2. This is just a fancy way of saying that both of these two points are on the line, nothing unusual. I think the point of this problem is to get you familiar with function notation, for you to not get intimidated if you see something like this. If you evaluate the function at 1.5, you get negative 3. So that's the coordinate if you imagine that y is equal to f of x. So this would be the y-coordinate. It would be equal to negative 3 when x is 1.5. Anyway, I've said it multiple times. Let's figure out the slope of this line. The slope which is change in y over change in x is equal to, let's start with 2 minus this guy, negative 3-- these are the y-values-- over, all of that over, negative 1 minus this guy. Let me write it this way, negative 1 minus that guy, minus 1.5. I do the colors because I want to show you that the negative 1 and the 2 are both coming from this, that's why I use both of them first. If I used these guys first, I would have to use both the x and the y first. If I use the 2 first, I have to use the negative 1 first. That's why I'm color-coding it. So this is going to be equal to 2 minus negative 3. That's the same thing as 2 plus 3. So that is 5. Negative 1 minus 1.5 is negative 2.5. 5 divided by 2.5 is equal to 2. So the slope of this line is negative 2. Actually I'll take a little aside to show you it doesn't matter what order I do this in. If I use this coordinate first, then I have to use that coordinate first. Let's do it the other way. If I did it as negative 3 minus 2 over 1.5 minus negative 1, this should be minus the 2 over 1.5 minus the negative 1. This should give me the same answer. This is equal to what? Negative 3 minus 2 is negative 5 over 1.5 minus negative 1. That's 1.5 plus 1. That's over 2.5. So once again, this is equal the negative 2. So I just wanted to show you, it doesn't matter which one you pick as the starting or the endpoint, as long as you're consistent. If this is the starting y, this is the starting x. If this is the finishing y, this has to be the finishing x. But anyway, we know that the slope is negative 2. So we know the equation is y is equal to negative 2x plus some y-intercept. Let's use one of these coordinates. I'll use this one since it doesn't have a decimal in it. So we know that y is equal to 2. So y is equal to 2 when x is equal to negative 1. Of course you have your plus b. So 2 is equal to negative 2 times negative 1 is 2 plus b. If you subtract 2 from both sides of this equation, minus 2, minus 2, you're subtracting it from both sides of this equation, you're going to get 0 on the left-hand side is equal to b. So b is 0. So the equation of our line is just y is equal to negative 2x. Actually if you wanted to write it in function notation, it would be that f of x is equal to negative 2x. I kind of just assumed that y is equal to f of x. But this is really the equation. They never mentioned y's here. So you could just write f of x is equal to 2x right here. Each of these coordinates are the coordinates of x and f of x. So you could even view the definition of slope as change in f of x over change in x. These are all equivalent ways of viewing the same thing.
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15304 | https://math.stackexchange.com/questions/3007719/number-of-increasing-sequences-alternating-between-odd-and-even-involving-only-i | combinatorics - Number of increasing sequences alternating between odd and even involving only integers from the set $[1, 2, 3, ..., n]$ - Mathematics Stack Exchange
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Number of increasing sequences alternating between odd and even involving only integers from the set [1,2,3,...,n][1,2,3,...,n]
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I am new to this forum so do correct me if I am doing anything wrong.
(Problem 6 of British Mathematics Olympiad1 - 1994)
An increasing sequence of integers is said to be alternating if it starts with an odd term, the second term is even, the third term is odd, the fourth is even, and so on. The empty sequence (with no term at all!) is considered to be alternating. Let A(n)A(n) denote the number of alternating sequences which only involve integers from the set [1,2,...,n][1,2,...,n]. Find the value of A(20)A(20) and prove that your value is correct.
I believe a method is to set up a recursion.
What I have done is as follows:
(I've slightly changed the definition of A(n) to be the set of all possible sequences)
Let O(n)O(n) be the number of increasing alternating sequences which only involve integers from the set [1,2,...,n][1,2,...,n] that end in an odd number and E(n)E(n) be the number of increasing alternating sequences which only involve integers from the set [1,2,...,n][1,2,...,n] that end in an even number.
For every even n n, O(n+2)=O(n)+E(n)+1 O(n+2)=O(n)+E(n)+1.
Proof: For every sequence that ends in an odd number in A(n)A(n), it will also be a sequence of A(n+2)A(n+2), hence add O(n)O(n). For every sequence that ends in an even number in A(n)A(n), n+1 n+1, which is odd, can be added to the end of that sequence, hence add E(n)E(n). The sequence with only n+1 n+1 also satisfies the conditions hence add 1 1.
For every even n n, E(n+2)=2 E(n)+O(n)+1 E(n+2)=2 E(n)+O(n)+1.
Proof: For every sequence that ends in an odd number in A(n)A(n), n+2 n+2, which is even, can be added to the end of that sequence, hence add O(n)O(n). For every sequence that ends in an even number in A(n)A(n), nothing or both n+1 n+1 and n+2 n+2 can be added to the end of that sequence, hence add 2 E(n)2 E(n). The sequence with only n+2 n+2 also satisfies the conditions hence add 1 1.
A(n)=E(n)+O(n)+1 A(n)=E(n)+O(n)+1
Proof: Every sequence either ends in an even number or an odd number or is an empty sequence.
Starting with E(0)=0 E(0)=0 and O(0)=0 O(0)=0, we get E(20)=10945 E(20)=10945 and O(20)=6765 O(20)=6765, meaning A(20)=17711 A(20)=17711.
Is there anything wrong with my method? If not is there a more efficient way to get to the answer? Thanks.
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asked Nov 21, 2018 at 13:18
E FresherE Fresher
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I think there's a small mistake. You say the sequence consisting only of n+2 n+2 is alternating, but this is not true unless n n is odd, because one of the conditions is that an alternating sequence start with an odd term. Since you eventually compute A(20),A(20), you need n n to be even. Your general approach looks fine, but you should review the details.saulspatz –saulspatz 2018-11-21 13:51:21 +00:00 Commented Nov 21, 2018 at 13:51
I believe I have avoided this by stating for every even n, or is this not what you mean.E Fresher –E Fresher 2018-11-21 14:12:16 +00:00 Commented Nov 21, 2018 at 14:12
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If a sequence starts with 1, the rest of the sequence can be found by adding 1 every term of a sequence from A(n−1)A(n−1).
If it doesn't begin with 1, you get it by adding 2 to every term of a sequence from A(n−2)A(n−2).
So A(n)=A(n−1)+A(n−2)A(n)=A(n−1)+A(n−2).
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answered Nov 21, 2018 at 13:48
Empy2Empy2
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And you need to start things off with A(0)=1,A(1)=2 A(0)=1,A(1)=2.TonyK –TonyK 2018-11-21 14:04:43 +00:00 Commented Nov 21, 2018 at 14:04
1 I recognized 6765 and 17711, and looked for that recursion.Empy2 –Empy2 2018-11-21 14:17:05 +00:00 Commented Nov 21, 2018 at 14:17
1 If I had not provided those values do you think you could have got there?E Fresher –E Fresher 2018-11-21 14:18:02 +00:00 Commented Nov 21, 2018 at 14:18
1 I might have come up with your method instead, which is a good method. Perhaps i might have used one equation to eliminate O(n)O(n) from the other equation. That gives a connection between E(n),E(n+2)E(n),E(n+2) and E(n+4)E(n+4) which I have been taught how to solve.Empy2 –Empy2 2018-11-21 14:23:11 +00:00 Commented Nov 21, 2018 at 14:23
1 The empty sequence is part of A(n−2)A(n−2). The empty sequence from A(n−1)A(n−1) contributes to the sequence {1}{1}.Empy2 –Empy2 2018-11-21 14:25:25 +00:00 Commented Nov 21, 2018 at 14:25
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15305 | https://people.tamu.edu/~yvorobets//MATH304-2010C/Lect2-10web.pdf | MATH 304 Linear Algebra Lecture 20: Linear transformations.
Range and kernel.
Linear mapping = linear transformation = linear function Definition.
Given vector spaces V1 and V2, a mapping L : V1 →V2 is linear if L(x + y) = L(x) + L(y), L(rx) = rL(x) for any x, y ∈V1 and r ∈R.
A linear mapping ℓ: V →R is called a linear functional on V .
If V1 = V2 (or if both V1 and V2 are functional spaces) then a linear mapping L : V1 →V2 is called a linear operator.
Linear mapping = linear transformation = linear function Definition.
Given vector spaces V1 and V2, a mapping L : V1 →V2 is linear if L(x + y) = L(x) + L(y), L(rx) = rL(x) for any x, y ∈V1 and r ∈R.
Remark.
A function f : R →R given by f (x) = ax + b is a linear transformation of the vector space R if and only if b = 0.
Properties of linear mappings Let L : V1 →V2 be a linear mapping.
• L(r1v1 + · · · + rkvk) = r1L(v1) + · · · + rkL(vk) for all k ≥1, v1, . . . , vk ∈V1, and r1, . . . , rk ∈R.
L(r1v1 + r2v2) = L(r1v1) + L(r2v2) = r1L(v1) + r2L(v2), L(r1v1 + r2v2 + r3v3) = L(r1v1 + r2v2) + L(r3v3) = = r1L(v1) + r2L(v2) + r3L(v3), and so on.
• L(01) = 02, where 01 and 02 are zero vectors in V1 and V2, respectively.
L(01) = L(001) = 0L(01) = 02.
• L(−v) = −L(v) for any v ∈V1.
L(−v) = L((−1)v) = (−1)L(v) = −L(v).
Examples of linear mappings • Scaling L : V →V , L(v) = sv, where s ∈R.
L(x + y) = s(x + y) = sx + sy = L(x) + L(y), L(rx) = s(rx) = r(sx) = rL(x).
• Dot product with a fixed vector ℓ: Rn →R, ℓ(v) = v · v0, where v0 ∈Rn.
ℓ(x + y) = (x + y) · v0 = x · v0 + y · v0 = ℓ(x) + ℓ(y), ℓ(rx) = (rx) · v0 = r(x · v0) = rℓ(x).
• Cross product with a fixed vector L : R3 →R3, L(v) = v × v0, where v0 ∈R3.
• Multiplication by a fixed matrix L : Rn →Rm, L(v) = Av, where A is an m×n matrix and all vectors are column vectors.
Linear mappings of functional vector spaces • Evaluation at a fixed point ℓ: F(R) →R, ℓ(f ) = f (a), where a ∈R.
• Multiplication by a fixed function L : F(R) →F(R), L(f ) = gf , where g ∈F(R).
• Differentiation D : C 1(R) →C(R), L(f ) = f ′.
D(f + g) = (f + g)′ = f ′ + g ′ = D(f ) + D(g), D(rf ) = (rf )′ = rf ′ = rD(f ).
• Integration over a finite interval ℓ: C(R) →R, ℓ(f ) = Z b a f (x) dx, where a, b ∈R, a < b.
Properties of linear mappings • If a linear mapping L : V →W is invertible then the inverse mapping L−1 : W →V is also linear.
• If L : V →W and M : W →X are linear mappings then the composition M ◦L : V →X is also linear.
• If L1 : V →W and L2 : V →W are linear mappings then the sum L1 + L2 is also linear.
Linear differential operators • an ordinary differential operator L : C ∞(R) →C ∞(R), L = g0 d2 dx2 + g1 d dx + g2, where g0, g1, g2 are smooth functions on R.
That is, L(f ) = g0f ′′ + g1f ′ + g2f .
• Laplace’s operator ∆: C ∞(R2) →C ∞(R2), ∆f = ∂2f ∂x2 + ∂2f ∂y 2 (a.k.a. the Laplacian; also denoted by ∇2).
Range and kernel Let V , W be vector spaces and L : V →W be a linear mapping.
Definition.
The range (or image) of L is the set of all vectors w ∈W such that w = L(v) for some v ∈V . The range of L is denoted L(V ).
The kernel of L, denoted ker L, is the set of all vectors v ∈V such that L(v) = 0.
Theorem (i) The range of L is a subspace of W .
(ii) The kernel of L is a subspace of V .
Example.
L : R3 →R3, L x y z = 1 0 −1 1 2 −1 1 0 −1 x y z .
The kernel ker L is the nullspace of the matrix.
L x y z = x 1 1 1 + y 0 2 0 + z −1 −1 −1 The range f (R3) is the column space of the matrix.
Example.
L : R3 →R3, L x y z = 1 0 −1 1 2 −1 1 0 −1 x y z .
The range of L is spanned by vectors (1, 1, 1), (0, 2, 0), and (−1, −1, −1). It follows that L(R3) is the plane spanned by (1, 1, 1) and (0, 1, 0).
To find ker L, we apply row reduction to the matrix: 1 0 −1 1 2 −1 1 0 −1 → 1 0 −1 0 2 0 0 0 0 → 1 0 −1 0 1 0 0 0 0 Hence (x, y, z) ∈ker L if x −z = y = 0.
It follows that ker L is the line spanned by (1, 0, 1).
More examples f : M2(R) →M2(R), f (A) = A + AT.
f a b c d = 2a b + c b + c 2d .
ker f is the subspace of anti-symmetric matrices, the range of f is the subspace of symmetric matrices.
g : M2(R) →M2(R), g(A) = 0 1 0 0 A.
g a b c d = c d 0 0 .
The range of g is the subspace of matrices with the zero second row, ker g is the same as the range = ⇒g(g(A)) = O.
P: the space of polynomials.
Pn: the space of polynomials of degree less than n.
D : P →P, (Dp)(x) = p′(x).
p(x) = a0 + a1x + a2x2 + a3x3 + · · · + anxn = ⇒(Dp)(x) = a1 + 2a2x + 3a3x2 + · · · + nanxn−1 The range of D is the entire P, ker D = P1 = the subspace of constants.
D : P4 →P4, (Dp)(x) = p′(x).
p(x) = ax3+bx2+cx+d = ⇒(Dp)(x) = 3ax2+2bx+c The range of D is P3, ker D = P1. |
15306 | https://engineeringlibrary.org/reference/conduction-heat-transfer-doe-handbook | Conduction Heat Transfer | Engineering Library
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Conduction Heat Transfer
This page provides the chapter on conduction heat transfer from the "DOE Fundamentals Handbook: Thermodynamics, Heat Transfer, and Fluid Flow," DOE-HDBK-1012/2-92, U.S. Department of Energy, June 1992.
Other related chapters from the "DOE Fundamentals Handbook: Thermodynamics, Heat Transfer, and Fluid Flow" can be seen to the right.
DOE Handbook: Heat Transfer
Heat Transfer Terminology
Conduction Heat Transfer
Convection Heat Transfer
Radiant Heat Transfer
Heat Exchangers
Boiling Heat Transfer
Heat Generation
Decay Heat
Conduction Heat Transfer
Conduction heat transfer is the transfer of thermal energy by interactions between adjacent atoms and molecules of a solid.
Conduction
Conduction involves the transfer of heat by the interaction between adjacent molecules of a material. Heat transfer by conduction is dependent upon the driving "force" of temperature difference and the resistance to heat transfer. The resistance to heat transfer is dependent upon the nature and dimensions of the heat transfer medium. All heat transfer problems involve the temperature difference, the geometry, and the physical properties of the object being studied.
In conduction heat transfer problems, the object being studied is usually a solid. Convection problems involve a fluid medium. Radiation heat transfer problems involve either solid or fluid surfaces, separated by a gas, vapor, or vacuum. There are several ways to correlate the geometry, physical properties, and temperature difference of an object with the rate of heat transfer through the object. In conduction heat transfer, the most common means of correlation is through Fourier's Law of Conduction. The law, in its equation form, is used most often in its rectangular or cylindrical form (pipes and cylinders), both of which are presented below.
Rectangular Q˙=k A(Δ T Δ x)Q˙=k A(Δ T Δ x)
(2-4)
Cylindrical Q˙=k A(Δ T Δ r)Q˙=k A(Δ T Δ r)
(2-5)
where:
Q˙Q˙=rate of heat transfer (Btu/hr)
A=cross-sectional area of heat transfer (ft 2)
Δx=thickness of slab (ft)
Δr=thickness of cylindrical wall (ft)
ΔT=temperature difference (°F)
k=thermal conductivity of slab (Btu/ft-hr-°F)
The use of Equations 2-4 and 2-5 in determining the amount of heat transferred by conduction is demonstrated in the following examples.
Conduction-Rectangular Coordinates
Example:
1000 Btu/hr is conducted through a section of insulating material shown in Figure 1 that measures 1 ft 2 in cross-sectional area. The thickness is 1 in. and the thermal conductivity is 0.12 Btu/hr-ft-°F. Compute the temperature difference across the material.
Figure 1: Conduction Through a Slab
Solution:
Using Equation 2-4:
Q˙=k A(Δ T Δ x)Q˙=k A(Δ T Δ x)
Solving for ΔT:
Δ T Δ T===Q˙(Δ x k A)(1000 Btu hr)(1 12 ft)(0.12 Btu hr-ft-∘F)(1 ft 2)694∘F Δ T=Q˙(Δ x k A)=(1000 Btu hr)(1 12 ft)(0.12 Btu hr-ft-∘F)(1 ft 2)Δ T=694∘F
Example:
A concrete floor with a conductivity of 0.8 Btu/hr-ft-°F measures 30 ft by 40 ft with a thickness of 4 inches. The floor has a surface temperature of 70°F and the temperature beneath it is 60°F. What is the heat flux and the heat transfer rate through the floor?
Solution:
Using Equations 2-1 and 2-4:
Q˙′′===Q˙A=k(Δ T Δ x)(0.8 Btu hr-ft-∘F)(10∘F 0.333 ft)24 Btu hr-ft 2 Q˙″=Q˙A=k(Δ T Δ x)=(0.8 Btu hr-ft-∘F)(10∘F 0.333 ft)=24 Btu hr-ft 2
Using Equation 2-3:
Q˙===k A(Δ T Δ x)=Q˙′′A(24 Btu hr-ft 2)(1200 ft 2)28,800 Btu hr Q˙=k A(Δ T Δ x)=Q˙″A=(24 Btu hr-ft 2)(1200 ft 2)=28,800 Btu hr
Equivalent Resistance Method
It is possible to compare heat transfer to current flow in electrical circuits. The heat transfer rate may be considered as a current flow and the combination of thermal conductivity, thickness of material, and area as a resistance to this flow. The temperature difference is the potential or driving function for the heat flow, resulting in the Fourier equation being written in a form similar to Ohm's Law of Electrical Circuit Theory. If the thermal resistance term Δx/k is written as a resistance term where the resistance is the reciprocal of the thermal conductivity divided by the thickness of the material, the result is the conduction equation being analogous to electrical systems or networks. The electrical analogy may be used to solve complex problems involving both series and parallel thermal resistances. The student is referred to Figure 2, showing the equivalent resistance circuit. A typical conduction problem in its analogous electrical form is given in the following example, where the "electrical" Fourier equation may be written as follows.
Q˙′′=Δ T R t h Q˙″=Δ T R t h
(2-6)
where:
Q˙′′Q˙″=Heat Flux (Q˙/A Q˙/A) (Btu/hr-ft 2)
ΔT=Temperature Difference (°F)
R th=Thermal Resistance (Δx/k) (hr-ft 2-°F/Btu)
Figure 2: Equivalent Resistance
Electrical Analogy
Example:
A composite protective wall is formed of a 1 in. copper plate, a 1/8 in. layer of asbestos, and a 2 in. layer of fiberglass. The thermal conductivities of the materials in units of Btu/hr-ft-°F are as follows: k Cu = 240, k asb = 0.048, and k fib = 0.022. The overall temperature difference across the wall is 500°F. Calculate the thermal resistance of each layer of the wall and the heat transfer rate per unit area (heat flux) through the composite structure.
Solution:
R C u===Δ x C u k C u 1 in(1 ft 12 in)240 Btu hr-ft-∘F 0.000347 hr-ft 2-∘F Btu R C u=Δ x C u k C u=1 in(1 ft 12 in)240 Btu hr-ft-∘F=0.000347 hr-ft 2-∘F Btu
R a s b===Δ x a s b k a s b 0.125 in(1 ft 12 in)0.048 Btu hr-ft-∘F 0.2170 hr-ft 2-∘F Btu R a s b=Δ x a s b k a s b=0.125 in(1 ft 12 in)0.048 Btu hr-ft-∘F=0.2170 hr-ft 2-∘F Btu
R f i b===Δ x f i b k f i b 2 in(1 ft 12 in)0.022 Btu hr-ft-∘F 7.5758 hr-ft 2-∘F Btu R f i b=Δ x f i b k f i b=2 in(1 ft 12 in)0.022 Btu hr-ft-∘F=7.5758 hr-ft 2-∘F Btu
Q˙A===(T i−T o)(R C u+R a s b+R f i b)500∘F(0.000347 0.2170+7.5758)hr-ft 2-∘F Btu 64.2 Btu hr-ft 2 Q˙A=(T i−T o)(R C u+R a s b+R f i b)=500∘F(0.000347 0.2170+7.5758)hr-ft 2-∘F Btu=64.2 Btu hr-ft 2
Conduction-Cylindrical Coordinates
Heat transfer across a rectangular solid is the most direct application of Fourier's law. Heat transfer across a pipe or heat exchanger tube wall is more complicated to evaluate. Across a cylindrical wall, the heat transfer surface area is continually increasing or decreasing. Figure 3 is a cross-sectional view of a pipe constructed of a homogeneous material.
Figure 3: Cross-sectional Surface Area of a Cylindrical Pipe
The surface area (A) for transferring heat through the pipe (neglecting the pipe ends) is directly proportional to the radius (r) of the pipe and the length (L) of the pipe.
A = 2πrL
As the radius increases from the inner wall to the outer wall, the heat transfer area increases.
The development of an equation evaluating heat transfer through an object with cylindrical geometry begins with Fourier's law Equation 2-5.
Q˙=k A(Δ T Δ r)Q˙=k A(Δ T Δ r)
From the discussion above, it is seen that no simple expression for area is accurate. Neither the area of the inner surface nor the area of the outer surface alone can be used in the equation. For a problem involving cylindrical geometry, it is necessary to define a log mean cross-sectional area (A lm).
A l m=A o u t e r−A i n n e r ln(A o u t e r A i n n e r)A l m=A o u t e r−A i n n e r ln(A o u t e r A i n n e r)
(2-7)
Substituting the expression 2πrL for area in Equation 2-7 allows the log mean area to be calculated from the inner and outer radius without first calculating the inner and outer area.
A l m==2 π r o u t e r L−2 π r i n n e r L ln(2 π r o u t e r L 2 π r i n n e r L)2 π L(r o u t e r−r i n n e r ln r o u t e r r i n n e r)A l m=2 π r o u t e r L−2 π r i n n e r L ln(2 π r o u t e r L 2 π r i n n e r L)=2 π L(r o u t e r−r i n n e r lnr o u t e r r i n n e r)
This expression for log mean area can be inserted into Equation 2-5, allowing us to calculate the heat transfer rate for cylindrical geometries.
Q˙Q˙===k A l m(Δ T Δ r)k2 π L(r o−r i ln r o r i)2 π k L(Δ T)ln(r o/r i)Q˙=k A l m(Δ T Δ r)=k2 π L(r o−r i lnr o r i)Q˙=2 π k L(Δ T)ln(r o/r i)
(2-8)
where:
L=length of pipe (ft)
r i=inside pipe radius (ft)
r o=outside pipe radius (ft)
Example:
A stainless steel pipe with a length of 35 ft has an inner diameter of 0.92 ft and an outer diameter of 1.08 ft. The temperature of the inner surface of the pipe is 122°F and the temperature of the outer surface is 118°F. The thermal conductivity of the stainless steel is 108 Btu/hr-ft-°F.
Calculate the heat transfer rate through the pipe.
Calculate the heat flux at the outer surface of the pipe.
Solution:
Q˙===2 π k L(T h−T c)ln(r o/r i)6.28(108 Btu hr-ft-∘F)(35 ft)(122∘F−118∘F)ln 0.54 ft 0.46 ft 5.92×10 5 Btu hr Q˙=2 π k L(T h−T c)ln(r o/r i)=6.28(108 Btu hr-ft-∘F)(35 ft)(122∘F−118∘F)ln0.54 ft 0.46 ft=5.92×10 5 Btu hr
Q˙′′====Q˙A Q˙2 π r o L 5.92×10 5 Btu hr 2(3.14)(0.54 ft)(35 ft)4985 Btu hr-ft 2 Q˙″=Q˙A=Q˙2 π r o L=5.92×10 5 Btu hr 2(3.14)(0.54 ft)(35 ft)=4985 Btu hr-ft 2
Example:
A 10 ft length of pipe with an inner radius of 1 in and an outer radius of 1.25 in has an outer surface temperature of 250°F. The heat transfer rate is 30,000 Btu/hr. Find the interior surface temperature. Assume k = 25 Btu/hr-ft-°F.
Solution:
Q˙=2 π k L(T h−T c)ln(r o/r i)Q˙=2 π k L(T h−T c)ln(r o/r i)
Solving for T h:
T h===Q˙ln(r o/r i)2 π k L+T c(30,000 Btu hr)(ln 1.25 in 1 in)2(3.14)(25 Btu hr-ft-∘F)(10 ft)+250∘F 254∘F T h=Q˙ln(r o/r i)2 π k L+T c=(30,000 Btu hr)(ln1.25 in 1 in)2(3.14)(25 Btu hr-ft-∘F)(10 ft)+250∘F=254∘F
The evaluation of heat transfer through a cylindrical wall can be extended to include a composite body composed of several concentric, cylindrical layers, as shown in Figure 4.
Figure 4: Composite Cylindrical Layers
Example:
A thick-walled nuclear coolant pipe (k s = 12.5 Btu/hr-ft-°F) with 10 in. inside diameter (ID) and 12 in. outside diameter (OD) is covered with a 3 in. layer of asbestos insulation (k a = 0.14 Btu/hr-ft-°F) as shown in Figure 5. If the inside wall temperature of the pipe is maintained at 550°F, calculate the heat loss per foot of length. The outside temperature is 100°F.
Figure 5: Pipe Insulation Problem
Solution:
Q˙L===2 π(T i n−T o)[ln(r 2/r 1)k s+ln(r 3/r 2)k a]2 π(550∘F−100∘F)[ln(6 in 5 in)12.5 Btu hr-ft-∘F+ln(9 in 6 in)0.14 Btu hr-ft-∘F]971 Btu hr-ft Q˙L=2 π(T i n−T o)[ln(r 2/r 1)k s+ln(r 3/r 2)k a]=2 π(550∘F−100∘F)[ln(6 in 5 in)12.5 Btu hr-ft-∘F+ln(9 in 6 in)0.14 Btu hr-ft-∘F]=971 Btu hr-ft
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15307 | https://pregatirematematicaolimpiadejuniori.wordpress.com/wp-content/uploads/2018/05/two_special_points__1_.pdf | ON TWO SPECIAL POINTS IN TRIANGLE KAPIL PAUSE 1. Introduction Today we will learn about two intriguing special points in triangle that seem to have many interesting properties. You will see many examples where spotting these hidden points often makes the problem very easy. These points do not seem to have a special name in the mathematical folklore. Being informal we take the liberty to call these two points as ’Humpty-dumpty points’. As you wil see, since these points are vertex dependent, we shall call them X -Humpty Dumpty points whenever they correspond to vertex X.
2. Some basic facts about Humpty-Dumpty points 2.1. Humpty point.
Definition 1. In triangle ABC the A Humpty point PA is defined to be a point inside triangle such that ∠PABC = ∠PAAB and ∠PACB = ∠PAAC A B C PA Figure 1. A-Humpty Point Facts about PA 1) lies on A median of ABC 2)lies on A appolonius circle of ABC that is, AB AC = PAB PAC 1 2 KAPIL PAUSE 3) B, PA, H, C are concyclic, where H is the orthocentre of ABC 4)HPA ⊥APA 2.2. Dumpty point.
Definition 2. In triangle ABC, the A Dumpty point QA is defined to be a point inside triangle such that ∠QABA = ∠QAAC and ∠QAAB = ∠QACA A B C QA Figure 2. A-Dumpty point facts about QA 1) lies on A symmedian 2) It is the centre of spiral symilarity sending ∆AQAC to ∆CQAB, ie sending AC to BA 3)B, QA, O, C are concyclic where O is circumcentre of ∆ABC.
4) OQA ⊥AQA.
Also notice that the humpty Dumpty points are isogonal conjugates of each other.
We recommend you to try to prove these facts by yourself. This will help you in befriending the humpty-dumpty points well.
If you are done, then lets move forward!
3. Examples Problem 1 (ELMO 2014). In triangle ABC H, O are respectively the orthocentre and circumcentre respectively. Circle ⊙BOC intersects a circle with diameter AO at M. AM intersects ⊙BOC again at X. Similarly ⊙BHC intersects circle with diameter AH at N and AN intersects ⊙BHC second time at Y . Prove that MN ∥ XY ON TWO SPECIAL POINTS IN TRIANGLE 3 A B C O H M N X Y Figure 3. ELMO 2014 Proof. We immediately realize that N, M are A Humpty dumpty points of triangle ABC.Indeed, HM ⊥AM and M lies on ⊙BHC implies its Humpty point and similarly for the other. Using the fact that they are isogonal conjugates we chase some angles and see that ∆AMB ∼∆ACY and ∆ANB ∼∆ACX. This implies AM.AY = AB.AC = AN.AX that is , AM AX = AN AY and we are done.
□ Problem 2 (USAMO 2008). ABC is a triangle. M, N are respectively the midpts of AB, AC.
The perpendicular bisector of AB, AC meet the A median at D, E respectively. BD and CE meet at F. Prove that A, M, F, N are concyclic.
Proof. Ah, yes F is none other than the A−dumpty point of ABC. To reach the desired conclusion, see that OF ⊥AF, ON ⊥AC, OM ⊥AB directly implies A, M, O, N, F are concyclic with AO as diameter. Another way to finish would have been to see that since a spiral similiarity centered at F maps AC to BA, it also maps the midpt of AB to midpoint of AC that is, M to N. The angle of spiral similarity is ∠AFC which is 180 −A, and this must also be the ∠MFN and we finish.
□ 4 KAPIL PAUSE A B C M N D E F Figure 4. USAMO 2008 Problem 3 (USA TSTST 2015). ABC is a scalene triangle. Ka, La, Ma are respec-tively the intersections BC with the interior angle bisector,exterior angle bisector and median from A. Circle ⊙AKaLa intersects AMa again at Xa. Similarly define Xb, Xc. Prove that the circumcentre of XaXbXc lies on euler line of ABC.
A B C H Xa Xb Xc G Figure 5. USA TSTST 2015 Proof. The problem looks really monstrous at first sight, but we will see how in-nocent it is. Clearly, ⊙AKaLa is the A applonius circle of ABC and so we have Xa as the A Humpty point of ABC. let us invoke another property of Xa, namely ON TWO SPECIAL POINTS IN TRIANGLE 5 HXa ⊥AMa, similarly for b, c. what do we get? Since AMa, BMb, CMc are con-current at centroid G, which implies Xa, Xb, Xc lies on a circle with diameter HG.
So the centre is midpt of HG, lies on euler line indeed.
□ Problem 4 (USA TST 2005). P is a point inside triangle ABC such that ∠PAB = ∠PBC, ∠PAC = ∠PCB. The perpendicular bisector of AP meets BC at Q. If O is the circumcentre of ABC then prove that ∠AQP = 2∠OQB A B C M O Q P Figure 6. USA TST 2005 Proof. Sure enough, it shouldnt take us much time to realize that P is the A Humpty point of ABC, hence AP is A median. Let AP intersect BCat M. The condition of problem requires 90 −1 2∠AQP = 90 −∠OQB ⇔∠QAM = ∠QOM ⇔AQMO is cyclic, that is ∠QAO = 90 or QA tangent to ⊙ABC.
This suggests that we should start working by taking Q′ such that Q′A tangent to ⊙ABC and try to prove that Q′A = Q′P. We know that AB AC = BQ′2 CQ′2 . also we know AB AC = PB PC , from properties of Humpty point. Hence we have PB PC = BQ′2 CQ′2 . We show that this property leads to Q′P tangent to ⊙PBC. Indeed, if tangent from P to ⊙PBC meets BC at Q” then PB PC = BQ”2 CQ”2 so BQ”2 CQ”2 = BQ′2 CQ′2 which implies Q′ = Q”.
Now by power of point, Q′A2 = Q′B.Q′C = Q′P 2.
Hence Q′A = Q′P and we are done.
□ 6 KAPIL PAUSE 4. excercises Given are some problems for the reader to try.
Exercise 1. The A symmedian of ABC meets its circumcircle at K. Reflection of K in BC is K′. Prove that AK′ is a median.
Exercise 2. P is a variable point on side BC of triangle ABC. M, N are respec-tively on AB, AC such that PM ∥AC, PN ∥AB. Prove that as P varies on BC, ⊙AMN passes through a fixed point.
Exercise 3. M, N are points on semicircle with diameter AB and centre O. NM meets AB at X. circles ⊙MBO, ⊙NAO meet at K. Prove that XK ⊥KO Exercise 4. QA is the A dumpty point of ABC. AD is altitude from A on BC.
Prove that DQA bisects the line segment joining the midpoints of AB, AC Exercise 5. P is point on A symmedian of ABC. O1, O2 are respectively cir-cumcentres of APB, CAP. If O is circumcentre of ABC, prove that AO bisects O1O2 Exercise 6. AD is an altitude from A to BC in triangle ABC. A circle with centre on AD is tangent externally to ⊙BOC at X, where O iis circumcentre of ABC. Prove that AX is the A symmedian. |
15308 | https://blog.chemondis.com/8283/know-your-chemical-uses-for-citric-acid/ | Visit Marketplace
Know Your Chemicals: Citric Acid (C6H8O7)
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Citric Acid’s Structure, Uses and Supply Landscape
Citric acid (C6H8O7) is a naturally occurring organic acid found in citrus fruits like lemons, limes, and oranges. Known for its tart taste, citric acid is widely used in food, cleaning products, cosmetics, and various industrial applications. It is highly versatile and safe, making it a staple in both household and commercial products. This article delves into the chemical structure, uses of citric acid, and the importance of it in multiple industries. It also gaves insights into its role in cleaning and skincare.
Chemical Structure and Function
The chemical formula of citric acid is C₆H₈O₇. Structurally, it consists of three carboxyl groups (-COOH) and a hydroxyl group (-OH) attached to a six-carbon skeleton. These functional groups give citric acid its characteristic acidic properties and allow it to act as a chelating agent, binding with metal ions like calcium and magnesium. Citric acid (C6H8O7) is also known by its CAS number which is (CAS: 77-92-9). You can search with this number to find Suppliers of citric acid at CheMondis.
This ability to bind with metal ions is what makes citric acid effective in softening water and removing mineral deposits, making it a key ingredient in cleaning agents. Citric acid also serves as a preservative by lowering pH levels. Which prevents the growth of bacteria and other microorganisms in various products. In food and beverages, uses of citric acid allowes the food to have flavors, stabilize ingredients, and prolong shelf life.
The Use of Citric Acid in Industries
Citric acid is a major player in multiple industries, particularly the food and beverage sector, which consumes about 70% of the global supply. It acts as a flavor enhancer, preservative, and pH regulator in products like soft drinks, jams, candies, and canned foods.
In addition to food production, uses of citric acid has significantly high reach in household cleaning products, cosmetics, pharmaceuticals, and water treatment. It is widely used in the textile and detergent industries to balance pH levels and soften fabrics. Its non-toxic, biodegradable nature has also made it a preferred choice in environmentally friendly cleaning solutions.
Citric Acid for Cleaning
One of the most notable uses of citric acid in daily life is its effectiveness in cleaning products. Citric acid is a natural, biodegradable alternative to harsher chemical cleaners and is known for its ability to dissolve calcium deposits and other mineral buildups. It acts as a descaler and can be used to remove hard water stains from faucets, glassware, and appliances.
Citric acid for cleaning is commonly found in eco-friendly cleaning products, as it is non-toxic and safe for use around the home. It effectively cleans surfaces, removes limescale, neutralizes odors, and acts as a disinfectant. Due to its ability to break down grease and grime, citric acid is a popular ingredient in multipurpose cleaning solutions. Whether used for cleaning bathrooms, kitchens, or laundry, citric acid provides a safe and effective solution for maintaining cleanliness without harmful chemicals.
In Skincare: What Does Citric Acid Do to Your Skin?
In addition to its industrial and cleaning applications, uses of citric acid in skincarehas gained popularity, particularly for its exfoliating and skin-brightening properties. As an alpha hydroxy acid (AHA), citric acid is used in skincare products such as toners, masks, and creams to promote healthy skin turnover.
But what does citric acid do to your skin? Here are a few key benefits:
Exfoliation: Citric acid helps to slough off dead skin cells, revealing fresher, smoother skin underneath. This promotes a brighter complexion and can help unclog pores, reducing the occurrence of breakouts.
Brightening: By reducing the appearance of dark spots and hyperpigmentation, citric acid contributes to an even skin tone, making it ideal for people with age spots or sun damage.
Anti-aging: Citric acid’s exfoliating properties promote the production of new skin cells, which can reduce the appearance of fine lines and wrinkles.
pH Regulation: Citric acid helps maintain the skin’s natural pH balance, protecting the skin’s moisture barrier and keeping it healthy and hydrated.
While citric acid (C6H8O7) offers numerous skincare benefits, it is important to use it properly to avoid irritation, especially for sensitive skin types. Using sunscreen is also critical, as exfoliated skin can be more susceptible to UV damage.
Understanding the Landscape of Citric Acid Suppliers
With increasing demand for citric acid across various industries, the need for reliable Suppliers is crucial. China leads in the production of citric acid, benefiting from both raw material availability and cost-effective manufacturing processes. The global supply chain for citric acid caters to industries ranging from food and beverages to pharmaceuticals, personal care, and household cleaning products.
CheMondis and Citric Acid
CheMondis, an online B2B platform that connects Buyers and Suppliers of chemicals, including citric acid. The platform simplifies the sourcing process by offering transparent procurement, technical data, and easy purchasing options. For companies looking to source high-quality citric acid, CheMondis provides a reliable marketplace that ensures quality and timely delivery.
Citric acid is a versatile compound with a wide range of applications, from food preservation to cleaning and skincare. Its unique chemical structure allows it to function as a preservative, pH regulator, and chelating agent. As industries continue to seek safer, more eco-friendly ingredients, citric acid’s demand will only rise. Platforms like CheMondis make it easier for businesses to source citric acid from trusted Suppliers. Ensuring they get the best quality for their specific needs.
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15309 | https://portal.tpu.ru/SHARED/k/KONVAL/Sites/English_sites/Site3_M/5/5_02.htm | Linear Equations
Algebraic Equations and Inequalities
Basic Conceptions Properties of Equations and Inequalities Graphical Interpretation of Solutions Linear Equations and Inequalities Linear Equations Linear Inequalities Linear Equations and Inequalities Involving Absolute Values Linear Equations Involving Absolute Value Linear Equations Involving a Few Absolute Values Linear Inequalities Involving Absolute Value Quadratic Equations and Inequalities Quadratic Equations Quadratic Equations and Quadratic Functions Extreme Value of Quadratic Function Quadratic Formula Solving Quadratic Equations by Factoring Quadratic InequalitiesLinear Equations Key Topics Remaining: Linear Inequalities » Linear Equations Involving Absolute Values » Linear Inequalities Involving Absolute Values » Quadratic Equations » Quadratic Equations and Quadratic Functions » Extreme Value of Quadratic Function » Quadratic Formula » Solving Quadratic Equations by Factoring » Quadratic Inequalities The simplest algebraic equation is a linear equation, k x + b = 0, where x is the variable; k and b are constants (k ≠ 0). A graph of a linear function, y = k x + b = 0(k ≠ 0), is a straight line with the only one x-intercept, x = – b / a. So a linear equation has a single solution, . In mathematical literature, the equation k x + b = 0 is known as an equation of a line in slope-intercept form, since b is the y-intercept, and k is the slope of the line. . |
15310 | https://www.cliffsnotes.com/study-notes/9887490 | Lab1B C6 Formal bomb calorimetry (docx) - CliffsNotes
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Lab1B C6 Formal bomb calorimetry
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CHEMICAL AND BIOMOLECULAR ENGINEERING LAB / BIOENGINEERING LAB CH1802 Bomb Calorimetry Location: N1.2 B3-12 Name: Matric Number: Group: CBE02 Date of experiment: 10/03/2023
SUMMARY AND OBJECTIVES This bomb calorimeter experiment is used to gain important thermodynamic properties such as standard enthalpy of combustion and standard enthalpy of formation of compounds. It does so by measuring the changes in internal energy (ΔU) during the combustion of a chemical substance. The objectives of this experiment: 1.Utilize bomb calorimetry to demonstrate materials and energy balance of a combustion process in a closed system. 2.Understand the use of bomb calorimetry to obtain heats of combustion for benzoic acid. 3.Calculate the standard internal energy of combustion and standard enthalpy of combustion of a substance. 4.Calculate the standard enthalpy of formation of a substance. 5.Practice unit conversions. THEORY The sample (or fuel) is introduced into a bomb and the latter charged with oxygen gas. Then the bomb is immersed in a water bath which is continuously stirred to ensure that thermal equilibrium is maintained between the bomb and the water bath. A thermometer is also immersed in the water. The water bath is thermally insulated from the external environment via a jacket. The bomb is then ignited by means of a fuse wire. The heat of combustion of the sample in oxygen will cause an increase in the water temperature, which is recorded. The stoichiometry of the combustion reaction (materials balance) is: C x H y O z+ (x+y 4− z 2)O 2(g) = xCO 2(g) + y 2 H 2 O The energy balance for the closed system comprising of the whole calorimeter is: ΔU + ΔE K+ ΔE p = Q + W where ΔU, ΔE K and ΔE p are the changes in internal energy, kinetic energy, and potential energy of the system respectively, Q is heat and W is work. For this experiment, the terms ΔE k, ΔE p , Q, and W can be neglected. Equation of energy balance then becomes: ΔU = 0 = ΔU sample+ ΔU fuse+ ΔU calorimeter The heat generated by the combustion process is absorbed by the calorimeter comprising of the bomb, water, and water bucket: (-ΔU c°)N + (-ΔU fuse) = C v Δt = (-ΔU c°)M + (-ΔU fuse ) ΔU fuse = -2.3cal/cm x length of fuse wire burnt in cm Where ΔU c° is the standard internal energy of combustion in cal/mol, or ΔU c ° is the standard internal energy of combustion in cal/g. N is the number of moles of the benzoic acid sample, M is the mass of the fuel, ΔU fuse is the change in thermal energy of the fuse wire, C v is the heat capacity of the calorimeter and
the Δt is the net corrected temperature rise of the calorimeter. Note that both ΔU c° and ΔU fuse represent exothermic processes and therefore have negative values. The standard enthalpy of combustion, ΔH c ° can be obtained from ΔU c °: ΔH c° = ΔU c° + Δ(PV) = ΔU c° + Δ(nRT) = ΔU c ° + RTΔn Where Δn is the change in the number of moles of gas during combustion based on the stoichiometry. Finally, the standard enthalpy of formation of the chemical compound, ΔH f ° can then be calculated from ΔH° using Hess's law.
MATERIALS - GENERAL LAB EQUIPMENT 1.Chemical balance (0.1mg sensitivity) 2.Solution balance (capacity of 2.0kg with 0.1g sensitivity) 3.Oxygen tank with CBA no. 540 outlet 4.1341 Parr oxygen bomb calorimetry 5.Parr 45C10 nickel chromium fuse wire 6.2 electrical outlets (one for ignition and one for the stirrer) 7.Tweezers 8.Stopwatch 9.Ruler MATERIALS - CONSUMABLES AND REAGENTS 1.2000g distilled water (or tap water containing less than 250ppm of dissolved solids) 2.Fuel samples (3x benzoic acid) 3.Pipette (1 ml) 4.Distilled water container 5.Paper towel
EXPERIMENTAL PROCEDURES A)Prepare the sample: 1)Set the bomb head on the A38A support stand (the smaller stand) and fasten a 10cm length of fuse wire between the 2 electrodes. The fuse wire is furnished on cards from which uniform 10cm lengths can be cut out without further measurement. 2)To attach the fuse quick-grip electrodes, insert the ends of the wire into the eyelet at the end of each stem and push cap downward to pinch the wire into place. 3)Weigh the samples (benzoic acid) up to an accuracy of 0.1 mg and place it in the fuel capsule. Use tweezers or gloves to handle the sample. Do not touch the sample with your bare hands. 4)Place the fuel capsule with its sample in the electrode loop and bend the wire downward toward the surface of the charge. When using pelleted samples, bend the wire so that the loop bears against the top of the pellet firmly enough to keep it from sliding against the side of the capsule (See Figure 2). Tilt the capsule slightly to one side so that the flame emerging from it will not impinge directly on the tip of the straight electrode. B)Close the bomb: 1)Add 1.0 ml of distilled water from a pipette to the bottom of the bomb (NOT the capsule). 2)Care must be taken not to disturb the sample when moving the bomb head from the support stand to the bomb cylinder. 3)Moisten the sealing ring with bit of water so that it will slide freely into the cylinder; then slide the head into the cylinder and push it down as far as it will go. For easy insertion, push the head straight down without twisting and leave the gas release valve open during this operation. 4)Set the screw cap on the cylinder and turn it down firmly by hand to a stop. Use hand pressure ONLY. When properly closed, no threads on the cylinder should be exposed. If the screw cap binds to the cylinder at this point, indicating that it might be difficult to open the bomb after it has been fired, turn the screw cap back slightly - but only a few degrees - enough to release the binding, since the bottom thread must remain fully engaged. 5)Hand tightens the gas release valve. C)Fill the bomb: 1)Carry the bomb carefully with both hands to where the oxygen cylinder is located. Care must be taken not to disturb the sample. Place the bomb on the support provided. 2)Slide the connector from the oxygen hose onto the gas inlet valve on the bomb head and push it down as far as it will go. 3)Open the filling connection control valve slowly and watch the gauge as the bomb pressure rises to 30atm. Then close the control valve. 4)Release the residual pressure in the filling hose by pushing on the lever attached to the relief valve. The gauge should now return to zero. Return the lever to the original position. 5)Remove the oxygen hose connector from the gas inlet valve. Carry the bomb back the bench VERY carefully with both hands as the bomb contains oxygen gas at HIGH PRESSURE. 6)If too much oxygen is accidentally introduced into the bomb (i.e. more than 40atm), DO NOT PROCEED WITH THE COMBUSTION. Ask for assistance. Detach the filling connection;
Page 1 of 18
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15311 | https://saylordotorg.github.io/text_general-chemistry-principles-patterns-and-applications-v1.0/s14-04-the-ideal-gas-law.html | Previous Section
Table of Contents
Next Section
10.4 The Ideal Gas Law
Learning Objective
To use the ideal gas law to describe the behavior of a gas.
In Section 10.3 "Relationships among Pressure, Temperature, Volume, and Amount", you learned how the volume of a gas changes when its pressure, temperature, or amount is changed, as long as the other two variables are held constant. In this section, we describe how these relationships can be combined to give a general expression that describes the behavior of a gas.
Deriving the Ideal Gas Law
Any set of relationships between a single quantity (such as V) and several other variables (P, T, and n) can be combined into a single expression that describes all the relationships simultaneously. The three individual expressions derived in Section 10.3 "Relationships among Pressure, Temperature, Volume, and Amount" are as follows:
Boyle’s law
Charles’s law
Avogadro’s law
Combining these three expressions gives
Equation 10.9
which shows that the volume of a gas is proportional to the number of moles and the temperature and inversely proportional to the pressure. This expression can also be written as
Equation 10.10
By convention, the proportionality constant in Equation 10.10 is called the gas constantA proportionality constant that is used in the ideal gas law., which is represented by the letter R. Inserting R into Equation 10.10 gives
Equation 10.11
Clearing the fractions by multiplying both sides of Equation 10.11 by P gives
Equation 10.12
PV = nRT
This equation is known as the ideal gas lawA law relating pressure, temperature, volume, and the amount of an ideal gas..
An ideal gasA hypothetical gaseous substance whose behavior is independent of attractive and repulsive forces. is defined as a hypothetical gaseous substance whose behavior is independent of attractive and repulsive forces and can be completely described by the ideal gas law. In reality, there is no such thing as an ideal gas, but an ideal gas is a useful conceptual model that allows us to understand how gases respond to changing conditions. As we shall see, under many conditions, most real gases exhibit behavior that closely approximates that of an ideal gas. The ideal gas law can therefore be used to predict the behavior of real gases under most conditions. As you will learn in Section 10.8 "The Behavior of Real Gases", the ideal gas law does not work well at very low temperatures or very high pressures, where deviations from ideal behavior are most commonly observed.
Note the Pattern
Significant deviations from ideal gas behavior commonly occur at low temperatures and very high pressures.
Before we can use the ideal gas law, however, we need to know the value of the gas constant R. Its form depends on the units used for the other quantities in the expression. If V is expressed in liters (L), P in atmospheres (atm), T in kelvins (K), and n in moles (mol), then
Equation 10.13
R = 0.082057 (L·atm)/(K·mol)
Because the product PV has the units of energy, as described in Chapter 5 "Energy Changes in Chemical Reactions", Section 5.1 "Energy and Work" and Essential Skills 4 (Chapter 5 "Energy Changes in Chemical Reactions", Section 5.6 "Essential Skills 4"), R can also have units of J/(K·mol) or cal/(K·mol):
Equation 10.14
R = 8.3145 J/(K·mol) = 1.9872 cal/(K·mol)
Scientists have chosen a particular set of conditions to use as a reference: 0°C (273.15 K) and 1 atm pressure, referred to as standard temperature and pressure (STP)The conditions 0°C (273.15 K) and 1 atm pressure for a gas.. We can calculate the volume of 1.000 mol of an ideal gas under standard conditions using the variant of the ideal gas law given in Equation 10.11:
Equation 10.15
Thus the volume of 1 mol of an ideal gas at 0°C and 1 atm pressure is 22.41 L, approximately equivalent to the volume of three basketballs. The quantity 22.41 L is called the standard molar volumeThe volume of 1 mol of an ideal gas at STP (0°C and 1 atm pressure), which is 22.41 L. of an ideal gas. The molar volumes of several real gases at STP are given in Table 10.3 "Molar Volumes of Selected Gases at Standard Temperature (0°C) and Pressure (1 atm)", which shows that the deviations from ideal gas behavior are quite small. Thus the ideal gas law does a good job of approximating the behavior of real gases at STP. The relationships described in Section 10.3 "Relationships among Pressure, Temperature, Volume, and Amount" as Boyle’s, Charles’s, and Avogadro’s laws are simply special cases of the ideal gas law in which two of the four parameters (P, V, T, and n) are held fixed.
Table 10.3 Molar Volumes of Selected Gases at Standard Temperature (0°C) and Pressure (1 atm)
| Gas | Molar Volume (L) |
--- |
| He | 22.434 |
| Ar | 22.397 |
| H2 | 22.433 |
| N2 | 22.402 |
| O2 | 22.397 |
| CO2 | 22.260 |
| NH3 | 22.079 |
If n, R, and T are all constant in Equation 10.11, the equation reduces to
Equation 10.16
which is exactly the same as Boyle’s law in Equation 10.6.
Similarly, Charles’s law states that the volume of a fixed quantity of gas is directly proportional to its temperature at constant pressure. If n and P in Equation 10.11 are fixed, then
Equation 10.17
which is exactly the same as Equation 10.7.
Applying the Ideal Gas Law
The ideal gas law allows us to calculate the value of the fourth variable for a gaseous sample if we know the values of any three of the four variables (P, V, T, and n). It also allows us to predict the final state of a sample of a gas (i.e., its final temperature, pressure, volume, and amount) following any changes in conditions if the parameters (P, V, T, and n) are specified for an initial state. Some applications are illustrated in the following examples. The approach used throughout is always to start with the same equation—the ideal gas law—and then determine which quantities are given and which need to be calculated. Let’s begin with simple cases in which we are given three of the four parameters needed for a complete physical description of a gaseous sample.
Example 5
The balloon that Charles used for his initial flight in 1783 was destroyed, but we can estimate that its volume was 31,150 L (1100 ft3), given the dimensions recorded at the time. If the temperature at ground level was 86°F (30°C) and the atmospheric pressure was 745 mmHg, how many moles of hydrogen gas were needed to fill the balloon?
Given: volume, temperature, and pressure
Asked for: amount of gas
Strategy:
A Solve the ideal gas law for the unknown quantity, in this case n.
B Make sure that all quantities are given in units that are compatible with the units of the gas constant. If necessary, convert them to the appropriate units, insert them into the equation you have derived, and then calculate the number of moles of hydrogen gas needed.
Solution:
A We are given values for P, T, and V and asked to calculate n. If we solve the ideal gas law (Equation 10.12) for n, we obtain
BP and T are given in units that are not compatible with the units of the gas constant [R = 0.082057 (L·atm)/(K·mol)]. We must therefore convert the temperature to kelvins and the pressure to atmospheres:
Substituting these values into the expression we derived for n, we obtain
Exercise
Suppose that an “empty” aerosol spray-paint can has a volume of 0.406 L and contains 0.025 mol of a propellant gas such as CO2. What is the pressure of the gas at 25°C?
Answer: 1.5 atm
In Example 5, we were given three of the four parameters needed to describe a gas under a particular set of conditions, and we were asked to calculate the fourth. We can also use the ideal gas law to calculate the effect of changes in any of the specified conditions on any of the other parameters, as shown in Example 6.
Example 6
Suppose that Charles had changed his plans and carried out his initial flight not in August but on a cold day in January, when the temperature at ground level was −10°C (14°F). How large a balloon would he have needed to contain the same amount of hydrogen gas at the same pressure as in Example 5?
Given: temperature, pressure, amount, and volume in August; temperature in January
Asked for: volume in January
Strategy:
A Use the results from Example 5 for August as the initial conditions and then calculate the change in volume due to the change in temperature from 86°F to 14°F. Begin by constructing a table showing the initial and final conditions.
B Rearrange the ideal gas law to isolate those quantities that differ between the initial and final states on one side of the equation, in this case V and T.
C Equate the ratios of those terms that change for the two sets of conditions. Making sure to use the appropriate units, insert the quantities and solve for the unknown parameter.
Solution:
A To see exactly which parameters have changed and which are constant, prepare a table of the initial and final conditions:
| August (initial) | January (final) |
--- |
| T | 30°C = 303 K | −10°C = 263 K |
| P | 0.980 atm | 0.980 atm |
| n | 1.23 × 103 mol H2 | 1.23 × 103 mol H2 |
| V | 31,150 L | ? |
Thus we are asked to calculate the effect of a change in temperature on the volume of a fixed amount of gas at constant pressure.
B Recall that we can rearrange the ideal gas law to give
Both n and P are the same in both cases, which means that nR/P is a constant. Dividing both sides by T gives
This is the relationship first noted by Charles.
C We see from this expression that under conditions where the amount (n) of gas and the pressure (P) do not change, the ratio V/T also does not change. If we have two sets of conditions for the same amount of gas at the same pressure, we can therefore write
where the subscripts 1 and 2 refer to the initial and final conditions, respectively. Solving for V2 and inserting the given quantities in the appropriate units, we obtain
It is important to check your answer to be sure that it makes sense, just in case you have accidentally inverted a quantity or multiplied rather than divided. In this case, the temperature of the gas decreases. Because we know that gas volume decreases with decreasing temperature, the final volume must be less than the initial volume, so the answer makes sense. We could have calculated the new volume by plugging all the given numbers into the ideal gas law, but it is generally much easier and faster to focus on only the quantities that change.
Exercise
At a laboratory party, a helium-filled balloon with a volume of 2.00 L at 22°C is dropped into a large container of liquid nitrogen (T = −196°C). What is the final volume of the gas in the balloon?
Answer: 0.52 L
Example 6 illustrates the relationship originally observed by Charles. We could work through similar examples illustrating the inverse relationship between pressure and volume noted by Boyle (PV = constant) and the relationship between volume and amount observed by Avogadro (V/n = constant). We will not do so, however, because it is more important to note that the historically important gas laws are only special cases of the ideal gas law in which two quantities are varied while the other two remain fixed. The method used in Example 6 can be applied in any such case, as we demonstrate in Example 7 (which also shows why heating a closed container of a gas, such as a butane lighter cartridge or an aerosol can, may cause an explosion).
Example 7
Aerosol cans are prominently labeled with a warning such as “Do not incinerate this container when empty.” Assume that you did not notice this warning and tossed the “empty” aerosol can in Exercise 5 (0.025 mol in 0.406 L, initially at 25°C and 1.5 atm internal pressure) into a fire at 750°C. What would be the pressure inside the can (if it did not explode)?
Given: initial volume, amount, temperature, and pressure; final temperature
Asked for: final pressure
Strategy:
Follow the strategy outlined in Example 6.
Solution:
Prepare a table to determine which parameters change and which are held constant:
| Initial | Final |
--- |
| V | 0.406 L | 0.406 L |
| n | 0.025 mol | 0.025 mol |
| T | 25°C = 298 K | 750°C = 1023 K |
| P | 1.5 atm | ? |
Once again, two parameters are constant while one is varied, and we are asked to calculate the fourth. As before, we begin with the ideal gas law and rearrange it as necessary to get all the constant quantities on one side. In this case, because V and n are constant, we rearrange to obtain
Dividing both sides by T, we obtain an equation analogous to the one in Example 6, P/T = nR/V = constant. Thus the ratio of P to T does not change if the amount and volume of a gas are held constant. We can thus write the relationship between any two sets of values of P and T for the same sample of gas at the same volume as
In this example, P1 = 1.5 atm, T1 = 298 K, and T2 = 1023 K, and we are asked to find P2. Solving for P2 and substituting the appropriate values, we obtain
This pressure is more than enough to rupture a thin sheet metal container and cause an explosion!
Exercise
Suppose that a fire extinguisher, filled with CO2 to a pressure of 20.0 atm at 21°C at the factory, is accidentally left in the sun in a closed automobile in Tucson, Arizona, in July. The interior temperature of the car rises to 160°F (71.1°C). What is the internal pressure in the fire extinguisher?
Answer: 23.4 atm
In Example 10.6 and Example 10.7, two of the four parameters (P, V, T, and n) were fixed while one was allowed to vary, and we were interested in the effect on the value of the fourth. In fact, we often encounter cases where two of the variables P, V, and T are allowed to vary for a given sample of gas (hence n is constant), and we are interested in the change in the value of the third under the new conditions. If we rearrange the ideal gas law so that P, V, and T, the quantities that change, are on one side and the constant terms (R and n for a given sample of gas) are on the other, we obtain
Equation 10.18
Thus the quantity PV/T is constant if the total amount of gas is constant. We can therefore write the relationship between any two sets of parameters for a sample of gas as follows:
Equation 10.19
This equation can be solved for any of the quantities P2, V2, or T2 if the initial conditions are known, as shown in Example 8.
Example 8
We saw in Example 5 that Charles used a balloon with a volume of 31,150 L for his initial ascent and that the balloon contained 1.23 × 103 mol of H2 gas initially at 30°C and 745 mmHg. Suppose that Gay-Lussac had also used this balloon for his record-breaking ascent to 23,000 ft and that the pressure and temperature at that altitude were 312 mmHg and −30°C, respectively. To what volume would the balloon have had to expand to hold the same amount of hydrogen gas at the higher altitude?
Given: initial pressure, temperature, amount, and volume; final pressure and temperature
Asked for: final volume
Strategy:
Follow the strategy outlined in Example 6.
Solution:
Begin by setting up a table of the two sets of conditions:
| Initial | Final |
--- |
| P | 745 mmHg = 0.980 atm | 312 mmHg = 0.411 atm |
| T | 30°C = 303 K | −30°C = 243 K |
| n | 1.23 × 103 mol H2 | 1.23 × 103 mol H2 |
| V | 31,150 L | ? |
Thus all the quantities except V2 are known. Solving Equation 10.19 for V2 and substituting the appropriate values give
Does this answer make sense? Two opposing factors are at work in this problem: decreasing the pressure tends to increase the volume of the gas, while decreasing the temperature tends to decrease the volume of the gas. Which do we expect to predominate? The pressure drops by more than a factor of two, while the absolute temperature drops by only about 20%. Because the volume of a gas sample is directly proportional to both T and 1/P, the variable that changes the most will have the greatest effect on V. In this case, the effect of decreasing pressure predominates, and we expect the volume of the gas to increase, as we found in our calculation.
We could also have solved this problem by solving the ideal gas law for V and then substituting the relevant parameters for an altitude of 23,000 ft:
Except for a difference caused by rounding to the last significant figure, this is the same result we obtained previously. There is often more than one “right” way to solve chemical problems.
Exercise
A steel cylinder of compressed argon with a volume of 0.400 L was filled to a pressure of 145 atm at 10°C. At 1.00 atm pressure and 25°C, how many 15.0 mL incandescent light bulbs could be filled from this cylinder? (Hint: find the number of moles of argon in each container.)
Answer: 4.07 × 103
Using the Ideal Gas Law to Calculate Gas Densities and Molar Masses
The ideal gas law can also be used to calculate molar masses of gases from experimentally measured gas densities. To see how this is possible, we first rearrange the ideal gas law to obtain
Equation 10.20
The left side has the units of moles per unit volume (mol/L). The number of moles of a substance equals its mass (m, in grams) divided by its molar mass (M, in grams per mole):
Equation 10.21
Substituting this expression for n into Equation 10.20 gives
Equation 10.22
Because m/V is the density d of a substance, we can replace m/V by d and rearrange to give
Equation 10.23
The distance between particles in gases is large compared to the size of the particles, so their densities are much lower than the densities of liquids and solids. Consequently, gas density is usually measured in grams per liter (g/L) rather than grams per milliliter (g/mL).
Example 9
Calculate the density of butane at 25°C and a pressure of 750 mmHg.
Given: compound, temperature, and pressure
Asked for: density
Strategy:
A Calculate the molar mass of butane and convert all quantities to appropriate units for the value of the gas constant.
B Substitute these values into Equation 10.23 to obtain the density.
Solution:
A The molar mass of butane (C4H10) is
(4)(12.011) + (10)(1.0079) = 58.123 g/mol
Using 0.082057 (L·atm)/(K·mol) for R means that we need to convert the temperature from degrees Celsius to kelvins (T = 25 + 273 = 298 K) and the pressure from millimeters of mercury to atmospheres:
B Substituting these values into Equation 10.23 gives
Exercise
Radon (Rn) is a radioactive gas formed by the decay of naturally occurring uranium in rocks such as granite. It tends to collect in the basements of houses and poses a significant health risk if present in indoor air. Many states now require that houses be tested for radon before they are sold. Calculate the density of radon at 1.00 atm pressure and 20°C and compare it with the density of nitrogen gas, which constitutes 80% of the atmosphere, under the same conditions to see why radon is found in basements rather than in attics.
Answer: radon, 9.23 g/L; N2, 1.17 g/L
A common use of Equation 10.23 is to determine the molar mass of an unknown gas by measuring its density at a known temperature and pressure. This method is particularly useful in identifying a gas that has been produced in a reaction, and it is not difficult to carry out. A flask or glass bulb of known volume is carefully dried, evacuated, sealed, and weighed empty. It is then filled with a sample of a gas at a known temperature and pressure and reweighed. The difference in mass between the two readings is the mass of the gas. The volume of the flask is usually determined by weighing the flask when empty and when filled with a liquid of known density such as water. The use of density measurements to calculate molar masses is illustrated in Example 10.
Example 10
The reaction of a copper penny with nitric acid results in the formation of a red-brown gaseous compound containing nitrogen and oxygen. A sample of the gas at a pressure of 727 mmHg and a temperature of 18°C weighs 0.289 g in a flask with a volume of 157.0 mL. Calculate the molar mass of the gas and suggest a reasonable chemical formula for the compound.
Given: pressure, temperature, mass, and volume
Asked for: molar mass and chemical formula
Strategy:
A Solve Equation 10.23 for the molar mass of the gas and then calculate the density of the gas from the information given.
B Convert all known quantities to the appropriate units for the gas constant being used. Substitute the known values into your equation and solve for the molar mass.
C Propose a reasonable empirical formula using the atomic masses of nitrogen and oxygen and the calculated molar mass of the gas.
Solution:
A Solving Equation 10.23 for the molar mass gives
Density is the mass of the gas divided by its volume:
B We must convert the other quantities to the appropriate units before inserting them into the equation:
The molar mass of the unknown gas is thus
C The atomic masses of N and O are approximately 14 and 16, respectively, so we can construct a list showing the masses of possible combinations:
The most likely choice is NO2 which is in agreement with the data. The red-brown color of smog also results from the presence of NO2 gas.
Exercise
You are in charge of interpreting the data from an unmanned space probe that has just landed on Venus and sent back a report on its atmosphere. The data are as follows: pressure, 90 atm; temperature, 557°C; density, 58 g/L. The major constituent of the atmosphere (>95%) is carbon. Calculate the molar mass of the major gas present and identify it.
Answer: 44 g/mol; CO2
Summary
The empirical relationships among the volume, the temperature, the pressure, and the amount of a gas can be combined into the ideal gas law, PV = nRT. The proportionality constant, R, is called the gas constant and has the value 0.08206 (L·atm)/(K·mol), 8.3145 J/(K·mol), or 1.9872 cal/(K·mol), depending on the units used. The ideal gas law describes the behavior of an ideal gas, a hypothetical substance whose behavior can be explained quantitatively by the ideal gas law and the kinetic molecular theory of gases. Standard temperature and pressure (STP) is 0°C and 1 atm. The volume of 1 mol of an ideal gas at STP is 22.41 L, the standard molar volume. All of the empirical gas relationships are special cases of the ideal gas law in which two of the four parameters are held constant. The ideal gas law allows us to calculate the value of the fourth quantity (P, V, T, or n) needed to describe a gaseous sample when the others are known and also predict the value of these quantities following a change in conditions if the original conditions (values of P, V, T, and n) are known. The ideal gas law can also be used to calculate the density of a gas if its molar mass is known or, conversely, the molar mass of an unknown gas sample if its density is measured.
Key Takeaway
The ideal gas law is derived from empirical relationships among the pressure, the volume, the temperature, and the number of moles of a gas; it can be used to calculate any of the four properties if the other three are known.
Key Equations
Ideal gas law
Equation 10.12: PV = nRT
Relationship between initial and final conditions
Equation 10.19:
Density of a gas
Equation 10.23:
Conceptual Problems
For an ideal gas, is volume directly proportional or inversely proportional to temperature? What is the volume of an ideal gas at absolute zero?
What is meant by STP? If a gas is at STP, what further information is required to completely describe the state of the gas?
For a given amount of a gas, the volume, temperature, and pressure under any one set of conditions are related to the volume, the temperature, and the pressure under any other set of conditions by the equation Derive this equation from the ideal gas law. At constant temperature, this equation reduces to one of the laws discussed in Section 10.3 "Relationships among Pressure, Temperature, Volume, and Amount"; which one? At constant pressure, this equation reduces to one of the laws discussed in Section 10.3 "Relationships among Pressure, Temperature, Volume, and Amount"; which one?
Predict the effect of each change on one variable if the other variables are held constant.
If the number of moles of gas increases, what is the effect on the temperature of the gas?
If the temperature of a gas decreases, what is the effect on the pressure of the gas?
If the volume of a gas increases, what is the effect on the temperature of the gas?
If the pressure of a gas increases, what is the effect on the number of moles of the gas?
What would the ideal gas law be if the following were true?
volume were proportional to pressure
temperature were proportional to amount
pressure were inversely proportional to temperature
volume were inversely proportional to temperature
both pressure and volume were inversely proportional to temperature
Given the following initial and final values, what additional information is needed to solve the problem using the ideal gas law?
| Given | Solve for |
--- |
| V1, T1, T2, n1 | n 2 |
| P1, P2, T2, n2 | n 1 |
| T1, T2 | V 2 |
| P1, n1 | P 2 |
7. Given the following information and using the ideal gas law, what equation would you use to solve the problem?
| Given | Solve for |
--- |
| P1, P2, T1 | T 2 |
| V1, n1, n2 | V 2 |
| T1, T2, V1, V2, n2 | n 1 |
8. Using the ideal gas law as a starting point, derive the relationship between the density of a gas and its molar mass. Which would you expect to be denser—nitrogen or oxygen? Why does radon gas accumulate in basements and mine shafts?
9. Use the ideal gas law to derive an equation that relates the remaining variables for a sample of an ideal gas if the following are held constant.
amount and volume
pressure and amount
temperature and volume
temperature and amount
pressure and temperature
Tennis balls that are made for Denver, Colorado, feel soft and do not bounce well at lower altitudes. Use the ideal gas law to explain this observation. Will a tennis ball designed to be used at sea level be harder or softer and bounce better or worse at higher altitudes?
Answer
P/T = constant
V/T = constant (Charles’ law)
P/n = constant
PV = constant (Boyle’s law)
V/n = constant (Avogadro’s law)
Numerical Problems
Calculate the number of moles in each sample at STP.
1580 mL of NO2
847 cm3 of HCl
4.792 L of H2
a 15.0 cm × 6.7 cm × 7.5 cm container of ethane
Calculate the number of moles in each sample at STP.
2200 cm3 of CO2
1200 cm3 of N2
3800 mL of SO2
13.75 L of NH3
Calculate the mass of each sample at STP.
36 mL of HI
550 L of H2S
1380 cm3 of CH4
Calculate the mass of each sample at STP.
3.2 L of N2O
65 cm3 of Cl2
3600 mL of HBr
Calculate the volume in liters of each sample at STP.
1.68 g of Kr
2.97 kg of propane (C3H8)
0.643 mg of (CH3)2O
Calculate the volume in liters of each sample at STP.
3.2 g of Xe
465 mg of CS2
5.34 kg of acetylene (C2H2)
Calculate the volume of each gas at STP.
1.7 L at 28°C and 96.4 kPa
38.0 mL at 17°C and 103.4 torr
650 mL at −15°C and 723 mmHg
Calculate the volume of each gas at STP.
2.30 L at 23°C and 740 mmHg
320 mL at 13°C and 97.2 kPa
100.5 mL at 35°C and 1.4 atm
A 8.60 L tank of nitrogen gas at a pressure of 455 mmHg is connected to an empty tank with a volume of 5.35 L. What is the final pressure in the system after the valve connecting the two tanks is opened? Assume that the temperature is constant.
At constant temperature, what pressure in atmospheres is needed to compress 14.2 L of gas initially at 25.2 atm to a volume of 12.4 L? What pressure is needed to compress 27.8 L of gas to 20.6 L under similar conditions?
One method for preparing hydrogen gas is to pass HCl gas over hot aluminum; the other product of the reaction is AlCl3. If you wanted to use this reaction to fill a balloon with a volume of 28,500 L at sea level and a temperature of 78°F, what mass of aluminum would you need? What volume of HCl at STP would you need?
An 3.50 g sample of acetylene is burned in excess oxygen according to the following reaction:
2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l)
At STP, what volume of CO2(g) is produced?
13. Calculate the density of ethylene (C2H4) under each set of conditions.
7.8 g at 0.89 atm and 26°C
6.3 mol at 102.6 kPa and 38°C
9.8 g at 3.1 atm and −45°C
Determine the density of O2 under each set of conditions.
42 g at 1.1 atm and 25°C
0.87 mol at 820 mmHg and 45°C
16.7 g at 2.4 atm and 67°C
At 140°C, the pressure of a diatomic gas in a 3.0 L flask is 635 kPa. The mass of the gas is 88.7 g. What is the most likely identity of the gas?
What volume must a balloon have to hold 6.20 kg of H2 for an ascent from sea level to an elevation of 20,320 ft, where the temperature is −37°C and the pressure is 369 mmHg?
What must be the volume of a balloon that can hold 313.0 g of helium gas and ascend from sea level to an elevation of 1.5 km, where the temperature is 10.0°C and the pressure is 635.4 mmHg?
A typical automobile tire is inflated to a pressure of 28.0 lb/in.2 Assume that the tire is inflated when the air temperature is 20°C; the car is then driven at high speeds, which increases the temperature of the tire to 43°C. What is the pressure in the tire? If the volume of the tire had increased by 8% at the higher temperature, what would the pressure be?
The average respiratory rate for adult humans is 20 breaths per minute. If each breath has a volume of 310 mL of air at 20°C and 0.997 atm, how many moles of air does a person inhale each day? If the density of air is 1.19 kg/m3, what is the average molecular mass of air?
Kerosene has a self-ignition temperature of 255°C. It is a common accelerant used by arsonists, but its presence is easily detected in fire debris by a variety of methods. If a 1.0 L glass bottle containing a mixture of air and kerosene vapor at an initial pressure of 1 atm and an initial temperature of 23°C is pressurized, at what pressure would the kerosene vapor ignite?
Answers
7.05 × 10−2 mol
3.78 × 10−2 mol
0.2138 mol
3.4 × 10−2 mol
0.21 g HI;
840 g H2S;
0.988 g CH4
0.449 L Kr
1510 L C3H8
3.13 × 10−4 L (CH3)2O
1.5 L
4.87 mL
650 mL
281 mmHg
20.9 kg Al, 5.20 × 104 L HCl
1.0 g/L
1.1 g/L
4.6 g/L
2174 L
Previous Section
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15312 | https://www.govinfo.gov/content/pkg/GOVPUB-C13-049896b7756a1eb71d691a56e46f21f4/pdf/GOVPUB-C13-049896b7756a1eb71d691a56e46f21f4.pdf | A11101 31flT c13 I / NBSIR 83 Calculation of Compressibility Factor for Air Over the Ranges of Pressure, Temperature, and Relative Humidity of Interest in Flowmeter Calibration -2652 U.S. DEPARTMENT OF COMMERCE National Bureau of Standards Center for Chemical Engineering Chemical Process Metrology Division Washington, DC 20234 March 1 983 c.
4 00 156 >3-2652 j L983 i ' .2_ U.S. DEPARTMENT OF COMMERCE national bureau of standards WATI' >irAL mutAv: op rrAifrrjiDP ’ r-MAY 2 ?983 HO ’ CA_Lr ^ NBSIR 83-2652 CALCULATION OF COMPRESSIBILITY FACTOR FOR AIR OVER THE RANGES OF PRESSURE, TEMPERATURE, AND RELATIVE HUMIDITY OF INTEREST IN FLOWMETER CALIBRATION Frank E. Jones U.S. DEPARTMENT OF COMMERCE National Bureau of Standards Center for Chemical Engineering Chemical Process Metrology Division Washington, DC 20234 March 1983 U.S. DEPARTMENT OF COMMERCE, Malcolm Baldrige, Secretary NATIONAL BUREAU OF STANDARDS, Ernest Ambler, Director Calculation of Compressibility Factor for Air Over the Ranges of Pressure, Temperature, and Relative Humidity of Interest in Flowmeter Calibration Frank E.
Jones Fluid Engineering Division Center for Chemical Engineering National Engineering Laboratory National Bureau of Standards Washington, D.C.
20234 A simple yet precise equation has been developed to enable calculation (using programmable calculators) of the compressi-bility factor, Z, for air from measurements of pressure, tempera-ture, and humidity.
The compressibility factor, a factor which accounts for the non-ideality of air in real-gas equations of state, is conventionally computed using virial coefficients.
In the present paper, an equation is fitted to tabulated values of Z.
The deviation between calculated and tabulated values is of the order of 0.01% or less; this does not imply, however, that the accuracy of calculated values is of this order.
Key Words: Air; compressibility factor; extrapolation formulas; flowmeter calibration; pressure; relative humidity; temperature.
1.
Introduction In the calculation of the flowrate of air through various metering devices (Venturi tubes, orifice plates, and nozzles, for example), the density of air enters.
Since air is not an "ideal" gas, the compressi-bility factor (Z) a measure of the nonideality is introduced in the real gas equation for density (p).
The compressibility factor is conventionally computed using the virial equation of state of an air-water vapor mixture (1,2).
In this paper a simple interpolation formula is developed which enables the engineer to conveniently make precise calculations of Z using small readily available programmable calculators.
2.
Real Gas Equation for Density of Air The real gas equation for the density, p, of dry air is PM a RTZ 5 1 p = ( 1 ) where P is the total pressure; M is the apparent molecular weight of dry air; R is the molar gas constant; T is absolute temperature; and Z is the compressibility factor.
The magnitude of the non-ideality of the gas is reflected in the departure of Z from 1.
For moist air, a mixture of dry air and water vapor, the real gas equation for p is PM p -[1 + (e ' 1) • (2) where e is the ratio of the molecular weight of water vapor to the apparent molecular weight of dry air, and e‘ is the effective vapor pressure of water in moist air.
The value of Z depends on P, T, and relative humidity of the air.
3.
Development of the Equations To develop the equation for Z, unpublished data for dry air provided by Wexler and Hyland and data for moist air in table 2 of reference were used.
The equation was developed in two parts: the first includes the dependence of Z on pressure and temperature; the second part is a correction due to the effect of relative humidity (RH).
At each of nine values of P (covering the range 1 to 40 atmo-spheres), values of (1-Z) were fitted by least squares to an equation quadratic in temperature, t, in °C.
At each value of pressure, 19 values of (1-Z) covering the temperature range 0°C to 54°C at inter-vals of three degrees were used.
The resulting equations are: at P = 1 atm, (1-Z) x 10 6 = 581.98 -12.471 t + at P = 5 atm, (1-Z) x 10 6 = 2860.3 -62.056 t + at P = 10 atm, (1-Z) x 10 6 = 5594.3 -123.32 t at P = 15 atm, (1-Z) x 10 6 = 8195.8 -183.24 t at P = 20 atm, (1-Z) x 10 6 = 10670.6 -243.22 t at P = 25 atm, (1-Z) x 10 6 = 13006.4 -301.64 t at P = 30 atm, (1-Z) x 10 6 = 15204.0 -358.89 t at P = 35 atm, (1-Z) x 10 6 = 17261.7 - 415.00 t 063156 t 2 , (3) 31 525 1.2 5 (4) .62894 t 2 .
(5) .93145 t 2 .
(6) 1.2517 t 2 .
(7) 1.5587 t 2 .
(8) 1.8613 t 2 , (9) 2.1607 t 2 .
(10) 2.4534 t 2 .
(ID The general form of eqs (3) through (11) is (1-Z) x 10 6 = a(P) + b(P) t + c(P)t 2 , ( 12 ) 2 where the dependence of Z on P is assigned to the coefficients, a, b, and c.
Each of these coefficients was fitted by least squares to an equation quadratic in P (in atm).
The resulting equations are a ( P ) = -10.864 + 588.26 P -2.7106 P 2 , (13) b(P) = 0.33297 -12.585 P + 2.0659 x 10" 2 P 2 , (14) c(P) = -2.4925 x 10" 3 + 6.3706 x 10" 2 P -5.5619 x 10" 5 P 2 (15) An equation expressing the dependence of Z for dry air on t and P is synthesized by substituting eqs (13) through (15) in eq (12): (1-Z) x 10 6 = -10.864 + 588.26 P -2.7106 P 2 + 0.33297 t -12.585 Pt + 2.0659 x 10" 2 P 2 t -2.4925 x 10“ 3 t 2 + 6.3706 x 10" 2 Pt 2 -5.5619 x 10" 5 P 2 t 2 (16) To develop an expression for the correction for the effect of RH on Z, values of Z for temperatures 19°C to 26°C at 1-degree intervals for RH of 0, 25, 50, 75, and 100 percent were used.
The five mean values of the difference (one corresponding to each value of RH) between Z for moist air and Z for dry air, a( 1-Z), were fitted by least squares to an equation quadratic in RH.
The intercept of the equation was forced sufficiently close to 0 by weighting the 0 percent RH point.
The resulting equation is (1-Z) x 10 6 = -0.35 (RH) -5.0 x 10" 3 (RH) 2 , (17) where RH is in percent.
By adding eq (17) to eq (16) and rearranging, the equation expressing the dependence of Z on P, t, and RH results: Z = 1.00001 -5.8826 x 10" 4 P + 2.7106 x 10" 6 P 2 -3.3297 x 10‘ 7 t + 1.2585 x 10' 5 Pt -2.0659 x 10' 8 P 2 t + 2.4925 x 10“ 9 t 2 -6.3706 x 10' 8 Pt 2 + 5.5619 x 10" 11 P 2 t 2 -3.5 x 10“ 7 (RH) -5.0 x 10" 9 (RH) 2 , (18) where P is in atm, t is in °C, RH is in percent, and Z is a pure number.
For P in PSI, t in °C, and RH in percent.
3 Z = 1.00001 -4.0029 x 10" 5 P + 1.2551 x 10" 8 P 2 + 3.3297 x 10" 7 t + 8.5636 x 10“ 7 Pt - 9.5827 x 2.4925 x 10' 9 t 2 - 4.3349 x 10“ 9 Pt 2 + 2.5753 3.5 x 10" 7 (RH) -5.0 x 10" 9 (RH) 2 10" 11 P 2 t x 10' 13 P 2 t 2 (19) For P in MPa, t in °C, and RH in percent, Z = 1.00001 -5.8057 x 10" 3 P + 2.6402 x 10" 4 P 2 -3.3297 x 10" 7 t + 1.2420 x 10“ 4 Pt - 2.0158 x 10“ 6 P 2 t + 2.4925 x 10“ 9 t 2 -6.2873 x 10" 7 Pt 2 + 5.4174 x 10' 9 P 2 t 2 -3.5 x 10“ 7 (RH) -5.0 x 10' 9 (RH) 2 (20) The values of Z for dry air calculated using eq (18) have been compared with the data used to develop the equation.
The mean of the 171 differences between calculated and data values is 6 x 10" , and the estimate of standard deviation is 2.6 x 10" 3 which corresponds to 0.0026%.
At a pressur deviation is 2.1 x 10" of 1 atmosphere the estimate of standard which corresponds to 0.00021%.
The estimate of overall uncertainty for the data on which the calculation of Z is based ranges from 0.00093% to 0.0011% at a pressure of 0.987 atmosphere over the temperature range of interest; at a pressure of 49.3 atmospheres the estimate of overall uncertainty ranges from 0.054% to 0.075%.
The estimate of standard deviation for the differ-ence between the calculated RH correction (the last two terms in eq (18)^ and that inferred from the data on which it is based is 1 x 10~ 3 which corresponds to 0.001% in Z.
Since the RH correction was fitted over the range 19°C to 26°C it will have a greater devia-tion outside this range.
However at the extremes of the temperature range of interest here, approximately 0°C to 50°C, the deviation is only of the order of 0.01% in Z.
4.
Conclusions A simple yet precise equation has been developed to enable calculation of the compressibility factor, Z, for air from measure-ments of pressure, temperature, and humidity.
The deviation between calculated and data values of Z is of the order of 0.01% or less; this does not imply, however, that the accuracy of calculated values is of this order.
5.
Acknowledgment The author gratefully acknowledges the assistance of R.
W.
Hyland in providing access to the compressibility factor data, the computa-tional assistance of C.
P.
Reeve, and the typing of the manuscript by Susan Johnson.
4 6.
References 1.
Jones, F.E.
The Air Density Equation and the Transfer of the Mass Unit, J.
Res.
Nat.
Bur.
Stand.
(U.S.).
83 (5) : 41 9-428 ; 1978 September-October.
2.
Hyland, R.W., and Wexler, A.
The Second Interaction (Cross) Virial Coefficient for Moist Air, J.
Res.
Nat.
Bur.
Stand.
(U.S.).
77A ( 1 ) : 1 33-1 47 ; 1973 January-February .
3.
Hyland, R.W.
Correlation for the Second Interaction Virial Coefficients and Enhancement Factors for Moist Air, J.
Res.
Nat.
Bur.
Stand.
(U.S.).
79A (4):551-560; 1975 July-August.
4.
Wexler, A., and Hyland, R.W.
Formulations for the Thermodynamic Properties of Dry Air from 173.15 to 473.15 K, and of Saturated Moist Air from 173.15 to 372.15 K, at Pressures to 5 MPa, May 1980, to be published by ASHRAE.
5 NBS-114A irev. 2-80 U.S. DEPT. OF COMM.
BIBLIOGRAPHIC DATA SHEET (See instructions) 1. PUBLICATION OR REPORT NO.
NBSIR 83-2652 2. Performing Organ. Report No 3. Publication Date March 1983 4. TITLE AND SUBTITLE Calculation of Compressibility Factor for Air Over the Ranges of Pressure, Temperature , and Relative Humidity of Interest in Flowmeter Calibration 5. AUTHOR(S) Frank E.
Jones 6. PERFORMING ORGANIZATION (If joint or other than N BS. see in struction s) 7. Contract/Grant No.
NATIONAL bureau of standards DEPARTMENT OF COMMERCE 8. Type of Report & Period Covered WASHINGTON, D.C.
20234 Final 9. SPONSORING ORGANIZATION NAME AND COMPLETE ADDRESS (Street. City. State. ZIP) 10.
SUPPLEMENTARY NOTES ~| Document describes a computer program; SF-185, FIPS Software Summary, is attached.
11.
ABSTRACT (A 200-word or less factual summary of most significant information.
If document includes a significant bibliography or literature survey, mention it here) A simple yet precise equation has been developed to enable calculation (using programmable calculators) of the compressibility factor, Z, for air from measure-ments of pressure, temperature, and humidity.
The compressibility factor, a factor which accounts for the non-ideality of air in real-gas equations of state, is conventionally computed using virial coefficients.
In the present paper, an equation is fitted to tabulated values of Z.
The deviation between calculated and tabulated values is of the order of 0.01% or less; this does not imply, however, that the accuracy of calculated values is of this order.
12.
KEY WORDS (Six to twelve entries; alphabetical order; capitalize only proper names; and separate key words by semi colon s) Air; compressibility factor; extrapolation formulas; flowmeter calibration; pressure; relative humidity; temperature.
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Qxj Order From National Technical Information Service (NTIS), Springfield, VA.
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15313 | http://www.stat.ucla.edu/~vlew/stat130/WEEK7/dalgaard13.pdf | 13 Logistic regression Sometimes you wish to model binary outcomes, variables that can have only two possible values: diseased or nondiseased, and so forth. For in-stance, you want to describe the risk of getting a disease depending on various kinds of exposures. Chapter 8 discusses some simple techniques based on tabulation, but you might also want to model dose-response re-lationships (where the predictor is a continuous variable) or model the effect of multiple variables simultaneously. It would be very attractive to be able to use the same modelling techniques as for linear models.
However, it is not really attractive to use additive models for probabili-ties since they have a limited range and regression models could predict off-scale values below zero or above 1. It makes better sense to model the probabilities on a transformed scale; this is what is done in logistic regression analysis.
A linear model for transformed probabilities can be set up as logit p = β0 + β1x1 + β2x2 + . . . βkxk in which logit p = log[p/(1 −p)] is the log odds. A constant additive ef-fect on the logit scale corresponds to a constant odds ratio. The choice of the logit function is not the only one possible, but it has some mathemat-ically convenient properties. Other choices do exist; the probit function (the quantile function of the normal distribution) or log(−log p), which has a connection to survival analysis models.
P. Dalgaard, Introductory Statistics with R, DOI: 10.1007/978-0-387-79054-1_13, © Springer Science+Business Media, LLC 2008 228 13. Logistic regression One thing to notice about the logistic model is that there is no error term as in linear models. We are modelling the probability of an event directly, and that in itself will determine the variability of the binary outcome. There is no variance parameter as in the normal distribution.
The parameters of the model can be estimated by the method of maximum likelihood. This is a quite general technique, similar to the least-squares method in that it finds a set of parameters that optimizes a goodness-of-fit criterion (in fact, the least-squares method itself is a slightly modified maximum-likelihood procedure). The likelihood function L(β) is simply the probability of the entire observed data set for varying parameters.
The deviance is the difference between the maximized value of −2 log L and the similar quantity under a “maximal model” that fits data perfectly.
Changes in deviance caused by a model reduction will be approximately χ2-distributed with degrees of freedom equal to the change in the number of parameters.
In this chapter, we see how to perform logistic regression analysis in R.
There naturally is quite a large overlap with the material on linear models since the description of models is quite similar, but there are also some special issues concerning deviance tables and the specification of models for pretabulated data.
13.1 Generalized linear models Logistic regression analysis belongs to the class of generalized linear models.
These models are characterized by their response distribution (here the binomial distribution) and a link function, which transfers the mean value to a scale in which the relation to background variables is described as linear ans additive. In a logistic regression analysis, the link function is logit p = log[p/(1 −p)].
There are several other examples of generalized linear models; for in-stance, analysis of count data is often handled by the multiplicative Poisson model, where the link function is log λ, with λ the mean of the Poisson-distributed observation. All of these models can be handled using the same algorithm, which also allows the user some freedom to define his or her own models by defining suitable link functions.
In R generalized linear models are handled by the glm function. This function is very similar to lm, which we have used many times for lin-ear normal models. The two functions use essentially the same model formulas and extractor functions (summary, etc.), but glm also needs to have specified which generalized linear model is desired. This is done via 13.2 Logistic regression on tabular data 229 the family argument. To specify a binomial model with logit link (i.e., logistic regression analysis), you write family=binomial("logit").
13.2 Logistic regression on tabular data In this section, we analyze the example concerning hypertension from Altman (1991, p. 353). First, we need to enter data, which is done as follows: > no.yes <- c("No","Yes") > smoking <- gl(2,1,8,no.yes) > obesity <- gl(2,2,8,no.yes) > snoring <- gl(2,4,8,no.yes) > n.tot <- c(60,17,8,2,187,85,51,23) > n.hyp <- c(5,2,1,0,35,13,15,8) > data.frame(smoking,obesity,snoring,n.tot,n.hyp) smoking obesity snoring n.tot n.hyp 1 No No No 60 5 2 Yes No No 17 2 3 No Yes No 8 1 4 Yes Yes No 2 0 5 No No Yes 187 35 6 Yes No Yes 85 13 7 No Yes Yes 51 15 8 Yes Yes Yes 23 8 The gl function to “generate levels” was briefly introduced in Section 7.3.
The first three arguments to gl are, respectively, the number of levels, the repeat count of each level, and the total length of the vector. A fourth argument can be used to specify the level names of the resulting factor.
The result is apparent from the printout of the generated variables. They were put together in a data frame to get a nicer layout. Another way of generating a regular pattern like this is to use expand.grid: > expand.grid(smoking=no.yes, obesity=no.yes, snoring=no.yes) smoking obesity snoring 1 No No No 2 Yes No No 3 No Yes No 4 Yes Yes No 5 No No Yes 6 Yes No Yes 7 No Yes Yes 8 Yes Yes Yes R is able to fit logistic regression analyses for tabular data in two different ways. You have to specify the response as a matrix, where one column is 230 13. Logistic regression the number of “diseased” and the other is the number of “healthy” (or “success” and “failure”, depending on context).
hyp.tbl <- cbind(n.hyp,n.tot-n.hyp) > hyp.tbl n.hyp [1,] 5 55 [2,] 2 15 [3,] 1 7 [4,] 0 2 [5,] 35 152 [6,] 13 72 [7,] 15 36 [8,] 8 15 The cbind function (“c” for “column”) is used to bind variables together, columnwise, to form a matrix. Note that it would be a horrible mistake to use the total count for column 2 instead of the number of failures.
Then, you can specify the logistic regression model as > glm(hyp.tbl~smoking+obesity+snoring,family=binomial("logit")) Actually, "logit" is the default for binomial and the family argument is the second argument to glm, so it suffices to write > glm(hyp.tbl~smoking+obesity+snoring,binomial) The other way to specify a logistic regression model is to give the proportion of diseased in each cell: > prop.hyp <- n.hyp/n.tot > glm.hyp <- glm(prop.hyp~smoking+obesity+snoring, + binomial,weights=n.tot) It is necessary to give weights because R cannot see how many observations a proportion is based on.
As output, you get in either case (except for minor details) Call: glm(formula = hyp.tbl ~ smoking + obesity + snoring, ...
Coefficients: (Intercept) smokingYes obesityYes snoringYes -2.37766 -0.06777 0.69531 0.87194 Degrees of Freedom: 7 Total (i.e. Null); 4 Residual Null Deviance: 14.13 Residual Deviance: 1.618 AIC: 34.54 13.2 Logistic regression on tabular data 231 which is in a minimal style similar to that used for printing lm objects.
Also in the result of glm is some nonvisible information, which may be extracted with particular functions. You can, for instance, save the result of a fit of a generalized linear model in a variable and obtain a table of regression coefficients and so forth using summary: > glm.hyp <- glm(hyp.tbl~smoking+obesity+snoring,binomial) > summary(glm.hyp) Call: glm(formula = hyp.tbl ~ smoking + obesity + snoring, family ...
Deviance Residuals: 1 2 3 4 5 6 -0.04344 0.54145 -0.25476 -0.80051 0.19759 -0.46602 7 8 -0.21262 0.56231 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -2.37766 0.38018 -6.254 4e-10 smokingYes -0.06777 0.27812 -0.244 0.8075 obesityYes 0.69531 0.28509 2.439 0.0147 snoringYes 0.87194 0.39757 2.193 0.0283 ---Signif. codes: 0 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 14.1259 on 7 degrees of freedom Residual deviance: 1.6184 on 4 degrees of freedom AIC: 34.537 Number of Fisher Scoring iterations: 4 In the following, we go through the components of summary output for generalized linear models: Call: glm(formula = hyp.tbl ~ smoking + obesity + snoring, family = ...
As usual, we start off with a repeat of the model specification. Obviously, more interesting is when the output is not viewed in connection with the function call that generated it.
Deviance Residuals: 1 2 3 4 5 6 -0.04344 0.54145 -0.25476 -0.80051 0.19759 -0.46602 7 8 -0.21262 0.56231 232 13. Logistic regression This is the contribution of each cell of the table to the deviance of the model (the deviance corresponds to the sum of squares in linear normal models), with a sign according to whether the observation is larger or smaller than expected. They can be used to pinpoint cells that are par-ticularly poorly fitted, but you have to be wary of the interpretation in sparse tables.
Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -2.37766 0.38018 -6.254 4e-10 smokingYes -0.06777 0.27812 -0.244 0.8075 obesityYes 0.69531 0.28509 2.439 0.0147 snoringYes 0.87194 0.39757 2.193 0.0283 ---Signif. codes: 0 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) This is the table of primary interest. Here, we get estimates of the re-gression coefficients, standard errors of same, and tests for whether each regression coefficient can be assumed to be zero. The layout is nearly identical to the corresponding part of the lm output.
The note about the dispersion parameter is related to the fact that the bino-mial variance depends entirely on the mean. There is no scale parameter like the variance in the normal distribution.
Null deviance: 14.1259 on 7 degrees of freedom Residual deviance: 1.6184 on 4 degrees of freedom AIC: 34.537 “Residual deviance” corresponds to the residual sum of squares in ordi-nary regression analyses which is used to estimate the standard deviation about the regression line. In binomial models, however, the standard devi-ation of the observations is known, and you can therefore use the deviance in a test for model specification. The AIC (Akaike information criterion) is a measure of goodness of fit that takes the number of fitted parameters into account.
R is reluctant to associate a p-value with the deviance. This is just as well because no exact p-value can be found, only an approximation that is valid for large expected counts. In the present case, there are actually a couple of places where the expected cell count is rather small.
The asymptotic distribution of the residual deviance is a χ2 distribution with the stated degrees of freedom, so even though the approximation may be poor, nothing in the data indicates that the model is wrong (the 5% significance limit is at 9.49 and the value found here is 1.62).
13.2 Logistic regression on tabular data 233 The null deviance is the deviance of a model that contains only the in-tercept (that is, describes a fixed probability, here for hypertension, in all cells). What you would normally be interested in is the difference from the residual deviance, here 14.13 −1.62 = 12.51, which can be used for a joint test for whether any effects are present in the model. In the present case, a p-value of approximately 0.6% is obtained.
Number of Fisher Scoring iterations: 4 This refers to the actual fitting procedure and is a purely technical item.
There is no statistical information in it, but you should keep an eye on whether the number of iterations becomes too large because that might be a sign that the model is too complex to fit based on the available data. Nor-mally, glm halts the fitting procedure if the number of iterations exceeds 25, but it is possible to configure the limit.
The fitting procedure is iterative in that there is no explicit formula that can be used to compute the estimates, only a set of equations that they should satisfy. However, there is an approximate solution of the equations if you supply an initial guess at the solution. This solution is then used as a starting point for an improved solution, and the procedure is repeated until the guesses are sufficiently stable.
A table of correlations between parameter estimates can be obtained via the optional argument corr=T to summary (this also works for linear models). It looks like this: Correlation of Coefficients: (Intercept) smokingYes obesityYes smokingYes -0.1520 obesityYes -0.1361 -9.499e-05 snoringYes -0.8965 -6.707e-02 -0.07186 It is seen that the correlation between the estimates is fairly small, so that it may be expected that removing a variable from the model does not change the coefficients and p-values for other variables much. (The correlations between the regression coefficients and intercept are not very informative; they mostly relate to whether the variable in question has many or few observations in the “Yes” category.) The z test in the table of regression coefficients immediately shows that the model can be simplified by removing smoking. The result then looks as follows (abbreviated): > glm.hyp <- glm(hyp.tbl~obesity+snoring,binomial) > summary(glm.hyp) ...
234 13. Logistic regression Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -2.3921 0.3757 -6.366 1.94e-10 obesityYes 0.6954 0.2851 2.440 0.0147 snoringYes 0.8655 0.3967 2.182 0.0291 13.2.1 The analysis of deviance table Deviance tables correspond to ANOVA tables for multiple regression analyses and are generated like these with the anova function: > glm.hyp <- glm(hyp.tbl~smoking+obesity+snoring,binomial) > anova(glm.hyp, test="Chisq") Analysis of Deviance Table Model: binomial, link: logit Response: hyp.tbl Terms added sequentially (first to last) Df Deviance Resid. Df Resid. Dev P(>|Chi|) NULL 7 14.1259 smoking 1 0.0022 6 14.1237 0.9627 obesity 1 6.8274 5 7.2963 0.0090 snoring 1 5.6779 4 1.6184 0.0172 Notice that the Deviance column gives differences between models as variables are added to the model in turn. The deviances are approximately χ2-distributed with the stated degrees of freedom. It is necessary to add the test="chisq" argument to get the approximate χ2 tests.
Since the snoring variable on the last line is significant, it may not be removed from the model and we cannot use the table to justify model reductions. If, however, the terms are rearranged so that smoking comes last, we get a deviance-based test for removal of that variable: > glm.hyp <- glm(hyp.tbl~snoring+obesity+smoking,binomial) > anova(glm.hyp, test="Chisq") ...
Df Deviance Resid. Df Resid. Dev P(>|Chi|) NULL 7 14.1259 snoring 1 6.7887 6 7.3372 0.0092 obesity 1 5.6591 5 1.6781 0.0174 smoking 1 0.0597 4 1.6184 0.8069 From this you can read that smoking is removable, whereas obesity is not, after removal of smoking.
13.2 Logistic regression on tabular data 235 For good measure, you should also set up the analysis with the two re-maining explanatory variables interchanged, so that you get a test of whether snoring may be removed from a model that also contains obesity: > glm.hyp <- glm(hyp.tbl~obesity+snoring,binomial) > anova(glm.hyp, test="Chisq") ...
Df Deviance Resid. Df Resid. Dev P(>|Chi|) NULL 7 14.1259 obesity 1 6.8260 6 7.2999 0.0090 snoring 1 5.6218 5 1.6781 0.0177 An alternative method is to use drop1 to try removing one term at a time: > drop1(glm.hyp, test="Chisq") Single term deletions Model: hyp.tbl ~ obesity + snoring Df Deviance AIC LRT Pr(Chi) 1.678 32.597 obesity 1 7.337 36.256 5.659 0.01737 snoring 1 7.300 36.219 5.622 0.01774 ---Signif. codes: 0 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1 Here LRT is the likelihood ratio test, another name for the deviance change.
The information in the deviance tables is fundamentally the same as that given by the z tests in the table of regression coefficients. The results may differ due to the use of different approximations, though. From theoretical considerations, the deviance test is preferable, but in practice the differ-ence is often small because of the large-sample approximation χ2 ≈z2 for tests with a single degree of freedom. However, to test factors with more than two categories, you have to use the deviance table because the z tests only relate to some of the possible group comparisons. Also, the small-sample situation requires special attention; see the next section.
13.2.2 Connection to test for trend In Chapter 8, we considered tests for comparing relative frequencies using prop.test and prop.trend.test, in particular the example of cae-sarean section versus shoe size. This example can also be analyzed as a logistic regression analysis on a “shoe score”, which — for want of a bet-ter idea — may be chosen as the group number. This gives essentially the same analysis in the sense that the same models are involved.
236 13. Logistic regression > caesar.shoe <4 4 4.5 5 5.5 6+ Yes 5 7 6 7 8 10 No 17 28 36 41 46 140 > shoe.score <- 1:6 > shoe.score 1 2 3 4 5 6 > summary(glm(t(caesar.shoe)~shoe.score,binomial)) ...
Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -0.87058 0.40506 -2.149 0.03161 shoe.score -0.25971 0.09361 -2.774 0.00553 ---Signif. codes: 0 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 9.3442 on 5 degrees of freedom Residual deviance: 1.7845 on 4 degrees of freedom AIC: 27.616 ...
Notice that caesar.shoe had to be transposed with t(...), so that the matrix was “stood on its end” in order to be used as the response variable by glm.
You can also write the results in a deviance table > anova(glm(t(caesar.shoe)~shoe.score,binomial)) ...
Df Deviance Resid. Df Resid. Dev NULL 5 9.3442 shoe.score 1 7.5597 4 1.7845 from the last line of which you see that there is no significant deviation from linearity (1.78 on 4 degrees of freedom), whereas shoe.score has a significant contribution.
For comparison, the previous analyses using standard tests are repeated: > caesar.shoe.yes <- caesar.shoe["Yes",] > caesar.shoe.no <- caesar.shoe["No",] > caesar.shoe.total <- caesar.shoe.yes+caesar.shoe.no > prop.trend.test(caesar.shoe.yes,caesar.shoe.total) Chi-squared Test for Trend in Proportions ...
X-squared = 8.0237, df = 1, p-value = 0.004617 > prop.test(caesar.shoe.yes,caesar.shoe.total) 13.3 Likelihood profiling 237 6-sample test for equality of proportions without continuity correction ...
X-squared = 9.2874, df = 5, p-value = 0.09814 ...
Warning message: In prop.test(caesar.shoe.yes, caesar.shoe.total) : Chi-squared approximation may be incorrect The 9.29 from prop.test corresponds to the 9.34 in residual deviance from a NULL model, whereas the 8.02 in the trend test corresponds to the 7.56 in the test of significance of shoe.score. Thus, the tests do not give exactly the same result but generally almost the same. Theoretical considerations indicate that the specialized trend test is probably slightly better than the regression-based test. However, testing the linearity by subtracting the two χ2 tests is definitely not as good as the real test for linearity.
13.3 Likelihood profiling The z tests in the summary output are based on the Wald approximation, which calculates what the approximate standard error of the parameter estimate would be if the true values of the parameters were equal to the es-timates. In large data sets, this is fine because the result is nearly the same for all parameter values that fit the data reasonably well. In smaller data sets, however, the difference between the Wald tests and the likelihood ratio test can be considerable.
This also affects the calculation of confidence intervals since these are based on inverting the tests, giving a set of parameter values that are not rejected by a statistical test. As an alternative to the Wald-based ±1.96 × s.e. technique, the MASS package allows you to compute inter-vals that are based on inverting the likelihood ratio test. In practice, this works like this > confint(glm.hyp) Waiting for profiling to be done...
2.5 % 97.5 % (Intercept) -3.2102369 -1.718143 obesityYes 0.1254382 1.246788 snoringYes 0.1410865 1.715860 The standard type of result can be obtained using confint.default.
The difference in this case is not very large, although visible in the lines relating to snoring and the intercept: 238 13. Logistic regression −4.0 −3.0 −2.0 −1.0 −4 −2 0 2 4 (Intercept) tau −0.5 0.0 0.5 1.0 1.5 −4 −2 0 2 obesityYes tau −0.5 0.5 1.0 1.5 2.0 2.5 −4 −2 0 2 snoringYes tau Figure 13.1. Profile plot for hypertension model.
confint.default(glm.hyp) 2.5 % 97.5 % (Intercept) -3.12852108 -1.655631 obesityYes 0.13670388 1.254134 snoringYes 0.08801498 1.642902 The way this works is via likelihood profiling. For a set of trial values of the parameter, the likelihood is maximized over the other parameters in the model. The result can be displayed in a profile plot as follows: > library(MASS) > plot(profile(glm.hyp)) Notice that we need to load the MASS package at this point. (The function was used by confint earlier on, but without putting it on the search path.) The plots require a little explanation. The quantity on the y-axis, labelled tau, is the signed square root of the likelihood ratio test.
τ(β) = sgn(β −ˆ β) q −2(ℓ(β) −ℓ( ˆ β)) 13.4 Presentation as odds-ratio estimates 239 Here ℓdenotes the profile log-likelihood. The main idea is that when the profile likelihood function is approximately quadratic, τ(β) is approxi-mately linear. Conversely, likelihood functions not well approximated by a quadratic show up as nonlinear profile plots.
One important thing to notice, though, is that although the profiling method will capture nonquadratic behaviour of the likelihood function, confidence intervals based on the likelihood ratio test will always be limited in accuracy by the approximation of the distribution of the test.
13.4 Presentation as odds-ratio estimates In parts of the epidemiological literature, it has become traditional to present logistic regression analyses in terms of odds ratios. In the case of a quantitative covariate, this means odds ratio per unit change in the covariate. That is, the antilogarithm (exp) of the regression coefficients is given instead of the coefficients themselves. Since standard errors make little sense after the transformation, it is also customary to give confidence intervals instead. This can be obtained quite easily as follows: > exp(cbind(OR=coef(glm.hyp), confint(glm.hyp))) Waiting for profiling to be done...
OR 2.5 % 97.5 % (Intercept) 0.09143963 0.04034706 0.1793989 obesityYes 2.00454846 1.13364514 3.4791490 snoringYes 2.37609483 1.15152424 5.5614585 The (Intercept) is really the odds of hypertension (for the not snoring non-obese) and not an odds ratio.
13.5 Logistic regression using raw data In this section, we again use Anders Juul’s data (see p. 85). For easy ref-erence, here is how to read data and convert the variables that describe groupings into factors (this time slightly simplified): > juul$menarche <- factor(juul$menarche, labels=c("No","Yes")) > juul$tanner <- factor(juul$tanner) In the following, we look at menarche as the response variable. This vari-able indicates for each girl whether or not she has had her first period. It is coded 1 for “no” and 2 for “yes”. It is convenient to look at a subset of data consisting of 8–20-year-old girls. This can be extracted as follows: 240 13. Logistic regression > juul.girl <- subset(juul,age>8 & age<20 & + complete.cases(menarche)) > attach(juul.girl) For obvious reasons, no boys have a nonmissing menarche, so it is not necessary to select on gender explicitly.
Then you can analyze menarche as a function of age like this: > summary(glm(menarche~age,binomial)) Call: glm(formula = menarche ~ age, family = binomial) Deviance Residuals: Min 1Q Median 3Q Max -2.32759 -0.18998 0.01253 0.12132 2.45922 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -20.0132 2.0284 -9.867 <2e-16 age 1.5173 0.1544 9.829 <2e-16 ---Signif. codes: 0 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 719.39 on 518 degrees of freedom Residual deviance: 200.66 on 517 degrees of freedom AIC: 204.66 Number of Fisher Scoring iterations: 7 The response variable menarche is a factor with two levels, where the last level is considered the event. It also works to use a variable that has the values 0 and 1 (but not, for instance, 1 and 2!).
Notice that from this model you can estimate the median menarcheal age as the age where logit p = 0. A little thought (solve −20.0132 + 1.5173 × age = 0) reveals that it is 20.0132/1.5173 = 13.19 years.
You should not pay too much attention to the deviance residuals in this case since they automatically become large in every case where the fitted probability “goes against” the observations (which is bound to happen in some cases). The residual deviance is also difficult to interpret when there is only one observation per cell.
A hint of a more complicated analysis is obtained by including the Tan-ner stage of puberty in the model. You should be warned that the exact interpretation of such an analysis is quite tricky and qualitatively different from the analysis of menarche as a function of age. It can be used for pre-diction purposes (although asking the girl whether she has had her first 13.6 Prediction 241 period would likely be much easier than determining her Tanner stage!), but the interpretation of the terms is not clear-cut.
summary(glm(menarche~age+tanner,binomial)) ...
Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -13.7758 2.7630 -4.986 6.17e-07 age 0.8603 0.2311 3.723 0.000197 tanner2 -0.5211 1.4846 -0.351 0.725609 tanner3 0.8264 1.2377 0.668 0.504313 tanner4 2.5645 1.2172 2.107 0.035132 tanner5 5.1897 1.4140 3.670 0.000242 ...
Notice that there is no joint test for the effect of tanner. There are a cou-ple of significant z-values, so you would expect that the tanner variable has some effect (which, of course, you would probably expect even in the absence of data!). The formal test, however, must be obtained from the deviances: > drop1(glm(menarche~age+tanner,binomial),test="Chisq") ...
Df Deviance AIC LRT Pr(Chi) 106.599 118.599 age 1 124.500 134.500 17.901 2.327e-05 tanner 4 161.881 165.881 55.282 2.835e-11 ...
Clearly, both terms are highly significant.
13.6 Prediction The predict function works for generalized linear models, too. Let us first consider the hypertension example, where data were given in tabular form: > predict(glm.hyp) 1 2 3 4 5 6 -2.3920763 -2.3920763 -1.6966575 -1.6966575 -1.5266180 -1.5266180 7 8 -0.8311991 -0.8311991 Recall that smoking was eliminated from the model, which is why the expected values come in identical pairs.
These numbers are on the logit scale, which reveals the additive structure.
Notice that 2.392 −1.697 = 1.527 −0.831 = 0.695 (except for roundoff er-242 13. Logistic regression ror), which is exactly the regression coefficient to obesity. Likewise, the regression coefficient to snoring is obtained by looking at the differences 2.392 −1.527 = 1.697 −0.831 = 0.866.
To get predicted values on the response scale (i.e., probabilities), use the type="response" argument to predict: > predict(glm.hyp, type="response") 1 2 3 4 5 6 0.08377892 0.08377892 0.15490233 0.15490233 0.17848906 0.17848906 7 8 0.30339158 0.30339158 These may also be obtained using fitted, although you then cannot use the techniques for predicting on new data, etc.
In the analysis of menarche, the primary interest is probably in seeing a plot of the expected probabilities versus age (Figure 13.2). A crude plot could be obtained using something like plot(age, fitted(glm(menarche~age,binomial))) (it will look better if a different plotting symbol in a smaller size, using the pch and cex arguments, is used) but here is a more ambitious plan: > glm.menarche <- glm(menarche~age, binomial) > Age <- seq(8,20,.1) > newages <- data.frame(age=Age) > predicted.probability <- predict(glm.menarche, + newages,type="resp") > plot(predicted.probability ~ Age, type="l") This is Figure 13.2. Recall that seq generates equispaced vectors, here ages from 8 to 20 in steps of 0.1, so that connecting the points with lines will give a nearly smooth curve.
13.7 Model checking For tabular data it is obvious to try to compare observed and fitted proportions. In the hypertension example you get > fitted(glm.hyp) 1 2 3 4 5 6 0.08377892 0.08377892 0.15490233 0.15490233 0.17848906 0.17848906 7 8 0.30339158 0.30339158 > prop.hyp 13.7 Model checking 243 8 10 12 14 16 18 20 0.0 0.2 0.4 0.6 0.8 1.0 Age predicted.probability Figure 13.2. Fitted probability of menarche having occurred.
0.08333333 0.11764706 0.12500000 0.00000000 0.18716578 0.15294118 0.29411765 0.34782609 The problem with this is that you get no feeling for how well the rela-tive frequencies are determined. It can be better to look at observed and expected counts instead. The former can be computed as > fitted(glm.hyp)n.tot 1 2 3 4 5 6 5.0267351 1.4242416 1.2392186 0.3098047 33.3774535 15.1715698 7 8 15.4729705 6.9780063 and to get a nice print for the comparison, you can use > data.frame(fit=fitted(glm.hyp)n.tot,n.hyp,n.tot) fit n.hyp n.tot 1 5.0267351 5 60 2 1.4242416 2 17 3 1.2392186 1 8 4 0.3098047 0 2 5 33.3774535 35 187 6 15.1715698 13 85 244 13. Logistic regression 8 10 12 14 16 18 20 0.0 0.2 0.4 0.6 0.8 1.0 Age predicted.probability Figure 13.3. Fitted probability for menarche having occurred and observed proportion in age groups.
7 15.4729705 15 51 8 6.9780063 8 23 Notice that the discrepancy in cell 4 between 15% expected and 0% ob-served really is that there are 0 hypertensives out of 2 in a cell where the model yields an expectation of 0.3 hypertensives!
For complex models with continuous background variables, it becomes more difficult to perform an adequate model check. It is especially a hindrance that nothing really corresponds to a residual plot when the observations have only two different values.
Consider the example of the probability of menarche as a function of age.
The problem here is whether the relation can really be assumed linear on the logit scale. For this case, you might try subdividing the x-axis in a number of intervals and see how the counts in each interval fit with the expected probabilities. This is presented graphically in Figure 13.3. Notice that the code adds points to Figure 13.2, which you are assumed not to have deleted at this point.
age.group <- cut(age,c(8,10,12,13,14,15,16,18,20)) 13.7 Model checking 245 > tb <- table(age.group,menarche) > tb menarche age.group No Yes (8,10] 100 0 (10,12] 97 4 (12,13] 32 21 (13,14] 22 20 (14,15] 5 36 (15,16] 0 31 (16,18] 0 105 (18,20] 0 46 > rel.freq <- prop.table(tb,1)[,2] > rel.freq (8,10] (10,12] (12,13] (13,14] (14,15] (15,16] 0.00000000 0.03960396 0.39622642 0.47619048 0.87804878 1.00000000 (16,18] (18,20] 1.00000000 1.00000000 > points(rel.freq ~ c(9,11,12.5,13.5,14.5,15.5,17,19),pch=5) The technique used above probably requires some explanation. First, cut is used to define the factor age.group, which describes a grouping into age intervals. Then a crosstable tb is formed from menarche and age.group. Using prop.table, the numbers are expressed relative to the row total, and column 2 of the resulting table is extracted. This con-tains the relative proportion in each age group of girls for whom menarche has occurred. Finally, a plot of expected probabilities is made, overlaid by the observed proportions.
The plot looks reasonable on the whole, although the observed proportion among 12–13-year-olds appears a bit high and the proportion among 13– 14-year-olds is a bit too low.
But how do you evaluate whether the deviation is larger than what can be expected from the statistical variation? One thing to try is to extend the model with a factor that describes a division into intervals. It is not prac-tical to use the full division of age.group because there are cells where either none or all of the girls have had their menarche.
We therefore try a division into four groups, with cutpoints at 12, 13, and 14 years, and add this factor to the model containing a linear age effect.
age.gr <- cut(age,c(8,12,13,14,20)) > summary(glm(menarche~age+age.gr,binomial)) ...
Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -21.5683 5.0645 -4.259 2.06e-05 age 1.6250 0.4416 3.680 0.000233 age.gr(12,13] 0.7296 0.7856 0.929 0.353024 age.gr(13,14] -0.5219 1.1184 -0.467 0.640765 246 13. Logistic regression 8 10 12 14 16 18 20 0.0 0.2 0.4 0.6 0.8 1.0 Age predicted.probability Figure 13.4. Logit-cubical fit of menarche data.
age.gr(14,20] 0.2751 1.6065 0.171 0.864053 ...
anova(glm(menarche~age+age.gr,binomial)) ...
Df Deviance Resid. Df Resid. Dev NULL 518 719.39 age 1 518.73 517 200.66 age.gr 3 8.06 514 192.61 > 1-pchisq(8.058,3) 0.04482811 That is, the addition of the grouping actually does give a significantly better deviance. The effect is not highly significant, but since the devia-tion concerns the ages where “much happens”, you should probably be cautious about postulating a logit-linear age effect.
Another possibility is to try a polynomial regression model. Here you need at least a third-degree polynomial to describe the apparent stagna-tion of the curve around 13 years of age. We do not look at this in great detail, but just show part of the output and in Figure 13.4 a graphical presentation of the model.
13.8 Exercises 247 > anova(glm(menarche~age+I(age^2)+I(age^3)+age.gr,binomial)) ...
Df Deviance Resid. Df Resid. Dev NULL 518 719.39 age 1 518.73 517 200.66 I(age^2) 1 0.05 516 200.61 I(age^3) 1 8.82 515 191.80 age.gr 3 3.34 512 188.46 Warning messages: 1: In glm.fit(x = X, y = Y, weights = weights, .... : fitted probabilities numerically 0 or 1 occurred 2: In method(x = x[, varseq <= i, drop = FALSE], .... : fitted probabilities numerically 0 or 1 occurred > glm.menarche <- glm(menarche~age+I(age^2)+I(age^3), binomial) Warning message: In glm.fit(x = X, y = Y, weights = weights, start = start, .... : fitted probabilities numerically 0 or 1 occurred > predicted.probability <-+ predict(glm.menarche, newages, type="resp") > plot(predicted.probability ~ Age, type="l") > points(rel.freq~c(9,11,12.5,13.5,14.5,15.5,17,19), pch=5) The warnings about fitted probabilities of 0 or 1 occur because the cubic term makes the logit tend much faster to ±∞than the linear model did.
There are two occurrences for the anova call because two of the models include the cubic term.
The thing to note in the deviance table is that the cubic term gives a sub-stantial improvement of the deviance, but once that is included, the age grouping gives no additional improvement. The plot should speak for itself.
13.8 Exercises 13.1 In the malaria data set, analyze the risk of malaria with age and log-transformed antibody level as explanatory variables.
13.2 Fit a logistic regression model to the graft.vs.host data set, pre-dicting the gvhd response. Use different transformations of the index variable. Reduce the model using backwards elimination.
13.3 In the analyses of the malaria and graft.vs.host data, try us-ing the confint function to find improved confidence intervals for the regression coefficients.
13.4 Following up on Exercise 8.2 about “Rocky Mountain spotted fever”, splitting the data by age groups gives the table below. Does this 248 13. Logistic regression confirm the earlier analysis?
Western Type Eastern Type Age Group Total Fatal Total Fatal Under 15 108 13 310 40 15–39 264 40 189 21 40 or above 375 157 162 61 747 210 661 122 13.5 A probit regression is just like a logistic regression but uses a differ-ent link function. Try the analysis of the menarche variable in the juul data set with this link. Does the fit improve? |
15314 | https://www.reddit.com/r/desmos/comments/ox0n4n/float_point_how_to_get_desmos_to_display_answers/ | Float Point, How to get Desmos to Display Answers in higher number of Decimal Places Globally : r/desmos
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15315 | http://apps.lonestar.edu/blogs/vindang/files/2017/06/1324-5.2.pdf | 1 Copyright © 2015, 2011, and 2008 Pearson Education, Inc. Chapter 5 Linear Inequalities and Linear Programming Section 2 Systems of Linear Inequalities in Two Variables 2 Copyright © 2015, 2011, and 2008 Pearson Education, Inc. Learning Objectives for Section 5.2 The student will be able to solve systems of linear inequalities graphically.
The student will be able to solve applications involving systems of linear inequalities. Systems of Linear Inequalities in Two Variables 3 Copyright © 2015, 2011, and 2008 Pearson Education, Inc. Solving Systems of Linear Inequalities Graphically We now consider systems of linear inequalities such as x + y > 6 2x – y > 0 We wish to solve such systems graphically, that is, to find the graph of all ordered pairs of real numbers (x, y) that simultaneously satisfy all the inequalities in the system.
The graph is called the solution region for the system (or feasible region.) To find the solution region, we graph each inequality in the system and then take the intersection of all the graphs. 4 Copyright © 2015, 2011, and 2008 Pearson Education, Inc. Graphing a System of Linear Inequalities: Example To graph a system of linear inequalities such as we proceed as follows: Graph each inequality on the same axes. The solution is the set of points whose coordinates satisfy all the inequalities of the system. In other words, the solution is the intersection of the regions determined by each separate inequality. 1 2 2 4 y x x y − < + − ≤ 5 Copyright © 2015, 2011, and 2008 Pearson Education, Inc. Graph of Example The graph of the first inequality y < –(1/2)x + 2 consists of the region shaded yellow. It lies below the dotted line y = –(1/2)x + 2.
The graph of the second inequality is the blue shaded region is above the solid line x – 4 = y.
The graph is the region which is colored both blue and yellow.
6 Copyright © 2015, 2011, and 2008 Pearson Education, Inc. Corner Points A corner point of a solution region is a point in the solution region that is the intersection of two boundary lines. In the previous example, the solution region had a corner point of (4,0) because that was the intersection of the lines y = –1/2 x + 2 and y = x – 4.
Corner point 7 Copyright © 2015, 2011, and 2008 Pearson Education, Inc. Bounded and Unbounded Solution Regions A solution region of a system of linear inequalities is bounded if it can be enclosed within a circle. If it cannot be enclosed within a circle, it is unbounded. The previous example had an unbounded solution region because it extended infinitely far to the left (and up and down.) We will now see an example of a bounded solution region. 8 Copyright © 2015, 2011, and 2008 Pearson Education, Inc. Graph of More Than Two Linear Inequalities To graph more than two linear inequalities, the same procedure is used. Graph each inequality separately. The graph of a system of linear inequalities is the area that is common to all graphs, or the intersection of the graphs of the individual inequalities.
Example: 9 Copyright © 2015, 2011, and 2008 Pearson Education, Inc. Application Suppose a manufacturer makes two types of skis: a trick ski and a slalom ski. Suppose each trick ski requires 8 hours of design work and 4 hours of finishing. Each slalom ski requires 8 hours of design and 12 hours of finishing. Furthermore, the total number of hours allocated for design work is 160, and the total available hours for finishing work is 180 hours. Finally, the number of trick skis produced must be less than or equal to 15.
How many trick skis and how many slalom skis can be made under these conditions? How many possible answers? Construct a set of linear inequalities that can be used for this problem. 10 Copyright © 2015, 2011, and 2008 Pearson Education, Inc. Application Solution Let x represent the number of trick skis and y represent the number of slalom skis. Then the following system of linear inequalities describes our problem mathematically. Actually, only whole numbers for x and y should be used, but we will assume, for the moment that x and y can be any positive real number. x and y must both be positive Number of trick skis has to be less than or equal to 15 Constraint on the total number of design hours Constraint on the number of finishing hours 11 Copyright © 2015, 2011, and 2008 Pearson Education, Inc. Application Graph of Solution The intersection of all graphs is the yellow shaded region.
The solution region is bounded and the corner points are (0,15), (7.5, 12.5), (15, 5), and (15, 0) The origin satisfies all the inequalities, so for each of the lines we use the side that includes the origin. |
15316 | https://www.genome.gov/Genetic-Disorders/Phenylketonuria | Published Time: 2019-03-09T00:14:40-0500
About Phenylketonuria
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About Phenylketonuria
Phenylketonuria is an inherited disorder of metabolism that causes an increase in the blood of a chemical known as phenylalanine.
What is phenylketonuria (PKU)?
Phenylketonuria (PKU) is an inherited disorder of metabolism that causes an increase in the blood of a chemical known as phenylalanine. Phenylalanine comes from a person's diet and is used by the body to make proteins. Phenylalanine is found in all food proteins and in some artificial sweeteners. Without dietary treatment, phenylalanine can build up to harmful levels in the body, causing mental disability and other serious problems.
Women who have high levels of phenylalanine during pregnancy are at high risk for having babies born with mental disability, heart problems, small head size (microcephaly) and developmental delay. This is because the babies are exposed to their mother's very high levels of phenylalanine before they are born.
In the United States, PKU occurs in 1 in 10,000 to 1 in 15,000 newborn babies. Newborn screening has been used to detect PKU since the 1960's. As a result, the severe signs and symptoms of PKU are rarely seen.
What is phenylketonuria (PKU)?Phenylketonuria (PKU) is an inherited disorder of metabolism that causes an increase in the blood of a chemical known as phenylalanine. Phenylalanine comes from a person's diet and is used by the body to make proteins. Phenylalanine is found in all food proteins and in some artificial sweeteners. Without dietary treatment, phenylalanine can build up to harmful levels in the body, causing mental disability and other serious problems.
Women who have high levels of phenylalanine during pregnancy are at high risk for having babies born with mental disability, heart problems, small head size (microcephaly) and developmental delay. This is because the babies are exposed to their mother's very high levels of phenylalanine before they are born.
In the United States, PKU occurs in 1 in 10,000 to 1 in 15,000 newborn babies. Newborn screening has been used to detect PKU since the 1960's. As a result, the severe signs and symptoms of PKU are rarely seen.
What are the symptoms of PKU?
Symptoms of PKU range from mild to severe. Severe PKU is called classic PKU. Infants born with classic PKU appear normal for the first few months after birth. However, without treatment with a low-phenylalanine diet, these infants will develop mental disability and behavioral problems. Other common symptoms of untreated classic PKU include seizures, developmental delay, and autism. Boys and girls who have classic PKU may also have eczema of the skin and lighter skin and hair than their family members who do not have PKU.
Babies born with less severe forms of PKU (moderate or mild PKU) may have a milder degree of mental disability unless treated with the special diet. If the baby has only a very slight degree of PKU, often called mild hyperphenylalaninemia, there may be no problems and the special dietary treatment may not be needed.
What are the symptoms of PKU?Symptoms of PKU range from mild to severe. Severe PKU is called classic PKU. Infants born with classic PKU appear normal for the first few months after birth. However, without treatment with a low-phenylalanine diet, these infants will develop mental disability and behavioral problems. Other common symptoms of untreated classic PKU include seizures, developmental delay, and autism. Boys and girls who have classic PKU may also have eczema of the skin and lighter skin and hair than their family members who do not have PKU.
Babies born with less severe forms of PKU (moderate or mild PKU) may have a milder degree of mental disability unless treated with the special diet. If the baby has only a very slight degree of PKU, often called mild hyperphenylalaninemia, there may be no problems and the special dietary treatment may not be needed.
How is PKU diagnosed?
PKU is usually diagnosed through newborn screening testing that is done shortly after birth on a blood sample (heel stick). However, PKU should be considered at any age in a person who has developmental delays or mental disability. This is because, rarely, infants are missed by newborn screening programs.
How is PKU diagnosed?PKU is usually diagnosed through newborn screening testing that is done shortly after birth on a blood sample (heel stick). However, PKU should be considered at any age in a person who has developmental delays or mental disability. This is because, rarely, infants are missed by newborn screening programs.
What is the treatment for PKU?
PKU is treated by limiting the amount of protein (that contains phenylalanine) in the diet. Treatment also includes using special medical foods as well as special low-protein foods and taking vitamins and minerals. People who have PKU need to follow this diet for their lifetime. It is especially important for women who have PKU to follow the diet throughout their childbearing years.
What is the treatment for PKU?PKU is treated by limiting the amount of protein (that contains phenylalanine) in the diet. Treatment also includes using special medical foods as well as special low-protein foods and taking vitamins and minerals. People who have PKU need to follow this diet for their lifetime. It is especially important for women who have PKU to follow the diet throughout their childbearing years.
Is PKU inherited?
PKU is inherited in families in an autosomal recessive pattern. Autosomal recessive inheritance means that a person has two copies of the gene that is altered. Usually, each parent of an individual who has PKU carries one copy of the altered gene. Since each parent also has a normal gene, they do not show signs or symptoms of PKU.
Gene alterations (mutations) in the PAH gene cause PKU. Mutations in the PAH gene cause low levels of an enzyme called phenylalanine hydroxylase. These low levels mean that phenylalanine from a person's diet cannot be metabolized (changed), so it builds up to toxic levels in the bloodstream and body. Having too much phenylalanine can cause brain damage unless diet treatment is started.
Additional Resources for PKU
NICHD: Phenylketonuria
Medline Plus: Phenylketonuria
Children's PKU Network
Genetic Testing Registry: Phenylketonuria
GARD: Phenylketonuria
Is PKU inherited?
PKU is inherited in families in an autosomal recessive pattern. Autosomal recessive inheritance means that a person has two copies of the gene that is altered. Usually, each parent of an individual who has PKU carries one copy of the altered gene. Since each parent also has a normal gene, they do not show signs or symptoms of PKU.
Gene alterations (mutations) in the PAH gene cause PKU. Mutations in the PAH gene cause low levels of an enzyme called phenylalanine hydroxylase. These low levels mean that phenylalanine from a person's diet cannot be metabolized (changed), so it builds up to toxic levels in the bloodstream and body. Having too much phenylalanine can cause brain damage unless diet treatment is started.
Additional Resources for PKU
NICHD: Phenylketonuria
Medline Plus: Phenylketonuria
Children's PKU Network
Genetic Testing Registry: Phenylketonuria
GARD: Phenylketonuria
Last updated: August 21, 2014
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15317 | https://www.geeksforgeeks.org/engineering-mathematics/newton-raphson-method/ | Newton Raphson Method
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Newton Raphson Method or Newton Method is a powerful technique for solving equations numerically. It is most commonly used for approximation of the roots of the real-valued functions.
It is a numerical technique for approximating the roots of real-valued functions.
It starts with initial guess of root and iteratively refines the result using a formula that involves derivative of the function.
Compared to other root-finding methods like bisection and secant methods, the Newton-Raphson method stands out due to its significantly faster convergence rate (quadratic while other have linear).
It requires computation of derivative and preferred over other methods when this computation easier and we can find good estimate of root.
Newton Raphson Method or Newton's Method is an algorithm to approximate the roots of zeros of the real-valued functions, using guess for the first iteration (x0) and then approximating the next iteration(x1) which is close to roots, using the following formula.
x1=x0−f′(x0)f(x0)
where,
x0is the initial value of x,
f(x0) is the value of the equation at initial value, and
f'(x0) is the value of the first order derivative of the equation or function at the initial value x0.
Note: f'(x0) should not be zero else the fraction part of the formula will change to infinity which means f(x) should not be a constant function.
Newton Raphson Method Formula
In the general form, the Newton-Raphson method formula is written as follows:
xn=xn−1−f′(xn−1)f(xn−1)
Where,
xn-1is the estimated (n-1)th root of the function
f(xn-1) is the value of the equation at (n-1)th estimated root
f'(xn-1) is the value of the first order derivative of the equation or function at xn-1
Newton Raphson Method Calculation
Assume the equation or functions whose roots are to be calculated as f(x) = 0.
In order to prove the validity of Newton Raphson method following steps are followed:
Step 1: Draw a graph of f(x) for different values of x as shown below:
Step 2: A tangent is drawn to f(x) at x0. This is the initial value.
Step 3:This tangent will intersect the X- axis at some fixed point (x1, 0) if the first derivative of f(x) is not zero i.e. f'(x0) ≠ 0.
Step 4: As this method assumes iteration of roots, this x1 is considered to be the next approximation of the root.
Step 5: Now steps 2 to 4 are repeated until we reach the actual root x.
Now we know that the slope-intercept equation of any line is represented as y = mx + c,
Where m is the slope of the line and c is the x-intercept of the line.
Using the same formula we, get
y = f(x0) + f'(x0) (x - x0)
Here f (x0) represents the c and f' (x0) represents the slope of the tangent m. As this equation holds true for every value of x, it must hold true for x1. Thus, substituting x with x1, and equating the equation to zero as we need to calculate the roots, we get:
0 = f(x0) + f'(x0) (x1 - x0)
x1=x0−f′(x0)f(x0)
Which is the Newton Raphson method formula.
Thus, Newton Raphson's method was mathematically proved and accepted to be valid.
Convergence of Newton Raphson Method
The Newton-Raphson method tends to converge if the following condition holds true:
| f(x). f''(x) | < | f'(x) |2
It means that the method converges when the modulus of the product of the value of the function at x and the second derivative of a function at x is lesser than the square of the modulo of the first derivative of the function at x. The Newton-Raphson Method has a convergence of order 2 which means it has a quadratic convergence.
Applications of Newton Raphson Method
Root Finding in Mathematics: The primary use of the Newton-Raphson method is to find the roots (or zeros) of functions. Given an equation f(x)=0, the method iteratively approximates the solution by refining guesses.
Solving Non-linear Equations: In engineering and physics, many real-world problems are modeled by non-linear equations. The Newton-Raphson method is used to find solutions to these equations efficiently.
Optimization Problems: In optimization, the method is used to find the maximum or minimum of a function. By iterating towards a point where the derivative of the function equals zero, the method can identify critical points in functions.
Machine Learning: It is used in some optimization techniques, such as for training models by finding the minimum of a loss function using the method’s iterative approach to minimizing error.
Engineering (Structural Analysis): In structural engineering, the method is used to solve complex systems of equations, such as those that arise when analyzing stresses and strains in materials.
Newton Raphson Method Example
Let's consider the following example to learn more about the process of finding the root of a real-valued function.
Example 1: For the initial value x0 = 3, approximate the root of f(x)=x3+3x+1.
Solution:
Given, x0 = 3 and f(x) = x3+3x+1
f'(x) = 3x2+3
f'(x0) = 3(9) + 3 = 30
f(x0) = f(3) = 27 + 3(3) + 1 = 37
Using Newton Raphson method:
x1 =x0 −f′(x0)f(x0)
= 3 - 37/30
= 1.767
Example 2: For the initial value x0= 1, approximate the root of f(x)=x2-5x+1.
Solution:
Given, x0= 1 and f(x) = x2-5x+1
f'(x) = 2x-5
f'(x0) = 2 - 5 = -3
f(x0) = f(1) = 1 - 5 + 1 = -3
Using Newton Raphson method:
⇒ x1 = 1 - (-3)/-3
⇒ x1 = 1 -1
⇒ x1 = 0
Problem 3: For the initial value x0= 2, approximate the root of f(x)=x3-6x+1.
Solution:
Given, x0= 2 and f(x) = x3-6x+1
f'(x) = 3x2 - 6
f'(x0) = 3(4) - 6 = 6
f(x0) = f(2) = 8 - 12 + 1 = -3
Using Newton Raphson method:
⇒ x1 = 2 - (-3)/6
⇒ x1 = 2 + 1/2
⇒ x1 = 5/2 = 2.5
Problem 4: For the initial value x0= 3, approximate the root of f(x)=x2-3.
Solution:
Given, x0= 3 and f(x) = x2-3
f'(x) = 2x
f'(x0) = 6
f(x0) = f(3) = 9 - 3 = 6
Using Newton Raphson method:
⇒ x1 = 3 - 6/6
⇒ x1 = 2
Problem 5: Find the root of the equation f(x) = x3 - 5x + 3 = 0, if the initial value is 3.
Solution:
Given x0 = 3 and f(x) = x3 - 5x + 3 = 0
f'(x) = 3x2 - 5
f'(x0 = 3) = 3 × 9 - 5 = 22
f(x0 = 3) = 27 - 15 + 3 = 15
Using Newton Raphson method
⇒ x1 = 3 - 15/22
⇒ x1 = 2.3181
Using Newton Raphson method again:
x2 = 1.9705
x3 = 1.8504
x4 = 1.8345
x5 = 1.8342
Therefore, the root of the equation is approximately x = 1.834.
Articles related to Newton Raphson Method:
Difference between Newton Raphson Method and Regular Falsi Method
Difference between Bisection Method and Newton Raphson Method
Newton Raphson Method: Practice Problems
Problem 1: Find the root of f(x) = x2-2 using the Newton-Raphson method starting with x0=1.
Problem 2: Find the root of f(x) = x3-2x+1 using the Newton-Raphson method starting with x0=0.
Problem 3: Find the root of f(x) = cos(x)-x using the Newton-Raphson method starting with x0=0.5.
Problem 4: Find the root of f(x) = ex-3x using the Newton-Raphson method starting with x0=1.
Problem 5: Find the root of f(x) = x3-4x2+6 using the Newton-Raphson method starting with x0=2.
Problem 6: Find the root of f(x) = ln(x)-1 using the Newton-Raphson method starting with x0=2.
Problem 7: Find the root of f(x) = x4-8x2+16 using the Newton-Raphson method starting with x0=2.5.
Problem 8: Find the root of f(x) = xsin(x)-1 using the Newton-Raphson method starting with x0=1.
Problem 9: Find the root of f(x)=x5-3x3+2 using the Newton-Raphson method starting with x0=1.
Problem 10: Find a root of f(x) = x3-6x2+11x-6 using the Newton-Raphson method starting with x0=3.
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15318 | https://ceur-ws.org/Vol-3702/paper23.pdf | The implementation of Montgomery modular reduction to speed up of modular exponentiation⋆ Ihor Prots’ko 1, Oleksandr Gryshchuk2 1 Lviv Polytechnic National University, S.Bandery, 12, Lviv, 79013, Ukraine 2 LtdС “SoftServe”, Sadova, 2d, Lviv, 79021, Ukraine Abstract Modular exponentiation over large integers involves multiple modular multiplications, which is very computationally expensive. Many processing systems use the Montgomery modular multiplication method, which reduces the latency of software and hardware implementations. The main directions of software development and outlines of the parts of Montgomery modular multiplication for the implementation are presented. The class Montgomery Arithmetic over large integers is implemented using four methods for Montgomery modular multiplication. We present the computation of modular exponentiation using the right-to-left binary exponentiation method for a fixed basis with a developed pre-computation of a reduced set of remainders using modular Montgomery multiplication. A comparison of the runtimes of three variants of functions for computing the modular exponentiation over large integers is performed. The algorithm with pre-computation of residues for fixed base provides a faster computation of modular exponentiation using Montgomery modular multiplication compared to the functions of modular exponentiation of the MPIR, OpenSSL libraries for large number more then 1K bits.
Keywords Montgomery modular multiplication, modular exponentiation, multithreading, large numbers 1. Introduction Modular reduction is the computation of x mod m. A basic operation in processing systems is computations in Zm integers modulo m, where m is a large positive integer, which may or may not be a prime. Modular reductions are normally used to create finite groups, rings, or fields. The most common usage for performance-driven modular reductions is in modular exponentiation algorithms. An efficient implementation of the modular reduction x mod m of large numbers is the key to high performance. The classical algorithm of modular reduction has no restriction on the size of x, m and can easily be adapted to a division algorithm with quotient and remainder. The formalization consists of estimating the quotient digit as accurately as possible. This is justified by the fact that using multiplication and division are the most time-consuming operations in the inner loops of algorithms, especially when calculating Modular reduction over multi-bit numerical data. Among the modular reduction algorithms: classical, Barrett, and Montgomery's, the Montgomery reduction is relatively simple and very efficient . The baseline Montgomery reduction algorithm will produce the residue for any size input. Montgomery reduction is a common algorithm used for modulus reduction. The unique property of this algorithm is that it does not compute the modulus directly, but instead, the modulus multiplied by a constant. The further development using Montgomery reduction for computing modular multiplication is much faster and does not require any division by m. This method is referred as Montgomery CMIS-2024: Seventh International Workshop on Computer Modeling and Intelligent Systems, May 3, 2024, Zaporizhzhia, Ukraine ihor.o.protsko@lpnu.ua (I. Prots’ko); ocr@ukr.net (O. Gryshchuk) 0000-0002-3514-9265 (I. Prots’ko); 0000-0001-8744-4242 (O. Gryshchuk) © 2024 Copyright for this paper by its authors. Use permitted under Creative Commons License Attribution 4.0 International (CC BY 4.0). CEUR Workshop Proceedings ceur-ws.org ISSN 1613-0073 modular multiplication and combines Montgomery reduction and multiple-precision multiplication. The scientific problem of speeding up modular reduction for processing systems is relevant for the present stage of the development of information and computer technologies. The software implementations of modular multiplication over large integers on general-purpose processors are an important target for optimization. The further increase in the speed of the computational implementation of the modular reduction operation and then the full multiplication part can be achieved only by using the multithreading of multi-core processor architectures. The paper is structured as follows: after the Introduction in Section 1 is described Montgomery Reduction as a common algorithm used for modulus reduction, and outlines the parts and basic stages of the Montgomery modular multiplication algorithm. Section 2 describes the developed software implementation of efficient Montgomery multiplication over large integers using the Multiple Precision Integers and Rationals library. For performance analysis in Section 3, the experiments and discussion of the software implementation of Montgomery modular multiplication for the computation of developed modular exponentiation are presented. As a result, the developed software implementation provides a faster computation of modular exponentiation using Montgomery modular multiplication compared to the general-purpose functions of modular exponentiation of the MPIR and OpenSSL libraries over large integers. Literature review There are different contemporary variations of Barrett and Montgomery algorithms, which have advantages and mines. Barrett reduction is a reduction algorithm proposed in 1986 by P.D. Barrett , designed to optimize the integers modulo m operation assuming m is constant and, divisions are replaced by multiplications. P. Barrett offered the idea of estimating the quotient x div m with operations that are less expensive in time than a classical multi-precision division by m. The only pre-computation [22n/m] required for successful modular reduction use of Barrett’s algorithm, where 2n is a number of bits. The computation of modular exponentiation based on Barrett's algorithm is better than the other known ones for small numerical values. Montgomery reduction uses on the changing of the original reduction modulo by some other convenient modulo. By representing the residue classes modulo Montgomery's algorithm replaces a division by m with a multiplication followed by a division by a power of radix r. In computer applications, b is usually defined as the power of 2, when m = 2k, k – the processor’s word-size, this operation is very easy and inexpensive. The idea developed by P. Montgomery's method suggests that the operations of addition and subtraction are practically unchanged, but multiplication changes slightly in a simple procedure without using reductions modulo m. Montgomery's algorithm (only for modulo m for which gcd(m, r) = 1) is faster than both the classical and Barrett's one and as fast as multiplication almost. There are different implementation algorithms of Montgomery reduction, which are improving to simpler and higher regularity. The paper proposes new residue number system Montgomery reduction algorithms, which achieve less number of unit multiplications. Traditional Montgomery approaches are combined with multiply-reduce methods at the bit-level in hardware implementations or based on the processor’s word-size level for software implementations . The parallel execution of modular operations “square and multiplications” based on Montgomery algorithm are described in the papers . The implementation of the Montgomery algorithm has been improved over the years, both at the software and hardware levels . 3. Montgomery Reduction and Modular Multiplication The Montgomery reduction of number T is defined as , mod 1 m R T (1) where m is a positive integer, T and R are integers such that R > m, gcd (m, R) =1, and 0 ≤ T < mR. The formula (1) is called a Montgomery reduction of number T modulo m with respect to R. Using Montgomery reduction easy to carry out modular reduction in the residue number system. The residue number system is a method for representing an integer as an n-tiple of its residues with respect to a given base. Montgomery Reduction i R−1mod m is a one-to-one mapping defined from Z/mZ to Z/mZ , for 0 ≤ i < m. To compute the Montgomery reduction, it is necessary to determine the value of R−1 that meets the condition R R−1 mod m =1. To find the inverse modulo, you can use the extended Euclidean algorithm. Indeed, if gcd(m, R) =1, then the following integers will be found u and v, that .
1 = mv Rv (2) If we pass to congruence modulo m in the last equality, then we obtain Ru ≡ 1 (mod m), which gives (R −1) ≡ u (mod m). For working with large numbers, where Montgomery multiplication is implemented, is common to write the Montgomery radix R as , 2 = k k r R (3) where k is the word-size of the computer architecture. Higher radices may be used but the radix-2 provides a simple algorithmic and hardware implementation. The algorithm to compute Montgomery constant μ=-m-1mod R for odd values m and R=2k is presented in the Fig. 1. Algorithm. Compute Montgomery constant μ=-m-1mod R Input : Odd integer m and R=2k Output : μ = - m -1 mod R_____ y ← 1 for i =2 to k do if (m y mod 2i ) ≠ 1 then y ← y + 2i-1 end if end for return μ ← R – y ; Figure 1: Algorithm of Computation of the Montgomery constant -m-1mod R There are different fast Modular Reduction Methods to implementing Montgomery modular reduction. The algorithm Montgomery Reduction for radix 2, which does not require some pre-computation is presented in Fig. 2. Algorithm Montgomery Reduction X R−1 mod m Input : X, m and R=2k Output : X 2-k mod m______ x = X for i= 1 to k do if x is odd then x = x + m; x = x/2; return x; Figure 2: Algorithm of Computation of the Montgomery Reduction for radix 2 This algorithm is based on scanning the bit of a large number X from the right (the least significant bit) to the left (the most significant bit). In the paper is described the efficiently computes Montgomery reduction. Let m’ = −m−1 mod R, if U = Tm’ mod R, m m−1 mod R =1, then .
/ ) ( mod 1 R m U T m R T (4) Taking the remainder modulo m was replaced by division by R, and also taking the remainder modulo R in the numerator of the formula (4). As a result, we can choose such R that truncation can be used instead of division. If we have long arithmetic with some radix r, then the degree of this radix ri. That is, modulo residues and divisions will turn into shifts and throw out extra numbers. In the chapter 14.3.2 Montgomery reduction are presented the algorithms and examples of Montgomery reduction based on formula (4). The algorithm does not require m’ = −m−1 mod R, but rather m’ = −m−1 mod r. Most processing systems are implemented by repetition of a modular multiplication with a large modulus m, that is, .
mod m y x z (5) where m is usually a large prime or a product of two large primes x = (xn−1 ... x1x0)r and y = (yn−1 ... y1y0)r , which are non-negative integers in a radix r representation such that x < m and y < m. Let us represent x’ and y’ of a number x and y in the Montgomery space as follows x’ = x R mod m and y’ = y R mod m. The Montgomery reduction of multiplication x’y’ is: .
mod mod / ) mod mod ( mod 1 1 ' ' m R y x m R m R y m R x m R y x (6) This means that, after doing the multiplication of two numbers in the Montgomery space, we need to reduce the result by multiplying it by R−1 and taking of modulo m. There is an efficient way to use Montgomery reduction. This operation called the Montgomery modular multiplication. Montgomery modular multiplication itself is fast, but it requires some pre-computation. Montgomery multiplication algorithm involves three basic stages: 1. The conversion of operands from integer domain to Montgomery space; 2. The multiplication of operands in the Montgomery space; 3. The conversion of operands back from Montgomery space to integer domain. The Montgomery multiplication needs to convert x and y into Montgomery space and their product out of Montgomery space (Fig. 3). In this method the costly division operation usually needed to perform modular reduction is replaced by simple shift operations by conversing the operands into the reduced number system domain before the operation and re-conversing the result after the operation. Montgomery modular multiplication involves: first conversion of operands into the Montgomery space, multiplication and then after the result is re-conversed into the Montgomery space. Conversion to Montgomery space to Montgomery space Fast multiplication in Montgomery space x y from Montgomery space x’ y’ x y R mod m x y mod m Conversion Conversion Figure 3: Computation of modular reduction using Montgomery modular multiplication For practical (Fig. 4) interest the R=rn will suffice when there can be a power of 2 and R=2n . The condition R > m is clearly satisfied, but gcd (m, R) =1 needs to be relatively prime i.e. must not have any common non-trivial divisors which will hold only if gcd (m, r) =1. Montgomery modular multiplication algorithm X Y (R-1) mod m Input : X, Y, m and R=2k, Output : X Y 2-k mod m S0 : LSB of S, xi∈ (xn−1 ... x1x0)2_________ S=0 for i= 0 to n do S = S + xi Y ; S = S + S0 m ; S = S/2 ; if S ≥ m then S = S-m; return S; Figure 4: Algorithm of computation of the Montgomery modular multiplication There are different implementations of Montgomery modular multiplication: the digit-serial architectures , special purpose circuits, what perform multiplication and reduction simultaneously , and parallel execution of modular multiplication . In practice at the software and hardware levels, Montgomery multiplication is the most efficient method when is used a very regular structure, which speeds up the implementation [13, 14]. The software implementations of modular multiplication over large integers on general-purpose processors are an important target and has been improved over the years. In the next Section, we describe the software implementation of efficient Montgomery multiplication over large integers using the Multiple Precision Integers and Rationals library. 4. The software implementation of Montgomery reduction to modular multiplication The software implementations of Montgomery modular multiplication on the general purpose processors are an important target for optimization. Important focus is on the software implementation of the full multiplication parts including the efficient reduction. Many works improve the performance of a Montgomery Multiplication [15, 16]. Almost all the implementations of modular multiplication in many processing systems are performed in assembly languages to take advantage of the specific architectural properties of the processor . In this section, we describe software implementations of modular multiplication on the basis of the realization of Montgomery modular multiplication, which includes the efficient modular reduction and multiplication parts. The modular multiplication is implemented in C++ language. The developed class MontgomeryArithmetic (Fig. 5) implements the Montgomery modular multiplication and reduction using the Multiple Precision Integers and Rationals library (MPIR) . Class MontgomeryArithmetic private: const mpzclass mod_; size_t mod_size_; mpz_class inv_; const size_t limbs_; const size_t bits_; public: explicit MontgomeryArithmetic(const mpz_class& mod); mpz_class init(const mpz_class& x) const; void multiply(mpz_class& a, const mpz_class& b) const; void reduce(mpz_class& x) const; Figure 5: The MontgomeryArithmetic class According to the markings in Fig. 5, the member variables of the Class MontgomeryArithmetic are: size_t mod_size_ is a divisor size in MPIR limbs (64-bit integers); mpz_class inv_ is a pre-computed inverse factor for the Montgomery reduction; const size_t limbs_ is the same as size_, but a more convenient name; const size_t bits_ is a bit count for the modular arithmetic. The parameters of the methods are: mod is a divisor for modular arithmetic; x is a number for the conversion; a and b are the first and second numbers converted to the Montgomery space. The constructor MontgomeryArithmetic(const mpz_class& mod) computes a modular inverse factor for the Montgomery reduction and initializes other member variables, where the argument mod is a divisor for modular arithmetic. For computing the inverse factor m’=m-1mod R efficiently, we can use the mathematical dependence, which is inspired by Newton's method. The algorithm for calculating the inverse factor is described and proved in : .
2 mod 1 ) 2 ( 2 mod 1 2k k x m x m x m (7) This means we can start with x =1, as the inverse of m modulo 21, apply the trick of power times and in each iteration we double the number . This algorithm uses only shifts, subtractions and multiplication of large numbers in each iteration and has the same computational complexity as the algorithm, which is shown in Fig. 1. The method init mpz_class init(const mpz_class& x) const converts a number to the Montgomery space. It is required to convert all numbers before applying the Montgomery multiplication. The algorithm for the conversion is described in , where the relation is used , / mod 2 2 R x R R x m R x (8) where x is a number for the conversion. Converting the number into the space is just a multiplication inside the space of the number with R 2. Therefore, we can pre-compute R 2mod m and just perform a multiplication instead of shifting the number. This algorithm uses the shifts and the subtractions and multiplications of large numbers in each bits_ iteration. The method returns the converted value, which can be used for the Montgomery multiplication. The method void multiply(mpz_class& a, const mpz_class& b) const multiplies two numbers, where a, b are the numbers converted to the Montgomery space. The method returns the result via first argument in place and then performs the Montgomery reduction. It modifies the first argument in place to improve efficiency and avoid copying. For multiplication, it uses regular multiplication provided by the MPIR library, which is optimized using AVX2 SIMD instructions. The method void reduce(mpz_class& x) const, where argument x is a number for the reduction in place, computes the Montgomery reduction in place. Any number from the Montgomery space can be converted back using this method. This is one of the most performance-critical methods. The MPIR library offers a few low-level implementations, which can be further optimized for specific use cases. This method calls the mpn_redc1() function provided by MPIR to compute the Montgomery reduction in place. The methods and initialized member variables in the developed class MontgomeryArithmetic provide an implementation of Montgomery modular multiplication corresponding to Fig. 3. The operations of multiplication and division by R=2k are very fast in the methods of class, as they are just bit shifts. Thus, Montgomery's algorithm is faster than the usual (a·b) mod m, which contains division by m. However, the computation R-1, m-1 and conversion of numbers to the remains and vice versa are time-intensive operations, as a result, of which it is inefficient to use the product for a single computation. Montgomery reduction is the fastest in computing a reasonably long series of modular reductions, for instance in computing exponential function. This algorithm is a time critical step in the computation of the modular exponentiation operation. 5. Experiments and discussions of the software implementation of Montgomery modular multiplication for the computation of modular exponentiation Modular exponentiation over large integers involves multiple modular multiplications, which is very computationally expensive. Modular exponentiation of large numbers is extremely necessary for providing high crypto capability of information data, for finding the discrete logarithm, in number-theoretic transforms and many other applications. Considerable attention is paid to the development of effective methods of modular exponentiation aimed at effective computation and reduction of the execution time of the modular exponentiation operations [20, 21]. One of the ways to speed up computations of modular exponentiation is parallelization of computations using modern software technologies for universal computer systems or creation of specialized computing tools. The software implementation of the Montgomery multiplication and modular exponentiation computation is included in the software libraries Crypto++, OpenSSL, PARI/GP, MPIR designed for working with large numbers. The production-grade software library and full-featured toolkit popular on Linux and other systems is OpenSSL library. OpenSSL library contains a set of tools that implements the Secure Sockets Layer (SSL v2/v3) and Transport Layer Security (TLS v1) . The functions BN_mod_mul_montgomery, BN_MONT_CTX_new of OpenSSL library implement Montgomery multiplication. The library includes three functions to calculate the modular exponentiation using Montgomery multiplication: BN_mod_exp_mont(), which calculates A to the power of x modulo m, and BN_mod_exp_mont_consttime(), BN_mod_exp_mont_consttime_x2(). Let's compare the use of Montgomery modular multiplication with the usual modular multiplication operation on the example of an efficient computation of modular exponentiation of large numbers. Consider the basic iterative algorithm using pre-computation to form a shortened sequence of residues of the fixed base A for computing the modular exponentiation .
modm A y x (9) The central idea to calculate Ах mod m is to use the binary representation of the exponent x. For a fixed-base A of the modular exponentiation (9), which is equal to the product of the residuals r.0, r.1,…, r.k-1 of the exponent (A2i ) mod m, (i = 0, 1, 2, …, k-1). Modular exponentiation is implemented using the development of the right-to-left binary exponentiation method for a fixed base with pre-computation of a reduced set of residuals. That can speed up the process of computing the modular exponentiation by pre-computing (Fig. 6) the sequence of residuals, and repetitions with the period T' after the offset u in the unit Precomputation u, T' . The scheme (Fig. 6) for computing the modular exponentiation consists of the denotations: A is the input of the base number; m is the input of the module; x is the input of an exponent with binary digits x.(k-1), x.(k-2),…,x.2, x.1, x.0; (A^2i)m are blocks of computation of the integer exponent of exponent 2i of the number A by the module m, i = 0,1,2,…, (k-1); – r.0, r.1,…, r.k-1 are residuals A^2i mod m, (i = 0, 1, 2, …, k-1), (X) mod m is the block of modular multiplication; y is the output of the modular exponentiation. Thus, applying the parallel execution of the computation of modular exponentiation with the pre-computation, threads are created during the software execution of the modular multiplication of residual values r.i, where i ≤ T', in the block of modular multiplication. These residual values r.i are determined in the process of computing of residual exponents (A^2i) mod m, (i = 0, 1, 2, ..., k-1). The only difficulty in organizing computations with such threads is the need to synchronize the streams and the unit of Precomputation u, T' to ensure the correct computation of the final value y of modular exponentiation. To implement the algorithm for computing the integer power of a number Ах by modulus m, the MPIR library is used, which is written in C and assembler and provides the ability to compile its functions in Visual Studio C++. Accordingly, in the MPIR library, the data type mpz_t represents large numbers of arbitrary length, which are chosen for the power of the number base and mod with the number of bits from 256 to 2048 bits for testing. However, using the function mpn_redc1() implement Montgomery multiplication is not efficient enough in the process of modular exponentiation. . . .
. . . x. (k-1) . . .
. . .
. . .
y . . .
. . .
x.0 .1 .2 .(k-2) .(k - 1 ) r.(k -1) . . . r.i r. j r. l .0 A m x ( A 2 0 ) m ^ ( A 21)m ^ (A 2 2)m ^ (A 2 k-2)m ^ (A 2 k-1)m ^ r r r r r .1 1 r r. 2 x x x x r.i r . j Precomputation u ,T’ x modular multiplication Figure 6: The scheme for computation of modular exponentiation y = Ах mod m with pre-computation The algorithm consists of precompute() and precompute_parallel() functions. The precompute() function determines the sequence of a reduced set of residues. The precompute() function calculates the sequence of remainders for fixed numbers base and mod for exp = 2i (i = 0, 1, 2, …) and analyzes the periodicity with the appearance of each defined remainder r.i, which are calculated by the find_remainders() function. The pre-computation has been made in a separate find_remainders() function to optimize multiple remainder searches (A^2i) mod m. The function update_remainers() reduces the length of the sequence of remainders as a result of fixing the periodicity T', taking into account the offset u. The precompute_parallel() function aim to compare the performance execution with the use of Montgomery modular multiplication and usual modular multiplication operation. To implement the algorithm, the mpz_init_set (mul, base), mpz_sizeinbase (exp, 2), mpz_tstbit (exp, i), mpz_mul (r, r, mul) functions from the MPIR library are used, the parameters of which can be multi-bit data limited to bit size 2048 bits. To organize efficient multithreading computation of modular exponentiation according to the precompute_parallel() function, the thread_function() and parallel() are implemented. The developed precompute_parallel() function uses multiple threads for the computation of the modular exponentiation. The method run() runs parallel exponentiation using multiple threads. It has the following steps: 1) creates a collection of the active exponent bits; 2) splits the exponent bits among the defined number of threads; 3) waits for every thread execution. 4) calculates the final result by multiplying partial results calculated by the threads. The final result of the function is written to the variable s_thread_result, and the computation time is fixed and averaged to output. We compare the time of calculating the modular exponent using the usual modular multiplication with the Montgomery modular multiplication based on the developed functions precompute_modulo(), precompute_parallel_modulo() and precompute_montgomery(), precompute_parallel_montgomery(), respectively. Testing of the calculation of modular exponentiation were carried out on a computer system with a multi-core microprocessor Intel Core i9-10980XE (18 cores, 36 threads, 3.0GHz) with shared memory in a 64-bit Windows. According to hyper-threading technology, each physical core of 18 consists of two virtual 36 ones. The numerical results are presented in Figure 7, which contains the values of average execution time (μs microseconds) for 500 and 250 trials of computing the modular exponentiation for pseudo-random data base, exp, mod for 1024 bits and 2048 bits. Figure 7: The results of testing the functions of computing the modular exponentiation on a computer system with an Intel Core i9-10980XE processor with a chosen number of threads of 12 The pre-computation time to determine the sequence of a reduced set of residues is taken into account, therefore the total average time for computing the modular exponentiation modexp() is equal to: 1) for the usual modular multiplication operation: modexp() = precompute_modulo() time + precompute_parallel_modulo() time. In accordance with the result of testing (Fig. 7) average time are equal to: modexp()=(301+153)μs=454μs; modexp()=(1290+411)μs =1701μs; for pseudo-random data the base, exp, mod of 1024 bits and 2048 bits respectively. 2) for the Montgomery modular multiplication: modexp()= precompute_montgomery() time + precompute_parallel_montgomery() time. In accordance with the result of testing average time (Fig. 7) are equal to: modexp()= (1290+71)μs =251 μs; modexp()= (1050+173)μs=1223μs; for pseudo-random data the base, exp, mod of 1024 bits and 2048 bits respectively. Therefore, the implementation of the Montgomery modular multiplication is based on the developed class MontgomeryArithmetic for computing the modular exponentiation speed up 454μs/251μs=1,8 and1701μs/1223μs=1,4 times for pseudo-random data the base, exp, mod of 1024 bits and 2048 bits. A highly optimized modification of the well-known GMP or GNU Multiple Precision Arithmetic Library the MPIR library contains the function mpz_powm () to realize the computation of modular exponentiation. The MPIR library uses an optimized version a floating-window algorithm of the modular exponentiation with Montgomery multiplication/reduction, which reduces the average number of multiplication operations. The function of the MPIR library mpz_powm(expected_result, base, exp, mod) better performs modular exponentiation than function BN_mod_exp_mont() of the OpenSSL and Crypto++ libraries in accordance with the results received in , therefore we chose the function mpz_powm() for comparison (Fig. 5). At testing results, the total average time of mpz_powm () function for computing the modular exponentiation modexp() is 301μs and 2088μs and is greater than the average time for computing the modular exponentiation the Montgomery modular multiplication 251μs and 1223μs for pseudo-random data the base, exp, mod of 1024 bits and 2048 bits respectively. The closest scientific work for comparing research results is work , where an approach that uses vector SIMD instructions for parallel computation of multiple Montgomery multiplications is applied. This work describes the fact of the comparison of a parallel version of Montgomery multiplication using vector SIMD instructions to the implementation of the function of modular exponentiation in the OpenSSL library. The parallel version of Montgomery multiplication using vector SIMD instructions performance increases by more than a factor of 1.5 compared to the implementation in the OpenSSL library in the classical arithmetic logic unit on the Atom platform for 2048-bit moduli. Our implementation of the modular Montgomery multiplication to compute the modular exponentiation has factors 1.8 and 1.4 for the pseudorandom data the base, exp, mod of 1024 bits and 2048 bits compared to the sequential implementation in MPIR library. According to the obtained results of modular exponentiation , the MPIR library is faster for large numbers than OpenSSL. The values of an average execution time of modular exponentiation depend on the computing capabilities in universal computer systems. Testing results was received on two computer systems with different computing capabilities with processors an Intel Core i9-10980XE (18 cores, 36 threads, 3.0GHz) and Intel Core і9-13900К (24 cores, 32 threads, 3.0GHz). The results are presented in Table 1, which contains the values of average execution time (μs microseconds) for 500 trials of the functions modexp() and montgomery_modexp() using developed Montgomery modular multiplication for computing the modular exponentiation with pseudo-random data of 1024 bits. Table 1 The average execution time (μs) of the functions of computing the modular exponentiation Release/x86 Intel Core i9-10980XE Intel Core і9-13900К Data bits / trials 1024 / 500 1024 / 500 precompute_ parallel modulo modexp() 554 225 precompute_parallel_montgomery_modexp() 255 153 The optimal number of threads is 12...16 for fast computation of modular exponentiation for universal computer systems . Therefore, based on the developed Montgomery modular multiplication software the further implementation of the computation of modular exponentiation using multithreaded technologies will provide an opportunity for the efficient computation of modular exponentiation with a fixed base. Conclusions In the work is compared and analysed the developed software implementation of the class MontgomeryArithmetic in modular exponentiation function. The main directions of software development and outline of the parts of Montgomery modular multiplication for the implementation are presented. Modular exponentiation with a fixed base is implemented using the development of the right-to-left binary exponentiation method with pre-computation of a reduced set of residuals with the use of Montgomery modular multiplication or the usual modular multiplication The average run time of the computation on multi-core microprocessors of universal computer systems have been defined. As a result, an algorithm with pre-computation of residues for fixed base provides faster computation in average 1,5 times of modular exponentiation using Montgomery modular multiplication compared to the functions of modular exponentiation using the usual modular multiplication.
The scientific novelty of obtained results lies in the implementation of parallelism using multithreading in the function of computing the modular exponentiation based on Montgomery modular multiplication, which is the best among the known modular exponentiation functions of Crypto++, OpenSSL and MPIR libraries for large numbers more than 1K bits. The practical significance of the work lies in the fact that the obtained results can be successfully applied in modern asymmetric cryptography, for efficient computation of number-theoretic transforms and other computational problems. Prospects for further research are the parallel implementation of Montgomery Modular Multipliers in the developed function of the modular exponentiation for large numbers using the computation on the video car Acknowledgements The authors are grateful to Roman Rykmas of the team leader of Uniservice LtdC for participation in testing and discussing the results obtained by the functions of computing the Montgomery modular multiplication. The work was carried out within the framework of the state-budget scientific-research work of DB "Neuroruh" of the Lviv Polytechnic National University. References M. J. Ferrao, K. Kiran V. G, N., M. Megha, Implementation of Modular Reduction and Modular Multiplication Algorithms. IOSR Journal of VLSI and Signal Processing (IOSR-JVSP), 8(6), Ver. I (2018). doi: 10.9790/4200-0806013438. Barrett reduction, 2023. URL: Montgomery Reduction Scheme Functions, 2022. URL: -2 /montgomery-reduction-scheme-functions.html. S. Kawamura, Y. Komano, H. Shimizu, T. Yonemura, RNS Montgomery reduction algorithms using quadratic residuosity. Journal of Cryptographic Engineering, 9 (2019). doi: 10.1007/s13389-018-0195-8. Bing Li, Jinley Wang, Guocheng Ding, Haisheng Fu, Bingjie Lei, Haitao Yang, Jiangang Bi, Shaochong Lei. “A high-performance and low-cost Montgomery modular multiplication based on redundant binary representation.” IEEE Trans. Circuits Syst. II Express Briefs 68.7(2021): 2660- 2664. doi:10.1109/tcsii.2021.3053630. I. Boiarshyn, O. Markovskyi, B. Ostrovska, Organization of parallel execution of modular multiplication to speed up the computational implementation of public-key cryptography, Information, Computing and Intelligent systems 3 (2022). doi:10.20535/2708-4930.3.2022.265418. J. Bos, S. Friedberger, Faster modular arithmetic for isogeny-based crypto on embedded devices, Journal of Cryptographic Engineering, 10.2 (2020). doi: 10.1007/s13389-019-00214-6. A. Menezes, J. Oorschot, A. Vanstone, Handbook of Applied Cryptography, Taylor & Francis excl. spl reprint, India, 2018. R. K. Venkata, S. C. Simranjeet, V. Desalphine, D. Selvakumar, A Low Latency Montgomery Modular Exponentiation, Procedia Computer Science, 171 (2020). doi:10.1016/j.procs.2020.04.087. S. Fatemi, M. Zare, A. Khavari, M. Maymandi-Nejad, Efficient implementation of digit-serial Montgomery modular multiplier architecture, IET Circuits Devices Syst., 13.7 (2019). doi:10.1049/iet-cds.2018.5182. S. Srinitha, S. Niveda, S. Rangeetha, V. Kiruthika, A High Speed Montgomery Multiplier used in Security Applications. In: Proceedings of the 3rd International Conference on Signal Processing and Communication (ICPSC), Coimbatore, India, 2021, pp. 299–303, URL: webtoc.pdf. B. Buhrow, B. Gilbert, C. Haider, Parallel modular multiplication using 512-bit advanced vector instructions: RSA fault-injection countermeasure via interleaved parallel multiplication, Journal of Cryptographic Engineering 2 (2021). doi: 10.1007/s13389-021-00256-10. Jinan Ding, Shuguo Li. “A low-latency and low-cost Montgomery modular multiplier based on NLP multiplication.” IEEE Trans. Circuits Syst. II Exp. Briefs, 67.7 (2020): 1319-1323. doi:10.1109/TCSII.2019.2932328. Z. Zhang, P. Zhang, A Scalable Montgomery Modular Multiplication Architecture with Low Area-Time Product Based on Redundant Binary Representation, Electronics 11.3712 (2022). doi: 10.3390/electronics11223712. M. Issad, B. Boudraa, M. Anane, A. M. Bellemou, Efficient PSoC Implementation of Modular Multiplication and Exponentiation Based on Serial-Parallel Combination, Journal of Circuits, Systems and Computers 28.13 (2019). doi: 10.1142/s0218126619502293. D. Mukhopadhyay, Improvement over Montgomery Modular Multiplication. Information Systems Security. In: Proceedings of 17th International Conference (ICISS 2021), Patna, India, 2021, pp. 212–217. doi: 10.1007/978-3-030-92571-0_14N. Drucker, S. Gueron, Fast modular squaring with AVX512IFMA, Report 2018/335, Cryptology ePrint Archive, Preprint. MINOR revision (2018). URL: MPIR: Multiple Precision Integers and Rationals. 2021. URL: Montgomery Multiplication, 2022. URL: montgomery_ multiplication.html#implementation. J. Robert, C. Negre, T. Plantard, Efficient Fixed Base Exponentiation and Scalar Multiplication based on a Multiplicative Splitting Exponent Recoding, Journal of Cryptographic Engineering, 9.2 (2019). doi: 10.1007/s13389-018-0196-7. N. Emmart, F. Zhengt, C. Weems, Faster modular exponentiation using double precision floating point arithmetic on the GPU. In: Proceedings IEEE 25th Symposium on Computer Arithmetic (ARITH), Amherst, MA, USA, 2018, pp. 130–137. doi: 10.1109/ARITH.2018.8464792. OpenSSL . Cryptography and SSL/TLS Toolkit, 2024. URL: I. Prots’ko O. Gryshchuk, The Modular Exponentiation with precomputation of reduced set of resedues for fixed-base, Radio Electronics, Computer Science, Control 1 (2022). doi: 10.15588/1607-3274-2022-1-7. I. Prots'ko, O. Gryshchuk, V. Riznyk, Efficient Multithreading Computation of Modular Exponentiation with Pre-computation of Residues for Fixed-base. In: Proceedings of Sixth International Workshop on Computer Modeling and Intelligent Systems (CMIS 2023), Zaporizhzhia, Ukraine, 2023, pp.224-234. doi:10.32782/cmis/3392-19. J. Bos, P. Montgomery, Montgomery arithmetic from a software perspective, in: J. W. Bos, A. K. Lenstra, (Ed.), Topics in Computational Number Theory Inspired by Peter L. Montgomery, 1st. ed., Cambridge University Press, 2017, pp.10-39. doi: 10.1017/9781316271575. |
15319 | https://www.youtube.com/watch?v=OA5DE2QCZMY | Transformations of Quadratic Functions: Lesson
Robert Sutton
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Posted: 9 Apr 2025
Topic: Transformations of Quadratic Functions
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Hello there. Mr. Sutton here bringing you the algebra 2 21 2 lesson on transformations of quadratic functions. Uh for algebra 2, this is a two-part lesson. For the honors, this is one part. Here's what we're doing. We're going to be describing and graphing transformations of quadratic functions. So, we've already done transformations with absolute value and linear functions. We're just applying that to quadratics now. And we're also going to be writing functions that represent these transformations of quadratic functions. So let's get to it. To uh start things off here, we're going to give you the function f ofx= x^2. Uh pause the video and take a moment and use this table with these x values from -2 to 2 to see if you can graph this out and then figure out the domain and range based off of that. Give it a shot. All right. So going through and getting each of these -2 squared is going to be pos4 - 1^2 is 1 0 2 is 0. And then 1^2 is 1 and 2^2 is 4. So putting those on my graph, we've got -24 right here. Uh here's one 1 0 0 1 1. And up here we have 2, 4. Connecting those dots, we get this nice little curve here. Um so in general, something with an x² that is called a quadratic function. And you might be thinking, well quad usually means four. Um, but the quad they're talking about is like a college quad, which is a square. Uh, so it's talking about this squared piece if you were ever wondering about the name. And the shape that we get here, uh, we can think of this as a U shape. It's kind of like the absolute value has a Vshape. It's very similar, but we also call this specific shape a parabola. So, it's good to know both words there. Now, for the domain of this, we see that we're going forever to the left and forever to the right. And in fact, you can plug in any x value you want and square it and you'll get a real number back. So, this has a domain of all reals just like the absolute value and the linear functions. Um, but we also see for now the range, the y-v values we can get back. We can't get back any of the negative values here. The lowest we can go is zero, at least with the parent function here. And we can get all the positive numbers. So our range just like the absolute value is going to have uh y greater than or equal to zero. So now one key feature of these uh types of functions here is that unlike the linear functions but but similar to the absolute value we see that we've got this turnaround point. We call that the vertex the point at which the parabola reverses directions. And there's actually a specific form of a parabola called the vertex form. And here it is. y equals or f ofx= a x - h ^2 + k. This form is kind of like the form that you were used to graphing the absolute values in where you could do transformations of the function by you know changing uh the the multiplier outside the function and adding and subtracting things inside and outside. So all the transformations we did with absolute value we can do with parabas or quadratics in this form. Um so just a little review of uh and this also gives us the vertex by the way. That's why we call this the vertex form. Uh because this h and this k are the vertex that we can pick right out of there. Um so just a little review now of transformations. Remember that there was inside and outside changes that we could do to a function. And by inside in this instance we mean inside this squared parenthesis that it's the thing being squared. Outside is outside the squared stuff. So remember that inside changes are horizontal. They affect the left right and they are inverse of how they appear. They behave oppositely from how they appear to behave. And then the outside changes things outside the square now are going to be your up down vertical changes. And they're going to do exactly what they appear to be doing. So with that in mind, let's apply some of these transformations for our first example. Now we're going to describe the transformation of uh x^2 that this g function is performing and then we're going to graph it. So what are we doing here with this this +4 inside and this minus1 outside? Well, remember that inside changes are left right and behave oppositely inversely of what they look like. So adding four to the x inside really means we're subtracting four. So we're shifting everything to the left four. Uh this minus one outside is a vertical change and it's exactly what it looks like. We're just shifting down by one. So we have it then a translation left four and down one. So we'll start by writing our parent table. Uh that's the one you generated in the beginning with uh the x squar. So -2 through 2 giving us 4 1 0 1 and 4. and then we'll just come up with some rules for X and Y based off of what we said here. So, you probably know enough to to take it the next step here. See if you can figure out the rules for XY. Pause the video uh and and see if you can then graph this thing out. All right, so we've got uh left four. That means we need to do x minus 4. And then down one means we're going to be doing y minus one. So, we'll subtract four from all the x values. So that's going to be -6, -5, -4, -3, and finally -2. That's all our new x values. And then for the y values, we're going to be doing, let's see, 4 - 1 is 3. 1 - 1 is 0. Then we've got - 1. And then these two y values are the same. So we get the same kind of going in the opposite direction. So 0 and 3. Uh let's put on our xy axis now and graph these. So we have -6 comma 3 right up here. And then we have -50 is right here. -41 down here. And then -3 0 and -2 3. And then graph the parabola through that. For this next example, we're going to be transforming x^2 into - 12x^2. So, pause the video and see if you can figure out what this transformation is doing and come up with the new x and y rules and then graph it. All right, let's see how that went. So, because we've got stuff happening outside the function on this one, these are all going to be vertical changes. A negation here gives us a reflection. So, a vertical reflection. Multiplying by 1/2 is going to give us a stretch or a shrink. in this case a shrink because it's less than one. So we've got uh a vertical reflection and shrink. Now if this stuff had happened inside the function that would have been a horizontal reflection and multiplying by 1/2 inside would have given us a stretch instead of a shrink. Remember uh stuff is inverse of what it looks like if it happens on the inside. So going back now, we've got our parent table again for x squ. We've got these these five points that work very well. Um, one reason I like these five points is because they include the vertex in them. Uh, the vertex is probably the most important part of the parabola. If you graph a parabola without showing the vertex, it's not really going to have that nice parabola shape to it. So now, what does the G rule look like for each of the x and the y's here? We didn't do anything horizontally. So x is still just going to be x. Nothing's changing with these x values. Vertically though, we reflected and shrunk things. We're essentially multiplying the y values by - one/2 here. So - 1/2 y. Let me go back and write my x values. Now we've just got all the original x values -2 -1 0 1 and 2. For the y values, though, we're multiplying all of these y values by - 1/2. So 4 -2 is -2. Then we have - 12 0 is still 0 - 12 and 4 - 1/2 is -2. One thing that is kind of worth noting here and you you might have noticed this also when we did the absolute value transformations is that this turnaround spot at 0 0 uh because zero is unaffected by multiplication. this spot is unaffected by reflections and stretches and shrinks. It still stays in the same spot. That's part of the reason the vertex form of a parabola actually works. Uh because if you just look at the the stuff that's happening in terms of plus and minus inside and outside the function, that's the only thing that's going to change these zero values. Uh multiplying all the shifting and reflecting doesn't do anything to it. So now let's put this on our graph. We've got -2 -2 right there. And then we have -1 - one/2 is right about here. 0 0 1 comma - one/2 and 2 comma -2. And I should this actually should be a -2 here. There we go. That was an easy fix. Now let's put it at 2, -2. I mean -2, -2 was a point, but we we don't need to put the same point twice. So, let's now draw the parabola through those. And there we go. Um, so just to kind of do a sanity check here, did we reflect this vertically? We did. Did we shrink it horizontally? It's kind of hard to tell with parabas unless you look at where the points started and where they ended up. But all of these points uh moved half the distance to the x-axis. So that's what a shrink is going to look like with these. Now, if you are doing the uh the algebra 2 uh non-honors, then this is the end of part one for you. Uh feel free to stop here and do whatever homework assignment you need to do. Uh if you were doing the honors version, then we're just going to keep on going. So, if you're doing the uh the non-honors algebra 2, then welcome back to part two of our presentation. Uh in this one, we're going to be writing rules for these quadratic transformations. If you're honors, then hey, you never left, so let's just keep going. So they want us to write a rule uh described by the transformations that they've listed here for this f function which in this case is x^2 but it's not necessarily always going to have to be just x^ squ and then they want us based off of this to identify the vertex. So when you're doing these uh my advice just like we did them with the absolute value and everything else is if there's multiple transformations happening do one of them at a time. create a new function for each transformation just so you don't lose anything. And I also recommend uh kind of using the uh do do like a general like big picture sort of transformation of the f function before you actually plug anything in. Um so for example like vertical stretch by a factor of two. Okay, what would that look like with the f function? Well, it's vertical so it would be outside. Stretching means we're multiplying. So I could write a g function that would equal 2 f ofx. And then I could go ahead and just replace uh the f function with this x squar here. And this is a function that now does this but with a vertical stretch of two. Uh so pause the video and see if you can go the rest of the way with these transformations to get to the final function. And then uh we'll we'll bring it together again before we talk about the vertex. All right. So, the next thing they wanted us to do was a reflection. Now, actually the reflection and the stretch could happen at the same time. Um, but let's just do one at a time. So, if we're reflecting in the yaxis, that's a left right reflection. That means I'm going to have to change the input to the g function to be a negative x. So, I'm going to write a new function h that's going to equal g ofx. And since the g function is just 2 x^2, this is 2 xant^ 2. And you could simplify this. When you take x and multiply by itself, you just get positive x^2. Um, so that the g function and the h function, that actually makes no difference. And that does kind of make sense for a parabola. If you're on the y ais, reflecting it over the y ais, you're going to get the same shape because it's symmetrical. All right, so we've got one more to do. That's a translation of three units down. So, I'm going to write another function. I don't like using I as a function. Um, so I'm going to use J instead. J of X. I'm going to call that H of X. And then 3 units down, I'm going to have to subtract 3 from the uh H function. So, H of X - 3. And since the h ofx function is just 2x^2, that's just 2x^2 - 3. So, that's what my transformed function looks like. Now, this is in vertex form. Um, there there's kind of like an x - 0 that we're not writing here, but you could think of this as x - 0 quantity squared. Uh, so pause the video for a little bit and see if you can figure out the vertex based off of this. All right, since we're doing x - 0, that means that uh 0 is the x value of our vertex. Subtracting three outside the squared stuff means we have a -3 for the y-value. So this is going to be a vertex of 0, -3. We're going to do one more example of writing rules of quadratics here based on transformations. In this one, you'll notice we didn't just give you like plain old boring x². We gave you a little bit more, but the still the principles are going to be the same in how we apply these things. So pause the video, see if you can give this one a shot. Uh write the new function and see if you can find the vertex based off of that. All right, let's see how that went. So we have a translation two units up. That means that my g function I'm going to write to do this has to take the f function and add 2 to that. Uh so that's just going to be all of this x - 2^ 2 - 6 stuff plus two. And uh all that gets affected is this minus 6 out here. Adding two to that just makes that a minus4. So not a huge change there, but we're not looking for big world changes here. Uh, the next thing they needed was a reflection in the x-axis. Since this is a vertical up down kind of change, if we're going over the x-axis, we need to put a negative outside whatever we're changing. Um, so I'm going to write the h function now, and I'm going to call that negative g ofx. That negative goes outside. G of x is all of this x - 2^2 - 4 stuff. So, I'm just going to put a negative outside of all of that. uh but I do have to distribute this now. So this is going to be - x -2^2 and then - gives us a +4 which is going to be important. Now if I want to pick the vertex out of this, this is in vertex form. There's actually two other forms of a parabola that we're going to see of a quadratic, but we're not there yet. Uh so the thing being subtracted inside again, that's the x value of the vertex. This minus2 inside means we have a positive2 for the x value of the vertex. Remember inside changes are inverse of what they look like. And then the plus4 really is a four for the y value. So 2a 4 is going to be our vertex. So that's it for this lesson on quadratic transformations. Hopefully upon reflection this was a transformative experience for you. Till next time, Mr. Sutton signing off. |
15320 | https://journals.sagepub.com/doi/10.1177/03946320241313426 | Skip to main content
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Micronutrient deficiencies in patients with celiac disease: A systematic review and meta-analysis
Saad Lamjadli saad.lamjadli@edu.uca.ma, Ider Oujamaa, […], Ikram Souli, Fatima ezzohra Eddehbi, Nadia Lakhouaja, Bouchra M’raouni, Abdelmouine Salami, Morad Guennouni, Moulay Yassine Belghali, Raja Hazime, and Brahim Admou+8-8View all authors and affiliations
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Abstract
This study aimed to characterize micronutrient deficiencies, including iron, ferritin, folic acid, vitamin D, zinc (Zn), vitamin B12, and copper, in patients with celiac disease, and evaluated the effects of these deficiencies on selected hematological parameters, including hemoglobin and mean corpuscular volume (MCV). Celiac disease (CeD), an immune-mediated disorder affecting the small bowel, is associated with genetic factors and micronutrient deficiencies. This meta-analysis was performed in accordance with the PRISMA guidelines. Literature searches of multiple databases retrieved 4140 studies, of which 45 were selected. Risk of Bias was performed in accordance with the STROBE checklist. Meta-analysis revealed a significant difference in hemoglobin levels between patients with CeD and controls (standardized mean difference (SMD) −0.59 (95% confidence interval (CI) −0.8459 to −0.3382); P = 0.0003). Iron levels were lower in patients with CeD (SMD ≈ −0.4 (95% CI −0.7385 to −0.0407); P = 0.0334), as were ferritin (SMD −0.6358 (95% CI −0.8962 to −0.3755); P = 0.0002), folic acid (SMD −0.5446 (95% CI −0.9749 to −0.1142); P = 0.0187), and vitamin D (SMD −0.4011 (95% CI −0.8020 to −0.0001); P = 0.0499) levels, while Zn levels were significantly reduced (SMD −1.1398 (95% CI −2.0712 to −0.2084); P = 0.0242). No significant differences were found in MCV, or copper or vitamin B12 levels between patients with CeD and controls. This study highlighted significantly higher micronutrient deficiencies in patients diagnosed with CeD than in controls, underscoring the importance of systematic nutritional assessment and multidisciplinary management to address micronutrient deficiencies and minimize negative health impact(s).
Introduction
Celiac disease (CeD) is a multifactorial condition influenced by genetic and environmental factors.1,2 More than 90% of patients with CeD carry the human leukocyte antigen (HLA)-DQ2 haplotype, whereas the remainder harbor HLA-DQ8. Although these genetic markers are necessary, they are insufficient for CeD genesis.3,4 Furthermore, the presence of single or double copies of HLA-DQB102 has been associated with an increased risk for developing CeD.5 Gluten consumption is the most important environmental factor in the pathological development of CeD; nevertheless, other factors, including viral infections, gut microbiota, and smoking, have also been implicated in its pathogenesis.3,4 CeD can occur at any age and has a wide range of symptoms. As a result, it is critical to make a diagnosis, not only in individuals experiencing conventional gastrointestinal symptoms, but also in those with extraintestinal clinical features because both forms can have significant and severe implications.1 The only effective treatment for CeD is strict adherence to gluten-free diet (GFD). Noncompliance with GFD increases the risk for morbidity and mortality due to related illnesses such as infertility, bone ailments, and cancer. According to the 2013 American College of Gastroenterology guidelines, one of the most common complications of CeD at diagnosis is micronutrient deficiency, primarily caused by chronic malabsorption due to villous atrophy (VA) in the small intestine, leading to a significant reduction in the absorptive surface area.6 Therefore, micronutrient deficiency should be identified and assessed in patients with newly diagnosed CeD.7 Although a GFD generally leads to mucosal recovery, some patients with CeD undergoing long-term treatment may experience persistent VA on follow-up, with or without ongoing or recurrent symptoms.4,8 This persistent enteropathy appears to be more common in individuals >45 years of age,9 as indicated by recent findings in which age ≥45 years was established as one of the independent variables predicting the persistence of VA,10 although it has also been described in 19% of younger patients.11 Furthermore, the persistence of enteropathy may be exacerbated by poor adherence to a GFD, lack of nutrient fortification in gluten-free products, or generally lower nutritional value of these products,12 which may further contribute to the persistence of micronutrient deficiencies in patients with CeD.13 Micronutrient deficiencies in patients with CeD are highly debated. As such, this study aimed to identify anomalies in micronutrient levels, including iron, ferritin, folic acid, vitamin D, zinc (Zn), vitamin B12, and copper, in patients with CeD, and to assess the effects of these deficiencies on specific hematological parameters, such as hemoglobin and mean corpuscular volume (MCV), which are indirectly influenced by deficiencies in iron, folate, and vitamin B12.
Material and methods
Search strategy and study selection
This meta-analysis was performed in accordance with the Preferred Reporting Items for Systematic Reviews and Meta-Analyses (i.e. “PRISMA”) statement.14 A comprehensive literature search of the PubMed, Scopus, Google Scholar, and Web of Science databases for relevant studies, published between 1963 and December 2023, was conducted. The search was repeated until March 2024 to identify the most recent studies. The main search terms included “celiac disease,” “celiac,” “coeliac disease,” “micronutrient,” and “deficiency.” Synonyms and alternative spelling methods were used in this study. A full search strategy is presented in Appendix 1. The results from all databases were aggregated using Zotero version 6.0.36 (Zotero.org). Five researchers independently reviewed titles and abstracts to identify potentially eligible studies for full-text review. All reviewers assessed the full text of the articles in detail. Disagreements were resolved by collaborative discussion, when required. In cases of persistent disagreement, the senior author made the final decision. The reference lists of the selected studies were also examined to identify additional, potentially eligible studies.
Inclusion and exclusion criteria
This meta-analysis included all studies reporting raw micronutrient values (mean and standard deviation (SD)) in both the CeD and control groups. Studies were included if they reported raw values for hemoglobin, MCV, iron, ferritin, folic acid, vitamin D, Zn, vitamin B12, and copper in patients diagnosed with CeD according to the American College of Gastroenterology guidelines15) or in the control group. Case reports, case series, commentaries, letters to the editor, and studies that did not report data regarding the research question were excluded. The literature search had no language or data restrictions.
Data extraction
Data were independently extracted by five investigators and entered into spreadsheet software (Excel, Professional Office LTSC Plus, Microsoft Corporation, Redmond, WA, USA). Additionally, we employed various packages from R software to support further data analysis and processing.16–26 Relevant data were extracted independently by five investigators and conflicts were resolved by consensus discussion. The following data were extracted from each study: country; author; year of publication; study design; total number of patients included in the CeD and control groups; mean age in the CeD and control groups; mean and SD values for hemoglobin, MCV, iron, ferritin, folic acid, vitamin D, Zn, vitamin B12, and copper in the CeD and control groups.
Statistical analysis
Meta-analyses were based on a single effect size of the standardized mean. Values were transformed from available statistics (mean and SD) to determine a standardized effect size (Hedges’ g statistic) using a comprehensive meta-analysis software packages.27–33 Hedges’ g is related to Cohen’s d and can be interpreted using the same conventions for effect size, as follows: small (0.2), medium (0.5), and large (0.8).34,35 An added benefit of Hedges’ g is the correction of the biases found in small sample sizes.34,35 The random-effects model was applied in the present meta-analysis, thereby adopting a conservative approach that assumes that the true effect size may vary from study to study, enabling the results to be generalized to populations beyond the study samples.35,36 The Q statistic was used to measure the homogeneity of effect sizes across the studies.35,37 A significant Q statistic indicates dissimilar effect sizes across studies, suggesting that differences in methodology or population samples could introduce variance in the results between studies.37 To complement the Q test, the I2 statistic was also calculated, which provides an index of the degree of heterogeneity across studies, in which I2 signifies the percentage of the total variability in effect sizes due to the variability between studies and not due to sampling errors within studies.38 Percentages of approximately 25% (I2 = 25), 50% (I2 = 50), and 75% (I2 = 75) were interpreted as low, medium, and high heterogeneity, respectively.35,39 Egger’s regression test was used to assess publication bias.40 Rucker’s Limit was used to adjust for suspected publication bias using a random-effects model.35,41 Sensitivity tests (right-skewness and flatness tests) were used to correct for publication bias.35,42 Outliers were addressed by considering studies as outliers if their confidence interval (CIs) did not overlap with those of the pooled effects.35,43
Systematic review registration
This review has been registered on PROSPERO: CRD42024544466. Available from:
Results
Study selection
The initial literature search retrieved 4140 studies, of which 145 were assessed by full-text review, and 45 were eligible for inclusion, with perfect agreement between investigators. The study selection process is illustrated in Figure 1.
Study characteristics
The studies selected were from North America (n = 7), South America (n = 3), Northern Europe (n = 2), Western Europe (n = 19), Southern Europe (n = 4), Eastern Europe (n = 2), Southern Asia (n = 4), Western Asia (n = 3), and Northern Africa (n = 1) (Appendices 2 and 3).
Pooled effect size of hemoglobin in the CeD versus control groups
Twelve studies including 545 patients with CeD and 915 controls were included in this meta-analysis. The pooled results revealed that the SMD of hemoglobin level in CD patients was—0.59 (95% CI −0.8459 to −0.3382]; P = 0.0003) compared with the controls (Table 1; Figure 2). Publication bias was not observed (Table 1.b in Supplemental material). The corrected real effect size estimate was −0.7684 (95% CI −1.2658 to −0.2709) (Table 1.c in Supplemental material). The sensitivity (p-curve test) of the estimated SMD was significant (Table 1.d in Supplemental material). (Detailed data for Tables 1.b to 1.d are provided in the Supplemental material).
Table 1. Pooled effect size (SMD) results (Hemoglobin).
| Type of Meta-Analyses | g (SMD) | 95% CI | P | I² (%) | 95% CI |
--- --- --- |
| Main meta-analysis (k = 18) | −1.1020 | −1.8346, −0.3695 | 0.0055 | 95.8 | 94.5–96.8 |
| Influencing cases removed (k = 12) | −0.5920 | −0.8459, −0.3382 | 0.0003 | 62.2 | 29.3–79.8 |
k = number of studies.
Removed as outliers: Ballestero et al.,44 Nestares et al.,45 Işikay et al.,46 Kalayci et al.,47 Caterina et al.,48 Kapur et al.49
Pooled effect size of iron in the CeD versus control groups
Eight studies including 519 patients with CeD and 14,566 controls were assessed. The pooled results of the meta-analysis revealed that the SMD of iron level in CD patients was −0.4 (95% CI −0.7385 to −0.0407; P = 0.0334) compared with the controls (Table 2; Figure 3). Publication bias was not observed (Table 2.b in Supplemental material). The estimated corrected true effect size was −0.0837 (95% CI −0.7317 to 0.5643) (Table 2.c in Supplemental material). The sensitivity (p-curve test) of the estimated SMD was significant (Table 2.d in Supplemental material). (Detailed data for Tables 2.b to 2.d are provided in the Supplemental material).
Table 2. Pooled effect size (SMD) results (Iron).
| Type of Meta-Analyses | g (SMD) | 95% CI | P | I² (%) | 95% CI |
--- --- --- |
| Main meta-analysis (k = 10) | −0.7639 | −1.3829, −0.1448 | 0.0210 | 91.6 | 86.7–94.7 |
| Influencing cases removed (k = 8) | −0.3896 | −0.7385, −0.0407 | 0.0334 | 74.4 | 48.2–87.4 |
k = number of studies.
Removed as outliers: Nestares et al.,45 Caterina et al.48
Pooled effect size of ferritin in the CeD versus control groups
Twelve studies, including 799 patients with CeD and 1442 controls, were included. The pooled results of meta-analysis revealed that the SMD of ferritin levels in CeD patients was −0.6358 (95% CI −0.8962 to −0.3755; P = 0.0002) compared with the controls (Table 3; Figure 4). No publication bias was observed (Table 3.b in Supplemental material). The estimated corrected true effect size was −0.3885 (95% CI −0.8173 to 0.0403) (Table 3.c in Supplemental material). The sensitivity (p-curve test) of the estimated SMD was significant (Table 3.d in Supplemental material). (Detailed data for Tables 3.b to 3.d are provided in the Supplemental material).
Table 3. Pooled effect size (SMD) results (Ferritin).
| Type of Meta-Analyses | g (SMD) | 95% CI | P | I² (%) | 95% CI |
--- --- --- |
| Main meta-analysis (k = 13) | −0.9097 | −1.5621, −0.2573 | 0.0103 | 86.4 | 78.5–91.4 |
| Influencing cases removed (k = 12) | −0.6358 | −0.8962, −0.3755 | 0.0002 | 70.3 | 46.4–83.6 |
k = number of studies.
Removed as outliers: Caterina et al.48
Pooled effect size of folic acid in the CeD versus control groups
Ten studies, including 834 patients with CeD and 16,378 controls, were included in this meta-analysis. The pooled results revealed that the SMD of folic acid in patients with CeD was −0.5446 (95% CI −0.9749 to −0.1142; P = 0.0187) compared with the controls (Table 4; Figure 5). No publication bias was observed (Table 4.b in Supplemental material). The estimate of the corrected true effect size was −0.2540 (95% CI −0.7134 to 0.2055) (Table 4.c in Supplemental material). The sensitivity (p-curve test) of the estimated SMD was significant (Table 4.d in Supplemental material). (Detailed data for Tables 4.b to 4.d are provided in the Supplemental material).
Table 4. Pooled effect size (SMD) results (Folic acid).
| Type of Meta-Analyses | g (SMD) | 95% CI | P | I² (%) | 95% CI |
--- --- --- |
| Main meta-analysis (k = 11) | −0.3981 | −0.9547, 0.1585 | 0.1421 | 93.4 | 90–95.6 |
| Influencing cases removed (k = 10) | −0.5446 | −0.9749, −0.1142 | 0.0187 | 88.3 | 80.6–93.0 |
k = number of studies.
Removed as outliers: Ballestero-Fernández et al.44
Pooled effect size of vitamin D in the CeD versus control groups
Fifteen studies were analyzed, including 655 patients with CeD and 14,717 controls. The pooled results revealed that the SMD of vitamin D in patients with CeD was −0.4011 (95% CI −0.8020 to −0.0001; P = 0.0499) compared with the controls (Table 5; Figure 6). Publication bias was not observed (Table 5.b in Supplemental material). The sensitivity test (right-skewness) for the estimated SMD was significant (Table 5.c in Supplemental material). (Detailed data for Tables 5.b to 5.c are provided in the Supplemental material).
Table 5. Pooled effect size (SMD) results (Vitamin D).
| Type of Meta-Analyses | g (SMD) | 95% CI | P | I² (%) | 95% CI |
--- --- --- |
| Main meta-analysis (k = 19) | −1.3599 | −3.0166, 0.2968 | 0.1017 | 99.2 | 99.1–99.4 |
| Influencing cases removed (k = 15) | −0.4011 | −0.8020, −0.0001 | 0.0499 | 85.8 | 78.2–90.8 |
k = number of studies.
Removed as outliers: Bayrak et al.,50 Stein et al.,51 Jamnik et al.,52 Björck et al.53
Pooled effect size of Zn in the CeD versus control groups
Eight studies were analyzed, including 343 patients with CeD and 14,250 controls. The pooled results of the meta-analysis revealed that the SMD of Zn in patients with CeD was −1.1398 (95% CI −2.0712 to −0.2084; P = 0.0242) compared with the controls (Table 6; Figure 7). There was publication bias (Table 6.b in Supplemental material). However, the sensitivity test (right-skewness) for the estimated SMD was significant (Table 6.c in Supplemental material). (Detailed data for Tables 6.b to 6.c are provided in the Supplemental material).
Table 6. Pooled effect size (SMD) results (Zn).
| Type of Meta-Analyses | g (SMD) | 95% CI | P | I² (%) | 95% CI |
--- --- --- |
| Main meta-analysis (k = 9) | −1.1398 | −2.0712; −0.2084 | 0.0242 | 92.7% | 87.4−95.7% |
| Influencing cases removed (k = 8) | −1.4092 | −2.4145; −0.4039 | 0.0129 | 94.6% | 91.4−96.6% |
k = number of studies.
Removed as outlier: Idris et al.54
Pooled effect size of MCV in the CeD versus control groups
Six studies comprising 134 patients with CeD and 806 controls were included in this meta-analysis. The pooled results of the meta-analysis revealed that the SMD of MCV in patients with CeD was −0.16 (95% CI−0.8 to 0.47; P >0.05) compared with the controls (Table 7; Figure 8). Publication bias was not observed (Table 7.b in Supplemental material). The sensitivity test (right skewness) for the estimated SMD was significant (Table 7.d in Supplemental material). (Detailed data for Tables 7.b to 7.d are provided in the Supplemental material).
Table 7. Pooled effect size (SMD) results (MCV).
| Type of Meta-Analyses | g (SMD) | 95% CI | P | I² (%) | 95% CI |
--- --- --- |
| Main meta-analysis (k = 8) | −1.12 | −2.84, 0.59 | > 0.05 | 95.5 | 93.2–97.0 |
| Influencing cases removed (k = 6) | −0.16 | −0.8, 0.47 | > 0.05 | 79 | 53.4–88.3 |
k = number of studies.
Removed as outliers: Kalayci et al.,47 Kapur et al.49
Pooled effect size of copper in the CeD versus control groups
Six studies comprising 189 patients with CeD and 3396 controls. The pooled results revealed that the SMD of copper in patients with CeD was −0.6429 (95% CI −1.5264 to 0.2407; P > 0.05) compared with controls (Table 8; Figure 9). No publication biases were observed.
Table 8. Pooled effect size (SMD) results (Copper).
| Type of Meta-Analyses | g (SMD) | 95% CI | P | I² (%) | 95% CI |
--- --- --- |
| Main meta-analysis (k = 7) | −0.6429 | −1.5264, 0.2407 | 0.1253 | 88.6 | 79–93.8 |
| Influencing cases removed (k = 6) | −0.3221 | −1.0000, 0.3557 | 0.2763 | 67.5 | 22.8–86.3 |
k = number of studies.
Removed as outlier: Guerrieri et al.55
Pooled effect size of vitamin B12 in the CeD versus control groups
Ten studies were conducted to assess the vitamin B12 levels, including 838 patients with CeD and 16,437 controls. The pooled results of meta-analysis revealed that the SMD of vitamin B12 in patients with CeD was 0.01 (95% CI −0.0121 to 0.15; P > 0.05) compared with controls (Table 9; Figure 10). No publication bias was observed (Table 9.b in Supplemental material). The estimated corrected true effect size was 0.1563 (95% CI −0.0205 to 0.3331; P = 0.0831) (Table 9.c in Supplemental material). (Detailed data for Tables 9.b to 9.c are provided in the Supplemental material).
Table 9. Pooled effect size (SMD) results (Vitamin B12).
| Type of Meta-Analyses | g (SMD) | 95% CI | P | I² (%) | 95% CI |
--- --- --- |
| Main meta-analysis (k = 10) | 0.01 | −0.12, 0.15 | > 0.05 | 20 | 0.0–65.7 |
Risk of bias
Risk of bias was calculated using STROBE.56 Using this tool, the studies were assessed using a 22-point checklist and grouped into low, moderate, and high risks of bias. Studies with a score <50 were considered to be poor, 50–70 as fair, 70–85 as good, and ≥85 as excellent. Studies with a high risk of bias were excluded (Supplemental Table 1).
Subgroup analysis
A subgroup analysis was used to investigate the sources of heterogeneity in the meta-analyses. The included studies were separated into ≥2 subgroups and the pooled effect sizes observed in these subgroups were examined to determine whether they differed significantly from one subgroup to another. The results of this subgroup analysis revealed significant unexplained heterogeneity within each subgroup as well as smaller and/or unequal data points. Consequently, the validity of the effect estimate for each subgroup is questionable, implying that the subgroup analysis is unlikely to yield valuable results (results not shown).
Discussion
Results of the present meta-analysis revealed that the pooled global effect of hemoglobin, ferritin, iron, and MCV was reduced in patients with CeD compared with the control group (−0.6 (95% CI −0.8459 to −0.3382), P = 0.0003; −0.6358 (95% CI −0.8962 to −0.3755), P = 0.0002; −0.4 (95% CI −0.7385 to −0.0407), P = 0.0334; and −0.16 (95% CI −0.8 to 0.4), P > 0.05, respectively). Furthermore, iron deficiency anemia (IDA) is not unusual because it is one of the main manifestations of CeD and is the predominant abnormality in approximately 45% of individuals with subclinical forms.57 Notably, Simon et al.58 reported that IDA could be the sole presenting feature in 39% of patients with CeD, underscoring its significance in the clinical assessment of this disorder. Furthermore, patients with CeD presenting with anemia at diagnosis have more advanced disease and a slower dietary response than those without anemia. This observation was reported in an excellent review of the extraintestinal manifestations of CeD, highlighting that when anemia is the primary reason for presentation of the disease, patients exhibit higher anti-transglutaminase levels, lower serum cholesterol, and higher degrees of VA than those presenting with diarrhea alone.59 Therefore, CeD should be considered as a possible cause of IDA in all patients. The underlying reason why some patients with CeD develop IDA while others do not remains poorly understood. However, it may be associated with deficiencies in specific regulatory proteins that are crucial for iron absorption at the enterocyte level, reflecting an imbalance between iron loss and absorption.59,60 Several disorders can affect the upper gastrointestinal tract, which is crucial for the absorption of dietary iron.61 GFD is recognized as the primary intervention for managing mild cases of IDA in CeD patients.62 However, the recovery of iron levels through GFD alone can be slow, particularly in severe cases.63 To accelerate the restoration of iron stores, oral iron supplementation may be implemented, which is especially beneficial for patients with mild enteropathy or those with inconsistent adherence to the GFD.59 In cases of advanced enteropathy, oral iron supplementation may lead to adverse effects, necessitating alternative strategies such as intravenous iron administration or methods to improve tolerability.59 Therefore, while dietary measures, such as a high-iron diet, can complement therapy, they are insufficient as standalone treatments and cannot replace the essential role of iron supplementation in managing anemia in CeD.59
CeD is a well-known cause of duodenal intraepithelial lymphocytosis, inflammation, and VA. It is mostly observed in the duodenum and upper jejunum. This may explain why folate deficiency has been reported in 8–85% of adult patients with CeD.64 The disparity in prevalence may be explained, in part, by the technical problems of measuring “folate” and “folic acid” because the bioavailability of folic acid is twice that of folate.65 Furthermore, patients with CeD exhibit megaloblastic anemia and neurological symptoms, and their chance of acquiring this deficiency is >5 times higher than that of healthy individuals. This was most likely caused by loss of villi in the proximal small intestine. As a result, the greater the degree of VA, the greater the folate insufficiency.66 Furthermore, a GFD appears to improve or even normalize folic acid levels in those affected by CeD.6 Our results are consistent with those of previous reports, given that the global pooled effect of folic acid in our study was −0.5446 (95% CI −0.9749 to −0.1142; P = 0.0187) in the CeD group compared with that in the control group.
The small intestine plays a critical role in Zn homeostasis. Zn deficiency in patients with CeD can be caused by an increased endogenous loss of this mineral rather than by abnormal Zn absorption.67 This cumulative loss can occur through several mechanisms, including the formation of insoluble Zn complexes with fat and phosphate, exudation of Zn protein complexes into the intestinal lumen, massive loss of intestinal secretions, and impaired Zn absorption resulting from damage to the intestinal epithelial cell membrane.68 Some CeD symptoms (e.g. anorexia and slow growth rate) may be linked to Zn deficiency. In recent years, Zn has emerged as a critical micronutrient for maintaining the integrity of the intestinal mucosa, immunity, and growth. Moreover, patients with CeD have been shown to have lower plasma Zn concentrations.68–73 Similar results were found in our meta-analysis, in which the pooled effect of Zn was −1.1398 (95% CI −2.0712 to −0.2084; P = 0.0242) in the CeD group compared with that in the control group.
CeD is linked to a wide range of endocrine concerns,74,75 the most prevalent of which are low bone mineral density (BMD), osteopenia, and osteoporosis,76 resulting in a high risk for bone fracture(s). Therefore, BMD measurements in adult patients are recommended.77 Although BMD was not considered in this meta-analysis, we found that the pooled effect of vitamin D in the CeD group was −0.4011 (95% CI −0.8020 to −0.0001; P = 0.0499) compared with that in the control group. In light of these results, the P-value analysis and p-curve results (P-Full and P-half ≤ 0.05), revealed that the pooled effect is not completely spurious; it is not merely a “mirage” produced by selective reporting.35 Our results suggest that vitamin D levels are low in patients with CeD. In addition to repairing and protecting the skeletal system during calcium metabolism, other roles of vitamin D have recently been reported. Vitamin D plays an important modulatory role in inflammation, immunological processes, and mucosal barrier control. In this context, vitamin D can cause immunological disorders and the role of vitamin D in immune regulation may be a major element in the initiation of CeD.78 Nonetheless, the results of studies investigating vitamin D levels and screening for vitamin D deficiency in patients are conflicting.79 Most vitamin D investigations on adult CeD have demonstrated that 25(OH) D insufficiency improves with a GFD, regardless of supplementation.80 The active form of 1,25 (OH) vitamin D was within normal range at the time of CeD diagnosis. It has been suggested that a GFD can boost vitamin D levels without the need for supplementation.81 However, our results provide evidence that patients with CeD should undergo nutritional assessment and receive nutritional counseling, as well as a strict GFD, and that dietary supplements should be recommended for those with severe deficiencies.
True deficits are difficult to demonstrate due to the complicated interplay between the elements. For example, folate requires vitamin B12 activation; therefore, low intracellular folate levels may result from vitamin B12 deficiency.82 Vitamin B12 deficiency appears to be rare in patients with CeD because it binds to intrinsic factors in the duodenum and the complex is absorbed in the terminal ileum, which is supposed to be protected from harm in CeD. Although the precise etiology of vitamin B12 deficiency in CeD remains unclear, potential contributing factors, such as reduced gastric acid production, small intestinal bacterial overgrowth (SIBO), autoimmune gastritis, and subtle dysfunction of the distal small intestine, have been suggested.67 Moreover, Dahele and Gosh,83 reported that 41% of adults with untreated CeD exhibited vitamin B12 deficiency despite the absence of intrinsic factor antibodies in all patients, with only one-third experiencing concurrent folate deficiency. We found no evidence of compromised vitamin B12 status (the pooled effect of vitamin B12 in the CD group was 0.01 (95% CI −0.12 to 0.15); P > 0.05). This could be due to higher dietary intake. Some patients with CeD have been reported to use vitamin and mineral supplements (vitamin B-complex) before being diagnosed with CeD.84
Primary dietary copper deficit is uncommon and is mostly caused by malabsorption syndrome. In our study, the pooled effect of copper in the CeD group was −0.6429 (95% CI −1.5264 to 0.2407)]. Although our results were not statistically significant, the trend was toward copper deficiency, which is consistent with many previous studies.54,71,85,86 An Iranian study reported in 2013 that the mean levels of Zn in patients with CeD were significantly lower than those in control group (75.97 ± 12 vs 92.83 ± 18, P < 0.0001).68 Similarly, Singhal et al. noted that serum Zn levels in patients with newly diagnosed CeD were significantly reduced (0.64 ± 0.34 mg/mL vs 0.94 ± 0.14 mg/mL in controls (95% CI −0.44 to −1.4)).73 Similarly, a recent study by Adam et al. showed that Zn levels were decreased in 59.4% of patients with CeD compared with 33.2% in controls.70
Micronutrient deficiencies observed in patients with CeD can be attributed to several factors related to disease pathophysiology. The CeD pathway is characterized by alterations in the small intestine, including intraepithelial lymphocytosis, crypt hyperplasia, and VA, which reduce nutrient absorption.87 Moreover, inflammation and small intestinal mucosal damage lead to loss of absorptive surfaces and nutrient malabsorption.88 Refined flours used in GFDs often lack fortification, potentially contributing to nutritional deficiencies in this population.89,90 In addition, GFDs commonly followed by patients with CeD are characterized by reduced intake of cereals, fruits, and vegetables, along with increased consumption of meat and meat-derived products.91 Whole-grain barley, rye, and wheat products are typically replaced by specialized gluten-free alternatives, which have been shown to possess lower nutritional value compared with their gluten-containing counterparts.91 These gluten-free products are often associated with higher levels of fats, particularly saturated and trans-fats, as well as refined sugars, phosphorus, and salt, which can reduce the intake of fibers, complex carbohydrates, and proteins.91 Furthermore, the inadequacy of dietary habits specific to this group may exacerbate the issue.92,93 In addition, the low demand for nutritional counseling from registered dietitians may foster insufficient food intake, particularly in rural areas. Despite being straightforward, GFD implementation poses significant challenges for patients and their families,94 one of which is the risk for cross-contamination, often leading to unintentional gluten transgression. These inadvertent exposures can perpetuate VA and contribute to ongoing nutritional deficiencies in individuals with CeD even when they adhere to a strict GFD.95 Registered dietitians play a critical role in guiding patients with CeD by adopting a GFD that is not only healthy but also interesting and practical, helping to mitigate these challenges.96 While it is acknowledged that a GFD entails dietary restrictions, patients who receive nutritional counseling from a registered dietitian can achieve a well-balanced and healthy diet. The dietary recommendations for a healthy GFD should align closely with those of a regular healthy diet, emphasizing nutritious and safe alternatives to cereal-based foods while avoiding excessive consumption of highly processed products. Such a diet should prioritize the intake of fresh, unprocessed, and naturally gluten-free foods, including a variety of fruits, vegetables, and proteins, preferably from plant sources such as legumes, whole grains, pseudocereals, tubers, and nuts.91 As such, a tailored diet could be beneficial in restoring a balanced gut microbiota.67 Our study has the merit of using a standardized meta-analytical methodology (with random-effects analyses) to assess the impact of CeD on different micronutrient categories. However, there were several limitations, including the lack of sample size calculation, the high level of heterogeneity observed among the included studies, and the disproportionate number of studies addressing the nutrients analyzed.
Conclusion
The present analysis revealed substantial differences in micronutrient levels between patients with CeD and controls. Decreases in hemoglobin, ferritin, iron, folic acid, Zn, and vitamin D levels highlight the multidimensional characteristics of nutritional deficits in CeD. These findings highlight the crucial role of a thorough nutritional evaluation and intervention techniques in CeD care to address a wide range of micronutrient deficits. Thus, it is critical to use a multidisciplinary strategy that includes registered dietitian counseling, supplementation when needed, and continued monitoring to reduce the negative health effects of micronutrient deficiencies in patients with CeD. Furthermore, additional studies should focus on identifying the underlying processes that contribute to micronutrient deficits in patients with CeD, as well as investigating novel techniques to improve nutrient absorption and overall nutritional status in this susceptible group.
Acknowledgments
We gratefully acknowledge Professor Daniel S. Quintana for his significant contributions to this study. His insightful feedback and dedication to rigorous scientific inquiry greatly enhanced the quality of this research.
Ethics approval
Not applicable.
Informed consent
Not applicable.
Declaration of conflicting interests
The author(s) declared no potential conflicts of interest with respect to the research, authorship, and/or publication of this article.
Funding
The author(s) received no financial support for the research, authorship, and/or publication of this article.
ORCID iD
Saad Lamjadli
Footnote
Trial registration Not applicable.
Data availability statement
All data supporting the findings of this meta-analysis are available in the Appendix.
Appendix
Appendix 2. Articles involving Hemoglobin included in the meta-analysis.
| ID | YEAR | COUNTRY | StudyType | AGE | AGE | Nc | Nd | AGEc | AGEd | HBc | HBd | SdHBc | SdHBd | AdherenceScore |
--- --- --- --- --- --- ---
| Karaman et al.97 | 2016 | Turkey | Prospective | Children | 6 | 238 | 12 | — | 6.2 | 8.7 | 7.3 | 1.6 | 1.4 | Good |
| Rafet Mete et al.98 | 2021 | Turkey | Prospective | Adult | 41.955 | 56 | 14 | 45.05 | 38.86 | 12.22 | 12.39 | 2.09 | 2.51 | Good |
| Botero-Lopez et al.99 | 2020 | Chile | Cross-sectional | Adult | 21.31 | 36 | 73 | 17.8 | 24.82 | 13.6 | 12.6 | 1.4 | 2.18 | Excellent |
| Bayrak et al.50 | 2019 | Turkey | Cross-sectional | Children | 12.845 | 135 | 228 | 12.77 | 12.92 | 13.56 | 12.09 | 1.15 | 1.71 | Excellent |
| Weintraub et al.100 | 2019 | Israel | Prospective | Children | 11.6 | 66 | 47 | 15 | 8.2 | 13.1 | 12.4 | 1.1 | 1.4 | Excellent |
| Tokgoz et al.101 | 2018 | Turkey | Case-control | Children | 8.85 | 50 | 52 | 8.7 | 9 | 12.3 | 11.3 | 1.1 | 1.7 | Excellent |
| Piatek-Guziewicz et al.102 | 2017 | Poland | Cross-sectional | Adult | 36 | 25 | 29 | 38 | 34 | 13.4 | 12.6 | 1.4 | 1.8 | Excellent |
| Armagan et al.103 | 2004 | Turkey | Case-control | Adult | 35.935 | 72 | 9 | 35.87 | 36 | 12.02 | 11.72 | 1.39 | 1.28 | Good |
| Narang et al.104 | 2016 | India | Cross-sectional | Children | — | 146 | 6 | — | — | 7.8 | 6.4 | 1.8 | 1.13 | Good |
| Xavier-Valente et al.105 | 2015 | Brazil | Cross-sectional | Adult | 36 | 39 | 20 | 36 | 36 | 13.4 | 13.6 | 1.3 | 1.2 | Good |
| Ince et al.86 | 2007 | Turkey | Cross-sectional | Adult | 0 | 32 | 35 | — | | 14 | 12.2 | 1.3 | 1.7 | Good |
| Troch et al.106 | 2019 | Mexico | Cross-sectional | Adult | — | 20 | 20 | — | — | 13.7 | 12.5 | 1.62 | 1.1 | Excellent |
Articles involving Ferritin included in the meta-analysis.
| ID | YEAR | COUNTRY | StudyType | AGE | Nc | Nd | AGEc | AGEd | FERRITINEc | FERRITINEd | SdFERRIc | SdFERRId | AdherenceScore |
--- --- --- --- --- --- --- |
| Emami et al.107 | 2011 | Iran | Cross-sectional | Adult | 117 | 13 | 36 | 35.3 | 8.41 | 7.2 | 3.2 | 0.7 | Excellent |
| Baghbanian et al.108 | 2015 | Iran | Cross-sectional | Adult | 378 | 24 | — | 25.76 | 11.13 | 11.13 | 9.06 | 8.71 | Excellent |
| Karaman et al.97 | 2016 | Turkey | Prospective | Children | 238 | 12 | — | 6.2 | 5.2 | 2 | 3.2 | 0.7 | Good |
| Rafet Mete et al.98 | 2021 | Turkey | Prospective | Adult | 56 | 14 | 45.05 | 38.86 | 28.34 | 13.23 | 35.87 | 11.23 | Good |
| Botero-Lopez et al.99 | 2020 | Chile | Cross-sectional | Adult | 36 | 73 | 17.8 | 24.82 | 28.7 | 14.25 | 41.925 | 23.9 | Excellent |
| Bayrak et al.50 | 2019 | Turkey | Cross-sectional | Children | 135 | 228 | 12.77 | 12.92 | 28.23 | 21.61 | 16.76 | 20.72 | Excellent |
| Weintraub et al.100 | 2019 | Israel | Prospective | Children | 66 | 47 | 15 | 8.2 | 28.2 | 15.3 | 17.62 | 12.14 | Excellent |
| Volkan et al.109 | 2017 | Turkey | Prospective | Children | 30 | 72 | — | 10.9 | 39.6 | 25.2 | 38.4 | 24.8 | Good |
| Işikay et al.46 | 2018 | Turkey | Cross-sectional | Children | 268 | 226 | 13.62 | 13.2 | 24.11 | 16.48 | 5.78 | 12.57 | Good |
| Nestares et al.45 | 2020 | Spain | Cross-sectional | Children | 68 | 43 | 10.3 | 8.5 | 50.3 | 43.1 | 6.5 | 7.7 | Excellent |
| Haapalahti et al.110 | 2004 | Finland | Prospective | Children | 29 | 26 | — | — | 27 | 14 | 30.5 | 19.5 | Good |
| Kapur et al.49 | 2003 | New Delhi | Prospective | Children | 21 | 21 | — | — | 88.3 | 38.25 | 36.8 | 14.4 | Good |
Articles involving Iron included in the meta-analysis.
| ID | YEAR | COUNTRY | StudyType | AGE | Nc | Nd | AGEc | AGEd | IRON_C | IRON_D | SdIRONc | SdIRONd | AdherenceScore |
--- --- --- --- --- --- --- |
| Karaman et al.97 | 2016 | Turkey | Prospective | Children | 238 | 12 | — | 6.2 | 17.4 | 13.4 | 6.6 | 4.6 | Good |
| Botero-Lopez et al.99 | 2020 | Chile | Cross-sectional | Adult | 36 | 73 | 17.8 | 24.82 | 145 | 150 | 52.5 | 51.875 | Excellent |
| Bayrak et al.50 | 2019 | Turkey | Cross-sectional | Children | 135 | 228 | 12.77 | 12.92 | 71.39 | 67.07 | 37.7 | 38.02 | Excellent |
| UnalpArida et al.64 | 2022 | USA | Cross-sectional | Adult | 14000 | 26 | — | — | 86.5 | 90.1 | 62.23 | 40.28 | Excellent |
| Ballestero-Fernández et al.44 | 2021 | Spain | Cross-sectional | Adult | 74 | 64 | 38 | 39 | 107.5 | 100.5 | 11.825 | 10.125 | Excellent |
| Ince et al.86 | 2007 | Turkey | Cross-sectional | Adult | 32 | 35 | — | — | 75 | 46 | 30 | 26 | Good |
| Kapur et al.49 | 2003 | New Delhi | Prospective | Children | 21 | 21 | — | — | 90.89 | 84.69 | 23 | 24.5 | Good |
| Karnani et al.85 | 2022 | India | Case control | Children | 30 | 60 | — | — | 80.63 | 58.24 | 21.18 | 29.63 | Excellent |
Articles involving MCV included in the meta-analysis.
| ID | YEAR | COUNTRY | StudyType | AGE | Nc | Nd | AGEc | AGEd | MCVc | MCVd | SdMCVc | SdMCVd | AdherenceScore |
--- --- --- --- --- --- --- |
| Armagan et al.103 | 2004 | Turkey | Case-control | Adult | 72 | 9 | 35.87 | 36 | 87.66 | 86.59 | 4.92 | 4.1 | Good |
| Ballestero-Fernández et al.44 | 2021 | Spain | Cross-sectional | Adult | 74 | 64 | 38 | 39 | 90.4 | 91.1 | 1.675 | 1.5 | Excellent |
| Hallert et al.111 | 1981 | Sweden | Prospective | Adult | 13 | 5 | — | — | 92 | 95 | 8 | 16 | Fair |
| Baghbanian et al.108 | 2015 | Iran | Cross-sectional | Adult | 378 | 24 | — | 25.76 | 72.76 | 72.57 | 6.64 | 6.22 | Excellent |
| Karaman et al.97 | 2016 | Turkey | Prospective | Children | 238 | 12 | — | 6.2 | 60.7 | 59.6 | 6.3 | 4.7 | Good |
| Caterina et al.48 | 2005 | Italy | Case-control | Children | 31 | 20 | — | — | 78 | 4 | 72 | 5 | Good |
Articles involving Folic Acid included in the meta-analysis.
| ID | YEAR | COUNTRY | StudyType | AGE | Nc | Nd | AGEc | AGEd | FOLATEc | FOLATEd | SdFOLATEc | SdFOLATEd | AdherenceScore |
--- --- --- --- --- --- --- |
| UnalpArida et al.64 | 2022 | USA | Cross-sectional | Adult | 15589 | 26 | — | 20 | 19.8 | 18.9 | 24.86 | 14.029 | Excellent |
| Işikay et al.46 | 2018 | Turkey | Cross-sectional | Children | 268 | 226 | 13.62 | 13.2 | 16.5 | 12.55 | 12.44 | 18.07 | Good |
| Wierdsma et al.84 | 2013 | Amsterdam | Prospective | Adult | 25 | 80 | 43 | 42.8 | 20.4 | 15.1 | 15 | 18.1 | Excellent |
| Volkan et al.109 | 2017 | Turkey | Prospective | Children | 30 | 72 | — | — | 8.17 | 7.7 | 8 | 6.2 | Good |
| Haapalahti et al.110 | 2004 | Finland | Prospective | Children | 29 | 26 | — | — | 109 | 91 | 55 | 40.25 | Good |
| Hallert et al.111 | 1981 | Sweden | Prospective | Adult | 13 | 5 | — | — | 4.9 | 2.2 | 1.5 | 0.8 | Fair |
| Bayrak et al.50 | 2019 | Turkey | Cross-sectional | Children | 135 | 228 | 12.77 | 12.92 | 8.18 | 8.6 | 2.69 | 5.33 | Excellent |
| Xavier Valente et al.105 | 2015 | Brazil | Cross-sectional | Adult | 39 | 20 | 36 | 36 | 29 | 17.5 | 9.4 | 8 | Good |
| Hadithi et al.112 | 2009 | Netherlands | Cross-sectional | Adult | 50 | 51 | — | — | 12.1 | 9.7 | 1.325 | 2.325 | Good |
| Dickey et al.113 | 2008 | Northern Ireland | Case-control | Adult | 200 | 100 | 54.7 | 55 | 24.1 | 12.9 | 10 | 14.2 | Good |
Articles involving vitamin B12 included in the meta-analysis.
| ID | YEAR | COUNTRY | StudyType | AGE | Nc | Nd | AGEc | AGEd | VITB12c | VITB12d | SdVITBc | SdVITBd | AdherenceScore |
--- --- --- --- --- --- --- |
| UnalpArida et al.64 | 2022 | USA | Cross-sectional | Adult | 15589 | 26 | — | 20 | 615.1 | 610 | 698.1 | 280.4 | Excellent |
| Armagan et al.103 | 2004 | Turkey | Case-control | Adult | 72 | 9 | 35.87 | 36 | 340.78 | 300.04 | 131.4 | 54.23 | Good |
| Xavier Valente et al.105 | 2015 | Brazil | Cross-sectional | Adult | 39 | 20 | 36 | 36 | 257.3 | 271.1 | 90.6 | 89 | Good |
| Haapalahti et al.110 | 2004 | Finland | Prospective | Children | 29 | 26 | — | — | 313 | 325 | 117.25 | 150.25 | Good |
| Dickey et al.113 | 2008 | Northern Ireland | Case-control | Adult | 200 | 100 | 55 | 54.7 | 248.6 | 274.1 | 104.6 | 170.9 | Good |
| Bayrak et al.50 | 2019 | Turkey | Cross-sectional | Children | 135 | 228 | 12.77 | 12.92 | 363.2 | 344.32 | 204.18 | 185.58 | Excellent |
| Volkan et al.109 | 2017 | Turkey | Prospective | Children | 30 | 72 | — | — | 393 | 384 | 228 | 172 | Good |
| Hadithi et al.112 | 2009 | Netherlands | Cross-sectional | Adult | 50 | 51 | — | — | 234.5 | 230.5 | 21.875 | 30.625 | Good |
| Wierdsma et al.84 | 2013 | Amsterdam | Prospective | Adult | 25 | 80 | 42.8 | 43 | 272.6 | 231.2 | 117.4 | 104.3 | Excellent |
| Işikay et al.46 | 2018 | Turkey | Cross-sectional | Children | 268 | 226 | 13.62 | 13.2 | 327.55 | 350.72 | 124.32 | 137.04 | Good |
Articles involving vitamin D included in the meta-analysis.
| ID | YEAR | COUNTRY | StudyType | AGE | Nc | Nd | AGEc | AGEd | VITDc | VITDd | SdVITDc | SdVITDd | AdherenceScore |
--- --- --- --- --- --- --- |
| Weintraub et al.100 | 2019 | Israel | Prospective | Children | 66 | 47 | 15 | 8.2 | 27 | 26 | 10.37 | 8.14 | Excellent |
| Szymczak et al.114 | 2012 | Poland | Cross-sectional | Adult | 36 | 35 | — | 41.5 | 39.5 | 29.9 | 19 | 18.3 | Good |
| Volkan et al.109 | 2017 | Turkey | Prospective | Children | 30 | 72 | — | 10.9 | 15.8 | 17.2 | 10.575 | 11.35 | Good |
| Unalp-Arida et al.64 | 2022 | USA | Cross-sectional | Adult | 14000 | 26 | — | 20 | 69.2 | 79.7 | 99.68 | 26.5 | Excellent |
| Villanueva et al.115 | 2012 | USA | Retrospective | Children | 50 | 24 | — | 9.42 | 26.2 | 27.04 | 10.45 | 9.91 | Good |
| Nwosu et al.116 | 2015 | USA | Cross-sectional | Children | 49 | 25 | — | 8.5 | 65.4 | 70.6 | 26.1 | 25.7 | Excellent |
| Setty-Shah et al.117 | 2014 | USA | Prospective | Children | 49 | 18 | 7.95 | 8.92 | 65.3 | 74.7 | 26 | 27.2 | Excellent |
| Uyanıkoglu et al.118 | 2021 | Turkey | Case-control | Adult | 40 | 40 | 40 | 40 | 118.43 | 134.33 | 48 | 44.35 | Good |
| Ballestero-Fernández et al.44 | 2021 | Spain | Cross-sectional | Adult | 74 | 64 | 38 | 39 | 33.7 | 34.7 | 11.15 | 8.85 | Excellent |
| Piatek-Guziewicz A et al.102 | 2017 | Poland | Cross-sectional | Adult | 25 | 29 | 38 | 34 | 29.7 | 19.4 | 5.1 | 9 | Excellent |
| Corazza et al.119 | 1995 | Milan | Cross-sectional | Adult | 15 | 23 | — | 37.5 | 27 | 15.1 | 4.75 | 5.87 | Good |
| Karnani et al.85 | 2022 | India | Case control | Children | 30 | 60 | — | — | 33.3 | 20.29 | 10.94 | 8.97 | Excellent |
| Tokgoz et al.101 | 2018 | Turkey | Case-control | Children | 50 | 52 | 8.7 | 9 | 27.6 | 19.8 | 10.4 | 7.9 | Excellent |
| Lionetti et al.120 | 2020 | Italy | Case-control | Children | 131 | 131 | 8.2 | 8.1 | 31.6 | 25.3 | 13.7 | 8 | Excellent |
| Armagan et al.103 | 2004 | Turkey | Case-control | Adult | 72 | 9 | 35.87 | 36 | 17.07 | 12.11 | 5.22 | 1.97 | Good |
Articles involving Copper included in the meta-analysis.
| ID | YEAR | COUNTRY | StudyType | AGE | Nc | Nd | AGEc | AGEd | Copperc | Copperd | SdCopperc | SdCopperd | AdherenceScore |
--- --- --- --- --- --- --- |
| UnalpArida et al.64 | 2022 | USA | Cross-sectional | Adult | 3274 | 26 | — | 20 | 118.4 | 116.5 | 62.94 | 7.07 | Excellent |
| Jameso et al.121 | 1985 | Netherlands | Case-control | Adult | 10 | 8 | — | — | 20.7 | 19 | 7.4 | 4.6 | Fair |
| Idris et al.54 | 2019 | Sudan | Case-control | Adult | 40 | 40 | — | — | 0.7 | 0.612 | 0.279 | 0.279 | Excellent |
| Solomons et al.71 | 1976 | USA | Cross-sectional | Adult | 10 | 20 | — | — | 74.3 | 51.1 | 9.7 | 13 | Fair |
| Karnani et al.85 | 2022 | India | Case control | Children | 30 | 60 | — | — | 90.95 | 90.95 | 17.62 | 17.62 | Excellent |
| Ince et al.86 | 2007 | Turkey | Cross-sectional | Adult | 32 | 35 | — | — | 105 | 105 | 16 | 16 | Good |
Articles involving Zinc included in the meta-analysis.
| ID | YEAR | COUNTRY | StudyType | AGE | Nc | Nd | ZINCc | ZINCd | SdZINCc | SdZINCd | AdherenceScore |
--- --- --- --- --- --- |
| BoteroLopez et al.99 | 2020 | Chile | Cross-sectional | Adulte | 36 | 73 | 90 | 82.5 | 16.25 | 20 | Excellent |
| UnalpArida et al.64 | 2022 | USA | Cross-sectional | Adulte | 14000 | 26 | 82.2 | 75.1 | 34.33 | 10.97 | Excellent |
| Rawal et al.122 | 2010 | India | Prospective | Children | 48 | 48 | 74.9 | 71.9 | 29.2 | 19.3 | Good |
| Ince et al.86 | 2007 | Turkey | Cross-sectional | Adulte | 35 | 32 | 101 | 70 | 20 | 14 | Good |
| Naveh et al.123 | 1983 | Israel | Prospective | Children | 31 | 34 | 100 | 62 | 15 | 11 | Fair |
| Fathi et al.68 | 2013 | Tehran | Case-control | Adulte | 30 | 30 | 92.83 | 75.97 | 18 | 12 | Good |
| Karnani et al.85 | 2022 | India | Case-control | Children | 30 | 60 | 102.13 | 16.52 | 80.63 | 21.18 | Good |
| Idris et al.54 | 2019 | Sudan | Case-control | Adulte | 40 | 40 | 1 | 0.285 | 0.245 | 0.1776 | Excellent |
References
Saturni L, Ferretti G, Bacchetti T (2010) The gluten-free diet: Safety and nutritional quality. Nutrients 2: 16–34.
Crossref
PubMed
Web of Science
Google Scholar
a [...] by genetic and environmental factors.
b [...] have significant and severe implications.
Beas R, Altamirano-Farfan E, Izquierdo-Veraza D, et al. (2024) Prevalence of celiac disease in systemic lupus erythematosus, sjogren syndrome and systemic sclerosis: A systematic review and meta-analysis. Digestive and Liver Disease 56(9): 1475–1482.
Go to Reference
Crossref
Web of Science
Google Scholar
Lindfors K, Ciacci C, Kurppa K, et al. (2019) Coeliac disease. Nature Reviews Disease Primers 5: 3.
Crossref
PubMed
Web of Science
Google Scholar
a [...] they are insufficient for CeD genesis.
b [...] also been implicated in its pathogenesis.
Lebwohl B, Sanders DS, Green PHR (2018) Coeliac disease. Lancet 391: 70–81.
Crossref
PubMed
Web of Science
Google Scholar
a [...] they are insufficient for CeD genesis.
b [...] also been implicated in its pathogenesis.
c [...] or without ongoing or recurrent symptoms.
Capittini C, De Silvestri A, Rebu C, et al. (2019) Relevance of HLA-DQB102 allele in the genetic predisposition of children with celiac disease: Additional cues from a meta-analysis. Medicina (B. Aires) 55: 190.
Go to Reference
Crossref
Web of Science
Google Scholar
Kreutz JM, Adriaanse MPM, van der Ploeg EMC, et al. (2020) Narrative review: Nutrient deficiencies in adults and children with treated and untreated celiac disease. Nutrients 12(2): 500.
Crossref
PubMed
Web of Science
Google Scholar
a [...] reduction in the absorptive surface area.
b [...] folic acid levels in those affected by CeD.
Rubio-Tapia A, Hill ID, Kelly CP, et al. (2013) Clinical guidelines: Diagnosis and management of celiac disease. American Journal of Gastroenterology 108: 656–676.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Ciacci C, Ciclitira P, Hadjivassiliou, et al. (2015) The gluten-free diet and its current application in coeliac disease and dermatitis herpetiformis. United European Gastroenterology Journal 3: 121–135.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Lebwohl B, Murray JA, Rubio-Tapia, et al. (2014) Predictors of persistent villous atrophy in coeliac disease: A population-based study. Alimentary Pharmacology & Therapeutics 39: 488–495.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Schiepatti A, Maimaris S, Raju SA, et al. (2023) Persistent villous atrophy predicts development of complications and mortality in adult patients with coeliac disease: A multicentre longitudinal cohort study and development of a score to identify high-risk patients. Gut 72(11): 2095–2102.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Leonard MM, Weir DC, DeGroote M, et al. (2017) Value of IgA TTG in predicting mucosal recovery in children with celiac disease on a gluten-free diet. Journal of Pediatric Gastroenterology and Nutrition 64: 286–291.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Abdi F, Zuberi S, Blom JJ, et al. (2023) Nutritional considerations in celiac disease and non-celiac gluten/wheat sensitivity. Nutrients 15(6): 1475.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Vici G, Belli L, Biondi M, et al. (2016) Gluten free diet and nutrient deficiencies: A review. Clinical Nutrition 35: 1236–1241.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Page MJ, McKenzie JE, Bossuyt PM, et al. (2021) The PRISMA 2020 statement: An updated guideline for reporting systematic reviews. BMJ 372:n71.
Go to Reference
Web of Science
Google Scholar
Rubio-Tapia A, Hill ID, Semrad C, et al. (2023) American college of gastroenterology guidelines update: Diagnosis and management of celiac disease. American Journal of Gastroenterology 118(1): 59–76.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Wickham H, Averick M, Bryan J, et al. (2019) Welcome to the tidyverse. Journal of Open Source Software 4(43): 1686.
Go to Reference
Crossref
Google Scholar
Wickham H, Bryan J (2023) readxl: Read excel files. R package version 1.4.3. Available at:
Google Scholar
Thompson C, Antal D, Parry J, et al. (2024) Spotifyr: R wrapper for the “Spotify” web API. R package version 2.2.5. Available at:
Google Scholar
Iannone R, Cheng J, Schloerke B, et al. (2024) gt: Easily create presentation-ready display tables. R package version 0.11.1. Available at:
Google Scholar
Mock T (2023) gtExtras: Extending “gt” for beautiful HTML tables. R package version 0.5.0. Available at:
Crossref
Google Scholar
Wickham H, Pedersen T, Seidel D (2023) Scales: Scale functions for visualization. R package version 1.3.0. Available at:
Google Scholar
Chang W (2023) Webshot2: Take screenshots of web pages. R package version 0.1.1. Available at:
Crossref
Google Scholar
Pebesma E, Bivand R (2023) Spatial Data Science: With Applications in R. Chapman and Hall/CRC: Boca Raton.
Crossref
Google Scholar
Massicotte P, South A (2023) Rnaturalearth: World map data from natural earth. R package version 1.0.1. Available at:
Google Scholar
Arel-Bundock V, Enevoldsen N, Yetman C (2018) Countrycode: An R package to convert country names and country codes. Journal of Open Source Software 3(28): 848.
Crossref
Google Scholar
Slowikowski K (2024) ggrepel: Automatically position non-overlapping text labels with “ggplot2”. R package version 0.9.5. Available at:
Go to Reference
Google Scholar
Balduzzi S, Rücker G, Schwarzer G (2019) How to perform a meta-analysis with R: A practical tutorial. Evidence-Based Mental Health 22: 153–160.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Harrer M, Cuijpers P, Furukawa, et al. (2019) dmetar: Companion R package for the guide “doing meta-analysis in R. R package version 0.0.9000. Available at:
Google Scholar
Viechtbauer W (2010). Conducting meta-analyses in R with the metafor package. Journal of Statistical Software 36(3): 1–48.
Crossref
Web of Science
Google Scholar
Peterson BG, Carl P (2020) Performance analytics: Econometric tools for performance and risk analysis. R package version 2.0.4. Available at:
Google Scholar
Schwarzer G, Carpenter J, Rücker G (2023) metasens: Statistical methods for sensitivity analysis in meta-analysis. R package version 1.5-2. Available at:
Google Scholar
McGuinness LA (2019). robvis: An R package and web application for visualising risk-of-bias assessments. Available at:
Google Scholar
Borenstien M, Hedges LV, Higgins J, et al. (2000) Comprehensive Meta-Analysis. Englewood, NJ: Biostat.
Go to Reference
Google Scholar
Cohen J (1988) Set correlation and contingency tables. Applied Psychological Measurement 12(4): 425–434.
Crossref
Web of Science
Google Scholar
a [...] small (0.2), medium (0.5), and large (0.8).
b [...] of the biases found in small sample sizes.
Harrer M, Cuijpers P, Furukawa TA (2021) Doing Meta-Analysis with R: A Hands-On Guide. Boca Raton, FL and London: Chapman & Hall/CRC Press.
Crossref
Google Scholar
a [...] small (0.2), medium (0.5), and large (0.8).
b [...] of the biases found in small sample sizes.
c [...] to populations beyond the study samples.
d [...] of effect sizes across the studies.
e [...] and high heterogeneity, respectively.
f [...] bias using a random-effects model.
g [...] were used to correct for publication bias.
h [...] overlap with those of the pooled effects.
i [...] a “mirage” produced by selective reporting.
Hedges LV, Olkin I (1958). Statistical Methods for Meta-Analysis. Orlando: Academic Press.
Go to Reference
Google Scholar
Shaddish WR, Haddock CK (1994) Combining estimates of effect size. In: Cooper H, Hedges LV (eds) The Handbook of Research Synthesis. New York, NY: Russel Sage Foundation, pp.261–285.
Google Scholar
a [...] of effect sizes across the studies.
b [...] variance in the results between studies.
Quintana DS, McGregor IS, Guastella AJ (2013) A meta-analysis on the impact of alcohol dependence on short-term resting-state heart rate variability: Implications for cardiovascular risk. Alcoholism: Clinical and Experimental Research 37: E23–E29.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Higgins JP, Thompson SG, Deeks JJ, et al. (2003) Measuring inconsistency in meta-analyses. BMJ 327: 557–560.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Egger M, Smith GD (1998) Bias in location and selection of studies. BMJ 316: 61–66.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Rücker G, Schwarzer G, Carpenter JR, et al. (2011) Treatment-effect estimates adjusted for small-study effects via a limit meta-analysis. Biostatistics 12: 122–142.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Simonsohn U, Nelson LD, Simmons JP (2014) P-curve and effect size: Correcting for publication bias using only significant results. Perspectives on Psychological Science 9(6): 666–681.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Viechtbauer W, Cheung MW (2010) Outlier and influence diagnostics for meta-analysis. Research Synthesis Methods 1(2): 112–125.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Ballestero-Fernández C, Varela-Moreiras G, Úbeda N, et al. (2021) Nutritional status in Spanish adults with celiac disease following a long-term gluten-free diet is similar to non-celiac. Nutrients 13: 1626.
Crossref
PubMed
Web of Science
Google Scholar
a [...] Removed as outliers: Ballestero et al.,
b [...] as outliers: Ballestero-Fernández et al.
c [...] Ballestero-Fernández et al.
d [...] Ballestero-Fernández et al.
e [...] Ballestero-Fernández et al.
Nestares T, Martín-Masot R, Labella A, et al. (2020) Is a gluten-free diet enough to maintain correct micronutrients status in young patients with celiac disease? Nutrients 12: 844.
Crossref
PubMed
Web of Science
Google Scholar
a [...] Nestares et al.,
b [...] Removed as outliers: Nestares et al.,
c [...] Nestares et al.
IŞıkay S, IŞıkay N, Per H, et al. (2018) Restless leg syndrome in children with celiac disease. The Turkish Journal of Pediatrics 60(1): 70–75.
Crossref
PubMed
Web of Science
Google Scholar
a [...] Işikay et al.,
b [...] Işikay et al.
c [...] Işikay et al.
d [...] Işikay et al.
Kalayci AG, Kanber Y, Birinci A, et al. (2005) The prevalence of coeliac disease as detected by screening in children with iron deficiency anaemia. Acta Paediatrica 94: 678–681.
Crossref
PubMed
Web of Science
Google Scholar
a [...] Kalayci et al.,
b [...] Removed as outliers: Kalayci et al.,
De Caterina M, Grimaldi E, Di Pascale G, et al. (2005) The soluble transferrin receptor (sTfR)-ferritin index is a potential predictor of celiac disease in children with refractory iron deficiency anemia. Clinical Chemistry and Laboratory Medicine 43(1): 38–42.
Crossref
PubMed
Web of Science
Google Scholar
a [...] Caterina et al.,
b [...] Caterina et al.
c [...] Removed as outliers: Caterina et al.
d [...] Caterina et al.
Kapur G, Patwari AK, Narayan S, et al. (2003) Iron supplementation in children with celiac disease. Indian Journal of Pediatrics 70: 955–958.
Crossref
PubMed
Google Scholar
a [...] Kapur et al.
b [...] Kapur et al.
c [...] Kapur et al.
d [...] Kapur et al.
Bayrak NA, Volkan B, Haliloglu B, et al. (2020) The effect of celiac disease and gluten-free diet on pubertal development: A two-center study. Journal of Pediatric Endocrinology and Metabolism 33(3): 409–415.
Crossref
PubMed
Web of Science
Google Scholar
a [...] Removed as outliers: Bayrak et al.,
b [...] Bayrak et al.
c [...] Bayrak et al.
d [...] Bayrak et al.
e [...] Bayrak et al.
f [...] Bayrak et al.
Stein EM, Rogers H, Leib A, et al. (2015) Abnormal skeletal strength and microarchitecture in women with celiac disease. Journal of Clinical Endocrinology and Metabolism 100(6): 2347–2353.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Jamnik J, Jenkins DJ, El-Sohemy A (2018) Biomarkers of cardiometabolic health and nutritional status in individuals with positive celiac disease serology. Nutrition and Health 24(1): 37–45.
Go to Reference
Crossref
PubMed
Google Scholar
Björck S, Brundin C, Karlsson M, et al. (2017) Reduced bone mineral density in children with screening-detected celiac disease. Journal of Pediatric Gastroenterology and Nutrition 65(5): 526–532.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Idris H, Elzein AM, Elxein HO (2019) Evaluation of zinc and copper levels in sudanese patients with celiac disease in red sea state. Pakistan Journal of Medical and Health Sciences 13(4): 1120–1123.
Google Scholar
a [...] Removed as outlier: Idris et al.
b [...] is consistent with many previous studies.
c [...] Idris et al.
d [...] Idris et al.
Guerrieri A, Catassi C, Pasquini E, et al. (1986) Plasma zinc levels in children with chronic diarrhoea. European Journal of Pediatrics 145(6): 563–564.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Cuschieri S (2019) The STROBE guidelines. Saudi Journal of Anaesthesia 13(Suppl 1): S31–S34.
Go to Reference
Crossref
Web of Science
Google Scholar
Bottaro G, Cataldo F, Rotolo N, et al. (1999) The clinical pattern of subclinical/silent celiac disease: An analysis of 1026 consecutive cases. The American Journal of Gastroenterology 94: 691–696.
Go to Reference
PubMed
Web of Science
Google Scholar
Simón E, Molero-Luis M, Fueyo-Díaz R, et al. (2023) The gluten-free diet for celiac disease: Critical insights to better understand clinical outcomes. Nutrients 15(18): 4013.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Montoro-Huguet MA, Santolaria-Piedrafita S, Cañamares-Orbis P (2021) Iron deficiency in celiac disease: Prevalence, health impact, and clinical management. Nutrients 13(10): 3437.
Crossref
PubMed
Web of Science
Google Scholar
a [...] than those presenting with diarrhea alone.
b [...] imbalance between iron loss and absorption.
c [...] with inconsistent adherence to the GFD.
d [...] or methods to improve tolerability.
e [...] supplementation in managing anemia in CeD.
Brittenham GM, Hoffman R, Benz EJ Jr, et al. (1995) Disorders of iron metabolism: Iron deficiency and overload. In: Hoffman R, Benz EJ, Shattil SJ (eds.) Haematology: Basic Principles and Practice, 2nd ed. Edinburgh: Churchill Livingstone, pp.492–523.
Go to Reference
Google Scholar
Lombard M, Chua E, O’Toole P (1997) Regulation of intestinal non-heme iron absorption. Gut 40: 435–439.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Saukkonen J, Kaukinen K, Koivisto AM, et al. (2017) Clinical characteristics and the dietary response in celiac disease patients presenting with or without anemia. Journal of Clinical Gastroenterology 51: 412–416.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Stefanelli G, Viscido A, Longo S, et al. (2020) Persistent iron deficiency anemia in patients with celiac disease despite a gluten-free diet. Nutrients 12: 2176.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Unalp-Arida A, Liu R, Ruhl CE (2022) Nutrient intake differs among persons with celiac disease and gluten-related disorders in the United States. Scientific Reports 12: 5566.
Crossref
PubMed
Web of Science
Google Scholar
a [...] in 8–85% of adult patients with CeD.
b [...] UnalpArida et al.
c [...] UnalpArida et al.
d [...] UnalpArida et al.
e [...] Unalp-Arida et al.
f [...] UnalpArida et al.
g [...] UnalpArida et al.
Suitor CW, Bailey LB (1999) Food folate vs. synthetic folic acid: A comparison. Journal of the American Dietetic Association 99: 285.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Kemppainen TA, Kosma VM, Janatuinen EK, et al. (1998) Nutritional status of newly diagnosed celiac disease patients before and after the institution of a celiac disease diet–association with the grade of mucosal villous atrophy. American Journal of Clinical Nutrition 67: 482–487.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Melini V, Melini F (2019) Gluten-free diet: Gaps and needs for a healthier diet. Nutrients 11: 170.
Crossref
PubMed
Web of Science
Google Scholar
a [...] rather than by abnormal Zn absorption.
b [...] small intestine, have been suggested.
c [...] in restoring a balanced gut microbiota.
Fathi F, Ektefa F, Tafazzoli M, et al. (2013) The concentration of serum zinc in celiac patients compared to healthy subjects in Tehran. Gastroenterology and Hepatology from Bed to Bench 6(2): 92–95.
PubMed
Google Scholar
a [...] to the intestinal epithelial cell membrane.
b [...] to have lower plasma Zn concentrations.
c [...] < 0.0001).
d [...] Fathi et al.
Altuntal B, Filik B, Ensari A, et al. (2000) Can zinc deficiency be used as a marker for the diagnosis of celiac disease in Turkish children with short stature? Pediatrics International 42: 682–684.
Crossref
PubMed
Web of Science
Google Scholar
Bledsoe AC, King KS, Larson JJ, et al. (2019) Micronutrient deficiencies are common in contemporary celiac disease despite lack of overt malabsorption symptoms. Mayo Clinic Proceedings 94(7):1253-1260. 31248695.
Go to Reference
Crossref
Google Scholar
Solomons NW, Rosenberg IH, Sandstead HH (1976) Zinc nutrition in celiac sprue. The American Journal of Clinical Nutrition 29: 371–375.
Crossref
PubMed
Web of Science
Google Scholar
a [...] is consistent with many previous studies.
b [...] Solomons et al.
MacMahon RA, Parker ML, McKinnon MC (1968) Zinc treatment in malabsorption. Medical Journal of Australia 2: 210–212.
Crossref
PubMed
Google Scholar
Singhal N, Alam S, Sherwani R, et al. (2008) Serum zinc levels in celiac disease. Indian Pediatrics 45(4): 319.
PubMed
Web of Science
Google Scholar
a [...] to have lower plasma Zn concentrations.
b [...] in controls (95% CI −0.44 to −1.4)).
Uyanikoglu A, Eren MA, Aydoğan T, et al. Celiac disease prezenting with osteoporosis: A case report. Euroasian Journal of Hepato-Gastroenterology 3: A054,21.
Go to Reference
Google Scholar
Uyanıkoğlu A, Aydoğan T, Nar H, et al. (2014) Demographic and laboratory features of celiac patients in Sanliurfa region. Journal of Current Gastroenterology 183: 339–341.
Go to Reference
Google Scholar
Krupa-Kozak U (2014) Pathologic bone alterations in celiac disease: Etiology, epidemiology, and treatment. Nutrition 30: 16–24.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Lucendo AJ, García-Manzanares A (2013) Bone mineral density in adult coeliac disease: An updated review. Revista Española De Enfermedades Digestivas 105: 154–162.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Vici G, Camilletti D, Polzonetti V (2020) Possible role of vitamin D in celiac disease onset. Nutrients 12: 1051.
Go to Reference
Google Scholar
Dos Santos S, Lioté F (2017) Osteoarticular manifestations of celiac disease and non-celiac gluten hypersensitivity. Joint Bone Spine 84: 263–266.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Lemieux B, Boivin M, Brossard JH, et al. (2001) Normal parathyroid function with decreased bone minerl density in treated celiac disease. Canadian Journal of Gastroenterology and Hepatology 15: 302–307.
Go to Reference
Google Scholar
Zingone F, Ciacci C (2018) The value and significance of 25(OH) and 1,25(OH) vitamin D serum levels in adult coeliac patients: A review of the literature. Digestive and Liver Disease 50: 757–760.
Go to Reference
Crossref
Web of Science
Google Scholar
Cuskelly GJ, Mooney KM, Young IS (2007) Folate and vitamin B12: Friendly or enemy nutrients for the elderly: Symposium on “Micronutrients through the life cycle.” Proceedings of the Nutrition Society 66(4): 548–558.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Dahele A, Ghosh S (2001) Vitamin B12 deficiency in untreated celiac disease. The American Journal of Gastroenterology 96: 745–750.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Wierdsma NJ, van Bokhorst-de MA, Berkenpas M, et al. (2013) Vitamin and mineral deficiencies are highly prevalent in newly diagnosed celiac disease patients. Nutrients 5(10): 3975–3992.
Crossref
PubMed
Web of Science
Google Scholar
a [...] B-complex) before being diagnosed with CeD.
b [...] Wierdsma et al.
c [...] Wierdsma et al.
Karnani P, Barupa R, Vishnoi SK, et al. Serum vitamin D, zinc, iron and copper levels in children with newly diagnosed coeliac disease. Sri Lanka Journal of Child Health 51(4): 519–524.
Crossref
Google Scholar
a [...] is consistent with many previous studies.
b [...] Karnani et al.
c [...] Karnani et al.
d [...] Karnani et al.
e [...] Karnani et al.
Ince AT, Kayadibi H, Soylu A, et al. (2008) Serum copper, ceruloplasmin and 24-h urine copper evaluations in celiac patients. Digestive Diseases and Sciences 53: 1564–1572.
Crossref
PubMed
Web of Science
Google Scholar
a [...] is consistent with many previous studies.
b [...] Ince et al.
c [...] Ince et al.
d [...] Ince et al.
e [...] Ince et al.
Ferretti G, Bacchetti T, Masciangelo S, et al. (2012) Celiac disease, inflammation and oxidative damage. A nutrigenetic approach. Nutrients 4: 243–257.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Green PH, Jabri B (2006) Celiac disease. Annual Review of Medicine 57: 207–221.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Hallert C, Grant C, Grehn S, et al. (2002) Evidence of poor vitamin status in coeliac patients on a gluten-free diet for 10 years. Alimentary Pharmacology & Therapeutics 16: 1333–1339.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Pagano AE (2006) Whole grains and the gluten-free diet. Practical Gastroenterology 29: 66–78.
Go to Reference
Google Scholar
Cardo A, Churruca I, Lasa A, et al. (2021) Nutritional imbalances in adult celiac patients following a gluten-free diet. Nutrients 13(8): 2877.
Crossref
PubMed
Web of Science
Google Scholar
a [...] of meat and meat-derived products.
b [...] with their gluten-containing counterparts.
c [...] complex carbohydrates, and proteins.
d [...] grains, pseudocereals, tubers, and nuts.
Kupper C (2005) Dietary guidelines and implementation for celiac disease. Gastroenterology 128: S121–S127.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Raymond N (2006) The gluten-free diet: An update for health professionals. Practical Gastroenterology 9: 73–91.
Go to Reference
Google Scholar
See J, Murray JA (2006) Gluten-free diet: The medical and nutrition management of celiac disease. Nutrition in Clinical Practice 21: 1–15.
Go to Reference
Crossref
PubMed
Google Scholar
Guennouni M, Admou B, El khoudri N, et al. (2022) Gluten contamination in labelled gluten-free, naturally gluten-free and meals in food services in low-, middle- and high-income countries: A systematic review and meta-analysis. British Journal of Nutrition 127(10): 1528–1542.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Garcia-Manzanares A, Lucendo AJ (2011) Nutritional and dietary aspects of celiac disease. Nutrition in Clinical Practice 26: 163–173.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Karaman K, Akbayram S, Kar S, et al. (2016) Prevalence of celiac disease in children with iron deficiency anemia in Van Lake Region of Turkey. Pediatric Hematology and Oncology 38: 143–146.
Crossref
Web of Science
Google Scholar
a [...] Karaman et al.
b [...] Karaman et al.
c [...] Karaman et al.
d [...] Karaman et al.
Mete R, Oran M, Avcı BA, et al. (2021) The diagnostic utility of flow cytometry in celiac disease presented isolated iron deficiency anemia. Turkish Journal of Gastroenterology 32(11): 932–936.
Crossref
PubMed
Web of Science
Google Scholar
a [...] Rafet Mete et al.
b [...] Rafet Mete et al.
Botero-Lopez J, Araya M, Parada A, et al. (2011) Micronutrient deficiencies in patients with typical and atypical celiac disease. Journal of Pediatric Gastroenterology and Nutrition 53: 265–270.
Crossref
PubMed
Web of Science
Google Scholar
a [...] Botero-Lopez et al.
b [...] Botero-Lopez et al.
c [...] Botero-Lopez et al.
d [...] BoteroLopez et al.
Weintraub Y, Ben-Tov A, Dotan G, et al. (2019) Vitamin A levels are comparable between children with newly diagnosed coeliac disease and non-coeliac controls. Acta Paediatrica 108(11): 2095–2099.
Crossref
PubMed
Web of Science
Google Scholar
a [...] Weintraub et al.
b [...] Weintraub et al.
c [...] Weintraub et al.
Tokgoz Y, Terlemez S, Karul A (2018) Fat soluble vitamin levels in children with newly diagnosed celiac disease, a case control study. BMC Pediatrics 18: 130.
Crossref
PubMed
Web of Science
Google Scholar
a [...] Tokgoz et al.
b [...] Tokgoz et al.
Piatek-Guziewicz A, Ptak-Belowska A, Przybylska-Felus M, et al. (2017) Intestinal parameters of oxidative imbalance in celiac adults with extraintestinal manifestations. World Journal of Gastroenterology 23: 7849–7862.
Crossref
PubMed
Web of Science
Google Scholar
a [...] Piatek-Guziewicz et al.
b [...] Piatek-Guziewicz A et al.
Armagan O, Uz T, Tascioglu F, et al. (2005) Serological screening for celiac disease in premenopausal women with idiopathic osteoporosis. Clinical Rheumatology 24: 239–243.
Crossref
PubMed
Web of Science
Google Scholar
a [...] Armagan et al.
b [...] Armagan et al.
c [...] Armagan et al.
d [...] Armagan et al.
Narang M, Natarajan R, Shah D, et al. (2018) Celiac disease in children with moderate-to-severe iron-deficiency anemia. Indian Pediatrics 15: 31–34.
Go to Reference
Crossref
Web of Science
Google Scholar
Valente FX, Campos Tdo N, Moraes LF, et al. (2015) B vitamins related to homocysteine metabolism in adults celiac disease patients: A cross-sectional study. Nutrition Journal 14: 110.
Crossref
PubMed
Web of Science
Google Scholar
a [...] Xavier-Valente et al.
b [...] Xavier Valente et al.
c [...] Xavier Valente et al.
Remes-Troche JM, Cobos-Quevedo ODJ, Rivera-Gutiérrez X, et al. (2020) Efectos de una dieta libre de gluten (DLG) durante 6 meses sobre el metabolismo en pacientes conenfermedad celíaca, sensibilidad al gluten no celíaca y controles asintomáticos. Revista de Gastroenterología de México 85(2): 109–117.
Go to Reference
Crossref
Web of Science
Google Scholar
Emami MH, Karimi S, Kouhestani S (2012) Is routine duodenal biopsy necessary for the detection of celiac disease in patients presenting with iron deficiency anemia? International Journal of Preventive Medicine 4: 273–277.
Go to Reference
Google Scholar
Baghbanian M, Farahat A, Vahedian HA, et al. (2015) The prevalence of celiac disease in patients with iron-deficiency anemia in Center and South area of Iran. Arquivos de gastroenterologia 52(4): 278–282.
Crossref
PubMed
Google Scholar
a [...] Baghbanian et al.
b [...] Baghbanian et al.
Volkan B, Fettah A, İşlek A, et al. (2018) Bone mineral density and vitamin K status in children with celiac disease: Is there a relation? Turkish Journal of Gastroenterology 29: 215–220.
Crossref
PubMed
Web of Science
Google Scholar
a [...] Volkan et al.
b [...] Volkan et al.
c [...] Volkan et al.
d [...] Volkan et al.
Haapalahti M, Kulmala P, Karttunen TJ, et al. (2005) Nutritional status in adolescents and young adults with screen-detected celiac disease. Journal of Pediatric Gastroenterology and Nutrition 40(5): 566–570.
Crossref
PubMed
Web of Science
Google Scholar
a [...] Haapalahti et al.
b [...] Haapalahti et al.
c [...] Haapalahti et al.
Hallert C, Tobiasson P, Walan A (1981) Serum folate determinations in tracing adult coeliacs. Scandinavian Journal of Gastroenterology 16: 263–267.
Crossref
PubMed
Web of Science
Google Scholar
a [...] Hallert et al.
b [...] Hallert et al.
Hadithi M, Mulder CJ, Stam F, et al. (2009) Effect of B vitamin supplementation on plasma homocysteine levels in celiac disease. World Journal of Gastroenterology 15: 955–960.
Crossref
PubMed
Web of Science
Google Scholar
a [...] Hadithi et al.
b [...] Hadithi et al.
Dickey W, Ward M, Whittle CR, et al. (2008) Homocysteine and related B-vitamin status in coeliac disease: Effects of gluten exclusion and histological recovery. Scandinavian Journal of Gastroenterology 43: 682–688.
Crossref
PubMed
Web of Science
Google Scholar
a [...] Dickey et al.
b [...] Dickey et al.
Szymczak J, Bohdanowicz-Pawlak A, Waszczuk E, et al. (2012) Low bone mineral density in adult patients with coeliac disease. Endokrynologia Polska 63(4): 270–276.
Go to Reference
PubMed
Web of Science
Google Scholar
Villanueva J, Maranda L, Nwosu BU (2012) Is vitamin D deficiency a feature of pediatric celiac disease? Journal of Pediatric Endocrinology and Metabolism 25(5–6): 607–610.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Nwosu BU, Maranda L (2015) Vitamin D status and adiposity in pediatric malabsorption syndromes. Digestion 92(1): 1–7.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Setty-Shah N, Maranda L, Nwosu BU (2014) Increased risk for vitamin d deficiency in obese children with both celiac disease and type 1 diabetes. Gastroenterology Research and Practice 2014(1): 561351.
Go to Reference
PubMed
Web of Science
Google Scholar
Uyanikoglu A, Cindioglu C, Ciftci A, et al. (2021) The value of 25 (OH) and 1,25 (OH) vitamin D serum levels in newly diagnosed or on diet adult celiac patients: A casecontrol study. International Medicine 3(2): 37–42.
Go to Reference
Crossref
Google Scholar
Corazza GR, Di Sario A, Cecchetti L, et al. (1996) Influence of pattern of clinical presentation and of gluten-free diet on bone mass and metabolism in adult coeliac disease. Bone 18: 525–530.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Lionetti E, Galeazzi T, Dominjanni V, et al. (2021) Lower level of plasma 25-Hydroxyvitamin D in children at diagnosis of celiac disease compared with healthy subjects: A case-control study. The Journal of Pediatrics 228: 132–137.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Jameson S, Hellsing K, Magnusson S (1985) Copper malabsorption in coelia disease. Science of the Total Environment 42: 29–36.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Rawal P, Thapa BR, Prasad R, et al. (2010) Zinc supplementation to patients with celiac disease—is it required? Journal of Tropical Pediatrics 56: 391–397.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
Naveh Y, Lightman A, Zinder O (1983) A prospective study of serum zinc concentration in children with celiac disease. Journal of Pediatrics 102(5): 734–736.
Go to Reference
Crossref
PubMed
Web of Science
Google Scholar
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International Journal of Immunopathology and Pharmacology
Volume 39
Article first published online: February 17, 2025
Issue published: January-December 2025
Keywords
gastrointestinal disorders
gluten-related disorders
celiac disease
micronutrient deficiencies
meta-analysis
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PubMed: 39959924
Authors
Saad Lamjadli
Laboratory of Immunology, Center of Clinical Research, Mohammed VI University Hospital, Marrakech, Morocco
saad.lamjadli@edu.uca.ma
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Ider Oujamaa
Biosciences Research Laboratory, Faculty of Medicine and Pharmacy, Cadi Ayyad University, Marrakech, Morocco
View all publications by this author
Ikram Souli
Laboratory of Immunology, Center of Clinical Research, Mohammed VI University Hospital, Marrakech, Morocco
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Fatima ezzohra Eddehbi
Laboratory of Immunology, Center of Clinical Research, Mohammed VI University Hospital, Marrakech, Morocco
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Nadia Lakhouaja
Laboratory of Immunology, Center of Clinical Research, Mohammed VI University Hospital, Marrakech, Morocco
View all publications by this author
Bouchra M’raouni
Laboratory of Immunology, Center of Clinical Research, Mohammed VI University Hospital, Marrakech, Morocco
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Abdelmouine Salami
Laboratory of Immunology, Center of Clinical Research, Mohammed VI University Hospital, Marrakech, Morocco
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Morad Guennouni
Biosciences Research Laboratory, Faculty of Medicine and Pharmacy, Cadi Ayyad University, Marrakech, Morocco
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Moulay Yassine Belghali
Biosciences Research Laboratory, Faculty of Medicine and Pharmacy, Cadi Ayyad University, Marrakech, Morocco
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Raja Hazime
Laboratory of Immunology, Center of Clinical Research, Mohammed VI University Hospital, Marrakech, Morocco
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Brahim Admou
Laboratory of Immunology, Center of Clinical Research, Mohammed VI University Hospital, Marrakech, Morocco
Biosciences Research Laboratory, Faculty of Medicine and Pharmacy, Cadi Ayyad University, Marrakech, Morocco
View all publications by this author
Notes
Saad Lamjadli, Laboratory of Immunology, Center of Clinical Research, Mohammed VI University Hospital, Marrakech, Morocco. Email: saad.lamjadli@edu.uca.ma
Author contributions
Conceptualization: Saad Lamjadli and Ikram Souli; Methodology: Ider Oujamaa, Morad Guennouni, Moulay Yassine Belghali, Raja Hazime and Brahim Admou; Software: Saad Lamjadli and Morad Guennouni; Data curation: Ikram Souli, Fatima ezzohra Eddehbi, Nadia Lakhouaja, Bouchra M’raouni and Abdelmouine Salami; Writing—Original Draft, Saad Lamjadli, Ider Oujamaa, Ikram Souli; Validation: Ider Oujamaa, Moulay Yassine Belghali, Raja Hazime and Brahim Admou; Writing—Review and Editing, all authors; Supervision: Brahim Admou.
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Uncovering Hidden Gluten Exposure in Celiac Patients: A Case Study in Family-Based Management and the Role of Point-of-Care Urine Testing and Psychological Assessment
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Figures
Figure 1. Flowchart of study selection process.
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Figure 2. Forest plot of hemoglobin levels.
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Figure 3. Forest plot of iron levels.
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Figure 4. Forest plot of ferritin levels.
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Figure 5. Forest plot of folic acid levels.
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Figure 6. Forest plot of vitamin D levels.
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Figure 7. Forest plot of zinc levels.
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Figure 8. Forest plot of MCV levels.
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Figure 9. Forest plot of copper levels.
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Figure 10. Forest plot of vitamin B12 levels.
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Appendix 1
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Appendix 3. Published articles included in this meta-analysis, categorized by region.
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Media
Tables
Table 1. Pooled effect size (SMD) results (Hemoglobin).
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Table 2. Pooled effect size (SMD) results (Iron).
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Table 3. Pooled effect size (SMD) results (Ferritin).
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Table 4. Pooled effect size (SMD) results (Folic acid).
Go to Table
Table 5. Pooled effect size (SMD) results (Vitamin D).
Go to Table
Table 6. Pooled effect size (SMD) results (Zn).
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Table 7. Pooled effect size (SMD) results (MCV).
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Table 8. Pooled effect size (SMD) results (Copper).
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Table 9. Pooled effect size (SMD) results (Vitamin B12).
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Appendix 2. Articles involving Hemoglobin included in the meta-analysis.
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Articles involving Ferritin included in the meta-analysis.
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Articles involving Iron included in the meta-analysis.
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Articles involving MCV included in the meta-analysis.
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Articles involving Folic Acid included in the meta-analysis.
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Articles involving vitamin B12 included in the meta-analysis.
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Articles involving vitamin D included in the meta-analysis.
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Articles involving Copper included in the meta-analysis.
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Articles involving Zinc included in the meta-analysis.
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15321 | https://www.youtube.com/watch?v=dBIlCD_JF9Q | Circle Theorems - GCSE Higher Maths
1st Class Maths
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This video is for students aged 14+ studying GCSE Maths.
A video explaining how to use and understand circle theorems for GCSE Higher Maths.
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0:00 Introduction
0:36 Angles in the same segment theorem
2:05 Angle in a semi circle theorem
2:52 Angle at the centre theorem
3:31 Opposite angles in a cyclic quadrilateral theorem
4:31 A tangent meets a radius theorem
4:55 Tangents from a point
5:55 Alternate Segment Theorem
6:44 All theorems on one page
6:56 Worked example
8:06 Second example
11:06 Third example
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Introduction [Music] foreign hello and welcome to this video where we're going to learn about Circle theorems before we do our first theorem we're going to need to know the names of some key points of a circle so if we draw a circle and then draw a straight line that goes from one side of the circle to the other this line is known as a chord a chord will always split a circle into two segments here we have a segment at the top and also a segment here at the bottom Angles in the same segment theorem now let's have a look at our first Circle theorem the first theorem States the angles in the same segment are equal this means if we use this chord that we've drawn to create an angle but keep that angle within the same segment all of those angles will be the same size so let's draw an angle in the upper segment using this chord like this if we draw a second angle like this one or even a third like this one then all of these angles must be the same size because they come from this chord and they're in the same segment so if the purple one was 68 degrees the green one would be as well and so would the blue one sometimes this theorem is drawn without the red chord so we can remove that and it may look something like this and the property still holds it's often drawn with only just two of the angles so if we remove one of them like this we end up with this picture where both of the purple angles here are the same size you can also apply this theorem by drawing a chord here if you do the same thing as we just did but upside down you can show that these two angles are also the same and again we don't need to draw that chord so very often you'll see this theorem drawn like this it's sometimes informally referred to as the bow tie theorem because this looks a little bit like the shape of a bow tie in an exam though you want to avoid that terminology and say that the angles in the same segment are equal but if the bow tie helps you recognize when this is true then that's fine also Angle in a semi circle theorem for the next theorem we're going to need to mark on the center of the circle we're going to draw a chord that goes straight through the center this is known as a diameter if we now do the same thing as we did for the first theorem so draw angles that are in the same segment we know they're going to be equal so if we draw one angle like this a second like this and even a third like this we know all three of these angles must be the same size but it turns out that if the chord that you draw happens to be the diameter of the circle going through the center all of these angles will always be right angles so the second theorem is that the angle in a semicircle is 90 degrees we say this is the angles in a semicircle because that diameter splits the circle in half Angle at the centre theorem for the next theorem we're going to draw a chord once again and draw an angle at the circumference also but rather than drawing another angle at the circumference we're now going to draw an angle at the center but using the same chord so something that looks like this if the green angle at the circumference of the circle was 50 degrees then the blue one would be twice this 100 degrees so for this theorem we say the angle at the center is twice the angle at the circumference you could also say in reverse the angle at the circumference is half of the angle at the center once again this theorem still applies if we remove the chord so sometimes you see it drawn like this Opposite angles in a cyclic quadrilateral theorem for the next theorem we need to learn some more terminology if you draw a quadrilateral where all four of the sides touch the circumference of the circle this is known as a cyclic quadrilateral now it's only true if all four of the corners of the shape are on the circumference for example this shape here is not a cyclic quadrilateral because this corner here isn't on the circumference of the circle so let's return to our original one this is a cyclic quadrilateral the theorem states that the opposite angles in a cyclic quadrilateral had to make 180 degrees for example if we look at these two angles here they're opposite each other and they're in a cyclic quadrilateral so if we knew that this one was 86 degrees we'd know the other one must be 94 degrees since 86 plus 94 makes 180. similarly these two angles must add up to 180 as well if we knew this one at the top was 110 the bottom one must be 70. since 110 plus 70 is 180. A tangent meets a radius theorem for the next theorem we're going to draw a tangent to the circle a tangent is a straight line that touches the circle in one place we're then going to draw a radius from the center of the circle to the point where the tangent also touches the circle in this diagram this angle here will always be a right angle so we can say for this theorem a tangent meets a radius at 90 degrees Tangents from a point now if we stick with the same diagram we can do some more theorems if we create a point at the end of this tangent on the right hand side and call that P and then if we draw a second tangent from P to the circle it would look something like this if you measure this distance here from P to the point where the tangent touches the circle at the top and the same distance at the bottom these two lines would be the same length we don't normally show this with arrows though we usually draw a line across both of those lines to show they're the same length like this if you are going to write this theorem down you could say that tangents from the same point are equal in length we can also add a bit more information to this diagram if you draw the angle at p then draw a line from the center of the circle to P then this line will split that angle in half or we could say it bisects the angle so if the whole angle here was 30 degrees then we'd have 15 degrees on the top and also 15 degrees on the bottom Alternate Segment Theorem now we're going to move on to the final theorem if we draw a tangent to the circle and then draw a chord from the point where the tangent touches the circle we form an angle here like this next we're going to use the cord to create an angle at the circumference but not in the same segment in the other segment so something that looks like this it turns out that the angle that the tangent makes with the chord is equal to this angle that the chord makes at the circumference the same happens with the chord on the other side as well so if you look at the chord on the right hand side the angle it makes with the tangent is this green one here but this chord will create an angle at the circumference in the other segment over here so these two angles are also equal we call this the alternate segment theorem and you'll need to know this name All theorems on one page here's a quick sketch of all of those theorems we've covered on one page you may find this useful for your revision now we're going to have a look at how we can use these theorems to solve some problems Worked example in this question we've been asked to find the angle ABD but we've also been asked to give reasons for our answers so every time we find an angle we need to explain what we did this will often involve just writing down the theorem that we've used so in this one we're looking for the angle ABD we're first of all going to find the angle ACD you should notice that this angle here ACD goes with the angle Ace to make a straight line so these two angles must add to make 180. so we could say that 180 minus 105 equals 75 so we know that this red angle here is 75 degrees and remember we need to give reasons for our answer so we would state that the angles on a straight line add up to 180 degrees now we can use one of our Circle theorems to find the angle ABD this one looks like the bow tie theorem the first one we did so the red angle we just found is actually the same as the angle ABD so this one is also 75 degrees so we could say that angle ABD equals 75 degrees and once again we need to give a reason because the question asked us for one and we would give the first theorem which was the angles in the same segment are equal Second example in this next question we've been asked to find angle BCF but this time we don't need to give reasons because it hasn't asked for them we do need to show working out still though we're going to start by looking at the triangle ACD which is this one here we know this is an isosceles triangle because we can see the a d and DC are the same length this means that the angle ACD down here is 35 degrees it's a good idea to Mark any angles that you find directly onto the diagram but we're also going to write down the angle ACD equals 35 degrees now we're going to stay inside this triangle and work out the angle ADC which is this one here we know angles in a triangle add to make 180 degrees so if we add up the two angles we have 35 at 35 that gives you 70 and then if we subtract this from 180 we get 110. so we can add this to our diagram as well and we're going to write down the angle ADC equals 110 degrees next we're going to look at the whole quadrilateral here you might remember from before that one of the theorems was about cyclic quadrilaterals we can be sure this is a cyclic quadrilateral because a b c and d are all on the circumference the theorem was that the opposite angles in a cyclic quadrilateral make 180 degrees so if we look at the angle that's opposite the 110 that's this one here ABC we can work this one out by subtracting 110 from 180. so if we do 180 take away 110 you get 70 degrees so we can mark this onto the diagram and also write down the angle ABC is 70 degrees next notice how the line a b goes through the center of the circle this means it's the diameter the second theorem we looked at was that the angle in a semicircle is always 90 degrees this means that the angle BCA which is here is a right angle so we can write down that BCA equals 90 degrees now if we stay inside this triangle ABC there's only one more angle to find the one at the top here you can find this by subtracting the other two from 180 so if we do 90 plus 70 that's 160 and then 180 take away 160 is 20 degrees so the angle at the top here is 20 degrees and we can write down the angle c a b is 20 degrees and there's only one more step to go to complete the question we're looking for the angle b c f we can use the alternate segment theorem to show that this is the same as the angle we just found this 20 degrees here is the same as 20 degrees here so the angle BCF equals 20 degrees now we didn't write down any of the worded reasons for this question because it didn't ask us to give reasons for our answers however you can see we have put down quite a substantial amount of working out and we've also labeled every angle that we found onto the diagram it's worth noting that this isn't the only way to solve this question in many Circle theorems questions there are different approaches as long as you get the right answer and show your method clearly you will get full marks Third example to begin this question we're going to look at the line ECF which is a tangent to the circle and also the line OC which is a radius one of the theorems said that a tangent meets a radius at 90 degrees so we know this angle here is 90 degrees so let's write that down angle ocf equals 90 degrees now we're going to look at the whole triangle ocf this one here we have two of the angles in this triangle we have the 32 degrees and the 90 degrees so we can find the missing one if we add together 90 and 32 you get 122 and if you subtract this from 180 you find the missing angle is 58 degrees so we can write the angle C of f equals 58 degrees and label this onto the diagram next we're going to look at the angle c a b which is this one here you should remember that the angle at the center is twice the angle at the circumference we can see the angle at the center from the chord CB is 58 degrees so the angle the chord CB makes up the circumference which is the angle I've marked must be half of 58. so if we do 58 divided by 2 which is 29 we know that this angle must be 29 degrees next we're going to look at the smaller triangle cob this one here sometimes in questions there's information that's not necessarily marked onto the diagram but it is useful it's useful for example to notice that OC and OB are both the radius of the circle this means they're the same length we can mark that onto the diagram but is very often not given to you you need to be able to spot this information yourself this means the triangle cob is an isosceles triangle so two of the angles must be the same so if we subtract 58 from 180 we get 122 and then if you half 122 you get 61. this means both of the base angles in the isosceles triangle must be 61 degrees the one that's important to us is the angle OBC so angle OBC is 61 degrees we can add this to the diagram as well now finally if we look at the triangle ABC we're only one step away from finding the angle which was required in the question so the angle is this one here all of these four angles must make 180 since they're the interior angles of a triangle so if we add up the angles we have 40 plus 61 plus 29 makes 130 and if we subtract this from 180 you get 50 which is the answer to the question and angle above thank you for watching this video I hope you found it useful check out the one I think you should watch next And subscribe so you don't miss out on future videos also check out the exam questions I put in this video's description |
15322 | https://pmc.ncbi.nlm.nih.gov/articles/PMC6082311/ | The multiple roles of titin in muscle contraction and force production - PMC
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The multiple roles of titin in muscle contraction and force production
Walter Herzog
Walter Herzog
1 Human Performance Laboratory, Faculty of Kinesiology, University of Calgary, Calgary, AB T2N 1N4 Canada
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1 Human Performance Laboratory, Faculty of Kinesiology, University of Calgary, Calgary, AB T2N 1N4 Canada
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Received 2017 Dec 21; Accepted 2017 Dec 29; Collection date 2018 Aug.
© International Union for Pure and Applied Biophysics (IUPAB) and Springer-Verlag GmbH Germany, part of Springer Nature 2018
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PMCID: PMC6082311 PMID: 29353351
Abstract
Titin is a filamentous protein spanning the half-sarcomere, with spring-like properties in the I-band region. Various structural, signaling, and mechanical functions have been associated with titin, but not all of these are fully elucidated and accepted in the scientific community. Here, I discuss the primary mechanical functions of titin, including its accepted role in passive force production, stabilization of half-sarcomeres and sarcomeres, and its controversial contribution to residual force enhancement, passive force enhancement, energetics, and work production in shortening muscle. Finally, I provide evidence that titin is a molecular spring whose stiffness changes with muscle activation and actin–myosin-based force production, suggesting a novel model of force production that, aside from actin and myosin, includes titin as a “third contractile” filament. Using this three-filament model of sarcomeres, the stability of (half-) sarcomeres, passive force enhancement, residual force enhancement, and the decrease in metabolic energy during and following eccentric contractions can be explained readily.
Keywords: Titin, Molecular spring, Mechanical functions, Active/passive force regulation, Muscle shortening, Force production, Cross-bridge theory, Three filament sarcomere model, Mechanisms of muscle contraction, Muscle energetics
Background
Since the 1950s, muscle contraction and force production have been associated with the relative sliding of the two contractile filaments, actin and myosin (referred to as the sliding filament theory) (Huxley and Hanson 1954; Huxley and Niedergerke 1954), and the cyclic interaction of myosin-based cross-bridges with specialized attachment sites on the actin filaments (referred to as the cross-bridge theory) (Huxley 1957a). However, from the very onset, the cross-bridge theory could not predict well some of the experimentally observed properties in skeletal muscles (Huxley 1957a). For example, the well-recognized and generally accepted residual force enhancement and residual force depression properties of muscles, observed well before the development of the cross-bridge theory (Abbott and Aubert 1952), cannot be predicted without making fundamental changes to the cross-bridge theory (Walcott and Herzog 2008). Furthermore, the stability of myosin filaments in the center of sarcomeres (Iwazumi 1979; Horowits and Podolsky 1987) and that of sarcomeres on the so-called descending limb of the force–length relationship (Zahalak 1997; Novak and Truskinovsky 2014) cannot be predicted with the cross-bridge theory (Iwazumi and Noble 1989; Zahalak 1997), and the forces and metabolic cost predicted by the original cross-bridge theory were much too high for eccentric contractions (Huxley 1957a).
Andrew Huxley, the father of the cross-bridge theory (Huxley 1957a), recognized the shortcomings of his approach. For example, in order to account for the excessive metabolic cost of eccentric contractions, he proposed that there might be multiple cross-bridge cycles for each energy unit hydrolyzed [adenosine triphosphate (ATP)], while for concentric and isometric contractions, the hydrolysis of one ATP molecule was tightly linked to one cross-bridge cycle (Huxley 1957a, 1969; Huxley and Simmons 1971; Rayment et al. 1993). Also, the excessive eccentric forces could be reduced to experimentally observed values by assuming that attached cross-bridges are torn from actin prior to the full completion of the cross-bridge cycle (Huxley 1957a). However, for the residual force enhancement properties of skeletal muscle, Huxley had no solution to offer. In his 1980 book, “Reflections on Muscle,” Huxley acknowledged the insufficiencies of current cross-bridge thinking in eccentric muscle contraction (Huxley 1980). Specifically, he mentions that special features must have evolved that allow the elongation of active muscles to take place without damaging muscles, that these special features allow for explanations of the mechanics and energetics of eccentric contractions, and that force regulation in eccentric contractions bears little relation to what happens in concentric muscle action (Huxley 1980).
In the mid- and late 1970s, just prior to the publication of Huxley’s book on muscle contraction, titin (initially also referred to as connection) was discovered (Maruyama 1976; Maruyama et al. 1977; Wang et al. 1979). Titin is a filamentous protein spanning the half-sarcomere from the M-band to Z-band (Fig. 1). While thought to be essentially rigidly attached to myosin in the A-band region (except possibly for extreme sarcomere excursions beyond the normal range encountered in typical everyday movements), titin’s I-band structure allows for large elongations and passive force production, and thus has been termed a “spring-like” molecule. Just prior to inserting into the Z-band, titin binds to actin along its most proximal 50 nm, thereby establishing a “permanent” bridge between actin and myosin (Trombitas and Pollack 1993; Linke et al. 1997; Trombitas and Granzier 1997): a bridge that is in parallel with attached cross-bridges and in series with the myosin filament in the passive muscle. With an estimated six titin filaments per half myosin (Granzier and Irving 1995; Cazorla et al. 2000; Liversage et al. 2001; Granzier and Labeit 2007), there is one titin for each actin filament in vertebrate skeletal muscles where actin filaments surround myosin in a hexagonal array (Huxley 1953b, 1957).
Fig. 1.
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Schematic two-dimensional illustration of a sarcomere bordered by Z-bands at either end. Thick, myosin-based filaments are in the center of the sarcomere (green), thin, actin-based filaments insert into the Z-band at either end of the sarcomere (red), and titin filaments (blue) run from the M-line in the middle of the sarcomere to the Z-band. Adapted from Granzier and Labeit (2007) with permission
Ever since its discovery, titin’s functions have been questioned, and titin’s recently proposed roles in active force regulation and mechanical work in muscle shortening are current topics of intense debate in the scientific community. At the recent European Muscle Conference (September 2017), debates on the functional role of titin ended inconclusively. Here, I will attempt to summarize both the acknowledged and the controversial aspects of titin’s mechanical functions, with an emphasis on titin’s proposed role in active force regulation and mechanical work.
Titin’s proposed functions
Titin has been associated with a variety of functions, including mechanical roles in active and passive force regulation in cardiac and skeletal muscles (e.g., Linke et al. 1994, 1996; Granzier and Labeit 2002, 2007; Linke and Fernandez 2002; Herzog et al. 2006; LeWinter and Granzier 2010; Herzog 2014a), structural and developmental roles in sarcomere organization (e.g., Linke and Fernandez 2002; Granzier and Labeit 2007), and functions associated with mechano-sensing and signaling (Schwarz et al. 2008; Kruger and Linke 2009; Linke and Kruger 2010; Granzier et al. 2014). Here, I primarily focus on the proposed mechanical functions of titin in skeletal muscles, although some comparisons with cardiac muscles will be made. However, many extensive reviews on titin’s properties in cardiac muscle have been published recently (Granzier and Labeit 2002, 2007; Granzier et al. 2002), whereas comparatively little has been said on titin’s mechanical role in skeletal muscle.
Passive force contributions of titin
It is generally accepted that titin contributes to the passive forces in skeletal and cardiac muscles. Passive force is defined here as any force that arises from structural elements of muscle, is not associated with metabolic energy consumption, and is not part of the actin–myosin-based cross-bridge forces. The primary passive force contributors in cardiac and skeletal muscles are collagen filaments embedded in the various connective tissue layers of muscles, and the sarcomeric filament titin. In isolated myofibril preparations, titin is the primary passive force contributor (Maruyama 1976; Funatsu et al. 1990; Bartoo et al. 1997; Colomo et al. 1997; Linke and Fernandez 2002; Joumaa et al. 2008b; Leonard and Herzog 2010; Herzog et al. 2012), and the elimination of titin abolishes virtually all passive force (e.g., Leonard and Herzog 2010) .
In skinned single fibers and myofibrils, passive force and titin isoforms are tightly related. Increasing molecular weight, and thus increasing titin subunits and length, are associated with decreasing passive forces. Prado et al. (2005) determined the molecular weights of titin in 37 rabbit skeletal muscles and compared the molecular weights to the passive forces in myofibrils, skinned fibers, and intact and skinned fiber bundles. These authors found a strong inverse relationship between the size of titin and passive force in myofibrils and skinned fibers, i.e., the greater the molecular weight of titin, the smaller the corresponding passive force for a given sarcomere length. However, titin size was not associated in any systematic manner with the passive force in intact fiber bundles (and thus the entire muscle), and the contribution of titin to the total passive force in fiber bundles varied considerably between muscles, ranging from a high of 57% in rabbit psoas (Granzier et al. 2002) to a low of 24% for soleus in the range of 2.0 to 3.2 μm/sarcomere (Prado et al. 2005).
Prado et al. (2005) also found that the contribution of titin to passive muscle force depends on the length of the muscle (i.e., the average sarcomere length). This result agrees with observations in cardiac muscle where titin is thought to contribute more substantially to the passive forces at short (average sarcomere length 2.0–2.2 μm) compared to long (> 2.3 μm/sarcomere) sarcomere length (Cazorla et al. 2000; Freiburg et al. 2000; Granzier et al. 2002). Since the physiologic cardiac cycle occurs between sarcomere lengths that range from approximately 1.9 to 2.3 μm (Ter Keurs et al. 1980), titin plays a significant role in the beating heart.
Whether titin plays an equally important role within the functional range of skeletal muscles has not been determined systematically. It has been observed that in rabbit psoas myofibrils, passive, titin-based forces start to emerge at average sarcomere lengths of approximately 2.6–2.7 μm (Linke et al. 1996; Bartoo et al. 1997; Joumaa et al. 2007; Leonard and Herzog 2010). However, our group measured the shortest (hip fully flexed) and longest (hip fully extended) sarcomere length for rabbit psoas muscles as 1.9 and 2.6 μm, respectively. Knowing that rabbits never fully extend their hip, the maximal sarcomere lengths are probably never reached in the live animal. Furthermore, our measurements were performed on the passive muscle in the anesthetized animal, while in the active muscle, a substantial fiber and sarcomere length shortening would be expected with force production, as elastic elements are stretched and the contractile machinery shortens (Fukunaga et al. 1997; Ichinose et al. 1997; Vaz et al. 2012; de Brito Fontana and Herzog 2016). Therefore, maximal sarcomere lengths in the rabbit psoas likely never exceed about 2.3–2.4 μm, and thus are below the sarcomere lengths where passive titin forces have been first observed. A similar argument could be made for the rabbit soleus and medial gastrocnemius muscles. Thus, it appears that titin passive force does not play a functional role in many skeletal muscles. Whether this statement can be generalized is not yet known, as the functional sarcomere length of most animal muscles are not known. However, should skeletal muscle functional sarcomere length reach values in excess of 2.6 μm, then titin would likely contribute to the passive force of intact muscles (Prado et al. 2005). Also, in the following text I discuss titin’s possible role in shifting its slack length upon muscle activation to shorter sarcomere lengths than those observed in the passive muscle, which could potentially change the argument made here, with titin possibly emerging as a powerful passive force contributor in active skeletal muscles after all (Herzog 2014b; Herzog et al. 2015, 2016).
Titin’s stiffness, and thus passive force at a given sarcomere length, can be modulated in a variety of ways, such as calcium binding to titin upon muscle activation (Labeit et al. 2003; Joumaa et al. 2008b; DuVall et al. 2013), titin phosphorylation (Yamasaki et al. 2002; Borbely et al. 2009; Anderson et al. 2010; Perkin et al. 2015), and titin interactions with actin and other sarcomeric proteins (Li et al. 1995; Linke et al. 1997, 2002; Trombitas and Granzier 1997; Astier et al. 1998; Kulke et al. 2001; Yamasaki et al. 2001; Nagy et al. 2004; Bianco et al. 2007; Leonard and Herzog 2010; Chung et al. 2011), to just name the most common mechanisms. Some of these mechanisms will be discussed below in the section on residual force enhancement properties of skeletal muscles in general (Abbott and Aubert 1952; Edman et al. 1982; Herzog et al. 2006) and the passive residual force enhancement specifically (Herzog and Leonard 2002). Excellent reviews on the regulation of titin’s stiffness in cardiac muscle are abundant (e.g., Granzier and Labeit 2002; Granzier et al. 2002; Linke and Fernandez 2002; LeWinter and Granzier 2010), and these results will not be repeated here, except for comparative purposes.
Titin as a stabilizer of sarcomeres and half-sarcomeres
In the two-filament (actin and myosin) cross-bridge model of muscle contraction, half-sarcomeres and sarcomeres are unstable (Morgan 1990, 1994; Allinger et al. 1996; Zahalak 1997; Epstein and Herzog 1998; Morgan et al. 2000; Novak and Truskinovsky 2014). The half-sarcomere is unstable because small differences in half-sarcomere forces, caused by the stochastic interaction of cross-bridges with actin, will cause an initially centered myosin filament to be displaced from its mid-point position in the sarcomere. This perturbation is unstable as the overlap between actin and myosin will increase in the half-sarcomere of initial myosin drift and thus make this half-sarcomere increasingly stronger, while in the other half-sarcomere, the actin–myosin filament overlap is lost, and force continuously decreases (Iwazumi 1979). An analogous argument can be made for the instability of serially arranged sarcomeres (for example, as they occur in a myofibril; Gordon et al. 1966; Campbell 2009; Stoecker et al. 2009) that are located on the descending limb of the force–length relationship (Hill 1953).
A-band shifts to one end of the sarcomere were observed after prolonged activation in rabbit psoas skinned fibers with an average sarcomere length of 2.6 μm and zero passive force (Horowits and Podolsky 1987, 1988). These shifts, as well as non-uniformities in associated half-sarcomere lengths, were abolished at average sarcomere lengths of about 3.0 μm and a passive stress of approximately 2 N/cm 2, corresponding to approximately 20% of the maximum, active, isometric force of these fibers at optimal sarcomere length and 7°C. The A-band shifts observed in these studies were interpreted as indicating myosin instability in the center of the sarcomere when titin forces were zero or small, while myosin was stabilized in the center of the sarcomere once titin forces had reached a certain passive force level corresponding to approximately 2 N/cm 2 (e.g., Horowits 1992). The same observations were made for rabbit soleus fibers, with the notable difference that A-band shifts were greater in soleus than in psoas fibers at corresponding sarcomere lengths and that half-sarcomere uniformity and stability (absence of A-band shifts) were only observed at average sarcomere lengths of about 3.6 μm when passive, titin-based tension had reached 2 N/cm 2. The explanation for this result was based on the smaller size of the titin isoforms for rabbit psoas (3.3 and 3.4 MDa in a 70:30% ratio) compared to that of the single titin isoform observed in rabbit soleus (3.6 MDa) (Prado et al. 2005), resulting in reduced titin-based passive forces at a given sarcomere length for rabbit soleus compared to psoas.
Instability of muscles and sarcomeres on the descending limb of the force–length relationship has been used to explain specific muscle properties for more than half a century (Hill 1953; Gordon et al. 1966; Julian et al. 1978; Julian and Morgan 1979). For example, the so-called “creep” property, which is a slow change in isometric force for muscles/fibers on the descending limb of the force–length relationship (Hill 1953; Gordon et al. 1966), and the residual force enhancement and residual force depression properties (Morgan 1990, 1994) have been thought to be caused by instabilities in sarcomere length and the associated development of sarcomere length non-uniformities in active muscles, particularly if the muscles were stretched onto the descending limb of the force–length relationship. Indeed, vast sarcomere length non-uniformities have been observed in entire muscle preparations (Llewellyn et al. 2008; Moo et al. 2016), single fibers (Huxley and Peachey 1961), and isolated myofibrils (Rassier et al. 2003a; Joumaa et al. 2008a; Johnston et al. 2016). However, these sarcomere length non-uniformities occur at all lengths (ascending, plateau, and descending portions of the force–length relationship; Moo et al. 2016) and similarly in isometric, shortening, and stretched muscles (Johnston et al. 2016); thus, they do not seem to be associated with the proposed instability of the negative slope of the descending limb of the force–length relationship.
Surprisingly, Rassier et al. (2003a, b) observed that strictly serially arranged sarcomeres in single, active myofibrils stretched onto the descending limb of the force–length relationship are highly non-uniform, but perfectly stable. Sarcomeres half-way down the descending limb of the force–length relationship, and thus at half of the maximal actin–myosin filament overlap, were perfectly stable and remained at a constant length in the presence of sarcomeres on the plateau of the force–length relationship with maximal overlap between actin and myosin (Fig. 2). It would appear, therefore, that sarcomere length, and thus actin–myosin filament overlap, alone does not predict the isometric force of a sarcomere for a given level of activation within a myofibril. Rather, since the force in serially arranged sarcomeres must be identical (neglecting any inertial effects, which can safely be done), there must be another way other than just actin–myosin filament overlap to determine the isometric, steady-state force of a sarcomere within its natural environment of a myofibril. Active or passive stabilizing mechanisms must ensure that the weakening behavior of sarcomeres on the descending limb of the force–length relationship is compensated for in some (as of yet unknown) manner to ensure stability of the muscle.
Fig. 2.
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Rabbit psoas myofibril comprised of six sarcomeres that is stretched while activated from an average sarcomere length of about 2.4 μm to about 3.0 μm. After active stretching, all individual sarcomeres are on the descending limb of the force–length relationship, but there is no apparent overstretching or popping (quick sarcomere elongations beyond actin–myosin filament overlap: 3.9 μm) as has been proposed by proponents of the sarcomere length non-uniformity theory. Rather, sarcomeres seem to remain at an essentially constant length following the active stretch
Titin has been implicated as a stretch sensor that activates so-called “super-relaxed” cross-bridges, thereby providing a mechanism by which a disadvantage caused by reduced actin–myosin filament overlap (long sarcomere length) can be compensated for by an increased proportion of attached cross-bridges per unit length of myofilament overlap (Fusi et al. 2016). Titin would be an ideal candidate for regulating cross-bridge kinetics in sarcomeres of different lengths, thereby guaranteeing stability. Also, since titin in adjacent half-sarcomeres overlap in the M-band and the Z-band regions (Granzier and Labeit 2007), it is easy to imagine that force transmission across the M-band and Z-band is continuous, thus providing possibilities for force transfer within half-sarcomeres and between sarcomeres. Such a passive, structural force transmission between half-sarcomeres would also provide sarcomere length/force stability and would eliminate the rather odd notion that sarcomeres are the smallest “independent” unit of force production. Muscle mechanics would be simplified greatly if serial sarcomeres in a myofibril were indeed not “independent” but rather mutually dependent force producers, such that forces can be transmitted and “re-distributed” along serially arranged sarcomeres. Aside from structural evidence for such mutual dependence among sarcomeres, there is also functional support for this notion (e.g., Yasuda et al. 1996).
Titin’s role in the residual force enhancement property of skeletal muscle
When an active muscle is stretched, its isometric, steady-state force following the stretch is greater than the corresponding (same length, same activation) purely isometric contraction. This increase in force caused by active stretching has been termed residual force enhancement (RFE) (Edman et al. 1982). RFE can produce forces twice as high as purely isometric contractions (Edman et al. 1982; Leonard and Herzog 2010; Leonard et al. 2010). Force in the enhanced state can easily exceed the isometric force at the plateau of the force–length relationship (Abbott and Aubert 1952; Rassier et al. 2003c; Peterson et al. 2004; Leonard et al. 2010). RFE increases with increasing stretch magnitude (Bullimore et al. 2007; Hisey et al. 2009) but is independent of stretch speed (Edman et al. 1982), is associated with a substantial decrease in the metabolic cost of force production (Joumaa and Herzog 2013), is long-lasting (Abbott and Aubert 1952; Herzog and Leonard 2002; Leonard et al. 2010), and has a passive component that contributes substantially to the enhanced force (Fig. 3a, b) (Herzog and Leonard 2002, 2005; Herzog et al. 2003; Joumaa et al. 2007, 2008b).
Fig. 3.
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Residual force enhancement observed in whole muscle (cat soleus; a), single myofibrils (rabbit psoas; b), and single sarcomeres (rabbit psoas; c). Note the increase in force enhancement (FE; a) with increasing stretch magnitude, and the increased passive force [passive force enhancement (PFE)] following deactivation of the muscles after an active stretch (a, b). Note also the vast FE in a single sarcomere (c) and the substantially greater force after active stretching compared to the isometric, steady-state force prior to stretching which occurred at the plateau of the force–length relationship (2.4 μm)
RFE was first studied and described systematically in 1952 (Abbott and Aubert 1952), and it has subsequently been observed consistently across all structural levels of muscle, from entire muscles (Fig. 3a) (Abbott and Aubert 1952; Lee and Herzog 2002; Oskouei and Herzog 2005; Hahn et al. 2010; Seiberl et al. 2012; Fortuna et al. 2016), to single fibers (Edman et al. 1982; Sugi and Tsuchiya 1988; Peterson et al. 2004; Lee and Herzog 2008) and myofibrils (Fig. 3b) (Rassier et al. 2003a, b; Leonard and Herzog 2010; Leonard et al. 2010; Pun et al. 2010; Rassier and Pavlov 2012), and even in single sarcomeres (Fig. 3c) (Leonard et al. 2010).
For most of the twentieth century, RFE was explained by the instability of sarcomere lengths and the associated development of sarcomere length non-uniformities when muscles are stretched onto the descending limb of the force–length relationship (Morgan 1990, 1994; Edman and Tsuchiya 1996; Morgan et al. 2000; Morgan and Proske 2004, 2006). The concepts of the so-called sarcomere length non-uniformity theory have been well discussed (for review, see Morgan 1994; Herzog et al. 2016) and will not be repeated here, with the exception of experimental results that demonstrate that this theory cannot explain most of the fundamental RFE properties of skeletal muscles.
The most basic predictions of the sarcomere length non-uniformity theory that have been shown to be violated are that:
RFE cannot occur on the ascending limb of the force–length relationship—but it actually does (Abbott and Aubert 1952; Rassier et al. 2003c; Peterson et al. 2004; Lee and Herzog 2008);
Forces in the RFE state cannot exceed the purely isometric forces measured in a muscle at optimal length on the plateau of the force–length relationship—but they do (Fig. 4) (Rassier et al. 2003c; Lee and Herzog 2008; Leonard et al. 2010);
RFE (by definition of the sarcomere length non-uniformity theory) cannot occur in a single sarcomere—but it does (Figs. 3c, 4) (Leonard et al. 2010);
The isometric reference contractions are achieved with essentially uniform sarcomere lengths—but they are not (Fig. 2) (Llewellyn et al. 2008; Johnston et al. 2016; Moo et al. 2016);
There is a distinct increase in sarcomere length non-uniformity, resulting in two distinct groups of sarcomere lengths when a muscle is stretched actively—but that is not the case (Fig. 2) (Joumaa et al. 2008a; Johnston et al. 2016)
Fig. 4.
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Steady-state isometric forces obtained in single, mechanically isolated sarcomeres (rabbit psoas) at sarcomere lengths of 2.4 μm [optimal length = 100% force (filled brown circle)] and 3.4 μm [approximately 50% of maximal isometric force at the plateau length (filled blue diamonds and filled black square = mean force). Also shown are the isometric steady-state forces of these isolated sarcomeres following a stretch from 2.4 to 3.4 μm (filled green triangles and filled black circle). FE Mean force enhancement observed in these sarcomeres, OFE the mean force above the maximal, isometric plateau forces for these sarcomeres. Note the enormous force enhancement and the consistently greater forces in the enhanced state compared to the plateau force. Adapted from Leonard et al. (2010) with permission
For these reasons, it would appear that the sarcomere length non-uniformity theory has little direct support. It will not be discussed further in this review, but interested readers may want to consult the following references for a more in depth treatment of this theory (Morgan 1990, 1994; Herzog 2014a, b; Herzog et al. 2015).
RFE increases with increasing stretch magnitude (Edman et al. 1982; Bullimore et al. 2007; Hisey et al. 2009), is independent of stretch speed (Edman et al. 1982), and is long lasting (Abbott and Aubert 1952; Leonard and Herzog 2010); however, it can be abolished instantaneously if activation of the muscle is interrupted for long enough for the force to drop to zero (Morgan et al. 2000). These properties gave early rise to the notion that RFE might be caused by the engagement of an elastic structural element upon muscle activation (Forcinito et al. 1998). Because of its location, attachment, and mechanical properties, titin was an early favorite for this role (Noble 1992). Simple modeling of passive structural element engagements upon activation allowed for excellent qualitative predictions of the RFE properties (Forcinito et al. 1998). However, direct proof of a passive element playing a role in RFE was missing for a long time.
In 2002, our research group discovered in experiments with cat soleus muscle that RFE was associated with a distinct increase in passive force (Herzog and Leonard 2002). Specifically, we demonstrated that an active stretch resulted in a substantial increase in the passive force (following the active stretch and following deactivation of the muscle) compared to the passive forces measured when the muscle was activated isometrically at the corresponding length or stretched passively to the final length (Fig. 5) (Herzog and Leonard 2002). We termed this increase in passive force following active muscle stretching “passive force enhancement” (PFE) and showed that PFE was long-lasting and increased with stretch magnitude and with final muscle length, but that it could be abolished instantaneously by a quick release of the muscle to a short (prior to stretch) length (Herzog et al. 2003). These studies provided the first direct evidence that a passive component was likely contributing to the RFE property of skeletal muscles.
Fig. 5.
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Force–time histories of cat soleus muscle stretched passively (lowest trace at 6 s), stretched actively (highest trace at 6 s), and activated isometrically at the final stretch length (middle trace at 6 s). Note the increased passive force following muscle deactivation (at 12 s) for the actively stretched muscle, compared to the passively stretched muscle and the muscle activated isometrically at the final stretch length. The increase in passive force following active muscle stretching (here seen at 12 s) was termed passive force enhancement (PFE). Adapted from Herzog and Leonard (2002) with permission
Subsequently, PFE was also observed in isolated myofibrils and sarcomeres (Fig. 3b) (Joumaa et al. 2007, 2008b). Since titin is the primary passive force producing structure in myofibrils (it produces in excess of 95% of the passive force), titin became directly implicated in the mechanisms causing PFE and contributing substantially to the total RFE (Herzog et al. 2006; Leonard and Herzog 2010; Powers et al. 2014).
How might titin contribute to the residual force enhancement?
If titin were to contribute to the PFE and the RFE, its force for a given muscle/sarcomere length would need to be greater when a muscle is stretched actively compared to when it is stretched passively. Such an increase in force could be achieved if titin was stiffer in an active muscle than in a passive muscle. There are two basic mechanisms by which titin (or any molecular spring) can increase its stiffness: (1) it can increase its inherent stiffness (a change in material property) or (2) it can shorten its spring length (resting length) while its material property remains unaltered.
Changes in the inherent stiffness of titin upon activation may occur if calcium binds to titin, thereby increasing its stiffness and force upon stretching. Labeit et al. (2003) showed that there was an approximately 20% increase in non-cross-bridge-based force in skinned mouse soleus fibers activated with calcium ([PCa 4.0]) whose cross-bridge forces were inhibited by 2,3-butanedione monoxime (BDM) compared to fibers stretched passively ([PCa 9.0]). Single molecule experiments with PEVK fragments of titin suggested that specific E-rich motifs in PEVK can bind calcium, thereby becoming stiffer. Our research group showed, using fluorescence spectroscopy, that I27 cardiac immunoglobulin domains also bind calcium in a dose-dependent and reversible manner and demonstrated, with atomic force microscopy, that unfolding of these domains required 20–25% more force in the presence of physiologically relevant concentrations of calcium compared to the passive state with low calcium concentration [PCa 8] (Fig. 6) (DuVall et al. 2013). Further experiments in which single myofibrils were activated with calcium but whose cross-bridge forces were inhibited (with BDM and/or troponin C deletion) also showed an increase in titin-based force (Joumaa et al. 2008b; Leonard and Herzog 2010) compared to low calcium (passive) conditions, as did experiments in which myofibrils were stretched actively and passively beyond the actin–myosin filament overlap (Leonard and Herzog 2010; Powers et al. 2014). These experiments all led to the conclusion that titin is indeed a spring whose stiffnessis changed by calcium, and thus, muscle activation. Aside from calcium activation, titin’s spring stiffness can also be changed with phosphorylation (Yamasaki et al. 2002; Borbely et al. 2009; Anderson et al. 2010; Hudson et al. 2010; Perkin et al. 2015) and disulfide bridging (Scott et al. 2002), among others. In summary, titin is a molecular spring whose inherent stiffness can be modulated by muscle activation. However, changes in the inherent stiffness of titin only seem to account for up to about 20% of the maximal RFE observed experimentally under optimal conditions (Granzier 2010; Leonard and Herzog 2010). Therefore, there must be other mechanisms to explain the remainder of the RFE.
Fig. 6.
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Unfolding force of the first five (out of 8 identical) cardiac I27 immunoglobulin (Ig) domains of titin. Note that unfolding of the I27 Ig domains in the absence of calcium (Control) required about 20% less force than in the presence of calcium (Calcium). Adapted from DuVall et al. (2013) with permission
Another way of increasing titin’s spring stiffness that may potentially account for the full RFE observed experimentally, is a change in titin’s resting or free spring length. A decrease in free spring length could be achieved theoretically if some of titin’s extensible domains were to become inextensible—for example, by binding to a more rigid structure. In vitro, titin fragments have been found to bind to actin (Linke et al. 1997, 2002; Yamasaki et al. 2001; Nagy et al. 2004; Li et al. 1995; Trombitas and Granzier 1997; Astier et al. 1998; Kulke et al. 2001; Bianco et al. 2007; Chung et al. 2011;), thereby slowing the progress of actin–myosin filament sliding in motility assays. However, functionally relevant binding seems to be restricted to cardiac titin’s PEVK domain, which has been found to interact with actin in a calcium-dependent manner (Kulke et al. 2001; Yamasaki et al. 2001). Specifically, calcium seems to release actin from titin, thereby reducing titin-based passive force and stiffness and facilitating the cardiac cycle (Yamasaki et al. 2001). Permanent titin–actin binding has been shown to occur in the most proximal titin segments near the Z-band (Trombitas and Pollack 1993; Trombitas and Granzier 1997). Together with the relatively rigid association of titin with myosin, this titin link to actin results in a passive molecular spring that is arranged in parallel with attached cross-bridges, while it is in series with the myosin filament and the actin filament near its insertion into the Z-band in the relaxed muscle. However, in vitro assays have largely excluded strong binding of skeletal titin sub-fragments to actin (Kulke et al. 2001; Yamasaki et al. 2001).
Antibody labeling of titin demonstrates that titin is extensible along its entire I-band length (except for titin’s most proximal 50–100 nm) in passive muscle (Horowits et al. 1989; Trombitas et al. 2000; Minajeva et al. 2001; Linke and Fernandez 2002). Horowits et al. (1989) used monoclonal antibodies that bind in the I-band region of titin to observe distal and proximal (relative to the antibody) titin segment elongation in passive and active rabbit psoas fibers. Passive fibers were analyzed for sarcomere lengths ranging from 2.1 to 3.8 μm, while activated fibers were kept at a constant sarcomere length of 2.6 μm (fibers were activated for 5 min to produce A-band shifts to one side of the sarcomere, thus compressing one-half of the sacromere and extending the other half). These authors found that in both passive and active fibers, the lengths of the proximal and distal titin segments depended only on the length of the half-sarcomere, leading them to conclude that activation did not change the mechanical properties of titin (Horowits et al. 1989).
In contrast, DuVall et al. (2017), using a variety of I-band-specific titin antibodies, found that stretching of rabbit psoas myofibrils resulted in different segmental elongations in the active and passive conditions. While these authors replicated previous results for segmental titin elongations for passive myofibril stretching (Horowits et al. 1989), they observed that proximal segments of titin stopped elongating after a short stretch amplitude in active conditions (Fig. 7). They interpreted their results with an activation- and stretch-induced entanglement of titin with myosin or cross-bridges that rendered some of the extensible distal regions of titin inextensible. However, an equally valid explanation would be that titin’s proximal segments bind to actin after a short stretch, thereby rendering the proximal segments inextensible. Since elongations of proximal titin segments stop occurring at sarcomere lengths as short as 2.3 μm, which is a sarcomere length at which titin is known to be unstrained in rabbit psoas myofibrils, this latter explanation seems the more feasible of the two. If correct, such titin–actin interactions in actively stretched muscles change titin’s free spring length, thereby increasing titin’s stiffness and consequently its force upon sarcomere elongation (Herzog 2014a). Theoretical modeling of titin–actin interactions, as observed experimentally (DuVall et al. 2017), demonstrate that even the largest RFE observed experimentally (e.g., Leonard et al. 2010) can be explained using titin binding to actin in actively stretched muscle (Schappacher-Tilp et al. 2015).
Fig. 7.
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Passive (a) and active (b) stretching of proximal titin segments labeled using an antibody [F146 that binds to the PEVK region (diamond symbols)] region that allows for measurements of proximal and distal titin segment elongations during passive and active stretching of single rabbit psoas myofibrils. Figures on the left show elongation of the half-sarcomere [top traces (circular symbols) using an M-line label) and elongations of the proximal titin segment (bottom traces: from X-band to F146 label) for two representative sarcomeres from two different myofibrils. Note in a (passive stretching) that the two proximal titin segments elongate continuously with half-sarcomere elongations, reaching final lengths of approximately 0.95 μm (at a sarcomere length of 4.0 μm) and about 0.6 μm (at a sarcomere length of about 3.5 μm). In contrast, when the myofibrils are stretched while activated, the proximal segments elongate similarly to the elongations observed in the passive condition, but then stop elongating and remain substantially shorter than in the passive case (i.e., with a length of about 0.6 and 0.35 μm, respectively). The panels on the right illustrate schematically what we believe might be happening. In the passive stretch (a), the proximal and distal titin segments elongate in accordance with their stiffness properties. In the active stretch (b), titin is thought to attach to actin at some point, thereby shortening titin’s free spring length, increasing its stiffness,eliminating elongation of proximal titin, and increasing titin-based force
The results by DuVall et al. (2017) do not agree with those observed by Horowits et al. (1989). However, in the DuVall study (DuVall et al. 2017), some sarcomere stretching was required prior to the loss of proximal titin segment elongation, while in the Horowits study (Horowits et al. 1989), sarcomeres were kept at a constant isometric length. Also, in the DuVall study, segmental elongations were measured continuously during the active and passive stretching, while in the Horowits study, the active and passive fibers were fixed at specific lengths, and thus measurements of titin segment lengths were not continuous, but were made at a single (average) sarcomere length following fixation. It is not known how fixation might affect possible titin-to-actin binding in active fibers, and if indeed some sarcomere stretching is required to initiate titin–actin interaction; both these aspects might have been missed in the Horowits experiments.
Linke et al. (1996) used epitope tracking to determine the elongations of specific titin segments in passive rabbit cardiac myofibrils and in rabbit soleus and psoas myofibrils. Their results using an N2A-based epitope do not mimic those by DuVall et al. (2017) for passive stretching of titin segments in rabbit psoas. While they found that proximal titin segment elongation stopped at average sarcomere lengths of about 2.5 μm with little if any further elongation, DuVall et al. (2017) found continuous elongations of proximal titin segments up to a sarcomere length of 4.0 μm (Fig. 7). Leonard and Herzog (2010) showed that titin-based forces were much greater in psoas myofibrils stretched actively than in those stretched passively from optimal lengths to lengths beyond the actin–myosin filament overlap (Fig. 8). When cross-bridge interaction in the active [pCa 3.5] myofibrils was inhibited by BDM or troponin C deletion (Joumaa et al. 2008b), titin-based forces were still greater than in myofibrils under passive [pCa 8.5] conditions but much smaller than those when cross-bridge interactions with actin were allowed to occur normally (Leonard and Herzog 2010; Leonard et al. 2010). Starting active stretching from a longer length (3.4 μm) than optimal (2.3–2.5 μm) also decreased the titin-based forces compared to when stretches were initiated at optimal length (Leonard and Herzog 2010). Although the molecular mechanisms associated with segment-specific changes in titin’s mechanical properties need further elucidation, it appears that changes in the titin’s mechanical properties that occur with muscle activation and stretching contribute substantially to the experimentally observed RFE and PFE in skeletal muscles. Furthermore, the results by Leonard and Herzog (2010) suggest that it is not calcium activation that produces these substantial changes in passive/titin-based forces; rather, active cross-bridge binding is required to observe substantial force enhancement. A possible explanation could be that strong cross-bridge binding causes a movement of the regulatory proteins (troponin and tropomyosin) that may expose titin binding sites on actin that cannot be accessed by titin in the passive muscle or in calcium-activated muscle in which strong cross-bridge binding to actin is inhibited.
Fig. 8.
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Stress (force/cross-sectional area) versus sarcomere length relationship for single rabbit psoas myofibrils stretched from an average sarcomere length of 2.0 μm to 6.0 μm. Myofibrils were stretched passively (Passive), actively (Active), actively from an average sarcomere length of 3.4 μm (Half-Force), and after deletion of titin (No Titin). Actin myosin filament overlap is lost at an average sarcomere length of about 4.0 μm (indicated by the shaded area). According to the cross-bridge theory, one would expect the forces beyond actin myosin filament overlap (non-shaded area) to be purely passive and the same for all conditions with intact titin filaments. However, the forces in the actively stretched myofibrils were substantially greater than those for the passively stretched myofibrils in the area where myofilament overlap was lost. Deactivation of selected myofibrils at an average sarcomere length of 5.0 μm did not result in a loss of force (results not shown), indicating that there was no remnant cross-bridge-based force at these lengths. Elimination of titin from the myofibrils abolished all passive and all active force in myofibrils, indicating that titin is not only an essential protein for passive force production but is absolutely essential for active force transmission from the cross-bridges to the Z-bands and for centering the myosin filaments in the sarcomere. Adapted from Leonard and Herzog (2010) with permission
Stretching psoas sarcomeres to very long lengths, i.e., in excess of 5.0 μm, has been criticized for its potential to damage sarcomeres permanently by dislodging titin from its attachment to the Z-band or the myosin filament. However, Herzog et al. (2012) found that stretching of myofibrils to an average sarcomere length of > 5 μm (and individual sarcomeres within these myofibrils up to 6 μm) was not associated with permanent damage (i.e., loss of force). Rather, given sufficient time (about 10 min) and recovery at a very short average sarcomere length (1.8–2.0 μm), full recovery of titin-based passive force was observed, indicating that for stretches of sarcomeres up to 5.0 μm, there does not appear to be permanent damage to the structural network of (rabbit psoas) sarcomeres that would prevent full force recovery (Fig. 9). The fact that force recovery required several minutes and needed to be done at short sarcomere lengths (recovery for < 10 min or at sarcomere lengths averaging ≥ 2.6 μm was always incomplete) was interpreted as indicating that titin immunoglobulin unfolding, which would have taken place at these long sarcomere lengths (Kellermayer et al. 1997; Herzog et al. 2012), is slow and only occurs when forces acting on titin are essentially zero (Kellermayer et al. 1997).
Fig. 9.
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Examples of two separate rabbit psoas myofibrils that were repeatedly stretched to sarcomere lengths beyond actin myosin filament overlap. Note that in both cases repeat stretches did not result in a decrease in peak force or loading energy, thus indicating that even stretching to lengths up to 5.0 μm did not result in permanent damage and loss of force. a Myofibril stretched to an average sarcomere length of approximately 5.2 μm (stretch 1), then shortened and rested for 10 min at an average sarcomere length of 2.6 μm and re-stretched (stretch 2).Two sets of three stretch-shortening cycles were performed and the third stretch of the first set(stretch 1), and the first stretch of the second set(stretch 2)are shown b Myofibril stretched to an average sarcomere length of approximately 4.2 μm, then shortened and rested for 10 min at an average sarcomere length of approximately 1.8 μm. Two sets of three stretch-shortening cycles were performed (with a 10-min rest in between) and the first cycles (stretch 1) of the first and second set (stretch 2) are shown. Adapted from Herzog et al. (2014) with permission
This interpretation has been challenged recently where immunoglobulin refolding has been thought to occur at force levels acting on titin of about 8 pN, a relatively large force corresponding to the force exerted by about two to four attached cross-bridges (Rivas-Pardo et al. 2016). If these latest results receive further support, then the notion of unfolding and subsequent refolding of titin’s immunoglobulin domain having physiological relevance may have to be revisited.
Conclusion
Titin is a multi-purpose spring with many acknowledged mechanical functions, including the provision of passive force, stability of the myosin filaments, and stability of sarcomeres on the descending limb of the force–length relationship (Granzier et al. 1996, 2002; Linke et al. 1998; Granzier and Labeit 2002, 2007; Anderson et al. 2010; Linke and Kruger 2010). Less clear are its possible functions in explaining the RFE property of skeletal and cardiac muscle and the molecular details of its function as an adaptable molecular spring (Herzog et al. 2006, 2016; Herzog 2014a). Specifically, many unexplained phenomena in eccentric muscle action (increased force, RFE, decreased energetic requirements) can easily be accounted for within the framework of the cross-bridge theory, assuming that titin is an adaptable spring (Schappacher-Tilp et al. 2015). It will be an exciting challenge in this upcoming decade to elucidate the detailed molecular properties of titin in actively elongating muscle.
Compliance with ethical standards
Walter Herzog declares that he has no conflict of interest.
Ethics approvals for all experiments described in this study were obtained by the Life Sciences and Animal Research Ethics Commitee of the University of Calgary.
References
Abbott BC, Aubert XM. The force exerted by active striated muscle during and after change of length. J Physiol. 1952;117:77–86. [PMC free article] [PubMed] [Google Scholar]
Allinger TL, Epstein M, Herzog W. Stability of muscle fibers on the descending limb of the force- length relation. A theoretical consideration. J Biomech. 1996;29:627–633. doi: 10.1016/0021-9290(95)00087-9. [DOI] [PubMed] [Google Scholar]
Anderson BR, Bogomolovas J, Labeit S, Granzier HLM. The effects of PKCalpha phosphorylation on the extensibility of titin’s PEVK element. J Struct Biol. 2010;170:270–277. doi: 10.1016/j.jsb.2010.02.002. [DOI] [PMC free article] [PubMed] [Google Scholar]
Astier C, Raynaud F, Lebart MC, Roustan C, Benyamin Y. Binding of a native titin fragment to actin is regulated by PIP2. FEBS Lett. 1998;429:95–98. doi: 10.1016/s0014-5793(98)00572-9. [DOI] [PubMed] [Google Scholar]
Bartoo ML, Linke WA, Pollack GH. Basis of passive tension and stiffness in isolated rabbit myofibrils. Am J Phys. 1997;273:C266–C276. doi: 10.1152/ajpcell.1997.273.1.C266. [DOI] [PubMed] [Google Scholar]
Bianco P, Nagy A, Kengyel A et al (2007) Interaction forces between F-actin and titin PEVK domain measured with optical tweezers. Biophys J 93:2102–2109 [DOI] [PMC free article] [PubMed]
Borbely A, Falcao-Pires I, van Heerebeek L et al (2009) Hypophosphorylation of the stiff N2B titin isoform raises cardiomyocyte resting tension in failing human myocardium. Circ Res 104(6):780–786 [DOI] [PubMed]
Bullimore SR, Leonard TR, Rassier DE, Herzog W. History-dependence of isometric muscle force: effect of prior stretch or shortening amplitude. J Biomech. 2007;40:1518–1524. doi: 10.1016/j.jbiomech.2006.06.014. [DOI] [PubMed] [Google Scholar]
Campbell KS. Interactions between connected half-sarcomeres produce emergent mechanical behavior in a mathematical model of muscle. PLoS Comput Biol. 2009;5:e1000560. doi: 10.1371/journal.pcbi.1000560. [DOI] [PMC free article] [PubMed] [Google Scholar]
Cazorla O, Freiburg A, Helmes M et al (2000) Differential expression of cardiac tintin isoforms and modulation of cellular stiffness. Circ Res 86:59–67 [DOI] [PubMed]
Chung CS, Bogomolovas J, Gasch A et al (2011) Titin-actin interaction: PEVK-actin-based viscosity in a large animal. J Biomed Biotechnol 2011:310791 [DOI] [PMC free article] [PubMed]
Colomo F, Piroddi N, Poggesi C, Te KG, Tesi C. Active and passive forces of isolated myofibrils from cardiac and fast skeletal muscle of the frog. J Physiol. 1997;500(Pt 2):535–548. doi: 10.1113/jphysiol.1997.sp022039. [DOI] [PMC free article] [PubMed] [Google Scholar]
de Brito Fontana H, Herzog W. Vastus lateralis maximum force-generating potential occurs at optimal fascicle length regardless of activation level. Eur J Appl Physiol. 2016;116:1267–1277. doi: 10.1007/s00421-016-3381-3. [DOI] [PubMed] [Google Scholar]
DuVall MM, Gifford JL, Amrein M, Herzog W. Altered mechanical properties of titin immunoglobulin domain 27 in the presence of calcium. Eur Biophys J. 2013;42:301–307. doi: 10.1007/s00249-012-0875-8. [DOI] [PubMed] [Google Scholar]
DuVall M, Jinha A, Schappacher-Tilp G, Leonard T, Herzog W. Differences in Titin segmental elongation between passive and active stretch in skeletal muscle. J Exp Biol. 2017;220(Pt 23):4418–4425. doi: 10.1242/jeb.160762. [DOI] [PubMed] [Google Scholar]
Edman KAP, Tsuchiya T. Strain of passive elements during force enhancement by stretch in frog muscle fibres. J Physiol. 1996;490(1):191–205. doi: 10.1113/jphysiol.1996.sp021135. [DOI] [PMC free article] [PubMed] [Google Scholar]
Edman KAP, Elzinga G, Noble MIM. Residual force enhancement after stretch of contracting frog single muscle fibers. J Gen Physiol. 1982;80:769–784. doi: 10.1085/jgp.80.5.769. [DOI] [PMC free article] [PubMed] [Google Scholar]
Epstein M, Herzog W. Theoretical models of skeletal muscle:biological and mathematical considerations. New York: John Wiley & Sons Ltd.; 1998. [Google Scholar]
Forcinito M, Epstein M, Herzog W. Can a rheological muscle model predict force depression/enhancement? J Biomech. 1998;31:1093–1099. doi: 10.1016/s0021-9290(98)00132-8. [DOI] [PubMed] [Google Scholar]
Fortuna R, Power GA, Mende E, Seiberl W, Herzog W. Residual force enhancement following shortening is speed-dependent. Sci Rep. 2016;5:21513. doi: 10.1038/srep21513. [DOI] [PMC free article] [PubMed] [Google Scholar]
Freiburg A, Trombitas K, Hell W et al (2000) Series of exon-skipping events in the elastic spring region of titin as the structural basis for myofibrillar elastic diversity. Circ Res 86:1114–1121 [DOI] [PubMed]
Fukunaga T, Ichinose Y, Ito M, Kawakami Y, Fukashiro S. Determination of fascicle length and pennation in a contracting human muscle in vivo. J Appl Physiol. 1997;82:354–358. doi: 10.1152/jappl.1997.82.1.354. [DOI] [PubMed] [Google Scholar]
Funatsu T, Higuchi H, Ishiwata S. Elastic filaments in skeletal muscle revealed by selective removal of thin filaments with plasma gelsolin. J Cell Biol. 1990;110:53–62. doi: 10.1083/jcb.110.1.53. [DOI] [PMC free article] [PubMed] [Google Scholar]
Fusi L, Brunello E, Yan Z, Irving M. Thick filament mechano-sensing is a calcium-independent regulatory mechanism in skeletal muscle. Nat Commun. 2016;7:13281. doi: 10.1038/ncomms13281. [DOI] [PMC free article] [PubMed] [Google Scholar]
Gordon AM, Huxley AF, Julian FJ. The variation in isometric tension with sarcomere length in vertebrate muscle fibres. J Physiol. 1966;184:170–192. doi: 10.1113/jphysiol.1966.sp007909. [DOI] [PMC free article] [PubMed] [Google Scholar]
Granzier HLM. Activation and stretch-induced passive force enhancement—are you pulling my chain? Focus on “regulation of muscle force in the absence of actin-myosin-based cross-bridge interaction”. Am J Physiol Cell Physiol. 2010;299:C11–C13. doi: 10.1152/ajpcell.00147.2010. [DOI] [PMC free article] [PubMed] [Google Scholar]
Granzier HLM, Irving TC. Passive tension in cardiac muscle: contribution of collagen, titin, microtubules, and intermediate filaments. Biophys J. 1995;68:1027–1044. doi: 10.1016/S0006-3495(95)80278-X. [DOI] [PMC free article] [PubMed] [Google Scholar]
Granzier HLM, Labeit S. Cardiac titin: an adjustable multi-functional spring. J Physiol. 2002;541(2):335–342. doi: 10.1113/jphysiol.2001.014381. [DOI] [PMC free article] [PubMed] [Google Scholar]
Granzier HLM, Labeit S. Structure-function relations of the giant elastic protein titin in striated and smooth muscle cells. Muscle Nerve. 2007;36:740–755. doi: 10.1002/mus.20886. [DOI] [PubMed] [Google Scholar]
Granzier HLM, Trombitas K, Kellermayer MSZ, Helmes M, Stockman B (1996) Titin: a bi-directional spring and modulator of filament sliding. Proc Can Soc Biomech 9:10–11
Granzier HLM, Labeit D, Wu Y, Labeit S. Titin as a modular spring: emerging mechanisms for elasticity control by titin in cardiac physiology and pathophysiology. J Muscle Res Cell Motil. 2002;23:457–471. doi: 10.1023/a:1023458406346. [DOI] [PubMed] [Google Scholar]
Granzier HLM, Hutchinson KR, Tonino P et al (2014) Deleting titin’s I-band/A-band junction reveals critical roles for titin in biomechanical sensing and cardiac function. Proc Natl Acad Sci USA 111:14589–14594 [DOI] [PMC free article] [PubMed]
Hahn D, Seiberl W, Schmidt S, Schweizer K, Schwirtz A. Evidence of residual force enhancement for multi-joint leg extension. J Biomech. 2010;43:1503–1508. doi: 10.1016/j.jbiomech.2010.01.041. [DOI] [PubMed] [Google Scholar]
Herzog W. Mechanisms of enhanced force production in lengthening (eccentric) muscle contractions. J Appl Physiol. 2014;116:1407–1417. doi: 10.1152/japplphysiol.00069.2013. [DOI] [PubMed] [Google Scholar]
Herzog W. The role of titin in eccentric muscle contraction. J Exp Biol. 2014;217:2825–2833. doi: 10.1242/jeb.099127. [DOI] [PubMed] [Google Scholar]
Herzog W, Leonard TR. Force enhancement following stretching of skeletal muscle: a new mechanism. J Exp Biol. 2002;205:1275–1283. doi: 10.1242/jeb.205.9.1275. [DOI] [PubMed] [Google Scholar]
Herzog W, Leonard TR. The role of passive structures in force enhancement of skeletal muscles following active stretch. J Biomech. 2005;38:409–415. doi: 10.1016/j.jbiomech.2004.05.001. [DOI] [PubMed] [Google Scholar]
Herzog W, Schachar R, Leonard TR. Characterization of the passive component of force enhancement following active stretching of skeletal muscle. J Exp Biol. 2003;206:3634–3643. doi: 10.1242/jeb.00645. [DOI] [PubMed] [Google Scholar]
Herzog W, Lee EJ, Rassier DE. Residual force enhancement in skeletal muscle. J Physiol Lond. 2006;574:635–642. doi: 10.1113/jphysiol.2006.107748. [DOI] [PMC free article] [PubMed] [Google Scholar]
Herzog JA, Leonard TR, Jinha A, Herzog W. Are titin properties reflected in single myofibrils? J Biomech. 2012;45:1893–1899. doi: 10.1016/j.jbiomech.2012.05.021. [DOI] [PubMed] [Google Scholar]
Herzog JA, Leonard TR, Jinha A, Herzog W. Titin (visco-) elasticity in skeletal muscle myofibrils. MCB. 2014;11:1–17. [PubMed] [Google Scholar]
Herzog W, Powers K, Johnston K, DuVall M. A new paradigm for muscle contraction: review. Front Physiol. 2015;6:174–185. doi: 10.3389/fphys.2015.00174. [DOI] [PMC free article] [PubMed] [Google Scholar]
Herzog W, Schappacher G, DuVall M, Leonard TR, Herzog JA. Residual force enhancement following eccentric contractions: a new mechanism involving Titin. Physiology. 2016;31:300–312. doi: 10.1152/physiol.00049.2014. [DOI] [PubMed] [Google Scholar]
Hill AV. The mechanics of active muscle. Proc R Soc Lond. 1953;141:104–117. doi: 10.1098/rspb.1953.0027. [DOI] [PubMed] [Google Scholar]
Hisey B, Leonard TR, Herzog W. Does residual force enhancement increase with increasing stretch magnitudes? J Biomech. 2009;42:1488–1492. doi: 10.1016/j.jbiomech.2009.03.046. [DOI] [PubMed] [Google Scholar]
Horowits R. Passive force generation and titin isoforms in mammalian skeletal muscle. Biophys J. 1992;61:392–398. doi: 10.1016/S0006-3495(92)81845-3. [DOI] [PMC free article] [PubMed] [Google Scholar]
Horowits R, Podolsky RJ. The positional stability of thick filaments in activated skeletal muscle depends on sarcomere length: evidence for the role of titin filaments. J Cell Biol. 1987;105:2217–2223. doi: 10.1083/jcb.105.5.2217. [DOI] [PMC free article] [PubMed] [Google Scholar]
Horowits R, Podolsky RJ. Thick filament movement and isometric tension in activated skeletal muscle. Biophys J. 1988;54:165–171. doi: 10.1016/S0006-3495(88)82941-2. [DOI] [PMC free article] [PubMed] [Google Scholar]
Horowits R, Maruyama K, Podolsky RJ. Elastic behaviour of connecting filaments during thick filament movement in activated skeletal muscle. J Cell Biol. 1989;109:2169–2176. doi: 10.1083/jcb.109.5.2169. [DOI] [PMC free article] [PubMed] [Google Scholar]
Hudson BD, Hidalgo CG, Gotthardt M, Granzier HLM. Excision of titin’s cardiac PEVK spring element abolishes PKCalpha-induced increases in myocardial stiffness. J Mol Cell Cardiol. 2010;48:972–978. doi: 10.1016/j.yjmcc.2009.12.006. [DOI] [PMC free article] [PubMed] [Google Scholar]
Huxley HE. Electron microscope studies of the organization of the filaments in striated muscle. Biochim Biophys Acta. 1953;12:387–394. doi: 10.1016/0006-3002(53)90156-5. [DOI] [PubMed] [Google Scholar]
Huxley AF. Muscle structure and theories of contraction. Prog Biophys Biophys Chem. 1957;7:255–318. [PubMed] [Google Scholar]
Huxley HE. The double array of filaments in cross-striated muscle. Biochem Biophys Acta. 1957;12:387–394. doi: 10.1083/jcb.3.5.631. [DOI] [PMC free article] [PubMed] [Google Scholar]
Huxley HE. The mechanism of muscular contraction. Science. 1969;164:1356–1366. [PubMed] [Google Scholar]
Huxley AF. Reflections on muscle. Liverpool: Liverpool University Press; 1980. [Google Scholar]
Huxley HE, Hanson J. Changes in cross-striations of muscle during contraction and stretch and their structural implications. Nature. 1954;173:973–976. doi: 10.1038/173973a0. [DOI] [PubMed] [Google Scholar]
Huxley AF, Niedergerke R. Structural changes in muscle during contraction. Interference microscopy of living muscle fibres. Nature. 1954;173:971–973. doi: 10.1038/173971a0. [DOI] [PubMed] [Google Scholar]
Huxley AF, Peachey LD. The maximum length for contraction in vertebrate striated muscle. J Physiol Lond. 1961;156:150–165. doi: 10.1113/jphysiol.1961.sp006665. [DOI] [PMC free article] [PubMed] [Google Scholar]
Huxley AF, Simmons RM. Proposed mechanism of force generation in striated muscle. Nature. 1971;233:533–538. doi: 10.1038/233533a0. [DOI] [PubMed] [Google Scholar]
Ichinose Y, Kawakami Y, Ito M, Fukunaga T. Estimation of active force-length characteristics of human vastus lateralis muscle. Acta Anat (Basel) 1997;159:78–83. doi: 10.1159/000147969. [DOI] [PubMed] [Google Scholar]
Iwazumi T. In: Crossbridge mechanism in muscle contraction. Sugi H, Pollack GH, editors. Tokyo: University of Tokyo Press; 1979. pp. 611–632. [Google Scholar]
Iwazumi T, Noble M. An electrostatic mechanism of muscular contraction. Int J Cardiol. 1989;24:267–275. doi: 10.1016/0167-5273(89)90003-x. [DOI] [PubMed] [Google Scholar]
Johnston K, Jinha A, Herzog W. The role of sarcomere length non-uniformities in residual force enhancement of skeletal muscle myofibrils. Royal Soc Open Sci. 2016;3:150657. doi: 10.1098/rsos.150657. [DOI] [PMC free article] [PubMed] [Google Scholar]
Joumaa V, Herzog W. Energy cost of force production is reduced after active stretch in skinned muscle fibres. J Biomech. 2013;46:1135–1139. doi: 10.1016/j.jbiomech.2013.01.008. [DOI] [PubMed] [Google Scholar]
Joumaa V, Rassier DE, Leonard TR, Herzog W. Passive force enhancement in single myofibrils. Pflügers Arch. 2007;455:367–371. doi: 10.1007/s00424-007-0287-2. [DOI] [PubMed] [Google Scholar]
Joumaa V, Leonard TR, Herzog W. Residual force enhancement in myofibrils and sarcomeres. Proc R Soc B. 2008;275:1411–1419. doi: 10.1098/rspb.2008.0142. [DOI] [PMC free article] [PubMed] [Google Scholar]
Joumaa V, Rassier DE, Leonard TR, Herzog W. The origin of passive force enhancement in skeletal muscle. Am J Physiol Cell Physiol. 2008;294:C74–C78. doi: 10.1152/ajpcell.00218.2007. [DOI] [PubMed] [Google Scholar]
Julian FJ, Morgan DL. The effects of tension on non-uniform distribution of length changes applied to frog muscle fibres. J Physiol. 1979;293:379–392. doi: 10.1113/jphysiol.1979.sp012895. [DOI] [PMC free article] [PubMed] [Google Scholar]
Julian FJ, Sollins MR, Moss RL. Sarcomere length non-uniformity in relation to tetanic response of stretched skeletal muscle fibres. Proc R Soc Lond B. 1978;200:109–116. doi: 10.1098/rspb.1978.0009. [DOI] [PubMed] [Google Scholar]
Kellermayer MSZ, Smith SB, Granzier HLM, Bustamante C. Folding-unfolding transitions in single titin molecules characterized with laser tweezers. Science. 1997;276:1112–1116. doi: 10.1126/science.276.5315.1112. [DOI] [PubMed] [Google Scholar]
Kruger M, Linke WA. Titin-based mechanical signaling in normal and failing myocardium. J Mol Cell Cardiol. 2009;46(4):490–498. doi: 10.1016/j.yjmcc.2009.01.004. [DOI] [PubMed] [Google Scholar]
Kulke M, Fuijita-Becker S, Rostkova E et al (2001) Interaction between PEVK-titin and actin filaments: origin of a viscous force component in cardiac myofibrils. Circ Res 89:874–881 [DOI] [PubMed]
Labeit D, Watanabe K, Witt C et al (2003) Calcium-dependent molecular spring elements in the giant protein titin. Proc Natl Acad Sci USA 100:13716–13721 [DOI] [PMC free article] [PubMed]
Lee HD, Herzog W. Force enhancement following muscle stretch of electrically and voluntarily activated human adductor pollicis. J Physiol. 2002;545:321–330. doi: 10.1113/jphysiol.2002.018010. [DOI] [PMC free article] [PubMed] [Google Scholar]
Lee EJ, Herzog W. Residual force enhancement exceeds the isometric force at optimal sarcomere length for optimized stretch conditions. J Appl Physiol. 2008;105:457–462. doi: 10.1152/japplphysiol.01109.2006. [DOI] [PubMed] [Google Scholar]
Leonard TR, Herzog W. Regulation of muscle force in the absence of actin-myosin based cross-bridge interaction. Am J Physiol Cell Physiol. 2010;299:C14–C20. doi: 10.1152/ajpcell.00049.2010. [DOI] [PubMed] [Google Scholar]
Leonard TR, Duvall M, Herzog W. Force enhancement following stretch in a single sarcomere. Am J Physiol Cell Physiol. 2010;299(6):C1398–C1401. doi: 10.1152/ajpcell.00222.2010. [DOI] [PubMed] [Google Scholar]
LeWinter MM, Granzier HLM. Cardiac titin: a multifunctional giant. Circulation. 2010;121:2137–2145. doi: 10.1161/CIRCULATIONAHA.109.860171. [DOI] [PMC free article] [PubMed] [Google Scholar]
Li Q, Jin J-P, Granzier HLM. The effect of genetically expressed cardiac titin fragments on in vitro actin motility. Biophys J. 1995;69:1508–1518. doi: 10.1016/S0006-3495(95)80021-4. [DOI] [PMC free article] [PubMed] [Google Scholar]
Linke WA, Fernandez JM. Cardiac titin: molecular basis of elasticity and cellular contribution to elastic and viscous stiffness components in myocardium. J Muscle Res Cell Motil. 2002;23:483–497. doi: 10.1023/a:1023462507254. [DOI] [PubMed] [Google Scholar]
Linke WA, Kruger M. The giant protein titin as an integrator of myocyte signaling pathways. Physiology (Bethesda) 2010;25:186–198. doi: 10.1152/physiol.00005.2010. [DOI] [PubMed] [Google Scholar]
Linke WA, Popov VI, Pollack GH. Passive and active tension in single cardiac myofibrils. Biophys J. 1994;67:782–792. doi: 10.1016/S0006-3495(94)80538-7. [DOI] [PMC free article] [PubMed] [Google Scholar]
Linke WA, Ivemeyer M, Olivieri N et al (1996) Towards a molecular understanding of the elasticity of titin. J Mol Biol 261:62–71 [DOI] [PubMed]
Linke WA, Ivemeyer M, Labeit S et al (1997) Actin-titin interaction in cardiac myofibrils: probing a physiological role. Biophys J 73:905–919 [DOI] [PMC free article] [PubMed]
Linke WA, Ivemeyer M, Mundel P, Stockmeier MR, Kolmerer B. Nature of PEVK-titin elasticity in skeletal muscle. Proc Natl Acad Sci U S A. 1998;95:8052–8057. doi: 10.1073/pnas.95.14.8052. [DOI] [PMC free article] [PubMed] [Google Scholar]
Linke WA, Kulke M, Li H et al (2002) PEVK domain of titin: an entropic spring with actin-binding properties. J Struct Biol 137:194–205 [DOI] [PubMed]
Liversage AD, Holmes D, Knight PJ, Tskhovrebova L, Trinick J. Titin and the sarcomere symmetry paradox1. J Mol Biol. 2001;305:401–409. doi: 10.1006/jmbi.2000.4279. [DOI] [PubMed] [Google Scholar]
Llewellyn ME, Barretto RPJ, Delp SL, Schnitzer MJ. Minimally invasive high-speed imaging of sarcomere contractile dynamics in mice and humans. Nature. 2008;454:784–788. doi: 10.1038/nature07104. [DOI] [PMC free article] [PubMed] [Google Scholar]
Maruyama K. Connectin, an elastic protein from myofibrils. J Biochem. 1976;80:405–407. doi: 10.1093/oxfordjournals.jbchem.a131291. [DOI] [PubMed] [Google Scholar]
Maruyama K, Kimura S, Kuroda M, Handa S. Connectin, an elastic protein of muscle. J Biochem. 1977;82:347–350. [PubMed] [Google Scholar]
Minajeva A, Kulke M, Fernandez JM, Linke WA. Unfolding of Titin domains explains the Viscoelastic behavior of skeletal myofibrils. Biophys J. 2001;80(3):1442–1451. doi: 10.1016/S0006-3495(01)76116-4. [DOI] [PMC free article] [PubMed] [Google Scholar]
Moo EK, Fortuna R, Sibole SC, Abusara Z, Herzog W. In vivo sarcomere lengths and sarcomere elongations are not uniform across an intact muscle. Front Physiol. 2016;7:1–9. doi: 10.3389/fphys.2016.00187. [DOI] [PMC free article] [PubMed] [Google Scholar]
Morgan DL. New insights into the behavior of muscle during active lengthening. Biophys J. 1990;57:209–221. doi: 10.1016/S0006-3495(90)82524-8. [DOI] [PMC free article] [PubMed] [Google Scholar]
Morgan DL. An explanation for residual increased tension in striated muscle after stretch during contraction. Exp Physiol. 1994;79:831–838. doi: 10.1113/expphysiol.1994.sp003811. [DOI] [PubMed] [Google Scholar]
Morgan DL, Proske U. Popping sarcomere hypothesis explains stretch-induced muscle damage. Clin Exp Pharmacol Physiol. 2004;31:541–545. doi: 10.1111/j.1440-1681.2004.04029.x. [DOI] [PubMed] [Google Scholar]
Morgan DL, Proske U. Can all residual force enhancement be explained by sarcomere non-uniformities? J Physiol. 2006;578(2):613–615. doi: 10.1113/jphysiol.2006.125039. [DOI] [PMC free article] [PubMed] [Google Scholar]
Morgan DL, Whitehead NP, Wise AK, Gregory JE, Proske U. Tension changes in the cat soleus muscle following slow stretch or shortening of the contracting muscle. J Physiol. 2000;522(3):503–513. doi: 10.1111/j.1469-7793.2000.t01-2-00503.x. [DOI] [PMC free article] [PubMed] [Google Scholar]
Nagy A, Cacciafesta P, Grama L et al (2004) Differential actin binding along the PEVK domain of skeletal muscle titin. J Cell Sci 117:5781–5789 [DOI] [PubMed]
Noble MIM. Enhancement of mechanical performance of striated muscle by stretch during contraction. Exp Physiol. 1992;77:539–552. doi: 10.1113/expphysiol.1992.sp003618. [DOI] [PubMed] [Google Scholar]
Novak I, Truskinovsky L. Nonaffine response of skeletal muscles on the ‘descending limb’. Math Mech Solids. 2014;20:1–24. [Google Scholar]
Oskouei AE, Herzog W. Observations on force enhancement in sub-maximal voluntary contractions of human adductor pollicis muscle. J Appl Physiol. 2005;98:2087–2095. doi: 10.1152/japplphysiol.01217.2004. [DOI] [PubMed] [Google Scholar]
Perkin J, Slater R, Del Favero G et al (2015) Phosphorylating Titin’s cardiac N2B element by ERK2 or CaMKII delta lowers the single molecule and cardiac muscle force. Biophys J 109:2592–2601 [DOI] [PMC free article] [PubMed]
Peterson D, Rassier DE, Herzog W. Force enhancement in single skeletal muscle fibres on the ascending limb of the force-length relationship. J Exp Biol. 2004;207:2787–2791. doi: 10.1242/jeb.01095. [DOI] [PubMed] [Google Scholar]
Powers K, Schappacher-Tilp G, Jinha A et al (2014) Titin force is enhanced in actively stretched skeletal muscle. J Exp Biol 217:3629–3636 [DOI] [PubMed]
Prado LG, Makarenko I, Andresen C et al (2005) Isoform diversity of giant proteins in relation to passive and active contractile properties of rabbit skeletal muscles. J Gen Physiol 126:461–480 [DOI] [PMC free article] [PubMed]
Pun C, Syed A, Rassier DE. History-dependent properties of skeletal muscle myofibrils contracting along the ascending limb of the force–length relationship. Proc Biol Sci. 2010;277:475–484. doi: 10.1098/rspb.2009.1579. [DOI] [PMC free article] [PubMed] [Google Scholar]
Rassier DE, Pavlov I. Force produced by isolated sarcomeres and half-sarcomeres after an imposed stretch. Am J Physiol Cell Physiol. 2012;302:C240–C248. doi: 10.1152/ajpcell.00208.2011. [DOI] [PubMed] [Google Scholar]
Rassier DE, Herzog W, Pollack GH. Stretch-induced force enhancement and stability of skeletal muscle myofibrils. Adv Exp Med Biol. 2003;538:501–515. doi: 10.1007/978-1-4419-9029-7_45. [DOI] [PubMed] [Google Scholar]
Rassier DE, Herzog W, Pollack GH. Dynamics of individual sarcomeres during and after stretch in activated single myofibrils. Proc R Soc Lond B. 2003;270:1735–1740. doi: 10.1098/rspb.2003.2418. [DOI] [PMC free article] [PubMed] [Google Scholar]
Rassier DE, Herzog W, Wakeling JM, Syme D. Stretch-induced, steady-state force enhancement in single skeletal muscle fibers exceeds the isometric force at optimal fibre length. J Biomech. 2003;36:1309–1316. doi: 10.1016/s0021-9290(03)00155-6. [DOI] [PubMed] [Google Scholar]
Rayment I, Holden HM, Whittaker M et al (1993) Structure of the actin-myosin complex and its implications for muscle contraction. Science 261:58–65 [DOI] [PubMed]
Rivas-Pardo JA, Eckels EC, Popa I et al (2016) Work done by Titin protein folding assists muscle contraction. Cell Rep 14:1–9 [DOI] [PMC free article] [PubMed]
Schappacher-Tilp G, Leonard T, Desch G, Herzog W. A novel three-filament model of force generation in eccentric contraction of skeletal muscles. PLoS One. 2015;10:e0117634. doi: 10.1371/journal.pone.0117634. [DOI] [PMC free article] [PubMed] [Google Scholar]
Schwarz ML, Witt SH, Schneider-Wald B et al (2008) Titin expression in human articular cartilage and cultured chondrocytes: a novel component in articular cartilage biomechanical sensing? Biomed Pharmacother 62:339–347 [DOI] [PubMed]
Scott KA, Steward A, Fowler SB, Clarke J. Titin: a multidomain protein that behaves as the sum of its parts. J Mol Biol. 2002;315:819–829. doi: 10.1006/jmbi.2001.5260. [DOI] [PubMed] [Google Scholar]
Seiberl W, Hahn D, Herzog W, Schwirtz A. Feedback controlled force enhancement and activation reduction of voluntarily activated quadriceps femoris during sub-maximal muscle action. J Electromyogr Kinesiol. 2012;22:117–123. doi: 10.1016/j.jelekin.2011.10.010. [DOI] [PubMed] [Google Scholar]
Stoecker U, Telley IA, Stüssi E, Denoth J. A multisegmental cross-bridge kinetics model of the myofibril. J Theor Biol. 2009;259:714–726. doi: 10.1016/j.jtbi.2009.03.032. [DOI] [PubMed] [Google Scholar]
Sugi H, Tsuchiya T. Stiffness changes during enhancement and deficit of isometric force by slow length changes in frog skeletal muscle fibres. J Physiol Lond. 1988;407:215–229. doi: 10.1113/jphysiol.1988.sp017411. [DOI] [PMC free article] [PubMed] [Google Scholar]
Ter Keurs HE, Rijnsburger WH, van Heuningen R, Nagelsmit MJ. Tension development and sarcomere length in rat cardiac trabeculae: evidence of length-dependent activation. Circ Res. 1980;46:703–714. doi: 10.1161/01.res.46.5.703. [DOI] [PubMed] [Google Scholar]
Trombitas K, Granzier HLM. Actin removal from cardiac myocytes shows that near Z line titin attaches to actin while under tension. Am J Phys. 1997;273:C662–C670. doi: 10.1152/ajpcell.1997.273.2.C662. [DOI] [PubMed] [Google Scholar]
Trombitas K, Pollack GH. Elastic properties of the titin filament in the z-line region of vertebrate striated-muscle. J Muscle Res Cell Motil. 1993;14:416–422. doi: 10.1007/BF00121293. [DOI] [PubMed] [Google Scholar]
Trombitas K, Redkar A, Centner T et al (2000) Extensibility of isoforms of cardiac titin: variation in contour length of molecular subsegments provides a basis for cellular passive stiffness diversity. Biophys J 79:3226–3234 [DOI] [PMC free article] [PubMed]
Vaz MA, de la Rocha FC, Leonard T, Herzog W. The force-length relationship of the cat soleus muscle. Muscles Ligaments Tendons J. 2012;2:79–84. [PMC free article] [PubMed] [Google Scholar]
Walcott S, Herzog W. Modeling residual force enhancement with generic cross-bridge models. Math Biosci. 2008;216:172–186. doi: 10.1016/j.mbs.2008.10.005. [DOI] [PubMed] [Google Scholar]
Wang K, Mcclure J, Tu A. Titin: major myofibrillar components of striated muscle. Proc Natl Acad Sci USA. 1979;76:3698–3702. doi: 10.1073/pnas.76.8.3698. [DOI] [PMC free article] [PubMed] [Google Scholar]
Yamasaki R, Berri M, Wu Y et al (2001) Titin-Actin interaction in mouse myocardium: passive tension modulation and its regulation by calcium/S100A1. Biophys J 81:2297–2313 [DOI] [PMC free article] [PubMed]
Yamasaki R, Wu Y, McNabb M et al (2002) Protein kinase a phosphorylates titin’s cardiac-specific N2B domain and reduces passive tension in rat cardiac myocytes. Circ Res 90:1181–1188 [DOI] [PubMed]
Yasuda K, Shindo Y, Ishwata S. Synchronous behavior of spontaneous oscillations of sarcomeres in skeletal myofibrils under isotonic conditions. Biophys J. 1996;70:1823–1829. doi: 10.1016/S0006-3495(96)79747-3. [DOI] [PMC free article] [PubMed] [Google Scholar]
Zahalak GI. Can muscle fibers be stable on the descending limbs of their sarcomere length–tension relations? J Biomech. 1997;30:1179–1182. doi: 10.1016/s0021-9290(97)00079-1. [DOI] [PubMed] [Google Scholar]
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15323 | https://www.youtube.com/watch?v=VeFF0_WLlA4 | An introduction to coordination compounds | Chapter 7 - Shriver & Atkins’ Inorganic Chemistry (5th)
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Chapter 7 of Shriver & Atkins’ Inorganic Chemistry (Fifth Edition) introduces the fundamental language, structures, and thermodynamics of coordination chemistry. It begins by defining complexes as central metal atoms or ions bonded to ligands, distinguishing between inner-sphere complexes (directly bonded ligands) and outer-sphere associations. Representative ligands are presented, including monodentate (NH₃, H₂O, Cl⁻), polydentate (en, bpy, edta⁴⁻), ambidentate (SCN⁻, NO₂⁻), and chelating agents that form stable rings. The chapter also reviews nomenclature rules established by IUPAC, showing how ligands are named in alphabetical order, how oxidation states are indicated, and how bridging ligands are denoted with μ. Structural classification follows, covering low coordination numbers such as linear [AgCl₂]⁻, trigonal [Pt(PCy₃)₃], and tetrahedral [CoCl₄]²⁻, as well as square-planar d⁸ complexes like [Pt(NH₃)₄]²⁺ and porphyrins. Five-coordinate species (trigonal bipyramidal, square pyramidal) and six-coordinate complexes (primarily octahedral with cis/trans, fac/mer, and distorted variants) are discussed in detail, along with higher coordination numbers (7–12) common in d- and f-block metals. Polymetallic complexes are introduced as clusters (with M–M bonds) and cages (ligand-bridged), with examples ranging from Hg₂Cl₂ to Fe₄S₄ biological cofactors. Isomerism is a major focus: linkage, ionization, hydrate, coordination, geometric (cis/trans, fac/mer), and optical isomerism are explained, with illustrations of tetrahedral chirality, square-planar cis/trans arrangements, octahedral enantiomers like [Co(en)₃]³⁺, and Δ/Λ stereochemistry. Ligand chirality and resolution of enantiomers are also covered. The thermodynamics of complex formation introduces stepwise and overall formation constants (Kf), dissociation constants (Kd), and their relation to Gibbs free energy. Trends in successive constants are analyzed statistically and with examples such as [Ni(NH₃)₆]²⁺ and [Fe(bpy)₃]²⁺. The chelate effect and macrocyclic effect are explained as entropic and enthalpic stabilizations of multidentate ligands, with practical importance in biology and analytical chemistry. Steric effects, bite angles, and electronic delocalization in chelating diimine ligands (bpy, phen) are emphasized, along with the stability of complexes such as [Ru(bpy)₃]²⁺. By covering ligands, geometries, isomerism, and thermodynamics, this chapter provides a complete framework for understanding the diversity and stability of coordination compounds.
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Shriver and Atkins Inorganic Chemistry Chapter 7 summary, introduction to coordination compounds explained, ligands monodentate polydentate ambidentate chelating, en edta bipyridine common ligands, coordination number geometry complexes, linear trigonal tetrahedral square planar octahedral trigonal bipyramidal square pyramidal, higher coordination numbers d-block f-block chemistry, polymetallic complexes clusters cages metal-metal bonds, Werner coordination chemistry history, linkage ionization hydrate coordination isomerism, cis trans fac mer isomers coordination complexes, optical isomerism chirality in octahedral complexes Δ Λ stereochemistry, chelate effect entropy stabilization, macrocyclic effect crown ethers porphyrins, bite angle steric effects ligand design, electronic delocalization diimine bpy phen ligands, thermodynamics formation constants Kf Kd stepwise constants, statistical trends in successive Kf values, stability of [Fe(bpy)3]2+ [Ni(NH3)6]2+, coordination chemistry summary inorganic
Transcript:
Welcome to the Deep Dive, your ultimate shortcut to truly understanding complex topics. Today, we're unlocking one of the most foundational and frankly pretty fascinating areas in chemistry, coordination compounds. Mhm. If you've ever wondered how metals do more than just, you know, form simple salts, or maybe you're navigating your way through inorganic chemistry, this deep dive is definitely for you. Absolutely. We're going to use a chapter from Shrivever and Atkins's Inorganic Chemistry fifth edition as our guide today. Basically, we're turning that dense textbook stuff into clear, hopefully captivating insights. Yeah. The goal is really to break it down. Exactly. Our mission to help you grasp these intricate ideas step by step without needing a single visual aid. And you know, this isn't just abstract theory. Coordination compounds are well, they're everywhere. Think catalysts in industry or even hemoglobin carrying oxygen in your blood, right? Understanding their structures, how they bond, it's like getting a secret decoder ring for a huge chunk of the chemical world, especially uh for those DB block elements, the transition metals. Okay, so let's start right at the beginning then. What is a coordination compound fundamentally and why should we really care about them? Great place to start. So at its core, a complex in chemistry, well, it describes a central metal atom or ion, often a transition metal that's surrounded by a group of other ions or molecules, okay? And these surrounding partners. We call them lians. Now, a lian is special because sure, it can exist on its own, but it also has at least one atom with a pair of electrons just ready to bond with that central metal. Ah, so it's like the ligan is bringing the electrons to the party. Exactly. Think of the metal as a lis acid and electron acceptor and the lian is the Louis bas the electron donor. the specific atom on the lian doing the donating like the nitrogen and ammonia or the oxygen and water that's the donor atom and a coordination compound itself right so that's either a neutral complex like n4 nickel tetracarbonel or it's an ionic compound where at least one of the ions is a complex like co- NH3 the cobalt bit inside the brackets is the complexation it's amazing how much of this basic picture was pieced together by Alfred Verer what over a century ago it really is remarkable I mean figuring out these 3D structures using just things like how they reacted or if they had isomers, same formula, different structure, or even just measuring electrical conductance all before we could really see atoms. Yeah, his deductive work was groundbreaking. Today, we've got powerful tools like X-ray defraction for precise structures, NMR for solution behavior, but Verer laid that foundation using pure chemical reasoning. Okay, so let's unpack the language a bit more. You mentioned a complex. Is it always about direct chemical bonds? Good question. Mostly when we talk about complexes, we mean inner sphere complexes. That's where the lians are directly chemically bonded to the metal. This forms the uh primary coordination sphere, right? The number of these direct bonds. That's the coordination number. And it can range quite a bit, you know, from two up to maybe 12 or even more. That's where a lot of the structural variety comes from. And the alternative, sometimes you get outer sphere complexes. This is more like well a charged complex ion just hanging out near its counterion due to electrostatic attraction. No direct bond just proximity. Most methods measure the sum of both. So it's a subtle point but worth knowing. Okay so coordination number tells us how many lians but what kind of lians are there? What kinds of partners do metals like? Lyans are often classified by how many points of attachment or uh donor atoms they use to bind to the metal. Mhm. The simplest are monotate one tooth like ammonia or chloride using just one atom just one connection. Yep. But many are polyate med teeth. If it has two donor atoms it's bidate. A classic example is ethylene adam and which we often just call N. Three points it's tridentate. So sense and then there are ambidentate lians. These are pretty clever. They have more than one potential donor atom but usually only use one at a time. Like the thioinate ion NCS. It could bind via the nitrogen which we'd write using this kappa notation like keem or it can flip around and bind through the sulfur heat mess. Same lian different connection point potentially different properties for the complex that kappa thing is neat tells you exactly how it's connected precisely. Yeah. Now when a polyentate lian uses more than one of its donor atoms to bind to the same metal ion forming a ring structure that includes the metal that's called a tillite like a claw you mentioned. Exactly from the Greek for claw. Ethylinine N is a perfect example. It's two nitrogens grab onto the metal forming a nice stable five membered ring. Ah, I see the ring structure. And some lians are amazing. Chillers. Think EDTA ethyleneate. It's hexodate. Six donor atoms. It can wrap around a metal ion completely forming multiple rings. It's why it's so good at say removing metal ions from hard water. Wow. Six points of contact. Yeah. And the geometry matters too. The angle formed by the lian metal lian bonds within that chillet ring. That's called the bite angle. If it's too strained, too tight, or too wide, the chill might not be as stable. Okay, this complexity definitely needs a clear naming system. How on earth do we name these things consistently? It seems like it could get messy fast. It can seem daunting, but there's a logic to it governed by IUCK rules. The key is clarity. First rule, just like simple salts, you name the IUP pay first, then the onion. Standard ionic compound rule. Okay. Then when you're naming the complex ion itself, you list the lians alphabetically. And here's a key point. You ignore prefixes like dri tetra when alphabetizing. Ah, so damine comes before chloro. Okay. Right. After naming all the lians, you state the name of the central metal and immediately after that you put its oxidation state in Roman numerals in parenthesis like cobalt 3. Got it. Ligan's alphabetical then metal then charge number. Exactly. Now there's a twist. If the entire complex ion has a negative charge, an overall negative charge, you modify the metal's name by adding the suffix 8. Sometimes you even use the Latin root like iron becomes ferate, copper becomes cuprate. Okay. So, FCN64 would be hexacyopate 2. Perfect. Hexa for six cyanol lians for because the complex is an annion and two for irons plus two oxidation state. What about multiple identical lians? For simple lians, you use the standard prefixes driet penta hexa. But if the lian name itself is complicated or already contains a prefix like ethylene edamine, we use different prefixes. bis for two, tries for three, to tryusets for four, and you put the liyan name in parenthesis. Okay, so tresthylamine cobalt 3. You got it. One last thing. If a lyan bridges between two metal centers, you use the prefix moo before its name. And always, always use square brackets in the formula to enclose the metal and the lians directly bonded to it. Metal symbol first, then lians alphabetically by symbol. Right, let me try one. What about pt2 NH342 plus lian? Okay, let's break it down. You've got four amine lians, NH3, and two chloro lians, CLA. Alphabetically, amine comes before chloro. The metal is platinum. To get the plus2 overall charge with two negative chlorides, the platinum must be plus four. Right? So, putting it together, tetramine 4 NH3, the chlorido 2 Clin tetrain chloro platinum, tetramini to chloro platinum. Okay, the system works. It does. Now, let's think about the actual shapes. What determines the constitution in geometry? That coordination number isn't always obvious just from a formula, right? Sometimes solvent is around but not bonded, right? Well, there are basically three main factors that influence the coordination number and thus the shape. First, the size of the central metal ion. Generally, larger ions can accommodate more lians around them. Makes sense, right? More space. Yeah. Physically larger can fit more stuff. Second, steric interactions. How bulky the lians themselves are. If you have really big bushy lians, they're going to crowd each other out and that favors a lower coordination number. They just can't all fit. Like trying to park too many big cars in a small lot. Exactly. And third, electronic factors. This relates to the metal's d electron count and its ability to accept electron density from the lagans. Metals with fewer d electrons, like those on the left of the dlock, often prefer higher coordination numbers than electron-rich metals on the right. Okay. size, crowding, and electronics. So, taking those into account, what are the common shapes these complexes actually adopt? Great question. The geometry is directly tied to the coordination number. Makes sense. For two coordination, it's pretty rare for isolated complexes, but when it happens, it's almost always linear. Yeah. Think of the metal in the middle and the two lians stretched out at 180° like AgCl2, a straight line. Got it. Three coordination is even rare. usually needs really bulky lians to prevent more from attaching. Typically, it's trional plane or flat like a triangle with the metal at the center. Four coordination is where things get really common and interesting. Two main geometries here. First is tetrahedral. Metal in the center, four lians pointing to the corners of a tetrahedrin. Think methane but with a metal. Very common for non-transition metals and some transition metal ions like M4, ZN42, the classic tetrahedral shape, right? But the other major four coordinate geometry is square planer. Here the metal and the four lians are all in the same plane forming a square. This is super important especially for D8 metal ions like platinum 2, palladium 2, gold 3. Think of the pferin and ring in heem or chlorophyll that metal sits in a square planer environment. Ah so flat versus 3D tetrahedral big difference huge difference. Then comes five coordination. It's less common than four or six and often structurally well flexible. The two main shapes are square pyramidal like a pyramid with a square base and trional bip pyramidal like two triangular pyramids joined at the base. And the interesting thing is these two shapes often interconvert really easily. It's a phenomenon called flexctionality. At room temperature the lians might seem to swap places rapidly between say the axial and equatorial positions in the trigonal by pyramid. It's like a molecular dance. We call one specific pathway the berry pseuda rotation. So they don't just sit still. Fascinating. What about six? Ah, six coordination. This is the most common coordination number by far. And the geometry is overwhelmingly ocahedral. Picture the metal at the center and six lians sitting at the vertices of an ocahedron like one pointing up, one down, and four around the equator. So workhorse geometry. Absolutely. Now even octahedra aren't always perfect. They can get distorted. A common one is tetraonal distortion where maybe the two lians along one axis are closer or further away than the four in the equatorial plane. There's also the famous John Teller distortion which must happen for certain electron configurations like D9 copper to remove electronic degeneracy. It has real chemical consequences. So even the most common shape has variations. Definitely. And while octaedral rule very rarely you might find a trigonal prismatic geometry for six coordination too. What about higher numbers? You mentioned up to 12. Yeah, higher coordination numbers 7 8 9 and even up to 12 become more common for larger metal ions especially the lanthnides and actonides the flock elements their size just allows more lians to pack around them so size is key there it really is seven coordination might be a pentagonal by pyramid or a cap ocahedron eight can be a square antiprism or a dodcied nine is actually very common for the early lanthnides in water like endoh293 plus compen 10 and 12 are rarer but known. Incredible range of shapes. Now, what if you have more than one metal atom involved? Good point. Polytallic complexes. We generally distinguish between two main types. Metal clusters have direct metaltometal bonds. Think of multiple metal atoms huddled together bonded to each other as well as to lians. Hg22+ is a simple example. Okay. Metals bonded to metals. All right. The other type is cage complexes. Here you have multiple metal atoms, but they're held together only by bridging lians. lians that bond to two or more metals simultaneously. There are no direct metal metal bonds. Many biological systems like iron sulfur proteins involved in electron transfer use these cage structures. Okay. Clusters have mm bonds. Cages don't. Got it. So clearly just knowing the formula like co-nh352 isn't enough. How do chemists deal with compounds that have the same formula but different structures? That's the whole concept of isomeism. Same formula, different arrangement of atoms. And in coordination chemistry there are several important types. Let's break them down. Okay, first remember those ambidentate lians. They lead to linkage isomeism. The same lian can attach through different atoms. The classic example is co- NH35 NO2 plus. If the NO2 binds through nitrogen nitro, it's yellow. If it binds through oxygen, nitro, it's red. Same formula, different color, different properties. Just depends on which atom connects. Cool. Then there's ionization isomerisism. This happens when a lian inside the coordination sphere swaps places with a counterarion outside the sphere. So ptl2, NHV over Br2 and PTBR2 NH24 CL2 are ionization isomers. They give different ions when dissolved in water. Ah inside versus outside the brackets matters definitely. Hydrate isomeism is a special case of that involving water molecules whether water is acting as a lian inside the sphere or just sitting outside as water of crystallization. Different isomers of CRCL 3.6 H2O actually have different colors because the number of coordinated waters changes. Okay. And coordination isomeism happens when both the canon and the annion are complex ions. You can swap the liyans between the metal centers. So the co NH36 CR CN6 is an isomer of CR NH36 co6. Wow. Okay. Several ways the connections can differ. But what about the 3D arrangement itself? That sounds crucial. Absolutely. That brings us to stereoism and specifically geometric isomeism. This arises because lians can occupy different positions relative to each other. For square planer complexes type MX2L2, this is huge, right? The flat ones. You can have the two X lians next to each other. That's the cis isomer. Or you can have them directly across from each other. That's the trans isomer. Cis and trans. I've heard of those. The most famous example is probably cis platin cis diameter chloroplatinum. It's a vital anti-cancer drug. The trans isomer, same formula, totally different shape, is inactive. Burner's work distinguishing cis and trans isomers of platinum was key evidence for the square planer geometry. So geometry is literally life or death in that case. What about tetrahedral? Tetrahedral complexes generally don't have geometric isomers because all four positions are equivalent relative to each other. However, if a tetrahedral complex has four different lians attached or certain unsymmetrical collides, it can be chyro. Chyro like handed. Exactly. It exists as a pair of non-s superimposable mirror images called enantiomers like your left and right hands. They rotate plain polarized light in opposite directions. That's optical activity. Okay. So tetrahedral can be chyal but not cyr. Correct. For five coordinate complexes the trigonal by pyramidal and square pyramidal shapes do have distinct positions like axial versus equatorial. But because of that flexionality we mentioned isomers are often hard or impossible to separate at room temperature. They intercon convert too fast pretty much. Now ocaedral complexes offer rich possibilities for geometric isomeism. For type MA4B2 you get cis and trans isomers again depending on whether the two bleans are adjacent 90° apart or opposite 180° apart to and trans again. Okay. For type MA3B3 you get two different arrangements. If the three A lians and the three B's occupy the corners of one triangular face of the ocahedron that's the fascial a facial isomer. If they occupy three positions around the equator or meridian of the octahedron that's the meridian isomer feck and merr. Got it. More lians more possibilities I assume. Definitely something like ma2 bdc2 can have five geometric isomers and one of those pairs is actually chyal andr which brings us back to kirality. You mentioned tetrahedral can be chyal. What about ocahedral? That seems more complex. Oh, ocaedral complexes are frequently chyal. A classic case is when you have three bid and tate lians like in coen 33 plus sand. Imagine those three n lians wrapping around the cobalt. The whole thing looks like a propeller. A propeller. Okay, I can visualize that it can twist one way like a right-handed screw or the other way like a left-handed screw. These are nons superimposable mirror images in anti-mer. We use the symbols a delta for the right-handed twist and lambda for the left-handed twist to describe their absolute configuration. Delta and lambda different from cyrs or fakma. Yes, this describes the overall handedness. And remember, this is different from DLL or plusa, which just tells you experimentally which way they rotate polarized light. We can often synthesize specific isomers or separate mixtures using chyro resolving agents. Okay, one last twist on structure. Can the lian itself be chyro? Yes. Or sometimes a lian that isn't chyro on its own becomes chyro just because of the way it has to twist to coordinate to the metal. The metal center essentially forces the lian into a specific chyro confirmation. So much structural subtlety. Let's shift gears slightly beyond how they're built. How strongly are they built? What governs the stability of these complexes? Why do some form so readily? That's all about thermodynamics. We measure the strength of liken binding using formation constants. usually symbolized as KF. Often we measure it relative to water lians being displaced. A large KF value means the lian binds strongly forming a stable complex. High KF strong bond. Exactly. And because these K values can span an enormous range. We often talk about them in logarithmic terms. Log FF. When lians add one by one, we have step-wise formation constants. K1 for the first lian, K2 for the second and so on. Okay. The overall formation constant aka to beta n represents the formation of the final complex mln from the bare metal ion and n lians. It's simply the product of all the stepwise constants k1 ak2sk kn. The inverse of kf kd is the dissociation constant. How easily it falls apart. How do those step-wise constants usually behave? Does K2 tend to be smaller than K1? Generally yes. You usually see KF1, KF2, KF3 and so on. Statistically, there are fewer water molecules left to replace. And sometimes adding negatively charged lians makes the metal less positive and thus less attractive to the next lian. Makes sense. Less attraction, lower K. But the really interesting situations are when this trend breaks down. If you see a sudden jump in a later KF value or a sudden drop, it signals something significant is happening electronically or structurally. Like what? Maybe adding a specific lian causes the metal to change its spin state becoming unexpectedly stable. Or maybe adding the say third chloride to mercury suddenly makes it much less favorable to add a fourth because it prefers to change geometry from tetrahedral HGCL42 back towards linear HGCL2. These deviations tell a story. Okay. So are there general principles that make certain lian metal combinations extra stable like a cheat code for stability? There absolutely is. The biggest one is the cal effect. Remember those chilating lians? The ones with multiple donor atoms that form rings. Yeah. The claws, right? Complexes formed with chilating lians are almost always significantly more stable than complexes formed with a comparable set of separate monodentate lians. Why is that? Stronger individual bonds. Not necessarily stronger bonds. The primary driving force is entropy. Think about it. If one bidentate lian like N replaces two monodentate water lians, you go from having say the metal complex plus one N molecule to the chilated complex plus two free water molecules. You've increased the number of independent particles floating around. Ah more molecules, more randomness, more entropy. Exactly. The universe favors increased entropy. So the reaction is thermodynamically more favorable. That positive entropy change makes the overall Gibbs free energy change more negative. Hence a larger kyop. So it's mostly an entropy game. Clever it is. And building on that is the macrocyclic effect. If your polyintentate lian is already pre-organized into a large ring, a macrocycle like a pferin ring or a crown ether, binding is often even more favorable than with the non-cyclic chiladine lian. Why the extra boost? Here you get an added enthalpic bonus. The lian is already set up in roughly the right confirmation to bind the metal. It doesn't have to twist and turn as much minimizing strain. This pre-organization makes the binding energy itself more favorable on top of the entropy gain. These effects are crucial in biology and analytical chemistry and faster binding too. Often yes there's a kinetic aspect too. Once one donor atom of a chile binds the others are already held close by making the subsequent ring closing steps much faster than waiting for separate monodentate lians to diffuse in. Makes sense. Any other factors? Sure. Steric effects matter. The size of the chlet ring is important. Five and six membered chilate rings tend to be the most stable because they have low ring strain like in cycllohexane or cyclopentane. Smaller or larger rings are usually less favorable. Okay, ring size matters. And finally, electron deoization can play a role. Some lians, particularly those with double bonds like bipur by bipantholane orphan are acceptors. They can accept electron density back from filled metal biz orbitals into their own empty orbitals. This backbonding strengthens the overall metal lian interaction, especially stabilizing metals in low oxidation states. Sometimes a metal can even act as a template, organizing precursor molecules around itself to facilitate the synthesis of a complex macrocyclic lian that might be hard to make otherwise. Wow. Okay. What a journey. We've gone from the basic idea of a metal surrounded by lians to intricate 3D shapes, a whole naming system, the subtleties of isomeism and kirality and finally the thermodynamics governing how stable these things are. It really is a whole interconnected world. It truly is. And I think the key takeaway is just how linked structure, bonding and reactivity are in coordination chemistry. Understanding the shape helps you understand the bonding which helps you predict stability in reactions. It forms a core part of inorganic chemistry for a reason. So, as you're listening, what stands out to you from this deep dive? Maybe you'll see the copper and wiring or the iron and hemoglobin a little differently now. What other questions does this spark for your own learning? Thank you so much for joining us on this deep dive into coordination compounds. We hope this breakdown has given you a clearer, more engaging picture of this vital area of chemistry. And a warm thank you from the entire last minute lecture |
15324 | https://www.geeksforgeeks.org/dsa/nth-non-square-number/ | Nth non-Square number - GeeksforGeeks
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Nth non-Square number
Last Updated : 20 Mar, 2023
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Given n, find the nth number which is not a perfect square among natural numbers (1, 2, 3, 4, 5, 6, ... )
Examples:
Input : 3
Output : 5
First three non-square numbers are 2, 3
and 5
Input : 5
Output : 7
Input : 16
Output : 20
Looking at the problem statement we can come up to a straight-forward brute-force approach. We can start from n = 1, and start to check if each of them is a perfect square or not. So we can come up to the nth non-square number.
However, the above approach is very slow as it searches for each in every number smaller than the target each time.
We can observe that the series under consideration is 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, ... .
We can come upto the constant time formula for the nth number in this sequence, by inspection.
a n s(n)=n+⌊1 2+n⌋ans(n) = n + \left \lfloor \frac{1}{2} + \sqrt{n}\right \rfloor an s(n)=n+⌊2 1+n⌋.
The correctness of the formula can be proved by the Principle of Mathematical Induction.
The implementation of the above formula is given below.
C++
```CPP
// CPP program to find n-th non-square number.
include
using namespace std;
// function to find the nth Non-Square Number
int findNthNonSquare(int n)
{
// conversion from int to long double is
// necessary in order to preserve decimal
// places after square root.
long double x = (long double)n;
// calculating the result
long double ans = x + floor(0.5 + sqrt(x));
return (int)ans;
}
// Driver code
int main()
{
// initializing the term number
int n = 16;
// Print the result
cout << "The " << n << "th Non-Square number is ";
cout << findNthNonSquare(n);
return 0;
}
```
// CPP program to find n-th non-square number.
include using namespace std;// function to find the nth Non-Square Number int findNthNonSquare(int n){ // conversion from int to long double is // necessary in order to preserve decimal // places after square root. long double x = (long double)n; // calculating the result long double ans = x + floor(0.5 + sqrt(x)); return (int)ans;}// Driver code int main(){ // initializing the term number int n = 16; // Print the result cout << "The " << n << "th Non-Square number is "; cout << findNthNonSquare(n); return 0;}
Java
```Java
// Java program to find
// n-th non-square number.
import java.io.;
import java.util.;
import java.lang.;
class GFG
{
// function to find the
// nth Non-Square Number
static int findNthNonSquare(int n)
{
// conversion from int to
// long double is necessary
// in order to preserve decimal
// places after square root.
double x = (double)n;
// calculating the result
double ans = x + Math.floor(0.5 +
Math.sqrt(x));
return (int)ans;
}
// Driver code
public static void main(String[] args)
{
// initializing
// the term number
int n = 16;
// Print the result
System.out.print("The " + n +
"th Non-Square number is ");
System.out.print(findNthNonSquare(n));
}
}
Python3Python3
Python3 program to find n-th
non-square number.
import math
function to find the nth
Non-Square Number
def findNthNonSquare(n):
# conversion from int to long
# double is necessary in order
# to preserve decimal places
# after square root.
x = n;
# calculating the result
ans = x + math.floor(0.5 + math.sqrt(x));
return int(ans);
Driver code
initializing the term number
n = 16;
Print the result
print("The", n, "th Non-Square number is",
findNthNonSquare(n));
This code is contributed by mits
C#CSHARP
// C# program to find
// n-th non-square number.
using System;
class GFG
{
// function to find the
// nth Non-Square Number
static int findNthNonSquare(int n)
{
// conversion from int
// to long double is
// necessary in order
// to preserve decimal
// places after square
// root.
double x = (double)n;
// calculating the result
double ans = x + Math.Floor(0.5 +
Math.Sqrt(x));
return (int)ans;
}
// Driver code
public static void Main()
{
// initializing
// the term number
int n = 16;
// Print the result
Console.Write("The " + n +
"th Non-Square " +
"number is ");
Console.Write(findNthNonSquare(n));
}
}
// This code is contributed
// by anuj_67.
PHPphp
php
// PHP program to find n-th
// non-square number.
// function to find the nth
// Non-Square Number
function findNthNonSquare($n)
{
// conversion from int to long
// double is necessary in order
// to preserve decimal places
// after square root.
$x = $n;
// calculating the result
$ans = $x + floor(0.5 + sqrt($x));
return (int)$ans;
}
// Driver code
// initializing the term number
$n = 16;
// Print the result
echo "The " . $n .
"th Non-Square number is ";
echo findNthNonSquare($n);
// This Code is Contributed by mits
?
JavaScriptjavascript
// Javascript program to find
// n-th non-square number.
// Function to find the
// nth Non-Square Number
function findNthNonSquare(n)
{
// Conversion from var to
// var var is necessary
// in order to preserve decimal
// places after square root.
var x = n;
// Calculating the result
var ans = x + Math.floor(0.5 + Math.sqrt(x));
return parseInt(ans);
}
// Driver code
// Initializing
// the term number
var n = 16;
// Print the result
document.write("The " + n + "th Non-Square number is ");
document.write(findNthNonSquare(n));
// This code is contributed by todaysgaurav
```
Output:
The 16th Non-Square number is 20
Time ComplexityO(log(n))
Space ComplexityO(1)
Comment
More info
A
AayushChaturvedi
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15325 | https://mathworld.wolfram.com/QuadraticIntegral.html | Quadratic Integral -- from Wolfram MathWorld
TOPICS
AlgebraApplied MathematicsCalculus and AnalysisDiscrete MathematicsFoundations of MathematicsGeometryHistory and TerminologyNumber TheoryProbability and StatisticsRecreational MathematicsTopologyAlphabetical IndexNew in MathWorld
Calculus and Analysis
Calculus
Integrals
Indefinite Integrals
Quadratic Integral
To compute an integral of the form
(1)
complete the square in the denominator to obtain
(2)
Let . Then define
(3)
where
(4)
is the negative of the polynomial discriminant. If , then
(5)
Now use partial fraction decomposition,
(6)
(7)
so and . Plugging these in,
(8)
for . Note that this integral is also tabulated in Gradshteyn and Ryzhik (2000, equation 2.172), where it is given with a sign flipped.
See also
Quadratic
Explore with Wolfram|Alpha
More things to try:
double integral
indefinite integrals
5-ary Lyndon words of length 12
References
Gradshteyn, I.S. and Ryzhik, I.M. Tables of Integrals, Series, and Products, 6th ed. San Diego, CA: Academic Press, 2000.
Referenced on Wolfram|Alpha
Quadratic Integral
Cite this as:
Weisstein, Eric W. "Quadratic Integral." From MathWorld--A Wolfram Resource.
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Calculus
Integrals
Indefinite Integrals
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15326 | https://chem.libretexts.org/Courses/Lubbock_Christian_University/LCU%3A_CHE_1305_-_Introductory_Chemistry/02%3A_Atoms/2.4%3A_Elements%3A_Defined_by_Their_Number_of_Protons | Skip to main content
2.4: Elements: Defined by Their Number of Protons
Last updated
: Nov 24, 2018
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2.3: The Properties of Protons, Neutrons, and Electrons
2.5: Counting Nails by the Pound
Page ID
: 118783
( \newcommand{\kernel}{\mathrm{null}\,})
Learning Objectives
Define atomic number.
Define mass number.
Determine the number of protons, neutrons, and electrons in an atom.
It's important to be able to distinguish atoms of one element from atoms of another element. Elements are pure substances that make up all other matter, so each one is given a unique name. The names of elements are also represented by unique one- or two-letter symbols, such as H for hydrogen, C for carbon, or He for helium. However, it would more powerful if these names could be used to identify the numbers of protons and neutrons in the atoms. That's where atomic number and mass number are useful.
Atomic Number
Scientists distinguish between different elements by counting the number of protons in the nucleus (Table 2.4.1). If an atom has only one proton, we know that it's a hydrogen atom. An atom with two protons is always a helium atom. If scientists count four protons in an atom, they know it's a beryllium atom. An atom with three protons is a lithium atom, an atom with five protons is a boron atom, an atom with six protons is a carbon atom . . . the list goes on.
Since an atom of one element can be distinguished from an atom of another element by the number of protons in its nucleus, scientists are always interested in this number, and how this number differs between different elements. The number of protons in an atom is called its atomic number (Z). This number is very important because it is unique for atoms of a given element. All atoms of an element have the same number of protons, and every element has a different number of protons in its atoms. For example, all helium atoms have two protons, and no other elements have atoms with two protons.
| Name | Protons | Neutrons | Electrons | Atomic Number (Z) | Mass Number (A) |
--- --- --- |
Table 2.4.1: Atoms of the First Six Elements
| Hydrogen | 1 | 0 | 1 | 1 | 1 |
| Helium | 2 | 2 | 2 | 2 | 4 |
| Lithium | 3 | 4 | 3 | 3 | 7 |
| Beryllium | 4 | 5 | 4 | 4 | 9 |
| Boron | 5 | 6 | 5 | 5 | 11 |
| Carbon | 6 | 6 | 6 | 6 | 12 |
Of course, since neutral atoms have to have one electron for every proton, an element's atomic number also tells you how many electrons are in a neutral atom of that element. For example, hydrogen has an atomic number of 1. This means that an atom of hydrogen has one proton, and, if it's neutral, one electron as well. Gold, on the other hand, has an atomic number of 79, which means that an atom of gold has 79 protons, and, if it's neutral, 79 electrons as well.
Neutral Atoms
Atoms are neutral in electrical charge because they have the same number of negative electrons as positive protons (Table 2.4.1). Therefore, the atomic number of an atom also tells you how many electrons the atom has. This, in turn, determines many of the atom's chemical properties.
Mass Number
The mass number (A) of an atom is the total number of protons and neutrons in its nucleus. The mass of the atom is a unit called the atomic mass unit (amu). One atomic mass unit is the mass of a proton, or about 1.67×10−27 kilograms, which is an extremely small mass. A neutron has just a tiny bit more mass than a proton, but its mass is often assumed to be one atomic mass unit as well. Because electrons have virtually no mass, just about all the mass of an atom is in its protons and neutrons. Therefore, the total number of protons and neutrons in an atom determines its mass in atomic mass units (Table 2.4.1).
Consider helium again. Most helium atoms have two neutrons in addition to two protons. Therefore the mass of most helium atoms is 4 atomic mass units (2amu for the protons + 2amu for the neutrons). However, some helium atoms have more or less than two neutrons. Atoms with the same number of protons but different numbers of neutrons are called isotopes. Because the number of neutrons can vary for a given element, the mass numbers of different atoms of an element may also vary. For example, some helium atoms have three neutrons instead of two (these are called isotopes and are discussed in detail later on).
Why do you think that the "mass number" includes protons and neutrons, but not electrons? You know that most of the mass of an atom is concentrated in its nucleus. The mass of an atom depends on the number of protons and neutrons. You have already learned that the mass of an electron is very, very small compared to the mass of either a proton or a neutron (like the mass of a penny compared to the mass of a bowling ball). Counting the number of protons and neutrons tells scientists about the total mass of an atom.
mass numberA=(number of protons)+(number of neutrons)
An atom's mass number is very easy to calculate, provided that you know the number of protons and neutrons in an atom.
Example 4.5.1
What is the mass number of an atom of helium that contains 2 neutrons?
Solution
(number of protons)=2 (Remember that an atom of helium always has 2 protons.)
(number of neutrons)=2
mass number=(number of protons)+(number of neutrons)
mass number=2+2=4
A chemical symbol is a one- or two-letter designation of an element. Some examples of chemical symbols are O for oxygen, Zn for zinc, and Fe for iron. The first letter of a symbol is always capitalized. If the symbol contains two letters, the second letter is lower case. The majority of elements have symbols that are based on their English names. However, some of the elements that have been known since ancient times have maintained symbols that are based on their Latin names, as shown in Table 2.4.2.
| Chemical Symbol | Name | Latin Name |
---
Table 2.4.2: Symbols and Latin Names for Elements
| Na | Sodium | Natrium |
| K | Potassium | Kalium |
| Fe | Iron | Ferrum |
| Cu | Copper | Cuprum |
| Ag | Silver | Argentum |
| Sn | Tin | Stannum |
| Sb | Antimony | Stibium |
| Au | Gold | Aurum |
| Pb | Lead | Plumbum |
Summary
Elements are pure substances that make up all matter, so each one is given a unique name.
The names of elements are also represented by unique one- or two-letter symbols.
Each element has a unique number of protons. An element's atomic number is equal to the number of protons in the nuclei of any of its atoms.
The mass number of an atom is the sum of the protons and neutrons in the atom.
Isotopes are atoms of the same element (same number of protons) that have different numbers of neutrons in their atomic nuclei.
2.3: The Properties of Protons, Neutrons, and Electrons
2.5: Counting Nails by the Pound |
15327 | https://www.teacherspayteachers.com/Product/Time-Zone-Conversions-Worksheets-11450864 | Time Zone Conversions Worksheets
Description
A packet of 8 different worksheets on time zone conversions. Each worksheet includes a chart with all the different US time zones. There are four worksheets where students will fill out a chart, converting between two different US time zones. There are four worksheets with 10-12 questions where students are given a time in one time zone and then they have to convert it to the second time zone. Each worksheet has an answer key following the worksheet.
Time Zone Conversions Worksheets
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15328 | https://w3.cs.jmu.edu/spragunr/CS228/lectures/intro_graphs/Chapter10.pdf | Graphs and Graph Models Section 10.1 Graphs Definition: A graph G = (V, E) consists of a nonempty set V of vertices (or nodes) and a set E of edges. Each edge has either one or two vertices associated with it, called its endpoints. An edge is said to connect its endpoints.
Some T erminology In a simple graph each edge connects two different vertices and no two edges connect the same pair of vertices.
Multigraphs may have multiple edges connecting the same two vertices. When m different edges connect the vertices u and v, we say that {u,v} is an edge of multiplicity m. An edge that connects a vertex to itself is called a loop.
A pseudograph may include loops, as well as multiple edges connecting the same pair of vertices.
Directed Graphs Definition: An directed graph (or digraph) G = (V, E) consists of a nonempty set V of vertices (or nodes) and a set E of directed edges (or arcs). Each edge is associated with an ordered pair of vertices. The directed edge associated with the ordered pair (u,v) is said to start at u and end at v. Remark: Graphs where the end points of an edge are not ordered are said to be undirected graphs.
Graph T erminology: Summary Graph Terminology and Special Types of Graphs Section 10.2 Basic T erminology Definition 1. T wo vertices u, v in an undirected graph G are called adjacent (or neighbors) in G if there is an edge e between u and v. Such an edge e is called incident with the vertices u and v and e is said to connect u and v. Definition 2. The set of all neighbors of a vertex v of G = (V, E), denoted by N(v), is called the neighborhood of v. If A is a subset of V, we denote by N(A) the set of all vertices in G that are adjacent to at least one vertex in A. So, Definition 3. The degree of a vertex in a undirected graph is the number of edges incident with it, except that a loop at a vertex contributes two to the degree of that vertex. The degree of the vertex v is denoted by deg(v).
Degrees of Vertices Theorem 1 (Handshaking Theorem): If G = (V,E) is an undirected graph with m edges, then Proof: Each edge contributes twice to the degree count of all vertices. Hence, both the left-hand and right-hand sides of this equation equal twice the number of edges. Think about the graph where vertices represent the people at a party and an edge connects two people who have shaken hands.
Directed Graphs Definition: An directed graph G = (V, E) consists of V, a nonempty set of vertices (or nodes), and E, a set of directed edges or arcs. Each edge is an ordered pair of vertices. The directed edge (u,v) is said to start at u and end at v.
Definition: Let (u,v) be an edge in G. Then u is the initial vertex of this edge and is adjacent to v and v is the terminal (or end) vertex of this edge and is adjacent from u. The initial and terminal vertices of a loop are the same.
Recall the definition of a directed graph.
Directed Graphs (continued) Definition: The in-degree of a vertex v, denoted deg−(v), is the number of edges which terminate at v. The out-degree of v, denoted deg+(v), is the number of edges with v as their initial vertex.
Example: In the graph G we have deg−(a) = 2, deg−(b) = 2, deg−(c) = 3, deg−(d) = 2, deg−(e) = 3, deg−(f) = 0.
deg+(a) = 4, deg+(b) = 1, deg+(c) = 2, deg+(d) = 2, deg+(e) = 3, deg+(f) = 0.
Directed Graphs (continued) Theorem 3: Let G = (V, E) be a graph with directed edges. Then: Proof: The first sum counts the number of outgoing edges over all vertices and the second sum counts the number of incoming edges over all vertices. It follows that both sums equal the number of edges in the graph.
Special T ypes of Simple Graphs: Complete Graphs A complete graph on n vertices, denoted by Kn, is the simple graph that contains exactly one edge between each pair of distinct vertices. Special T ypes of Simple Graphs: Cycles and Wheels A cycle Cn for n ≥ 3 consists of n vertices v1, v2 ,⋯ , vn, and edges {v1, v2}, {v2, v3} ,⋯ , {vn-1, vn}, {vn, v1}.
A wheel Wn is obtained by adding an additional vertex to a cycle Cn for n ≥ 3 and connecting this new vertex to each of the n vertices in Cn by new edges.
Special T ypes of Simple Graphs: n-Cubes An n-dimensional hypercube, or n-cube, Qn, is a graph with 2n vertices representing all bit strings of length n, where there is an edge between two vertices that differ in exactly one bit position. Bipartite Graphs Definition: A simple graph G is bipartite if V can be partitioned into two disjoint subsets V1 and V2 such that every edge connects a vertex in V1 and a vertex in V2. In other words, there are no edges which connect two vertices in V1 or in V2.
It is not hard to show that an equivalent definition of a bipartite graph is a graph where it is possible to color the vertices red or blue so that no two adjacent vertices are the same color. G is bipartite H is not bipartite since if we color a red, then the adjacent vertices f and b must both be blue.
Complete Bipartite Graphs Definition: A complete bipartite graph Km,n is a graph that has its vertex set partitioned into two subsets V1 of size m and V2 of size n such that there is an edge from every vertex in V1 to every vertex in V2.
New Graphs from Old Definition: A subgraph of a graph G = (V,E) is a graph (W,F), where W ⊂ V and F ⊂ E. A subgraph H of G is a proper subgraph of G if H ≠ G.
Example: Here we show K5 and one of its subgraphs.
Definition: Let G = (V, E) be a simple graph. The subgraph induced by a subset W of the vertex set V is the graph (W,F), where the edge set F contains an edge in E if and only if both endpoints are in W. Example: Here we show K5 and the subgraph induced by W = {a,b,c,e}. New Graphs from Old (continued) Definition: The union of two simple graphs G1 = (V1, E1) and G2 = (V2, E2) is the simple graph with vertex set V1 ⋃ V2 and edge set E1 ⋃ E2. The union of G1 and G2 is denoted by G1 ⋃ G2.
Example: |
15329 | https://math.stackexchange.com/questions/4626114/how-to-know-how-to-sketch-cos2-x-graphs-from-cos-x | graphing functions - How to know how to sketch $\cos^2 x$ graphs from $\cos x$? - Mathematics Stack Exchange
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How to know how to sketch cos 2 x cos 2x graphs from cos x cosx?
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How to know how to sketch cos 2 x cos 2x graphs from cos x cosx?
I am trying to understand spherical harmonics, but I need to brush up on my fundamentals to understand it faster and better.
For example, cos x cosx graphs peaks at 0 and π π, but what about cos 2 x cos 2x graphs?
How do I understand where it peaks for cos 2 x cos 2x ? I've been told to visualise it as $2 cosine graphs, but what does that mean?
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asked Jan 26, 2023 at 8:42
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For every x x-value, take the y y-value of the cos cos graph and square it. cos 2 cos 2 then has a maximum wherever cos cos is furthest away from the x x-axis, and a minimum wherever cos cos is closest to the x x-axis.Zuy –Zuy 2023-01-26 08:48:24 +00:00 Commented Jan 26, 2023 at 8:48
2 cos 2 x=1+cos(2 x)2 cos 2x=1+cos(2 x)2.Kavi Rama Murthy –Kavi Rama Murthy 2023-01-26 08:52:42 +00:00 Commented Jan 26, 2023 at 8:52
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Genetics: .
Chapter1 Genes, Genomes, and Genetic Analysis
Chapter2 DNA Structure and Genetic Variation
Chapter3 Transmission Genetics: The Principle of Segregation
Chapter4 Chromosomes and Sex-Chromosome Inheritance
Chapter5 Genetic Linkage and Chromosome Mapping
Chapter6 Molecular Biology of DNA Replication and Recombination
Chapter7 Molecular Organization of Chromosomes
Chapter8 Human Karyotypes and Chromosome Behavior
Chapter9 Genetics of Bacteria and Their Viruses
Chapter10 Molecular Biology of Gene Expression
Chapter11 Molecular Mechanisms of Gene Regulation
Chapter12 Genomics, Proteomics, and Transgenics
Chapter13 Genetic Control of Development
Chapter14 Molecular Mechanisms of Mutation and DNA Repair
Chapter15 Molecular Genetics of the Cell Cycle and Cancer
Chapter16 Mitochondrial DNA and Extranuclear Inheritance
Chapter17 Molecular Evolution and Population Genetics
Chapter18 The Genetic Basis of Complex Traits
Chapter19 Human Evolutionary Genetics
1126808731
Genetics: .
Chapter1 Genes, Genomes, and Genetic Analysis
Chapter2 DNA Structure and Genetic Variation
Chapter3 Transmission Genetics: The Principle of Segregation
Chapter4 Chromosomes and Sex-Chromosome Inheritance
Chapter5 Genetic Linkage and Chromosome Mapping
Chapter6 Molecular Biology of DNA Replication and Recombination
Chapter7 Molecular Organization of Chromosomes
Chapter8 Human Karyotypes and Chromosome Behavior
Chapter9 Genetics of Bacteria and Their Viruses
Chapter10 Molecular Biology of Gene Expression
Chapter11 Molecular Mechanisms of Gene Regulation
Chapter12 Genomics, Proteomics, and Transgenics
Chapter13 Genetic Control of Development
Chapter14 Molecular Mechanisms of Mutation and DNA Repair
Chapter15 Molecular Genetics of the Cell Cycle and Cancer
Chapter16 Mitochondrial DNA and Extranuclear Inheritance
Chapter17 Molecular Evolution and Population Genetics
Chapter18 The Genetic Basis of Complex Traits
Chapter19 Human Evolutionary Genetics
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Overview
Chapter1 Genes, Genomes, and Genetic Analysis
Chapter2 DNA Structure and Genetic Variation
Chapter3 Transmission Genetics: The Principle of Segregation
Chapter4 Chromosomes and Sex-Chromosome Inheritance
Chapter5 Genetic Linkage and Chromosome Mapping
Chapter6 Molecular Biology of DNA Replication and Recombination
Chapter7 Molecular Organization of Chromosomes
Chapter8 Human Karyotypes and Chromosome Behavior
Chapter9 Genetics of Bacteria and Their Viruses
Chapter10 Molecular Biology of Gene Expression
Chapter11 Molecular Mechanisms of Gene Regulation
Chapter12 Genomics, Proteomics, and Transgenics
Chapter13 Genetic Control of Development
Chapter14 Molecular Mechanisms of Mutation and DNA Repair
Chapter15 Molecular Genetics of the Cell Cycle and Cancer
Chapter16 Mitochondrial DNA and Extranuclear Inheritance
Chapter17 Molecular Evolution and Population Genetics
Chapter18 The Genetic Basis of Complex Traits
Chapter19 Human Evolutionary Genetics
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Product Details
About the Author
Table of Contents
Product Details
| ISBN-13: | 9781449635961 |
| Publisher: | Jones & Bartlett Learning |
| Publication date: | 08/19/2011 |
| Edition description: | Older Edition |
| Pages: | 804 |
| Product dimensions: | 8.70(w) x 11.00(h) x 1.30(d) |
About the Author
Harvard University
Harvard University
Table of Contents
Chapter1 Genes, Genomes, and Genetic Analysis
Chapter2 DNA Structure and Genetic Variation
Chapter3 Transmission Genetics: The Principle of Segregation
Chapter4 Chromosomes and Sex-Chromosome Inheritance
Chapter5 Genetic Linkage and Chromosome Mapping
Chapter6 Molecular Biology of DNA Replication and Recombination
Chapter7 Molecular Organization of Chromosomes
Chapter8 Human Karyotypes and Chromosome Behavior
Chapter9 Genetics of Bacteria and Their Viruses
Chapter10 Molecular Biology of Gene Expression
Chapter11 Molecular Mechanisms of Gene Regulation
Chapter12 Genomics, Proteomics, and Transgenics
Chapter13 Genetic Control of Development
Chapter14 Molecular Mechanisms of Mutation and DNA Repair
Chapter15 Molecular Genetics of the Cell Cycle and Cancer
Chapter16 Mitochondrial DNA and Extranuclear Inheritance
Chapter17 Molecular Evolution and Population Genetics
Chapter18 The Genetic Basis of Complex Traits
Chapter19 Human Evolutionary Genetics
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15331 | https://jnccn.org/abstract/journals/jnccn/23/5/article-p132.xml | NCCN Continuing Education Target Audience: This journal article is designed to meet the educa-tional needs of oncologists, nurses, pharmacists, and other healthcare professionals who manage patients with cancer.
Accreditation Statements
In support of improving patient care, National Comprehensive Cancer Network (NCCN) is jointly accredited by the Accreditation Council for Continuing Medical Education (ACCME), the Accreditation Council for Pharmacy Edu-cation (ACPE), and the American Nurses Credentialing Center (ANCC), to provide continuing education for the healthcare team.
Physicians: NCCN designates this journal-based CME activity for a maximum of 1.0 AMA PRA Category 1 Credit TM . Physicians should claim only the credit commensurate with the extent of their par-ticipation in the activity.
Nurses: NCCN designates this educational activity for a maximum of 1.0 contact hour.
Pharmacists: NCCN designates this knowledge-based continuing education activity for 1.0 contact hour (0.1 CEUs) of continuing education credit. UAN: JA4008196-0000-25-012-H01-P
PAs: NCCN has been authorized by the American Academy of PAs (AAPA) to award AAPA Category 1 CME credit for activities planned in accordance with AAPA CME Criteria. This activity is des-ignated for 1.0 AAPA Category 1 CME credit. Approval is valid until May 10, 2026. PAs should only claim credit commensurate with the extent of their participation. All clinicians completing this activity will be issued a certificate of participation. To participate in this journal CE activity: (1) review the educational content; (2) take the posttest with a 66% minimum passing score and complete the evaluation at nccn.org/May2025 ; and (3) view/print certificate.
Pharmacists: You must complete the posttest and evaluation within 30 days of the activity. Continuing pharmacy education credit is re-ported to the CPE Monitor once you have completed the posttest and evaluation and claimed your credits. Before completing these re-quirements, be sure your NCCN profile has been updated with your NAPB e-profile ID and date of birth. Your credit cannot be reported without this information. If you have any questions, please email education@nccn.org. Release date: May 10, 2025; Expiration date: May 10, 2026
Learning Objectives:
Upon completion of this activity, participants will be able to:
Integrate into professional practice the updates to the NCCN Guidelines for Multiple Myeloma
Describe the rationale behind the decision-making process for de-veloping the NCCN Guidelines for Multiple Myeloma
Disclosure of Relevant Financial Relationships
None of the planners for this educational activity have relevant financial relationship(s) to disclose with ineligible companies whose primary business is producing, marketing, selling, re-selling, or distributing healthcare products used by or on patients.
Individuals Who Provided Content Development and/or Authorship Assistance:
The faculty listed below have no relevant financial relationship(s) with ineligible companies to disclose.
Natalie S. Callander, MD, Panel Vice Chair
Emily Kovach, Guidelines Layout Specialist, NCCN
Rashmi Kumar, PhD, Senior Director, Clinical Content, NCCN The faculty listed below have the following relevant financial relationship(s) with ineligible companies to disclose. All of the relevant financial relationships listed for these individuals have been mitigated.
Shaji K. Kumar, MD, Panel Chair, has disclosed receiving grant/research support from AbbVie Inc., Amgen Inc., Bristol Myers Squibb, CARsgen Therapeutics, GSK plc, Janssen Pharmaceutica Products, LP, Oricell Therapeutics, Regeneron Pharmaceuticals, Inc., Roche, Sanofi US, and Takeda Pharmaceuticals; receiving consulting fees from Moderna, Inc.; and serving as a scientific advisor for AbbVie Inc., Amgen Inc., Bristol Myers Squibb, GSK plc, Janssen Pharmaceutica Products, LP, Regeneron Pharmaceuticals, Inc., Roche, Sanofi US, and Takeda Pharmaceuticals.
Muhamed Baljevic, MD, has disclosed receiving consulting fees from AbbVie Inc., Johnson & Johnson, and Pfizer; and serving as a scientific advisor for AbbVie Inc., Bristol Myers Squibb/Celgene Corporation, Janssen Biotech, Parexel International Corporation, Pfizer, Prothena Corporation, and Sanofi Genzyme. To view disclosures of external relationships for the NCCN Guidelines panel, go to NCCN.org/guidelines/guidelines-panels-and-disclosure/disclosure-panels This activity is supported by educational grants from AstraZeneca, Coherus BioSciences, Geron, Janssen Biotech, Inc., administered by Janssen Scientific Affairs, LLC, Novartis, SpringWorks Therapeutics, Inc., and Taiho Oncology, Inc. This activity is supported by an independent educational grant from Rigel Pharmaceuticals, Inc.
CE NCCN GUIDELINES ® INSIGHTS Multiple Myeloma, Version 1.2025 132 © JNCCN —Journal of the National Comprehensive Cancer Network | Volume 23 Issue 5 | May 2025 Multiple Myeloma, Version 1.2025
Featured Updates to the NCCN Guidelines ®
Shaji K. Kumar, MD 1 ,; Natalie S. Callander, MD 2 ,; Kehinde Adekola, MD, MSCI 3 ; Larry D. Anderson Jr, MD, PhD 4 ;Muhamed Baljevic, MD 5 ,; Rachid Baz, MD 6 ; Erica Campagnaro, MD 7 ; Caitlin Costello, MD 8 ; Christopher D ’Angelo, MD 9 ;Benjamin Derman, MD 10 ; Srinivas Devarakonda, MD 11 ; Noura Elsedawy, MD 12 ; Amandeep Godara, MD 13 ; Kelly Godby, MD 14 ;Jens Hillengass, MD, PhD 15 ; Leona Holmberg, MD, PhD 16 ; Myo Htut, MD 17 ; Carol Ann Huff, MD 18 ; Malin Hultcrantz, MD, PhD 19 ;Yubin Kang, MD 20 ; Sarah Larson, MD 21 ; Hans C. Lee, MD 22 ; Michaela Liedtke, MD 23 ; Thomas Martin, MD 24 ; James Omel, MD 25 ;Timothy Robinson, MD, PhD 26 ; Aaron Rosenberg, MD 27 ; Mark A. Schroeder, MD 28 ; Daniel Sherbenou, MD, PhD 29 ;Attaya Suvannasankha, MD 30 ; Jason Valent, MD 31 ; Asya Nina Varshavsky-Yanovsky, MD, PhD 32 ; Dan Vogl, MD 33 ;Emily Kovach 34 ,; and Rashmi Kumar, PhD 34 ,
Abstract
The NCCN Guidelines for Multiple Myeloma (MM) provide recommendations for diagnosis, initial workup, treatment, follow-up, and sup-portive care for patients with MM. These NCCN Guidelines Insights highlight the important updates and changes speci fi c to systemic ther-apy for patients with newly diagnosed as well as previously treated MM included in Version 1.2025 of the NCCN Guidelines for MM.
J Natl Compr Canc Netw 2025;23(5):132 –140 doi:10.6004/jnccn.2025.0023
Overview
Multiple myeloma (MM) is a malignant neoplasm of plasma cells that typically accumulate in bone marrow, leading to bone destruction and anemia as well as renal damage secondary to the secreted monoclonal protein. MM is most frequently diagnosed in individuals aged 65 to 74 years, with a median age of 69 years. 1The American Cancer Society estimates 36,100 new MM cases and 12,030 deaths will occur in the United States in 2025. 2 Glob-ally, age-standardized prevalence and mortality rates, as well as disability-adjusted life-years have all increased between 1990 and 2021. The number of MM cases, deaths, and disability-adjusted life-years is higher in males than in females. 3The NCCN Guidelines Panel for Plasma Cell Neoplasms has developed guidelines for managing patients with various plasma cell neoplasms, including monoclonal gammopathies of clinical signi fi cance, POEMS syndrome, solitary plasmacy-toma, smoldering myeloma, MM, systemic light-chain amy-loidosis, and Waldenstr €om macroglobulinemia. These guidelines are updated annually or more frequently if new high-quality clinical data become available. Signi fi cant updates have been made in Version 1.2025 of the NCCN Guidelines for MM. These NCCN Guidelines Insights focus only on important updates speci fi c to treatment options for patients with newly diagnosed MM (NDMM) who are eligible or ineligible for hematopoietic cell transplant (HCT), recom-mendations for maintenance therapy, and treatment options for previously treated MM. For a complete list of updates in Version 1.2025, visit NCCN.org.
Updates to Treatment Options for Patients With NDMM
During the annual NCCN Guidelines panel meeting, the panel reviewed the list of regimens and the preference categories and added new regimens, reclassi fi ed the preference category, and updated categories of evidence and consensus based on new or updated evidence and panel consensus. The panel added the 4-drug regimen daratumumab/ bortezomib/lenalidomide/dexamethasone ( D -VRd) as a pre-ferred regimen for patients with NDMM who are eligible for transplant (Figure 1). The data for this regimen are derived from 2 trials. The phase III PERSEUS trial evaluated the ef fi cacy and safety of adding daratumumab to VRd in patients with NDMM who are eligible for transplant (n 5709) and demonstrated a clear bene fi t of adding the monoclonal antibody. 4 Patients were ran-domized to receive treatment induction with 4 cycles of D -VRd or VRd, followed by autologous HCT, 2 cycles of consolidation with the same regimen as induction, and then maintenance
1
Mayo Clinic Comprehensive Cancer Center; 2University of Wisconsin Carbone Cancer Center; 3Robert H. Lurie Comprehensive Cancer Center of Northwestern University; 4UT Southwestern Simmons Comprehensive Cancer Center; 5Vanderbilt-Ingram Cancer Center; 6Mof fi tt Cancer Center; 7University of Michigan Rogel Cancer Center; 8UC San Diego Moores Cancer Center; 9Fred & Pamela Buffett Cancer Center; 10 The UChicago Medicine Comprehensive Cancer Center; 11 The Ohio State University Comprehensive Cancer Center –James Cancer Hospital and Solove Research Institute; 12 St. Jude Children ’s Research Hospital/The University of Tennessee Health Science Center; 13 Huntsman Cancer Institute at the University of Utah; 14 O’Neal Comprehensive Cancer Center at UAB; 15 Roswell Park Comprehensive Cancer Center; 16 Fred Hutchinson Cancer Center; 17 City of Hope National Medical Center; 18 Johns Hopkins Kimmel Cancer Center; 19 Memorial Sloan Kettering Cancer Center; 20 Duke Cancer Institute; 21 UCLA Jonsson Comprehensive Cancer Center; 22 The University of Texas MD Anderson Cancer Center; 23 Stanford Cancer Institute; 24 UCSF Helen Diller Family Comprehensive Cancer Center; 25 Patient Advocate; 26 Yale Cancer Center/Smilow Cancer Hospital; 27 UC Davis Comprehensive Cancer Center; 28 Siteman Cancer Center at Barnes-Jewish Hospital and Washington University School of Medicine; 29 University of Colorado Cancer Center; 30 Indiana University Melvin and Bren Simon Comprehensive Cancer Center; 31 Case Comprehensive Cancer Center/University Hospitals Seidman Cancer Center and Cleveland Clinic Taussig Cancer Institute; 32 Fox Chase Cancer Center; 33 Abramson Cancer Center at the University of Pennsylvania; and 34 National Comprehensive Cancer Network.
Provided content development and/or authorship assistance.
The full and most current version of these NCCN Guidelines is available at NCCN.org.
JNCCN.org | Volume 23 Issue 5 | May 2025 133 CE NCCN GUIDELINES ® INSIGHTS therapy with lenalidomide in the VRd group and daratumumab 1
lenalidomide in the D -VRd group. The addition of daratumumab to VRd signi fi cantly improved progression-free survival (PFS). Median PFS was not reached in either arm; at 47.5 months, the PFS rate was 84.3% (95% CI, 79.5% –88.1%) in the D -VRd group compared with 67.7% (95% CI, 62.2% –72.6%) in the VRd group, with a hazard ratio (HR) of 0.42 (95% CI, 0.30 –0.59; P ,.001), in-dicating a 58% reduction in the risk of progression or death. 4 The D-VRd group also demonstrated a higher rate of complete response (CR) or better (87.9% vs 70.1%; P ,.001). 4 A substantial proportion of patients in the D-VRd group achieved minimal re-sidual disease (MRD) at the ,10 25 threshold (75.2% vs 47.5%;
P ,.001), suggesting deeper responses and potentially longer duration of remission. After at least 24 months of maintenance therapy, daratumumab was discontinued in patients who had achieved CR or better and had sustained MRD-negative status for at least 12 months. These patients continued to receive main-tenance lenalidomide until disease progression or unacceptable
NCCN CATEGORIES OF EVIDENCE AND CONSENSUS Category 1: Based upon high-level evidence ( $1 randomized phase 3 trials or high-quality, robust meta-analyses), there is uniform NCCN consensus ($85% support of the Panel) that the intervention is appropriate.
Category 2A: Based upon lower-level evidence, there is uniform NCCN consensus ( $85% support of the Panel) that the intervention is appropriate.
Category 2B: Based upon lower-level evidence, there is NCCN consensus ($50%, but ,85% support of the Panel) that the intervention is appropriate.
Category 3: Based upon any level of evidence, there is major NCCN disagreement that the intervention is appropriate.
All recommendations are category 2A unless otherwise indicated. NCCN CATEGORIES OF PREFERENCE Preferred intervention: Interventions that are based on superior ef fi cacy, safety, and evidence; and, when appropriate, affordability.
Other recommended intervention: Other interventions that may be somewhat less ef fi cacious, more toxic, or based on less mature data; or signi fi cantly less affordable for similar outcomes.
Useful in certain circumstances: Other interventions that may be used for selected patient populations (de fi ned with recommendation).
All recommendations are considered appropriate. NCCN recognizes the importance of clinical trials and encourages participation when applicable and available. Trials should be designed to maximize inclu-siveness and broad representative enrollment. PLEASE NOTE
The NCCN Guidelines ® are a statement of evidence and consensus of the authors regarding their views of currently accepted approaches to treatment.
The NCCN Guidelines ® Insights highlight important changes in the NCCN Guidelines ® recommendations from previous versions. Colored markings in the algorithm show changes and the discussion aims to further understanding of these changes by summarizing salient portions of the panel ’s discus-sion, including the literature reviewed.
The NCCN Guidelines Insights do not represent the full NCCN Guidelines; further, the National Comprehensive Cancer Network ® (NCCN ® ) makes no representations or warranties of any kind regarding the content, use, or application of the NCCN Guidelines and NCCN Guidelines Insights and disclaims any responsibility for their application or use in any way.
Version 1.2025 © 2025 National Comprehensive Cancer Network ®(NCCN ®). All rights reserved. The NCCN Guidelines ®and this illustration may not be reproduced in any form without the express written permission of NCCN.
MYEL-G 1 OF 5 PRIMARY THERAPY FOR TRANSPLANT CANDIDATES a-d Preferred Regimens • Daratumumab/lenalidomide/bortezomib/dexamethasone (catgory 1)
Other Recommended Regimens • Bortezomib/lenalidomide/dexamethasone (category 1)
• Carfilzomib/lenalidomide/dexamethasone
• Isatuximab-irfc/bortezomib/lenalidomide/dexamethasone
Useful In Certain Circumstances • Bortezomib/cyclophosphamide/dexamethasone e
• Carfilzomib/cyclophosphamide/dexamethasone e,f
• Daratumumab/bortezomib/cyclophosphamide/dexamethasone
• Daratumumab/carfilzomib/lenalidomide/dexamethasone
• Dexamethasone/thalidomide/cisplatin/doxorubicin/cyclophosphamide/etoposide/ bortezomib g (VTD-PACE)
• Isatuximab-irfc/carfilzomib/lenalidomide/dexamethasone
MAINTENANCE THERAPY Preferred Regimens • Lenalidomide h (category 1)
Other Recommended Regimens • Carfilzomib/lenalidomide i,h
• Daratumumab/lenalidomide i,h
Useful In Certain Circumstances • Bortezomib ± lenalidomidei, h
• Ixazomib (category 2B) a Selected, but not inclusive of all regimens. The regimens under each preference category are listed by order of NCCN Category of Evidence and Consensus alphabetically. b Supportive Care Treatment for Multiple Myeloma (MYEL-H). c General Considerations for Myeloma Therapy (MYEL-F). d Management of Renal Disease in Multiple Myeloma (MYEL-K). e Preferred primarily as initial treatment in patients with acute renal insufficiency or those who have no access to proteasome inhibitor (PI)/lenalidomide/dexamethasone. Consider switching to PI/lenalidomide/dexamethasone after renal function improves. f Treatment option for patients with renal insufficiency and/or peripheral neuropathy. g Generally reserved for the treatment of aggressive MM. h There appears to be an increased risk for secondary cancers, especially with lenalidomide maintenance following transplant. The benefits and risks of maintenance therapy vs. secondary cancers should be discussed with patients. i Two drug maintenance recommended for high-risk MM.
Figure 1. MYEL-G, 1 of 5. NCCN Clinical Practice Guidelines in Oncology for Multiple Myeloma, Version 1.2025.
CE NCCN GUIDELINES ® INSIGHTS Multiple Myeloma, Version 1.2025 134 © JNCCN —Journal of the National Comprehensive Cancer Network | Volume 23 Issue 5 | May 2025 toxicity. The percentage of patients with MRD-negative status for at least 12 months was higher in the D -VRd group than in the VRd group (64.8% vs 29.7%). 4 The percentage of patients with MRD-negative status assessed at a sensitivity threshold of 10 26was 65.1% in the D -VRd group and 32.2% in the VRd group. The safety pro fi le of D-VRd was consistent with known pro-
fi les of the individual agents. Common grade 3 or 4 adverse events observed more frequently in the daratumumab arm compared with the control arm included neutropenia (62.1% vs 51%) and thrombocytopenia (29.1% and 17.3%). Serious adverse events occurred in 57.0% of patients treated with D -VRd compared with 49.3% of those treated with VRd. 4The fi nal analysis of the GRIFFIN trial also supports the clinical bene fi t of adding daratumumab to VRd, demonstrating a signi fi cant improvement in depth of response (stringent CR of 67% vs 48% with VRd) and a 4-year PFS rate of 87.2% versus 70% with VRd (HR, 0.45; 95% CI, 0.21 –0.95). 5Lastly, a subgroup analysis comparing D-VRd versus VRd from a systematic review and meta-analysis of daratumumab-based quadruplet versus triplet induction regimens in patients with NDMM who are eligible for transplant (n 53,327) from 3 randomized controlled trials and 1 nonrandomized controlled study further con fi rmed signi fi cant improvements in both over-all survival (OS) (pooled HR, 0.68; 95% CI, 0.48 –0.97; P5.03;
I 250%) and PFS (pooled HR, 0.41; 95% CI, 0.31 –0.54; P ,.00001;
I 250%) with the addition of daratumumab. 6Therefore, because adding daratumumab to VRd showed signi fi cant improvement in PFS compared with VRd, the triple-drug regimen VRd (category 1) was moved from “Preferred ”
to “Other Recommended Regimens. ” Based on a similar ratio-nale, because the ENDURANCE trial 7 demonstrated compara-ble PFS between car fi lzomib/lenalidomide/dexamethasone and bortezomib/lenalidomide/dexamethasone, car fi lzomib/ lenalidomide/dexamethasone was also moved from “Preferred ”
to “Other Recommended Regimens ” (Figure 1). Isatuximab/bortezomib/lenalidomide/dexamethasone (Isa-VRd) was also added to the list of “Other Recommended Regimens ” for patients who are HCT-eligible based on data from the phase III GMMG-HD7 trial (Figure 1). In this trial, pa-tients with NDMM (n 5660) eligible for autologous HCT were randomly assigned to receive three 42-day cycles of VRd in-duction therapy with or without isatuximab. An initial analy-sis found that Isa-VRd led to deeper responses and higher rates of MRD negativity ( ,10 25 by next-generation fl ow [NGF]) after induction (50% vs 36%; odds ratio [OR], 1.82; 95% CI, 1.33 –2.48). The addition of isatuximab resulted in higher inci-dences of grade 3 or 4 neutropenia (23% vs 7%) and grade 3 or 4 infections (12% vs 10%). One treatment-related death (from sep-tic shock) occurred in the isatuximab group, whereas 4 were re-ported in the VRd group (one due to cardiac decompensation, one from hepatic and renal failure, one from cardiac arrest, and one from drug-induced enteritis). 8 The fi nal analysis of outcomes following the fi rst randomization to HCT has now been pub-lished. At the time of the publication of Version 1.2025 of the NCCN Guidelines, these data were not yet available. The NGF MRD-negative rates continued to deepen, and after HCT, they were 66% with Isa-VRd versus 48% with VRd. 9 Importantly, at a median follow-up of 4 years from the fi rst random assign-ment, the data showed that 18-week induction with Isa-VRd signi fi cantly prolonged PFS compared with VRd, regardless of maintenance therapy (HR, 0.70; 95% CI, 0.52 –0.95; P 5.0184), with PFS not yet reached in either arm. 9Another anti –CD-38 regimen, isatuximab/car fi lzomib/ lenalidomide/dexamethasone (Isa-KRd), was added by the panel to the list of regimens “Useful in Certain Circumstances ” based on data from the phase II GMMG-CONCEPT trial and the phase III IsKia trial (Figure 1). The GMMG-CONCEPT trial eval-uated Isa-KRd for the treatment of high-risk NDMM in patients who were HCT-eligible and HCT-ineligible (n 5125). High-risk disease was de fi ned by ISS stage II or III and speci fi c cytogenetic abnormalities, such as del(17p), t(4;14), t(14;16), or .3 copies of 1q21. Patients received induction therapy followed by consoli-dation with Isa-KRd plus maintenance therapy with Isa-KR. Patients eligible for transplant received high-dose melphalan followed by HCT, whereas those not eligible for transplant re-ceived 2 additional cycles of Isa-KRd after initial treatment. The primary endpoint was a MRD negativity rate of ,10 25 by NGF after consolidation, with PFS as the secondary endpoint. After consolidation, the MRD negativity rates were 67.7% in patients eligible for HCT and 54.2% in those who were ineligible. In total, 81.8% of patients who were HCT-eligible achieved MRD nega-tivity, with 62.6% sustaining it for at least 1 year. At the time of publication, the median PFS had not yet been reached in either group. 10 The phase III IsKia trial assessed the ef fi cacy and safety of Isa-KRd compared with KRd in patients with NDMM who are eligible for transplant, fi nding that Isa-KRd signi fi cantly increased MRD negativity rates after induction and consolidation phases, with 77% of patients in the Isa-KRd group achieving MRD negativity by next-generation sequencing (NGS) at the ,10 25threshold after consolidation, compared with 67% in the KRd group. Isa-KRd also demonstrated higher MRD negativity rates across all subgroups, including patients with high-risk MM. 11 Additionally, regimens that are no longer used as fi rst-line therapy in the United States and were previously listed as
“regimens useful in certain circumstances ” were removed from the guidelines in Version 1.2025. These include bortezomib/ doxorubicin/dexamethasone and the thalidomide-containing 4-drug regimen daratumumab/bortezomib/thalidomide/dexa-methasone (Figure 1). For patients ineligible for HCT, Isa-VRd was included as a category 1 option for those aged ,80 years who are not frail, based on data from the following trials (Figure 2). The registra-tional, phase III, open-label IMROZ trial randomized patients (n 5446) aged #80 years with NDMM who were not eligible for autologous stem cell transplantation to receive either Isa-VRd or VRd in four 6-week cycles, followed by continuous treatment in 4-week cycles of Isa-Rd versus Rd, respectively. At a median follow-up of 59.7 months, the estimated PFS was 63.2% in the Isa-VRd group compared with 45.2% in the VRd group, with a 40% re-duction in risk of disease progression or death (HR, 0.60; 95% CI, 0.44 –0.81; P5.0009). The rate of CR or better was signi fi cantly higher in the isatuximab group (74.7% vs 64.1%; P5.01), and the proportion of patients achieving both MRD-negative status ( ,10 25by NGS) and CR was also higher (55.5% vs 40.9%; P 5.003). 12 No-tably, patients aged $65 years were deemed ineligible for HCT, and those aged .80 years were ineligible for the study. The inci-dence of peripheral sensory neuropathy was not higher with Isa-VRd than with VRd in this trial (all grades, 54.4% vs 60.8%, respectively); this high rate re fl ects the use of twice-weekly bortezomib dosing.
Multiple Myeloma, Version 1.2025
NCCN GUIDELINES ® INSIGHTS CE JNCCN.org | Volume 23 Issue 5 | May 2025 135 The nonregistrational phase III BENEFIT/IFM2020-05 study demonstrated the ef fi cacy and safety of isatuximab combined with weekly VRd compared with the triplet combination of isatuximab 1 lenalidomide/dexamethasone in patients with NDMM who are ineligible for HCT. 13 The NGS MRD negativity rate at 10 25 at 18 months was signi fi cantly higher in the quadruplet arm, with 71 patients (53%; 95% CI, 44% –61%) achieving MRD negativity compared with 35 patients (26%; 95% CI, 19% –34%) in the triplet arm. The OR for MRD negativity in the quadruplet group, compared with the triplet group, was 3.16 (95% CI, 1.89 –5.28; P ,.0001). 13 The weekly and bimonthly schedule of subcutaneous bortezomib (cycles 1 –12 and 13 –18, respectively) in the Isa-VRd arm led to a higher rate of peripheral neuropathy: grade $2 occurred in 37 (27%) patients (4 had grade 3 neuropa-thy) in the Isa-VRd arm versus 13 (10%) patients (1 had grade 3 neuropathy) in the Isa-Rd arm. However, as expected, the inci-dence was lower than with twice-weekly dosing. The panel also voted to move daratumumab/cyclophos-phamide/bortezomib/dexamethasone, which was previously included based on the LYRA study, 14 from the list of “Other Recom-mended Regimens ” to “Useful in Certain Circumstances, ” be-cause it is mainly used as a treatment option for patients with renal insuf fi ciency or acute kidney injury at the time of presentation (Figure 2). Although randomized controlled efforts are still lack-ing, mounting evidence highlights the feasibility and impor-tance of early initiation of daratumumab-based treatment in patients with MM who present with acute kidney injury in order to induce rapid and signi fi cant reductions in disease burden, improve renal outcomes, and provide a plasmaphe-resis-free approach. 15 Isa-KRd was added to the list of regimens “Useful in Certain Circumstances ” for patients ineligible for HCT who have high-risk cytogenetics based on the results of the GMMG-CONCEPT trial (Figure 2). However, because this trial included very few patients who were ineligible for HCT, the data supporting its inclusion for non-HCT candidates are based more on extrapola-tion from the HCT-eligible population. As a result, there was no uniform consensus on the use of this regimen in this setting, given the availability of other treatment options, and therefore it is a category 2B recommendation. Ixazomib-containing regimens are currently FDA-indicated for the treatment of patients with MM who have received at least one prior therapy. In the updated version, the panel has added a footnote noting that in patients with NDMM who are receiving regimens containing bortezomib or car fi lzomib, these drugs may be substituted with ixazomib only in select patients with intolerance to bortezomib/car fi lzomib 16 or for logistical reasons. 17 The panel discussed examples of the logis-tical reasons for considering the use of ixazomib, which included patients who are assessed as very frail with many comorbid con-ditions, those unable to travel to a treatment site, or situations arising during public health emergencies.
Updates to Maintenance Therapy Recommendations
The NCCN panel clari fi ed in the algorithm section that the maintenance regimens are appropriate for patients who have received autologous HCT and classi fi ed them as either Pre-ferred, Other Recommended, or Useful in Certain Circumstan-ces. For patients who are not eligible for autologous HCT, in general, primary therapy is continued until progression, with de-escalation of therapy or modi fi cations to dose and duration based on treatment response and toxicity, as needed. Therefore, given the nuances needed in this setting, such as needing to con-tinue induction therapy, de-escalate therapy as appropriate, make dose modi fi cation, or discontinue treatment, the panel re-moved the list of maintenance options previously included for those not eligible for autologous HCT. Based on the results of the phase III PERSEUS trial 4 and the phase II GRIFFIN trial, 5 the panel moved daratumumab/ lenalidomide as a maintenance therapy option from “Useful in Certain Circumstances ” to “Other Recommended Regimens ”
(Figure 1). It is important to note that these trials did not random-ize the maintenance therapy after induction plus consolidation, and therefore it is unclear whether the MRD negativity and PFS bene fi t observed with daratumumab were due to the induction
Other Recommended Regimens
Version 1.2025 © 2025 National Comprehensive Cancer Network ®(NCCN ®). All rights reserved. The NCCN Guidelines ®and this illustration may not be reproduced in any form without the express written permission of NCCN.
MYEL-G 2 OF 5
aSelected, but not inclusive of all regimens. The regimens under each preference category are listed by order of NCCN Category of Evidence and Consensus alphabetically. bSupportive Care Treatment for Multiple Myeloma (MYEL-H). cGeneral Considerations for Myeloma Therapy (MYEL-F). dManagement of Renal Disease in Multiple Myeloma (MYEL-K). ePreferred primarily as initial treatment in patients with acute renal insufficiency or those who have no access to PI/lenalidomide/dexamethasone. Consider switching to PI/lenalidomide/dexamethasone after renal function improves.
PRIMARY THERAPY FOR NON-TRANSPLANT CANDIDATES a-d,j In general, continue primary therapy until progression with de-escalation of therapy (modification of dose and duration) as needed. Preferred Regimens • Daratumumab/lenalidomide/dexamethasone (category 1)
• Isatuximab-irfc/bortezomib/lenalidomide/dexamethasone (for patients <80 years old who are not frail)(category 1)
• Lenalidomide/bortezomib/dexamethasone (category 1)
• Carfilzomib/lenalidomide/dexamethasone
Useful In Certain Circumstances • Lenalidomide/low-dose dexamethasone (category 1)
• Bortezomib/cyclophosphamide/dexamethasone e
• Bortezomib/dexamethasone
• Bortezomib/lenalidomide/dexamethasone (VRD-lite) for patients assessed as being frail
• Carfilzomib/cyclophosphamide/dexamethasone e,f
• Daratumumab/cyclophosphamide/bortezomib/dexamethasone k
• Isatuximab-irfc/carfilzomib/lenalidomide/dexamethasone (category 2B)
• Lenalidomide/cyclophosphamide/dexamethasone f Treatment option for patients with renal insufficiency and/or peripheral neuropathy. j Bortezomib or carfilzomib may be substituted with ixazomib in select patients in case of intolerance/logistical reasons. k Treatment option for patients with renal insufficiency.
Figure 2. MYEL-G, 2 of 5. NCCN Clinical Practice Guidelines in Oncology for Multiple Myeloma, Version 1.2025.
CE NCCN GUIDELINES ® INSIGHTS Multiple Myeloma, Version 1.2025 136 © JNCCN —Journal of the National Comprehensive Cancer Network | Volume 23 Issue 5 | May 2025 followed by consolidation therapy with the daratumumab-containing regimen, or whether adding daratumumab to the maintenance regimen provided additional bene fi t. The panel strongly encourages enrolling patients in clinical trials in gen-eral, and especially to clarify remaining questions around maintenance therapy. Based on the results from the FORTE 18 and ATLAS 19 trials, the panel moved car fi lzomib/lenalidomide as maintenance op-tion from “Useful in Certain Circumstances ” to “Other Recom-mended Regimens ” (Figure 1). The FORTE trial demonstrated improved PFS with car fi lzomib/lenalidomide maintenance compared with lenalidomide maintenance (3-year PFS from randomization to maintenance, 75% vs 65%; HR, 0.64; 95% CI, 0.44 –0.94; P 5.023). The randomized phase III ATLAS trial eval-uated car fi lzomib/lenalidomide/dexamethasone versus single-agent lenalidomide as maintenance therapy following autologous HCT in patients with newly diagnosed MM. After a median follow-up of 33.8 months, the car fi lzomib/lenalidomide/dexamethasone group demonstrated a median PFS of 59.1 months, compared with 41.6 months in the lenalidomide group, representing a 49% reduction in the risk of disease progression or death. Bortezomib monotherapy is included as a maintenance ther-apy option based on the results of the HOVON 20 and UPFRONT 21 study data. However, the isolated bene fi t of bortezomib mono-therapy in the maintenance setting is unclear. Therefore, given the availability of other maintenance options, the NCCN Multiple Myeloma Panel members moved bortezomib monotherapy as a maintenance therapy option to “Useful in Certain Circum-stances, ” such as when a patient is unable to tolerate lenalidomide due to toxicity (Figure 1).
Updates to Treatment Options for Previously Treated MM
Editor ’s Note: Updates included in Version 2.2025 were not available at press time, these can be found at NCCN.org. The choice of appropriate therapy for a speci fi c patient with relapsed/refractory disease depends on the context of the clinical relapse, such as prior treatment and duration of response. As a general principle, to maximize the bene fi t of systemic therapy, if disease relapse occurs at least 6 months after stopping therapy, previous regimens may be reconsidered or repeated. For patients still sensitive to daratumumab, bortezomib, and/or lenalidomide, any of the regimens listed in guidelines for previously treated MM may be appropriate. However, because daratumumab-, bortezomib-, and lenalidomide-containing regimens are often given as induction therapy, it is likely that the disease may not be sensitive to one or more of these agents at relapse or if relapse is well within 6 months of primary treatment completion. In such cases, regimens that do not include previously used agents are preferred. Most of the updates in this setting in Version 1.2025 were for disease relapses after 1 to 3 prior therapies. Considering that anti-CD38 –containing regimens are listed as “Preferred ” for primary therapy in patients with MM, the NCCN panel has provided a list of regimens for anti-CD38 –
refractory disease in Version 1.2025. In the previous version of the guidelines, the panel included a list of regimens for bortezomib-refractory and lenalidomide-refractory disease (Figure 3). The preferred regimens for anti-CD38 –refractory disease include the following non –anti-CD38 –containing regimens: car fi lzomib/pomalidomide/dexamethasone, car fi lzomib/ lenalidomide/dexamethasone, pomalidomide/bortezomib/ dexamethasone, and elotuzumab/pomalidomide/dexamethasone. Car fi lzomib/lenalidomide/dexamethasone has been speci-
fi ed as a category 1 option (assigned to situations where patients are refractory to anti-CD38 antibody but not to lenalidomide treatment) based on the results of the randomized, multicenter, phase III ASPIRE trial. 22 This trial studied the combination of lenalidomide and dexamethasone with or without car fi lzomib in patients (n 5792) with relapsed/refractory MM who had received 1 to 3 prior lines of therapy and were sensitive to both lenalidomide and dexamethasone. The primary endpoint of the study was PFS. The results showed that adding car fi lzomib to lenalidomide 1 dexamethasone signi fi cantly improved PFS by 8.7 months (26.3 months for the car fi lzomib arm vs 17.6 months for the lenalidomide 1 low-dose dexamethasone arm; HR for progression or death, 0.69; 95% CI, 0.57 –0.83; P 5.0001). The median duration of treatment was longer in the car fi lzomib group (88 vs 57 weeks). The incidence of peripheral neuropathy was identical in both arms (17%). Nonhematologic adverse effects (grade $3) that were more common in the car fi lzomib group compared with lenalidomide 1 dexamethasone included dyspnea (2.8% vs 1.8%), cardiac failure (3.8% vs 1.8%), and hypertension (4.3% vs 1.8%). Fewer discontinuations due to side effects occurred in the car fi lzomib arm (15.3% vs 17.7%). Patients in the car fi lzomib arm reported superior health-related quality of life compared with those receiving lenalidomide 1 dexamethasone alone. Car fi lzomib/pomalidomide/dexamethasone has been in-cluded as a treatment option for patients who are refractory to anti-CD38 antibody treatment, based on data from several phase II trials. 23,24 The fi rst was a phase II trial that investigated car fi lzomib/pomalidomide/dexamethasone, followed by con-tinuous pomalidomide/dexamethasone, as second-line therapy for relapsed/refractory MM in patients whose disease progres-sion during lenalidomide maintenance therapy and in those eligi-ble for HCT who had not yet received it. On this regimen, 75% of patients experienced at least a very good partial response (VGPR), and 37% reached a CR or better. At 40-months of follow-up, the median PFS was 26 months for patients who received therapy with HCT, and 17 months for those who received car fi lzomib/ pomalidomide/dexamethasone therapy without HCT. The me-dian OS was 67 months, with the most common grade 3 and 4 ad-verse events related to treatment including hematologic toxicity (41%), cardiovascular events (6%), respiratory (3%) events, and in-fections (17%). 23 Because its inclusion is based on the results of a phase II study, it is included as a category 2A recommendation. Pomalidomide/bortezomib/dexamethasone has also been in-cluded as a category 1 option based on the results of the OPTIMISMM study, a phase III open-label, multicenter, randomized trial. 25 The trial compared pomalidomide/bortezomib/dexamethasone (n 5281) versus bortezomib/dexamethasone in patients (n 5278) with relapsed or refractory MM who had previously received lenalidomide. 25 After a median follow-up of 15.9 months, the pomalidomide arm showed a signi fi cantly improved PFS (me-dian, 11.20 vs 7.10 months; HR, 0.61; 95% CI, 0.49 –0.77; P ,.0001). The most common grade 3/4 treatment-related adverse events in the pomalidomide arm were neutropenia, infections, and throm-bocytopenia. 25 A post hoc subgroup analysis of the OPTIMISMM trial evaluated outcomes in 226 patients at fi rst relapse who had received only one prior line of therapy. Analyses were
Multiple Myeloma, Version 1.2025
NCCN GUIDELINES ® INSIGHTS CE JNCCN.org | Volume 23 Issue 5 | May 2025 137 conducted based on lenalidomide-refractory status, prior bor-tezomib exposure, and prior HCT. Statistically signi fi cant im-provements in PFS were observed in both patients with lenalidomide-refractory disease (17.8 vs 9.5 months; P 5.0276) and those with lenalidomide-nonrefractory disease (22.0 vs 12.0 months; P 5.0491). There were also statistically signi fi cant improvements in PFS among patients who had received prior bortezomib (17.8 vs 12 months) and those with (22 vs 13.8 months) and without (16.5 vs 9.5 months) prior HCT. 26 Elotuzumab/pomalidomide/dexamethasone (EPd) is in-cluded as an option for patients who have received at least 2 prior therapies including an immunomodulatory agent (IMiD) and a proteasome inhibitor (PI). In a phase II study, 117 patients with refractory/relapsed MM refractory to lenalidomide and a PI were randomized to receive either pomalidomide/dexametha-sone or elotuzumab/pomalidomide/dexamethasone. 27 After a follow-up of 9.1 months, the median PFS and overall response rate (ORR) were both more than double with elotuzumab (PFS, 10.3 vs 4.7 months; ORR, 53% vs 26%). 27 A more recent follow-up analy-sis showed that median OS was also signi fi cantly improved with elotuzumab/pomalidomide/dexamethasone compared with pomalidomide/dexamethasone (29.8 vs 17.4 months; HR, 0.59; 95% CI, 0.37 –0.93; P 5.0217). 27 Of note, daratumumab had not yet been approved for use in earlier lines of therapy at the time of this study (only 3 patients received daratumumab as prior ther-apy). Therefore, further exploration of the use of EPd in patients who are refractory to anti-CD38 antibodies is warranted. 28 Ixazomib/pomalidomide/dexamethasone has been included as an option for patients who experience relapse after 2 prior therapies, including an immunomodulatory drug and a PI, with disease progression on/within 60 days of completing the last therapy. This is based on data from the following trials. In the phase I/II Alliance A061202 study (n 529), patients with lenalidomide/PI-refractory MM were treated with ixazomib/ pomalidomide/dexamethasone. In this trial, 51.7% of patients achieved at least a partial response (PR), with a median PFS of 4.4 months, a median response duration of 16.8 months, and a median OS of 34.3 months. Common adverse events included hematologic toxicity and gastrointestinal events. 29 Another phase I/II study evaluated the safety and ef fi cacy of ixazomib/ pomalidomide/dexamethasone in patients who received multiple prior therapies. 30 With a median follow-up of 11.9 months, at an established ixazomib dose of 4 mg, 48% of patient achieved at least a PR, with 20% achieving a VGPR and 76% having stable disease. The most common grade $2 adverse events reported were anemia, neutropenia, thrombocytopenia, and infections. 31
Updates to CAR-T Cell Therapy for Previously Treated MM
The panel discussed the inclusion of CAR T-cell therapies for early relapse, based on the FDA approvals and the results of the CARTITUDE-4 trial 32 for ciltacabtagene autoleucel (cilta-cel) in patients who have received at least one prior line of therapy, including a PI and an IMiD, and are refractory to lenalidomide. The panel also considered the results of the KarMMa-3 trial 33 for idecabtagene vicleucel (ide-cel) in patients with triple-class –exposed relapsed/refractory MM who had received 2 to 4 prior lines of therapy.
Version 1.2025 © 2025 National Comprehensive Cancer Network ®(NCCN ®). All rights reserved. The NCCN Guidelines ®and this illustration may not be reproduced in any form without the express written permission of NCCN.
MYEL-G 3 OF 5
aSelected, but not inclusive of all regimens. The regimens under each preference category are listed by order of NCCN Category of Evidence and Consensus alphabetically. b Supportive Care Treatment for Multiple Myeloma (MYEL-H). c General Considerations for Myeloma Therapy (MYEL-F). d Management of Renal Disease in Multiple Myeloma (MYEL-K). l Regimens included under 1–3 prior therapies can also be used later in the disease course. Attempt should be made to use drugs/drug classes the patients have not been exposed to or exposed to >1 line prior. mAutologous HCT should be considered in patients who are eligible and have not previously received HCT or had a prolonged response to initial HCT. nIn order to maximize benefit of systemic therapy, agents/regimens may be reconsidered or repeated if relapse is after at least 6 months of stopping therapy. oAlkylating agents can impact the ability to collect T cells for CAR T-cell therapy. See NCCN Guideline for Management of Immunotherapy-Related Toxicities.
For Other Recommended Regimens and for regimens Useful in Certain Circumstances for Relapsed/Refractory Disease After 1–3 Prior Therapies, see MYEL-G 4 of 5
THERAPY FOR PREVIOUSLY TREATED MULTIPLE MYELOMA a-d,l-o Relapsed/Refractory Disease After 1–3 Prior Therapies Preferred Regimens
Order of regimens does not indicate comparative efficacy
Anti-CD-38 Refractory Bortezomib-Refractory Lenalidomide-Refractory • Carfi lzomib/lenalidomide/dexamethasone (category 1)
• Carfi lzomib/pomalidomide/dexamethasone
• Pomalidomide/bortezomib/dexamethasone (category 1)
After two prior therapies including lenalidomide and a PI
Elotuzumab/pomalidomide/dexamethasone
After two prior therapies including an IMiD and a PI and with disease progression on/within 60 days of completion of last therapy
Ixazomib/pomalidomide/dexamethasone • Carfilzomib/lenalidomide/dexamethasone (category 1) • Daratumumab/carfilzomib/dexamethasone (category 1) • Daratumumab/lenalidomide/dexamethasone (category 1) • Isatuximab-irfc/carfilzomib/dexamethasone (category 1) • Carfilzomib/pomalidomide/dexamethasone
After one prior therapy including lenalidomide and a PI
Daratumumab/pomalidomide/dexamethasone (category 1)
After two prior therapies including lenalidomide and a PI
Isatuximab-irfc/pomalidomide/dexamethasone (category 1) Elotuzumab/pomalidomide/dexamethasone
• Daratumumab/bortezomib/dexamethasone (category 1)
• Daratumumab/carfilzomib/dexamethasone (category 1)
• Isatuximab-irfc/carfilzomib/dexamethasone (category 1)
• Pomalidomide/bortezomib/dexamethasone (category 1)
• Carfilzomib/pomalidomide/dexamethasone
After one prior therapy including lenalidomide and a PI
Daratumumab/pomalidomide/dexamethasone (category 1)
After two prior therapies including lenalidomide and a PI
Isatuximab-irfc/pomalidomide/dexamethasone (category 1) Elotuzumab/pomalidomide/dexamethasone
After two prior therapies including an IMiD and a PI and with disease progression on/within 60 days of completion of last therapy
Ixazomib/pomalidomide/dexamethasone
CAR T-Cell Therapy
After one prior line of therapy including IMiD and a PI, and refractory to lenalidomide
Ciltacabtagene autoleucel (category 1)
After two prior lines of therapies including an IMiD, an anti-CD38 monoclonal antibody and a PI
Idecabtagene vicleucel (category 1)
Figure 3. MYEL-G, 3 of 5. NCCN Clinical Practice Guidelines in Oncology for Multiple Myeloma, Version 1.2025.
CE NCCN GUIDELINES ® INSIGHTS Multiple Myeloma, Version 1.2025 138 © JNCCN —Journal of the National Comprehensive Cancer Network | Volume 23 Issue 5 | May 2025 In the phase III CARTITUDE-4 study, patients treated with cilta-cel therapy (n 5208) were compared with those treated with standard regimens (n 5211). 32 In the cilta-cel arm, the rate of CR or better was 73.1%, compared with 21.8% in the standard regi-men arm (OR, 10.3; 95% CI, 6.5 –16.4). The ORRs in the cilta-cel and standard-regimen arms were 84.6% and 67.3%, respectively (OR, 3.0; 95% CI, 1.8 –5.0). MRD negativity ( ,10 25 determined by NGS) was higher in the cilta-cel arm, at 60.6%, compared with 15.6% in the standard regimen arm (OR, 8.7; 95% CI, 5.4 –13.9). At a median follow-up of 15.9 months (range, 0.1 –27.3), the me-dian PFS was not yet reached with cilta-cel and was 11.8 months with standard regimens (HR, 0.26; 95% CI, 0.18 –0.38; P ,.001). 32 Based on these data, the panel included cilta-cel as an option for patients after one prior line of therapy, including an IMiD and a PI, and whose disease is refractory to lenalidomide (Figure 3). Importantly, after median follow-up 33.6 months, cilta-cel sig-ni fi cantly improved OS, with a 30-month OS of 76.4% versus 63.8% in the standard regimen arm (HR, 0.55; 95% CI, 0.39 –0.79;
P 5.0009), with consistent OS bene fi t observed across prespeci-
fi ed subgroups. 34 The peer-reviewed publication of this key follow-up analysis is awaited. In the phase III KarMMa-3 trial, patients treated with ide-cabtagene vicleucel (ide-cel; n 5254) were compared with those treated with standard regimens (n 5132). 33 In this trial, 66% of the patients had triple-class –refractory disease, and 95% had daratumumab-refractory disease. At a median follow-up of 18.6 months, the median PFS was 13.3 months (95% CI, 11.8 –16.1) in the ide-cel arm, compared with 4.4 months (95% CI, 3.4 –5.9) in the standard-regimen arm, leading to a 51% reduction in the risk of disease progression or death (HR for disease progression or death, 0.49; 95% CI, 0.38 –0.65; P ,.001). Ide-cel also demonstrated an improvement in ORR versus stan-dard regimens, with 71% (95% CI, 66% –77%) of patients achiev-ing a response, compared with 42% (95% CI, 33% –50%) of those who received standard regimens ( P ,.0001). 33 Based on these results, the panel included ide-cel as an option for patients after 2 prior lines of therapy, including an IMiD, an anti-CD38 mono-clonal antibody, and a PI (Figure 3).
Conclusions
The NCCN Guidelines are in continuous evolution, with updates occurring annually or more frequently if new, high-quality clinical data become available. These NCCN Guidelines In-sights highlight the important updates in Version 1.2025 of the NCCN Guidelines for MM speci fi c to the systemic therapy op-tions for newly diagnosed and previously treated MM. The rec-ommendations in the NCCN Guidelines are based on evidence from clinical trials. Expert clinical judgment is required when applying these guidelines in the context of individual clinical circumstances to provide optimal care. The physician and the patient have the responsibility to jointly explore and select the most appropriate treatment from among the available options. Consistent with NCCN philosophy, when possible, the panel strongly encourages participation in prospective clinical trials when available and applicable.
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National Cancer Institute. SEER cancer stat facts: myeloma. Accessed March 31, 2025. Available at mulmy.html 2. Siegel RL, Kratzer TB, Giaquinto AN, et al. Cancer statistics, 2025. CA Cancer J Clin 2025;75:10 –45. 3. Diao X, Ben T, Cheng S, et al. Global, regional, and national multiple myeloma burden from 1990 to 2021: a systematic analysis for of the Global Burden of Disease Study 2021. BMC Public Health 2025;25: 1054. 4. Sonneveld P, Dimopoulos MA, Boccadoro M, et al. Daratumumab, bortezomib, lenalidomide, and dexamethasone for multiple myeloma. N Engl J Med 2024;390:301 –313. 5. Voorhees PM, Sborov DW, Laubach J, et al. Addition of daratumumab to lenalidomide, bortezomib, and dexamethasone for transplantation-eligible patients with newly diagnosed multiple myeloma (GRIFFIN):
fi nal analysis of an open-label, randomised, phase 2 trial. Lancet Haematol 2023;10:e825 –837. 6. Souto Filho JTD, Cantadori LO, Crusoe EQ, et al. Daratumumab-based quadruplet versus triplet induction regimens in transplant-eligible newly diagnosed multiple myeloma: a systematic review and meta-analysis. Blood Cancer J 2025;15:37. 7. Kumar SK, Jacobus SJ, Cohen AD, et al. Car fi lzomib or bortezomib in combination with lenalidomide and dexamethasone for patients with newly diagnosed multiple myeloma without intention for immediate autologous stem-cell transplantation (ENDURANCE): a multicentre, open-label, phase 3, randomised, controlled trial. Lancet Oncol 2020; 21:1317 –1330. 8. Goldschmidt H, Mai EK, Bertsch U, et al. Addition of isatuximab to le-nalidomide, bortezomib, and dexamethasone as induction therapy for newly diagnosed, transplantation-eligible patients with multiple mye-loma (GMMG-HD7): part 1 of an open-label, multicentre, randomised, active-controlled, phase 3 trial. Lancet Haematol 2022;9:e810 –821. 9. Mai EK, Bertsch U, Pozek E, et al. Isatuximab, lenalidomide, bortezomib, and dexamethasone induction therapy for transplant-eligible newly diagnosed multiple myeloma: fi nal part 1 analysis of the GMMG-HD7 trial. J Clin Oncol 2024:JCO2402266. 10. Leypoldt LB, Tichy D, Besemer B, et al. Isatuximab, car fi lzomib, lenalido-mide, and dexamethasone for the treatment of high-risk newly diagnosed multiple myeloma. J Clin Oncol 2024;42:26 –37. 11. Gay F, Roeloffzen W, Dimopoulos MA, et al. Results of the phase III ran-domized IsKia trial: isatuximab-car fi lzomib-lenalidomide-dexamethasone vs car fi lzomib-lenalidomide-dexamethasone as pre-transplant induction and post-transplant consolidation in newly diagnosed multiple myeloma patients. Blood 2023;142(Suppl 1):Abstract 4. 12. Facon T, Dimopoulos MA, Leleu XP, et al. Isatuximab, bortezomib, lenali-domide, and dexamethasone for multiple myeloma. N Engl J Med 2024; 391:1597 –1609. 13. Leleu X, Hulin C, Lambert J, et al. Isatuximab, lenalidomide, dexametha-sone and bortezomib in transplant-ineligible multiple myeloma: the ran-domized phase 3 BENEFIT trial. Nat Med 2024;30:2235 –2241. 14. Yimer H, Melear J, Faber E, et al. Daratumumab, bortezomib, cyclophos-phamide and dexamethasone in newly diagnosed and relapsed multiple myeloma: LYRA study. Br J Haematol 2019;185:492 –502. 15. Kim EB, Malespini JE, Lei M, et al. Early daratumumab therapy improves renal outcomes in newly diagnosed myeloma patients admitted with kidney injury. Blood Adv. Published online March 14, 2025. doi:10.1182/bloodadvances.2025015901 16. Rifkin RM, Costello CL, Birhiray RE, et al. In-class transition from bortezo-mib-based therapy to IRd is an effective approach in newly diagnosed multiple myeloma. Future Oncol 2024;20:131 –143. 17. Rifkin RM, Girnius SK, Noga SJ, et al. In-class transition (iCT) of protea-some inhibitor-based therapy: a community approach to multiple mye-loma management. Blood Cancer J 2023;13:147. 18. Gay F, Musto P, Rota-Scalabrini D, et al. Car fi lzomib with cyclophospha-mide and dexamethasone or lenalidomide and dexamethasone plus autologous transplantation or car fi lzomib plus lenalidomide and dexamethasone, followed by maintenance with car fi lzomib plus lenalidomide or lenalidomide alone for patients with newly diagnosed
Multiple Myeloma, Version 1.2025 NCCN GUIDELINES ® INSIGHTS CE JNCCN.org | Volume 23 Issue 5 | May 2025 139 multiple myeloma (FORTE): a randomised, open-label, phase 2 trial. Lancet Oncol 2021;22:1705 –1720. 19. Dytfeld D, Wrobel T, Jamroziak K, et al. ATLAS: a phase 3 randomized trial of car fi lzomib, lenalidomide, and dexamethasone versus lenalido-mide alone after stem-cell transplant for multiple myeloma. J Clin Oncol 2022;40(16 Suppl):Abstract 8001. 20. Sonneveld P, Schmidt-Wolf IG, van der Holt B, et al. Bortezomib induc-tion and maintenance treatment in patients with newly diagnosed multiple myeloma: results of the randomized phase III HOVON-65/ GMMG-HD4 trial. J Clin Oncol 2012;30:2946 –2955. 21. Niesvizky R, Flinn IW, Rifkin R, et al. Community-based phase IIIB trial of three UPFRONT bortezomib-based myeloma regimens. J Clin Oncol 2015;33:3921 –3929. 22. Stewart AK, Rajkumar SV, Dimopoulos MA, et al. Car fi lzomib, lenalido-mide, and dexamethasone for relapsed multiple myeloma. N Engl J Med 2015;372:142 –152. 23. Sonneveld P, Zweegman S, Cavo M, et al. Car fi lzomib, pomalidomide, and dexamethasone as second-line therapy for lenalidomide-refractory multiple myeloma. Hemasphere 2022;6:e786. 24. Perrot A, Delimpasi S, Spanoudakis E, et al. An open-label phase 2 study treating patients with fi rst or second relapse of multiple myeloma with car fi lzomib, pomalidomide, and dexamethasone (KPd): SELECT study. Leuk Lymphoma 2024;65:833 –842. 25. Richardson PG, Oriol A, Beksac M, et al. Pomalidomide, bortezomib, and dexamethasone for patients with relapsed or refractory multiple myeloma previously treated with lenalidomide (OPTIMISMM): a randomised, open-label, phase 3 trial. Lancet Oncol 2019;20:781 –794. 26. Dimopoulos M, Weisel K, Moreau P, et al. Pomalidomide, bortezomib, and dexamethasone for multiple myeloma previously treated with lenalidomide (OPTIMISMM): outcomes by prior treatment at fi rst relapse. Leukemia 2021;35:1722 –1731. 27. Dimopoulos MA, Dytfeld D, Grosicki S, et al. Elotuzumab plus pomalido-mide and dexamethasone for multiple myeloma. N Engl J Med 2018; 379:1811 –1822. 28. Dimopoulos MA, Dytfeld D, Grosicki S, et al. Elotuzumab plus pomalido-mide and dexamethasone for relapsed/refractory multiple myeloma: fi nal overall survival analysis from the randomized phase II ELOQUENT-3 trial. J Clin Oncol 2023;41:568 –578. 29. Voorhees PM, Suman VJ, Tuchman SA, et al. A phase I/II study of ixazo-mib, pomalidomide, and dexamethasone for lenalidomide and protea-some inhibitor refractory multiple myeloma (Alliance A061202). Am J Hematol 2021;96:1595 –1603. 30. Krishnan AY, Kapoor P, Palmer J, et al. A phase I/II study of ixazomib (Ix) pomalidomide (POM) dexamethasone (DEX) in relapsed refractory (R/R) mul-tiple myeloma: initial results. J Clin Oncol 2016;34(Suppl):Abstract 8008. 31. Krishnan A, Kapoor P, Palmer JM, et al. Phase I/II trial of the oral regimen ixazomib, pomalidomide, and dexamethasone in relapsed/refractory mul-tiple myeloma. Leukemia 2018;32:1567 –1574. 32. San-Miguel J, Dhakal B, Yong K, et al. Cilta-cel or standard care in lenali-domide-refractory multiple myeloma. N Engl J Med 2023;389:335 –347. 33. Rodriguez-Otero P, Ailawadhi S, Arnulf B, et al. Ide-cel or standard regi-mens in relapsed and refractory multiple myeloma. N Engl J Med 2023; 388:1002 –1014. 34. Mateos MV, San-Migue J, Dhakal B, et al. Overall survival (OS) with cilta-cabtagene autoleucel (cilta-cel) versus standard of care (SoC) in lenalido-mide (len)-refractory multiple myeloma (MM): phase 3 CARTITUDE-4 study update. Presented at the 2024 International Myeloma Society Annual Meeting; September 25 –28, 2024; Rio de Janeiro, Brazil.
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REVIEW article
Front. Mol. Neurosci., 12 June 2025
Sec. Molecular Signalling and Pathways
Volume 18 - 2025 |
Roles of ERK signaling pathway in regulating myelination of the peripheral nervous system
Di LiuJingwei Zhou
Department of Endocrinology and Nephrology, Dongzhimen Hospital, The First Affiliated Hospital of Beijing University of Chinese Medicine, Beijing, China
Myelination of Schwann cells is a complex biological process that plays a crucial role in peripheral nervous system (PNS) development and repair. Recent studies have indicated that the extracellular signal-related kinase (ERK) signaling pathway participates in both developmental PNS myelination and remyelination. This review focuses on recent evidence identifying the roles of the ERK signaling pathway in regulating Schwann cell differentiation, myelination, and remyelination. In addition, the crosstalk between the ERK signaling pathway and other cellular signaling pathways that control Schwann cell myelination, such as c-Jun and Notch, are discussed. This review provides an overview of recent studies, revealing that dysregulated expression of the ERK signaling pathway participated in the pathogenesis of hereditary and acquired peripheral neuropathies.
Introduction
To form the myelin sheath and insulation, which speeds up the transmission of electrical signals and offers metabolic and trophic support to nerve axons, Schwann cells spirally wrap around large-diameter axons at a 1:1 ratio in the peripheral nervous system (Feltri et al., 2016; Salzer, 2015). This fine-tuned process plays a central role in PNS development, as emphasized by inherited demyelinating neuropathies caused by malformation of the myelin sheath. Moreover, regeneration following the peripheral nerve injury also requires remyelination of the regenerated axons, which is critical for the successful functional restoration of PNS. Functionally impaired intact axons with abnormal myelination are a key mechanism in many human peripheral neuropathies (Höke, 2006). Therefore, understanding the underlying mechanisms that regulate myelination holds significance in elucidating the pathogenesis of peripheral neuropathies and developing new therapeutic strategies.
Recent studies have reported that the extracellular signal-related kinase (ERK) signaling pathway actively participates in developmental PNS myelination and peripheral nerve regeneration (Napoli et al., 2012; Sheean et al., 2014). Therefore, this review gives attention to the roles of the ERK signaling pathway in regulating PNS myelin development as well as the remyelination process after damage. The crosstalk between the ERK signaling pathway and other cellular signaling pathways that control Schwann cell myelination is discussed. Finally, the roles of the ERK signaling pathway in developing both inherited and acquired peripheral neuropathies are reviewed.
The ERK1/2 signaling pathway
ERK1/2, as constituents of the mitogen-activated protein kinase (MAPK) family, are extensively expressed hydrophilic non-receptor proteins that contribute to the Ras–Raf–MEK–ERK signaling pathway (Roskoski, 2012). Ras-GTP initiates the activation of Raf kinases, which then catalyze MEK1 and MEK2 phosphorylation and stimulation (Raman et al., 2007). MEK1/2 are dual-specificity protein kinases that promote tyrosine and threonine phosphorylation in ERK1 and ERK2, their sole recognized physiological substrates (Raman et al., 2007). The ERK signaling pathway has long been acknowledged as a regulator of numerous processes, such as cell survival, differentiation, proliferation, adhesion, cycle progression, and metabolism (Wortzel and Seger, 2011; Cargnello and Roux, 2011; Lefloch et al., 2009; Gonsalvez et al., 2016). The disruption of this pathway can potentially result in a range of diseases, such as neurodegenerative disorders, metabolic disease, cancer, cardiac disorders, and inflammatory disease (Pylayeva-Gupta et al., 2011; Subramaniam and Unsicker, 2010; Steelman et al., 2011; Kim and Choi, 2010).
Previous investigations showed that various cellular stressors, such as cytokines, bradykinin, insulin, epidermal growth factor (EGF), fibroblast growth factor, insulin-like growth factor-1, platelet-derived growth factor, and others, can stimulate ERK signaling (Raman et al., 2007). Furthermore, for myelination, the ERK signaling pathway participated in controlling the promyelinating effects of numerous growth factors, such as brain-derived neurotrophic factor (BDNF), fibroblast growth factor, and EGF family proteins like neuregulin 1 (NRG-1) type III (Gonsalvez et al., 2016; Newbern and Birchmeier, 2010; Xiao et al., 2012; Cui and Almazan, 2007; Syed et al., 2010).
ERK signaling pathway in Schwann cell differentiation
In the initial stages of embryonic development, a small, temporary group of cells known as neural crest cells, isolated from the neural tube at the time of neural tube closure, have been identified to originate Schwann cell precursors (SCPs) (Scherer, 1997). At the initial transitional stage in the Schwann cell lineage, SCPs serve as the glial cells of early embryonic neurons and provide the immature Schwann cells (Monk et al., 2015). Schwann cells, which are present in mature nerve trunks, undergo differentiation into myelinating and non-myelinating varieties during birth (Salzer, 2015). The earliest commitment toward myelin synthesis is expected to occur during the shift from immature Schwann cells to promyelinating Schwann cells, which is why this phase is crucial. Schwann cell differentiation and maturation are precisely controlled, relying on the synchronized expression of differentiation-enhancing genes and inhibition of negative regulators (Monk et al., 2015).
One of the most important regulators of Schwann cell differentiation and proliferation is ERK signaling. In primary rat Schwann cells, strong activation of ERK was found to suppress promyelinating markers of Schwann cells, while this suppressive effect was entirely abolished by blocking ERK signaling, suggesting that the ERK pathway suppresses Schwann cell differentiation (Harrisingh et al., 2004). Concurrently, another study also revealed that ERK pathway activation can block Schwann cell differentiation and actively drive the dedifferentiation process (Ogata et al., 2004). Moreover, in mouse primary Schwann cells, the ERK1/2 phosphorylation was found to be linked to elevated mRNA levels of markers for undifferentiated Schwann cells such as Krox-24 and Scip (Nadra et al., 2008). A previous study reported that the suppression effect of adenosine on PDGF-induced Schwann cell proliferation was mediated through the stimulation of the ERK signaling pathway (Stevens et al., 2004). Collectively, these outcomes provide convincing evidence that ERK signaling exerts negative effects on Schwann cell differentiation and proliferation (Figure 1).
Figure 1
Figure 1. Roles of ERK signaling pathway in regulating Schwann cell differentiation and myelination. Schwann cell precursors (SCPs) are the first transitional stage in the Schwann cell lineage derived from neural crest cells. SCPs cease migration and develop into immature Schwann cells. Immature Schwann cells envelop single large-diameter axons and differentiate into myelinating Schwann cells. During the developmental process, basal levels of ERK pathway activity are indispensable for the differentiation of precursors and myelin maintenance, while high levels of ERK activation negatively regulate Schwann cell differentiation and myelination.
However, recent research has shown that the ERK pathway is essential for Schwann cell differentiation during development (Figure 1). A significant reduction of SCPs was detected in the peripheral nerve of ERK1 null allele embryos, suggesting that ERK1/2 is critical for SCP development of the peripheral nerve in vivo (Newbern et al., 2011). The neuregulin/ErbB axis has an essential function in Schwann cell differentiation (Newbern and Birchmeier, 2010). This study also stated that ERK1/2 is essential for mediating the impacts of NRG-1 on Schwann cells in vivo. Furthermore, conditional deletion of Shp2, an upstream ERK/MAPK stimulator, in neural crest cells leads to defects in Schwann cell development, including cell proliferation, migration, and differentiation (Grossmann et al., 2009). Nevertheless, the results from a recent study indicated that small molecule inhibitors targeting MEK and ERK did not influence the expression levels of Oct6 and EGR2, which are essential transcription factors of Schwann cell differentiation (Park et al., 2021). Nonetheless, the roles of the ERK pathway on Schwann cell differentiation remain incompletely understood.
Different studies have concluded that ERK signaling has both positive and negative effects on Schwann cell differentiation. The reconciling explanation could be that distinct levels or duration of ERK activity would define the state of Schwann cell differentiation. Low or basal activity of ERK would be required for Schwann cell differentiation while high ERK levels would drive dedifferentiation and proliferation (Newbern and Snider, 2012). ERK signaling seems to play a pro-differentiating role during development and a pro-dedifferentiating role after nerve injury. This dualistic role may depend on the lasting of activation. Following peripheral nerve injury, the levels of phosphorylated ERK/MAPK in the distal nerve rise more than threefold and remain elevated in the Büngner bands for up to a month (Sheu et al., 2000). Transient activation of ERK signaling during developmental stages may be linked to positive differentiation modulation, while long-term activation ERK signaling following nerve injury may lead to its negative modulation (Castelnovo et al., 2017). Another possibility is that ERK signaling pathway may interact with other pathways that regulate Schwann cell differentiation. The development of Schwann cells follows a structured sequence from the neural crest to fully differentiated Schwann cells, involving integration of both cell autonomous and non-cell autonomous signaling mechanisms. For example, during nerve development, cAMP/PKA signaling directs ERK signaling to facilitate Schwann cell differentiation, while in damaged adult nerves, lower cAMP levels link ERK signaling to Schwann cell dedifferentiation (Arthur-Farraj et al., 2011). Nonetheless, the roles of the ERK pathway on Schwann cell differentiation remain incompletely understood.
ERK signaling pathway in myelination by Schwann cells
Schwann cell myelination involves the recognizing of crucial external signals, the activation of intracellular pathways, and the transcriptional and epigenetic programs that synergistically promote the myelinating phenotype (Salzer, 2015). Growing evidence supports the role of the ERK signaling pathway in Schwann cell myelination (Figures 1, 2). Activation of the ERK signaling pathway was observed to downregulate myelin proteins in Schwann cell cultures expressing myelin proteins induced by cAMP, as well as cause demyelination in DRG neuron/Schwann cell co-cultures (Harrisingh et al., 2004). Another study investigating neuron/Schwann cell co-cultures also indicated that the endogenous Mek/ERK action in co-cultures is likely to collaborate with soluble Nrg1 type II (GGF) to offer extra inhibitory signals for myelination. GGF inhibitory effect on myelination is linked to the activation of c-Jun induced by the MEK/ERK signaling pathway (Syed et al., 2010). This study also demonstrated that a high concentration of soluble Nrg1 type III suppresses myelination depending on MEK/ERK. Furthermore, in vivo evidence indicates that a significant rise in the MEK/ERK/c-Jun pathway is associated with a decrease in the myelin gene expression as well as a delay in Schwann cell differentiation (Scapin et al., 2020). Collectively, these researches indicate that high levels of phosphorylation of the ERK signaling pathway negatively regulate Schwann cell myelination.
Figure 2
Figure 2. Schematic of ERK signaling pathway that regulate Schwann cell myelination. Schwann cell myelination is a process strongly dependent on instructive signals provided by the axons. The major axonal signals include type III NRG1, which signal via erbB receptors. Following NGF binding to tropomyosin, receptor kinase activates Ras. The MEK1/2-dependent phosphorylation of ERK1/2 induces their nuclear translocation, enabling ERK1/2 to potentially phosphorylate and/or stabilize transcription factors and proteins that then alter gene expression. MEK, the kinase directly upstream of ERK1/2 induce the phosphorylation of YY1. YY1 drive peripheral myelination transcription, such as Krox20. ERK1/2 phosphorylation can upregulate Krox-24 expression. The phosphatase Shp2, which activates the ERK1/2 signaling pathway, is necessary for Schwann cell differentiation and myelination. In addition, MEK/ERK can collaborate with soluble Nrg1 type II (GGF) to offer extra inhibitory signals for myelination. During the myelination process, basal levels of ERK activity are necessary for differentiation of precursors while high levels of ERK activity negatively regulate Schwann cell differentiation and myelination.
Conversely, ERK signaling is required for PNS myelination. Newbern et al. (2011) reported that the absence of ERK 1/2 in SCPs resulted in differentiation disruption and significant hypomyelination of axons, highlighting the importance of ERK 1/2 for the progression of the myelinating Schwann cell lineage following initial specification. Moreover, the MEK-dependent transcription factor Yy has been proven to be critical in the process of peripheral myelination (He et al., 2010). In addition, conditional mutation of Shp2 in myelinating Schwann cells can cause hypomyelination and is found to be linked to decreased ERK1/2 phosphorylation (Grossmann et al., 2009). Finally, recent studies indicate that sustained stimulation of the ERK signaling pathway can override the signals that terminate myelination, leading to continuous myelin growth (Sheean et al., 2014; Ishii et al., 2013). It has been observed that the stimulation of ERK signaling can compensate for the lack of ErbB3/Shp2 signaling pathway during Schwann cell growth and myelination (Sheean et al., 2014). These outcomes indicate that the ERK signaling pathway is a conserved mechanism that promotes PNS developmental myelination (Figures 1, 2).
At first, different levels or duration of ERK activity may elicit distinct responses, potentially explaining these seemingly paradoxical observations. The basal levels of ERK pathway is necessary for the differentiation of precursors and myelin maintenance, whereas high levels of ERK activation negatively regulate Schwann cell differentiation and myelination (Newbern and Snider, 2012). Therefore, the state of Schwann cell development might be defined by levels of ERK activity. Secondly, there are differences in the regulation of developmental myelination and remyelination. The ERK signaling is activated by NRG-1 and promotes myelination during development, but sustain activation of ERK by NRG-1 inhibits myelination (Scherer, 1997). It is also possible that the impact of the ERK signaling pathway on neural repair is related to the intensity and duration of its expression. In the early stages of nerve injury, the ERKsignaling can promote myelin clearance. However, during the advanced stages of nerve repair, the ERK signaling can inhibit myelination, which hinders the remyelination of regenerating axons and delays the restoration of neural function (Cervellini et al., 2018). Thirdly, it has also been suggested that other signaling pathways may interact with ERK signaling to influence the myelination by Schwann cells. For example, during active myelination, ERK signaling is dependent on mTOR signaling to drive the growth of the myelin sheath and regulate myelin thickness (Ishii et al., 2021). Furthermore, the transition into the myelinating Schwann cell stage requires accurate asymmetrical Schwann cell polarization created upon axo-glial contact. Axon-glial communications play a vital role in regulating the myelination process and axonal cytoskeleton by altering important nodes within intracellular signaling pathways and the transcriptional network of neuron–glia (Rao and Pearse, 2016). Axon derived NRG1-type III can bind the ErbB2/3 receptor on Schwann cells and supresses Nrg1-type I expression by MEK/ERK signaling. In the context of Charcot–Marie-Tooth (CMT) disease, the disrupted axon-glia communication reduces PI3K/AKT signaling activity, which in turn causes MEK/ERK hyperactivation. The imbalanced activity of PI3K/AKT and MEK/ERK signaling pathways triggers the transcription of Schwann cell immature and repair markers, resulting in impaired Schwann cell differentiation and myelination. When the activity of PI3K/AKT and MEK/ERK signaling pathways is balanced, thereby maintaining Schwann cells in their differentiation state. In addition, the differentiation of Schwann cells is also influenced by signals coming from the extracellular matrix (ECM) externally. ECM-derived collagen VI negatively ERK activation and inhibits myelination (Chen et al., 2013). However, clarifying the roles of the ERK signaling pathway and determining how ERK signaling operate with other signaling pathways control changes in the transcriptional network that regulates Schwann cell behavior will be challenging.
ERK signaling pathway in remyelination by Schwann cells
Following axotomy or nerve crush injury, the distal stump of the peripheral nerve undergoes Wallerian degeneration. Wallerian degeneration involves several phases of structural changes: alterations of initial axons, recruitment of macrophages, removal of myelin and axon debris, responses of Schwann cells, and actions of the other neural cells (Chen et al., 2007). When a nerve is transected, the axons at the distal stump experience granular degeneration, which is then followed by the recruitment of circulating macrophages into the endoneurium to phagocytize debris of myelin and axons. At injury sites, Schwann cells dedifferentiate and proliferate, forming Büngner bands to facilitate a supportive environment for axon regeneration and promote remyelination (Jessen et al., 2015). The growth of regenerating axons occurs along Büngner bands, leading to an increase in their diameter. The process of remyelination starts once Schwann cells encounter the regrowing axons. Hereditary and acquired peripheral neuropathies are characterized by demyelination accompanied by incomplete remyelination (Höke, 2006).
In vivo and in vitro experiments reveal that the ERK signaling pathway has a central function in peripheral nerve regeneration, including regulating immune responses, Schwann cell dedifferentiation and proliferation, neurotrophic factors, and axon regeneration and remyelination (Figure 3).
Figure 3
Figure 3. Roles of the ERK signaling pathway in regulating remyelination. After nerve crush injury or axotomy, the distal stump of the peripheral nerve undergoes Wallerian degeneration, which includes early degenerative axonal changes, immune cell recruitment, Schwann cell dedifferentiation and proliferation, and axon regeneration and remyelination. Nerve injury results in a rapid activation of ERK signaling in Schwann cells. A number of cytokines are upregulated, and a variety of immune cells are recruited following the activation of the ERK signaling pathway. ERK activation results in Schwann cell dedifferentiation and proliferation. The ERK signaling pathway also participates in regulating neurotrophic factor expressions and axonal outgrowth. After phosphorylated ERK levels return to basal levels, dedifferentiated Schwann cells redifferentiate and remyelinate.
Immune responses
The recruitment of immune cells plays an essential role in clearing axon and myelin debris and supporting subsequent revascularization of damaged nerves. The ERK signaling pathway has been found to participate in regulating inflammatory responses after nerve injury. In vivo, following the stimulation of the ERK signaling pathway, there was a significant rise in the count of macrophages, neutrophils, mast cells and T cells in the nerve, which are all recruited after nerve injury (Napoli et al., 2012). Furthermore, in injured nerves, the process of immune cell recruitment can be blocked by highly selective ERK signaling inhibitors. Similarly, conditioned media from Schwann cells treated with an ERK signaling activator also effectively recruited immune cells. A number of cytokines, including c-kit ligand, MCP-1, IL11, and Cxcl10, were upregulated after stimulation of ERK signaling in dedifferentiated Schwann cells. In addition, in the nerves of a mouse model for Charcot–Marie Tooth neuropathy, the activation of ERK1/2 was found to be directly associated with the elevated secretion of the macrophage-attachment cytokine MCP-1 (Fischer et al., 2008). Clearance of myelin debris following acute demyelination is crucial for functional recovery after nerve injury. Functioning as “amateur” phagocytes, microvascular endothelial cells have been proven to engulf and clear myelin debris, which enhances inflammatory response, vigorous angiogenesis, and ongoing fibrosis. Furthermore, ERK signaling is also involved in regulating myelin debris that triggers endothelial-to-mesenchymal transition in a dose-dependent way and supports the migration of endothelial cells (Yao et al., 2023).
Schwann cells dedifferentiation and proliferation
Schwann cell dedifferentiation and proliferation are significantly associated with successful nerve regeneration (Jessen et al., 2015). In addition, it was observed in vivo axotomy and in vitro culture of nerve segments that there was a rapid rise in ERK1/2 phosphorylation in Schwann cells, which subsequently caused a corresponding elevation in Schwann cell proliferation at ERK1/2 stimulation sites, which can be inhibited by ERK1/2 inhibitor (Martensson et al., 2007). Likewise, active ERK was mainly located in Schwann cells that form Büngner bands in the distal sciatic nerve segments (Agthong et al., 2006). Following a crush injury to the sciatic nerve, Schwann cells exhibit increased expressions of Epo and EpoR (Li et al., 2005). The application of exogenous Epo to damaged sciatic nerves or primary cultures of Schwann cells promotes Schwann cell proliferation by activating ERK/MAPK. Conversely, metalloproteinase-9, production of which was increased markedly by Schwann cells after nerve damage, was demonstrated to suppress Schwann cell proliferation and mitogenic activity through selective activation of ERK1/2 (Chattopadhyay and Shubayev, 2009). Furthermore, following the transfection of DRG neuron/Schwann cell co-cultures, active ERK1/2 led to Schwann cell dedifferentiation (Harrisingh et al., 2004). In a recent in vivo study, it was found that 3 days of increased ERK/MAPK phosphorylation notably induced Schwann cell dedifferentiation, and proliferating Schwann cell progenitors in the nerve was markedly increased (Napoli et al., 2012). This study provides convincing evidence that the RAF/MEK/ERK pathway activation alone can successfully induce Schwann cell dedifferentiation in vivo, even in nerves lacking damaged axons. Concurrently, a study reported that rapamycin repaired damaged nerve cells and neurological function by controlling Schwann cell proliferation via the ERK signaling pathway (Liu et al., 2020). Following peripheral nerve damage, the protein kinase C α (PKCα) expression, a serine/threonine kinase in SCs, was markedly increased. In vivo and in vitro culture trials further illustrated that PKCα significantly promoted Schwann cell migration by inducing the stimulation of the ERK signaling pathway. These outcomes suggest a potential mechanism for ERK to control Schwann cell proliferation after peripheral nerve injury (Li et al., 2020).
Neurotrophic factors
Following peripheral nerve damage, the release of neurotrophic factors is crucial for neuronal survival and efficient nerve regeneration (Richner et al., 2014). Nerve damage triggers the neurotrophic factors’ expression, including NGF, GDNF, and BDNF. The sequential and overlapping induction of NGF, GDNF, and BDNF expression in the distal stump correlates with the sustained activation of ERK, indicating the essential role of ERK activation in establishing an extracellular environment that facilitates regeneration at the distal end of transected nerves (Abe et al., 2001). In contrast, markedly lower levels of CNTF expression were found in nerves that were injured and regenerating (Friedman et al., 1992). Experiments further revealed that blocking the ERK signaling pathway caused a significant elevation in CNTF expression in cultured IMS32 cells; moreover, there was an inverse association between CNTF expression and ERK activity in the Schwann cells of the sciatic nerve (Abe et al., 2001). As a part of the neurotrophin family, neurotrophin-3 (NT-3) is crucial for the development, upkeep, and survival of neurons within the nervous system. A recent study reported that NT-3 supported the regeneration of peripheral nerves by sustaining Schwann cells in a repair state after prolonged denervation through the TrkC/ERK/c-Jun signaling pathway (Xu et al., 2023). Collectively, these outcomes indicate that the ERK signaling pathway participates in regulating neurotrophic factors during peripheral nerve repair.
Axon regeneration
Axon regeneration is also a key process for effective nerve regeneration. Increasing data indicates that the ERK signaling pathway has a function in controlling axon degeneration, sprouting, and regrowth after nerve injury. Research indicates that the suppression of the ERK pathway through the proteasome is involved in Wallerian degeneration of injured axons and axonal pruning in the absence of local NGF (Macinnis and Campenot, 2005). ERK signaling pathway not only promotes axon growth but also triggers the anti-apoptotic protein Bcl-2 through CREB to support axonal survival (Akram et al., 2022). Moreover, an in vitro study has revealed that PD98059, an inhibitor of MEK1/ERK1/2, almost fully inhibited neuron sprouting in damaged PC12 cells, indicating that MEK1/ERK1/2 has a crucial function in neurite re-formation (Waetzig and Herdegen, 2005). After peripheral nerve injury, the inhibitory microenvironment associated with myelin restricts nerve regeneration. A recent study revealed that ERK signaling is required for the regulation of axonal microenvironment. Blocking the EGFR-ERK pathway was shown to enhance TRIM32 expression, which can counteract the inhibitory effects of the microenvironment on the neuronal differentiation of neural stem cells and support neurogenesis (Ma et al., 2017). Following peripheral nerve damage, administering EGFR-ERK inhibitors enhances the neurogenesis in nestin(+) cells activated by the injury and improves functional recovery (Xue et al., 2020). Consistent with the pivotal role of ERK signaling in axon regeneration, a study from Tsuda revealed lower levels of p-ERK 1/2 in unrepaired transected nerves compared to the immediate or delayed repair groups (Tsuda et al., 2011). Notably, the study also proved that there was a correlation between axonal outgrowth length and p-ERK 1/2 expression at the lesion site and in the distal nerve segment. Moreover, the outcomes from in vitro and in vivo trials validated that the ERK pathway could promote myelinated axonal regrowth (Yang et al., 2023). A recent study from Tsuda showed that In chicken DRG, the non-integrin laminin receptor (LamR) promoted axonal outgrowth by regulating the Akt and ERK pathways, both of which are important for controlling axonal outgrowth (Mrówczyńska et al., 2024). These studies suggest that ERK plays an essential role in axon regeneration after nerve damage.
Peripheral nerve remyelination
Functional peripheral nerve regeneration necessitates remyelination of the regenerated axons by Schwann cells. Growing evidence indicates that the ERK signaling pathway plays negative regulatory roles in remyelination. A study using DRG neuron/Schwann cell co-cultures indicated that ERK signaling pathway activation resulted in a marked decrease in mRNA levels of Protein zero (P0), peripheral myelin protein-22 (PMP22), and myelin basic protein (MBP), along with the myelin sheath breakdown (Harrisingh et al., 2004). Consistent with this report, a study employing an innovative transgenic mouse model that facilitates Schwann cell-specific, reversible ERK/MAPK stimulation also indicated that ERK signaling activation resulted in a rapid reduction in expressing myelin genes, P0, MBP, and periaxin, causing substantial nerve demyelination and motor/proprioceptive impairments during behavioral testing (Napoli et al., 2012). The observation of remyelination and motor recovery following the normalization of ERK signaling activity implies that a subsequent decrease in ERK/MAPK activity may be essential for remyelination (Napoli et al., 2012). Moreover, an in vivo study revealed that autocrine or paracrine NRG1 type I signaling by Schwann cells promotes efficient remyelination after damage, whereas axonal NRG1 type III signaling to myelinating Schwann cells continuously suppresses Nrg1 type I transcription via the ERK1/2 signaling pathway (Stassart et al., 2013).
Crosstalk between ERK signaling pathway and other myelination regulators
Recently, the possibility that the ERK signaling pathway may interact with other pathways that control myelination has been elevated in investigations on Schwann cell development, myelination, and peripheral nerve repair (Figure 4). C-Jun, an important constituent of the AP-1 transcription factor complex, has been recognized as a negative regulator of myelination (Jessen and Mirsky, 2008; Heinen et al., 2013). A previous study in co-cultures revealed that induction of myelin genes is suppressed in Jun-enforced Schwann cells, whereas the expression of myelin genes is elevated in c-Jun null Schwann cells (Parkinson et al., 2008). U0126, a highly selective ERK signaling inhibitor, was found to decrease c-Jun upregulation in the sciatic nerve, suggesting an interaction between the ERK1/2 and JNK pathways during SC transcription (Lindwall Blom et al., 2014). In differentiated PC12 cells after injury, both ERK1/2 and JNK suppression could suppress c-Jun expression and phosphorylation, implying that c-Jun can be controlled by the Ras/ERK1/2 signaling pathway (Waetzig and Herdegen, 2005). Furthermore, a study in DRG neuron/Schwann cell co-cultures found that by inhibiting the ERK pathway, both demyelination and c-jun expression induced by NRG1 were suppressed (Syed et al., 2010). In Schwann cells, NT-3 was found to upregulate c-Jun, mainly through the ERK pathway (Xu et al., 2023). Notch, a transmembrane receptor protein, also functions as a negative regulator of myelination (Jessen and Mirsky, 2008; Heinen et al., 2013). In vitro studies showed that upregulating the myelin proteins periaxin and P0 through enforced Krox20 expression was blocked by simultaneous infection with Ad-Notch intracellular domain (NICD). Another in vivo experiment revealed that in mice, elevating Schwann cell NICD expression temporarily around birth causes a delay in myelination (Woodhoo et al., 2009). Strong upregulation of c-Jun and the Notch ligand jagged-1 occurs in Schwann cells after Raf stimulation, indicating that the Notch pathway and c-Jun are downstream of the ERK signaling pathway (Napoli et al., 2012) (Figure 4). Concurrently, increased ERK1/2 phosphorylation in sciatic nerves was shown to be associated with a substantial increase in Notch1 receptor gene expression (Nasser et al., 2022). In summary, strong upregulation of c-Jun and the Notch ligand jagged-1 occurs in Schwann cells after Raf stimulation. C-Jun and Notch pathway inhibit Schwann cell differentiation and myelination by interfering with Krox20 activity, thus repressing peripheral myelination transcription such as periaxin and P0 (Figure 4). Furthermore, a study indicated that eIF2α phosphorylation exerts a protective function in CMT1B Schwann cells by limiting ERK/c-Jun hyperactivation (Scapin et al., 2020). However, the precise mechanism still needs to be further studied.
Figure 4
Figure 4. Crosstalk between ERK signaling pathway and other myelination regulators. The ERK signaling pathway may interact with other pathways that control myelination. C-Jun has been recognized as a negative regulator of myelination. Notch signaling, via its intracellular domain (NICD), also functions as a negative regulator of myelination. Strong upregulation of c-Jun and the Notch ligand jagged-1 occurs in Schwann cells after Raf fstimulation, indicating that c-Jun and Notch pathway are downstream of the ERK signaling pathway. C-Jun and Notch pathway inhibit Schwann cell differentiation and myelination by interfering with Krox20 activity, thus repressing peripheral myelination transcription such as periaxin and P0.
ERK signaling pathway and peripheral neuropathies
ERK signaling pathway has a function in the progression of both hereditary and acquired peripheral neuropathies, including inherited disorders that cause demyelination or Schwann cell plasticity in damaged or diseased nerves. Charcot–Marie Tooth neuropathy is an inherited demyelinating neuropathy (Berger et al., 2006). In the mouse model of Charcot–Marie Tooth neuropathy, the ERK signaling pathway was activated within the nuclei of certain myelinating Schwann cells, which is directly related to elevated levels of the macrophage-attracting cytokine MCP-1 in the nerves, identifying ERK signaling pathway as a crucial intracellular pathway linking the Schwann cell mutations to the activation of macrophages that are pathogenetically significant in the peripheral nerves (Fischer et al., 2008). Neurofibromatosis type 1 (NF1) is a prevalent genetic condition of the nervous system, inherited in an autosomal dominant manner, impacting one in every 3,500 individuals globally (Rubin and Gutmann, 2005). The loss of neurofibromin was found to induce Ras hyperactivation and its downstream effectors (Cichowski and Jacks, 2001). In NF1−/− and malignant peripheral nerve sheath tumor (MPNST)-derived Schwann cells, the expressions of active Ras were increased and proven to be vital for sustaining the transformed phenotype in these cells (Declue et al., 1992; Basu et al., 1992; Kim et al., 1995). In primary co-culture systems, NF1 loss was found to hinder Schwann cells from connecting with axons and cause already connected Schwann cells to detach from axons by upregulating the Ras/Raf/ERK signaling pathway. Moreover, Normal axonal interactions were restored by treating NF1−/-Schwann cells with the MEK inhibitor U0126 after losing axonal interactions (Parrinello et al., 2008). In a three-dimensional in vitro culture, the findings demonstrated that hyperactivation of focal adhesion kinase (FAK) leads to the aberrant activation of ERK and AKT downstream of Ras, allowing human NF1-deficient and mouse Nf1−/− cells to proliferate inside a three-dimensional matrix. Furthermore, the study demonstrated that combining both the MEK inhibitor Selumetinib and the FAK inhibitor defactinib entirely inhibits their transformational potential, offering novel approaches for treating plexiform neurofibroma (Errico et al., 2021). These findings not only clarify a molecular mechanism for tumorigenesis but also present a novel strategy for developing NF1 therapies. Moreover, the ERK signaling pathway is associated with immune-mediated peripheral nerve injury caused by leprosy bacilli. In research on the response of human primary Schwann cells to prolonged Mycobacterium leprae infection, it was found that intracellular M. leprae directly stimulated ERK1/2 via a PKC-dependent and MEK-independent signaling pathway, thereby inducing continuous proliferation (Fang et al., 2015). Diabetic peripheral neuropathy is a kind of acquired demyelinating disease. The DRG and sciatic nerves showed elevated ERK phosphorylation in diabetic animal models (Stavniichuk et al., 2013). In vitro, high glucose levels also triggered the activation of ERK in cultured SCs in a dose-dependent way (Zhu et al., 2012). Furthermore, in primary-culture Schwann cells subjected to elevated glucose levels and in Schwann cell-dorsal root ganglion (DRG) neuron co-cultures under similar conditions, the inhibition of the ERK signaling pathway prompted Schwann cell differentiation, increased CNTF release, augmented both protein and mRNA expressions of myelin, and improved the quantity and length of myelin segments (Liu et al., 2020). These findings suggest that the ERK signaling pathway may play a role in the development of diabetic peripheral neuropathy. In addition, the contribution of the ERK signaling pathway activation to neuropathic pain has been observed in mice after partial sciatic nerve ligation, implying that p-ERK and mediators associated with ERK could be targeted for treating neuropathic pain (Kiguchi et al., 2009). A study conducted in a TRPV4-mediated trigeminal neuralgia rat model observed that the ERK antagonist treatment could reverse the mechanical hyperalgesia threshold, nerve fiber abnormalities, myelin degradation, and Schwann cell proliferation (Liu et al., 2021). Furthermore, a recent study reported that miR-30b-5p contained within skin-derived precursor Schwann cells (SKP-SC-EVs) promoted the regeneration of injured sciatic nerves and supported axon growth and myelination in dogs by activating the phosphorylation of ERK, suggesting that SKP-SC-EVs-incorporating TENGs represent an innovative bioactive material that could be used for peripheral nerve repair in clinical practice (Yu et al., 2024).
Conclusions and future perspectives
Myelination is a complex process and is determined by a balance between positive and negative factors. Recent studies have demonstrated that the ERK signaling pathway is essential for developmental PNS myelination and post-injury remyelination, encompassing the regulation of Schwann cell differentiation, myelination, dedifferentiation, proliferation, remyelination, inflammatory responses, neurotrophic factors, and axon regeneration. Nonetheless, the findings must be contextualized, as varying degrees of ERK activity delineate the state of Schwann cell differentiation. In addition, the ERK signaling pathway may potentially interact with other pathways that control Schwann cell myelination, including c-Jun and Notch. Understanding how these signaling pathways control alterations in the transcriptional network that regulates Schwann cell activity will be tough. The dysregulation of the ERK signaling pathway has been demonstrated to contribute to the development of peripheral neuropathies. Nonetheless, the role of the ERK signaling pathway in human neuropathic disorders is not yet fully understood. Additional investigation in suitable animal models and humans is essential to assess the function of the ERK signaling pathway in the pathogenesis of hereditary and acquired peripheral nerve disorders and to determine if the restoration of ERK signaling pathway regulation could facilitate effective remyelination and regeneration. Enhanced comprehension of the molecular mechanisms governing the unique functions of the ERK signaling pathway in modulating Schwann cell myelination would aid in the formulation of novel therapeutic approaches for severe disorders associated with dysregulated myelination.
Author contributions
DL: Funding acquisition, Resources, Writing – original draft. JZ: Supervision, Writing – review & editing.
Funding
The author(s) declare that financial support was received for the research and/or publication of this article. This work was supported by the National High Level Hospital of TCM SM Funding of China (Grant no. CZ015) and the National Natural Sciences Foundation of China (Grant no. 82305153).
Conflict of interest
The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.
Generative AI statement
The authors declare that no Gen AI was used in the creation of this manuscript.
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References
Abe, K., Namikawa, K., Honma, M., Iwata, T., Matsuoka, I., Watabe, K., et al. (2001). Inhibition of Ras extracellular-signal-regulated kinase (ERK) mediated signaling promotes ciliary neurotrophic factor (CNTF) expression in Schwann cells. J. Neurochem. 77, 700–703. doi: 10.1046/j.1471-4159.2001.00286.x
PubMed Abstract | Crossref Full Text | Google Scholar
Agthong, S., Kaewsema, A., Tanomsridejchai, N., and Chentanez, V. (2006). Activation of MAPK ERK in peripheral nerve after injury. BMC Neurosci. 7:45. doi: 10.1186/1471-2202-7-45
PubMed Abstract | Crossref Full Text | Google Scholar
Akram, R., Anwar, H., Javed, M. S., Rasul, A., Imran, A., Malik, S. A., et al. (2022). Axonal regeneration: underlying molecular mechanisms and potential therapeutic targets. Biomedicines 10:3186. doi: 10.3390/biomedicines10123186
PubMed Abstract | Crossref Full Text | Google Scholar
Arthur-Farraj, P., Wanek, K., Hantke, J., Davis, C. M., Jayakar, A., Parkinson, D. B., et al. (2011). Mouse schwann cells need both NRG1 and cyclic AMP to myelinate. Glia 59, 720–733. doi: 10.1002/glia.21144
PubMed Abstract | Crossref Full Text | Google Scholar
Basu, T. N., Gutmann, D. H., Fletcher, J. A., Glover, T. W., Collins, F. S., and Downward, J. (1992). Aberrant regulation of ras proteins in malignant tumour cells from type 1 neurofibromatosis patients. Nature 356, 713–715. doi: 10.1038/356713a0
PubMed Abstract | Crossref Full Text | Google Scholar
Berger, P., Niemann, A., and Suter, U. (2006). Schwann cells and the pathogenesis of inherited motor and sensory neuropathies (Charcot-Marie-tooth disease). Glia 54, 243–257. doi: 10.1002/glia.20386
PubMed Abstract | Crossref Full Text | Google Scholar
Cargnello, M., and Roux, P. P. (2011). Activation and function of the MAPKs and their substrates, the MAPK-activated protein kinases. Microbiol. Mol. Biol. Rev. 75, 50–83. doi: 10.1128/MMBR.00031-10
PubMed Abstract | Crossref Full Text | Google Scholar
Castelnovo, L., Bonalume, V., Melfi, S., Ballabio, M., Colleoni, D., and Magnaghi, V. (2017). Schwann cell development, maturation and regeneration: a focus on classic and emerging intracellular signaling pathways. Neural Regen. Res. 12, 1013–1023. doi: 10.4103/1673-5374.211172
PubMed Abstract | Crossref Full Text | Google Scholar
Cervellini, I., Galino, J., Zhu, N., Allen, S., Birchmeier, C., and Bennett, D. L. (2018). Sustained MAPK/ERK activation in adult Schwann cells impairs nerve repair. J. Neurosci. 38, 679–690. doi: 10.1523/JNEUROSCI.2255-17.2017
PubMed Abstract | Crossref Full Text | Google Scholar
Chattopadhyay, S., and Shubayev, V. I. (2009). MMP-9 controls Schwann cell proliferation and phenotypic remodeling via IGF-1 and ErbB receptor-mediated activation of MEK/ERK pathway. Glia 57, 1316–1325. doi: 10.1002/glia.20851
PubMed Abstract | Crossref Full Text | Google Scholar
Chen, P., Cescon, M., Megighian, A., and Ronaldo, P. (2013). Collagen VI regulates peripheral nerve myelination and function. FASEB J. 28, 1145–1156. doi: 10.1096/fj.13-239533
PubMed Abstract | Crossref Full Text | Google Scholar
Chen, Z. L., Yu, W. M., and Strickland, S. (2007). Peripheral regeneration. Annu. Rev. Neurosci. 30, 209–233. doi: 10.1146/annurev.neuro.30.051606.094337
PubMed Abstract | Crossref Full Text | Google Scholar
Cichowski, K., and Jacks, T. (2001). NF1 tumor suppressor gene function: narrowing the GAP. Cell 104, 593–604. doi: 10.1016/s0092-8674(01)00245-8
PubMed Abstract | Crossref Full Text | Google Scholar
Cui, Q. L., and Almazan, G. (2007). IGF-I-induced oligodendrocyte progenitor proliferation requires PI3K/Akt, MEK/ERK, and Src-like tyrosine kinases. J. Neurochem. 100, 1480–1493. doi: 10.1111/j.1471-4159.2006.04329.x
PubMed Abstract | Crossref Full Text | Google Scholar
Declue, J. E., Papageorge, A. G., Fletcher, J. A., Diehl, S. R., Ratner, N., Vass, W. C., et al. (1992). Abnormal regulation of mammalian p21ras contributes to malignant tumor growth in von Recklinghausen (type 1) neurofibromatosis. Cell 69, 265–273. doi: 10.1016/0092-8674(92)90407-4
Crossref Full Text | Google Scholar
Errico, A., Stocco, A., Riccardi, V. M., Gambalunga, A., Bassetto, F., Grigatti, M., et al. (2021). Neurofibromin deficiency and extracellular matrix cooperate to increase transforming potential through FAK-dependent signaling. Cancers 13:2329. doi: 10.3390/cancers13102329
PubMed Abstract | Crossref Full Text | Google Scholar
Fang, F., Liu, C., Zhang, J., Zhu, L., Qian, Y., Yi, J., et al. (2015). Involvement of MAPK ERK activation in upregulation of water channel protein aquaporin 1 in a mouse model of bell’s palsy. J. Mol. Neurosci. 56, 164–176. doi: 10.1007/s12031-014-0477-2
PubMed Abstract | Crossref Full Text | Google Scholar
Feltri, M. L., Poitelon, Y., and Previtali, S. C. (2016). How Schwann cells Sort axons: new concepts. Neuroscientist 22, 252–265. doi: 10.1177/1073858415572361
PubMed Abstract | Crossref Full Text | Google Scholar
Fischer, S., Weishaupt, A., Troppmair, J., and Martini, R. (2008). Increase of MCP-1 (CCL2) in myelin mutant Schwann cells is mediated by MEK-ERK signaling pathway. Glia 56, 836–843. doi: 10.1002/glia.20657
PubMed Abstract | Crossref Full Text | Google Scholar
Friedman, B., Scherer, S. S., Rudge, J. S., Helgren, M., Morrisey, D., Mcclain, J., et al. (1992). Regulation of ciliary neurotrophic factor expression in myelin-related Schwann cells in vivo. Neuron 9, 295–305. doi: 10.1016/0896-6273(92)90168-D
PubMed Abstract | Crossref Full Text | Google Scholar
Gonsalvez, D., Ferner, A. H., Peckham, H., Murray, S. S., and Xiao, J. (2016). The roles of extracellular related-kinases 1 and 2 signaling in CNS myelination. Neuropharmacology 110, 586–593. doi: 10.1016/j.neuropharm.2015.04.024
PubMed Abstract | Crossref Full Text | Google Scholar
Grossmann, K. S., Wende, H., Paul, F. E., Cheret, C., Garratt, A. N., Zurborg, S., et al. (2009). The tyrosine phosphatase Shp2 (PTPN11) directs Neuregulin-1/ErbB signaling throughout Schwann cell development. Proc. Natl. Acad. Sci. USA 106, 16704–16709. doi: 10.1073/pnas.0904336106
PubMed Abstract | Crossref Full Text | Google Scholar
Harrisingh, M. C., Perez-Nadales, E., Parkinson, D. B., Malcolm, D. S., Mudge, A. W., and Lloyd, A. C. (2004). The Ras/Raf/ERK signalling pathway drives Schwann cell dedifferentiation. EMBO J. 23, 3061–3071. doi: 10.1038/sj.emboj.7600309
PubMed Abstract | Crossref Full Text | Google Scholar
He, Y., Kim, J. Y., Dupree, J., Tewari, A., Melendez-Vasquez, C., Svaren, J., et al. (2010). Yy1 as a molecular link between neuregulin and transcriptional modulation of peripheral myelination. Nat. Neurosci. 13, 1472–1480. doi: 10.1038/nn.2686
PubMed Abstract | Crossref Full Text | Google Scholar
Heinen, A., Lehmann, H. C., and Küry, P. (2013). Negative regulators of Schwann cell differentiation—novel targets for peripheral nerve therapies? J. Clin. Immunol. 33, 18–26. doi: 10.1007/s10875-012-9786-9
PubMed Abstract | Crossref Full Text | Google Scholar
Höke, A. (2006). Mechanisms of disease: what factors limit the success of peripheral nerve regeneration in humans? Nat. Clin. Pract. Neurol. 2, 448–454. doi: 10.1038/ncpneuro0262
PubMed Abstract | Crossref Full Text | Google Scholar
Ishii, A., Furusho, M., and Bansal, R. (2013). Sustained activation of ERK1/2 MAPK in oligodendrocytes and schwann cells enhances myelin growth and stimulates oligodendrocyte progenitor expansion. J. Neurosci. 33, 175–186. doi: 10.1523/JNEUROSCI.4403-12.2013
PubMed Abstract | Crossref Full Text | Google Scholar
Ishii, A., Furusho, M., and Bansal, R. (2021). Mek/ERK1/2-MAPK and PI3K/Akt/mTOR signaling plays both independent and cooperative roles in Schwann cell differentiation, myelination and dysmyelination. Glia 69, 2429–2446. doi: 10.1002/glia.24049
PubMed Abstract | Crossref Full Text | Google Scholar
Jessen, K. R., and Mirsky, R. (2008). Negative regulation of myelination: relevance for development, injury, and demyelinating disease. Glia 56, 1552–1565. doi: 10.1002/glia.20761
PubMed Abstract | Crossref Full Text | Google Scholar
Jessen, K. R., Mirsky, R., and Lloyd, A. C. (2015). Schwann cells: development and role in nerve repair. Cold Spring Harb. Perspect. Biol. 7:a20487. doi: 10.1101/cshperspect.a020487
Crossref Full Text | Google Scholar
Kiguchi, N., Maeda, T., Kobayashi, Y., Fukazawa, Y., and Kishioka, S. (2009). Activation of extracellular signal-regulated kinase in sciatic nerve contributes to neuropathic pain after partial sciatic nerve ligation in mice. Anesth. Analg. 109, 1305–1311. doi: 10.1213/ane.0b013e3181b1dca2
PubMed Abstract | Crossref Full Text | Google Scholar
Kim, E. K., and Choi, E. J. (2010). Pathological roles of MAPK signaling pathways in human diseases. Biochim. Biophys. Acta 1802, 396–405. doi: 10.1016/j.bbadis.2009.12.009
PubMed Abstract | Crossref Full Text | Google Scholar
Kim, H. A., Rosenbaum, T., Marchionni, M. A., Ratner, N., and Declue, J. E. (1995). Schwann cells from neurofibromin deficient mice exhibit activation of p21ras, inhibition of cell proliferation and morphological changes. Oncogene 11, 325–335.
Google Scholar
Lefloch, R., Pouyssegur, J., and Lenormand, P. (2009). Total ERK1/2 activity regulates cell proliferation. Cell Cycle 8, 705–711. doi: 10.4161/cc.8.5.7734
PubMed Abstract | Crossref Full Text | Google Scholar
Li, X., Gonias, S. L., and Campana, W. M. (2005). Schwann cells express erythropoietin receptor and represent a major target for Epo in peripheral nerve injury. Glia 51, 254–265. doi: 10.1002/glia.20202
PubMed Abstract | Crossref Full Text | Google Scholar
Li, J., Zhang, Z., Wang, J., Du, S., Yao, D., Cao, R., et al. (2020). Protein kinase Calpha promotes proliferation and migration of Schwann cells by activating ERK signaling pathway. Neuroscience 433, 94–107. doi: 10.1016/j.neuroscience.2020.03.007
PubMed Abstract | Crossref Full Text | Google Scholar
Lindwall Blom, C., Mårtensson, L. B., and Dahlin, L. B. (2014). Nerve injury-induced c-Jun activation in Schwann cells is JNK independent. Biomed. Res. Int. 2014, 1–11. doi: 10.1155/2014/392971
PubMed Abstract | Crossref Full Text | Google Scholar
Liu, D., Jia, S., Sun, D., Wang, S. Y., Meng, F. C., and Guo, W. C. (2020). Rapamycin repairs damaged nerve cells and neurological function in rats with spinal cord injury through ERK signaling pathway. J. Biol. Regul. Homeost. Agents 34, 865–873. doi: 10.23812/20-122-L-45
PubMed Abstract | Crossref Full Text | Google Scholar
Liu, D., Liang, X. C., Sun, Y., Wu, Y. N., and Zhang, H. (2020). Combination of quercetin, Hirudin and Cinnamaldehyde promotes Schwann cell differentiation and myelination against high glucose by inhibiting ERK signaling pathway. Chin. J. Integr. Med. 26, 591–598. doi: 10.1007/s11655-020-2721-7
PubMed Abstract | Crossref Full Text | Google Scholar
Liu, W., Pu, B., Liu, M., Zhang, X., and Zeng, R. (2021). Down-regulation of MAPK pathway alleviates TRPV4-mediated trigeminal neuralgia by inhibiting the activation of histone acetylation. Exp. Brain Res. 239, 3397–3404. doi: 10.1007/s00221-021-06194-6
PubMed Abstract | Crossref Full Text | Google Scholar
Ma, Z., Zang, T., Birnbaum, S. G., Wang, Z., Johnson, J. E., Zhang, C., et al. (2017). TrkB dependent adult hippocampal progenitor differentiation mediates sustained ketamine antidepressant response. Nat. Commun. 8, 1614–1668. doi: 10.1038/s41467-017-01709-8
Crossref Full Text | Google Scholar
Macinnis, B. L., and Campenot, R. B. (2005). Regulation of Wallerian degeneration and nerve growth factor withdrawal-induced pruning of axons of sympathetic neurons by the proteasome and the MEK/Erk pathway. Mol. Cell. Neurosci. 28, 430–439. doi: 10.1016/j.mcn.2004.10.003
PubMed Abstract | Crossref Full Text | Google Scholar
Martensson, L., Gustavsson, P., Dahlin, L. B., and Kanje, M. (2007). Activation of extracellular-signal-regulated kinase-1/2 precedes and is required for injury-induced Schwann cell proliferation. Neuroreport 18, 957–961. doi: 10.1097/WNR.0b013e32819f8f27
PubMed Abstract | Crossref Full Text | Google Scholar
Monk, K. R., Feltri, M. L., and Taveggia, C. (2015). New insights on Schwann cell development. Glia 63, 1376–1393. doi: 10.1002/glia.22852
PubMed Abstract | Crossref Full Text | Google Scholar
Mrówczyńska, E., Machalica, K., and Mazur, A. J. (2024). Non-integrin laminin receptor (LamR) plays a role in axonal outgrowth from chicken DRG via modulating the Akt and Erk signaling. Front. Cell Dev. Biol. 12:1433947. doi: 10.3389/fcell.2024.1433947
PubMed Abstract | Crossref Full Text | Google Scholar
Nadra, K., de Preux, C. A., Medard, J. J., Hendriks, W. T., Han, G. S., Gres, S., et al. (2008). Phosphatidic acid mediates demyelination in Lpin1 mutant mice. Genes Dev. 22, 1647–1661. doi: 10.1101/gad.1638008
Crossref Full Text | Google Scholar
Napoli, I., Noon, L. A., Ribeiro, S., Kerai, A. P., Parrinello, S., Rosenberg, L. H., et al. (2012). A central role for the ERK-signaling pathway in controlling Schwann cell plasticity and peripheral nerve regeneration in vivo. Neuron 73, 729–742. doi: 10.1016/j.neuron.2011.11.031
PubMed Abstract | Crossref Full Text | Google Scholar
Nasser, A. H., Gendy, A. M., El-Yamany, M. F., and El-Tanbouly, D. M. (2022). Upregulation of neuronal progranulin mediates the antinociceptive effect of trimetazidine in paclitaxel-induced peripheral neuropathy: role of ERK1/2 signaling. Toxicol. Appl. Pharmacol. 448:116096. doi: 10.1016/j.taap.2022.116096
PubMed Abstract | Crossref Full Text | Google Scholar
Newbern, J., and Birchmeier, C. (2010). Nrg1/ErbB signaling networks in Schwann cell development and myelination. Semin. Cell Dev. Biol. 21, 922–928. doi: 10.1016/j.semcdb.2010.08.008
PubMed Abstract | Crossref Full Text | Google Scholar
Newbern, J. M., Li, X., Shoemaker, S. E., Zhou, J., Zhong, J., Wu, Y., et al. (2011). Specific functions for ERK/MAPK signaling during PNS development. Neuron 69, 91–105. doi: 10.1016/j.neuron.2010.12.003
PubMed Abstract | Crossref Full Text | Google Scholar
Newbern, J. M., and Snider, W. D. (2012). Bers-ERK Schwann cells coordinate nerve regeneration. Neuron 73, 623–626. doi: 10.1016/j.neuron.2012.02.002
PubMed Abstract | Crossref Full Text | Google Scholar
Ogata, T., Iijima, S., Hoshikawa, S., Miura, T., Yamamoto, S., Oda, H., et al. (2004). Opposing extracellular signal-regulated kinase and Akt pathways control Schwann cell myelination. J. Neurosci. 24, 6724–6732. doi: 10.1523/JNEUROSCI.5520-03.2004
PubMed Abstract | Crossref Full Text | Google Scholar
Park, K., Shin, Y., Lee, G., Park, H., and Choi, Y. (2021). Dabrafenib promotes Schwann cell differentiation by inhibition of the MEK-ERK pathway. Molecules 26:2141. doi: 10.3390/molecules26082141
PubMed Abstract | Crossref Full Text | Google Scholar
Parkinson, D. B., Bhaskaran, A., Arthur-Farraj, P., Noon, L. A., Woodhoo, A., Lloyd, A. C., et al. (2008). C-Jun is a negative regulator of myelination. J. Cell Biol. 181, 625–637. doi: 10.1083/jcb.200803013
PubMed Abstract | Crossref Full Text | Google Scholar
Parrinello, S., Noon, L. A., Harrisingh, M. C., Digby, P. W., Rosenberg, L. H., Cremona, C. A., et al. (2008). NF1 loss disrupts Schwann cell-axonal interactions: a novel role for semaphorin 4F. Genes Dev. 22, 3335–3348. doi: 10.1101/gad.490608
PubMed Abstract | Crossref Full Text | Google Scholar
Pylayeva-Gupta, Y., Grabocka, E., and Bar-Sagi, D. (2011). RAS oncogenes: weaving a tumorigenic web. Nat. Rev. Cancer 11, 761–774. doi: 10.1038/nrc3106
PubMed Abstract | Crossref Full Text | Google Scholar
Raman, M., Chen, W., and Cobb, M. H. (2007). Differential regulation and properties of MAPKs. Oncogene 26, 3100–3112. doi: 10.1038/sj.onc.1210392
PubMed Abstract | Crossref Full Text | Google Scholar
Rao, S. N. R., and Pearse, D. D. (2016). Regulating axonal responses to injury: the intersection between signaling pathways involved in axon myelination and the inhibition of axon regeneration. Front. Mol. Neurosci. 9:33. doi: 10.3389/fnmol.2016.00033
Crossref Full Text | Google Scholar
Richner, M., Ulrichsen, M., Elmegaard, S. L., Dieu, R., Pallesen, L. T., and Vaegter, C. B. (2014). Peripheral nerve injury modulates neurotrophin signaling in the peripheral and central nervous system. Mol. Neurobiol. 50, 945–970. doi: 10.1007/s12035-014-8706-9
PubMed Abstract | Crossref Full Text | Google Scholar
Roskoski, R. (2012). ERK1/2 MAP kinases: structure, function, and regulation. Pharmacol. Res. 66, 105–143. doi: 10.1016/j.phrs.2012.04.005
PubMed Abstract | Crossref Full Text | Google Scholar
Rubin, J. B., and Gutmann, D. H. (2005). Neurofibromatosis type 1 - a model for nervous system tumour formation? Nat. Rev. Cancer 5, 557–564. doi: 10.1038/nrc1653
PubMed Abstract | Crossref Full Text | Google Scholar
Salzer, J. L. (2015). Schwann cell myelination. Cold Spring Harb. Perspect. Biol. 7:a20529. doi: 10.1101/cshperspect.a020529
Crossref Full Text | Google Scholar
Scapin, C., Ferri, C., Pettinato, E., Bianchi, F., Feltri, M. L., et al. (2020). Phosphorylation of eIF2alpha promotes Schwann cell differentiation and myelination in CMT1B mice with activated UPR. J. Neurosci. 40, 8174–8187. doi: 10.1523/JNEUROSCI.0957-20.2020
Crossref Full Text | Google Scholar
Scherer, S. S. (1997). The biology and pathobiology of Schwann cells. Curr. Opin. Neurol. 10, 386–398. doi: 10.1097/00019052-199710000-00006
PubMed Abstract | Crossref Full Text | Google Scholar
Sheean, M. E., Mcshane, E., Cheret, C., Walcher, J., Muller, T., Wulf-Goldenberg, A., et al. (2014). Activation of MAPK overrides the termination of myelin growth and replaces Nrg1/ErbB3 signals during Schwann cell development and myelination. Genes Dev. 28, 290–303. doi: 10.1101/gad.230045.113
PubMed Abstract | Crossref Full Text | Google Scholar
Sheu, J. Y., Kulhanek, D. J., and Eckenstein, F. P. (2000). Differential patterns of ERK and STAT3 phosphorylation after sciatic nerve transection in the rat. Exp. Neurol. 166, 392–402. doi: 10.1006/exnr.2000.7508
PubMed Abstract | Crossref Full Text | Google Scholar
Stassart, R. M., Fledrich, R., Velanac, V., Brinkmann, B. G., Schwab, M. H., Meijer, D., et al. (2013). A role for Schwann cell-derived neuregulin-1 in remyelination. Nat. Neurosci. 16, 48–54. doi: 10.1038/nn.3281
PubMed Abstract | Crossref Full Text | Google Scholar
Stavniichuk, R., Obrosov, A. A., Drel, V. R., Nadler, J. L., Obrosova, I. G., and Yorek, M. A. (2013). 12/15-lipoxygenase inhibition counteracts MAPK phosphorylation in mouse and cell culture models of diabetic peripheral neuropathy. J. Diabetes Mellitus 3:33015. doi: 10.4236/jdm.2013.33015
PubMed Abstract | Crossref Full Text | Google Scholar
Steelman, L. S., Chappell, W. H., Abrams, S. L., Kempf, R. C., Long, J., Laidler, P., et al. (2011). Roles of the Raf/MEK/ERK and PI3K/PTEN/Akt/mTOR pathways in controlling growth and sensitivity to therapy-implications for cancer and aging. Aging 3, 192–222. doi: 10.18632/aging.100296
PubMed Abstract | Crossref Full Text | Google Scholar
Stevens, B., Ishibashi, T., Chen, J. F., and Fields, R. D. (2004). Adenosine: an activity-dependent axonal signal regulating MAP kinase and proliferation in developing Schwann cells. Neuron Glia Biol. 1, 23–34. doi: 10.1017/S1740925X04000055
PubMed Abstract | Crossref Full Text | Google Scholar
Subramaniam, S., and Unsicker, K. (2010). ERK and cell death: ERK1/2 in neuronal death. FEBS J. 277, 22–29. doi: 10.1111/j.1742-4658.2009.07367.x
PubMed Abstract | Crossref Full Text | Google Scholar
Syed, N., Reddy, K., Yang, D. P., Taveggia, C., Salzer, J. L., Maurel, P., et al. (2010). Soluble neuregulin-1 has bifunctional, concentration-dependent effects on Schwann cell myelination. J. Neurosci. 30, 6122–6131. doi: 10.1523/JNEUROSCI.1681-09.2010
PubMed Abstract | Crossref Full Text | Google Scholar
Tsuda, Y., Kanje, M., and Dahlin, L. B. (2011). Axonal outgrowth is associated with increased ERK 1/2 activation but decreased caspase 3 linked cell death in Schwann cells after immediate nerve repair in rats. BMC Neurosci. 12:12. doi: 10.1186/1471-2202-12-12
PubMed Abstract | Crossref Full Text | Google Scholar
Waetzig, V., and Herdegen, T. (2005). MEKK1 controls neurite regrowth after experimental injury by balancing ERK1/2 and JNK2 signaling. Mol. Cell. Neurosci. 30, 67–78. doi: 10.1016/j.mcn.2005.06.001
PubMed Abstract | Crossref Full Text | Google Scholar
Woodhoo, A., Alonso, M. B., Droggiti, A., Turmaine, M., D'Antonio, M., Parkinson, D. B., et al. (2009). Notch controls embryonic Schwann cell differentiation, postnatal myelination and adult plasticity. Nat. Neurosci. 12, 839–847. doi: 10.1038/nn.2323
PubMed Abstract | Crossref Full Text | Google Scholar
Wortzel, I., and Seger, R. (2011). The ERK Cascade: distinct functions within various subcellular organelles. Genes Cancer 2, 195–209. doi: 10.1177/1947601911407328
PubMed Abstract | Crossref Full Text | Google Scholar
Xiao, J., Ferner, A. H., Wong, A. W., Denham, M., Kilpatrick, T. J., and Murray, S. S. (2012). Extracellular signal-regulated kinase 1/2 signaling promotes oligodendrocyte myelination in vitro. J. Neurochem. 122, 1167–1180. doi: 10.1111/j.1471-4159.2012.07871.x
PubMed Abstract | Crossref Full Text | Google Scholar
Xu, X., Song, L., Li, Y., Guo, J., Huang, S., Du, S., et al. (2023). Neurotrophin-3 promotes peripheral nerve regeneration by maintaining a repair state of Schwann cells after chronic denervation via the TrkC/ERK/c-Jun pathway. J. Transl. Med. 21:733. doi: 10.1186/s12967-023-04609-2
PubMed Abstract | Crossref Full Text | Google Scholar
Xue, W., Zhao, Y., Xiao, Z., Wu, X., Ma, D., Han, J., et al. (2020). Epidermal growth factor receptor-extracellular-regulated kinase blockade upregulates TRIM32 signaling cascade and promotes neurogenesis after spinal cord injury. Stem Cells 38, 118–133. doi: 10.1002/stem.3097
PubMed Abstract | Crossref Full Text | Google Scholar
Yang, Q., Su, S., Liu, S., Yang, S., Xu, J., Zhong, Y., et al. (2023). Exosomes-loaded electroconductive nerve dressing for nerve regeneration and pain relief against diabetic peripheral nerve injury. Bioact Mater 26, 194–215. doi: 10.1016/j.bioactmat.2023.02.024
PubMed Abstract | Crossref Full Text | Google Scholar
Yao, F., Luo, Y., Chen, Y., Li, Y., Hu, X., You, X., et al. (2023). Myelin debris impairs tight junctions and promotes the migration of microvascular endothelial cells in the injured spinal cord. Cell. Mol. Neurobiol. 43, 741–756. doi: 10.1007/s10571-022-01203-w
PubMed Abstract | Crossref Full Text | Google Scholar
Yu, M., Shen, M., Chen, D., Li, Y., Zhou, Q., Deng, C., et al. (2024). Chitosan/PLGA-based tissue engineered nerve grafts with SKP-SC-EVs enhance sciatic nerve regeneration in dogs through miR-30b-5p-mediated regulation of axon growth. Bioact Mater 40, 378–395. doi: 10.1016/j.bioactmat.2024.06.011
PubMed Abstract | Crossref Full Text | Google Scholar
Zhu, H., Yu, W. J., Le, Y., Wang, W. J., Li, F., Gui, T., et al. (2012). High glucose levels increase the expression of neurotrophic factors associated with p-p42/p44 MAPK in Schwann cells in vitro. Mol. Med. Rep. 6, 179–184. doi: 10.3892/mmr.2012.896
PubMed Abstract | Crossref Full Text | Google Scholar
Keywords: ERK, myelination, remyelination, Schwann cell, peripheral nervous system
Citation: Liu D and Zhou J (2025) Roles of ERK signaling pathway in regulating myelination of the peripheral nervous system. Front. Mol. Neurosci. 18:1617976. doi: 10.3389/fnmol.2025.1617976
Received: 25 April 2025; Accepted: 28 May 2025;
Published: 13 June 2025.
Edited by:
Barbara Hausott, Innsbruck Medical University, Austria
Reviewed by:
Jose Javier Miguel-Hidalgo, University of Mississippi Medical Center, United States
Raquel Oliveira, King’s College London, United Kingdom
Vincenzo Macaione, University of Messina, Italy
Copyright © 2025 Liu and Zhou. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms.
Correspondence: Jingwei Zhou, 13910634708@163.com
Disclaimer: All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article or claim that may be made by its manufacturer is not guaranteed or endorsed by the publisher. |
15333 | https://tass.ru/ekonomika/18861569 | Минфин ожидает в 2024 году роста нефтегазовых доходов почти на треть
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28 сентября 2023, 07:46,
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Минфин ожидает в 2024 году роста нефтегазовых доходов почти на треть
Они могут увеличиться до 11,5 трлн рублей
МОСКВА, 28 сентября. /ТАСС/. Нефтегазовые доходы федерального бюджета РФ могут увеличиться в 2024 году почти на 30% по сравнению с 2023 годом - с 8,86 трлн рублей до 11,5 трлн рублей. Это следует из Основных направлений бюджетной, налоговой и таможенно-тарифной политики на 2024 год и плановый период 2025 и 2026 годов.
В 2025 году министерство ожидает дальнейшего роста нефтегазовых доходов до 11,8 трлн рублей, а в 2026 году - снижения до 11,4 трлн рублей.
В документе также подчеркивается, что в 2024 году доля нефтегазовых доходов прогнозируется на уровне 6,4% от общего объема ВВП против 5,3% в 2023 году с учетом динамики обменного курса и корректировок налогового законодательства. При этом к 2026 году доля нефтегазовых доходов в ВВП, как ожидает Минфин, сократится до 5,6% вследствие стабилизации ценовой конъюнктуры и увеличения доли добычи нефти на месторождениях с льготным режимом налогообложения.
Как следует из документа, Минфин также планирует поэтапно сократить предельный дисконт российской нефти Urals для целей налогообложения с $15 за баррель в 2024 году до $10 за баррель в 2025 году и $6 за баррель в 2026 году.
В материалах к проекту бюджета РФ на 2024 год и плановый период 2025-2026 годов, которые имеются в распоряжении ТАСС, НДПИ на нефть по итогам 2023 года оценивается в 7,7 трлн рублей, в 2024 году он может вырасти до 9,7 трлн рублей, а в 2025-2026 годах будет держаться на уровне 9,8-9,9 трлн рублей. При этом НДПИ на газ в 2023 году прогнозируется на уровне 1,2 трлн рублей, в 2024 году - 1,6 трлн рублей, в 2025 году - 1,76 трлн рублей, в 2026 году - 1,3 трлн рублей.
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LSAT Grouping Games
The document provides a lesson on solving LSAT grouping games, emphasizing the importance of understanding the setup, identifying and organizing rules, and eliminating invalid options. It ou…
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LSA T Grouping Games: A Short Lesson for Presentation
Introduction
Grouping games on the LSAT require organizing elements into distinct groups based on given conditions. Mastering these games involves identifying key rules, making deductions, and eliminating invalid options.
Step 1: Understand the Setup
● Identify the total number of elements and groups. ● Determine whether groups have fixed sizes.
Example:
"Seven people (Aiden, Bella, Carlos, Daisy, Evan, Fiona, George) are split into three teams: Team 1 (3 members), Team 2 (2 members), Team 3 (2 members)."
Step 2: Identify and Organize the Rules
● Some elements must be together (
Aiden & Carlos
). ● Some elements must be apart (
Bella & Daisy
). ● Some must be placed in specific ways (
Evan & Fiona together
). ● Some restrict ions apply (
George cannot be with Aiden
).
Step 3: Apply Rules to Eliminate Invalid Options
● Check if all mandatory pairings are satisfied. ● Ensure restricted pairs are not together. ● Verify that all groups are correctly sized.
Example of an Invalid Teaming:
Team 1: Aiden, Carlos, Daisy
Team 2: Bella, Evan
Team 3: Fiona, George
❌
Fails because
Evan & Fiona are separated
.
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Step 4: Find the Valid Option
If no answer choices meet all conditions,
none of the above
is the correct response.
Pro Tip:
● Use diagrams to visualize groupings. ● Look for chains of dependency (e.g., If A is with B and B is with C, then A must be with C).
Practice Question
Six people (Anna, Ben, Claire, David, Emma, Fred) must be divided into two teams of three, following these rules:
Anna must be with David.
Ben cannot be with Claire.
Emma must be with Claire.
❓
Which of the following is a valid grouping?
A) Anna, Ben, Cla ire | David, E mma, Fred B) Anna, David, Fred | Ben, Claire, Emma C) Anna, David, Claire | Ben, Emma, Fred D) None of the above
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15335 | https://math.stackexchange.com/questions/332167/proper-methods-of-solving-parametric-equations | Skip to main content
Proper methods of solving parametric equations
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I'm learning parametric equations in this section. Although I understand why the following works, I'm having difficulty understanding why the method employed for solving it is the correct one.
I'm instructed to derive a Cartesian equation for the parametric equations given and describe the path the particle traverses. Describing the path amounts to simply graphing the path, i.e. the Cartesian function or parametric equation, and superimpose arrows indicating direction.
So, without further ado, the parametric equations give are:
xy==cos2tsin2t
Over the interval: 0≤t≤π
The solution involves observing that x2+y2=1=>cos2(2t)+sin2(2t)=1
Squaring these functions to make them produce something convenient seems like "fixing" the problem to do what I'd like it to do. Why is this a valid approach? I hope I'm asking my question well. Squaring both of these functions seems like turning the problem into something it isn't, but yet, it's the right thing to do. Why?
Thanks,
Andy
parametric
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edited Mar 16, 2013 at 18:55
Adi Dani
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asked Mar 16, 2013 at 18:49
Andrew FalangaAndrew Falanga
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In general, we want to use the two equations to "eliminate" t. That cannot always be done explicitly, and the strategy is quite equations-dependent.
– André Nicolas
Commented
Mar 16, 2013 at 18:54
You are not "fixing" anything "to do what you want" (??) but realizing some relation exists between the parametric functions. The trigonometric Pythagoras Theorem is hardly something out of the blue in this particular case, and in other cases you will have to come up with soem other ideas.
– DonAntonio
Commented
Mar 16, 2013 at 18:54
I apologize for the lateness of approving an answer or commenting. I didn't see the application of the Pythagorean Theorem as something out of the blue. Rather, it is simply that x=cos2t and y=sin2t weren't squared. By squaring them, I was considering that as "changing" what was there to something that wasn't. However, I think I see that it's because the parametric equations aren't a single Cartesian relationship but two. I'm trying to make a single relationship from them. Am I getting warm?
– Andrew Falanga
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Mar 25, 2013 at 21:01
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It is the right approach.
Take a candidate cartesian equation, replace x and y withe their expression x(t), y(t) and massage the formula until the parameter t disappears.
Then you are sure that every point of your parametric curve is also on the curve with the candidate cartesian equation.
You may have points of the parametric curve which are not on the cartesian curve. Like the equations (x,y)=(et,e−t) which gives only one branch of the hyperbola xy=1, the one for which x,y>0.
A part of that, finding the cartesian equation is some kind of a guess work.
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answered Mar 16, 2013 at 19:07
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15336 | https://en.wikipedia.org/wiki/Plant_hormone | Published Time: 2003-07-02T11:08:40Z
Plant hormone - Wikipedia
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Contents
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(Top)
1 Characteristics
2 ClassesToggle Classes subsection
2.1 Abscisic acid
2.2 Auxins
2.3 Brassinosteroids
2.4 Cytokinins
2.5 Ethylene
2.6 Gibberellins
2.7 Jasmonates
2.8 Salicylic acid
2.9 Strigolactones
2.10 Other known hormones
3 Use in horticultureToggle Use in horticulture subsection
3.1 Seed dormancy
4 Human useToggle Human use subsection
4.1 Salicylic acid
4.2 Jasmonic acid
5 See also
6 References
7 External links
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Plant hormone
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From Wikipedia, the free encyclopedia
Chemical compounds that regulate plant growth and development
Lack of the plant hormone auxin can cause abnormal growth (right)
Plant hormones (or phytohormones) are signal molecules, produced within plants, that occur in extremely low concentrations. Plant hormones control all aspects of plant growth and development, including embryogenesis, the regulation of organ size, pathogen defense, stress tolerance and reproductive development. Unlike in animals (in which hormone production is restricted to specialized glands) each plant cell is capable of producing hormones. Went and Thimann coined the term "phytohormone" and used it in the title of their 1937 book.
Phytohormones occur across the plant kingdom, and even in algae, where they have similar functions to those seen in vascular plants ("higher plants"). Some phytohormones also occur in microorganisms, such as unicellular fungi and bacteria, however in these cases they do not play a hormonal role and can better be regarded as secondary metabolites.
Characteristics
[edit]
Phyllody on a purple coneflower (Echinacea purpurea), a plant development abnormality where leaf-like structures replace flower organs. It can be caused by hormonal imbalance, among other reasons.
The word hormone is derived from Greek, meaning set in motion. Plant hormones affect gene expression and transcription levels, cellular division, and growth. They are naturally produced within plants, though very similar chemicals are produced by fungi and bacteria that can also affect plant growth. A large number of related chemical compounds are synthesized by humans. They are used to regulate the growth of cultivated plants, weeds, and in vitro-grown plants and plant cells; these manmade compounds are called plant growth regulators (PGRs). Early in the study of plant hormones, "phytohormone" was the commonly used term, but its use is less widely applied now.
Plant hormones are not nutrients, but chemicals that in small amounts promote and influence the growth, development, and differentiation of cells and tissues. The biosynthesis of plant hormones within plant tissues is often diffuse and not always localized. Plants lack glands to produce and store hormones, because, unlike animals—which have two circulatory systems (lymphatic and cardiovascular) —plants use more passive means to move chemicals around their bodies. Plants utilize simple chemicals as hormones, which move more easily through their tissues. They are often produced and used on a local basis within the plant body. Plant cells produce hormones that affect even different regions of the cell producing the hormone.
Hormones are transported within the plant by utilizing four types of movements. For localized movement, cytoplasmic streaming within cells and slow diffusion of ions and molecules between cells are utilized. Vascular tissues are used to move hormones from one part of the plant to another; these include sieve tubes or phloem that move sugars from the leaves to the roots and flowers, and xylem that moves water and mineral solutes from the roots to the foliage.
Not all plant cells respond to hormones, but those cells that do are programmed to respond at specific points in their growth cycle. The greatest effects occur at specific stages during the cell's life, with diminished effects occurring before or after this period. Plants need hormones at very specific times during plant growth and at specific locations. They also need to disengage the effects that hormones have when they are no longer needed. The production of hormones occurs very often at sites of active growth within the meristems, before cells have fully differentiated. After production, they are sometimes moved to other parts of the plant, where they cause an immediate effect; or they can be stored in cells to be released later. Plants use different pathways to regulate internal hormone quantities and moderate their effects; they can regulate the amount of chemicals used to biosynthesize hormones. They can store them in cells, inactivate them, or cannibalise already-formed hormones by conjugating them with carbohydrates, amino acids, or peptides. Plants can also break down hormones chemically, effectively destroying them. Plant hormones frequently regulate the concentrations of other plant hormones. Plants also move hormones around the plant diluting their concentrations.
The concentration of hormones required for plant responses are very low (10−6 to 10−5 mol/L). Because of these low concentrations, it has been very difficult to study plant hormones, and only since the late 1970s have scientists been able to start piecing together their effects and relationships to plant physiology. Much of the early work on plant hormones involved studying plants that were genetically deficient in one or involved the use of tissue-cultured plants grown in vitro that were subjected to differing ratios of hormones, and the resultant growth compared. The earliest scientific observation and study dates to the 1880s; the determination and observation of plant hormones and their identification was spread out over the next 70 years.
Synergism in plant hormones refers to the how of two or more hormones result in an effect that is more than the individual effects. For example, auxins and cytokinins often act in cooperation during cellular division and differentiation. Both hormones are key to cell cycle regulation, but when they come together, their synergistic interactions can enhance cell proliferation and organogenesis more effectively than either could in isolation.
Classes
[edit]
Different hormones can be sorted into different classes, depending on their chemical structures. Within each class of hormone, chemical structures can vary, but all members of the same class have similar physiological effects. Initial research into plant hormones identified five major classes: abscisic acid, auxins, brassinosteroids, cytokinins and ethylene. This list was later expanded, and brassinosteroids, jasmonates, salicylic acid, and strigolactones are now also considered major plant hormones. Additionally there are several other compounds that serve functions similar to the major hormones, but their status as bona fide hormones is still debated.
Abscisic acid
[edit]
Abscisic acid
Abscisic acid (also called ABA) is one of the most important plant growth inhibitors. It was discovered and researched under two different names, dormin and abscicin II, before its chemical properties were fully known. Once it was determined that the two compounds are the same, it was named abscisic acid. The name refers to the fact that it is found in high concentrations in newly abscissed or freshly fallen leaves.
This class of PGR is composed of one chemical compound normally produced in the leaves of plants, originating from chloroplasts, especially when plants are under stress. In general, it acts as an inhibitory chemical compound that affects bud growth, and seed and bud dormancy. It mediates changes within the apical meristem, causing bud dormancy and the alteration of the last set of leaves into protective bud covers. Since it was found in freshly abscissed leaves, it was initially thought to play a role in the processes of natural leaf drop, but further research has disproven this. In plant species from temperate parts of the world, abscisic acid plays a role in leaf and seed dormancy by inhibiting growth, but, as it is dissipated from seeds or buds, growth begins. In other plants, as ABA levels decrease, growth then commences as gibberellin levels increase. Without ABA, buds and seeds would start to grow during warm periods in winter and would be killed when it froze again. Since ABA dissipates slowly from the tissues and its effects take time to be offset by other plant hormones, there is a delay in physiological pathways that provides some protection from premature growth. Abscisic acid accumulates within seeds during fruit maturation, preventing seed germination within the fruit or before winter. Abscisic acid's effects are degraded within plant tissues during cold temperatures or by its removal by water washing in and out of the tissues, releasing the seeds and buds from dormancy.
ABA exists in all parts of the plant, and its concentration within any tissue seems to mediate its effects and function as a hormone; its degradation, or more properly catabolism, within the plant affects metabolic reactions and cellular growth and production of other hormones. Plants start life as a seed with high ABA levels. Just before the seed germinates, ABA levels decrease; during germination and early growth of the seedling, ABA levels decrease even more. As plants begin to produce shoots with fully functional leaves, ABA levels begin to increase again, slowing down cellular growth in more "mature" areas of the plant. Stress from water or predation affects ABA production and catabolism rates, mediating another cascade of effects that trigger specific responses from targeted cells. Scientists are still piecing together the complex interactions and effects of this and other phytohormones.
In plants under water stress, ABA plays a role in closing the stomata. Soon after plants are water-stressed and the roots are deficient in water, a signal moves up to the leaves, causing the formation of ABA precursors there, which then move to the roots. The roots then release ABA, which is translocated to the foliage through the vascular system and modulates potassium and sodium uptake within the guard cells, which then lose turgidity, closing the stomata.
Auxins
[edit]
The auxin, indole-3-acetic acid
Auxins are compounds that positively influence cell enlargement, bud formation, and root initiation. They also promote the production of other hormones and, in conjunction with cytokinins, control the growth of stems, roots, and fruits, and convert stems into flowers. Auxins were the first class of growth regulators discovered. A Dutch Biologist Frits Warmolt Went first described auxins. They affect cell elongation by altering cell wall plasticity. They stimulate cambium, a subtype of meristem cells, to divide, and in stems cause secondary xylem to differentiate.
Auxins act to inhibit the growth of buds lower down the stems in a phenomenon known as apical dominance, and also to promote lateral and adventitious root development and growth. Leaf abscission is initiated by the growing point of a plant ceasing to produce auxins. Auxins in seeds regulate specific protein synthesis, as they develop within the flower after pollination, causing the flower to develop a fruit to contain the developing seeds.
In large concentrations, auxins are often toxic to plants; they are most toxic to dicots and less so to monocots. Because of this property, synthetic auxin herbicides including 2,4-dichlorophenoxyacetic acid (2,4-D) and 2,4,5-trichlorophenoxyacetic acid (2,4,5-T) have been developed and used for weed control by defoliation. Auxins, especially 1-naphthaleneacetic acid (NAA) and indole-3-butyric acid (IBA), are also commonly applied to stimulate root growth when taking cuttings of plants. The most common auxin found in plants is indole-3-acetic acid (IAA).
Brassinosteroids
[edit]
Brassinolide, a major brassinosteroid
Brassinosteroids (BRs) are a class of polyhydroxysteroids, the only example of steroid-based hormones in plants. Brassinosteroids control cell elongation and division, gravitropism, resistance to stress, and xylem differentiation. They inhibit root growth and leaf abscission. Brassinolide was the first brassinosteroid to be identified and was isolated from extracts of rapeseed (Brassica napus) pollen in 1979. Brassinosteroids are a class of steroidal phytohormones in plants that regulate numerous physiological processes. This plant hormone was identified by Mitchell et al. who extracted ingredients from Brassica pollen only to find that the extracted ingredients’ main active component was Brassinolide. This finding meant the discovery of a new class of plant hormones called Brassinosteroids. These hormones act very similarly to animal steroidal hormones by promoting growth and development.
In plants these steroidal hormones play an important role in cell elongation via BR signaling. The brassinosteroids receptor brassinosteroid insensitive 1 (BRI1) is the main receptor for this signaling pathway. This BRI1 receptor was found by Clouse et al. who made the discovery by inhibiting BR and comparing it to the wildtype in Arabidopsis. The BRI1 mutant displayed several problems associated with growth and development such as dwarfism, reduced cell elongation and other physical alterations. These findings mean that plants properly expressing brassinosteroids grow more than their mutant counterparts. Brassinosteroids bind to BRI1 localized at the plasma membrane which leads to a signal cascade that further regulates cell elongation. This signal cascade however is not entirely understood at this time. What is believed to be happening is that BR binds to the BAK1 complex which leads to a phosphorylation cascade. This phosphorylation cascade then causes BIN2 to be deactivated which causes the release of transcription factors. These released transcription factors then bind to DNA that leads to growth and developmental processes and allows plants to respond to abiotic stressors.
Cytokinins
[edit]
Zeatin, a cytokinin
Cytokinins (CKs) are a group of chemicals that influence cell division and shoot formation. They also help delay senescence of tissues, are responsible for mediating auxin transport throughout the plant, and affect internodal length and leaf growth. They were called kinins in the past when they were first isolated from yeast cells. Cytokinins and auxins often work together, and the ratios of these two groups of plant hormones affect most major growth periods during a plant's lifetime. Cytokinins counter the apical dominance induced by auxins; in conjunction with ethylene, they promote abscission of leaves, flower parts, and fruits.
Among the plant hormones, the three that are known to help with immunological interactions are ethylene (ET), salicylates (SA), and jasmonates (JA), however more research has gone into identifying the role that cytokinins play in this. Evidence suggests that cytokinins delay the interactions with pathogens, showing signs that they could induce resistance toward these pathogenic bacteria. Accordingly, there are higher CK levels in plants that have increased resistance to pathogens compared to those which are more susceptible. For example, pathogen resistance involving cytokinins was tested using the Arabidopsis species by treating them with naturally occurring CK (trans-zeatin) to see their response to the bacteria Pseudomonas syringa. Tobacco studies reveal that over expression of CK inducing IPT genes yields increased resistance whereas over expression of CK oxidase yields increased susceptibility to pathogen, namely P. syringae.
While there’s not much of a relationship between this hormone and physical plant behavior, there are behavioral changes that go on inside the plant in response to it. Cytokinin defense effects can include the establishment and growth of microbes (delay leaf senescence), reconfiguration of secondary metabolism or even induce the production of new organs such as galls or nodules. These organs and their corresponding processes are all used to protect the plants against biotic/abiotic factors.
Ethylene
[edit]
Main article: Ethylene as a plant hormone
Ethylene
Unlike the other major plant hormones, ethylene is a gas and a very simple organic compound, consisting of just six atoms. It forms through the breakdown of methionine, an amino acid which is in all cells. Ethylene has very limited solubility in water and therefore does not accumulate within the cell, typically diffusing out of the cell and escaping the plant. Its effectiveness as a plant hormone is dependent on its rate of production versus its rate of escaping into the atmosphere. Ethylene is produced at a faster rate in rapidly growing and dividing cells, especially in darkness. New growth and newly germinated seedlings produce more ethylene than can escape the plant, which leads to elevated amounts of ethylene, inhibiting leaf expansion (see hyponastic response).
As the new shoot is exposed to light, reactions mediated by phytochrome in the plant's cells produce a signal for ethylene production to decrease, allowing leaf expansion. Ethylene affects cell growth and cell shape; when a growing shoot or root hits an obstacle while underground, ethylene production greatly increases, preventing cell elongation and causing the stem to swell. The resulting thicker stem is stronger and less likely to buckle under pressure as it presses against the object impeding its path to the surface. If the shoot does not reach the surface and the ethylene stimulus becomes prolonged, it affects the stem's natural geotropic response, which is to grow upright, allowing it to grow around an object. Studies seem to indicate that ethylene affects stem diameter and height: when stems of trees are subjected to wind, causing lateral stress, greater ethylene production occurs, resulting in thicker, sturdier tree trunks and branches.
Ethylene also affects fruit ripening. Normally, when the seeds are mature, ethylene production increases and builds up within the fruit, resulting in a climacteric event just before seed dispersal. The nuclear protein Ethylene Insensitive2 (EIN2) is regulated by ethylene production, and, in turn, regulates other hormones including ABA and stress hormones. Ethylene diffusion out of plants is strongly inhibited underwater. This increases internal concentrations of the gas. In numerous aquatic and semi-aquatic species (e.g. Callitriche platycarpus, rice, and Rumex palustris), the accumulated ethylene strongly stimulates upward elongation. This response is an important mechanism for the adaptive escape from submergence that avoids asphyxiation by returning the shoot and leaves to contact with the air whilst allowing the release of entrapped ethylene. At least one species (Potamogeton pectinatus) has been found to be incapable of making ethylene while retaining a conventional morphology. This suggests ethylene is a true regulator rather than being a requirement for building a plant's basic body plan.
Gibberellins
[edit]
Gibberellin A1
Gibberellins (GAs) include a large range of chemicals that are produced naturally within plants and by fungi. They were first discovered when Japanese researchers, including Eiichi Kurosawa, noticed a chemical produced by a fungus called Gibberella fujikuroi that produced abnormal growth in rice plants. It was later discovered that GAs are also produced by the plants themselves and control multiple aspects of development across the life cycle. The synthesis of GA is strongly upregulated in seeds at germination and its presence is required for germination to occur. In seedlings and adults, GAs strongly promote cell elongation. GAs also promote the transition between vegetative and reproductive growth and are also required for pollen function during fertilization.
Gibberellins breaks the dormancy (in active stage) in seeds and buds and helps increasing the height of the plant. It helps in the growth of the stem[citation needed]
Jasmonates
[edit]
Jasmonic acid
Jasmonates (JAs) are lipid-based hormones that were originally isolated from jasmine oil. JAs are especially important in the plant response to attack from herbivores and necrotrophic pathogens. The most active JA in plants is jasmonic acid. Jasmonic acid can be further metabolized into methyl jasmonate (MeJA), which is a volatile organic compound. This unusual property means that MeJA can act as an airborne signal to communicate herbivore attack to other distant leaves within one plant and even as a signal to neighboring plants. In addition to their role in defense, JAs are also believed to play roles in seed germination, the storage of protein in seeds, and root growth.
JAs have been shown to interact in the signalling pathway of other hormones in a mechanism described as “crosstalk.” The hormone classes can have both negative and positive effects on each other's signal processes.
Jasmonic acid methyl ester (JAME) has been shown to regulate genetic expression in plants. They act in signalling pathways in response to herbivory, and upregulate expression of defense genes. Jasmonyl-isoleucine (JA-Ile) accumulates in response to herbivory, which causes an upregulation in defense gene expression by freeing up transcription factors.
Jasmonate mutants are more readily consumed by herbivores than wild type plants, indicating that JAs play an important role in the execution of plant defense. When herbivores are moved around leaves of wild type plants, they reach similar masses to herbivores that consume only mutant plants, implying the effects of JAs are localized to sites of herbivory. Studies have shown that there is significant crosstalk between defense pathways.
Salicylic acid
[edit]
Salicylic acid
Salicylic acid (SA) is a hormone with a structure related to benzoic acid and phenol. It was originally isolated from an extract of white willow bark (Salix alba) and is of great interest to human medicine, as it is the precursor of the painkiller aspirin. In plants, SA plays a critical role in the defense against biotrophic pathogens. In a similar manner to JA, SA can also become methylated. Like MeJA, methyl salicylate is volatile and can act as a long-distance signal to neighboring plants to warn of pathogen attack. In addition to its role in defense, SA is also involved in the response of plants to abiotic stress, particularly from drought, extreme temperatures, heavy metals, and osmotic stress.
Salicylic acid (SA) serves as a key hormone in plant innate immunity, including resistance in both local and systemic tissue upon biotic attacks, hypersensitive responses, and cell death. Some of the SA influences on plants include seed germination, cell growth, respiration, stomatal closure, senescence-associated gene expression, responses to abiotic and biotic stresses, basal thermo tolerance and fruit yield. A possible role of salicylic acid in signaling disease resistance was first demonstrated by injecting leaves of resistant tobacco with SA. The result was that injecting SA stimulated pathogenesis related (PR) protein accumulation and enhanced resistance to tobacco mosaic virus (TMV) infection. Exposure to pathogens causes a cascade of reactions in the plant cells. SA biosynthesis is increased via isochorismate synthase (ICS) and phenylalanine ammonia-lyase (PAL) pathway in plastids. It was observed that during plant-microbe interactions, as part of the defense mechanisms, SA is initially accumulated at the local infected tissue and then spread all over the plant to induce systemic acquired resistance at non-infected distal parts of the plant. Therefore with increased internal concentration of SA, plants were able to build resistant barriers for pathogens and other adverse environmental conditions
Strigolactones
[edit]
5-deoxystrigol, a strigolactone
Strigolactones (SLs) were originally discovered through studies of the germination of the parasitic weed Striga lutea. It was found that the germination of Striga species was stimulated by the presence of a compound exuded by the roots of its host plant. It was later shown that SLs that are exuded into the soil also promote the growth of symbiotic arbuscular mycorrhizal (AM) fungi. More recently, another role of SLs was identified in the inhibition of shoot branching. This discovery of the role of SLs in shoot branching led to a dramatic increase in the interest in these hormones, and it has since been shown that SLs play important roles in leaf senescence, phosphate starvation response, salt tolerance, and light signalling.
Other known hormones
[edit]
Other identified plant growth regulators include:
Plant peptide hormones – encompasses all small secreted peptides that are involved in cell-to-cell signaling. These small peptide hormones play crucial roles in plant growth and development, including defense mechanisms, the control of cell division and expansion, and pollen self-incompatibility. The small peptide CLE25 is known to act as a long-distance signal to communicate water stress sensed in the roots to the stomata in the leaves.
Polyamines – are strongly basic molecules with low molecular weight that have been found in all organisms studied thus far. They are essential for plant growth and development and affect the process of mitosis and meiosis. In plants, polyamines have been linked to the control of senescence and programmed cell death.
Nitric oxide (NO) – serves as signal in hormonal and defense responses (e.g. stomatal closure, root development, germination, nitrogen fixation, cell death, stress response). NO can be produced by a yet undefined NO synthase, a special type of nitrite reductase, nitrate reductase, mitochondrial cytochrome c oxidase or non enzymatic processes and regulate plant cell organelle functions (e.g. ATP synthesis in chloroplasts and mitochondria).
Karrikins – are not plant hormones as they are not produced by plants themselves but are rather found in the smoke of burning plant material. Karrikins can promote seed germination in many species. The finding that plants which lack the receptor of karrikin receptor show several developmental phenotypes (enhanced biomass accumulation and increased sensitivity to drought) have led some to speculate on the existence of an as yet unidentified karrikin-like endogenous hormone in plants. The cellular karrikin signalling pathway shares many components with the strigolactone signalling pathway.
Triacontanol – a fatty alcohol that acts as a growth stimulant, especially initiating new basal breaks in the rose family. It is found in alfalfa (lucerne), bee's wax, and some waxy leaf cuticles.
Use in horticulture
[edit]
Synthetic plant hormones or PGRs are used in a number of different techniques involving plant propagation from cuttings, grafting, micropropagation and tissue culture. Most commonly they are commercially available as "rooting hormone powder".
The propagation of plants by cuttings of fully developed leaves, stems, or roots is performed by gardeners utilizing auxin as a rooting compound applied to the cut surface; the auxins are taken into the plant and promote root initiation. In grafting, auxin promotes callus tissue formation, which joins the surfaces of the graft together. In micropropagation, different PGRs are used to promote multiplication and then rooting of new plantlets. In the tissue-culturing of plant cells, PGRs are used to produce callus growth, multiplication, and rooting.
When used in field conditions, plant hormones or mixtures that include them can be applied as biostimulants.
Seed dormancy
[edit]
Main article: Seed dormancy
Plant hormones affect seed germination and dormancy by acting on different parts of the seed.
Embryo dormancy is characterized by a high ABA:GA ratio, whereas the seed has high abscisic acid sensitivity and low GA sensitivity. In order to release the seed from this type of dormancy and initiate seed germination, an alteration in hormone biosynthesis and degradation toward a low ABA/GA ratio, along with a decrease in ABA sensitivity and an increase in GA sensitivity, must occur.
ABA controls embryo dormancy, and GA embryo germination. Seed coat dormancy involves the mechanical restriction of the seed coat. This, along with a low embryo growth potential, effectively produces seed dormancy. GA releases this dormancy by increasing the embryo growth potential, and/or weakening the seed coat so the radical of the seedling can break through the seed coat. Different types of seed coats can be made up of living or dead cells, and both types can be influenced by hormones; those composed of living cells are acted upon after seed formation, whereas the seed coats composed of dead cells can be influenced by hormones during the formation of the seed coat. ABA affects testa or seed coat growth characteristics, including thickness, and effects the GA-mediated embryo growth potential. These conditions and effects occur during the formation of the seed, often in response to environmental conditions. Hormones also mediate endosperm dormancy: Endosperm in most seeds is composed of living tissue that can actively respond to hormones generated by the embryo. The endosperm often acts as a barrier to seed germination, playing a part in seed coat dormancy or in the germination process. Living cells respond to and also affect the ABA:GA ratio, and mediate cellular sensitivity; GA thus increases the embryo growth potential and can promote endosperm weakening. GA also affects both ABA-independent and ABA-inhibiting processes within the endosperm.
Human use
[edit]
Salicylic acid
[edit]
Willow bark has been used for centuries as a painkiller. The active ingredient in willow bark that provides these effects is the hormone salicylic acid (SA). In 1899, the pharmaceutical company Bayer began marketing a derivative of SA as the drug aspirin. In addition to its use as a painkiller, SA is also used in topical treatments of several skin conditions, including acne, warts and psoriasis. Another derivative of SA, sodium salicylate has been found to suppress proliferation of lymphoblastic leukemia, prostate, breast, and melanoma human cancer cells.
Jasmonic acid
[edit]
Jasmonic acid (JA) can induce death in lymphoblastic leukemia cells. Methyl jasmonate (a derivative of JA, also found in plants) has been shown to inhibit proliferation in a number of cancer cell lines, although there is still debate over its use as an anti-cancer drug, due to its potential negative effects on healthy cells.
See also
[edit]
Look up plant hormone in Wiktionary, the free dictionary.
Plants portal
Forchlorfenuron
Phytoestrogen
Phytoandrogen
Chlormequat
Glyphosine
References
[edit]
^ Méndez-Hernández HA, Ledezma-Rodríguez M, Avilez-Montalvo RN, Juárez-Gómez YL, Skeete A, Avilez-Montalvo J, et al. (2019). "Signaling Overview of Plant Somatic Embryogenesis". Frontiers in Plant Science. 10: 77. doi:10.3389/fpls.2019.00077. PMC 6375091. PMID 30792725.
^ Shigenaga AM, Argueso CT (August 2016). "No hormone to rule them all: Interactions of plant hormones during the responses of plants to pathogens". Seminars in Cell & Developmental Biology. 56: 174–189. doi:10.1016/j.semcdb.2016.06.005. PMID 27312082.
^ Bürger M, Chory J (August 2019). "Stressed Out About Hormones: How Plants Orchestrate Immunity". Cell Host & Microbe. 26 (2): 163–172. doi:10.1016/j.chom.2019.07.006. PMC 7228804. PMID 31415749.
^ Ku YS, Sintaha M, Cheung MY, Lam HM (October 2018). "Plant Hormone Signaling Crosstalks between Biotic and Abiotic Stress Responses". International Journal of Molecular Sciences. 19 (10): 3206. doi:10.3390/ijms19103206. PMC 6214094. PMID 30336563.
^ Ullah A, Manghwar H, Shaban M, Khan AH, Akbar A, Ali U, et al. (November 2018). "Phytohormones enhanced drought tolerance in plants: a coping strategy". Environmental Science and Pollution Research International. 25 (33): 33103–33118. Bibcode:2018ESPR...2533103U. doi:10.1007/s11356-018-3364-5. PMID 30284160. S2CID 52913388.
^ Pierre-Jerome E, Drapek C, Benfey PN (October 2018). "Regulation of Division and Differentiation of Plant Stem Cells". Annual Review of Cell and Developmental Biology. 34: 289–310. doi:10.1146/annurev-cellbio-100617-062459. PMC 6556207. PMID 30134119.
^ "Plant hormones". NCS Pearson. Archived from the original on 2021-11-27. Retrieved 2018-08-14.
^ "Plant Hormones". Archived from the original on 2019-12-18. Retrieved 2018-08-14.
^ Went FW, Thimann KV (1937). Phytohormones. New York: The Macmillan Company.
^ Tarakhovskaya ER, Maslov Y, Shishova MF (2007). "Phytohormones in algae". Russian Journal of Plant Physiology. 54 (2): 163–170. doi:10.1134/s1021443707020021. S2CID 27373543.
^ Rademacher W (1994). "Gibberellin formation in microorganisms". Plant Growth Regulation. 15 (3): 303–314. doi:10.1007/BF00029903. S2CID 33138732.
^ Srivastava LM (2002). Plant growth and development: hormones and environment. Academic Press. p. 140. ISBN 978-0-12-660570-9.
^ Öpik H, Rolfe SA, Willis JA, Street HE (2005). The physiology of flowering plants (4th ed.). Cambridge University Press. p. 191. ISBN 978-0-521-66251-2.[permanent dead link]
^ Swarup R, Perry P, Hagenbeek D, Van Der Straeten D, Beemster GT, Sandberg G, et al. (July 2007). "Ethylene upregulates auxin biosynthesis in Arabidopsis seedlings to enhance inhibition of root cell elongation". The Plant Cell. 19 (7): 2186–96. doi:10.1105/tpc.107.052100. PMC 1955695. PMID 17630275.
^ Srivastava 2002, p. 143
^ Botany: a brief introduction to plant biology. New York: Wiley. 1979. pp. 155–170. ISBN 978-0-471-02114-8.
^ Feurtado JA, Ambrose SJ, Cutler AJ, Ross AR, Abrams SR, Kermode AR (February 2004). "Dormancy termination of western white pine (Pinus monticola Dougl. Ex D. Don) seeds is associated with changes in abscisic acid metabolism". Planta. 218 (4): 630–9. Bibcode:2004Plant.218..630F. doi:10.1007/s00425-003-1139-8. PMID 14663585. S2CID 25035678.
^ Kermode AR (December 2005). "Role of Abscisic Acid in Seed Dormancy". J Plant Growth Regul. 24 (4): 319–344. doi:10.1007/s00344-005-0110-2.
^ Ren H, Gao Z, Chen L, Wei K, Liu J, Fan Y, et al. (2007). "Dynamic analysis of ABA accumulation in relation to the rate of ABA catabolism in maize tissues under water deficit". Journal of Experimental Botany. 58 (2): 211–9. doi:10.1093/jxb/erl117. PMID 16982652.
^ Else MA, Coupland D, Dutton L, Jackson MB (January 2001). "Decreased root hydraulic conductivity reduces leaf water potential, initiates stomatal closure, and slows leaf expansion in flooded plants of castor oil (Ricinus communis) despite diminished delivery of ABA from the roots to shoots in xylem sap". Physiologia Plantarum. 111 (1): 46–54. doi:10.1034/j.1399-3054.2001.1110107.x.
^ Yan J, Tsuichihara N, Etoh T, Iwai S (October 2007). "Reactive oxygen species and nitric oxide are involved in ABA inhibition of stomatal opening". Plant, Cell & Environment. 30 (10): 1320–5. doi:10.1111/j.1365-3040.2007.01711.x. PMID 17727421.
^ Osborne DJ, McManus MT (2005). Hormones, signals and target cells in plant development. Cambridge University Press. p. 158. ISBN 978-0-521-33076-3.
^ Tomic S, Gabdoulline RR, Kojic-Prodic B, Wade RC (1998). "Classification of auxin related compounds based on similarity of their interaction fields: Extension to a new set of compounds". Internet Journal of Chemistry. 1 (26): CP1 – U21.
^ Walz A, Park S, Slovin JP, Ludwig-Müller J, Momonoki YS, Cohen JD (February 2002). "A gene encoding a protein modified by the phytohormone indoleacetic acid". Proceedings of the National Academy of Sciences of the United States of America. 99 (3): 1718–23. Bibcode:2002PNAS...99.1718W. doi:10.1073/pnas.032450399. PMC 122257. PMID 11830675.
^ Grove MD, Spencer GF, Rohwedder WK, Mandava N, Worley JF, Warthen JD, et al. (1979). "Brassinolide, a plant growth-promoting steroid isolated from Brassica napus pollen". Nature. 281 (5728): 216–217. Bibcode:1979Natur.281..216G. doi:10.1038/281216a0. S2CID 4335601.
^ a b Tang J, Han Z, Chai J (December 2016). "Q&A: what are brassinosteroids and how do they act in plants?". BMC Biology. 14 (1): 113. doi:10.1186/s12915-016-0340-8. PMC 5180403. PMID 28007032.
^ Yamagami A, Saito C, Nakazawa M, Fujioka S, Uemura T, Matsui M, et al. (July 2017). "Evolutionarily conserved BIL4 suppresses the degradation of brassinosteroid receptor BRI1 and regulates cell elongation". Scientific Reports. 7 (1): 5739. Bibcode:2017NatSR...7.5739Y. doi:10.1038/s41598-017-06016-2. PMC 5515986. PMID 28720789.
^ Yamagami A, Nakazawa M, Matsui M, Tujimoto M, Sakuta M, Asami T, Nakano T (February 2009). "Chemical genetics reveal the novel transmembrane protein BIL4, which mediates plant cell elongation in brassinosteroid signaling". Bioscience, Biotechnology, and Biochemistry. 73 (2): 415–421. doi:10.1271/bbb.80752. PMID 19202280. S2CID 35568960.
^ a b c Planas-Riverola A, Gupta A, Betegón-Putze I, Bosch N, Ibañes M, Caño-Delgado AI (March 2019). "Brassinosteroid signaling in plant development and adaptation to stress". Development. 146 (5): dev151894. doi:10.1242/dev.151894. PMC 6432667. PMID 30872266.
^ Ahammed GJ, Li X, Liu A, Chen S (19 March 2020). "Brassinosteroids in Plant Tolerance to Abiotic Stress". Journal of Plant Growth Regulation. 39 (4): 1451–1464. doi:10.1007/s00344-020-10098-0. ISSN 0721-7595. S2CID 213166792.
^ Sipes DL, Einset JW (August 1983). "Cytokinin stimulation of abscission in lemon pistil explants". J Plant Growth Regul. 2 (1–3): 73–80. doi:10.1007/BF02042235. S2CID 43997977.
^ Akhtar SS, Mekureyaw MF, Pandey C, Roitsch T (2020). "Role of Cytokinins for Interactions of Plants With Microbial Pathogens and Pest Insects". Frontiers in Plant Science. 10: 1777. doi:10.3389/fpls.2019.01777. PMC 7042306. PMID 32140160.
^ "Cytokinin - an overview | ScienceDirect Topics". www.sciencedirect.com. Retrieved 2021-06-10.
^ Wang Y, Liu C, Li K, Sun F, Hu H, Li X, et al. (August 2007). "Arabidopsis EIN2 modulates stress response through abscisic acid response pathway". Plant Molecular Biology. 64 (6): 633–44. doi:10.1007/s11103-007-9182-7. PMID 17533512. S2CID 42139177.
^ Jackson MB (1985). "Ethylene and Responses of Plants to Soil Waterlogging and Submergence". Annual Review of Plant Physiology. 36 (1): 145–174. doi:10.1146/annurev.pp.36.060185.001045. ISSN 0066-4294.
^ Jackson MB (January 2008). "Ethylene-promoted elongation: an adaptation to submergence stress". Annals of Botany. 101 (2): 229–48. doi:10.1093/aob/mcm237. PMC 2711016. PMID 17956854.
^ Jackson MB, Ram PC (January 2003). "Physiological and molecular basis of susceptibility and tolerance of rice plants to complete submergence". Annals of Botany. 91 Spec No (2): 227–41. doi:10.1093/aob/mcf242. PMC 4244997. PMID 12509343.
^ Voesenek LA, Benschop JJ, Bou J, Cox MC, Groeneveld HW, Millenaar FF, et al. (January 2003). "Interactions between plant hormones regulate submergence-induced shoot elongation in the flooding-tolerant dicot Rumex palustris". Annals of Botany. 91 Spec No (2): 205–11. doi:10.1093/aob/mcf116. PMC 4244986. PMID 12509341.
^ Summers JE, Voesenek L, Blom C, Lewis MJ, Jackson MB (July 1996). "Potamogeton pectinatus Is Constitutively Incapable of Synthesizing Ethylene and Lacks 1-Aminocyclopropane-1-Carboxylic Acid Oxidase". Plant Physiology. 111 (3): 901–908. doi:10.1104/pp.111.3.901. PMC 157909. PMID 12226336.
^ Grennan AK (June 2006). "Gibberellin metabolism enzymes in rice". Plant Physiology. 141 (2): 524–6. doi:10.1104/pp.104.900192. PMC 1475483. PMID 16760495.
^ Tsai FY, Lin CC, Kao CH (January 1997). "A comparative study of the effects of abscisic acid and methyl jasmonate on seedling growth of rice". Plant Growth Regulation. 21 (1): 37–42. doi:10.1023/A:1005761804191. S2CID 34099546.
^ Demole E, Lederer E, Mercier D (1962). "Isolement et détermination de la structure du jasmonate de méthyle, constituant odorant caractéristique de l'essence de jasmin" [Isolation and determination of the structure of methyl jasmonate, a fragrant constituent characteristic of jasmine oil]. Helvetica Chimica Acta (in French). 45 (2): 675–685. doi:10.1002/hlca.19620450233.
^ a b Browse J (2005). "Jasmonate: an oxylipin signal with many roles in plants". Plant Hormones. Vitamins & Hormones. Vol. 72. pp. 431–56. doi:10.1016/S0083-6729(05)72012-4. ISBN 9780127098722. PMID 16492478.
^ Katsir L, Chung HS, Koo AJ, Howe GA (August 2008). "Jasmonate signaling: a conserved mechanism of hormone sensing". Current Opinion in Plant Biology. 11 (4): 428–35. doi:10.1016/j.pbi.2008.05.004. PMC 2560989. PMID 18583180.
^ Lorenzo O, Solano R (October 2005). "Molecular players regulating the jasmonate signalling network". Current Opinion in Plant Biology. 8 (5): 532–40. doi:10.1016/j.pbi.2005.07.003. PMID 16039901.
^ Wasternack C (October 2007). "Jasmonates: an update on biosynthesis, signal transduction and action in plant stress response, growth and development". Annals of Botany. 100 (4): 681–97. doi:10.1093/aob/mcm079. PMC 2749622. PMID 17513307.
^ a b Howe GA, Jander G (2008). "Plant Immunity to Insect Herbivores". Annu. Rev. Plant Biol. 59: 41–66. doi:10.1146/annurev.arplant.59.032607.092825. PMID 18031220.
^ Paschold A, Halitschke R, Baldwin IT (July 2007). "Co(i)-ordinating defenses: NaCOI1 mediates herbivore- induced resistance in Nicotiana attenuata and reveals the role of herbivore movement in avoiding defenses". The Plant Journal: For Cell and Molecular Biology. 51 (1): 79–91. doi:10.1111/j.1365-313X.2007.03119.x. PMID 17561925.
^ Zarate SI, Kempema LA, Walling LL (February 2007). "Silverleaf whitefly induces salicylic acid defenses and suppresses effectual jasmonic acid defenses". Plant Physiology. 143 (2): 866–75. doi:10.1104/pp.106.090035. PMC 1803729. PMID 17189328.
^ Rivas-San Vicente M, Plasencia J (June 2011). "Salicylic acid beyond defence: its role in plant growth and development". Journal of Experimental Botany. 62 (10): 3321–38. doi:10.1093/jxb/err031. PMID 21357767.
^ Dempsey DA, Klessig DF (March 2017). "How does the multifaceted plant hormone salicylic acid combat disease in plants and are similar mechanisms utilized in humans?". BMC Biology. 15 (1): 23. doi:10.1186/s12915-017-0364-8. PMC 5364617. PMID 28335774.
^ Kumar D (November 2014). "Salicylic acid signaling in disease resistance". Plant Science. 228: 127–34. doi:10.1016/j.plantsci.2014.04.014. PMID 25438793.
^ Ding P, Ding Y (June 2020). "Stories of Salicylic Acid: A Plant Defense Hormone". Trends in Plant Science. 25 (6): 549–565. doi:10.1016/j.tplants.2020.01.004. PMID 32407695. S2CID 213666131.
^ Xie X, Yoneyama K, Yoneyama K (2010-07-01). "The strigolactone story". Annual Review of Phytopathology. 48 (1): 93–117. doi:10.1146/annurev-phyto-073009-114453. PMID 20687831. S2CID 27305711.
^ Akiyama K, Matsuzaki K, Hayashi H (June 2005). "Plant sesquiterpenes induce hyphal branching in arbuscular mycorrhizal fungi". Nature. 435 (7043): 824–7. Bibcode:2005Natur.435..824A. doi:10.1038/nature03608. PMID 15944706. S2CID 4343708.
^ Gomez-Roldan V, Fermas S, Brewer PB, Puech-Pagès V, Dun EA, Pillot JP, et al. (September 2008). "Strigolactone inhibition of shoot branching". Nature. 455 (7210): 189–94. Bibcode:2008Natur.455..189G. doi:10.1038/nature07271. PMID 18690209. S2CID 205214618.
^ Saeed W, Naseem S, Ali Z (2017-08-28). "Strigolactones Biosynthesis and Their Role in Abiotic Stress Resilience in Plants: A Critical Review". Frontiers in Plant Science. 8: 1487. doi:10.3389/fpls.2017.01487. PMC 5581504. PMID 28894457.
^ Lindsey K, Casson S, Chilley P (February 2002). "Peptides: new signalling molecules in plants". Trends in Plant Science. 7 (2): 78–83. doi:10.1016/S0960-9822(01)00435-3. PMID 11832279. S2CID 5064533.
^ Takahashi F, Suzuki T, Osakabe Y, Betsuyaku S, Kondo Y, Dohmae N, et al. (April 2018). "A small peptide modulates stomatal control via abscisic acid in long-distance signalling". Nature. 556 (7700): 235–238. Bibcode:2018Natur.556..235T. doi:10.1038/s41586-018-0009-2. PMID 29618812. S2CID 4598494.
^ Pandey S, Ranade SA, Nagar PK, Kumar N (September 2000). "Role of polyamines and ethylene as modulators of plant senescence". Journal of Biosciences. 25 (3): 291–9. doi:10.1007/BF02703938. PMID 11022232. S2CID 21925829.
^ Moschou PN, Roubelakis-Angelakis KA (March 2014). "Polyamines and programmed cell death". Journal of Experimental Botany. 65 (5): 1285–96. doi:10.1093/jxb/ert373. PMID 24218329.
^ Shapiro AD (2005). "Nitric oxide signaling in plants". Plant Hormones. Vitamins & Hormones. Vol. 72. pp. 339–98. doi:10.1016/S0083-6729(05)72010-0. ISBN 9780127098722. PMID 16492476. S2CID 13308295.
^ Roszer T (2012). "Nitric Oxide Synthesis in the Chloroplast". In Roszer T (ed.). The Biology of Subcellular Nitric Oxide. New York, London, Heidelberg: Springer. ISBN 978-94-007-2818-9.
^ Chiwocha SD, Dixon KW, Flematti GR, Ghisalberti EL, Merritt DJ, Nelson DC, et al. (2009-10-01). "Karrikins: A new family of plant growth regulators in smoke". Plant Science. 177 (4): 252–256. doi:10.1016/j.plantsci.2009.06.007.
^ Li W, Nguyen KH, Chu HD, Ha CV, Watanabe Y, Osakabe Y, et al. (November 2017). "The karrikin receptor KAI2 promotes drought resistance in Arabidopsis thaliana". PLOS Genetics. 13 (11): e1007076. doi:10.1371/journal.pgen.1007076. PMC 5703579. PMID 29131815.
^ Yakhin OI, Lubyanov AA, Yakhin IA, Brown PH (2017-01-26). "Biostimulants in Plant Science: A Global Perspective". Frontiers in Plant Science. 7: 2049. doi:10.3389/fpls.2016.02049. PMC 5266735. PMID 28184225.
^ Leubner G (2000). "Seed Dormancy". The Seed Biology Place. Royal Holloway University of London.
^ Diarmuid J (December 2008). Aspirin : the remarkable story of a wonder drug. New York, NY. ISBN 9781596918160. OCLC 879610692.{{cite book}}: CS1 maint: location missing publisher (link)
^ Madan RK, Levitt J (April 2014). "A review of toxicity from topical salicylic acid preparations". Journal of the American Academy of Dermatology. 70 (4): 788–792. doi:10.1016/j.jaad.2013.12.005. PMID 24472429.
^ a b Fingrut O, Flescher E (April 2002). "Plant stress hormones suppress the proliferation and induce apoptosis in human cancer cells". Leukemia. 16 (4): 608–16. doi:10.1038/sj.leu.2402419. PMID 11960340.
^ Zhang M, Zhang MW, Zhang L, Zhang L (2015-07-24). "Methyl jasmonate and its potential in cancer therapy". Plant Signaling & Behavior. 10 (9): e1062199. Bibcode:2015PlSiB..10E2199Z. doi:10.1080/15592324.2015.1062199. PMC 4883903. PMID 26208889.
External links
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Wikimedia Commons has media related to Plant hormones.
Simple plant hormone table with location of synthesis and effects of application — this is the format used in the description templates at bottom of Wikipedia articles about plant hormones.
Hormonal Regulation of Gene Expression and Development — Detailed introduction to plant hormones, including genetic information.
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Plant cellsCell wallPhragmoplastPlastidPlasmodesmaVacuoleTissuesCorkGround tissueMesophyllMeristemStorage organsVascular tissueVascular bundleWoodVegetativeBulbRootRhizoidRhizomeShootBudLeafCataphyllPetioleSessilityStemReproductive(incl. Flower)ArchegoniumAntheridiumAndroeciumPollenStamenAntherFilamentStaminodeTapetumFlowerAestivationFlower developmentFloral diagramFloral formulaFloral symmetryWhorlFruitAnatomyBerryCapsuleNutPyrenaSeedDispersalEndospermGametophyteGynandriumGynoeciumCarpelOvaryLoculeOvuleStigmaStyleHypanthium (Floral cup)InflorescenceBractPedicellateRacemeUmbelPerianthTepalPetalSepalPlant embryoReceptacleSporophyllSporophyteSurface structuresCuticleEpicuticular waxEpidermisNectarStomaThorns, spines, and pricklesTrichome
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Plant hormone
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15337 | https://api.pageplace.de/preview/DT0400.9780192524836_A30545946/preview-9780192524836_A30545946.pdf | Acute abdomen 606 Acute kidney injury 298 Addisonian crisis 836 Anaphylaxis 794 Aneurysm, abdominal aortic 654 intracranial/extradural 78, 482 gastrointestinal 256, 820 rectal 629 variceal 257, 820 Antidotes, poisoning 842 Arrhythmias, broad complex 128, 804 narrow complex, SVT 126, 806 Asthma 810 Asystole 895 Atrial fl utter/fi brillation Bacterial shock 790 Blast injury 851 Bradycardia 124 Burns 846 Cardiac arrest 894 (Fig A3) Cardiogenic tamponade 802 Cardioversion, DC 770 Central line insertion (CVP line) 774 Cerebral oedema 830 Chest drain 766 Coma 786 Cricothyrotomy 772 Cyanosis 186–9 Cut-down 761 Defi brillation 770, 894 (Fig A3) Diabetes emergencies 832–4 Disseminated intravascular coagulopathy (DIC) 352 Disaster, major 850 Encephalitis 824 Epilepsy, status 826 Extradural haemorrhage 482 Fluids, IV 666, 790 Haematemesis 256–7 Haemorrhage 790 Hyperthermia 790, 838 Hypoglycaemia 214, 834 Hypothermia 848 Intracranial pressure, raised 830 Ischaemic limb 656 Malaria 416 Malignant hyperpyrexia 572 Index to emergency topics ‘Don’t go so fast: we’re in a hurry!’—Talleyrand to his coachman.
Malignant hypertension 140 Meningitis 822 Meningococcaemia 822 Myocardial infarction 796 Needle pericardiocentesis 773 Neutropenic sepsis 352 Obstructive uropathy 641 Oncological emergencies 528 Opioid poisoning 842 Overdose 838–44 Pacemaker, temporary 776 Pericardiocentesis 773 Phaeochromocytoma 837 Pneumonia 816 Pneumothorax 814 Poisoning 838–44 Potassium, hyperkalaemia 674 hypokalaemia 674 Pulmonary embolism 818 Respiratory arrest 894 (Fig A3) Respiratory failure 188 Resuscitation 894 (Fig A3) Rheumatological emergencies 538 Shock 790 Smoke inhalation 847 Sodium, hypernatraemia 672 hyponatraemia 672 Spinal cord compression 466, 543 Status asthmaticus 810 Status epilepticus 826 Stroke 470 Superior vena cava obstruction 528 Supraventricular tachycardia (SVT) 806 Testicular torsion 652 Thrombotic thrombocytopenic purpura (TTP) 315 Thyroid storm 834 Transfusion reaction 349 Varices, bleeding 257, 820 Vasculitis, acute systemic 556 Venous thromboembolism, leg 656 pulmonary 818 Ventricular arrhythmias 128, 804 Ventricular failure, left 800 Ventricular fi brillation 894 (Fig A3) Ventricular tachycardia 128, 804 Common haematology values Haemoglobin men: 130–180g/L p324 women: 115–160g/L p324 Mean cell volume, MCV 76–96fL p326; p332 Platelets 150–400 ≈ 10 9/L p364 White cells (total) 4–11 ≈ 10 9/L p330 neutrophils 2.0–7.5 ≈ 10 9/L p330 lymphocytes 1.0–4.5 ≈ 10 9/L p330 eosinophils 0.04–0.4 ≈ 10 9/L p330 Blood gases pH 7.35–7.45 p670 P aO2 >10.6kPa p670 P aCO2 4.7–6kPa p670 Base excess ± 2mmol/L p670 U&ES (urea and electrolytes) Sodium 135–145mmol/L p672 Potassium 3.5–5.3mmol/L p674 Creatinine 70–100μmol/L p298–301 Urea 2.5–6.7mmol/L p298–301 eGFR >60 p669 LFTS (liver function tests) Bilirubin 3–17μmol/L p272, p274 Alanine aminotransferase, ALT 5–35IU/L p272, p274 Aspartate transaminase, AST 5–35IU/L p272, p274 Alkaline phosphatase, ALP 30–130IU/L (non-pregnant adults) p272, p274 Albumin 35–50g/L p686 Cardiac enzymes Troponin T <99th percentile of upper reference limit: value depends on local assay p119 Other biochemical values Cholesterol <5mmol/L p690 Triglycerides Fasting: 0.5–2.3mmol/L p690 Amylase 0–180 IU/dL p636 C-reactive protein, CRP <10mg/L p686 Corrected calcium 2.12–2.60mmol/L p676 Glucose, fasting 3.5–5.5mmol/L p206 Thyroid stimulating hormone, TSH 0.5–4.2mU/L p216 For all other reference intervals, see p750–7 Reading tests Hold this chart (well-illuminated) 30cm away, and record the smallest type read (eg N12 left eye, N6 right eye, spectacles worn) or object named accurately.
all the brightest gems N. 24 He moved N. 48 faster and faster towards the N. 18 ever-growing bucket of lost hopes; had there been just one more year N. 14 of peace the battalion would have made a floating system of perpetual drainage.
N. 12 A silent fall of immense snow came near oily remains of the recently eaten supper on the table.
N. 10 We drove on in our old sunless walnut. Presently classical eggs ticked in the new afternoon shadows.
N. 8 We were instructed by my cousin Jasper not to exercise by country house visiting unless accompanied by thirteen geese or gangsters.
N. 6 The modern American did not prevail over the pair of redundant bronze puppies.
The worn-out principle is a bad omen which I am never glad to ransom in August.
N. 5 OXFORD HANDBOOK OF CLINICAL MEDICINE TENTH EDITION Ian B. Wilkinson Tim Raine Kate Wiles Anna Goodhart Catriona Hall Harriet O’Neill 3 Oxford University Press, Great Clarendon Street, Oxford OX2 6DP Oxford University Press is a department of the University of Oxford. It furthers the University’s objective of excellence in research, scholarship, and education by publishing worldwide. Oxford is a registered trade mark of Oxford University Press in the UK and in certain other countries.
Published in the United States by Oxford University Press Inc., New York © Oxford University Press, 2017 The moral rights of the authors have been asserted Database right Oxford University Press (maker) First published 1985 Fifth edition 2001 Tenth edition 2017 (RA Hope & JM Longmore) (JM Longmore & IB Wilkinson) (IB Wilkinson, T Raine & K Wiles) Second edition 1989 Sixth edition 2004 Third edition 1993 Seventh edition 2007 Fourth edition 1998 Eighth edition 2010 Ninth edition 2014 Translations: Chinese French Hungarian Polish Russian Czech German Indonesian Portuguese Spanish Estonian Greek Italian Romanian All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without the prior permission in writing of Oxford University Press, or as expressly permitted by law, or under terms agreed with the appropriate reprographics rights organization. Enquiries concerning reproduction outside the scope of the above should be sent to the Rights Department, Oxford University Press, at the address above.
You must not circulate this book in any other binding or cover and you must impose the same condition on any acquirer.
British Library Cataloguing in Publication Data Data available Library of Congress Control Number: 2017939060 Typeset by GreenGate Publishing Services, Tonbridge, UK; printed in China by C&C Off set Printing Co. Ltd.
ISBN 978-0-19-968990-3 Except where otherwise stated, recommendations are for the non-pregnant adult who is not breastfeeding and who has reasonable renal and hepatic function.
We have made every eff ort to check this text, but it is still possible that drug or other errors have been missed. OUP makes no representation, express or implied, that doses are correct. Readers are urged to check with the most up to date prod-uct information, codes of conduct, and safety regulations. The authors and the publishers do not accept responsibility or legal liability for any errors in the text, or for the misuse or misapplication of material in this work.
For updates/corrections, see Drugs Contents Each chapter’s contents are detailed on its fi rst page Prefaces to the fi rst and tenth editions iv Acknowledgements v Symbols and abbreviations vi 1 Thinking about medicine 0 2 History and examination 24 3 Cardiovascular medicine 92 4 Chest medicine 160 5 Endocrinology 202 6 Gastroenterology 242 7 Renal medicine 292 8 Haematology 322 9 Infectious diseases 378 10 Neurology 444 11 Oncology and palliative care 518 12 Rheumatology 538 13 Surgery 564 14 Clinical chemistry 662 15 Eponymous syndromes 694 16 Radiology 718 17 Reference intervals, etc.
750 18 Practical procedures 758 19 Emergencies 778 20 References 852 Index 868 Early warning score 892 Cardiac arrest 894 Preface to the tenth edition This is the fi rst edition of the book without either of the original authors—Tony Hope and Murray Longmore. Both have now moved on to do other things, and enjoy a well-earned rest from authorship. In this book, I am joined by a Nephrologist, Gas-troenterologist, and trainees destined for careers in Cardiology, Dermatology, and General Practice. Five physicians, each with very diff erent interests and approaches, yet bringing their own knowledge, expertise, and styles. When combined with that of our specialist and junior readers, I hope this creates a book that is greater than the sum of its parts, yet true to the original concept and ethos of the original authors. Life and medicine have moved on in the 30 years since the fi rst edition was published, but medicine and science are largely iterative; true novel ‘ground-breaking’ or ‘prac-tice-changing’ discoveries are rare, to quote Isaac Newton: ‘If I have seen further, it is by standing on the shoulders of giants’. Therefore, when we set about writing this edition we drew inspiration from the original book and its authors; updating, adding, and clarifying, but trying to retain the unique feel and perspective that the OHCM has provided to generations of trainees and clinicians.
IBW, 2017 We wrote this book not because we know so much, but because we know we remember so little…the problem is not simply the quantity of information, but the diversity of places from which it is dispensed. Trailing eagerly behind the surgeon, the student is admonished never to forget alcohol withdrawal as a cause of post-operative confusion. The scrap of paper on which this is written spends a month in the pocket before being lost for ever in the laundry. At diff erent times, and in inconvenient places, a number of other causes may be presented to the student. Not only are these causes and aphorisms never brought together, but when, as a surgical house offi cer, the former student faces a confused patient, none is to hand.
We aim to encourage the doctor to enjoy his patients: in doing so we believe he will prosper in the practice of medicine. For a long time now, house offi cers have been encouraged to adopt monstrous proportions in order to straddle the diverse pinnacles of clinical science and clinical experience. We hope that this book will make this endeavour a little easier by moving a cumulative memory burden from the mind into the pocket, and by removing some of the fears that are naturally felt when starting a career in medicine, thereby freely allowing the doctor’s clinical acumen to grow by the slow accretion of many, many days and nights.
RA Hope and JM Longmore, 1985 Preface to the fi rst edition Acknowledgements Heart-felt thanks to our advisers on specifi c sections—each is acknowledged on the chapter’s fi rst page. Thanks also to our junior readers, Charles Badu-Boateng, Clare Coggins, and Luke Walls. We especially thank all our mentors and teach-ers, and patients who provide our inspiration and remind us that one never stops learning. We acknowledge the Department of Radiology at both the Leeds Teach-ing Hospitals NHS Trust and the Norfolk and Norwich University Hospital for their kind help in providing many images, particularly Dr Edmund Godfrey, whose tire-less hunt for perfect images has improved so many chapters.
Readers’ comments These have formed a vital part of our endeavour to provide an accurate, comprehensive, and up-to-date text. We sincerely thank the many stu-dents, doctors, and other health professionals who have found the time and the generosity to write to us on our Reader’s Comments Cards, in editions past, or, in more recent times, via the web. These have now become so numerous for past edi-tions that they cannot all be listed. See oxhmed/links for a full list, and our very heart-felt tokens of thanks.
3rd-party web addresses We disclaim any responsibility for 3rd-party content.
Symbols and abbreviations ..........this fact or idea is important .......don’t dawdle!—prompt action saves lives 1 ...........reference :......male-to-female ratio. :=2:1 means twice as common in males .........therefore ~ ..........approximately –ve ......negative (+ve is positive) ........increased or decreased .......normal (eg serum level) 1° ........primary 2° ........secondary ..........diagnosis ........diff erential diagnosis A:CR ......albumin to creatinine ratio (mg/mmol) A2 .........aortic component of the 2nd heart sound Ab ......antibody ABC ......airway, breathing, and circulation ABG .....arterial blood gas: P aO2, P aCO2, pH, HCO3 ABPA ....allergic bronchopulmonary aspergillosis ACE-i .....angiotensin-converting enzyme inhibitor ACS .......acute coronary syndrome ACTH ....adrenocorticotrophic hormone ADH .....antidiuretic hormone AF ........atrial fi brillation AFB ......acid-fast bacillus Ag .......antigen AIDS ....acquired immunodefi ciency syndrome AKI ........acute kidney injury ALL ......acute lymphoblastic leukaemia ALP ......alkaline phosphatase AMA ....antimitochondrial antibody AMP .....adenosine monophosphate ANA .....antinuclear antibody ANCA ...antineutrophil cytoplasmic antibody APTT ....activated partial thromboplastin time AR ........aortic regurgitation ARB .....angiotensin II receptor ‘blocker’ (antagonist) ARDS ...acute respiratory distress syndrome ART ......antiretroviral therapy AS ........aortic stenosis ASD .....atrial septal defect AST ......aspartate transaminase ATN ......acute tubular necrosis ATP ......adenosine triphosphate AV ........atrioventricular AVM .....arteriovenous malformation(s) AXR .....abdominal X-ray (plain) Ba ........barium BAL ......bronchoalveolar lavage bd .......bis die (Latin for twice a day) BKA .....below-knee amputation BNF ......British National Formulary BNP ......brain natriuretic peptide BP ........blood pressure BPH ......benign prostatic hyperplasia bpm ....beats per minute ca ........cancer CABG ...coronary artery bypass graft cAMP ...cyclic adenosine monophosphate (AMP) CAPD ...continuous ambulatory peritoneal dialysis CCF ......congestive cardiac failure (ie left and right heart failure) CCU ......coronary care unit CDT ......Clostridium diffi cile toxin CHB ......complete heart block CHD ......coronary heart disease CI .........contraindications CK ........creatine (phospho)kinase CKD ......chronic kidney disease CLL ......chronic lymphocytic leukaemia CML .....chronic myeloid leukaemia CMV .....cytomegalovirus CNS ......central nervous system COC ......combined oral contraceptive pill COPD ....chronic obstructive pulmonary disease CPAP ....continuous positive airway pressure CPR ......cardiopulmonary resuscitation CRP ......c-reactive protein CSF ......cerebrospinal fl uid CT ........computed tomography CVA ......cerebrovascular accident CVP ......central venous pressure CVS ......cardiovascular system CXR ......chest x-ray d ..........day(s); also expressed as /7; months are /12 DC ........direct current DIC ......disseminated intravascular coagulation DIP ......distal interphalangeal dL .......decilitre DM .......diabetes mellitus DOAC ...direct oral anticoagulant DU ........duodenal ulcer D&V .....diarrhoea and vomiting DVT ......deep venous thrombosis DXT ......deep radiotherapy EBV ......Epstein–Barr virus ECG ......electrocardiogram Echo ...echocardiogram EDTA ....ethylene diamine tetra-acetic acid (anticoagulant coating, eg in FBC bottles) EEG ......electroencephalogram eGFR ....estimated glomerular fi ltration rate (in mL/ min/1.73m 2) ELISA ...enzyme-linked immunosorbent assay EM .......electron microscope EMG .....electromyogram ENT ......ear, nose, and throat ERCP ....endoscopic retrograde cholangiopancreatography ESR ......erythrocyte sedimentation rate ESRF ....end-stage renal failure EUA ......examination under anaesthesia FBC ......full blood count FDP ......fi brin degradation products FEV1 .....forced expiratory volume in 1st sec FiO2 ....partial pressure of O2 in inspired air FFP ......fresh frozen plasma FSH ......follicle-stimulating hormone FVC ......forced vital capacity g ..........gram G6PD ....glucose-6-phosphate dehydrogenase GA .......general anaesthetic GCS ......Glasgow Coma Scale GFR ......glomerular fi ltration rate GGT ......gamma-glutamyl transferase GH ........growth hormone GI ........gastrointestinal GN ........glomerulonephritis GP ........general practitioner GPA ......granulomatosis with polyangiitis (formerly Wegener’s granulomatosis) GTN ......glyceryl trinitrate GTT ......glucose tolerance test GU(M) ..genitourinary (medicine) h ..........hour HAV .....hepatitis A virus Hb .......haemoglobin HbA1c .glycated haemoglobin HBSAg ..hepatitis B surface antigen HBV .....hepatitis B virus HCC ......hepatocellular cancer HCM .....hypertrophic obstructive cardiomyopathy Hct ......haematocrit HCV ......hepatitis C virus HDV .....hepatitis D virus HDL ......high-density lipoprotein HHT ......hereditary haemorrhagic telangiectasia HIV ......human immunodefi ciency virus HLA ......human leucocyte antigen HONK ...hyperosmolar non-ketotic (coma) HPV ......human papillomavirus HRT ......hormone replacement therapy HSP ......Henoch–Schönlein purpura HSV ......herpes simplex virus HUS ......haemolytic uraemic syndrome IBD ......infl ammatory bowel disease IBW .....ideal body weight ICD ......implantable cardiac defi brillator ICP .......intracranial pressure IC(T)U ..intensive care unit IDDM ...insulin-dependent diabetes mellitus IFN- ..interferon alpha IE .........infective endocarditis Ig ........immunoglobulin IHD ......ischaemic heart disease IM ........intramuscular INR ......international normalized ratio IP .........interphalangeal IPPV ....intermittent positive pressure ventilation ITP .......idiopathic thrombocytopenic purpura IU ........international unit IVC ......inferior vena cava IV(I) ....intravenous (infusion) IVU ......intravenous urography JVP ......jugular venous pressure K ..........potassium kg .......kilogram KPa ......kiloPascal L ..........litre LAD ........left axis deviation on the ECG LBBB ....left bundle branch block LDH ......lactate dehydrogenase LDL ......low-density lipoprotein LFT ......liver function test LH ........luteinizing hormone LIF .......left iliac fossa LKKS ....liver, kidney (R), kidney (L), spleen LMN .....lower motor neuron LMWH ..low-molecular-weight heparin LOC ......loss of consciousness LP ........lumbar puncture LUQ ......left upper quadrant LV ........left ventricle of the heart LVF .......left ventricular failure LVH ......left ventricular hypertrophy MAI .....Mycobacterium avium intracellulare MALT ...mucosa-associated lymphoid tissue mane ..morning (from Latin) MAOI ...monoamine oxidase inhibitor MAP .....mean arterial pressure MC&S ...microscopy, culture, and sensitivity mcg ....microgram MCP .....metacarpo-phalangeal MCV .....mean cell volume MDMA ..3,4-methylenedioxymethamphetamine ME .......myalgic encephalomyelitis mg ......milligram MI ........myocardial infarction min(s) minute(s) mL .......millilitre mmHg millimetres of mercury MND .....motor neuron disease MR .......modifi ed release or mitral regurgitation MRCP ...magnetic resonance cholangiopancreatography MRI ......magnetic resonance imaging MRSA ...meticillin-resistant Staph. aureus MS .......multiple sclerosis MSM ....men who have sex with men MSU .....midstream urine N&V .....nausea and/or vomiting NAD .....nothing abnormal detected NBM .....nil by mouth ND ........notifi able disease NEWS ..National Early Warning Score ng .......nanogram NG ........nasogastric NHS .....National Health Service (UK) NICE ....National Institute for Health and Care Excellence, NIDDM .non-insulin-dependent diabetes mellitus NMDA ..N-methyl-D-aspartate NNT .....number needed to treat nocte ..at night NR ........normal range (=reference interval) NSAID ..non-steroidal anti-infl ammatory drug OCP ......oral contraceptive pill od .......omni die (Latin for once daily) OGD .....oesophagogastroduodenoscopy OGTT ....oral glucose tolerance test OHCS ....Oxford Handbook of Clinical Specialties om ......omni mane (in the morning) on .......omni nocte (at night) OPD ......outpatients department OT ........occupational therapist P:CR .....protein to creatinine ratio (mg/mmol) P2 .........pulmonary component of 2nd heart sound PaCO2 ...partial pressure of CO2 in arterial blood PAN ......polyarteritis nodosa PaO2 .....partial pressure of O2 in arterial blood PBC ......primary biliary cirrhosis PCR ......polymerase chain reaction PCV ......packed cell volume PE ........pulmonary embolism PEEP ....positive end-expiratory pressure PEF(R) ..peak expiratory fl ow (rate) PERLA ..pupils equal and reactive to light and accommodation PET ......positron emission tomography PID ......pelvic infl ammatory disease PIP .......proximal interphalangeal (joint) PMH .....past medical history PND .....paroxysmal nocturnal dyspnoea PO ........per os (by mouth) PPI .......proton pump inhibitor, eg omeprazole PR ........per rectum (by the rectum) PRL ......prolactin PRN ......pro re nata (Latin for as required) PRV ......polycythaemia rubra vera PSA ......prostate-specifi c antigen PTH ......parathyroid hormone PTT ......prothrombin time PUO ......pyrexia of unknown origin PV ........per vaginam (by the vagina, eg pessary) PVD ......peripheral vascular disease QDS ......quater die sumendus; take 4 times daily qqh .....quarta quaque hora: take every 4h R ..........right RA ........rheumatoid arthritis RAD .....right axis deviation on the ECG RBBB ...right bundle branch block RBC ......red blood cell RCT ......randomized controlled trial RDW ....red cell distribution width RFT ......respiratory function tests Rh ........Rhesus status RIF .......right iliac fossa RRT ......renal replacement therapy RUQ .....right upper quadrant RV ........right ventricle of heart RVF ......right ventricular failure RVH ......right ventricular hypertrophy .........recipe (Latin for treat with) s/sec ...second(s) S1, S2 ....fi rst and second heart sounds SBE ......subacute bacterial endocarditis SC ........subcutaneous SD ........standard deviation SE ........side-eff ect(s) SIADH ..syndrome of inappropriate anti-diuretic hormone secretion SL ........sublingual SLE ......systemic lupus erythematosus SOB ......short of breath SOBOE .short of breath on exertion SpO2 ....peripheral oxygen saturation (%) SR ........slow-release Stat ....statim (immediately; as initial dose) STD/I ...sexually transmitted disease/infection SVC ......superior vena cava SVT ......supraventricular tachycardia T° .........temperature t½ ........biological half-life T3 ........tri-iodothyronine T4 ........thyroxine TB ........tuberculosis TDS ......ter die sumendus (take 3 times a day) TFT ......thyroid function test (eg TSH) TIA ......transient ischaemic attack TIBC ....total iron-binding capacity TPN ......total parenteral nutrition TPR ......temperature, pulse, and respirations count TRH ......thyrotropin-releasing hormone TSH ......thyroid-stimulating hormone TTP ......thrombotic thrombocytopenic purpura U ..........units UC ........ulcerative colitis U&E .....urea and electrolytes and creatinine UMN .....upper motor neuron URT(I) ..upper respiratory tract (infection) US(S) ....ultrasound (scan) UTI ......urinary tract infection VDRL ....Venereal Diseases Research Laboratory VE ........ventricular extrasystole VF ........ventricular fi brillation VHF ......viral haemorrahgic fever VMA ....vanillylmandelic acid (HMMA) V/Q .......ventilation/perfusion scan VRE ......vancomycin resistant enterococci VSD ......ventricular-septal defect VT ........ventricular tachycardia VTE ......venous thromboembolism WBC ....white blood cell WCC ....white blood cell count wk(s) ..week(s) yr(s) ...year(s) ZN ........Ziehl–Neelsen stain, eg for mycobacteria ‘He who studies medicine without books sails an unchartered sea, but he who studies medicine without patients does not go to sea at all’ William Osler 1849–1919 The word ‘patient’ occurs frequently throughout this book.
Do not skim over it lightly.
Rather pause and doff your metaphorical cap, off ering due respect to those who by the opening up of their lives to you, become your true teachers.
Without your patients, you are a technician with a useless skill.
With them, you are a doctor.
1 Thinking about medicine Contents The Hippocratic oath 1 Medical care 2 Compassion 3 The diagnostic puzzle 4 Being wrong 5 Duty of candour 5 Bedside manner and communication skills 6 Prescribing drugs 8 Surviving life on the wards 10 Death 12 Medical ethics 14 Psychiatry on medical and surgical wards 15 The older person 16 The pregnant woman 17 Epidemiology 18 Randomized controlled trials 19 Medical mathematics 20 Evidence-based medicine (EBM) 22 Medicalization 23 Fig 1.1 Asclepius, the god of healing and his three daughters, Meditrina (medicine), Hygieia (hy-giene), and Panacea (healing). The staff and single snake of Asclepius should not be confused with the twin snakes and caduceus of Hermes, the dei-fi ed trickster and god of commerce, who is viewed with disdain. Plate from Aubin L Millin, Galerie Mythologique (1811) We thank Dr Kate Mansfi eld, our Specialist Reader, for her contribution to this chapter. Thinking about medicine 1 The Hippocratic oath ~4th century BC I swear by Apollo the physician and Asclepius and Hygieia and Panacea and all the gods and goddesses, making them my witnesses, that I will fulfi l according to my ability and judgement this oath and this covenant.
T o hold him who has taught me this art as equal to my parents and to live my life in partnership with him, and if he is in need of money to give him a share of mine, and to regard his off spring as equal to my own brethren and to teach them this art, if they desire to learn it, without fee and covenant. I will impart it by pre-cept, by lecture and by all other manner of teaching, not only to my own sons but also to the sons of him who has taught me, and to disciples bound by covenant and oath according to the law of physicians, but to none other.
T he regimen I shall adopt shall be to the benefi t of the patients to the best of my power and judgement, not for their injury or any wrongful purpose.
I will not give a deadly drug to anyone though it be asked of me, nor will I lead the way in such counsel.
1 And likewise I will not give a woman a pessary to pro-cure abortion.
2 But I will keep my life and my art in purity and holiness. I will not use the knife, 3 not even, verily, on suff erers of stone but I will give place to such as are craftsmen therein.
W hatsoever house I enter, I will enter for the benefi t of the sick, refraining from all voluntary wrongdoing and corruption, especially seduction of male or female, bond or free.
W hatsoever things I see or hear concerning the life of men, in my attend-ance on the sick, or even apart from my attendance, which ought not to be blabbed abroad, I will keep silence on them, counting such things to be as reli-gious secrets.
I f I fulfi l this oath and do not violate it, may it be granted to me to enjoy life and art alike, with good repute for all time to come; but may the contrary befall me if I transgress and violate my oath.
Paternalistic, irrelevant, inadequate, and possibly plagiarized from the followers of Pythagoras of Samos; it is argued that the Hippocratic oath has failed to evolve into anything more than a right of passage for physicians. Is it adequate to address the scientifi c, political, social, and economic realities that exist for doctors today? Certainly, medical training without a fee appears to have been confi ned to history. Yet it remains one of the oldest binding documents in history and its principles of commitment, ethics, justice, professionalism, and confi dentiality transcend time.
The absence of autonomy as a fundamental tenet of modern medical care can be debated. But just as anatomy and physiology have been added to the doctor’s repertoire since Hippocrates, omissions should not undermine the oath as a para-digm of self-regulation amongst a group of specialists committed to an ideal. And do not forget that illness may represent a temporary loss of autonomy caused by fear, vulnerability, and a subjective weighting of present versus future. It could be argued that Hippocratic paternalism is, in fact, required to restore autonomy.
Contemporary versions of the oath often fail to make doctors accountable for keeping to any aspect of the pledge. And beware the oath that is nothing more than historic ritual without accountability, for then it can be superseded by per-sonal, political, social, or economic priorities: ‘In Auschwitz, doctors presided over the murder of most of the one million victims…. [They] did not recall being especially aware in Auschwitz of their Hippocratic oath, and were not surprisingly, uncomfortable discussing it…The oath of loyalty to Hitler…was much more real to them.’ Robert Jay Lifton, The Nazi Doctors.
The endurance of the Hippocratic oath 1 This is unlikely to be a commentary on euthanasia (easeful death) as the oath predates the word. Rather, it is believed to allude to the common practice of using doctors as political assassins.
2 Abortion by oral methods was legal in ancient Greece. The oath cautions only against the use of pessaries as a potential source of lethal infection.
3 The oath does not disavow surgery, merely asks the physician to cede to others with expertise.
2 Thinking about medicine Medical care Advice for doctors • Do not blame the sick for being sick. • Seek to discover your patient’s wishes and comply with them. • Learn. • Work for your patients, not your consultant. • Respect opinions. • Treat a patient, not a disease. • Admit a person, not a diagnosis. • Spend time with the bereaved; help them to shed tears. • Give the patient (and yourself) time: for questions, to refl ect, and to allow healing. • Give patients the benefi t of the doubt. • Be optimistic. • Be kind to yourself: you are not an inexhaustible resource. • Question your conscience. • Tell the truth. • Recognize that the scientifi c approach may be fi nite, but experience and empathy are limitless.
The National Health Service ‘The resources of medical skill and the apparatus of healing shall be placed at the disposal of the patient, without charge, when he or she needs them; that medical treatment and care should be a communal responsibility, that they should be made available to rich and poor alike in accordance with medical need and by no other criteria...Society becomes more wholesome, more serene, and spiritually healthier, if it knows that its citizens have at the back of their con-sciousness the knowledge that not only themselves, but all their fellows, have ac-cess, when ill, to the best that medical skill can provide...You can always ‘pass by on the other side’. That may be sound economics. It could not be worse morals.’ Aneurin Bevan, In Place of Fear, 1952.
In 2014, the Commonwealth Fund presented an overview of international healthcare systems examining fi nancing, governance, healthcare quality, effi ciency, evidence-based practice, and innovation. In a scoring system of 11 nations across 11 catego-ries, the NHS came fi rst overall, at less than half the cost per head spent in the USA.1 The King’s Fund debunks the myth that the NHS is unaff ordable in the modern era,2 although funding remains a political choice. Bevan prophesied, ‘The NHS will last as long as there are folk left with the faith to fi ght for it.’ Guard it well.
Decision and intervention are the essence of action, refl ection and conjecture are the essence of thought; the essence of medicine is combining these in the ser-vice of others. We off er our ideals to stimulate thought and action: like the stars, ideals are hard to reach, but they are used for navigation. Orion (fi g 1.2) is our star of choice. His constellation is visible across the globe so he links our readers everywhere, and he will remain recognizable long after other constellations have dis-torted.
Medicine and the stars Fig 1.2 The const ellation of Orion has three superb stars: Bel-latrix (the stetho scope’s bell), Betel geuse (B), and Rigel (R). The three stars at the cross over (Orion’s Belt) are Alnitak, Alnilam, and Mint a ka.
©JML and David Malin.
Thinking about medicine 3 QALYS and resource rationing ‘There is a good deal of hit and miss about general medicine. It is a profession where exact measurement is not easy and the absence of it opens the mind to endless conjecture as to the effi cacy of this or that form of treatment.’ Aneurin Bevan, In Place of Fear, 1952.
A QALY is a quality-adjusted life year. One year of healthy life expectancy = 1 QALY, whereas 1 year of unhealthy life expectancy is worth <1 QALY, the precise value falling with progressively worsening quality of life. If an intervention means that you are likely to live for 8 years in perfect health then that intervention would have a QALY value of 8. If a new drug improves your quality of life from 0.5 to 0.7 for 25 years, then it has a QALY value of (0.7 Ω 0.5)≈25=5. Based on the price of the intervention, the cost of 1 QALY can be calculated. Healthcare priorities can then be weighted towards low cost QALYs. The National Institute for Health and Care Excellence (NICE) consid-ers that interventions for which 1 QALY=<£30 000 are cost-eff ective. However, as a practical application of utilitarian theory, QALYs remain open to criticism (table 1.1). Remember that although for a clinician, time is unambiguous and quantifi able, time experienced by patients is more like literature than science: a minute might be a chapter, a year a single sentence.3 Table 1.1 The advantages and disadvantages of QALYs Advantages Disadvantages Transparent societal decision making Focuses on slice (disease), not pie (health) Common unit for diff erent interventions Based on a value judgement that living longer is a measure of success Allows cost-eff ectiveness analysis Quality of life assessment comes from general public, not those with disease Allows international comparison Potentially ageist—the elderly always have less ‘life expectancy’ to gain Focus on outcomes, not process ie care, compassion The inverse care law, equity, and distributive justice: The inverse care law states that the availability of good medical care varies inversely with the need for it. This arises due to poorer quality services, barriers to service ac-cess, and external disadvantage. By focusing on the benefi t gained from an interven-tion, the QALY system treats everyone as equal. But is this really equality? Distributive justice is the distribution of ‘goods’ so that those who are worst off become better off . In healthcare terms, this means allocation of resources to those in greatest need, regardless of QALYs.
The importance of compassion4,5 in medicine is undisputed. It is an emotional re-sponse to negativity or suff ering that motivates a desire to help. It is more than ‘pity’, which has connotations of inferiority; and diff erent from ‘empathy’, which is a vicarious experience of the emotional state of another. It requires imaginative indwelling into another’s condition. The fi ctional Jules Henri experiences a loss of sense of the second person; another person’s despair alters his perception of the world so that they are ‘connected in some universal, though unseen, pattern of humanity’.
4 With compassion, the pain of another is ‘intensifi ed by the imagina-tion and prolonged by a hundred echoes’.
5 Compassion cannot be taught; it re-quires engagement with suff ering, cultural understanding, and a mutuality, rather than paternalism. Adverse political, excessively mechanical, and managerial envi-ronments discourage its expression. When compassion (what is felt) is diffi cult, etiquette (what is done) must not fail: refl ection, empathy, respectfulness, atten-tion, and manners count: ‘For I could never even have prayed for this: that you would have pity on me and endure my agonies and stay with me and help me’.
6 Compassion 4 Sebastian Faulkes, Human Traces, 2005.
5 Milan Kundera, The Unbearable Lightness of Being, 1984.
6 Philoctetes by Sophocles 409 BC (translation Phillips and Clay, 2003).
4 Thinking about medicine The diagnostic puzzle How to formulate a diagnosis Diagnosing by recognition: For students, this is the most irritating method. You spend an hour asking all the wrong questions, and in waltzes a doctor who names the disease before you have even fi nished taking the pulse. This doctor has simply recognized the illness like he recognizes an old friend (or enemy).
Diagnosing by probability: Over our clinical lives we build up a personal database of diagnoses and associated pitfalls. We unconsciously run each new ‘case’ through this continuously developing probabilistic algorithm with increasing speed and ef-fortlessness.
Diagnosing by reasoning: Like Sherlock Holmes, we must exclude each diff erential, and the diagnosis is what remains. This is dependent on the quality of the diff erential and presupposes methods for absolutely excluding diseases. All tests are statistical rather than absolute (5% of the population lie outside the ‘normal’ range), which is why this method remains, like Sherlock Holmes, fi ctional at best.
Diagnosing by watching and waiting: The dangers and expense of exhaustive tests may be obviated by the skilful use of time.
Diagnosing by selective doubting: Diagnosis relies on clinical signs and investiga-tive tests. Yet there are no hard signs or perfect tests. When diagnosis is diffi cult, try doubting the signs, then doubting the tests. But the game of medicine is unplayable if you doubt everything: so doubt selectively.
Diagnosis by iteration and reiteration: A brief history suggests looking for a few signs, which leads to further questions and a few tests. As the process reiterates, various diagnostic possibilities crop up, leading to further questions and further tests. And so history taking and diagnosing never end.
Heuristic pitfalls Heuristics are the cognitive shortcuts which allow quick decision-making by focus-ing on relevant predictors. Be aware of them so you can be vigilant of their traps.7 Representativeness: Diagnosis is driven by the ‘classic case’. Do not forget the atypical variant.
Availability: The diseases that we remember, or treated most recently, carry more weight in our diagnostic hierarchy. Question whether this more readily available information is truly relevant.
Overconfi dence: Are you overestimating how much you know and how well you know it? Probably.
Bias: The hunt for, and recall of, clinical information that fi ts with our expectations. Can you disprove your own diagnostic hypothesis?
Illusory correlation: Associated events are presumed to be causal. But was it treat-ment or time that cured the patient?
Consider three wise men:6 Occam’s razor: Entia non sunt multiplicanda praeter necessitatem translates as ‘entities must not be multiplied unnecessarily’. The physician should therefore seek to achieve diagnostic parsimony and fi nd a single disease to explain all symptoms, rather than proff er two or three unrelated diagnoses.
Hickam’s dictum: Patients can have as many diagnoses as they damn well please. Signs and symptoms may be due to more than one pathology. Indeed, a patient is statistically more likely to have two common diagnoses than one unify-ing rare condition.
Crabtree’s bludgeon: No set of mutually inconsistent observations can exist for which some human intellect cannot conceive a coherent explanation however complicated. This acts as a reminder that physicians prefer Occam to Hickam: a unifying diagnosis is a much more pleasing thing. Confi rmation bias then ensues as we look for supporting information to fi t with our unifying theory. Remember to test the validity of your diagnosis, no matter how pleasing it may seem. A razor, a dictum, and a bludgeon Thinking about medicine 5 It is always possible to be wrong8 because you remain unaware of it while it is happening. Such error-blindness is why ‘I am wrong’ is a statement of impos-sibility. Once you are aware that you are wrong, you are no longer wrong, and can therefore only declare ‘I was wrong’. It is also the reason that fallibility must be accepted as a universally human phenomenon. Conversely, certainty is the convic-tion that we cannot be wrong because our biases and beliefs must be grounded in fact. Certainty produces the comforting illusion that the world (and medicine) is knowable. But be cautious of certainty for it involves a shift in perspective inwards, towards our own convictions. This means that other people’s stories can cease to matter to us. Certainty becomes lethal to empathy.
In order to determine how and why mistakes are made, error must be acknowl-edged and accepted. Defensiveness is bad for progress. ‘I was wrong, but...’ is rarely an open and honest analysis of error that will facilitate diff erent and better action in the future. It is only with close scrutiny of mistakes that you can see the possibility of change at the core of error. And yet, medical practice is littered with examples of resistance to disclosure, and reward for the concealment of error. This must change.4 Remember error blindness and protect your whistle-blowers. Listen. It is an act of humility that acknowledges the position of others, and the possibility of error in yourself. Knowledge persists only until it can be disproved. Better to aspire to the aporia of Socrates: ‘At fi rst, he didn’t know...just as he doesn’t yet know the answer now either; but he still thought he knew the answer then, and he was answering con-fi dently, as if he had knowledge. He didn’t think he was stuck before, but now he appreciates that he is stuck...At any rate, it would seem that we’ve increased his chances of fi nding out the truth of the matter, because now, given his lack of knowledge, he’ll be glad to undertake the investigation...Do you think he’d have tried to enquire or learn about this matter when he thought he knew it (even though he didn’t), until he’d become bogged down and stuck, and had come to appreciate his ignorance and to long for knowledge?’ Plato: Meno and other dialogues, 402 BC; Waterfi eld translation, 2005.
Being wrong In a world in which a ‘mistake’ can be redefi ned as a ‘complication’, it is easy to conceal error behind a veil of technical language. In 2014, a professional duty of candour became statutory in England for incidents that cause death, severe or moderate harm, or prolonged psychological harm. As soon as practicable, the pa-tient must be told in person what happened, given details of further enquiries, and off ered an apology. But this should not lead to the proff ering of a ‘tick-box’ apology of questionable value. Be reassured that an apology is not an admission of liability. Risks and imperfections are inherent to medicine and you have the freedom to be sorry whenever they occur. Focus not on legislation, but on transparency and learning. The ethics of forgiveness require a complete response in which the pa-tient’s voice is placed at the heart of the process.9 Duty of candour Error provides a link between medicine and the humanities. Both strive to bridge the gap between ourselves and the world. Medicine attempts to do this in an ob-jective manner, using disproved hypotheses (error) to progress towards a ‘truth’. Art, however, accepts the unknown, and celebrates transience and subjectivity. By seeing the world through someone else’s eyes, art teaches us empathy. It is at the point where art and medicine collide that doctors can re-attach themselves to the human race and feel those emotions that motivate or terrify our patients. ‘Unknowing’ drives medical theory, but also stories and pictures. And these are the hallmark of our highest endeavours.
‘We all know that Art is not truth. Art is a lie that makes us realise the truth, at least the truth that is given to us to understand.’ Pablo Picasso in Picasso Speaks, 1923.
Medicine, error, and the humanities 6 Thinking about medicine Bedside manner and communication skills A good bedside manner is dynamic. It develops in the light of a patient's needs and is grounded in honesty, humour, and humility, in the presence of human weakness. But it is fragile: `It is unsettling to fi nd how little it takes to defeat success in medicine... You do not imagine that a mere matter of etiquette could foil you. But the social dimension turns out to be as essential as the scientifi c... How each interaction is negotiated can determine whether a doctor is trusted, whether a patient is heard, whether the right diagnosis is made, the right treatment given. But in this realm there are no perfect formulas.' (Atul Gawande, Better: A Surgeon's Notes on Performance, 2008) A patient may not care how much you know, until they know how much you care. Without care and trust, there can be little healing. Pre-set formulas off er, at best, a guide: Introduce yourself every time you see a patient, giving your name and your role.
‘Introductions are about making a human connection between one human being who is suffering and vulnerable, and another human being who wishes to help. They begin therapeutic relationships and can instantly build trust’ Kate Granger, hellomynameis.org.uk, #hellomynameis Be friendly. Smile. Sit down. Take an interest in the patient and ask an unscripted question. Use the patient’s name more than once.
Listen. Do not be the average physician who interrupts after 20–30 seconds.
‘Look wise, say nothing, and grunt. Speech was given to conceal thought.’ William Osler (1849–1919).
Increase the wait-time between listening and speaking. The patient may say more.
Pay attention to the non-verbal. Observe gestures, body language, and eye contact. Be aware of your own.
Explain. Consider written or drawn explanations. When appropriate, include rela-tives in discussions to assist in understanding and recall.
Adapt your language. An explanation in fl uent medicalese may mean nothing to your patient.
Clarify understanding. ‘Acute’, ‘chronic’, ‘dizzy’, ‘jaundice’, ‘shock’, ‘malignant’, ‘re-mission’: do these words have the same meaning for both you and your patient?
Be polite. It requires no talent.
‘Politeness is prudence and consequently rudeness is folly. To make enemies by being...unnecessarily rude is as crazy as setting one’s house on fi re.’ Arthur Schopenhauer (1788–1860).
Address silent fears. Give patients a chance to raise their concerns: ‘What are you worried this might be?’, ‘Some people worry about...., does that worry you?’ Consider the patient’s disease model. Patients may have their own explanations for their symptoms. Acknowledge their theories and, if appropriate, make an eff ort to explain why you think them unlikely.
‘A physician is obligated to consider more than a diseased organ, more even than the whole man - he must view the man in his world.’ Harvey Cushing (1869–1939).
Keep the patient informed. Explain your working diagnosis and relate this to their understanding, beliefs, and concerns. Let them know what will happen next, and the likely timing. ‘Soon’ may mean a month to a doctor, but a day to a patient. Apologize for any delay.
Summarize. Is there anything you have missed?
Communication, partnership, and health promotion are improved when doctors are trained to KEPe Warm:10 • Knowing—the patient’s history, social talk. • Encouraging—back-channelling (hmmm, aahh). • Physically engaging—hand gestures, appropriate contact, lean in to the patient. • Warm up—cooler, professional but supportive at the start of the consultation, making sure to avoid dominance, patronizing, and non-verbal cut-off s (ie turning away from the patient) at the end.
Thinking about medicine 7 Open questions ‘How are you?’, ‘How does it feel?’ The direction a patient chooses off ers valuable information. ‘Tell me about the vomit.’ ‘It was dark.’ ‘How dark?’ ‘Dark bits in it.’ ‘Like…?’ ‘Like bits of soil in it.’ This information is gold, although it is not cast in the form of coff ee grounds.
Patient-centred questions Patients may have their own ideas about what is causing their symptoms, how they impact, and what should be done. This is ever truer as patients frequently consult Dr Google before their GP. Unless their ideas, concerns, and expectations are elucidated, your patient may never be fully satisfi ed with you, or able to be fully involved in their own care.
Considering the whole Humans are not self-suffi cient units; we are complex relational beings, constantly reacting to events, environments, and each other. To understand your patient’s concerns you must understand their context: home-life, work, dreams, fears. Information from family and friends can be very helpful for identifying triggering and exacerbating factors, and elucidating the true underlying cause. A headache caused by anxiety is best treated not with analgesics, but by helping the patient access support.
Silence and echoes Often the most valuable details are the most diffi cult to verbalize. Help your patients express such thoughts by giving them time: if you interrogate a robin, he will fl y away; treelike silence may bring him to your hand.
‘Trade Secret: the best diagnosticians in medicine are not internists, but pa-tients. If only the doctor would sit down, shut up, and listen, the patient will eventually tell him the diagnosis.’ Oscar London, Kill as Few Patients as Possible, 1987.
Whilst powerful, silence should not be oppressive—try echoing the last words said to encourage your patient to continue vocalizing a particular thought.
Try to avoid Closed questions: These permit no opportunity to deny assumptions. ‘Have you had hip pain since your fall?’ ‘Yes, doctor.’ Investigations are requested even though the same hip pain was also present for many years before the fall!
Questions suggesting the answer: ‘Was the vomit black—like coff ee grounds?’ ‘Yes, like coff ee grounds, doctor.’ The doctor’s expectations and hurry to get the evidence into a pre-decided format have so tarnished the story as to make it useless.
Asking questions Shared decision-making: no decision about me, without me Shared decision-making aims to place patients’ needs, wishes, and preferences at the centre of clinical decision-making. • Support patients to articulate their understanding of their condition. • Inform patients about their condition, treatment options, benefi ts, and risk. • Make decisions based on mutual understanding.
Consider asking not, ‘What is the matter?’ but, ‘What matters to you?’.
Consider also your tendency towards libertarian paternalism or ‘nudge’. This is when information is given in such a way as to encourage individuals to make a particular choice that is felt to be in their best interests, and to correct apparent ‘reasoning failure’ in the patient. This is done by framing the information in either a positive or negative light depending on your view and how you might wish to sway your audi-ence. Consider the following statements made about a new drug which off ers 96% survival compared to 94% with an older drug: • More people survive if they take this drug. • This new drug reduces mortality by a third. • This new drug benefi ts only 2% of patients. • There may be unknown side-eff ects to the new drug.
How do you choose?
8 Thinking about medicine Prescribing drugs Consult the BNF or BNF for Children or similar before giving any drug with which you are not thoroughly familiar.
Check the patient’s allergy status and make all reasonable attempts to qualify the reaction (table 1.2). The burden of iatrogenic hospital admission and avoidable drug-related deaths is real. Equally, do not deny life-saving treatment based on a mild and predictable reaction.
Check drug interactions meticulously.
Table 1.2 Drug reactions Type of reaction Examples True allergy Anaphylaxis: oedema, urticaria, wheeze (p794–5) Side-eff ect All medications have side-eff ects. The most common are rash, itch, nausea, diarrhoea, lethargy, and headache Increased eff ect/ toxicity Due to inter-individual variance. Dosage regimen normally cor-rects for this but beware states of altered drug clearance such as liver and renal (p305) impairment Drug interaction Reaction due to drugs used in combination, eg azathioprine and allopurinol, erythromycin and warfarin Remember primum non nocere: fi rst do no harm. The more minor the illness, the more weight this carries. Overall, doctors have a tendency to prescribe too much rather than too little.
Consider the following when prescribing any medication: 1 The underlying pathology. Do not let the amelioration of symptoms lead to failure of investigation and diagnosis.
2 Is this prescription according to best evidence?
3 Drug reactions. All medications come with risks, potential side-eff ects, incon-venience to the patient, and expense.
4 Is the patient taking other medications?
5 Alternatives to medication. Does the patient really need or want medication? Are you giving medication out of a sense of needing to do something, or because you genuinely feel it will help the patient? Is it more appropriate to off er infor-mation, reassurance, or lifestyle modifi cation?
6 Is there a risk of overdose or addiction?
7 Can you assist the patient? Once per day is better than four times. How easy is it to open the bottle? Is there an intervention that can help with medicine manage-ment, eg a multi-compartment compliance aid, patient counselling, an IT solution such as a smartphone app?
8 Future planning. How are you going to decide whether the medication has worked? What are the indications to continue, stop, or change the prescribed regimen?
Pain is often seen as an unequivocally bad thing, and certainly many patients dream of a life without pain. However, without pain we are vulnerable to ourselves and our behaviours, and risk ignorance of underlying conditions.
While most children quickly learn not to touch boiling water as their own body disciplines their behaviour with the punishment of pain; children born with con-genital insensitivity to pain (CIPA) can burn themselves, break bones, and tear skin without feeling any immediate ill eff ect. Their health is constantly at risk from unconsciously self-mutilating behaviours and unnoticed trauma. CIPA is very rare but examples of the human tendency for self-damage without the protective fac-tor of pain are common. Have you ever bitten your tongue or cheek after a dental anaesthetic? Patients with diabetic neuropathy risk osteomyelitis and arthropa-thy in their pain-free feet.
If you receive a message of bad news, you do not solve the problem by hiding the message. Listen to the pain as well as making the patient comfortable.
In appreciation of pain Thinking about medicine 9 The placebo eff ect The placebo eff ect is a well-recognized phenomenon whereby patients improve af-ter undergoing therapy that is believed by clinicians to have no direct eff ect on the pathophysiology of their disease. The nature of the therapy (pills, rituals, massages) matters less than whether the patient believes the therapy will help.
Examples of the placebo eff ect in modern medicine include participants in the pla-cebo arm of a clinical trial who see dramatic improvements in their refractory illness, and patients in severe pain who assume the saline fl ush prior to their IV morphine is opioid and reporting relief of pain before the morphine has been administered. It is likely that much of the symptomatic relief experienced from ‘active’ medicines in fact results from a placebo eff ect.
The complementary therapy industry has many ingenious ways of utilizing the placebo eff ect. These can give great benefi ts to patients, often with minimal risk; but there remains the potential for signifi cant harm, both fi nancially and by dissuading patients from seeking necessary medical help.
Why evolution has given us bodies with a degree of self-healing ability in response to a belief that healing will happen, and not in response to a desire for healing, is unclear. Perhaps the belief that a solution is underway ‘snoozes’ the internal alarm systems that are designed to tell us there is a problem, and so improve the symptoms that result from the body’s perception of harm.
Many patients who receive therapies are unaware of their intended eff ects, thus missing out on the narrative that may give them an expectation of improvement. Try to fi nd time to discuss with your patients the story of how you hope treatment will address their problems.
Compliance embodies the imbalance of power between doctor and patient: the doctor knows best and the patient’s only responsibility is to comply with that monopoly of medical knowledge. Devaluing of patients and ethically dubious, the term ‘compliance’ has been relegated from modern prescribing practice. Con-cordance is now king: a prescribing agreement that incorporates the beliefs and wishes of the patient.
Only 50–70% of patients take medicines as prescribed to them. This leads to concern over wasted resources and avoidable illness. Interventions that increase concordance are promoted using the mnemonic: Educating Patients Enhances Care Received • Explanation: discuss the benefi ts and risks of taking and not-taking medication. Some patients will prefer not to be treated and, if the patient has capacity and understands the risks, such a decision should be respected. • Problems: talk through the patient’s experience of their treatment—have they suff ered side-eff ects which have prompted non-concordance? • Expectations: discuss what they should expect from their treatment. This is im-portant especially in the treatment of silent conditions where there is no symp-tomatic benefi t, eg antihypertensive treatment. • Capability: talk through the medication regimen with them and consider ways to reduce its complexity. • Reinforcement: reproduce your discussion in written form for the patient to take home. Check how they are managing their medications when you next see them.
But remember that there is little evidence that increasing information improves concordance. And if concordance is increased solely by the ‘education’ of the pa-tient then it starts to look a lot like compliance.11 A truly shared agreement will not always ‘comply’ or ‘concord’ with the prescriber. The capacity of the informed individual to consent or not, means that in some cases, concordance looks more like informed divergence.
Compliance and concordance 10 Thinking about medicine Surviving life on the wards The ward round • All entries on the patient record must have: date, time, the name of the clinician leading the interaction, the clinical fi ndings and plan, your signature, printed name, and contact details. Make sure the patient details are at the top of every side of paper. Write legibly—this may save more than the patient. • A problem list will help you structure your thoughts and guide others. • BODEX: Blood results, Observations, Drug chart, ECG, X-rays. Look at these. If you think there is something of concern, make sure someone else looks at them too. • Document what information has been given to the patient and relatives.
Handover • Make sure you know when and where to attend. • Make sure you understand what you need to do and why. ‘Check blood results’ or ‘Review warning score’ is not enough. Better to: ‘Check potassium in 4 hours and discuss with a senior if it remains >6.0mmol/L’ .
On call • Write it down. • The ABCDE approach (p779) to a sick patient is never wrong. • Try and establish the clinical context of tasks you are asked to do. Prioritize and let staff know when you are likely to get to them. • Learn the national early warning score (NEWS) (p892, fi g A1). • Smile, even when talking by phone. Be polite. • Eat and drink, preferably with your team.
Making a referral • Have the clinical notes, observation chart, drug chart, and investigation results to hand. Read them before you call. • Use SBAR: Situtation (who you are, who the patient is, the reason for the call), Background, Assessment of the patient now, Request. • Anticipate: urine dip for the nephrologist, PR exam for the gastroenterologist.
With the going down of the sun we can momentarily cheer ourselves up by the thought that we are one day nearer to the end of life on earth—and our respon-sibility for the unending tide of illness that fl oods into our corridors, and seeps into our wards and consulting rooms. Of course you may have many other quiet satisfactions, but if not, read on and wink with us as we hear some fool telling us that our aim should be to produce the greatest health and happiness for the great-est number. When we hear this, we don’t expect cheering from the tattered ranks of on-call doctors; rather, our ears detect a decimated groan, because these men and women know that there is something at stake in on-call doctoring far more elemental than health or happiness: namely survival.
Within the fi rst weeks, however brightly your armour shone, it will now be smeared and spattered, if not with blood, then with the fallout from the many decisions that were taken without suffi cient care and attention. Force majeure on the part of Nature and the exigencies of ward life have, we are suddenly stunned to realize, taught us to be second-rate; for to insist on being fi rst-rate in all areas is to sign a death warrant for ourselves and our patients. Don’t keep re-polishing your armour, for perfectionism does not survive untarnished in our clinical world. Rather, to fl ourish, furnish your mind and nourish your body. Regular food makes midnight groans less intrusive. Drink plenty: doctors are more likely to be oliguric than their patients. And do not voluntarily deny yourself the restorative power of sleep, for it is our natural state, in which we were fi rst created, and we only wake to feed our dreams.
We cannot prepare you for fi nding out that you are not at ease with the person you are becoming, and neither would we dream of imposing a specifi c regimen of exercise, diet, and mental fi tness. Finding out what can lead you through adversity is the art of living.
Living with blood spattered armour Thinking about medicine 11 Resilience and coping ‘Burnout’ is common in clinical medicine. It is a syndrome of lost enthusiasm, re-duced empathy, increased cynicism, and a decrease in the meaningfulness of work. Coping styles and resilience can protect doctors and better equip them to meet, and learn from, the challenges of clinical practice:12 • Self-directedness correlates strongly with resilience. A personal sense of responsi-bility allows learning from mistakes and moving on. • Cooperativeness is the ability to work with opinions and behaviours diff erent to your own, preventing them becoming a source of stress. • Clinicians who are low in harm avoidance are better able to accept uncertainty and a degree of risk. This facilitates decision-making as it is unclouded by anxiety and pessimism about potential problems. Supervised experience outside your comfort zone may help you deal better with uncertainty. • Be persistent but set realistic goals. Perfectionism can be detrimental. • Task-orientated coping occurs when a situation is seen as changeable. This is as-sociated with less burnout than emotion-orientated coping when situations are considered unchangeable: don’t just do something, stand there. • Be self-aware. Development or modifi cation of your personality traits may reduce your vulnerability.
Dr Corrigan of Dublin was: ‘tall, erect, of commanding fi gure...He had the countenance of an intellectual...
and his face “beamed with kindness”...In temperament his distinguishing traits were kindness and tenderness towards the sick, and the ability to make a bold decision.’ E. O’Brien, Conscience and Confl ict: A Biography of Sir Dominic Corrigan 1802–1880, 1983 Was he busy? At the start of his professional life he was advised that the best way to get business was to pretend to have it. It was suggested that a note marked ‘Im-mediate and pressing’ should be ostentatiously handed to him at the dinner table, but always at a suitable time so as not to miss the best food. Such advice was not taken. Corrigan aspired to hard work and taught his students the value of ‘never doing nothing’. The city in which he practised had a ‘degree of fi lth, stench and dark-ness, inconceivable by those who have not experienced them’, and ‘not enough hos-pital beds to care for the great numbers in need.’ And so the story of a secret door, made in his consulting room, to escape the ever growing queue of eager patients.
In times of chaos, fi lled with competing, urgent, simultaneous demands, excessive paperwork, too few beds, eff ort–reward imbalance, personal sacrifi ce, and despair; we all need Corrigan to take us by the shadow of our hand, and walk with us through a secret door into a calm inner world. Our metaphorical door has fi ve parts: 1 However lonely you feel, you are not usually alone. Do not pride yourself on not asking for help. If a decision is a hard one, share it with a colleague.
2 Take any chance you get to sit down and rest. Have a cup of tea with other members of staff , or with a friendly patient (patients are sources of renewal, not just devourers of your energies).
3 Do not miss meals. If there is no time to go to the canteen, ensure that food is put aside for you to eat when you can: hard work and sleeplessness are twice as bad when you are hungry.
4 Avoid making work for yourself. It is too easy for doctors, trapped in their image of excessive work, and blackmailed by misplaced guilt, to remain on the wards re-clerking patients, re-writing notes, or re-checking results at an hour when the priority should be caring for themselves.
5 Look to the future. Plan for a good time after a bad rota.
The origins of the story of Corrigan’s secret door are unknown. It may never have existed other than in these hallowed pages. But when the legend becomes fact, print the legend.
7 On being busy: Corrigan’s secret door 7 The Man Who Shot Liberty Valance, 1962: Ransom Stoddard (Jimmy Stewart) becomes a legend after killing Liberty Valance in a duel. It does not matter that the real shooter was Tom Doniphon (John Wayne) all along.
12 Thinking about medicine Death Diagnosing dying Would you be surprised if your patient were to die in the next few days, weeks, or months? If the answer is ‘no’ then end-of-life choices, decisions, and care should be addressed.
Consider: decline in functional performance, eg in bed or chair >50% of day, in-creasing dependence, weight of co-morbidity, unstable or deteriorating symptom burden, decreased treatment response, weight loss >10% in 6 months, crisis admis-sions, serum albumin <25g/L, sentinel event, eg fall, transfer to nursing home.
Death is nature’s cruel master stroke, allowing genotypes space to try new pheno-types. The time comes in the life of every organism when it is better to start from scratch, rather than carry on with the weight and muddle of endless accretions. Our bodies and minds are the perishable phenotypes on the wave of our genes. But our genes are not really our genes. It is we who belong to them for a few decades. And death is nature’s great insult, that she should prefer to put all her eggs in the basket of a defenceless, incompetent neonate; rather than in the tried and tested custody of our own superb minds. But as our neurofi brils begin to tangle, and that neonate walks to a wisdom that eludes us, we are forced to give nature credit for her daring idea. Of course, nature, in her careless way, can get it wrong: people often die in the wrong order and one of our chief roles is to prevent this misorder-ing of deaths, not the phenomenon of death itself. With that exception, we must admit that dying is a brilliant idea, and one that it is most unlikely we would ever have devised ourselves.
The wisdom of death Death13 is the irreversible loss of the essential characteristics which are necessary for the existence of a human being.
Death following cessation of cardiorespiratory function: Simultaneous and irreversible onset of apnoea, absence of circulation, and un-consciousness.
Cardiorespiratory arrest is confi rmed by observation of the following: • Absence of central pulse on palpation. • Absence of heart sounds on auscultation. • After 5 minutes of cardiorespiratory arrest absence of brainstem activity is con-fi rmed by the absence of pupillary responses to light, an absent corneal refl ex, and no motor response to supra-orbital pressure.
The time of death is the time at which these criteria are fulfi lled.
Brainstem death: Brainstem pathology causing irreversible damage to its integrative functions including neural control of cardiorespiratory function and consciousness.
Diagnosed by an absence of brainstem refl exes: • No pupil light response. • No corneal refl ex (blink to touch). • Absent oculovestibular refl exes (no eye movements seen with injection of ice-cold water into each external auditory meatus, tympanic membranes visualized). • No motor response to stimulation within the cranial nerve distribution (supra-orbital pressure). • No cough/gag refl ex. • No respiratory response to hypercarbia: oxgenation is maintained (SpO2>85%) but ventilation is reduced to achieve P aCO2 ≥6.0kPa with pH ≤7.40. No respiratory response is seen within 5 minutes and P aCO2 rises by >0.5kPa.
Diagnosis is made by two competent doctors registered for >5 years testing to-gether completely and successfully on two separate occasions. Diagnosing death Thinking about medicine 13 Death may be regarded as a medical failure rather than an inevitable consequence of life. But when medical treatments can no longer off er a cure, and a patient enters the end of life, active management of death is vital. Remember that the focus of medicine is narrow and concerned more with the repair of health, rather than the sustenance of the soul. This medical imperative may fail in its duty to make life-in-death better. Priorities at the end of life include freedom from pain, achieving a sense of completeness, being treated as a whole person, and fi nding peace with God.14 Swift death due to a catastrophic event is rare. Most death is the end product of a struggle with long-term, progressive disease: cancer, COPD, vascular disease, neu-rological deterioration, frailty, or dementia. Although death is inevitable, prognos-tication is diffi cult and inaccurate with remarkable variation in time to death. The patient in front of you may be the median, mean, or on the 99th centile. Dare to hope, but prepare for the worst. Prioritize preferences and aim to meet individual needs.15 • Seek help from experienced members of staff including palliative care teams. • Elicit needs: physiological, psychological, social, and spiritual. Discuss fears. • Establish the wishes of the patient. What trade-off s are they willing to accept, eg treatment toxicity for potential time gained? What is unacceptable to them? • Consider the views of those important to the patient. • Hydration: give support to allow the dying to drink, off er mouth care. Consider clinically assisted hydration (parenteral, enteral, intravenous) according to wish-es and if distressing signs/symptoms of dehydration are possible. Stop according to wishes and harm. • Manage pain promptly and eff ectively. Treat any reversible causes of pain. • Consider a syringe pump if symptom control medications are required more than twice in 24 hours (see p536). • Anticipate likely symptoms: the PRN side of the drug chart should cover all pos-sibilities (see p536).
Death may or may not come with peace and acceptance. Patients may rage might-ily against the dying of the light. Bear witness for them: listen and hold their hand.
Managing death Organ donation Over 6000 people are waiting for an organ transplant in the UK and approximately 1000 people in need of a transplant will die each year (see p308).
Any patient who is a potential donor can be referred to a specialist organ donation service. That service will provide advice as to suitability for transplantation and will coordinate the approach to families. They are contactable 24 hours a day and their details will be held in your A&E and/or ITU departments.
Organs can be retrieved from: • donor after brainstem death or heart-beating donor. • donor after cardiac death or non-heart-beating donor. Includes death following unsuccessful CPR and patients for whom death is inevitable but do not meet the criteria for brainstem death.
There are two legislative frameworks for organ donation: • Opt-in. Donors give their explicit consent • Opt-out. Anyone who has not actively refused consent is a donor.
The association between an opt-out system and higher organ donation rates is complicated by the presence of multiple cofounding factors. Non-legislative change including national coordination, support and training of clinicians, routine discussion as part of end-of-life care, and effi cient organ retrieval also increase donation rates. The ethics of presumed consent should also be considered: the absence of an objec-tion would not be an acceptable substitute for informed consent in other areas of clinical practice.
In the UK, although consent for transplantation rests with the deceased, if the patient’s family or representative cannot support donation then it will not go ahead. Register your decision on the NHS Organ Donor Register (
nhs.uk/register-to-donate/register-your-details/) and more importantly, let your family know your wishes.
14 Thinking about medicine To force an ethical problem to fi t a framework may be inadequate, reduction-istic, and inconsistent.18 It is potentially biased towards Western culture, dis-counts the non-autonomous, and is vulnerable to poorly considered emphasis and error. ‘Doing’ ethics can become a check-list exercise where thinking is lost. But doctors are not moral philosophers. They are clinicians. A framework there-fore provides a starting point from which to work. It is the toe which tests the water of moral deliberation. Be aware of the cultural setting of your dilemma and consider carefully the weight of synthesis. Be prepared to wade deeper if needed. But acknowledge that moral wisdom may well be out of your depth. Beyond the ethical framework Medical ethics ‘Our clinical practice is steered by ethical principles. They guide the decisions we make in our clinics and ward rounds, what we tell our patients, and what we omit to tell them.’ Tony Lopez, Journal of Royal Society of Medicine 2001;94:603–4.
In the silences of our consultations it is we who are under the microscope, and we cannot escape our destiny in the sphere of ethics. To give us courage in this enterprise, we can recall the law of the aviator and seagull: it is only by facing the prevailing wind that we can become airborne, and achieve a new vantage point from which to survey our world. We hope for moral perception: to be able to visualize the morally salient features of a situation. For without this, ethical issues may fl oat past never to be resolved. Be alert to words which may carry hidden assumptions: ‘futil-ity’, ‘consent’, ‘best interests’.16 Consider WIGWAM in your routine patient review: • Wishes of the patient: are they known or unknown? • Issues of confi dentiality/disclosure. • Goals of care: are they clear? Whose are they: yours or the patient’s? • Wants: to decline treatment or discharge against advice. • Arguments between family/friends/doctors. • Money: concerns of the patient, concerns of the healthcare provider.
Ethical frameworks Off er structure, comprehensiveness, and transparency in deliberation.16,17 Four principles Autonomy: Self governance: the ability of a patient to make a choice based on their own values and beliefs. Benefi cence: The obligation to benefi t patients. Links with autonomy as benefi t is dependent upon the view of the patient. Non-malefi cence: Do not harm. Or more appropriately, do no overall harm: you should stick a needle into someone when they need dialysis. Justice: A collection of obligations including legality, human rights, fairness, and resource distribution.
Four quadrants method Medical indications: Identify the clinical problem, treatment options, goals of treat-ment, and likelihood of success. Patient preferences: What is the patient’s autono-mous decision? (And is the patient capable of making one? If not, look for previously expressed wishes from advanced directives, family, friends, GP.) Quality of life: How will the proposed treatment aff ect quality of life? This is subjective: recognize your own biases and accommodate those of the patient. Contextual factors: The wider context: legal, cultural, religious, familial, and anything else that may impact.
These frameworks describe individual voices within the ethics choir. Sometimes there is a beautiful harmony, but how should you act when there is discordance? There is no hierarchy within the frameworks. Each component is binding unless it is trumped by a stronger principle. How you weigh up and balance the ethical com-ponents of a situation is not easy, but it should be clear and justifi ed. Know the patient. Consult others, especially those who hold diff erent opinions to yourself. Can you adequately defend your decision to the patient? Their family? Your consultant? Another consultant? A lawyer? If an investigative journalist were to sit on a sulcus of yours, having full knowledge of all thoughts and actions, would he be composing vitriol for tomorrow’s newspapers? If so, can you answer him, point for point?
Thinking about medicine 15 Psychiatry on medical and surgical wards ‘Body and soul cannot be separated for purposes of treatment, for they are one and indivisible. Sick minds must be healed as well as sick bodies.’ C Jeff Miller, 1931.
Mental state examination: ASEPTIC • Appearance and behaviour: dress, hygiene, eye contact, rapport. • Speech: volume, rate, tone. • Emotion: mood (subjective and objective), aff ect (how mood is expressed with behaviour—appropriate or incongruent?). • Perception: hallucinations—auditory (in the second or third person)?, visual? • Thought: • Form: block, insertion, broadcast, fl ight of ideas, knight’s move. • Content: delusions, obsessions, phobias, preoccupations, self-harm, suicide. • Insight: ask the patient why they have presented today. • Cognition: orientation, registration, recall, concentration, knowledge.
Do not be afraid to ask about suicidal thoughts and plans. Remove yourself from the situation if you feel threatened.
Depression Two questions can be used to identify depression:19 1 During the last month, have you been bothered by feeling down, depressed, or hopeless?
2 During the last month, have you often been bothered by having little interest or pleasure in doing things?
If a person answers ‘yes’ to either question they should undergo mental health as-sessment including a risk assessment of self-harm and suicide. Appropriate treat-ments include psychosocial intervention (guided self-help, cognitive behavioural therapy, structured physical activity) and medication. Treatment choice depends on disease severity, previous psychiatric history, response to treatment, and patient preference. If medication is indicated, a generic SSRI should be considered fi rst line after consideration of GI bleeding risk, drug interactions, toxicity, overdose, and dis-continuation symptoms. The full eff ect of medication is gradual, over 4–6 weeks.
Capacity The Mental Capacity Act (MCA) 2005 has a two-stage test for lack of capacity: 1 There is an impairment or disturbed functioning of the mind.
2 The patient is unable to make a decision.
Decision-making is impaired if the patient is unable to: understand the relevant information, retain it for long enough to make a decision, weigh up the informa-tion, communicate their decision. Capacity is decision-specifi c not patient-specifi c. When treatment is proposed to those who lack capacity, a capacity advocate should be provided. Even patients without capacity should be as involved as possible in decision-making.
Mental Health Act (MHA) and common law A patient can be detained under common law (subject to a test of reasonableness) or under the MHA, only if they lack capacity to remain informally and are a danger to themselves or others. You will have more experience in verbal and non-verbal communication, than in detention under the MHA, so use these skills fi rst to try and de-escalate the situation. If rapid tranquillization is needed, be familiar with dosage, side-eff ects, and the need for ongoing observation. If there is no history to guide choice of medication, intramuscular lorazepam can be used.20 Doctors and mental health Suicide rates are three times higher in doctors compared to the general population. Up to 7% of doctors will have a substance abuse problem within their lifetime. Do not ignore feeling low, poor concentration, and reduced energy levels. Do not self-diagnose and manage. Avoid ‘corridor consultations’. Trust your GP. Seek support: • British Medical Association: www.bma.org.uk/doctorsfordoctors. • Doctors’ Support Network: www.dsn.org.uk. • Doctors’ Support Line: 0844 395 3010. • Sick Doctors Trust: www.sick-doctors-trust.co.uk.
16 Thinking about medicine The older person ‘To know how to grow old is the master-work of wisdom, and one of the most diffi cult chapters in the great art of living.’ Henri Amiel, Journal Intime, 21 Sept 1874.
Ageing is an inevitable and irreversible decline in organ function that occurs with time, in the absence of injury or illness, and despite the existence of complex path-ways of maintenance and repair.
Healthy ageing is the maintenance of physical and mental abilities that enable well-being and independence in older age.
Do not presume ageing. Look for preventable and reversible pathology. Old age does not cause disease (although it can increase vulnerability and recovery time).
Look for ways to reduce disability and support older people in their own homes.
Diff erences in the evaluation of the older person 1 Multiple pathologies: Elderly patients have, on average, six diagnosable disor-ders. Eff ects may be multiplicative. Treatment must be integrated.
2 Multiple aetiologies: One problem may have several causes, eg falls. Treating each alone may do little good, treating all may be of great benefi t.
3 Non-specifi c/atypical presentation: Delirium, dizziness, falls, mobility prob-lems, weight loss, and incontinence can be due to disorders in more than one organ system. Typical signs and symptoms may be absent. Ask about functional decline in activities of daily living—this may be the only symptom.
4 Missed or delayed diagnosis: The older person may decline quickly if treat-ment is delayed. Complications are common. Use a collateral history: what is the patient usually like?
5 Pharmacy and polypharmacy: NSAIDS, anticoagulants, anti-parkinson drugs, hypoglycaemic drugs, and psychoactive drugs can pose a particular risk in the older patient. Double check for interactions. Consider body weight, liver and re-nal function—drug doses may need to be modifi ed. The STOPP/START criteria detail >100 potentially inappropriate prescriptions and prescribing omissions relevant to the older patient.21 6 Prolonged recovery time: Anticipate and plan for this. Don’t forget nutrition.
7 Rehabilitation and social factors: Essential for healthy ageing.
Falls 50% aged >80 will fall at least once per year. Falls23 lead to injury, pain, distress, loss of confi dence, loss of independence, and mortality. Cost to the NHS is £2.3bn/year. • History: frequency, context and circumstances, severity, injuries. • Multifactorial risk assessment: gait, balance, muscle strength, osteoporosis risk, perceived functional ability, fear of falls, vision, cognition, neurological examina-tion, continence, home and hazards, cardiovascular examination, medication re-view. • Interventions: strength and balance training, home hazard intervention, correct vision, modifi cation/withdrawal of medication (cardiovascular, psychotropic), in-tegrated management of contributing morbidities. Consider barriers to change, eg fear, patient preference.
History: In addition to routine elements, include function in activities of daily liv-ing, continence, and social support. Ask if there is an advanced care directive and nominated proxy healthcare decision maker.
Examination: • Appearance and aff ect: hygiene, nutrition, hydration. Briefl y assess mood. • Senses: vision, hearing, assess swallowing with 20mL of water. • Cognition: brief screening test, eg AMTS (p64), 2-step command. • Pulse and blood pressure: lying/sitting and standing. • Peripheral neurological exam: tone, power, wasting, active range of movement. • Other periphery: pulses, oedema, skin integrity, pressure areas. • Walking: stand patient, balance, transfers, observe gait (be ready to assist). • Other systems: CV, respiratory, abdomen (don’t forget to palpate for bladder).22 A quick ward assessment of the older person Thinking about medicine 17 The pregnant woman Pre-existing conditions and non-obstetric disease cause more maternal deaths in the UK than obstetric complications.24 Pregnant women should receive the same investigations and treatment as non-preg-nant patients, with avoidance of harm/potential harm to the fetus whenever possible.
Most mistakes made in the medical management of pregnant women are due to acts of omission caused by inappropriate weighting of risk and benefi t.
Physiological changes in pregnancy Clinical assessment in pregnancy requires knowledge of the physiological changes associated with the gravid state. Expected changes and guidance on when to inves-tigate for possible underlying pathology is given in table 1.3.
Table 1.3 Physiology and pathology in pregnancy System Normal pregnancy Consider pathology Cardiovascular BP before 20 weeks’ gestation Diastolic BP >80mmHg in 1st trimester Heart rate Sustained tachycardia >100/min Respiratory Compensated respiratory alkalosis Serum bicarbonate <18mmol/L No change in PEFR Decrease in PEFR Respiratory rate by 10% Respiratory rate >20/min Renal GFR and creatinine clearance Creatinine >85μmol/L (eGFR not valid in pregnancy) Protein excretion Protein:creatinine ratio >30mg/mmol Endocrine Altered glucose handling Fasting glucose >5.0mmol/L Haematology Haemodilution Hb <10.5g/dL, platelets <100x10 9/L Radiology If the uterus is positioned outside the imaging fi eld of view, the radiation dose to the conceptus is minimal. Exposure from the following investigations is well below the threshold of risk to the fetus: • Plain radiograph: chest, extremities, spine. • CT: head, chest (but consider radiation to maternal breast in pregnancy/lactation).
Ultrasound and MRI are preferentially used when imaging the abdomen.
Reassure your pregnant patient that a chest x-ray is safe. It is the equivalent of 3 days of background radiation. Do not presume it is not required—how else will you pick up the widened mediastinum as a cause for her chest pain?
Drugs For drugs prescribed in pregnancy, benefi t must be balanced against risk (table 1.4). For information on drugs in lactation see:
Table 1.4 Drugs in pregnancy Considered safe Contraindicated Penicillins Tetracycline/doxycycline Macrolides Ciprofl oxacin Low-molecular-weight heparin Trimethoprim (1st trimester) Aspirin NSAIDS (3rd trimester) Labetalol ACE-i Nifedipine ARA Adenosine Mycophenolate Prednisolone Warfarin Treatment for asthma: salbutamol, ipratropium, aminophylline, leukotriene antagonists Live vaccines (MMR, BCG, Varicella) Sepsis Do not underestimate sepsis in pregnancy. Septic shock can be rapid. Do not ignore tachypnoea. All pregnant women should receive the infl uenza vaccine.
18 Thinking about medicine Epidemiology ‘The work of epidemiology is related to unanswered questions, but also to un-questioned answers.’ Patricia Buffl er, North American Congress of Epidemiology, 2011.
Who, what, when, where, why, and how?
Epidemiology is the study of the distribution of clinical phenomena in populations.
It analyses disease in terms of host, agent, and environment (the ‘epidemiologist’s triad’). It elucidates risks and mechanisms for the development of disease, and re-veals potential targets for disease prevention and treatment. Epidemiology does not look at the individual patient, but examines a defi ned population. How applicable its fi ndings are depend upon how well the sample population mirrors the study popula-tion, which must, in turn, mirror the target population. Does your patient fi t in this ‘target’? If ‘yes’, then the epidemiological fi ndings may be applicable.
Measures of disease frequency Incidence proportion is the number of new cases of disease as a proportion of the population. Synonyms include probability of disease, cumulative incidence, risk.
Incidence rate is the number of new cases per unit of person-time, ie one person observed for 5 years contributes 5 person-years of follow-up.
Prevalence is the number of cases that exist at a given time (point prevalence) or time-frame (period prevalence), divided by the total population being studied. For example, the lifetime prevalence of hiccups is ~100% and incidence is millions/year. However, the point prevalence at 3am may be 0 if no one is actually having hiccups.
Comparisons of outcome frequency Diff erences in outcome rates between populations point to an association between the outcome and factors distinguishing the populations (eg a smoking population compared to a non-smoking population). Challenges arise as populations tend to diff er from each other in many ways, so it may not be clear which factor(s) aff ect outcome frequency. This leads to confounding. For example, we might fi nd that heart disease is more common in those who use walking sticks. But we cannot con-clude that walking sticks cause heart disease as age is a confounding factor: age is causal, not walking sticks.
Ways of accounting for associations: A may cause B (antacids cause cancer), B may cause A (cancer causes antacid use), a 3rd unknown agent X (eg age) may cause A and B, or the association may be a chance fi nding. When considering the options, it is useful to bear in mind the Bradford Hill ‘criteria’ for causation (NB he did not claim any were essential): 1 Consistency of fi ndings: among diff erent populations, studies, time periods.
2 Temporality: the eff ect must occur after the cause.
3 Biological gradient: a dose response whereby more exposure = more eff ect.
4 Specifi city: exposure causes a single outcome (smoking does not conform!).
5 Strength of association: strong associations are more likely to be causal.
6 Biological plausibility: there is a mechanism linking cause and eff ect.
7 Coherence: the relationship is supported by current disease knowledge.
8 Experiment: does removal of exposure reduce outcome frequency?
Epidemiological studies Studies should be designed to give an adequate answer to a specifi c research ques-tion. Samples need to be representative and of suffi cient size to answer the question.
Ecological studies: Outcome rates are examined in diff erent populations, eg trend over time, geographically distinct groups, social class. Populations rather than indi-viduals are the unit of study.
Longitudinal (cohort) studies: Subjects are followed over time with measurement of exposure and outcome.
Case–control studies: Patients with the outcome of interest are identifi ed and past exposure is assessed in comparison to ‘controls’ who did not develop the outcome. Cases and controls should be adequately matched for other factors that may aff ect outcome, or these diff erences should be corrected for (mathematical assumption).
Experimental studies: Exposure is allocated to a study group and compared to those who are not exposed, eg randomized controlled trials.
Thinking about medicine 19 Randomized controlled trials In a randomized controlled trial (RCT), participants are allocated to an intervention/ exposure (eg new drug treatment) or no intervention (eg placebo, standard care) by a process which equates to the fl ip of a coin, ie all participants have an equal chance of being in either arm of the study. The aim is to minimize bias and attempt to get at the truth as to whether the intervention is any good or not. Both groups are followed up and analysed against predefi ned end-points.
Randomizing Done with the aim of eliminating the eff ects of non-studied factors. With randomization (and suffi cient study size) the two arms of the study will be identical (on average), with the exception of the intervention of interest.
Blinding There is a risk that factors during the trial may aff ect the outcome, eg participant or clinician optimism if they know the patient is on active treatment, or an unwillingness to expose more severe disease to placebo. If the subject does not know which intervention they are having, the trial is single-blind. Ideally, the experi-menter should not know either, and the study should be double-blind.
In a good trial, the blind lead the blind.
When a randomized controlled trial might not be the best method • Generating new ideas beyond current paradigms (case reports). • Researching causes of illnesses and prognoses (cohort studies). • Evaluating diagnostic tests (cohort study and decision model). • Where the researcher has no idea of the eff ective dose of a drug (dose-ranging adaptive design). • When recruiting of patients would be impossible or unethical. • When personalized medicine is the aim, eg treatments matched to patients’ bio-marker profi les (adaptive design, cohort study).
In the end, all randomized trials have to submit to the ultimate test when the sta-tistical collides with the personal: ‘Will this treatment help me?’, ‘Will this procedure help you?’ No randomized trial is complete until real-life decisions taken in the light of its fi ndings are scrutinized. Remember Osler: ‘no two individuals react alike and behave alike under the abnormal conditions which we know as disease. This is the fundamental diffi culty of the physician’. Do not ask for defi nitive trials: everything is provisional. • Does the study answer a useful question? Does it add to current literature: big-ger, better, diff erent target population? • Does the target population in the study include your patient(s)? Check the inclu-sion and exclusion criteria including age and comorbidity. • Is the intervention well described so it can be replicated in clinical practice? • Was the sample size big enough to detect an eff ect? Can you fi nd a sample size calculation? Watch out for sub-group analyses for which the sample size was not calculated. • Were outcome measures predefi ned? • Is randomization adequate? Look at the baseline data for each group—are there signifi cant diff erences? Are any parameters of interest (that might aff ect out-come) not included? • Who was blinded and how blind were they? • Are statistical methods reported and appropriate? There should be a measure of the eff ect size and its precision (confi dence interval, see p20). • Is the eff ect clinically signifi cant? Watch out for surrogate end-points which do not directly measure benefi t, harm, or the treatment response of interest. • How long was the follow-up? Was it long enough to determine outcome? • How complete was the follow-up? How many patients were left at the end of the follow-up period? Were those who left the study included in the analysis (intention-to-treat)?25 Journal club: how good is this RCT?
20 Thinking about medicine Medical mathematics ‘When you can measure what you are speaking about, and express it in numbers, you know something about it; but when you cannot measure it, when you cannot express it in numbers, your knowledge is of a meagre and unsatisfactory kind.’ Lord Kelvin, 1883.
Comparison measures Comparisons between ‘exposed’ and ‘unex-posed’ populations are made in terms of the risk or likelihood of an outcome This can be appreci-ated by plotting a 2≈2 table (table 1.5). • Absolute risk difference (attributable risk) = disease frequency in exposed minus the disease frequency in unexposed. Example (table 1.5): (70/90) Ω (120/570) = 0.57 exposure increases risk by 57%. • Relative risk = ratio of outcome in exposed population compared to unexposed. Relative risk of 1 means risk is same in both populations. Relative risk >1 means ex-posure increases risk. Relative risk <1 means exposure lessens risk, eg vaccination. Example (table 1.5): (70/90)÷(120/570) = 3.69 risk is 3.69 ≈ higher with exposure. • Odds ratio = ratio of the probability of an outcome occurring compared to the probability of an outcome not occurring. Example (table 1.5): (70/20)÷(120/450) = 13.13 odds of outcome are 13.13 ≈ higher with exposure.
Relative risk is easier to interpret than odds ratio but relies on a meaningful preva-lence/incidence. For the individual, absolute risk diff erence may be most relevant.
Table 1.5 2≈2 table analysis Outcome Exposure Event No event Total Yes 70 20 90 No 120 450 570 Total 190 470 In a study of two groups (eg new treatment versus placebo), it is possible that there is no diff erence (ie new treatment has no benefi t). This is the null hypothesis. A p-value measures the strength of evidence in relation to the null hypothesis: • Low p-value: data unlikely if null hypothesis is true. • High p-value: data likely is null hypothesis is true.
A p-value is not the probability that your results occurred by chance, and it can-not tell you how good a study is. There will be many assumptions in the statistical model. Look at the details: have confounding or bias aff ected the result? Do not consider p <0.05 as ‘statistically signifi cant’: a small p-value just fl ags the data as unusual.26 You need to question why and decide if this is clinically important.
Confi dence intervals (CI) give a guide to the eff ect size and direction (eg benefi t/ harm). They give a margin of error that indicates the amount of uncertainty in the statistical estimate.
P-values and confi dence intervals Assessing validity The validity of a test which dichotomizes study participants can be assessed by examining the results from the test against a standard reference (or outcome: did the participant actually have the disease?) (see table 1.6).
Sensitivity TP/(TP + FN) = of those with the condition, how many test positive? A sensi-tive test is able to correctly identify those with the disease.
Specifi city TN/(TN + FP) = of those who do not have the condition, how many test negative? A specifi c test is able to correctly identify those without a disease.
‘Do they have abdominal pain?’ as a test for appendicitis will have sensitivity (most cases have pain), but specifi city (many patients with pain do not have appendicitis).
Positive predictive value (TP/(TP + FP) indicates how likely it is that someone with a positive test result has the condition.
Negative predictive value (TN/(TN + FN) indicates how likely it is that someone with a negative test result does not have the condition. When you receive a test result, you need to know how likely it is to be correct.
Table 1.6 Table of possible test results Test result Patient has condition Patient does not have condition Positive True positive (TP) False positive (FP) Negative False negative (FN) True negative (TN) Thinking about medicine 21 Number needed to treat Number needed to treat (NNT) is a useful way of reporting the results of randomized clinical trials. It is the reciprocal of the absolute risk diff erence: 1 ÷ ARR.
A large treatment eff ect means that fewer patients need to receive treatment in order for one to benefi t. It is specifi c to the chosen comparator (eg placebo or usual care), the measured outcome (eg death, blood pressure fall), and the duration of treatment follow-up used in the study. Look carefully at the details of the question that the NNT is attempting to quantify. • Advantages: easily calculated, single numerical value for effi cacy, can be used to examine harm (becomes the number needed to harm). • Disadvantages: confi dence intervals are diffi cult when the diff erences between treatments are not signifi cant. Yes or No? Your tutor asks whether Gobble’s disease is commoner in women or men. You have no idea, and make a guess. What is the chance of getting it right? Common sense decrees that you have a 50:50 chance. Sod’s law predicts that whatever you guess, you will always be wrong. Somewhere between the two is Damon Runyon’s view that ‘all life is 6 to 5 against’: Will you pick the right answer? Perhaps, but don’t bet on it!
New or existing disease? Suppose singultus is a rare symptom of Gobble’s dis-ease (seen in 5% of patients), but that it is a very common symptom of Kobble’s disease (seen in 90%). If we have a man whom we already know has Gobble’s disease, who goes on to develop singultus, is it more likely to be due to Kobble’s, rather than Gobble’s disease? The answer is usually no: it is generally the case that most symptoms are due to a disease that is already known, and do not imply a new disease (Occam’s razor, p4). The ‘odds ratio’ makes this clearer, ie the ratio of [the probability of the symptom, given the known disease] to [the probability of the symptom due to new disease ≈ the probability of developing the new dis-ease]. Usually this is vastly in favour of the symptom being due to the old disease, because of the prior odds of the two diseases. This will work until Kobble’s dis-ease increases in prevalence so as to increase the odds of a second disease (then Hickam trumps Occam, p4).
How to play the odds It is distasteful to think that doctors can gamble with patients’ lives. It is also distasteful to think of serious diseases being ‘missed’, and invasive procedures being done unnecessarily. Yet we do not have an evidence base or an experience base which can tell us defi nitively which cough or lethargy or sore toe is just ‘one of those things’, and which is the result of undiagnosed cancer or HIV or osteomyelitis. And so we gamble.
Medicine is not for pessimists—almost anything can be made to seem fatal, so that a pessimistic doctor would never get any sleep at night due to worry about the meaning of their patients’ symptoms. Medicine is not for blind optimists either, who too easily embrace a fool’s paradise of false reassurance. Rather, medicine is for informed gamblers: gamblers who are happy to use subtle clues to change their outlook from pessimism to optimism and vice versa. Sometimes the gambling is scientifi c, rational, methodical, and reproducible (odds ratio); sometimes it instinc-tual, due to clinical intuition (vital but ill-defi ned).
Of course, gambling inevitably results in losses, and in medicine the chips are not just fi nancial. They betoken the health of your patient, your reputation, and your confi dence. Perhaps the hardest part of medicine is the inevitability of making mistakes whilst attempting to help (see ‘Being wrong’, p5). But do not worry about gambling: gambling is your job. If you cannot gamble, you cannot walk the thin line between successfully addressing health needs, and causing over-medicalization (p23). But try hard to assemble suffi cient evidence to maximize the chances of being lucky. Lucky gambling is a requisite for successful doctoring and the casino of medical practice celebrates the card counter. But the cardinal clinical virtue is courage: without it we would not follow our hunches and take justifi ed risks.
The doctor as a gambler 22 Thinking about medicine Evidence-based medicine (EBM) EBM is the conscientious and judicious use of current, best research evidence to opti-mize management plans and integrate them with patients’ values by: 1 Asking answerable questions.
2 Finding the best information.
3 Appraising the information for quality, validity, and relevance.
4 Dialogue to fi nd out what the patient wants.
5 Applying data to patient care.
6 Evaluation.
The amount of evidence More than 2 million new biomedical papers are published each year including >20 000 new randomized trials. Patients benefi t directly from a tiny fraction of these papers. How do we fi nd them? • A hierarchy of evidence (fi g 1.3) is used to identify the best research available to answer our question. • Specialist EBM journals, eg Evidence-based Med-icine, appraise published information for quality, relevance, and interest on our behalf. • The Cochrane Collabora-tion gathers and summa-rizes best evidence, free from commercial spon-sorship and confl icts of interest. >37 000 research-ers from 130 countries contribute.
Problems: • The concept of scientifi c rigour is opaque. What do we want? The science, the rigour, the truth, or what will be most useful to patients? These may overlap, but they are not the same. Can average cohort results inform clinical decisions on an individual level (especially in the context of comorbidity)? • Can we really appraise ALL the evidence? 27 We are hindered by publication bias. Around half of all clinical trials remain unpublished. See www.alltrials.net for the cam-paign to register all trials, and ensure methods and summary results are available. • Evidence can be expensive. Who paid the bill and what is their vested interest? • Is the result clinically signifi cant? What is the level of benefi t to the individual, as opposed to the population? Is the EBM tail wagging the clinical dog? • How is our innate hierarchy of evidence constructed? Do we maintain the same standard of the evidence for all changes to our practice? • Have you checked the correspondence columns in journals from which winning papers are extracted? It may take years for unforeseen fl aws to surface. • There is a danger that by always asking, ‘What is the evidence?’, we will divert resources from hard-to-prove areas (eg psychosocial interventions). • EBM is never 100% up to date and reworking meta-analyses takes time and mon-ey. Specialists may ostensibly reject a new trial due to a tiny fl aw, when the real dread is that it might fl ip their once-perfect formulation. • EBM lies uncomfortably in a world of clinical intuition and instinctual premoni-tion. Yet these instincts may be vital. • If EBM is prescriptive, patient choice declines. Does our zeal for EBM make us ar-rogant, mechanical, and defensive? Where is the shared decision-making (p7)? • By focusing on answerable questions, EBM can distract us from our patients’ un-answerable questions; questions that still require time and acknowledgement.
The practice of EBM must be informed by clinical judgement and compassion.
Fig 1.3 Hierarchy of evidence.
EBM Pyramid and EBM Page Generator, copyright 2006 Trustees of Dartmouth College and Yale University.
Thinking about medicine 23 Medicalization Using Illness as a Metaphor, 8 Susan Sontag describes two kingdoms: that of the well, and that of the sick. She describes our dual citizenship, and the use of a pass-port to travel from one kingdom to the other. But medicalization blurs this distinc-tion. The boundary between the ‘Kingdom of the Sick’ and the ‘Kingdom of Well’ is lost and there is an anschluss of healthy people annexed into the potentially preda-tory and frightening kingdom of the sick from which there may well be no escape.
‘Too much medicine’ occurs as a result of: • Overdiagnosis: Labelling an (asymptomatic) person as ‘sick’ despite the fact that subsequent treatment, lifestyle advice, or monitoring provides no benefi t to their outcome (and potentially causes harm), eg non-progressive breast cancer. • Overdetection: Increasingly sensitive tests identify pathology that is indolent or non-progressive, eg subsegmental pulmonary emboli diagnosed on CT angiography. • Overdefi niton: Expansion of disease defi nitions or lowering of disease thresholds, eg an eGFR diagnosis of chronic kidney disease now means that 1 in 8 adults are labelled with the disease, many of whom will never progress to symptomatic kid-ney failure; 15% of pregnant women now have subclinical hypothyroidism without evidence that thyroxine replacement is benefi cial (2016). • Disease mongering: The creation of pseudodiseases which pose no threat to health, eg restless legs, sexual health dysfunction, multiple chemical sensitivity. • Overutilization: Healthcare practice that provides no net benefi t, eg routine MRI for lower back pain. • Overtreatment: Treatment that is of no benefi t (and may cause harm), eg antibiot-ics for viral infections, polypill for the population.
Too much medicine arises from the fear of missing a diagnosis, and concern about avoidable morbidity or mortality. A punitive society means there is a perceived need for more tests, to seek more certainty. But certainty is the holy grail of myth and legend. The individual patient is a unique set of symptoms, stoicism, experience, and need. And by the nature of life, all cure can only ever be temporary.
Choosing wisely CHOOSING WISELY is an initiative to change doctors’ practice away from interventions that are not: • supported by evidence • free from harm • truly necessary (including duplicative tests).
The top 5–10 interventions that should not be used routinely are given for each spe-cialty. Search for those relevant to your current post at: www.choosingwisely.org/doctor-patient-lists/.
Consider medicalization when screening for disease. Remember all screening programmes do harm, some do good. The Wilson criteria for screening lists the important features necessary for a screening programme and the mnemonic IATROGENIC reminds of our pressing duty to do no harm: 1 The condition screened for should be an important one.
2 There should be an acceptable treatment for the disease.
3 Diagnostic and treatment facilities should be available.
4 A recognizable latent or early symptomatic stage is required.
5 Opinions on who to treat must be agreed.
6 The test must be good: high discriminatory power, valid, and reproducible with safety guaranteed.
7 The examination must be acceptable to the patient.
8 The untreated natural history of the disease must be known.
9 It should be inexpensive.
10 Screening must be continuous (ie not a ‘one-off ’ aff air).
Screening 8 Susan Sontag, Illness as a Metaphor, 1978 2 History and examination Contents Taking a history 26 Symptoms 28 Systemic enquiry 30 Physical examination 32 Signs 34 The cardiovascular system: History 36 Examination 38–41 Pulses 42 The jugular venous pressure (JVP) 43 The heart sounds 44 Cardiac murmurs 46 The respiratory system: History 48 Examination 50–53 Important presentations 54 The gastrointestinal system: History 56 Gastrointestinal symptoms 58 Examination of the abdomen 60 The gastrointestinal system: examination 62 The neurological system: History 64 Neurological examination of the upper limbs 66 Neurological examination of the lower limbs 68 Cranial nerve examination 70 Cranial nerve lesions of the eye 72 Musculoskeletal hand examination 74 The peripheral vascular system: Examination 78 Arterial 79 Venous 79 The genitourinary system: History 80 The breast: History 82 Examination 83 The thyroid: Examination 84 Speech and higher mental function 86 Movement disorders 87 Psychiatric assessment 88 Method and order for routine examination 90 Fig 2.1 William Osler (1849–1919) was a great medical educationalist who loved practical jokes. He introduced many nov-elties to the classroom, including, on one occasion, a gaggle of geese. We can all identify with his geese, because these birds show exceptional learning ability and resilience.
Osler did not agree with gavage, a method whereby geese (and medical students) are forcibly stuff ed by funnel to fatten them for the delight of gluttons. We are too familiar with the three Rs of medical education: RamRememberRegurgitate, a sequence that turns once-bright medi-cal students into tearful wrecks. Luckily in the realm of History & Examination we can fl ee the library and alight at the bedside, bearing in mind another of Osler’s aphor-isms: ‘He who studies medicine without books sails an uncharted sea, but he who studies medicine without patients does not go to sea at all.’ We thank Dr Petra Sulentic, our Specialist Reader, for her contribution to this chapter. 25 History and examination 1 The way to learn physical signs is at the bedside, with guidance from a senior doc-tor or an experienced colleague. This chapter is not a substitute for this process: it is simply an aide-memoire both on the wards and when preparing for exams.
2 We ask questions to get information to help with diff erential diagnosis. But we also ask questions to fi nd out about the lives our patients live so that we can respect them as individuals. The patient is likely to notice and reciprocate this respect, and the rapport that you build with your patient in this way is a key com-ponent to diagnosing and managing their disease.
3 Patients (and diseases) rarely read textbooks, so don’t be surprised that some symptoms are ambiguous, and others meaningless. Get good at recognizing pat-terns, but not so good that you create them when none exist. We all fall into this trap!
4 Signs can be easy to detect, or subtle. Some will be found by all the new medical students, others require experienced ears or eyes. Remember, you can be a fi ne doctor without being able to elicit every sign.1 However, fi nding signs and put-ting together the clues they give us to fi nd a diagnosis is one of the best parts of being a doctor. It is also essential that we learn those signs that highlight diseases we should never miss. However, in an exam, if you cannot fi nd a sign, never be tempted to make up something you think should be there. If the examiner is push-ing you to describe something you cannot see, be honest and admit you cannot see it. Learning is a lifelong process, and nobody becomes a consultant overnight.
Advice and experience Having a template for the all-important history and examination is no more than a rough guide and you must fl esh it out with your own learning. We start out nerv-ous of missing some question or sign, but what we should really be nervous about is losing our humanity in the hurly-burly of a time-pressed interview. Here is how one student put some fl esh on the bones—for a man in a wheelchair: she asked all about the presenting complaint, and how it fi tted in with his CNS condition and life at home—and then found out that his daughter had had a nervous breakdown at the start of his illness, 5 years ago. ‘How is she now?’ she asked. ‘Fine—I’ve got two lovely grand children…Jim is just learning to walk…’ ‘Oh…you must be so busy!’ the student said with a joyful smile. This man had not been busy for 5 years, and was fed up with his passive dependency. The thought of being busy again made his face light up—and when the student left he rose up out of his wheelchair to shake her by the hand, a movement we doctors thought was impossible. Jim and his grandfather were learning to walk, but this student was up and running—far ahead of her teachers.
An insightful student While on the acute medical or surgical take you will ‘clerk’ countless numbers of patients. This involves taking a full history: history-taking may seem deceptively easy, as if the patient knew the hard facts and the only problem was extracting them; but what a patient says is a mixture of hearsay (‘She said I looked very pale’), innuendo (‘You know, doctor, down below’), legend (‘I suppose I bit my tongue; it was a real fi t, you know’), exaggeration (‘I didn’t sleep a wink’), and improbabilities (‘The Pope put a transmitter in my brain’). The great skill (and pleasure) in taking a history lies not in ignoring these garbled messages, but in making sense of them. Next you will likely perform all the core examinations (cardiovascular, respiratory, abdominal, and neurological) and any relevant additional ones (eg breast, thyroid, peripheral vascular). No two doctors will have identical examination techniques. Relish this variation as it helps you craft your own routine.
Developing your own routine 26 History and examination Taking a history Taking a good history is an art and an essential skill: 80% of diagnoses should be made on history alone, with the signs you elicit adding an extra 10% and tests only giving the fi nal 5% or so. Do not rely on signs or investigations for your diagnosis, but use them rather to confi rm what you suspected. Try to put the patient at ease: a good rapport may relieve distress. Introduce yourself and check whether the patient is comfortable. Be conversational rather than interrogative. Start with open ques-tions, allow the patient to tell their story, but if they stray off topic, gently steer them back towards the important points.
Presenting complaint (PC) Open questions: ‘Why have you come to see me today?’ Record the patient’s own words rather than medical terms.
History of presenting complaint (HPC) When did it start? What was the fi rst thing noticed? Progress since then. Ever had it before? ‘SOCRATES’ questions: site; onset (gradual, sudden); character; radiation; associations (eg nausea, sweating); timing of pain/duration; exacerbating and alleviating factors; severity (eg scale of 1–10, compared with worst ever previous pain). Direct questioning (to narrow list of possible diagnoses). Specifi c or ‘closed’ questions about the diff erential diagnoses you have in mind (+risk factors, eg travel—p414) and a review of the relevant system.
Past medical history (PMH) Ever in hospital? Illnesses? Operations? Ask specifi -cally about MIJTHREADS: MI, jaundice, TB, high BP, rheumatic fever, epilepsy, asthma, diabetes, stroke, anaesthetic problems.
Drug history (DH) Any tablets, injections, ‘over-the-counter’ drugs, herbal remedies, oral contraceptives? Ask about allergies and what the patient experienced, eg may be an intolerance (nausea, diarrhoea), or may have been a minor reaction of sensiti-zation (eg rash and wheeze) before full-blown anaphylaxis.
Social history (SH) Probe without prying. ‘Who else is there at home?’ Job. Marital status. Spouse’s job and health. Housing—any stairs at home? Who visits—relatives, neighbours, GP, nurse? Are there any dependants at home? Mobility—any walking aids needed? Who does the cooking and shopping? What can the patient not do be-cause of the illness? Ask about occupation, hobbies, sport, exercise, and ethnic origin.
The social history is all too often seen as a dispensable adjunct but vital clues may be missed about the quality of life and it is too late to ask when the surgeon’s hand is deep in the belly and they are wondering how radical a procedure to perform. Utilize the GP’s knowledge of the patient: they may have known them and/or their family for decades. He or she may even hold a ‘living will’ or advance directive if they cannot speak for themselves. Tactfully ask about alcohol, tobacco, and recreational drugs. How much? How long? When stopped? 1 unit = 8g of ethanol = 1 spirits measure = 1/2 glass of wine = 1/3 pint of beer. The CAGE questionnaire is a useful screening test for alcoholism (p281). Quantify smoking in terms of pack-years: 20 cigarettes/day for 1 year equals 1 pack-year. We all like to present ourselves well, so be inclined to double stated quantities (Holt’s ‘law’).
Family history (FH) Areas of the family history may need detailed questioning, eg to determine if there is a signifi cant family history of heart disease you need to ask about the health of the patient’s grandfathers and male siblings, smoking, tendency to hypertension, hyperlipidaemia, and claudication before they were 60 years old, as well as ascertaining the cause of death. Ask about TB, diabetes, and other relevant diseases. Draw a family tree (see BOX). Be tactful when asking about a family history of malignancy.
Systemic enquiry (See p30.) Helps uncover undeclared symptoms. Some of this may already have been incorporated into the history.
Always enquire, without sounding robotic, if your patient has any ideas of what the problem might be, if he/she has any particular concerns or expectations, and give him/her an opportunity to ask you questions or tell you anything you may have missed.
Don’t hesitate to review the history later: recollections change (as you will fi nd, often on the post-take ward round when the Consultant is asking the questions!).
27 History and examination Advances in genetics are touching all branches of medicine. It is increasingly im-portant for doctors to identify patients at high risk of genetic disease, and to make appropriate referrals. The key skill is drawing a family tree to help you structure a family history as follows: 1 Start with your patient. Draw a square for a male and a circle for a female. Add a small arrow (see fi g 2.2) to show that this person is the propositus (the person through whom the family tree is ascertained).
2 Add your patient’s parents, brothers, and sisters. Record basic information only, eg age, and if alive and well (a&w). If dead, note age and cause of death, and pass an oblique stroke through that person’s symbol.
3 Ask the key question ‘Has anybody else in your family had a similar problem as yourself?’, eg heart attack/angina/stroke/cancer. Ask only about the family of diseases that relate to your patient’s main problem. Do not record a potted medical history for each family member: time is too short.
4 Extend the family tree upwards to include grandparents. If you haven’t revealed a problem by now, go no further—you are unlikely to miss important familial disease. If your patient is elderly it may be impossible to obtain good informa-tion about grandparents. If so, fi ll out the family tree with your patient’s uncles and aunts on both the mother’s and father’s sides.
5 Shade those in the family tree aff ected by the disease. • = an aff ected female; = an aff ected male. This helps to show any genetic problem and, if there is one, will help demonstrate the pattern of inheritance.
6 If you have identifi ed a familial susceptibility, or your patient has a recognized genetic disease, extend the family tree down to include children, to identify others who may be at risk and who may benefi t from screening. You should fi nd out who is pregnant in the family, or may soon be, and arrange appropriate genetic counselling (OHCS p154). Refer for genetics opinion.
The family tree (fi g 2.2) shows these ideas at work and indicates that there is evidence for genetic risk of colon cancer, meriting referral to a geneticist. N.B: use a diff erent approach in paediatrics, and for autosomal or sex-linked disease. Ask if parents are related (consanguinity risk of recessive diseases).
Drawing family trees to reveal dominantly inherited disease Acknowlegement The box in this section owes much to Dr Helen Firth, who we thank.
Fig 2.2 Genetic risk of colon cancer in a family tree.
28 History and examination Common causes of admission in the elderly, and can lead to loss of confi dence and independence. Causes are often multifactorial: Intrinsic: Typically osteo- or rheumatoid arthritis, but remember fractured neck of femur, CNS disease, vision, cognitive impairment, depression, postural hypoten-sion, periph eral neuropathy, medication (eg antihypertensives, sedatives), pain, eg arthritis, parkinsonism (eg drugs: prochlorperazine, neuroleptics, metoclopra-mide), muscle weakness (consider vitamin D defi ciency), incontinence, UTI, pneu-monia, anaemia, hypothyroidism, renal impairment, hypothermia, and alcohol.
Environ ment: Poor lighting, uneven walking surface. Treatment includes address-ing injuries, reducing risk factors, and reducing the risk of injury, eg treat osteopo-rosis (p682). A multidisciplinary multifactorial approach alongside occupational therapists and physiotherapists is likely to be benefi cial. See gait disorders, p467.
If there is ataxia, the cause is not always alcohol: other chemicals may be involved (eg cannabis or prescribed sedatives). There may be a metastatic or non-meta-static manifestation of malignancy, or a cerebellar lesion.
Bilateral weak legs may suggest a cord lesion: see p466. If there is associated urinary or faecal incontinence ± saddle anaesthesia or lower limb sensory loss, urgent imaging (MRI) and treatment for cord compression may well be needed.
‘Off -legs’—falls and diffi culty walking Itching (pruritus) Common and, if chronic, most unpleasant.
Table 2.1 Aetiology of pruritus Local causes Systemic (do FBC, ESR, glucose, LFT, U&E, ferritin, TFT) Eczema, atopy, urticaria Liver disease (bile salts, eg PBC) Old age; pregnancy Scabies Uraemia (eg CKD) Drugs (eg morphine) Lichen planus Malignancy (eg lymphoma) Diabetes mellitus Dermatitis herpetiformis Polycythaemia rubra vera Thyroid disease Spinal cord tumours (rare)2 Iron defi ciency anaemia HIV infection Questions: Wheals (urticaria)? Worse at night? Others aff ected (scabies)? What provokes it? After a bath ≈ polycythaemia rubra vera (p366). Exposure, eg to ani-mals (atopy?) or fi bre glass (irritant eczema?).
See table 2.1. Look for local causes: Scabies burrows in fi nger webs, lice on hair shafts, knee and elbow blisters (dermatitis herpetiformis). Systemic: Splenomeg-aly, nodes, jaundice, fl ushed face, or thyroid signs? : Treat causes; try soothing bland emollients ± emollient bath oils ± sedative antihistamines at night, eg chlor-phenamine 4mg PO.
Itch Symptoms Symptoms are features which patients report. Physical signs are elicited at the bedside. Together, they constitute the features of the condition in that patient. Their evolution over time and interaction with the physical, psychological, and social spheres comprise the natural history of any disease. Throughout this chapter, we discuss symptoms in isolation and attempt to classify them into a ‘system’ or present them in the following BOXES as ‘non-specifi c’. This is unnatural but a good fi rst step in learning how to diagnose. All doctors have to know about symptoms and their relief. Part of becoming a good doctor is learning to link symptoms together, to identify those that may be normal, and those that are worrying. There are many online tools and books that can help with this, but there is no substitute for experience. If you aren’t sure, ask a specialist in that area for advice.
The following are common ‘non-specifi c’ presentations.
29 History and examination So common that it is a variant of normality. Only 1 in 400 episodes of fatigue leads to visiting the doctor. Don’t miss depression (p15). Even if depressed, still rule out common treatable causes—eg anaemia, hypothyroidism, diabetes. After history and examination: FBC, ESR, U&E, plasma glucose, TFT, ± CXR. Follow up to see what develops, and to address emotional problems. Take a sleep history.
Fatigue While some night sweating is common in anxiety, drenching sweats requiring changes of night-clothes are a more ominous symptom associated with infection (eg TB, brucellosis), lymphoproliferative disease, or other malignancies. Patterns of fever may be relevant (see p442).
Rigors are uncontrolled paroxysms of shivering which occur as a patient’s tem-perature rises rapidly.
Sweating excessively (hyperhidrosis) may be primary (eg hidradenitis sup-purativa may be very distressing to the patient)—or secondary to fever, pain or anxiety (cold & sweaty) or a systemic condition: the menopause, hyperthyroidism (warm & sweaty), acromegaly, malignancy, phaeochromocytoma, amyloidosis, or neuroleptic malignant syndrome (+hyper thermia). Or it may refl ect gabapentin or opiate withdrawal, or a cholin er gic or parasympathomimetic side-eff ect (ami-triptyline, bethanechol, distigmine, spider bites)—also hormonal drugs, eg levo-thyroxine, gonadorelin or somatostatin analog ues, vasopressin, and ephedrine. Also amiodarone, ciprofl oxacin, levodopa, lisinopril, rivastigmine, ritonavir, pioglitazone, venlafaxine. At the bedside: ask about all drugs, examine all over for nodes; any signs of hyperthyroidism? Any spleno megaly? Test the urine; do T°, ESR, TSH, FBC, & blood culture. : Antiperspirants (aluminium chloride 20%=Driclor®), sympathec-tomy, or iontophoresis may be tried.
Fevers, rigors, sweats This is trivial—until we ourselves have a few sleepless nights. Then sleep becomes the most desirable thing imaginable, and bestowing it the best thing we can do, like relieving pain. But don’t give drugs without looking for a cause. • Self-limiting: Jet lag; stress; shift work; in hospital. We need less sleep as we age. • Psychic: Depression; anxiety; mania; grief; psychomotor agitation/psychosis. • Organic: Drugs (many; eg caff eine; mefl oquine; nicotine withdrawal); nocturia; alcohol; pain (eg acid refl ux—worse on lying down); itch; tinnitus; asthma; dys-tonias; obstructive sleep apnoea (p194); dementia; restless leg syndrome (p698, check ferritin). Rarer: encephalitis (eg West Nile virus) and encephalopathy (Whipple’s; pellagra; HIV; prion diseases, eg CJD, p696, and fatal familial insomnia).
:Sleep hygiene. No daytime naps; don’t turn in till you feel sleepy; regular bed-time routines. Keep a room for sleep; don’t eat or work in it (not viable for much of the world). Less caff eine, nicotine, late exercise (but sexual activity may give excellent torpor!), and alcohol (its abuse causes paradoxical pro-adrenergic trem-or and insomnia). Try monitoring quality with a sleep diary (unless already over-obsessive). Music and relaxation may make sleep more restorative and augment personal resources.
Hypnotic drugs. Give for a few nights only (addictive and cause daytime somno-lence ± rebound insomnia on stopping). Warn about driving/machine use. Exam-ple: zopiclone 3.75–7.5mg. Obstructive sleep apnoea, p194. Parasomnias, sleep paralysis, etc. OHCS p371. Narcolepsy, p700.
Insomnia 30 History and examination Systemic enquiry Just as skilled acrobats are happy to work without safety nets, so experienced clini-cians may operate without the functional enquiry. But to do this you must be experi-enced enough to understand all the nuances of the presenting complaint.
General questions May be the most signifi cant, eg in TB, endocrine problems, or cancer: • Weight loss. • Night sweats. • Any lumps. • Fatigue/malaise/lethargy. • Sleeping pattern.
1 • Appetite. • Fevers. • Itch or rash. • Recent trauma.
Cardiorespiratory symptoms • Chest pain (p94). • Exertional dyspnoea (=breathlessness): quantify exercise tolerance and how it has changed, eg stairs climbed, or distance walked, before onset of breathlessness. • Paroxysmal nocturnal dyspnoea (PND). Orthopnoea, ie breathlessness on lying fl at (a symptom of left ventricular failure): quantify in terms of number of pillows the patient must sleep on to prevent dyspnoea. • Oedema: ankles, legs, lower back (dependent areas). • Palpitations (awareness of heartbeats): can they tap out the rhythm? • Cough: sputum, haemoptysis (coughing up blood). • Wheeze.
Gastrointestinal symptoms • Abdominal pain (constant or colicky, sharp or dull; site; radiation; duration; onset; severity; relationship to eating and bowel action; alleviating or exacerbating, or associated features). • Other questions—think of symptoms throughout the GI tract, from mouth to anus: • Swallowing (p250). • Indigestion (p252). • Nausea/vomiting; blood? (p250). • Bowel habit (p258 & p260). • Stool: colour, consistency, blood, mucus; diffi culty fl ushing away (p266); tenes-mus or urgency.
Tenesmus is the feeling of incomplete evacuation of the bowels (eg due to a tumour or irritable bowel syndrome). Haematemesis is vomiting blood. Melaena is altered (black) blood passed PR (p256), with a characteristic off ensive smell and tar like appearance.
Genitourinary symptoms • Incontinence (stress or urge, p648). • Dysuria (painful micturition). • Urinary abnormalities: colour? Haematuria (streaks or pink urine?) Frothy? • Nocturia (needing to micturate at night). • Frequency (frequent micturition) or polyuria (the frequent passing of large vol-umes of urine). • Hesitancy (diffi culty starting micturition). • Terminal dribbling. • Vaginal discharge (colour, odour); pain on intercourse (dyspareunia) (p412). • Menses: frequency, regularity, heavy or light, duration, painful? First day of last menstrual period (LMP). Number of pregnancies and births. Menarche. Menopause. Any chance of pregnancy now?
1 Too sleepy? Think of myxoedema or narcolepsy. Early waking? Think of depression. Being woken by pain is always a serious sign. For the signifi cance of the other questions listed here, see Chapter 3.
31 History and examination Neurological symptoms • Special senses: sight, hearing, smell, and taste. • Seizures, faints, ‘funny turns’. • Headache. • ‘Pins and needles’ (paraesthesiae) or numbness. • Limb weakness (‘Are your arms and legs weaker than normal?’), poor balance. • Speech problems (p86). • Sphincter disturbance. • Higher mental function and psychiatric symptoms (p86–p89). The important thing is to assess function: what the patient can and cannot do at home, work, etc.
Musculoskeletal symptoms • Pain, stiff ness, swelling of joints. • Diurnal variation in symptoms (ie worse in mornings). • Functional defi cit. • Signs of systemic disease: rashes, mouth ulcers, nasal stuffi ness, malaise, and con-stitutional symptoms.
Thyroid symptoms • Hyperthyroidism: Prefers cold weather, bad tempered, sweaty, diarrhoea, oli-gomenorrhoea, weight (though often appetite), tremor, palpitations, visual prob-lems. • Hypothyroidism: Depressed, slow, tired, thin hair, croaky voice, heavy periods, con-stipation, dry skin, prefers warm weather.
32 History and examination Physical examination The physical examination is not so much an extension of the history, but more of the fi rst investigation, to confi rm, exclude, defi ne, or show the progress of the provi-sional diagnosis as revealed in the history. Even in the emergency department where the history may be brief, eg ‘trauma’, the examination is to confi rm a fracture, or to decide that a fracture is less likely. The examination sheds further light on the his-tory. As you get better, your physical examination gets briefer. Establish your own routine—practice is the key.
End of the bed • Look at the patient—are they well or in extremis? What makes you think this? Are they in pain? If so, does it make them lie still (eg peritonitis) or writhe about (eg colic)? What is the pattern of breathing: laboured; rapid; shallow; irregular; distressed? Are they obese or cachectic? Is their behaviour appropriate? Can you detect any unusual smell, eg hepatic fetor (p274), cigarettes, alcohol? • Also take a moment to look around the bed for other clues, eg inhalers, insulin administration kit, walking aids, etc.
Face and body habitus • Does the patient’s appearance suggest any particular diseases, eg acromegaly, thy-rotoxicosis, myxoedema, Cushing’s syndrome, or hypopituitarism? See p202. • Is there an abnormal distribution of body hair (eg bearded , or hairless ) sug-gestive of endocrine disease? • Is there anything about the patient to trigger thoughts about Paget’s disease, Mar-fan’s, myotonia, or Parkinson’s syndrome? Look for rashes, eg the malar fl ush of mitral disease and the butterfl y rash of SLE.
Peripheral stigmata of disease Specifi c signs are associated with diff erent diseases: consider the nails (koilonychia = iron defi ciency), subcutaneous nodules (rheumatoid, neurofi broma?), and look for lymph nodes (cervical, axillary, inguinal). See specifi c systems for features to assess for, but for all systems consider: Skin colour: • Blue/purple = cyanosis (can also be central only, p34). • Yellow = jaundice (yellow skin can also be caused by uraemia, pernicious anaemia, carotenaemia—check the sclera: if they are also yellow it is jaundice). • Pallor: this is non-specifi c; anaemia is assessed from the palmar skin creases (when spread) and conjunctivae (fi g 8.3)—usually pale if Hb <80–90g/L: you cannot con-clude anything from normal conjunctival colour, but if they are pale, the patient is probably anaemic. • Hyperpigmentation: Addison’s, haemo chromatosis (slate-grey) and amiodarone, gold, silver, and minocycline therapy.
Charts: • Temperature: varies during the day; a morning oral temperature >37.2°C or evening >37.7°C constitutes a fever.3 Rectal temperatures are generally 0.6°C above oral temperatures. Remember that temperatures are generally lower in elderly patients and therefore fevers may not be as pronounced.4 A core temperature <35°C indi-cates hypothermia; special low-reading thermometers may be required. • Blood pressure and pulse—trends are more important than one-off values; repeat if concerned. • Urine: check urinalysis and input/output charts if available.
Fluid status When admitting an unwell patient, don’t forget to assess their hydra-tion, check skin turgor and mucous membranes, look for sunken eyes, and check capillary refi ll (if well perfused <2s) and JVP.
33 History and examination When you don’t know: ask. If you are wondering if you should ask: ask.
Frequently, the skills needed for diagnosis or treatment will lie beyond the team you are working for, so, during ward rounds, agree who should be asked for an opinion. You will be left with the job of making the arrangements, so check before your senior leaves exactly what their question is. Don’t be intimidated, but follow these simple rules: • Know the history and examination fi ndings (ideally your own), and have the pa-tient’s notes, observations, recent test results, and drug charts to hand (table 2.2). • At the outset, state if you are just looking for advice or if you are asking if the patient could be seen. Make it clear exactly what the question is that you want addressed, allowing the listener to focus their thoughts and ask relevant questions. • Give the patient’s age and run through a brief history including relevant past medical history. If you would like the patient to be seen, give warning if they will be leaving the ward for a test at a particular time. • The visiting doctor may be unfamiliar with your ward. When he or she arrives introduce yourself, get the notes and charts, and give your contact details in case they have further questions.
Table 2.2 Referring for a specialist opinion Team Key questions Anaesthetics Previous anaesthetic? Reaction? Last ate/drank?
Cardiology Known IHD? BP? ECG fi ndings? Echo fi ndings? Murmurs? Troponin? Temperature/possibility of endocarditis? (ESR, microscopic haematuria, etc. p150) Dermatology Site, onset, and appearance of rash? Drugs? Systemic disease? History of atopy? Endocrinology Diabetes: blood glucose, usual insulin regimen, complications. Other: blood results? Stable/unstable—eg Addisonian crisis. Usual steroid dose?
Gastroenterology/ Hepatology Bleeding: Rockall score (p257)? Shock? Diarrhoea: blood? Foreign travel? Frequency per day? Liver disease: signs of decompensation (p274)? Ascites? Encephalopathy grade?
Gynaecology/ Obstetrics LMP? Possibility of pregnancy? Previous pregnancies? Vaginal discharge? Hormonal contraceptives? STIS?
Haematology Blood results? Splenomegaly? Fever? Lymphadenopathy? Bleeding: anticoagulants? Clotting results? Infectious diseases/ Microbiology Possible source? Antibiotics (current/recent/previous)? Foreign travel? Risk factors for HIV? Nephrology Creatinine (current, old)? Clotting? Urine output? Potassium? BP? Fluid status? Drugs? Known renal disease?
Neurology/Stroke Neurological examination? CT/MRI scan fi ndings? Radiology See p720. Contrast or not? Creatinine? Clotting? Cannula in situ? Metallic implants?
Respiratory O2 sats? Respiratory rate? ABG? CXR? Inhalers/nebs? Home O2? Respiratory support, eg NIV/CPAP?
Surgery (general) Pain? Scan fi ndings? Acutely unwell? Clotting? Urology History of LUTS (lower urinary tract symptoms) p642? Catheter? Haematuria? History of stones? Scan fi ndings (ultrasound, CT)?
You would be amazed at how many people refer to neurology/stroke without having done a neurologi-cal examination! Don’t be one of them...
Unexplained signs and symptoms: how to refer for an opinion 34 History and examination Signs The following signs are not specifi c to a particular system: Cyanosis Dusky blue skin (peripheral—of the fi ngers) or mucosae (central—of the tongue), representing 50g/L of Hb in its reduced (hence hypoxic) form, it occurs more readily in polycythaemia than anaemia.
Causes: • Lung disease with inadequate oxygen transfer, eg luminal obstruction, asthma, COPD, pneumonia, PE, pulmonary oedema—may be correctable by inspired O2. • Congenital cyanotic heart disease, where there is a mixture, eg transposition of the great arteries or right-to-left shunt (eg VSD with Eisenmenger’s syndrome; see p156)—cyanosis is not reversed by increasing inspired oxygen. • Rare causes—methaemoglobinaemia, a congenital or acquired red cell disorder.
Acute cyanosis is an emergency. Is there asthma, an inhaled foreign body, a pneu-mothorax (p749, fi g 1) or pulmonary oedema? See p814.
Peripheral cyanosis will occur in causes of central cyanosis, but may also be induced by changes in the peripheral and cutaneous vascular systems in patients with nor-mal oxygen saturations. It occurs in the cold, in hypovolaemia, and in arterial disease, and is, therefore, not a specifi c sign.
Pallor May be racial or familial—or from anaemia, shock/faints, Stokes–Adams attack (p460, pale fi rst, then fl ushing), hypothyroid ism, hypopituitarism, and albinism.
Anaemia is haemoglobin concentration <130g/L in men and <120g/L in non-pregnant women (p324). It may be assessed from the conjunctivae and skin creases. Koilo-nychia and stomatitis (p32) suggest iron defi ciency. Anaemia with jaundice suggests haemolysis. If pallor just one limb or digit, think of emboli.
Skin discolouration Generalized hyperpigmentation may be genetic (racial) or due to radiation; ACTH (cross-reacts with melanin receptors, eg Addison’s disease (p226), Nelson’s syn-drome (p76), ectopic ACTH in bronchial carcinoma); chronic kidney disease (urea, p302); malabsorption; chloasma (seen in pregnancy or with the oral contraceptive pill); biliary cirrhosis; haemochromatosis (‘bronzed diabetes’); carotenaemia; or drugs (eg chlorpromazine, busulfan, amiodarone, gold).
Obesity Defi ned by the World Health Organization as a BMI of over 30kg/m 2. A higher waist to hip ratio, indicating central fat distribution, is commoner in and is associated with greater health risks, which include type 2 diabetes mellitus, IHD, dyslipidaemia, BP, osteoarthritis of weight-bearing joints, and cancer (breast and bowel); see p206. The majority of cases are not due to specifi c metabolic disorders. Lifestyle change is key to treatment, to increase energy expenditure and reduce intake (p244). Medica-tion ± surgery may be considered if the patient fulfi ls strict criteria (BMI of 40 kg/m 2 or more, or between 35 kg/m 2 and 40 kg/m 2 and other signifi cant disease that could improve with weight loss, non-surgical measures have been tried and failed, patient receives intensive management in a tier 3 service, and fi t for anaesthesia and sur-gery). Conditions associated with obesity include: genetic (Prader–Willi syndrome, Lawrence–Moon syndrome), hypothyroidism, Cushing’s syndrome, and hypothalamic damage (eg tumour or trauma damage to satiety regions).
35 History and examination Lymphadenopathy Causes of lymphadenopathy are either reactive or infi ltrative: Reactive: Infective: • Bacterial: eg pyogenic, TB, brucella, syphilis. • Viral: EBV, HIV, CMV, infectious hepatitis. • Others: toxoplasmosis, trypanosomiasis.
Non-infective: sarcoidosis, amyloidosis, berylliosis, connective tissue disease (eg rheumatoid, SLE), dermatological (eczema, psoriasis), drugs (eg phenytoin).
Infi ltrative: Benign histiocytosis—OHCS p644, lipoidoses.
Malignant: • Haematological: lymphoma or leukaemia: ALL, CLL, AML (p356). • Metastatic carcinoma: from breast, lung, bowel, prostate, kidney, or head and neck cancers.
Oedema (See p579.) Pitting oedema: Fluid can either be squeezed out of the veins (increased hydrostatic pressure, eg DVT, right heart failure) or diff use out because of reduced oncotic pres-sure (low plasma proteins, eg cirrhosis, nephrotic syndrome, protein-losing enter-opathy) leading to an osmotic gradient with the tissues (fi g 2.9, p39, p579). The cause of oedema is still not completely understood.5 Periorbital oedema: Oedema around the face has a very diff erent diff erential; the eyelid skin is very thin so periorbital oedema is usually the fi rst sign—think of al-lergies (contact dermatitis, eg from eye make-up, stings), angioedema (can be hereditary), infection (orbital cellulitis can be life-threatening, refer to hospital immediately if concerned, other infections include EBV and sinusitis); if there is prop-tosis (p219) think Graves’ disease, connective tissue diseases (eg dermatomyositis, SLE, sarcoid, amyloid); and many others. Assess for systemic disease before putting it down to allergies.
Non-pitting oedema: Ie non-indentable, is lymphoedema due to poor lymphatic drainage. Can be due to radiotherapy, malignant infi ltration, infection, fi lariasis, or rarely primary lymphoedema (Milroy’s syndrome p706).
Weight loss Weight loss can be both a symptom (ie reported by the patient) and a sign (identifi ed by physician). A feature of chronic disease, depression, malnutrition, malignancy, chronic infections (eg TB, HIV/enteropathic AIDS), diabetes mellitus, and hyperthy-roidism (typically in the presence of increased appetite). Severe generalized muscle wasting is also seen as part of a number of degenerative neurological diseases and in cardiac failure (cardiac cachexia), although in the latter, right heart failure may not make weight loss a major complaint. Do not forget ano-rexia nervosa (OHCS p382) as an underlying cause of weight loss.
Rule out treatable causes, eg diabetes is easy to diagnose—TB can be very hard. For example, the CXR may look like cancer so don’t forget to send bronchoscopy samples for ZN stain and TB culture. Unintentional weight loss should always ring alarm bells, so assess patients carefully.
Cachexia General muscle wasting from famine, or eating (dementia; stroke; MND, p506; ano-rexia nervosa), malabsorption (entero pathic AIDS/slim disease/Crypto spor idium; Whipple’s) or catabolism (neoplasia; CCF; TB; chronic kidney disease; leptin).6 36 History and examination The cardiovascular system: history Table 2.3 Presenting symptoms and questions to ask Presenting symptoms Direct questions Chest pain (see pp94–5 and p784) Site? Central?
Onset? (Sudden? What was the patient doing?) Character? Ask patient to describe pain (Crushing? Heavy?).
Radiation? Ask specifi cally if moves to arm, neck, or jaw?
Associations? Ask specifi cally about shortness of breath, nausea, sweating.
Timing? Duration?
Exacerbating and alleviating factors? Worse with respiration or move-ment (less likely to be angina)? Relieved by GTN? Worse on inspiration and better when sitting forwards (pericarditis)?
Severity: out of 10?
Is patient known to have angina or chest pain; better/worse/same as usual pain; is it more frequent? Decreasing exercise tolerance?
NB: ‘heartburn’ more likely if ‘burning’, onset after eating/drinking, worse lying fl at, or associated with dysphagia.
Palpitations ‘Ever aware of your own heartbeat’? When and how did it start/stop? Duration? Onset sudden/gradual? Associated with blackout (how long)? Chest pain? Dyspnoea? Food related (eg caff eine)?
Regular fast palpitations may refl ect paroxysmal supraventricular tachycardia (SVT) or ventricular tachycardia (VT).
Irregular fast palpitations are likely to be paroxysmal AF, or atrial fl ut-ter with variable block.
Dropped or missed beats related to rest, recumbency, or eating are likely to be atrial or ventricular ectopics.
Regular pounding may be due to anxiety.
Slow palpitations are likely to be due to drugs such as -blockers, or bigeminus (fi g 3.34, p129).
Reassurance is vital and can be therapeutic. Check a TSH and consider a 24h ECG (Holter monitor, p125). An event recorder, if available, is better than 24h ECGS.
Dyspnoea (see p52, and p782) Duration? At rest? On exertion? Determine exercise tolerance (and any other reason for limitation, eg arthritis). NYHA classifi cation (p135)? Worse when lying fl at, how many pillows does the patient sleep with (orthopnoea)? Does the patient ever wake up in the night gasping for breath (paroxysmal nocturnal dyspnoea), and how often? Any ankle swelling?
Dizziness/ blackouts (see pp460–3) Dizziness is a loose term, so try to clarify if your patient means: did patient lose consciousness, and for how long (short duration suggests cardiac while longer duration suggests a neurological cause)? Any warning (pre-syncope)? What was patient doing at the time? Sudden/ gradual? Associated symptoms? Any residual symptoms, eg confusion? How long did it take for patient to return to ‘normal’? Tongue biting (pp460–1), seizure, incontinence? Witnessed? Memory loss pre/post event?
Vertigo (p462), the illusion of rotation of either the patient or their sur-roundings ± diffi culty walking/standing, patients may fall over.
Imbalance, a diffi culty in walking straight but without vertigo, from peripheral nerve, posterior column, cerebellar, or other central pathway failure.
Faintness ie ‘light-headedness’, seen in anaemia, BP, postural hypoten-sion, hypoglycaemia, carotid sinus hypersensitivity, and epilepsy.
Claudication SOCRATES? Foot/calf/thigh/buttock? ‘Claudication distance’, ie how long can patient walk before onset of pain? Rest pain?
37 History and examination Screen for presenting symptoms (table 2.3) before proceeding to past history: Past history Ask specifi cally about: angina, any previous heart attack or stroke, rheumatic fe-ver, diabetes, hypertension, hypercholesterolaemia, previous tests/procedures (ECG, angiograms, angioplasty/stents, echocardiogram, cardiac scintigraphy, coronary artery bypass grafts (CABGS)).
Drug history Particularly note aspirin/GTN/-blocker/diuretic/ACE-i/digoxin/statin/anticoagulant use.
Family history Enquire specifi cally if any 1st-degree relatives having cardiovascular events (espe-cially if <60yrs).
Social history Smoking, impact of symptoms on daily life, alcohol (clarify number of units), hob-bies, exercise. • Hypertension. • Smoking. • Diabetes mellitus. • Family history (1st-degree relative <60yrs old with IHD). • Hyperlipidaemia.
Ischaemic heart disease risk factors 38 History and examination The cardiovascular system: examination 1 Introduce yourself, obtain consent to examine, and position the patient appropri-ately: lying on a bed, sitting up at 45°. Expose them to the waist (for female patients, delay until examining the praecordium). Explain what you are doing throughout.
5 Neck • JVP: Ask patient to turn head to the left and look at the supraclavicular fossa (see fi g 2.6 and p43). Comment on the height of the JVP and waveform. Press on the abdomen to check the abdomino-jugular refl ex. • Carotid pulse: inspect (visible carotid = Cor-rigan's sign of aortic regurgitation), and palpate volume and character on one side then the other.
4 Blood pressure • Hyper- or hypotensive? • Pulse pressure (wide = aortic regurgitation, ar-teriosclerosis, narrow = aortic stenosis, dry) 3 Radial and brachial pulses • Radial: Rate, rhythm; radio–radial delay (palpate pulse bilaterally simultaneously), radiofemoral delay (palpate ipsilateral pulses simultaneously), collapsing pulse (identify radial pulse (fi g 2.5), then wrap your fi ngers around wrist. Before el-evating arm from the elbow check for pain in arm/ shoulder. Lift arm straight up: collapsing pulse, felt as ‘waterhammer’ pulsation. • Brachial: (Just medial to tendinous insertion of biceps.) Waveform character.
2 Hands • Temperature: Capillary refi ll time • Inspect: Skin: tobacco staining, peripheral cyanosis (fi g 2.4), tendon xanthomata, Janeway lesions, Osler's nodes (signs of infective endocarditis) Nails: clubbing, splinter haemorrhages, nail bed pulsation (Quincke's sign of aortic regurgitation) 1 General inspection • Assess general state (ill/well) • Look for clues (oxygen, GTN spray) • Colour (pale, cyanosed, fl ushed) • Short of breath? • Scars on chest wall (fi g 2.3)?
Fig 2.6 The JVP.
Reproduced from Thomas J, et al. (eds). Oxford Handbook of Clinical Examination and Practical Skills (2014), with permission from Oxford University Press.
Fig 2.3 CABG scar.
Fig 2.4 Peripheral cyanosis.
Reproduced from Ball G, et al. (eds). Oxford Textbook of Vasculitis (2014), with permission from Oxford University Press.
Fig 2.5 Radial pulse.
Reproduced from Thomas J, et al. (eds). Oxford Handbook of Clinical Examination and Practical Skills (2014), with permission from Oxford University Press.
39 History and examination Fig 2.7 Corneal arcus.
8 To complete the examination • Palpate for sacral and ankle oedema (fi g 2.9). • Auscultate the lung bases for inspiratory crackles. • Examine the abdomen for a pulsatile liver and aortic aneurysm. • Check peripheral pulses, observation chart for temperature and O2, sats, dip urine, perform fundoscopy.
Fig 2.9 Pitting oedema, apply fi rm pressure for a few seconds.
7 The praecordium Inspect: • Scars—midline sternotomy, lateral thoracotomy (mitral stenosis valvotomy).
Palpate: • Apex beat (lowermost lateral pulsation)— usually 5th intercostal space in mid-clavicular line; measure position by counting intercostal spaces (sternal notch = 2nd intercostal space). Undisplaced/displaced? Character: impalpable (?dextrocardia/COPD), tapping (palpable S1), dou-ble impulse, sustained/strong. Count rate if pulse irregular (AF, p130). • 'Heaves’ and ‘thrills’—place the heel of the hand fl at on chest to left then right of sternum. Heave: sustained, thrusting usually felt at left sternal edge (= right ventricular enlargement). Thrill: palpable murmur felt as a vibration beneath your hand.
Auscultate: (palpate carotid pulse simultaneously) • Apex (mitral area)—listen with bell and dia-phragm. Identify 1st and 2nd heart sounds: are they normal? Listen for added sounds (p44) and murmurs (p46); with the diaphragm listen for a pansystolic murmur radiating to the axilla—mitral regurgitation (see fi g 2.8). • At apex with bell, ask the patient to ‘Roll over onto your left side, breathe out, and hold it there’ (a rumbling mid-diastolic murmur—mitral stenosis). • Lower left sternal edge (tricuspid area) and pulmonary area (left of manubrium in the 2nd intercostal space): if suspect right-sided mumur, listen with patient’s breath held in inspiration. • Right of manubrium in 2nd intercostal space (aortic area)—ejection systolic murmur radiating to the carotids—aortic stenosis. • Sit the patient up and listen at the lower left sternal edge with patient held in expiration (early diastolic murmur: aortic regurgitation?).
6 Face • Colour: Pale, fl ushed, central cyanosis • Features: Corneal/senile arcus (fi g 2.7), xanthe-lasma (see fi g 2.29, p60) • Pallor of the conjunctiva (anaemia) • Malar fl ush (mitral stenosis) • Dental hygiene Fig 2.8 Praecordium/heart sounds.
Reproduced from Thomas J, et al. (eds). Oxford Handbook of Clinical Examination and Practical Skills (2014), with permission from Oxford University Press.
40 History and examination The cardiovascular system: examination 2 General inspection Ill or well? In pain? Dyspnoeic? Are they pale, cold, and clammy? Can you hear the click of a prosthetic valve? Inspect for scars: median sternotomy (CABG; valve replacement; congenital heart disease). Inspect for any pacemakers/ internal cardiac defi brillators (ICDS). Look around the bed for oxygen and GTN spray.
Hands Finger clubbing occurs in congenital cyanotic heart disease and endocardi-tis. Splinter haemorrhages, Osler’s nodes (tender nodules, eg in fi nger pulps) and Janeway lesions (red macules on palms, fi g 3.38, p151) are signs of infective endocar-ditis. If found, examine the fundi for Roth’s spots (retinal infarcts, p560). Are there nail fold infarcts (vasculitis, p556) or nailbed capillary pulsation (Quincke’s sign in aortic regurgitation)? Is there arachnodactyly (Marfan’s) or polydactyly (ASD)? Are there tendon xanthomata (see BOX ‘Hyperlipidaemia’)?
Pulse See p42. Feel for radio-femoral delay (coarctation of the aorta) and radio-radial delay (eg from aortic arch aneurysm).
Blood pressure (see BOX ‘An unusual BP measurement’) Systolic BP is the pressure at which the pulse is fi rst heard as on cuff defl ation (Korotkoff sounds); the diastolic is when the heart sounds disappear or become muffl ed (eg in the young). The pulse pressure is the diff erence between systolic and diastolic pressures. It is narrow in aortic stenosis and hypovolaemia, and wide in aortic regurgitation, arteriosclerosis, and septic shock. Defi n-ing hypertension: see p138. Examine the fundi for hypertensive changes (p138). Shock may occur if systolic <90mmHg (p790). Postural hypotension is defi ned as a drop in systolic >20mmHg or diastolic >10mmHg on standing for 3–5 min (see BOX ‘Postural hypotension’).
Carotid pulse (See p42.) Jugular venous pressure (See p43.) Face Is there corneal arcus (fi g 2.7, p39) or xanthelasma (fi g 2.29, p60, signifying dyslipidaemia, p690)? Is there a malar fl ush (mitral stenosis, low cardiac output)? Are there signs of Graves’ disease, eg bulging eyes (exophthalmos) or goitre—p218)? Is the face dysmorphic, eg Down’s syndrome, Marfan’s syndrome (p706)—or Turner’s, Noonan’s, or Williams syndromes (p149)?
Praecordium Palpate the apex beat. Normal position: 5th intercostal space in the mid-clavicular line. Is it displaced laterally? Is it abnormal in nature: heaving (caused by outfl ow obstruction, eg aortic stenosis or systemic hypertension), thrust-ing (caused by volume overload, eg mitral or aortic incompetence), tapping (mitral stenosis, essentially a palpable 1st heart sound), diffuse (LV failure, dilated cardio-myopathy) or double impulse (H(O)CM, p152)? Is there dextrocardia? Feel for left parasternal heave (RV enlargement, eg in pulmonary stenosis, cor pulmonale, ASD) or thrills (transmitted murmurs).
Auscultating the heart Also auscultate for bruits over the carotids and elsewhere, particularly if there is inequality between pulses or absence of a pulse. Causes: ath-erosclerosis (elderly), vasculitis (young, p556).
Lungs Examine the bases for creps & pleural eff usions, indicative of cardiac failure.
Oedema Examine the ankles, legs, sacrum, and torso for pitting oedema. (You may prefer to examine ankles while standing at the foot of the bed as it is a good early clue that there may be further pathology to be found.) Abdomen Hepatomegaly and ascites in right-sided heart failure; pulsatile hepato-megaly with tricuspid regurgitation; splenomegaly with infective endocarditis.
Fundoscopy Roth spots (infective endocarditis).
Urine dipstick Haematuria.
1 Signs of heart failure?
2 Clinical evidence of infective endocarditis?
3 Sinus/abnormal rhythm?
4 Heart sounds normal, abnormal, or additional?
5 Murmurs?
Presenting your fi ndings 41 History and examination Don’t interpret a BP value in isolation (p138). We cannot diagnose hypertension (or hypotension) on one BP reading. Take into account pain, the ‘white coat’ eff ect (BP higher in a medical setting), and equipment. Getting cuff size right is vital. Optimal cuff width is 40% of the arm circumference. If you suspect a BP read-ing to be anomalous, check the equipment and review the observation chart for previous readings and other vital signs. Consider taking a manual reading with a diff erent set yourself.
Often a quiet chat will bring the BP down (yours and your patient’s: keep your ears open, and the patient may reveal some new tangential but vital fact that the offi cial history glossed over). Many things aff ect BP readings from background noise to how much you touch the patient. If BP, eg 150/90, check both arms. If the systolic diff erence is >20mmHg, consider peripheral vascular disease, and if the patient could have a thoracic aortic aneurysm or co arctation (rare). NB: right arm diastolic is normally 2.4–5mmHg higher than left.
An unusual BP measurement Xanthomata are localized deposits of fat under the skin, occurring over joints, tendons, hands, and feet. Xanthel asma refers to xanthoma on the eyelid (p691, fi g 14.13). Corneal arcus (fi g 2.7, p39) is a crescentic-shaped opacity at the periph-ery of the cornea. Common in those over 60yrs, can be normal, but may represent hyperlipidaemia, especially in those under this age.
Hyperlipidaemia This is an important cause of falls and faints in the elderly. It is defi ned as a drop in systolic BP >20mmHg or diastolic >10mmHg after standing for 3min vs lying.
Causes: Hypo volaemia (early sign); drugs, eg nitrates, diuretics, antihyperten-sives, antipsychotics; Addison’s (p226); hypopituitarism (ACTH); autonomic neu-ropathy (p505, DM, multisystem atrophy, p494); after a marathon run (peripheral resistance is low for some hours); idiopathic.
Treatment: • Lie down if feeling faint. • Stand slowly (with escape route: don’t move away from the chair too soon!). • Consider referral to a ‘falls clinic’, where special equipment is available for moni-toring patient under various tilts. • Manage autonomic neuropathy, p505. • Water and salt ingestion can help (eg 150mmol Na +/d), but Na + has its problems. • Physical measures: leg crossing, squatting, elas tic compression stockings (check dorsalis pedis pulse is present), and careful exercise may help. • If post-prandial dizziness, eat little and often; carbohydrate and alcohol intake. • Head-up tilt of the bed at night renin release, so fl uid loss and standing BP. • 1st-line drugs: fl udrocortisone (retains fl uid) 50mcg/d; go up to 300mcg/24h PO only if tolerated. Monitor weight; beware if CCF, renal impairment, or albumin as fl udrocortisone worsens oedema. • 2nd-line drugs: sympathomimetics, eg midodrine (not always available) or ephed-rine; pyridostigmine (eg if detrusor under-activity too). • If these fail, turn things on their head and ask: is this really supine hypertension?
Postural hypotension The hand can be used as a manometer to estimate JVP/CVP if you cannot see the neck properly (eg central line in situ). Hold the hand palm down below the level of the heart until the veins dilate (patient must be warm!), then lift slowly, keeping the arm horizontal. The veins should empty as the hand is raised. Empty veins below the level of the heart suggests a low CVP, if they remain full it suggests a normal/high CVP.
Top tips 42 History and examination The waterhammer was a popular toy that consisted of a vacuum tube half-fi lled with water. On inversion, the whoosh of water produced an intriguing hammer-blow as it rushed from end to end. This is the alternative name for Corrigan’s collapsing pulse—ie one in which the upstroke is abrupt and steep, whose peak is reached early and with abnormal force—before a rapid downstroke (as blood whooshes back into the left ventricle through an incompetent aortic valve).
Waterhammer pulse Pulses Assess the radial pulse to determine rate and rhythm. Character and volume are best assessed at the brachial or carotid arteries. A collapsing pulse may also be felt at the radial artery when the patient’s arm is elevated above their head. See fi g 2.10.
Rate Is the pulse fast (100bpm, p126) or slow (60bpm, p124)?
Rhythm An irregularly irregular pulse occurs in AF or multiple ectopics. A regularly irregular pulse occurs in 2° heart block and ventricular bigeminus.
Character and volume • Bounding pulses are caused by CO2 retention, liver failure, and sepsis. • Small volume pulses occur in aortic stenosis, shock, and pericardial eff usion. • Collapsing (‘waterhammer’) pulses are caused by aortic incompetence, AV mal-formations, and a patent ductus arteriosus. • Anacrotic (slow-rising) pulses occur in aortic stenosis. • Bisferiens pulses occur in combined aortic stenosis and regurgitation. • Pulsus alternans (alternating strong and weak beats) suggests LVF, cardio-myopathy, or aortic stenosis. • Jerky pulses occur in H(O)CM. • Pulsus paradoxus (systolic pressure weakens in inspiration by >10mmHg) occurs in severe asthma, pericardial constriction, or cardiac tamponade.
Peripheral pulses (See p36.) See p771 for arterial blood gas (ABG) sampling.
Fig 2.10 Arterial pulse waveforms.
Reproduced from Thomas J, et al. (eds). Oxford Handbook of Clinical Examina-tion and Practical Skills (2014), with permission from Oxford University Press.
43 History and examination The jugular venous pressure (JVP) The internal jugular vein acts as a capricious manometer of right atrial pressure. Ob-serve the height and the waveform of the pulse. JVP observations are often diffi cult. so do not be downhearted if the skill seems to elude you. Examine necks, and the patterns you see may slowly start to make sense—see fi g 2.11 for the local venous anatomy. Concomitantly palpate the arterial pulse to help decipher patterns.
The height Observe the patient at 45°, with their head turned slightly to the left and neck relaxed. Good lighting and correct positioning are key. Look for the right internal jugular vein as it passes just medial to the clavicular head of the sterno cleidomastoid up behind the angle of the jaw to the earlobes. The JVP is assessed by measuring the vertical height from the manubriosternal angle (not the sternal notch) to the top of the pulse. Pressure at zero (at the sternal angle) is 5cm, so add the height of the JVP with 5cm to obtain the right heart fi lling pressure in cm of water. A pressure above 9cm (4cm above the sternal angle at 45°) is elevated.
Is the pulse venous (and not arterial)? • Usually impalpable, and obliterated by fi nger pressure on the vessel. • Rises transiently with pressure on abdomen (abdominojugular refl ux) 2 or on liver (hepatojug ular refl ux), and alters with posture and respiration (disappears when patient sits from lying fl at). • Usually has a double pulse for every arterial pulse. See fi g 2.12.
Abnormalities of the JVP • Raised JVP with normal waveform: Fluid overload, right heart failure. • Fixed raised JVP with absent pulsation: SVC obstruction (p528). • Large a wave: Pulmonary hypertension, pulmonary stenosis. • Cannon a wave: When the right atrium contracts against a closed tricuspid valve, large ‘cannon’ a waves result. Causes—complete heart block, single chamber ven-tricular pacing, ventricular arrhythmias/ectopics. • Absent a wave: Atrial fi brillation. • Large v waves: Tricuspid regurgitation—look for earlobe movement. • Constrictive pericarditis: High plateau of JVP (which rises on inspiration—Kuss-maul’s sign) with deep x and y descents. • Absent JVP: When lying fl at, the jugular vein should be fi lled. If there is reduced circulatory volume (eg dehydration, haemorrhage) the JVP may be absent.
2 This sign was fi rst described by Pasteur in 1885 in the context of tricuspid incompetence.
Posterior auricular vein Superficial temporal vein Maxillary vein Facial vein Thyroid cartilage Anterior jugular vein Retromandibular vein External jugular vein Internal jugular vein Fig 2.11 The jugular venous system.
Fig 2.12 The jugular venous pressure wave. The JVP drops as the X descent during ventricular systole because the right atrium is no longer contracting. This means that the pressure in the right atrium is dropping and this is refl ected by the JVP.
After Clinical Examination, Macleod, Churchill and Aids to Undergraduate Medicine, J Burton, Churchill.
44 History and examination The heart sounds Listen systematically: sounds then murmurs. While listening, palpate the carotid artery: S1 is synchronous with the upstroke.
Heart sounds See fi g 2.13. The 1st and 2nd sounds are usually clear. Confi dent pro-nouncements about other sounds and soft murmurs may be diffi cult. Even senior colleagues disagree with one another about the more diffi cult sounds and murmurs.
The 1st heart sound (S1) Represents closure of mitral (M1) and tricuspid (T1) valves. Splitting in inspiration may be heard and is normal. • Loud S1 In mitral stenosis, because the narrowed valve orifi ce limits ventricular fi lling, there is no gradual decrease in fl ow towards the end of diastole. The valves are, therefore, at their maximum excursion at the end of diastole, and so shut rap-idly leading to a loud S1 (the ‘tapping’ apex). S1 is also loud if diastolic fi lling time is shortened, eg if the PR interval is short, and in tachycardia. • Soft S1 occurs if the diastolic fi lling time is prolonged, eg prolonged PR interval, or if the mitral valve leafl ets fail to close properly (ie mitral incompetence).
The intensity of S1 is variable in AV block, AF, and nodal or ventricular tachycardia.
The 2nd heart sound (S2) Represents aortic (A2) and pulmonary valve (P2) closure. The most important abnormality of A2 is softening in aortic stenosis. • A2 is said to be loud in tachycardia, hypertension, and transposition, but this is probably not a useful clinical entity. • P2 is loud in pulmonary hypertension and soft in pulmonary stenosis. • Splitting of S2 in inspiration is normal and is mainly due to the variation of right heart venous return with respiration, delaying the pulmonary component. • Wide splitting occurs in right bundle branch block (BBB), pulmonary stenosis, deep inspiration, mitral regurgitation, and VSD. • Wide fi xed splitting occurs in atrial septal defect (ASD). • Reversed splitting (ie A2 following P2, with splitting increasing on expiration) occurs in left bundle branch block, aortic stenosis, PDA (patent ductus arterio-sus), and right ventricular pacing. • A single S2 occurs in Fallot’s tetralogy, severe aortic or pulmonary stenosis, pulmonary atresia, Eisenmenger’s syndrome (p156), large VSD, or hypertension.
NB: splitting and P2 are heard best in the pulmonary area.
Additional sounds 3rd heart sound (S3) may occur just after S2. It is low pitched and best heard with the bell of the stethoscope. S3 is pathological over the age of 30yrs. A loud S3 occurs in a dilated left ventricle with rapid ventricular fi lling (mitral regurgitation, VSD) or poor LV function (post MI, dilated cardiomyopathy). In constrictive pericarditis or restrictive cardiomyopathy it occurs early and is more high pitched (‘pericardial knock’).
4th heart sound (S4) occurs just before S1. Always abnormal, it represents atrial contraction against a ventricle made stiff by any cause, eg aortic stenosis or hyper-tensive heart disease.
Triple and gallop rhythms A 3rd or 4th heart sound occurring with a sinus tachycar-dia may give the impression of galloping hooves. An S3 gallop has the same rhythm as ‘Ken-tucky’, whereas an S4 gallop has the same rhythm as ‘Tenne-ssee’. When S3 and S4 occur in a tachycardia, eg with pulmonary embolism, they may summate and appear as a single sound, a summation gallop.
An ejection systolic click is heard early in systole with bicuspid aortic valves, and if BP. The right heart equivalent lesions may also cause clicks.
Mid-systolic clicks occur in mitral valve prolapse (p144).
An opening snap precedes the mid-diastolic murmur of mitral (and tricuspid) steno-sis. It indicates a pliable (non-calcifi ed) valve.
Prosthetic sounds are caused by non-biological valves, on opening and closing: rum-bling sounds ≈ ball and cage valves (eg Starr–Edwards); single clicks ≈ tilting disc valve (eg single disc: Bjork Shiley; bileafl et: St Jude—often quieter). Prosthetic mitral valve clicks occur in time with S1, aortic valve clicks in time with S2.
45 History and examination Fig 2.13 The cardiac cycle. 46 History and examination Cardiac murmurs Always consider other symptoms and signs before auscultation and think: ‘What do I expect to hear?’ But don’t let your expectations determine what you hear.
Use the stethoscope correctly: remember that the bell is good for low-pitched sounds (eg mitral stenosis) and should be applied gently. The diaphragm fi lters out low pitches, making higher-pitched murmurs easier to detect (eg aortic regurgita-tion). NB: a bell applied tightly to the skin becomes a diaphragm.
Consider any murmur in terms of character, timing, loudness, area where loudest, radiation, and accentuating manoeuvres.
When in doubt, rely on echocardiography rather than disputed sounds. (But still enjoy trying to fi gure out the clinical conundrum!) Character and timing (See fi g 2.14.) • An ejection-systolic murmur (ESM, crescendo–decrescendo) usually originates from the outfl ow tract and waxes and wanes with the intraventricular pressures. ESMS may be innocent and are common in children and high-output states (eg tachycardia, pregnancy). Organic causes include aortic stenosis and sclerosis, pul-monary stenosis, and H(O)CM. • A pansystolic murmur (PSM) is of uniform intensity and merges with S2. It is usu-ally organic and occurs in mitral or tricuspid regurgitation (S1 may also be soft in these), or a ventricular septal defect (p156). Mitral valve prolapse may produce a late systolic murmur ± midsystolic click. • Early diastolic murmurs (EDMs) are high pitched and easily missed: listen for the ‘absence of silence’ in early diastole. An EDM occurs in aortic and, though rare, pul-monary regurgitation. If the pulmonary regurgitation is secondary to pulmonary hypertension resulting from mitral stenosis, then the EDM is called a Graham Steell murmur. • Mid-diastolic murmurs (MDMs) are low pitched and rumbling. They occur in mitral stenosis (accentuated presystolically if heart still in sinus rhythm), rheumatic fever (Carey Coombs’ murmur: due to thickening of the mitral valve leafl ets), and aortic regurgitation (Austin Flint murmur: due to the fl uttering of the anterior mitral valve cusp caused by the regurgitant stream).
Intensity All murmurs are graded on a scale of 1–6 (see table 2.4), though in practice diastolic murmurs, being less loud, are only graded 1–4. Intensity is a poor guide to the severity of a lesion—an ESM may be inaudible in severe aortic stenosis.
Area where loudest Though an unreliable sign, mitral murmurs tend to be loudest over the apex, in contrast to the area of greatest intensity from lesions of the aortic (right 2nd intercostal space), pulmonary (left 2nd intercostal space), and tricuspid (lower left sternal edge) valves.
Radiation The ESM of aortic stenosis classically radiates to the carotids, in contrast to the PSM of mitral regurgitation, which radiates to the axilla.
Accentuating manoeuvres • Movements that bring the relevant part of the heart closer to the stethoscope ac-centuate murmurs (eg leaning forward for aortic regurgitation, left lateral position for mitral stenosis). • Expiration increases blood fl ow to the left side of the heart and therefore accentu-ates left-sided murmurs. Inspiration has the opposite eff ect. • Valsalva manoeuvre (forced expiration against a closed glottis) decreases sys-temic venous return, accentuating mitral valve prolapse and H(O)CM, but softening mitral regurgitation and aortic stenosis. Squatting has exactly the opposite eff ect. Exercise accentuates the murmur of mitral stenosis.
Non-valvular murmurs A pericardial friction rub may be heard in pericarditis. It is a superfi cial scratching sound, not confi ned to systole or diastole. Continuous murmurs are present throughout the cardiac cycle and may occur with a patent ductus arteriosus, arteriovenous fi stula, or ruptured sinus of Valsalva.
47 History and examination The following grading is commonly used for murmurs—systolic murmurs from 1 to 6 and diastolic murmurs from 1 to 4, never being clinically >4/6.
Table 2.4 Grading of heart murmurs.
Grade Description 1/6 Very soft, only heard after listening for a while 2/6 Soft, but detectable immediately 3/6 Clearly audible, but no thrill palpable 4/6 Clearly audible, palpable thrill 5/6 Audible with stethoscope only partially touching chest 6/6 Can be heard without placing stethoscope on chest Grading intensity of heart murmurs • de Musset’s sign—head nodding in time with the pulse. • Müller’s sign—systolic pulsations of the uvula. • Corrigan’s sign—visible carotid pulsations. • Quincke’s sign—capillary nailbed pulsation in the fi ngers. • Traube’s sign—‘pistol shot’ femorals, a booming sound heard over the femorals. • Duroziez’s sign—to and fro diastolic murmur heard when compressing the femo-rals proximally with the stethoscope.
Eponymous signs of aortic regurgitation Fig 2.14 Typical waveforms of common heart murmurs.
Prosthetic valves: Created either from synthetic material (mechanical prosthesis) or from biological tissue (bioprosthesis). The choice of prosthesis is determined by the anticipated longevity of the patient and the patient’s ability to tolerate antico-agulation. Three mechanical valve designs exist: the caged ball valve, the tilting disc (single leafl et) valve, and the bileafl et valve. Tissue valves are made from porcine valves or bovine pericardium.
Prosthetic aortic valves: All types produce a degree of outfl ow obstruction and thus have an ESM. The intensity of this murmur increases as the valve fails. Ball and cage valves (eg Starr–Edwards) and tissue valves do close completely in diastole and so any diastolic murmur implies valve failure.
Prosthetic mitral valves: Ball and cage valves project into the left ventricle and can cause a low-intensity ESM as they interfere with the ejected stream. Tissue valves and bileafl et valves can have a low-intensity diastolic murmur. Consider any systolic murmur of loud intensity to be a sign of regurgitation and failure.
Prosthetic valve murmurs 48 History and examination The respiratory system: history Table 2.5 Presenting symptoms and questions to ask Presenting symptoms Direct questions Cough (see BOX ‘Charac-teristic coughs’) Duration? Character (eg barking/hollow/dry)? Nocturnal (≈asthma, ask about other atopic symptoms, ie eczema, hay fever)? Exacerbat-ing factors? Sputum (colour? How much?). Any blood/haemoptysis?
Haemoptysis (see table 2.6 and BOX ‘Haemop-tysis’) Always think about TB (recent foreign travel?) and mal ignancy (weight loss?). Mixed with sputum? (Blood not mixed with sputum suggests pulmonary embolism, trauma, or bleeding into a lung cav-ity. ) Melaena? (Occurs if enough coughed-up blood is swallowed.) Dyspnoea (see table 2.7 and BOX ‘Dyspnoea’ and p782) Duration? Steps climbed/distance walked before onset? NYHA clas-sifi cation (p135)? Diurnal variation (≈asthma)?
Ask specifi cally about circumstances in which dyspnoea occurs (eg occupational allergen exposure).
Hoarseness (OHCS p568) Eg due to laryngitis, recurrent laryngeal nerve palsy, Singer’s nodules, or laryngeal tumour.
Wheeze (p52) Fever/night sweats (p29) Chest pain (p94 & p784) SOCRATES (see p36), usually ‘pleuritic’ if respiratory (ie worse on inspiration?).
Stridor (see BOX ‘Stridor’) History Ask about current symptoms (table 2.5) and past history: pneumonia/bron-chitis; TB; atopy 3 (asthma/eczema/hay fever); previous CXR abnormalities; lung surgery; myopathy; neurological disorders. Connective tissue disorders, eg rheumatoid, SLE.
Drug history Respiratory drugs (eg steroids, bronchodilators)? Any other drugs, esp -ecially with respiratory SE (eg ACE inhibitors, cytotoxics, -blockers, amiodarone)?
Family history Atopy?
3 Emphysema? TB?
Social history Quantify smoking in ‘pack-years’ (20 cigarettes/day for 1 year = 1 pack-year). Occupational exposure (farming, mining, asbestos) has possible compen-satory implications. Pets at home (eg birds)? Recent travel/TB contacts?
Inspiratory sound due to partial obstruction of upper airways. Obstruction may be due to something within the lumen (eg foreign body, tumour, bilateral vocal cord palsy), within the wall (eg oedema from anaphylaxis, laryngospasm, tumour, croup, acute epiglottitis, amyloidosis), or extrinsic (eg goitre, oesophagus, lymph-adenopathy, post-op stridor, after neck surgery). It’s an emerg ency (p772) if gas exchange is compromised. NB: wheeze is an expiratory sound.
Characteristic coughs Coughing is relatively non-specifi c, resulting from irritation anywhere from the pharynx to the lungs. The character of a cough may, however, give clues as to the underlying cause: • Loud, brassy coughing suggests pressure on the trachea, eg by a tumour. • Hollow, ‘bovine’ coughing is associated with recurrent laryngeal nerve palsy. • Barking coughs occur in croup. • Chronic cough Think of pertussis, TB, foreign body, asthma (eg nocturnal). • Dry, chronic coughing may occur following acid irritation of the lungs in oe-sophageal refl ux, and as a side-eff ect of ACE inhibitors.
Do not ignore a change in character of a chronic cough; it may signify a new problem, eg infection, malignancy.
Stridor 3 Atopy implies predisposition to, or concurrence of, asthma, hay fever and eczema with production of specifi c IgE on exposure to common allergens (eg house dust mite, grass, cats).
49 History and examination Blood is coughed up, eg frothy, alkaline, and bright red, often in a context of known chest disease (vomited blood is acidic and dark).
Table 2.6 Respiratory causes of haemoptysis.
1 Infective TB; bronchiectasis; bronchitis; pneumonia; lung abscess; COPD; fungi (eg aspergillosis); viruses (from pneumonitis, cryo globulinaemia, eg with hepatitis viruses, HIV-associ-ated pneumocystosis, or MAI, p400). Helminths: paragon-imiasis; hydatid (p435); schistosomiasis.
2 Neoplastic Primary or secondary.
3 Vascular Lung infarction (PE); vasculitis (ANCA-associated; RA; SLE); hereditary haemorrhagic telangiectasia; AV malforma-tion; capillaritis.
4 Parenchymal Diff use interstitial fi brosis; sarcoidosis; haemosid erosis; Goodpasture’s syndrome; cystic fi brosis.
5 Pulmonary hypertension Idiopathic, thromboembolic, congenital cyanotic heart disease (p156), pulmonary fi brosis, bronchiectasis.
6 Coagulopathies Any—eg thrombocytopenia, p344; DIC; warfarin excess.
7 Trauma/foreign body Eg post-intubation, or an eroding implanted defi brillator.
8 Pseudo-haemoptysis Munchausen’s (p706); aspirated haematemesis; red pig ment (prodigiosin) from Serratia marcescens (Gram-negative bacteria) in sputum.7 Rare causes refuse to be classifi ed neatly: vascular causes may have infective origins, eg hydatid cyst may count as a foreign body, and infection, and vascular if it fi stulates with the aorta; ditto for infected (mycotic) aneurysm rupture, or TB aortitis. Infective causes entailing coag ulopathy: dengue; leptospirosis. In monthly haemoptysis, think of lung endometriosis.
: Haemoptysis may need treating in its own right, if mass ive (eg trauma, TB, hydatid cyst, cancer, AV malformation): call chest team, consider interventional radiology input (danger is drowning: lobe resection, endobronchial tamponade, or arterial embolization may be needed). Set up IVI, do CXR, blood gases, FBC, INR/APTT, crossmatch. If distressing, give prompt IV morphine, eg if inoperable malignancy.
Dyspnoea Subjective sensation of shortness of breath, often exacerbated by exertion. • Lung—airway and interstitial disease. May be hard to separate from cardiac causes; asthma may wake patient, and cause early morning dyspnoea & wheeze. • Cardiac—eg ischaemic heart disease or left ventricular failure (LVF), mitral ste-nosis, of any cause. LVF is associated with orthopnoea (dyspnoea worse on lying; ‘How many pillows?’) and paroxysmal nocturnal dyspnoea (PND; dyspnoea waking one up). Other features include ankle oedema, lung crepitations, and JVP. • Anatomical—eg diseases of the chest wall, muscles, pleura. Ascites can cause breathlessness by splinting the diaphragm, restricting its movement. • Others Any shocked patient may also be dyspnoeic (p790 & p607)—dyspnoea may be shock’s presenting feature. Also anaemia or metabolic acidosis caus-ing respiratory compensation, eg ketoacidosis, aspirin poisoning. Look for other clues—dyspnoea at rest unassociated with exertion, may be psychogenic: pro-longed hyperventilation causes respiratory alkalosis. This causes a fall in ion ized calcium leading to apparent hypocalcaemia. Features include peripheral and perioral paraesthesiae ± carpopedal spasm. Speed of onset helps diagnosis: Table 2.7 Aetiology of dyspnoea by timing of onset.
Acute Subacute Chronic Foreign body Pneumothorax (p749, fi g 16.43) Pulmonary embolus Acute pulmonary oedema Psychogenic Asthma Parenchymal disease, eg alveolitis pneumonia Eff usion Psychogenic COPD and chronic parenchymal diseases Non-respiratory causes, eg cardiac failure, anaemia Haemoptysis 50 History and examination Fig 2.18 Horner’s syndrome.
The respiratory system: examination 1 Begin by introducing yourself, obtaining consent to examine and position the patient appropriately: lying on a bed, sitting up at 45°. Expose them to the waist (for female patients, delay until examining the chest). Explain what you are doing throughout.
5 Face • Inspect: For signs of Horner’s (fi g 2.18), conjuncti-val pallor, central cyanosis (ask patient to stick out tongue), pursed lip breathing 3 Arms • Time pulse rate, with fi ngers still on the pulse, check respiratory rate (this can increase if the pa-tient is aware you are timing it)—and pattern (p53) • Bounding pulse (CO2 retention)? • Check blood pressure 2 Hands • Inspect: Tobacco staining (fi g 2.16), peripheral cyanosis, clubbing, signs of systemic disease (systemic scle-rosis, rheumatoid arthritis) • Asterixis: Ask the patient to hold their hands out and cock their wrists back 1 General inspection • Assess general state (ill/well/cachexic) • Look for clues (oxygen, inhalers, nebulizers, venturi mask) • Colour (pale, cyanosed (fi g 2.15), fl ushed) • Short of breath? Accessory muscle use? • Scars on chest wall?
Ask the patient to take a deep breath in, watch chest movement and symmetry, any coughing?
4 Neck • Trachea: Feel in sternal notch (fi g 2.17, deviated?), assess cricosternal distance in fi nger-breadths and feel for tracheal tug • Lymphadenopathy: From behind with patient sat forward palpate lymph nodes of head and neck • JVP: Raised in cor pulmonale, fi xed and raised in superior vena cava obstruction Fig 2.15 Cyanosis.
Fig 2.16 Tar stains.
Fig 2.17 Sternal notch.
Reproduced from Thomas et al. (eds), Oxford Handbook of Clini-cal Examination and Practical Skills (2014) with permission from Oxford University Press.
51 History and examination Fig 2.19 Placement of the hands for testing chest expan-sion: anchor with the fi ngers and leave the thumbs free-fl oating.
Reproduced from Thomas J, et al., Oxford Handbook of Clinical Examination and Practical Skills (2014) with permission from Oxford University Press.
7 Back of chest • Expansion • Tactile vocal fremitus • Percussion • Auscultation • Vocal resonance 8 To complete the examination • Palpate for sacral and ankle oedema (fi g 2.20) • Check peripheral pulses, observation chart for temperature and O2 sats • Examine the sputum pot and check PEFR 6 Front of chest • Apex beat. • Expansion: Ask patient to ‘breathe all the way out’, place hands as in fi g 2.19, ‘now a deep breath in’, and note distance of thumbs to midline, is expan-sion equal? Repeat with hands laid on upper chest. • Tactile vocal fremitus: Palpate the chest wall with your fi ngertips and ask the patient to repeat ‘99’, each time they feel your hand, comparing right to left. This is rarely used. • Percussion: Percuss over diff erent respiratory seg-ments, comparing right and left (see fi g 2.21, p53). • Auscultation: Ask patient to ‘take steady breaths in and out through your mouth’ and listen with diaphragm from apices to bases, comparing right and left (see table 2.8, p52). • Vocal resonance: Repeat auscultation, asking pa-tient to repeat ‘99’ each time they feel the stetho-scope. If marked resonance heard, repeat with asking patient to whisper ‘99’; if clearly heard this is termed ‘whispering pectoriloquy’ and is a sensi-tive sign for consolidation. Outside of exams, the choice of vocal resonance or tactile vocal fremitus is a personal preference. Many clinicians prefer vo-cal resonance as it provides more information than tactile vocal fremitus.
Fig 2.20 Ankle oedema. • Whispering pectoriloquy is a classic and specifi c sign of consolidation. • If you don’t adequately expose the chest you may miss small scars, eg from video thoracoscopy. • If you see Horner’s syndrome, check for wasting of the small muscles of the hand; see p702 and p708.
Top tips 52 History and examination The respiratory system: examination 2 General inspection ‘Comfortable at rest’ or unwell? Cachectic? Respiratory dis-tress? (if high negative intrathoracic pressures are needed to generate air entry). Stridor? Respiratory rate, breathing pattern (see BOX ‘Breathing patterns’). Look for chest wall and spine deformities (see p55). Inspect for scars of past surgery, chest drains, or radiotherapy (skin thickening, tattoos for radiotherapy). Chest wall move-ment: symmetrical? (if not, pathology on restricted side). Paradoxical respiration? (abdomen sucked in with inspiration; seen in diaphragmatic paralysis, see p502).
Hands Clubbing, peripheral cyanosis, tar stains, fi ne tremor (-agonist use), wast-ing of in trinsic muscles (T1 lesions, eg Pancoast’s tumour, p708). Tender wrists (hy-pertrophic pulmonary osteo arthropathy—cancer). Aster ixis (CO2 retention). Pulse: paradoxical (respiratory distress), bounding (CO2 retention).
Face Ptosis and constricted pupil (Horner’s syndrome, eg Pancoast’s tumour, p708)? Bluish tongue and lips (central cyanosis, p34)? Conjunctival pallor (anaemia)?
Neck Trachea: Central or displaced? (towards collapse or away from large pleural eff usion/tension pneumothorax; slight deviation to right is normal). Cricosternal dis-tance <3cm is hyperexpansion. Tracheal tug: descent of trachea with inspiration (se-vere airfl ow limitation). Lymphadenopathy: TB/Ca? JVP: in cor pulmonale.
Palpation Apex beat: Impalpable? (COPD/pleural eff usion/dextrocardia?) Ex-pansion: <5cm on deep inspiration is abnormal. Symmetry? Tactile vocal fremitus: implies consolidation.
Percussion Dull percussion note: Collapse, consolidation, fi brosis, pleural thicken-ing, or pleural eff usion (‘stony dull’). Cardiac dullness usually detectable over the left side. Liver dullness usually extends up to 5th rib, right mid-clavicular line; below this, resonant chest is a sign of lung hyperexpansion (eg asthma, COPD). Hyper-resonant percussion note: Pneumothorax or hyperinfl ation (COPD).
Table 2.8 Auscultation Breath sounds Description Pathology Vesicular Rustling quality Normal Bronchial breathing Harsh with gap between inspiration and expiration. Increased vocal reso-nance and whispering pectoriloquy Consolidation, localized fi bro-sis, above pleural/percardial eff usion (Ewart’s sign, p154) Diminished breath sounds Diffi cult to hear Pleural eff usion, pleural thickening, pneumothorax, bronchial obstruction, asthma, or COPD Silent chest Inaudible breath sounds Life-threatening asthma Wheeze (rhonchi) Air expired through narrow airways • Monophonic (single note, partial obstruction one airway) • Polyphonic (multiple notes, wide-spread airway narrowing) • Tumour occluding airway • Asthma, cardiac wheeze (LVF) Crackles (crepi-tations) Reopening of small airways on inspiration • Fine and late in inspiration • Coarse and mid inspiratory • Early inspiratory • Pulmonary oedema • Bronchiectasis • Small airway disease • Late/pan inspiratory • Disappear post cough • Alveolar disease • Insignifi cant Pleural rub Movement of visceral pleura over parietal when both are roughened (eg due to infl ammatory exudate) • Pneumonia • Pulmonary infarction Pneumothorax click Shallow left pneumo thorax between layers of parietal pleura overlying heart, heard during cardiac systole 53 History and examination Fig 2.21 The respiratory segments supplied by the segmental bronchi.
Further examination—sputum, temperature charts, O2 sats, PEFR: Inspect spu-tum and send suspicious sputum for microscopy (Gram stain and auramine/ZN stain, if indicated), culture, and cytology. • Black carbon specks suggests smoking: commonest cause of increased sputum. • Yellow/green sputum suggests infection, eg bronchiectasis, pneumonia. • Pink frothy sputum suggests pulmonary oedema. • Bloody sputum (haemoptysis) may be due to malignancy, TB, infection, or trau-ma, and requires investigation for these causes. See p49. • Clear sputum is probably saliva.
Sputum examination Hyperventilation: Tachypnoea (ie >20 breaths/min) or deep (hyper pnoea, ie tidal volume). Hyperpnoea is not unpleasant, unlike dyspnoea. It may cause respiratory alkalosis, hence paraesthesiae ± muscle spasm (Ca 2+). The main cause is anxiety: associated dizziness, chest tightness/pain, palpit ations, and panic. Rare causes: response to metabolic acidosis; brainstem lesions. • Kussmaul respiration is deep, sighing breaths in severe metabolic acidosis (blowing off CO2), eg diabetic or alcoholic ketoacidosis, renal impairment. • Neurogenic hyperventilation is produced by pontine lesions. • The hyperventilation syndrome involves panic attacks associated with hyper-ventilation, palpitations, dizziness, faintness, tinnitus, alarming chest pain/ tightness, perioral and peripheral tingling (plasma Ca 2+). Treatment: relaxation techniques and breathing into a paper bag (inspired CO 2 corrects the alkalosis).
NB: the anxious patient in A&E with hyperventilation and a respiratory alkalosis may actually be presenting with an aspirin overdose (p844).
Cheyne–Stokes breathing: Breaths get deeper and deeper, then shallower (±epi-sodic apnoea) in cycles. Causes—brainstem lesions or compression (stroke, ICP). If the cycle is long (eg 3min), the cause may be a long lung-to-brain circulation time (eg in chronic pulmonary oedema or cardiac output). It is enhanced by opioids.
Breathing patterns • Tachypnoea. • Nasal fl aring. • Tracheal tug (pulling of thyroid cartilage towards sternal notch in inspiration). • Use of accessory muscles (sternocleidomastoid, platysma, infrahyoid). • Intercostal, subcostal, and sternal recession. • Pulsus paradoxus (p42).
Signs of respiratory distress 54 History and examination The respiratory system: important presentations Some physical signs (fig 2.22).
(There may be bronchial breathing at the top of an effusion) PLEURAL EFFUSION Expansion: Percussion: (stony dull) Air entry: Vocal resonance: Trachea + mediastinum central (shift away from affected side only with massive effusions ≥ 1000mL) CONSOLIDATION Expansion Percussion note Vocal resonance Bronchial breathing ± coarse crackles (with whispering pectoriloquy) Trachea + mediastinum central SPONTANEOUS PNEUMOTHORAX/ EXTENSIVE COLLAPSE ( LOBECTOMY/ PNEUMONECTOMY) Expansion Percussion note Breath sounds Trachea + mediastinum shift towards the affected side TENSION PNEUMOTHORAX (See fig 16.43, p749 for chest X-ray image) Expansion Percussion note Breath sounds Trachea + mediastinum shift away from the affected side FIBROSIS Expansion Percussion note Breath sounds bronchial ± crackles Trachea + mediastinum central or pulled towards the area of fibrosis Fig 2.22 Physical signs on chest examination. 55 History and examination Fig 2.23 Pectus carinatum (pigeon chest). Prominent sternum, from lung hyperinfl a-tion while the bony thorax is still develop-ing, eg in chronic childhood asthma. Often seen with Harrison’s sulcus, a groove de-formity caused by indrawing of lower ribs at the diaphragm attachment site. This usually has little functional signifi cance in terms of respiration but can have signifi -cant psychological eff ects: see BOX.
Image courtesy of Prof Eric Fonkalsrud.
Fig 2.24 Pectus excavatum; the term for funnel or sunken chest. It is often asymptomatic, but may cause dis-placement of the heart to the left, and restricted ventilatory capacity ± mild air-trapping. Associ ations: scoliosis; Mar fan’s; Ehlers–Danlos syndrome.
Image courtesy of Prof Eric Fonkalsrud. • Barrel chest: AP diameter, tracheal descent and chest expansion , seen in chronic hyperinfl ation (eg asthma/COPD). • Pigeon chest (pectus carinatum): See fi g 2.23. • Funnel chest (pectus excavatum): Develop mental defect involving local ster-num depression (lower end). See fi g 2.24. • Kyphosis: ‘Hump back’ from AP thoracic spine curvature. • Scoliosis: Lateral curvature (OHCS p674); all of these may cause a restrictive ven-tilatory defect. Chest deformities Chest wall deformities such as pectus excavatum are quite common, often appearing during adolescent growth spurts. Exercise intolerance is the main symptom (from heart comp ression—consider CXR/CT). Indications for surgical correction (rarely needed): ≥2: a severe, symptomatic deformity; progression of deformity; paradoxical respiratory chest wall motion; pectus index >3.25 on CT; cardiac or lung compression; restrictive spirometry; cardiac pathology that might be from compression of the heart.
Psychological eff ects are interesting and not to be dismissed as their eff ects may be greater than any physical eff ects.8 Because these people hate exposing their chests they may become introverted, and never learn to swim, so don’t let them sink without trace. Be sympathetic, and remember Herr Minty, who inau-gurated Graham Greene’s theory of compensation: wherever a defect exists we must look for a compensating perfection to account for how the defect survives. In Minty’s case, although ‘crooked and yellow and pigeon-chested he had his deep refuge, the in exhaustible ingenuity of his mind.’ Herr Minty and his pigeon chest 56 History and examination The gastrointestinal system: history See table 2.9 for direct questions to ask regarding presenting symptoms.
Table 2.9 Presenting symptoms and questions to ask Presenting symptoms Direct questions Abdominal pain (see p57 and p606) SOCRATES (p36) Distension (see p57) Nausea, vomiting (see table 2.10) Timing? Relation to meals? Amount? Content (liquid, solid, bile, blood)? Frequency? Fresh (bright red)/dark/‘coff ee grounds’? Consider neoplasia (weight loss, dysphagia, pain, mel-aena?), NSAIDS/warfarin? Surgery? Smoking?
Haematemesis (pp256–7) Dysphagia (p250) Level? Onset? Intermittent? Progressive? Painful swallow (odynophagia)?
Indigestion/dyspepsia/refl ux (p252) Timing (relation to meals)?
Recent change in bowel habit Consider neoplasia (weight loss, dysphagia, pain, melaena?) Diarrhoea (p258), constipation (p260) Rectal bleeding (p629) or melaena (p246) Pain on defecation? Mucus? Fresh/dark/black? Mixed with stool/on surface/on paper/in the pan?
Appetite, weight change Intentional? Quantify. Dysphagia? Pain?
Jaundice (p272) Pruritus? Dark urine? Pale stools?
Past history Peptic ulcer disease, carcinoma, jaundice, hepatitis, blood transfusions, tattoos, previous operations, last menstrual period (LMP), dietary changes.
Drug history Especially steroids, NSAIDS, antibiotics, anticoagulants (eg clopidogrel with SSRI—see BOX ‘SSRIS and upper GI bleeding risk’).
Family history Irritable bowel syndrome (IBS), infl ammatory bowel disease (IBD), peptic ulcer disease, polyps, cancer, jaundice.
Social history Smoking, alcohol (quantify units/week), recreational drug use, travel history, tropical illnesses, contact with jaundiced persons, occupational exposures, sexual history, blood transfusions, surgery over-seas.
Vomiting History is vital. Associated symptoms and past medical history often indi-cate cause (table 2.10). Examine for dehydration, distension, tenderness, abdominal mass, succussion splash in children (pyloric stenosis), or tinkling bowel sounds (in-testinal obstruction).
Table 2.10 Causes of vomiting Gastrointestinal CNS Metabolic/endocrine • Gastroenteritis • Peptic ulceration • Pyloric stenosis • Intestinal obstruction • Paralytic ileus • Acute cholecystitis • Acute pancreatitis • Meningitis/encephalitis • Migraine • Intracranial pressure • Brainstem lesions • Motion sickness • Ménière’s disease • Labyrinthitis • Uraemia • Hypercalcaemia • Hyponatraemia • Pregnancy • Diabetic ketoacidosis • Addison’s disease Alcohol and drugs Psychiatric Others • Antibiotics • Opiates • Cytotoxics • Digoxin • Self-induced • Psychogenic • Bulimia nervosa • Myocardial infarction • Autonomic neuropathy • Sepsis (UTI; meningitis) How to remember the chief non-GI causes of vomiting? Try ABCDEFGHI: Acute kidney injury Addison’s disease; Brain (eg ICP); Cardiac (myocardial infarct); Diabetic ketoacidosis; Ears (eg labyrinthitis, Ménière’s disease); Foreign substances (alcohol; drugs, eg opiates); Gravidity (eg hyperemesis gravidarum); Hypercal-caemia/Hyponatraemia; Infection (eg UTI, meningitis).
57 History and examination Abdominal pain Character depends on underlying cause. Examples: irritation of the mucosa (acute gastritis), smooth muscle spasm (acute enterocolitis), capsular stretching (liver con-gestion in CCF), peritoneal infl ammation (acute appendicitis), and direct splanchnic nerve stimulation (retro peritoneal extension of tumour). The character (constant or colicky, sharp or dull), duration, and frequency depend on the mechanism of production. The location and distribution of referred pain depend on the anatomi-cal site. Time of occurrence and aggravating or relieving factors such as meals, defecation, and sleep also have special signifi cance related to the underlying disease process. The site of the pain may provide a clue: • Epigastric: Pancreatitis, gastritis/duodenitis, peptic ulcer, gallbladder disease, aortic aneurysm. • Left upper quadrant: Peptic ulcer, gastric or colonic (splenic fl exure) cancer, splenic rupture, subphrenic or perinephric abscess, renal (colic, pyelonephritis). • Right upper quadrant: Cholecystitis, biliary colic, hepatitis, peptic ulcer, colonic cancer (hepatic fl exure), renal (colic, pyelonephritis), subphrenic/perinephric ab-scess. • Loin: (lateral ⅓ of back between thorax and pelvis—merges with the fl ank, p565) Renal colic, pyelonephritis, renal tumour, perinephric abscess, pain referred from vertebral column. Causes of fl ank pain are similar (see index for fuller list). • Left iliac fossa: Diverticulitis, volvulus, colon cancer, pelvic abscess, infl am m atory bowel disease, hip pathology, renal colic, urinary tract infection (UTI), cancer in undescended testis; zoster—wait for the rash! (p454). Gynae: torsion of ovarian cyst, salpingitis, ectopic pregnancy. • Right iliac fossa pain: All causes of left iliac fossa pain plus appendicitis and Crohn’s ileitis, but usually excluding diverticulitis. • Pelvic: Urological: UTI, retention, stones. Gynae: menstruation, pregnancy, endome-triosis (OHCS p288), salpingitis, endometritis (OHCS p274), ovarian cyst torsion. • Generalized: Gastroenteritis, irritable bowel syndrome, peritonitis, constipation. • Central: Mesenteric ischaemia, abdominal aneurysm, pancreatitis.
Remember referred pain: Myocardial infarct epigastrium; pleural pathology.
Abdominal distension (masses and the ‘famous fi ve’ Fs) Enid Blyton’s Famous Five characters can generally solve any crime or diag nos tic problem using 1950s methodologies steeped in endless school holi days, copious confection-laden mid night feasts, and lashings of homemade ginger beer.
Let’s give them the problem of abdominal distension. The sweets and drinks used by the Famous Five actually contribute to the distension itself: fat, fluid, faeces, flatus, and fetus. If you think it far-fetched to implicate ginger beer in the genesis of fetuses, note that because it was homemade, like the fun, there was no limit to its intoxicating powers in those long-gone vintage summers. The point is to think to ask ‘When was your last period?’ whenever confronted by a distended abdomen.
Flatus will be resonant on percussion. Fluid will be dull, and can be from ascites (eg from malignancy or cirrhosis: look for shifting dullness), distended bladder (cannot get below it) or an aortic aneurysm (expansile). Masses can be pelvic (think of uter-ine fi broids or ovarian pathology) or tumours from colon, stomach, pancreas, liver, or kidney. Also see causes of ascites with portal hypertension (p604), hepatomegaly (p61), splenomegaly, and other abdominal masses (p604).
SSRIs have been associated with an increased risk of bleeding.9 SSRIs are thought to increase gastric acidity and serotonin is thought to play a role in platelet ag-gregation. This may lead to an increased risk of ulcers and bleeding, particularly when co-prescribed with anticoagulants and drugs aff ecting intestinal lining (eg NSAIDS). NICE10 recommends cautious concomitant use of SSRIs with anticoagu-lants or NSAIDS and recommends gastroprotection (eg PPI) for older patients tak-ing NSAIDs or aspirin.
SSRIS and upper GI bleeding risk 58 History and examination Gastrointestinal symptoms Faecal incontinence This is common in the elderly. Do your best to help, and get social services involved if concerned. Continence depends on many factors—mental function, stool (volume and consistency), anatomy (sphincter function, rectal distensibility, anorectal sensa-tion and refl exes). Defects in any area can cause loss of faecal continence.
Causes: Often multifactorial. Is it passive faecal soiling or urgency-related stool loss? Consider the following: • Sphincter dysfunction: • Vaginal delivery is the commonest cause due to sphincter tears or pudendal nerve damage. • Surgical trauma, eg following procedures for fi stulas, haemorrhoids, fi ssures. • Impaired sensation—diabetes, MS, dementia, any spinal cord lesions (consider cord compression if acute faecal incontinence). • Faecal impaction—overfl ow diarrhoea, extremely common, especially in the el-derly, and very easily treated. • Idiopathic—although there is often no clear cause found, especially in elderly women, this is usually multifactorial, including a combination of poor sphincter tone and pudendal damage leading to poor sensation.
Assessment: Do PR (overfl ow incontinence? poor tone?) and assess neurological function of legs, particularly checking sensation.
Refer to a specialist (esp. if rectal prolapse, anal sphincter injury, lumbar disc disease, or alarm symptoms for colon ca exist). Consider anorectal manometry, pelvic ultra-sound or MRI, and pudendal nerve testing may be needed.
Treat according to cause and to promote dignity: Never let your own embarrassment stop you from off ering help. Knowledge and behaviour are key factors: • Ensure toilet is in easy reach. Plan trips in the knowledge of toilet locations. • Obey call-to-stool impulses (esp. after meal, ie the gastro-colic refl ex). • Ensure access to latest continence aids and advice on use, refer to continence nurse specialist for assessment. • Pelvic fl oor rehabilitation: eg can help faecal incontinence, squeeze pressure, and maximal tolerated volume. • Loperamide 2–4mg 45min before social engagements may prevent accidents out-side home. An anal cotton plug may help isolated internal sphincter weakness. Skin care. Support agencies.
If all sensible measures fail, try a brake-and-accelerator approach: enemas to empty the rectum (twice weekly) and codeine phosphate, eg 15mg/12h, on non-enema days to constipate. It’s not a cure, but makes the incontinence manageable.
Flatulence Normally, 400–1300mL of gas is expelled PR in 8–20 discrete (or indiscrete) episodes per day. If this, with any eructation (belching) or distension, seems excessive to the patient, they may complain of fl atulence. Eructation occurs in hiatus hernia—but most patients with ‘fl atulence’ have no GI disease. Air swal lowing (aerophagy) is the main cause of fl atus; here N2 is the chief gas. If fl atus is mostly methane, H2 and CO2, then fermentation by bowel bacteria is the cause, and reducing carbohydrate intake (eg less lactose and wheat) may help.
59 History and examination Tenesmus This is a sensation in the rectum of incomplete emptying after defecat ion. It’s com-mon in irritable bowel syndrome (p266), but can be caused by tumours.
Regurgitation Gastric and oesophageal contents are regurgitated eff ortlessly into the mouth— without contraction of abdominal muscles and diaphragm (so distinguishing it from true vomiting). It may be worse on lying fl at, and can cause cough and nocturnal asthma. Regurgitation is rarely preceded by nausea, and when due to gastro-oe-sophageal refl ux, it is often associated with heartburn. An oesophageal pouch may cause regurgitation. Very high GI obstructions (eg gastric volvulus, p611) cause non-productive retching rather than true regurgitation.
Steatorrhoea These are pale stools that are diffi cult to fl ush, and are caused by malabsorption of fat in the small intestine and hence greater fat content in the stool.
Causes: Ileal disease (eg Crohn’s or ileal resection), pancreatic disease, and obstruc-tive jaundice (due to excretion of bile salts from the gallbladder).
Dyspepsia Dyspepsia and indigestion (p252) are broad terms. Dyspepsia is defi ned as one or more of post-prandial fullness, early satiety (unable to fi nish meal), and/or epigastric or retrosternal pain or burning. ‘Indigestion’ reported by the patient can refer to dyspepsia, bloating, nausea, and vomiting. Try to fi nd out exactly what your pa-tient means and when these symptoms occur in relation to meals, eg the classic symptoms of peptic ulcers occur 2–5 hours after a meal and on an empty stomach. Look for alarm symptoms (see p248); these have high negative predictive value. If all patients with dyspepsia undergo endoscopy, <33% have clinically signifi cant fi ndings.11 Myocardial infaction may present as ‘indigestion’.
Halitosis Halitosis (fetor oris, oral malodour) results from gingivitis (rarely severe enough to cause Vincent’s angina, p712), metabolic activity of bacteria in plaque, or sulfi de-yield-ing food putrefaction, eg in gingival pockets and tonsillar crypts. Patients can often be anxious and convinced of halitosis when it is not present (and vice versa!).
Contributory factors: Smoking, drugs (disulfi ram; isosorbide), lung disease, hangovers.
: Try to eliminate anaerobes: • Good dental hygiene, dental fl oss, tongue scraping. • 0.2% aqueous chlorhexidine gluconate.
The very common halitosis arising from the tongue’s dorsum is secondary to over-populated volatile sulfur compound-producing bacteria. Locally retained bacteria metabolize sulfur-containing amino acids to yield volatile ( smelly) hydrogen sulfi de and methylmercaptane, which perpetuate periodontal disease. At night and be-tween meals, conditions are optimal for odour production—so eating regularly may help. Treat by mechanical cleansing/scraping using tongue brushes or scrapes plus mouthwashes. Oral care products containing metal ions, especially Zn, inhibit odour formation, it is thought, because of affi nity of the metal ion to sulfur. It is possible to measure the level of volatile sulfur-containing compounds in the air in the mouth directly by means of a portable sulfi de monitor.
60 History and examination Examination of the abdomen Begin by introducing yourself, obtaining consent to examine, and position the patient appropriately; lie the patient down as fl at as possible, ideally exposing from ‘nipples to knees’. In practice, keep the groin covered and examine separately for hernias, etc. Fig 2.28 Cervical and supraclavicular nodes.
Reproduced from Thomas J, et al. (eds). Oxford Handbook of Clinical Examination and Practi-cal Skills (2014), with permission from Oxford University Press.
5 Face • Skin and eyes: Jaundice, conjunc-tival pallor, Kayser–Fleischer rings, xanthelasma (see fi g 2.29), sunken eyes (dehydration) • Mouth: Angular stomatitis, pigmenta-tion, telangiectasia, ulcers, glossitis 1 General inspection • Assess general state (ill/well/cachexic) • Clues (vomit bowl, stoma bags, cath-eter, urine colour) • Colour (pale, jaundiced, uraemic) • Body mass index? • Scars on the abdomen? Stomas (fi g 2.25)?
Ask the patient to lift their head off the bed, or cough, looking for bulges, disten-sion or pain.
4 Neck • Examine cervical and supraclavicular lymph nodes (see fi g 2.28) • JVP raised in fl uid overload (renal dys-function, liver dysfunction), tricuspid regurgitation (may cause pulsatile he-patomegaly) • Scars from tunnelled haemodialysis lines (see p303) or other central venous access 3 Arms • Check pulse and blood pressure • Look in the distribution of the SVC (arms, upper chest, upper back) for spider naevi (fi g 2.27) • Check for track marks, bruising, pigmentation, scratch marks, arterio-venous fi stulae (see p303 for signs seen in patients with chronic kidney disease) 2 Hands • Inspect: Clubbing, koilonychia, leuconychia, Muehrcke’s lines, palmar erythema, Dupuytren’s contracture (fi g 2.26), pigmentation of the palmer creases • Asterixis: (See p50) Fig 2.25 Stoma.
Reproduced from MacKay G, et al. (eds). Oxford Specialist Handbook of Colorectal Surgery (2010), with permission from Oxford University Press.
Fig 2.26 Dupuytren’s contracture.
Fig 2.27 Spider naevi.
Fig 2.29 Xanthelasma.
61 History and examination 7 To complete the examination • Palpate for ankle oedema, examine the hernial orifi ces, external genitalia, and perform a rectal examination. Check the observation chart and dipstick urine.
6 Abdomen Inspection: • Scars—previous surgery, transplant, stoma • Visible masses, hernias, or pulsation of AAA • Visible veins suggesting portal hypertension • Gynaecomastia, hair loss, acanthosis nigricans Palpation: Squat by the bed so that the patient’s abdomen is at your eye level. Ask if there is any pain and examine this part last. Watch the patient’s face for signs of discomfort. Palpate the entire abdomen (see p565): • Light palpation—if this elicits pain, check for re-bound tenderness. Any involuntary tension in mus-cles (‘guarding’)? See p606. • Deep palpation—to detect masses. • Liver—using the radial border of the index fi nger aligned with the costal margin start palpation from the RIF. Press down and ask patient to take a deep breath. Continue upwards towards the costal mar-gin until you feel the liver edge. • Spleen—start palpation in RIF and work towards the left costal margin asking the patient to take a deep breath in and feeling for edge of the spleen. • Kidneys—for each kidney: place one hand behind patient’s loin, press down on the abdomen with your other hand and ‘ballot’ the kidney up with your lower hand against your upper hand (fi g 2.30). Un-less slim or pathology present, may not be palpable. • Aorta—palpate midline above umbilicus, is it ex-pansile? (fi g 2.49, p79).
Percussion: • Liver—percuss to map upper & lower border of liver. • Spleen—percuss from border of spleen as palpat-ed, around to mid-axillary line. • Bladder—if enlarged, suprapubic region will be dull. • Ascites—shifting dullness: percuss centrally to lat-erally until dull, keep your fi nger at the dull spot and ask patient to lean onto opposite side. If the dull-ness was fl uid, this will now have moved by gravity and the previously dull area will be resonant.
Auscultation: • Bowel sounds—listen just below the umbilicus. • Bruits—listen over aorta and renal arteries (either side of midline above umbilicus).
Fig 2.30 Ballottement of the kidneys.
Reproduced from Thomas J, et al. (eds). Oxford Handbook of Clini-cal Examination and Practical Skills (2007), with permission from Oxford University Press. • If you think there is a spleen tip, roll the patient onto their right side and feel again. This tips the spleen forward and allows you to percuss around to the back. • Check the back for spider naevi, even if the chest appears clear (look for nephrec-tomy scars as you do this). • Light palpation really should be light, to check for tenderness and very large masses, watching the patient’s face throughout. • If you suspect voluntary guarding, use the diaphragm of your stethoscope to as-sist with palpation and distract the patient who will think you are auscultating!
Top tips 62 History and examination The gastrointestinal system: examination Inspection Does your patient appear comfortable or in distress? Look for abnormal contours/ distension. Tattoos? Cushingoid appearance may suggest steroid use post-trans-plant or IBD. Inspect (and smell) for signs of chronic liver disease: • Hepatic fetor on breath (p274). • Gynaecomastia. • Clubbing (rare). • Purpura (purple-stained skin, p344). • Scratch marks. • Muscle wasting. • Spider naevi (fi g 2.27, p60). • Palmar erythema. • Jaundice. • Asterixis.
Look for signs of malignancy (cachexia, masses), anaemia, jaundice, Virchow’s node. From the end of the bed inspect the abdomen for: • Visible pulsation (aneurysm, p654). • Peristalsis. • Scars. • Masses. • Striae (stretch marks, eg pregnancy). • Distension. • Genitalia. • Herniae.
If abdominal wall veins look dilated, assess direction of fl ow. In inferior vena caval (IVC) obstruction, below the umbilicus blood fl ows up; in portal hypertension (caput medusae), fl ow radiates out from the umbilicus.
The cough test: While looking at the face, ask the patient to cough. If this causes abdominal pain, fl inching, or a protective movement of hands towards the abdomen, suspect peritonitis.
Hands Clubbing, leuconychia (whitening of the nails due to hypoalbuminaemia), koilo-nychia (‘spooning’ of the nails due to iron, B12, or folate defi ciency), Muehrcke’s lines (transverse white lines due to hypoalbuminaemia), blue lunulae (bluish discoloura-tion seen in Wilson’s disease). Palmar erythema (chronic liver disease, pregnancy), Dupuytren’s contracture (thickening and fi brous contraction of palmar fascia (see fi g 2.26, p60; alcoholic liver disease)). Hepatic fl ap/asterixis (hepatic encephalopa-thy, uraemia from renal disease), check pulse and respiratory rate (infection/sepsis?), palpate for AV fi stulae in the forearm (haemodialysis access in renal failure).
Face Assess for jaundice, anaemia, xanthelasma (PBC, chronic obstruction), Kayser–Fleis-cher rings (green-yellow ring at corneal margin seen in Wilson’s disease). Inspect mouth for angular stomatitis (thiamine, B12, iron defi ciency), pigmentation (Peutz– Jeghers syndrome, p709, fi g 15.14), telangiectasia (Osler–Weber–Rendu syndrome/ hereditary haemorrhagic telangiectasia, p709, fi g 15.12), ulcers (IBD), glossitis (iron, B12, or folate defi ciency).
Cervical lymph nodes Palpate for enlarged left supraclavicular lymph node (Virchow’s node/Troisier’s sign) (gastric carcinoma?).
Abdomen Inspect: Look around to the fl anks for nephrectomy scars.
Palpate: Note any masses, tenderness, guarding (involuntary tensing of abdominal muscles—pain or fear of it), or rebound tenderness (greater pain on removing hand than on gently depressing abdomen—peritoneal infl ammation); Rovsing’s sign (ap-pendicitis, p608); Murphy’s sign (cholecystitis, p634). Palpating the liver: Assess size (see BOX ‘Causes of hepatomegaly’), regularity, smoothness, and tenderness. Pulsa-tile (tricuspid regurgitation)? The scratch test is another way to fi nd the lower liver edge (if it is below the costal margin): start with diaphragm of stethoscope at right costal margin. Gently scratch the abdominal wall, starting in the right lower quad-rant, working towards the liver edge. A sharp increase in trans mission of the scratch is heard when the border of the liver is reached. Palpating the spleen: If suspect splenomegaly but cannot detect it, assess patient in the right lateral position with your left hand pulling forwards from behind the rib cage. Palpating the kidneys: See fi g 2.30, p61. May be non palpable unless slim. Enlarged? Nodular? Palpating the aorta: Normally palpable transmitted pulsation in thin individuals.
63 History and examination Percussion Confi rm the lower border and defi ne the upper border of the liver and spleen (dull in the mid-axillary line in the 10th intercostal space). Percuss all regions of abdomen. If this induces pain, there may be peritoneal infl ammation below (eg an infl amed ap-pendix). Some experts percuss fi rst, before palpation, because even anxious patients do not expect this to hurt—so, if it does hurt, this is a very valuable sign. Percuss for the shifting dullness of ascites (p61 & p604) but ultrasound is a more reliable way of detecting ascites.
Auscultation Bowel sounds: absence implies ileus; they are enhanced and tink ling in bowel ob-struction. Listen for bruits in the aorta, renal and femoral arteries.
Further examination Check for hernias (p612), perform a PR examination see BOX ‘Examination of the rectum and anus’. • Cannot get above it (ribs overlie the upper border of the spleen). • Dull to percussion (kidney is usually resonant because of overlying bowel). • Moves towards RIF with inspiration (kidney tends to move downwards). • May have palpable notch on its medial side.
Features of the spleen diff erentiating it from an enlarged kidney (For hepatosplenomegaly, see p604.) Malignancy: Metastatic or primary (usually craggy, irregular edge).
Hepatic congestion: Right heart failure—may be pulsatile in tricuspid incompe-tence, hepatic vein thrombosis (Budd–Chiari syndrome, p696).
Anatomical: Riedel’s lobe (normal variant).
Infection: Infectious mononucleosis (glandular fever), hepatitis viruses, malaria, schistosomiasis, amoebic abscess, hydatid cyst.
Haematological: Leukaemia, lymphoma, myeloproliferative disorders (eg my-elofi brosis), sickle-cell disease, haemolytic anaemias.
Others: Fatty liver, porphyria, amyloidosis, glycogen storage disorders.
Splenomegaly • Abnormally large spleen.
Causes: See p604. If massive, think of: chronic myeloid leukaemia, myelofi brosis, malaria (or leishmaniasis).
Causes of hepatomegaly It is necessary to have a chaperone present for the examination. Explain what you are about to do. Make sure curtains are pulled. Have the patient lie on their left side, with knees brought up towards the chest. Use gloves and lubricant. Part the buttocks and inspect the anus: •A gaping anus suggests a neuropathy or megarectum. •Symmetry (a tender unilateral bulge suggests an abscess). •Pro-lapsed piles. •A subanodermal clot may peep out. •Prolapsed rectum (descent of >3cm when asked to strain, as if to pass a motion). •Anodermatitis (from frequent soiling). The anocutaneous refl ex tests sensory and motor innervation—on lightly stroking the anal skin, does the external sphincter briefl y contract?
Press your index fi nger against the side of the anus. Ask the patient to breathe deeply and insert your fi nger slowly. Feel for masses (haemorrhoids are not palpable) or impacted stool. Twist your arm so that the pad of your fi nger is feeling anteriorly. Feel for the cervix or prostate. Note consistency, size, and symmetry of the prostate. If there is faecal incontinence or concern about the spinal cord, ask the patient to squeeze your fi nger and note the tone. This is best done with your fi nger pad facing posteriorly. Note stool or blood on the glove and test for occult blood.
Wipe the anus. Consider proctoscopy (for the anus) or sigmoidoscopy (which mainly inspects the rectum).
Examination of the rectum and anus 64 History and examination The neurological system: history History This should be taken from the patient and if possible from a close friend or relative as well for corroboration/discrepancies. The patient’s memory, perception, or speech may be aff ected by the disorder, making the history diffi cult to obtain. Note the progression of the symptoms and signs: gradual deterioration (eg tumour) vs intermittent exacerbations (eg multiple sclerosis) vs rapid onset (eg stroke). Ask about age, occupation, and ethnic origin. Right- or left-hand dominant?
Presenting symptoms • Headache: (p456 & p780.) Diff erent to usual headaches? Acute/chronic? Speed of onset? Single/recurrent? Unilateral/bilateral? Associated symptoms (eg aura with migraine, p458)? Any meningism (p822)? Worse on waking (ICP)? Decreased conscious level? Take a ‘worst-ever’ headache very seriously. (See p749.) • Muscle weakness: (p466.) Speed of onset? Muscle groups aff ected? Sensory loss? Any sphincter disturbance? Loss of balance? Associated spinal/root pain? • Visual disturbance: (OHCS p410.) eg blurring, double vision (diplopia), photophobia, visual loss. Speed of onset? Any preceding symptoms? Pain in eye? • Change in other senses: Hearing (p464), smell, taste? Abnormalities are not al-ways due to neurological disease, consider ENT disease. • Dizziness: (p462.) Illusion of surroundings moving (vertigo)? Hearing loss/tinnitus? Any loss of consciousness? Positional? • Speech disturbance: (p86.) Diffi culty in expression, articulation, or comprehension (can be diffi cult to determine)? Sudden onset or gradual? • Dysphagia: (p250.) Solids and/or liquids? Intermittent or constant? Diffi culty in coordination? Painful (odynophagia)? • Fits/faints/‘funny turns’/involuntary movements: (p468.) Frequency? Durat ion? Mode of onset? Preceding aura? Loss of consciousness? Tongue biting? Inconti-nence? Any residual weakness/confusion? Family history? • Abnormal sensations: Eg numbness, ‘pins & needles’ (paraesthesiae), pain, odd sensations. Distribution? Speed of onset? Associated weakness? • Tremor: (p65.) Rapid or slow? Present at rest? Worse on deliberate movement? Taking -agonists? Any thyroid problems? Any family history? Fasciculations?
Cognitive state If there is any doubt about the patient’s cognition, cognitive testing should be undertaken. There are a number of tools including MMSE (subject to strict copyright), GPCOG, TYM, and 6-CIT. The Abbreviated Mental Test Score (AMTS) is a commonly used screening questionnaire for cognitive impairment:12 1 Tell patient an address to recall at the end (eg 42 West Street) A score of ≤6 suggests poor cognit-ion, acute (delirium), or chronic (de-mentia). AMTS correlates well with the more detailed Mini-Mental State Examination (MMSE™) NB: deaf, dys-phasic, depressed, and uncoop erative patients, as well as those who do not understand English, will also get low scores.13 2 Age 3 Time (to nearest hour) 4 What year is it?
5 Recognize 2 people (eg doctor & nurse) 6 Date of birth 7 Dates of the Second World War 8 Name of current monarch/prime minister 9 Where are you now? (Which hospital?) 10 Count backwards from 20 to 1 Past medical history Ask about meningitis/encephalitis, head/spine trauma, sei-zures, previous operations, risk factors for vascular disease (p470, AF, hypertension, hyperlipidaemia, diabetes, smoking), and recent travel, especially exotic destina-tions. Is there any chance that the patient is pregnant (eclampsia, OHCS p48)?
Drug history Any anticonvulsant/antipsychotic/antidepressant medication? Any psychotropic drugs (eg ecstasy)? Any medication with neurological side-eff ects (eg isoniazid which can cause a peripheral neuropathy)?
Social and family history What can the patient do/not do, ie activities of daily living (ADLs)? What’s the Barthel Index score? Any family history of neurological or psychiatric disease? Any consanguinity? Consider sexual history, eg syphilis.
65 History and examination Cramp This is painful muscle spasm. Leg cramps are com mon at night or after heavy exer-cise, and in patients with renal impairment or on dialysis. Cramp can signify salt de-pletion, and rarely: muscle ischaemia (claudication, DM), myopathy (McArdle, p704), or dystonia (writer’s cramp, p469). Forearm cramps suggest motor neuron disease. Night cramps may respond to quinine bisulfate 300mg at night PO.
Drugs causing cramp: Diuretics (? from K +), domperidone, salbutamol/terbutaline IVI, ACE-i, telmisartan, celecoxib, lacidipine, ergot alkaloids, levothyroxine.
Paraesthesiae ‘Pins and needles’, numbness/tingling, which can hurt or ‘burn’ (dysaes th esia).
Causes: Metab olic, Ca 2+ (perioral); P aCO2; myxoedema; neurotoxins (tick bite; sting). Vascular, arterial emboli; Raynaud’s; DVT; high plasma viscosity. Anti body-mediated, paraneo-plastic; SLE; ITP. Infection, rare: Lyme; rabies. Drugs, ACE-i. Brain,thalamic/parietal le-sions. Cord, MS; myelitis/HIV; B12; lumbar fract ure. Plexopathy/mono neuropathy, see p502, cervical rib; carpal tun nel; sciatica. Peripheral neuropathy, glove & stocking, p504, eg DM; CKD. If paroxysmal, migraine; epilepsy; phaeochromocytoma. If wander-ing, take travel history, consider infection, eg strongyloides.
Tremor Tremor is rhythmic oscillation of limbs, trunk, head, or tongue. Three types: 1 Resting tremor—worst at rest—eg from parkinsonism (±bradykinesia and rigidity; tremor is more resistant to treatment than other symptoms). It is usually a slow tremor (frequency of 3–5Hz), typically ‘pill-rolling’ of the thumb over a fi nger.
2 Postural tremor—worst if arms are outstretched. Typically rapid (8–12Hz). May be exaggerated physiological tremor (eg anxiety, hyperthyroidism, alcohol, drugs), due to brain damage (eg Wilson’s disease, syphilis) or benign essential tremor (BET). This is often familial (autosomal dominant) tremor of arms and head pre-senting at any age. Cogwheeling may occur but there is no brady kinesia. It is sup-pressed by alcohol, and patients may self-medicate rather than admit problems. Rarely progressive (unless onset is unilateral). Propranolol (40–80mg/8–12h PO) can help, but not in all patients.
3 Intention tremor—worst on movement, seen in cerebellar disease, with past-pointing and dysdiadochokinesis (see p499). No eff ective drug has been found.
Facial pain CNS causes: Migraine, trigeminal, or glossopharyngeal neuralgia (p457) or from any other pain-sensitive structure in the head or neck. Post-herpetic neuralgia: nasty burning-and-stabbing pain involves dermatomal areas aff ected by shingles (p404); it may aff ect cranial nerves V and VII in the face. It all too often becomes chronic and intractable (skin aff ected is exquisitely sensitive). Treatment is hard. Always give strong psychological support. Transcutaneous nerve stimulation, capsaicin oint-ment, and infi ltrating local anaesthetic are tried. Neuropathic pain agents, such as amitriptyline, eg 10–25mg/24h at night, or gabapentin (p504) may help. NB: famciclo-vir or valaciclovir given in acute shingles may duration of neuralgia.14 Vascular and non-neurological causes: • Neck—cervical disc pathology. • Bone/sinuses—sinusitis; neoplasia. • Eye—glaucoma; iritis; orbital cellulitis; eye strain; AVM. • Temporomandibular joint—arthritis or idiopathic dysfunction (common). • Teeth/gums—caries; broken teeth; abscess; malocclusion. • Ear—otitis media; otitis externa. • Vascular/vasculitis—arteriovenous fi stula; aneurysm; or AVM at the cerebellopon-tine angle; giant cell arteritis; SLE.
66 History and examination Neurological examination of the upper limbs The neurological system is usually the most daunting examination, so learn at the bedside from a senior colleague, preferably a neurologist. Keep practising. Be aware that books present ideal situations: often one or more signs are equivocal or even contrary to expectation; consider signs in the context of the history and try re-ex-amining the patient, as signs may evolve over time. The only essential point is to distinguish whether weakness is upper (UMN) or lower (LMN) motor neuron (p446).
Position the patient comfortably, sitting up at 45° and with arms exposed. The order of examination should be Inspection, Tone, Power, Refl exes, Coordination, Sensation (fi g 2.31). 3 Power Direct patient to adopt each position and follow commands while you as the examiner stabilize the joint above and resist movements as appropriate to grade power (see BOX ‘Muscle weakness grading’ on p446). Test each muscle group bilaterally before moving on to the next position. See p452–3 for myotomes. • ‘Shrug your shoulders and don’t let me push down; push your arms out to the side against me; try to pull them back in.’ • ‘Hold your arms up like this and pull me towards you, now push me away.’ • ‘Hold your hand out fl at, don’t let me push it down; now don’t let me push it up.’ • Off er the patient two (crossed) fi ngers of yours and ask them to ‘squeeze my fi ngers.’ • Ask patient to ‘spread your fi n gers and stop me pushing them back together’, then hand the pa-tient a piece of paper to grip between two fi ngers. You as the examiner should grip the paper with your corresponding fi ngers while asking patient to ‘grip the paper and don’t let me pull it away.’ 2 Tone Ask patient to ‘relax/go fl oppy like a rag-doll’. Ask if patient has any pain in hands/arm/shoulder before passively fl exing and extending limb while also pro-nating and supinating the forearm. Any spasticity or rigidity?
1 General inspection Abnormal posturing, asymmetry, abnormal move-ments (fasciculation/tremor/dystonia/athetosis), muscle wasting (especially small muscles of the hand)—symmetrical/asymmetrical? Local/general?
67 History and examination • Use the tendon hammer like a pendulum, let it drop, don’t grip it too tightly. • Ensure you are testing light touch, not stroke sensation.
Top tips 6 Sensation • Light touch: Use cotton wool, touch (not rub) it to sternum fi rst—‘this is what it should feel like, tell me where you feel it and if it feels diff erent’. Pro-ceed to test with cotton wool in all dermatomes (see p454), comparing left and right. • Pin prick: Repeat as above using a neurological pin, asking patient to tell you if it feels sharp or dull. • Temperature: Repeat as above, alternating hot and cold probes. Can the patient tell hot from cold? • Vibration: Using a 128Hz tuning fork (128 vibrate!) confi rm with patient that they ‘can feel a buzzing’ when you place the tuning fork on their sternum. Proceed to test at the most distal bony promi-nence and move proximally by placing the buzzing fork on the bony prominence, then stopping it with your fi ngers. Ask the patient to tell you when the buzzing stops. • Proprioception: With the patient’s eyes closed grasp distal phalanx of the index fi nger at the sides, not on top. Stabilize the rest of the fi nger. Flex and extend the joint, stopping at intervals to ask whether the fi nger tip is up or down.
5 Coordination Holding your fi nger in front of the patient instruct ‘touch my fi nger then your nose…as fast as you can’. Look for intention tremor, and ‘past pointing’. • Test for dysdiadokokinesis: ask patient to repeatedly pronate and supinate forearm, tapping hands each time. Test both limbs. You may have to demonstrate. Failure to perform rapidly alternating movements is dysdiadokokinesis. • Test for pronator drift: with patient’s eyes closed and arms outstretched, tap down on their up-facing palms and look for a failure to maintain supination.
4 Refl exes For each refl ex, test right, then left and com-pare. If absent, attempt to elicit with ‘rein-forcement’ by asking patient to clench their teeth on a count of three, at which time you strike (Jendrassik manoeuvre). Are refl exes absent/present (with reinforcement)/nor-mal/brisk/exaggerated? •Biceps (C5,6) •Tri-ceps (C7) •Supinator (C6). Fig 2.31 Sensory dermatomes.
Reproduced from Harrison (ed) Revision Notes for MCEM Part A (2011), with permission from Oxford University Press.
68 History and examination 1 General inspection and gait Gait: Ask patient to walk a few metres, turn, and walk back to you. Note use of walking aids, symmetry, size of paces, arm swing. Ask patient to ‘walk heel-to-toe as if on a tightrope’ to exaggerate any instability. Ask patient to walk on tiptoes, then on heels. Inability to walk on tiptoes indicates S1 or gastroc nemius lesion. Inability to walk on heels indicates L4,5 lesion or foot drop.
Romberg’s test: Ask patient to stand unaided with arms by their sides and close their eyes (be ready to support them). If they sway/lose balance the test is positive and indicates posterior column disease/sensory ataxia.
Inspect: Abnormal posturing, muscle wasting, fasciculation (LMN lesion?), deformities of the foot (eg pes cavus of Frie-dreich’s ataxia or Charcot–Marie–Tooth disease). Is one leg smaller than the other (old polio, infantile hemiplegia)?
2 Tone Ask patient to ‘relax/go fl oppy like a rag-doll’. Ask if they have any pain in feet/legs/hips before passively fl exing and extend-ing each limb while also internally and externally rotating. Hold the patient’s knee and roll it from side to side. Put your hand behind the knee and raise it quickly. The heel should lift slightly from the bed if tone is normal. Any spasticity/rigidity?
Clonus: Plantar fl ex the foot then quickly dorsifl ex and hold. More than 3 ‘beats’ of plantar fl exion is sustained clonus and is abnormal. Clonus can also be elicited at the patella with rapid downward movement of patella. Hypertonia and clonus suggest an upper motor neuron lesion.
Neurological examination of the lower limbs If the patient is able, begin your examination by asking the patient to remove their lower garments down to underwear, and to walk across the room. Gait analysis (p467) gives you more information than any other test. If they aren’t able to walk, start with them lying down, legs fully exposed. Then, Inspection, Tone, Refl exes, Power, Coordination, Sensation (fi g 2.32).
3 Refl exes For each refl ex, test right, then left and compare. If absent, attempt to elicit with ‘reinforcement’. Decide whether refl exes are absent/present (with reinforcement)/normal/brisk/exag-gerated. • Knee: (L3,4.) Strike on the patella tendon, just below the patella. • Ankle: (L5,S1.) Several accepted methods; ideally ask the patient to slightly bend the knee, then drop it laterally, grasp the foot and dorsifl ex, then strike the Achilles tendon. If hip pain limits mobility, dorsifl ex the foot with a straight leg and strike your hand, feeling for an ankle jerk. • Plantar refl exes: (L5, S1, S2.) Stroke the patient’s sole with an orange stick or similar. The normal refl ex is downward movement of the great toe. Babinski’s sign is positive if there is dorsifl exion of the great toe (this is abnormal (up-per motor neuron lesion) if patient age >6 months).
69 History and examination • If you are limited for time, gait is the most useful test to start with. • Make sure you test each muscle group individually by stabilizing above the joint you are testing. • Test vibration by putting a buzzing tuning fork on the bony part of a joint (most distal point) with the patient’s eyes closed then ask them to tell you when the buzzing stops (pinch the tuning fork to stop it) to distinguish vibration from pressure sensation.
Top tips 4 Power Direct patient to adopt position and follow the following commands while you as the examiner resist movements as appropriate to grade power (p446). Test each muscle group bilaterally before moving on to the next position. See pp452–3 for myotomes. • Hip fl exion: ‘Keeping your leg straight, can you lift your leg off the bed, don’t let me push it down.’ • Hip extension: ‘And now using your leg, push my hand into the bed.’ • Hip abduction: Position hands on outer thighs—‘push your legs out to the sides.’ • Hip adduction: Position hands on inner thighs—‘and push your legs together.’ • Knee fl exion and extension: ‘Bend your knee and bring your heel to your bottom, don’t let me pull it away… and now kick out against me and push me away.’ • Ankle plantar fl exion: With your hand on the underside of the patient’s foot ask them to ‘bend your foot down, pushing my hand away.’ • Ankle dorsifl exion: Put your hand on the dorsum of the foot and ask them to ‘lift up your foot, point your toes at the ceiling, don’t let me push your foot down.’ 6 Sensation As upper limbs (p67). • Light touch: Lower limb dermatomes (p454). • Pin prick • Temperature • Vibration • Joint position sense: With the patient’s eyes closed grasp distal phalanx of the great toe at the sides. Stabilize the rest of the toe. Move the joint up and tell patient ‘this is up’, and down, saying ‘this is down’. Flex and extend the joint, stopping at intervals to ask whether the toe is up or down.
5 Coordination Heel–shin test: Using your fi nger on the patient’s shin to demonstrate, instruct patient to ‘put your heel just below your knee then run it smoothly down your shin, lift it up and place it back on your knee, now run it down again’, etc. Repeat on the other side. Also, fast alternate foot tapping onto examiners’s hands with patient lying down.
Fig 2.32 Dermatomes of lower limb.
Reproduced from Harrison (ed) Revision Notes for MCEM Part A (2011), with permission from Oxford University Press.
70 History and examination Cranial nerve examination Approach to examining the cranial nerves Where is the lesion? Think systematically. Is it in the brain-stem (eg MS) or outside, pressing on the brainstem? Is it the neuromuscular junction (myasthenia) or the muscles (eg a dystrophy)? Cranial nerves may be af-fected singly or in groups. Face the patient (helps spot asymmetry). For causes of lesions see BOX ‘Caus-es of cranial nerve lesions’. • I: Smell—test ability of each nostril (separately) to distinguish familiar smells, eg coff ee. • II: Acuity—test each eye separately, and its correct-ability with glasses or pin-hole; use Snellen chart, or the one inside the cover of this book. Visual fi elds— compare with your own fi elds or formally via perim-etry testing. Any losses/inattention? Sites of lesions: OHCS p428. Pupils (p72)—size, shape, symmetry, reac-tion to light (direct and consensual) or accommoda-tion. Swinging light test for relative aff erent pupillary defect. Ophthalmoscopy (OHCS, p414)—best learnt from an ophthalmologist and dilat-ing drops help! Darken the room, warn the patient you will need to get close to their face. Focus the lens on the optic disc (pale? swollen?). Follow vessels outwards to view each quadrant. If the view is obscured, examine the red refl ex, with your focus on the margin of the pupil, to look for a cataract. Try to get a view of the fovea by asking the patient to look directly at the ophthalmoscope Pathology here needs prompt oph-thalmic review. If in doubt, ask for slit lamp examination or photography of the retina. • III p, IV, & VI—eye movements. Ask the patient to keep their head still and follow your fi nger as you trace an imaginary ‘H’. IIIrd nerve palsy—ptosis, large pupil, eye down and out. IVth nerve palsy—diplopia on looking down and in (often no-ticed on descending stairs)—head tilting compensates for this (ocular torticollis). VIth nerve palsy—horizontal diplopia on looking out. Nystagmus is involuntary, often jerky, eye oscillations. Horizontal nystagmus is often due to a vestibular le-sion (acute: nystagmus away from lesion; chronic: towards lesion), or cerebellar lesion (unilateral lesions cause nystagmus towards the aff ected side). If it is more in whichever eye is abducting, MS may be the cause (internuclear ophthalmoplegia, see fi g 2.34). If also deafness/tinnitus, suspect a peripheral cause (eg VIIIth nerve lesion, barotrauma, Ménière’s, p462). If it varies with head position, suspect benign positional vertigo (p462). If it is up-and-down, ask a neurologist to review—upbeat nystagmus classically occurs with lesions in the midbrain or at the base of the 4th ventricle, downbeat nystagmus in foramen magnum lesions. Nystagmus lasting ≤2 beats is normal, as is nystagmus at the extremes of gaze. • V: Motor palsy—‘Open your mouth’; jaw deviates to side of lesion, muscles of mas-tication (temporalis, masseter and pterygoids). Sensory—check all three divisions. Consider corneal refl ex (lost fi rst). • VII: p Facial nerve lesions cause droop and weakness. As the forehead has bilateral representation in the brain, only the lower two-thirds is aff ected in UMN lesions, but all of one side of the face in LMN lesions. Ask to ‘raise your eyebrows’, ‘show me your teeth’, ‘puff out your cheeks’. Test taste with salt/sweet solutions (supplies anterior two-thirds of tongue). • VIII: Hearing—p464. Ask to repeat a number whispered in an ear while you block the other. Perform Weber’s and Rinne’s tests (p464). Balance/vertigo—p462. • IX p & X: p Gag refl ex—ask the patient to say ‘Ah’. Xth nerve lesions also cause the pal-ate to be pulled to the normal side on saying ‘Ah’, uvula deviates away. Ask them to swallow a sip of water. Consider gag refl ex—touch the back of the soft palate with an orange stick. The aff erent arm of the refl ex involves IX; the eff erent arm involves X. • XI: Trapezii—‘Shrug your shoulders’ against resistance. Sternocleidomastoid: ‘Turn your head to the left/right’ against resistance. • XII: Tongue movement—the tongue deviates to the side of the lesion.
I olfactory II optic III oculomotor IV trochlear V1 ophthalmic division V2 maxillary division V3 mandibular division VI abducens VII facial VIII vestibulo cochlear IX glossopharyngeal X vagus XI accessory XII hypoglossal Cranial nerve names 71 History and examination Any cranial nerve may be aff ected by diabetes mellitus; stroke; MS; tumours; sar-coidosis; vasculitis (p556), eg PAN (p556), SLE (p554); syphilis. Chronic menin gitis (malignant, TB, or fungal) tends to pick off the lower cranial nerves one by one. • I: Trauma; respiratory tract infection; meningitis; frontal lobe tumour. • II: Field defects may start as small areas of visual loss (scotomas, eg in glaucoma). Monocular blindness—lesions of one eye or optic nerve eg MS, giant cell arteritis. Bilateral blindness 4—any cause of mononeuritis, eg diabetes, MS; rarely methanol, neurosyphilis. Field defects—bitemporal hemianopia—optic chiasm compres-sion, eg pituitary adenoma, craniopharyngioma, internal carotid artery aneurysm (fi g 10.3, p451). Homonymous hemianopia—aff ects half the visual fi eld contralat-eral to the lesion in each eye. Lesions lie beyond the chiasm in the tracts, radiation, or occipital cortex, eg stroke, abscess, tumour. Optic neuritis (pain on moving eye, loss of central vision, relative aff erent pupillary defect, disc swelling from papillitis 5)—causes demyelination (eg MS); rarely sinusitis, syphilis, collagen vas-cular disorders. Ischaemic papillopathy—swelling of optic disc due to stenosis of the posterior ciliary artery (eg in giant cell arteritis). Papilloedema (bilateral-ly swollen discs, fi g 12.20, p560)—most commonly ICP (tumour, abscess, enceph-alitis, hydrocephalus, idiopathic intracranial hypertension); rarer: retro-orbital lesion (eg cavernous sinus thrombosis, p480). Optic atrophy (pale optic discs and reduced acuity)—MS; frontal tumours; Friedreich’s ataxia; retinitis pigmen-tosa; syphilis; glaucoma; Leber’s optic atrophy; chronic optic nerve compression. • III: c Alone c—‘medical’ causes (pupillary sparing): diabetes; HTN; giant cell arteri-tis; syphilis; idiopathic. ‘Surgical’ causes (early pupil involvement due to external compression of nerve damaging parasympathetic fi bres): posterior communi-cating artery aneurysm (+ surgery) ICP (if uncal herniation through the tento-rium compresses the nerve); tumours. • IV:c Alone—rare and usually due to trauma to the orbit. • V: c Sensory—trigeminal neuralgia (pain but no sensory loss, p457); herpes zoster, nasopharyngeal cancer, acoustic neuroma (p462). Motor—rare. • VI: c Alone—MS, Wernicke’s encephalopathy, false localizing sign in ICP, pontine stroke (presents with fi xed small pupils ± quadriparesis). • VII: LMN—Bell’s palsy (p500), polio, otitis media, skull fracture; cerebellopontine angle tumours, eg acoustic neuroma, malignant parotid tumours, herpes zoster (Ramsay Hunt syndrome p501, OHCS p652). UMN—(spares the forehead, because of its bilateral cortical representation), stroke, tumour. • VIII: (p462 & p464.) Noise damage, Paget’s disease, Ménière’s disease, herpes zoster, acoustic neuroma, brainstem CVA, drugs (eg aminoglycosides). • IX, X, XI: Trauma, brainstem lesions, neck tumours. • XII: Rare. Polio, syringomyelia, tumour, stroke, bulbar palsy, trauma, TB.
Groups of cranial nerves: VIII, then V, VI, IX, & X: cerebellopontine angle tumours, eg acoustic neuroma (p462; facial weakness is not a prominent sign). III, IV & VI: stroke, tumours, Wernicke’s enceph alopathy; aneurysms, MS. III, IV, Va, & VI: cavernous sinus thrombosis, supe-rior orbital fi ssure lesions (Tolosa–Hunt syndrome, OHCS p654). IX, X, & XI: jugular foramen lesion. : myasthenia gravis, muscular dystrophy, myotonic dystrophy, mononeuritis multiplex (p502).
Causes of cranial nerve lesions 4 Remember the commonest cause of monocular or binocular blindness is not a cranial nerve lesion but a problem with the eye itself (cataracts, retinal problems). Neurological disorders more commonly cause loss of part of the visual fi eld.
5 Unilateral disc swelling = papillitis, bilateral papillitis/disc swelling = papilloedema. Check both eyes! c= structures passing through the cavernous sinus; see BOX ‘Psychiatric symptoms’, p89. NB: Va is the only division of V to do so.
p= Remember that these cranial nerves carry parasympathetic fi bres. Sympathetic fi bres originate from the thoracic chain and run with the arterial supply to distribute about the body (see also OHCS, fi g 9.6, p621).
If the patient is able to shake their head, there is no meningism.
Top tips 72 History and examination Fig 2.33 Light refl ex. Action potentials go along optic nerve (red), traversing optic chiasm, passing synapses at pre-tectal nu-cleus, en route to Edinger–Westphal nuclei of CNIII. These send fi bres to both irises’ ciliary muscles (so both pupils constrict) via ciliary ganglion (also relays accom-modation and corneal sensation, and gets sympathetic roots from C8–T2, carrying fi bres to dilate pupil).
Cranial nerve lesions of the eye Pupillary abnormalities Key questions: • Equal, central, circular, dilated, or constricted? • React to light, directly and consensually? • Constrict normally on convergence/accommodation?
Irregular pupils: Anterior uveitis (iritis), trauma to the eye, syphilis.
Dilated pupils: CN III lesions (inc. ICP, p830) and mydriatic drugs. Always ask: is this pupil dilated, or is it the other that is constricted?
Constricted pupils: Old age, sympathetic nerve damage (Horner’s, p702, and ptosis, p73), opiates, miotics (pilocarpine drops for glaucoma), pontine damage.
Unequal pupils (anisocoria): May be due to unilateral lesion, eye-drops, eye surgery, syphilis, or Holmes–Adie pupil. Some inequality is normal.
Light reaction: Test: cover one eye and shine light into the other obliquely. Both pupils should constrict, one by direct, other by consensual light refl ex (fi g 2.33). The lesion site is deduced by knowing the pathway: from the retina the message passes up the optic nerve (CNII) to the superior colliculus (midbrain) and thence to the CNIII nuclei on both sides. The IIIrd cranial nerve causes pupillary constriction. If a light in one eye causes only contralateral constriction, the defect is ‘eff erent’, as the aff erent pathways from the retina being stimulated must be intact. Test for relative afferent pupillary defect: move torch quickly from pupil to pupil. If there has been incomplete damage to the aff erent pathway, the aff ected pupil will paradoxically dilate when light is moved from the normal eye to the abnormal eye. This is because, in the face of reduced aff erent input from the aff ected eye, the consensual pupillary relaxation response from the normal eye predominates. This is the Marcus Gunn sign, and may occur after apparent complete recovery from the initial lesion.
Reaction to accommodation/convergence: If the patient fi rst looks at a distant object and then at the examiner’s fi nger held a few inches away, the eyes will con-verge and the pupils constrict. Aff erent fi bres in each optic nerve pass to the lateral geniculate bodies. Impulses then pass to the pre-tectal nucleus and then to the para-sympathetic nuclei of the IIIrd cranial nerves, causing pupillary constriction. • Holmes–Adie (myotonic) pupil: The aff ected pupil is normally moderately dilated and is poorly reactive to light, if at all. It is slowly reactive to accommodation; wait and watch carefully: it may eventually constrict more than a normal pupil. It is often associated with diminished or absent ankle and knee refl exes, in which case the Holmes–Adie syndrome is present. Usually a benign incidental fi nding. Rare causes: Lyme disease, syphilis, parvovirus B19, HSV, autoimmunity. >. • Argyll Robertson pupil: This occurs in neurosyphilis. The pupil is constricted and unreactive to light, but reacts to accommodation. Other possible causes: Lyme disease; HIV; zoster; diabetes mellitus; sarcoidosis; MS; paraneoplastic; B12. The iris may be patchily atrophied, irregular, and depigmented. The lesion site is not always near the Edinger–Westphal nucleus or even in the midbrain. Pseudo-Argyll Robertson pupils occur in Parinaud’s syndrome (p708). • Hutchinson pupil: This is the sequence of events resulting from rapidly rising uni-lateral intracranial pressure (eg in intracerebral haemorrhage). The pupil on the side of the lesion fi rst constricts then widely dilates. The other pupil then goes through the same sequence. See p830.
73 History and examination Get ophthalmology help. See OHCS p434–p455. Consider: • Is the eye red? (Glaucoma, uveitis p561.) • Pain? Giant cell arteritis: severe temporal headache, jaw claudication, scalp ten-derness, ESR: urgent steroids (p556). Optic neuritis: eg in MS. • Is the cornea cloudy: corneal ulcer (OHCS p435), glaucoma (OHCS p433)? • Is there a contact lens problem (infection)? • Any fl ashes/fl oaters? (TIA, migraine, retinal detachment?) • Is there a visual fi eld problem (stroke, space-occupying lesion, glaucoma)? • Are there any focal CNS signs? • Any valvular heart disease/carotid bruits (emboli)? Hyperlipidaemia (p690)? • Is there a relative aff erent pupillary defect (p72)? • Any past history of trauma, migraine, hypertension, cerebrovascular disease, MS, diabetes or connective tissue disease? • Any distant signs: eg HIV (causes retinitis), SLE, sarcoidosis?
Sudden: • Acute glaucoma • Retinal detachment • Vitreous haemorrhage (eg in diabetic proliferative retinopathy) • Central retinal artery or vein occlusion • Migraine • CNS: TIA (amaurosis fugax), stroke, space-occupying lesion • Optic neu-ritis (eg MS) • Temporal arteritis • Drugs: quinine/methanol • Pituitary apoplexy.
Gradual: • Optic atrophy • Chronic glaucoma • Cataracts • Macular degeneration • Tobacco amblyopia.
Visual loss Drooping of the upper eyelid. Best observed with patient sitting up, with head held by examiner. Oculomotor nerve (CN III) innervates main muscle concerned (levator palpebrae), but nerves from the cervical sympathetic chain innervate superior tar-sal muscle, and a lesion of these nerves causes mild ptosis which can be overcome on looking up. Causes: 1 CN III lesions cause unilateral complete ptosis: look for other evidence of a CN III lesion: ophth almoplegia with ‘down and out’ deviation of the eye, pupil dil ated and unreactive to light or accommodation. If eye pain too, suspect infi ltration (eg by lymphoma or sarcoidosis). If T° or consciousness, suspect infection (any tick bites?).
2 Sympathetic paralysis usually causes unilateral partial ptosis. Look for other evidence of a sympathetic lesion, as in Horner’s syndrome (p702): constricted pupil = miosis, lack of sweating on same side of the face (=anhidrosis).
3 Myopathy, eg dystrophia myotonica, myasthenia gravis (cause bilateral partial ptosis).
4 Congenital; usually partial and without other CNS signs.
Ptosis Fig 2.34 Internuclear ophthal-moplegia (INO) and its causes. To produce synchronous eye move-ments, cranial nerves III, IV, and VI communicate through medial long it udinal fasciculus in midbrain. In INO, a lesion disrupts commu n-i cation, causing weakness in ad-duction of the ipsilat eral eye with nystagmus of the contralateral eye only when abducting. There may be incomplete or slow abduction of the ipsilateral eye during lateral gaze. Convergence is preserved. Causes MS or vascular (rarely: HIV; syphilis; Lyme disease; brainstem tumours; phenothiazine toxicity).
74 History and examination Skin On both the palm and the dorsum start by inspecting the skin for: 1 Colour—pigmentation of creases, jaundice, palmar erythema (fi g 2.44).
2 Consistency—tight (sclerodactyly), thick (DM, acromegaly) (fi g 2.39).
3 Characteristic lesions—pulp infarcts, rashes, purpura, spider naevi, telangiecta-sia, tophi (fi g 2.35), scars (eg carpal tunnel release).
Musculoskeletal hand examination 1 Begin by introducing yourself, obtaining consent to examine and position the patient appropriately. Expose the arms, then ask the patient to rest their hands on a pillow. Start by examining the dorsal surface and then turn the hands over. Always ask about pain or tender areas. Follow the ‘look, ask the patient to move, then feel’ to avoid causing pain.
Nails Look for the same skin changes as the palm, plus tendon xan-thomata, plaques, and joint re-placement scars, and examine the nails for: • Pitting and onycholysis (p76). • Clubbing (p77). • Nail fold infarcts and splinter haemorrhages. • Other lesions, eg Beau’s lines (fi g 2.41, p76), koilonychia, leuconychia (fi g 2.36).
Muscles Examine the muscles for wasting and fascicu-lations; on the dorsal surface look for wasting, particularly of dorsal interossei. On the palm look particularly at the thenar and hypothenar eminences. • Thenar wasting (fi g 2.37) = median nerve lesion. • Generalized wasting, particularly of the interossei on the dorsum, but sparing of the thenar eminence = ulnar nerve lesion.
Also look for Dupuytren's contracture and perform Tinel's test (percuss over the distal skin crease of the wrist). Phalen's test (patient holds dorsal surfaces of both hands together for 60 seconds). Both tests are positive if tin-gling reported, suggesting carpal tunnel syn-drome.
Fig 2.35 Gouty tophi.
Fig 2.36 Leuconychia.
Fig 2.37 Thenar wasting.
75 History and examination Joints Examine for acute infl ammation (swol-len, red joints) as well as the charac-teristic deformities of chronic arthritis, eg rheumatoid, osteoarthritis (fi g 2.38). • Ulnar deviation at the wrist. • Z deformity of the thumb. • Swan-neck (fl exed DIP, hyperextended PIP—fi g 12.2, p540). • Boutonnière (hyperextended DIP, fl exed PIP). • Heberden’s nodes (DIP joints, p77). • Bouchard’s nodes (PIP joints).
Move and feel By this point, you should know the likely diagnosis, so assess neurological func-tion looking at power, function, and sen-sation: • Wrist and forearm: Extension (prayer position) and fl exion (reverse prayer), supination and pronation. Look at the elbows. • Small muscles: Pincer grip, power grip (squeeze my two fi ngers), abduction of the thumb, abduction (spread your fi n-gers), and adduction (grip this piece of paper between your fi ngers) of the fi n-gers. NB Froment's sign = fl exion of the thumb during grip as ulnar nerve lesion prevents adduction (p453). • Function: Write a sentence, undo a but-ton, pick up a coin. • Sensation: Test little fi nger (ulnar), index fi nger (median), and anatomi-cal snuff box (radial) using light touch/ pinprick.
When you have clinched the diagnosis and functional status, examine each joint, palpating for tenderness, eff usions, and crepitus. Test sensation (see p67) and ex-amine the elbows. Consider examination of upper limbs and face. • Cross your fi ngers before the patient grips them, it hurts less! • Don’t forget to palpate the radial pulse. • Don’t forget to look at the elbows for plaques of psoriasis and rheumatoid nodules.
Top tips Fig 2.38 Osteoarthritis.
Fig 2.39 Sclerosis.
76 History and examination Musculoskeletal hand examination 2 The hands can give you a wealth of information about a patient. Shaking hands can tell you about thyroid disease (warm, sweaty, tremor), anxiety (cold, sweaty), and neurological disease (myotonic dystrophy patients have diffi culty relaxing their grip, a weak grip may suggest muscle wasting or peripheral neuropathy). The nails and skin can inform about systemic disease: Nail abnormalities • Koilonychia (spoon-shaped nails, fi g 2.40) suggests iron defi ciency, haemochromatosis, infection (eg fungal), endocrine disorders (eg acromegaly, hypo-thyroidism), or malnutrition. • Onycholysis (detachment of the nail from the nail-bed) is seen with hyper thyroidism, fungal infection, and psoriasis. • Beau’s lines (fi g 2.41) are transverse furrows from temporary arrest of nail growth at times of biologi-cal stress: severe infection. Nails grow at ~0.1mm/d, the furrow’s distance from the cuticle allows dating of the stress. • Mees’ lines are single white transverse bands clas-sically seen in arsenic poisoning, chronic kidney dis-ease, and carbon monoxide poisoning among others. • Muehrcke’s lines are paired white parallel trans-verse bands (without furrowing of the nail itself, distinguishing them from Beau’s lines) seen, eg, in chronic hypoalbuminaemia, Hodgkin’s disease, pel-lagra (p268), chronic kidney disease. • Terry’s nails: Proximal portion of nail is white/ pink, nail tip is red/brown (causes include cirrhosis, chronic kidney disease, congestive cardiac failure). • Pitting is seen in psoriasis and alopecia areata. • Splinter haemorrhages (fi g 2.42) are fi ne longitudi-nal haemorrhagic streaks (under the nails), which in the febrile patient may suggest infective endocardi-tis. They may be microemboli, or be normal—eg due to gardening. • Nail-fold infarcts: Embolic, typically seen in vascu-litic disorders (OHCS, p452). • Nail clubbing See p77. • Chronic paronychia is a chronic infection of the nail-fold and presents as a painful swollen nail with intermittent discharge (fi g 2.43).
Skin changes • Palmar erythema (fi g 2.44) is associated with cir-rhosis, pregnancy, hyperthyroidism, rheumatoid arthritis, polycythaemia; also chronic liver disease— via inactivation of vasoactive endotoxins by the liver. Also chemotherapy-induced palmar/plantar erythrodysaesthesia. • Pallor of the palmar creases suggests anaemia. • Pigmentation of the palmar creases is normal in people of African-Caribbean or Asian origin but is also seen in Addison’s disease and Nelson’s syn-drome (increased ACTH after removal of the adrenal glands in Cushing’s disease).15 • Gottron’s papules (purple rash on the knuckles) with dilated end-capillary loops at the nail fold sug-gests dermatomyositis (p552). Fig 2.40 Koilonychia.
Fig 2.44 Palmar erythema.
Fig 2.41 Beau’s lines, here due to chemotherapy, a new line is seen with each cycle. See p525.
Fig 2.42 Splinter haemor-rhages.
Fig 2.43 Paronychia.
Reproduced from Burge et al. Oxford Handbook of Medical Dermatology 2016, with permission from Oxford University Press.
77 History and examination Fingernails (± toenails) have increased curvature in all directions and loss of the angle between nail and nail fold and feel boggy (fi gs 2.46, 2.47). Pathogenesis is unclear although the platelet theory was developed in 1987.16 Megakaryocytes are normally fragmented into platelets in the lungs, and the original theory was that any disruption to normal pulmonary circulation (infl ammation, cancer, cardiac right-to-left shunting) would allow large megakaryocytes into the systemic cir-culation. They become lodged in the capillaries of the fi ngers and toes, releasing platelet-derived growth factor and vascular endothelial growth factor, which lead to tissue growth, vascular permeability, and recruitment of infl ammatory cells. Evidence showing platelet microthrombi in clubbed fi ngers, and high levels of PDGF and VEGF in patients with hypertrophic osteoarthropathy, support the theory. This does not explain the changes in patients with unilateral clubbing, usually seen in neurological disorders.
Causes Thoracic: GI: Cardiovascular: • Bronchial cancer (clubbing is twice as common in women); us u ally not small cell cancer • Chronic lung suppuration: • Empyema, abscess • Bronchiectasis • Cystic fi brosis • Fibrosing alveolitis • Mesothelioma • TB. • Infl ammatory bowel dis-ease (especially Crohn’s) • Cirrhosis • GI lymphoma • Malabsorption, eg coeliac.
Rare: • Familial • Thyroid acropachy (p562). • Cyanotic cong enital heart disease • Endocarditis • Atrial myxoma • Aneurysms • Infected grafts.
Unilateral clubbing: • Hemiplegia • Vascular lesions, eg upper-limb artery aneurysm, Takayasu’s arteritis, brachial arteriovenous malformations (including iatrogenic— haemodialysis fi stulas).
Clubbing Nodules and contractures • Dupuytren’s contracture (see fi g 2.26, p60) fi brosis and contrac ture of palmar fascia, p698) is seen in liver disease, trauma, epilepsy, and ageing. • Look for Heberden’s (DIP) fi g 2.45 and Bouchard’s (PIP) ‘nodes’—osteophytes (bone over-growth at a joint) seen with osteoarthritis.
Fig 2.45 Heberden’s (DIP).
Reproduced from Watts et al. (eds) Oxford Textbook of Rheumatology (2013), with permission from Oxford University Press.
Fig 2.46 Finger clubbing.
Fig 2.47 Testing for fi nger clubbing. |
15338 | https://www.epilepsy.va.gov/Library/Lecture3diagnostictests2.pdf | 1 Rizwana.Rehman@va.gov Statistics in Evidence Based Medicine II Lecture 2: Linear Correlation and Regression Statistics in Evidence Based Medicine II Lecture 3: Summary Statistics for Diagnostic Tests Statistics in Evidence Based Medicine IILecture 3: Summary Statistics for Diagnostic Tests Audio Information: Dial 1-855-767-1051 Conference ID 61304911 Text Books Main: Statistics at Square One 12th edition (2010) M J Campbell & T D V Swinscow Secondary: Basic and Clinical Biostatistics (2004) Beth Dawson, Robert G. Trapp eID=62 For more information, program materials, and to complete evaluation for CME credit visit www.epilepsy.va.gov/Statistics 2 Audio Information: Dial 1-855-767-1051 Conference ID 61304911 Overview Vocabulary Prevalence Sensitivity and Specificity Positive predictive value Likelihood ratios Comparison Reading an article about a diagnostic test Reporting the results for publication 3 Evaluation of a Diagnostic Test A gold standard is needed to evaluate the performance of a test. Gold standard or reference standard is a definite tool that identifies if a person has a particular condition. Why not use gold standard all the time? Expensive Difficult to administer 4 Binary Outcomes Gold Standard Positive Negative Diagnostic test Positive a b Negative c d Total a+c b+d n=a+b+c+d 5 Statistics At Square One 11th edition a=number of true positives, d=number of true negatives. c= number of false negatives, b=number of false positives Example of Binary Outcomes Diagnosis of Generalized Anxiety Disorder (GAD) by Mental Health Professionals Positive Negative GAD 2 ≥3(+ve) 63 152 215 <3(-ve) 10 740 750 Total 73 892 965 6 Statistics At Square One 11th edition Prevalence of a Disease 7 Prevalence is the proportion of people diagnosed by the gold standard. Prevalence=(a+c)/n Prevalence=73/965=0.076=7.6% Sensitivity of the Diagnostic Test 8 Given a person has the disease, sensitivity is the proportion of people who have a positive diagnostic test. Sensitivity=a/(a+c) Sensitivity=63/73=0.86=86% Specificity of a Diagnostic Test Given a person does not have the disease, specificity is the proportion of people who have a negative diagnostic test. Specificity=d/(b+d) Specificity=740/892=0.83=83% 9 Useful Mnemonic SeNsitivity=1-proportion of false negatives (n in each sides) SPecificity=1-proportion of false positives (p in each side) 10 Things to Remember For a test with high sensitivity, a negative result rules out the disease. mnemonic=SnNout For a test with high specificity, a positive result rules in the disease. mnemonic=SpPin 11 Positive Predictive Value 12 Proportion of truly positive cases among the positive cases detected by the test. “If I have a positive test, what are the chances I have the disease?” Positive Predictive Value=PPV=a/(a+b) PPV=63/215=0.29=29% Negative Predictive Value 13 Proportion of truly negative cases among the negative cases detected by the test. “If I have a negative test, what are the chances I don’t have the disease?” Negative Predictive Value=NPV=d/(c+d) NPV=740/750=0.99=99% Comparison of Sensitivity, Specificity and Positive Predictive Value Sensitivity and specificity are independent of prevalence of a disease. Positive predicative value depends upon prevalence. Increasing the prevalence increases the positive predictive value. 14 Receiver Operating Characteristics Curve (ROC) For tests that produce results on continuous or ordinal scale, we need a cut-off value to calculate sensitivity and specificity. We plot a graph of sensitivity vs. 1-specificity for different cut-off values. ROC curves are used to compare the results of different diagnostic tests. 15 Likelihood Ratios A single summary statistic Positive Likelihood Ratio (LR+) (prob. of positive test given the disease)/(prob. of positive test without the disease)=sensitivity/(1-specificity) Negative Likelihood Ratio (LR-) (prob. of negative test given the disease)/(prob. of negative test without the disease)=(1-sensitivity)/specificity 16 Likelihood Ratios 17 LR(+)=[a/(a+c)]/[b/(b+d)] LR(-)=[c/(a+c)]/[d/(b+d)] 18 Example of Likelihood Ratios LR(+)=0.86/(1-0.83)=5.06 A high GAD2 score is 5.1 times as likely to occur in a patient with, as opposed to a patient without, GAD. LR(-)=(1-0.86)/0.83=0.17 Advantages of Likelihood Ratio Knowing the likelihood of a disease gives a way to estimate how likely is someone to have a disease, if one knows the prevalence or probability of the disease before the test. The LR indicates by how much a given diagnostic test result will raise or lower the pretest probability of the target disease. 19 A Rough Guide for Interpretation of LR from Pretest to Posttest Probabilities >10 or <0.1 imply conclusive change from pretest probability to post test probability. Between 5 and 10 and 0.1 to 0.2 imply moderate shifts. Between 2 and 5 and 0.5 to 0.2 generate small changes. Between 1 and 2 and 0.5 to 1 rarely alter the probability 20 User’s Guide to the Medical Literature :III. How to Use an Article About a Diagnostic Test: B. What are the Results and Will They Help Me in Caring for My Patients. JAMA 1994 Guide-Diagnosis_PartII.pdf Bayes’ Theorem Posttest odds=pretest odds × LR Pretest odds=73/892=0.082 LR(+)=5.06 Posttest odds=5.06 × 0.082=0.4149 Posttest probability=Posttest odds/(1+Posttest odds) Posttest Probability=0.4149/1.4149=0.293=29% Posttest probability is the Positive Predictive Value 21 Another Example: PIOPED Study Pulmonary Embolism Present Absent V/Q Scan Results No. Proportion No. Proportion Likelihood Ratios High Probability 102 102/251=0.408 14 14/630=0.022 18.3 Intermediate Probability 105 105/251=0.418 217 217/630=0.344 1.2 Low Probability 39 39/251=0.155 273 273/630=0.433 0.36 Normal/near normal 5 5/251=0.020 126 126/630=0.200 0.10 Total 251 630 22 User’s Guide to the Medical Literature :III. How to Use an Article About a Diagnostic Test: B. What are the Results and Will They Help Me in Caring for My Patients. JAMA 1994 Guide-Diagnosis_PartII.pdf Fagan’s Nomogram 23 r =0.846 Fagan’s Nomogram for PIOPED Example 24 r =0.846 Computing Sensitivity and Specificity 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0 2 4 6 8 10 BAC Beers Beers and BAC r = 0.894 25 Changing Threshold of Positive vs. Negative 26 Comparison of Likelihood Ratios with Sensitivity and Specificity 27 With sensitivity and specificity we loose important information. We have to calculate sensitivity and specificity with every cut point. For likelihood ratio method just need to know one number. Likelihood ratios can be used on individual’s level. Assumptions 28 All summary statistics are based upon: The diseases or diagnoses being considered are mutually exclusive and include the actual diagnosis. The result of each diagnostic test are independent from the results of all other tests. Reading and Reporting Diagnostic Tests Always report confidence intervals of measures. Always report the prevalence of the condition. Report how the subjects were selected. Question whether treatment would be changed depending on the results of a test. 29 Helpful Articles 30 Users' Guides to the Medical Literature: III. How to Use an Article About a Diagnostic Test: A. Are the Results of the Study Valid? User’s Guide to the Medical Literature :III. How to Use an Article About a Diagnostic Test: B. What are the Results and Will They Help Me in Caring for My Patients. JAMA 1994 Guide-Diagnosis_PartII.pdf Statistical Guidance on Reporting Results from Studies Evaluating Diagnostic Tests dGuidance/GuidanceDocuments/ucm071287.pdf Thank you! Questions/Comments Rizwana.Rehman@va.gov (919) 286-0411 ext: 5024 31 For more information, program materials, and to complete evaluation for CME credit visit www.epilepsy.va.gov/Statistics |
15339 | https://www.youtube.com/watch?v=aw7uLsWexOs | The turn ratio of a transformer is given as 2: 3. If the current through the primary coil is 3 A...
PW Solutions
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Posted: 14 Oct 2022
The turn ratio of a transformer is given as 2: 3. If the current through the primary coil is 3 A, thus calculate the current through load resistance?
a. 1 A
b. 4.5 A
c. 2 A
d. 1.5 A
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Transcript:
हेलो बच्चों लेट्स सी दिस क्वेश्चन द टर्न रेशो ऑफ़ अ ट्रांसफॉर्मर इज गिवन बाय 2/3 ओके सो n p अप n की वैल्यू दे दी 2/3 इफ द करंट थ्रू द प्राइमरी कॉइल ओके इज हाउ मच 3 एयर्स देन कैलकुलेट द करंट थ्रू द लोड रेजिस्टेंस व्हिच इज आ तो यहां पे कितना आ जाएगा 2/3 3 2/3 3 = टू कितना हो गया 2 एंपियर हो गया यह मेरा फाइनल आंसर हो जाएगा लुकिंग एट द ऑप्शंस ए गलत बी गलत डी गलत सी सही हेंस द फाइनल आंसर इ ऑप्शन सी ऑल द बेस्ट |
15340 | https://www.youtube.com/watch?v=8vElQO9CWV4 | Exploring points of inflection and concavity using GeoGebra Graphing Calculator app on iPad
Mark Willis
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Posted: 23 Jun 2019
A-Level 26 Differentiation
Transcript:
hi in this video I'm going to show you how to use the geogebra graphing calculator on the iPad to explore the idea of concavity and finding a point of inflection okay so first of all I'm going to open up the graphing calculator so in my school apps here graphing calculators quite a lot of jihad for apps so make sure you've got the right one doesn't this is like the finished product so I'm going to try to recreate this product for you so if I press the free arrows and clear also I go back to start a starting point again so a couple things to begin with if you press a cog and press the labels you can put the X and white labels in okay and although obviously I've probably going to need to make the font a little bit bigger so you probably need to do this last bit so I'm gonna make this a little bit bigger so you can actually see what I'm doing now in order to get the equation that I want I'm just going to log in to my sorry just open that again and what I wanted to do is do split through so hold on to the thing and then move it over here like that and then you'll get a quick view so this is the if it's a function I want to try and recreate so just going to recreate that so when you press this you need to make sure your in number so I'm going to want to X to the power of three we need to press the arrow key now bring it back down plus three X this is automatically a squared button minus four X minus three and then I don't need the Safari file anymore so just come all about that now I with two fingers on the x-axis if you now pull apart you'll be able to make the axes a bit bigger so maybe we'll make that bit smaller than we're going to now create a slider so to get slide that you see you've got a calculator then you've got a circle and a triangle press the circle and the triangle and press slider and put where you want it okay I want to go from minus five to five step 0.05 is okay press ok now I can move it around to where I want it to be and now if I go back to the calculator it's there if I press the three dots and press the settings I could perhaps change the color if I wanted to right right now I want to what I want to do is put a moving point on to the curve so the curve is defined as f of X so if I press import and now press bracket now to get the letter A because it needs to correspond with us like that need to press of the ABC press a comma F and then to get the brackets unfortunate you need to go back to that one and then go back to that one little bit cumbersome okay press ENTER right if I now move you can see now got a moving point okay so if we now select the point three dots go to the Settings will change the color to red so it stands out it's a little personal taste and I'm gonna rename the points at the top it's good to rename it as a capital P okay you can see it's all changed okay so the next step is to put the tangent in so press the circle and the triangle and scroll down here and then in construct you'll see tangent pressing with your finger the arrow come up will say tangent hit the point hit the curve and then you've got the tangent go back to the calculator press the three dots put the settings and let's change the color to something like blue okay so now what I've got is a tangent no what you'll notice is when you put this across this curve so if you want to label the curve that's another thing we should do is if we press the free dots and go to settings set caption style and then just put down the not value it name of value at this time it will add in the name of the function it does seem to do it right at the very bottom so it's a bit difficult to see but never mind okay so if you move the slight and what happens you'll know that here there's a horizontal tangent and I hear there's a horizontal tangent but what you will notice if we just move that a little bit what you'll notice is that here sorry starting from here the tangent is above the curse okay and it's still above the curve and then at some point just down here it will change from going from one side to the other which is actually if we actually at minus 0.5 let's just do that okay at that particular point it will just change from one side to the other okay and then then it will be underneath okay so the point where it changes from one side to the other is called the point of inflection it's called a non-stationary point of inflection and the above the curve we got concave downwards and below the curve we got concave upwards so if we go back to this and we asked it to put in F and then you have to press the alpha beta gamma and press the apostrophe twice okay it will put in the second derivative okay and you will notice at that point minus 0.5 that the second derivative is actually equal to zero so there's this don't change the settings so let's go for style and change it to a dust dashed line okay ceilings and tricks here settings and then we'll put in the that we will put the value in won't do that we'll just keep that hidden okay it doesn't look very nice okay anyhow so there we've got we can see at the point inflection at the moment that the second derivative is actually equal to zero okay so if we now ask it to put the curve for X comma go back to numbers X less than minus 0.5 enter that's not something I've got the X button so backspace and then just press X okay it will pretend so we'll just change the color of that again go to settings and change the color orange okay where the curve is orange it's where the curve is concave downwards okay and then if we put in here F comma back to 1 2 3 X is greater than minus 0.5 and so okay we're going to change the color again so settings and make that one red so it stands out okay so where the graph is red the curve is concave upwards the tangent is below the curve and where the curve is orange where the curve is orange is concave downwards and if we just plot the point comma minus 0.5 comma F of minus 0.5 okay and we see that point up here when the point a is is the actual point of inflection that's the point where the tangent crosses from one side to the other okay so I hope you find this useful you can move this back there and then you've got the whole thing if you don't want to see where it's called caved you can we move move anything like this pressing absolutely it was the original curve there's where it's concave downwards and that's where it's concave upwards we're just changing colors it's quite good way of demonstrating just one last thing if you prep the back to the original curve and we go to table values and let's start from minus 3 to 3 we press ok that gives you a table of values for the original curve - free going up in steps of minus 2 which is also a useful fact feature of geogebra ok so this has been a video to show you how to use the graph of geogebra graphing calculator to draw a curve and to explore the idea concavity and finding a point of inflection I hope you've understood and I thank you very much for watching |
15341 | https://docs.gap-system.org/doc/ref/chap16.html | [MathJax on]
16 Combinatorics
This chapter describes functions that deal with combinatorics. We mainly concentrate on two areas. One is about selections, that is the ways one can select elements from a set. The other is about partitions, that is the ways one can partition a set into the union of pairwise disjoint subsets.
16.1 Combinatorial Numbers
16.1-1 Factorial
| | |
--- |
| ‣ Factorial( n ) | ( function ) |
‣ Factorial
returns the factorial n! of the positive integer n, which is defined as the product 1 ⋅ 2 ⋅ 3 ⋯ n.
n! is the number of permutations of a set of n elements. 1 / n! is the coefficient of x^n in the formal series exp(x), which is the generating function for factorial.
PermutationsList (16.2-12) computes the set of all permutations of a list.
PermutationsList
16.1-2 Binomial
| | |
--- |
| ‣ Binomial( n, k ) | ( function ) |
‣ Binomial
returns the binomial coefficient {n choose k} of integers n and k. This is defined by the conditions {n choose k} = 0 for k < 0, {0 choose k} = 0 for k ≠ 0, {0 choose 0} = 1 and the relation {n choose k} = {n-1 choose k} + {n-1 choose k-1} for all n and k.
There are many ways of describing this function. For example, if n ≥ 0 and 0 ≤ k ≤ n, then {n choose k} = n! / (k! (n-k)!) and for n < 0 and k ≥ 0 we have {n choose k} = (-1)^k {-n+k-1 choose k}.
If n ≥ 0 then {n choose k} is the number of subsets with k elements of a set with n elements. Also, {n choose k} is the coefficient of x^k in the polynomial (x + 1)^n, which is the generating function for {n choose .}, hence the name.
NrCombinations (16.2-3) is the generalization of Binomial for multisets. Combinations (16.2-1) computes the set of all combinations of a multiset.
NrCombinations
Binomial
Combinations
16.1-3 Bell
| | |
--- |
| ‣ Bell( n ) | ( function ) |
‣ Bell
returns the Bell number B(n). The Bell numbers are defined by B(0) = 1 and the recurrence B(n+1) = ∑_{k = 0}^n {n choose k} B(k).
B(n) is the number of ways to partition a set of n elements into pairwise disjoint nonempty subsets (see PartitionsSet (16.2-16)). This implies of course that B(n) = ∑_{k = 0}^n S_2(n,k) (see Stirling2 (16.1-6)). B(n)/n! is the coefficient of x^n in the formal series exp( exp(x)-1 ), which is the generating function for B(n).
PartitionsSet
Stirling2
16.1-4 Bernoulli
| | |
--- |
| ‣ Bernoulli( n ) | ( function ) |
‣ Bernoulli
returns the n-th Bernoulli number B_n, which is defined by B_0 = 1 and B_n = -∑_{k = 0}^{n-1} {n+1 choose k} B_k/(n+1).
B_n / n! is the coefficient of x^n in the power series of x / (exp(x)-1). Except for B_1 = -1/2 the Bernoulli numbers for odd indices are zero.
16.1-5 Stirling1
| | |
--- |
| ‣ Stirling1( n, k ) | ( function ) |
‣ Stirling1
returns the Stirling number of the first kind S_1(n,k) of the integers n and k. Stirling numbers of the first kind are defined by S_1(0,0) = 1, S_1(n,0) = S_1(0,k) = 0 if n, k ne 0 and the recurrence S_1(n,k) = (n-1) S_1(n-1,k) + S_1(n-1,k-1).
S_1(n,k) is the number of permutations of n points with k cycles. Stirling numbers of the first kind appear as coefficients in the series n! {x choose n} = ∑_{k = 0}^n S_1(n,k) x^k which is the generating function for Stirling numbers of the first kind. Note the similarity to x^n = ∑_{k = 0}^n S_2(n,k) k! {x choose k} (see Stirling2 (16.1-6)). Also the definition of S_1 implies S_1(n,k) = S_2(-k,-n) if n, k < 0. There are many formulae relating Stirling numbers of the first kind to Stirling numbers of the second kind, Bell numbers, and Binomial coefficients.
Stirling2
16.1-6 Stirling2
| | |
--- |
| ‣ Stirling2( n, k ) | ( function ) |
‣ Stirling2
returns the Stirling number of the second kind S_2(n,k) of the integers n and k. Stirling numbers of the second kind are defined by S_2(0,0) = 1, S_2(n,0) = S_2(0,k) = 0 if n, k ne 0 and the recurrence S_2(n,k) = k S_2(n-1,k) + S_2(n-1,k-1).
S_2(n,k) is the number of ways to partition a set of n elements into k pairwise disjoint nonempty subsets (see PartitionsSet (16.2-16)). Stirling numbers of the second kind appear as coefficients in the expansion of x^n = ∑_{k = 0}^n S_2(n,k) k! {x choose k}. Note the similarity to n! {x choose n} = ∑_{k = 0}^n S_1(n,k) x^k (see Stirling1 (16.1-5)). Also the definition of S_2 implies S_2(n,k) = S_1(-k,-n) if n, k < 0. There are many formulae relating Stirling numbers of the second kind to Stirling numbers of the first kind, Bell numbers, and Binomial coefficients.
PartitionsSet
Stirling1
16.2 Combinations, Arrangements and Tuples
16.2-1 Combinations
| | |
--- |
| ‣ Combinations( mset[, k] ) | ( function ) |
‣ Combinations
returns the set of all combinations of the multiset mset (a list of objects which may contain the same object several times) with k elements; if k is not given it returns all combinations of mset.
A combination of mset is an unordered selection without repetitions and is represented by a sorted sublist of mset. If mset is a proper set, there are {|mset| choose k} (see Binomial (16.1-2)) combinations with k elements, and the set of all combinations is just the power set of mset, which contains all subsets of mset and has cardinality 2^{|mset|}.
Binomial
To loop over combinations of a larger multiset use IteratorOfCombinations (16.2-2) which produces combinations one by one and may save a lot of memory. Another memory efficient representation of the list of all combinations is provided by EnumeratorOfCombinations (16.2-2).
IteratorOfCombinations
EnumeratorOfCombinations
16.2-2 Iterator and enumerator of combinations
| | |
--- |
| ‣ IteratorOfCombinations( mset[, k] ) | ( function ) |
‣ IteratorOfCombinations
| | |
--- |
| ‣ EnumeratorOfCombinations( mset ) | ( function ) |
‣ EnumeratorOfCombinations
IteratorOfCombinations returns an Iterator (30.8-1) for combinations (see Combinations (16.2-1)) of the given multiset mset. If a non-negative integer k is given as second argument then only the combinations with k entries are produced, otherwise all combinations.
IteratorOfCombinations
Iterator
Combinations
EnumeratorOfCombinations returns an Enumerator (30.3-2) of the given multiset mset. Currently only a variant without second argument k is implemented.
EnumeratorOfCombinations
Enumerator
The ordering of combinations from these functions can be different and also different from the list returned by Combinations (16.2-1).
Combinations
16.2-3 NrCombinations
| | |
--- |
| ‣ NrCombinations( mset[, k] ) | ( function ) |
‣ NrCombinations
returns the number of Combinations(mset,k).
Combinations(mset,k)
The function Arrangements (16.2-4) computes ordered selections without repetitions, UnorderedTuples (16.2-6) computes unordered selections with repetitions, and Tuples (16.2-8) computes ordered selections with repetitions.
Arrangements
UnorderedTuples
Tuples
16.2-4 Arrangements
| | |
--- |
| ‣ Arrangements( mset[, k] ) | ( function ) |
‣ Arrangements
returns the set of arrangements of the multiset mset that contain k elements. If k is not given it returns all arrangements of mset.
An arrangement of mset is an ordered selection without repetitions and is represented by a list that contains only elements from mset, but maybe in a different order. If mset is a proper set there are |mset|! / (|mset|-k)! (see Factorial (16.1-1)) arrangements with k elements.
Factorial
16.2-5 NrArrangements
| | |
--- |
| ‣ NrArrangements( mset[, k] ) | ( function ) |
‣ NrArrangements
returns the number of Arrangements(mset,k).
Arrangements(mset,k)
As an example of arrangements of a multiset, think of the game Scrabble. Suppose you have the six characters of the word "settle" and you have to make a four letter word. Then the possibilities are given by
"settle"
Can you find the five proper English words, where "lets" does not count? Note that the fact that the list returned by Arrangements (16.2-4) is a proper set means in this example that the possibilities are listed in the same order as they appear in the dictionary.
"lets"
Arrangements
The function Combinations (16.2-1) computes unordered selections without repetitions, UnorderedTuples (16.2-6) computes unordered selections with repetitions, and Tuples (16.2-8) computes ordered selections with repetitions.
Combinations
UnorderedTuples
Tuples
16.2-6 UnorderedTuples
| | |
--- |
| ‣ UnorderedTuples( set, k ) | ( function ) |
‣ UnorderedTuples
returns the set of all unordered tuples of length k of the set set.
An unordered tuple of length k of set is an unordered selection with repetitions of set and is represented by a sorted list of length k containing elements from set. There are {|set| + k - 1 choose k} (see Binomial (16.1-2)) such unordered tuples.
Binomial
Note that the fact that UnorderedTuples returns a set implies that the last index runs fastest. That means the first tuple contains the smallest element from set k times, the second tuple contains the smallest element of set at all positions except at the last positions, where it contains the second smallest element from set and so on.
UnorderedTuples
16.2-7 NrUnorderedTuples
| | |
--- |
| ‣ NrUnorderedTuples( set, k ) | ( function ) |
‣ NrUnorderedTuples
returns the number of UnorderedTuples(set,k).
UnorderedTuples(set,k)
As an example for unordered tuples think of a poker-like game played with 5 dice. Then each possible hand corresponds to an unordered five-tuple from the set { 1, 2, ..., 6 }.
The function Combinations (16.2-1) computes unordered selections without repetitions, Arrangements (16.2-4) computes ordered selections without repetitions, and Tuples (16.2-8) computes ordered selections with repetitions.
Combinations
Arrangements
Tuples
16.2-8 Tuples
| | |
--- |
| ‣ Tuples( set, k ) | ( function ) |
‣ Tuples
returns the set of all ordered tuples of length k of the set set.
An ordered tuple of length k of set is an ordered selection with repetition and is represented by a list of length k containing elements of set. There are |set|^k such ordered tuples.
Note that the fact that Tuples returns a set implies that the last index runs fastest. That means the first tuple contains the smallest element from set k times, the second tuple contains the smallest element of set at all positions except at the last positions, where it contains the second smallest element from set and so on.
Tuples
16.2-9 EnumeratorOfTuples
| | |
--- |
| ‣ EnumeratorOfTuples( set, k ) | ( function ) |
‣ EnumeratorOfTuples
This function is referred to as an example of enumerators that are defined by functions but are not constructed from a domain. The result is equal to that of Tuples( set, k ). However, the entries are not stored physically in the list but are created/identified on demand.
Tuples( set, k )
16.2-10 IteratorOfTuples
| | |
--- |
| ‣ IteratorOfTuples( set, k ) | ( function ) |
‣ IteratorOfTuples
For a set set and a positive integer k, IteratorOfTuples returns an iterator (see 30.8) of the set of all ordered tuples (see Tuples (16.2-8)) of length k of the set set. The tuples are returned in lexicographic order.
IteratorOfTuples
Tuples
16.2-11 NrTuples
| | |
--- |
| ‣ NrTuples( set, k ) | ( function ) |
‣ NrTuples
returns the number of Tuples(set,k).
Tuples(set,k)
Tuples(set,k) can also be viewed as the k-fold cartesian product of set (see Cartesian (21.20-15)).
Tuples(set,k)
Cartesian
The function Combinations (16.2-1) computes unordered selections without repetitions, Arrangements (16.2-4) computes ordered selections without repetitions, and finally the function UnorderedTuples (16.2-6) computes unordered selections with repetitions.
Combinations
Arrangements
UnorderedTuples
16.2-12 PermutationsList
| | |
--- |
| ‣ PermutationsList( mset ) | ( function ) |
‣ PermutationsList
PermutationsList returns the set of permutations of the multiset mset.
PermutationsList
A permutation is represented by a list that contains exactly the same elements as mset, but possibly in different order. If mset is a proper set there are |mset| ! (see Factorial (16.1-1)) such permutations. Otherwise if the first elements appears k_1 times, the second element appears k_2 times and so on, the number of permutations is |mset| ! / (k_1! k_2! ...), which is sometimes called multinomial coefficient.
Factorial
16.2-13 NrPermutationsList
| | |
--- |
| ‣ NrPermutationsList( mset ) | ( function ) |
‣ NrPermutationsList
returns the number of PermutationsList(mset).
PermutationsList(mset)
The function Arrangements (16.2-4) is the generalization of PermutationsList (16.2-12) that allows you to specify the size of the permutations. Derangements (16.2-14) computes permutations that have no fixed points.
Arrangements
PermutationsList
Derangements
16.2-14 Derangements
| | |
--- |
| ‣ Derangements( list ) | ( function ) |
‣ Derangements
returns the set of all derangements of the list list.
A derangement is a fixpointfree permutation of list and is represented by a list that contains exactly the same elements as list, but in such an order that the derangement has at no position the same element as list. If the list list contains no element twice there are exactly |list|! (1/2! - 1/3! + 1/4! - ⋯ + (-1)^n / n!) derangements.
Note that the ratio NrPermutationsList( [ 1 .. n ] ) / NrDerangements( [ 1 .. n ] ), which is n! / (n! (1/2! - 1/3! + 1/4! - ⋯ + (-1)^n / n!)) is an approximation for the base of the natural logarithm e = 2.7182818285..., which is correct to about n digits.
NrPermutationsList( [ 1 .. n ] ) / NrDerangements( [ 1 .. n ] )
16.2-15 NrDerangements
| | |
--- |
| ‣ NrDerangements( list ) | ( function ) |
‣ NrDerangements
returns the number of Derangements(list).
Derangements(list)
As an example of derangements suppose that you have to send four different letters to four different people. Then a derangement corresponds to a way to send those letters such that no letter reaches the intended person.
The function PermutationsList (16.2-12) computes all permutations of a list.
PermutationsList
16.2-16 PartitionsSet
| | |
--- |
| ‣ PartitionsSet( set[, k] ) | ( function ) |
‣ PartitionsSet
returns the set of all unordered partitions of the set set into k pairwise disjoint nonempty sets. If k is not given it returns all unordered partitions of set for all k.
An unordered partition of set is a set of pairwise disjoint nonempty sets with union set and is represented by a sorted list of such sets. There are B( |set| ) (see Bell (16.1-3)) partitions of the set set and S_2( |set|, k ) (see Stirling2 (16.1-6)) partitions with k elements.
Bell
Stirling2
16.2-17 NrPartitionsSet
| | |
--- |
| ‣ NrPartitionsSet( set[, k] ) | ( function ) |
‣ NrPartitionsSet
returns the number of PartitionsSet(set,k).
PartitionsSet(set,k)
Note that PartitionsSet (16.2-16) does currently not support multisets and that there is currently no ordered counterpart.
PartitionsSet
16.2-18 Partitions
| | |
--- |
| ‣ Partitions( n[, k] ) | ( function ) |
‣ Partitions
returns the set of all (unordered) partitions of the positive integer n into sums with k summands. If k is not given it returns all unordered partitions of n for all k.
An unordered partition is an unordered sum n = p_1 + p_2 + ⋯ + p_k of positive integers and is represented by the list p = [ p_1, p_2, ..., p_k ], in nonincreasing order, i.e., p_1 ≥ p_2 ≥ ... ≥ p_k. We write p ⊢ n. There are approximately exp(π sqrt{2/3 n}) / (4 sqrt{3} n) such partitions, use NrPartitions (16.2-21) to compute the precise number.
NrPartitions
If you want to loop over all partitions of some larger n use the more memory efficient IteratorOfPartitions (16.2-19).
IteratorOfPartitions
It is possible to associate with every partition of the integer n a conjugacy class of permutations in the symmetric group on n points and vice versa. Therefore p(n) :=NrPartitions(n) is the number of conjugacy classes of the symmetric group on n points.
NrPartitions
Ramanujan found the identities p(5i+4) = 0 mod 5, p(7i+5) = 0 mod 7 and p(11i+6) = 0 mod 11 and many other fascinating things about the number of partitions.
16.2-19 IteratorOfPartitions
| | |
--- |
| ‣ IteratorOfPartitions( n ) | ( function ) |
‣ IteratorOfPartitions
For a positive integer n, IteratorOfPartitions returns an iterator (see 30.8) of the set of partitions of n (see Partitions (16.2-18)). The partitions of n are returned in lexicographic order.
IteratorOfPartitions
Partitions
16.2-20 IteratorOfPartitionsSet
| | |
--- |
| ‣ IteratorOfPartitionsSet( set[, k[, flag]] ) | ( function ) |
‣ IteratorOfPartitionsSet
IteratorOfPartitionsSet returns an iterator (see 30.8) for all unordered partitions of the set set into pairwise disjoint nonempty sets (see PartitionsSet (16.2-16)). If k given and flag is omitted or equal to false, then only partitions of size k are computed. If k is given and flag is equal to true, then only partitions of size at most k are computed.
IteratorOfPartitionsSet
PartitionsSet
false
true
16.2-21 NrPartitions
| | |
--- |
| ‣ NrPartitions( n[, k] ) | ( function ) |
‣ NrPartitions
returns the number of Partitions(set,k).
Partitions(set,k)
The function OrderedPartitions (16.2-22) is the ordered counterpart of Partitions (16.2-18).
OrderedPartitions
Partitions
16.2-22 OrderedPartitions
| | |
--- |
| ‣ OrderedPartitions( n[, k] ) | ( function ) |
‣ OrderedPartitions
returns the set of all ordered partitions of the positive integer n into sums with k summands. If k is not given it returns all ordered partitions of set for all k.
An ordered partition is an ordered sum n = p_1 + p_2 + ... + p_k of positive integers and is represented by the list [ p_1, p_2, ..., p_k ]. There are totally 2^{n-1} ordered partitions and {n-1 choose k-1} (see Binomial (16.1-2)) ordered partitions with k summands.
Binomial
Do not call OrderedPartitions with an n much larger than 15, the list will simply become too large.
OrderedPartitions
16.2-23 NrOrderedPartitions
| | |
--- |
| ‣ NrOrderedPartitions( n[, k] ) | ( function ) |
‣ NrOrderedPartitions
returns the number of OrderedPartitions(set,k).
OrderedPartitions(set,k)
The function Partitions (16.2-18) is the unordered counterpart of OrderedPartitions (16.2-22).
Partitions
OrderedPartitions
16.2-24 PartitionsGreatestLE
| | |
--- |
| ‣ PartitionsGreatestLE( n, m ) | ( function ) |
‣ PartitionsGreatestLE
returns the set of all (unordered) partitions of the integer n having parts less or equal to the integer m.
16.2-25 PartitionsGreatestEQ
| | |
--- |
| ‣ PartitionsGreatestEQ( n, m ) | ( function ) |
‣ PartitionsGreatestEQ
returns the set of all (unordered) partitions of the integer n having greatest part equal to the integer m.
16.2-26 RestrictedPartitions
| | |
--- |
| ‣ RestrictedPartitions( n, set[, k] ) | ( function ) |
‣ RestrictedPartitions
In the first form RestrictedPartitions returns the set of all restricted partitions of the positive integer n into sums with k summands with the summands of the partition coming from the set set. If k is not given all restricted partitions for all k are returned.
RestrictedPartitions
A restricted partition is like an ordinary partition (see Partitions (16.2-18)) an unordered sum n = p_1 + p_2 + ... + p_k of positive integers and is represented by the list p = [ p_1, p_2, ..., p_k ], in nonincreasing order. The difference is that here the p_i must be elements from the set set, while for ordinary partitions they may be elements from [ 1 .. n ].
Partitions
[ 1 .. n ]
16.2-27 NrRestrictedPartitions
| | |
--- |
| ‣ NrRestrictedPartitions( n, set[, k] ) | ( function ) |
‣ NrRestrictedPartitions
returns the number of RestrictedPartitions(n,set,k).
RestrictedPartitions(n,set,k)
The last example tells us that there are 451 ways to return 50 pence change using 1, 2, 5, 10, 20 and 50 pence coins.
16.2-28 SignPartition
| | |
--- |
| ‣ SignPartition( pi ) | ( function ) |
‣ SignPartition
returns the sign of a permutation with cycle structure pi.
This function actually describes a homomorphism from the symmetric group S_n into the cyclic group of order 2, whose kernel is exactly the alternating group A_n (see SignPerm (42.4-1)). Partitions of sign 1 are called even partitions while partitions of sign -1 are called odd.
SignPerm
16.2-29 AssociatedPartition
| | |
--- |
| ‣ AssociatedPartition( pi ) | ( function ) |
‣ AssociatedPartition
AssociatedPartition returns the associated partition of the partition pi which is obtained by transposing the corresponding Young diagram.
AssociatedPartition
16.2-30 PowerPartition
| | |
--- |
| ‣ PowerPartition( pi, k ) | ( function ) |
‣ PowerPartition
PowerPartition returns the partition corresponding to the k-th power of a permutation with cycle structure pi.
PowerPartition
Each part l of pi is replaced by d = gcd(l, k) parts l/d. So if pi is a partition of n then pi^k also is a partition of n. PowerPartition describes the power map of symmetric groups.
PowerPartition
16.2-31 PartitionTuples
| | |
--- |
| ‣ PartitionTuples( n, r ) | ( function ) |
‣ PartitionTuples
PartitionTuples returns the list of all r-tuples of partitions which together form a partition of n.
PartitionTuples
r-tuples of partitions describe the classes and the characters of wreath products of groups with r conjugacy classes with the symmetric group on n points, see CharacterTableWreathSymmetric (71.20-6) and CharacterValueWreathSymmetric (71.20-7).
CharacterTableWreathSymmetric
CharacterValueWreathSymmetric
16.2-32 NrPartitionTuples
| | |
--- |
| ‣ NrPartitionTuples( n, r ) | ( function ) |
‣ NrPartitionTuples
returns the number of PartitionTuples( n, r ).
PartitionTuples( n, r )
16.2-33 BetaSet
| | |
--- |
| ‣ BetaSet( alpha ) | ( function ) |
‣ BetaSet
For a list alpha that describes a partition of a nonnegative integer (see Partitions (16.2-18)), BetaSet returns the list of integers obtained by reversing the order of alpha and then adding the sequence [ 0, 1, 2, ... ] of the same length, cf. [JK81, Section 2.7].
Partitions
BetaSet
[ 0, 1, 2, ... ]
16.3 Fibonacci and Lucas Sequences
16.3-1 Fibonacci
| | |
--- |
| ‣ Fibonacci( n ) | ( function ) |
‣ Fibonacci
returns the nth number of the Fibonacci sequence. The Fibonacci sequence F_n is defined by the initial conditions F_1 = F_2 = 1 and the recurrence relation F_{n+2} = F_{n+1} + F_n. For negative n we define F_n = (-1)^{n+1} F_{-n}, which is consistent with the recurrence relation.
Using generating functions one can prove that F_n = ϕ^n - 1/ϕ^n, where ϕ is (sqrt{5} + 1)/2, i.e., one root of x^2 - x - 1 = 0. Fibonacci numbers have the property gcd( F_m, F_n ) = F_{gcd(m,n)}. But a pair of Fibonacci numbers requires more division steps in Euclid's algorithm (see Gcd (56.7-1)) than any other pair of integers of the same size. Fibonacci(k) is the special case Lucas(1,-1,k) (see Lucas (16.3-2)).
Gcd
Fibonacci(k)
Lucas(1,-1,k)
Lucas
16.3-2 Lucas
| | |
--- |
| ‣ Lucas( P, Q, k ) | ( function ) |
‣ Lucas
returns the k-th values of the Lucas sequence with parameters P and Q, which must be integers, as a list of three integers. If k is a negative integer, then the values of the Lucas sequence may be nonintegral rational numbers, with denominator roughly Q^k.
Let α, β be the two roots of x^2 - P x + Q then we define Lucas( P, Q, k ) = U_k = (α^k - β^k) / (α - β) and Lucas( P, Q, k ) = V_k = (α^k + β^k) and as a convenience Lucas( P, Q, k ) = Q^k.
Lucas( P, Q, k )
Lucas( P, Q, k )
Lucas( P, Q, k )
The following recurrence relations are easily derived from the definition U_0 = 0, U_1 = 1, U_k = P U_{k-1} - Q U_{k-2} and V_0 = 2, V_1 = P, V_k = P V_{k-1} - Q V_{k-2}. Those relations are actually used to define Lucas if α = β.
Lucas
Also the more complex relations used in Lucas can be easily derived U_2k = U_k V_k, U_{2k+1} = (P U_2k + V_2k) / 2 and V_2k = V_k^2 - 2 Q^k, V_{2k+1} = ((P^2-4Q) U_2k + P V_2k) / 2.
Lucas
Fibonacci(k) (see Fibonacci (16.3-1)) is simply Lucas(1,-1,k). In an abuse of notation, the sequence Lucas(1,-1,k) is sometimes called the Lucas sequence.
Fibonacci(k)
Fibonacci
Lucas(1,-1,k)
Lucas(1,-1,k)
16.4 Permanent of a Matrix
16.4-1 Permanent
| | |
--- |
| ‣ Permanent( mat ) | ( attribute ) |
‣ Permanent
returns the permanent of the matrix mat. The permanent is defined by ∑_{p ∈ Sym(n)} ∏_{i = 1}^n mat[i][i^p].
Note the similarity of the definition of the permanent to the definition of the determinant (see DeterminantMat (24.4-4)). In fact the only difference is the missing sign of the permutation. However the permanent is quite unlike the determinant, for example it is not multilinear or alternating. It has however important combinatorial properties.
DeterminantMat
generated by GAPDoc2HTML |
15342 | https://www.youtube.com/watch?v=3C5g4zkJV5g | Array Division with remainder ( Model Division )
MooMooMath and Science
583000 subscribers
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Description
27685 views
Posted: 1 Mar 2018
Array Division with a remainder.
Arrays are objects divided into rows and columns
Rows are horizontal and columns are vertical.
The dividend is the number being divided and divisor tells you how many numbers are in the groups.
Sometimes when you divide it does not divide evenly and numbers are left over.
These leftover numbers are remainders.
When dividing with arrays you can see the remainders and can help you visualize the math.
14 comments
Transcript:
[Music] welcome to moomoomath and science in this video I'd like to show how to divide using an array in these problems have remainders before we get started with division let's learn the difference between a row and a column a row is horizontal and moves from left to right much like on a farm the crops are in a row from left to right a column on the other hand goes up and down or vertical much like the columns on a house go up and down or vertical ok let's look at this array how many rows are there well there are 1 2 horizontal rows next how many columns are there there are 3 columns 1 2 3 so first we know the difference now between a row and a column in this first example we will divide 8 divided by 3 8 is our dividend so we will use circles to represent the 8 and we have 8 circles the 3 is our divisor and it tells us how many are in each group I will create columns so each column will have 3 circles in it then I will count the number of columns and that will give us our quotient or answer so now I count them out in columns so we have 1 2 3 4 5 6 and I have 2 more left over 7 8 so now I count the number of columns for the quotient and that is 1 2 and then I have a remainder of 2 1 2 so our answer 8 divided by 3 is 2 with a remainder of 2 okay let's look at our next problem thirteen divided by three the 13 is our dividend and we have 13 circles the divisor is three so I'll put them in groups of three and I'm going to create columns again so let's start counting one two three four five six seven eight nine ten eleven twelve and then I'll place the other one here so now I count the number of columns for the answer I have one two three four columns and then a remainder of one so 13 divided by 3 is 4 with a remainder of one I hope that helps in dividing with arrays thanks for watching and moomoomath uploads a new math and science video everyday please subscribe and share |
15343 | https://www.onelook.com/?loc=dmapirel&w=hackneyed | "hackneyed": Unoriginal due to frequent overuse - OneLook
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15344 | https://www.khanacademy.org/science/in-in-class10th-physics/in-in-electricity/in-in-circuits-ohms-law-resistance/e/ohm-s-law-and-resistance | Use of cookies
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15345 | https://visionofthepalantir.com/2023/08/02/the-passing-of-the-grey-company/ | The Passing of the Grey Company – Vision Of The Palantir
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The Passing of the Grey Company
mortendall88Lord of the Rings Saga2 August 2023
“_“The Dead are following,” said Legolas. “I see shapes of Men and of horses, and pale banners like shreds of cloud, and spears like winter-thickets on a misty night. The Dead are following.”
“Yes, the Dead ride behind. They have been summoned,” said Elladan._”
– The Return of the King, The Passing of the Grey Company
The Flame of the West Saga Expansion is probably best known amongst fans for the third scenario, the legendary Battle of Pelennor Fields. However, before we arrive at the final confrontation with the Witch-King, Aragorn must undergo a journey, both physical and metaphorical, through the Paths of the Dead. In contrast to how they were depicted in the Peter Jackson films (seriously, who came up with theskull trap?), the dead were never really a physical threat in the book. They cause the Corsairs to flee, but they do not kill them. How do you make an enemy that isn’t a physical threat? You cause them to attack something different than the hero’s health…
The Passing of the Grey Company
Found inThe Flame of the West Saga Expansion, Scenario 1
Official Difficulty: –
Community Difficulty: 6.5
Encounter sets:The Passing of the Grey Company
Quest cards: 3
Play if:You like puzzle scenarios that require a specific deck. You like fighting the Army of the Dead. You like playing with threat. You want to play a quest with a lot of progress required. You like cards with heavy metal sounding names.
What is different about this quest?:Enemies rarely attack your health but your threat level. The amount of progress required to advance to stage 2 increases with your threat level. You get a special setup turn to get you going.
Solo or Multiplayer?:In solo, you need to put a lot of progress down and manage our threat effectively. In multiplayer, you can better have one player taking care of the Dead, but everyone needs to manage their threat somehow – otherwise, we will never be done!
Can I bring Side Quests for this scenario?:You can, but since a lot of progress is required and your threat is under constant attack, this can stall you to the point where you get too far behind. If you have discarded Overcome by Fear, then Double Back can be a good side-quest to lower threat and reduce the quest points on the main quest at the same time!
The Quest
Setup
Setup is pretty straightforward. Because this is a Saga quest, the first player takes control of the fellowship sphere version ofAragorn– this version has the classical Aragorn statline (2/3/2 – 5 hp) and rewards you for putting an artifact on him. If he has an artifact attached, he gains Sentinel and does not exhaust to quest; very versatile. You then continue with the quest setup. You start by settingThe Stone of ErechandThe Army of the Deadaside, out of play. Then, each player attaches a copy ofOvercome By Fearto their threat dial. Yes, you read that right. Threat dials can have attachments now. Additional copies of Overcome by Fear are removed from the game, so there is one copy per player in the game, and each threat dial can have a maximum of one. Since Overcome by Fear is central to the mechanics of this scenario, it is worth spending some time on before we complete additional setup.
Throughout the game, Overcome by Fear may bounce on and off your threat dial. While it is attached, your threat level cannot be reduced. This is a problem because, as we will see, the encounter deck attacks your threat relentlessly, and your threat level determines how easy it is to progress from stage 2. So you want it discarded as soon as possible. However, to do that, you need to spend a Fellowship resource from Aragorn and raise your threat by one. This can only be done at the end of the round. In a solo game, this is obviously much easier to do than in a four-player game, where you may have to wait for Aragorn to get back to you before you can discard it. This can lead to some tense moments when your threat gets high. As a general rule, you want to discard it when you can. There is generally little reason to keep it around. When discarded, it is set aside, out of play, so it can return later. You will usually get plenty of opportunities to get it back into play from encounter cards, but unless the cost of refusing it is excessively high (threating out or costing you a hero), you should generally avoid taking it whenever you can. Trust me. You want to keep that threat low.
Having dealt with our dial attachment, we continue with setup. If you play Saga mode, you will have some additional setup steps to do before you are ready to flip the quest card. First of all, you are allowed to switch heroes without incurring the +1 starting threat penalty. Then, the first player has a choice to make – they may raise the threat of every player by 1 to search their deck for the boonAnduril, a juicy 3-cost restricted attachment that only goes on Noble heroes or Aragorn, gives +1 to all stats and gives the wielder the response that after the wielder defends an attack, you can exhaust Anduril to strike back – and with Aragorn that is for 4 attack with Anduril, which is definitely not trivial. +1 extra threat is a minor price to pay, so make sure that the first player is the one who has Anduril in their deck in multiplayer. The Keen-eyed Tooks among you may also notice that Anduril is an artifact – and hence you can play him down on Fellowship Aragorn and get his juicy bonuses from the start of the game. It’s a gift that keeps on giving. When you are done drooling over Anduril, you proceed with setting aside burdens from the Frodo side of the story – A Shadow of the Past, Flight to the Ford, The Ring Goes South, The Passage of the Marshes, Shelob’s Lair. You are now ready to proceed!
The Forbidden Door (- Quest Points)
When flipped, stage 1B provides us with a choice – each player may add an additional resource to each hero they control! Each player who does so must raise their threat by 3. This tempting offer is not just granted out of the goodness of the designer’s heart – if you read on, you will see why it is a tempting offer. At the end of the first planning phase, each player discards their hand and advances to stage 2A. So you have this planning phase to play cards from your opening hand. Make the most of it. It is, of course, dependent on what you have in your opening hand whether this is a good idea. But generally, I take the extra resource to get my board state set. Because I will lose what remaining cards I have, card draw such asDaeron’s Runesand heroGaladriel, and cards that recur things from your discard pile such asDwarven Tombare all advantageous to have in this scenario.Galdor of the Havenspractically shines here as he lets you draw a new opening hand. Decks withErestorwill also be well-suited for this stage, as they tend to build around losing their hand every turn. This time it happens a bit faster than usual, but at least you will draw a lot of cards next turn! Tricks such as playing a couple of allies down (perhaps withTimely Aid) followed by a cheekyVery Good Taleto get even more allies out are the kind of tricks you can use to get over the setback of losing your hand. And as we will discuss later when we talk strategy, Silvans are particularly useful in overcoming this scenario. But more on that later. Once we have played our cards, we move on to stage 2A.
Stage 2 – The Paths of the Dead (X Quest Points)
When advancing to stage 2 at the end of the first planning phase, we see another “when revealed” effect. Each player must search the encounter deck for a different location and adds it to the staging area. We will cover each location in more detail but note that the encounter set only contains four different locations, so if you play four-player, you will get a copy of each. At this stage in the game’s life, it was common for locations to prevent specific strong player strategies such as location control and treachery cancellation. These locations certainly fit that description. Hence, my choice may vary depending on what my deck is trying to accomplish. In general, my order of choice will be:
Haunted Path – low threat, but more threat when players have a threat score over 35 – should not be a problem at the game’s start.
Deadly Road – punishes cancellation by raising your threat when cancels are played, requires fellowship resources for traveling, but low threat.
Trysting Place – prevents sniping of enemies in staging area, and brings out an Oathbreaker enemy as a traveling cost.
Dark Door – Mild effect as it “only” negates location control, by preventing progress placement in the staging area. However, it is a 4 threat location and triggers an encounter card reveal as a Travel cost. I do not like the uncertainty of extra encounter cards.
Once relevant locations have been added, stage 2A is flipped over. The first thing you will notice is that the quest points are not fixed but are equal to the threat of the player with the highest threat – so the number will keep on going up unless threat is managed – and remember, if you are Overcome by Fear, you cannot reduce your threat… This is the main strategy of the encounter deck. Drive the player’s threat up, and keep them at stage 2B forever until they threat out. And it can be relentlessis not optional here – I needed to side-board threat management cards into my decks to stand a chance against this quest when I did my playthroughs for research. So find out what kind of threat control you can include in your deck and swap it in! Threat is both dealt out through encounter deck effects and also because all enemies in the encounter deck have the Phantom keyword – a keyword invented for the occasion! The Phantom keyword states that “during the ‘determine combat damage’ step of an attack made by an enemy with Phantom, if any damage would be dealt by the attack, it is canceled, and the defending player raises his Threat by an equal amount instead.” So instead of taking damage, your threat goes up. This also means that undefended attacks (usually) are safe to take if you can handle the threat. So a dedicated blocker can help you keep the threat low but is less necessary than in other quests. But this also means that chump blocking is less effective – because you will just receive the damage as threat instead.
OK, so it’s all just a matter of getting our threat levels super low and power quest through, right? Well, there are some additional complications here. First of all, your threat cannot go lower than your starting threat – so no excessive lowering withSpirit MerryandThe Shirefolk. Additionally, at the beginning of the planning phase, each player has a tough choice to make – either they must raise their threat by one, or they cannot play any cards this phase. This latter part is a devilishly clever design, as you must constantly make tough decisions on whether you play something or wait to keep your threat low. This is why you ideally want to be setup with the added resources at the end of the first planning phase – even with massive card draw, playing new allies down each planning phase will cost with the increased tax of a threat bump, potentially making it harder to reach the end of Stage 2B if you are the player with the highest threat. And since you need to make a lot of progress to clear locations and accumulate the required progress to advance, you need to be able to quest right out of the gate. There really is no alternative.
So how do you build up a board state with the added threat tax? You have a few different options. One is to just power through it. Take one threat, spam out allies, and have enough to just power through. This, of course, works best if you are doing Outlands, Dwarf, or Gondor swarms, where you generate massive economy and play cards like there is no tomorrow. But this can be tough to keep up with, and most likely, you will need threat management, or you will be at serious risk of threating out. Plus, the discarding of your hand is punishing for swarm decks unless they have strong card draw. If you have a somewhat stable board state, you can make the call on a round-to-round basis. If you have a few good cards to play, take the threat and play them. Otherwise, hold on to them, try to make some progress, and build up a strong hand and resources. Ideally, you want to work towards a somewhat cleared staging area (which is, of course, much more likely in lower-player games). Then you can gradually add more and more cards in bursts. The third option, which is my favorite, is to play cards outside the planning phase. This is where Silvans are very helpful.Thranduillets you play an ally in the combat phase.The Tree Peoplelets you put allies into play outside of the planning phase too. And if you throwGaladrielinto the mix, you also have the option to draw cards and lower your threat. It is a strong tool against this quest. Make a plan for how you will address this added threat. Because it is just a small drip of threat, but as we all know, those drips tend to add up…
As mentioned above, your goal should be to keep the staging area as clear as possible. You need to quest for a low, so get some strong willpower characters in there, get a clear staging area by dealing with locations and enemies as they appear, and place as much progress as you can. Once the end is in sight, you should have a board state strong enough to face the final battle: the Army of the Dead.
Stage 3: The Dead Are Following (5 progress)
When you finally overcome the marathon that is stage 2, your board state is probably in a good place (otherwise, you won’t put down that much progress), and you hopefully have a relatively low threat, and the staging area is somewhat clear, so you are ready for the final confrontation with the Oath-breakers. You begin by addingThe Stone of Erechand theArmy of the Dead(enemy side face-up) to the staging area. Then, each player can raise their threat by 3 to detach Overcome by Fear from their threat dial and set it aside. Unless you are close to threating out or have no threat reduction in hand, and you can see threat is climbing, this is definitely worth doing if you have the attachment. This is also worth remembering if you are about to advance from 2B – if an encounter deck effect lets you attach Overcome by Fear, you know you have an easy way to discard it in stage 3. With everyone free of fear, we can look at how we win this damn thing.
Stage 3B tells us that when theStone of Erechleaves play as an explored location, we win! So this seems pretty straightforward. The stone itself has 5 threat and require 5 progress to clear, so it is adding a substantial amount of threat to the staging area. It is immune to player card effects (no surprise there), and to travel there, you must remove 5 progress from stage 3B. So you cannot travel to it immediately after advancing to stage 3A. And you need to add sufficient progress if you want to travel to it after the next quest phase. Traveling there also means enemies: After The Stone of Erech becomes the active location, each player searches the encounter deck and discard pile for an Oathbreaker enemy and adds it to the staging area, and shuffles the encounter deck. I prefer eitherFaithless DeadorGhosts of Menbecause they only attack by 3 and because you bypass their “when revealed” effects by adding them to the staging area. This is all the text on the actual location. However, the quest card has some additional effects to add: While Stone of Erech is the active location, each Oathbreaker enemy loses thePhantomkeyword and gets -20 engagement cost. So until we clear it, the Oathbreakers are ANGRY! This includes the scenario boss: The Army of the Dead.
With X threat, where X is twice the number of players in the game, 6 attack, 2 defense, and 8 hit points, Army of the Dead is a beefy enemy. However, it “only” takes 10 damage in one swing to bring them down. At this point, we have fought Balrogs and Witch-Kings – it takes more to impress us! It has the Phantom keyword and is immune to player card effects. It also has a Forced effect: When Army of the Dead attacks, you must either attach a set aside Overcome by Fear to your threat dial or discard an ally you control. If you play Saga, you do not want to have Overcome by Fear attached at the end of the game. So I prefer tossing a cheap ally to the effect. It has an engagement cost of 50, so once it gets out, you do not necessarily have to engage it until the Stone is the active location. In multiplayer, it will contribute a pretty hefty threat amount to the staging area, so if you can clear it, it is a good idea to do so straight away. Use a chump blocker or a defender with many hit points (Aragorn is a decent choice here if you dare) and throw your attack power after them. Once you deal 10 damage, they go to the victory display.
Ideally your stage 3 will consist of the following 3 rounds:
Round 1: Advance. Optionally engage and kill the Army of the Dead
Round 2:Quest over The Stone of Erech and place enough progress here that you can travel there. Be ready for combat with remaining enemies
Round 3: Clear The Stone of Erech and win!
You might have to linger for a few rounds if you get swarmed with enemies, or you are location locked, but this can be a vicious cycle. In Solo, the Stone and the Army will add 7 threat, and this number goes up by player count. If you are swarmed with locations, it can be difficult to quest sufficiently and have enough for combat. But if you manage to go through, you will have won! If you play Saga and have Overcome by Fear attached, you add it to the campaign pool as a burden. You will begin subsequent plays with Overcome by Fear attached to your threat dial. If Army of the Dead is in the victory display, add the Army of the Dead boon to the campaign pool. You are now ready to proceed in the Saga!
The Encounter Deck
Global
The encounter deck consists of 32 cards on Normal and 25 cards on Easy.
On Normal mode, 20 cards have shadow effects, giving you a risk of 62.5% for triggering a shadow effect during combat. On Easy, that risk is 60% with 15 cards.
Average threat is 1.7 per card on Normal and 1.6 on Easy.
Surge is present on 2 cards on both difficulty levels (The Way is Shut)
Peril is present on 10 cards on Normal and 9 cards on Easy (Faithless Dead, Ghosts of Men, The Way is Shut and Whispers in the Dark)
Doomed 1 is present on 3 cards on normal and 1 on easy (Blood Runs Chill). There are plenty of other ways that the encounter deck raises your threat, so don’t bring any Doomed yourself.
Immunity:
The Stone of Erech and the Army of the Dead are immune to player card effects.
While Overcome by Fear is attached to your threat dial, your threat cannot be reduced.
While Dark Door is in the staging area, progress cannot be placed on locations in the staging area.
While Trysting Place is in the staging area, each enemy in the staging area cannot take damage.
Enemies
First off, we have theDead Man of Dunharrow, a small enemy that wants to slam Overcome by Fear back on that juicy threat dial of yours. With four copies in the deck on both difficulties, it is an enemy you will encounter. With 2 threat, 2 attack, 1 defense, and 3 hit points, it is pretty straightforward to deal with, and most dedicated attackers should be able to take it down easily. It has an engagement cost of 30, meaning that most decks will have to engage it from early game (especially due to the many threat increases). It has the Phantom keyword, so it deals its damage as threat. The truly annoying part comes from the Forced effect when you engage it: you must either attach a set aside Overcome by Fear to your threat dial or raise your threat by 2 (which more or less constitutes an undefended attack). As this is an engagement effect and not a When Revealed effect, this is not something you can cheese with some of the locations that fish out enemies. As a shadow card, it can also be inconvenient or even deadly in the rare situations when Phantom is turned off, as it adds a +1 attack to the attacking enemy. In summary, this is an enemy that slows you down but is relatively easy to clear again. With 2 threat, this is actually a great target for Ranger Spikes, allowing you to keep it in the staging area forever and not have to worry about it.
Faithless Dead is not only a fantastic metal band name but also the name of an enemy that comes in three copies on normal and two copies on easy. 2 threat, 3 attack, 1 defense, and 4 hitpoints make it a step up from the Dead Men, but it should still be something a dedicated attacking duo can take down easily. It has an engagement cost of 35, which in this scenario still means you will face it early. It has Phantom and Peril for its When Revealed effect: Either raise each player’s threat by 2 or reveal an additional encounter card. I usually go for the threat increase, but this might be a solo bias. But additional encounter cards are usually bad for business. I like to take the threat. If two threat kicks you over the edge, things are not going great. I prefer taking this enemy from the encounter deck withTrysting Placeas I can then bypass the effect. It also removes a scary shadow effect: +1 attack and Phantom is negated for the turn. So that threat increase you planned to shrug off might suddenly cost you a hero. Ouch.
Ghosts of Men is another great band name, though I do believe they would play a different kind of music than Faithless Dead. Perhaps some more industrial vibes? This enemy has 3 threat, 3 attack, 1 defense, and 5 hit points and comes in two copies. So now we are definitely moving into territory where you need multiple attackers to take it down. It has an engagement cost of 40, so this is one you can leave to attack in the staging area for some time if this is your modus operandi. It has Phantom and Peril for its When Revealed effect: Either attach a set aside Overcome by Fear to your threat dial, or Ghosts of Men makes an immediate attack against you. My choice here varies a bit based on board state. If I have plenty of resources on Aragorn and I have few available defenders (or if defending would make me unable to take it down in one attack), then I might take Overcome by Fear (with the intention of discarding it as soon as possible). But if my threat is well-managed or I have plenty of defenders out, I will not sweat too much about an extra attack. It will likely only be a threat bump. You can fish this out with Trysting Place if you would rather not have to worry about this effect. As a shadow card, it has no effect, so that is a nice place to see it.
WithShadow Host,we are definitely back in metal territory. This is some dark and slow metal, like Tiamat. In our case, it is a mini-boss that is present in two copies on normal and one copy on easy. 4 threat, 4 attack, 2 defense, and 6 hit points make it a threat that you might not be able to deal with right out of the gate. 45 engagement means you can likely engage it on your own terms. It has Phantom and Forced: When Shadow Host engages you, either attach a set aside Overcome by Fear to your threat dial, or it makes an immediate attack, so same thing as Ghosts of Men, but every time they engage you, which might happen multiple times due to card effects we will see in a bit. It is an enemy you cannot ignore, especially because of the threat tag. Unless you get it late game, it will probably have to be your focus for a round or two. So keep an eye out for it as a shadow card (it has no shadow effect) and plan accordingly if it is still in your deck. It can make your threat go up very rapidly if unmanaged.
Locations
The first location of the encounter deck isDark Door– which would be a pretty good metal track name. I sense this is going to be a recurring theme in this section. But to return to the actual location, with 4 threat and 4 progress required to clear it, this is a tough one to see during the quest phase. It is the kind of location you might as well travel to because clearing it eats up as much of your willpower than leaving it in the staging area. There is only one problem – the travel cost. To go there, you must reveal a card from the encounter deck. This is, most of the time, bad news, especially at lower player counts. You might get a location to replace it, which can contribute to location lock. Or it can bring out a nasty enemy or treachery. You cannot use mostlocation controlto deal with it, as while it is in the staging area, progress cannot be placed on locations in the staging area. However, you can use some of the tried and true location swap shenanigans, such asWestroad TravelerorThrór’s Map, or blank it withThròr’s Key. It has no shadow effect, so it is nice to see it during combat. The encounter deck contains two copies on Normal and one on Easy.
Trysting Place is not a catchy name for a song, but a location that comes in two copies on easy and three on normal. With 3 threat and 3 progress to clear it, it can be somewhat bothersome to leave it around for too long, but the low progress threshold makes it an easy target for location control effects. While it is in the staging area, it protects enemies in the staging area from Gandalf slaps and other direct damage effects. To travel there, the first player must search the encounter deck and discard pile for an enemy and put it into play engaged with them. Use it to fish out some of the enemies described above with When Revealed effects! It has no shadow effects. Overall, it is a location that can be a problem if you are getting location locked, but otherwise is a somewhat convenient way to take the sting out of the encounter deck.
WithDeadly Road, we are back in metal territory – I can basically hear the Dragonforce riff that accompanies it! It comes in three copies, no matter your difficulty. It is a fairly benign location to encounter with 2 threat and 3 progress required to clear it, with the usual anti-Test of Will tech that later scenarios were so keen to implement: While Deadly Road is in the staging area, it gains:“Forced:After a ‘when revealed’ effect is canceled, raise each player’s threat by 1.” So another threat increase. This should not keep you from canceling treacheries but maybe think twice before doing it. The travel cost is that you must spend one fellowship resource to go there. It is mostly an issue at higher player counts, where Overcome by Fear will bounce in and out of play a lot more, and you, therefore, will need more resources. However, this is such a small price to pay that I don’t spend my location control bypassing it. It has no shadow effect, and it is a location that I usually breathe a sign of relief when I see it.
Lastly, we haveHaunted Path. Here we are definitely in goth rock territory. The encounter deck contains 3 copies, no matter your difficulty. With 1 threat and 4 progress required, it is a softball in the early game. At higher player counts, this can get a bit more troublesome, as it gets +1 threat pr. A player with threat over 35. I would argue that this is unlikely yo be the straw that breaks the pony’s back, but of course, it can location-lock you at an inconvenient time. Usually, it is a card I am happy to see. However, the travel cost can be a bit steep, as traveling there requires you to raise each player’s threat by 1. As a shadow card, this one can actually be pretty bad: You must either raise your threat by 2 or return the attacking enemy to the staging area after this attack. This means you can get some nasty engagement effects from some of the stronger Oathbreaker enemies. This is one you potentially cancel if you can.
Treacheries
We were just getting started with the metal song titles in the location section, and luckily the treacheries did not let us down.Blood Runs Chillsounds like it would be written by Mastodon. In our case, it is present in 3 copies on normal and 1 on easy. It has Doomed 1, and when revealed, each player with a threat of 35 or higher deals 1 damage to each exhausted character he controls. This treachery is a good reason to keep your threat low by any means necessary. Especially decks like Silvan or Gondor decks are prone to board wipes when a treachery like this fires. If your threat runs high, this is the kind of treachery you cancel if you can. As a shadow card, it punishes the defender readying. If your threat is 35 or higher, defending character cannot ready until the end of the round. This can be a problem if you have a dedicated defender, and you should therefore be aware of how many copies of this are in the deck if this is your main strategy for combat.
The Way is Shut sounds a little more light than Blood Runs Chill – perhaps it could be a song by Kamelot? It has peril and surge, and when it is revealed, either attach a set aside Overcome by Fear to your threat dial, or progress cannot be placed on the current quest until the end of the round. As we have discussed previously, you should generally avoid Overcome by Fear unless your threat is very low and you have Aragorn around to clear it easily. The second effect stalls you and can be problematic in the final part of the quest, where you are trying to blast through. To my understanding, you can still clear locations, so it is not a completely wasted effort. But you are racing against the clock in this quest, so it is an annoying effect. It comes in two copies regardless of difficulty and has no shadow effect, so combat is where I prefer seeing it.
Whispers in the Dark sounds like a creepy track. Perhaps we are in Dimmu Borgir territory? Here it is present in three copies, for both difficulties. It has Peril and When revealed: Either attach a set aside Overcome by Fear to your threat dial or each character you control gets -1 to each of the three stats until the end of the round. Usually, this will lead to a round of lost questing, potentially forcing you to eat some threat if you had problems questing hard enough. If this is the case, Overcome by Fear might be a better option. It is also annoying if you are in the process of dealing with one of the beefier enemies the encounter deck has to offer. This rewards questing with a few dedicated questers rather than a big swarm of weenie Dwarf or Gondor questers. However, I would usually save my cancellations for worse treacheries. As a shadow card, it punishes undefended attacks, granting +1 attack or +2 if the attack is undefended. This usually leads to a pretty substantial threat increase in the early game or a dead hero in the rare cases where Phantom is lost. I would keep this in mind when I decide where to use any shadow cancellation.
No Turning Back is a difficult one to review because it is not very metal. This could even be some Nickelback stuff or other soft rock… I am sorry to disappoint you, dear reader. I had hoped for a stronger ending! No turning back is present in two copies on normal and one copy on easy. It completes the trifecta of treacheries trying to stick overcome by fear back to your threat dial. However, this one triggers for ALL players – not just the one who drew it! When revealed, each player must choose: either attach a set aside Overcome by Fear to your threat dial or discard an ally you control. Usually, this is not too bad. Because of Phantom, chump blockers will have a tendency to stick around, and you can easily toss a cheap ally two it. There are, of course, niche cases where you play Fellowship or Three Hunters, where you either do not want to discard any allies or you simply have no allies in play. It is not one I often cancel unless my board state is in a tricky spot. Keep in mind that it does not have Peril – you are free to discuss with the table whether you should cancel it! I would be more prone to cancel it as a shadow card: Either attach a set aside Overcome by Fear to your threat dial, or the attacking enemy makes an additional attack after this one. If you have the blockers, this is fine, but otherwise, it can lead to a lot of threat or dead heroes. I prefer to see this one in the quest phase!
Tips and Tricks
Galadriel and other heroes with consistent threat-lowering effects, such asBeregond, are really solid choices for this quest. You want all the threat lowering you can get, and especially the tax of 1 threat for playing anything in the planning phase can really sneak up on you.
Card effects that put cards into play outside the planning phase are really valuable.Thranduilis a good example, as he allows you to play an ally in the combat phase.A very good talecan also be utilized to good effect.
Threat controlis not really optional, which means you will likely have to bring some spirit cards into your deck. Plan accordingly. Remember, you are allowed to switch heroes in Saga mode.
As mentioned above, placing Overcome by Fear on your threat dial is really dependent on the situation you find yourself in. It often depends on your threat and whether you have Aragorn with 1 resource available to clear it. One thing I often look at is how much threat I would get from not taking it. Sometimes eating the +1 threat and a fellowship resource is a better alternative. You generally want to clear it as quickly as you can, so you have the option of putting it back on if the alternative is worse on a treachery or if you plan on playing threat reduction.
While his ability would be perfect for this scenario, Lore Aragorn, unfortunately, cannot be played because of Fellowship Aragorn. And remember, you cannot playThorongil on fellowship-Aragorn either.
If you play standalone or do not want to include Anduril for some reason, consider adding one or more Artifact attachments to your deck for Aragorn to unlock his bonus. It is well worth it to get his free questing each turn.
Playthroughs
Cardboard of the Rings has quite a few playthroughs of this quest. Chad played through it in hisThematic Saga. He also played through it in hisOriginal Saga. Chad also participated in aTwitch playthroughwith Relic of the First Age Sean and community members bigfomlof and Stokesbook.
One Stop co-op shop also has a nice video and goes through some detail about setup and the Phantom keyword in the beginning. It is also made with real cards (like Chad’s Thematic Saga) if you prefer that compared to DragnCards and/or OCTGN.
Another set of videos with real cards was uploaded recently by user Carl Swanson (There and back again). You can find the videos here:video 1,video 2andvideo 3. I am not really sure about the order of the videos, so please excuse me if I mixed them up.
LOTRLCGSolo made a progression-style video with the quest.
Bronco’s Gaming World also made a video with a playthrough recently.
Card Talk also has a couple of videos where they played this quest as a live stream:Video 1andVideo 2. They also did a theatrical campaign.
Our brave companions made it through the Paths of the Dead! But the Corsairs await on the other side… When the Saga campaign returns, it will be all hands on deck! I can already promise a lot of stupid nautical wordplay…
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15346 | https://www.apollo247.com/blog/article/typhoid-test-normal-range-and-interpretation | Buy Medicines
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Widal Test for Typhoid Fever – Normal Range, Purpose, Procedure, and Test Results
7 min read
By Apollo 24|7, Published on- 03 January 2025
What Is Typhoid Fever?Understanding the Widal TestPurpose of the Widal TestPre-Test PreparationWhen is the Widal Test Recommended?Widal Test ProcedureInterpreting Widal Test ResultsNormal Range and Clinical ImplicationsInterpreting Widal Test TitersAlternative Tests for Typhoid DiagnosisPost-Test Care and Follow-UpConclusion
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The Widal test is a widely used blood test to diagnose typhoid fever, a bacterial infection caused by Salmonella Typhi and Salmonella Paratyphi. It works by detecting antibodies that your immune system produces in response to these bacteria. Typhoid fever remains a serious health concern, particularly in areas with poor sanitation and contaminated water supplies.
By measuring antibody levels (agglutination) against the bacterial O (somatic) and H (flagellar) antigens, the Widal test helps healthcare providers confirm the diagnosis and begin timely treatment.
What Is Typhoid Fever?
Typhoid fever is a severe bacterial infection caused by Salmonella Typhi. The disease spreads primarily through contaminated food, water, or close contact with infected individuals. If not treated promptly, it can lead to life-threatening complications such as intestinal perforation, organ damage, and sepsis.
Symptoms of Typhoid Fever
The common symptoms of typhoid fever include:
High fever: Gradually rising, lasting for more than a week.
Abdominal pain: Discomfort, often felt in the lower right abdomen.
Fatigue and headache: Persistent weakness, tiredness, and general malaise.
Gastrointestinal issues: Nausea, loss of appetite, diarrhoea, or constipation.
Skin rashes: Flat, rose-coloured spots on the chest or abdomen.
What is the Typhoid Normal Range?
The typhoid normal range indicates the acceptable levels of antibodies present in the blood when a person is free from typhoid infection. Generally, the typhoid normal range for the Widal test is less than 1:160.
Typhoid Normal Range in Male and Female
Typhoid normal range in male remains consistent with the general population. Similarly,
Typhoid normal range in female individuals is the same as that for males.
Understanding the Widal Test
The Widal test was developed in 1896 by French physician Georges Widal. For over a century, it has been a cornerstone diagnostic tool for typhoid fever, particularly in resource-limited settings.
The Widal test is a serological blood test that detects the presence of antibodies (agglutinins) in response to Salmonella Typhi and Salmonella Paratyphi. It targets two primary antigens:
O antigen (somatic): Found in the outer cell wall of the bacteria.
H antigen (flagellar): Found in the flagella (tail-like structure) of the bacteria.
Purpose of the Widal Test
The Widal test helps confirm typhoid fever in individuals showing prolonged fever, gastrointestinal issues, and other related symptoms. It is particularly useful in regions where typhoid is endemic or where patients have a history of recent travel to affected areas.
Many febrile illnesses, such as malaria, dengue, and other bacterial infections, present similar symptoms. The Widal test helps differentiate typhoid fever from these illnesses, enabling appropriate treatment.
Pre-Test Preparation
No special preparation is required for the Widal test:
Fasting: Not necessary.
Medications: Inform your doctor about any medicines you are taking for ongoing treatments, as antibiotics may interfere with test accuracy.
When is the Widal Test Recommended?
Doctors may recommend the Widal test if you experience:
Fever lasting 3-5 days or more.
Abdominal pain or discomfort.
Persistent fatigue and headaches.
Loss of appetite, nausea, or gastrointestinal issues.
Rose-coloured rashes on the chest or abdomen.
Additionally, if you’ve recently travelled to areas where typhoid is common or have been exposed to poor sanitation, the test may be advised.
Widal Test Procedure
The widal test procedure includes:
Blood Sample Collection: A small blood sample is drawn from a vein in your arm by a healthcare practitioner. The sample is then sent to a laboratory for testing.
Agglutination Reaction: The lab mixes your blood sample with antigens (O and H).
Result Analysis: The presence of antibodies in your blood sample causes visible clumping (agglutination). The strength of this reaction indicates the antibody levels (titers).
Report Generation: Results are typically available within 24-48 hours.
Interpreting Widal Test Results
Widal test results can be interpreted in the following ways:
1. Understanding Titers
In the Widal test, titers refer to the measurement of the concentration of antibodies in the blood that react to specific antigens of Salmonella bacteria (O and H antigens). The test measures the highest dilution of the patient's serum that still causes agglutination (clumping) of the bacterial antigens.
2. Limitations in Interpretation
The Widal test is not always definitive. False positives and false negatives can occur due to:
Cross-reaction with other bacteria (e.g., malaria, other Salmonella strains).
Early-stage infections, when antibodies are not yet detectable.
Prior typhoid vaccination, which can cause elevated titers.
Doctors may order repeat tests (7-10 days apart) or additional diagnostics like blood cultures to confirm the diagnosis.
Normal Range and Clinical Implications
Understanding the Widal test normal range is essential for the accurate diagnosis and management of typhoid fever.
Widal Test Normal Range
A result of less than 1:160 is considered normal, indicating no significant presence of Salmonella bacteria antibodies.
| | | |
---
| Antigen | Negative(Normal Range) | Positive Range |
| O | Less than 1:80 | 1:160 or higher |
| H | Less than 1:160 | 1:160 or higher |
Interpreting Widal Test Titers
Widal Test Positive Values
When the Widal test shows antibody titers greater than 1:160, it is considered a Widal test positive values. The interpretation of titers helps determine whether a person has an active infection, has been previously exposed to the bacteria, or is likely unaffected.
A negative result in the Widal test is indicated by O antigen titers less than 1:80 and H antigen titers less than 1:160, suggesting no active infection.
A positive result occurs when titers for both antigens are 1:160 or higher, indicating a current or recent infection.
A four-fold rise in titers between two blood samples taken 7 to 10 days apart is a strong indicator of active infection, particularly for diagnosing acute typhoid fever.
Alternative Tests for Typhoid Diagnosis
If the Widal test results are inconclusive, doctors may recommend:
1. Blood Culture
A blood culture Test is considered one of the most accurate tests for diagnosing typhoid fever. It involves taking a blood sample from the patient and incubating it in a laboratory to allow any Salmonella bacteria to grow. Since Salmonella enters the bloodstream early during infection, blood cultures are especially useful during the first week of illness.
This test typically takes 2 to 5 days to yield results, as the bacteria require time to multiply. Despite this, it remains one of the most reliable methods for confirming a typhoid diagnosis and determining the best course of treatment.
2. Stool and Urine Cultures
A stool culture is a test used to detect Salmonella bacteria in a patient’s stool sample. This test is most useful in diagnosing typhoid after the infection has progressed and the bacteria moves to the intestines. The stool sample is incubated in a lab to allow the bacteria to grow, which helps confirm the infection.
Stool cultures usually take 3 to 5 days to provide results. It is less useful in the early stages of infection. However, they are valuable for confirming typhoid and detecting carriers of the bacteria.
3. PCR Tests
The Polymerase Chain Reaction (PCR) test is a quick and highly accurate way to diagnose typhoid fever. It works by detecting the genetic material of Salmonella bacteria in blood, stool, or other body fluids.
Unlike culture tests, PCR can provide results in just a few hours. This test is especially useful in the early stages of infection when bacterial levels are still low and harder to detect. With its speed and accuracy, the PCR test has become an essential tool for diagnosing typhoid.
Post-Test Care and Follow-Up
If the Widal test confirms typhoid fever, your doctor will prescribe antibiotics such as ciprofloxacin or azithromycin to clear the infection.
After completing your antibiotic treatment for typhoid, it’s important to rest, stay hydrated, and eat a balanced diet to help your body recover. Most people start feeling better within a few days, but full recovery can take several weeks. Continue following your doctor’s instructions and complete the entire course of antibiotics to ensure the infection is fully cleared, even if you feel better.
Regular follow-ups are important to avoid complications like a relapse of the infection. Again, practising good hygiene, such as washing your hands and drinking safe water, is essential.
Conclusion
The Widal test plays a crucial role in diagnosing typhoid fever by measuring the presence of antibodies against Salmonella bacteria in the blood. Understanding the normal range, purpose, procedure, and test results is essential for accurate diagnosis and effective management of the disease. This test, when used alongside clinical assessment and other diagnostic methods, provides valuable insights into the presence and progression of typhoid fever.
If you notice symptoms such as prolonged fever, abdominal pain, or any other signs of typhoid, make sure to see your doctor without delay. A simple Widal test can confirm the diagnosis and help you start the right treatment on time. Book your Widal test with Apollo 24|7 and get quick and accurate results.
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15347 | https://byjus.com/maths/rhombus/ | In Euclidean geometry, a rhombus is a type of quadrilateral. It is a special case of a parallelogram, whose all sides are equal and diagonals intersect each other at 90 degrees. This is the basic property of rhombus. The shape of a rhombus is in a diamond shape. Hence, it is also called a diamond. Check lines of symmetry in a rhombus.
You must have seen the diamond shape in the playing cards. All the rhombi are parallelograms and kites. If all the angles of the rhombus are 90 degrees, then it is a square.
Table of Contents:
Definition
Formulas
Area
Perimeter
Properties
Examples
Now, before we discuss rhombus and its properties, let us know what a quadrilateral is? A quadrilateral is a closed polygon containing 4 sides and 4 vertices enclosing 4 angles. The sum of the interior angles of a quadrilateral is equal to 360 degrees. The quadrilateral is basically of 6 types such as:
Parallelogram
Trapezium
Square
Rectangle
Kite
Rhombus
Rhombus Definition
A rhombus is a special case of a parallelogram. In a rhombus, opposite sides are parallel and the opposite angles are equal. Moreover, all the sides of a rhombus are equal in length, and the diagonals bisect each other at right angles. The rhombus is also called a diamond or rhombus diamond. The plural form of a rhombus is rhombi or rhombuses.
In the above figure, you can see a rhombus ABCD, where AB, BC, CD and AD are the sides of a rhombus and AC and BD are the diagonals of a rhombus.
Properties of Quadrilaterals
Types Of Quadrilaterals
Area of Rhombus
Difference Between Rhombus And Parallelogram
Quadrilaterals for Class 9
Is Square a Rhombus?
Rhombus has all its sides equal and so does a square. Also, the diagonals of the square are perpendicular to each other and bisect the opposite angles. Therefore, a square is a rhombus.
Angles of Rhombus
Below are some important facts about the rhombus angles:
Rhombus has four interior angles.
The sum of interior angles of a rhombus adds up to 360 degrees.
The opposite angles of a rhombus are equal to each other.
The adjacent angles are supplementary.
In a rhombus, diagonals bisect each other at right angles.
The diagonals of a rhombus bisect these angles.
Rhombus Formulas
The formulas for rhombus are defined for two major attributes, such as:
Area
Perimeter
Get more:Mathematics formulas
Area of Rhombus
The area of the rhombus is the region covered by it in a two-dimensional plane. The formula for the area is equal to the product of diagonals of the rhombus divided by 2. It can be represented as:
Area of Rhombus, A = (d1x d2)/2 square units
where d 1 and d 2 are the diagonals of a rhombus.
Perimeter of Rhombus
The perimeter of a rhombus is the total length of its boundaries. Or we can say the sum of all the four sides of a rhombus is its perimeter. The formula for its perimeter is given by:
The perimeter of Rhombus, P = 4a units
Where the diagonals of the rhombus are d 1& d 2 and ‘a’ is the side.
Properties of Rhombus
Some of the important properties of the rhombus are as follows:
All sides of the rhombus are equal.
The opposite sides of a rhombus are parallel.
Opposite angles of a rhombus are equal.
In a rhombus, diagonals bisect each other at right angles.
Diagonals bisect the angles of a rhombus.
The sum of two adjacent angles is equal to 180 degrees.
You will get a rectangle when you join the midpoint of the sides.
You will get another rhombus when you join the midpoints of half the diagonal.
Around a rhombus, there can be no circumscribing circle.
Within a rhombus, there can be no inscribed circle.
You will get a rectangle where the midpoints of the 4 sides are joined together, and the length and width of the rectangle will be half the value of the main diagonal so that the area of the rectangle will be half of the rhombus.
When the shorter diagonal is equal to one of the sides of a rhombus, two congruent equilateral triangles are formed.
You will get a cylindrical surface having a convex cone at one end and concave cone at another end when the rhombus is revolved about any side as the axis of rotation.
You will get a cylindrical surface having concave cones on both the ends when the rhombus is revolved about the line joining the midpoints of the opposite sides as the axis of rotation.
You will get solid with two cones attached to their bases when the rhombus is revolving about the longer diagonal as the axis of rotation. In this case, the maximum diameter of the solid is equal to the shorter diagonal of the rhombus.
You will get solid with two cones attached to their bases when the rhombus is revolving about the shorter diagonal as the axis of rotation. In this case, the maximum diameter of the solid is equal to the longer diagonal of the rhombus.
Rhombus Solved Problems
The sample example for the rhombus is given below.
Question 1:
The two diagonal lengths d 1 and d 2 of a rhombus are 6cm and 12 cm, respectively. Find its area.
Solution:
Given:
Diagonal d 1 = 6cm
Diagonal d 2= 12 cm
Area of the rhombus, A = (d 1 x d 2)/2 square units
A = ( 6 x 12)/2
A = 72/2
A = 36 cm 2
Therefore, the area of rhombus = 36 cm 2
Question 2:
Find the diagonal of a rhombus if its area is 121 cm 2 and length measure of longest diagonal is 22 cm.
Solution:
Given: Area of rhombus = 121 cm 2 and Lets say d 1 = 22 cm.
Using Area of the rhombus formula, A = (d 1 x d 2)/2 square units, we get
121 = (22 x d 2)/2
121 = 11 x d 2
or 11 = d 2
Therefore, the Length of another diagonal is 11 cm.
Question 3:
What are the basic properties of rhombus?
Solution:
The basic properties of the rhombus are:
The opposite angles are congruent.
The diagonals intersect each other at 90 degrees.
The diagonals bisect the opposite interior angles.
The adjacent angles are supplementary.
Question 4:
What is the perimeter of a rhombus whose sides are all equal to 6 cm?
Solution:
Given, the side of rhombus = 6cm
Since all the sides are equal, therefore,
Perimeter = 4 x side
P = 4 x 6
P = 24cm
For more such interesting information on properties of the quadrilateral, register with BYJU’S – The Learning App and also watch videos to learn with ease.
Frequently Asked Questions on Rhombus – FAQs
Q1
Is a rhombus a square?
No, rhombus is not a square but a square is a rhombus.
Q2
Why is a rhombus not a square?
Rhombus is not a square since for a square all the sides are equal and all the interior angles are right angles. However, in rhombus all the interior angles are not equal even though they have equal sides.
Q3
Does a rhombus have 4 right angles?
No, a rhombus does not have four right angles.
Q4
Are all angles of a rhombus equal?
No, in rhombus only the opposite angles are equal.
Q5
Do rhombus angles add up to 360?
We know that the sum of all the interior angles of a quadrilateral is equal to 360 degrees. Hence, the angles of a rhombus add up to 360 degrees.
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15348 | https://math.stackexchange.com/questions/1610334/name-of-famous-triangle-area-formula-a-frac12ab-sinc | geometry - Name of Famous Triangle Area Formula? $A=\frac{1}{2}ab\sin(C)$ - Mathematics Stack Exchange
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Name of Famous Triangle Area Formula? A=1 2 a b sin(C)A=1 2 a b sin(C)
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Asked 9 years, 8 months ago
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I'm sorry for this short question, but I really could not find the answer to this anywhere.
Is there a name for the formula A=1 2 a b sin(C)A=1 2 a b sin(C)?
geometry
area
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edited Jan 13, 2016 at 4:47
JedJed
asked Jan 13, 2016 at 4:35
JedJed
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This particular method has been used in some form since the year 500 500 A.D. in India by Indian mathematician Aryabhata. In practice though, I often simply hear it referred to as the "one-half base times height" formula.JMoravitz –JMoravitz 2016-01-13 04:42:10 +00:00 Commented Jan 13, 2016 at 4:42
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I've heard it called the SAS (side-angle-side) area formula.
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answered Jan 13, 2016 at 4:45
gariepygariepy
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Yup, a google search tells me that seems to be right. I'll accept when the time limit goes away.Jed –Jed 2016-01-13 04:47:03 +00:00 Commented Jan 13, 2016 at 4:47
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I think it is just called the sine formula for the area of a triangle. In any case, it is easily derivable. Given a triangle A B C A B C
we can say that the area of the rectangle is h b h b and thus that the area of the triangle is 1 2 h b 1 2 h b. But we can also say that h=a sin C h=a sinC giving rise to the formula A=1 2 a b sin C A=1 2 a b sinC.
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answered Jan 13, 2016 at 4:43
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15349 | https://arxiv.org/pdf/2201.01351 | QUADRATIC EMBEDDING CONSTANTS OF PATH GRAPHS
WOJCIECH MLOTKOWSKI
Abstract. We characterize positive definiteness for some family of matrices. As an application, we derive the explicit value of the quadratic embedding constants of the path graphs.
Introduction
Let G = ( V, E ) be a graph, with V as the set of vertices and E as the set of edges, i.e. two-element subsets of V . We assume that G is connected , which means that for every
x, y ∈ V there exists a walk connecting x and y: a finite sequence x0, x 1, . . . , x n ∈ V
such that n ≥ 0, x0 = x, x n = y and {xi−1, x i} ∈ E for i = 1 , . . . , n . The distance
d(x, y ) is defined as the smallest possible length n of such a walk. For some connected graphs the distance matrix (d(x, y )) x,y ∈V is conditionally negative definite, equivalently, there exists a function φ which maps V into a Hilbert space H and satisfies
d(x, y ) = ‖φ(x) − φ(y)‖2
for all x, y ∈ V . This motivated the authors of to introduce and study the quadratic embedding constant , defined as (1) QEC( G) := sup
{ ∑
x,y ∈V
d(x, y )f (x)f (y) : f ∈ F 0,1(V )
}
,
where F0,1(V ) denotes the set of all finitely supported functions f : V → R satisfying ∑
x∈V
f (x) = 0 and ∑
x∈V
f (x)2 = 1. In particular, the distance matrix ( d(x, y )) x,y ∈V
is conditionally negative definite if and only if QEC( G) ≤ 0. Several examples and properties were furnished in [1, 2, 3, 4, 5]. Applying the min-max theorem and the Perron-Frobenius theorem, one can observe that if V is finite and λ1(G) ≥ λ2(G) ≥
. . . ≥ λ|V |(G) are the eigenvalues of the distance matrix of G, then
λ2(G) ≤ QEC( G) < λ 1(G).
In this paper we will study finite path graphs , i.e. graphs of the form Pn := ( V, E ), where
V := {1, 2, . . . , n }, E := {{1, 2}, {2, 3}, . . . , {n − 1, n }}.
The eigenvalues λi(Pn) of the distance matrix of Pn were found in . In particular,
λ2(Pn) =
−11 + cos( π/n ) if n is even,
−11 − cos θ∗ if n is odd,
2010 Mathematics Subject Classification. Primary 05C50; Secondary 05C12, 15A15.
Key words and phrases. Positive definite matrix, conditionally negative definite matrix, distance matrix of a graph, path graph, quadratic embedding constant.
1
arXiv:2201.01351v2 [math.CO] 19 Mar 2022 2WOJCIECH MLOTKOWSKI
where θ∗ is the maximal solution of the equation: tan( θ/ 2) tan( nθ/ 2) = −1/n, θ ∈ (0 , π ).
It was observed in [3, Proposition 5.4] that QEC( Pn) is equal to the minimal t such that the matrix (2) [2 min {i, j } + t + t · δi,j ]n−1
i,j =1
is positive definite, i.e. all the eigenvalues of this matrix are nonnegative. This led to inequality: QEC( Pn) ≤ − 1/2, see [3, Theorem 5.6]. Our aim is to provide the exact value:
Theorem 1.1. For n ≥ 2 we have
(3) QEC( Pn) = −11 + cos( π/n ).
Consequently, if n is even then λ2(Pn) = QEC( Pn) and if n is odd then λ2(Pn) <
QEC( Pn).
As an immediate consequence we obtain the quadratic embedding constant for N
and Z regarded as infinite path graphs, with edges {i, i + 1 }, i ∈ N or i ∈ Z, c.f. [3, Theorem 5.7].
Corollary 1.2. QEC( N) = QEC( Z) = −1/2.
The paper is organized as follows. First we examine a family of auxiliary polynomials
Sn(a, b ; t), a, b ∈ R, n ≥ 0. Section 3 is devoted to the study of a family An(s, t ) of matrices, a two-parameter version of (2). We provide formula for det An(s, t ), n < ∞,in terms of polynomials Sn(a, b ; t), and characterize these s, t ∈ R for which An(s, t ), 1 ≤ n ≤ ∞ , is positive definite. Finally we prove Theorem 1.1. 2. A family of polynomials
Now we are going to study a family of polynomials defined by the following recurrence:
S0(a, b ; t) := 1, S1(a, b ; t) := at + b and (4) Sn(a, b ; t) = (1 + 2 t)Sn−1(a, b ; t) − t2Sn−2(a, b ; t)for n ≥ 2, a, b, t ∈ R.
Proposition 2.1. For n ≥ 0 we have deg Sn(a, b ; t) ≤ n and the coefficient at tn in
Sn(a, b ; t) is equal to an − n + 1 . Moreover,
(5) Sn(a, b ; t) = ( at + b)
n−1
∑
k=0
(2n − k − 1
k
)
tk −
n−2
∑
k=0
(2n − k − 3
k
)
tk+2
= 12n+1 √1 + 4 t
[(
2b − 1 + 2( a − 1) t + √1 + 4 t
) (
1 + 2 t + √1 + 4 t
)n
−
(
2b − 1 + 2( a − 1) t − √1 + 4 t
) (
1 + 2 t − √1 + 4 t
)n]
.
The former formula holds for n ≥ 1, while the latter for n ≥ 0 and t 6 = −1/4. For
t = −1/4, n ≥ 0 we have
(6) Sn(a, b ; −1/4) = 14n (4 nb − na − n + 1) .QUADRATIC EMBEDDING CONSTANTS OF PATH GRAPHS 3
Proof. The first statement can be easily proved by induction. Now define: (7) Wn(t) :=
n
∑
k=0
(2n − k + 1
k
)
tk
(see A172431 in OEIS ). Then W0(t) = 1, W1(t) = 2 t + 1, and one can check that
Wn(t) = (1 + 2 t)Wn−1(t) − t2Wn−2(t)for n ≥ 2, so that Wn(t) = Sn(2 , 1; t). Putting Q0(t) := 1 and
Qn(t) := ( at + b)Wn−1(t) − t2Wn−2(t)for n ≥ 1 ( W−1(t) := 0) we have Q1(t) = at + b and Qn(t) = (1 + 2 t)Qn−1(t) − t2Qn−2(t)for n ≥ 1, so Qn(t) coincides with Sn(a, b ; t) and the first formula in (5) holds. Moreover, it is easy to verify by induction that Wn(−1/4) = ( n + 1) /4n, which leads to (6). For n ≥ 0, a, b ∈ R, t 6 = −1/4 put
Tn(t) := 12n+1 √1 + 4 t
[(
2b − 1 + 2( a − 1) t + √1 + 4 t
) (
1 + 2 t + √1 + 4 t
)n
−
(
2b − 1 + 2( a − 1) t − √1 + 4 t
) (
1 + 2 t − √1 + 4 t
)n]
.
Then T0(t) = 1, T1(t) = at + b and from the identity
(
1 + 2 t ± √1 + 4 t
)2
= 2(1 + 2 t)
(
1 + 2 t ± √1 + 4 t
)
− 4t2
we have
Tn(t) = (1 + 2 t)Tn−1(t) − t2Tn−2(t)for n ≥ 2, which implies that Tn(t) = Sn(a, b ; t) for n ≥ 0, t 6 = −1/4.
For later use, we record the following identities:
Proposition 2.2. For Wn(t) defined in (7) we have:
(8) Wn(t) = Sn(2 , 1; t) = Sn(1 , t + 1; t).
Moreover,
(9) Wn(t) = 12n+1 √1 + 4 t
[(
1 + 2 t + √1 + 4 t
)n+1
−
(
1 + 2 t − √1 + 4 t
)n+1 ]
.
Proof. The former equality in (8) was noted in the previous proof, the latter, as well as (9), is a consequence of (5).
In some particular cases we are able to find the roots of Sn(a, b ; t). 4 WOJCIECH MLOTKOWSKI
Proposition 2.3. For n ≥ 1 we have
1
n + 1 Sn(2 , 1; t) =
n
∏
k=1
(
t + 12 + 2 cos ( kπ n+1
))
,(10)
Sn(1 , 1/2; t) =
n
∏
k=1
(
t + 12 + 2 cos ((2 k−1) π
2n
))
,(11)
Sn(1 , 1; t) =
n
∏
k=1
(
t + 12 + 2 cos ( 2kπ
2n+1
))
,(12) 12n + 1 Sn(3 , 1; t) =
n
∏
k=1
(
t + 12 + 2 cos ((2 k−1) π
2n+1
))
.(13)
Proof. First we note that in view of the first part of Proposition 2.1, if a ≥ 1 then deg Sn(a, b ; t) = n and the leading term is ( an − n + 1) tn.Fix a, b, t ∈ R, with 1 + 4 t < 0, and put
w := 2 b − 1 + 2( a − 1) t + √1 + 4 t, z := 1 + 2 t + √1 + 4 t,
with Im( w) > 0, Im( z) > 0, α := arg( w), β := arg( z), 0 < α, β < π . Then cos β = 1 + 2 t
−2t and hence t = −12 + 2 cos β .
In view of (5), we have Sn(a, b ; t) = 0 if and only if w · zn = w · zn, equivalently,
α + nβ = kπ for some k = 1 , 2, . . . , n . Now we consider four cases. 1. If a = 2, b = 1 then w = z, α = β and therefore β = kπ/ (n + 1). 2. If a = 1, b = 1 /2 then α = π/ 2 and then β = (2 k − 1) π/ (2 n). 3. If a = b = 1 then cos α = 1
√−4t =
√1 + cos β
2 = cos( β/ 2) ,
which implies α = β/ 2 and β = 2 kπ/ (2 n + 1). 4. Finally, if a = 3, b = 1 then cos α = 1 + 4 t
√4t(1 + 4 t) = −
√1 + 4 t
4t = −
√1 − cos β
2= − sin( β/ 2) = cos( π/ 2 + β/ 2) ,
therefore α = π/ 2 + β/ 2 and, consequently, β = (2 k − 1) π/ (2 n + 1).
Lemma 2.4. For n ≥ 1, a, b ∈ R we have
(14) Sn(2 , 1; t) · Sn(a, b ; t) − Sn−1(2 , 1; t) · Sn+1 (a, b ; t) = t2n.
Proof. Putting
u± := 2 b − 1 + 2( a − 1) t ± √1 + 4 t, v± := 1 + 2 t ± √1 + 4 tQUADRATIC EMBEDDING CONSTANTS OF PATH GRAPHS 5
we have
(vn+1 + − vn+1
−
) ( u+vn
+
− u−vn
−
) − (vn
+
− vn
−
) ( u+vn+1 + − u−vn+1
−
)
= ( u+ − u−)( v+ − v−)( v+v−)n = 4(1 + 4 t) (4t2)n ,
and, by (5), the formula follows.
Define (15) tn :=
−∞ if n = 1,
−12 + 2 cos( π/n ) if n ≥ 2.
Lemma 2.5. If tn < t < t n+1 then Sn−1(2 , 1; t) > 0, Sn(2 , 1; t) < 0 and if t > t n+1 then
Sn−1(2 , 1; t) > 0, Sn(2 , 1; t) > 0.Proof. First note that the function α 7 → − 1/(2 + 2 cos α) is decreasing with α ∈ (0 , π ). Therefore the statement is a consequence of (10), because for n ≥ 2 we have
−12 + 2 cos (2π/ (n + 1) ) < −12 + 2 cos (π/n ) < −12 + 2 cos (π/ (n + 1) ).
Now we collect properties of polynomials of the form Sn(1 , s + 1; t), which will be applied in the next section.
Proposition 2.6. For s ∈ R, n ≥ 1, we have
(16) Sn(1 , s + 1; t) = Sn(1 , 1; t) + s · Sn−1(2 , 1; t)=
n
∑
k=0
(2n − kk
)
tk + s
n−1
∑
k=0
(2n − 1 − kk
)
tk
= 12n+1 √1 + 4 t
[(
1 + 2 s + √1 + 4 t
) (
1 + 2 t + √1 + 4 t
)n
−
(
1 + 2 s − √1 + 4 t
) (
1 + 2 t − √1 + 4 t
)n]
.
The latter formula is valid for t 6 = −1/4, while
(17) Sn(1 , s + 1; −1/4) = 4ns + 2 n + 1 4n .
Proof. The first equality in (16) can be verified by induction: putting Pn(t) := Sn(1 , 1; t)+
s·Sn−1(2 , 1; t) we have P0(t) = 1, P1(t) = t+1+ s and Pn(t) = (1+2 t)Pn−1(t)−t2Pn−2(t)for n ≥ 2, consequently Pn(t) = Sn(1 , s + 1; t) for all n ≥ 0. Now (16) and (17) are consequences of (5) and (6).
A family of matrices
Define a family of matrices (18) An(s, t ) := [min {i, j } + s + tδ i,j ]ni,j =1 ,
where s, t ∈ R, 1 ≤ n ≤ ∞ . These matrices seem interesting on their own, for the sake of Theorem 1.1 we are particularly interested in the case s = t. We are going to study 6 WOJCIECH MLOTKOWSKI
their determinants and positive definiteness. Note that if s1 ≤ s2, t1 ≤ t2, n1 ≥ n2 and
An1 (s1, t 1) is positive definite then so is An2 (s2, t 2).
Theorem 3.1. For s, t ∈ R, n ∈ N, we have
(19) det An(s, t ) = Sn(1 , s + 1; t) = Sn(1 , 1; t) + s · Sn−1(2 , 1; t).
With the notation of (15), the matrix An(s, t ) is positive definite if and only if
(20) t > t n and Sn(1 , s + 1; t) ≥ 0.
Proof. It is easy to verify (19) for n = 1 , 2. Now assume that n ≥ 3 and let kj denote the jth column of An(s, t ). Then det An(s, t ) = det( k1, . . . , kn) = det( k1, . . . , kn−1, kn − kn−1)and, denoting the transposition by “T”, we have
kn − kn−1 = (0 , . . . , 0, −t, 1 + t)T.
Expanding the determinant along the last column kn − kn−1, we get det An(s, t ) = (1 + t) det An−1(s, t ) + t det B,
where
B = (k′
1
, . . . , k′
n−2
, k′
n−1
η) ,
k′
j
is the jth column of An−1(s, t ) and η := (0 , . . . , 0, −t)T. This yields det B = det An−1(s, t ) − t det An−2(s, t ),
and, consequently, det An(s, t ) = (1 + 2 t) det An−1(s, t ) − t2 det An−2(s, t ),
which completes the proof of (19). Fix t′ ∈ R and put
`k(s) := det Ak(s, t ′) := bk + s · ak.
Then, by Lemma 2.4, we have (21) akbk−1 − ak−1bk = ( t′)2k−2.
With the notation of (15), if tk−1 < t ′ ≤ tk then ak−1 > 0, ak ≤ 0. Consequently, the lines k−1, k cross at some point ( s′, y ′), i.e. k−1(s′) =k(s′) = y′, and, by (21),
y′ = akbk−1 − ak−1bk
ak − ak−1
< 0.
This implies, that for every s ∈ R we have either k−1(s) < 0 ork(s) < 0, therefore the matrix An(s, t ′) is not positive definite for n ≥ k.Now, if we assume that t′ > t n then a1, . . . , a n > 0. Defining sk by `k(sk) = 0, we have s1 ≤ s2 ≤ . . . ≤ sn, by (21). Therefore, if det An(s, t ′) > 0 then s > s n,det Ak(s, t ′) > 0 for all k ≤ n, and hence An(s, t ′) is positive definite. If det An(s, t ′) = 0 then An(s, t ′) is positive definite as pointwise limit of positive definite matrices. The proof is complete.
Note, that in particular, (22) det An(t, t ) = Sn(1 , t + 1; t) = Sn(2 , 1) = Wn(t).
Since Sn(1 , s 0 + 1; t) is the characteristic polynomial of a real symmetric matrix, we have QUADRATIC EMBEDDING CONSTANTS OF PATH GRAPHS 7
Corollary 3.2. For every fixed s0 ∈ R, n ≥ 1, the polynomial Sn(1 , s 0 + 1; t) has only real roots.
Now we consider some particular cases.
Theorem 3.3.
• The matrix An(t, t ) is positive definite if and only if
t ≥ −12 + 2 cos (π/ (n + 1) ).
• The matrix An(−1/2, t ) is positive definite if and only if
t ≥ −12 + 2 cos (π/ (2 n)).
• The matrix An(0 , t ) is positive definite if and only if
t ≥ −12 + 2 cos (2π/ (2 n + 1) ).
• The matrix An(2 t, t ) is positive definite if and only if
t ≥ −12 + 2 cos (π/ (2 n + 1) ).
Proof. First note that by (5) we have Sn(1 , 2t + 1; t) = Sn(3 , 1; t). It is easy to check the statements for n = 1. For 0 < k < m put
tkm := −12 + 2 cos( kπ/m ).
Then, for n > 1, we have t2
n+1
< t 1
n
< t 1
n+1
, t32n < t 1
n
< t 12n, t42n+1 < t 1
n
< t 22n+1 ,
t32n+1 < t 1
n
< t 12n+1 , and the statements follow from Theorem 3.1 and Proposition 2.3.
Now we are able to describe the case n = ∞.
Theorem 3.4. The infinite matrix A∞(s, t ) is positive definite if and only if
1 + 4 t ≥ 0 and 1 + 2 s + √1 + 4 t ≥ 0.
Proof. If 1 + 4 t ≥ 0 and 1 + 2 s + √1 + 4 t ≥ 0 then all the matrices An(s, t ), 1 ≤ n < ∞,are positive definite by (19), (16) and (17), hence so is A∞(s, t ). Now assume that A∞(s, t ) is positive definite. Then t > −1/(2 + 2 cos( π/n )) for every n ∈ N, which implies t ≥ − 1/4. If t = −1/4 then s ≥ − 1/2 in view of (17). If 1 + 4 t > 0 then, putting
q := 1 + 2 t − √1 + 4 t
1 + 2 t + √1 + 4 t ,
we have 0 < q < 1 and, in view of the last formula in (16), the inequality 1 + 2 s + √1 + 4 t ≥
(
1 − 2s + √1 + 4 t
)
qn
holds for every n ≥ 1. This implies that 1 + 2 s + √1 + 4 t ≥ 0. 8 WOJCIECH MLOTKOWSKI -2 -1 1 2 s
-1
1
2
3
t
Figure 1. The range of positive definiteness of A∞(s, t )For example, the following matrix is positive definite: 4A∞(−1/2, −1/4) =
1 2 2 2 2 . . .
2 5 6 6 6 . . .
2 6 9 10 10 . . .
2 6 10 13 14 . . .
2 6 10 14 17 . . .
... ... ... ... ... . . .
.
Path graphs
Now we are ready to prove the main result of this paper.
Proof of Theorem 1.1. In view of [3, Proposition 5.4], QEC( Pn) is equal to the mini-mal t such that An−1(t/ 2, t/ 2) is positive definite. It remains to apply the first part of Theorem 3.3.
Now we indicate explicitly a vector xn on the vertices of Pn for which the supremum in (1) is attained. Put (23) xn := ( xn, 1, x n, 2, . . . , x n,n ) , with xn,i := ( −1) i sin
(2i − 12n π
)
.
If n is even then xn is the eigenvector corresponding to the second eigenvalue λ2(Pn), see .
Proposition 4.1. For n ≥ 2 we have
(24)
n
∑
i=1
xn,i = 0 ,
n
∑
i=1
x2
n,i
= n
2
and
(25)
n
∑
i,j =1
|i − j|xn,i · xn,j = −11 + cos( π/n ) · n
2 .QUADRATIC EMBEDDING CONSTANTS OF PATH GRAPHS 9
Proof. Denoting by “ i” the imaginary unit, we have
n
∑
i=1
(−1) i exp
(2i − 12n πi
)
= − exp ( πi/(2 n)) 1 + ( −1) n
1 + exp ( πi/n ) = −1 − (−1) n
2 cos (π/ (2 n)).
Taking the imaginary part we get the first equation in (24). Similarly, since
n
∑
i=1
exp
((2 i − 1) πn i
)
= 0 ,
we get 2
n
∑
i=1
sin 2
((2 i − 1) π
2n
)
=
n
∑
i=1
[
1 − cos
((2 i − 1) πn
)]
= n,
which completes the proof of (24). Now we will prove (25). For 1 ≤ k < n we have 2
n−k
∑
i=1
sin
(2i − 12n π
)
sin
(2i + 2 k − 12n π
)
=
n−k
∑
i=1
[cos( kπ/n ) − cos ((2 i + k − 1) π/n )]
= ( n − k) cos( kπ/n ) + sin( kπ/n )sin( π/n ) .
Therefore (26)
n
∑
i,j =1
|i − j|xn,i · xn,j = 2
n−1
∑
k=1
k
n−k
∑
i=1
xn,i · xn,i +k
=
n−1
∑
k=1
(−1) kk(n − k) cos( kπ/n ) +
n−1
∑
k=1
(−1) k k sin( kπ/n )sin( π/n ) .
Now applying elementary formulas:
n−1
∑
k=1
kq k = q(1 − qn) − nq n(1 − q)(1 − q)2 ,
n−1
∑
k=1
k(n − k)qk = nq (1 − q)(1 + qn) − q(1 + q)(1 − qn)(1 − q)3
to q := − exp( πi/n ), so that qn = −(−1) n, we obtain
n−1
∑
k=1
k(−1) k exp( kπ i/n ) = −1 − (−1) n + ( −1) nn( exp( −πi/n ) + 1 )
4 cos 2 (π/ (2 n)) ,10 WOJCIECH MLOTKOWSKI
n−1
∑
k=1
k(n − k)( −1) k exp( kπ i/n )= 2n cos (π/ (2 n))( (−1) n − 1) + 2 sin (π/ (2 n))( 1 + ( −1) n)i
8 cos 3 (π/ (2 n)) .
Consequently,
n
∑
k=0
(−1) kk sin( kπ/n )sin( π/n ) = −(−1) nn
4 cos 2(π/ (2 n)) ,
n
∑
k=0
(−1) kk(n − k) cos( kπ/n ) = (( −1) n − 1) n
4 cos 2(π/ (2 n)) ,
which, together with (26), leads to (25).
Acknowledgements
The author is grateful to Nobuaki Obata for useful discussions, in particular for pointing out reference , and also to the anonymous referee for careful reading the paper and useful comments, in particular for simplifying the proof of Theorem 3.1.
References
E. T. Baskoro, N. Obata, Determining finite connected graphs along the quadratic embedding constants of paths, Electron. J. Graph Theory Appl. (EJGTA) 9 (2021), no. 2, 539–560. Zhenzhen Lou, Nobuaki Obata, Qiongxiang Huang, Quadratic Embedding Constants of Graph Joins, arXiv 2020. W. Mlotkowski, N. Obata, On quadratic embedding constants of star product graphs, Hokkaido Math. J. 49 (2020), no. 1, 129–163. N. Obata, Quadratic embedding constants of wheel graphs, Interdiscip. Inform. Sci. 23 (2017), no. 2, 171–174. N. Obata, A. Y. Zakiyyah, Distance matrices and quadratic embedding of graphs, Electron. J. Graph Theory Appl. (EJGTA) 6 (2018), no. 1, 37–60. The On-Line Encyclopedia of Integer Sequences (OEIS), S. N. Ruzieh, D. L. Powers, The distance spectrum of the path Pn and the first distance eigenvector of connected graphs, Linear and Multilinear Algebra, 28 (1990), 75–81.
Instytut Matematyczny, Uniwersytet Wroclawski, Plac Grunwaldzki 2/4, 50-384 Wroclaw, Poland
Email address : mlotkow@math.uni.wroc.pl |
15350 | https://link.springer.com/book/10.1007/978-0-387-44899-2 | Skip to main content
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Advanced Organic Chemistry
Part A: Structure and Mechanisms
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Richard J. Sundberg
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Francis A. Carey
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Parts A and B may stand alone; together, they provide a comprehensive foundation for study in organic chemistry
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Since its original appearance in 1977, Advanced Organic Chemistry has maintained its place as the premier textbook in the field, offering broad coverage of the structure, reactivity and synthesis of organic compounds. As in the earlier editions, the text contains extensive references to both the primary and review literature and provides examples of data and reactions that illustrate and document the generalizations. While the text assumes completion of an introductory course in organic chemistry, it reviews the fundamental concepts for each topic that is discussed.
The two-part fifth edition has been substantially revised and reorganized for greater clarity. Among the changes: Updated material reflecting advances in the field since 2001’s Fourth Edition, especially in computational chemistry; A companion Web site provides digital models for study of structure, reaction and selectivity; Solutions to the exercises provided to instructors online.
The material in Part Ais organized on the basis of fundamental structural topics such as structure, stereochemistry, conformation and aromaticity and basic mechanistic types, including nucleophilic substitution, addition reactions, carbonyl chemistry, aromatic substitution and free radical reactions. Together with Part B: Reaction and Synthesis, the two volumes are intended to provide the advanced undergraduate or beginning graduate student in chemistry with a sufficient foundation to comprehend and use the research literature in organic chemistry.
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Keywords
Aromaticity
Nucleophilic substitution
Pericyclic reaction
bonding
carbon
organic chemistry
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structure
synthesis
Table of contents (12 chapters)
Front Matter
Pages I-XXI
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2. ### Chemical Bonding and Molecular Structure
Francis A. Carey, Richard J. Sundberg
Pages 1-117
3. ### Stereochemistry, Conformation, and Stereoselectivity
Francis A. Carey, Richard J. Sundberg
Pages 119-251
4. ### Structural Effects on Stability and Reactivity
Francis A. Carey, Richard J. Sundberg
Pages 253-388
5. ### Nucleophilic Substitution
Francis A. Carey, Richard J. Sundberg
Pages 389-472
6. ### Polar Addition and Elimination Reactions
Francis A. Carey, Richard J. Sundberg
Pages 473-577
7. ### Carbanions and Other Carbon Nucleophiles
Francis A. Carey, Richard J. Sundberg
Pages 579-628
8. ### Addition, Condensation and Substitution Reactions of Carbonyl Compounds
Francis A. Carey, Richard J. Sundberg
Pages 629-711
9. ### Aromaticity
Francis A. Carey, Richard J. Sundberg
Pages 713-770
10. ### Aromatic Substitution
Francis A. Carey, Richard J. Sundberg
Pages 771-831
Concerted Pericyclic Reactions
Francis A. Carey, Richard J. Sundberg
Pages 833-964
12. ### Free Radical Reactions
Francis A. Carey, Richard J. Sundberg
Pages 965-1071
13. ### Photochemistry
Francis A. Carey, Richard J. Sundberg
Pages 1073-1153
14. ### Back Matter
Pages 1155-1203
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Authors and Affiliations
Department of Chemistry, University of Virginia, Charlottesville
Francis A. Carey,
Richard J. Sundberg
About the authors
Francis A. Carey is a native of Pennsylvania, educated in the public schools of Philadelphia, at Drexel University (B.S. in chemistry, 1959), and at Penn State (Ph.D. 1963). Following postdoctoral work at Harvard and military service, he was appointed to the chemistry faculty of the University of Virginia in 1966. Prior to retiring in 2000, he regularly taught the two-semester lecture courses in general chemistry and organic chemistry. With his students, Professor Carey has published over forty research papers in synthetic and mechanistic organic chemistry.
Professor Sundberg is primarily engaged in teaching and chemical education. Along with Francis A. Carey he is the author of “Advanced Organic Chemistry. Professor Sundberg is also interested in synthetic methodology in heterocyclic chemistry and is the author of “Indoles” in the Best Synthetic Methods Series (Academic Press, 1996).
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Book Title: Advanced Organic Chemistry
Book Subtitle: Part A: Structure and Mechanisms
Authors: Francis A. Carey, Richard J. Sundberg
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Advanced Organic Chemistry
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Origin and history of embezzle
embezzle(v.)
early 15c., "make away with money or property of another, steal," from Anglo-French enbesiler "to steal, cause to disappear" (c. 1300), from Old French em- (see en- (1)) + besillier "torment, destroy, gouge," which is of unknown origin. Sense of "dispose of fraudulently to one's own use," is first recorded 1580s. Related: Embezzled; embezzling.
also from early 15c.
Entries linking to embezzle
variant of embezzle (q.v.).
"theft or misappropriation of funds placed in one's trust or belonging to one's employer," 1540s, from embezzle + -ment. An earlier noun was embezzling (early 15c.).
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15352 | https://arxiv.org/pdf/2403.02627 | Eight-Partitioning Points in 3D, and Efficiently Too ∗
Boris Aronov # Ñ
Department of Computer Science and Engineering, Tandon School of Engineering, New York University, Brooklyn, NY 11201 USA
Abdul Basit # Ñ
School of Mathematics, Monash University, VIC 3800, Australia
Indu Ramesh
Department of Computer Science and Engineering, Tandon School of Engineering, New York University, Brooklyn, NY 11201 USA.
Gianluca Tasinato # Ñ
Institute of Science and Technology Austria, Am Campus 1, 3400 Klosterneuburg, Austria
Uli Wagner # Ñ
Institute of Science and Technology Austria, Am Campus 1, 3400 Klosterneuburg, Austria
Abstract
An eight-partition of a finite set of points (respectively, of a continuous mass distribution) in R3
consists of three planes that divide the space into 8 octants, such that each open octant contains at most 1/8 of the points (respectively, of the mass). In 1966, Hadwiger showed that any mass distribution in R3 admits an eight-partition; moreover, one can prescribe the normal direction of one of the three planes. The analogous result for finite point sets follows by a standard limit argument. We prove the following variant of this result: Any mass distribution (or point set) in R3 admits an eight-partition for which the intersection of two of the planes is a line with a prescribed direction. Moreover, we present an efficient algorithm for calculating an eight-partition of a set of n points in R3 (with prescribed normal direction of one of the planes) in time O∗(n7/3).
2012 ACM Subject Classification Theory of computation → Randomness, geometry and discrete structures → Computational geometry; Mathematics of computing → Discrete mathematics
Keywords and phrases Mass partitions, partitions of points in three dimensions, Borsuk-Ulam Theorem, Ham-Sandwich Theorem
Funding Boris Aronov : Work has been supported by NSF grants CCF 15-40656 and CCF 20-08551, and by grant 2014/170 from the US-Israel Binational Science Foundation. Part of this research was conducted while BA was visiting ISTA in the summers of 2022 and 2023. The visit of BA to ISTA in the summer of 2022 was supported by an ISTA Visiting Professorship. Research of BA also partially supported by ERC grant no. 882971, “GeoScape,” and by the Erdős Center.
Abdul Basit : Work has been supported by Australian Research Council grant DP220102212.
Indu Ramesh : Work supported by a Tandon School of Engineering Fellowship and by NSF Grant CCF-20-08551.
Acknowledgements BA and AB would like to thank William Steiger for insightful initial discussions of the problems addressed in this work.
1 Introduction
Geometric methods for partitioning space, point sets, or other geometric objects are a central topic in discrete and computational geometry. Partitioning results are often proved using
∗
A preliminary version of this work appeared in SoCG’24 .
arXiv:2403.02627v4 [cs.CG] 16 May 2025
2 Eight-partitioning points in 3D
topological methods and also play an important role in topological combinatorics [ 8, 19 , 21 , 24 ]. A classical example is the famous Ham-Sandwich Theorem , which goes back to the work of Steinhaus, Banach, Stone, and Tukey (see [21, Sec. 1] for more background and references). A “discrete” version of this theorem asserts that, given any d finite point sets P1, . . . , P d
in Rd, there is an (affine) hyperplane H that simultaneously bisects all Pi, i.e., each of the two open half-spaces determined by H contains at most |Pi|/2 points, 1 ≤ i ≤ d. This follows (by a standard limit argument, see [ 19 , Sec. 3.1]) from the following “continuous” version: Let μ1, . . . , μ d be mass distributions in Rd, i.e., finite measures such that every open set is measurable and every hyperplane has measure zero. Then there exists a hyperplane H
such that μi(H+) = μi(H−) = 12 μi(Rd) for 1 ≤ i ≤ d, where H+ and H− are the two open half-spaces bounded by H.In this paper, we are interested in another classical equipartitioning problem, first posed by Grünbaum [ 12 ] in 1960: Given a mass distribution (respectively, a finite point set) in Rd,can one find d hyperplanes that subdivide Rd into 2d open orthants, each of which contains exactly 1/2d of the mass (respectively, at most 1/2d of the points)? We call such a d-tuple of hyperplanes a 2d-partition of the mass distribution (respectively, of the point set). For d = 2 , it is an easy consequence of the planar Ham-Sandwich theorem that any mass distribution (or point set) in R2 admits a four-partition; moreover, the four-partition can be chosen such that one of the lines has a prescribed direction (indeed, start by choosing a first line in the prescribed direction that bisects the given mass distribution; by the Ham-Sandwich Theorem, there exists a second line that simultaneously bisects the two parts of the mass on either side of the first line). Alternatively, one can also show that there is always a four-partition such that the two lines are orthogonal. Intuitively, the reason that we can impose such additional conditions is that the four-partitioning problem in the plane is underconstrained : A line in the plane can be described by two independent parameters, so a pair of lines have four degrees of freedom, while the condition that the four quadrants have the same mass can be expressed by three equations, leaving one degree of freedom; either one of the additional constraints uses this extra degree of freedom. In 1966, Hadwiger [ 13 ] gave an affirmative answer to Grünbaum’s question for d = 3 and showed that any mass distribution in R3 admits an eight-partition; moreover, the normal vector of one of the planes can be prescribed arbitrarily. This result was later re-discovered by Yao, Dobkin, Edelsbrunner, and Paterson .
▶ Theorem 1.1 ([ 13 , 25 ]) . Let μ be a mass distribution on R3, and let v ∈ S2. Then there exists a triple of planes (H1, H 2, H 3) that form an eight-partition for μ and such that the normal vector of H1 is v.
More recently, Blagojević and Karasev [ 6] gave a different proof for the existence of eight-partitions and showed the following variant:
▶ Theorem 1.2 ([ 6]) . Let μ be a mass distribution on R3. Then there exists an eight-partition
(H1, H 2, H 3) of μ such that the plane H1 is perpendicular to both H2 and H3.
Our first result is the following alternative version of eight-partitioning, which to the best of our knowledge is new:
▶ Theorem 1.3. Given a mass distribution μ in R3 and a vector v ∈ S2, there exists an eight-partition (H1, H 2, H 3) of μ such that the intersection of the two planes H1 and H2 is a line in the direction of v.B. Aronov, A. Basit, I. Ramesh, G. Tasinato, and U. Wagner 3
As in the case of the Ham-Sandwich Theorem, each of the three theorems above also implies the existence of the corresponding type of eight-partition for finite point sets, again by a standard limit argument (see Lemma A.1). We remark that, in general, d hyperplanes in Rd are described by d2 independent parameters, while the condition that 2d orthants have equal mass can be expressed by
2d − 1 equations. For d = 3 , this leaves 9 − 7 = 2 degrees of freedom, which allows for any one of the additional conditions imposed in Theorems 1.1, 1.2, and 1.3, respectively. On the other hand, for d ≥ 5, we have d2 < 2d − 1, so intuitively Grünbaum’s problem is overconstrained. Avis [ 4 ] made this precise and constructed explicit counterexamples using the well-known moment curve γ = {(t, t 2, . . . , t d) : t ∈ R} in Rd. The crucial fact is that any hyperplane intersects the moment curve γ in at most d points ([ 19 , Lemma 1.6.4]). Thus, for
d ≥ 5, a mass distribution supported on γ admits no 2d-partition because any d hyperplanes intersect γ in at most d2 points, which subdivide γ into at most d2 + 1 intervals, hence there are always at least 2d − d2 − 1 > 0 orthants that do not intersect γ and hence contain no mass. The last remaining case d = 4 of Grünbaum’s problem, i.e., the question whether any mass distribution in R4 admits a 16 -partition by four hyperplanes, remains stubbornly open (see [ 5], [ 8, Conjecture 7.2], [ 19 , pp. 50–51], and [ 21 , Problem 2.1.4] for more background and related open problems). We now turn to the algorithmic question of computing eight-partitions in R3.
▶ Problem 1. Given a set P of n points in R3, in sufficiently general position, compute three planes H1, H 2, H 3 that form an eight-partition of the points. As noted above, the corresponding problem of computing a four-partition of a planar point set can be reduced to finding a Ham-Sandwich cut of two planar point sets that are separated by a line. Megiddo showed that this can be done in linear time. To characterize the complexity of Problem 1, we introduce the following concept. A
halving line (resp., halving plane ) for an n-point set in R2 (resp., R3) in general position is a line (resp., plane) that passes through two (resp., three) of the points and divides the remaining ones as equally as possible. Let h2(n) (resp., h3(n)) denote the maximum number of halving lines (resp., planes) for an n-point set in R2 (resp., R3). The best known upper and lower bounds for h2(n) are O(n4/3), due to Dey [ 9 ], and Ω( ne
√log n), due to Tóth [ 23 ], respectively. For h3(n), the best-known bounds are are O(n5/2), due to Sharir, Smorodinsky, and Tardos , and Ω( n2e
√log n), due to Tóth . By a result of Lo, Matoušek, and Steiger [ 18 , Proposition 2], a Ham-Sandwich cut of
n points in R3 can be computed in time O∗(h2(n)) = O∗(n4/3) (see also [ 14 ]), where the
O∗(·)-notation suppresses polylogarithmic factors. However, computing eight-partitions in R3
appears to be significantly more difficult. Unlike the planar four-partition problem, there is no known way of reducing it to the computation of a Ham-Sandwich cut. In particular, given two planes H1 and H2 that four-partition a finite point set P in R3 (in the sense that every one of the four open quadrants determined by H1 and H2 contains at most |P |/4 points), there generally need not exist a third plane H3 such that H1, H 2, H 3 form an eight-partition. We note that, for fixed dimension d ≥ 3, the best known algorithm for computing Ham-Sandwich cuts in Rd runs in time O(nd−1−αd ) where αd > 0 is a constant depending only on d [18 ]. When the dimension is part of the input, a decision variant of the problem becomes computationally hard, see, e.g., . A brute-force algorithm that checks every triple of halving planes solves Problem 1 in time comparable to O(h3(n)3) = O(n15 /2). Yao et al. [ 25 ] and Edelsbrunner [ 10 ] gave a O(n6)-time algorithm that computes an eight-partition (with a prescribed normal direction for one of 4 Eight-partitioning points in 3D
the planes, as in Theorem 1.1) by an exhaustive search, using the fact that only two planes need to be identified. Fixing one plane and performing a brute-force search for the remaining two would yield an algorithm with a running time comparable to O(h3(n)2) = O(n5).Here, we present, to our knowledge, the fastest known algorithm for Problem 1. Roughly speaking, our algorithm runs in time near-linear in h3(n) rather than quadratic in it. Slightly more precisely, our algorithm runs in time near-linear in nh 2(n), which is not known to be
o(h3(n)) , but for which the best known upper bound is strictly stronger; see Theorem 4.2 and Fact 3.2:
▶ Theorem 1.4 (Algorithm) . An eight-partition of n points in general position in R3, with a prescribed normal vector for one of the planes, can be computed in time O∗(nh 2(n)) , hence
O∗(n7/3); here, the O∗(·)-notation suppresses polylogarithmic factors.
Our algorithm can be seen as a constructive version of Hadwiger’s proof [ 13 ]. We start by bisecting the point set by a plane with a fixed normal direction, which partitions the initial point set into two subsets of “red” and “blue” points, respectively, of equal size. After that, our algorithm finds two more planes that simultaneously four-partition both the red and the blue points. It remains an open question whether Theorem 1.2 or our own Theorem 1.3 can also be used to obtain an efficient algorithm for Problem 1. It would also be interesting to decide whether there is an algorithm for Problem 1 with running time o(nh 2(n)) .
2 The topological result 2.1 Notation and preliminaries
In what follows, it will often be convenient to assume that the mass distributions we work with have connected support , where the support of a mass distribution μ is Supp (μ) := {x ∈
R3 : μ(Br (x)) > 0 for every r > 0} and Br (x) denotes the ball of radius r > 0 centered at x.By a standard limit argument (see Lemma A.3), the existence of eight-partitions for mass distributions with connected support implies the existence of eight-partitions for the general case. Hereafter, unless stated otherwise, we assume, without loss of generality, that every mass distribution has connected support. We denote the scalar product of two vectors x, y ∈ R3 by x · y := ∑3
i=1
xiyi. A vector
v ∈ R3 \ { 0} and a scalar a ∈ R determine an (affine) plane
H = Hv (a) := {x ∈ R3 : x · v = a},
together with an orientation of H (given by the direction of the normal vector v). We denote by −H := H−v (−a) the affine plane with the same equation as H but with opposite orientation. The oriented plane H determines two open half-spaces, denoted by
H+ := {x ∈ R3 : x · v > a } and H− := {x ∈ R3 : x · v < a }.
More generally, let H = ( H1, . . . , H k) be an ordered k-tuple of (oriented) planes in R3, k ≤ 3.In what follows, it will be convenient to identify the set {+, −} with the group Z2 (where the group operation is multiplication of signs). Elements of {+, −} k = Zk
2
are strings of signs of length k, and we will denote by + = + · · · + the identity element of Zk
2
.For α = ( α1, . . . , α k) ∈ Zk
2
= {+, −} k, we define the open orthant determined by H and α
as OH
α
:= Hα1
1
∩ · · · ∩ Hαk
k
. Given a mass distribution μ in R3, we say that an ordered k-tuple
H = ( H1, . . . , H k) of planes ( k ≤ 3) forms a 2k-partition of μ if every orthant contains 1/2kB. Aronov, A. Basit, I. Ramesh, G. Tasinato, and U. Wagner 5
of the mass, i.e., μ(OH
α
) = μ(R3)/2k for every α ∈ { +, −} k. For k = 1 , 2, 3, this corresponds to the notions of bisecting, four-partitioning, and eight-partitioning μ as mentioned in the introduction. Analogously, we say that H forms a 2k-partition of a finite point set P in R3 if
|P ∩ O H
α
| ≤ |P |
2k
for all α.We will parameterize oriented planes in R3 by S3, where the north pole e4 and the south pole −e4 map to the plane at infinity with opposite orientations. For this we embed R3
into R4 via the map (x1, x 2, x 3) 7 → (x1, x 2, x 3, 1) . An oriented plane in R3 is mapped to an oriented affine 2-dimensional subspace of R4 and is extended (uniquely) to an oriented linear hyperplane. The unit normal vector on the positive side of the linear hyperplane defines a point on the sphere S3. Hence, there is a one-to-one correspondence between points v in
S3 \ { e4, −e4} and oriented affine planes Hv in R3. The positive side of the plane at infinity is
R3 for v = e4 and ∅ for v = −e4. Hence H+
−v
= H−
v
for every v. Note that planes at infinity cannot arise as solutions to the measure partitioning problem, since they produce empty orthants. Therefore we do not need to worry about the fact that the sphere includes these. We parameterize triples of planes (called plane configurations ) in R3 by (S3)3, and denote by Hv the triple corresponding to v ∈ (S3)3. Given a mass distribution μ on R3, for each
v ∈ (S3)3 and α ∈ Z32 \ { +}, we set
Fα(v, μ ) = ∑
β∈Z32
(−1) p(α,β )μ(OHv
β
).
where p(α, β ) is the number of coordinates where both α and β are −. The functions Fα
were also utilized in the proof of Theorem 1.1 in . As an example, with H := Hv = ( H1, H 2, H 3) and α = − − + ∈ Z32 \ { +}, we obtain
F−− +(H, μ ) = ∑
β∈Z32
(−1) p(α,β )μ(OH
β
) = ∑
β∈Z32:p(α,β )=0
μ(OH
β
) − ∑
β∈Z32:p(α,β )=1
μ(OH
β
)= (μ(OH
+++
) + μ(OH
++ −
) + μ(OH−− +) + μ(OH−−− ))
− (μ(OH−++ ) + μ(OH−+−) + μ(OH
+−+
) + μ(OH
+−−
))
= μ(H+1 ∩ H+2 ) + μ(H−
1
∩ H−
2
) − μ(H−
1
∩ H+2 ) − μ(H+1 ∩ H−
2
).
When μ is clear from context, we write Fα(H) instead of Fα(H, μ ). The definitions of alternating sums for a pair of planes or a single plane are analogous. The alternating sums have the following properties which will play an important role in the proof below.
▶ Observation 2.1. Let μ be a mass distribution and fix k = 2 , 3.
(i) Let α ∈ Zk−12 \ { +} and let H = ( H1, . . . , H k) be a k-tuple of planes. Then F+α(H) =
Fα(( H2, . . . , H k)) (the equivalent statement holds for any other entry of a k-tuple
(α1, · · · , α k) instead of just for α1).
(ii) A k-tuple H of planes 2k-partitions if and only if Fα(H) = 0 for every α ∈ Zk
2
\ { +}.
Proof. (i) Since every hyperplane has null measure it follows that, for any β ∈ Zk−12
μ(O(H2,...,H k )
β
) == μ(OH
+β
) + μ(OH−β ).
If ˜α = + α then, from the definition of F˜α, for any β ∈ Zk−12 the two orthants OH
+β
and
OH−β are counted with the same sign in the sum, therefore
F˜α(H) = ∑
β∈Zk−12
p(α,β )=0
(μ(OH
+β
) + μ(OH−β )) − ∑
β∈Zk−12
p(α,β )=1
(μ(OH
+β
) + μ(OH−β )) = Fα(( H2, . . . , H k)) .6 Eight-partitioning points in 3D
(ii) It is clear that, if H is a 2k-partition, then all the alternating sums are 0. We will prove the other implication. Suppose first that k = 1 and that H = H. The only alternating sum is F−(H) =
μ(H+) − μ(H−) and F−(H) = 0 implies that H bisects μ.Suppose now that k = 2 and that H = ( H1, H 2). By (i) and the statement for a single plane, F+−(H) = 0 and F−+(H) = 0 imply that both H1 and H2 bisect. Therefore, if
λ := μ(OH
++
), we have that
0 = F−− (H) = μ(OH
++
)+ μ(OH−− )−μ(OH−+)−μ(OH
+−
) = λ+λ−( 12 −λ)−( 12 −λ) = 4 λ−1;
hence λ = 14 as desired. Finally, suppose that k = 3 and that H = ( H1, H 2, H 3). By (i) and the statement for single planes and for pairs of planes, we have that all planes Hi bisect and all pairs (Hi, H j )
four-partition. Therefore, if λ := μ(OH
+++
), we have that
0 = F−−− (H) = μ(OH
+++
) + μ(OH
+−−
) + μ(OH−+−) + μ(OH−− +)
− μ(OH−++ ) − μ(OH
+−+
) − μ(OH
++ −
) − μ(OH−−− )= λ + λ + λ + λ − ( 14 − λ) − ( 14 − λ) − ( 14 − λ) − ( 14 − λ) = 8 λ − 1;
hence λ = 18 as claimed. ◀
2.2 The main topological result
Our goal is to prove the following result, which is a more precise statement of Theorem 1.3:
▶ Theorem 2.2. Given a mass distribution μ and a direction u ∈ S2, there exists a triple
H = ( H1, H 2, H 3) of oriented planes that eight-partition μ so that the oriented direction of the intersection H1 ∩ H2 is u.
By Lemma A.3, it is sufficient to prove Theorem 2.2 for mass distributions with connected support. We require the following technical lemma about partitioning a mass distribution on
R2, due to Blagojević and Karasev [ 6 ]. For completeness, the proof is given in Appendix B.
▶ Lemma 2.3 (Four-partitioning a mass distribution in R2 [ 6]) . Let μ# be a mass distribution (with connected support) on R2 and v ∈ S1. Then there exists a pair (ℓ1, ℓ 2) of lines in R2
that four-partitions μ# and such that v bisects the angle between ℓ1 and ℓ2.Moreover, if we orient ℓ1 and ℓ2 so that ℓ1 is in the first direction clockwise from v, and
ℓ2 is in the first direction counterclockwise, the oriented pair is unique and the lines depend continuously on v.
Proof of Theorem 2.2. Without loss of generality, let u = (0 , 0, 1) . Our proof proceeds in two steps. In the first step, we construct a map Φ : S1 × S3 → R4 whose zeros codify eight-partitions of μ; then we prove that Φ is equivariant with respect to a suitable choice of actions of G := Z4 × Z2 on the two spaces. In the second step we show that any continuous
G-equivariant map Ψ : S1 × S3 → R4 has to have a zero.
Step 1: The key step in constructing the map Φ is to show that we can parameterize pairs of planes that have intersection direction u and four-partition μ, by a vector in S1.We project μ onto u⊥, the plane orthogonal to u, to obtain a mass distribution μ# on
R2. Specifically, identifying the plane with R2, let A ⊆ R2 and set A × R to be the cylinder over A in the u-direction. Then μ#(A) = μ(A × R).B. Aronov, A. Basit, I. Ramesh, G. Tasinato, and U. Wagner 7ℓ2(v)
ℓ1(v)
v
μ#
g1·
ℓ1(g1·v)
ℓ2(g1·v)
g1·v
μ#
Figure 1 Example of the action of g1.
Let v ∈ S1 ⊆ R2. By Lemma 2.3, there are two oriented lines ℓ1(v) and ℓ2(v) (that we can interpret as points (ℓi
1
, ℓ i
2
, a i) ∈ S2) in the plane u⊥ such that v bisects the angle between the two and ℓ1, ℓ 2 four-partition the projected measure μ#. Define Hi(v) :=(ℓi
1
(v), ℓ i
2
(v), 0, a i(v)) ∈ S3 to be ℓi(v) × R the (oriented) span of ℓi(v) and u; the two planes now four-partition μ and have the desired intersection direction. Now let g1 be a generator of Z4×{ +} ⊆ G and define its action on S1 by a counterclockwise rotation by π
2
. We use g1 · v to denote the action of g1 on v. Then, by the uniqueness in Lemma 2.3, we have that (see Figure 1): ⃗ℓ1(g1 · v) = ⃗ ℓ2(v) and ⃗ ℓ2(g1 · v) = −⃗ ℓ1(v). (1) Using this construction, we can define a function S1 → S3 × S3 by v 7 → (H1(v), H 2(v)) .It follows from eq. (1) that g1 · v is mapped to (H2(v), −H1(v)) . Therefore, if we fix the corresponding action 1 of Z4 on S3 × S3, the map is Z4-equivariant. The group {e} × Z2 acts by antipodality on S3; therefore, if G acts on (S3 × S3) × S3
component-wise, the map Φ : S1×S3 → (S3 × S3)×S3 defined as Φ( v, w ) := ( H1(v), H 2(v), H w)
is G-equivariant. By construction, the first two planes are always a four-partition of the mass distribution, therefore by Observation 2.1, a configuration Φ( v, w ) is an eight-partition if and only if the four alternating sums with α3 = − (i.e., α = + + −, − + −, + − − and − − − ) are 0.To compute the action of G on the alternating sums, it is enough to specify what happens on g1 (a generator of Z4 × { e}) and g2 (the generator of {e} × Z2). Recalling that
g1 · Φ( v, w ) = ( H2(v), −H1(v), H w) and applying Observation 2.1(i), we obtain
F+−− (g1 · Φ( v, w )) = F+−− (( H2(v), −H1(v), H w)) = F−− (( −H1(v), H w)) = μ(−H1(v)+ ∩ H+
w
) + μ(−H1(v)− ∩ H−
w
) − μ(−H1(v)− ∩ H+
w
) − μ(−H1(v)+ ∩ H−
w
)= μ(H1(v)− ∩ H+
w
) + μ(H1(v)+ ∩ H−
w
) − μ(H1(v)+ ∩ H+
w
) − μ(H1(v)− ∩ H−
w
)= −F−+−(Φ( v, w ))
1Formally, for any (x, y )∈S3×S3the generator g1of Z4× { +} ⊆ Gacts by g1·(x, y ) = ( y, −x).
8 Eight-partitioning points in 3D
Similar computations imply that, if we act with g1, we obtain
F++ −(g1 · Φ( v, w )) = F++ −(Φ( v, w )) ,F−+−(g1 · Φ( v, w )) = F+−− (Φ( v, w )) , and
F−−− (g1 · Φ( v, w )) = −F−−− (Φ( v, w )) ,
while acting with g2 produces
F++ −(g2 · Φ( v, w )) = −F++ −(Φ( v, w )) ,F+−− (g2 · Φ( v, w )) = −F−+−(Φ( v, w )) ,F−+−(g2 · Φ( v, w )) = −F+−− (Φ( v, w )) , and
F−−− (g2 · Φ( v, w )) = −F−−− (Φ( v, w )) ,
for every (v, w ) ∈ S1 × S3.Finally, we can choose a linear G-action on R4 that is consistent with the previous equations. In particular, if we define
g1 · (x, y, z, u ) = ( x, −z, y, −u) and g2 · (x, y, z, u ) = ( −x, −y, −z, −u),
then the map Ψ : S1 × S3 → R4, given by
(v, w ) 7 → (F++ −(v, w ), F +−− (v, w ), F −+−(v, w ), F −−− (v, w ))
is G-equivariant, i.e., Ψ( g1 ·v, w ) = g1 ·Ψ( v, w ) and Ψ( v, −w) = −Ψ( v, w ). By Observation 2.1, the zeros of Ψ are exactly the configurations of planes that eight-partition the measure and have the desired intersection property.
Step 2: Suppose now, for a contradiction, that Ψ does not have a zero. This means that it is possible to define a G-equivariant map Ψ : S1 × S3 → S3 by Ψ( v, w ) := Ψ( v,w )
∥Ψ( v,w )∥
.Denote by Ψa, for a ∈ S1, the map Ψa : S3 → S3, Ψa(w) = Ψ( a, w ); this function has two key properties:
(i) for any a ∈ S1, Ψa is antipodal;
(ii) for any a, b ∈ S1, let γ : [0 , 1] → S1 be a parametrization of the arc between γ(0) = a
and γ(1) = b. Then Ψa and Ψb are homotopic via H : S3 × [0 , 1] → S3 with H(t, x ) = Ψγ(t)(x).For any n ≥ 1, the group of orthogonal matrices O(n) contains exactly two connected components, distinguished by the sign of the determinant. Since the map g1 : S3 → S3 is induced by a matrix with determinant −1, it is homotopic to any other orthogonal linear map with determinant −1. In particular, it is homotopic to the reflection r along the last coordinate and, thus, deg( g1) = deg( r) = −1. Combining everything together we have:
deg(Ψ a) = deg(Ψ g1·a) = deg( g1 · Ψa) = deg( g1) deg(Ψ a) = − deg(Ψ a).
Hence deg(Ψ a) = 0 , contradicting the Borsuk-Ulam theorem (see [19, Theorem 2.1.1]). ◀
Theorem 2.2, along with Lemma A.1, immediately implies the following.
▶ Theorem 2.4. Let P ⊆ R3 be a finite set of points and p ∈ S2 a fixed direction. Then there exists a triple H = ( H1, H 2, H 3) of oriented planes that eight-partitions P , so that the oriented direction of the intersection H1 ∩ H2 is p.B. Aronov, A. Basit, I. Ramesh, G. Tasinato, and U. Wagner 9
3 Levels in arrangements of planes
Let P ⊆ R3 be a set of n points in general position. Specifically, we assume that the points in P satisfy the following: no four in a plane, no three in a vertical plane, and no two in a horizontal plane. Recall that a halving plane for a point set in R3 in general position is a plane that passes through three of the points and divides the remaining points as equally as possible; h3(n) is the maximum number of halving planes for an n-point set in R3.The duality transform maps points in R3 to planes in R3 and vice versa. Specifically, the point p = ( p1, p 2, p 3) ∈ R3 is mapped to the non-vertical plane p∗ : z = p1x + p2y − p3 in R3,and vice versa. See [15, Chapter 25.2] for standard properties of the duality transform. Let H be a set of planes in R3 in general position. Specifically, we assume that the planes are non-vertical, every triple of planes in H meets in a unique point, and no point in R3 is incident with more than three of the planes. The planes in H partition R3 into a complex of convex cells, called the arrangement of H and denoted by A(H). The k-level in A(H) is defined as the closure of the set of all points which lie on a unique plane of the arrangement and have exactly k − 1 planes below it. Note that the k-level is a piecewise linear surface in R3 whose faces are contained in planes of H. The complexity of a level is the total number of vertices, edges and faces contained in the level. When k = ⌊(|H| + 1) /2⌋, the k-level is called the median level of the arrangement. Duality, k-levels, and complexity of levels are defined analogously in R2. Let g2(n) (resp.,
g3(n)) be the maximum complexity of any k-level in an arrangement of n planes in R2 (resp.,
R3). It is well-known that h2(n) = Θ( g2(n)) and that h3(n) = Θ( g3(n)) (see [ 2, Theorem 3], [11, Theorem 3.3]). The main object of interest in bounding the complexity of our algorithm is the intersection of median levels of two arrangements of disjoint sets of planes. We first show that the complexity of this is proportional to h3(n), in the worst case.
▶ Fact 3.1. Let ℓ(n) be the maximum complexity of the intersection of a level in the arrangement A(R) and a level in the arrangement A(B) of disjoint sets R and B of planes in R3, with H := A ∪ B in general position and n := |H|. Then ℓ(n) = Θ( h3(n)) .
Proof. Let L be the intersection of level k of A(R) and level k′ of A(B). As the planes of H
are in general position, L is (a disjoint union of) a collection of (open) edges and vertices in A(R ∪ B). In fact, L is a collection of cycles and bi-infinite curves so its complexity is asymptotically determined by the number of its edges. A point on an edge of L has the property that it lies on one plane of R and one plane of B, and has exactly k + k′ − 2 planes of R ∪ B below it. Hence L is contained in the
(k + k′ − 1) -level of A(R ∪ B). It follows that the complexity of L is bounded above by the complexity of a level in an arrangement of n planes, implying ℓ(n) ≤ g3(n) = O(h3(n)) . This proves the upper bound. For the lower bound, suppose first that n is of the form 4k + 6 and let P ⊆ R3 be a set
n/ 2 = 2 k + 3 points in general position achieving the maximum number h3(2 k + 3) of halving planes. Let P ′ be a copy of P translated by a sufficiently small distance ε > 0 in a generic direction. We then slightly perturb the points of P ∪ P ′ to ensure general position. For a point p ∈ P , we denote by p′ its copy (or twin ) in P ′. Consider the sets R∗ (red) and B∗
(blue) of k points each obtained as follows: for each pair (p, p ′), we assign one point to R∗
and the other to B∗ uniformly and independently at random. Recall that a halving plane is defined by a triple of points and divides the remaining points as evenly as possible. Let π1 be a halving plane of P defined by a triple (a, b, c ), so that π1 has exactly k points of P on each side. Consider the set S := {a, b, c, a ′, b ′, c ′}. We 10 Eight-partitioning points in 3D
claim that, with constant probability, there exists a plane π2 that passes through a red point and a blue point of S that are not twins, and has precisely one red and one blue point of S
on each side. As our perturbation ε is arbitrarily small, π2 partitions the remaining points of R∗ ∪ B∗
in the same manner as π1 partitions P . Let R and B be the sets of planes dual to points in R∗ and B∗, respectively. In the dual, the plane π2 corresponds to a point π∗
2
on an edge of the arrangement A(R ∪ B) that lies on level k of A(R) and of A(B). Therefore it lies on the curve L of intersection of two k-levels and this curve contains all the points π∗
2
(as we range over halving planes π1 of P ). Since different halving planes correspond to partitioning P in different ways, the planes
π2 partition P ∪ P ′ in different ways and, hence, points π∗
2
arising from different halving planes of P lie on different edges of A(R ∪ B). By construction, the number of edges of L is
Ω( h3(n)) in expectation, completing the lower bound proof for n = 4 k + 6 .Finally, for a general n, we write n = 4 k + 6 + c, for 0 ≤ c ≤ 5, we apply the above construction to n − c and, at the end, add c planes in general position to the set R lying above all vertices of A(R ∪ B). It is easily checked that this addition does not reduce the size of L. ◀
Note, however, that in our application, the sets of points R∗ and B∗ are strictly separated, which is not the case in the above lower-bound construction. We show that, with this additional constraint, the complexity of L is O(nh 2(n)) and that there exist pairs of separated point sets R∗ and B∗ that achieve this bound.
▶ Fact 3.2. Let ℓ∗(n) be the maximum complexity of the intersection of two levels in A(R)
and A(B) as in Fact 3.1, with the additional constraint that the dual sets R∗ and B∗ are strictly separated by a plane. Then ℓ∗(n) = Θ( nh 2(n)) .
Proof. Upper bound : Let L be the intersection of level k in A(R) and level k′ in A(B).Without loss of generality, we consider vertices of L that are the intersection of one plane in
R and two planes in B. Such a vertex has k − 1 planes of R and either k − 1 or k − 2 planes of B below it. We work with the dual point sets R∗ (red) and B∗ (blue), strictly separated by the plane π. Here, a vertex as above corresponds to a plane passing through one red point and two blue points, with k − 1 red points and k − 1 or k − 2 blue points above it. We count all such planes σa passing through a fixed red point a ∈ R∗.Let B′ be a set of points obtained by radially projecting points of B∗ onto π with center a.Let ℓa be the line π ∩ σa. Since R∗ and B∗ are separated by π, ℓa has k − 1 or k − 2 points of B′ on one side of it. Hence, the dual of ℓa is a vertex on level k (or its complement) in the planar arrangement of the dual of B′ in π. Therefore, there are at most 2g2(|B′|) choices for the plane σa, and at most |R| · 2g2(|B|) = O(nh 2(n)) such planes overall. Planes passing through exactly one blue point and two red points are handled symmetri-cally, completing the proof of the upper bound.
Lower bound : Once again, it is sufficient to make the argument for n of the form 4k + 3
for a positive integer k; the general case is handled as in the lower bound proof of Fact 3.1. Consider a set R∗ of 2k + 2 points realizing h2(2 k + 2) lying in the xy -plane, scaled to fit in the rectangle (0 , 1) × (0 , ε ), for a sufficiently small ε > 0.Let B∗ be a set of 2k + 1 points equally spaced on a unit circle in the plane x = 0 , centered at the origin, so that no point lies in the xy -plane. Note that R∗ and B∗ are separated by the plane x = δ, for a sufficiently small δ > 0.B. Aronov, A. Basit, I. Ramesh, G. Tasinato, and U. Wagner 11
By making ε small enough, we can ensure that any halving line of R∗ passes arbitrarily close to the origin. In particular, any pair of a halving line of R∗ and a point of B∗ define a halving plane that passes through two points of R∗ and one of B∗, and has exactly 2k points of R∗ ∪B∗ on each side. The number of such halving planes is (2 k +1) h2(2 k +2) = Ω( nh 2(n)) .Finally, the points of R∗ ∪ B∗ can be perturbed to satisfy the general position assumption without reducing this number. So we have constructed two separated sets of points R∗ and B∗ with the property that there are Ω( nh 2(n)) halving planes spanned by three of the points that simultaneously bisect both sets. The lower bound follows by considering the dual sets R and B. ◀
4 The algorithm
We can deduce the existence of eight-partitions of a finite point set P ⊂ R3 of a certain advantageous form from Theorem 1.1.
▶ Observation 4.1. Let k > 0 be an integer and P ⊆ R3 be a set of n = 8 k + 7 points in general position. Then, there exists a triple of planes (H1, H 2, H 3) that eight-partitions P
with the following properties:
(i) H1 is horizontal (i.e., parallel to the xy -plane) and passes through the z-median point of P . From here on, we refer to the 4k + 3 points that lie below (resp., above) H1 as
red (resp., blue ) points and denote the sets R (resp. B).
(ii) H2 and H3 each contain exactly three points, and each open octant contains exactly k
points.
(iii) H2, H 3 each bisect R and B, and the pair (H2, H 3) four-partitions both R and B.Furthermore, H2 and H3 contain at least one point of each color.
Proof. Since the set X := {(H1, H 2, H 3) : H1 is horizontal } ⊂ (S3)3 is compact, by Theo-rem 1.1 and Lemma A.1, there exists a configuration H∞ = ( H1, H 2, H 3) that eight-partitions the point set with H1 horizontal. Along with the general position assumption, this implies that H1 contains only the z-median point. This proves (i). To see (ii), note that any eight-partition has at most k points of P in each of the eight open octants, one point in H1, and at most three points in each of H2 and H3, by general position, for a total of at most 8k + 1 + 2 · 3 = 8 k + 7 = n points. So, in fact, all the inequalities are equalities: there must be exactly k points in each open quadrant and exactly three points of R ∪ B in each of H2 and H3.It remains to show (iii). By the preceding paragraph, it is straightforward to see that
(H2, H 3) four-partitions both R and B. By Corollary A.2, we have that any pair (Hi, H j )
four-partitions P . Since (H1, H 2) four-partitions P , each quadrant formed by (H1, H 2) has at most ⌈(8 k + 7) /4⌉ = 2 k + 1 points. H1 has 4k + 3 points on each side. Hence, we obtain that H2 bisects R and B, and, in particular, contains at least one point of each color. A symmetric argument shows that H3 bisects both R and B, and contains at least one point of each color. This completes the proof. ◀▶ Theorem 4.2 (Computation of an eight-partition) . Let P ⊆ R3 be a set of n > 0 points in general position and v ∈ S2. An eight-partition (H1, H 2, H 3) of P , with v being the normal vector of H1, can be computed in time O∗(n + ℓ∗(n)) .
▶ Remark. Since ℓ∗(n) = Θ( nh 2(n)) = O(n7/3) by Fact 3.2, we can compute an eight-partition in time O∗(n7/3).12 Eight-partitioning points in 3D
The rest of this section is devoted to the proof of Theorem 4.2. We assume, without loss of generality, that v = e3 = (0 , 0, 1) is the vertical vector, so H1 is required to be horizontal. If n ≤ 7, the statement holds trivially — set H1 to be the horizontal plane containing any point of P , and H2, H 3 to contain at most three distinct points each, so that the octants do not contain any points. From here on, we will assume that n = 8 k + 7 , for an integer
k > 0. If n is not of this form, we may add dummy points to P (in general position) until the number of points is of the required form and run the algorithm. Once the algorithm terminates, we discard the dummy points, resulting in an eight-partition with at most k
points in each octant. We now describe the algorithm to construct an eight-partition of P satisfying the properties in Observation 4.1. Let H1 be the horizontal plane containing the z-median point of P , and, without loss of generality, identify H1 with the xy -plane. Now consider the sets R and B of
4k + 3 points each lying below and above, respectively, H1. We further assume, without loss of generality, that B is contained in the half-space x < 0 and R is contained in the half-space
x > 0. Otherwise, since no point in R ∪ B has z = 0 by the general position assumption, there exists a plane H containing the y-axis and with sufficiently small negative slope in the
x direction such that all red points are below H and all blue points are above H. Applying a generic sheer transformation (so as not to violate the general position assumption) that fixes the xy -plane and maps H to the plane x = 0 , we obtain point sets with the required properties. Let R∗ = {p∗ : p ∈ R} be the set of red planes dual to points in R and set A(R) := A(R∗)
to be the arrangement formed by the set R∗. The set of blue planes B∗ and the blue arrangement A(B) are defined analogously. We will write A := A(R ∪ B) for the arrangement formed by the planes in R∗ ∪ B∗. For a (dual) point p ∈ R3, we set R+
p
, R −
p
⊆ R∗ to be the set of red planes lying strictly above and below p, respectively. For a pair p, q of (dual) points, put
X(p, q ) := |R+
p
∩ R+
q
| − | R+
p
∩ R−
q
| − | R−
p
∩ R+
q
| + |R−
p
∩ R−
q
|.
The sets B+
p
, B −
p
⊆ B∗ and the function Y (p, q ) are defined analogously for B∗.Let L be the intersection of the median levels of A(B) and A(R). Let m be the complexity, i.e., the number of vertices and edges, of L. By the following lemma, we have that L is a connected y-monotone polygonal curve and is an alternating sequence of edges and vertices of A terminated by half-lines.
▶ Lemma 4.3. L is a connected y-monotone curve.
Proof. Recall that B lies in the quadrant x < 0, z < 0 and R lies in the quadrant x > 0, z > 0.Hence, the dual planes in B∗ and R∗ have equations of the form z = ax + by + c with a < 0
and a > 0, respectively. Consider the intersection of R∗ ∪ B∗ with the plane Πd : y = d. The intersection of the plane z = ax + by + c is the line z = ax + ( bd + c), so planes in B∗ correspond to lines with negative slope and planes in R∗ correspond to lines with positive slope. In particular, the median levels of lines corresponding to B∗ and R∗ are graphs of piecewise-linear total functions that are decreasing and increasing, respectively. It follows that the two curves intersect exactly once. This intersection point corresponds to the intersection of L with the plane Πd.By general position, L is a union of vertex-disjoint cycles and bi-infinite paths composed of edges and vertices of A, since incident to every vertex of A contained in L are precisely two edges belonging to L. By monotonicity in y, L must be a single bi-infinite chain. ◀B. Aronov, A. Basit, I. Ramesh, G. Tasinato, and U. Wagner 13
In fact, L can be computed efficiently using standard tools [ 1, 7], which we outline now.
▶ Lemma 4.4 (Computing the intersection of two levels [ 1 , 7]) . The intersection curve L of the median level of A(B) and the median level of A(R) can computed in time O∗(n + m),where m is the complexity of the curve and O∗(·) notation hides polylogarithmic factors.
Proof. We use the standard dynamic data structure for ray-shooting queries in the intersection of half-spaces; the currently fastest algorithm is due to Chan [ 7], see also earlier work of Agarwal and Matoušek . A starting ray of L can be computed by computing the intersection of the median levels in the vertical plane Πd : y = d for a small enough d, defined as in the proof of Lemma 4.3, in linear time, using an algorithm of Megiddo . Consider a point p on the initial edge of L (infinite in the −y-direction). It lies on one plane of π ∈ B and one plane π′ ∈ R. Let ℓ be the intersection line of π and π′, and consider the half line ρ of ℓ starting at p and infinite in the +y-direction. The planes of B ∪ R (besides
π and π′) are classified into those lying above p and those lying below it. Call the first set U
and the second D. We preprocess the intersection of the set of lower half-spaces defined by the planes of U and the intersection of the set of upper half-spaces defined by the planes of D
for dynamic ray shooting and shoot with ρ. The earlier of the two intersections identifies the first plane π2 of (B ∪ R) \ { π, π ′} that ρ meets. This is the next vertex of A(B ∪ R) on L;
L turns here. If π2 belongs to B, L now follows the intersection line of π2 and π′. Otherwise it follows the intersection line of π and π2. Past the intersection, the sets U and D need to be updated and we continue, following the next ray, until we trace all of L.The only cost besides the initial computation of p are identifying U and D in O(n) time, initializing the dynamic structure, in O∗(n) time, and performing two ray shots and O(1)
updates on U and D per vertex of L, each at a cost of O∗(1) . ◀▶ Note. As we construct L, we can store it as a sequence of vertices and edges. Each edge is associated with the red-blue pair of planes containing it. An endpoint of an edge is contained in an additional plane. For each consecutive edge/vertex pair (e, v ), in either direction, we record which new plane contains v together with its color. We now return to the computation of the eight-partition (H1, H 2, H 3). By the general position assumption, H2 and H3 cannot be vertical, so H2 and H3 correspond to vertices in A,by Observation 4.1. With the above setup, we can reformulate the problem of computing H2
and H3 as follows.
▷ Claim 4.5 (The dual alternating sign functions). Computing H2 and H3 is equivalent to identifying a pair of vertices p, q ∈ L such that Y (p, q ) = X(p, q ) = 0 .
Proof. By Observation 4.1(ii), the eight-partition (H1, H 2, H 3) has exactly k points in each of the eight open octants. Setting p := H∗
2
and q := H∗
3
, we obtain that |R±
p
∩ R±
q
| =
|B±
p
∩ B±
q
| = k for all combinations of signs. Therefore Y (p, q ) = X(p, q ) = 0 , as claimed. We now argue the other direction. Let p, q ∈ L be vertices such that X(p, q ) = Y (p, q ) = 0 .Since p and q lie on L, H2 := p∗ and H3 := q∗ bisect both R and B and contain exactly three points each, at least one of each color. Hence, it suffices to show that (H2, H 3) is a four-partition of both R and B, i.e., |R±
p
∩ R±
q
|, |B±
p
∩ B±
q
| ≤ k for all combinations of signs. Indeed, this implies that each octant formed by (H1, H 2, H 3) contains exactly k points, completing the proof. Let ar := |R+
p
∩ R+
q
|, br := |R+
p
∩ R−
q
|, cr := |R−
p
∩ R+
q
|, and dr := |R−
p
∩ R−
q
|. Define
ab, b b, c b, d b analogously for the blue planes. Without loss of generality, for a contradiction, suppose ar > k .14 Eight-partitioning points in 3D
We first consider the case ar ≥ k + 2 . Since p lies on the median level of A(R), we have
ar + br ≤ | R+
p
| = 2 k + 1 , implying br ≤ k − 1. Similarly, since q lies on the median level of A(R), we have cr ≤ k − 1. Recall that, by assumption, X(p, q ) = ar + dr − br − cr = 0 ,implying dr = br + cr − ar ≤ k − 4. Hence, ar + br + cr + dr ≤ 4k − 4, so p and q together are contained in 4k + 3 − (ar + br + cr + dr ) ≥ 7 red planes, contradicting the general position assumption. We may now assume ar = k + 1 . Following the same reasoning we obtain br ≤ k, cr ≤ k,and dr = br + cr − ar ≤ k − 1. This implies ar + br + cr + dr ≤ 4k, and, in particular, that p
and q together are contained in at least 3 red planes. Now consider the blue planes and note that ab + bb + cb + db ≤ 4k — this is clear if each of sets B±
p
∩ B±
q
contains at most k blue planes, otherwise it follows by the same argument as above. Hence, p and q together are contained in 4k + 3 − (ab + bb + cb + db) ≥ 3 blue planes. By Observation 4.1(ii), p and q are contained in at most 6 planes of R∗ ∪ B∗. Combined with the argument above, this implies p and q together are contained in exactly 3 planes of each color. It follows that ar + br + cr + dr = ab + bb + cb + db = 4 k, which, by the assumption
ar = k + 1 , implies br = cr = k and dr = k − 1. Since |R−
p
| = 2 k + 1 and br + dr = 2 k − 1,there are exactly 2 red planes containing q below p. Similarly, since |R−
q
| = 2 k + 1 and
br + dr = 2 k − 1, there are exactly 2 red planes containing p below q. But then p and q are contained in a total of 4 red planes, a contradiction. This exhausts all possibilities and, hence, |R±
p
∩ R±
q
|, |B±
p
∩ B±
q
| ≤ k for all combinations of signs, completing the proof. ◀
To summarize, once we construct L in time O∗(n + m), to compute an eight-partition, it is sufficient, by Claim 4.5, to find two vertices p, q ∈ L satisfying X(p, q ) = Y (p, q ) = 0 . In particular, it is possible to construct an eight-partition by enumerating all the Θ( m2) pairs of vertices in L; the exact running time depends on how efficiently one can check candidate pairs. Below, we describe how to reduce the amount of remaining work to O(( m + n) log m).
Speed up For simplicity of later calculations, we orient L in the y-direction (which is possible by Lemma 4.3) and view it as an alternating sequence of edges and vertices, starting and ending with a half-line. We denote these elements by x1, x 2, . . . , x m. Recall that the goal is to identify i, j ∈ [m] so that xi, x j are vertices and X(xi, x j ) = Y (xi, x j ) = 0 .We extend the definition of X, Y as follows. If xi, x j are both edges, we pick arbitrary points p and q in the open edges xi and xj , respectively, and set X(xi, x j ) := X(p, q ) and
Y (xi, x j ) := Y (p, q ); the cases where xi is an edge or xj is an edge, but not both, are handled analogously. Note that specifying the (open) edges containing p and q is sufficient to determine X and Y , hence the definition is unambiguous. Define π : [ m]2 → Z2 by
π(i, j ) := ( X(xi, x j ), Y (xi, x j )) .
With this setup, our goal is to identify a point (i, j ) ∈ [m]2 (corresponding to a pair of vertices on L) such that π(i, j ) = 0.We define a grid curve C to be a sequence of points (i1, j 1), . . . , (it, j t) in Z2 such that
(iℓ+1 , j ℓ+1 ) ∈ { (iℓ, j ℓ), (iℓ±1, j ℓ), (iℓ, j ℓ±1)} for each ℓ ∈ [t − 1] . In words, a grid curve is a walk in Z2 which, at each step, does not move at all or moves by exactly one unit up/down/left/right. A curve is closed if (i1, j 1) = ( it, j t). A grid curve is simple if non-consecutive points are distinct (we think of the start and end points as consecutive) — so the curve does not revisit a point after it moves away from the point. To each grid curve C, we associate a piecewise linear curve C in R2, consisting of line segments connecting consecutive points (iℓ, j ℓ), (iℓ+1 , j ℓ+1 ) of C for each ℓ ∈ [t − 1] . For a B. Aronov, A. Basit, I. Ramesh, G. Tasinato, and U. Wagner 15
curve C not passing through the origin 0, the winding number w(C) about 0 is defined in the standard way. Slightly abusing notation, we set w(C) := w(C). In particular, provided
C misses the origin,
w(C) = w(C) =
0 if C does not wind around 0,
n > 0 if C winds around 0 n times counterclockwise,
n < 0 if C winds around 0 −n times clockwise. We omit the rigorous definition of w(C) as a contour integral in the complex plane since it does not add to the discussion and, instead, refer the reader to [17, Chapter 4.4.4]. Our algorithm proceeds as follows:
Step 1 Set C := T (see Definition 4.6). If π(C) meets 0, then stop — we have found a point that maps to 0 (see Lemma 4.8). Otherwise π(C) has odd winding number, by Lemma 4.9.
Step 2 Construct two simple closed curves C1, C2 so that (a) C = C1 + C2, (b) at least one of π(C1), π (C2) has odd winding number (unless they meet 0), (c) the regions enclosed by C1 and C2 partition the region enclosed by C, and (d) the area enclosed by each of C1, C 2 is a fraction of that enclosed by C (see Lemma 4.11).
Step 3 If π(C1) or π(C2) meets 0, then stop — we found a point that maps to 0, by Lemma 4.8.
Step 4 Compute w(π(C1)) and w(π(C2)) , and replace C with the one with the odd winding number. Goto Step 2. We now proceed to fill in the details, starting with the definition of the initial curve T .
▶ Definition 4.6 (The triangular grid curve T ). The simple closed grid curve T traverses a
triangular path defined as follows: Starting with the bottom horizontal side of the grid [m]2, T traverses the points
(x1, x 1), (x2, x 1), . . . , (xm, x 1),
continuing along the right vertical side of the grid [m]2 along the points
(xm, x 1), (xm, x 2), . . . , (xm, x m),
finally, traversing back diagonally along
(xm, x m), (xm−1, x m), (xm−1, x m−1), (xm−2, x m−1), . . . , (x1, x 2), (x1, x 1).
Along the diagonal side of T , we are really only interested in points of the form (xℓ, x ℓ)
with ℓ ∈ [m]. However, since this doesn’t give a grid curve, we “patch” it up by introducing intermediate points. Fortunately, this does not change the desired properties of T .
▶ Lemma 4.7. If C is a grid curve, then π(C) is a grid curve. Moreover, if L has already been computed, π(C) can be computed in time O(n + |C|).
Proof. Consider a step in C from (xi, x j ) to (xi+1 , x j ), where xi is an edge of L and xi+1 is a vertex. Then xi+1 is contained in the planes that contain xi and one additional plane H.Suppose, without loss of generality, that H is red. This means that the cardinality of one of the sets R±
p
changes by one. Hence, the cardinality of R±
p
∩ R±
q
, for each combination of signs, changes by at most one — if H contains q, nothing changes. It follows that the function X changes by at most one, and the function Y remains unchanged. 16 Eight-partitioning points in 3D
Note that, up to symmetry, only one such transition or its reverse occurs in a single step of C. We’ve shown that each step causes either X or Y (but not both) to change by at most one, and, hence, π(C) is a grid curve. The computation can be carried out in constant time per incident edge-vertex pair of C,since L has been already computed, after a O(n)-time initialization that computes X, Y at an arbitrary starting point of C by brute force. ◀
Lemma 4.7 immediately implies the following.
▶ Lemma 4.8. If π(C) meets 0, then some point of C is mapped to 0.
A key property of the triangular grid curve T is the following.
▶ Lemma 4.9. If 0̸ ∈ π(T ), then w(T ) is odd.
Proof. Let N := 4 k + 2 , and let H, V, D be the images (under π) of the horizontal, vertical, diagonal sides of T , respectively. Note that π(T ) is the concatenation of H, V, and D in that order. Observe that if p = q = xi with i ∈ [m], then |R+
p
∩ R−
q
| = |R−
p
∩ R+
q
| = 0 . Hence,
X(xi, x i) ∈ { 4k + 1 , 4k + 2 } depending on whether xi is contained in one or two red planes. Similarly, Y (xi, x i) ∈ { 4k + 1 , 4k + 2 }. Hence π(xi, x i) ∈ { (N, N ), (N − 1, N − 1) } and, in particular, π(xi, x i) = ( N, N ) if xi is an edge. Along with Lemma 4.7, this implies that the grid curve D is a closed walk on the points in {N − 2, N − 1, N, N + 1 }2 \ { 0} starting and ending at the point (N, N ).Noting that x1 and xm are half-lines contained in the same red plane, and that every red plane that lies above x1 lies below xm and vice versa, we obtain π(xm, x 1) = ( −N, −N ).Hence, H is a grid curve from the point (N, N ) to (−N, −N ) and V is a grid curve from the point (−N, −N ) to (N, N ).The discussion above implies that w(T ) is equal to the winding number of the concatena-tion of V and H. We argue below that V is the image of H under a rotation by 180 ° around the origin, i.e., the map (x, y ) 7 → (−x, −y). Since, by assumption, neither H nor V contain
0, the concatenation of H and V has odd winding number as claimed. Specifically, we need to show that π(xi, x 1) = −π(xm, x i). Since π is symmetric in the two arguments, it suffices to show that π(x1, x i) = −π(xm, x i). As mentioned before, every plane that lies above x1 lies below xm and vice versa. That is, R+
x1
= R−
xm
and R−
x1
= R+
xm
,and similarly B+
x1
= B−
xm
and B−
x1
= B+
xm
. The claim is now obvious from the definition of the functions X and Y . ◀▶ Lemma 4.10. If w(π(C)) is odd, then there is a point (i, j ) ∈ Z2 enclosed by C with
π(i, j ) = 0.
Proof. A grid square S is a simple closed grid curve of the form
(i, j ), (i + 1 , j ), (i + 1 , j + 1) , (i, j + 1) , (i, j )
with (i, j ) ∈ Z2. A square is S for some grid square S. If there is a grid square S enclosed by C such that π(S) meets 0, then we are done by Lemma 4.8. Otherwise, note that π(C) is the sum of the images of the corresponding squares. Hence, there is a grid square S with
w(π(S)) odd. By Lemma 4.7, π(S) is a grid curve. By enumerating all possibilities (see Fig. 2), we conclude that w(π(S)) cannot be odd. ◀B. Aronov, A. Basit, I. Ramesh, G. Tasinato, and U. Wagner 17 π(i, j )π(i + 1 , j )π(i + 1 , j + 1) π(i, j + 1) π(i, j )π(i + 1 , j + 1) π(i, j + 1) π(i + 1 , j )π(i + 1 , j )π(i + 1 , j + 1) π(i, j )π(i, j + 1) π(i, j )π(i + 1 , j )π(i + 1 , j + 1) π(i, j + 1) (i + 1 , j )(i + 1 , j + 1) (i, j )(i, j + 1) π(i, j )π(i + 1 , j )π(i + 1 , j + 1) π(i, j + 1)
Figure 2 Up to symmetries, the different possibilities for the image under π of a grid square S,which is always always a grid curve in Z2, by Lemma 4.7. Note that the image cannot have odd winding number.
Next, we show how to decompose a curve C. We restrict our attention to “trapezoidal” curves: Such a curve is the boundary of the intersection of the region bounded by the initial triangle T with a grid-aligned rectangle. This property is maintained inductively.
▶ Lemma 4.11. Given a trapezoidal curve C whose image misses 0, with w(π(C)) odd, we can construct two trapezoidal curves C1 and C2 so that
(i) the region R surrounded by C is partitioned into region R1 surrounded by C1 and region
R2 surrounded by C2.
(ii) area( R1), area( R2) ≤ c · area( R), for an absolute constant c < 1.
(iii) either 0 is in the image of C1 and C2 or w(π(C)) = w(π(C1)) + w(π(C2)) .
Proof. We already noted that the image of a grid square cannot have odd winding number, therefore R is not a grid square. As long as R is at least two units high, divide it by a horizontal grid chord into two near-equal-height pieces (that is, the two parts have equal height, or the lower one is one smaller) producing two regions R1 and R2. The curves C1
and C2 are the boundaries of the regions (refer to Fig. 3). If the height of R is one, perform a similar partition by a vertical chord into to near-equal-width pieces. Property (i) is satisfied by construction. If the image of the new chord misses 0, then both C1 and C2 avoid 0 and property (iii) follows from the properties of the winding number on the plane. Finally, an easy calculation shows that, if the split height/width is even, then each part contains at most 3/4 of the original area; this fraction can rise to 5/6 if R has odd width or length (the extreme case is achieved at width/length of three), which proves property (ii). ◀▶ Lemma 4.12. Given a simple closed grid curve C in [m]2 we can determine whether π(C)
contains a zero. If not, we can compute w(π(C)) , all in time O(|C| + n).
Proof. By Lemma 4.7, we can trace π(C) step by step and, in particular, detect whether
0 ∈ π(C). So suppose this is not the case. Consider the (open) ray ρ from the origin directed to the right in Z2. To determine the winding number of the curve π(C) not passing through the origin, it’s sufficient to count how many times the curve crosses the ray ρ. We can compute the number of times π(C)
crosses ρ by computing π for every vertex of C in order and counting the number of times
(X, 0) occurs along it, with X > 0.18 Eight-partitioning points in 3D C
C1
C2
T
(x1, x 1)
(xm, x m)
Figure 3 Curve C; the blue region is bounded by C1, and the red by C2, with the horizontal dividing chord drawn dashed.
As π(C) may partially overlap ρ, we need to check whether π(C) arrives at (X, 0) with
X > 0 from below the X-axis and (possibly after staying on the axis for a while) departs into the region above X-axis, or vice versa. That would count as a signed crossing. Arriving from below and returning below, or arriving from above and returning above, does not count as a crossing. All of the above calculations can be done in time O(1) per step of π(C), after proper initialization, by Lemma 4.7. ◀
Running time We now analyze the running time of the algorithm we described. We can traverse a length-O(m) closed grid curve C, compute its image π(C), and check whether it passes through the origin in time O(m + n) by Lemma 4.7. One can check whether π(C)
winds around the origin an odd number of times by Lemma 4.12, also in time O(m + n).The number of rounds of the main loop of the algorithm is O( log m), as the starting curve cannot enclose an area larger than O(m2) and areas shrink by a constant factor in every iteration, by Lemma 4.11. Combining everything together, we conclude that L can be computed in O∗(n + m) time, and the algorithm can then identify the pair of vertices of L corresponding to an eight-partition in at most O( log m) rounds, each costing at most
O(m + n). This concludes the proof of Theorem 4.2.
References
1Pankaj K. Agarwal and Jirí Matousek. Dynamic half-space range reporting and its applications.
Algorithmica , 13(4):325–345, 1995.
2Artur Andrzejak, Boris Aronov, Sariel Har-Peled, Raimund Seidel, and Emo Welzl. Results on k-sets and j-facets via continuous motion. In Proceedings of the 14th Annual Symposium on Computational Geometry , pages 192–199, 1998.
3Boris Aronov, Abdul Basit, Indu Ramesh, Gianluca Tasinato, and Uli Wagner. Eight-Partitioning Points in 3D, and Efficiently Too. In 40th International Symposium on Computa-tional Geometry (SoCG 2024) , pages 8:1–8:15, 2024.
4David Avis. Non-partitionable point sets. Information Processing Letters , 19(3):125–129, 1984.
B. Aronov, A. Basit, I. Ramesh, G. Tasinato, and U. Wagner 19
5 Pavle Blagojević, Florian Frick, Albert Haase, and Günter Ziegler. Topology of the Grünbaum– Hadwiger–Ramos hyperplane mass partition problem. Transactions of the American Mathe-matical Society , 370(10):6795–6824, 2018.
6 Pavle V. M. Blagojević and Roman Karasev. Partitioning a measure in R3. Manuscript, 2016.
7 Timothy M. Chan. A dynamic data structure for 3-D convex hulls and 2-D nearest neighbor queries. J. ACM , 57(3):16:1–16:15, 2010.
8 Jesús A. De Loera, Xavier Goaoc, Frédéric Meunier, and Nabil H. Mustafa. The discrete yet ubiquitous theorems of Carathéodory, Helly, Sperner, Tucker, and Tverberg. Bulletin of the American Mathematical Society , 56(3):415–511, 2019.
9 Tamal K Dey. Improved bounds for planar k-sets and related problems. Discrete & Computa-tional Geometry , 19:373–382, 1998.
10 Herbert Edelsbrunner. Edge-skeletons in arrangements with applications. Algorithmica ,1:93–109, 1986.
11 Herbert Edelsbrunner. Algorithms in Combinatorial Geometry , volume 10 of EATCS Mono-graphs on Theoretical Computer Science . Springer, 1987.
12 Branko Grünbaum. Partitions of mass-distributions and of convex bodies by hyperplanes.
Pacific Journal of Mathematics , 10(4):1257–1261, 1960.
13 H. Hadwiger. Simultane Vierteilung zweier Körper. Archiv der Mathematik (Basel) , 17:274–278, 1966.
14 Dan Halperin and Micha Sharir. Arrangements. In Jacob E. Goodman, Joseph O’Rourke, and Csaba D. Tóth, editors, Handbook of Discrete and Computational Geometry , pages 551–580. CRC Press LLC, 3rd edition, 1997.
15 Sariel Har-Peled. Geometric Approximation Algorithms . American Mathematical Society, USA, 2011.
16 Christian Knauer, Hans Raj Tiwary, and Daniel Werner. On the computational complexity of Ham-Sandwich cuts, Helly sets, and related problems. In STACS’11 , pages 649–660, 2011.
17 Steven George Krantz. Handbook of Complex Variables . Springer, 1999.
18 Chi-Yuan Lo, Jiří Matoušek, and William Steiger. Algorithms for ham-sandwich cuts. Discrete & Computational Geometry , 11(4):433–452, 1994.
19 Jiří Matoušek. Using the Borsuk–Ulam Theorem . Springer Berlin Heidelberg, 2003.
20 Nimrod Megiddo. Partitioning with two lines in the plane. Journal of Algorithms , 6(3):430–433, 1985.
21 Edgardo Roldán-Pensado and Pablo Soberón. A survey of mass partitions. Bulletin of the American Mathematical Society , 2021.
22 Micha Sharir, Shakhar Smorodinsky, and Gábor Tardos. An improved bound for k-sets in three dimensions. Discrete & Computational Geometry , 26(2):195–204, 2001.
23 Géza Tóth. Point sets with many k-sets. Discrete & Computational Geometry , 26(2):187–194, 2001.
24 Rade T. Živaljević. Topological methods. In Jacob E. Goodman, Joseph O’Rourke, and Csaba D. Tóth, editors, Handbook of Discrete and Computational Geometry , pages 551–580. CRC Press LLC, 3rd edition, 1997.
25 F. Frances Yao, David P. Dobkin, Herbert Edelsbrunner, and Mike Paterson. Partitioning space for range queries. SIAM Journal on Computing , 18(2):371–384, 1989.
A Limit arguments
We prove some standard facts using limit arguments.
▶ Lemma A.1 (Limit argument for finite point sets) . Let X ⊆ (S3)3 be a compact subset such that, for all μ mass distributions (with connected support) on R3 there is a plane configuration
H ∈ X that eight-partitions μ; then for any set P of points in R3, there is a configuration
H∞ ∈ X that eight-partitions the point set. 20 Eight-partitioning points in 3D
Proof. Let n be the number of points in P . Let μi be the measure defined as follows. At every point in P , place a ball of radius εi = 12i with uniform density and total mass (1 − εi)/n ;finally, add a normal Gaussian distribution “on the background,” with total mass εi/n . Note that the total measure μi of the complement of the union of balls is less than εi.By choosing i large enough, we can assume that a plane can intersect a collection of balls only if their centres are coplanar; hence, without loss of generality, we can assume that this happens for i = 1 .By assumption, there is a plane configuration Hi that eight-partitions the mass μi for each i; by compactness of X, there is a subsequence Hij that converges to some limit H∞;up to reindexing we can assume that the original sequence Hi does. The obtained limit point eight-partitions the original point set P : in fact, there is a i0 big enough such that for every orthant α ∈ Z32 and any m ≥ i0, every point p ∈ P ∩ O H∞
α
is “far away” (e.g., at least 1/2i0 )from the planes in the configuration Hm; hence
1 − εm
n |P ∩ O H∞
α
| ≤ μm(OHm
α
) = 18 .
Taking the limit in m we obtain the desired result. ◀▶ Corollary A.2. Let X and P as above. If H∞ = ( H1, H 2, H 3) is the configuration constructed in the proof of Lemma A.1, then any plane Hi in H∞ bisects P and any pair
(Hi, H j ) four-partitions the points.
Proof. For simplicity we show the result for the first plane H1 and the pair (H1, H 2), all the other cases are identical. First, construct μi and Hi = ( Hi, 1, H i, 2, H i, 3) converging to the limit H∞ as in the proof of Lemma A.1. Again, by choosing i0 big enough we obtain that, for any m ≥ i0 and any sign α ∈ Z2
every point p ∈ P ∩ Hα
1
is sufficiently far form Hm, 1 hence
1 − εm
n |P ∩ Hα
1
| ≤ μm(Hαm, 1) = 12 .
Similarly, for any pair of signs (α1, α 2) ∈ Z22, any point p ∈ P ∩ Hα1
1
∩ Hα2
2
is sufficiently far from both Hm, 1 and Hm, 2, therefore
1 − εm
n |P ∩ Hα1
1
∩ Hα2
2
| ≤ μm(Hα1
m, 1
∩ Hα2
m, 2
) = 14 .
By taking the limit we obtain the desired result. ◀▶ Lemma A.3 (Limit argument for mass distributions with possibly disconnected support) . Let
X be a compact set in (S3)3 such that, for any mass distribution with connected support there is a configuration H ∈ X that eight-partitions the measure. Let μ be a “general” mass distribution. Then there is a plane arrangement H∞ ∈ X that eight-partitions μ.
Proof. Define εi := 12i and let μi the measure defined, on a measurable set A ⊆ R3, as
μi(A) := (1 − εi) μ(A) + εiN (A),
where N is a normal Gaussian distribution on R3. Then, μi is a mass distribution with connected support hence there is a configuration Hi that eight-partitions μi. By compactness, up to taking a subsequence, Hi converges to a configuration H∞.Now, for any α ∈ Z32, we have that
(1 − εi) μ(OHi
α
) ≤ μi(OHi
α
) = 18 .B. Aronov, A. Basit, I. Ramesh, G. Tasinato, and U. Wagner 21
For any fixed measure ˜μ, the map H → ˜μ(OH
α
) is continuous; hence by taking the limit in i we obtain that, for all α ∈ Z32,
μ(OH∞
α
) ≤ 18 .
However, since
∑
α∈Z32
μ(OH∞
α
) = μ(R3) = 1 ,
it follows that all the previous inequalities are equalities. ◀
B Four-partitioning in the plane with a prescribed bisector
This section is devoted to the proof of Lemma 2.3. For convenience, we restate the lemma here. Both the lemma and proof are due to .
▶ Lemma 2.3 (Four-partitioning a mass distribution in R2 [ 6]) . Let μ# be a mass distribution (with connected support) on R2 and v ∈ S1. Then there exists a pair (ℓ1, ℓ 2) of lines in R2
that four-partitions μ# and such that v bisects the angle between ℓ1 and ℓ2.Moreover, if we orient ℓ1 and ℓ2 so that ℓ1 is in the first direction clockwise from v, and
ℓ2 is in the first direction counterclockwise, the oriented pair is unique and the lines depend continuously on v.
Proof. Suppose, without loss of generality, that v = (0 , 1) . We first prove existence. Let
α ∈ [0 , π/ 2] , and rotate v counterclockwise and clockwise by angle α to obtain uα and wα,respectively. Then, since the measure has connected support, there exist unique lines ℓα and
mα that are perpendicular to uα and wα, respectively, and bisect μ. Note that v bisects the angle between ℓα and mα.The (oriented) lines ℓα and mα partition the plane into four octants, which we label
PN , P E , P S , P W (north, east, south, west) in the obvious manner. Since both lines are bisecting, we have
μ(PN ) = μ(PS ) = x, μ(PW ) = μ(PE ) = μ(R2)/2 − x = y.
When α tends to 0, PW and PE tend to empty sets and evidently x > y for α sufficiently close to 0. When α tends to π/ 2, PN and PS tend to empty sets and then x < y for α
sufficiently close to π/ 2. Since x depends continuously on α, we must have x = y somewhere in between, by the intermediate value theorem. Thus, we have existence. We now show uniqueness. Assume we have a partition PN , P E , P S , P W with angle α and another partition QN , Q E , Q S , Q W with angle α′. Assume without loss of generality that
α′ ≤ α. Moreover, since for α′ = α the partition is defined uniquely, we may assume α′ < α .Let p = ℓα ∩ mα and consider the following cases:
p ∈ QN : In this case PN ⊂ QN and both sets have the same measure. This contradicts connectivity of the set where the density is positive, since the density is positive in QS
and in PN , it must be positive somewhere in QN \ PN , implying μ(QN ) > μ (PN ).
p ∈ QE : In this case PW ⊂ QW and we obtain a similar contradiction.
p ∈ QS and p ∈ QW are similar to considered cases. Since the lines ℓα, m α, ℓ α′ , and mα′ are distinct, this covers all cases. In each case, we obtain a contradiction, hence, we have uniqueness. 22 Eight-partitioning points in 3D
As for continuity, it follows from the standard fact that a map with compact codomain and closed graph must in fact be continuous. The codomain is compact since we are only interested in directions of the halving lines that afterwards produce halving lines continuously, thus working with S1 × S1 as the space of parameters. ◀ |
15353 | https://bostonpads.com/real-estate-glossary/point-of-beginning-pob/ | Point of Beginning (POB) Definition - Real Estate Glossary - Boston Pads
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In a metes-and-bounds legal description, the starting point of the survey, situated in one corner of the parcel; all metes-and-bounds descriptions must follow the boundaries of the parcel back to the point of beginning.
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15354 | https://www.diva-portal.org/smash/get/diva2:1453911/FULLTEXT01.pdf | INOM EXAMENSARBETE SAMHÄLLSBYGGNAD, AVANCERAD NIVÅ, 30 HP , STOCKHOLM SVERIGE 2020 Numerical study of the effect of thermal ice loads on concrete dams WASSEEM KORDOGHLY IMAL ZUWAK KTH SKOLAN FÖR ARKITEKTUR OCH SAMHÄLLSBYGGNAD Numerical study of the effect of thermal ice loads on concrete dams Imal Zuwak, Wasseem Kordoghly Master of Science Thesis Stockholm, Sweden 2020 TRITA-ABE-MBT-20318 ISBN: 978-91-7873-579-2 KTH School of ABE SE-100 44 Stockholm SWEDEN © IMAL ZUWAK, WASSEEM KORDOGHLY 2020 Royal Institute of Technology (KTH) Department of Civil and Architectural Engineering Division of Concrete Structures i Abstract It is essential to understand the mechanics of ice load and how it affects concrete dams located in a cold climate, such as Sweden, where the temperature becomes sufficiently cold to freeze the surface of the reservoir. The purpose of this thesis is to study ice load distribution along concrete dams, and its response during the application of an ice load. Two types of concrete dams were analysed, an arch dam and a buttress dam. For these dams, the influence from different parameters on the ice load distribution along the dams is studied. In addition to this, a study on how the ice load affects dam stability had also been performed. Stability analyses based on the finite element method were performed using both linear and nonlinear formulation of the interaction behaviour between the base of the dam and the underlying rock. A parametric study of ice sheet expansion on different dam types and geometries were performed. The expansion of the ice sheet was assumed to either be caused by a constant temperature 15 ˚C uniformly distributed over the ice thickness, or by a temperature gradient from 15 ˚C at the top surface of the ice sheet and 0 ˚C at the bottom. The parametric study also includes an investigation about influence of the shape of the reservoir beaches, where it either had a perpendicular shape towards the surface of the dam, or it had an angle of 30˚ with the dam surface. In the linear stability analysis, the structure continued to deform with increasing of the resultant pressure until it reached nonlinearity. The dam deflection had a linear relation with the applied ice load force until it reached the point when structure behaviour was nonlinear. The structure failed due to sliding, overturning or combination of both sliding and overturning. A material failure can also occur if the nonlinear material behaviour is considered, however this was not considered in this study. The parametric study showed that the ice load distribution was less near the beaches, and the distribution of the load on the concrete dam was higher near the top surface of the ice sheet. It was also shown in the study that the distribution of the ice load along the dam was as a cosine function where it had the maximum value at the buttress and the minimum at monolith connections. The result also showed that the load distribution over the thickness of the ice sheet was the same along the dam, regardless of the shape of the beaches or the length of the ice sheet. Keywords: Arch dam, Buttress dam, Concrete, Finite element analysis, ice load. iii Sammanfattning Det är viktigt att förstå hur islasten beter sig och hur den påverkar betongdammar som är belägna i kallt klimat, som t.ex. Sverige, där temperaturen blir tillräckligt låg för att frysa ytvattnet i en flod. Syftet med detta examensarbete är att studera isbelastningsfördelningen längs en betongdamm och dess respons under en belastningen. Två olika typer av betongdammar har analyserats, vilka är valvdamm samt lamelldamm. För dessa, studerades det hur olika parametrar påverkar lastfördelningen från istrycket längsmed dessa dammar. Slutligen har det studerats hur islasten påverkar dammsäkerheten och risken för dammbrott. Inverkan från interaktionen mellan dammen och det underliggande berget som linjär eller olinjärt har studerats i stabilitetsanalyser baserade på finita elementmetoden. En parameterstudie har också genomförts för olika dammtyper och geometrier där islasten orsakades av en expansion av isytan. Denna expansion antogs vara orsakad av antingen av en jämn fördelad temperatur över istjockleken på 15 ˚C, eller en temperaturgradient över istjockleken med +15 ˚C på den övre ytan och 0 °C vid isens bottenyta. Den parametriska studien beaktar även inverkan från utformningen av stränderna, där den har definierats som antingen vinkelrät mot dammen eller med en lutande vinkel på 30 grader. I fallet med linjära stabilitetsanalyser kommer konstruktionen att fortsätta att deformeras som ett resultat av ökande resulterande tryckkraft. Dammens deformation har ett linjärt förhållande med den applicerade islasten till dess att den når en punkt då strukturens beteende övergå till olinjärt. Strukturens brottmod kan uppstå på grund av glidning, stjälpning eller i en kombination av både glidning och stjälpning. materialbrott kan uppstå om icke-lineariteterna beaktas. Dammen gick till brott på grund av glidning, vältning eller i kombinationen av dessa då. Materialbrott kan uppstå om icke-linjära materialmodeller inkluderas, men detta beaktades dock ej i denna studie. Den parametriska studien visar att isbelastningen är mindre nära stränderna och att belastningen på betongdammen är högre vid isens ovanyta. Studien visar att islastfördelningen längsmed dammen liknar en cosinusfunktion som når sitt maximum vid stödskivan och sitt minimum vid monolitanslutningen. Resultatet visar även att islastfördelningen genom islastens tjocklek har samma form längsmed dammen oavsett utformningen av stränder eller istäckets längd. Nyckelord: Valvdamm, Lamelldamm, Betong, Finit elementanalys, Islast. iv v Preface This master thesis has been performed at the division of concrete structures and department of Civil and Architectural Engineering at KTH, the Royal Institute of Technology. The authors are thankful to Docent Richard Malm for his supervision, support and guidance throughout this thesis. During the thesis, Docent Richard Malm has shared knowledge with authors and authors has learned much from that. In addition to that, the authors also want to show an appreciation for Assoc. Prof. Bert Norlin, who has taken his time and helped us when needed. Stockholm, June 2020 Imal Zuwak Wasseem Kordoghly vii Table of content Abstract ................................................................................................................... i Sammanfattning ..................................................................................................... iii Preface .................................................................................................................... v 1 Introduction ................................................................................................... 1 1.1 Background ................................................................................................. 1 1.1 Aim and purpose ......................................................................................... 2 1.2 Limitations .................................................................................................. 3 1.3 Thesis structure........................................................................................... 3 2 Concrete dams .................................................................................................. 5 2.1 Gravity dams .............................................................................................. 5 2.1.1 Mass dams ....................................................................................... 5 2.1.2 Buttress dams .................................................................................. 6 2.2 Arch dams ................................................................................................... 7 2.3 Loads ........................................................................................................ 11 2.3.1 Gravity .......................................................................................... 12 2.3.2 Uplift pressure ............................................................................... 12 2.3.3 Hydrostatic pressure ...................................................................... 12 2.3.4 Ice load .......................................................................................... 13 2.3.5 Thermal loads ................................................................................ 13 2.4 Material properties .................................................................................... 14 2.4.1 Concrete ........................................................................................ 14 2.5 Ice load measurements............................................................................... 15 2.5.1 Thermal effects on ice load ............................................................ 15 2.5.2 Reservoir water level effect on ice load ........................................... 17 2.5.3 Other factors that effects the ice load ............................................ 18 viii 3 Finite element analyses of concrete dams .......................................................... 19 3.1 Geometrical model .................................................................................... 19 3.2 Meshing ..................................................................................................... 21 3.2.1 Shape function ............................................................................... 22 3.2.2 The discretisation of the elements .................................................. 23 3.3 Boundary conditions and interactions ....................................................... 23 3.3.1 Interactions between the surfaces .................................................. 24 3.3.2 Modelling of the rock ..................................................................... 25 4 Case studies .................................................................................................... 27 4.1 Dam geometries ......................................................................................... 27 4.1.1 Concrete buttress dams ................................................................. 28 4.1.2 Concrete arch dam ........................................................................ 30 4.2 Material properties .................................................................................... 31 4.3 Mesh ......................................................................................................... 32 4.4 Loads and loading procedure ..................................................................... 36 4.4.1 Case 1 ............................................................................................ 36 4.4.2 Case 2 ............................................................................................ 38 4.5 Boundary conditions and interactions ....................................................... 38 4.5.1 Case 1 ............................................................................................ 38 4.5.2 Case 2 ............................................................................................ 40 5 Results ............................................................................................................ 43 5.1 Ice load force and deformation relation on dams ........................................ 43 5.1.1 Linear behaviour of the concrete dam ............................................ 43 5.1.2 Nonlinear behaviour of the concrete dam ....................................... 45 5.2 Variation in ice load due to dam geometry ................................................ 47 5.2.1 Arch Dam ...................................................................................... 47 5.2.2 Buttress Dams ............................................................................... 49 5.3 Influence of reservoir length ...................................................................... 55 5.4 The inclination of the beaches ................................................................... 57 5.5 Pressure distribution over the ice thickness ............................................... 59 6 Discussion ....................................................................................................... 61 6.1 Influence of ice loads on dams .................................................................... 61 6.2 Variation in ice load along dams ................................................................ 62 6.3 Influence of the inclination of the beaches ................................................. 63 ix 6.4 Pressure distribution over the ice thickness ............................................... 64 7 Conclusions and further research ...................................................................... 67 7.1 Conclusions ............................................................................................... 67 7.2 Further research ........................................................................................ 69 Bibliography ........................................................................................................... 71 Appendix A ............................................................................................................ 75 1.1. BACKGROUND 1 1 Introduction 1.1 Background Following the increased demand for electrical energy and water supply, which are essential for any growing society. Societies have used dams constructed from different materials as a solution to ensure a steady supply of water and for flood control. At the beginning of the 20th century, the use of generating electricity was introduced to the dam’s function. There are 10 000 dams of all types in Sweden, and 2 000 of those are used for energy production. The purpose of dams is to store water, and then a powerhouse is built to produce energy from the kinetic energy of the water through turbines. (Malm, 2020) In Sweden, when it comes to hydropower concrete structures design, the engineers should perform the design in accordance with the Swedish dam safety guideline, RIDAS (Swedish hydropower companies guidelines for dam safety). RIDAS refers to the Eurocodes for the design of concrete structures with some modifications to increase the safety requirements. In addition, RIDAS is to a large extent based on the international guidelines developed by ICOLD (International Commission of Large Dams). RIDAS, thereby define how to evaluate hydraulic structures including, the type of loads to consider and their magnitude, type of design scenarios to consider and design requirements for dam stability and cross-sectional design. Yet during the lifetime of the structure, the loads may not reach to the ultimate state load, but it will vary and cause deflections and deformations in the structure. The deformation may lead to cracks, and these cracks will affect the strength capability of the structure. This, together with inadequate assessment and maintenance, may lead to the total failure of the structure. In cold regions, such as Sweden, the temperature variation is significant between summer and winter and can reach up to 70 °C. The winter is usually sufficiently cold that the surface of the river or a lake freezes and any movement of the ice sheet applies pressure to the dams. It is an essential factor for the safety of the dams’ structure to have a better understanding of the ice loads effect on dams and use accurate descriptions of the ice load. (Malm et al.,2018) Chapter CHAPTER 1. INTRODUCTION 2 The designing ice loads were considered as a unit load, and the value for the load varies between 50-200 kN/m, according to RIDAS (2017). The value for the design load depends on the geographical location (Gebre et al., 2013), water depth and inclination of the waterside (Comfort et al., 2003). Adolfi och Eriksson (2013) has collected the previous ice load data and derived a lognormal distribution of the maximum annual ice load with an average value of 81 kN/m, and a standard deviation of 86 kN/m. In Norway, the ice load is assumed as a line load as well, but the values varies between 100-150 kN/m according to the Norwegian water resources and energy directorate, (NVE, 2003). The seasonal temperature variation during a year is fairly similar from one year to another, which means that the seasonal ice load on a concrete dam should be quite similar from one year to another. The background is unknown for the theoretical values of the design codes (Jeppsson, 2003). As mentioned previously, the assumptions about the ice load in the design codes introduces uncertainties to the calculations and it is unclear if the ice load can be considered as a constant line load along a dam. This assumption differs from the real distribution along the dam that has been investigated in this thesis. There have been previous researches conducted to measure the ice load on concrete dams, for example, Malm et al. (2017), Comfort et al. (2003), Carter et al. (1998), Petrich et al. (2015). These studies aimed to determine the actual size of the ice load and the parameters that impact its magnitude. However, there is a lack of studies on the load distribution along different dam geometries and the structural response due to this ice load. 1.1 Aim and purpose The purpose of this thesis is to study the ice load distribution along a concrete dam and its response during an ice load. The main focus is on the influence of thermal ice load caused by the expansion of the ice sheet due to an uneven or constant temperature profile over the ice thickness. The ice load distribution will be investigated for different dam geometries, to study the influence of the dam stiffness. These investigations will be performed with the finite element method. The research questions for this project are presented below. • Is it possible to perform structural analyses that can capture the distribution in thermal ice loads acting on a concrete dam? • How is the ice load distribution along the dam influenced by different dam types and geometries? • What is the effect of the beaches shape on the ice load distribution along arch and buttress dams? • Will the length of the ice sheet affect the results produced in the parametric study? • Can a simplified finite element study also be used to capture the variation in ice pressure over the ice thickness? 1.2. LIMITATIONS 3 1.2 Limitations The performed studies in the thesis are limited to only two types of dams, gravity buttress dams and arch dam. The properties of freshwater ice have been taken into consideration and not saline ice. The properties of concrete and its behaviour are considered regarding Eurocodes. In the studies of the pressure distribution over the ice thickness, only three monoliths section of the dam will be considered. Many parameters may influence the ice load effect on dams. However, this thesis focuses on the parameters about the dam geometry, thermal expansion of the ice, temperature variation, ice sheet length and influence from the beaches (different angles of the reservoir with respect to the upstream of the dam). 1.3 Thesis structure In Chapter 2, the theoretical background of the subjects of this study is summarised. Different types of dams will be described, and typical mechanical loads that act on dams together with a description of ice load and the parameters that influence its size and spatial distribution against a concrete dam. The theory behind the finite element (FE)-method regarding modelling and analysing concrete structures is explained in Chapter 3. The FE concepts about modelling concrete dams and identifying the boundary conditions and the contact between the dam structure and the foundations are described. Chapter 4 is focused on describing the study case for this thesis, where the studied models are described including their geometries, properties, loads and the underlying assumptions and approximations that have been considered in this study. Chapter 5 presents the results from the FE-analysis, in the first part, the results from analyses based on linear and nonlinear interaction between the dam and rock is presented. In the rest of this chapter, the results of the parametric study are presented consisting of displacements, stresses and moments at the contact between the ice sheet and the dam surface. In Chapter 6, a discussion of the results that were presented in chapter 5 is given. This chapter also contains reflections about the results of this thesis. Chapter 8 contains the procedure and the outcome of this work and offer some recommendations for further research. CHAPTER 1. INTRODUCTION 4 2.1. GRAVITY DAMS 5 2 Concrete dams 2.1 Gravity dams Concrete gravity dams depend on the material weight of the structure to counter the forces that applied to it. It is thereby designed so that its self-weight provide sufficient stability and prevent the risk of sliding or overturning failure caused by external loads, i.e. hydrostatic pressure, ice load, uplift pressure. The main characteristic of the gravity dams is that they are made of individually stable sections (monoliths). Each of the monoliths can be considered as a fixed cantilever at the foundation, the horizontal forces acting on the dam cause a moment which generates tensile stresses on the upstream surface of the dam and compressive stresses on the downstream (Chen, 2015). The monoliths are joined with vertical joints, and these monoliths can be constructed of either filled triangular uniform cross-section along the dam, or to save the use of materials the monoliths compose from parts which are the front plate and the buttress. The dams built of the previously mentioned two types of monoliths are called massive dams and buttress dams, respectively (Bergh, 2014). 2.1.1 Mass dams Mass dams are defined by its characteristically filled triangular cross-section along a dam with a wide base that distributes the loads to the foundation ground and works on stabilising the structure using the structure weight (Bergh, 2014). The dam is constructed by casting vertical independent parts called monoliths, with a uniform cross-section see Figure 2.1. Each of the monoliths is stable individually and joined together by vertical sealed joints. These contraction joints counteract the thermal dilation effect, i.e. the contraction and expansion of the concrete dam monoliths due to temperature variations after the construction phase is done (Bergh, 2014). One of the disadvantages of this type of dams appears in the case of uneven foundations. Which leads to uneven settlements of the monoliths, therefore leakage occurs in the contracting joints and cracks form. Leakage and cracking results in increased moisture transport in the concrete and deterioration of concrete and reinforcement through processes such as leaching, frost-damage, carbonisation, corrosion, etc. In addition to this, the bulk of concrete yields low dissipation of the heat generated by the hydration process, which makes the concrete more likely to crack, which may result in deterioration of the dam strength. Another disadvantage of this type of dams is that large amounts of concrete are required to satisfy the stability conditions, which makes it a costly choice economically and environmentally because of high carbon emissions (Chen, 2015). CHAPTER 2. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 6 Figure 2.1 Cross-section for a massive dam 2.1.2 Buttress dams Concrete buttress dams are a type of gravity dams that have the advantage of saving a significant amount of concrete compared to massive dams. Buttress dams also have reduced footprint compared to massive dams, which reduce the area subjected to uplift pressure. Finally, to increase the stability, an inclined upstream surface is often used in buttress dams (Chen, 2015). A buttress dam is classified as a gravity dam since it relies on its self-weight for stability, and is constructed of monoliths, similar to massive dams. The monoliths consist of two parts, an upstream plate which is supported by a triangular buttress see Figure 2.2. These two parts have the following properties; • The front plate is a relatively thin plate and could be constructed as a vertical or inclined plate. A vertical plate resists the water pressure and transfers the loads to the buttress, whereas an inclined plate is not only relying on the self-weight for stabilization but also take advantage of the vertical resultant of the hydrostatic pressure to help stabilize the structure. • The buttress (or the support), is placed downstream the front-plate. It transfers the loads from the front plate and distributes them on the foundations. Concrete buttress dams are often used to create a transition from an embankment dam to the spillway section. Buttress dams require to be constructed over relatively strong 2.2. ARCH DAMS 7 foundations due to the higher concentration of forces that the buttress transfers to the foundation, which requires a higher quality of the rock. Buttress dams can be distinguished according to the monolith shapes into three types; massive head, flat slab and multi-arch buttress dams (Chen, 2015). Figure 2.2 Buttress dams: (a) Massive head buttress type, (b) flat slab buttress type, (c) multi-arch buttress type (Akıntuğ, 2012). 2.2 Arch dams Arch dams are curved on the upstream side, and partially transfers the hydrostatic pressure by arch action in the horizontal direction to the abutments. Arch dams can be classified into; • Single curvature arch dams are generally constructed from thick parts joined with a single curve; i.e. horizontal curve see Figure 2.3a. • Double curvature arch dams are in general slenderer and more constructed as vertical parts joined to form a shell or dome construction. The dam then is curved vertically and horizontally, which improve the dam efficiency of transferring the loads, see Figure 2.3b. CHAPTER 2. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 8 (a) (b) Figure 2.3 (a) Cross-section of a single curvature arch dam, (b) Cross-section of a double curvature arch dam. Arch dams are most suitable to be built in narrow canyons and are in general built with a small ratio of span/height up to five. This value may be exceeded, but for higher values, other types of dams are in general, more suitable. Arch dams have been constructed in different geographies, hydrology and climate conditions, which mean that these are not conditional factors (Pedro, 1999). Other factors to be considered in the choice of arch dams- like any other type of dams- are economical and reliability factors. Arch dams are durable, reliable and have a significantly smaller volume, especially in comparison to mass gravity dams, yet cast on situ arch dams require a more expensive framework and usage of high-quality concrete in addition to the layout challenges. To overcome the economical high cost and maintain the durability for arch dams (Pedro, 1999) concludes that roller-compacted arch dams optimise the economy/reliability balance. Arch dams transfer loads horizontally to the abutments and vertically to the foundation, hence a large pressure obtained in the foundation, this arises the awareness of the water tightness and foundation stability problems, which were the cause of the dramatic accidents like Malpasset dam (1958), and the Vajont reservoir (1962) (Pedro, 1999) see Figures 2.4 to 2.7. Pedro (1999) concluded that arch dams require rock foundation of high quality with special characteristic regarding drainage, consolidation and water tightness. In addition to this, other essential aspects are making the interface between the dam and the rock as smooth as possible and extend the dam into the foundations, which will prevent the dam from sliding but gives it the ability for rotation vertically in the front toe which make it more flexible, and improve the water tightness (USACE, 1994). 2.2. ARCH DAMS 9 Figure 2.4 Malpasset dam after the end of construction (Marinos, 1958). Figure 2.5 Malpasset dam after failure, end 1959 (Duffault, 2013). CHAPTER 2. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 10 Figure 2.6 Vajont reservoir dam after the end of construction (Casagrande, 2014). Figure 2.7 Vajont reservoir dam after failure, 1962 (Lio, 2019). 2.3. LOADS 11 2.3 Loads Loads acting on a dam could be divided into static and dynamic loads and permanent and time-dependent loads. The loads that a dam is subjected to are: • The gravity load, i.e. the dead weight of the dam structure and facilitating parts. • Hydrostatic pressure, i.e. static water pressure from the headwater acting on the upstream and the tailwater is acting on the downstream part of the dam. • Uplift, i.e. pore water pressure acting on the surface of contact between the dam foot and the foundation ground. • Thermal loads, i.e. temperature-related loads from ambient temperature and chemical reactions in concrete during the hydration. • Seismic loads, i.e. earthquakes. • Ice loads, i.e. load acting on the contact surface between the ice sheet and dam upstream at winter. • Wind pressure is a mechanical load due to wind pressure on the dam structure. • Wave pressure, i.e. the pressure loads from streams in the water acting on the dam upstream. • Forces in the concrete due to internal reactions. In this section, a brief description of the static loads will be presented. Figure 2.8 is an illustration for some of the loads mentioned above. Figure 2.8 Loads acting on a dam CHAPTER 2. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 12 2.3.1 Gravity Gravity load is the equivalent downward gravity force due to the total weight of the concrete dam and additional appurtenances such as gates, bridges and the rest of the dam components. The weight of the accessories such as gates is negligible compared to the dam weight; therefore, the gravity load can, in general, be calculated using the following equation. 𝐺𝐺= 𝛾𝛾c ∗V (1) where, 𝐺𝐺 is the gravity load [kg] 𝛾𝛾𝑐𝑐 is the density of concrete [kg/m3] V is the volume of the concrete dam [m3] 2.3.2 Uplift pressure Stored water behind the upstream face penetrates the concrete dam through the pores. The water can also be transported through cracks of the foundation material, cracks in the concrete structure and joints between the dam toe and the foundation. The penetrated water results in an upward hydrostatic pressure underneath the dam, which is denoted uplift pressure (Ghanaat, 1993). The uplift pressure counteracts the vertical downward force, thus lower the resistance to sliding failure. The uplift has a considerable effect on the sliding stability of gravity dams and thick arch dams but can usually be neglected in thin arch dams (Ghanaat, 1993). 2.3.3 Hydrostatic pressure The hydrostatic pressure is a triangular distributed pressure load that acts perpendicular to the upstream face, where the pressure intensity depends on the water elevation. The pressure force increases linearly from the top to the bottom of the reservoir. In some cases, the presence of water on the downstream face of the dam results in a pressure that works as stabilising force. For safety reasons, the hydrostatic downstream pressure is usually neglected. However, downstream water pressure results in increased uplift pressure. Therefore it is always considered in the uplift load. 2.3. LOADS 13 The hydrostatic pressure is calculated using the following equation: 𝑝𝑝= 𝜌𝜌w. 𝑔𝑔. ℎ (2) where, 𝑝𝑝 is the hydrostatic pressure [N/m2] 𝜌𝜌𝑤𝑤 is the water density [kg/m3] 𝑔𝑔 is the gravitational constant [m/s2] ℎ is the height of the hydraulic head [m] 2.3.4 Ice load The influence of ice load should be considered in cold climates regions. According to RIDAS Sweden is divided into three regions; • South of Sweden with a horizontal ice pressure of magnitude 50 kN/m and an ice sheet thickness of 0.6 m. • The middle part of Sweden with a horizontal ice pressure of 100 kN/m and an ice sheet thickness of 0.6 m. • The north of Sweden with a horizontal ice pressure of 200 kN/m and an ice sheet thickness of 1.0 m. The ice pressure is assumed to act horizontally on the contact surface of the structure. The ice pressure is caused by several mechanisms, acting separately or combined. The most important causes for the ice load pressure are; temperature variations which lead to thermal expansion and contraction, water level variations and drag forces from wind and currents (Johansson et al.,2013). The horizontal ice pressure caused by a thermal variation is represented by triangular distributed load degraded from the maximum at the top surface to zero at the bottom surface. The resultant force located one-third of the ice sheet thickness from the top surface (Malm, 2020). 2.3.5 Thermal loads Temperature variations cause the concrete dams to expand or contract, and if this volume change is restrained, stresses are introduced. The expansion and contraction may vary depending on the type and amount of aggregates and varies typically between ±50 mm, Contraction joints or dilation joints are placed between monoliths or between a dam sections to help the concrete expand and contract without the risk of cracking (Chen, 2015). CHAPTER 2. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 14 If the resulting tensile stresses are equal to the concrete tensile strength, cracks will occur. The restrained volume change may induce cracks in hydropower plants, especially in members with thinner thickness subjected to a temperature gradient from possible inner heat from producing electricity and the outdoor climate. In this dams, since the reservoir water not used in the hydration of concrete, hence the Concrete dries out faster, and as a result, shrinkage occurs much faster. If the design of joints and reinforcement is not sufficient enough to limit the width of the cracks caused by the shrinkage, then deterioration of the concrete dam strength occurs (Chen, 2015). 2.4 Material properties 2.4.1 Concrete Three parameters can describe the linear behaviour of the concrete in mechanical analyses; elastic modulus poisons ratio and the density of the concrete. The density is only required for calculating the gravity loads in static analyses. The mean value for concrete that is 28 days old can be calculated by Eq. (2), according to Eurocode 2 (2008). 𝐸𝐸𝑐𝑐𝑐𝑐= 22 ∗(𝑓𝑓 𝑐𝑐𝑐𝑐 10 )0.3 (3) where, 𝐸𝐸𝑐𝑐𝑐𝑐 is the mean value of elastic modulus of concrete 𝑓𝑓 𝑐𝑐𝑐𝑐 is the mean value of cylindrical compressive strength of concrete The elastic modulus, i.e. the stiffness of concrete, increases with hardening while creep or cracking cause a reduction in the elastic modulus. According to Eurocode 1 (2011), the density of the concrete is assumed to 24 kN/m3. In the case of the concrete dams, it is used 23 kN/m3 RIDAS (2011), the reason for this is gravity load is beneficial in the stability analyses. According to Eurocode 2 poisons ratio is 0.2 for uncracked concrete and 0 for cracked concrete. FE-analysis also requires the thermal expansion coefficient a which varies between 0.74 ∙ 10−5 ≤ 𝛼𝛼 ≤ 1.3 ∙ 10−5 [𝐾𝐾−1] but according to Eurocode 2 the assumption of alfa can be 𝛼𝛼 = 1.0 ∙ 10−5 [𝐾𝐾−1]. 2.5. ICE LOAD MEASUREMENTS 15 2.5 Ice load measurements The ice load varies depending on external variables and may be caused by different factors. According to Johansson et al. (2013), the ice load can be influenced by the following mechanisms: • Temperature variations. • Water level variations. • Water currents and winds. These mechanisms can affect the ice load separately or as a combination of two or more mechanisms at the same time, which makes it hard to study the influence of one factor on the ice load without enough data. That data needs to be extracted during a long period of time in different locations of the dam, to describe the variation of one or more factors and record their effect on the ice load. Therefore, the complexity of the experiments that have been made varied depending on the resources that were available for researchers. When considering combined factors for the ice load, the complexity of data analysis increases significantly. Therefore, a higher risk of interpretations and decrease of precision will occur, which will make it more difficult to draw conclusions about the ice load, according to Comfort et al. (2003). The size of ice loads has been found in Previous measurements were between 100 kN/m to 460 kN/m (Adolfi and Eriksson, 2013; Saether, 2012). These measurements show a significant difference compared to the design ice line-loads between 50 kN/m and 200 kN/m that RIDAS (2017) recommends. 2.5.1 Thermal effects on ice load The influence of thermal effects on the ice load is governed by the thermal expansion of the ice. The temperature of the bottom of the ice sheet remains constant 0˚C, while the top surface varies due to the ambient air temperature see Figure 2.9 (Johansson et al.,2013). CHAPTER 2. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 16 Figure 2.9 Thermal loads (Malm et al., 2017) According to Hellgren et al. (2019), the ice load due to temperature gradient along the ice thickness has an inverse relation with the ambient air temperature. The daily measurement of ice load and air temperature shows that the ice load peaks occur when the temperature drops. This relation is inverted compared to the normal expectation that ice load increases due to the expansion from increasing temperature. Hellgren et al. (2019) presents a figure showing how the ice load differs daily with the change of the temperature during one spring see Figure 2.10. Figure 2.10 Ice loads and temperature variation during one spring (Hellgren et al., 2019) Tensile stress is generated in the ice either while the ice thickness grows or when the ice upper surface temperature changes rapidly. For instance, when the top surface of the ice sheet is cooling while the bottom surface temperature remains constant, tensile stress is formed as the top surface is shrinking. An ice sheet tolerates these stresses until it reaches the cracking level. Water will fill these cracks and freeze, which lead to further growth 2.5. ICE LOAD MEASUREMENTS 17 and expansion of the ice sheet. Due to the expansion of the ice sheet, the ice load pressure on the concrete dam increase (Johansson et al., 2013). When the top surface of the ice sheet expands and contracts, a bending moment occur over the ice thickness. This bending moment cannot be carried by the ice sheet since the ice sheet is floating on the water surface. Hence, the ice sheet cannot bend upwards nor downwards, since the ice is restrained against bending by its self-weight. Considering that the ice sheet is restrained from bending while a low resistance to the tensile forces caused by bending moments, the ice sheet will crack (Bergdahl, 1977). Another cause of the ice sheet to crack is when the temperature changes unevenly along the ice sheet. For instance, if the ambient temperature rises, causing the heated surface to expand. The expanding part of the ice sheet will face resistance from the surrounding boundaries, and this resistance leads to rising in tension and the ice sheet cracks to relieve this tension (Johansson et al., 2013) There are more parameters that affect the thermal effect on an ice sheet. Some examples are; the reservoir surface shape and its boundary, if an ice sheet is free to grow in one or more directions, is an ice sheet constrained by the surrounding geometry, and the snow cover over an ice sheet (Malm et al., 2018 ). The stress distribution within an ice cover is quite nonlinear; however, for simplicity, the stress distribution is often considered linear in previous researches. The expected stresses at the top of the sheet are to be -200 kPa and +200 kPa at the bottom for a typically cold winter. For severe cold peaks, the tension may go up to -600 kPa at the top surface, and the compressions in the bottom surface also reach +600 kPa (Kharik et al., 2017). 2.5.2 Reservoir water level effect on ice load Water level variation results in tensile and bending stresses near the bond in the interface between an ice sheet and a dam due to sag. The reservoir water level effect varies depending on some factors : • The difference in water level variation. • The sequence of the water level variation. • The duration takes for the water level to change. In an eight-year research program, Comfort et al. (2000) derived the effect of these factors into three main classes: • Negligible effect: when the water height change is small (10 - 15 cm ) with a low rate for changing (0 - 0,5 times/day). • Medium effect: the water height change between (10 - 30 cm) with relatively frequent changes (1 - 2 times/day). • Large effect: with water height change between (35 - 45 cm ) and frequent. The results from seven measured points on different dams show that an ice load reaches as high as 300 kN/m due to the variation in reservoir water level. CHAPTER 2. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 18 A study by Stander (2006) supports the assumption of an ice sheet bonded to the dam upstream face would crack when the reservoir water level is lowered. These cracks will be filled with water and freeze, which will lead to the expansion of the ice sheet. This expansion will cause pressure on the dam face under some condition that Stander (2006). To mention one of the conditions and not limited to, an ice sheet should be limited from expanding to the free water direction. Otherwise, the expansion will cause negligible compressive stresses on the dam surface, to read more see Stander (2006). The previous condition was one of the assumptions that had been considered while modelling the ice sheet in this thesis. 2.5.3 Other factors that effects the ice load The effect of dam geometries will be studied in this thesis since it was encouraged to investigate ice load distribution along with dams due to the lack of studies in this regard. An example of previous studies there is the studies of Morse et al. (2011) and Taras et al. (2011), where the stress distribution studied with sensors placed three meters apart along the dam. The outcomes of these two studies show that the ice load acting on the dam varies significantly along a few meters of the dam and has high and the variation is described as chaotic. In addition, when at one sensor high load is measured, the magnitude measured in that sensor is always higher than the average measured stress over two or more sensors. Another property that influences ice load on a dam which has not been investigated to any larger extent is the influence of reservoir beaches. Therefore in this thesis, the influence of the angle of the reservoir beaches with respect to the dam surface was studied, i.e. the shape of the reservoir. As mentioned in Hellgren ( 2019 ), the only reference is found that studied the influence of reservoir beaches is a study by Ko et al. (1994). Ko et al. ( 1994 ) study the influence of beach slope not beaches angle, the study divides thermal ice load into three categories according to the slope of the beaches. Steep beaches which have the slope smaller than 45° the design load is increased by 1.5 times compared to flatter beaches with a slope higher than 20°. 3.1. GEOMETRICAL MODEL 19 3 Finite element analyses of concrete dams When it comes to using FEM software like Abaqus, the general procedure consists of three steps, which are; pre-processing, analysis, and post-processing. The first step, pre-processing, is defining the required parameters such as geometry, material properties, boundary condition, and loads. The next stage is the analysis where, for instance, the defined geometry or model in the software can be analysed depending on the purpose of the study. The last step is post-processing, which is evaluating the obtained results (Malm, 2016). 3.1 Geometrical model For continuous gravity dams, it is better to utilise the plane strain theory and use 2D analyses. If material nonlinearity is included in the analyses, 2D analyses prevent the out of plane cracking in the model. Figures 3.1 and 3.2 below are shown as an example of a situation where plane and strain theory can be preferred to utilise (Malm, 2016). Figure 3.1 Sliding of a gravity dam (Linsbauer and Bhattacharjee, 1999). CHAPTER 3. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 20 Figure 3.31 Gravity dam with a fracture in the rock at the upstream toe (Ruggeri, 2004). For arch dams, where the cross-section various along the width of the dam, or when the load is carried in the lateral direction of the dam, i.e. through arch action, then Malm (2016) recommends performing the analyses with 3D solid elements. Concrete buttress dams can be analysed by using either 2D or 3D solid elements. If the 2D analysis is used, then the lateral bending of the front plates is disregarded. Therefore, 3D solid or shell elements are a better option for even the buttress dams. Figure 3.3 below shows the example of where 3D solid elements are used for the analyses (Malm, 2016). Figure 3.3 Geometric model of a dam (Malla and Wieland, 1999). Shell elements can be utilized for the slender dams, for instance, ambursen and buttress dams. Shell elements have the advantage of lower computational effort and that the sectional forces can be obtained directly. The disadvantage is that the complicated corners are difficult to constrain, which leads to wrong stiffness value. To tackle this 3.2. MESHING 21 problem, the shell elements may have to be defined with an offset to represent the real geometry, and its real stiffness can be represented (Malm, 2016). 3.2 Meshing There are several element types to choose from to mesh their model, as an engineer, it is important to know about them so that these can be used where appropriate. Different element types have a different degree of freedom; below is a list of the common element types used in the FEM (Malm, 2016). The simplification of a real structure is the hardest part of the numerical analysis for an inexperienced engineer, due to the lack of understanding the fundamental behaviour. For instance, to translate the real structural behaviour into boundary conditions, loads, interaction etc. Figure 3.4 is showing the most common element types in structural analysis. Most common types of elements and their degrees of freedom in FEM are : • Truss elements (three degrees of freedom, translations, at the nodes) • Solid elements (three degrees of freedom, translations, at the nodes) • Beam elements (six degrees of freedom, translations and rotations, at the nodes) • Shell elements (six degrees of freedom, translations and rotations, at the nodes) Figure 3.4 Common elements for structural analyses (Malm, 2016). There are few more differences than only in degrees of freedom in elements. For the design of the beam, the cross-sectional forces are interesting. In the Euler Bernoulli theory, the shear deformations are neglected, but the theory of Timoshenko considers the shear deformation. Similarly, for shell elements, the option is between Kirchhoff that neglect the shear and Mindlin theory includes it (Malm, 2016). When connecting a shell or a beam to a solid element, only the translation can be restrained. This means that the rotations in the beam or shell are unrestrained and acts as a hinge. In order to solve this problem, another vertical shell element has to be created CHAPTER 3. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 22 in the connection between the elements almost like a plate. This solution gives a distributed pressure over the face of a solid element, see Figure 6.2 for an illustration. 3.2.1 Shape function FE-theory is based on nodes which are discrete values that describe the behaviour of a structure. Between these nodes are the solution which is interpolated either by linear shape function or quadratic. Interpolation is made by shape function, and an illustration is presented in Figure 3.5 below (Malm, 2016). The line that is straight between the nodes in the linear and the line that is bent is the quadratic shape function. Figure 3.5 Shape function (Malm, 2016). To describe the bending of the structure, a linear shape function is a poor choice. A better option for bending is to choose a quadratic shape function. If the serendipity elements are excluded, the quadratic shapes have twice as many nodes as the linear shape functions; this doubles the total degree of freedom for the structure and leads to expensive calculations. An example can be seen in Figure 3.6, where it presents a 2D plane stress or plane strain element. The left one is four noded elements, and the right one is eight noded elements (Malm, 2016). Figure 3.6 2D plane stress elements, left one is four noded, and the right one is eight noded elements (Malm, 2016). 3.3. BOUNDARY CONDITIONS AND INTERACTIONS 23 3.2.2 The discretisation of the elements The definition of the mesh should be on a way where skew elements are avoided. Figure 3.7 below shows the error in percentage when a point load is applied to a cantilever beam. One important thing to have in mind is to perform convergence test, means that analysis should be done with both higher and lower grade elements and it is preferred to have lower grade linear elements in case material nonlinearity includes which requires smaller elements. Smaller element size obtains a better representation of the crack pattern. To increase the accuracy of the FE analysis, either the linear upgrade elements to quadratic elements or increase the total number of the elements (Malm, 2016). Figure 3.7 Error of elements (Eriksson 2002 and Malm 2016). 3.3 Boundary conditions and interactions In the definition of boundary conditions and interactions, one of the many factors that influence the accuracy of the numerical model is the boundaries, and how large part of the actual dam is included in the analyses. The models must capture the mechanical modes of action for the part that is analysed. If the simplifications of the model are made correctly, then there is practically no significant difference between the model and the real structure. This means that engineering judgment is an important factor when evaluating the results. In some cases, it is necessary to model the whole structure, for instance, in the case of arch dams, according to Malm (2016). When analysing a whole dam, special consideration may be needed at the contraction joints. A valid assumption could be in this case to includes the joints and not allow force transmission over the joints. Those joints often consist of PVC, steel, or bitumen. Another important factor in FE modelling is the interaction between the rock and the dam. Interaction can be handled by three approaches, including the large part of the rock, using spring elements, defining the boundary condition at the bottom of the dams (Malm, 2016). Special consideration also has to be taken into account when combining solid elements with shell elements. A direct connection between a shell and solid elements results in a CHAPTER 3. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 24 high displacement and stress at the connection node, which works like a hinge and no bending moment can be transmitted from solid to shell. To transfer the bending moment from solid to shell, it should be connected through a rigid link. Figure 3.8 illustrates both types of connections on how to couple solid to shell elements. Figure 3.8 To the left it is a direct connection between shell and solid element. To the left a common displacement restraint between the nodes is shown. To the right, a rigid link connection is shown which also ensures moment restraint between the solid and the shell strutures. 3.3.1 Interactions between the surfaces One way to simplify the contact between the rock and concrete when analysing dams is by to assume complete bond. This can be a fair assumption in some cases if the concrete has some bond strength to the foundation, for instance, when the rock surface is blasted and cleaned from loose particles. However, it must be ensured to not subject the dam body that is closest to the rock for high tensile stresses since the bond and shear strength is lower than the tensile shear strength (Malm, 2016). In order to make the assumption of the FE-model more realistic, a contact formulation between the rock and the dam can be used. This contact formulation should include the frictional resistance of the horizontal plane and high stiffness for compressive forces in the normal direction. It is important that the contact formulation also allows separation between the rock and the dam when subjected to tensile forces, which may occur in case of an overturning failure. According to RIDAS (2011), the friction coefficient can be set as μ= 1.0 for concrete dams built on rock (Malm, 2016). In contact formulations, it is important to consider which nodes that are defined as master and slave. The surfaces with coarser mesh are recommended to be defined as a master surface, according to Malm (2016). The reason for this is that the mid-nodes on the slave surface can be adjusted to remove penetration based on a linear interpolation between the nodes of the master surface. This increases the convergency and the convergency rate (Malm, 2016). There are several ways to define the stiffness in the normal direction of the contact surface in Abaqus, soft contact and hard contact. The hard contact allows the physical surfaces to interact and separate. The soft contact keeps the contact even after 3.3. BOUNDARY CONDITIONS AND INTERACTIONS 25 separation and the normal pressure is decides the pressure-overclosure relationship (Dassault Systèmes, 2014). Figure 3.9 and Figure 3.10 below shows the normal pressure when using these two types of contact. Figure 3.9 Normal pressure when using soft contact (Dassault Systèmes, 2014). Figure 3.10 Normal pressure when using hard contact (Dassault Systèmes, 2014). 3.3.2 Modelling of the rock An option for the boundary condition is to include the rock mass into the numerical model, but then there are some recommendations for this type of boundary condition that needs to be considered in the FE-analysis (Malm, 2016). The boundary condition should be in the outer part of the rock mass, so that the displacements of the rock is restricted in the direction perpendicular to each external surface. CHAPTER 3. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 26 4.1. DAM GEOMETRIES 27 4 Case studies In this chapter, the two different case studies used to analyse the influence of ice load on concrete dams are described. Case 1 involved stability analyses of two different concrete dams to assess the influence of an ice load on the dam behaviour. The first dam was an arch dam and the second dam was a monolith from a buttress dam. These geometries had been modelled in the FEM-software Abaqus. The stability analyses were performed using two different approaches for the concrete-rock interface, where both complete bond and a contact formulation had been studied. Case 2 was consisting of a parametric study regarding the loads acting on the two different concrete dams caused by the expansion of an ice sheet. The parametric study also included investigations of the shape of the reservoir beaches, where it either had a perpendicular shape towards the surface of the dam, or it had an angle on 30 degrees. In both cases that are mentioned above, the analyses are performed with linear elastic material behaviour. In the analyses with linear elastic material behaviour, the reinforcement had a minor influence on the results. The reason for this is that the stresses of embedded reinforcement are small in uncracked concrete. Therefore, in instability analyses, the reinforcement had a minor effect of the behaviour of the dam, and only influence the gravity load. This can, however, easily be compensated by defining a density of the concrete that includes the weight of reinforcement (Malm, 2020). For this reason, the modelling of reinforcement had been not taken into consideration since it is not going to affect the stability of the dam. 4.1 Dam geometries Five different geometries of the concrete buttress dam and one arch dam had been studied. The buttress dam consists of a concrete buttress dam built out of 3 and 9 monoliths as seen in Figure 4.2 – 4.6 in the geometry consisting of 9 monoliths, three of the monoliths on each side were placed with an angle of 30 degrees, see Figure 4.6. The last one is the same arch dam that was studied previously. It is connected to a spillway and abutment on sides, see Figure 4.7. All the different geometries studied in case 2 are summarised in Table 4.1. CHAPTER 4. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 28 4.1.1 Concrete buttress dams The concrete dam consists of one single monolith of a small typical Swedish concrete buttress dam. Figure 4.1 illustrates the model that had been analysed in Case 1. Figure 4.1 FE-Model consist of one monolith of a buttress dam considered the foundation to rock. All the geometries that had been studied in case 2 are included in table 4.1. Table 4.1 Dams geometries that are considered in the study of case 2 Geometries that had been analysed in case 2 The numbers of figures that are presenting the model Three monoliths dam with perpendicular ice sheet Figure 4.2 Three monoliths dam with inclined ice sheet Figure 4.3 Nine monoliths dam with perpendicular ice sheet Figure 4.4 Nine monoliths dam with inclined ice sheet Figure 4. 5 Rotated Monoliths dam Figure 4.6 Arch dam with perpendicular ice sheet Figure 4.8 Arch dam with inclined ice sheet Figure 4.9 Monolith Ground 4.1. DAM GEOMETRIES 29 Figure 4.2 Three monoliths dam with the perpendicular ice sheet. Figure 4.3 Three monoliths dam with the inclined ice sheet. Figure 4.4 Nine monoliths dam with the perpendicular ice sheet. Figure 4.5 Nine monoliths dam with the inclined ice sheet. CHAPTER 4. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 30 Figure 4.6 Rotated Monoliths dam. 4.1.2 Concrete arch dam The dam is a typical Swedish arch dam previously used in the 14th ICOLD International Benchmark Workshop on Numerical analysis of Dams, see Malm et al. (2018). In Figure 4.7, the arch dam is illustrated. It is supported by abutment and spillway on each side, from underneath there is a rock support foundation for the arch dam. Figure 4.7 FE-Model for an existing arch dam in Sweden. Arch Dam Whole Rock 4.2. MATERIAL PROPERTIES 31 Figure 4.8 Arch dam with the perpendicular ice sheet. Figure 4.9 Arch dam with inclined ice sheet. 4.2 Material properties The material properties that were used for concrete in this thesis are presented in Table 4.2, and Table 4.3 presents the ice sheet properties that were used in the models. The values specified for the arch dam can be referred to the 14th ICOLD Benchmark proceedings (Malm et al. 2018) CHAPTER 4. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 32 Table 4.2 Material properties for concrete. Material properties Unit Concrete Rock Quality C30/37 Density (Monolith) kg / m3 2300 2700 Density (Arch dam) kg / m3 2400 2700 Poisson's ratio 0.2 0.15 Young's modulus GPa 33 40 Table 4.3 Material properties for the ice sheet. Coefficient Value Youngs modulus 8.710^9 Expansion Coeff. 110^-5 Poissons ratio 0.31 Mass density 918.9 Conductivity 2.3 All the models were analysed with the fixed time steps using 10 increments. The only load that was applied to dams’ models was from the ice sheet; the rest of the loads were deactivated. The water level was assumed to 1.5 below the crest. In order to give analyse the stresses at the edges, an inclination on 30 degrees were given to the edges. 4.3 Mesh A mesh in Abaqus was created to analyse the monolith model. The used elements are four noded tetrahedral(C3D4). The details for the mesh are presented in Table 4.4 and shown in Figure 4.10. 4.3. MESH 33 Table 4.4 Mesh properties for one monolith model. Part Element type Element length (m) Number of nodes Number of elements Ground C3D4 1.2 1707 7535 Monolith C3D4 0.35 9481 45514 Figure 4.10 The mesh that is used for one monolith model. Mesh details for three monoliths and the ice sheet is presented below in Table 4.5. Element types are linear four noded tetrahedral that are used for monoliths, and four noded quadrilateral elements are used for the ice sheet. Table 4.5 Mesh details for three monoliths and the ice sheet. Part Element type Element length (m) Number of nodes Number of elements Ice sheet, perpendicular S4R 0.75 1089 1024 Ice sheet, inclined S4R, S3 0.75 2566 2484 Monoliths C3D4 0.75 3870 15012 CHAPTER 4. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 34 Mesh details for nine monoliths and the ice sheet is presented below in Table 4.6. Element types are linear four noded tetrahedral that are used for all monoliths, and four noded quadrilateral elements are used for the ice sheet. Table 4.6 Mesh details for nine monoliths and the ice sheet. Part Element type Element length (m) Number of nodes Number of elements Ice sheet, perpendicular S4R 0.75 3201 3072 Ice sheet, inclined S3,S4R 0.75 22420 22405 All Monoliths C3D4 0.75 54450 45036 Mesh details for the arch and the ice sheet is presented below in Table 4.7. Element types are linear four noded tetrahedral that are used for all monoliths, and four noded quadrilateral elements are used for the ice sheet. 4.3. MESH 35 Table 4.7 Mesh details for the arch dam and the ice sheet. Part Element type Element length (m) Number of nodes Number of elements Ice sheet, perpendicular S4R 0.75 3201 3072 Ice sheet, inclined S4R 0.75 22420 22405 Arch C3D4 0.75 66697 330384 Foundation C3D4 0.75 50551 259038 Spillway C3D4 0.75 33493 174343 Abutment C3D4 0.75 4441 21832 Small Rock C3D4 3 8165 36945 Large rock C3D4 3 58517 306085 Mesh details for rotated monoliths and the ice sheet is presented below in Table 4.8. Element types are linear four noded tetrahedral that are used for all monoliths, and four noded quadrilateral elements are used for the ice sheet. Table 4.8 Mesh details for the model with rotated monoliths. Part Element type Element length (m) Number of nodes Number of elements The ice sheet, straight S4R 0.75 3201 3072 Monolith C3D4 0.75 53250 53036 CHAPTER 4. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 36 4.4 Loads and loading procedure 4.4.1 Case 1 The loads that had been taken into account in the analyses are; gravity load, hydrostatic pressure, ice load and uplift load from the hydrostatic press. The models were defined with three load steps, where the gravity load was applied as a first step. In the second step, the effect from the gravity load is propagated, and the hydrostatic pressure is applied with the default hydrostatic distribution in Abaqus. In the last step, the ice load is applied as pressure with a triangular distribution. The ice load is defined as a height of one meter, starting from the point that is just below the crest and applied along the whole dam length. An illustration of the ice load distribution is seen in Figure 4.11. The maximum ice load pressure was defined as 400 kN/m2, which results in a pressure of 200 kN/m along the dam crest for both dams. With this distribution, the resultant force is applied 1/3 from the top surface in accordance with the specifications in RIDAS. Figure 4.11 The ice load distribution on the point just below the crest. How monolith is loaded for in analysis model is illustrated in Figure 4.12. The ice load is acting at the top of the monolith; the surface it is acting on is also marked in Figure 4.12 with a rectangle. The rest of the surface is subjected to hydrostatic pressure, and an arrow that is pointing downwards is the gravity load. 4.4. LOADS AND LOADING PROCEDURE 37 Figure 4.12 Loads illustration for monolith in case 1. Illustration for loads acting on the arch dam is represented in Figure 4.13. The gravity load in the figure is pointing downwards and at the top of the dam is the ice load, and the rest of the surface is subject to hydrostatic pressure. Figure 4.13 Loads illustration for the arch in case 1. CHAPTER 4. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 38 4.4.2 Case 2 The load that is acting on the concrete dams in case 2 was caused by thermal expansion of the ice sheet through the thickness. This expansion had been defined as a predefined field function in Abaqus. It had either been set as a constant temperature on the ice sheet or as a temperature gradient where the top surface and the bottom surface had two different temperatures. One of the expansions had been set at a time to analyse all models in case 2. The top surface temperature is set to + 15 ˚C and the bottom to 0 ˚C for the gradient. The constant temperature on the ice sheet was set to +15 ˚C. The pressure in the models had been extracted as a nodal force. The unit for these nodal forces are kN/m, so it had been divided by the height of the contact surface, i.e. the ice thickness of 1 m, to convert it to pressure. 4.5 Boundary conditions and interactions 4.5.1 Case 1 The interaction between the monolith and rock was defined with a surface to surface contact using nonlinear interaction. The behaviour of the contact in the normal direction, was set as a type of hard contact, see Section 3.3.1. In the tangential behaviour, the friction value in the buttress dam was defined as 1.0 according to RIDAS, and the contact thereby allows for sliding behaviour. The arch dam friction value was defined as a 40 to represent the reality, where the dam was going through the foundation, and therefore, the sliding of the dam was restricted. See figure 4.14 for illustration how the bottom surface of the monolith is connected to the foundation below, where the highlighted surfaces had the hard contact formulation. Figure 4.14 The contact surfaces between the monolith and foundation in case 1. 4.5. BOUNDARY CONDITIONS AND INTERACTIONS 39 The boundary condition was set for monolith in case 1 is represented below by Figure 4.15. The arrows are showing the restricted directions for translations in the model. Figure 4.15 Boundary condition for monolith in case 1. The boundary condition was set for the arch dam in case 1 is represented below by Figure 4.16. The arrows are showing the restricted directions for translations in the model. Figure 4.16 Boundary condition for the arch dam in case 1. CHAPTER 4. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 40 4.5.2 Case 2 In case 2, the interaction between the rock and the concrete dams surfaces had been considered as fixed boundary conditions see Figure 4.17. This means that the rock was neglected in the FE analysis, and only the concrete dams had been considered. Figure 4.18 shows an example of the connection between the monoliths in studied buttress dams. The connection between the monoliths surfaces had been set to tie connection. This means that all the monoliths were considered as one structure. Figure 4.17 Boundary condition for the arch dam in case 2. Figure 4.18 Tie connection between the monoliths The connection between the concrete dams and the ice sheet had been defined as a tie connection in Abaqus for all the models that were included in case 2, see Figure 4.19 for illustration. This tie connection is a direct connection between the ice sheet and the dam. One of the models that belonged to case 2 had been connected; differently. A vertical shell element was created; this vertical shell element was connecting the dam surface and 4.5. BOUNDARY CONDITIONS AND INTERACTIONS 41 the ice sheet as a rigid link see Figure 4.20. A tie connection between an ice sheet and a concrete dam is presented in Figure 4.19. Figure 4.19 Direct tie connection between the monoliths and ice sheet. Figure 4.20 Rigid link connection between the monoliths and ice sheet. Figure.4.21 represents how the boundary conditions for perpendicular ice sheet and buttress dam that are set in the models with constant temperature on the ice sheet. The ice sheet has been restricted from bending upwards or downwards by setting a constrain. CHAPTER 4. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 42 Figure 4.21 Tie connection between the monoliths and perpendicular ice sheet, including all the boundary conditions. Figure 4.22 shows the boundary condition for the buttress dam and inclined ice sheet with a temperature gradient; the bending in the ice sheet had been allowed. Figure 4.22 Tie connection between the monoliths and ice sheet, including all the boundary conditions.
5.1. ICE LOAD FORCE AND DEFORMATION RELATION ON DAMS 43 5 Results In this chapter, the most important results that had been generated from the numerical analyses are presented. Two main aims of interest for this project are presented in the results; one is to model the ice load as a simple triangular load and study the relation between ice load and deflection of the dam, second is to model the ice sheet and study the variation in ice load due to different dam geometries. Data of the deflection was taken at the dam crest since the maximum deflection occurred at the crest, for more results see Appendix A. 5.1 Ice load force and deformation relation on dams To establish an idea about the force and displacement relation for the effect of ice load on a concrete dam, models mentioned in chapter 4.1.1 had been used (one monolith dam and arch dam). Depending on the dam foundation contact condition, the deflection-force relation had either linear or nonlinear behaviour. 5.1.1 Linear behaviour of the concrete dam When the interaction between rock and studied dams assumed as fixed, i.e., the translations and rotations were prevented in all directions. As a result, Linear relationship between the ice load pressure applied and the deflection was produced. The structure would continue to deform with the increase of the resultant pressure force caused by the triangular ice load until it reached the maximum applied load, As shown in Figures 5.1 and 5.2 for buttress and arch dams respectively. Figure 5.3 is presenting the deformed shape of the arch dam with a linear relation between displacement and ice load force. CHAPTER 5. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 44 Figure 5.1 Linear displacement-force graph in one monolith of a buttress dam. Figure 5.2 Linear displacement-force graph in the arch dam. 0 50 100 150 200 250 300 350 400 1,19 1,26 1,33 1,39 1,46 1,53 1,60 1,67 1,74 1,81 1,88 Force (kN) Displacement (mm) 0 50 100 150 200 250 300 350 400 17,20 17,54 17,89 18,24 18,59 18,94 19,29 19,64 19,99 20,34 20,69 Force (kN) Displacement (mm) 5.1. ICE LOAD FORCE AND DEFORMATION RELATION ON DAMS 45 Figure 5.3 Deformed shape of the arch dam with linear displacement-force relation. 5.1.2 Nonlinear behaviour of the concrete dam When the interaction between rock and dam was considered more complex than the first case, i.e. there was hard contact between the dam and the rock with friction coefficient. The dam deflection had a linear relation with the applied ice load force until it reached the nonlinear relation state. The structure either slide where the deformation started to increase significantly even for a constant load, as presented in Figure 5.4 and Figure 5.6 for the monolith and arch dam respectively. To reach the nonlinear behaviour, the ice pressure load had to be applied significantly higher than the maximum design load and cannot be reached in real-life situations. In Figure 5.5, the deformed shape of one monolith buttress dam after sliding failure is shown. CHAPTER 5. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 46 Figure 5.4 Nonlinear displacement-force graph in one monolith of a buttress dam. Figure 5.5 Deformed shape of one monolith of a buttress dam after sliding failure. 0 150 300 450 600 750 1 1,5 2 2,5 3 3,5 4 Force (kN) Displacement (mm) 5.2. VARIATION IN ICE LOAD DUE TO DAM GEOMETRY 47 Figure 5.6 Nonlinear displacement-force graph in the arch dam. 5.2 Variation in ice load due to dam geometry This parametric study was performed to analyse how the thermal expansion of an ice sheet caused displacements and normal forces along the dam for different dam geometries. The studied geometries are presented in the following for two different dam types; arch dams and buttress dams. The normal forces acting on the concrete dam were caused by thermal expansion of the ice sheet. This ice sheet was subjected to either constant temperature or a temperature gradient between the top and the bottom surface of the ice sheet. 5.2.1 Arch Dam • Constant temperature The displacement along the crest of the arch dam and the variation in normal forces along the crest is presented in Figure 5.7 for a constant temperature increase of +15 ˚C in the ice sheet (the positive displacement sign is towards the downstream direction, the origin was placed at the spillway). The maximum displacement, and subsequently the lowest force, appeared almost in the middle of the arch. The displacement was zero at both ends, and the normal forces at the ends of the arch were maximum due to the boundary conditions that were set in the FEM model. Figure 5.8 is presenting the deformed shape of the arch dam due to constant temperature expansion of the ice sheet. 0 150 300 450 600 16,5 17 17,5 18 18,5 19 19,5 20 20,5 Force (kN) Displacement (mm) CHAPTER 5. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 48 Figure 5.7 Normal force and displacement along the arch dam due to constant temperature increase in the ice sheet. Figure 5.8 Deformed shape of the arch dam due to constant temperature expansion of the ice sheet. • Thermal gradient The displacement and normal force in the arch dam due to the expansion of the ice sheet subjected from a thermal gradient is shown in Figure 5.9. The thermal gradient in the ice sheet was defined to be + 15 ˚C at the top surface and 0 ˚C at the bottom (the positive displacement sign is towards the downstream direction, the origin placed at the spillway). The maximum displacement in the arch dam occurred in the connection between the spillway and arch then gradually decreased along the arch. Figure 5.10 is presenting the deformed shape of the arch dam due to thermal gradient. Spillway Aboutment 0,0055 0,011 0,0165 0,022 0,0275 0,033 0,0385 0,044 0,0495 0,055 0% 10% 20% 30% 40% 50% 60% 70% 80% 90% 100% 0 17 34 51 68 85 102 119 136 153 170 Displacement [m] Normal force (N/Nmax)% Arch length (m) Normal force% Displacement[M] 5.2. VARIATION IN ICE LOAD DUE TO DAM GEOMETRY 49 Figure 5.9 Normal force and displacement along the arch dam due to thermal gradient. Figure 5.10 Deformed shape of the arch dam due to a thermal gradient. 5.2.2 Buttress Dams • Constant temperature The displacements and forces when the thermal expansion occurred due to constant applied temperature in the ice sheet are presented in Figures 5.11, 5.15 and 5.16 (the positive displacement sign is towards the downstream direction). In Figure 5.11 displacement for models -three and nine monoliths- is compared, in these models, the ice sheet was with straight edges. The displacement for both dams was zero at the dam boundaries and increased to follow a harmonic shaped curve with the maximum displacement between two monoliths and the lowest at the buttresses. Figures 5.12 to 5.14 are presenting deformed shape of three and nine monoliths buttress dams due to constant temperature expansion of the ice sheet. Spillway -0,00005 -0,00004 -0,00003 -0,00002 -0,00001 0 0,00001 0,00002 0,00003 0,00004 -40% -20% 0% 20% 40% 60% 80% 100% 0 17 34 51 68 85 102 119 136 153 170 Displacement [m] Normal force(N/Nmax) % Arch lenghth (m) Normal force% Displacement[M] CHAPTER 5. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 50 Figure 5.11 Displacement for three and nine monoliths buttress dam due to the constant temperature. Figure 5.12 Section view for the deformed shape of three monoliths buttress dam due to constant temperature. 0 0,002 0,004 0,006 0,008 0 10 20 30 40 50 60 70 Displacement (m) Dam length (m) 9 Monoliths 3 Monoliths 5.2. VARIATION IN ICE LOAD DUE TO DAM GEOMETRY 51 Figure 5.13 Plane view for the deformed shape of three monoliths buttress dam due to constant temperature. Figure 5.14 Deformed shape of nine monoliths buttress dam due to constant temperature. The normal force acting on the contact surface between the ice sheet and dams upstream surface is shown in Figure 5.15 below. The normal force affecting both models -three and nine monoliths models- was minimum at the dam boundaries then followed a harmonic shaped curve with the lowest values at the connection between monoliths and greatest at the location of buttresses support. CHAPTER 5. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 52 Figure 5.15 Normal force for three and nine monoliths buttress dam due to the constant temperature. The previous results had been presented for buttress dams with a straight dam axis. Therefore, an analysis was also performed for a dam where the monoliths were placed to create an inclination in the dam axis. In Figure 5.16, the displacement and normal force along the dam crest are shown for the buttress dam defined with 30 degrees rotation of the outer monoliths. Also, in this analysis, the displacement was zero at the boundaries where the normal force was at its maximum. The displacement increased to follow then the same pattern as the studied dams with a straight dam axis. The significant difference between these results was that at the inclined monoliths model, it was noticeable that the values increased from the edges towards the meeting between the straight and the 30 degrees rotated parts of the dam. At the connection between the 30 degrees rotated edges and the straight part of the dam, the joint displacement was reduced compared to the adjacent monoliths. The maximum displacement of the whole dam occurred at the joints of the central monoliths. Figures 5.17 is presenting the deformed shape of rotated monoliths buttress dams due to constant temperature expansion of the ice sheet. Figure 5.16 Normal force & displacement along the rotated monoliths dam due to constant temperature. 20% 40% 60% 80% 100% 0 10 20 30 40 50 60 70 Normal force (N/Nmax)% Dam length (m) 9 Monoliths 3 Monoliths 0 0,0002 0,0004 0,0006 0,0008 0,001 0,0012 0,0014 0,0016 0,0018 0,002 0% 10% 20% 30% 40% 50% 60% 70% 80% 90% 100% 0 20 40 60 80 100 120 Displacement (m) Normal force(N/Nmax) % Dam length (m) Normal force% Displacement[M] 5.2. VARIATION IN ICE LOAD DUE TO DAM GEOMETRY 53 Figure 5.17 Deformed shape of the buttress dam with rotated monoliths due to constant temperature. • Thermal gradient Due to temperature variation between the top and bottom surface of the ice sheet, an upward bending of the ice sheet occurred. Therefore, it caused stresses on the concrete dam. The variation in displacement and normal force are shown in Figure 5.18 and 5.19 for the straight buttress dams consisting of three and nine monoliths, respectively. The displacement pattern is the same for both dams. The effect was concentrated on the first three monoliths from each edge and reduced to be negligible in the rest of the monoliths. The ice pressure primarily occurs to the exterior monoliths and the beaches. As shown in Figure 5.18 and 5.19 (Where the positive displacement sign was towards the downstream direction), The highest displacement value occurs at the joints between the first and second monoliths both at the start and the end of the dam. The lowest deflection was zero in the middle of the dam if the number of monoliths is more than three. Thereby, the thermal ice load due to thermal gradient was small on a longer dam. CHAPTER 5. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 54 Figure 5.18 Displacement for three and nine monoliths buttress dam due to thermal gradient. Figure 5.19 Normal forces for three and nine monoliths buttress dam due to thermal gradient. In the buttress dam with 30 degrees angled sides, the displacement along the dam is shown in Figure 5.20 (Where the positive displacement sign is towards the downstream direction). The maximum displacement for the 30 degrees rotated monoliths part occurred in the joint between the first and second monoliths. The maximum displacement of the whole dam was in the corner connection between the straight part and the 30 degrees rotating parts. Figures 5. 21 is presenting the deformed shape of rotated monoliths buttress dams due to thermal gradient. 0 0,000002 0,000004 0,000006 0,000008 0,00001 0 0,00002 0,00004 0,00006 0,00008 0,0001 0,00012 0 10 20 30 40 50 60 70 Displacement-9monoliths (m) Displacement-3monoliths (m) Dam length (m) 3 Monoliths 9 Monoliths -100% -80% -60% -40% -20% 0% 20% 40% 60% 80% 100% 0 10 20 30 40 50 60 70 Normal force(N/Nmax) % Dam length (m) 9 Monoliths 3 Monolioths 5.3. INFLUENCE OF RESERVOIR LENGTH 55 Figure 5.20 Displacement for rotated monoliths dam due to thermal gradient. Figure 5.21 Deformed shape of rotated monoliths buttress dams due to thermal gradient. 5.3 Influence of reservoir length Due to constant applied temperature in the ice sheet, displacements and normal forces data had been extracted from two models, each of them consisted of three monoliths. The difference between the models was, in one model, the ice sheet made three times longer than the other. In Figure 5.22 and 5.23 (Where the positive displacement sign is towards the downstream direction) the comparison between the data extracted from the two models is presented. The variation in normal force and the displacement along the crest are shown for the comparison between long and short ice sheet. 0,000000 0,000001 0,000002 0,000003 0,000004 0,000005 0,000006 0,000007 0,000008 0 7 13 21 30 37 49 64 79 87 92 100 108 116 Displacement (m) Dam length (m) CHAPTER 5. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 56 Figure 5.22 Normal forces for two models, one with small ice sheet and the other with three times larger ice sheet in the normal direction of the dam, the values normalized for each model individually. Figure 5.23 Displacement for two models, one with small ice sheet and the other with three times bigger ice sheet in the normal direction of the dam. 20% 40% 60% 80% 100% 0 4 8 12 16 20 24 Normal force (N/Nmax)% Dam length (m) 3Bigger normal 0,000 0,001 0,002 0,003 0,004 0,005 0,006 0,007 0 4 8 12 16 20 24 Displacement (m) Dam length (m) 3Bigger Normal 5.4. THE INCLINATION OF THE BEACHES 57 5.4 The inclination of the beaches This study focused on the influence of the shape of the beaches. It involved a change of the shape of the beaches, from straight edges, perpendicular to the dam axis, to where the beaches were inclined with 30 degrees from the orthogonal axis. In the dams consisting of three monoliths, the normal force along the dam crest due to constant temperature expansion is presented in Figure 5.24 for perpendicular and inclined beaches. For both models -perpendicular and inclined ice sheet- the maximum normal forces occurred at the boundaries, the reduction of 35 % in the values when the beaches were inclined was noticed. In the middle of the dam, the values were the same in both cases. Figure 5.24 Normal forces for three monoliths dam for two different reservoir beaches, the values normalized for each model individually. In the longer dam, consisting of nine monoliths, similar results were found as in the shorter dam as seen in Figure 5.25. For the model with inclined beaches, the maximum normal forces reduction was around 32% near the beaches. In the centre, the values from the inclined beaches were higher than the model with perpendicular beaches. The peak to peak difference between the highest and the lowest values, at the buttresses and the joints respectively, were the same for the two studied configurations of the beaches. 0% 20% 40% 60% 80% 100% 0 2 3 5 6 8 9 11 12 14 15 17 18 20 21 23 24 Normal force(N/Nmax) % Dam length (m) Normal forces for perpendicular beaches Normal forces for inclined beaches CHAPTER 5. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 58 Figure 5.25 Normal forces for nine monoliths buttress dam for two different reservoir beaches; the values normalized for each model individually. The normal forces in the models of the arch dam with inclined and perpendicular beaches are presented in Figure 5.26. In this figure, a singularity had been excluded to improve the quality of reading the graph, yet the original results are included in the appendix. The maximum normal force occurred at the contact between the spillway and the arch. It can be seen from the figure that the model with perpendicular beaches resulted in significantly lower force in this region. For both studied configurations of the beaches, the force was constant in the central part of the dam and increased towards the abutment. Figure 5.26 Normal forces for an arch dam for two different reservoir beaches, the values normalized for each model individually. 0% 20% 40% 60% 80% 100% 0 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60 63 66 69 72 Normal force(N/Nmax) % Dam length (m) Normal forces for perpendicular beaches Normal forces for inclined beaches 0% 20% 40% 60% 80% 100% 0 7 15 22 30 37 45 52 60 67 75 82 90 97 105112120127135142150157165 Normal force(N/Nmax) % Dam Length (m) Normal forces for perpendicular beaches Normal forces for inclined beaches 5.5. PRESSURE DISTRIBUTION OVER THE ICE THICKNESS 59 5.5 Pressure distribution over the ice thickness To have an understanding of the distribution of the contact force at the contact surface between the ice sheet and the concrete dam. The study performed on two buttress dams’ models; both of them consisted of three monoliths. In the first model, the ice sheet was modelled as a planar sheet and was connected to the concrete dam along its edge, i.e. similarly as modelled in the previous sections. As an alternative, another model defined with a vertical sheet along the edge was added. This vertical sheet was defined with a height of 1.0 m, i.e. the ice thickness, it was included to ensure surface to surface contact between the ice sheet and the concrete dam. The distribution of the ice pressure over the thickness of the ice sheet that had a direct connection with the dam is presented in Figure 5.27. The ice sheet had a perpendicular beach. As seen in the figure, the maximum pressure occurred in the middle of the ice thickness. This result thereby shows that modelling an edge contact between the ice sheet and the concrete surface resulted in a triangular pressure with its maximum at the middle of the ice thickness. This showed that the local pressure thereby deviates from the assumption of the ice pressure give in RIDAS. Figure 5.27 Pressure distribution over the ice thickness from direct connection The pressure over the ice thickness is illustrated in Figure 5.28, where a more realistic connection between the ice sheet and the concrete dam was considered through the vertical ice sheet, i.e. rigid connection, The pressure was distributed linearly, where the top is subjected to tension, and the bottom is subjected to compression. The pressure distribution indicated the bending of the ice sheet where the resulting force seems to be close to zero. -200 4800 9800 14800 19800 24800 0 0,2 0,4 0,6 0,8 1 Force (N) Vertical Distance on the contact surface (m) CHAPTER 5. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 60 Figure 5.28 Pressure distribution over the ice thickness from a rigid connection -15000 -10000 -5000 0 5000 10000 15000 0 0,2 0,4 0,6 0,8 1 Force (N) Vertical Distance on the contact surface (m) 6.1. INFLUENCE OF ICE LOADS ON DAMS 61 6 Discussion This chapter contains the discussion for the results that had been presented in chapter five; the discussion is divided into four subcategories: • Influence of ice loads on dams. • Evaluation of the parametric study regarding the dam crest deflection. • Influence of the inclination of the beaches. • Pressure distribution over the ice thickness. Each part includes the author's interpretations of the results regarding the discussed subcategory. 6.1 Influence of ice loads on dams In chapter 5.1, the results of the one monolith buttress dam and an arch dam were presented. These models had been studied in order to gain an insight of the force and displacement relationship for the effect of ice load on concrete dams. In the models of buttress dam and arch dam, the ice load was applied as a triangular distributed load with a maximum resultant force of 640 kN/m. When the contact between dam and rock foundation was fixed, the resultant load caused deformation in the dam crest that reached 2 mm when the ice load reached 640 kN/m. This deflection is small in comparison to deflection due to temperature variation that had been studied in the “Proceedings of the 14th ICOLD International Benchmark Workshop on Numerical Analysis of Dams” (Malm et al., 2018). The maximum deflection due to temperature variation is 80 mm in cold winter and -22 mm in warm summer where the positive deflection sign is towards the downstream (Malm et al., 2018). Due to the interaction between concrete and rock, the buttress dam behaved as a cantilever beam after applied ice load, where it deformed, and stresses were introduced. When the concrete and rock was defined with contact interaction using a friction coefficient of µ = H/N = 1, the one monolith buttress dam started to slide as the ice load resultant force reached 640 kN. This value is significantly higher than typical design code values for an 1 m thick ice sheet. Hence this situation is not likely to occur in reality. The real forces are lower than this and typically less than 253 kN/m according to Adolfi and Eriksson (2013). CHAPTER 6. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 62 In the model of the arch dam, the friction coefficient was defined as the high value (40) to consider that the base of the dam is excavated into the rock.This high coefficient of friction, thereby makes it more or less impossible for sliding to occur. Instead, the dam reached stability failure when the ice load reached around 550 kN/m, which is also a higher value than real-life situations. The uplift pressure was not included, which leads to that the ice load that is required to obtain a failure should be lower than this. As mentioned above, the ice load effect on the dam stability in the studied cases was small compare to other dominant forces like hydrostatic pressure, gravity and temperature variation thermal effect on the dam. This might not be the case for all dams, especially smaller dams (Malm, 2020). Yet according to Ridas (2017) in the ultimate state design, the dominant variable load should be increased with a factor of 1.95, and at the same time, other loads should be increased but with a smaller factor. The displacement was small in comparison to deflections from other forces, and the sliding failure needed a higher value of the ice load for a longer time compared to the real-life situation. There are not many cases in the literature that clearly showed that a dam ever had failed due to an ice load. In some cases, the failure cause had been inconclusive, where ice load may have been one of the primary factors causing the failure (Hellgren, 2019). 6.2 Variation in ice load along dams The results for the displacements and the normal forces along different dam geometries was presented in Chapter 5.2. These displacements and normal forces were due to expansion that caused by either a constant temperature expansion in an ice sheet or due to a thermal gradient. The variation in displacement and normal forces along the dams shows that the maximum ice loads occur at the dam supports, i.e. near the beaches or spillway and abutments. At these locations, the dam had high stiffness resulting in small displacement, which initiates these large ice load. The reason for this was the interactions, which resulted that the deformation of the dam was restrained at these locations. The assumption for the boundary conditions is that it is completely anchored to the rock at the sides of the dam. In all the studied concrete buttress dams, it had been found that the maximum field displacement (hence the minimum field normal force) occurred at the connection between two monoliths. The minimum field displacement and thus maximum field normal forces occur at the location where the monolith had the highest stiffness, i.e. over the buttresses support. The explanation is the rigidity of the structure since it is low at the connection between two monoliths and higher at the buttresses supports. Because the thermal ice pressure is a restraint force similar to shrinkage, it wants to expand as the temperature increase. This expansion induces stresses if this expansion is not allowed to occur (i.e. there is some form of restraint). The higher degree of restraint, the higher the forces will be. In normal design, the assumption is that the ice load is at its maximum value along the full length of the dam. As the results of this study showed, this will not occur since the dam will deform and thereby reduce the ice load. 6.3. INFLUENCE OF THE INCLINATION OF THE BEACHES 63 In the arch dam, when the ice sheet was subjected to a thermal gradient that caused the ice sheet to bend; therefore, the maximum displacement and minimum normal forces occurred at the connection between the spillway and the arch. When the ice sheet is expanding, due to constant temperature effect, the normal forces from the ice load and the distribution of displacements along the dam had the same magnitude as long as the dam consisted of three or more monoliths. If the dam had a curvature or horizontal inclination, where some monoliths were installed with an angle to the straight monoliths, the corner connection had the minimum displacement. This was due to that this part of the dam had the highest stiffness. In addition, the maximum displacement will occur at the straight part of the dam. Although the ice expands in all directions, the reasons why the inclined part is subjected to lower ice loads is likely because it was stiffer due to its support. In the studied buttress dams, when the ice sheet was subjected to a thermal gradient, this caused the ice sheet to bend. The ice sheet was restrained at the edges, and it caused forces on to the first and last monoliths of the dam because of the rigid boundary conditions. In the central parts of the dam, the ice load was almost zero due to the bending of the ice sheet which unloaded the monoliths in this region. 6.3 Influence of the inclination of the beaches The results for the study were made to investigate the influence of the reservoir beaches inclination on the ice load acting on a dam. These results were presented in Chapter 5.4, where the studied models were modelled with two different beaches. The first one was a perpendicular beach to the dam surface and the second one was inclined beach with 30 degrees from the dam surface towards the upstream direction. In the studied models with perpendicular beaches, the ice sheet was restrained along its edges, which caused the maximum ice load to occur on the edges of the dams. Due to the difference in the stiffness between the joints that connected the monoliths and the parts adjacent to the buttress, the ice load variation was as a harmonic shaped curve. In the studied models with inclined beaches, the distribution of the normal forces along the dam was similar to the distribution in the models with perpendicular beaches. The results also showed for the buttress dams having inclined beaches, that the normal forces had a lower magnitude at the edges in comparison to the same dams having perpendicular beaches. The reduction was 35 % in the models with three monoliths meanwhile for the nine monoliths it was 31 %. The arch model behaved differently than buttress dams’ models at its edges, and obtained increased forces with 30 % in comparison to the same dam that had perpendicular beaches. The reason for this reduction in the buttress dams was that the inclined edges had a fixed boundary condition where it was defined as rigid edges as in the mountain and rock reservoir beaches. These rigid edges carried a high amount of the ice load and reduced the ice load that was acting on the dam. CHAPTER 6. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 64 High normal forces were obtained on the dam, if the edges of the ice sheet were not rigid like the reservoir beaches. Figure 6.1 illustrates the studied arch dam, where the hatched triangular area is an extra amount of ice sheet that acts on the dam in compared to the ice sheet with perpendicular beaches. This extra amount area had caused about 30 % higher value of normal forces in the connection between the spillway and arch in comparison to the perpendicular beaches. The same phenomenon applies to the model with nine monoliths, wherein the central part of the dam the inclined beaches normal forces were higher than the forces in the model with perpendicular beaches. Figure 6.1 Top view of the arch dam. 6.4 Pressure distribution over the ice thickness The distribution of the contact forces, at the contact surface between the ice sheet and dams, were presented in Chapter 5.5 for two different studied models. One of the models had a direct connection between the ice sheet and the dam and the second one had a rigid link connection. The ice sheet was modelled as a shell member, while the concrete dam was modelled as a solid member. This was creating a problem to define suitable contact formulations and to obtain the contact pressure, representing the ice load (see Chapter 3.4). A comparison was made between two models with a different type of connection between solid (dam) and shell (ice sheet) elements. Figure 6.2 represents how the pressure was distributed through the nodes at the connection. Figure 6.2a represents a direct connection between solid and shell element, and Figure 6.2b represents a rigid link connection. The model that had a rigid link connection showed a different distribution of the pressure over the ice thickness than the one that had a direct connection. See results from the model with a direct connection in Figure 5.27, and the model with a rigid connection in Figure 5.28. The results were reasonable since the rigid link distributed the pressure from several nodes to the dam while the direct connection distributed the pressure from one node to the dam, see Figure 6.2 for illustration 6.4. PRESSURE DISTRIBUTION OVER THE ICE THICKNESS 65 ( a ) ( b ) Figure 6.2 (a) Shell planar nodes connected to a solid surface, (b).planar shell connection to another shell surface which in its turn connected to the solid surface. 7.1. CONCLUSIONS 67 7 Conclusions and further research 7.1 Conclusions A parametric study was performed for different dam geometries and shapes of the ice sheet in the FE software Abaqus. The analyses were performed to study the response of different dams subjected to thermal ice loads, through investigations of the displacements, forces and distribution of the ice load. All analyses were performed based on the linear elastic properties for the concrete, rock and ice. In this study, the distribution of thermal ice load acting on concrete dams was investigated by using FE analysis, and this distribution was indicated by extracting the normal forces caused by the expansion of the ice sheet. The results showed that it was possible to extract the distribution of thermal ice load by using FE analysis. The ice load influenced the global stability of the studied concrete dams, but for these dams, the ice loads gave rise to quite small displacements compared to the thermal gradient that can be expected from for instance seasonal temperature variations. The size of the ice load normally consider in design was not large enough to introduce failures in the studied dams. The ice loads had to be increased with 252 % and 213 % before failure was initiated in the buttress dam and arch dam respectively. The deflection was 55 mm for the arch dam and 8 mm for the buttress dam. According to “Proceedings of the 14th ICOLD International Benchmark Workshop on Numerical Analysis of Dams” the maximum deflection due to temperature variation in an arch dam is 80 mm. This leads to the conclusion that the displacement from ice load in studied dams is small compared to temperature variation. For instance, in the arch dam, the deflection is reduced with 31.2 %. In the studied dams, the contribution of the ice load in the overturning is small, and other dominant forces had a bigger effect on the dam stability than the ice load. Yet there are several dams, especially low dams, that cannot satisfy the safety criteria due to the significant influence of the ice load. An arch dam in addition to buttress dams with different numbers and placement of monoliths had been studied regarding ice load distribution along with the dams. As a conclusion from the results of the thesis, it can be said that at the higher degree of restraint and the higher stiffness is the higher value of ice load will occur. For the arch dam with a constant temperature on the ice sheet, the maximum ice load occurred at the locations where the stiffness was high, for instance, dam supports, near the beaches or spillway and abutments. The thermal gradient caused a bending in the ice sheet, and thereby it resulted that the minimum forces occurred in the connection between the CHAPTER 7. FINITE ELEMENT ANALYSES OF CONCRETE DAMS 68 spillway and the arch. In the models of the concrete buttress dams using constant temperature in the ice sheet, the maximum ice load occurred at the supports. The distribution of the ice load was the same in all the models as far as the dam was consisted of at least three monoliths. In the models of the buttress dams where the ice was subjected to a thermal gradient, the ice sheet bent upwards due to the effect of the thermal gradient. The ice load occurred at the first and last monoliths due to the rigid boundary conditions, and the ice load was zero at the central parts of the dam. In the buttress dam with curvature or horizontal inclination of some monoliths, the maximum ice load occurred at the corner connection between the straight and angled monoliths. The minimum ice load occurred at the straight part of the dam. The corner connections were modelled as a tie connection and had a high stiffness. To study the effect of the beaches shape on the ice load distribution along with the dams. A comparison between two different shapes of the edges of the ice sheet had been made. The edges were either perpendicular to the dams’ surface or had an inclination towards the surface of the dams. The results showed that the inclination will distribute the load into two-components, where one will act perpendicular to the dam and the other parallel to the dam. The inclination in the buttress dams model resulted in a lower normal force from the ice load at the edges in comparison to the buttress model subjected to the perpendicular beaches. If these beaches are rigid and start at the edges of the dam, a significant part of the ice load is transmitted to the boundary conditions which results in about 30 % lower ice loads near the edges and increase the loads at the field of the dam or the points that in the central part of the dam. In the arch dam, the inclination of the ice sheet resulted in 30 % higher forces in the connection between the arch dam and the spillway. This was due to that inclination increased the volume of the ice sheet and thereby the forces from ice load. The results showed that the length of the modelled ice sheet did not have a significant influence on the distributions of normal forces and displacement of the studied dams. Because the farther end of the ice sheet from the dam was set as a fixed constrained the length of the ice sheet had no effect on the produced results from the parametric study. Two models had been compared with a different type of connection between the modelled ice sheet and the concrete dam. One of them had a direct connection between the dam and the ice sheet, and the second had a rigid link connection. The direct connection between the ice sheet and the dam resulted in the pressure distribution as a point load, see Figure 6.2a. The pressure from the ice sheet was directly transmitted from a single node to multiple nodes at the dam surface. Therefore, a vertical shell element had been presented into the model to create a rigid link connection, see Figure 6.2b. The rigid link distributed the pressure from several nodes to the dam surface resulting in a better distribution of the contact pressure. The pressure distribution was linear; the top of the ice sheet was subjected to tension, and the bottom subjected to compression. This linear distribution indicated the bending of the ice sheet. Due to the differnce in the ice pressure distribution over the thickness of the ice sheet between the two models that was compared, it can be concluded that it was uncertain to capture the variation in ice pressure over the ice thickness. 7.2. FURTHER RESEARCH 69 7.2 Further research Further evaluation for the ice loads should be done with models that can capture the nonlinear behaviour of the ice sheet, where cracks can occur. Also, a nonlinear analysis can be done to the concrete dams, where the stiffness of the dams can vary due to the cracks, and it might influence the ice load distribution. Future studies are advised to investigate different thermal gradients and compare their effect on the ice load. It is interesting to know how the height of the concrete dams will affect the influence of the ice load on the dam. The models in this study captured only the linear behaviour of concrete dams and ice sheets. The study was focused on a thermal gradient that was same for all the models with different buttress dams geometries that have the same height. Another important topic for future research that has not been covered in this report is the influence of variations in water level. Future research should, therefore, be performed to analyse how the ice loads are influenced by variations in water level with different magnitudes. BIBLIOGRAPHY 71 Bibliography Adolfi E & Eriksson J (2013): “Islastens inverkan på brottsannolikheten för stjälpningen av betongdammar, ” MSc Thesis 13/01, ISSN 1652-599. KTH Royal Institute of Technology, Dept. Soil and Rock Mechanics , Stockholm, Sweden. Akıntuğ B (2012): Lecture notes from course CVE471 Water Resources Engineering, Middle East Technical University Northern Cyprus Campus. Bergdahl L (1977): “Physics of ice and snow as affects thermal pressure, ” MSc Thesis ISSN 0348-1050. Department Hydraulics, Chalmers University of Technology, Gothenburg, Sweden. Bergh H (2014): Hydraulic Engineering. KTH Royal Institute of Technology Dept. Civil and Architectural Engineering, Stockholm, Sweden. Carter D, Sodhi D, Stander E, Caron O & Quach T (1998): “Ice thrust in reservoirs,” Journal of Cold Regions Engineering, 12(4), PP. 169-183. Casagrande G (2014): Spaces and places fifty years after the Vajont tragedy, Geographic Reasearch and Application Laboratory ISSN: 2281-5694 vol. 1, PP. 51-62, , European University of Rome, Rome, Italy. Chen S-H (2015): “Hydraulic Structures”. Wuhan University, Wuhan, China. Comfort G, Gong Y, & Singh S (2000): “The factors controlling static ice loads on dams, ” Proceedings, 15th International Symposium on Ice, Gdansk, Vol 1, pp. 189–197. Comfort G, Gong Y Singh S & Abdelnour R (2003): “Static ice loads on dams, ” Canadian Journal of Civil Engineering, Canada, 30(1), PP. 42-68. Dassult Systèms (2014): Abaqus 6.14 Documentation [Online], Dassault Systèmes Simulia Corp, Available: [Accessed Dec 12 2019]. Duffault P (2013): “The traps behind the failure of Malpasset arch dam, France, in 1959,” Journal of Rock Mechanics and Geotechnical Engineering 5(5), Chinese Academy of Sciences, Institute of Rock and Soil Mechanics,. PP.335–341 Eriksson A (2002): Lecture notes from course 1C1117 “Finite Element Methods,” KTH Royal Institute of Technology Dept. Civil and Architectural Engineering, Stockholm, Sweden. BIBLIOGRAPHY 72 Eurocode 1 (2011): “Actions on structures.” Part 1-1: General actions – Densities, selfweight, imposed loads for buildings. SS EN 1991-1-1, SIS. Eurocode 2 (2008): Design of concrete structures. Part 1-1: General rules and rules for buildings. SS-EN 1992-1-1:2005: Swedish Standards Institute (SIS). Lio F (2019): “56 anni fa il disastro del vajont, quasi 2000 vittime”. Available at: (Accessed 2020-05-16). Gebre S , Alfredsen K , Lia L, Morten S & Einar T (2013): “Ice Effects on Hydropower Systems A Review,” Journal of cold regions engineering 27(4), Norway. Ghanaat Y & Clough R W (1993): “Concrete dams: Evaluation for seismic loading,” International Workshop on Dam Safety, Grindelwald, Switzerland, Vol. 4, PP.137-169. Hellgren R (2019): “Condition assessment of concrete dams in cold climate”. Licentiate Thesis, TRITA-ABE-DLT 1934, KTH Royal Institute of Technology, Dept. Civil and Architectural Engineering, Stockholm, Sweden. Hellgren R, Malm R, Fransson L, Johansson F, Nordström E & Westberg M (2020): “Measurement of ice pressure on a concrete dam with a prototype ice load panel”. Cold Regions Science and Technology, Vol, 170, February 2020, 102923. Jeppsson J (2003): “Reliability-based assessment procedures for existing concrete structures”. Doctoral Thesis, Rapport TVBK – 1026. LTH School of Engineering, Dept. Structural Engineering, Lund, Sweden. Johansson F, Spross J & Lennart F (2013): “Islast mot dammkonstruktioner: Sammanställning av kunskapsläget samt förslag till forskning och utveckling.” Report 13:56, Elforsk AB, Stockholm, Sweden. Kharik E, Morse B, Roubtsova V, Fafard M, Côté A & Comfort G (2017): “Numerical studies for a better understanding of static ice loads on dams,” Canadian Journal of Civil Engineering, Vol.45(1), PP.18-29. Ko P K, Ho M S & Smith G F (1994): “Thermal ice forces on concrete dams: Recent developments,” Proceedings of the Xth Canadian Dam Safety Conference, Winnipeg, Canada, PP.17-33. Linsbauer H & Bhattacharjee S (1999): “Dam safety assessment due to uplift pressure action in a dam-foundation interface crack,” Proceedings Fifth Benchmark Workshop on Numerical Analysis of Dams, USCOLD. Malla S & Wieland M (1999): “Analysis of an arch-gravity dam with a horizontal crack.” Computers & Structures (72), Zurich, Switzerland, pp.267 – 278. Malm R (2016): “Guideline for FE analyses of concrete dams.” Report 2016:270, Energiforsk AB, Stockholm, Sweden. BIBLIOGRAPHY 73 Malm R, Nordström E, Johansson F, Hellgren R & Westberg W M (2017): ”Lastförutsättningar avseende istryck.” Report 2017:439, Energiforsk AB, Stockholm, Sweden. Malm R, Ansell A, Gasch T & Eriksson D (2018): “Freezing of partially saturated air-entrained concrete: A multiphase description of the hygro-thermo-mechanical behaviour.” KTH Royal Institute of Technology, Dept. Civil and Architectural Engineering, Stockholm, Sweden. Malm R, Hassanzadeh M & Hellgren R (2018): “Proceedings of the 14th ICOLD International Benchmark Workshop on Numerical Analysis of Dams” TRITA-ABE-RPT,Vol. 1802001, KTH Royal Institute of Technology, Dept. Civil and Architectural Engineering, Stockholm, Sweden. Malm R (2020): RE: Personal communication: General information in arch dams, KTH Royal Institute of Technology, Dept. Civil and Architectural Engineering, Stockholm, Sweden. Marinos G P (1958): “A Field Trip for Civil Engineering students to Demonstrate the importance of EngineeringGeology.” Available at: (Accessed 2020-05-16). Morse B, Stander E, Coté A, Taras A, Santerre R & Prat Y (2011): “SpatialTemporal Variability of Static Ice Forces,” Proceedings of the 21st International Conference on Port and Ocean Engineering under Arctic Conditions, Montréal, Canada. NVE (2003): “Retningslinjer for laster och og dimensionering”, The Norwegian Water Resources and Energy Directorate, Oslo, Norway. Pedro J O (1999): “Arch dams: designing and monitoring for safety.” CISM Courses and lectures, international centre for mechanical sciences 367, Springer-Verlag Wien. Petrich C, Sand B, Arnsten B, Fransson L & Saether I (2015): “Time-dependent spatial distribution of thermal stresses in the ice cover of a small reservoir.” Cold Regions Science and Technology, 120:35–44, 2015. ISSN 0165232X. DOI:10.1016/j.coldregions.2015.09.003. RIDAS (2011): “Swedish Hydropower companies guidelines for dam safety,” Application guideline 7.3 Concrete dams, Svensk Energi, Stockholm, Sweden. RIDAS (2017): “Swedish Hydropower companies guidelines for dam safety,” Application guideline 7.3 Concrete dams, Svensk Energi, Stockholm, Sweden. Ruggeri G (2004): “Sliding safety of existing gravity dams: final report.” Working Group on Sliding Safety of Existing Gravity Dams, ICOLD European Club. Saether I (2012): “Methods for prediction of thermally induced ice loads on dams and hydro-electrical structures,” literature review, Technical Report, Report No. 2012/03, (Norut) Narvik, Norway. BIBLIOGRAPHY 74 Stander E (2006): “Ice stresses in reservoirs: effect of water level fluctuations.” Journal of Cold Regions Engineering, Vol. 20(2), pp.52–67. Taras A, Côté A, Comfort G, Thériault L & Morse B (2011): “Measurements of Ice Thrust at Arnprior and Barrett Chute Dams,” 16th Workshop on the Hydraulics of Ice Covered Rivers, Winnipeg, Manitoba, PP.317–328. USACE (1994): “Arch dam design.” EM 1110-2-2201, US Army Corps of Engineers (USACE), Washington DC, U.S.A.. 75 Appendix A Displacement due to constant temperature Figure A-1 Figure A-2 0 0,002 0,004 0,006 0,008 0 1 2 3 4 5 6 7 8 9 101112131415161718192021222324 Displacement (m) Dam length (m) 3 Monoliths dam with straight 3 times bigger ice sheet 0 0,001 0,002 0,003 0,004 0,005 0,006 0 1 2 3 4 5 6 7 8 8 9 111112131415161718192021222324 Displacement (m) Dam length (m) 3 Monoliths dam with straight ice sheet APPENDIX A 76 Figure A-3 Figure A-4 0 0,001 0,002 0,003 0,004 0,005 0,006 0 1 2 3 4 5 6 7 8 8 9 111112131415161718192021222324 Displacement (m) Dam length (m) 3 Monoliths dam with Angled ice sheet 0 0,002 0,004 0,006 0,008 0,01 0 3 6 9 12151720232528313437404346495255576063666972 Displacement (m) Dam length (m) 9 Monoliths dam with angled ice sheet 77 Figure A-5 Figure A-6 Figure A-7 0 0,002 0,004 0,006 0,008 0,01 0 3 6 9 12151720232528313437404346495255576063666972 Displacement (m) Dam length (m) 9 Monoliths dam with straight ice sheet 0 0,01 0,02 0,03 0,04 0,05 0,06 0 10 19 29 39 49 58 68 78 88 97 107117127136146156166 Displacement (m) Arch length (m) Arch dam with straight ice sheet 0 0,01 0,02 0,03 0,04 0,05 0 10 19 29 39 49 58 68 78 88 97 107117127136146156166 Displacement (m) Arch length (m) Arch dam with angled ice sheet APPENDIX A 78 Figure A-8 Displacement due to thermal gradient Figure A-9 0 0,0005 0,001 0,0015 0,002 0 8 15 22 29 37 44 51 59 66 73 80 88 95 103110117 Displacement (m) Dam length (m) Rotated Monoliths dam 0 0,00002 0,00004 0,00006 0,00008 0,0001 0,00012 0 1 3 4 6 7 9 10 12 13 15 16 17 19 20 22 23 Displacement (m) Dam length (m) 3 Monoliths dam with straight ice sheet 79 Figure A-10 Figure A-11 0 0,000001 0,000002 0,000003 0,000004 0,000005 0,000006 0,000007 0,000008 0 1 3 4 6 7 9 10 12 13 15 16 17 19 20 22 23 Displacement (m) Dam length (m) 3 Monoliths dam with angled ice sheet 0 0,000002 0,000004 0,000006 0,000008 0,00001 0,000012 0 3 6 9 121517202326293235384144474952555861646770 Displacement (m) Dam length (m) 9 Monoliths dam with straight ice sheet APPENDIX A 80 Figure A-12 Figure A-13 0 0,000002 0,000004 0,000006 0,000008 0,00001 0,000012 0,000014 0 3 6 9 121517202326293235384144474952555861646770 Displacement (m) Dam length (m) 9 Monoliths dam with angled ice sheet -0,00005 -0,00004 -0,00003 -0,00002 -0,00001 0 0,00001 0,00002 0,00003 0,00004 0,00005 0 10 21 31 42 52 63 73 84 94 105115126136147157168 Displacement (m) Dam length (m) Arch dam with straight ice sheet 81 Figure A-14 Normal forces due to constant temperature Figure A-15 0 0,000001 0,000002 0,000003 0,000004 0,000005 0,000006 0,000007 0,000008 0 7 11 18 23 31 38 49 61 73 83 88 94 100107114 Displacement (m) Dam length (m) Rotated Monoliths dam 0 1000000 2000000 3000000 4000000 5000000 6000000 7000000 0 2 3 5 6 8 9 11 12 14 15 17 18 20 21 23 24 NForce (N) Dam length (m) 3 Monoliths dams with straight ice sheet APPENDIX A 82 Figure A-16 Figure A-17 0 500000 1000000 1500000 2000000 2500000 3000000 3500000 4000000 4500000 0 2 3 5 6 8 9 11 12 14 15 17 18 20 21 23 24 NForce (N) Dam length (m) 3 Monoliths dam with Angled ice sheet 0 500000 1000000 1500000 2000000 2500000 3000000 3500000 4000000 4500000 5000000 0 3 6 9 121518212427303336394245485154576063666972 NForce (N) Dam length (m) 9 Monoliths dam with angled ice sheet 83 Figure A-18 Figure A-19 0 1000000 2000000 3000000 4000000 5000000 6000000 7000000 8000000 0 2 3 5 6 8 9 11 12 14 15 17 18 20 21 23 24 Nforce(N) Dam length (m) 3 Monoliths dams with straight ice sheet 3 times big size 0 1000000 2000000 3000000 4000000 5000000 6000000 7000000 8000000 0 3 6 9 121518212427303336394245485154576063666972 NForce(N) Dam length (m) 9 Monoliths dam with straight ice sheet APPENDIX A 84 Figure A-20 Figure A-21 0 200000 400000 600000 800000 1000000 1200000 1400000 1600000 1800000 2000000 0 1 2 3 4 5 6 7 8 9 101112131415161718192021222324 NForce(N) Dam length (m) 3 Monoliths dams with straight ice sheet with vertical shell element 0 1000000 2000000 3000000 4000000 5000000 6000000 0 10 21 31 42 52 63 73 84 94 105115126136147157168 NForce(N) Dam Length (m) Arch dam with angled ice sheet 85 Figure A-22 Figure A-23 -4000000 -3000000 -2000000 -1000000 0 1000000 2000000 3000000 4000000 5000000 0 11 22 34 45 56 67 79 90 101112124135146157169 NForce(N) Dam length (m) Arch dam with straight ice sheet 0 200000 400000 600000 800000 1000000 1200000 1400000 0 8 15 23 30 38 45 53 60 68 75 83 91 98 106113 NForce(N) Dam length (m) Rotated Monoliths dam APPENDIX A 86 Normal forces due to thermal gradient Figure A-24 Figure A-25 -1500 -1000 -500 0 500 1000 1500 0 2 3 5 6 8 9 11 12 14 15 17 18 20 21 23 24 NForce(N) Dam length (m) 3 Monoliths dam with Angled ice sheet -2500 -2000 -1500 -1000 -500 0 500 1000 1500 2000 2500 3000 NForce(N) Dam length (m) 3 Monoliths dam with straight ice sheet 87 Figure A-26 Figure A-27 -3000 -2000 -1000 0 1000 2000 3000 4000 0 3 6 9 121518212427303336394245485154576063666972 NForce(N) Dam length (m) 9 Monoliths dam with angled ice sheet -2500 -2000 -1500 -1000 -500 0 500 1000 1500 2000 2500 NForce(N) Dam length [m] 9 Monoliths dam with straight ice sheet APPENDIX A 88 Figure A-28 Figure A-29 -20000 -15000 -10000 -5000 0 5000 10000 15000 20000 25000 30000 NForce(N) Dam length (m) Arch dam with angled ice sheet -10000 0 10000 20000 30000 40000 50000 NForce(N) Dam length (m) Arch dam with straight ice sheet 89 Figure A-30 -4000 -3000 -2000 -1000 0 1000 2000 3000 NForce(N) Dam length (m) Rotated Monoliths dam TRITA-ABE-MBT-20318 www.kth.se |
15355 | https://www.engr.colostate.edu/ECE412/SP17/Lectures%2019-20.pdf | Stability Analysis of Digital Control Systems Digital Filter Design Digital Controls & Digital Filters Lectures 19 & 20 M.R. Azimi, Professor Department of Electrical and Computer Engineering Colorado State University Spring 2017 M.R. Azimi Digital Control & Digital Filters Stability Analysis of Digital Control Systems Digital Filter Design Stability Analysis (Chapter 7) Definition 1: Bounded Input, Bounded Output (BIBO) Stability An LTI system is BIBO stable iffa bounded input yields a bounded output (for any IC), i.e.
If |x(n)| < M < ∞= ⇒|y(n)| < N < ∞ Theorem 1: The necessary and sufficient condition for BIBO stability is that the impulse response h(n) is absolutely summable.
∞ P n=0 |h(n)| < P < ∞ Definition 2: Asymptotic Stability Consider autonomous (no input) LTI system: x(n + 1) = Ax(n) System is asymptotically stable ifffor any IC x(0) such that ||x(0)|| < δ = ⇒ lim n→∞||x(n)|| = 0 where ||x|| represents the Euclidean norm of vector x.
M.R. Azimi Digital Control & Digital Filters Stability Analysis of Digital Control Systems Digital Filter Design Stability Analysis-Cont.
Theorem 2: A discrete-time LTI system given before, is asymptotically stable iffall the eigenvalues of matrix A (poles of the system) lie strictly inside the unit circle.
Theorem 3: Asymptotic stability implies BIBO stability and vice versa.
Example: Consider the following LTI system x(n + 1) = 0 1 −1 0 x(n) + 1 0 u(n) y(n) = 1 0 x(n) The roots of CE or eigenvalues of matrix A are obtained by solving, ϱ(z) = |zI −A| = z −1 1 z = (z2 + 1) = 0 = ⇒z1,2 = ±j = e±jπ/2 i.e. roots on the unit circle and hence not asymptotically stable.
On the other hand, we can test for BIBO stability by finding h(n) and checking absolute summable requirement.
M.R. Azimi Digital Control & Digital Filters Stability Analysis of Digital Control Systems Digital Filter Design Stability Analysis-Cont.
State transition matrix, An = φ(n) = Z−1 (zI −A)−1z (zI −A)−1 = z 1 −1 z z2+1 An = Z−1 " z2 z2+1 z z2+1 −z z2+1 z2 z2+1 # = cos nπ 2 sin nπ 2 −sin nπ 2 cos nπ 2 Then the impulse response, h(n) = CAn−1B, becomes, h(n) = 1 0 " cos (n−1)π 2 sin (n−1)π 2 −sin (n−1)π 2 cos (n−1)π 2 # 1 0 = cos (n−1)π 2 ∞ P n=0 |h(n)| = ∞ P n=0 | cos (n−1)π 2 | →∞ i.e. not BIBO stable.
M.R. Azimi Digital Control & Digital Filters Stability Analysis of Digital Control Systems Digital Filter Design Jury’s Stability Test Let the closed-loop CE be, ϱ(z) = aNzN + aN−1zN−1 + . . . + a1z + a0 = 0 where a′ is are the coefficients with aN > 0 Form the Jury’s table: Row 0 z0 z1 . . .
zN−2 zN−1 zN Row 1 a0 a1 . . .
aN−2 aN−1 aN Row 2 aN aN−1 . . .
a2 a1 a0 Row 3 b0 b1 . . .
bN−2 bN−1 Row 4 bN−1 bN−2 . . .
b1 b0 Row 5 c0 c1 . . .
cN−2 Row 6 cN−2 cN−3 . . .
c0 .
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Row (2N −3) m0 m1 m2 where bk = a0 aN−k aN ak , ck = b0 bN−1−k bN−1 bk , dk = c0 cN−2−k cN−2 ck M.R. Azimi Digital Control & Digital Filters Stability Analysis of Digital Control Systems Digital Filter Design Jury’s Stability Test-Cont.
Theorem: Necessary and sufficient conditions for asymptotic stability are: 1 ϱ(1) > 0 2 (−1)Nϱ(−1) > 0 3 |a0| < aN 4 |b0| > |bN−1| 5 |c0| > |cN−2| 6 |d0| > |dN−3| .
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|m0| > |m2| Remarks: 1. For a second order system (N = 2) Jury’s table contains only one row and the conditions for stability are: 1 ϱ(1) > 0 2 ϱ(−1) > 0 3 |a0| < a2 M.R. Azimi Digital Control & Digital Filters Stability Analysis of Digital Control Systems Digital Filter Design Jury’s Stability Test-Cont.
2. For N = 3 conditions are: 1 ϱ(1) > 0 2 −ϱ(−1) > 0 3 |a0| < a3 4 |b0| > |b2| Example 1: Given the following digital control system, find the range of gain K for stability.
G(z) = (1 −z−1)Z h 6.4 s(s+1)(s+0.2) i = 0.16z+0.064 z2−1.78z+0.78 Closed-loop transfer function, T(z) = C(z) R(z) = KG(z) 1+0.03KG(z) M.R. Azimi Digital Control & Digital Filters Stability Analysis of Digital Control Systems Digital Filter Design Jury’s Stability Test-Cont.
Thus, the CE of the closed-loop system, ϱ(z) = 1 + 0.03KG(z) = z2 −1.78z + 0.78 + 0.03K(0.16z + 0.064) = 0 or, ϱ(z) = z2 −(1.78 −0.0048K)z + (0.78 + 0.0019K) = 0 For stability, 1 ϱ(1) > 0 = ⇒0.007 + 0.0067K > 0 = ⇒K > −1.04 2 ϱ(−1) > 0 = ⇒3.57 −0.0028K > 0 = ⇒K < 1.24 × 103 3 a0 < a2 = ⇒0.78 + 0.0019K < 1 = ⇒K < 111 Therefore, the range of K for stability is 0 < K < 111.
Example 2: The CE of a digital control system is given by, ϱ(z) = z3 + (111.6T 2 + 16.74T −3)z2 + (3 −33.48T + 1.395 × 10−4KT 3)z + (1.395 × 10−4KT 3 + 16.74T −111.6T 2 −1) = 0 find the range of T (sampling period) and gain K for stability.
M.R. Azimi Digital Control & Digital Filters Stability Analysis of Digital Control Systems Digital Filter Design Jury’s Stability Test-Cont.
1 ϱ(1) > 0 = ⇒KT 3 > 0 = ⇒K > 0 2 ϱ(−1) < 0 = ⇒−8 + 66.96T < 0 = ⇒T < 0.12 sec 3 |a0| < a3 = ⇒|1.395 × 10−4KT 3 + 16.74T −111.6T 2 −1| < 1 4 |b0| > |b2| = ⇒(1.395 × 10−4KT 3 + 16.74T −111.6T 2 −1)2 −1 > |(1.395 × 10−4KT 3 + 16.74T −111.6T 2 −1)(111.6T 2 + 16.74T −3) − (3 −33.48T + 1.395 × 10−4KT 3)| The stable region can be identified by plotting.
This example shows the utility of the Jury’s test for design purposes as well.
M.R. Azimi Digital Control & Digital Filters Stability Analysis of Digital Control Systems Digital Filter Design Jury’s Stability Test-Cont.
Remark: When some or all elements of a row in Jury’s table become zero, tabulation terminates.
Remedy: Expand or contract the unit circle by z →(1 ± ϵ)z where ϵ > 0 zN →(1 ± ϵ)NzN But since (1 ± ϵ)N ≈1 ± Nϵ we use zN →(1 ± Nϵ)zN.
i.e. we use (1 ± Nϵ)zN instead of zN in table and proceed.
Example: Let ϱ(z) = z3 + z2 + z + 1 Row z0 z1 z2 z3 1 1 1 1 1 2 1 1 1 1 3 0 0 0 Substitute z3 →(1 + 3ϵ)z3 , z2 →(1 + 2ϵ)z2, and z →(1 + ϵ)z M.R. Azimi Digital Control & Digital Filters Stability Analysis of Digital Control Systems Digital Filter Design Jury’s Stability Test-Cont.
Then, Jury’s table becomes Row z0 z1 z2 z3 1 1 1 + ϵ 1 + 2ϵ 1 + 3ϵ 2 1 + 3ϵ 1 + 2ϵ 1 + ϵ 1 3 −6ϵ −4ϵ −2ϵ 4 −2ϵ −4ϵ −6ϵ 1 ϱ(1) > 0 = ⇒ϱ(1) = 4 > 0 2 −ϱ(−1) > 0 = ⇒ϱ(1) = 0 3 |a0| < |a3| = ⇒1 < 1 + 3ϵ = ⇒ϵ > 0 4 |b0| > |b2| = ⇒6ϵ > 2ϵ Stable for ϵ > 0 and unstable for ϵ < 0 = ⇒poles must be on the unit circle.
M.R. Azimi Digital Control & Digital Filters Stability Analysis of Digital Control Systems Digital Filter Design Design of Digital Filters Comprises of 4 general steps: 1 Approximation: Process of generating a transfer function satisfying a set of desired specs (frequency domain).
2 Realization: Process of converting a transfer function into a filter structure.
3 Quantization effects: Process of studying finite word length effects in digital systems.
i Input Quantization (at A/D).
ii Coefficient Quantization (at multipliers).
iii Product Quantization (output of multipliers).
4 Implementation: Process of implementing the designed filter either in i Software, or ii Hardware.
M.R. Azimi Digital Control & Digital Filters Stability Analysis of Digital Control Systems Digital Filter Design Approximation 1. Approximation Methods for Recursive or IIR Digital Filters Consider a recursive digital filter described by difference equation (or transfer function): y(n) = M P i=0 bix(n −i) − N P j=0 ajy(n −j) x(n): input, y(n) output, N: order, a′ js, b′ is: filter coefficients Advantages of IIR filters: 1 Low order (small N) gives sharper cutofffrequencies than FIR filters of the same order.
2 Can be implemented easily in time-domain (recursive).
Disadvantages: 1 Nonlinear phase 2 Stability issues 3 Quantization issues M.R. Azimi Digital Control & Digital Filters Stability Analysis of Digital Control Systems Digital Filter Design Approximation Methods Indirect Methods: Based on the design of analog prototype.
Idea: Start with the design of an analog filter, HA(s), and map it to a digital transfer function, HD(z), i.e. HA(s) →HD(z) using an appropriate mapping: 1 Impulse Invariant 2 Bilinear z-Mapping 3 Matched z-Transform Direct Methods: Based on using optimization packages: 1 Linear programming.
2 Non-linear programming.
M.R. Azimi Digital Control & Digital Filters Stability Analysis of Digital Control Systems Digital Filter Design Indirect Method Conditions for Mapping 1 Should preserve all essential characteristics of analog filter (BW, PB loss, SB loss).
2 A stable analog filter should be mapped to a stable digital filter.
3 The transfer function HD(z) should be rational and proper with real coefficients.
ωc : Cutofffrequency ωr : Lower stopband frequency δ1 : Passband error δ2 : Stopband error M.R. Azimi Digital Control & Digital Filters Stability Analysis of Digital Control Systems Digital Filter Design Review of Analog Filters-Preliminaries Recall: HA(s) = Y (s) U(s) = ∞ ´ 0 h(t)e−stdt Define Magnitude-Squared Function: A(−s2) = HA(s)HA(−s) (Spectral Factorization) Alternatively by letting s = jω A(ω2) = HA(jω)HA(−jω) = HA(jω)H∗ A(jω) = |HA(jω)|2 Properties of A(−s2): 1 Even with real coefficients.
2 Poles and zeros of A(−s2) occur in quadruples i.e. if pi is a pole/zero then −pi, p∗ i , and −p∗ i are also poles/zeros.
Goal: Given A(−s2) (or A(ω2)), HA(s) should be chosen such that it contains all the poles and zeros of A(−s2) that lie in the LH of the s-plane (spectral factorization).
M.R. Azimi Digital Control & Digital Filters Stability Analysis of Digital Control Systems Digital Filter Design Review of Analog Filters-Preliminaries Example: Given A(ω2) = 2+ω2 1+ω4 , determine HA(s).
A(−s2) = 2−s2 1+s4 = ( √ 2−s)( √ 2+s) (s+α)(s+α∗)(s−α)(s+α∗) = HA(s)HA(−s) α = 1+j √ 2 HA(s) = s+ √ 2 (s+α)(s+α∗) i.e. HA(s) captures all poles/zeros that are in LH of s-plane.
M.R. Azimi Digital Control & Digital Filters |
15356 | https://www.comprehensionconnection.net/2017/04/using-close-reading-for-deeper-thinking.html | How To Use Close Reading For Deeper Thinking
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How to Use Close Reading for Deeper Thinking
Of all of the strategies I’ve used with the kids I’ve taught, close reading has been the most effective at helping my students with reading comprehension. Why is that? I believe it teaches children to not just read the words and get surface level understandings. Rather, it requires engagement and high level thinking. In this post, I will share the close reading process. I hope that these tips might provide support as many approach end of year assessments.
What is Close Reading?
Close reading is not moving your paper closer to your eyes (one of the kids suggested this today). It is not rereading alone.Close reading is active engagement with the text demonstrated by annotations that highlight important points.For successful use of close reading, the teacher should model with think aloud to model how to analyze the importance of information AS we read. Thinking aloud is very important and should be used with read alouds, articles, and even with anchor charts we use. This is especially imporant as we scaffold the instruction for your struggling readers (or when you first share it with your students for the first time). Gradually, you want them to take over the responsibility and independently do it.
If you are looking for a great teacher resource book about Close Reading, Chris Lehman’s book,Falling in Love with Close Readingoffers a thorough explanation of how and why it works. I heard Chris speak two years ago. During his presentation, he used song lyrics for the text for modeling with us, and that is an important point. Any text can be used for this strategy, and in fact, using a variety of texts is necessary for our kids. Here are a few tips I learned from his book…
Read through different lenses in order to select specific information.
According to Chris, Close Reading is“making careful observations”of something and then“developing interpretations from those observations”. In other words, we stop to look carefully at choices an author (or painter or musician or director or architect) has made, and then develop ideas from what we have noticed (stop and jot). This means students read with different lenses to match the purpose we give them and observe text evidence to fit that designated purpose.
Look for connections with the text and analyze the author’s craft.
We want our readers to be strategic in their reading and thinking. We want them to observe the author’s use of language to convey meaning and apply that learning in their own work. I tell my students all the time that reading and writing go hand-in-hand. When I share a read aloud with my kids to introduce a new writing assignment or to model a reading skill, I’m not simply reading the book. We areANALYZINGthe author’s craft to apply it to our own writing ideas or using that text to practice the skill.
The best way to become a strong reader or writer is to interact with strong writing that is filled with vivid vocabulary, includes varied sentence length and type, that’s well organized, and that shares a strong message or idea. As we break down the text, we are thinking, andtest taking is all about thinking strategies.
Read sections of text to see how small ideas connect to the bigger picture.
As students work with the close reading strategy, their level of understanding improves. With the first reading, my students aren’t as fluent. They get the gist of the reading, and observe basic information with a pencil in hand to mark it. They scratch the surface. With the second and third visits to the text, fluency improves. We hone in on specific skills, record annotations in the margins that support our thinking and that match the assigned purpose. We share our learning and opinions with one another. It is through group discussions that we quickly see the depths of understanding our students have achieved. Will your kids reread on your state assessment? Maybe a few. I think if we can slow them down by getting them to annotate, they will improve comprehension.
Choose a variety of texts including songs, poems, articles, ads, and even books.
Close reading is a strategy that can be used with all sorts of text types. We shouldn’t confine it’s practicality to just short fiction and nonfiction stories. It works well with video clips, song lyrics, poetry, television ads, and movies. Students in upper elementary through high school have a need to talk and crave controversy. Chris gives examples of how teachers can capitalize on that energy in studying point of view, argument, and text structure. Although Chris recommends close reading for grades 5-8, I believe this strategy can be adapted for use with younger students. We can use close reading with poetry as we think about the author’s choice of words. We can model with all types of texts.
Keep instruction clear with a purpose, engaging and active, so that the strategy is transferred to other texts.
With all of this in mind, there is a routine that is used with Close Reading. Chris talks at length about what Close Reading is and is not. It is not just answering questions, doing book reports, or filling out a worksheet. It is a strategic method of looking at and using text. The focus is on the interactions between reader and text, the ideas drawn, and the conclusions made. Here are the general steps I use when doing a Close Read with my students.
First Reading-Getting the Gist
Set the purpose for the reading.
Assign sections of the text to read and share your expectations for marking it.
Give time to read and a stopping point for discussion. Having kids turn and talk about the section may flush out confusions they may have.
Second Reading-Thinking More Deeply
Set a new purpose for the reading.
Again, read and interact with the text in sections.
Respond to the text (questions or an organizer with a skill focus)
Third Reading-Extending Ideas
Set a new purpose for the last reading.
With a third reading, students must be using high level thinking which requires them to analyze the ideas included in the text for the purpose.
The information learned from the reading is used in an extended response.
Resources for Close Reading
In working with a group of fourth graders this week who hadn’t used close reading much, I decided to begin with an anchor chart that walked through the process as we worked with a weather close reading set in one of my resources. Of course, it was new to them, and because they really hadn’t been taught to interact with the text in this way, there were a fair number of groans. However, they all expressed an understanding of how this would help them in future research projects, reading of complex materials, and with testing. Close reading would not be something to do every single day, but I think it definitely helps kids rethink their thinking about their reading.
For practice, there are certainly lots of wonderful free resources to get you started. If you[CLICK HERE], you will be taken to a list of free close reading sets. You can find close reading resource sets for all grade levels now too. I hope you’ll give them a try and see the benefits as we have at my school.
For a look at what’s available in my store, you can check out these two bundles for science. There are a few others too.
Light and Sound Close Reading Sets Science Activities$4.00 Add to cart
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Sale Product on sale Polar Animals Close Reading Bundle and Research Project~~$17.00~~Original price was: $17.00.$10.00 Current price is: $10.00. Add to cart
Related Posts:
Five Ways to Improve Deep Thinking in the Classroom
Teaching Tips Using Reading A to Z
Close reading strategies really helped my students with processing information from their reading. I hope these tips and strategies help your students too.
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Lesson Explainer: Logarithmic Equations with Different Bases Mathematics • Second Year of Secondary School
In this explainer, we will learn how to solve logarithmic equations involving logarithms with different bases.
Let’s first recall the relationship between logarithmic and exponential forms.
Definition: Relationship between Logarithmic and Exponential Forms
For and base , , the logarithmic form is equivalent to the exponential form , which allows us to convert from one form to another once we identify , , and .
An exponential function is the inverse of the logarithmic function . This means if you raise to the power of log of with base or if you raise to the power of first then take the log with base of the result, you get back :
The relationship between logarithmic and exponential forms allows us to deduce the properties satisfied by the logarithmic forms, known as the laws of logarithms, that follow from the laws of exponents. Let’s recall the laws of logarithms.
Definition: Laws of Logarithms
Suppose , , , and are positive numbers with . The laws of logarithms are
product: ,
division: ,
powers: ,
change of base: .
These will be useful to solving logarithmic equations with different bases, especially the last one, which allows us to convert a logarithm of one base to another. In order to see where this comes from, first note that which follow directly from the fact that exponents and logarithms are inverses. If we substitute the second expression into the first, we obtain
Using the law of exponents , we can rewrite this as
Taking the logarithm of both sides with base , we find as required. This formula along with the laws of logarithms and the equivalent exponential form allow us to solve logarithmic equations involving logarithms of different bases. Another fact that will be important for solving logarithmic equations is which follows directly from the relationship between logarithmic and exponential forms, in particular, from the fact that the logarithmic and exponential functions are strictly monotonic functions. We can show this directly using the laws of logarithms:
Converting the last expression to exponential form, we find and thus, , as required.
As an example, suppose we want to find the solutions to the logarithmic equation
We first convert the logarithm on the left-hand side to a logarithm of base 2 using the change in base formula: where we have used the fact that to obtain the last line. Thus, using this and the laws of logarithms, the given logarithmic equation becomes
Using the fact that we established, if , then , or converting to exponential form, we obtain
Thus, the only solution to the logarithmic equation is .
Now, let’s consider a few examples to practice and deepen our understanding of solving logarithmic equations. In the first example, we have two logarithms of different bases and the unknown appears inside the logarithm.
Example 1: Finding the Solution Set of a Logarithmic Equation over the Set of Real Numbers
Find the solution set of in .
Answer
In this example, we want to determine the solution set of a particular logarithmic equation with different bases and the unknown appearing inside the logarithm.
In order to solve the equation, we will make use of the change of base formula, and the power law,
On applying the change of base formula, we can write the right-hand side of the given logarithmic equation, , as a logarithm of base 3 using and :
Using this, the given equation becomes
Since , we have . Thus, the solution set is given by .
Now, let’s consider an example where a logarithmic equation contains two logarithms of different bases and an unknown appearing in each.
Example 2: Finding the Solution Set of a Logarithmic Equation over the Set of Real Numbers
Determine the solution set of the equation in .
Answer
In this example, we want to determine the solution set of a particular logarithmic equation with different bases and the unknown appearing inside two logarithms of different bases.
In order to solve the equation, we will make use of the change of base formula, and the power law,
On applying the change of base formula, we can write the second term on the left-hand side of the given logarithmic equation, , as a logarithm of base 3 using and :
On substituting this into the given logarithmic equation, we obtain
For and base , , the logarithmic form is equivalent to the exponential form .
Finally, converting to exponential form we have
Thus, the solution set is .
In the next example, we will find the solution to a logarithmic equation that contains the sum of three logarithms of different bases and the unknown appearing inside each of the logarithms.
Example 3: Finding the Solution Set of a Logarithmic Equation over the Set of Real Numbers
Find the solution set of in .
Answer
In this example, we want to determine the solution set of a particular logarithmic equation with different bases and the unknown appearing inside the logarithm.
In order to solve the equation, we will make use of the change of base formula, and the power law,
On applying the change of base formula, we can write the second term on the left-hand side of the given logarithmic equation, , as a logarithm of base 2 using and , and similarly, for the third term, , using as
Substituting these expressions into the logarithmic equation, we obtain
For and base , , the logarithmic form is equivalent to the exponential form .
Thus, converting to exponential form, we obtain
Therefore, the solution set is .
Now, let’s consider an example where we have to find the solution to a logarithmic equation that contains the sum of the reciprocal of three logarithms of different bases and the unknown appearing inside each of the logarithms.
Example 4: Finding the Solution Set of a Logarithmic Equation over the Set of Real Numbers
Find the solution set of in .
Answer
In this example, we want to determine the solution set of a particular logarithmic equation with different bases and the unknown appearing inside three logarithms of different bases.
In order to solve the equation, we will make use of the change of base formula, and the power law,
On applying the change of base formula, we can write the second term on the left-hand side of the given logarithmic equation, , as a logarithm of base 2 using and , and similarly, for the third term, , using as
Substituting these expressions into the logarithmic equation, we obtain
For and base , , the logarithmic form is equivalent to the exponential form .
Thus, converting to exponential form we find
Thus, the solution set is .
In the next example, we will solve a logarithmic equation that contains logarithms of different bases, including a fractional base, and an unknown appearing inside the logs as a linear and quadratic term.
Example 5: Solving Logarithmic Equations Involving Laws of Logarithms
Find the solution set of in .
Answer
In this example, we want to determine the solution set of a particular logarithmic equation with different bases and the unknown appearing inside two logarithms of different bases.
In order to solve the equation, we will make use of the change of base formula, and the power law,
On applying the change of base formula, we can write the second term on the left-hand side of the given logarithmic equation, , as a logarithm of base 3 using and :
Finally, substituting this expression into the given logarithmic equation, we have
For and base , , the logarithmic form is equivalent to the exponential form .
Finally, converting to exponential form, we have
Thus, the solution set is .
As we have seen in the previous examples, the change of base rule for logarithms also allows us to evaluate expressions of the form by rewriting this logarithm of base as a logarithm of base (i.e., ) along with the power law and the fact that :
Now, let’s consider an example where we have a logarithmic equation that contains a logarithm of the logarithm with two different bases and an unknown appearing inside the logarithm as a quadratic.
Example 6: Solving Logarithmic Equations Involving Laws of Logarithms and Quadratic Equations
Solve , where .
Answer
In this example, we want to determine the solution of a particular logarithmic equation with different bases and an unknown appearing inside the logarithm of a logarithm in the form of a quadratic equation.
For and base , , the exponential form is equivalent to the logarithmic form , which allows us to convert from one form to another once we identify , , and .
Converting to exponential form, we obtain
Repeating the process, we obtain
Therefore, we obtain or . The solution set is .
So far, the examples we have considered had the unknown variable , which we have to solve for, appearing inside the logarithm itself. We can also find solutions to logarithmic equations where the unknown can appear in the base of the logarithm.
In the next example, let’s consider a logarithmic equation with a triple logarithm of different bases and an unknown appearing as the base.
Example 7: Solving Logarithmic Equations over the Set of Real Numbers
Solve , where .
Answer
In this example, we want to determine the solution of a particular logarithmic equation with three different bases and an unknown appearing as a base of a logarithm.
Recall that, for and base , , the logarithmic form is equivalent to the exponential form . The common logarithm where no base is specified has base 10: .
Converting to exponential form, we obtain
Repeating this process for the resulting expression, we get
And repeating it lastly again, we obtain
Thus, the solutions are or , but we ignore the second solution since the logarithm of a negative base (i.e., ), is undefined.
Therefore, the solution to the logarithmic equation is .
Now, let’s consider an example where we have a logarithmic equation containing logarithms of different bases and an unknown appearing both inside the logarithm and the base of the same logarithm.
Example 8: Finding the Solution Set of Exponential Equations Involving Logarithms over the Set of Real Numbers
Find the solution set of in .
Answer
In this example, we want to determine the solution of a particular logarithmic equation with two different bases and an unknown appearing inside and appearing as a base of a logarithm.
In order to solve the given equation, we will make use of the power law:
On applying this and using , we obtain and
Substituting these into the given equation, we obtain
The solutions are and ; however, we will ignore the second as the logarithm of a negative number or with a negative base is undefined. The solution set is therefore .
Using the change of base formula, we can interchange the argument of the logarithm and the base. Changing the logarithm to base and using , we have
Suppose we want to find the solution of ; we want to determine the values of that satisfy this logarithmic equation. Since , this simplifies to
Thus, converting this to exponential form with base , we obtain .
Finally, let’s consider an example where the unknown that we have to solve for appears both inside the logarithm and as a base of another logarithm.
Example 9: Finding the Solution Set of Logarithmic Equations over the Set of Real Numbers
Determine the solution set of the equation in .
Answer
In this example, we want to determine the solution of a particular logarithmic equation with two different bases and an unknown appearing inside and appearing as a base of a logarithm.
In order to solve the logarithmic equation, we will make use of the change in base formula:
If we use the base , this formula allows us to interchange the argument of the logarithm and the base: where we used the fact that . Using this, the given equation becomes
So, if we let , then we have to solve
Multiplying both sides of this equation by and rearranging, we obtain
Thus,
For and base , , the logarithmic form is equivalent to the exponential form .
Thus, upon converting to exponential form, we have
Thus, the solution set is .
Let’s summarize what has been learned in this explainer.
Key Points
In order to solve the logarithmic equations, we made use of the laws of logarithms:
product: ,
division: ,
powers: ,
change of base: .
If we have a logarithm of base , we can also convert this to a logarithm of base using
We can also interchange the base and the argument of a logarithm using
We can also solve logarithmic equations by converting the logarithmic form to the exponential form .
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15358 | https://stats.stackexchange.com/questions/495912/chi2-test-for-the-variance-in-python | hypothesis testing - $\chi^2$ test for the variance in Python - Cross Validated
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χ 2 test for the variance in Python [closed]
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I am looking for a function in Python testing the hypothesis that the variance of a Gaussian sample is equal to a given value, to validate my own function.
I talk about this test:
I could find the χ 2 test for categorical variance, and the Levene and Bartlett tests to compare sample variances, but not this simple test. Anybody aware of such a function in Python?
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asked Nov 11, 2020 at 3:04
marcomarco
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```python
import numpy as np
from scipy.stats import chi2
def var_test(x, va0, direction = "two-tailed", alpha = 0.05):
n = len(x)
Q = (n - 1) np.var(x) / va0
if direction == "lower":
q = chi2.ppf(alpha, n - 1)
if Q <= q:
return "H_0 rejected"
else:
return "H_0 not rejected"
elif direction == "upper":
q = chi2.ppf(1 - alpha, n - 1)
if Q >= q:
return "H_0 rejected"
else:
return "H_0 not rejected"
else:
q1 = chi2.ppf(alpha / 2, n - 1)
q2 = chi2.ppf(1 - (alpha / 2), n - 1)
if Q <= q1 or Q >= q2:
return "H_0 rejected"
else:
return "H_0 not rejected"
n = 25
x = np.random.normal(0, 3, n)
var_test(x, va0 = 1)
```
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edited Nov 12, 2020 at 3:21
answered Nov 11, 2020 at 4:44
ZenZen
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15359 | https://www.electrical4u.com/inductance-in-power-transmission-line/ | Skip to content
Inductance in Transmission Line
by Electrical4U
Contents
💡
Key learnings:
Definition of Inductance: Inductance in a transmission line is the property that causes it to oppose changes in current, due to the magnetic flux created by the current.
Flux Linkage: Flux linkage in a conductor involves the interaction of magnetic flux with the current, influencing inductance.
Internal and External Inductance: Internal inductance is due to flux within the conductor, while external inductance arises from flux outside the conductor.
Two-Wire System Inductance: The inductance in a two-wire single-phase system depends on the currents in both wires and their distance apart.
Three-Phase System Inductance: In a three-phase transmission line, the inductance is influenced by the currents in all three conductors and their mutual interactions.
Reason of Transmission Line Inductance
Generally, electric power is transmitted through the transmission line with AC high voltage and current. High valued alternating current while flowing through the conductor sets up magnetic flux of high strength with alternating nature. This high valued alternating magnetic flux makes a linkage with other adjacent conductors parallel to the main conductor. Flux linkage in a conductor happens internally and externally. Internally flux linkage is due to self-current and externally flux linkage due to external flux. Now the term inductance is closely related to the flux linkage, denoted by λ. Suppose a coil with N number of turn is linked by flux Φ due to current I, then,
But for transmission line N = 1. We have to calculate only the value of flux Φ, and hence, we can get the transmission line inductance.
Calculation of Inductance of Single Conductor
Calculation of Internal Inductance due to Internal Magnetic Flux of a Conductor
Consider a conductor carrying current I along its length l. Let x be the variable internal radius, and r the original radius of the conductor. The cross-sectional area for radius x is πx2 square units, with current Ix flowing through it. The value of Ix can be expressed in terms of the original current I and the cross-sectional area πr2 square units.
Now, consider a small thickness dx along a 1m length of the conductor, where Hx is the magnetizing force due to current Ix in the area πx2.And magnetic flux density Bx = μHx, where μ is the permeability of this conductor. Again, µ = µ0µr. If it is considered that the relative permeability of this conductor µr = 1, then µ = µ0. Hence, here Bx = μ0 Hx.dφ for small strip dx is expressed by
Here entire cross-sectional area of the conductor does not enclose the above expressed flux. The ratio of the cross sectional area inside the circle of radius x to the total cross section of the conductor can be thought about as fractional turn that links the flux. Therefore the flux linkage is
Now, the total flux linkage for the conductor of 1m length with radius r is given byHence, the internal inductance is
External Inductance due to External Magnetic Flux of a Conductor
Assume that due to the skin effect, current I is concentrated near the conductor’s surface. Consider y as the distance from the center, representing the external radius of the conductor.Hy is the magnetizing force and By is the magnetic field density at y distance per unit length of the conductor.Let us assume magnetic flux dφ is present within the thickness dy from D1 to D2 for 1 m length of the conductor as per the figure.As the total current I is assumed to flow in the surface of the conductor, so the flux linkage dλ is equal to dφ.But we have to consider the flux linkage from conductor surface to any external distance, i.e. r to D
Inductance of Two Wire Single Phase Transmission Line
Suppose conductor A of radius rA carries a current of IA in opposite direction of current IB through the conductor B of radius rB. Conductor A is at a distance D from conductor B and both are of length l. They are in close vicinity with each other so that flux linkage takes place in both of the conductors due to their electromagnetic effects.
Let us consider the magnitude of current in both conductors are same and hence IA = – IB,Now, total flux linkage in conductor A = flux linkage by self-current of conductor A + flux linkage on conductor A due to current in the conductor B.Similarly, flux linkage in conductor B = flux linkage by self-current of conductor B + flux linkage on conductor B due to current through conductor A.Now if we consider a point P in close vicinity both conductor A and B, the flux linkage at point P would be, flux linkage at point P for current carrying conductor A + flux linkage at point P for current carrying conductor B i.e.
Now, ……… shown in the figure below in figure (a) and (b).
λAAP is the flux linkage at point P for conductor A due to current through conductor A itself.
λABP is the flux linkage at point P for conductor A due to current through conductor B.
λBAP is the flux linkage at point P for conductor B due to current through conductor A.
λBBP is the flux linkage at point P for conductor B due to current through conductor B itself.
λABP and λBAP are negative in value because the directions current are opposite with respect to each other.
If we consider that both conductor are with same radius, i.e. rA = rB = r and point P is shifted to infinite distance then we can write thatIf conductor A becomes bundled conductor, then its geometrical mean radius (GMR) will be calculated for n number of conductors per bundle.Where, d is the distance between the central axis of conductors within the bundle.
Inductance in Three Phase Transmission Line
In the three phase transmission line, three conductors are parallel to each other. The direction of the current is same through each of the conductors.
Let us consider conductor A produces magnetic flux φA,Conductor B produces magnetic flux φB,And conductor C produces magnetic flux φC.When they carry the current of the same magnitude “I”, they are in flux linkage with each other.Now, let us consider a point P near three conductors. So, flux linkage at point P due to current through conductor A is, Flux linkage at point P for conductor A due to current through conductor A =
Flux linkage at point P for conductor A due to current through conductor B = Flux linkage at point P for conductor A due to current through conductor C = Therefore, flux linkage at point P for conductor A,As, and in balanced system, then we can write that
If we arrange them in matrix form, then we get
Where, λA, λB, λC are the total flux linkages of conductor A, B and C.LAA, LBB and LCC are the self inductances of conductor A, B and C.LAB, LAC, LBC, LBA, LCA, LCB are the mutual inductances of between conductors A, B and C.
Again balanced systemAndIn balanced system, then we can write thatSimilarly,
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15361 | https://cs.ucsb.edu/~daniello/papers/dodgsonHard.pdf | Foundations of Software Technology and Theoretical Computer Science (2009) Submission Determining the Winner of a Dodgson Election is Hard Michael Fellows1, Bart Jansen2∗, Daniel Lokshtanov3, Frances A.
Rosamond1 and Saket Saurabh4 1Parameterized Complexity Research Unit, University of Newcastle, Callaghan, NSW Australia. {Michael.Fellows|Frances.Rosamond}@newcastle.edu.au 2Department of Information and Computing Sciences, Utrecht University, P.O. Box 80.089, 3508 TB, Utrecht, The Netherlands. bart@cs.uu.nl 3Department of Informatics, University of Bergen, N-5020 Bergen, Norway.
daniello@ii.uib.no 4Institute of Mathematical Sciences, CIT Campus, Taramani, 600 113 Chennai, India.
saket@imsc.res.in ABSTRACT. Computing the Dodgson Score of a candidate in an election is a hard computational problem, which has been analyzed using classical and parameterized analysis. In this paper we resolve two open problems regarding the parameterized complexity of DODGSON SCORE. We show that DODGSON SCORE parameterized by the target score value k has no polynomial kernel unless the polynomial hierarchy collapses to the third level; this complements a result of Fellows, Rosamond and Slinko who obtain a non-trivial kernel of exponential size for this problem. We also prove that DODGSON SCORE parameterized by the number n of votes is hard for W.
1 Introduction Complexity issues play an important role in the relatively new area of computational so-cial choice, especially in the area of election systems, which has applications in finance and economics (agreement on the winner of an auction), internet search engines (agree-ment on the order of web pages presented), web mining (consensus is the notion of “public opinion”), mechanism design (agreement by participants in large networks involving au-tonomous software agents), and computational biology (finding consensus in feature selec-tion), among many others. The involvement of increasingly larger numbers of participants and the increasing sophistication of the information objects of debate, have made election systems a vital area of computer science research.
In this paper we study the hard election problem DODGSON SCORE. We consider an election in which we allow each voter to specify a complete preference ranking of the candi-dates: each vote is a strict total order on the set of candidates, and a vote in an election with three candidates could be represented as a < b < c stating that candidate a is least preferred ∗Bart Jansen was supported by the Netherlands Organisation for Scientific Research (NWO), project “KER-NELS: Combinatorial Analysis of Data Reduction”.
NOT FOR DISTRIBUTION 2 DETERMINING THE WINNER OF A DODGSON ELECTION IS HARD and c is most preferred. Given the votes that were cast in an election, we can compare the relative ranking of two candidates a, b as follows: candidate a beats candidate b in pairwise comparison if a is ranked above b more often than below b. A candidate who beats every other candidate in pairwise comparison is said to be a Condorcet winner. If such a winner ex-ists then it must be unique, and it wins the election. But unfortunately a Condorcet winner may not always exist, as is shown by the following election with three candidates and three voters: a < b < c, b < c < a, c < a < b. This situation has a cyclic preference structure: candidate a beats b, candidate b beats c and c beats a (in pairwise comparison), so there is no candidate who beats all others. In 1876 the mathematician Charles Dodgson formulated a rule that defines the winner of an election even if there is no Condorcet winner. The idea is to measure how close a candidate is to being a Condorcet winner; the candidate who is closest then wins the election. This can be formalized as follows. The Dodgson score of a candidate c in an election, is defined to be the minimum number of swaps of adjacent can-didates in the voter’s preference orders that have to be made to ensure that c becomes a Condorcet winner. The candidates that have the minimum Dodgson score are the winners of the election. Dodgson’s rule is not the only voting scheme resulting from Condorcet’s criterion; similar schemes have been suggested by Young and Kemeny .
Unfortunately, DODGSON SCORE, YOUNG SCORE and KEMENY SCORE and many other election problems are NP-hard or worse, and finding an “approximate” winner of an elec-tion is usually not appropriate. Thus, election problems are well-suited for parameterized analysis because it offers an exact result, taking advantage of natural parameters to the prob-lems, such as the number of votes that were cast, the number of candidates, or the score of a candidate.
Earlier Work.
Bartholdi et al. initiated the study of the complexity of the Dodgson voting scheme in 1989 , when they showed that determining the winner of a Dodgson election is NP-hard.
They also proved that computing the Dodgson or Kemeny score of a given candidate is NP-complete. The complexity of the winner problem for Dodgson elections was later es-tablished exactly; Hemaspaandra et al. showed in 1997 that this problem is complete for PNP || (“parallel access to NP”).
McCabe-Dansted was the first to investigate DODGSON SCORE from a parameter-ized perspective, and observed that the problem is fixed-parameter tractable when parame-terized by the number m of candidates in the election, by refining an integer linear program from Bartholdi et al. and exploiting the fact that the number of variables in the ILP can be bounded by a function of m alone. The parameterization by the target score k was first studied in 2007, when Fellows and Rosamond showed that k-DODGSON SCORE is in FPT.
The group of Betzler et al. [5, 6] independently reached the same conclusion and obtained a dynamic programming algorithm with running time O(2k · nk + nm) where n is the number of votes and m the number of candidates. Fellows, Rosamond and Slinko considered a generalization of Dodgson’s rule where each possible preference ranking specifies a cost for every swap that can be made; a candidate wins the election if the minimum total cost of making that candidate a Condorcet winner is not higher than the minimum cost of mak-FELLOWS, JANSEN, LOKSHTANOV, ROSAMOND AND SAURABH FSTTCS 2008 3 ing any other candidate a Condorcet winner. They obtained a kernel of exponential size O(eO(k2)) for this k-GENERALIZED DODGSON SCORE problem.
The election problems KEMENY SCORE and YOUNG SCORE have also been studied from the parameterized perspective. The YOUNG SCORE problem is W-complete when pa-rameterized by the target score, and the same holds when using the dual of this parame-ter [5, 6]. The KEMENY SCORE problem admits several natural parameterizations that lead to fixed-parameter tractability. Results have been found for parameters ‘number of votes’, ‘average Kendall-Tau distance’, ‘maximum range (or maximum Kendall-Tau distance)’ and combined parameters of ‘number of votes and average KT-distance’, and ‘number of votes and maximum KT-distance’ .
Our Results.
The parameterized analysis of DODGSON SCORE by Betzler et al. left two open problems unanswered: 1) does k-DODGSON SCORE admit a polynomial kernel when parameterized by the target score, and 2) is the problem fixed-parameter tractable when parameterized by the number of votes? We answer both questions in this paper. We use the framework de-veloped by Bodlaender et al. to prove that there is no polynomial kernel for k-DODGSON SCORE unless the polynomial hierarchy collapses to the third level (denoted as PH = Σ3 p).
Our second result is a non-trivial reduction establishing that k-DODGSON SCORE parame-terized by the number of votes is hard for W.
2 Preliminaries In this section we formalize some notions that were introduced in Section 1. An election is a tuple (V, C) where V is a multiset of votes, and C is a set of candidates. A vote v ∈V is a preference list on the candidates, i.e. a strict total ordering. For candidates a, b ∈C the value na,b counts the number of votes in V that rank a above b. A Condorcet winner is a candidate x ∈C such that nx,y > ny,x for all y ∈C \ {x}.† To swap candidate x upwards in a vote v ∈V means to exchange the positions of x and the candidate immediately above it in the ranking; an upward swap operation is undefined if x is already the most preferred candidate in the vote. For example, if x < z < w < y is a vote, then swapping x upwards once results in the vote z < x < w < y. We say that the candidate x gains a vote on candidate z through this swap, since this swap increases nx,z by one and decreases nz,x by one. The Dodgson score of a candidate x ∈C is the minimum number of swaps needed to make x a Condorcet winner. It is not hard to verify that if x can be made a Condorcet winner by k swaps, then this can also be done by k swaps that only move candidate x upwards.
Consult [4, Lemma 4.0.5] for a formal proof of this claim. Therefore we may also define the Dodgson score as the minimum number of upwards swaps of x that are required to make x a Condorcet winner.
†The notion of Condorcet winner used by Fellows et al. slightly differs from the standard notion; under their definitions a Condorcet winner is a candidate who draws or defeats each candidate in a pairwise comparison; hence such a Condorcet winner need not be unique. The standard definition (and the one that we use here) requires a Condorcet winner to strictly beat each other candidate in pairwise comparison. It is not hard to see that the two variants are equivalent from a FPT perspective.
4 DETERMINING THE WINNER OF A DODGSON ELECTION IS HARD The theory of parameterized complexity offers a toolkit for the theoretical analy-sis of the structure of NP-hard problems. A parameterized decision problem is a language L ⊆Σ∗× N, where an instance (x, k) is composed of the classical input x and the pa-rameter value k that describes some property of x. A parameterized problem L is in the class (strongly uniform) FPT (for Fixed-Parameter Tractable) if there is an algorithm that decides L in f (k)p(|x|) time, where p is a polynomial and f is a computable function. A kernelization algorithm (kernel) is a mapping that transforms an instance (x, k) ∈Σ∗× N in p(|x| + k) time for some polynomial p, into an equivalent instance (x′, k′) such that (x, k) ∈L ⇔(x′, k′) ∈L and such that |x′|, k′ ≤f (k) for some computable function f.
The function f is called the size of the kernel. Recent developments in the theory of ker-nelization have yielded tools to show that certain problems are unlikely to have kernels of polynomial size .
The DODGSON SCORE problem is formally defined as follows: DODGSON SCORE Instance: A set C of candidates, a distinguished candidate x ∈C, a multiset V of votes and a positive integer k.
Question: Can x be made a Condorcet winner by making at most k swaps be-tween adjacent candidates?
We consider two different parameterizations in this work. When the problem is parameter-ized by the number of allowed swaps k then we will refer to it as k-DODGSON SCORE; the other variant considers a bounded number of votes n := |V| which we call n-DODGSON SCORE.
3 Kernelization Lower Bound for k-DODGSON SCORE In this section we prove that k-DODGSON SCORE has no polynomial kernel unless PH = Σ3 p.
To prove this result we need some notions related to parameterized reducibility.
DEFINITION 1. Let P and Q be parameterized problems. We say that P is polynomial parameter reducible to Q, written P ≤Ptp Q, if there exists a polynomial time computable function g : Σ∗× N 7→Σ∗× N and a polynomial p, such that for all (x, k) ∈Σ∗× N (a) (x, k) ∈P ⇔(x′, k′) = g(x, k) ∈Q and (b) k′ ≤p(k). The function g is called polynomial parameter transformation.
THEOREM 2. Let P and Q be parameterized problems and ˜ P and ˜ Q be the unparam-eterized versions of P and Q respectively. Suppose that ˜ P is NP-hard and ˜ Q is in NP. Fur-thermore if there is a polynomial parameter transformation from P to Q, then if Q has a polynomial kernel then P also has a polynomial kernel.
We use the following problem as the starting point for our transformation: SMALL UNIVERSE SET COVER Instance: A set family F ⊆2U of subsets of a finite universe U and a positive integer k ≤|F|.
Question: Is there a subfamily F ′ ⊆F with |F ′| ≤k such that ∪S∈F ′S = U?
Parameter: The value k + |U|.
FELLOWS, JANSEN, LOKSHTANOV, ROSAMOND AND SAURABH FSTTCS 2008 5 SMALL UNIVERSE SET COVER is a parameterized version of the NP-complete SET COVER problem [13, SP5]. We need the following incompressibility result for this problem [14, Theorem 2]: THEOREM 3. The problem SMALL UNIVERSE SET COVER parameterized by k + |U| does not admit a polynomial kernel unless PH = Σ3 p.
The transformation that we shall use to prove that k-DODGSON SCORE has no poly-nomial kernel (unless PH = Σ3 p) is similar in spirit to the original NP-completeness re-sult for DODGSON SCORE. Let (U, F, k) be an instance of SMALL UNIVERSE SET COVER.
We show how to construct an equivalent instance (V, C, x, k′) of k-DODGSON SCORE with k′ := k(|U| + 1) in polynomial time. Since the problem can be solved in polynomial time when |F| < 3, we may assume without loss of generality that the set family F contains at least 3 sets.
The set of candidates is composed of several parts. We create one candidate for each element u in the universe U; we will use an element u ∈U to refer both to the corresponding candidate and to the element of the finite universe, since the meaning will be clear from the context. We also take one candidate x to use as the distinguished candidate for whom the Dodgson score must be computed, one candidate y that will encode the fact that we must cover the universe with exactly k subsets, and finally we use three sets of dummy candidates D0, D1 and D2 that are needed for padding. These dummy candidates will ensure that we can make the distance between x and y in the total orders sufficiently large, i.e. that it takes a lot of swaps for x to gain a vote on y. We want to ensure that x beats all the dummy candidates in pairwise comparison in the initial situation, to ensure that the dummies do not interfere with the encoding of the set cover instance. Our three sets of dummies D0, D1 and D2 each contain |U| + 1 candidates. If we want to use some d ≤|U| + 1 dummy candidates that rank above x in the i-th vote that we create, then we use d candidates from the set Di mod 3; the other dummies are ranked below x. Since x beats every dummy candidate in at least two out of three votes, this ensures that x will beat all dummy candidates in pairwise comparison if we use at least 5 votes. Using this scheme we will from now on write Dj to denote a set of j ≤|U| + 1 dummy candidates that can be used in the vote we are constructing.
The set V of votes is built out of two parts, each containing |F| votes. Since |F| ≥3 this will ensure that we create at least 6 votes. Using the terminology of we create a set of swing votes corresponding to elements of F, and a set of equalizing votes that create the proper initial conditions. We introduce an abbreviation to write down total orders: if C′ ⊆C is a set of candidates, then by writing a total order a < C′ < b we mean a total order in which all candidates of C′ are ranked below b and above a. The relative ranking of the candidates in C′ among each other is not important. We now define the two parts of the vote set.
Swing votes For every S ∈F we make a vote (. . . < x < S < D|U|−|S| < y). All candi-dates that are not explicitly mentioned in the construction are ranked below x in arbitrary order. The set S in this vote represents the candidates corresponding to the universe ele-ments in S ⊆U.
6 DETERMINING THE WINNER OF A DODGSON ELECTION IS HARD The swing votes correspond to the sets in the family F.
Observe that it takes ex-actly |U| + 1 switches for x to gain a swing vote on y. The name “swing vote” comes from the fact that if x can become a Condorcet winner in k(|U| + 1) switches, then all those switches must be made in swing votes.
Equalizing votes The goal of the set of equalizing votes is to create initial conditions in which x must gain k votes on candidate y, and one vote on every candidate corresponding to some u ∈U in order to become the Condorcet winner. We construct the equalizing votes so that no switches made in them will allow x to become a winner in k(|U| + 1) steps. We do not give an explicit construction for the equalizing votes; instead we present the conditions that they must satisfy. It will be easy to see that such a set of votes exists, and can be constructed in polynomial time.
1. For every candidate corresponding to an element u ∈U, the number of equalizing votes in which x is ranked above u is equal to the number of swing votes in which x is ranked below u. This ensures that overall, every candidate u is ranked above x exactly as often as below x; hence x needs to gain one vote on every u to beat it in pairwise comparison.
2. There are |F| −k + 1 equalizing votes in which x is ranked above y. Since x is ranked below y in all swing votes, this implies that there are |F| −k + 1 votes in which x ranks above y, and 2|F| −(|F| −k + 1) = |F| + k −1 votes in which x ranks below y. The reader may verify that this means that x needs to gain at least k votes on y to beat y in a pairwise comparison.
3. Whenever x is ranked below y, then (by inserting dummies if necessary) there are at least |U| + 1 candidates between x and y. This ensures that at least |U| + 2 swaps are needed for x to gain an equalizing vote on y.
This concludes the construction of the instance (V, C, x, k′).
LEMMA 4. If the instance (U, F, k) has a set cover of size k, then candidate x can be made a Condorcet winner in the election (V, C, x, k′) by k′ swaps.
PROOF.
Suppose F ′ ⊆F is a set cover of size k. Every S ∈F ′ corresponds to a swing vote. Consider the effect of swapping x upwards for |U| + 1 steps in the swing votes cor-responding to the elements of F ′. Since there are exactly |U| candidates between x and y in every swing vote, this means that x gains these k votes on y. Since F ′ is a set cover of U it follows from the construction of the swing votes that we must have swapped x over every candidate u ∈U at least once. By the earlier observations this shows that after these k′ = k(|U| + 1) swaps the candidate x must be a Condorcet winner.
LEMMA 5. If candidate x can be made a Condorcet winner in the election (V, C, x, k′) by k′ swaps, then instance (U, F, k) has a set cover of size k.
PROOF.
Assume there is some series of k(|U| + 1) swaps that makes x a Condorcet win-ner. By the observations in the preliminaries we may assume that these swaps only move x upwards. Since x needs to gain k votes on y in order to become a Condorcet winner, we can conclude that at most |U| + 1 swaps on average can be used for every vote that x gains over y. But by construction it is impossible to improve over y using fewer than |U| + 1 FELLOWS, JANSEN, LOKSHTANOV, ROSAMOND AND SAURABH FSTTCS 2008 7 swaps per vote, which shows that none of the swaps can be made in equalizing votes since there it takes |U| + 2 swaps for x to improve over y. It follows that the swaps that make x a Condorcet winner in k(|U| + 1) steps must be composed of |U| + 1 swaps in k different swing votes. Since x had to gain one vote on every candidate corresponding to u ∈U in order to become a Condorcet winner, we may conclude that in these k swing votes every candidate u ∈U was ranked above x at least once. But this shows that the sets correspond-ing to the k swing votes form a set cover for U of size k, which shows that U has a set cover of the requested size.
It is not hard to verify that the transformation can be computed in polynomial time.
The transformation is a polynomial parameter transformation because the parameter k′ = k(|U| + 1) of the k-DODGSON SCORE instance is bounded by the square of the original pa-rameter k + |U|. By combining Theorem 2 with Theorem 3 the existence of this polynomial parameter transformation yields the following theorem.
THEOREM 6. k-DODGSON SCORE does not admit a polynomial kernel unless PH = Σ3 p.
4 Parameterized Hardness of n-DODGSON SCORE We now consider the parameterization by the number of votes n and show that this leads to W-hardness. We use a reduction from the following well-known problem .
MULTI-COLORED CLIQUE Instance: A simple undirected graph G = (V, E), a positive integer k and a coloring function c : V 7→{1, 2, . . . , k} on the vertices.
Question: Is there a clique in G that contains exactly one vertex from each color class?
Parameter: The value k.
We give a FPT-reduction from MULTI-COLORED CLIQUE to n-DODGSON SCORE. In par-ticular, given an instance (G = (V, E), c, k) of MULTI-COLORED CLIQUE we construct an instance (C′, V′, x′, k′) of n-DODGSON SCORE such that |V′| = n = 4((k 2) + k).
Let V1, . . . , Vk be the color classes of G, that is, for every v ∈Vi we have c(v) = i. For every pair of distinct integers 1 ≤i < j ≤k we define Ei,j to be the set of edges with one endpoint in Vi and one in Vj. We will assume without loss of generality that all color classes of G have the same number N of vertices, and that between every pair of color classes there are exactly M edges. We define the target score value k′ of the DODGSON SCORE instance as k′ := ((N + 1)(Mk + 1) + 2)k + (5M −3)(k 2). The set C′ of candidates is built out of five groups.
1. We have a distinguished candidate x′ for which we need to compute the Dodgson score.
2. We use 3k′ dummy candidates, just as in the proof of Theorem 6. This allows us to use up to k′ dummy candidates in each vote, while maintaining the property that the candidate x′ is ranked above every dummy candidate in more than half of the votes.
3. For every color class 1 ≤i ≤N there are candidates ap i for 0 ≤p ≤N + 2.
4. For every pair of color classes 1 ≤i < j ≤k there are candidates ap i,j for 1 ≤p ≤M + 1.
5. For every edge e ∈Ei,j there are candidates ei, e′ i, ej and e′ j.
8 DETERMINING THE WINNER OF A DODGSON ELECTION IS HARD From these definitions it is easy to verify that the number of candidates is polynomial in the size of the MULTI-COLORED CLIQUE instance. We now describe the vote set. As in the proof of Theorem 6 we will distinguish between swing votes and equalizing votes. There are 2((k 2) + k) votes of each type, and hence |V′| = n = 4((k 2) + k) from which it follows that the parameter n for the n-DODGSON SCORE instance is polynomial in the parameter k of the MULTI-COLORED CLIQUE instance.
Equalizing votes The equalizing votes create the right initial conditions for the election.
We build the equalizing votes such that in the resulting election the distinguished candi-date x′ must gain exactly one vote on each non-dummy candidate in order to win the elec-tion. We ensure that no swaps made in an equalizing vote can allow x′ to become a Con-dorcet winner in k′ steps, by ranking k′ dummy candidates ranked immediately above x′ in every equalizing vote. It is not hard to see that we do not need more equalizing votes than swing votes to encode these requirements.
Swing votes The swing votes encode the behavior of the MULTI-COLORED CLIQUE in-stance into the election. Every edge e gets an identification number ID(e) between 1 and M. Since the total number of edges is M(k 2) the identification of two edges may be the same, but we ensure that for two distinct edges e1, e2 both in Ei,j we always have ID(e1) ̸= ID(e2).
Similarly we give every vertex v ∈V an identification number ID(v) between 1 and N, and we ensure that distinct vertices in the same color class have different ID’s. As in the previous construction we know that only the part of the vote above x′ is relevant, so we do not show the remainder. None of the described candidates are dummies, unless specified otherwise. For every pair of integers 1 ≤i < j ≤k we make two swing votes, v1 i,j and v2 i,j as follows.
v1 i,j : x < a1 i,j < . . . < a2 i,j < . . . < a3 i,j < . . . < aM+1 i,j (1) v2 i,j : x < aM+1 i,j < . . . < aM i,j < . . . < aM−1 i,j < . . . < a1 i,j (2) The gaps between consecutive candidates ap i,j are filled as follows. For every edge e ∈Ei,j we insert ei, e′ i, ej and e′ j between aID(e) i,j and aID(e)+1 i,j in v1 i,j. Also, we insert ei, e′ i, ej and e′ j between aID(e)+1 i,j and aID(e) i,j in v2 i,j. Notice that between any consecutive ap i,j’s in v1 i,j and v2 i,j there are exactly 4 other candidates.
For every integer 1 ≤i ≤k we make two swing votes, v1 i and v2 i as follows.
v1 i : x < a0 i < . . . < a1 i < . . . < a2 i < . . . < a3 i < (. . .) < aN i (3) v2 i : x < aN+2 i < . . . < aN+1 i < . . . < aN i < . . . < aN−1 i < (. . .) < a2 i (4) For every edge e with one endpoint v in Vi we add ei between aID(v)−1 i and aID(v) i in v1 i and we add e′ i between aID(v)+2 i and aID(v)+1 i in v2 i . Having done this for every edge, we add dummy candidates between each consecutive pair of ap i ’s in v1 i and v2 i such that the total number of candidates between each consecutive pair of ap i ’s in v1 i and v2 i is exactly kM. This concludes the construction of (C′, V′, x′, k′).
FELLOWS, JANSEN, LOKSHTANOV, ROSAMOND AND SAURABH FSTTCS 2008 9 LEMMA 7. If G contains a colored k-clique, then x′ can be made a winner of the election (C′, V′, x′, k′) in k′ = ((N + 1)(kM + 1) + 2)k + (5M −3)(k 2) swaps.
PROOF.
Let C = c1, c2 . . . ck be a clique in G such that ci ∈Vi. For each vertex ci ∈C we move x′ in v1 i such that x beats aID(ci) i . We also move x′ in v2 i such that x beats aID(ci)+1 i . For each value of i this takes exactly (N + 1)(kM + 1) + 2 swaps: we need 1 swap to move over a0 i in v1 i and 1 swap to move over aN+2 i in v2 i , and for all N + 1 other candidates ap i we need to swap over the block of kM in front of them and over the candidates themselves, resulting in (N + 1)(kM + 1) more swaps. Thus in total there are ((N + 1)(kM + 1) + 2)k swaps in the v1 i and v2 i swing votes.
For every pair of distinct integers 1 ≤i < j ≤k we move x′ in v1 i,j such that x beats a ID(cicj) i,j , and move x′ in v2 i,j such that x beats a ID(cicj)+1 i,j . The number of swaps to do this is (5M −3)(k 2). Thus the total number of swaps is ((N + 1)(kM + 1) + 2)k + (5M −3)(k 2) = k′.
We show that x has gained a swing vote on each non-dummy candidate. It is easy to see that for every 1 ≤i ≤k, the candidate x′ has gained a swing vote on all ap i ’s and that for every 1 ≤i < j ≤k, x′ has gained a swing vote on all ap i,j’s. Observe that in the two swing votes v1 i,j and v2 i,j, x′ has gained a swing vote on all candidates ei, e′ i, ej and e′ j except for the four candidates corresponding to the edge e[i, j] = cicj. Let these four candidates be ei[i, j], e′ i[i, j], ej[i, j] and e′ j[i, j] respectively. However, x′ gains a swing vote on ei[i, j] in v1 i , on e′ i[i, j] in v2 i , on ej[i, j] in v1 j and on e′ j[i, j] in v2 j . This concludes the proof of the lemma.
LEMMA 8. If x′ can be made a winner of the election (C′, V′, x′, k′) in k′ = ((N + 1)(kM + 1) + 2)k + (5M −3)(k 2) swaps, then G contains a colored k-clique.
PROOF.
Observe that for any fixed i we need to perform at least (N + 1)(kM + 1) + 2 swaps in v1 i and v2 i in total in order to gain a swing vote on all ap i ’s. Similarly for any fixed 1 ≤i < j ≤k we need to perform at least 5M −3 swaps in v1 i,j and v2 i,j in total in order gain a swing vote on all ap i,j’s.
Thus, if strictly more than (N + 1)(kM + 1) + 2 swaps are performed in v1 i and v2 i in total for some i, or if more than 5M −3 swaps are performed in v1 i,j and v2 i,j in total for some i, j, then the total number of swaps performed must be greater than t′ if the swaps make x′ a Condorcet winner. So under the given assumptions for each i, exactly N + 3 + kM(N + 1) swaps are performed in v1 i and v2 i in total, and for each 1 ≤i < j ≤k, exactly (M + 1 + 4(M −1)) swaps are performed in v1 i,j and v2 i,j in total.
For a fixed i, if x′ has gained a swing vote for all the ap i ’s in v1 i and v2 i , then there must be some ci ∈Vi such that x′ has been moved right over aID(ci) i in v1 i and right over aID(ci)+1 i in v2 i . Similarly, for every 1 ≤i < j ≤k there must be some e[i, j] ∈Ei,j such that x′ has been moved right over aID(e[i,j]) i,j in v1 i,j and moved right over aID(e[i,j])+1 i,j in v2 i,j. Notice that x′ did not get any swing vote on ei[i, j], e′ i[i, j], ej[i, j] and e′ j[i, j]. The only other places x′ could have gotten a swing vote on these candidates are in v1 i , v2 i , v1 j and v2 j respectively. We prove that e[i, j] is incident to ci. Let vi be the vertex incident to e[i, j] in Vi. If ID(ci) < ID(vi) then x′ does not gain a swing vote on e[i, j]i. Hence ID(ci) ≥ID(vi). If, on the other hand, ID(ci) > ID(vi) then x′ does not gain a swing vote on e′[i, j]i, and therefore we must have 10 DETERMINING THE WINNER OF A DODGSON ELECTION IS HARD ID(ci) ≤ID(vi). But then vi = ci and therefore we know that for each i the vertex ci that is selected in the votes v1 i and v2 i is an endpoint of the edge that was selected in the votes v1 i,j and v2 i,j. Using e[i, j]j and e′[i, j]j one can similarly prove that e[i, j] is incident to cj. Hence C = {c1, c2, . . . , ck} forms a clique in G. This concludes the proof of the lemma.
The construction of (C′, V′, x′, k′) together with Lemmata 7 and 8 shows that there is a FPT-reduction from MULTI-COLORED CLIQUE to n-DODGSON SCORE. Since it is well-known that MULTI-COLORED CLIQUE is hard for W, we obtain the following result.
THEOREM 9. n-DODGSON SCORE is hard for W.
5 Conclusion and Discussions In this paper we answered two open problems with respect to the parameterized complexity of DODGSON SCORE. The parameterization k-DODGSON SCORE has no polynomial kernel unless the polynomial hierarchy collapses, and n-DODGSON SCORE is hard for W. The proof that k-DODGSON SCORE has no polynomial kernel unless PH = Σ3 p also implies that the exponential size kernel by Fellows et al. for k-GENERALIZED DODGSON SCORE can-not be improved to a polynomial kernel unless PH = Σ3 p.
In a natural variant of the DODGSON SCORE problem we are given a set of votes over a set of candidates C, together with an integer k, and asked whether any candidate can be made a Condorcet Winner by performing at most k swaps. A simple construction extends the hardness result of Theorems 6 and 9 to this problem as well.
References McLean, I., Urken, A.: Classics of Social Choice. University of Michigan Press, Ann Arbor, Michigan (1995) Bartholdi, J.J., Tovey, C.A., Trick, M.A.: Voting schemes for which it can be difficult to tell who won the election. Social Choice and Welfare 6 (1989) 157–165 Hemaspaandra, E., Hemaspaandra, L.A., Rothe, J.: Exact analysis of Dodgson elec-tions: Lewis Carroll’s 1876 voting system is complete for parallel access to NP. J. ACM 44 (1997) 806–825 McCabe-Dansted, J.C.: Approximability and computational feasibility of Dodgson’s rule. Master’s thesis, University of Auckland (2006) Betzler, N., Guo, J., Niedermeier, R.: Parameterized computational complexity of Dodgson and Young elections. In: Proc. 11th SWAT. (2008) 402–413 Betzler, N., Guo, J., Niedermeier, R.: Parameterized computational complexity of Dodgson and Young elections. Information and Computation 208 (2010) 165–177 Fellows, M., Rosamond, F., Slinko, A.: Sensing God’s will is fixed parameter tractable.
Technical report, ResearchSpace@Auckland 561 (2008) Betzler, N., Fellows, M.R., Guo, J., Niedermeier, R., Rosamond, F.A.: Fixed-parameter algorithms for Kemeny rankings. Theor. Comput. Sci. 410 (2009) 4554–4570 Bodlaender, H.L., Downey, R.G., Fellows, M.R., Hermelin, D.: On problems without polynomial kernels. J. Comput. Syst. Sci. 75 (2009) 423–434 FELLOWS, JANSEN, LOKSHTANOV, ROSAMOND AND SAURABH FSTTCS 2008 11 Downey, R., Fellows, M.R.: Parameterized complexity. Monographs in computer sci-ence. Springer, New York (1999) Niedermeier, R.: Invitation to Fixed-Parameter Algorithms. Oxford University Press (2006) Bodlaender, H.L., Thomass´ e, S., Yeo, A.: Kernel bounds for disjoint cycles and disjoint paths. In: Proc. 17th ESA. (2009) 635–646 Garey, M.R., Johnson, D.S.: Computers and Intractability, A Guide to the Theory of NP-Completeness. W.H. Freeman and Company, New York (1979) Dom, M., Lokshtanov, D., Saurabh, S.: Incompressibility through colors and ids. In: Proc. 36th ICALP. (2009) 378–389 Fellows, M.R., Hermelin, D., Rosamond, F.A., Vialette, S.: On the parameterized com-plexity of multiple-interval graph problems. Theor. Comput. Sci. 410 (2009) 53–61 |
15362 | https://stackoverflow.com/questions/14882530/python-coin-toss | random - Python Coin Toss - Stack Overflow
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Python Coin Toss
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Asked 12 years, 7 months ago
Modified1 year, 1 month ago
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I am VERY new to Python and I have to create a game that simulates flipping a coin and ask the user to enter the number of times that a coin should be tossed. Based on that response the program has to choose a random number that is either 0 or 1 (and decide which represents “heads” and which represents “tails”) for that specified number of times. Count the number of “heads” and the number of “tails” produced, and present the following information to the user: a list consisting of the simulated coin tosses, and a summary of the number of heads and the number of tails produced. For example, if a user enters 5, the coin toss simulation may result in [‘heads’, ‘tails’, ‘tails’, ‘heads’, ‘heads’]. The program should print something like the following: “ [‘heads’, ‘tails’, ‘tails’, ‘heads’, ‘heads’]
This is what I have so far, and it isn't working at all...
```python
import random
def coinToss():
number = input("Number of times to flip coin: ")
recordList = []
heads = 0
tails = 0
flip = random.randint(0, 1)
if (flip == 0):
print("Heads")
recordList.append("Heads")
else:
print("Tails")
recordList.append("Tails")
print(str(recordList))
print(str(recordList.count("Heads")) + str(recordList.count("Tails")))
```
python
random
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edited Aug 12, 2024 at 9:11
Penny Liu
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asked Feb 14, 2013 at 19:27
YeahScienceYeahScience
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3
There are better options in random but I presume you have to solve the problem long hand so to speak.sotapme –sotapme 2013-02-14 19:45:28 +00:00 Commented Feb 14, 2013 at 19:45
Can you give some more information? "it isn't working at all" isn't very descriptive. What specifically is the problem?Matthew –Matthew 2013-02-14 20:00:46 +00:00 Commented Feb 14, 2013 at 20:00
The program simply prints either 'Tails' or 'Heads' and 0 or 1 YeahScience –YeahScience 2013-02-14 20:06:16 +00:00 Commented Feb 14, 2013 at 20:06
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14 Answers 14
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You need a loop to do this. I suggest a for loop:
python
import random
def coinToss():
number = input("Number of times to flip coin: ")
recordList = []
heads = 0
tails = 0
for amount in range(number):
flip = random.randint(0, 1)
if (flip == 0):
print("Heads")
recordList.append("Heads")
else:
print("Tails")
recordList.append("Tails")
print(str(recordList))
print(str(recordList.count("Heads")) + str(recordList.count("Tails")))
I suggest you read this on for loops.
Also, you could pass number as a parameter to the function:
python
import random
def coinToss(number):
recordList, heads, tails = [], 0, 0 # multiple assignment
for i in range(number): # do this 'number' amount of times
flip = random.randint(0, 1)
if (flip == 0):
print("Heads")
recordList.append("Heads")
else:
print("Tails")
recordList.append("Tails")
print(str(recordList))
print(str(recordList.count("Heads")) + str(recordList.count("Tails")))
Then, you need to call the function in the end: coinToss().
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edited Feb 14, 2013 at 19:37
answered Feb 14, 2013 at 19:32
Rushy PanchalRushy Panchal
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You are nearly there:
1) You need to call the function:
python
coinToss()
2) You need to set up a loop to call random.randint() repeatedly.
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answered Feb 14, 2013 at 19:30
NPENPE
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This is possibly more pythonic, although not everyone likes list comprehensions.
```python
import random
def tossCoin(numFlips):
flips= ['Heads' if x==1 else 'Tails' for x in [random.randint(0,1) for x in range(numflips)]]
heads=sum([x=='Heads' for x in flips])
tails=numFlips-heads
```
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edited Jun 29, 2016 at 11:47
answered Oct 7, 2013 at 22:41
John PowellJohn Powell
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Abdullah Tariq
Abdullah TariqSep 18 at 14:58
Ap ko ager Game say paisy lany han muje bati ma apko paisa du ja
2025-09-18T14:58:09.88Z+00:00
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I'd go with something along the lines of:
python
from random import randint
num = input('Number of times to flip coin: ')
flips = [randint(0,1) for r in range(num)]
results = []
for object in flips:
if object == 0:
results.append('Heads')
elif object == 1:
results.append('Tails')
print results
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edited Feb 14, 2013 at 20:21
Donato Szilagyi
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answered Feb 14, 2013 at 19:58
AitchAitch
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```python
import random
import time
flips = 0
heads = "Heads"
tails = "Tails"
heads_and_tails = [(heads),
(tails)]
while input("Do you want to coin flip? [y|n]") == 'y':
print(random.choice(heads_and_tails))
time.sleep(.5)
flips += 1
else:
print("You flipped the coin",flips,"times")
print("Good bye")
```
You could try this, i have it so it asks you if you want to flip the coin then when you say no or n it tells you how many times you flipped the coin. (this is in python 3.5)
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answered May 3, 2018 at 1:54
BallisticZer0BallisticZer0
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Create a list with two elements head and tail, and use choice() from random to get the coin flip result. To get the count of how many times head or tail came, append the count to a list and then use Counter(list_name) from collections. Use uin() to call
```python
coin flip
import random
import collections
def tos():
a=['head','tail']
return(random.choice(a))
def uin():
y=[]
x=input("how many times you want to flip the coin: ")
for i in range(int(x)):
y.append(tos())
print(collections.Counter(y))
```
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edited Aug 13, 2019 at 18:59
answered Aug 13, 2019 at 18:45
Abhishek BahukhandiAbhishek Bahukhandi
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Fixing the immediate issues
The highest voted answer doesn't actually run, because it passes a string into range() (as opposed to an int).
Here's a solution which fixes two issues: the range() issue just mentioned, and the fact that the calls to str() in the print() statements on the last two lines can be made redundant. This snippet was written to modify the original code as little as possible.
python
def coinToss():
number = int(input("Number of times to flip coin: "))
recordList = []
heads = 0
tails = 0
for _ in range(number):
flip = random.randint(0, 1)
if (flip == 0):
recordList.append("Heads")
else:
recordList.append("Tails")
print(recordList)
print(recordList.count("Tails"), recordList.count("Heads"))
A more concise approach
However, if you're looking for a more concise solution, you can use a list comprehension. There's only one other answer that has a list comprehension, but you can embed the mapping from {0, 1} to {"Heads", "Tails"} using one, rather than two, list comprehensions:
python
def coinToss():
number = int(input("Number of times to flip coin: "))
recordList = ["Heads" if random.randint(0, 1) else "Tails" for _ in range(number)]
print(recordList)
print(recordList.count("Tails"), recordList.count("Heads"))
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answered Mar 3, 2022 at 5:13
BrokenBenchmarkBrokenBenchmark
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Instead of all that, you can do like this:
python
import random
options = ['Heads' , 'Tails']
number = int(input('no.of times to flip a coin : ')
for amount in range(number):
heads_or_tails = random.choice(options)
print(f" it's {heads_or_tails}")
print()
print('end')
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edited Jan 18, 2020 at 9:20
groenhen
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answered Jan 18, 2020 at 8:55
user12736398user12736398
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I did it like this. Probably not the best and most efficient way, but hey now you have different options to choose from. I did the loop 10000 times because that was stated in the exercise.
```python
Coinflip program
import random
numberOfStreaks = 0
emptyArray = []
for experimentNumber in range(100):
Code here that creates a list of 100 heads or tails values
headsCount = 0
tailsCount = 0
print(experimentNumber)
for i in range(100):
if random.randint(0, 1) == 0:
emptyArray.append('H')
headsCount +=1
else:
emptyArray.append('T')
tailsCount += 1
Code here that checks if the list contains a streak of either heads or tails of 6 in a row
heads = 0
tails = 0
headsStreakOfSix = 0
tailsStreakofSix = 0
for i in emptyArray:
if i == 'H':
heads +=1
tails = 0
if heads == 6:
headsStreakOfSix += 1
numberOfStreaks +=1
if i == 'T':
tails +=1
heads = 0
if tails == 6:
tailsStreakofSix += 1
numberOfStreaks +=1
print('\n' + str(headsStreakOfSix))
print('\n' + str(tailsStreakofSix))
print('\n' + str(numberOfStreaks))
print('\nChance of streak: %s%%' % (numberOfStreaks / 10000))
```
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answered Jan 3, 2021 at 10:40
R. Sp.R. Sp.
13 4 4 bronze badges
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```python
program to toss the coin as per user wish and count number of heads and tails
import random
toss=int(input("Enter number of times you want to toss the coin"))
tail=0
head=0
for i in range(toss):
val=random.randint(0,1)
if(val==0):
print("Tails")
tail=tail+1
else:
print("Heads")
head=head+1
print("The total number of tails is {} and head is {} while tossing the coin {} times".format(tail,head,toss))
```
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answered Apr 14, 2021 at 15:47
MADhuMADhu
1
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```python
import random
def coinToss(number):
heads = 0
tails = 0
for flip in range(number):
coinFlip = random.choice([1, 2])
if coinFlip == 1:
print("Heads")
recordList.append("Heads")
else:
print("Tails")
recordList.append("Tails")
number = input("Number of times to flip coin: ")
recordList = []
if type(number) == str and len(number)>0:
coinToss(int(number))
print("Heads:", str(recordList.count("Heads")) , "Tails:",str(recordList.count("Tails")))
```
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edited Dec 16, 2019 at 15:30
B. Go
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answered Dec 16, 2019 at 15:06
Panagiotis NtinisPanagiotis Ntinis
16 3 3 bronze badges
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All Possibilities in Coin Toss for N number of Coins
```python
def Possible(n, a):
if n >= 1:
Possible(n // 2, a)
z = n % 2
z = "H" if z == 0 else "T"
a.append(z)
return a
def Comb(val):
for b in range(2 N):
A = Possible(b, [])
R = N - len(A)
c = []
for x in range(R):
c.append("H")
Temp = (c + A)
if len(Temp) > N:
val.append(Temp[abs(R):])
else:
val.append(Temp)
return val
N = int(input())
for c in Comb([]):
print(c)
```
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edited Sep 20, 2021 at 18:18
clarj
1,053 6 6 silver badges 15 15 bronze badges
answered Sep 20, 2021 at 13:32
KrishnakumarKrishnakumar
1
1 Comment
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Community
CommunityOver a year ago
As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center.
2021-09-20T16:57:13.737Z+00:00
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```python
heads = 1
tails = 0
input("choose 'heads' or 'tails'. ").upper()
random_side = random.randint(0, 1)
if random_side == 1:
print("heads you win")
else:
print("sorry you lose ")
```
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edited Oct 31, 2021 at 20:05
Tomerikoo
19.5k 16 16 gold badges 57 57 silver badges 68 68 bronze badges
answered Oct 31, 2021 at 18:58
Peter_JrPeter_Jr
1 1 1 bronze badge
1 Comment
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nima
nimaOver a year ago
While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value. You can find more information on how to write good answers in the help center: stackoverflow.com/help/how-to-answer . Good luck 🙂
2021-10-31T20:25:19.043Z+00:00
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None of the other solutions worked for me. This is what I did and it worked:
```python
import random
random_side = random.randint(0, 1)
if random_side == 1:
print("Heads")
else:
print("Tails")
```
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edited Mar 25, 2024 at 6:12
L Tyrone
7,928 23 23 gold badges 32 32 silver badges 45 45 bronze badges
answered Mar 25, 2024 at 5:32
Gleb GuninGleb Gunin
1 1 1 bronze badge
1 Comment
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Vivek Atal
Vivek AtalOver a year ago
This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
2024-03-27T02:02:56.52Z+00:00
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15363 | http://hyperphysics.phy-astr.gsu.edu/hbase/Nuclear/raddec.html | Radioactive Half-Life
Radioactive Decay Calculation ============================= The radioactive half-life for a given radioisotope is a measure of the tendency of the nucleus to "decay" or "disintegrate" and as such is based purely upon that probability. Alternatively, this information can be expressed in terms of the decay constant λ. The number of atoms, the mass of the substance, and the level of activity all follow the same exponential decay form. The calculation below is stated in terms of the amount of the substance remaining, but can be applied to intensity of radiation or any other property proportional to it. If the radioactive half-life of an isotope is T = x10^ s = min = hr T = yr = x10^ yr and amount A 0 is present at time t=0 and amount A remains at time t, then = x10^ s = min t = hr = yr = x10^ yr the fraction remaining will be given by = = x10^ If the beginning amount was Ci then the amount remaining is Ci This calculation is designed for exploration. It is presumed that you will enter a half-life for a particular radioisotope. Then if you enter a time, then it will calculate the fraction remaining at that time. If you enter a fraction or amount remaining, it will calculate the time required (or age) required to achieve that fraction. The amount is listed in Ci (Curies), but the mass, intensity of radiation, number of counts or other ways of characterizing the radioisotpe will decrease by the same ratio, so the calculation can be used for any of those quantities. Graph of decayHalf-life discussionIndex
HyperPhysics NuclearR NaveGo Back |
15364 | https://www.chegg.com/homework-help/questions-and-answers/2consider-initial-value-problem-y-y-2-y-2-y-x0-y0-first-order-ordinary-differential-equati-q30960886 | Solved 2.Consider the initial value problem: y’=y^2+y-2 | Chegg.com
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2.Consider the initial value problem: y’=y^2+y-2 y(x0)=y0. This is a first order ordinary differential equation with dependent variable y and independent variable x. a) Verify that the hypotheses of the existence and uniqueness theorem are satisfied. b) Determine all equilibrium solutions to this differential equation, i.e., all solutions where y is a
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Question: 2.Consider the initial value problem: y’=y^2+y-2 y(x0)=y0. This is a first order ordinary differential equation with dependent variable y and independent variable x. a) Verify that the hypotheses of the existence and uniqueness theorem are satisfied. b) Determine all equilibrium solutions to this differential equation, i.e., all solutions where y is a
2.Consider the initial value problem: y’=y^2+y-2 y(x0)=y0. This is a first order ordinary differential equation with dependent variable y and independent variable x.
a) Verify that the hypotheses of the existence and uniqueness theorem are satisfied.
b) Determine all equilibrium solutions to this differential equation, i.e., all solutions where y is a constant.
c) using your answers to the last two questions, explain why any solution y=y(x) to this differential equation with -2<y0=y(x0)<1 must have -2<y(x)<1 for all x and have y be a decreasing function of x
d) Sketch a slope field for (a). Sketch representative curves for the three conditions y0<-2,-2<y0<1, and y0>1. Use the curve plotter to check your sketches.
e) For a solution y=y(x) to this differential equation with y0=y(x0)=0, what are lim x->infinity y(x) and lim x->-infinity y(x) equal to?
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y′=y 2+y−2 y(x 0)=0 View the full answer Step 2 UnlockAnswer Unlock Previous question
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mmmmmmmmmmlli mmmmmmmmmmlli mmmmmmmmmmlli |
15365 | https://www.nagwa.com/en/explainers/567105213104/ | Lesson Explainer: Perpendicular Bisector of a Chord Mathematics • Third Year of Preparatory School
Join Nagwa Classes
Attend live Mathematics sessions on Nagwa Classes to learn more about this topic from an expert teacher!
In this explainer, we will learn how to use the theory of the perpendicular bisector of a chord from the center of a circle and
its converse to solve problems.
Let’s begin by recalling how we can define a radius, a chord, and a diameter. A radius is a line segment that has one end
at the center of the circle and the other on the circumference.
We define a chord as any straight line segment whose end points both lie on the circumference of the same circle.
The diameter is a special type of chord, which passes through the center of the circle. As you can see in the diagram, it also
consists of two radii.
In this explainer, we will be considering the perpendicular bisectors of chords, such as the following.
There are three theorems related to perpendicular bisectors of chords, which we will be covering. Let’s start by supposing
that we have a chord in a circle, with a straight line that passes through the center of the circle, 𝐴, and also
bisects the chord, 𝐵𝐶, as shown in the diagram.
We will now prove by contradiction that the bisector of the chord, which passes through the center of the circle, is also
perpendicular to the chord. We will assume that ∠𝐴𝐷𝐵≠90∘ and ∠𝐴𝐷𝐶≠90∘.
In our diagram, we can see that we have a circle centered at 𝐴, the chord 𝐵𝐶,
the bisector ⃖⃗𝐸𝐹, and the two radii 𝐴𝐵 and
𝐴𝐶.
Now, since 𝐴𝐵 and 𝐴𝐶 are radii of the circle, they
will be equal in length. Also, since ⃖⃗𝐸𝐹 bisects 𝐶𝐵, we can
say that the lengths of 𝐶𝐷 and 𝐷𝐵 will be equal. Finally,
△𝐴𝐶𝐷 and △𝐴𝐷𝐵 share the side
𝐴𝐷, so we can see that all three side of these triangles are equal in length. Therefore,
by SSS congruence, we have that
△𝐴𝐶𝐷≅△𝐴𝐷𝐵.
Next, we consider the angles ∠𝐴𝐷𝐶 and ∠𝐴𝐷𝐵. We know that
𝐶𝐵 is a straight line because it is a chord within the circle. Angles on a straight
line sum to 180∘. Due to the congruence of
△𝐴𝐶𝐷 and △𝐴𝐷𝐵, we have that
∠𝐴𝐷𝐵=∠𝐴𝐷𝐶. Combining these two points, we have that
∠𝐴𝐷𝐵=∠𝐴𝐷𝐶=1802=90.∘∘
This contradicts our assumption at the beginning and, hence, completes our proof. Let’s summarize what we have just
proven with the following theorem.
Theorem: The Chord Bisector Theorem—Part 1
If we have a circle with center 𝐴 containing a chord 𝐵𝐶, then the
straight line that passes through 𝐴 and bisects chord 𝐵𝐶 is
perpendicular to 𝐵𝐶.
The second chord bisector theorem is very similar to this one. However, we start with a slightly different assumption. This
time, we have a circle containing a chord, with a straight line passing through the center of the circle, 𝐴,
which is also perpendicular to chord 𝐵𝐶 as shown in this diagram.
The theorem states that the line that passes through the center of the circle and is perpendicular to the chord also bisects
that chord. The proof of this theorem is very similar to the first proof, and so we will not be proving it here.
Theorem: The Chord Bisector Theorem—Part 2
If we have a circle with center 𝐴 containing a chord 𝐵𝐶, then the
straight line that passes through 𝐴 and is perpendicular to 𝐵𝐶 also
bisects 𝐵𝐶.
The final theorem we will be covering is often called the converse to the chord bisector theorem. It is very similar to the
other two theorems but again starts with slightly different assumptions. This theorem states that the perpendicular bisector
of a chord also passes through the center of the circle. The proof is also very similar to the proof of the first theorem,
and so this is left as an exercise for the reader.
Theorem: The Converse to the Chord Bisector Theorem
If we have a circle with center 𝐴 containing a chord 𝐵𝐶, then
the perpendicular bisector of 𝐵𝐶 passes through 𝐴.
Now that we have covered these three theorems, let us look at some examples of how we can use them to find missing lengths
and angles.
Example 1: Finding a Missing Length Using Perpendicular Bisectors of Chords
Given 𝐴𝑀=200cm and
𝑀𝐶=120cm, find
𝐴𝐵.
Answer
In the diagram, we can see that we have a circle with center 𝑀. There is a chord
𝐴𝐵, which is bisected by the line segment 𝑀𝐷
at point 𝐶. Here, we can apply the chord bisector theorem, which states that if we have a circle with
center 𝑀 containing a chord 𝐴𝐵, then the straight line that passes
through 𝑀 and bisects chord 𝐴𝐵 is perpendicular to
𝐴𝐵. Hence, we can say that
𝑚∠𝑀𝐶𝐵=90.∘
The question tells us that 𝐴𝑀=200cm. 𝐴𝑀 is a radius of the circle, and so is 𝐵𝑀;
therefore,
𝐵𝑀=200.cm
We have also been given that 𝑀𝐶=120cm, so
let’s add these to our diagram.
We can see in our diagram that △𝑀𝐶𝐵 is a right triangle of which we know two of the
sides. We can find the remaining side using the Pythagorean theorem:
𝐵𝐶=√200−120=160.cm
All that is left to do now is to use the fact that 𝐶 bisects 𝐴𝐵,
so 𝐴𝐵=2𝐵𝐶. Substituting our value for 𝐵𝐶 in, we obtain that
𝐴𝐵=320.cm
In the next example, we will see how we can apply the theorems to find the area of a triangle.
Example 2: Finding a Missing Length and the Area of a Triangle Using Perpendicular Bisectors of Chords
In the figure below, if 𝑀𝐴=17.2cm
and 𝐴𝐵=27.6cm, find the length of
𝑀𝐶 and the area of △𝐴𝐷𝐵 to the nearest tenth.
Answer
Since 𝑀 is the center of the circle and line 𝑀𝐷 bisects chord
𝐴𝐵 at 𝐶, we can apply the chord bisector theorem, which states
that if we have a circle with center 𝑀 containing a chord 𝐴𝐵,
then the straight line that passes through 𝑀 and bisects chord 𝐴𝐵
is perpendicular to 𝐴𝐵. Hence, we can say that
𝑚∠𝑀𝐶𝐴=90.∘
We have been given the length of 𝐴𝐵, and we also know that 𝐶
bisects this chord, so we can say that
𝐴𝐶=27.62=13.8.cm
Adding this information to our diagram, we can see that we have a right triangle, △𝑀𝐶𝐴,
for which we know two of the side lengths. We can use the Pythagorean theorem to find the length of
𝑀𝐶:
𝑀𝐶=√17.2−13.8=10.266….cm
To answer the first part of the question, we need to round this value to the nearest tenth. In doing this, we obtain
𝑀𝐶=10.3.cm
Next, we need to find the area of △𝐴𝐷𝐵. We know that 𝑀𝐷
is perpendicular to 𝐴𝐵, so we have that 𝑚∠𝐴𝐶𝐷=90∘. We also have that 𝐴𝐵=27.6cm, so all we need to find is the length of
𝐶𝐷.
𝑀𝐷 and 𝑀𝐴 are both radii of the circle, so
they will have the same length. Therefore, 𝑀𝐷=17.2cm. We can find the length of 𝐶𝐷 by subtracting the length we found previously for
𝑀𝐶 from the length of 𝑀𝐷. In doing this, we obtain
𝐶𝐷=17.2−10.266…=6.933….cm
We know that the area of a triangle is given by the formula
areaofatrianglebaseheight=12××.
Using this formula, we find that the area of this triangle is
areaofcm△𝐴𝐷𝐵=12×27.6×6.933…=95.683….
All we need to do now is to round to the nearest tenth, to find that our solution is
areaofcm△𝐴𝐷𝐵=95.7.
In our next example, we will see how we can use perpendicular bisectors of chords to find a missing angle.
Example 3: Finding a Missing Angle Using Perpendicular Bisectors of Chords
𝐴𝐵 and 𝐴𝐶 are two chords in a circle with center
𝑀 on two opposite sides of the center, where 𝑚∠𝐵𝐴𝐶=33∘. If 𝐷 and 𝐸
are the midpoints of 𝐴𝐵 and 𝐴𝐶, respectively,
find 𝑚∠𝐷𝑀𝐸.
Answer
We can start our solution by noting that both 𝑀𝐸 and
𝑀𝐷 pass through the center of the circle and bisect chords
𝐴𝐶 and 𝐴𝐵 respectively. Therefore, we can apply
the chord bisector theorem, which states that if we have a circle with center 𝑀 containing a chord
𝐴𝐵, then the straight line that passes through 𝑀 and bisects
chord 𝐴𝐵 is perpendicular to 𝐴𝐵. Using this
theorem, we can say that 𝑀𝐷⟂𝐴𝐵 and 𝑀𝐸⟂𝐴𝐶, so we have that
𝑚∠𝑀𝐸𝐴=90𝑚∠𝑀𝐷𝐴=90.∘∘and
Now, if we consider quadrilateral 𝐴𝐷𝑀𝐸, we can see that we now know the measures of three of the
interior angles. Since the interior angles of a quadrilateral always sum to
360∘, we can find the measure of the unknown angle as follows:
𝑚∠𝐷𝑀𝐸=360−90−90−33=147.∘
In the next example, we will find the perimeter of a triangle using perpendicular bisectors of chords.
Example 4: Finding the Perimeter of a Triangle Drawn inside Another Triangle Whose Vertices Touch a Circle
In a circle of center 𝑂, 𝐴𝐵=35cm, 𝐶𝐵=25cm, and 𝐴𝐶=40cm. Given that 𝑂𝐷⟂𝐵𝐶 and 𝑂𝐸⟂𝐴𝐶,
find the perimeter of △𝐶𝐷𝐸.
Answer
The first thing we notice about this question is that line segments 𝑂𝐸 and
𝑂𝐷 both pass through 𝑂 and meet chords
𝐴𝐶 and 𝐶𝐵, respectively, at right angles. Therefore, we can apply the theorem, which states that if we have a circle with center 𝑂 containing
a chord 𝐵𝐶, then the straight line that passes through 𝑂 and
is perpendicular to 𝐵𝐶 also bisects 𝐵𝐶.
Using this information, we can say that
𝐴𝐸=𝐶𝐸𝐶𝐷=𝐷𝐵.and
This tells us that 𝐸 is the midpoint of 𝐴𝐶 and 𝐷
is the midpoint of 𝐵𝐶.
Now we will use the midpoint theorem, which tells us that the line segment in a triangle joining the midpoints of two sides
of the triangle is said to be parallel to the third side and is also half the length of the third side. Hence, we have that
𝐸𝐷=12𝐴𝐵=352=17.5.cm
We also know that 𝐶𝐸=12𝐴𝐶=20cm
and 𝐶𝐷=12𝐶𝐵=12.5cm. By summing
these three lengths, we will reach our solution:
theperimeterofcm△𝐶𝐷𝐸=17.5+20+12.5=50.
In our final example, we will see how we can use the perpendicular bisectors to find the diameter of a circle, given the
length of a chord.
Example 5: Determining the Length of the Diameter of a Circle given the Length of a Chord in It
Using the figure and the fact that 𝐵𝐶=32√3cm, determine the diameter of the circle.
Answer
The question has asked us to find the length of the diameter of the circle. Since the diameter is double of the radius, a sensible
place to start will be to identify the circle’s radius.
The first thing we notice when looking at this question is that the line segment 𝑀𝐴
passes through the center of the circle and is also perpendicular to chord 𝐵𝐶. This
tells us that 𝑀𝐴 bisects 𝐵𝐶. Since
𝐵𝐶=32√3cm, we can say that
𝐵𝐷=16√3.cm
Next, let’s consider △𝑀𝐵𝐴. We know that 𝑀𝐵=𝐵𝐴, but we also
know that 𝑀𝐵=𝑀𝐴, since they are both radii of the circle. Hence, this triangle is an equilateral triangle. Using this, we can say that
𝑚∠𝐴𝑀𝐵=60.∘
The following diagram shows the information we have found so far.
We will now consider △𝑀𝐷𝐵. We can see that it is a right triangle, for which we know
∠𝐷𝑀𝐵=60∘ and 𝐵𝐷=16√3cm. There are two methods we can use to find
the length of 𝑀𝐵, which is also the radius of the circle. We will cover both methods.
Method 1
We will start by considering ∠𝑀𝐵𝐷. We know that the angles in a triangle sum to
180∘ and the other two angles in △𝑀𝐷𝐵
are 90∘ and 60∘. Therefore, we have that
∠𝑀𝐵𝐷=180−90−60=30.∘
Now we can use the fact that in a 30-60-90 triangle, the side that is opposite the angle of
30∘ is half the length of the hypotenuse, so 𝑀𝐵=2𝑀𝐷.
Since this is a right triangle, we can use the Pythagorean theorem, which tells us that the square of the hypotenuse is
equal to the sum of the squares of the other two sides. Applying this to our triangle gives us
𝑀𝐵=𝑀𝐷+𝐵𝐷.
Next, we can substitute in 𝑀𝐵=2𝑀𝐷 and 𝐵𝐷=16√3cm and then simplify:
(2𝑀𝐷)=𝑀𝐷+16√33𝑀𝐷=768𝑀𝐷=256𝑀𝐷=16.cm
We note that when we take the square root in the final step here, we can ignore the negative solution since the length,
𝑀𝐷, must be positive. Now we can find the length of 𝑀𝐵 using
𝑀𝐵=2𝑀𝐷. This give us that
𝑀𝐵=32.cm
This is the radius of the circle. However, the question asked us to find the diameter, so we need to double this to
reach our solution of
64.cm
Method 2
In the second method, we will be using right triangle trigonometry. We know that
sinOppHyp𝜃=.
In our case, the opposite side is 𝐵𝐷 and the hypotenuse is 𝑀𝐵. We can rearrange this
formula to obtain
𝑀𝐵=16√360=32.sincm∘
Now that we have found the length of the radius, all that is left to do is to double it to find the length of the diameter. This gives us a solution of
64.cm
We have now seen a variety of examples of how perpendicular bisectors of chords can be used to find missing lengths, angle
measures, and other unknowns in problems involving circles. Let’s recap some key points of the explainer.
Key Points
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15366 | https://www.quora.com/What-is-a10-of-the-geometric-sequence-if-the-first-term-is-35-and-the-sum-of-the-first-ten-terms-is-355 | What is a10 of the geometric sequence if the first term is -35 and the sum of the first ten terms is 355? - Quora
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What is a10 of the geometric sequence if the first term is -35 and the sum of the first ten terms is 355?
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Enrico Gregorio
Associate professor in Algebra · Author has 18.4K answers and 16M answer views
·2y
Let the sequence be a n,a n, where a 1=−35.a 1=−35. If r r is the constant ratio we know that
a 10=−35 r 9 a 10=−35 r 9
but we also know that
355=a 1+a 2+⋯+a 10=−35 r 10−1 r−1 355=a 1+a 2+⋯+a 10=−35 r 10−1 r−1
The equation is therefore
71(r−1)=−35(r 10−1)71(r−1)=−35(r 10−1)
that simplifies to
7 r 10+71 r−78=0 7 r 10+71 r−78=0
Consider the function f(x)=7 x 10+71 x−78.f(x)=7 x 10+71 x−78. We see that
f′(x)=70 x 9+71 f′(x)=70 x 9+71
vanishes only for x=−9√71/70 x=−71/70 9 so the function f f has an absolute minimum, which is negative because f(0)<0.f(0)<0. Hence f f only vanishes twice. One root is x=1,x=1, which is invalid for our problem. The other root is in the interval (−2,−1),(−2,−1), because f(−2)>0 f(−2)>0 and f(−1)<0.f(−1)<0.
An approximate value of t
Continue Reading
Let the sequence be a n,a n, where a 1=−35.a 1=−35. If r r is the constant ratio we know that
a 10=−35 r 9 a 10=−35 r 9
but we also know that
355=a 1+a 2+⋯+a 10=−35 r 10−1 r−1 355=a 1+a 2+⋯+a 10=−35 r 10−1 r−1
The equation is therefore
71(r−1)=−35(r 10−1)71(r−1)=−35(r 10−1)
that simplifies to
7 r 10+71 r−78=0 7 r 10+71 r−78=0
Consider the function f(x)=7 x 10+71 x−78.f(x)=7 x 10+71 x−78. We see that
f′(x)=70 x 9+71 f′(x)=70 x 9+71
vanishes only for x=−9√71/70 x=−71/70 9 so the function f f has an absolute minimum, which is negative because f(0)<0.f(0)<0. Hence f f only vanishes twice. One root is x=1,x=1, which is invalid for our problem. The other root is in the interval (−2,−1),(−2,−1), because f(−2)>0 f(−2)>0 and f(−1)<0.f(−1)<0.
An approximate value of the unique solution is −1.381−1.381
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Muthusamy Piramanayagam ( முத்துசாமி பிரமநாயகம்)
M.Sc Physics, Diploma in science teaching. H.O.D in Science · Author has 11K answers and 7.8M answer views
·2y
Sum = a(1 - r^n) / (1 - r), when r ≠ 1.
355 = (-35) ( 1- r ^10) / ( 1- r)
355 ( 1-r) = 35 ( r^10- 1)
r = -1.380532
10 th term = -35 ( - 1.380532) ^ 9
= 637.5
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To find the 10 t h 10 t h term of the geometric sequence with the first term a 1=−35 a 1=−35 and the sum of the first ten terms S 10=355 S 10=355, we can use the formula for the sum of the first n n terms of a geometric sequence:
S n=a 1 1−r n 1−r S n=a 1 1−r n 1−r
where a 1 a 1 is the first term, r r is the common ratio, and n n is the number of terms.
For our case:
S 10=355 S 10=355
a 1=−35 a 1=−35
n=10 n=10
Substituting these values into the formula gives us:
355=−35 1−r 10 1−r 355=−35 1−r 10 1−r
To simplify, we can first multiply both sides by (1−r)(1−r):
355(1−r)=−35(1−r 10)355(1−r)=−35(1−r 10)
Expanding both sides:
355−355 r=−35+35 r 10 355−355 r=−35+35 r 10
Rearrangin
Continue Reading
To find the 10 t h 10 t h term of the geometric sequence with the first term a 1=−35 a 1=−35 and the sum of the first ten terms S 10=355 S 10=355, we can use the formula for the sum of the first n n terms of a geometric sequence:
S n=a 1 1−r n 1−r S n=a 1 1−r n 1−r
where a 1 a 1 is the first term, r r is the common ratio, and n n is the number of terms.
For our case:
S 10=355 S 10=355
a 1=−35 a 1=−35
n=10 n=10
Substituting these values into the formula gives us:
355=−35 1−r 10 1−r 355=−35 1−r 10 1−r
To simplify, we can first multiply both sides by (1−r)(1−r):
355(1−r)=−35(1−r 10)355(1−r)=−35(1−r 10)
Expanding both sides:
355−355 r=−35+35 r 10 355−355 r=−35+35 r 10
Rearranging the equation:
355+35=35 r 10+355 r 355+35=35 r 10+355 r
This simplifies to:
390=35 r 10+355 r 390=35 r 10+355 r
Dividing the entire equation by 5 to simplify:
78=7 r 10+71 r 78=7 r 10+71 r
Now, we need to solve this equation for r r. This is a non-linear equation, and it may require numerical methods or graphing to find r r. However, we can also try some reasonable values for r r to see if we can find a solution.
Trial for r=1 r=1:
78=7(1)10+71(1)=7+71=78(True)78=7(1)10+71(1)=7+71=78(True)
Thus, r=1 r=1 is a solution.
Now, we can find the 10 t h 10 t h term a 10 a 10:
a n=a 1⋅r n−1 a n=a 1⋅r n−1
For n=10 n=10:
a 10=−35⋅1 10−1=−35 a 10=−35⋅1 10−1=−35
Therefore, the 10 t h 10 t h term of the geometric sequence is:
−35−35
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Related questions
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What is the sum of the first 35 terms of the sequence whose general term is An=5?
The second term of a geometric sequence is 35 and the fourth term is 875. What is the first term and the fifth term?
If the sum of the first 17 terms is 289, what is the sum of its first 9 terms?
If the 5th term of geometric progress is 162 and the 8th term is 4374, what is the sum of the first 10 terms?
Peter Groot
B.S. in Mathematics, Massachusetts Institute of Technology (Graduated 1971) · Author has 9.3K answers and 3.5M answer views
·2y
If the sum of the first 10 terms is 355, and the first term is -35, the multiplier must be negative in order to generate positive terms. I used a spreadsheet and got a multiplier of -1.38053 and the 10th term is 637.5
Edit: If it were an arithmetic sequence, the mean is 355/10 = 35.5. The sum of the first term -35 and the last term is 2 35.5 = 71, so the last term a10 = 71 - (-35) = 106.
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Miyuki Nanase
Author has 4.9K answers and 1M answer views
·2y
355 = S(10)
= (10 / 2) × ( T(1) + T(10) )
= 5 × ( —35 + T(10) )
T(10) — 35 = 355/5
= 71
T(10) = 71 + 35
= 106 if this was an AP
355 = S(10)
= –35 × (r^10 — 1) / (r — 1)
71r — 71 = 7 — 7r^10
7r^10 + 71r = 7 + 71
0 = 7r^10 + 71r — 78
Real solutions are r = 1 and r ≈ —1.38053 ( r ≈ -1.3805298797310627210 )
T(10) = ar^9
= (–35) × (–1.38053)^9
≈ 637.5
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Related questions
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In an AP sum of the first 10 terms is 25 and the sum of the first 25 terms is 10. What is the sum of 35 terms?
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The fifth term of a geometric sequence is -2 and the fourth term is 432. What is the first term?
The fourth term of a geometric sequence is 27 and the seventh term is 1.What is the first term?
What is the number of terms in the following GP 81, 27, 9…1/27?
Bruce Stevens
Author has 2.8K answers and 1.5M answer views
·1y
Related
What is the sum of the first 7 terms of a geometric sequence if the 3rd term is 108 and the 6th term is -32?
Set the question into GS equations:
t(3) = 108 = ( a (r^2) ),
t(6) = ( —32) = ( a (r^5) ).
Solve for the value, of the constant ratio, r, between consecutive terms:
((a (r^5)) / (a (r^2))) = (( — 32) / 108),
NOTE: The 1st term, a, in the numerator, and denominator, cancel each other. Both numerator,and denominator, can be simplified, by being divided by 4. Also, by the quotient rule of exponents, with the same base:
( (x^a) / (x^b) ) = ( x^(a — b) ).
So,
( r^(5 — 2) ) = ( — (8 / 27)),
( r^3 ) = ( — (8 / 27)),
NOTE: 8 = ( 2^3 ), and 27 = ( 3^3 ). By the power of a fraction rule:
( (x^a) / (y^a) )
Continue Reading
Set the question into GS equations:
t(3) = 108 = ( a (r^2) ),
t(6) = ( —32) = ( a (r^5) ).
Solve for the value, of the constant ratio, r, between consecutive terms:
((a (r^5)) / (a (r^2))) = (( — 32) / 108),
NOTE: The 1st term, a, in the numerator, and denominator, cancel each other. Both numerator,and denominator, can be simplified, by being divided by 4. Also, by the quotient rule of exponents, with the same base:
( (x^a) / (x^b) ) = ( x^(a — b) ).
So,
( r^(5 — 2) ) = ( — (8 / 27)),
( r^3 ) = ( — (8 / 27)),
NOTE: 8 = ( 2^3 ), and 27 = ( 3^3 ). By the power of a fraction rule:
( (x^a) / (y^a) ) = ( (x / y)^a ).
So,
( r^3 ) = ( — ((2^3) / (3^3))),
( r^3 ) = ( — ((2 / 3)^3)),
NOTE: The same exponents on both sides, of the equal sign, so, drop the exponents.
r = ( — (2 / 3)).
Solve for the value, of the 1st term, t(1):
t(3) = 108 = ( t(1) (( — (2 / 3))^2) ),
t(1) = ( 108 / (( — (2 / 3))^2) ),
t(1) = ( 108 / (4 / 9) ),
Flip the denominator of ( 4 / 9 ), so it becomes ( 9 / 4 ), and multiply:
t(1) = ( 108 (9 / 4) ), → t(1) = ((108 9) / 4),
t(1) = (972 / 4), → t(1) = 243.
Solve for the value, of the sum of the first seven terms, Sum(7):
By the geometric sequence summation formula:
Sum(n) = ( (a ((r^n) — 1)) / (r — 1) ),
where Sum(n) = ?, a = 243, n = 7, r = ( —(2/3)),
Sum(7) = ( (243 ((( —(2/3))^7) — 1)) / (( —(2/3)) — 1) ),
Sum(7) = ( (243 (( —(128/2187)) — 1)) / (( —(2/3)) — (3/3)) ),
Sum(7) = ( (243 (( —(128/2187)) — (2187/2187))) / ( —(5/3)) ),
Sum(7) = ( (243 ( —(2315/2187))) / ( —(5/3)) ),
Sum(7) = ( ( —((243 2315) / 2187)) / ( —(5/3)) ),
Sum(7) = ( ( —(562545/2187)) / ( —(5/3)) ),
Flip the denominator of ( —(5/3)), so it becomes ( —(3/5)), and multiply:
Sum(7) = ( ( —(562545/2187)) ( —(3/5)) ),
Sum(7) = ( (( —562545) ( —3)) / (2187 5) ),
Sum(7) = (1687635 / 10935),
Sum(7) = (463 / 3).
P.S.
The geometric sequence is:
243, —162, 108, —72, 48, —32, (64/3), —(128/9), (256/27), —(512/81), (1024/243), …
The general term rule, of this geometric sequence, when n = 1, 2, 3, …, is:
t(n) = ( 243 (( — (2 / 3))^(n — 1)) )
OR
t(n) = ( 243 / (( — (2 / 3))^(1 — n)) )
OR
t(n) = ( (( —2)^(n — 1)) (3^(6 — n)) )
OR
t(n) = ( (( —2)^(n — 1)) / (3^(n — 6)) ).
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Calvin L.
2nd year mathematics student · Author has 10K answers and 2.2M answer views
·2y
S 10=10 2[2 a+9 d]=355 S 10=10 2[2 a+9 d]=355
2 a+9 d=71 2 a+9 d=71
a+9 d=71+35=106 a+9 d=71+35=106
∴a 10=106∴a 10=106
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Maung Maung Sein
Advisor (Mech. Engg) · Author has 77 answers and 172K answer views
·5y
Related
The first, third, and sixth terms of an arithmetic sequence are the first 3 terms of a geometric sequence. If the fifteen term is 15, how do you determine the first four terms of the geometric sequence if r>1?
^Solution : Let the first, third and sixth terms of the AP be a, (a + 2d), and (a + 5d).
The first term of GP = the first term of AP.
Therefore the first term of GP = a.
So the first 3 terms of GP are a, ar, and ar^2.
The first, third and sixth terms of AP are the first three terms of GP (given). Therefore,
a = a, a + 2d = ar, and a + 5d = ar^2.
(a + 2d) - a = ar - a,
2d = a (r - 1)…….#1.
(a +5d) - (a + 2d) = ar^2 - ar,
3d = ar (r - 1)…..#2.
Dividing #2 by #1 gives, r = 3/2 (r > 1)….#3.
Substituting the value of “r” in #1, we have 2d = a (3/2 - 1),
2d = a/2,
4d = a…..#3.
It is given that the 15th. term of A
Continue Reading
^Solution : Let the first, third and sixth terms of the AP be a, (a + 2d), and (a + 5d).
The first term of GP = the first term of AP.
Therefore the first term of GP = a.
So the first 3 terms of GP are a, ar, and ar^2.
The first, third and sixth terms of AP are the first three terms of GP (given). Therefore,
a = a, a + 2d = ar, and a + 5d = ar^2.
(a + 2d) - a = ar - a,
2d = a (r - 1)…….#1.
(a +5d) - (a + 2d) = ar^2 - ar,
3d = ar (r - 1)…..#2.
Dividing #2 by #1 gives, r = 3/2 (r > 1)….#3.
Substituting the value of “r” in #1, we have 2d = a (3/2 - 1),
2d = a/2,
4d = a…..#3.
It is given that the 15th. term of AP is 15. So,
a + 14d = 15,…#4.
Substituting the value of “a” from #3 in #4 gives, 18d = 15, from which d = 5/6….#5.
Substituting the value of “d” from #5 in #4, gives, a + 14 (5/6) = 15,
a + 35/3 = 15,
a = 10/3….#6.
The first four terms of GP are 10/3, 10/3 × (3/2), 10/3 × (3/2)^2, 10/3 × (3/2)^3.
Answer : The first four terms of GP will be, “10/3, 5, 15/2, 45/4”.
Verification : The first, third and sixth terms of AP are 10/3, 10/3 + 2 × 5/6, 10/3 + 5( 5/6),
10/3, 5, 15/2.
Please upvote if my solution is acceptable.
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And honestly? Putting them to use was way easier than I expected. I bet you can knock out at least three or four of these right now—yes, even from your phone. Don’t wait like I did.
Cancel Your Car Insurance
You might not even realize it, but your car insurance company is probably overcharging you. In fact, they’re kind of counting on you not noticing. Luckily, this problem is easy to fix.
Don’t waste your time
Continue Reading
Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included.
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Cancel Your Car Insurance
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Matt Zoske
Lifelong lover of math
·4y
Related
The first term of a geometric sequence is 3 and the last term is 48. What is the common ratio if the sum of n terms is 93?
This is an easy one to guess and check or do without formulas, but if the numbers were not as easy, you can use the following steps:
Use the formulas for the sum of n terms of a geometric sequence and for the nth term of a geometric sequence,
Continue Reading
This is an easy one to guess and check or do without formulas, but if the numbers were not as easy, you can use the following steps:
Use the formulas for the sum of n terms of a geometric sequence and for the nth term of a geometric sequence,
Upvote ·
9 3
Bruce Stevens
Author has 2.8K answers and 1.5M answer views
·1y
Related
The 1st, 5th, and 10th term of linear sequence are geometric progression. If the sum of the 2nd and 8th terms of the linear sequence is 30, what is the 1st term?
Set the question into AS / GS equations:
AS_t(1) = a = GS_t(1) = a,
AS_t(5) = (a + 4d) = GS_t(2) = (a r),
AS_t(10) = (a + 9d) = GS_t(3) = (a (r^2)),
AS_Sum( t(2) + t(8) ) = 30 = ( (a + d) + (a + 7d) ).
Solve for the algebraic value, of the 1st term, a, from the arithmetic summation equation:
AS_Sum( t(2) + t(8) ) = 30 = (2a + 8d),
Divide throughout, by 2:
15 = (a + 4d), → t(5) = (a + 4d),
Therefore, t(5) = 15, → a = (15 — 4d).
Solve for the algebraic value, of the 1st term, a, from the arithmetic terms divided as geometric terms:
AS_t(1) = a, AS_t(5) = 15 = (a + 4d), AS_t(10) = (a + 9d) are in a geom
Continue Reading
Set the question into AS / GS equations:
AS_t(1) = a = GS_t(1) = a,
AS_t(5) = (a + 4d) = GS_t(2) = (a r),
AS_t(10) = (a + 9d) = GS_t(3) = (a (r^2)),
AS_Sum( t(2) + t(8) ) = 30 = ( (a + d) + (a + 7d) ).
Solve for the algebraic value, of the 1st term, a, from the arithmetic summation equation:
AS_Sum( t(2) + t(8) ) = 30 = (2a + 8d),
Divide throughout, by 2:
15 = (a + 4d), → t(5) = (a + 4d),
Therefore, t(5) = 15, → a = (15 — 4d).
Solve for the algebraic value, of the 1st term, a, from the arithmetic terms divided as geometric terms:
AS_t(1) = a, AS_t(5) = 15 = (a + 4d), AS_t(10) = (a + 9d) are in a geometric sequence. Therefore, to determine the constant difference between arithmetic terms, solve for the constant difference the same way, as the constant ratio, would be solved! That is:
Constant ratio = (term_2 / term_1) = (term_3 / term_2) = … = (term_n / term_(n — 1)),
So,
Constant difference = (term_2 / term_1) = (term_3 / term_2) = … = (term_n / term_(n — 1)),
( (a + 4d) / a ) = ( (a + 9d) / (a + 4d) ),
Cross multiply:
( (a + 4d) (a + 4d) ) = ( a (a + 9d) ),
( (a^2) + (4ad) + (4ad) + (16d^2) ) = ( (a^2) + (9ad) ),
( (a^2) + (8ad) + (16d^2) ) = ( (a^2) + (9ad) ),
The (a^2) on both sides of the equal sign, cancel each other:
( (8ad) + (16d^2) ) = (9ad),
Group like terms:
(16d^2) = ( (9ad) — (8ad) ),
(16d^2) = (ad),
Divide throughout, by d:
a = ( 16d ).
Solve for the value, of the constant difference, d, between consecutive terms:
a = (15 — 4d), → a = 16d,
16d = (15 — 4d), → 20d = 15,
d = (15 / 20), → d = (3 / 4).
Solve for the value, of the 1st term, a:
a = (15 — 4d), → a = (15 — 4(3/4)),
The 4, multiplying the numerator of 3, cancels the denominator, of 4:
a = (15 — 3), → a = 12.
ALTERNATIVELY:
a = 16d, → a = (16 (3/4)),
a = ((16 3) / 4), → a = (48 / 4),
a = 12.
P.S.
The arithmetic terms, from the question, are:
t(1) = 12, t(5) = 15, t(10) = (75/4).
The arithmetic sequence is:
12, (51/4), (54/4), (57/4), 15, (63/4), (66/4), (69/4), 18, (75/4), (78/4), (81/4), 21, (87/4), (90/4), (93/4), 24, …,
The general term rule, of this arithmetic sequence, when n = 1, 2, 3, …, is:
t(n) = ( 12 + ((n — 1) (3/4)) )
OR
t(n) = ( (3n + 45) / 4 ).
NOTE:
AS_t(1) = a = 12 = GS_t(1) = a = 12, AS_t(5) = 15 = GS_t(2), and, GS_t(3) = (a (r^2)),
The constant ratio is:
Constant ratio = (term_2 / term_1) = (term_3 / term_2) = … = (term_n / term_(n — 1)),
(15 / 12) = ( (12 (r^2)) / 15 ),
Cross multiply:
(15^2) = ((12^2) (r^2)),
Drop the square, throughout:
15 = (12 r), → r = ± (15/12),
r = ± (5 / 4).
NOTE: — (5 / 4) is discarded, since AS_t(5) = + 15 = GS_t(2), which means, the geometric sequence is not alternating!
The geometric terms, from the question, are:
t(1) = 12, t(2) = 15, t(3) = (75/4).
The general term rule, of this geometric sequence, when n = 1, 2, 3, …, is:
t(n) = ( 12 ((5/4)^(n — 1)) )
OR
t(n) = ( 12 ((4/5)^(1 — n)) )
OR
t(n) = ( 3 (2^(4 — 2n)) (5^(n — 1)) ).
Upvote ·
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Friendly Fire
Just a nerd with too much time · Author has 63 answers and 9K answer views
·1y
Related
What is the sum of the first 10 terms of the geometric sequence: 2, 6, 18?
tldr: answer is 59048
assuming that we are multiplying two by three each time, we can make the formula which will give us the value of the nth term in a sequence
Using summation notation, the sum of the first “N” terms of the geometric sequence is given by this equation
Now if we make N=10, then we would get an answer of 59048,
Or you can just add it up by hand
Continue Reading
tldr: answer is 59048
assuming that we are multiplying two by three each time, we can make the formula which will give us the value of the nth term in a sequence
Using summation notation, the sum of the first “N” terms of the geometric sequence is given by this equation
Now if we make N=10, then we would get an answer of 59048,
Or you can just add it up by hand
Upvote ·
Neil Morrison
I do all my own differentiation! · Author has 9.5K answers and 27.7M answer views
·6y
Related
The sum of the first three terms of a geometric sequence is 8 and the sum of the first six terms is 12. What is the common ratio?
Upvote ·
99 12
9 4
Bruce Stevens
Author has 2.8K answers and 1.5M answer views
·1y
Related
What is the geometric series of the first 10 terms of the sequence 4, 12, and 36?
Set the question into GS equations:
t(1) = 4 = a,
t(2) = 12 = ( a r ),
t(3) = 36 = ( a (r^2) ).
Solve for the constant ratio, r, between consecutive terms:
Constant ratio = (term_2 / term_1) = (term_3 / term_2) = … = (term_n / term_(n — 1)),
r = (12 / 4) = (36 / 12),
r = ( 3 ) = ( 3 ).
Show the next seven terms:
t(4) = ( a (r^3) ) = ( 4 (3^3) ) = ( 4 27 ) = 108,
t(5) = ( a (r^4) ) = ( 4 (3^4) ) = ( 4 81 ) = 324,
t(6) = ( a (r^5) ) = ( 4 (3^5) ) = ( 4 243 ) = 972,
t(7) = ( a (r^6) ) = ( 4 (3^6) ) = ( 4 729 ) = 2916,
t(8) = ( a (r^7) ) = ( 4 (3^7) ) = ( 4 2187 ) = 8748,
t(9)
Continue Reading
Set the question into GS equations:
t(1) = 4 = a,
t(2) = 12 = ( a r ),
t(3) = 36 = ( a (r^2) ).
Solve for the constant ratio, r, between consecutive terms:
Constant ratio = (term_2 / term_1) = (term_3 / term_2) = … = (term_n / term_(n — 1)),
r = (12 / 4) = (36 / 12),
r = ( 3 ) = ( 3 ).
Show the next seven terms:
t(4) = ( a (r^3) ) = ( 4 (3^3) ) = ( 4 27 ) = 108,
t(5) = ( a (r^4) ) = ( 4 (3^4) ) = ( 4 81 ) = 324,
t(6) = ( a (r^5) ) = ( 4 (3^5) ) = ( 4 243 ) = 972,
t(7) = ( a (r^6) ) = ( 4 (3^6) ) = ( 4 729 ) = 2916,
t(8) = ( a (r^7) ) = ( 4 (3^7) ) = ( 4 2187 ) = 8748,
t(9) = ( a (r^8) ) = ( 4 (3^8) ) = ( 4 6561 ) = 26244,
t(10) = ( a (r^9) ) = ( 4 (3^9) ) = ( 4 19683 ) = 78732.
P.S.
The general term rule, of this geometric sequence, when n = 1, 2, 3, …, ∞, is:
t(n) = ( 4 (3^(n — 1)) )
OR
t(n) = ( 4 / (3^(1 — n)) )
OR
t(n) = ( (4/3) (3^n) ).
Upvote ·
9 1
Related questions
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What is the sum of the first 30 terms of a geometric sequence?
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If the sum of the first 17 terms is 289, what is the sum of its first 9 terms?
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MEDLINE plus: building and maintaining the National Library of Medicine's consumer health Web service
Naomi Miller
Naomi Miller, M.L.S.
1 National Library of Medicine, 8600 Rockville Pike, Bethesda, Maryland 20894
Systems Librarian
Find articles by Naomi Miller
1, Eve-Marie Lacroix
Eve-Marie Lacroix, M.S.
1 National Library of Medicine, 8600 Rockville Pike, Bethesda, Maryland 20894
Chief, Public Services Division
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1, Joyce E B Backus
Joyce E B Backus, M.S.L.S.
1 National Library of Medicine, 8600 Rockville Pike, Bethesda, Maryland 20894
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Copyright and License information
1 National Library of Medicine, 8600 Rockville Pike, Bethesda, Maryland 20894
Roles
Naomi Miller: M.L.S., Systems Librarian
Eve-Marie Lacroix: M.S., Chief, Public Services Division
Joyce E B Backus: M.S.L.S., Systems Librarian
Received 1999 Jun; Accepted 1999 Aug.
Copyright © 2000, The Medical Library Association
PMC Copyright notice
PMCID: PMC35193 PMID: 10658959
Abstract
MEDLINE plus i s a Web-based consumer health information resource, made available by the National Library of Medicine (NLM). MEDLINE plus has been designed to provide consumers with a well-organized, selective Web site facilitating access to reliable full-text health information. In addition to full-text resources, MEDLINE plus directs consumers to dictionaries, organizations, directories, libraries, and clearinghouses for answers to health questions. For each health topic, MEDLINE plus includes a preformulated MEDLINE search created by librarians. The site has been designed to match consumer language to medical terminology. NLM has used advances in database and Web technologies to build and maintain MEDLINE plus, allowing health sciences librarians to contribute remotely to the resource. This article describes the development and implementation of MEDLINE plus, its supporting technology, and plans for future development.
INTRODUCTION
The Web has dramatically changed information seeking and nowhere is this more evident than in the field of medicine. Health information is one of the most frequently requested types of information on the Web. A recent study by Deloitte & Touche and VHA has found that 17.5 million adults in the United States, or 43% of the 40.6 million who use the Internet, are searching for health information . The number of health Web sites is now estimated at 15,000 . Clearly, Americans have both the need and the desire to find answers to their health questions. Many organizations who do not normally provide health information are forming partnerships to add health content to their sites. Health information is now featured at such disparate sites as the Washington Post, Netscape, and Cable News Network (CNN).
Although the National Library of Medicine (NLM) has traditionally focused its services on health professionals, the relatively recent explosive growth of the Internet and Web has made NLM resources increasingly attractive to the general public as well. NLM's first Web site, launched in October 1993, described selected NLM programs and provided links to samplers of activities such as the Visible Human Project. In 1996, the site was reorganized and expanded to cover NLM's programs more comprehensively for a wider general audience. Early examinations of the Web search logs showed that users most often searched for names of diseases or other medical terms, probably because they expected the Web site to contain full-text medical information .
When NLM made MEDLINE on the Web free of charge in June 1997, the general public immediately became an important NLM user group. Today 30% of MEDLINE searching performed from NLM's Web site is by students and the general public . MEDLINE contains more than nine million citations and abstracts to articles largely written by and for a health professional audience. In early 1998, NLM added twelve consumer health journals to MEDLINE to increase its coverage of information written specifically for the general public. In October 1998, NLM launched MEDLINE plus, a Web-based resource designed to provide consumers with immediate access to authoritative health information. The release of MEDLINE plus represented a significant step for NLM in using Web technology to provide consumer-level information. The purpose of this article is to describe MEDLINE plus: its goals, development, implementation, supporting technology, current status, and plans for the future.
GOALS
A multitude of Web sites provide consumer health information, but these sites vary greatly in the type, depth, and quality of information they provide . Determining whether the information is reliable and up to date is often difficult. In addition, the Web has spawned duplication of information, with many sites republishing documents from freely available information sources, such as the National Institutes of Health (NIH) and other federal agencies. The increasingly commercial nature of the Web further complicates the search for reliable, unbiased information.
NLM's primary goal in developing MEDLINE plus was to help consumers searching the Web find answers to health questions. MEDLINE plus was designed to be a well-organized, selective, Web resource that directs consumers to full-text health information available from NLM, NIH, and other reliable sources. In addition, MEDLINE plus was designed to assist consumers with searching MEDLINE through preformulated searches on selected health topics. The site's use of consumer language and simple search interface were designed to facilitate use by the general public.
DESIGN AND DEVELOPMENT
As a first step, the MEDLINE plus team drafted guidelines for selecting resources. These Selection Guidelines are included as a link on MEDLINE plus. NLM chose to link to relevant Web-based information from NLM, NIH, and other government health agencies as a first priority. Additional links to other organizations were selected if the sites included unique information that was primarily educational and not fee-based or commercial in nature.
The prototype design was developed by a small team within the NLM Public Services Division, with assistance from NLM's graphic artist. The focus of MEDLINE plus is on national (i.e., United States) resources, not state, local, or international information sources, though there may be exceptions in specialized areas. The sections are:
▪ Health Topics—a list of diseases, conditions, and other medical topics for which NLM has organized links to reliable health information.
▪ Dictionaries—general health-related dictionaries and lists of dictionaries to assist the public with spelling and definitions of medical terms.
▪ Databases—NLM databases and other federal databases that may be useful to the general public. This section also links to the home pages of select Web sites containing comprehensive and reliable information, such as NetWellness, healthfinder, and Mayo Clinic Health Oasis.
▪ Organizations—federal government resources such as the Food and Drug Administration (FDA) and Centers for Disease Control and Prevention (CDC), as well as national associations such as the American Academy of Pediatrics and the American Cancer Society, which have organized information targeted to the general public. Most disease-specific organizations are not included in this section, but are included in the specific Health Topic.
▪ Clearinghouses—federal government and government-sponsored clearinghouses, funded to provide health information to the general public. Other select clearinghouses such as the National Information Center on Deafness at Gallaudet University are also included. Though traditionally these clearinghouses have mailed pamphlets, videos, and other information upon request, they are currently in transition to providing full-text information and databases for direct access through the Web.
▪ Publications/News—federal government newsletters or magazines written mainly for the general public, as well as links to medical atlases, encyclopedias, and major news Web sites with health features such as CNN and MSNBC.
▪ Libraries—libraries offering services to the general public that also have Web sites, including consumer health libraries and selected public libraries; lists of other libraries or library services, such as the National Network of Libraries of Medicine (NN/LM) and U.S. state libraries.
▪ Directories—directories of health professionals and health care facilities.
Each of these eight sections organizes links to reputable Web sites with information useful to the general public. The major focus of MEDLINE plus, however, is to identify full-text sources on health topics for reading online, that is, article- or pamphlet-level information. The abundance of such information from the eighteen individual institutes of the NIH provides a solid foundation for MEDLINE plus. Other organizations such as the CDC, the FDA, and many national health-related government and professional organizations and associations have been rich sources of excellent information for the public.
DEVELOPING HEALTH TOPICS
As noted earlier, analysis of search terms used on the NLM home page has consistently shown that the majority of searches (90% or more) were for specific diseases, conditions, or other medical terms. Based on this finding, NLM staff anticipated that the Health Topics section would form the core of MEDLINE plus. The original list of health topics was developed from an analysis of a sample of search terms consumers used. Staff of NLM's Lister Hill National Center for Biomedical Communication (LHC) performed mapping and analysis on a five-week sample of searches. The search log contained 87,423 searches that the LHC staff mapped to 56,905 concept terms using the Unified Medical Language System (UMLS), Metathesaurus, SPECIALIST Lexicon, and lexical programs . Of these, over 8,446 occurred more than once and 2,676 could be mapped directly to MeSH terms. For example, shingles, zoster, and herpes zoster were mapped to herpes zoster; herniated disk and slipped disk were mapped to intervertebral disk displacement.
Once mapped, the terms were ranked by search frequency. The most frequently searched topics included diabetes, shingles, prostate, hypertension, asthma, lupus, fibromyalgia, multiple sclerosis, cancer, and other diseases. Drug topics such as Viagra and Zoloft, and alternative medicines such as St. John's Wort were also frequently searched. A target health topic list of more than 300 was developed based on the analysis of user searches to ensure that the health topics of interest to users were included in MEDLINE plus.
Each Health Topic page was organized into three major sections: NLM/NIH Resources, including one or more preformulated MEDLINE search; Other Federal Government Resources; and Government and Other Resources by Category. Each Government and Other Resources by Category section may contain up to twenty-one subcategories such as Treatment, Diagnosis, Dictionaries/Glossaries, and Español/Spanish.
NLM's reference librarians created preformulated searches of MEDLINE for each health topic using advanced features of the PubMed search interface. For example, for the diabetes health topic page, preformulated searches on subtopics such as “reviews, practice guidelines, and clinical trials,” “diagnosis,” “diabetic foot,” “diabetic retinopathy,” and “therapy” were available. These searches when run against PubMed produced a focused list of the most recent citations from English language journals with articles most likely to be readable by consumers. Using PubMed, users could use search techniques such as “find related” to locate other articles of interest similar to those presented.
Initially, each Health Topic was constructed as one HTML page. Very early in development, NLM staff realized that if hundreds of Health Topic pages were to be developed, maintaining them as static HTML pages would not be practical. Even with a single topic page, links broke regularly. Within the first eight topics developed, inconsistencies in format and approach by individual creators became evident. Individual creators pointed to the same resource on different topic pages—for example, heart attack and stroke—but used varying site names and descriptions or pointed to different uniform resource locators (URLs) for the same resource—for example, the home page versus the introduction section to a pamphlet. A more efficient way to build and maintain MEDLINE plus was essential if it was to continue to grow and maintain a high level of quality and consistency.
PROCESS AND TECHNOLOGY
The MEDLINE plus development team turned to NLM's Office of Computer and Communications Systems (OCCS) for assistance with developing a system for creation, review, and maintenance of MEDLINE plus. OCCS recommended using a Web browser compatible database. The database would help ensure consistency of terms and categories and would allow site records to be created and maintained centrally for display on multiple Web pages. The system OCCS implemented used technologies that combined an interactive Oracle database with a Cold Fusion Web forms-based input system. The architecture of the system is outlined in Figure 1.
Figure 1.
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MEDLINE plus system architecture
Using this system, health sciences librarians on contract to NLM contribute to MEDLINE plus using a Web browser. The technology allows them to work from any location with Internet access. Currently most MEDLINE plus-contributing librarians work in states outside Maryland, where NLM is located. The system architecture has allowed NLM to take advantage of expertise at other institutions but maintain the high degree of consistency and quality control afforded by the database.
SELECTION AND REVIEW
The process for creating MEDLINE plus records involves selectors and reviewers. Selectors use Web-based forms to contribute records; reviewers approve records for display or return them to selectors for editing. NLM staff and outside contractors are both selectors and reviewers.
Selectors use a Web form to enter records. The form combines free-text fields and fields using lists, radio buttons, and check boxes. Health topics and subcategories are chosen from lists; the more than 300 health topics currently under development can be searched as well as displayed alphabetically. Organization names are subject to authority control. Before the public release of a health topic, a staff member from NLM's cataloging section validates the names of all organizations associated with the records in the database. Selectors may search organization names directly or use a selection list; cross-references are provided to acronyms and variant forms of the organization name. Selectors may enter a new, unapproved organization name, but it will not be released for display on MEDLINE plus until it has passed cataloging review.
Successful completion of the input form is dependent upon a set of rules. For example, the system is able to detect a URL that is a duplicate of one already in the database. If a selector has created a record but has omitted a required field (e.g., name of site, type of site, URL), the system prompts the selector to return to the form to complete the required field. An example of a form is shown in Figures 2, 3, and 4.
Figure 2.
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Top of MEDLINE plus entry form: section of form showing site identification and display fields
Figure 3.
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MEDLINE plus entry form: section of form showing topics window, type of site, subcategory fields, organization window
Figure 4.
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MEDLINE plus entry form: section of form showing unapproved organization, special site type, and PubMed search fields
Reviewers are able to view a list of submitted forms and may view, edit, approve, or return individual records to a selector for reworking. Figure 5 shows a Web form of records pending review. Reviewers may also release topics for public display, move records among topics, manage records for organizations, manage topics by adding topics or changing the names of topics, and manage subcategories by adding subcategories or changing the names of subcategories. The complete administrative menu is shown in Figure 6.
Figure 5.
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MEDLINE plus pending records screen
Figure 6.
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MEDLINE plus site administration menu
The system used by the development team when reviewing individual records and pages uses the Oracle database. One advantage of this approach is that selectors and reviewers can immediately see the results of approving a site record for display or check on the status of a topic that is still in process. To ensure optimum performance on the public site, however, the public does not attach directly to the Oracle database. The system generates static HTML pages nightly from the Oracle database. These pages are indexed for searching and load quickly on a user's remote browser.
SITE MAINTENANCE
To date, the major effort has been the development of the internal structure and content of MEDLINE plus. Through systematic “combing” of major government and professional organization Web sites and through the serendipitous discoveries of the many librarians now building this site, the Web site topics have been continuously updated. An automatic link checker is in place to identify broken links, which are fixed daily. Automated Web page checkers that monitor specific URLs for changes to pages are also used to keep MEDLINE plus up to date. For example, the NIH “What's New” pages are monitored for new documents. Seven months after its introduction, a schedule for systematically updating all sections of MEDLINE plus is being implemented. Upon release of a health topic, a release date is associated with the record. These release dates are used to schedule periodic updating. All major sections of the site are continuously updated, but a complete review will be done every six months.
CURRENT STATUS
MEDLINE plus made its official debut on October 22, 1998, in conjunction with NLM's consumer health pilot project with public libraries. The pilot project was NLM's first major initiative to increase the public's awareness of and access to health information on the Internet. NLM has worked with thirty-nine public library organizations at some 200 sites in five states and the District of Columbia. NLM, in partnership with medical libraries, has provided training and support to librarians in searching MEDLINE and MEDLINE plus so that they can better assist patrons in finding health information.
NLM has obtained feedback on MEDLINE plus from a number of sources. Feedback from the public libraries participating in NLM's pilot project has been positive. NLM has implemented changes based on suggestions from users, such as shortening the Health Topics pages by listing some government Web sites only under the categorized list. The librarians and patrons who have used MEDLINE plus have consistently reported that as more topics are added, the site becomes more useful to them. At this writing, the health topics covered have increased from twenty-two to more than eighty, and this number is expected to grow quickly to complete the identified topics. A “feedback” button has recently been added to the site and dozens of topic suggestions have been received, along with numerous compliments and other suggestions.
CHALLENGES
Experience with and feedback on MEDLINE plus have supported NLM's plan to continue its development and to begin to promote this service nationwide. Though users are generally very pleased with the organization and authoritative information provided through MEDLINE plus, there is room for improvement. The most common recommendation from users is that MEDLINE plus should contain more topics. Completing over 300 health topics with preformulated MEDLINE searches by the end of 1999 is a priority.
Complicating development is the paucity of authoritative Web information on some of the topics most frequently searched by consumers. At the same time, avoiding duplication is difficult when multiple Web sites post the same information. Sometimes different Web servers from the same organization will publish the identical document with different URLs. Though NLM does not plan to create new information content, NLM does plan to encourage NIH and other appropriate health organizations to create information for access through MEDLINE plus.
NLM has continued to analyze the terms input by users of the NLM main Web site and the MEDLINE plus site to keep up on what users want. The MEDLINE plus search engine, ht://Dig , offers “fuzzy” searching and soundex options. A search for “Alzimers” presents the user with the link to the Alzheimer's Health Topic page. NLM staff are optimizing the ht://Dig search engine and testing other search engines to optimize retrieval. An analysis of the terms has provided the development team with an insight into the difficulty many Web users have with the medical language. In a follow-up study of over 300,000 queries of the NLM Web site, NLM's Lister Hill Center staff has reported numerous examples of misspelled words. Words such as fibromyalgia, gynecology, and prostate are often misspelled. Eponymic terms such as Crohn's disease are misspelled in a number of ways. NLM will explore the development of a terminology server to mediate between user terminology and the terminology used in medical sources .
NLM has begun to investigate whether the database methodology being developed for MEDLINE plus can be used by other organizations organizing health information on the Web for other audiences. Shared records, vocabulary, and other tools, as well as authority work done by NLM, may prove feasible to reduce redundancy among efforts and allow other organizations to combine what NLM has done on a national level with consumer health resources at the local level.
The major development for MEDLINE plus will be completed during 1999, but as with the Web itself, NLM expects MEDLINE plus to continue to expand as a dynamic, useful resource for the general public, health sciences librarians, and health professionals.
REFERENCES
Deloitte N, Touche. VHA, Inc. 43% of consumers use Internet for health information. [Web document].Irving,. TX: Deloitte & Touche, April 5, 1999. [rev 8 Apr 1998; cited 7 Sep 1999]. . [Google Scholar]
Grove AS. Internet health day keynote speech. [Web document]. Intel. Corporation, 1999. [rev 1999; cited 7 Sep 1999]. . [Google Scholar]
Miller N. Improving the NLM home page: from logs to links. Poster presented at Medical Library Association annual meeting, May 26–27, 1997. [Web document]. Bethesda, MD: National Library of Medicine, 1997. . [Google Scholar]
Online usage statistics smashed; free MEDLINE rewrites NLM record book. NLM Newsline. 1998;53(1):1–2. [Google Scholar]
MEDLINE to index selected consumer health publications. NLM Newsline 1998531 2 [Google Scholar]
Jadad AR, Gagliardi A. Rating health information on the Internet: navigating to knowledge or to Babel? JAMA. 1998 Feb 25;279(8):611–4. doi: 10.1001/jama.279.8.611. [DOI] [PubMed] [Google Scholar]
National Library of Medicine. MEDLINE plus: selection guidelines. [Web document]. Bethesda,. MD: The Library, 1998. [rev 22 Jan 1999; cited 7 Sep 1999]. . [Google Scholar]
National Library of Medicine. Unified Medical Language System. [Web document]. Bethesda, MD: The Library. . [Google Scholar]
ht://Dig. WWW search engine software. [Web document]. The ht//Dig Group, 1995–99. [rev 1 Jun 1999; cited 7 Sep 1999]. . [Google Scholar]
McCray AT, Loane RF, Browne AC, and Bangalore AK. Terminology issues in user access to Web-based medical information. AMIA 1999 Annual Symposium, Washington DC. 8November1999 Session S-16. [PMC free article] [PubMed] [Google Scholar]
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INTRODUCTION
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15368 | https://courses.physics.illinois.edu/phys436/sp2011/p436web/p436web/special_rel3-notes.pdf | 1 Special Relativity 3 • 4-momenta – Scalar derivative of a 4 vector interval – Use of proper time – Interpretation of zero component – Non-relativistic limits of energy and momenta • Rest mass as the 4-vector invariant “length” – 4 momenta from boosts • Subatomic energy units – Calculating velocity from kinetic energy – What determines the energy of a photon?
• Doppler shifts using photon 4-momenta • Conservation of 4-momenta – Deducing the mass of a parent from the energy and momenta of its daughters.
– Analysis of Compton Scattering • Center of momentum (cm) frame – Computing the cm energy using invariants – Computing the energy threshold for subatomic processes – Describing collisions in the cm frame • Computing the cm energies from s and m 2 What will we do in this chapter?
We introduce the idea of a photon with an energy proportional to its frequency and use this idea to derive a more general form for the Doppler shift by transforming the photon energy in an arbitrary frame using our invariant trick.
We discuss the conservation of 4-momenta in a particle collision and discuss how the mass of a short-lived subatomic parent can be deduced from the energy and momenta of its daughters. We analyze Compton scattering or the scattering of x-rays from electrons in metal.
We introduce the concept of the center of momentum frame and use it determine the minimum (threshold) energy required to allow subatomic collisions to take place. Finally we show how to boost into the cm frame and compute the incident and final particle cm energies and momenta from the cm energy and the particle masses.
In this chapter we discuss the concept of 4 momentum. We convert Newtonian 3-momentum into a 4 vector using a velocity based on differentiating the interval with respect to proper time rather than time. In the process we obtain a 4-vector with a zero component of energy and momenta as the 1,2,3 components. The relativistic energy includes a rest energy (mc2) as well as a kinetic energy term which goes over to the Newtonian form (T = mv2/2) as v/c →0.
We show that dot product of the 4 momentum with itself is the square of the particle’s rest mass and show that a Lorentz boost of the rest mass creates the relativistic energy and momentum of the particle.
We discuss eV based units and give the rest energy of the some common subatomic particles. We show how the velocity of a particle can be computed from its kinetic energy.
3 4-momentum ( ) A natural candidate for 4-momenta is p mv This result should raise some questions. Does p have anything to do with conventional momentum? And what's up with the zero component p m p mc mv p η γ γ = ⇒ = → = r r r % % % % ( ) ( ) 0 1/2 2 2 2 2 0 0 2 0 mc ? We can answer both by expanding the answer to second order in . 1 We begin with . Recall 1 1 / 2 = 1 / 2 1-1 / 2 Defining = , 2 we have c x x mv p mc mc mc p p p c c c p mc γ β γ γ β β γ β − = − ≈+ → ≈+ ⎛ ⎞ = ≈ + ≈ + + → = ⎜ ⎟ ⎝ ⎠ = ≈ + E E E r % L 2 and 2 mv p mv mv γ = ≈ r r r We see the usual Newtonian 3 momentum. The cp0 component contains the usual kinetic energy plus an “energy” of the form E=mc2. This is called the rest energy. We will show later that one can indeed convert rest energy into kinetic energy.
4 What is the length of 4-momentum?
( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 What is if ?
1 Hence the squared length of the momentum is the squared rest energy/c of the particle carrying the momentum. In this formulation of rel p p p mc mv p p mc mc p p mc mc γ γ γ γ β γ β • = • = − = • = − = r % % % % % % % ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 ativistic kinematics, We will usually write the 4 momentum as / mass is th and implies / ) or e important inv For 1 1 1 ariant.
2 p c p p p mc c p mc cp mc cp cp mc mc mc cp mc mc = • = − = = + ⎛ ⎞ << = + ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ ≈ + ⎜ ⎝ E (E E E E r % % % 2 2 2 2 p mc m ⎛ ⎞ ≈ + ⎜ ⎟ ⎟ ⎜ ⎟ ⎠ ⎝ ⎠ As a 4 vector, p can be transformed to another frame using the Lorentz boost. As an example, consider transforming p from its rest frame to the lab frame.
Let the primed frame be the rest frame of mas % % ( ) s .
In the mass's rest frame we have p'= mc 0 since in this frame the mass has no momentum but only rest energy. m r % v ' x x 0 o 1 x x 2 p /c p ' p p p ' mc mc mc 0 mc m In general / = mc = mc and p= m v c v γ βγ βγ γ γ βγ γ γ βγ γ γβ γ γ γ γ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ = = = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ = = ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠ → E E E r r We thus have obtained exactly the same form for 4-momentum by boosting the particle’s rest-mass to the lab.
5 Units and velocity calculation Relativity is often used to describe subatomic particles which often travel at relativistic velocities (i.e. velocities approaching the speed of light). It is much more convenient to use the rest energy (m c2) for these particles rather than their mass and specify this energy in units based on electron volts or eV. An electron (or any charged particle with a charge of |e| = 1.6×10-19 C acquires a kinetic energy of 1 eV when accelerated through a 1 volt electrical potential difference. 1eV = 1.6×10-19 Joules. If an electron with m c2 = 0.511 MeV is accelerated through 1 million Volts it acquires a lab energy of 1.511 MeV.
V 2 6 2 2 6 Accelerating an electron from rest through 1 million volts.
mc 0.511 10 eV If V=1 MeV = 1 10 eV T = 1 MeV and = 1.511 MeV T mc eV mc = × = + = + × E E ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 How fast does the 1 MeV electron travel? p= mv = mc Thus 1 For the case of 1.511 0.511 1 0.941 1.511 v cp c cp mc cp mc mc v cp mc c MeV v c γ γ → = − = → = − − ⎛ ⎞ = = = −⎜ ⎟ ⎝ ⎠ = ⎛ ⎞ = − = ⎜ ⎟ ⎝ ⎠ E E E E E E E E E We illustrate a typical kinematics calculation for the velocity of a 1 MeV electron (i.e. an electron w/ a kinetic energy of 1 MeV).
It is clear from our formula for v/c that an infinite amount of energy is required to accelerate a massive particle up to v=c.
6 More particles The electron is a real light weight. We have been introduced to the muon which has a rest energy of mμ c2 = 105 MeV. Another common particle is the proton with a rest energy of mp c2 = 940 MeV. At present protons and antiprotons are accelerated Tevatron at Fermilab with a kinetic energy of 1 TeV or 1000 GeV or 1 million MeV or T=1×1012 eV How fast is a 1 TeV proton ?
2 2 2 2 2 2 2 2 2 6 7 Very clos We go to the limit where T>> 1 1 1 2 1 1 940 1 1 2 2 1 10 1 4.4 10 e to c!
mc v mc mc c T mc v mc c T v c − ⎛ ⎞ ⎛ ⎞ = − ≈− ⎜ ⎟ ⎜ ⎟ + ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ ≈− = − ⎜ ⎟ ⎜ ⎟ × ⎝ ⎠ ⎝ ⎠ ≈− × E Another important particle is the photon which is the carrier of light waves. Our formula shows that the photon must be massless in order to travel at the speed of light with finite energy. Since the mass and velocity of the photon is set, something else must control the photon’s energy … this is the frequency of the photon.
( ) The energy of a photon is E= where is Planck's constant and is the it's angular frequency. Since 2 2 / , 2 / 1240 / . A visible light photon with 600 has E 2 eV. A photon is very f c c eV nm nm γ ω ω ω π π λ π λ λ λ = = = = = ≈ E h h h ( ) -1 -19 20 2 2 2 small -- a 60 Watt bulb puts out 60 J s /(2 1.6 10 J/ ) 2 10 /s Since the photon is massless / 0 ˆ ˆ and cp = , and p = / / where k is photon's direction. Boosting to a different fr c p c ck γ γ γ ω ω × × ≈ × − = E E % h h ame changes and thus changes . This is another way of looking at the Doppler shift.
ω E 7 The Doppler shift from 4 momenta Θ % A % ( ) 0 0 r ( ) ct r r In our first lecture on Special Relativity, we obtained the above formula for transforming the time ( or A0 component) of interval A into the Θ reference frame. The derivation of this formula only depended on dot product invariance so we can use it for any 4-vectors Here is a transcription for photons.
( ) c A Θ Θ• = Θ•Θ E % % % % Θ % pγ % ( ) 0 0 r ( ) ˆ / c p k c c ω ω ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ E h h r ( ) t A c Θ Θ• = Θ•Θ % % % % ( ) 2 2 2 ( ) is ideal choice for Recall ( ) (1 ) = where ˆ and c v c c p p c c p k c v c c η γ γ γ η η η η γ γ β η η η η ω ω η γ 2 2 Θ ⇒Θ = • = − = − = • • = • ⎛ ⎞ = = ⎜ ⎟ ⎝ ⎠ E % % % % % % % % % % % % r h h % % 8 Check of general Doppler formula ( ) ( ) ( ) ( ) =c 1 ; ˆ ˆ 1 ˆ ˆ 1 v c v p c p v k c c c c c v k v c k c c k c c c γ η γ η η η γ γ η ω ω ω γ γ ω ω γ ω ω γω ω ⎛ ⎞ = = ⎜ ⎟ ⎝ ⎠ • ⎛ ⎞ = = − • ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ ⎜ ⎟ ⎛ ⎞ • → = − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⇒ ⎝ ⎛ ⎞ • = − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎠ r r % % r % % h h h r h h r h h ( ) (rec) (send) 2 (rec) (send ( (rec) ( rec) (send) (send) (rec) (send s ) en ) d) ˆ ˆ ˆ , 1 1 1 1 1s ˆ ˆ If , 0 1 2nd orde t orde r r 2 ˆ 1 v v c c v k k k x v v x v c v k c v β ω ω γ ω β ω ω γω β ω ω ω ω β γ ω << << ⊥ • = = − + ⎛ ⎞ = + = ⎜ ⎟ − ⎝ ⎠ ⎯⎯⎯ ⎛ ⎞ • = − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ = → = ⎛ ⎞ ⎯⎯⎯ → ⎝ → ⎜ + + ⎟ ⎠ r r r r v ˆ x Θ v ˆ x ˆ k Θ ˆ k 9 Momentum Conservation We are use to the idea of conservation of momentum in all collision problems and the conservation of energy in elastic collision problems. In the subatomic world, all collisions conserve energy since there are no internal degrees of freedom which can absorb energy in the form of heat or friction. Thus in a subatomic collision, all 4 components of 4 momenta are conserved. A % B % C % D % A B C D = + + % % % % In subatomic physics, particles often change identities during an interaction and “matter” can turn to “energy”.
0 0 2 -17 0 An example is the decay where the has a rest energy of m 135 MeV and a mean lifetime of 8.4 10 . Lets find the photon energies in the rest frame.
c s π π γγ π π → = × ( ) 1 1 / c ω ω − h h ( ) 1 2 1 1 1 2 2 2 1 2 1 2 2 2 1 1 2 0 0 2 / 2 p p c mc mc mc mc ω ω ω ω ω ω ω ω ω − + + = → = → = + = → + = → = ∴ = = E E r h h r r h h h h h 0 π ( ) 0 mc 1 γ 2 γ ( ) 2 2 / c ω ω h h ˆ i We thus get two back-to-back 62.5 MeV gamma rays. Matter into energy? What if the π0 is moving to the right in the lab with a velocity v? Solve it with Doppler shifts. γ1 is red shifted; γ2 is blue shifted.
( ) ( ) ( ) 0 0 2 1 1 2 2 2 2 1 2 2 2 2 1 1 2 1 1 1 1 2 1 The energies sum to la 1 2 1 1 b 1 2 mc mc c mc mc m π π β β β β ω ω β β β ω ω β β γ β ω β ω γ ⎛ ⎞ − + + = + = ⎜ ⎟ ⎜ ⎟ + − ⎝ ⎠ + = + + − − = − + = − + = = ∴ E E h h h h h h 0 π 1 γ 2 γ v 10 Reconstructing the parent from the daughters ( ) 2 (2) (2) (2) (2) 2 (1) ( (1) (1 2 2 1) (1) (1) (2) 1) (2) ( 2 ) ˆ ˆ 1 2 1 1 1 ˆ ˆ 2 1 1 ˆ 1 1 ˆ 2 2 1 1 1 1 x x mc x mc p c p p p mc p c p p p x mcx mcx P P c γ γ γ γ γ γ γ γ γ γ γ γ γ β β β β β β β βγ β β β β β ⎛ ⎞ − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ + − − ⎜ ⎟ ⎜ ⎟ − + ⎝ ⎠ ⎛ ⎞ + − ⎜ ⎟ = − = ⎜ ⎟ − − ⎝ ⎠ ⎛ − = + = = + = − + + + ⎞ • = − • ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ + + E E E E E E r r r r r r r % r % ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 ( 2 ) 2 2 1 It works!
mc mc mc mc γ γ βγ γ β = − = − = 0 π 1 γ 2 γ v ˆ x The π0 lives such a short time, it hardly exists at all and must be deduced by measuring the direction and energy of its daughter photons. We can easily do this using conservation and the 4 vector length.
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 0 0 ( ) ( ) 1 2 2 ( ) ( ) ( ) ( ) 1 2 1 2 ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 1 2 2 ( ) ( ) ( ) ( ) 1 2 1 2 2 2 ( ) ( ) 2 ( ) ( ) ( ) ( ) 1 2 1 2 1 2 / / / / / o o p p p m c p p p p p p p p c p c p c c p p m c c p p p p γ γ π π π π γ γ γ γ γ γ γ γ γ γ γ γ γ γ π γ γ γ γ γ γ = + = • = + • + + = + = + + = + − + • + E E E E E E % % % % % % % % % r r % % r r r r r r Lets illustrate this technique by reconstructing the mass of a πο from the energies and directions of two photons which decay along the direction of the πο velocity.
11 Compton Scattering k % p % ' k % ' p % ( ) ( ) 2 4 0 0 vectors: electrons and ' ˆ ˆ ; ; ' ' ' ' photons and ' ; ' ' p p p mc k k k k c c c c p p mc p p k k k k k k ⎛ ⎞ ⎛ ⎞ = = = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ • = = • • = • = E E E' E' ) % % % % % % % % % % % % % % Metal electron –photon scattering ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 1 2 2 ' ' ' ' ; ' ˆ ˆ ' ' ' Eliminate ' ' since don't measure ' s e' ' ' co p mc p k k p p k k k k p k k m k k k k k k k k k k k k c c mc k c p k m m p θ ⎡ ⎤ = = + − = + • − + − • − = − ⎣ ⎦ − = • − • + • = − • = − − • = − − ∴ = + − = − + − E E' EE' EE' EE' E E' % % % % % % % % % % % % % % % % % % % % % % % % % % % ( ) 2 2 1 1 cos cos c mc θ θ − − − → = E E' EE' ( ) ( ) 2 2 6 1 1 1 1 1 1240 1 0 511 10 412 Compton Compton 2 ' cos Now some QM: = 2 2 ' cos cos (x-r ; .
ay) c c mc c mc ev nm nm ev π λ λ θ ω λ π π λ λ θ λ θ λ − − − ⎡ ⎤ ⎛ ⎞ = → = − = = ⎜ ⎟ ⎢ ⎥ ⎣ ⎦ ⎝ ⎠ → − = − = − = = × E E' E EE' E' E h h h h ' e Metal rod θ ' λ λ 12 Center of momentum in a collision A % B % D % C % There exists a reference frame where the vector sum of the 3-momenta of incoming and outgoing particles is zero. We will call this the center of momentum (cm) frame. It is useful to measure the total energy in the cm frame which is easy to do using dot product invariants. The square of E(cm) is often called s. s is the same for the incident and final state.
( ) ( ) ( ) 2 2 ( ) 2 cm In any frame we have: / In cm: 0 0 is called s We can use the cm to compute the minimum (threshold) energy re X A B i i i i cm cm cm i i X i i cm X X i i p p p c p p p c s c p p ⎛ ⎞ = + = ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = = ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ ≡ • = ⎜ ⎟ ⎝ ⎠ ∑ ∑ ∑ ∑ ∑ E E E E r % % % r r % % % quired to create particles. A famous example is the creation of antiprotons via p p p p p p. This is the simplest reaction which conserves baryon number and charge. Both particles have the same rest energy. It is easy to show t → ( ) ( ) (cm) 2 min 4 4 2 2 (cm) 2 (cm) 2 2 min 1 1 hat that E 4 . Use this to find the threshold kinetic energy required for an incident p to collide with a stationary p to create the final p p p p . 4 i i i mc mc cp mc mc = = = = + > > ∑ ∑ E ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 B Thus 4 We can simplify this a bit by expanding: 2 2 16 or 7 Let B be the projectile proton T+ with P = p A B A B A B A B A A A B B B A B A B s c p p p p mc p p p p p p p p p p mc p p mc mc p p mc mc c = + • + > + • + = • + • + • = + • + > • > ⎛ ⎝ % % % % % % % % % % % % % % % % % % r % ( ) ( ) ( ) 2 2 2 2 2 A 2 and A be at rest with P = 0 ; T+ 7 or 6 6(940) 5640 A B mc c p p mc mc mc T mc MeV ⎞ ⎜ ⎟ ⎠ • = > > > > r % % % The antiproton was discovered once such a powerful accelerator existed.
X 13 More on the CM frame ( ) ( ) ( ) ( ) It is easy to find a boost into the cm frame.
We want a frame where A+ / 0 .
A+ A+ / 0 . We boost along the direction of the lab momentum sum.
A B A B B s c B B P P c s c Λ Λ = + ⎛ ⎞ Λ Λ = Λ = Λ + ⎜ ⎟ ⎝ ⎠ = E E r % % r r % % % % r ( ) ( ) ( ) ( ) ( ) s c 0 0 A B A B A B A B A B A B c P P P P c c P P γ βγ βγ γ βγ γ β + ⎛ ⎞ ⎛ ⎞ − ⎛ ⎞⎜ ⎟ ⎜ ⎟= ⎜ ⎟⎜ ⎟ ⎜ ⎟ − ⎝ ⎠ ⎜ ⎟ ⎜ ⎟ + ⎝ ⎠ ⎝ ⎠ + = − + + + = + E E E E E E r r r r r r r A % B % D % C % In the cm frame, the above collision would look like this: A r B r C r D r The two entering momenta are equal and opposite. The two exiting momenta are equal and opposite. β ' x A B P P + r r A P r B P r ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) (cm) A (cm) (cm) 2 (cm) A A It is easy to get the energies and momenta of the particles in the cm frame using our old trick.
/ Since / 0 , / and P A B A A B A c s c A B A B A B s c A B A B s c A c + • + • = = + • + + = ⎛ + • + = = ⎜ ⎝ E E % % % % % % % % % % r % % r % % % % % ⎞ ⎟ ⎠ Entering and exiting |P| are often unequal 14 CM description ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 (cm) A 2 2 2 2 2 2 2 2 (cm) A 2 2 2 2 (cm) (cm) (cm) B A B = It is easy to get from the masses 2 2 And 2 We can easil A A B A B B A mc B A c A B A s s B A s A B A B A A B A B B c s mc mc s mc mc c B A s s mc mc s s + • + • = • = + • + = • + • + • − − + − • = → = + − = + = E E E E E % % % % % % % % % % % % % % % % % % % ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 (cm) (cm) 2 A 2 2 (cm) (cm) 2 B 2 2 2 2 2 2 2 2 (cm) (cm) C D y find momenta magnitudes using Analogous expressions exist for C and D 2 2 A A B B C D D C cp m c cp m c s mc mc s mc mc s s = − = − + − + − = = E E E E A r B r This problem can also be solved without invariant methods by realizing that two incoming momenta have equal magnitude A B p = = r r ( ) ( ) ( ) ( ) 2 2 (cm) (cm) 2 A B 2 2 2 We could solve for p using a quadratic equation.
A B s pc mc pc mc = + = + + + E E |
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Fuel Efficiency
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Sadie has a cousin Nanette in Germany. Both families recently bought new cars and the two girls are comparing how fuel efficient the two cars are. Sadie tells Nanette that her family’s car is getting 42 miles per gallon. Nanette has no idea how that compares to her family’s car because in Germany mileage is measured differently. She tells Sadie that her family’s car uses 6 liters per 100 km. Which car is more fuel efficient?
IM Commentary
The problem requires students to not only convert miles to kilometers and gallons to liters but they also have to deal with the added complication of finding the reciprocal at some point. In the USA we use distance per unit volume to measure fuel efficiency but in Europe we use volume per unit distance. Furthermore, the unit of distance is not simply 1 km but rather 100 km.
The required computation is not particularly long, but requires a firm understanding of the concepts and skills of unit conversion. The task would therefore be suitable for individual assessment, or alternatively, this task could be used as an activity where students work in groups and discuss their strategies. The problem also has elements of a modeling task since students have to decide on a solution method and find the appropriate conversion factors. Most likely, they have not been presented with a similar task before, so there won’t be a clear solution path to follow. To solve the problem it is particularly helpful to follow the units of the quantities involved.
Solutions
Solution: mpg to liters per 100 km
We need to convert miles per gallon to liters per 100 km. So we somehow have to get from miles to km and from gallons to liters. A quick Google search finds that 1 mile = 1.609 km and 1 gallon = 3.79 liters.
We can start by converting miles per gallon into km per liter.
42 miles 1 gallon=42 miles 1 gallon⋅1.609 km 1 mile⋅1 gallon 3.79 liters=67.578 km 3.79 liters
Since we want to find the answer in liters per unit distance, we need to take the reciprocal and convert the units in the denominator into 100 km to finish the computation:
3.79 liters 67.578 km=0.0561 liters per km=5.61 liters per 100 km
This means that 42 miles per gallon is equivalent to 5.61 liters per 100 km, and Sadie’s family’s car is slightly more fuel efficient than Nanette’s family’s car.
Solution: liters per 100 km to mpg
As an alternative solution, we would start with 6 liters per 100 km and converted it to miles per gallon:
6 liters 100 km=0.06 liters 1 km
This is equivalent to
1 km 0.06 liters=1 km 0.06 liters⋅1 mile 1.609 km⋅3.79 liters 1 gallon=39.26 miles 1 gallon.
Nanette’s family’s car gets 39.26 miles per gallon, which is slightly worse than Sadie’s family’s car at 42 miles per gallon.
Fuel Efficiency
Sadie has a cousin Nanette in Germany. Both families recently bought new cars and the two girls are comparing how fuel efficient the two cars are. Sadie tells Nanette that her family’s car is getting 42 miles per gallon. Nanette has no idea how that compares to her family’s car because in Germany mileage is measured differently. She tells Sadie that her family’s car uses 6 liters per 100 km. Which car is more fuel efficient?
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15370 | https://brainly.com/question/59998560 | [FREE] The derivation for the equation of a parabola with a vertex at the origin is started - brainly.com
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The derivation for the equation of a parabola with a vertex at the origin is started below:
(x−0)2+(y−p)2=(x−x)2+(y−(−p))2
(x)2+(y−p)2=(0)2+(y+p)2
x 2+y 2−2 p y+p 2=y 2+2 p y+p 2
If the equation is further simplified, which equation for a parabola does it become?
A. x 2=4 p y
B. x 2=2 y 2+2 p 2
C. y 2=4 p x
D. y 2=4 p y
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We start with the equation
x 2+(y−p)2=(0)2+(y+p)2.
Squaring both sides (which is valid because both square roots are nonnegative) gives
x 2+(y−p)2=(y+p)2.
Next, we expand the squares on both sides:
x 2+(y−p)2=x 2+(y 2−2 p y+p 2)=x 2+y 2−2 p y+p 2,
and
(y+p)2=y 2+2 p y+p 2.
Thus, our equation becomes:
x 2+y 2−2 p y+p 2=y 2+2 p y+p 2.
Notice that the y 2 and p 2 terms appear on both sides and cancel out, leaving
x 2−2 p y=2 p y.
Now, add 2 p y to both sides to collect the y-terms:
x 2=2 p y+2 p y=4 p y.
This is the standard form of a parabola with the vertex at the origin, hence the correct equation is
x 2=4 p y.
Thus, the correct option is the one that corresponds to
x 2=4 p y.
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The derived equation for a parabola with a vertex at the origin is x 2=4 p y. This corresponds to option A in the multiple-choice question. Therefore, the correct choice is option A.
Explanation
To derive the equation of a parabola with its vertex at the origin, we start from the given equation:
egin{aligned} \text{1. }\text{ } \ \ \text{1. } \text{ } \ \ \text{ }\text{1. }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \text{ }\text{ })=0. By simplifying, we obtain:
egin{aligned}
\text{2. }x^2 + (y - p)^2 = (y + p)^2\ \
\text{{At this point, we will expand both sides:}} \
egin{aligned} \text{a. Left Side: } & x^2 + (y^2 - 2py + p^2) \ & = x^2 + y^2 - 2py + p^2 \ \text{b. Right Side: } & 0 + (y^2 + 2py + p^2) \ & = y^2 + 2py + p^2 \ \text{After substituting, the equation becomes: } \x^2 + y^2 - 2py + p^2 = y^2 + 2py + p^2.
\text{Now we can cancel out } y^2 \text{ and } p^2 \text{ from both sides: }
\text{So, we are left with: }
egin{aligned}
x^2 - 2py = 2py.
\text{Next, we organize this so that all } y \text{-terms are on one side: }\
\rightarrow x^2 = 2py + 2py = 4py.
\text{This is the standard equation of a parabola modeled as: }
egin{aligned} \text{So the equation of a parabola is } x^2 = 4py, \ \text{corresponding to option A: } \textbf{Answer: A. }\ \textbf{Answer: A. } \text{Thus, we have found that the correct answer is: }
Examples & Evidence
For example, if p=2, the equation becomes x^2 = 8y, which shows how the graph of the parabola opens upwards and has its vertex at the origin. This illustrates how changing the value of p affects the shape and position of the parabola.
The standard form of a parabola with a vertex at the origin is widely recognized in algebra and geometry textbooks, confirming that x 2=4 p y characterizes such parabolas.
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Community Answer The derivation for the equation of a parabola with a vertex at the origin is started below: [tex][ \sqrt{(x-0)^2+(y-p)^2} = \sqrt{(x-x)^2+(y-(-p))^2} ][/tex] 1. tex^2 + (y-p)^2 = (0)^2 + (y+p)^2)[/tex] 2. tex[/tex] If the equation is further simplified, which equation for a parabola does it become? A. tex[/tex] B. tex[/tex] C. tex[/tex] D. tex[/tex]
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15371 | https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carboxylic_Acids/Reactivity_of_Carboxylic_Acids/Conversion_of_carboxylic_acids_to_acid_chlorides | Conversion of carboxylic acids to acid chlorides - Chemistry LibreTexts
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Reactivity of Carboxylic Acids
Carboxylic Acids
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Mon, 23 Jan 2023 07:35:57 GMT
Conversion of carboxylic acids to acid chlorides
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Contents
1. Home
2. Bookshelves
3. Organic Chemistry
4. Supplemental Modules (Organic Chemistry)
5. Carboxylic Acids
6. Reactivity of Carboxylic Acids
7. Conversion of carboxylic acids to acid chlorides
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Bookshelves
Introductory, Conceptual, and GOB Chemistry
General Chemistry
Organic Chemistry
Supplemental Modules (Organic Chemistry)
Acid Halides
Alkanes
Alkenes
Alkynes
Alcohols
Aldehydes and Ketones
Alkyl Halides
Amides
Amines
Anhydrides
Arenes
Aryl Halides
Azides
Carbohydrates
Carboxylic Acids
Nomenclature of Carboxylic Acids
Properties of Carboxylic Acids
Reactivity of Carboxylic Acids
Conversion of a Carboxylic Acid to an Amide
Conversion of carboxylic acids to acid chlorides
Conversion of carboxylic acids to alcohols using LiAlH4
Conversion of Carboxylic acids to amides using DCC as an activating agent
Fischer Esterification
Hell-Volhard-Zelinskii Reaction
Making Acyl Chlorides (Acid Chlorides)
Making Esters From Carboxylic Acids
Reactions of Carboxylic Acids
Reduction of Carboxylic Acids with (LiAlH_4)
Simple Reactions of Carboxylic Acids as Acids
The Decarboxylation of Carboxylic Acids and Their Salts
Synthesis of Carboxylic Acids/Carboxylic_Acids/Synthesis_of_Carboxylic_Acids)
Chirality/Chirality)
Conjugation/Conjugation)
Esters/Esters)
Ethers/Ethers)
Fundamentals/Fundamentals)
Hydrocarbons/Hydrocarbons)
Lipids/Lipids)
Nitriles/Nitriles)
Organo-phosphorus Compounds/Organo-phosphorus_Compounds)
Phenols/Phenols)
Phenylamine and Diazonium Compounds/Phenylamine_and_Diazonium_Compounds)
Polymers/Polymers)
Reactions/Reactions)
Thiols and Sulfides/Thiols_and_Sulfides)
Spectroscopy/Spectroscopy)
Organic Chemistry (OpenStax))
Organic Chemistry (Morsch et al.))
Organic Chemistry with a Biological Emphasis (Soderberg))
Organic Chemistry Lab Techniques (Nichols))
Basic Principles of Organic Chemistry (Roberts and Caserio))
Organic Chemistry Nomenclature Workbook (O'Donnell))
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Conversion of carboxylic acids to acid chlorides
Last updated Jan 23, 2023
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Conversion of a Carboxylic Acid to an Amide
Conversion of carboxylic acids to alcohols using LiAlH4
Page ID 5437
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. General Reaction
1. Example
Mechanism
Contributors
Carboxylic acids react with Thionyl Chloride (SOCl 2) to form acid chlorides. During the reaction the hydroxyl group of the carboxylic acid is converted to a chlorosulfite intermediate making it a better leaving group. The chloride anion produced during the reaction acts a nucleophile.
General Reaction
Example
Mechanism
1) Nucleophilic attack on Thionyl Chloride
2) Removal of Cl leaving group
3) Nucleophilic attack on the carbonyl
4) Leaving group removal
5) Deprotonation
Contributors
Prof. Steven Farmer (Sonoma State University)
Conversion of carboxylic acids to acid chlorides is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.
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Conversion of a Carboxylic Acid to an Amide
Conversion of carboxylic acids to alcohols using LiAlH4
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Conversion of carboxylic acids to alcohols using LiAlH4
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15372 | http://jilliancross.weebly.com/uploads/2/4/3/3/24334594/5.4_subtracting_polynomials_worksheet.pdf | Corner Brook Intermediate Math 9 Unit 5: Polynomials Name: ______ 5.3 Subtracting Polynomials - Worksheet 1. Use algebra tiles. Sketch your tile model. Record your answer symbolically. a) (4x + 2) – (2x + 1) b) (4x + 2) – (2x – 1) c) (2s2 + 3s + 6) – (s2 + s + 2) d) (2s2 + 3s – 6) – (s2 + s – 2) e) (–2s2 + 3s + 6) – (–s2 + s + 2) f) (2s2 – 3s + 6) – (s2 – s + 2) 2. Subtract. a) (2x + 3) – (5x + 4) b) (4 – 8w) – (7w + 1) c) (x2 + 2x – 4) – (4x2 + 2x – 2) d) (–9z2 – z – 2) – (3z2 – z – 3) 3. A student subtracted (3y2 + 5y + 2) – (4y2 + 3y + 2) like this: = 3y2 – 5y – 2 – 4y2 – 3y – 2 = 3y2 – 4y2 – 5y – 3y – 2 – 2 = –y2 – 8y – 4 a) Explain why the student’s solution is incorrect. b) What is the correct answer? Show your work. 4. The difference between two polynomials is (5x + 3). One of the two polynomials is (4x + 1 – 3x2). What is the other polynomial? Explain how you found your answer. 5. Subtract. a) (mn – 5m – 7) – (–6n + 2m + 1) b) (2a + 3b – 3a2 + b2) – (–a2 + 8b2 + 3a – b) c) (xy – x – 5y + 4y2) – (6y2 + 9y – xy) |
15373 | https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/21%3A_Acids_and_Bases/21.19%3A_Titration_Curves | Skip to main content
21.19: Titration Curves
Last updated
: Mar 20, 2025
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21.18: Titration Calculations
21.20: Indicators
Page ID
: 53949
( \newcommand{\kernel}{\mathrm{null}\,})
The x-y plot that we know of as a graph was the brainchild of the French mathematician-philosopher Rene Descartes (1596-1650). His studies in mathematics led him to develop what was known as "Cartesian geometry", including the concept of today's graphs. The coordinates are often referred to as Cartesian coordinates.
Titration Curves
As
base
is added to
acid
at the beginning of a
titration
, the
pH
rises very slowly. Nearer to the
equivalence point
, the
pH
begins to rapidly increase. If the
titration
is a strong
acid
with a strong
base
, the
pH
at the
equivalence point
is equal to 7. A bit past the
equivalence point
, the rate of change of the
pH
again slows down. A titration
curve is a graphical representation of the
pH
of a
solution
during a
titration
. The figure below shows two different examples of a strong
acid
-strong
base
titration
curve. On the left is a
titration
in which the
base
is added to the
acid
, and so the
pH
progresses from low to high. On the right is a
titration
in which the
acid
is added to the
base
. In this case, the
pH
starts out high and decreases during the
titration
. In both cases, the
equivalence point
is reached when the moles of
acid
and
base
are equal and the
pH
is 7. This also corresponds to the color change of the
indicator
.
Titration
curves can also be generated in the case of a
weak acid
-strong
base
titration
or a strong
acid
-
weak base
titration
. The general shape of the
titration
curve is the same, but the
pH
at the
equivalence point
is different. In a
weak acid
-strong
base
titration
, the
pH
is greater than 7 at the
equivalence point
. In a strong
acid
-
weak base
titration
, the
pH
is less than 7 at the
equivalence point
.
Summary
A
titration
curve is a graphical representation of the
pH
of a
solution
during a
titration
.
In a strong
acid
-strong
base
titration
, the
equivalence point
is reached when the moles of
acid
and
base
are equal and the
pH
is 7.
In a
weak acid
-strong
base
titration
, the
pH
is greater than 7 at the
equivalence point
.
In a strong
acid
-
weak base
titration
, the
pH
is less than 7 at the
equivalence point
.
21.18: Titration Calculations
21.20: Indicators |
15374 | https://allen.in/dn/qna/185017902 | If k (x) =5x +2, what is the vlaue of k (4) - k(1) ?
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15375 | https://www.reddit.com/r/calculus/comments/1d1kgw0/difference_bewteen_definte_and_indefinte_intergal/ | difference bewteen definte and indefinte intergal : r/calculus
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difference bewteen definte and indefinte intergal
Differential Calculus
I have been looking at the definition. It shows that the indefinite integral is F(x) + C. It is called an indefinite integral because we don't know C, but it is only definite when we know C (I could be way off).
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15376 | https://www.youtube.com/watch?v=Sntww-uEYmI | Prove: sin^2theta + cos^2theta = 1 | Trigonometric Identities | Class 10 Maths
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sin square theta + cos square theta. sin^2 theta+cos^2 theta=1 proof. sin2 theta cos2 theta 1 prove. How to prove sine squared theta plus cosine squared theta equals one. sin^2(x)+cos^2(x)=1 proof. Prove sin^2(x)+cos^2(x)=1. Prove sin^2 theta+cos^2 theta=1. sin^2 theta + cos^2 theta=1. Prove that sin^2 theta + cos^2 theta = 1. sin squared theta + cos squared theta is equal to 1 prove. Proof of sin^2(θ)+cos^2(θ)=1. How to prove sine squared θ plus cos squared θ equals 1. Sum of sine squared θ and cos squared θ equals 1.
sin2θ+cos2θ=1
Prove: sin^2theta + cos^2theta = 1 | Trigonometric Identities | Class 10 Maths
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हेलो हेलो फ्रेंड्स वेलकम टू माय YouTube चैनल आज हम इस वीडियो में क्लास टेंथ चैप्टर ट्रिग्नोमेट्री कवर मोस्ट इंपोर्टेंट टॉपिक कवर करेंगे डेबिट कॉल ट्रिगोनोमेट्रिक आईडेंटिटीज और आपको बताइए ट्रिगोनोमेट्रिक आईडेंटिटीज ही टोटली फ्री होती है बट इस वीडियो में हम पोस्ट आईडेंटिटी को अवरुद्ध करने वाले हैं और आज हम इस वीडियो में आपको फर्स्ट आईडेंटिटी कार्ड प्रॉपर्टी प्रूफ आपको दिखाएंगे वह भी इस एक्सप्लेनेशन के साथ तो स्टार्ट करते हैं वीडियो में [संगीत] मैं आपको मजा कि ट्रिगोनोमेट्रिक आईडेंटिटीज क्या कहती है कि साइन स्क्वायर थीटा प्लस कौस स्क्वायर थीटा बराबर होगा1 मतलब कि साइन का स्क्वायर करेंगे और कॉस्कास कट करेंगे दोनों का तिरंगा से मोबिलाइजर ठीक है यहां पर बिठा दें इन दोनों के स्क्वायर कसम किसके बराबर होता वन के वहीं आपको प्रूव करने पर यहां पर आपके टोटल ट्रिक ऐसी जुदाई सो सकते हैं यह अगर आपका जीरो डिग्री और सकता है या फिर यह गला पर नौकरी हो सकता है या फिर सिंगल एक्यूट एंगल हो सकता बिकॉज आपको बताया कि ट्रिग्नोमेट्री हमारी राइट एंगल ट्रायंगल में ही यूज होती है तो राइट एंगल ट्रायंगल यहां तो 100 सकता है या तो एंगल इन आंसू निकल एक्यूट एंगल है डिग्री से छोटा तो वह टीमों के सिस्टम लेकर चलेंगे और वन बाय वन टीमों के बीच में इस आइडिया को फॉलो करेंगे तो केस नंबर वन कि वैन थीटा बराबर है गुरु तेघ को यूज करेंगे तो एक टाइम में डाल प्लस कौस स्क्वायर थीटा थे साइंस को पर जीरो डिग्री प्लस कॉस स्क्वेयर कि आपने ट्रिगोनोमेट्रिक रेश्योस टेबल अपने लंच किया वहीं और आपको पता कि साइज जीरो की तरह जीरो तो जीरो का स्क्वायर प्लस पॉजिटिव विकास को 2010 प्लस वन अधिकारी अगर आप यहां पर ब्रिंग टू फ्रंट [प्रशंसा] यह हो गया नंबर दो कि वैन एक ही टाइम बराबर है नौकरी की प्रॉब्लम है कि यहां पर एंगल आपको लेने लगी तो थैंक्स कोई नौकरी कॉस्ट को नौकरी आपको पता ही नहीं तो 200 और यह पेज नंबर 66 होगा तो यहां पर हमने यहां पर आपने देख लिया है जो अलार्म्स मैंने अपने आप लिखा गया है कि साइंस को ठीक-ठाक प्लस पॉइंट तो अब हम क्या करेंगे साइन थीटा और कौन सी की रैली निकालेंगे आपको बताया कि साइज छोटा किसके बराबर होता है परपेंडिकुलर अपऑन थे विटल है अगर मैं सीता के हिसाब से परपेंडिकुलर देखो तो यह आपका क्या होगा परपेंडिकुलर हफ्ता होगा बेस और योग आपका है पिता ने यहां पर महाप्रबंधक है बल्कि यह भी अपऑन एस ई ई कि अगर हम टॉस निकाले तक व्हाट्सएप पर किसके बराबर होता है विवाह आज दिल्ली करोगे पॉइंट है यह सब क्या बीसीडी और कितना आप करेंगे तो आपको मिलेगा सब्सक्राइब दोनों लेंगे तो सब्सक्राइब करें आपको पता लगा सकते हैं लगा सकते है सब्सक्राइब कर दो ए फिल्म ओं कि अगर हम फ्रॉम फर्स्ट क्वेश्चन देखें तो यह विकास पर उपलब्धि थी कि इसको है और आपको बताइए दोनों के स्वयं किसके द्वारा बनाई गई थी कस्टमर तो इस विधि को यहां पर क्या करते हैं पॉइंट तो आपका एक्सप्रेस बनेगा फ्रॉम फर्स्ट यहां पर आप देखोगे ऐसी का स्कोर अपऑन इसी का त्यौहार और यह आपको क्या करना है प्रूफ मैं यहां पर आपने कहती थीं कि फॉर एनी रिलीफ ठीक-ठाक टाइम्स स्क्वेयर थीटा प्लस कौस स्क्वायर थीटा किसके बराबर आ जाएगा वन किया जाएगा और यह आपकी फेस आईडेंटिटी का इजी प्रूफ तो यह था आपका पोस्ट आईडेंटिटी का इजी प्रूफ आई हॉप आपको वीडियो अच्छी लगी हो तो प्लीज इस वीडियो को लाइक करें ज्यादा शेयर करें और पौधा वीडियो देखने के लिए प्लीज हमारे चैनल को सब्सक्राइब करें बटन को प्रेस करके जो रखी बुंदेली वीडियो अपलोड करें जिनकी नौकरी से ऑफिस जाकर मिलते जल्द आपको नेक्स्ट वीडियो में तब |
15377 | https://courses.lumenlearning.com/odessa-collegealgebra/chapter/multiply-and-divide-complex-numbers/ | Complex Numbers
Multiply and divide complex numbers
Multiplying Complex Numbers
Multiplying complex numbers is much like multiplying binomials. The major difference is that we work with the real and imaginary parts separately.
Example 4: Multiplying a Complex Number by a Real Number
Figure 5
Let’s begin by multiplying a complex number by a real number. We distribute the real number just as we would with a binomial. So, for example,
How To: Given a complex number and a real number, multiply to find the product.
Use the distributive property.
Simplify.
Example 5: Multiplying a Complex Number by a Real Number
Find the product [latex]4\left(2+5i\right)[/latex].
Solution
Distribute the 4.
[latex]\begin{cases}4\left(2+5i\right)=\left(4\cdot 2\right)+\left(4\cdot 5i\right)\hfill \ =8+20i\hfill \end{cases}[/latex]
Try It 4
Find the product [latex]-4\left(2+6i\right)[/latex].
Solution
Multiplying Complex Numbers Together
Now, let’s multiply two complex numbers. We can use either the distributive property or the FOIL method. Recall that FOIL is an acronym for multiplying First, Outer, Inner, and Last terms together. Using either the distributive property or the FOIL method, we get
[latex]\left(a+bi\right)\left(c+di\right)=ac+adi+bci+bd{i}^{2}[/latex]
Because [latex]{i}^{2}=-1[/latex], we have
[latex]\left(a+bi\right)\left(c+di\right)=ac+adi+bci-bd[/latex]
To simplify, we combine the real parts, and we combine the imaginary parts.
[latex]\left(a+bi\right)\left(c+di\right)=\left(ac-bd\right)+\left(ad+bc\right)i[/latex]
How To: Given two complex numbers, multiply to find the product.
Use the distributive property or the FOIL method.
Simplify.
Example 6: Multiplying a Complex Number by a Complex Number
Multiply [latex]\left(4+3i\right)\left(2 - 5i\right)[/latex].
Solution
Use [latex]\left(a+bi\right)\left(c+di\right)=\left(ac-bd\right)+\left(ad+bc\right)i[/latex]
[latex]\begin{cases}\left(4+3i\right)\left(2 - 5i\right)=\left(4\cdot 2 - 3\cdot \left(-5\right)\right)+\left(4\cdot \left(-5\right)+3\cdot 2\right)i\hfill \ \text{ }=\left(8+15\right)+\left(-20+6\right)i\hfill \ \text{ }=23 - 14i\hfill \end{cases}[/latex]
Try It 5
Multiply [latex]\left(3 - 4i\right)\left(2+3i\right)[/latex].
Solution
Dividing Complex Numbers
Division of two complex numbers is more complicated than addition, subtraction, and multiplication because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator. We need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary portion of the denominator so that we end up with a real number as the denominator. This term is called the complex conjugate of the denominator, which is found by changing the sign of the imaginary part of the complex number. In other words, the complex conjugate of [latex]a+bi[/latex] is [latex]a-bi[/latex].
Note that complex conjugates have a reciprocal relationship: The complex conjugate of [latex]a+bi[/latex] is [latex]a-bi[/latex], and the complex conjugate of [latex]a-bi[/latex] is [latex]a+bi[/latex]. Further, when a quadratic equation with real coefficients has complex solutions, the solutions are always complex conjugates of one another.
Suppose we want to divide [latex]c+di[/latex] by [latex]a+bi[/latex], where neither a nor b equals zero. We first write the division as a fraction, then find the complex conjugate of the denominator, and multiply.
[latex]\frac{c+di}{a+bi}\text{ where }a\ne 0\text{ and }b\ne 0[/latex]
Multiply the numerator and denominator by the complex conjugate of the denominator.
[latex]\frac{\left(c+di\right)}{\left(a+bi\right)}\cdot \frac{\left(a-bi\right)}{\left(a-bi\right)}=\frac{\left(c+di\right)\left(a-bi\right)}{\left(a+bi\right)\left(a-bi\right)}[/latex]
Apply the distributive property.
[latex]=\frac{ca-cbi+adi-bd{i}^{2}}{{a}^{2}-abi+abi-{b}^{2}{i}^{2}}[/latex]
Simplify, remembering that [latex]{i}^{2}=-1[/latex].
[latex]\begin{cases}=\frac{ca-cbi+adi-bd\left(-1\right)}{{a}^{2}-abi+abi-{b}^{2}\left(-1\right)}\hfill \ =\frac{\left(ca+bd\right)+\left(ad-cb\right)i}{{a}^{2}+{b}^{2}}\hfill \end{cases}[/latex]
A General Note: The Complex Conjugate
The complex conjugate of a complex number [latex]a+bi[/latex] is [latex]a-bi[/latex]. It is found by changing the sign of the imaginary part of the complex number. The real part of the number is left unchanged.
When a complex number is multiplied by its complex conjugate, the result is a real number.
When a complex number is added to its complex conjugate, the result is a real number.
Example 7: Finding Complex Conjugates
Find the complex conjugate of each number.
[latex]2+i\sqrt{5}[/latex]
[latex]-\frac{1}{2}i[/latex]
Solution
The number is already in the form [latex]a+bi[/latex]. The complex conjugate is [latex]a-bi[/latex], or [latex]2-i\sqrt{5}[/latex].
We can rewrite this number in the form [latex]a+bi[/latex] as [latex]0-\frac{1}{2}i[/latex]. The complex conjugate is [latex]a-bi[/latex], or [latex]0+\frac{1}{2}i[/latex]. This can be written simply as [latex]\frac{1}{2}i[/latex].
Analysis of the Solution
Although we have seen that we can find the complex conjugate of an imaginary number, in practice we generally find the complex conjugates of only complex numbers with both a real and an imaginary component. To obtain a real number from an imaginary number, we can simply multiply by i.
How To: Given two complex numbers, divide one by the other.
Write the division problem as a fraction.
Determine the complex conjugate of the denominator.
Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator.
Simplify.
Example 8: Dividing Complex Numbers
Divide [latex]\left(2+5i\right)[/latex] by [latex]\left(4-i\right)[/latex].
Solution
We begin by writing the problem as a fraction.
[latex]\frac{\left(2+5i\right)}{\left(4-i\right)}[/latex]
Then we multiply the numerator and denominator by the complex conjugate of the denominator.
[latex]\frac{\left(2+5i\right)}{\left(4-i\right)}\cdot \frac{\left(4+i\right)}{\left(4+i\right)}[/latex]
To multiply two complex numbers, we expand the product as we would with polynomials (the process commonly called FOIL).
[latex]\begin{cases}\frac{\left(2+5i\right)}{\left(4-i\right)}\cdot \frac{\left(4+i\right)}{\left(4+i\right)}=\frac{8+2i+20i+5{i}^{2}}{16+4i - 4i-{i}^{2}}\hfill & \hfill \ \text{ }=\frac{8+2i+20i+5\left(-1\right)}{16+4i - 4i-\left(-1\right)}\hfill & \text{Because } {i}^{2}=-1\hfill \ \text{ }=\frac{3+22i}{17}\hfill & \hfill \ \text{ }=\frac{3}{17}+\frac{22}{17}i\hfill & \text{Separate real and imaginary parts}.\hfill \end{cases}[/latex]
Note that this expresses the quotient in standard form.
Example 9: Substituting a Complex Number into a Polynomial Function
Let [latex]f\left(x\right)={x}^{2}-5x+2[/latex]. Evaluate [latex]f\left(3+i\right)[/latex].
Solution
Figure 6
Substitute [latex]x=3+i[/latex] into the function [latex]f\left(x\right)={x}^{2}-5x+2[/latex] and simplify.
Analysis of the Solution
We write [latex]f\left(3+i\right)=-5+i[/latex]. Notice that the input is [latex]3+i[/latex] and the output is [latex]-5+i[/latex].
Try It 6
Let [latex]f\left(x\right)=2{x}^{2}-3x[/latex]. Evaluate [latex]f\left(8-i\right)[/latex].
Solution
Example 10: Substituting an Imaginary Number in a Rational Function
Let [latex]f\left(x\right)=\frac{2+x}{x+3}[/latex]. Evaluate [latex]f\left(10i\right)[/latex].
Solution
Substitute [latex]x=10i[/latex] and simplify.
[latex]\begin{cases}\frac{2+10i}{10i+3}\hfill & \text{Substitute }10i\text{ for }x.\hfill \ \frac{2+10i}{3+10i}\hfill & \text{Rewrite the denominator in standard form}.\hfill \ \frac{2+10i}{3+10i}\cdot \frac{3 - 10i}{3 - 10i}\hfill & \text{Prepare to multiply the numerator and}\hfill \ \hfill & \text{denominator by the complex conjugate}\hfill \ \hfill & \text{of the denominator}.\hfill \ \frac{6 - 20i+30i - 100{i}^{2}}{9 - 30i+30i - 100{i}^{2}}\hfill & \text{Multiply using the distributive property or the FOIL method}.\hfill \ \frac{6 - 20i+30i - 100\left(-1\right)}{9 - 30i+30i - 100\left(-1\right)}\hfill & \text{Substitute }-1\text{ for } {i}^{2}.\hfill \ \frac{106+10i}{109}\hfill & \text{Simplify}.\hfill \ \frac{106}{109}+\frac{10}{109}i\hfill & \text{Separate the real and imaginary parts}.\hfill \end{cases}[/latex]
Try It 7
Let [latex]f\left(x\right)=\frac{x+1}{x - 4}[/latex]. Evaluate [latex]f\left(-i\right)[/latex].
Solution
Simplifying Powers of i
The powers of i are cyclic. Let’s look at what happens when we raise i to increasing powers.
[latex]\begin{cases}{i}^{1}=i\ {i}^{2}=-1\ {i}^{3}={i}^{2}\cdot i=-1\cdot i=-i\ {i}^{4}={i}^{3}\cdot i=-i\cdot i=-{i}^{2}=-\left(-1\right)=1\ {i}^{5}={i}^{4}\cdot i=1\cdot i=i\end{cases}[/latex]
We can see that when we get to the fifth power of i, it is equal to the first power. As we continue to multiply i by itself for increasing powers, we will see a cycle of 4. Let’s examine the next 4 powers of i.
[latex]\begin{cases}{i}^{6}={i}^{5}\cdot i=i\cdot i={i}^{2}=-1\ {i}^{7}={i}^{6}\cdot i={i}^{2}\cdot i={i}^{3}=-i\ {i}^{8}={i}^{7}\cdot i={i}^{3}\cdot i={i}^{4}=1\ {i}^{9}={i}^{8}\cdot i={i}^{4}\cdot i={i}^{5}=i\end{cases}[/latex]
Example 11: Simplifying Powers of i
Evaluate [latex]{i}^{35}[/latex].
Solution
Since [latex]{i}^{4}=1[/latex], we can simplify the problem by factoring out as many factors of [latex]{i}^{4}[/latex] as possible. To do so, first determine how many times 4 goes into 35: [latex]35=4\cdot 8+3[/latex].
[latex]{i}^{35}={i}^{4\cdot 8+3}={i}^{4\cdot 8}\cdot {i}^{3}={\left({i}^{4}\right)}^{8}\cdot {i}^{3}={1}^{8}\cdot {i}^{3}={i}^{3}=-i[/latex]
Q & A
Can we write [latex]{i}^{35}[/latex] in other helpful ways?
As we saw in Example 11, we reduced [latex]{i}^{35}[/latex] to [latex]{i}^{3}[/latex] by dividing the exponent by 4 and using the remainder to find the simplified form. But perhaps another factorization of [latex]{i}^{35}[/latex] may be more useful. The table below shows some other possible factorizations.
| | | | | |
--- ---
| Factorization of [latex]{i}^{35}[/latex] | [latex]{i}^{34}\cdot i[/latex] | [latex]{i}^{33}\cdot {i}^{2}[/latex] | [latex]{i}^{31}\cdot {i}^{4}[/latex] | [latex]{i}^{19}\cdot {i}^{16}[/latex] |
| Reduced form | [latex]{\left({i}^{2}\right)}^{17}\cdot i[/latex] | [latex]{i}^{33}\cdot \left(-1\right)[/latex] | [latex]{i}^{31}\cdot 1[/latex] | [latex]{i}^{19}\cdot {\left({i}^{4}\right)}^{4}[/latex] |
| Simplified form | [latex]{\left(-1\right)}^{17}\cdot i[/latex] | [latex]-{i}^{33}[/latex] | [latex]{i}^{31}[/latex] | [latex]{i}^{19}[/latex] |
Each of these will eventually result in the answer we obtained above but may require several more steps than our earlier method.
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15378 | http://dspace.mit.edu/bitstream/handle/1721.1/78251/6-003-spring-2010/contents/lecture-notes/MIT6_003S10_lec05_handout.pdf | ˙ x(t) x(t) X AX ˙ x(t) x(t) X AX ˙ x(t) x(t) X AX X 6.003: Signals and Systems Lecture 5 February 18, 2010 6.003: Signals and Systems Concept Map: Continuous-Time Systems Relations among representations. Laplace Transform Block Diagram System Functional Y Y 2A2 = X 2 + 3A + A2 + 1 + 1 2 X − − Impulse Response h(t) = 2(e−t/2 − e−t) u(t) Differential Equation System Function 2¨ y(t) + 3 ˙ y(t) + y(t) = 2x(t) Y (s) 2 February 18, 2010 X(s) = 2s2 + 3s + 1 Check Yourself Concept Map: Continuous-Time Systems How to determine impulse response from system functional? Today: new relations based on Laplace transform. Block Diagram Block Diagram System Functional System Functional X Y Y 2A2 Y Y 2A2 + + − − = + + − − = 1 X 2 + 3A + A2 1 X 2 + 3A + A2 1 2 1 2 Impulse Response Impulse Response h(t) = 2(e−t/2 − e−t) u(t) h(t) = 2(e−t/2 − e−t) u(t) Differential Equation System Function Differential Equation System Function 2¨ y(t) + 3 ˙ y(t) + y(t) = 2x(t) Y (s s ) ) = 2 2¨ y(t) + 3 ˙ y(t) + y(t) = 2x(t) Y (s s ) ) = 2 X( 2s2 + 3s + 1 X( 2s2 + 3s + 1 Laplace Transform: Definition Laplace Transforms Laplace transform maps a function of time t to a function of s. Example: Find the Laplace transform of x1(t): x1(t) X(s) = x(t)e −stdt x1(t) = e−t if t ≥ 0 0 otherwise There are two important variants: Unilateral (18.03) ∞ ∞ e−(s+1)t ∞ 1 ∞ X1(s) = −∞ x1(t)e −stdt = 0 e −t e −stdt = −(s + 1) = s + 1 X(s) = x(t)e −stdt 0 0 provided Re(s + 1) > 0 which implies that Re(s) > −1.
Bilateral (6.003) ∞ 0 t X(s) = x(t)e −stdt −∞ Both share important properties — will discuss differences later. 1 s + 1 ; Re(s) > −1 −1 s-plane ROC 1 6.003: Signals and Systems Lecture 5 February 18, 2010 Check Yourself x2(t) x2(t) = e−t − e−2t if t ≥ 0 0 otherwise t 0 Which of the following is the Laplace transform of x2(t)? 1. X2(s) = (s+1)( 1 s+2) ; Re(s) > −1 2. X2(s) = (s+1)( 1 s+2) ; Re(s) > −2 s 3. X2(s) = (s+1)(s+2) ; Re(s) > −1 s 4. X2(s) = (s+1)(s+2) ; Re(s) > −2 5. none of the above Regions of Convergence Left-sided signals have left-sided Laplace transforms (bilateral only). −1 ROC Example: x3(t) x3(t) = −e−t if t ≤ 0 t 0 otherwise −1 s) = ∞ t)e −stdt = 0 −e −t e −stdt = −e−(s+1)t 0 = 1 X3( −∞ x3( −∞ −(s + 1) −∞ s + 1 Re( + 1) 0 Re( ) 1 provided which implies that − < < s s . s-plane ; Re(s) < −1 s + 1 1 Left- and Right-Sided ROCs Laplace transforms of left- and right-sided exponentials have the same form (except −); with left- and right-sided ROCs, respectively. time function Laplace transform e−tu(t) 1 0 t s + 1 s-plane ROC −1 ROC −1 s-plane −e−tu(−t) 1 t s + 1 −1 Left- and Right-Sided ROCs Laplace transforms of left- and right-sided exponentials have the same form (except −); with left- and right-sided ROCs, respectively. time function Laplace transform e−t −e−t u(t) t 0 −1 s-plane ROC 1 s + 1 s-plane u(−t) t −1 −1 ROC 1 s + 1 Check Yourself Find the Laplace transform of x4(t). 0 t x4(t) x4(t) = e−|t| 1. X4(s) = 2 1−s2 ; −∞ < Re(s) < ∞ 2. X4(s) = 2 1−s2 ; −1 < Re(s) < 1 3. X4(s) = 2 1+s2 ; −∞ < Re(s) < ∞ 4. X4(s) = 2 1+s2 ; −1 < Re(s) < 1 5. none of the above s-plane s-plane −1 Time-Domain Interpretation of ROC ∞ X(s) = x(t) e −stdt −∞ x1(t) t −1 x2(t) t −2 −1 x3(t) s-plane t −1 x4(t) −1 1 s-plane t 2 6.003: Signals and Systems Lecture 5 February 18, 2010 Time-Domain Interpretation of ROC Time-Domain Interpretation of ROC ∞ ∞ X(s) = x(t) e −stdt X(s) = x(t) e −stdt −∞ −∞ x1(t) t s-plane −1 t x2(t) s-plane −1 −2 t x3(t) −1 s-plane −1 t x4(t) s-plane −1 1 t x1(t) s-plane −1 t x2(t) s-plane −1 −2 t x3(t) −1 s-plane −1 t x4(t) s-plane −1 1 Time-Domain Interpretation of ROC Check Yourself ∞ X(s) = x(t) e −stdt −∞ The Laplace transform 2 2 − s 4 corresponds to how many of t x1(t) s-plane −1 t x2(t) s-plane −1 −2 t x3(t) −1 −1 s-plane t x4(t) s-plane −1 1 s the following signals? 1. e−2tu(t) + e2tu(t) 2. e−2tu(t) −e2tu(−t) 3. −e−2tu(−t) + e2tu(t) 4. −e−2tu(−t) −e2tu(−t) Solving Differential Equations with Laplace Transforms Solve the following differential equation: y ˙(t) + y(t) = δ(t) Take the Laplace transform of this equation. L {y ˙(t) + y(t)} = L {δ(t)} The Laplace transform of a sum is the sum of the Laplace transforms (prove this as an exercise). L {y ˙(t)} + L {y(t)} = L {δ(t)} What’s the Laplace transform of a derivative? Laplace transform of a derivative Assume that X(s) is the Laplace transform of x(t): ∞ X(s) = x(t)e −stdt −∞ Find the Laplace transform of y(t) = ˙ x(t). ∞ ∞ Y (s) = y(t)e −stdt = x ˙(t) e − st dt −∞ −∞ ˙ v u = x(t) e −st ∞ − ∞ x(t)(−se −st)dt −∞ u v −∞ v u ˙ The first term must be zero since X(s) converged. Thus ∞ Y (s) = s x(t)e −stdt = sX(s) −∞ 3 6.003: Signals and Systems Solving Differential Equations with Laplace Transforms Back to the previous problem: L {y ˙(t)} + L {y(t)} = L {δ(t)} Let Y (s) represent the Laplace transform of y(t). Then sY (s) is the Laplace transform of y ˙(t). sY (s) + Y (s) = L {δ(t)} What’s the Laplace transform of the impulse function? Solving Differential Equations with Laplace Transforms Back to the previous problem: sY (s) + Y (s) = L {δ(t)} = 1 This is a simple algebraic expression. Solve for Y (s): 1 Y (s) = s + 1 We’ve seen this Laplace transform previously. y(t) = e −t u(t) (why not y(t) = −e−tu(−t) ?) Notice that we solved the differential equation y ˙(t)+y(t) = δ(t) without computing homogeneous and particular solutions. Solving Differential Equations with Laplace Transforms Recognizing the form ... Is there a more systematic way to take an inverse Laplace transform? Yes ... and no. Formally, 1 σ+j∞ x(t) = X(s)e stds 2πj σ−j∞ but this integral is not generally easy to compute. This equation can be useful to prove theorems. We will find better ways (e.g., partial fractions) to compute inverse transforms for common systems. Lecture 5 February 18, 2010 Laplace transform of the impulse function Let x(t) = δ(t). ∞ X(s) = δ(t)e −stdt −∞ ∞ = δ(t) e −st t=0 dt −∞ ∞ = δ(t) 1 dt −∞ = 1 Sifting property: δ(t) sifts out the value of e−st at t = 0. Solving Differential Equations with Laplace Transforms Summary of method. Start with differential equation: y ˙(t) + y(t) = δ(t) Take the Laplace transform of this equation: sY (s) + Y (s) = 1 Solve for Y (s): 1 Y (s) = s + 1 Take inverse Laplace transform (by recognizing form of transform): y(t) = e −t u(t) Solving Differential Equations with Laplace Transforms Example 2: y ¨(t) + 3 ˙ y(t) + 2y(t) = δ(t) Laplace transform: s 2Y (s) + 3sY (s) + 2Y (s) = 1 Solve: 1 1 1 Y (s) = (s + 1)(s + 2) = s + 1 − s + 2 Inverse Laplace transform: y(t) = e −t −e −2t u(t) These forward and inverse Laplace transforms are easy if • differential equation is linear with constant coefficients, and • the input signal is an impulse function. 4 ˙ x(t) x(t) X AX 6.003: Signals and Systems Properties of Laplace Transforms The use of Laplace Transforms to solve differential equations de pends on several important properties. Property x(t) X(s) ROC Linearity ax1(t) + bx2(t) aX1(s) + bX2(s) ⊃(R1 ∩R2) Delay by T x(t −T ) X(s)e −sT R Lecture 5 February 18, 2010 Concept Map: Continuous-Time Systems Where does Laplace transform fit in? Block Diagram System Functional X Y Y 2A2 + + − − = 1 2 1 X 2 + 3A + A2 dX(s) Multiply by t tx(t) − R ds Multiply by e−αt x(t)e −αt X(s + α) shift R by −α Impulse Response dx(t) h(t) = 2(e−t/2 −e−t) u(t) Differentiate in t sX(s) ⊃R dt Integrate in t t x(τ) dτ X(s) ⊃ R ∩ Re(s)>0 −∞ s ∞ Differential Equation System Function Convolve in t x1(τ)x2(t −τ) dτ X1(s)X2(s) ⊃(R1 ∩R2) Y (s) 2 −∞ 2¨ y(t) + 3 ˙ y(t) + y(t) = 2x(t) X(s) = 2s2 + 3s + 1 Initial Value Theorem Final Value Theorem If x(t) = 0 for t < 0 and x(t) contains no impulses or higher-order If x(t) = 0 for t < 0 and x(t) has a finite limit as t →∞ singularities at t = 0 then x(∞) = lim sX(s) . s→0 x(0+) = lim sX(s) . ∞ ∞ s→∞ ∞ ∞ Consider lim sX(s) = lim s x(t)e −stdt = lim x(t) se −stdt. s→0 s→0 −∞ s→0 0 Consider lim sX(s) = lim s x(t)e −stdt = lim x(t) se −stdt. s→∞ s→∞ −∞ s→∞ 0 As s →0 the function e−st flattens out. But again, the area under As s →∞ the function e−st shrinks towards 0. se−st is always 1. −st e −st e s = 1 s = 1 s = 25 s = 5 s = 25 s = 5 x(∞) t t Area under e−st is 1 → area under se−st is 1 → lim se −st = δ(t) ! As s →0, area under se−st monotonically shifts to higher values of t s s→∞ ∞ ∞ −st (e.g., the average value of se is 1 s which grows as s →0).
lim sX(s) = lim x(t)se −stdt → x(t)δ(t)dt = x(0+) s→∞ s→∞ 0 0 In the limit, lim sX(s) →x(∞) . s→0 (the 0+ arises because the limit is from the right side.) 5 MIT OpenCourseWare 6.003 Signals and Systems Spring 2010 For information about citing these materials or our Terms of Use, visit: |
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3.1.2: Breaking the octet rule with higher electron counts (hypervalent atoms)
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: Nov 6, 2022
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3.1.1: Resonance
3.1.3: Formal Charge
Page ID
: 167323
Kathryn Haas
Duke University
( \newcommand{\kernel}{\mathrm{null}\,})
The octet rule applies well to atoms in the second row of the periodic table, where a full valence shell includes eight electrons with an electron configuration of . Even elements in the third and fourth row are known to follow this rule sometimes, but not always. In larger atoms, where the valence shell contains additional subshells: the subshells. Therefore, atoms with can have higher valence shell counts by "expanding" into these additional subshells. When atoms contain more than eight electrons in their valence shell, they are said to be hypervalent. Hypervalency allows atoms with to break the octet rule by having more than eight electrons. This also means they can have five or more bonds; something that is nearly unheard of for atoms with . Complete the exercises below to see examples of molecules containing hypervalent atoms.
Exercise
Draw the Lewis structures for sulfur hexafluoride ().
Answer
: Each fluorine atom has one valence electron and will make one bond each. The sulfur has six valence electrons, and must make six bonds to form a molecule with the six fluorine atoms. The molecular structure has six bonds to sulfur, with twelve valence electrons. The sulfur is hypervalent.
Exercise
Draw the Lewis structure for chlorine trifluoride (.
Answer
: All atoms are halogens and each has seven valence electrons. Chlorine is capable of hypervalency because it is in the third row of the periodic table; however fluorine cannot have more than eight valence electrons in its valence because it is in the second row. The structure has the three fluorine atoms bonded to a central chlorine atom. The chlorine has a valence of ten electrons due to its three bonds and two lone pairs.
.
Is hypervalency real? Not exactly. Hypervalency is a concept associated with hybrid orbital theory and Lewis theory. It's useful for some simple things, like predicting how atoms are connected and predicting molecular shape. But the idea that the d-orbitals are involved in bonding isn't accurate according to wave mechanics.
For main group molecules, chemists (like Pauling) thought a long time ago that hypervalence is due to expanded s2p6 octets. The consensus is now clear that d orbitals are NOT involved in bonding in molecules like SF6 any more than they are in SF4 and SF2. In all three cases, there is a small and roughly identical participation of d-orbitals in the wavefunctions. This has been established in both MO and VB theory. However, using hybrid orbitals with d-orbital contributions equips us with a language which can pragmatically describe the geometries of highly coordinated substances.
While hybrid orbitals are a powerful tool to describe the geometries and shape of molecules and metal complexes, in "real" molecules their significance may be debated. Often a more realistic molecular orbitals approach is needed. However, from an epistemologically simple point of view, bonding theories can only be judged by their predictions. To the extent that hybridization can explain the shapes of PF5 and SF6, valence bond theory is a perfectly good theory. To the extent that if you write out the valence bond wavefunction using hybridized orbitals and calculate energies and other properties à la Pauling (i.e., ionization energy and electron affinities) and find them to be off from experimental results (by tens of kcals/mol), then valence bond theory is not accurate.
Bonding theories can only be judged by their predictions.
A simple explanation that can be given is that molecular wavefunctions constructed out of hybridized atomic orbitals are accurate enough to predict some things, but not others. Predictions of any theory must be compared with empirical evidence to assess when they work and when they fail. When a theory gives the wrong answer, at least one assumption must not hold. In this case, the valence bond wavefunction is not accurate enough to capture some important features of a system's electronic structure. It may not be the most intellectually satisfying answer, but to say more would result in a much more complicated answer and certainly far beyond the level reasonably expected from general chemistry discussions.
3.1.1: Resonance
3.1.3: Formal Charge |
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#13 Density of Brine
/ Tech Talks / By admin / November 30, 2020
Accurate correlations exist for the density of pure water. An example is the model of Keyes et al. , which is used in PVTsim Nova. Formation water produced together with gas and oil is not pure, but contains dissolved salts, and these salts can have a significant effect on the density.
The importance of water density for the design of pipelines and process equipment increases with the water cut. There are cases where oil is produced with a water cut as high as 90%. Also when designing water-based drilling fluids, it is of much importance to be able to calculate how much dissolved salts affect the density of the water phase.
In PVTsim, the salt concentration can either be input as mg/l or the concentration of each individual salt can be entered. In both cases, the model by Laliberté and Cooper is used to correct the density for salt content. When the amount of salt is given as mg/l, it is assumed that all salt is sodium chloride (NaCl).
The correlation of Laliberté and Cooper takes the form:
where
ρbrine Density of salt water (brine)wH2OWeight fraction of waterρH2O Density of pure waterwiWeight fraction of salt number iρApp,i Density of salt number i when dissolved in water (apparent salt density)
The apparent density (in kg/m3) of a salt dissolved in water is calculated from
where c0-c4 are salt specific constants and t the temperature in oC. PVTsim Nova handles the effect of the below salts on the water phase density:
– Sodium Chloride (NaCl)– Potassium Chloride (KCl)– Sodium Bromide (NaBr)– Potassium Bromide (KBr)– Calcium Chloride (CaCl2)– Calcium Bromide (CaBr2)– Zinc Bromide (ZnBr2)– Sodium Formate (HCOONa)– Potassium Formate (HCOOK)– Cesium Formate (HCOOCs)
Table 1 shows the constants to be used for NaCl.
| | |
--- |
| Constant | Value for NaCl |
| c0 | 0.00433 |
| c1 | 0.06471 |
| c2 | 1.0166 |
| c3 | 0.014624 |
| c4 | 3315.6 |
Figure 1 shows measured and simulated densities of NaCl brine at four different temperatures at 1 bar and at four different pressures at 50 oC. An almost perfect match is seen between the experimental densities and those simulated using the correlation of Laliberté and Cooper.
Figure 1Measured and simulated densities of NaCl salt water at four different temperatures at 1 bar and at four different pressures at 50 oC.
It has also been verified that the correlation of Laliberté and Cooper works well for mixed salts. Densities were simulated of water with NaCl and NaBr. The data covered NaCl weight%’s from 0 – 14.8 and NaBr weight%’s from 0 – 24.0 in several mixing ratios. The match of the data is shown in Figure 2 as Calculated versus Experimental densities. With a perfect match, all the points on the plot should follow the diagonal, which is very close to being the case.
Figure 2Calculated versus experimental densities of salt water solutions with NaCl and NaBr at 25 oC and 1 bar.
References
Keyes, F.G., Keenan, J.H., Hill, P.G. and Moore, J.G., ”A Fundamental Equation for Liquid and Vapor Water”, presented at the Seventh International Conference on the Properties of Steam, Tokyo, Japan, Sept. 1968.
Laliberté, M. and Cooper, W.E., “Model for Calculating the Density of Aqueous Electrolyte Solutions”, J. Chem. Eng. Data 49, 2004, pp. 1141-1151.
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15381 | https://www.ncbi.nlm.nih.gov/books/NBK540990/ | Mode - StatPearls - NCBI Bookshelf
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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-.
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Mode
Anthony A. Mercadante; Steven Tenny.
Author Information and Affiliations
Authors
Anthony A. Mercadante 1; Steven Tenny 2.
Affiliations
1 Hackensack Meridian Medical School
2 University of Nebraska Medical Center
Last Update: July 17, 2023.
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Definition/Introduction
The mode refers to the most frequently occurring number found in a group of numbers. The mode is determined by collecting data to count the frequency of each result. The result with the highest count of occurrences is known as the mode of the set, which is also commonly referred to as the modal value. If multiple results all have the same high number of counts, then there can be multiple modes in the data set.
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Issues of Concern
The mode of a list or group of events is the most frequent output in that grouping. Therefore the mode will is the most abundant outcome. For example, if there is a set of numbers 1,2,2,2,2,2,2,3,3, the mode will be 2 as this is the number appears six times while the next most frequent number is 3, with only two occurrences. Besides working with numbers mode can also compare the number of occurrences of categorical lists. If for example, a zoologist is listing the animals in the zoo, then the list may look like: bird, bird, tiger, tiger, tiger, lion, lion, lion, lion, tiger. This example is important as there is not just one mode but two separate modes. One can consider the tiger and lion the mode of this set since they both appear four different times, making this an example of how there can be multiple modes. In this way, one can use mode in various situations with either quantitative or qualitative categories, making the mode crucial to many data sets and a prominent statistical figure in many medical research papers.
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Clinical Significance
The mode is one of the significant statistical values that go hand in hand with the mean and median of data sets to compare number sets or data for medical studies. Even with this connection,the mode has little correlation to the median and mean and therefore changes with the addition or subtraction of data points. It is also an important distinction to make that there can be no mode for a data set if all the values only appear once. Here there would be no mode as no values appear most frequently. The last situation in which mode is useful is when there is a set of data that does not have a normal distribution, making it bimodal, with each peak representing a different mode of the data set.
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Review Questions
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References
1.
Phillips S, Wilson WH. Categorial compositionality III: F-(co)algebras and the systematicity of recursive capacities in human cognition. PLoS One. 2012;7(4):e35028. [PMC free article: PMC3325926] [PubMed: 22514704]
2.
Ner-Kluza J, Wawrzykowski J, Franczyk M, Siberring J, Kankofer M. Identification of protein patterns in bovine placenta at early-mid pregnancy - Pilot studies. Rapid Commun Mass Spectrom. 2019 Jun 30;33(12):1084-1090. [PubMed: 30912867]
3.
Messer-Hannemann P, Bätz J, Lampe F, Klein A, Püschel K, Campbell GM, Morlock M. The influence of cavity preparation and press-fit cup implantation on restoring the hip rotation center. Clin Biomech (Bristol). 2019 Mar;63:185-192. [PubMed: 30913461]
4.
Rigaud B, Klopp A, Vedam S, Venkatesan A, Taku N, Simon A, Haigron P, de Crevoisier R, Brock KK, Cazoulat G. Deformable image registration for dose mapping between external beam radiotherapy and brachytherapy images of cervical cancer. Phys Med Biol. 2019 May 31;64(11):115023. [PubMed: 30913542]
5.
Shine R. Reproductive mode may determine geographic distributions in Australian venomous snakes (Pseudechis, Elapidae). Oecologia. 1987 Mar;71(4):608-612. [PubMed: 28312236]
Disclosure:Anthony Mercadante declares no relevant financial relationships with ineligible companies.
Disclosure:Steven Tenny declares no relevant financial relationships with ineligible companies.
Definition/Introduction
Issues of Concern
Clinical Significance
Review Questions
References
Copyright © 2025, StatPearls Publishing LLC.
This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal.
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15382 | https://www.aimspress.com/article/doi/10.3934/steme.2021020?viewType=HTML | STEM Education
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2021, Volume 1, Issue 4: 309-329. doi: 10.3934/steme.2021020
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Case study
The Laplace transform as an alternative general method for solving linear ordinary differential equations
William Guo ,
School of Engineering and Technology, Central Queensland University, Bruce Highway, North Rockhampton, QLD 4702, Australia
Academic Editor: Zlatko Jovanoski
Received: 01 October 2021 Revised: 01 November 2021
Abstract
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The Laplace transform is a popular approach in solving ordinary differential equations (ODEs), particularly solving initial value problems (IVPs) of ODEs. Such stereotype may confuse students when they face a task of solving ODEs without explicit initial condition(s). In this paper, four case studies of solving ODEs by the Laplace transform are used to demonstrate that, firstly, how much influence of the stereotype of the Laplace transform was on student's perception of utilizing this method to solve ODEs under different initial conditions; secondly, how the generalization of the Laplace transform for solving linear ODEs with generic initial conditions can not only break down the stereotype but also broaden the applicability of the Laplace transform for solving constant-coefficient linear ODEs. These case studies also show that the Laplace transform is even more robust for obtaining the specific solutions directly from the general solution once the initial values are assigned later. This implies that the generic initial conditions in the general solution obtained by the Laplace transform could be used as a point of control for some dynamic systems.
Keywords:+ Laplace transform,
+ linear ordinary differential equations,
+ initial value problem,
+ generic initial conditions,
+ convolution,
+ engineering mathematics
Citation: William Guo. The Laplace transform as an alternative general method for solving linear ordinary differential equations[J]. STEM Education, 2021, 1(4): 309-329. doi: 10.3934/steme.2021020
### Related Papers:
### Abstract
The Laplace transform is a popular approach in solving ordinary differential equations (ODEs), particularly solving initial value problems (IVPs) of ODEs. Such stereotype may confuse students when they face a task of solving ODEs without explicit initial condition(s). In this paper, four case studies of solving ODEs by the Laplace transform are used to demonstrate that, firstly, how much influence of the stereotype of the Laplace transform was on student's perception of utilizing this method to solve ODEs under different initial conditions; secondly, how the generalization of the Laplace transform for solving linear ODEs with generic initial conditions can not only break down the stereotype but also broaden the applicability of the Laplace transform for solving constant-coefficient linear ODEs. These case studies also show that the Laplace transform is even more robust for obtaining the specific solutions directly from the general solution once the initial values are assigned later. This implies that the generic initial conditions in the general solution obtained by the Laplace transform could be used as a point of control for some dynamic systems.
### References
| | |
--- |
| | Kreyszig, E., Advanced Engineering Mathematics, 10th ed. 2011, USA: Wiley. |
| | Fatoorehchi, H. and Rach, R., A method for inverting the Laplace transform of two classes of rational transfer functions in control engineering. Alexandria Engineering Journal, 2020, 59: 4879-4887. |
| | Ha, W. and Shin C., Seismic random noise attenuation in the Laplace domain using SVD. IEEE Access, 2021, 9: 62037. |
| | Grasso, F., Manetti, S., Piccirilli, M.C. and Reatti, A., A Laplace transform approach to the simulation of DC-DC converters. International Journal of Numerical Modelling: Electronic Networks, Devices and Fields, 2019, 32(5): e2618. |
| | Han, H. and Kim, H., The solution of exponential growth and exponential decay by using Laplace transform. International Journal of Difference Equations, 2020, 15(2):191-195. |
| | Huang, L., Deng, L., Li, A., Gao, R., Zhang, L. and Lei, W., A novel approach for solar greenhouse air temperature and heating load prediction based on Laplace transform. Journal of Building Engineering, 2021, 44: 102682. |
| | Daci, A. and Tola, S., Laplace transform, application in population growth. International Journal of Recent Technology and Engineering, 2019, 8(2): 954-957. |
| | Etzweiler, G. and S. Steele., The Laplace transformation of the impulse function for engineering problems. IEEE Transactions on Education, 1967, 10: 171-173. |
| | Zill, D.G., A First Course in Differential Equations with Modeling Applications, 10th ed. 2013, Boston, USA: Cengage Learning. |
| | Nise, N.S. Control Systems Engineering, 8th ed. 2019, USA: John Wiley & Sons. |
| | AL-Khazraji, H., Cole, C. and Guo, W., Analysing the impact of different classical controller strategies on the dynamics performance of production-inventory systems using state space approach. Journal of Modelling in Management, 2018, 13(1): 211-235. doi: 10.1108/JM2-08-2016-0071 |
| | AL-Khazraji, H., Cole, C. and Guo, W., Optimization and simulation of dynamic performance of production–inventory systems with multivariable controls. Mathematics, 2021, 9(5): 568. doi: 10.3390/math9050568 |
| | Guo, W.W., Advanced Mathematics for Engineering and Applied Sciences, 3rd ed. 2016, Sydney, Australia: Pearson. |
| | Ngo, V. and Ouzomgi, S., Teaching the Laplace transform using diagrams. The College Mathematics Journal, 1992, 23(4): 309-312. |
| | Holmberg, M. and Bernhard. J., University teachers' perspectives on the role of the Laplace transform in engineering education. European Journal of Engineering Education, 2017, 42(4): 413-428. doi: 10.1080/03043797.2016.1190957 |
| | Croft, A. and Davison, R., Engineering Mathematics, 5th ed. 2019, Harlow, UK: Pearson. |
| | Greenberg, M.D., Advanced Engineering Mathematics. 2nd ed. 1998, Upper Saddle River, USA: Prentice Hall. |
| | James, G., Modern Engineering Mathematics, 2nd ed. 1996, Harlow, UK: Addison-Wesley Longman. |
| | Yeung, K.S. and Chung, W.D., On solving differential equations using the Laplace transform. International Journal of Electrical Engineering and Education. 2007, 44(4): 373-376. |
| | Robertson, R.L., Laplace transform without integration. PRIMUS, 2017, 27(6): 606–617. doi: 10.1080/10511970.2016.1235643 |
| | Srivastava, H.M., Masjed-Jamei, M. and Aktas, R., Analytical solutions of some general classes of differential and integral equations by using the Laplace and Fourier transforms. Filomat, 2020, 34(9): 2869-2876. doi: 10.2298/FIL2009869S |
| | Guo, W., Unification of the common methods for solving the first-order linear ordinary differential equations. STEM Education, 2021, 1(2): 127-140. doi: 10.3934/steme.2021010 |
| | Guo, W., Li, W. and Tisdell, C.C., Effective pedagogy of guiding undergraduate engineering students solving first-order ordinary differential equations. Mathematics, 2021, 9(14):1623. doi: 10.3390/math9141623 |
| | Garner, B. and Garner, L., Retention of concepts and skills in traditional and reformed applied calculus. Mathematics Education Research Journal, 2001, 13: 165-184. |
| | Kwon, O.N., Rasmussen, C. and Allen, K., Students' retention of mathematical knowledge and skills in differential equations. School Science and Mathematics, 2005,105: 227-239. |
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William Guo. The Laplace transform as an alternative general method for solving linear ordinary differential equations[J]. STEM Education, 2021, 1(4): 309-329. doi: 10.3934/steme.2021020
William Guo. The Laplace transform as an alternative general method for solving linear ordinary differential equations[J]. STEM Education, 2021, 1(4): 309-329. doi: 10.3934/steme.2021020
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Figure 1.
A system represented by a single block diagram in the state space
Figure 2.
The series RL circuit with input $ f(t) = E = {E_0}\sin \omega t $
Figure 3.
Electric currents in the RL circuit with R = 10 Ω, L = 5 H, ω = 2, and E0 = 10 V
Figure 4.
Plot of the output for Case 2 with y(0) = 0 and y′(0) = 0
Figure 5.
An example of student's work on solving the ODE in Case 2 by the Laplace transform
Figure 6.
Plots of the solution to Case 3 with fixed y(0) = 1 and different values for $ y'(0) = {v_1} $
Figure 7.
Electric currents of the RL circuit in Case 1 with R = 10 Ω, L = 5 H, ω = 2, and E0 = 10 V [v0 = i(0) for –10, 0, and 10 amperes, respectively]
Figure 8.
Plots of the output of the ODE in Case 2 by Laplace transform with different initial values [Black: v0 = –3, v1 = –3; Red: v0 = 0, v1 = 0; Blue: v0 = 5, v1 = 5]
Figure 9.
The mixing problem for Case 4
Figure 10.
Plots of the mixing processes in the solutions (34) and (35) V = 4000 liters, x0 = 600 kg, y0 = 0 kg; blue curves: q1 = 40 l/m; red curves: q2 = 100 l/m
Figure 11.
Plots of the mixing processes in the solutions (36) and (37) V = 4000 liters, x0 = 500 kg, y0 = 100 kg; blue curves: q1 = 40 l/m; red curves: q2 = 100 l/m |
15383 | https://www.dailywritingtips.com/dozen-singular-or-plural/ | DailyWritingTips
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Dozen: Singular or Plural?
Referring to a recent post, a reader wants to know why I wrote, “Here are a dozen common subordinating conjunctions” and not, “Here is a dozen common subordinating conjunctions.”
Because I was referring to what I regard as twelve distinct conjunctions with different uses, I treated dozen as a plural.
Dozen is a collective noun, like committee. Collective nouns name groups of people or items. If the group is seen as identical or as acting in unison, the noun is treated as singular. If individuals in the group do not act in unison, the collective noun is treated as plural. For example:
The committee has agreed to appropriate money for new sidewalk.
The committee are in disagreement as to the importance of a new sidewalk.
The same rule applies to dozen. If dozen is regarded as a group of undifferentiated items, it takes a singular verb and singular pronouns. If dozen refers to a collection of individual persons or things, it takes a plural verb and pronouns.
On the Google Ngram Viewer, the construction “Here are a dozen” far outnumbers “Here is a dozen,” but the reverse is true in a Web search.
Although common, the singular construction “here is a dozen” is unidiomatic when it is followed by what are clearly distinct items. The construction is often used to introduce lists, as in these examples:
Here is a dozen top aquariums around the country.
Here is a dozen resources for every student.
The decision to regard dozen as singular or plural ultimately lies with the writer.
If the dozen consists of items that differ from one another in some marked way, then dozen should be regarded as plural. For example, the aquariums are all in different cities; the resources are of different kinds.
Better:
Here are a dozen top aquariums around the country.
Here are a dozen resources for every student.
The writer’s decision should be made on the basis of the noun that follows dozen and not because dozen is preceded by the indefinite article a.
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13 thoughts on “Dozen: Singular or Plural?”
Rewording the sentence can often relieve the awkward sound: “The committee members are in disagreement as to the importance of a new sidewalk.” The members, after all, are the ones who are disagreeing.
Another common example is the use of “couple.” If the couple is acting in unison, then the singular is used: “The couple is on a date.” If they are acting independently, then the plural is used:
“The couple are arguing with each other over the bill.”
An alternative might be: “The diners are arguing with each other over the bill.” However, this only works if it doesn’t alter the meaning or intent of the context; the preceding example could be focusing either on the couple or on the argument that they’re having.
“Dozen” strikes me as a different kind of collective noun from “committee” because it refers to a specific number and is like using the word “twelve.” I can’t imagine saying “A dozen eggs is needed to make an omelette for six people.”
Also, I would rewrite “The committee are in disagreement,” which sounds awkward, as “Members of the committee are in disagreement,” or “There is disagreement in the committee.”
Since the sentence exhibits an expletive structure (Here is/Here are), the subject of the sentence follows the verb. In this particular sentence, I would suggest that the noun “dozen” functions as an adjective modifying/describing the plural noun “conjunctions,” that your verb “are” is actually agreeing in number with “conjunctions,” not “dozen.”
Totally agree w/Bill, and the example of eggs is exactly what came to mind. There are a dozen eggs in a carton. Here are a dozen eggs. Maybe the explanation given by Larry Barkley is more applicable here. I also agree that the “committee” example is awkward, and I agree with the recasting proposed by Bill. I think the same would apply to the word “staff,” and I would also revise a sentence from “The staff were in disagreement” to “The staff members were in disagreement.” It is not the staff, as an entity, that is in disagreement with itself; it is individual members who disagree with each other.
Dozen is possibly different in some cases because it refers to a specific number, but it is still “a” dozen. That is why it seems it should be singular. “Here is a dozen eggs” sounds fine and correct to me, as does, “Here is a pair of keets for your birdcage.” Likewise, “The committee is reconsidering its numbers”, and, “The committee is divided about its position.” And in all similar cases: the team is, the family is, the group is, the public is, the government is, etc.
Thanks for all the constructive criticism. I’m going to take another run at this one.
Ought not the plural of “aquarium” be “aquaria” and not “aquariums”? That is how it was taught to me at Eton, being told it was the queen’s English and all that.
Many thanks! I’m pretty good with grammar, but this was something I didn’t know.
@venqax: When the entity is functioning as a unit, is unanimous or acting/thinking as one, I agree with using the singular form (“The family is going on vacation” or The committee is recommending new guidelines.”). When the entity is internally divided, you now have at least two factions, making it plural (IMO). “The staff (or better, the staff members) were in disagreement as to how to handle the money.”
Four years later, still useful.
I think rather than “If dozen is regarded as a group of undifferentiated items, it takes a singular verb and singular pronouns”, it is more like “If dozen is regarded as a single entity, …”
because the twelve eggs you need for your (rather large) omelet aren’t exactly differentiated from each other, but like Bill says, you wouldn’t say “a dozen eggs is …”
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15384 | https://www.youtube.com/watch?v=nSto5HuDJf4 | How to Find the Perimeter of a Square | Math with Mr. J
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[Music] Welcome to Math with Mr. J. In this video, I'm going to cover how to find the perimeter of a square. Now, remember, perimeter is the distance around the outside of a shape. Let's jump into our examples starting with number one. So, let's find the perimeter of this square. And we can do this by adding the lengths of all of the sides. Now for this square each side is 8 in. So we have 8 in 8 in 8 8 in and 8 in. So perimeter equals 8 + 8 + 8 + 8. 8 + 8 is 16 + 8 is 24 + 8 is 32. So, the perimeter is 32 in. That's the distance around the outside of that square. Now, I do want to mention for squares, we can just multiply a side length by 4 since all of the sides are the same. Instead of 8 + 8 + 8 + 8, we can just do 8 4. We get the same thing either way. So we can use the formula perimeter equals 4 s 4 a side length for this square. Each side is 8 in. So let's plug in 8 for s. So 4 8. 4 8 is 32. So the perimeter equals 32 in that way as well. So for the perimeter of a square, we can either add all of the sides or we can multiply a side length by four. Either way will work. Let's move on to number two where we have a square with sides that are 11 m. And you'll notice that we only have one side labeled. That's because this is a square and we know all of the sides are the same. So all of the sides are 11 m. These tick marks right here show us that all of the sides are the same. We have one tick mark on each side indicating that they are all again the same. They are all 11 m. Then we have right angle symbols as well. 90° angle symbols. I wanted to include all of these symbols in case you come across them. That way you will know what they mean. So something to keep in mind. As far as perimeter, let's start by adding all of the sides. So we have perimeter equals. Again, all of the sides are 11 m. So 11 + 11 + 11 + 11. 11 + 11 is 22 + 11 is 33 + 11 is 44. So the perimeter is 44 m. Now we can also find perimeter of a square by multiplying a side length by 4. So we can use the formula perimeter equals 4s. Let's plug in 11 fors. So 4 11 4 11 gives us a perimeter of 44 m that way as well. So again, we can find the perimeter of a square by either adding all of the side lengths, so all four sides, or we can multiply a side length by four since it's a square and all of the sides are the same. So there you have it. There's how to find the perimeter of a square. I hope that helped. Thanks so much for watching. Until next time, peace. |
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Fluid Dynamics
Table of Contents
Unit 4 Overview: Incompressible inviscid flows
4.1 Irrotational flow
4.2 Velocity potential
4.3 Stream function
4.4 Circulation and vorticity
4.5 Kelvin's circulation theorem
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💨 fluid dynamics review
4.2 Velocity potential
Citation:
MLA
Velocity potential simplifies fluid flow analysis by describing irrotational flows with a scalar function. It reduces variables needed and satisfies Laplace's equation for incompressible flows. This powerful tool allows for easier calculation of velocity fields and other flow properties.
Understanding velocity potential is crucial for analyzing various flow scenarios. It enables the study of uniform flows, sources, sinks, and flow around objects like cylinders and spheres. Complex potential theory extends this concept to two-dimensional flows using complex analysis.
Definition of velocity potential
Velocity potential is a scalar function that describes the velocity field of an irrotational flow
It simplifies the analysis of fluid flow by reducing the number of variables needed to describe the flow
The velocity potential is denoted by the Greek letter $\phi$ and has units of $m^2/s$
Irrotational flow and velocity potential
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Irrotational flow is a type of fluid flow where the fluid particles do not rotate about their own axis
In irrotational flow, the curl of the velocity field is zero $(\nabla \times \vec{V} = 0)$
For irrotational flows, a velocity potential exists such that the velocity field can be expressed as the gradient of the velocity potential $(\vec{V} = \nabla \phi)$
Laplace's equation for velocity potential
For incompressible and irrotational flows, the velocity potential satisfies Laplace's equation $(\nabla^2 \phi = 0)$
Laplace's equation is a second-order partial differential equation that describes the spatial distribution of the velocity potential
Solving Laplace's equation with appropriate boundary conditions allows for the determination of the velocity potential and, consequently, the velocity field
Properties of velocity potential
The velocity potential is a powerful tool for analyzing irrotational flows due to its unique properties
These properties simplify the analysis and provide insights into the behavior of the flow
Relationship between velocity potential and velocity field
The velocity field can be obtained from the velocity potential by taking its gradient $(\vec{V} = \nabla \phi)$
The x-component of velocity is given by $u = \frac{\partial \phi}{\partial x}$, the y-component by $v = \frac{\partial \phi}{\partial y}$, and the z-component by $w = \frac{\partial \phi}{\partial z}$
This relationship allows for the determination of the velocity field once the velocity potential is known
Uniqueness of velocity potential
For a given irrotational flow, the velocity potential is unique up to an additive constant
This means that if two velocity potentials satisfy the same boundary conditions, they will differ only by a constant value
The uniqueness property ensures that the solution for the velocity potential is well-defined and consistent
Superposition principle for velocity potential
The velocity potential satisfies the superposition principle due to the linearity of Laplace's equation
If $\phi_1$ and $\phi_2$ are velocity potentials for two irrotational flows, then their sum $(\phi_1 + \phi_2)$ is also a valid velocity potential
The superposition principle allows for the construction of complex flow fields by combining simpler flow solutions
Boundary conditions for velocity potential
To solve for the velocity potential, appropriate boundary conditions must be specified
The boundary conditions describe the behavior of the flow at the boundaries of the domain and ensure a unique solution
Solid boundary conditions
At a solid boundary, the fluid velocity normal to the surface must be zero to satisfy the no-penetration condition
This condition is expressed as $\frac{\partial \phi}{\partial n} = 0$, where $n$ is the normal direction to the solid surface
For moving solid boundaries, the normal velocity of the fluid must match the normal velocity of the boundary
Free surface boundary conditions
At a free surface (e.g., the interface between a liquid and air), two conditions must be satisfied:
Kinematic condition: The fluid particles at the free surface must remain on the surface
Dynamic condition: The pressure at the free surface must be equal to the atmospheric pressure
These conditions lead to the free surface boundary condition $\frac{\partial \phi}{\partial n} = \frac{\partial \eta}{\partial t}$, where $\eta$ is the free surface elevation
Far-field boundary conditions
Far away from the region of interest, the flow is assumed to be undisturbed and uniform
The velocity potential should approach the uniform flow potential $(\phi_\infty)$ as the distance from the origin tends to infinity
This condition is expressed as $\lim_{r \to \infty} \phi = \phi_\infty$, where $r$ is the radial distance from the origin
Applications of velocity potential
Velocity potential is a powerful tool for analyzing various types of irrotational flows
By solving for the velocity potential, the velocity field and other flow properties can be determined
Uniform flow and velocity potential
For a uniform flow with velocity $U$ in the x-direction, the velocity potential is given by $\phi = Ux$
The velocity field is constant and equal to $\vec{V} = (U, 0, 0)$
Uniform flow is a fundamental building block for more complex flows and is often used as a far-field boundary condition
Source, sink, and doublet flow
A source is a point from which fluid emanates uniformly in all directions, while a sink is a point where fluid is uniformly absorbed
The velocity potential for a source or sink with strength $m$ is given by $\phi = \frac{m}{4\pi r}$, where $r$ is the radial distance from the source or sink
A doublet is formed by placing a source and a sink of equal strength infinitesimally close to each other
The velocity potential for a doublet with strength $\mu$ oriented along the x-axis is given by $\phi = \frac{\mu \cos \theta}{4\pi r^2}$, where $\theta$ is the angle between the radial direction and the x-axis
Flow around a cylinder using velocity potential
The flow around a circular cylinder can be modeled using a doublet and a uniform flow
The velocity potential for the flow around a cylinder of radius $a$ with a uniform flow $U$ in the x-direction is given by $\phi = U(r + \frac{a^2}{r})\cos \theta$
By taking the gradient of the velocity potential, the velocity field and streamlines can be obtained
Flow around a sphere using velocity potential
The flow around a sphere can be modeled using a doublet and a uniform flow, similar to the flow around a cylinder
The velocity potential for the flow around a sphere of radius $a$ with a uniform flow $U$ in the x-direction is given by $\phi = U(r + \frac{a^3}{2r^2})\cos \theta$
The velocity field and streamlines can be obtained by taking the gradient of the velocity potential
Complex potential theory
Complex potential theory is an extension of velocity potential theory that uses complex analysis to study two-dimensional irrotational flows
It provides a powerful framework for solving flow problems and visualizing flow patterns
Complex velocity potential definition
The complex velocity potential $w(z)$ is a complex-valued function that combines the velocity potential $\phi$ and the stream function $\psi$
It is defined as $w(z) = \phi(x, y) + i\psi(x, y)$, where $z = x + iy$ is the complex coordinate
The real part of the complex velocity potential represents the velocity potential, while the imaginary part represents the stream function
Relationship between complex potential and velocity field
The velocity field can be obtained from the complex velocity potential by taking its derivative with respect to the complex coordinate $z$
The complex velocity is given by $V(z) = \frac{dw}{dz} = u - iv$, where $u$ and $v$ are the x and y components of the velocity, respectively
This relationship allows for the determination of the velocity field from the complex velocity potential
Conformal mapping using complex potential
Conformal mapping is a technique that uses complex analysis to transform one flow domain into another while preserving local angles
By finding a suitable complex velocity potential, a flow in a complicated domain can be mapped to a simpler domain where the solution is known
Common conformal mappings include the Joukowski transformation, which maps the flow around a cylinder to the flow around an airfoil
Numerical methods for velocity potential
Analytical solutions for velocity potential are not always possible, especially for flows with complex geometries or boundary conditions
Numerical methods provide a way to approximate the solution of the velocity potential and the associated flow field
Finite difference method for velocity potential
The finite difference method discretizes the flow domain into a grid and approximates the derivatives in Laplace's equation using finite differences
The resulting system of linear equations is solved to obtain the velocity potential at the grid points
The velocity field can then be calculated from the velocity potential using finite difference approximations of the gradient
Boundary element method for velocity potential
The boundary element method (BEM) is a numerical technique that discretizes only the boundaries of the flow domain
It is based on the integral formulation of Laplace's equation and uses fundamental solutions (e.g., sources, sinks, and doublets) to represent the velocity potential
BEM reduces the dimensionality of the problem and is particularly useful for exterior flow problems with unbounded domains
Comparison of numerical methods for velocity potential
The choice of numerical method depends on the specific flow problem, the geometry of the domain, and the desired accuracy
Finite difference methods are straightforward to implement but may require a large number of grid points for accurate solutions
BEM is more efficient for exterior flow problems and can handle complex geometries, but it requires the evaluation of singular integrals and the solution of dense linear systems
Other numerical methods, such as finite element methods and spectral methods, can also be used to solve for the velocity potential
Limitations of velocity potential
While velocity potential theory is a powerful tool for analyzing irrotational flows, it has certain limitations that should be considered
Applicability to rotational flows
Velocity potential theory is strictly applicable only to irrotational flows, where the curl of the velocity field is zero
For rotational flows, such as those encountered in turbulence or flows with vorticity, velocity potential theory cannot be directly applied
In such cases, alternative formulations, such as the stream function or the vorticity-velocity formulation, may be more appropriate
Nonlinear effects and velocity potential
Velocity potential theory is based on the assumption of inviscid and irrotational flow, which leads to a linear governing equation (Laplace's equation)
However, many real-world flows exhibit nonlinear effects, such as flow separation, vortex shedding, and turbulence
These nonlinear effects cannot be captured by the linear velocity potential formulation, and more advanced models, such as the Navier-Stokes equations, are required
Compressibility effects and velocity potential
Velocity potential theory assumes that the fluid is incompressible, meaning that the density of the fluid remains constant
For flows with significant compressibility effects, such as high-speed gas flows or flows with large pressure variations, the incompressibility assumption breaks down
In such cases, the velocity potential formulation must be modified to account for the compressibility of the fluid, leading to more complex governing equations and boundary conditions
###### 4.3 Stream function |
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Elementary Algebra 2e
3.4 Solve Geometry Applications: Triangles, Rectangles, and the Pythagorean Theorem
Elementary Algebra 2e
3.4
Solve Geometry Applications: Triangles, Rectangles, and the Pythagorean Theorem
Search for key terms or text.
Learning Objectives
By the end of this section, you will be able to:
Solve applications using properties of triangles
Use the Pythagorean Theorem
Solve applications using rectangle properties
Be Prepared 3.12
Before you get started, take this readiness quiz.
Simplify: 12(6h).12(6h).
If you missed this problem, review Example 1.122.
Be Prepared 3.13
The length of a rectangle is three less than the width. Let w represent the width. Write an expression for the length of the rectangle.
If you missed this problem, review Example 1.26.
Be Prepared 3.14
Solve: A=12bhA=12bh for b when A=260A=260 and h=52.h=52.
If you missed this problem, review Example 2.61.
Be Prepared 3.15
Simplify: √144.144−−−√.
If you missed this problem, review Example 1.111.
Solve Applications Using Properties of Triangles
In this section we will use some common geometry formulas. We will adapt our problem-solving strategy so that we can solve geometry applications. The geometry formula will name the variables and give us the equation to solve. In addition, since these applications will all involve shapes of some sort, most people find it helpful to draw a figure and label it with the given information. We will include this in the first step of the problem solving strategy for geometry applications.
How To
Solve Geometry Applications.
Step 1.
Read the problem and make sure all the words and ideas are understood. Draw the figure and label it with the given information.
Step 2.
Identify what we are looking for.
Step 3.
Label what we are looking for by choosing a variable to represent it.
Step 4.
Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
Step 5.
Solve the equation using good algebra techniques.
Step 6.
Check the answer by substituting it back into the equation solved in step 5 and by making sure it makes sense in the context of the problem.
Step 7.
Answer the question with a complete sentence.
We will start geometry applications by looking at the properties of triangles. Let’s review some basic facts about triangles. Triangles have three sides and three interior angles. Usually each side is labeled with a lowercase letter to match the uppercase letter of the opposite vertex.
The plural of the word vertex is vertices. All triangles have three vertices. Triangles are named by their vertices: The triangle in Figure 3.4 is called △ABC.△ABC.
Figure
3.4
Triangle ABC has vertices A, B, and C. The lengths of the sides are a, b, and c.
The three angles of a triangle are related in a special way. The sum of their measures is 180°.180°. Note that we read m∠Am∠A as “the measure of angle A.” So in △ABC△ABC in Figure 3.4,
m∠A+m∠B+m∠C=180°
m∠A+m∠B+m∠C=180°
Because the perimeter of a figure is the length of its boundary, the perimeter of △ABC△ABC is the sum of the lengths of its three sides.
P=a+b+c
P=a+b+c
To find the area of a triangle, we need to know its base and height. The height is a line that connects the base to the opposite vertex and makes a 90°90° angle with the base. We will draw △ABC△ABC again, and now show the height, h. See Figure 3.5.
Figure
3.5
The formula for the area of △ABC△ABC is A=12bh,A=12bh, where b is the base and h is the height.
Triangle Properties
For △ABC△ABC
Angle measures:
m∠A+m∠B+m∠C=180
m∠A+m∠B+m∠C=180
The sum of the measures of the angles of a triangle is 180°.180°.
Perimeter:
P=a+b+c
P=a+b+c
The perimeter is the sum of the lengths of the sides of the triangle.
Area:
A=12bh,b=base,h=height
A=12bh,b=base,h=height
The area of a triangle is one-half the base times the height.
Example 3.34
The measures of two angles of a triangle are 55 and 82 degrees. Find the measure of the third angle.
Solution
| | |
--- |
| Step 1. Read the problem. Draw the figure and label it with the given information. | |
| Step 2. Identify what you are looking for. | the measure of the third angle in a triangle |
| Step 3. Name. Choose a variable to represent it. | Let x=x= the measure of the angle. |
| Step 4. Translate. | |
| Write the appropriate formula and substitute. | m∠A+m∠B+m∠C=180m∠A+m∠B+m∠C=180 |
| Step 5. Solve the equation. | 55+82+x=180137+x=180x=4355+82+x137+xx===18018043 |
| Step 6. Check. 55+82+43≟180180=180✓55+82+43180≟=180180✓ | |
| Step 7. Answer the question. | The measure of the third angle is 43 degrees. |
Try It 3.67
The measures of two angles of a triangle are 31 and 128 degrees. Find the measure of the third angle.
Try It 3.68
The measures of two angles of a triangle are 49 and 75 degrees. Find the measure of the third angle.
Example 3.35
The perimeter of a triangular garden is 24 feet. The lengths of two sides are four feet and nine feet. How long is the third side?
Solution
| | |
--- |
| Step 1. Read the problem. Draw the figure and label it with the given information. | |
| Step 2. Identify what you are looking for. | length of the third side of a triangle |
| Step 3. Name. Choose a variable to represent it. | Let c=c= the third side. |
| Step 4. Translate. | |
| Write the appropriate formula and substitute. | |
| Substitute in the given information. | |
| Step 5. Solve the equation. | |
| Step 6. Check. P=a+b+c24≟4+9+1124=24✓P2424=≟=a+b+c4+9+1124✓ | |
| Step 7. Answer the question. | The third side is 11 feet long. |
Try It 3.69
The perimeter of a triangular garden is 48 feet. The lengths of two sides are 18 feet and 22 feet. How long is the third side?
Try It 3.70
The lengths of two sides of a triangular window are seven feet and five feet. The perimeter is 18 feet. How long is the third side?
Example 3.36
The area of a triangular church window is 90 square meters. The base of the window is 15 meters. What is the window’s height?
Solution
| | |
--- |
| Step 1. Read the problem. Draw the figure and label it with the given information. | Area =90m2=90m2 |
| Step 2. Identify what you are looking for. | height of a triangle |
| Step 3. Name. Choose a variable to represent it. | Let h=h= the height. |
| Step 4. Translate. | |
| Write the appropriate formula. | |
| Substitute in the given information. | |
| Step 5. Solve the equation. | |
| Step 6. Check. A=12bh90≟12⋅15⋅1290=90✓A9090=≟=12bh12⋅15⋅1290✓ | |
| Step 7. Answer the question. | The height of the triangle is 12 meters. |
Try It 3.71
The area of a triangular painting is 126 square inches. The base is 18 inches. What is the height?
Try It 3.72
A triangular tent door has an area of 15 square feet. The height is five feet. What is the base?
The triangle properties we used so far apply to all triangles. Now we will look at one specific type of triangle—a right triangle. A right triangle has one 90°90° angle, which we usually mark with a small square in the corner.
Right Triangle
A right triangle has one 90°90° angle, which is often marked with a square at the vertex.
Example 3.37
One angle of a right triangle measures 28°.28°. What is the measure of the third angle?
Solution
| | |
--- |
| Step 1. Read the problem. Draw the figure and label it with the given information. | |
| Step 2. Identify what you are looking for. | the measure of an angle |
| Step 3. Name. Choose a variable to represent it. | Let x=x= the measure of an angle. |
| Step 4. Translate. | m∠A+m∠B+m∠C=180m∠A+m∠B+m∠C=180 |
| Write the appropriate formula and substitute. | x+90+28=180x+90+28=180 |
| Step 5. Solve the equation. | x+118=180x=62x+118x==18062 |
| Step 6. Check. 180≟90+28+62180=180✓180180≟=90+28+62180✓ | |
| Step 7. Answer the question. | The measure of the third angle is 62°. |
Try It 3.73
One angle of a right triangle measures 56°.56°. What is the measure of the other small angle?
Try It 3.74
One angle of a right triangle measures 45°.45°. What is the measure of the other small angle?
In the examples we have seen so far, we could draw a figure and label it directly after reading the problem. In the next example, we will have to define one angle in terms of another. We will wait to draw the figure until we write expressions for all the angles we are looking for.
Example 3.38
The measure of one angle of a right triangle is 20 degrees more than the measure of the smallest angle. Find the measures of all three angles.
Solution
| | |
--- |
| Step 1. Read the problem. | |
| Step 2. Identify what you are looking for. | the measures of all three angles |
| Step 3. Name. Choose a variable to represent it. | Let a=1sta=1st angle. a+20=2nda+20=2nd angle 90=3rd90=3rd angle (the right angle) |
| Draw the figure and label it with the given information | |
| Step 4. Translate | |
| Write the appropriate formula. Substitute into the formula. | |
| Step 5. Solve the equation. | 55 90 third angle |
| Step 6. Check. 35+55+90≟180180=180✓35+55+90180≟=180180✓ | |
| Step 7. Answer the question. | The three angles measure 35°, 55°, and 90°. |
Try It 3.75
The measure of one angle of a right triangle is 50° more than the measure of the smallest angle. Find the measures of all three angles.
Try It 3.76
The measure of one angle of a right triangle is 30° more than the measure of the smallest angle. Find the measures of all three angles.
Use the Pythagorean Theorem
We have learned how the measures of the angles of a triangle relate to each other. Now, we will learn how the lengths of the sides relate to each other. An important property that describes the relationship among the lengths of the three sides of a right triangle is called the Pythagorean Theorem. This theorem has been used around the world since ancient times. It is named after the Greek philosopher and mathematician, Pythagoras, who lived around 500 BC.
Before we state the Pythagorean Theorem, we need to introduce some terms for the sides of a triangle. Remember that a right triangle has a 90°90° angle, marked with a small square in the corner. The side of the triangle opposite the 90°90° angle is called the hypotenuse and each of the other sides are called legs.
The Pythagorean Theorem tells how the lengths of the three sides of a right triangle relate to each other. It states that in any right triangle, the sum of the squares of the lengths of the two legs equals the square of the length of the hypotenuse. In symbols we say: in any right triangle, a2+b2=c2,a2+b2=c2, where aandbaandb are the lengths of the legs and cc is the length of the hypotenuse.
Writing the formula in every exercise and saying it aloud as you write it, may help you remember the Pythagorean Theorem.
The Pythagorean Theorem
In any right triangle, a2+b2=c2.a2+b2=c2.
where a and b are the lengths of the legs, c is the length of the hypotenuse.
To solve exercises that use the Pythagorean Theorem, we will need to find square roots. We have used the notation √mm−−√ and the definition:
If m=n2,m=n2, then √m=n,m−−√=n, for n≥0.n≥0.
For example, we found that √2525−−√ is 5 because 25=52.25=52.
Because the Pythagorean Theorem contains variables that are squared, to solve for the length of a side in a right triangle, we will have to use square roots.
Example 3.39
Use the Pythagorean Theorem to find the length of the hypotenuse shown below.
Solution
| | |
--- |
| Step 1. Read the problem. | |
| Step 2. Identify what you are looking for. | the length of the hypotenuse of the triangle |
| Step 3. Name. Choose a variable to represent it. Label side c on the figure. | Let c = the length of the hypotenuse. |
| Step 4. Translate. | |
| Write the appropriate formula. | a2+b2=c2a2+b2=c2 |
| Substitute. | 32+42=c232+42=c2 |
| Step 5. Solve the equation. | 9+16=c29+16=c2 |
| Simplify. | 25=c225=c2 |
| Use the definition of square root. | √25=c25−−√=c |
| Simplify. | 5=c5=c |
| Step 6. Check. | |
| Step 7. Answer the question. | The length of the hypotenuse is 5. |
Try It 3.77
Use the Pythagorean Theorem to find the length of the hypotenuse in the triangle shown below.
Try It 3.78
Use the Pythagorean Theorem to find the length of the hypotenuse in the triangle shown below.
Example 3.40
Use the Pythagorean Theorem to find the length of the leg shown below.
Solution
| | | |
---
| Step 1. Read the problem. | | |
| Step 2. Identify what you are looking for. | | the length of the leg of the triangle |
| Step 3. Name. Choose a variable to represent it. | | Let b = the leg of the triangle. |
| Lable side b. | | |
| Step 4. Translate | | |
| Write the appropriate formula. | | a2+b2=c2a2+b2=c2 |
| Substitute. | | 52+b2=13252+b2=132 |
| Step 5. Solve the equation. | | 25+b2=16925+b2=169 |
| Isolate the variable term. | | b2=144b2=144 |
| Use the definition of square root. | | b2=√144b2=144−−−√ |
| Simplify. | | b=12b=12 |
| Step 6. Check. | | |
| Step 7. Answer the question. | | The length of the leg is 12. |
Try It 3.79
Use the Pythagorean Theorem to find the length of the leg in the triangle shown below.
Try It 3.80
Use the Pythagorean Theorem to find the length of the leg in the triangle shown below.
Example 3.41
Kelvin is building a gazebo and wants to brace each corner by placing a 10″10'' piece of wood diagonally as shown above.
If he fastens the wood so that the ends of the brace are the same distance from the corner, what is the length of the legs of the right triangle formed? Approximate to the nearest tenth of an inch.
Solution
| | |
--- |
| Step 1. Read the problem. | |
| Step 2. Identify what we are looking for. | the distance from the corner that the bracket should be attached |
| Step 3. Name. Choose a variable to represent it. | Let x=x= the distance from the corner. |
| Step 4. Translate Write the appropriate formula and substitute. | a2+b2=c2x2+x2=102a2+b2x2+x2==c2102 |
| Step 5. Solve the equation. Isolate the variable. Use the definition of square root. Simplify. Approximate to the nearest tenth. | 2x2=100x2=50x=√50x≈7.12x2x2xx===≈1005050−−√7.1 |
| Step 6. Check. a2+b2=c2(7.1)2+(7.1)2≈102Yes.a2+b2(7.1)2+(7.1)2=≈c2102Yes. | |
| Step 7. Answer the question. | Kelvin should fasten each piece of wood approximately 7.1" from the corner. |
Table
3.10
Try It 3.81
John puts the base of a 13-foot ladder five feet from the wall of his house as shown below. How far up the wall does the ladder reach?
Try It 3.82
Randy wants to attach a 17 foot string of lights to the top of the 15 foot mast of his sailboat, as shown below. How far from the base of the mast should he attach the end of the light string?
Solve Applications Using Rectangle Properties
You may already be familiar with the properties of rectangles. Rectangles have four sides and four right (90°)(90°) angles. The opposite sides of a rectangle are the same length. We refer to one side of the rectangle as the length, L, and its adjacent side as the width, W.
The distance around this rectangle is L+W+L+W,L+W+L+W, or 2L+2W.2L+2W. This is the perimeter, P, of the rectangle.
P=2L+2W
P=2L+2W
What about the area of a rectangle? Imagine a rectangular rug that is 2-feet long by 3-feet wide. Its area is 6 square feet. There are six squares in the figure.
A=6A=2·3A=L·W
A=6A=2⋅3A=L⋅W
The area is the length times the width.
The formula for the area of a rectangle is A=LW.A=LW.
Properties of Rectangles
Rectangles have four sides and four right (90°)(90°) angles.
The lengths of opposite sides are equal.
The perimeter of a rectangle is the sum of twice the length and twice the width.
P=2L+2W
P=2L+2W
The area of a rectangle is the product of the length and the width.
A=L·W
A=L⋅W
Example 3.42
The length of a rectangle is 32 meters and the width is 20 meters. What is the perimeter?
Solution
| | |
--- |
| Step 1. Read the problem. Draw the figure and label it with the given information. | |
| Step 2. Identify what you are looking for. | the perimeter of a rectangle |
| Step 3. Name. Choose a variable to represent it. | Let P = the perimeter. |
| Step 4. Translate. | |
| Write the appropriate formula. | |
| Substitute. | |
| Step 5. Solve the equation. | |
| Step 6. Check. P≟10420+32+20+32≟104104=104✓P20+32+20+32104≟≟=104104104✓ | |
| Step 7. Answer the question. | The perimeter of the rectangle is 104 meters. |
Try It 3.83
The length of a rectangle is 120 yards and the width is 50 yards. What is the perimeter?
Try It 3.84
The length of a rectangle is 62 feet and the width is 48 feet. What is the perimeter?
Example 3.43
The area of a rectangular room is 168 square feet. The length is 14 feet. What is the width?
Solution
| | |
--- |
| Step 1. Read the problem. Draw the figure and label it with the given information. | |
| Step 2. Identify what you are looking for. | the width of a rectangular room |
| Step 3. Name. Choose a variable to represent it. | Let W = the width. |
| Step 4. Translate. | |
| Write the appropriate formula. | A=LWA=LW |
| Substitute. | 168=14W168=14W |
| Step 5. Solve the equation. | 16814=14W1416814=14W14 12=W12=W |
| Step 6. Check. A=LW168≟14⋅12168=168✓ | |
| Step 7. Answer the question. | The width of the room is 12 feet. |
Try It 3.85
The area of a rectangle is 598 square feet. The length is 23 feet. What is the width?
Try It 3.86
The width of a rectangle is 21 meters. The area is 609 square meters. What is the length?
Example 3.44
Find the length of a rectangle with perimeter 50 inches and width 10 inches.
Solution
| | |
--- |
| Step 1. Read the problem. Draw the figure and label it with the given information. | |
| Step 2. Identify what you are looking for. | the length of the rectangle |
| Step 3. Name. Choose a variable to represent it. | Let L = the length. |
| Step 4. Translate. | |
| Write the appropriate formula. | |
| Substitute. | |
| Step 5. Solve the equation. | |
| Step 6. Check. P=5015+10+15+10≟5050=50✓ | |
| Step 7. Answer the question. | The length is 15 inches. |
Try It 3.87
Find the length of a rectangle with: perimeter 80 and width 25.
Try It 3.88
Find the length of a rectangle with: perimeter 30 and width 6.
We have solved problems where either the length or width was given, along with the perimeter or area; now we will learn how to solve problems in which the width is defined in terms of the length. We will wait to draw the figure until we write an expression for the width so that we can label one side with that expression.
Example 3.45
The width of a rectangle is two feet less than the length. The perimeter is 52 feet. Find the length and width.
Solution
| | |
--- |
| Step 1. Read the problem. | |
| Step 2. Identify what you are looking for. | the length and width of a rectangle |
| Step 3. Name. Choose a variable to represent it. Since the width is defined in terms of the length, we let L = length. The width is two feet less than the length, so we let L − 2 = width. | P=52 ft |
| Step 4. Translate. | |
| Write the appropriate formula. The formula for the perimeter of a rectangle relates all the information. | P=2L+2W |
| Substitute in the given information. | 52=2L+2(L−2) |
| Step 5. Solve the equation. | 52=2L+2L−4 |
| Combine like terms. | 52=4L−4 |
| Add 4 to each side. | 56=4L |
| Divide by 4. | 564=4L4 14=L The length is 14 feet. |
| Now we need to find the width. | The width is L−2. The width is 12 feet. |
| Step 6. Check. Since 14+12+14+12=52, this works! | |
| Step 7. Answer the question. | The length is 14 feet and the width is 12 feet. |
Try It 3.89
The width of a rectangle is seven meters less than the length. The perimeter is 58 meters. Find the length and width.
Try It 3.90
The length of a rectangle is eight feet more than the width. The perimeter is 60 feet. Find the length and width.
Example 3.46
The length of a rectangle is four centimeters more than twice the width. The perimeter is 32 centimeters. Find the length and width.
Solution
| | |
--- |
| Step 1. Read the problem. | |
| Step 2. Identify what you are looking for. | the length and the width |
| Step 3. Name. Choose a variable to represent the width. | |
| The length is four more than twice the width. | |
| Step 4. Translate | |
| Write the appropriate formula. | |
| Substitute in the given information. | |
| Step 5. Solve the equation. | 12 The length is 12 cm. |
| Step 6. Check. P=2L+2W32≟2⋅12+2⋅432=32✓ | |
| Step 7. Answer the question. | The length is 12 cm and the width is 4 cm. |
Try It 3.91
The length of a rectangle is eight more than twice the width. The perimeter is 64. Find the length and width.
Try It 3.92
The width of a rectangle is six less than twice the length. The perimeter is 18. Find the length and width.
Example 3.47
The perimeter of a rectangular swimming pool is 150 feet. The length is 15 feet more than the width. Find the length and width.
Solution
| | |
--- |
| Step 1. Read the problem. Draw the figure and label it with the given information. | P=150 ft |
| Step 2. Identify what you are looking for. | the length and the width of the pool |
| Step 3. Name. Choose a variable to represent the width. The length is 15 feet more than the width. | |
| Step 4. Translate | |
| Write the appropriate formula. | |
| Substitute. | |
| Step 5. Solve the equation. | |
| Step 6. Check. P=2L+2W150≟2(45)+2(30)150=150✓ | |
| Step 7. Answer the question. | The length of the pool is 45 feet and the width is 30 feet. |
Try It 3.93
The perimeter of a rectangular swimming pool is 200 feet. The length is 40 feet more than the width. Find the length and width.
Try It 3.94
The length of a rectangular garden is 30 yards more than the width. The perimeter is 300 yards. Find the length and width.
Section 3.4 Exercises
Practice Makes Perfect
Solving Applications Using Triangle Properties
In the following exercises, solve using triangle properties.
211.
The measures of two angles of a triangle are 26 and 98 degrees. Find the measure of the third angle.
The measures of two angles of a triangle are 61 and 84 degrees. Find the measure of the third angle.
213.
The measures of two angles of a triangle are 105 and 31 degrees. Find the measure of the third angle.
The measures of two angles of a triangle are 47 and 72 degrees. Find the measure of the third angle.
215.
The perimeter of a triangular pool is 36 yards. The lengths of two sides are 10 yards and 15 yards. How long is the third side?
A triangular courtyard has perimeter 120 meters. The lengths of two sides are 30 meters and 50 meters. How long is the third side?
217.
If a triangle has sides 6 feet and 9 feet and the perimeter is 23 feet, how long is the third side?
If a triangle has sides 14 centimeters and 18 centimeters and the perimeter is 49 centimeters, how long is the third side?
219.
A triangular flag has base one foot and height 1.5 foot. What is its area?
A triangular window has base eight feet and height six feet. What is its area?
221.
What is the base of a triangle with area 207 square inches and height 18 inches?
What is the height of a triangle with area 893 square inches and base 38 inches?
223.
One angle of a right triangle measures 33 degrees. What is the measure of the other small angle?
One angle of a right triangle measures 51 degrees. What is the measure of the other small angle?
225.
One angle of a right triangle measures 22.5 degrees. What is the measure of the other small angle?
One angle of a right triangle measures 36.5 degrees. What is the measure of the other small angle?
227.
The perimeter of a triangle is 39 feet. One side of the triangle is one foot longer than the second side. The third side is two feet longer than the second side. Find the length of each side.
The perimeter of a triangle is 35 feet. One side of the triangle is five feet longer than the second side. The third side is three feet longer than the second side. Find the length of each side.
229.
One side of a triangle is twice the shortest side. The third side is five feet more than the shortest side. The perimeter is 17 feet. Find the lengths of all three sides.
One side of a triangle is three times the shortest side. The third side is three feet more than the shortest side. The perimeter is 13 feet. Find the lengths of all three sides.
231.
The two smaller angles of a right triangle have equal measures. Find the measures of all three angles.
The measure of the smallest angle of a right triangle is 20° less than the measure of the next larger angle. Find the measures of all three angles.
233.
The angles in a triangle are such that one angle is twice the smallest angle, while the third angle is three times as large as the smallest angle. Find the measures of all three angles.
The angles in a triangle are such that one angle is 20° more than the smallest angle, while the third angle is three times as large as the smallest angle. Find the measures of all three angles.
Use the Pythagorean Theorem
In the following exercises, use the Pythagorean Theorem to find the length of the hypotenuse.
235.
236.
237.
238.
In the following exercises, use the Pythagorean Theorem to find the length of the leg. Round to the nearest tenth, if necessary.
239.
240.
241.
242.
243.
244.
245.
246.
In the following exercises, solve using the Pythagorean Theorem. Approximate to the nearest tenth, if necessary.
247.
A 13-foot string of lights will be attached to the top of a 12-foot pole for a holiday display, as shown below. How far from the base of the pole should the end of the string of lights be anchored?
Pam wants to put a banner across her garage door, as shown below, to congratulate her son for his college graduation. The garage door is 12 feet high and 16 feet wide. How long should the banner be to fit the garage door?
249.
Chi is planning to put a path of paving stones through her flower garden, as shown below. The flower garden is a square with side 10 feet. What will the length of the path be?
Brian borrowed a 20 foot extension ladder to use when he paints his house. If he sets the base of the ladder 6 feet from the house, as shown below, how far up will the top of the ladder reach?
Solve Applications Using Rectangle Properties
In the following exercises, solve using rectangle properties.
251.
The length of a rectangle is 85 feet and the width is 45 feet. What is the perimeter?
The length of a rectangle is 26 inches and the width is 58 inches. What is the perimeter?
253.
A rectangular room is 15 feet wide by 14 feet long. What is its perimeter?
A driveway is in the shape of a rectangle 20 feet wide by 35 feet long. What is its perimeter?
255.
The area of a rectangle is 414 square meters. The length is 18 meters. What is the width?
The area of a rectangle is 782 square centimeters. The width is 17 centimeters. What is the length?
257.
The width of a rectangular window is 24 inches. The area is 624 square inches. What is the length?
The length of a rectangular poster is 28 inches. The area is 1316 square inches. What is the width?
259.
Find the length of a rectangle with perimeter 124 and width 38.
Find the width of a rectangle with perimeter 92 and length 19.
261.
Find the width of a rectangle with perimeter 16.2 and length 3.2.
Find the length of a rectangle with perimeter 20.2 and width 7.8.
263.
The length of a rectangle is nine inches more than the width. The perimeter is 46 inches. Find the length and the width.
The width of a rectangle is eight inches more than the length. The perimeter is 52 inches. Find the length and the width.
265.
The perimeter of a rectangle is 58 meters. The width of the rectangle is five meters less than the length. Find the length and the width of the rectangle.
The perimeter of a rectangle is 62 feet. The width is seven feet less than the length. Find the length and the width.
267.
The width of the rectangle is 0.7 meters less than the length. The perimeter of a rectangle is 52.6 meters. Find the dimensions of the rectangle.
The length of the rectangle is 1.1 meters less than the width. The perimeter of a rectangle is 49.4 meters. Find the dimensions of the rectangle.
269.
The perimeter of a rectangle is 150 feet. The length of the rectangle is twice the width. Find the length and width of the rectangle.
The length of a rectangle is three times the width. The perimeter of the rectangle is 72 feet. Find the length and width of the rectangle.
271.
The length of a rectangle is three meters less than twice the width. The perimeter of the rectangle is 36 meters. Find the dimensions of the rectangle.
The length of a rectangle is five inches more than twice the width. The perimeter is 34 inches. Find the length and width.
273.
The perimeter of a rectangular field is 560 yards. The length is 40 yards more than the width. Find the length and width of the field.
The perimeter of a rectangular atrium is 160 feet. The length is 16 feet more than the width. Find the length and width of the atrium.
275.
A rectangular parking lot has perimeter 250 feet. The length is five feet more than twice the width. Find the length and width of the parking lot.
A rectangular rug has perimeter 240 inches. The length is 12 inches more than twice the width. Find the length and width of the rug.
Everyday Math
277.
Christa wants to put a fence around her triangular flowerbed. The sides of the flowerbed are six feet, eight feet and 10 feet. How many feet of fencing will she need to enclose her flowerbed?
Jose just removed the children’s playset from his back yard to make room for a rectangular garden. He wants to put a fence around the garden to keep out the dog. He has a 50 foot roll of fence in his garage that he plans to use. To fit in the backyard, the width of the garden must be 10 feet. How long can he make the other length?
Writing Exercises
279.
If you need to put tile on your kitchen floor, do you need to know the perimeter or the area of the kitchen? Explain your reasoning.
If you need to put a fence around your backyard, do you need to know the perimeter or the area of the backyard? Explain your reasoning.
281.
Look at the two figures below.
ⓐ Which figure looks like it has the larger area?
ⓑ Which looks like it has the larger perimeter?
ⓒ Now calculate the area and perimeter of each figure.
ⓓ Which has the larger area?
ⓔ Which has the larger perimeter?
Write a geometry word problem that relates to your life experience, then solve it and explain all your steps.
Self Check
ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?
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Annals of Mathematics and Artificial Intelligence
Article
An improvement of Random Node Generator for the uniform generation of capacities
Published: 01 December 2023
Volume 92,pages 1381–1406, (2024)
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Annals of Mathematics and Artificial IntelligenceAims and scopeSubmit manuscript
An improvement of Random Node Generator for the uniform generation of capacities
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Peiqi SunORCID: orcid.org/0000-0002-7037-35881,
Michel Grabisch2&
Christophe Labreuche3
96 Accesses
1 Citation
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Abstract
Capacity is an important tool in decision-making under risk and uncertainty and multi-criteria decision-making. When learning a capacity-based model, it is important to be able to generate uniformly a capacity. Due to the monotonicity constraints of a capacity, this task reveals to be very difficult. The classical Random Node Generator (RNG) algorithm is a fast-running speed capacity generator, however with poor performance. In this paper, we firstly present an exact algorithm for generating a n elements’ general capacity, usable when n<5. Then, we present an improvement of the classical RNG by studying the distribution of the value of each element of a capacity. Furthermore, we divide it into two cases, the first one is the case without any conditions, and the second one is the case when some elements have been generated. Experimental results show that the performance of this improved algorithm is much better than the classical RNG while keeping a very reasonable computation time.
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References
Angilella, S., Corrente, S., Greco, S.: Stochastic multiobjective acceptability analysis for the Choquet integral preference model and the scale construction problem. Eur. J. Oper. Res. 240, 172–182 (2015)
ArticleMathSciNetGoogle Scholar
Angilella, S., Corrente, S., Greco, S., Slowinski, R.: Robust ordinal regression and stochastic multiobjective acceptability analysis in multiple criteria hierarchy process for the Choquet integral preference model. Omega 63, 154–169 (2016)
ArticleGoogle Scholar
Bana e Costa, C.A., De Corte, J.M., Vansnick J.C.: MACBETH. Int. J. Inf. Technol. Decis. Mak. 11, 359–387 (2012)
Combarro, E.F., Díaz, I., Miranda, P.: On random generation of fuzzy measures. Fuzzy Sets and Systems 228, 64–77 (2013)
ArticleMathSciNetGoogle Scholar
Combarro, E.F., Hurtado de Saracho, J., Díaz I.: Minimals plus: an improved algorithm for the random generation of linear extensions of partially ordered sets. Inf. Sci. 501, 50–67 (2019)
Choquet, G.: Theory of capacities. Annales de l’Institut Fourier 5, 131–295 (1953)
ArticleMathSciNetGoogle Scholar
Grabisch, M.: Set Functions, Games and Capacities in Decision Making, volume 46 of Theory and Decision Library C, Springer (2016)
Grabisch, M., Kojadinovic, I., Meyer, P.: Using the Kappalab R package for Choquet integral based multi-attribute utility theory. In: Proc. Int. Conf. on Information Processing and Management of Unicertainty (IPMU’06), pp.1702–1709, Paris, France, July (2006)
Grabisch, M., Labreuche, Ch.: A decade of application of the Choquet and Sugeno integrals in multi-criteria decision aid. Ann. Oper. Res. 175, 247–286 (2010)
ArticleMathSciNetGoogle Scholar
Grabisch, M., Labreuche, Ch., Sun, P.: An approximation algorithm for random generation of capacities. To appear in Order (2023).
ArticleGoogle Scholar
Greco, S., Mousseau, V., Słowinski, R.: Ordinal regression revisited: multiple criteria ranking with a set of additive value functions. Eur. J. Oper. Res. 191, 416–436 (2008)
ArticleMathSciNetGoogle Scholar
Havens, T.C., Pinar, A.J.: Generating random fuzzy (capacity) measures for datafusion simulations. In: IEEE Symposium Series on Computational Intelligence (IEEE SSCI2017), pp 1–8 (2017)
Herin, M., Perny, P., Sokolovska N.: Learning sparse representations of preferences within Choquet expected utility theory. In: 38th conference on Uncertainty in Artificial Intelligence (UAI’2022), Eindhoven, Netherlands (2022)
Karzanov, A., Khachiyan, L.: On the conductance of order Markov chains. Order 8, 7–15 (1991)
ArticleMathSciNetGoogle Scholar
Keeney, R.L., Raiffa, H.: Decision with multiple objectives. Wiley, New York (1976)
Google Scholar
Kojadinovic, I., Marichal, J.-L., Roubens, M.: An axiomatic approach to the definition of the entropy of a discrete Choquet capacity. Inf. Sci. 172, 131–153 (2005)
ArticleMathSciNetGoogle Scholar
Krantz, D.H., Luce, R.D., Suppes, P., Tversky A.: Foundations of measurement, volume 1: Additive and Polynomial Representations. Academic Press (1971)
Lahdelma, R., Hokkanen, J., Salminen, P.: SMAA - Stochastic Multiobjective Acceptability Analysis. Eur. J. Oper. Res. 106, 137–143 (1998)
ArticleGoogle Scholar
De Loof, K., De Meyer, H., De Baets, B.: Exploiting the lattice of ideals representation of a poset. Fundamenta Informaticae 71, 309–321 (2006)
ArticleMathSciNetGoogle Scholar
Magoč, T., Modave, F.: Optimization of the Choquet Integral Using Genetic Algorithm, pp. 97–109. Springer International Publishing, Cham (2014)
Google Scholar
Mayag, B., Grabisch, M., Labreuche, Ch.: A characterization of the 2-additive Choquet integral through cardinal information. Fuzzy Sets Syst. 184, 84–105 (2011)
ArticleMathSciNetGoogle Scholar
Mayag, B., Grabisch, M., Labreuche, Ch.: A representation of preferences by the Choquet integral with respect to a 2-additive capacity. Theor. Decis. 71, 297–324 (2011)
ArticleMathSciNetGoogle Scholar
Stanley, R.: Two poset polytopes. Discret. Comput. Geom. 1, 9–23 (1986)
ArticleMathSciNetGoogle Scholar
Download references
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Authors and Affiliations
Université Paris I - Panthéon-Sorbonne, Paris, France
Peiqi Sun
Paris School of Economics, Université Paris I - Panthéon-Sorbonne, Paris, France
Michel Grabisch
Thales Research & Technology, Palaiseau, France
Christophe Labreuche
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1. Peiqi SunView author publications Search author on:PubMedGoogle Scholar
2. Michel GrabischView author publications Search author on:PubMedGoogle Scholar
3. Christophe LabreucheView author publications Search author on:PubMedGoogle Scholar
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Correspondence to Peiqi Sun.
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A:The Markov Chain method
A:The Markov Chain method
This method is due to Karzanov and Khachiyan . Consider a given poset P, and E the set of linear extensions of P. Two linear extensions X 1,X 2∈E are said to be neighbors, if X 2 can be obtained by a single transposition of two consecutive elements in X 1, that is X 1=(x 1,x 2,…,x i,x i+1,…,x|P|) and X 2=(x 1,x 2,…,x i+1,x i,…,x|P|) for i∈[1,|P|−1]. Consider a Markov chain with E being the set of states. The transition probability between two states X 1 and X 2 is given by:
p(X 1,X 2)={1/(2|P|−2)If X 1 and X 2 are neighbors 1−n(X 1)/(2|P|−2)If X 1=X 2 0 otherwise
where n(X 1) denotes the number of neighbors of X 1. The Markov chain with the above transition matrix describes a random walk through the simplices, and this random walk starts at an arbitrary simplex X 0 in the triangulation, then there is a move to a new simplex with probability 1 2 n−2. Karzanov and Khachiyan proved that the transition matrix induces an ergodic time-reversible Markov chain with uniform stationary distribution, that means for an arbitrary initial probability distribution on E, after T steps (T big enough), the distribution converges to the uniform distribution on E. Thus, for a given poset P and sufficiently large T, the following algorithm gives a nearly uniform generator of linear extensions of the poset.
Algorithm Classical Karzanov-Khachiyan chain
1.
Input X 0 as the initial state (an arbitrary linear extension).
2.
Randomly select r∈{0,1}.
3.
If r=1, randomly select i from {1,2,3,…,|P|−1}.
4.
Interchange the value of i th position and (i+1)th position of X 0, if the new X 0 belongs to the set of linear extensions E.
5.
Repeat T times Step 2,3,4, and output new X 0.
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Sun, P., Grabisch, M. & Labreuche, C. An improvement of Random Node Generator for the uniform generation of capacities. Ann Math Artif Intell92, 1381–1406 (2024).
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Keywords
Random generation
Capacity
Linear extension
Mathematics Subject Classification (2010)
28E10
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Angilella, S., Corrente, S., Greco, S.: Stochastic multiobjective acceptability analysis for the Choquet integral preference model and the scale construction problem. Eur. J. Oper. Res. 240, 172–182 (2015)
ArticleMathSciNetGoogle Scholar
Angilella, S., Corrente, S., Greco, S., Slowinski, R.: Robust ordinal regression and stochastic multiobjective acceptability analysis in multiple criteria hierarchy process for the Choquet integral preference model. Omega 63, 154–169 (2016)
ArticleGoogle Scholar
Bana e Costa, C.A., De Corte, J.M., Vansnick J.C.: MACBETH. Int. J. Inf. Technol. Decis. Mak. 11, 359–387 (2012)
Combarro, E.F., Díaz, I., Miranda, P.: On random generation of fuzzy measures. Fuzzy Sets and Systems 228, 64–77 (2013)
ArticleMathSciNetGoogle Scholar
Combarro, E.F., Hurtado de Saracho, J., Díaz I.: Minimals plus: an improved algorithm for the random generation of linear extensions of partially ordered sets. Inf. Sci. 501, 50–67 (2019)
Choquet, G.: Theory of capacities. Annales de l’Institut Fourier 5, 131–295 (1953)
ArticleMathSciNetGoogle Scholar
Grabisch, M.: Set Functions, Games and Capacities in Decision Making, volume 46 of Theory and Decision Library C, Springer (2016)
Grabisch, M., Kojadinovic, I., Meyer, P.: Using the Kappalab R package for Choquet integral based multi-attribute utility theory. In: Proc. Int. Conf. on Information Processing and Management of Unicertainty (IPMU’06), pp.1702–1709, Paris, France, July (2006)
Grabisch, M., Labreuche, Ch.: A decade of application of the Choquet and Sugeno integrals in multi-criteria decision aid. Ann. Oper. Res. 175, 247–286 (2010)
ArticleMathSciNetGoogle Scholar
Grabisch, M., Labreuche, Ch., Sun, P.: An approximation algorithm for random generation of capacities. To appear in Order (2023).
ArticleGoogle Scholar
Greco, S., Mousseau, V., Słowinski, R.: Ordinal regression revisited: multiple criteria ranking with a set of additive value functions. Eur. J. Oper. Res. 191, 416–436 (2008)
ArticleMathSciNetGoogle Scholar
Havens, T.C., Pinar, A.J.: Generating random fuzzy (capacity) measures for datafusion simulations. In: IEEE Symposium Series on Computational Intelligence (IEEE SSCI2017), pp 1–8 (2017)
Herin, M., Perny, P., Sokolovska N.: Learning sparse representations of preferences within Choquet expected utility theory. In: 38th conference on Uncertainty in Artificial Intelligence (UAI’2022), Eindhoven, Netherlands (2022)
Karzanov, A., Khachiyan, L.: On the conductance of order Markov chains. Order 8, 7–15 (1991)
ArticleMathSciNetGoogle Scholar
Keeney, R.L., Raiffa, H.: Decision with multiple objectives. Wiley, New York (1976)
Google Scholar
Kojadinovic, I., Marichal, J.-L., Roubens, M.: An axiomatic approach to the definition of the entropy of a discrete Choquet capacity. Inf. Sci. 172, 131–153 (2005)
ArticleMathSciNetGoogle Scholar
Krantz, D.H., Luce, R.D., Suppes, P., Tversky A.: Foundations of measurement, volume 1: Additive and Polynomial Representations. Academic Press (1971)
Lahdelma, R., Hokkanen, J., Salminen, P.: SMAA - Stochastic Multiobjective Acceptability Analysis. Eur. J. Oper. Res. 106, 137–143 (1998)
ArticleGoogle Scholar
De Loof, K., De Meyer, H., De Baets, B.: Exploiting the lattice of ideals representation of a poset. Fundamenta Informaticae 71, 309–321 (2006)
ArticleMathSciNetGoogle Scholar
Magoč, T., Modave, F.: Optimization of the Choquet Integral Using Genetic Algorithm, pp. 97–109. Springer International Publishing, Cham (2014)
Google Scholar
Mayag, B., Grabisch, M., Labreuche, Ch.: A characterization of the 2-additive Choquet integral through cardinal information. Fuzzy Sets Syst. 184, 84–105 (2011)
ArticleMathSciNetGoogle Scholar
Mayag, B., Grabisch, M., Labreuche, Ch.: A representation of preferences by the Choquet integral with respect to a 2-additive capacity. Theor. Decis. 71, 297–324 (2011)
ArticleMathSciNetGoogle Scholar
Stanley, R.: Two poset polytopes. Discret. Comput. Geom. 1, 9–23 (1986)
ArticleMathSciNetGoogle Scholar
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15388 | https://www.youtube.com/watch?v=uWKGVSpWcc8 | Using Function Notation - What is f(x)?
mroldridge
34700 subscribers
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Description
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Posted: 12 Sep 2011
What is f(x)? It is a different way of writing "y" in equations, but it's much more useful!
469 comments
Transcript:
Quick little note about function notation. What we mean by function notation is that we have a new way of writing y. And the way that we can write it is f of x. That's how we pronounce this f of x. Now it doesn't have to be f. It could be h. It could be m. It could be anything else you want except x. Obviously, it's just another way of writing y. If y = 2x - 3, you can just the same way say f ofx is 2x - 3. If y is x^2 + 2x - 8, then maybe h of x, if you want to call it that, is x^2 + 2x - 8. The only reason that I've ever found that we do this is because when you plug in different values for x, this allows us to show what number we did plug in. For example, if x is three, then we say f of three. See how we're plugging in 3 for x because x is 3 is 2 3. See 2 x is 2 3 - 3 turn 2 3 is 6 - 3 turns out to be 3. And so what we say is that f of 3 = 3. Rather than just saying y equ= 3. This way, we're able to also tell whoever's watching our math that the x that we plugged in was 3 to get y = 3. Maybe uh x turns out to be uh0.1. So h of0.1 is0.1^ 2ar - 2 0.1 - 8. See everywhere we have x we just plug in 0.1. Uh this turns out to be 0.01 + 0.2 - 8. And when you do all that math, it turns out to be -7.79, I believe. Forgive me if the math is wrong. We say that h of0.1 is -7.79. See, it's the exact same thing as y, but it's telling you what x you plugged in to get it. That's the only special thing about this. [Music] |
15389 | https://www.mometrix.com/academy/area-and-perimeter-of-a-trapezoid/ | Skip to content
Hi, and welcome to this video on finding the area and perimeter of a trapezoid!
What is a Trapezoid?
A trapezoid is a four-sided polygon, or “quadrilateral,” that has at least one set of parallel sides. There are two types of sides in a trapezoid: legs and bases. A trapezoid has two legs and two bases.
We can tell which sides are the bases because they are parallel to each other.
Here, we can see the top and bottom are parallel because of the matching arrows on those sides.
How to Find the Perimeter of a Trapezoid
When we know the lengths of the legs and the lengths of the bases we can find the perimeter of the trapezoid. The perimeter is the distance around an object. For instance, if we wanted to build a fence around a trapezoid-shaped yard, we’d need to know the perimeter of the yard to know how much fencing to buy.
Trapezoid Perimeter Formula
For a trapezoid, the formula for perimeter is P=b1+b2+l1+l2.
We don’t need to remember this formula though, because just like with every other type of polygon, it’s just a fancy way of saying add all of the sides together!
Let’s go ahead and find the perimeter of this trapezoid:
10+21+12+16=59m
That’s all there is to it! Let’s move on to area.
How to Find the Area of a Trapezoid
Here’s a trapezoid on some graph paper:
Remember that area is a measure of how many square units will fit inside a shape. How many squares are inside our trapezoid?
There are 24 full squares plus eight half squares, which means the area of the trapezoid is 28 square units. But what if we don’t have graph paper or a conveniently sized trapezoid? That’s why we need a formula!
Trapezoid Area Formula
The formula for finding the area of a trapezoid is A=h(b1+b22).
Note that dividing the sum of the bases by two is the average of those lengths. Because our sample problem is on a graph, we can see that the top base, which we’ll call base 1, is three units long. Our bottom base, base 2, is 11 units longs. The height of the trapezoid, which is the distance between the bases, is four units:
For area, we don’t need the measurements of the two legs, just the two bases and the height, which can also be called the altitude. Since we have all three we can plug them into our formula:
A=h(b1+b22)=4(3+112)=4(142)=4(7)=28 units2
That’s the same answer we got when we counted!
Let’s try another one:
Okay, it looks a bit different than the trapezoid we just did. But we can tell it’s a trapezoid because it has one set of parallel sides. We can use the formula, so now we just need to figure out which numbers go where. The parallel sides are the bases so we can set base one as 6 centimeters and base two as 3 centimeters. There’s no dashed or colored line inside the trapezoid connecting the bases that would clearly be the height, but the bottom side is connecting the bases and is perpendicular to them, as we can tell by the right angle symbol. So 4 centimeters is the height, even though it’s sideways! Let’s plug it all in:
A=h(b1+b22)=4(6+32)=4(92)=4(4.5)=18 cm2
This formula also works to find the area of parallelograms too. That’s because all parallelograms are trapezoids since they have at least one set of parallel sides. In fact, all parallelograms have two sets.
That’s about all there is to finding the perimeter and area of trapezoids.
Thanks for watching, and happy studying!
“Area of a Trapezoid. Definition, Formula and Calculator – Math Open Reference.” n.d.
“Trapezoid – Math Word Definition – Math Open Reference.” n.d.
Frequently Asked Questions
Q
How do you find the area of a trapezoid?
A
Each of the two parallel sides of a trapezoid is a base. The distance between the bases (measured perpendicular to each) is the height. To find the area of a trapezoid, we multiply the average length of the two bases by the height. In symbols, if the lengths of the bases are a and b and the height is h (see diagram), then the area, A, of the trapezoid is A=(a+b)2h, which we can also write as A=12(a+b)h.
For example, the trapezoid in the diagram with bases of length 5 cm and 9 cm and with height 3 cm has an area of A=5+92⋅3=142⋅3=7⋅3=21 square centimeters, which we may also write as 21 cm2.
Q
Why does the area formula for a trapezoid work?
A
The area formula for a trapezoid works because it comes from the formula for the area of a parallelogram. The trapezoid below (with solid sides) has bases of length a and b and height h. Suppose we make a copy of it, rotate it halfway around, and place it adjoining the original trapezoid so that legs (non-parallel sides) of the same length coincide (the shaded trapezoid with dashed sides). Together these figures form a parallelogram with a base of length a+b and height h. By the standard formula, the area of this parallelogram is area=base×height=(a+b)h. The area of the original trapezoid is half of this, namely 12(a+b)h, or, equivalently, (a+b)2h. This same procedure works for every trapezoid.
Q
What units do you use for the area of a trapezoid?
A
We measure the area of a trapezoid, like all areas, in square units–square feet, square inches, square meters, etc.—which are sometimes written ft2, in2, m2, etc. Usually, we use the square of the unit used to measure the bases and height of the trapezoid. For instance, if we measure the bases and height in centimeters, we usually give the area in square centimeters.
Q
How do you find the perimeter of a trapezoid?
A
The perimeter of a figure is the distance around it. We find the perimeter of a trapezoid by adding up the lengths of its four sides.
Q
How do you find the perimeter of a trapezoid using the Pythagorean theorem?
A
If we do not know the lengths of all four sides of a trapezoid, sometimes we have enough other information to find the lengths of the missing sides using the Pythagorean theorem. For example, in the trapezoid in the diagram, the bases are 2 cm and 9 cm, the height is 4 cm, and the longer base sticks out past the shorter base by 3 cm on the left and 4 cm on the right. This makes sides c and d hypotenuses of right triangles with sides whose lengths we know. By the Pythagorean theorem, c2=32+42=9+16=25, so c=25−−√=5 cm. Similarly, d2=42+42=16+16=32, so d=32−−√=16⋅2−−−−√=16−−√⋅2–√=42–√≈5.7 cm. Now we can find the perimeter, P, of the trapezoid by adding up the four sides: P=9+5+2+42–√=16+42–√≈16+5.7=21.7 cm. This is one example of finding the perimeter of a trapezoid using the Pythagorean theorem.
Q
How do you find the area of a trapezoid without the height?
A
If we do not know the height of a trapezoid, sometimes we have enough other information to find the height using the Pythagorean theorem. For example, in the trapezoid in the diagram, the bases are 2 cm and 9 cm, the longer base sticks out 3 cm past the shorter base on the left, and the left leg (the non-parallel side) is 5 cm. This makes the dashed line a side of a right triangle whose hypotenuse and other side we know. By the Pythagorean theorem, h2+32=52 or h2+9=25. So, h2=16 and h=16−−√=4 cm, which is the height of the trapezoid. Now we can apply the standard formula to find the area of this trapezoid: A=(a+b)2h=(2+9)2×4=112×4=111⋅2=22 cm2. This is an example of finding the area of a trapezoid without the height.
Area and Perimeter of a Trapezoid Practice Problems
Question #1:
What is the perimeter of this trapezoid?
74 in
86 in
142 in
300 in
Answer:
The correct answer is 74 in. To find the perimeter of a trapezoid, add all four side lengths together.
P=23+12+27+12=74 in
Question #2:
What is the area of this trapezoid?
96 cm2
81 cm2
57 cm2
41 cm2
Answer:
The correct answer is 81 cm2. The formula for area of a trapezoid is:
A=12(b1+b2)h
The length base 1 is 12 cm. The length of base 2 is 15 cm. The length of the height is 6 cm.
A=12(12+15)(6)=12(27)(6)=81 cm2
Question #3:
What is the perimeter of this trapezoid?
34 in
48 in
36 in
42 in
Answer:
The correct answer is 36 in. Find the perimeter of the trapezoid by adding the lengths of all four sides together.
P=11+6+13+6=36 in
Question #4:
What is the area of this trapezoid?
17 ft2
15 ft2
18 ft2
23 ft2
Answer:
The correct answer is 15 ft2. The formula for area of a trapezoid is:
A=12(b1+b2)h
The length of base 1 is 4 ft. The length of base 2 is 6 ft. The length of the height is 3 ft.
A=12(4+6)(3)=12(10)(3)=15 ft2
Question #5:
What is the perimeter of this trapezoid?
153 cm
140 cm
47 cm
55 cm
Answer:
The correct answer is 55 cm. To find the perimeter of a trapezoid, add the lengths of all four sides together.
P=15+8+20+12=55 cm
Return to Geometry Videos
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15390 | https://www.scribd.com/document/480175236/nt2-2 | Number Theory II: Congruences | PDF | Mathematical Concepts | Ring Theory
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Number Theory II: Congruences
This document provides an overview of congruences in number theory. It defines congruences modulo m as integers a and b such that their difference is divisible by m. Congruences can be added…
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15391 | https://askfilo.com/user-question-answers-smart-solutions/show-that-problem-1-93-implies-that-r-3-3-leq-6r-3-3-andgt-5-3333313936393031 | Question asked by Filo student
Show that R(3,3)=6.Problem 1.93 implies that R(3,3)≤6.ToshowthatR(3,3)>5, it is enough to consider a seating arrangement of 5 people about a round table in which each person knows only the 2 people on either side. In such a situation there is no set of 3 mutual acquaintances and no set of 3 people not known to one another.
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| Question Text | Show that R(3,3)=6.Problem 1.93 implies that R(3,3)≤6.ToshowthatR(3,3)>5, it is enough to consider a seating arrangement of 5 people about a round table in which each person knows only the 2 people on either side. In such a situation there is no set of 3 mutual acquaintances and no set of 3 people not known to one another. |
| Updated On | May 8, 2025 |
| Topic | All topics |
| Subject | Smart Solutions |
| Class | Grade 12 |
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15392 | https://www.sincerelyspain.com/blog/2019/03/07/thriving-is-just-striving/ | Thriving is Just Striving! – Sincerely, Spain
Sincerely, Spain
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March 7, 2019/No Comments
Dear Jamie,
I’ve been thinking a lot lately and I feel like so many of my thriving posts over the past year have been about how I’m taking stock of where I’m at and figuring out ways to readjust, to get back on track, and/or make changes in order to ‘thrive’ again. At times, this has felt like a confession of my lack of thriving, but now I realize that this process is exactly what thriving entails!
I believe that, quite often, thriving does not mean a state of actualization or success but rather the ongoing process of getting there. It means having goals and aspirations and working each day to move closer to them. More often than not, we are in the process of doing rather than having and that’s okay. In fact, it’s great!
As I mentioned in my post about the importance of mindful reflection, the majority of us spend so much of our time worried about where we want to go next and so little of our time actually celebrating our successes when we get there. If we look at thriving as a destination to reach rather than the journey, we are missing out on the larger part (and perhaps the BEST part) of the experience.
To be honest with you, before sitting down to write this post, I thought that this interpretation of thriving was something I had concocted on my own—a unique understanding I had reached through ongoing trial and error and compassion for myself and my journey. But guess what? The Cambridge dictionary literally defines ‘thriving’ as “growing, developing, or being successful.” Google’s first definition it churned out for me was “growing well or vigorously,” followed by a secondary definition of “prospering; flourishing.” It turns out, the TRUE definition of thriving is, first and foremost, the journey of striving for success.
When did we get it so backwards?! As a society, we’ve generally stopped recognizing growth as the definition and focused in instead on the final destination—which is part of it, but literally not even definition #1!! This is a huge realization for me and it’s utterly vindicating to see that the understanding I have reached in regards to what thriving is (which sincerely felt like swimming upstream against a general consensus that defines it otherwise) is in fact correct on more levels than I realized.
Hooray for thriving being striving!! It’s a lot easier for many of us to admit that we’re trying, that we’re working on something, or that we’re moving towards our goals. On the contrary, we often feel undeserving of the title ‘successful’ or ‘thriving’ because we think that we should have something tangible to show for it.
Well, friends, that’s simply not the case—thriving means developing, it means working on it! And so does striving. In fact, striving may be an even harder process at it generally involves “making great efforts to achieve something,” “especially for a long time or against difficulty.” If you feel like you’ve been striving a lot recently (I know I do!) then GOOD FOR YOU—you’ve truly been thriving!
If you’re loving this concept, be sure to check out a number of our other posts that get into this as well:
Earlier on in my recognition of this definition: Accepting Who I Am
Working on my obstacles: Six Strategies for Managing Unstructured Time
Recognizing one of my self-sabotaging quirks: I Have Non-Commitment Issues
A recent time a thriving post called for more striving in my real life: It’s Really Hard for me to Take a Day Off
Claudia’s reflections on her journey: Accepting my own Journey
How Claudia thrives through her workaholic ways: I’m a Workaholic
How Dani has come to define her “real life” in Spain: My Real Life in Spain
And an important reminder that it’s okay to seek support: Sometimes I Have to Ask for Help
I’m sure you’ll find something you can relate to on this list (and please reach out and share if any of them resonate with you; we love hearing from you all)!
Had you previously defined ‘thriving’ in this way? Are you feeling like it’s time for a change if not? I’m certainly going to move forward with my head held higher and spread the good news to all the striving people I meet. We’re thriving after all!
Sincerely,
Dani
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15393 | https://www.spanishdict.com/answers/242767/personal-a | personal a
On my recent Spanish test a few questions dealt with the "personal a", but i do not remember learning about it. What is it?
3 Answers
This lesson might help, it's all about the Spanish 'a' and its uses including as the personal 'a' - here's an excerpt:
5) The personal 'a'.
This term 'personal a' has been coined to help us understand this
untranslatable use of the preposition 'a' but you won't find that term
in many (if any) Spanish grammar books written in Spanish for Spanish
speakers.
To introduce a direct object that is a definite person (or treated
like a person - eg a pet), also used for groups of people (made up of
or including known people) and personified nouns:
Conozco a Pablo. I know Peter.
Encontré a mi perro Rover. I found my dog Rover.
Oigo a los estudiantes. I hear the students.
Respeta a los ancianos. Respect your elders.
The main exception to the rule about 'known people' is that certain pronouns, such as alguien and nadie, always require the personal a when used as direct objects, even when they aren't referring to a specific person.
Eg: No conozco a nadie. I don't know anybody.
And here's a more indepth article that deals only with the personal 'a'.
The presonal a is used when the action of the verb is directed at a person, rather a thing (this also works with pets that are considered part of someone´s family).
Estoy mirando el jardín .... I am looking at the garden
Estoy mirando a mi hermana ..... I am looking at my sister
Estoy mirando a mi perro ... I am looking at my dog
Estoy mirando un perro callejero ..... I am looking at a stray dog
The personal "a" as noted is used for a direct object that is a specific person and not a thing.
It is also used with words like alquien, nadie, alguno , ninguno, cualquiera and any other pronoun that refers to a person. It is used with numbers that refer to persons.
Personal "a" is not used with the verb tener.
The specfic person aspect can be tricky. Por ejemplo, "Este restaurante busca camareros.." This restaurant is looking for waiters. Waiters in this sentence are not considered to be "specific" person(s).
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15394 | https://metatutor.co.uk/wp-content/uploads/PDFs/Worksheets/Foundation/ALGEBRA-SUBSTITUTION.pdf | ALGEBRA – SUBSTITUTION – PRACTICE QUESTIONS CALCULATOR ALLOWED 1. a = 2b + 3c Find the value of a when b = 5 and c = 2. 2. d = 5e – 2f Find the value of d when e = 4 and f = 6. 3. g = 4h – i Find the value of g when h = 2 and i = -5. 4. z = x2 – 3y Find the value of z when x = 3 and y = -2. 5. t = 3s2 – 4u Find the value of t when s = -3 and u = 4. 6. Below are four algebraic expressions. 3x + y 2x2 5y – x y + 12 Which expression gives the largest value when x = 3 and y = 4? 7. Below are four algebraic expressions. 5p + 2q p2 – 5 3(p – q) q2 + 9 Which expression gives the largest value when p = 4 and q = -2? 8. Below are four algebraic expressions. a3 + 5 3a – 4b a(5a + b) ba Which expression gives the smallest value when a = 2 and b = -3? 9. p = r + at Find the value of p when r = 5, a = 2 and t = 6. 10. A = 5c – de Find the value of A when c = 4, d = 0.5 and e = 12. 11. w2 = u2 – 2vz Find the value of w when u = -6, v = 1 and z = 10. 12. x2 = y2 + 4wz Find the value of x when y = 3, w = 5 and z = 2. 13. L = 3m + n Find the value of n when L = 20 and m = 4. 14. A = 4b – 3d Find the value of b when A = 9 and d = 3. 15. w = 10z + 3t Find the value of z when w = 21 and t = -3. 16. r = 3p2 + 2q Find the value of q when r = 10 and p = 2. 17. The formula below can be used to work out the cost in pounds (C) of hiring a bike for h hours. 𝐶= 9ℎ+ 5 (a) Carla wants to hire a bike for 4 hours. Work out how much Carla will pay. (b) Darren also hired a bike. He paid £50. Work out how many hours Darren hired the bike for. 18. The formula below can be used to convert temperatures between ° Celsius (C) and ° Fahrenheit (F). 𝐹= 2𝐶+ 30 (a) The temperature in Athens is 28°C. Work out the temperature in Athens in ° Fahrenheit. (b) The temperature in Helsinki is 20°F. Work out the temperature in Helsinki in ° Celsius. 19. A taxi company uses this formula to calculate the cost (C) of a journey. 𝐶= 5 + 1.5𝑚+ 𝐵 where m is the number of miles travelled and B is a booking fee. (a) Mae booked a taxi with the company. She travelled 6 miles and paid £15.50. Work out the booking fee. (b) Noreen also booked a taxi with the company. She travelled 10 miles. Work out how much Noreen paid. 20. The formula below can be used to calculate the temperature in Celsius (T) at different heights on Mount Everest. 𝑇= 48 ℎ−36 where h is the height in kilometres. (a) A hiker has set up a base at a height of 1,500 metres. Work out the temperature at the base. (b) The temperature at another base on the mountain is –20°C. How high up the mountain is the base, in kilometres? 21. The formula below is used to calculate the number of points (P) a football team has. 𝑃= 3𝑊+ 𝐷 where W is the number of games won and D is the number of games drawn. (a) Liverpool have won 24 games and drawn 8 games. Everton have won 16 games and drawn 6 games. How many more points do Liverpool have than Everton? (b) Manchester United have played 38 games. They have drawn 13 games and have 79 points. How many games have Manchester United lost? 22. Aaron works for a company selling car and home insurances. His monthly pay in pounds (P) can be calculated using the formula below. 𝑃= 1250 + 𝐶 40 + 3𝐻 100 where C is the total value of the car insurances he sells and H is the total value of the home insurances he sells. In June, Aaron sold 6 car insurances worth £260 each and 2 home insurances worth £350 each. Work out Aaron’s pay in June. 23. Below are 5 numbers. a + b b2 2a – b a2 – 1 5b + 9 (a) Find the mode of the numbers when a = 5 and b = 3. (b) Find the median of the numbers when a = 3 and b = 4. 24. The formula below can be used to calculate the final velocity in metres per second (v) of a journey. 𝑣2 = 𝑢2 + 2𝑎𝑠 where u is the initial velocity in metres per second, a is the acceleration in metres per second per se (a) A car travelled from point A to point B. The distance between point A and point B is 88 metres. The car’s initial velocity was 2 metres per second. The car’s acceleration was 0.75 metres per second per second. Work out the car’s final velocity, to 1 decimal place. (b) Another car travelled from point C to point D. The car’s initial velocity was 3 metres per second. The car’s final velocity was 15 metres per second. The car’s acceleration was 1.02 metres per second. Work out the distance travelled, to the nearest metre. |
15395 | https://www.nagwa.com/en/videos/175149047283/ | Question Video: Finding the Intersection of Two Intervals Mathematics • Second Year of Preparatory School
Given that 𝑥 = [−8, 3) and 𝑦 = [−4, −3) find 𝑥 ∩ 𝑦.
Video Transcript
Given that 𝑥 is equal to the
left-closed, right-open interval from negative eight to three and 𝑦 is equal to the
left-closed, right-open interval from negative four to negative three, find the
intersection between 𝑥 and 𝑦.
In this question, we are given two
sets 𝑥 and 𝑦 in interval notation. We need to use this to find the
intersection between the two sets. To do this, let’s start by
recalling what we mean by the intersection of two sets.
We can recall that we say that 𝑎
is a member of the intersection between 𝑥 and 𝑦 if 𝑎 is a member of 𝑥 and 𝑎 is
a member of 𝑦. In other words, the intersection
between 𝑥 and 𝑦 includes all of the elements in both sets. To find the intersection between 𝑥
and 𝑦, let’s recall what is meant by the interval notation for 𝑥 and 𝑦.
Let’s start with 𝑥. In interval notation, the first
number is a lower bound of the set and the second number is an upper bound of the
set. We also note that we have a bracket
at negative eight and a parenthesis at three. This means that we want to include
negative eight but not include three. So 𝑥 is the set of all real
numbers between negative eight and three, where we include negative eight. In the same way, we can see that 𝑦
is the set of real values between negative four and negative three, where we include
negative four.
We now want to find the
intersection of these sets, that is, all of the values in both sets. To do this, it is a good idea to
represent these sets visually on a number line. Let’s start with 𝑥. We know that 𝑥 contains all of the
real values between negative eight and three. However, we need to include
negative eight and not three. We can show this on a number line
by drawing a solid circle at negative eight to show that it is included and a hollow
circle at three to show that it is not included in this set. We then connect these with a line
to show that all of the values between the endpoints are in the set.
We can follow the same process for
𝑦. We see that 𝑦 includes negative
four but does not include negative three. So we sketch a solid circle at
negative four and a hollow circle at negative three and connect them with a line to
represent the set 𝑦. The intersection of these two sets
includes every value in both of the sets. We can see in the diagram that the
lowest value in both sets is negative four. We can then see that all of the
values up to negative three are elements of both sets. So they are in the
intersection.
Importantly, we can note that
negative three is not a member of 𝑦. So it is not a member of the
intersection. There are no other elements in the
intersection. So the intersection is the
left-closed, right-open interval from negative four to negative three. We can represent this in interval
notation in two different ways: either using a parenthesis to show that the interval
is open on the right or by using a backwards bracket.
It is also worth noting that in our
diagram, we showed that every element of 𝑦 is an element of 𝑥. So 𝑦 is a subset of 𝑥. This means that 𝑥 intersect 𝑦
will just be equal to 𝑦. We can see this in our answer. The intersection between 𝑥 and 𝑦
is equal to 𝑦, which is the left-closed, right-open interval from negative four to
negative three.
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15396 | https://askfilo.com/user-question-answers-chemistry/on-passing-a-current-through-agno3-solution-5-4-g-ag-is-3234303436373135 | Question asked by Filo student
On passing a current through AgNO3 solution, 5.4 g Ag is deposited at cathode (assume 100% current efficiency). If same current is passed through CuSO4 for same time, only 0.8 g Cu is deposited, then current efficiency in second experiment is
Views: 5,533 students
Updated on: May 27, 2025
Text SolutionText solutionverified iconVerified
Concepts
Faraday's Laws of Electrolysis, Equivalent Weight, Current Efficiency Calculation
Explanation
Step-By-Step Solution
Step 1: Mass relation for silver (Ag) deposition at 100% efficiency
Given: mAg=5.4 g, efficiency = 100%, EAg=108 g/eq
From Faraday's law:
mAg=FEAg⋅Q
So:
Q=EAgmAg⋅F=1085.4×96500
Calculate:
Q=1085.4×96500=108521100≈4824.07 C
Step 2: Expected mass of Cu deposited with this Q, at 100% efficiency
Plug in values:
mCu, expected=9650031.75×4824.07
Calculate numerators and denominator:
31.75×4824.07=153174.22
Then,
mCu, expected=96500153174.22≈1.5886 g
Step 3: Given mass of Cu deposited = 0.8 g. Calculate current efficiency.
Efficiency =expected massactual mass×100
So:
Efficiency=1.58860.8×100≈50.38%
Final Answer
Current efficiency in the second experiment is:
50.4% (rounded to one decimal place)
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| Question Text | On passing a current through AgNO3 solution, 5.4 g Ag is deposited at cathode (assume 100% current efficiency). If same current is passed through CuSO4 for same time, only 0.8 g Cu is deposited, then current efficiency in second experiment is |
| Updated On | May 27, 2025 |
| Topic | All Topics |
| Subject | Chemistry |
| Class | Class 11 |
| Answer Type | Text solution:1 |
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15397 | https://www.etymonline.com/word/affluence | Want to remove ads? Log in to see fewer ads, and become a Premium Member to remove all ads.
Origin and history of affluence
affluence(n.)
mid-14c., "a plentiful flowing, an abundant supply," from Old French affluence, from Latin affluentia "affluence, abundance," literally "a flowing to," abstract noun from affluentem (nominative affluens) "flowing toward; abounding, rich, copious" (see affluent). The notion in the figurative Latin sense is of "a plentiful flow" of the gifts of fortune, hence "wealth, abundance of earthly goods," a sense attested in English from c. 1600. Latin affluentia is glossed in Ælfric's vocabulary (late Old English) by oferflowendnys.
also from mid-14c.
Entries linking to affluence
early-15c., "abounding in, copious" (of God's grace); mid-15c. "flowing to" (of liquids), both senses now obsolete, from Old French afluent (14c.) or directly from Latin affluentem (nominative affluens) "abounding, rich, copious," literally "flowing toward," present participle of affluere "flow toward," from assimilated form of ad "to" (see ad-) + fluere "to flow" (see fluent). The especial sense of "abounding in wealth or possessions" is from 1753.
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15398 | https://www.nature.com/scitable/knowledge/library/water-uptake-and-transport-in-vascular-plants-103016037/ | Water Uptake and Transport in Vascular Plants | Learn Science at Scitable
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Water Uptake and Transport in Vascular Plants
By:Andrew J. McElrone(U.S. Department of Agriculture, Agricultural Research Service, University of California, Davis),Brendan Choat(University of Western Sydney),Greg A. Gambetta(University of California, Davis)&Craig R. Brodersen(University of Florida)©2013 Nature Education
)
Citation:McElrone,A.J.,Choat,B.,Gambetta,G.A.&Brodersen,C.R.(2013)Water Uptake and Transport in Vascular Plants.Nature Education Knowledge 4(5):6
How does water move through plants to get to the top of tall trees? Here we describe the pathways and mechanisms driving water uptake and transport through plants, and causes of flow disruption.
Aa)Aa)Aa)
Why Do Plants Need So Much Water?
Water is the most limiting abiotic (non-living) factor to plant growth and productivity, and a principal determinant of vegetation distributions worldwide. Since antiquity, humans have recognized plants' thirst for water as evidenced by the existence of irrigation systems at the beginning of recorded history. Water's importance to plants stems from its central role in growth and photosynthesis, and the distribution of organic and inorganic molecules. Despite this dependence, plants retain less than 5% of the water absorbed by roots for cell expansion and plant growth. The remainder passes through plants directly into the atmosphere, a process referred to as transpiration. The amount of water lost via transpiration can be incredibly high; a single irrigated corn plant growing in Kansas can use 200 L of water during a typical summer, while some large rainforest trees can use nearly 1200 L of water in a single day!
If water is so important to plant growth and survival, then why would plants waste so much of it? The answer to this question lies in another process vital to plants — photosynthesis. To make sugars, plants must absorb carbon dioxide (CO 2) from the atmosphere through small pores in their leaves called stomata (Figure 1). However, when stomata open, water is lost to the atmosphere at a prolific rate relative to the small amount of CO 2 absorbed; across plant species an average of 400 water molecules are lost for each CO 2 molecule gained. The balance between transpiration and photosynthesis forms an essential compromise in the existence of plants; stomata must remain open to build sugars but risk dehydration in the process.
Figure 1:Rendering of an open stoma on the surface of a tobacco leaf.
Stomata are pores found on the leaf surface that regulate the exchange of gases between the leaf's interior and the atmosphere. Stomatal closure is a natural response to darkness or drought as a means of conserving water.
© 2013 Nature Education All rights reserved. ;)
From the Soil into the Plant
Essentially all of the water used by land plants is absorbed from the soil by roots. A root system consists of a complex network of individual roots that vary in age along their length. Roots grow from their tips and initially produce thin and non-woody fine roots. Fine roots are the most permeable portion of a root system, and are thought to have the greatest ability to absorb water, particularly in herbaceous (i.e., non-woody) plants (McCully 1999). Fine roots can be covered by root hairs that significantly increase the absorptive surface area and improve contact between roots and the soil (Figure 2). Some plants also improve water uptake by establishing symbiotic relationships with mycorrhizal fungi, which functionally increase the total absorptive surface area of the root system.
Figure 2:Root hairs often form on fine roots and improve water absorption by increasing root surface area and by improving contact with the soil.
© 2013 Nature Education All rights reserved. ;)
Roots of woody plants form bark as they age, much like the trunks of large trees. While bark formation decreases the permeability of older roots they can still absorb considerable amounts of water (MacFall et al. 1990, Chung & Kramer 1975). This is important for trees and shrubs since woody roots can constitute ~99% of the root surface in some forests (Kramer & Bullock 1966).
Roots have the amazing ability to grow away from dry sites toward wetter patches in the soil — a phenomenon called hydrotropism. Positive hydrotropism occurs when cell elongation is inhibited on the humid side of a root, while elongation on the dry side is unaffected or slightly stimulated resulting in a curvature of the root and growth toward a moist patch (Takahashi 1994). The root cap is most likely the site of hydrosensing; while the exact mechanism of hydrotropism is not known, recent work with the plant model Arabidopsis has shed some light on the mechanism at the molecular level (see Eapen et al. 2005 for more details).
Roots of many woody species have the ability to grow extensively to explore large volumes of soil. Deep roots (>5 m) are found in most environments (Canadell et al. 1996, Schenk & Jackson 2002) allowing plants to access water from permanent water sources at substantial depth (Figure 3). Roots from the Shepard's tree (Boscia albitrunca)have been found growing at depths 68 m in the central Kalahari, while those of other woody species can spread laterally up to 50 m on one side of the plant (Schenk & Jackson 2002). Surprisingly, most arid-land plants have very shallow root systems, and the deepest roots consistently occur in climates with strong seasonal precipitation (i.e., Mediterranean and monsoonal climates).
Figure 4:Tree roots at significant depths accessed via caves.
Plant scientists examine: deep roots of Juniperus asheii growing at 7m depth in a cave in Austin, TX USA (left); an extensive fine root network attached to a single ~1cm diameter tap root accessing a perennial underground stream at 20m depth in a cave in central TX, USA; and twisty roots in a cave located in southwest Western Australia below a forest dominated by Eucalyptus diversicolor — roots in this cave system are commonly found from 20-60m depth.
© 2013 Nature Education Images provided by W. T. Pockman (Univ of New Mexico), A. J. McElrone, and T. M. Bleby (Univ of Western Australia). All rights reserved. ;)
Through the Plant into the Atmosphere
Water flows more efficiently through some parts of the plant than others. For example, water absorbed by roots must cross several cell layers before entering the specialized water transport tissue (referred to as xylem) (Figure 4). These cell layers act as a filtration system in the root and have a much greater resistance to water flow than the xylem, where transport occurs in open tubes. Imagine the difference between pushing water through numerous coffee filters versus a garden hose. The relative ease with which water moves through a part of the plant is expressed quantitatively using the following equation:
Flow = Δψ /R,
which is analogous to electron flow in an electrical circuit described by Ohm's law equation:
i = V / R,
where R is the resistance, i is the current or flow of electrons, and V is the voltage. In the plant system, V is equivalent to the water potential difference driving flow (Δψ) and i is equivalent to the flow of water through/across a plant segment. Using these plant equivalents, the Ohm's law analogy can be used to quantify the hydraulic conductance (i.e., the inverse of hydraulic R) of individual segments (i.e., roots, stems, leaves) or the whole plant (from soil to atmosphere).
Upon absorption by the root, water first crosses the epidermis and then makes its way toward the center of the root crossing the cortex and endodermis before arriving at the xylem (Figure 4). Along the way, water travels in cell walls (apoplastic pathway) and/or through the inside of cells (cell to cell pathway, C-C) (Steudle 2001). At the endodermis, the apoplastic pathway is blocked by a gasket-like band of suberin — a waterproof substance that seals off the route of water in the apoplast forcing water to cross via the C-C pathway. Because water must cross cell membranes (e.g., in the cortex and at apoplastic barriers), transport efficiency of the C-C pathway is affected by the activity, density, and location of water-specific protein channels embedded in cell membranes (i.e., aquaporins). Much work over the last two decades has demonstrated how aquaporins alter root hydraulic resistance and respond to abiotic stress, but their exact role in bulk water transport is yet unresolved.
Figure 4:Representation of the water transport pathways along the soil-plant-atmosphere continuum (SPAC).
(A) Water moves from areas of high water potential (i.e. close to zero in the soil) to low water potential (i.e., air outside the leaves). Details of the Cohesion-Tension mechanism are illustrated with the inset panels (A), where tension is generated by the evaporation of water molecules during leaf transpiration (1) and is transmitted down the continuous, cohesive water columns (2) through the xylem and out the roots to the soil (3). The pathways for water movement out of the leaf veins and through the stomata (B) and across the fine roots (C) are detailed and illustrate both symplastic and apoplastic pathways.
© 2013 Nature Education All rights reserved. ;)
Once in the xylem tissue, water moves easily over long distances in these open tubes (Figure 5). There are two kinds of conducting elements (i.e., transport tubes) found in the xylem: 1) tracheids and 2) vessels (Figure 6). Tracheids are smaller than vessels in both diameter and length, and taper at each end. Vessels consist of individual cells, or "vessel elements", stacked end-to-end to form continuous open tubes, which are also called xylem conduits. Vessels have diameters approximately that of a human hair and lengths typically measuring about 5 cm although some plant species contain vessels as long as 10 m. Xylem conduits begin as a series of living cells but as they mature the cells commit suicide (referred to as programmed cell death), undergoing an ordered deconstruction where they lose their cellular contents and form hollow tubes. Along with the water conducting tubes, xylem tissue contains fibers which provide structural support, and living metabolically-active parenchyma cells that are important for storage of carbohydrates, maintenance of flow within a conduit (see details about embolism repair below), and radial transport of water and solutes.
Figure 5:Three dimensional reconstructions of xylem imaged at the Ghent microCT facility.
Differences in xylem structure and conduit distributions can be seen between Ulmus americana (left) and Fraxinus americana (right) xylem.
© 2013 Nature Education Images from S. Jansen, Ulm University. All rights reserved. ;)
When water reaches the end of a conduit or passes laterally to an adjacent one, it must cross through pits in the conduit cell walls (Figure 6). Bordered pits are cavities in the thick secondary cell walls of both vessels and tracheids that are essential components in the water-transport system of higher plants. The pit membrane, consisting of a modified primary cell wall and middle lamella, lies at the center of each pit, and allows water to pass between xylem conduits while limiting the spread of air bubbles (i.e., embolism) and xylem-dwelling pathogens. Thus, pit membranes function as safety valves in the plant water transport system. Averaged across a wide range of species, pits account for >50% of total xylem hydraulic resistance. The structure of pits varies dramatically across species, with large differences evident in the amount of conduit wall area covered by pits, and in the porosity and thickness of pit membranes (Figure 6).
Figure 6:Comparison of different types of wood from flowering and cone-bearing plants.
This features wider conduits from flowering plants (top), a cartoon reconstruction of vessels, tracheids and their pit membranes (middle), which are also shown in SEM images (bottom).
© 2013 Nature Education Image from Choat et al. 2008. All rights reserved. ;)
After traveling from the roots to stems through the xylem, water enters leaves via petiole (i.e., the leaf stalk) xylem that branches off from that in the stem. Petiole xylem leads into the mid-rib (the main thick vein in leaves), which then branch into progressively smaller veins that contain tracheids (Figure 7) and are embedded in the leaf mesophyll. In dicots, minor veins account for the vast majority of total vein length, and the bulk of transpired water is drawn out of minor veins (Sack & Holbrook 2006, Sack & Tyree 2005). Vein arrangement, density, and redundancy are important for distributing water evenly across a leaf, and may buffer the delivery system against damage (i.e., disease lesions, herbivory, air bubble spread). Once water leaves the xylem, it moves across the bundle sheath cells surrounding the veins. It is still unclear the exact path water follows once it passes out of the xylem through the bundle sheath cells and into the mesophyll cells, but is likely dominated by the apoplastic pathway during transpiration (Sack & Holbrook 2005).
Figure 7:An example of a venation pattern to illustrate the hydraulic pathway from petiole xylem into the leaf cells and out the stomata.
© 2013 Nature Education Image from Beerling and Franks 2010. All rights reserved. ;)
Mechanism Driving Water Movement in Plants
Unlike animals, plants lack a metabolically active pump like the heart to move fluid in their vascular system. Instead, water movement is passively driven by pressure and chemical potential gradients. The bulk of water absorbed and transported through plants is moved by negative pressure generated by the evaporation of water from the leaves (i.e., transpiration) — this process is commonly referred to as the Cohesion-Tension (C-T) mechanism. This system is able to function because water is "cohesive" — it sticks to itself through forces generated by hydrogen bonding. These hydrogen bonds allow water columns in the plant to sustain substantial tension (up to 30 MPa when water is contained in the minute capillaries found in plants), and helps explain how water can be transported to tree canopies 100 m above the soil surface. The tension part of the C-T mechanism is generated by transpiration. Evaporation inside the leaves occurs predominantly from damp cell wall surfaces surrounded by a network of air spaces. Menisci form at this air-water interface (Figure 4), where apoplastic water contained in the cell wall capillaries is exposed to the air of the sub-stomatal cavity. Driven by the sun's energy to break the hydrogen bonds between molecules, water evaporates from menisci, and the surface tension at this interface pulls water molecules to replace those lost to evaporation. This force is transmitted along the continuous water columns down to the roots, where it causes an influx of water from the soil. Scientists call the continuous water transport pathway the S oil P lant A tmosphere C ontinuum (SPAC).
Stephen Hales was the first to suggest that water flow in plants is governed by the C-T mechanism; in his 1727 book Hales states "for without perspiration the [water] must stagnate, notwithstanding the sap-vessels are so curiously adapted by their exceeding fineness, to raise [water] to great heights, in a reciprocal proportion to their very minute diameters." More recently, an evaporative flow system based on negative pressure has been reproduced in the lab for the first time by a ‘synthetic tree' (Wheeler & Stroock 2008).
When solute movement is restricted relative to the movement of water (i.e., across semipermeable cell membranes) water moves according to its chemical potential (i.e., the energy state of water) by osmosis — the diffusion of water. Osmosis plays a central role in the movement of water between cells and various compartments within plants. In the absence of transpiration, osmotic forces dominate the movement of water into roots. This manifests as root pressure and guttation — a process commonly seen in lawn grass, where water droplets form at leaf margins in the morning after conditions of low evaporation. Root pressure results when solutes accumulate to a greater concentration in root xylem than other root tissues. The resultant chemical potential gradient drives water influx across the root and into the xylem. No root pressure exists in rapidly transpiring plants, but it has been suggested that in some species root pressure can play a central role in the refilling of non-functional xylem conduits particularly after winter (see an alternative method of refilling described below).
Disruption of Water Movement
Water transport can be disrupted at many points along the SPAC resulting from both biotic and abiotic factors (Figure 8). Root pathogens (both bacteria and fungi) can destroy the absorptive surface area in the soil, and similarly foliar pathogens can eliminate evaporative leaf surfaces, alter stomatal function, or disrupt the integrity of the cuticle. Other organisms (i.e., insects and nematodes) can cause similar disruption of above and below ground plant parts involved in water transport. Biotic factors responsible for ceasing flow in xylem conduits include: pathogenic organisms and their by-products that plug conduits (Figure 8); plant-derived gels and gums produced in response to pathogen invasion; and tyloses, which are outgrowths produced by living plant cells surrounding a vessel to seal it off after wounding or pathogen invasion (Figure 8).
Figure 8:Sources of dysfunction in the xylem.
Left to right: (A) xylem-dwelling pathogens like Xylella fastidiosa bacteria; (B) tyloses (plant-derived); (C and D) conduit (in blue) implosion (Brodribb and Holbrook 2005, Pine needle tracheids); and (E) embolized conduits among water filled ones in a frozen plant samples (Choat unpublished figure, Cryo SEM).
© 2013 Nature Education All rights reserved. ;)
Abiotic factors can be equally disruptive to flow at various points along the water transport pathway. During drought, roots shrink and lose contact with water adhering to soil particles — a process that can also be beneficial by limiting water loss by roots to drying soils (i.e., water can flow in reverse and leak out of roots being pulled by drying soil). Under severe plant dehydration, some pine needle conduits can actually collapse as the xylem tensions increase (Figure 8).
Water moving through plants is considered meta-stable because at a certain point the water column breaks when tension becomes excessive — a phenomenon referred to as cavitation. After cavitation occurs, a gas bubble (i.e., embolism) can form and fill the conduit, effectively blocking water movement. Both sub-zero temperatures and drought can cause embolisms. Freezing can induce embolism because air is forced out of solution when liquid water turns to ice. Drought also induces embolism because as plants become drier tension in the water column increases. There is a critical point where the tension exceeds the pressure required to pull air from an empty conduit to a filled conduit across a pit membrane — this aspiration is known as air seeding (Figure 9). An air seed creates a void in the water, and the tension causes the void to expand and break the continuous column. Air seeding thresholds are set by the maximum pore diameter found in the pit membranes of a given conduit.
Figure 9:Air seeding mechanism.
Demonstrates how increasing tension in a functional water filled vessel eventually reaches a threshold where an air seed is pulled across a pit membrane from an embolized conduit. Air is seeded into the functional conduit only after the threshold pressure is reached.
© 2013 Nature Education Adapted from Tyree & Zimmermann 2002. All rights reserved. ;)
Fixing the Problem
Failure to re-establish flow in embolized conduits reduces hydraulic capacity, limits photosynthesis, and results in plant death in extreme cases. Plants can cope with emboli by diverting water around blockages via pits connecting adjacent functional conduits, and by growing new xylem to replace lost hydraulic capacity. Some plants possess the ability to repair breaks in the water columns, but the details of this process in xylem under tension have remained unclear for decades. Brodersen et al. (2010) recently visualized and quantified the refilling process in live grapevines (Vitis vinifera L.) using high resolution x-ray computed tomography (a type of CAT scan) (Figure 10). Successful vessel refilling was dependent on water influx from living cells surrounding the xylem conduits, where individual water droplets expanded over time, filled vessels, and forced the dissolution of entrapped gas. The capacity of different plants to repair compromised xylem vessels and the mechanisms controlling these repairs are currently being investigated.
Figure 10:Embolism repair documented in grapevines (Vitis vinifera L.) with X-ray micro-CT at the ALS facility at Lawrence Berkeley National Lab CA, USA.
(A) Longitudinal section showing a time series of cavitated vessels refilling in less than 4 hrs; (B) 3D reconstruction of four vessel lumen with water droplets forming on the vessel walls and growing over time to completely fill the embolized conduit.
© 2013 Nature Education Image from Brodersen et al. 2010. All rights reserved. ;)
References and Recommended Reading
Agrios, G. N. Plant Pathology. New York, NY: Academic Press, 1997.
Beerling, D. J. & Franks, P. J. Plant science: The hidden cost of transpiration. Nature464, 495-496 (2010).
Brodersen, C. R. et al. The dynamics of embolism repair in xylem: In vivo visualizations using high-resolution computed tomography Plant Physiology154, 1088-1095 (2010).
Brodribb, T. J. & Holbrook, N. M. Water stress deforms tracheids peripheral to the leaf vein of a tropical conifer. Plant Physiology137, 1139-1146 (2005)
Canadell, J. et al. Maximum rooting depth of vegetation types at the global scale. Oecologia108, 583-595 (1996).
Choat, B., Cobb, A. R. & Jansen, S. Structure and function of bordered pits: New discoveries and impacts on whole-plant hydraulic function. New Phytologist177, 608-626 (2008).
Chung, H. H. & Kramer, P. J. Absorption of water and "P through suberized and unsuberized roots of loblolly pine. Canadian Journal of Forest Research5, 229-235 (1975).
Eapen, D. et al. Hydrotropism: Root growth responses to water. Trends in Plant Science10, 44-50 (2005).
Hetherington, A. M. & Woodward, F. I. The role of stomata in sensing and driving environmental change. Nature424, 901-908 (2003).
Holbrook, N. M. & Zwieniecki, M. A. Vascular Transport in Plants. San Diego, CA: Elsevier Academic Press, 2005.
Javot, H. & Maurel, C. The role of aquaporins in root water uptake. Annals of Botany90, 1-13 (2002).
Kramer, P. J. & Boyer, J. S. Water Relations of Plants and Soils. New York, NY: Academic Press, 1995.
Kramer, P. J. & Bullock, H. C. Seasonal variations in the proportions of suberized and unsuberized roots of trees in relation to the absorption of water. American Journal of Botany53, 200-204 (1966).
MacFall, J. S., Johnson, G. A. & Kramer, P. J. Observation of a water-depletion region surrounding loblolly pine roots by magnetic resonance imaging. Proceedings of the National Academy of Sciences of the United States of America87, 1203-1207 (1990).
McCully, M. E. Roots in Soil: Unearthing the complexities of roots and their rhizospheres. Annual Review of Plant Physiology and Plant Molecular Biology50, 695-718 (1999).
McDowell, N. G. et al. Mechanisms of plant survival and mortality during drought: Why do some plants survive while others succumb to drought? New Phytologist178, 719-739 (2008).
Nardini, A., Lo Gullo, M. A. & Salleo, S. Refilling embolized xylem conduits: Is it a matter of phloem unloading? Plant Science180, 604-611 (2011).
Pittermann, J. et al. Torus-margo pits help conifers compete with angiosperms. Science310, 1924 (2005).
Sack, L. & Holbrook, N. M. Leaf hydraulics. Annual Review of Plant Biology57, 361-381 (2006).
Sack, L. & Tyree, M. T. "Leaf hydraulics and its implications in plant structure and function," in Vascular Transport in Plants, eds. N. M. Holbrook & M. A. Zwieniecki. (San Diego, CA: Elsevier Academic Press, 2005) 93-114.
Schenk, H. J. & Jackson, R. B. Rooting depths, lateral root spreads, and belowground/aboveground allometries of plants in water-limited environments. Journal of Ecology90, 480-494 (2002).
Sperry, J. S. & Tyree, M. T. Mechanism of water-stress induced xylem embolism. Plant Physiology88, 581-587 (1988).
Steudle, E. The cohesion-tension mechanism and the acquisition of water by plants roots. Annual Review of Plant Physiological and Molecular Biology52, 847-875 (2001).
Steudle, E. Transport of water in plants. Environmental Control in Biology40, 29-37 (2002).
Takahashi, H. Hydrotropism and its interaction with gravitropism in roots. Plant Soil165, 301-308 (1994).
Tyree, M. T. & Ewers, F. W. The hydraulic architecture of trees and other woody plants. New Phytologist119, 345-360 (1991).
Tyree, M. T. & Sperry, J. S. Vulnerability of xylem to cavitation and embolism. Annual Review of Plant Physiology and Molecular Biology40, 19-38 (1989).
Tyree, M. T. & Zimmerman, M. H. Xylem Structure and the Ascent of Sap. 2nd ed. New York, NY: Springer-Verlag, 2002.
Tyree, M. T. & Ewers, F. The hydraulic architecture of trees and other woody plants. New Phytologist119, 345-360 (1991).
Wheeler, T. D. & Stroock, A. D. The transpiration of water at negative pressures in a synthetic tree. Nature455, 208-212 (2008).
Wullschleger, S. D., Meinzer, F. C. & Vertessy, R. A. A review of whole-plant water use studies in trees. Tree Physiology18, 499-512 (1998).
Zimmerman, M. H. Xylem Structure and the Ascent of Sap. 1st ed. Berlin, Germany: Springer-Verlag, 1983.
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15399 | https://wayground.com/admin/quiz/63e50ea71368a6001ecbd41b/igcse-physics-waves-reflection-refraction-etc | IGCSE Physics - Waves (reflection, Refraction etc) 10th Grade Quiz | Wayground (formerly Quizizz)
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IGCSE Physics - Waves (reflection, Refraction etc)
Quiz
•
Physics
•
10th Grade
•
Medium
•
NGSS
HS-PS4-1, MS-PS4-2, HS-PS4-3
+1
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Claire Phillips
Used 37+ times
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26 questions
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[x]
1.
MULTIPLE CHOICE QUESTION
2 mins • 1 pt
Preview
Edit
The units of frequency are...
metres
seconds
Hertz
degrees
Tags
NGSS.HS-PS4-1
2.
MULTIPLE CHOICE QUESTION
2 mins • 1 pt
Preview
Edit
Light is an example of a transverse wave
True
False
Tags
NGSS.HS-PS4-3
3.
MULTIPLE CHOICE QUESTION
2 mins • 1 pt
Preview
Edit
Sound is an example of a transverse wave
True
False
4.
MULTIPLE CHOICE QUESTION
2 mins • 1 pt
Preview
Edit
A microwave has a frequency of 300,000 kHz. What is the wavelength of these waves? (speed of light = 300,000,000 m/s)
1000 m
100 m
10 m
1 m
5.
MULTIPLE CHOICE QUESTION
2 mins • 1 pt
Preview
Edit
The following image demonstrates...
Reflection
Refraction
Interference
Diffraction
Tags
NGSS.HS-PS4-5
6.
MULTIPLE CHOICE QUESTION
2 mins • 1 pt
Preview
Edit
θ 1=35°,θ 2=25°\theta1\ =35\degree,\ \theta2=25\degree θ 1=3 5°,θ 2=2 5°What is the refractive index, n, of water?
1.37
1.36
1.35
1.34
7.
MULTIPLE CHOICE QUESTION
2 mins • 1 pt
Preview
Edit
20 waves pass a point in 5 s. How is the frequency of these waves?
100 Hz
20 Hz
5 Hz
4 Hz
Tags
NGSS.HS-PS4-1
8.
MULTIPLE CHOICE QUESTION
2 mins • 1 pt
Preview
Edit
All waves can be reflected and refracted.
True
False
Tags
NGSS.MS-PS4-2
9.
MULTIPLE CHOICE QUESTION
2 mins • 1 pt
Preview
Edit
The critical angle in the diagram is 43 degrees. What is the angle of refraction at the critical angle?
Less than 43
Equal to 43
More than 43 but less than 90
Equal to 90 degrees
10.
MULTIPLE CHOICE QUESTION
2 mins • 1 pt
Preview
Edit
A ray of light passes through a window. Which path does it take?
A
B
C
D
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