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Graph Representations
Standard Form
Tangent Lines
Circle (shape)
Coordinate Plane
Standard Form of Circle E...
Geometry of Circles
5
What is the equation of the circle in standard form if the circle is tangent to the x- and y-axes and passes through the point (6, 3)?
Gordon M. Brown
Math Tutor at San Diego City College (2018-Present)
·
Author has 6.2K answers and 4.3M answer views
·
3y
As we’ll see shortly, there are two circles that meet these criteria. Regardless, since each circle is tangent to both axes, we know that the x- and y-coordinates of their respective centers must be equal, and further, they must equal the radii of the circles. So we’ll just set up the general equation accordingly, letting a represent the quantities involved:
(x - a)^2 + (y - a)^2 = a^2
Now substitute what we know about the given point (6, 3):
(6 - a)^2 + (3 - a)^2 = a^2
36 - 12a + a^2 + 9 - 6a + a^2 = a^2
2a^2 - 18a + 45 = a^2
a^2 - 18a + 45 = 0
(a - 3) (a - 15) = 0
a = { 3, 15 }
Therefore, the tw
As we’ll see shortly, there are two circles that meet these criteria. Regardless, since each circle is tangent to both axes, we know that the x- and y-coordinates of their respective centers must be equal, and further, they must equal the radii of the circles. So we’ll just set up the general equation accordingly, letting a represent the quantities involved:
(x - a)^2 + (y - a)^2 = a^2
Now substitute what we know about the given point (6, 3):
(6 - a)^2 + (3 - a)^2 = a^2
36 - 12a + a^2 + 9 - 6a + a^2 = a^2
2a^2 - 18a + 45 = a^2
a^2 - 18a + 45 = 0
(a - 3) (a - 15) = 0
a = { 3, 15 }
Therefore, the two equations are
(x - 3)^2 + (y - 3)^2 = 9 and (x - 15)^2 + (y - 15)^2 = 225
In the graph below, notice that both circles pass through the point (6, 3).
Haresh Sagar
Studied Science & Mathematics (Graduated 1988)
·
Author has 6.2K answers and 7M answer views
·
3y
If a circle is tangent to both the axes, its centre lies either on x+y=0 or x−y=0. Here circle passes through (6,3) so it's centre must lie on x−y=0. So arbitrary centre of circle can be written as (h,h).
Since tangent points of this circle lie on x as well as y axis, its radius length must be equal to numerical value of x and y coordinates of centre which is (h,h) so from point on circumference and arbitrary centre, we can find coordinates of centre of circle as well as radius of circle.
(h−6)2+(h−3)2=h2⟹
h2−12h+36+h2−6h+9=h2⟹
h2−18h+45=0⟹
(h−15)(h−3)=0⟹
h
If a circle is tangent to both the axes, its centre lies either on x+y=0 or x−y=0. Here circle passes through (6,3) so it's centre must lie on x−y=0. So arbitrary centre of circle can be written as (h,h).
Since tangent points of this circle lie on x as well as y axis, its radius length must be equal to numerical value of x and y coordinates of centre which is (h,h) so from point on circumference and arbitrary centre, we can find coordinates of centre of circle as well as radius of circle.
(h−6)2+(h−3)2=h2⟹
h2−12h+36+h2−6h+9=h2⟹
h2−18h+45=0⟹
(h−15)(h−3)=0⟹
h=3,h=15
C1=(3,3),C2=(15,15)
Equation of C1⟹
(x−3)2+(y−3)2=9
Equation of C2⟹
(x−15)2+(y−15)2=225
Assistant
Bot
·
1y
To find the equation of the circle in standard form that is tangent to the x- and y-axes and passes through the point (6, 3), we can follow these steps:
Understanding the Tangent Condition
:
If a circle is tangent to the x- and y-axes, its center must be at a distance equal to its radius
r
from both axes. Therefore, if the center of the circle is at
(
h
,
k
)
, we have
h
=
r
and
k
=
r
. Thus, the center can be expressed as
(
r
,
r
)
.
2. Circle Equation
:
- The standard form of the equation of a circle is given by:
(
x
−
h
)
2
+
(
y
−
k
)
2
=
r
2
- Substituting
h
=
r
and
k
=
r
:
(
x
−
r
)
2
+
(
y
To find the equation of the circle in standard form that is tangent to the x- and y-axes and passes through the point (6, 3), we can follow these steps:
Understanding the Tangent Condition:
If a circle is tangent to the x- and y-axes, its center must be at a distance equal to its radius r from both axes. Therefore, if the center of the circle is at (h,k), we have h=r and k=r. Thus, the center can be expressed as (r,r).
Circle Equation:
The standard form of the equation of a circle is given by:
(x−h)2+(y−k)2=r2
Substituting h=r and k=r:
(x−r)2+(y−r)2=r2
Using the Given Point (6, 3):
Since the circle passes through the point (6, 3), we can substitute x=6 and y=3 into the circle equation:
(6−r)2+(3−r)2=r2
Expanding the Equation:
Expanding both squares:
(6−r)2=36−12r+r2
(3−r)2=9−6r+r2
Combining these gives:
(36−12r+r2)+(9−6r+r2)=r2
Simplifying:
36+9−12r−6r+2r2=r2
45−18r+2r2=r2
Rearranging leads to:
r2−18r+45=0
Solving the Quadratic Equation:
Using the quadratic formula:
r=−b±√b2−4ac2a
Here, a=1, b=−18, and c=45:
r=18±√(−18)2−4⋅1⋅452⋅1
r=18±√324−1802
r=18±√1442
r=18±122
This gives two potential solutions for r:
r=302=15andr=62=3
Finding the Circle's Center:
The center of the circle is (r,r):
For r=15: the center is (15,15)
For r=3: the center is (3,3)
Circle Equations:
For r=15:
(x−15)2+(y−15)2=152⇒(x−15)2+(y−15)2=225
For r=3:
(x−3)2+(y−3)2=32⇒(x−3)2+(y−3)2=9
Final Answer:
The equations of the circles are:
1. (x−15)2+(y−15)2=225 (if r=15)
2. (x−3)2+(y−3)2=9 (if r=3)
Both circles satisfy the conditions given in the problem.
Related questions
A circle is tangent to y-axis, and its center is on the line x − 3y = 0. If the circle passes through point A (6, 1), what is the equation of the circles?
What is the standard equation of the circle that has center (3,4) and tangent to the y-axis?
What is the equation of a circle that passes through point (2,5) and tangent to the line y=3x-2 at the point (1,1)?
What is the equation of the circle if the center is (3,4) tangent to x-axis?
A circle has radius 10 units and passes through the point (5,-16). The x-axis is a tangent to the circle. What are the possible equations of the circle?
Armando Flores
Former Consultant/Senior Engineer, IBM/Lexmark Retired
·
Author has 2.1K answers and 1.6M answer views
·
3y
What is the equation of the circle in standard form if the circle is tangent to the x- and y-axes and passes through the point (6, 3)?
What is the equation of the circle in standard form if the circle is tangent to the x- and y-axes and passes through the point (6, 3)?
Pramodkumar Tandon
Retired as Prof. & Head at Institute of Engineering and Rural Technology (1965–present)
·
Author has 2.1K answers and 1.2M answer views
·
2y
Since the circle is tangent to both the axes, it’s center ( C ) will have both the coordinates numerically equal to each it’s radius (r). Hence center is ( r, r )
Again since it touches both the axes, whole of the circle must lie in the same quadrant. As it passes through (6,3), the whole circle will lie in the First quadrant.
Distance from center ( C ) to the given point (6,3) = Radius = r
=> Sqrt [ (r-6)^2 + (r-3)^2 ] = r
=> (r-6)^2 + (r-3)^2 = r^2
=> r^2 - 12 r + 36 + r^2 - 6 r + 9 = r^2
=> 2 r^2 - 18 r + 45 = r^2
=> r^2 - 18 r + 45= 0
=> r^2 - 15 r - 3 r + 45 = 0
=> r ( r - 15 ) - 3 ( r - 15 ) = 0
=
Since the circle is tangent to both the axes, it’s center ( C ) will have both the coordinates numerically equal to each it’s radius (r). Hence center is ( r, r )
Again since it touches both the axes, whole of the circle must lie in the same quadrant. As it passes through (6,3), the whole circle will lie in the First quadrant.
Distance from center ( C ) to the given point (6,3) = Radius = r
=> Sqrt [ (r-6)^2 + (r-3)^2 ] = r
=> (r-6)^2 + (r-3)^2 = r^2
=> r^2 - 12 r + 36 + r^2 - 6 r + 9 = r^2
=> 2 r^2 - 18 r + 45 = r^2
=> r^2 - 18 r + 45= 0
=> r^2 - 15 r - 3 r + 45 = 0
=> r ( r - 15 ) - 3 ( r - 15 ) = 0
=> ( r - 15 ) ( r - 3 ) = 0
=> r = 3, 15
Hence there will be two circles. One with Radius = 3 and the other with radius = 15
Equations will be —
(x-3)^2 + (y-3)^2 = 9 and (x-15)^2 + (y-15) = 225 ………………… Answer
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Dave Benson
trying to make maths easy.
·
Author has 6.1K answers and 2.1M answer views
·
3y
Do a sketch, there are 2 circles with centres on line at 45º or y = x
Standard equation of a circle (x-h)²+(y-k)² = r² & centre (h,k) & radius r
But h = k and h = r so plug into the equation with x = 6 and y = 3
(6-h)²+(3-h)² = h² expand
36 -12h+h²+9–6h+h² = h²
h²-18h+45 = 0 factorise
(h-30)(h-15) = 0
h = 15 = r = k for one circle and h = 3 = r = k for the other circle.
so equations for the circles are:
(x-3)²+(y-3)² = 3² and (x-15)²+(y-15)² = 15² Answer
Do a sketch, there are 2 circles with centres on line at 45º or y = x
Standard equation of a circle (x-h)²+(y-k)² = r² & centre (h,k) & radius r
But h = k and h = r so plug into the equation with x = 6 and y = 3
(6-h)²+(3-h)² = h² expand
36 -12h+h²+9–6h+h² = h²
h²-18h+45 = 0 factorise
(h-30)(h-15) = 0
h = 15 = r = k for one circle and h = 3 = r = k for the other circle.
so equations for the circles are:
(x-3)²+(y-3)² = 3² and (x-15)²+(y-15)² = 15² Answer
Related questions
What is the equation of a circle which has both axes at its tangents and passes through point (1;2)?
A circle passes through (-6,9), touches the x axis and its y intercept is 8. What is the equation of the circle?
What is the equation of the circle with the center at (-4, 3) and tangent to the y-axis?
What is the equation of a circle that has both axes as tangents and passes through (3,-6)?
A circle passes through the point (12,0), and is tangent to the y-axis at the point (0,3). What is the radius of the circle?
Khaliquzzaman Khan
Taught Secondary School level classes
·
Author has 1.2K answers and 1.1M answer views
·
3y
Equation of a circle with centre (h,,h) and radius r is:
(x-h)^2+(y-k)^2=r^2
It passes through the point (6,3)
(6-h)^2+(3-k)^2=r^2
Both Axis of x and y are tangents to the circle
so, h=k=r
(6-h)^2+(3-h)^2=h^2
Or, 36+h^2–12h+9+h^2-6h=h^2
Or, h^2–18h+45=0
h=15,3
Put h=k=r=15
Equation of circle is
(x-15)^2+(y-15)^2=15^2
Put h=k=r=3
Equation of circle is
(x-3)^2+(y-3)^2=3^2
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Pradeep Hebbar
Many years of Structural Engineering & Math enthusiasm
·
Author has 9.2K answers and 6.2M answer views
·
3y
Equation of circle with center (h,k) having radius r is given by,
(x−h)2+(y−k)2=r2…(1)
Circle is tangential to x-axis and y-axis. Hence its center is equidistant from x and y axis.
h=k=r
Given that the circle passes through the point (6,3)
(6−h)2+(3−k)2=r2
(6−r)2+(3−r)2=r2
36+r2–12r+9+r2–6r=r2
r2–18r+45=0
Roots of this equation are, r=15 and r=3
To get the equation of circle, for each case we substitute the values of h, k and r in eqn. (1).
Case 1: r=15⟹h=k=15
(x−15)2+(y−15)2=152
Case 2: r=3⟹h=k=3
(x−3)2+(y−3)2=32
Janet Heberling
Lives in San Francisco, CA (2022–present)
·
Author has 21.5K answers and 9.4M answer views
·
3y
the circle is tangent to the x- and y-axes
Draw a picture. The center lies on the line y=x in Quadrant I.
Radius r is the distance from center to x- or y-axis
(x−r)2+(y−r)2=r2
passes through the point (6, 3)
(6−r)2+(6−r)2=r2(r2−12r+36)+(r2−6r+9)=r2r2−18r+45=0r=15,3
two solutions:
(x−15)2+(y−15)2=152
(x−3)2+(y−3)2=32
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Mike Hirschhorn
Honorary Associate Professor of Mathematics at UNSW
·
Author has 8.1K answers and 2.7M answer views
·
3y
centre (a,a), radius a,
(x−a)2+(y−a)2=a2,
(6−a)2+(3−a)2=a2,
a2–18a+45=0,
(a−15)(a−3)=0,
a=15 or 3,
x2+y2−30x−30y+225=0 or x2+y2−6x−6y+9=0.
Haresh Sagar
Studied Science & Mathematics (Graduated 1988)
·
Author has 6.2K answers and 7M answer views
·
Sep 9
Related
A circle is tangent to y-axis, and its center is on the line x − 3y = 0. If the circle passes through point A (6, 1), what is the equation of the circles?
If a circle is tangent to y axis, the radius is absolute value of x coordinate of centre. So if we assume centre as (h,k), the radius is |h|.
Given that centre is on x-3y=0, it can be written as (h,h/3). One more thing given is A(6,1) is on circle so…
(6−h)2+(1−h3)2=h2
36−12h+h2+1−23h+h29=h2
h29−383h+37=0
h2−114h+333=0
h=111,3
Equations…
(x−3)2+(y−1)2=32
(x−111)2+(y−37)2=1112
If a circle is tangent to y axis, the radius is absolute value of x coordinate of centre. So if we assume centre as (h,k), the radius is |h|.
Given that centre is on x-3y=0, it can be written as (h,h/3). One more thing given is A(6,1) is on circle so…
(6−h)2+(1−h3)2=h2
36−12h+h2+1−23h+h29=h2
h29−383h+37=0
h2−114h+333=0
h=111,3
Equations…
(x−3)2+(y−1)2=32
(x−111)2+(y−37)2=1112
Pramodkumar Tandon
Retired as Prof. & Head at Institute of Engineering and Rural Technology (1965–present)
·
Author has 2.1K answers and 1.2M answer views
·
Sep 10
Related
A circle is tangent to y-axis, and its center is on the line x − 3y = 0. If the circle passes through point A (6, 1), what is the equation of the circles?
Roman Andronov
Solving technical problems is my day job.
·
Author has 427 answers and 13.4M answer views
·
Updated 9mo
Related
A circle passes through the point (-1,1), (0,6) and (5,5). What is the equation of the circle?
We instantly see a nice little C program in here: assume that in a certain plane
p
an orthogonal coordinate system is fixed and in that coordinate system the following three distinct points
P
1
(
x
1
,
y
1
)
,
P
2
(
x
2
,
y
2
)
and
P
3
(
x
3
,
y
3
)
are given.
If all three
x
-coordinates of these points are equal and all three
y
-coordinates of these points are equal then the given points are one and the same and we return a NON_DISTINCT_POINTS error. If any two
x
-coordinates are equal and the corresponding two
y
-coordinates are equal then the respective
two
points are one and the same and we return the same err
We instantly see a nice little C program in here: assume that in a certain plane p an orthogonal coordinate system is fixed and in that coordinate system the following three distinct points P1(x1,y1),P2(x2,y2) and P3(x3,y3) are given.
If all three x-coordinates of these points are equal and all three y-coordinates of these points are equal then the given points are one and the same and we return a NON_DISTINCT_POINTS error. If any two x-coordinates are equal and the corresponding two y-coordinates are equal then the respective two points are one and the same and we return the same error. In all of the above cases infinitely many circles pass through a single point or two distinct points. Either way, the input is invalid.
If the said three distinct points are collinear then there does not exist a circle that passes through them (and we return a COLLINEAR_POINTS error).
Assume that the three given distinct points are not collinear.
If any pair of such points PiPj defines a straight line that is perpendicular to the x-axis then there always exist the third input point Pk that can be used to construct the equations of the straight lines PkPi and PkPj none of which will be perpendicular to the x-axis, see below.
In that case there exists a unique circle c(C(cx,cy),r) in the plane p with the following equation in the chosen coordinate system:
(x−cx)2+(y−cy)2=r2(1)
(and there exist infinitely many spheres that pass through these three points; the centers of all such spheres will be located on the straight line that is perpendicular to the plane p and that passes through the circumcenter of the triangle P1P2P3 - a straight line that is equally distanced from the points P1,2,3)
Thus, our problem is reduced to recovering the center of the circle C(cx,cy) and the value of its radius r given three distinct non collinear points P1,2,3.
The Center
To that end, we observe that the sought-after center C is located at the intersection of the perpendicular bisectors a and b of the line segments, say, P1P2 and P2P3 respectively (Fig. 1):
Therefore, it is sufficient to construct the equations of the straight lines λa and λb passing through the points P1,2 and P2,3 respectively:
λa:ya=ma(x−x1)+y1(2)
λb:yb=mb(x−x2)+y2(3)
where the slopes ma and mb are readily computable as follows:
ma=y2−y1x2−x1(4)
mb=y3−y2x3−x2(5)
In other words, the real numbers ma and mb are known to us - ahead of time in their symbolic form.
But we all remember from our middle school days that the slope of a normal to a given straight line is a negative inverse of the slope of the parent straight line.
Hence, the slopes and [math]m'_b[/math] of our perpendicular bisectors [math]a[/math] and [math]b[/math] are, respectively:
[math]m'_a = -\,\dfrac{1}{m_a},\;m'_b = -\,\dfrac{1}{m_b} \tag{6}[/math]
and the equations of the said bisectors are given by:
[math]a\,:\;y'_a = -\,\dfrac{1}{m_a}\,\left(x-\dfrac{x_1+x_2}{2}\right)+\dfrac{y_1+y_2}{2} \tag{7}[/math]
[math]b\,:\;y'_b = -\,\dfrac{1}{m_b}\,\left(x-\dfrac{x_2+x_3}{2}\right)+\dfrac{y_2+y_3}{2} \tag{8}[/math]
Across the equations (7, 8) we have two equations and two unknowns: [math]c_x[/math] and [math]c_y[/math].
Because the center [math]C[/math] is located at the intersection of the straight lines [math]a[/math] and [math]b[/math], in order to recover [math]c_x[/math] and [math]c_y[/math] we, first, equate (7) to (8):
[math]-\,\dfrac{1}{m_a}\,\left(c_x-\dfrac{x_1+x_2}{2}\right)+\dfrac{y_1+y_2}{2} = -\,\dfrac{1}{m_b}\,\left(c_x-\dfrac{x_2+x_3}{2}\right)+\dfrac{y_2+y_3}{2} \tag{9}[/math]
and solve (9) for [math]c_x[/math] by multyplying (9) through by [math]-2m_am_b[/math]:
[math]c_x = \dfrac{m_am_b(y_1-y_3)+m_b(x_1+x_2)-m_a(x_2+x_3)}{2(m_b-m_a)} \tag{10}[/math]
and then we put [math]c_x[/math] from (10) into either (7) or (8) to recover [math]c_y[/math]:
[math]c_y = -\,\dfrac{1}{m_a}\,\left(c_x-\dfrac{x_1+x_2}{2}\right)+\dfrac{y_1+y_2}{2} \tag{11}[/math]
[math]c_y = -\,\dfrac{1}{m_b}\,\left(c_x-\dfrac{x_2+x_3}{2}\right)+\dfrac{y_2+y_3}{2} \tag{12}[/math]
Game over.
At this point it is never the case that [math]m_a=m_b[/math] because we already made sure that the three given points are not collinear and, hence, the respective perpendicular bisectors will never be parallel to each other.
If any of the above straight lines is strictly vertical and its slope is equal to infinity (or not defined) then we can always choose another pair of points that will eliminate this case.
Note also that all along we have been manipulating the linear quantities only.
The Radius
Once the location of the center [math]C[/math] of the circle [math]c[/math] in [math]p[/math] is known (Fig. 2):
then the value of the radius [math]r[/math] of the circle [math]c[/math] in [math]p[/math] is trivially known as the Euclidean metric or the Euclidean distance between the point [math]C[/math] and any of the given distinct points [math]P_{1,2,3}[/math]:
[math]r = \sqrt{(c_x-x_i)^2+(c_y-y_i)^2},\;i=1,2,3 \tag{13}[/math]
Pick your favorite [math]i[/math].
In your case we have:
[math]P_1(-1,1),\;P_2(0,6),\;P_3(5,5) \tag{}[/math]
Putting the respective values of [math]x_i[/math] and [math]y_i[/math] into the equations (4, 5, 10, 11, 13), we find the equation of the circle sought-after:
math^2 + (y - 3)^2 = (\sqrt{13})^2 \tag{14}[/math]
For more information on and ideas about puzzles and problem-solving in mathematics, physics and computer science please visit my YouTube channel ProbLemma.
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11601 | https://frrq.cvg.utn.edu.ar/pluginfile.php/29484/mod_resource/content/0/CIRCUITOS%20MAGNETICOS%20Rev2.pdf | CIRCUITOS MAGNETICOS CIRCUITOS MAGNETICOS < Núcleo Magnético 𝑰→ Φ → Φ = 𝐵⋅𝑆= 𝜇0 ⋅𝜇𝑟⋅𝐻⋅𝑆 Φ = 𝜇0 ⋅𝜇𝑟⋅𝑁⋅𝐼 𝐿𝑚 ⋅𝑆→ Φ = 𝑁⋅𝐼 𝐿𝑚 𝜇0 ⋅𝜇𝑟⋅𝑆 ℱ 𝑚𝑚= 𝑁⋅𝐼𝐴⋅𝑣→𝐹𝑢𝑒𝑟𝑧𝑎𝑚𝑎𝑔𝑛𝑒𝑡𝑜𝑚𝑜𝑡𝑟𝑖𝑧 ℜ= 𝐿𝑚 𝜇0 ⋅𝜇𝑟⋅𝑆 1 𝐻→𝑅𝑒𝑙𝑢𝑐𝑡𝑎𝑛𝑐𝑖𝑎 Circuitos Magnético Circuitos Eléctrico Φ = ℱ 𝑚𝑚 ℜ 𝐼= 𝐸 𝑅 ℱ 𝑚𝑚= 𝑁⋅𝐼= Φ ⋅ℛ𝑖= 𝐻𝑖⋅𝑙𝑖= Φ1 ⋅ℛ1 + Φ2 ⋅ℛ2 = 𝐻1 ⋅𝑙1 + 𝐻2 ⋅𝑙2 + 𝐻𝑒⋅𝑙𝑒 𝐷𝑖𝑠𝑡𝑖𝑛𝑡𝑜𝑠𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙𝑒𝑠𝑞𝑢𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑒𝑙𝑛ú𝑐𝑙𝑒𝑜 Ley de Hopkinsón Ley de Ohm de los Circuitos Magnético ℱ 𝑚𝑚= Φ ⋅ℜ Ley de Ohm De Circuitos Eléctrico 𝐸= 𝐼⋅𝑅 Analogía entre los circuitos eléctricos y magnéticos.
Núcleos Magnético a b • Núcleo Sección rectangular 𝐴𝑒= 𝑎+ 𝑒∗(𝑏+ 𝑒) • Núcleo Sección circular 𝑒 𝐷2 𝐴= 𝜋𝑅1 2 𝐴𝑒= 𝜋𝑅2 2 = 𝜋(𝑅2 + 𝑒2) 𝐷1 Cuando una bobina esta alimentada con C.C., la intensidad de corriente depende únicamente de la tensión aplicada, pero es absolutamente independiente de la naturaleza y características de o los materiales que esta fabricado el núcleo.
Sucesión de efectos para la bobina alimentada con C.C. Si se aumentara la ℜ, del circuito magnético, por ejemplo agregando un entrehierro, la intensidad de corriente absorbida por la bobina no cambia, lo que si cambia es el Φ, haciendo que este disminuya.
CIRCUITOS MAGNETICOS CIRCUITOS MAGNETICOS CIRCUITOS MAGNETICOS CIRCUITOS MAGNETICOS CIRCUITOS MAGNETICOS Maquina Para soldar Monofásica CIRCUITOS MAGNETICOS • PROBLEMA DIRECTO (Conocido el flujo determino la Fm.m.) • PROBLEMA INDIRECTO Conocida la f.m.m. determino el flujo. (Método de ensayo y error) -Luego reduzco B por caída en el hierro y con este nuevo valor determino -De la tabla saco H para el material y determino -Luego comparo con Si el error esta dentro del 5% acepto la aproximación, si no ajusto nuevamente el valor de B y repito los cálculos.
0 0 0 l B Fmm = m m mm l H F = i N 0 mm mm F F m + 𝐻= 𝐵 𝜇0 • Se tiene en cuenta el % de Error con el que se va a trabajar, Por Ej. 5%.
• Suponemos que toda la f.m.m cae en el entrehierro o en material con menor permeabilidad si no hay entrehierro.
• Si el núcleo está compuesto sin entrehierro y por dos materiales muy similares , puedo suponer una caída del 50% en cada uno.
Determino la caída de tensión magnética y comparo con la Fuerza magnetomotriz de la bobina Supongamos que tiene entrehierro: Elijo un valor de B y determino H ) 𝑈𝑒= 𝐻𝑒𝑙𝑒 Ejemplo de Aplicación CIRCUITOS MAGNETICOS • En Corriente Alterna Aplicando la Ley de Kirchhoff 𝑖(𝑡) Φ(𝑡) 𝑓𝑒𝑚 ↓ ↓ ↓ ↓ 𝑣= 𝑅𝑖+ 𝑁𝑑Φ 𝑑𝑡 𝑣= 𝑁𝑑Φ 𝑑𝑡 ↓ V(t) R N espiras Φ(𝑡) = 1 𝑁න𝑣𝑑𝑡= 2 𝑁𝜔𝑉𝑠𝑒𝑛𝜔𝑡 La constante de integración es nula Siempre que se considere que no hay magnetismo remanente en el núcleo.
Φ(𝑡) = Φ𝑚𝑠𝑒𝑛𝜔𝑡= Φ𝑚𝑐𝑜𝑠(𝜔𝑡−90°) Φ𝑚= 2 𝑁𝜔𝑉 𝑉= 2𝜋𝑓 2 𝑁Φ𝑚 𝑉= 4,44𝑓𝑁Φ𝑚 → 1 • La tensión aplicada es el valor eficaz y el flujo es el máximo.
• El flujo se retrasa 90° respecto de la tensión en la bobina.
• El valor del flujo queda definido por el valor de tensión aplicada y de su frecuencia, pero es absolutamente independiente de la naturaleza y características de o los materiales que está fabricado el núcleo.
Despejando → 𝑉 Si se aumentara la ℜ, del circuito magnético, por ejemplo agregando un entrehierro, la intensidad de corriente absorbida por la bobina cambia, aumenta, lo hace para poder mantener el flujo Φ constante, ya que este depende del valor de la tensión de alimentación.
Sucesión de efectos para la bobina alimentada con C.A. CIRCUITOS MAGNETICOS Se considerara que es circuito magnético es lineal, o sea, que la permeabilidad es cte.
Que el núcleo no tenga perdidas Que el núcleo tenga perdidas.
• Núcleo sin perdidas Circuito eléctrico equivalente de una bobina con núcleo de hierro alimentada con c.a.
Si la potencia activa consumida por la bobina es nula, esto hace de considerar que la resistencia de la bobina es despreciable.
Φ = ℱ 𝑚𝑚 ℜ = 𝑁⋅𝑖𝑒𝑥𝑐 𝑙𝑚 𝜇⋅𝑆 = 𝜇⋅𝑁⋅𝑖𝑒𝑥𝑐 𝑙𝑚 ⋅𝑆 𝑖𝑒𝑥𝑐𝐶𝑜𝑟𝑟𝑖𝑒𝑛𝑡𝑒𝑑𝑒𝑒𝑥𝑐𝑖𝑡𝑎𝑐𝑖𝑜𝑛𝑖𝑛𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑎 𝑣= 𝑁𝑑Φ 𝑑𝑡 𝑣= 𝜇⋅𝑁2 ⋅𝑆 𝑙 𝑑𝑖𝑒𝑥𝑐 𝑑𝑡 𝑣 Φ 𝑣= 𝜇⋅𝑁2 ⋅𝑆 𝑙 𝑑𝑖𝑒𝑥𝑐 𝑑𝑡 𝑣= 𝐿𝑑𝑖𝑒𝑥𝑐 𝑑𝑡 𝐿= 𝜇⋅𝑁2 ⋅𝑆 𝑙 Circuito equivalente Representado por una bobina de valor L Al no existir perdidas la 𝑖𝑒𝑥𝑐 está en fase con el Φ Bobina con núcleo de hierro.
• Núcleo con perdidas En el caso que el núcleo tenga perdidas en el hierro, la potencia activa absorbida de la red debe vencer estas perdidas, por lo tanto el ángulo formado entre la corriente de excitación Y la tensión ya no va a ser de 90 grados.
𝑃𝐹𝑒= 𝑉𝐼𝑒𝑥𝑐𝑐𝑜𝑠𝜑𝑣 𝐼𝑒𝑥𝑐 𝐼𝜇 → 𝐶𝑜𝑟𝑟𝑖𝑒𝑛𝑡𝑒𝑚𝑎𝑔𝑛𝑒𝑡𝑖𝑧𝑎𝑛𝑡𝑒 𝐼𝐹𝑒 →𝐶𝑜𝑟𝑟𝑖𝑒𝑛𝑡𝑒𝑑𝑒𝑝𝑒𝑟𝑑𝑖𝑑𝑎𝑒𝑛𝑒𝑙𝑛𝑢𝑐𝑙𝑒𝑜 Diagrama vectorial 𝐼𝐹𝑒= 𝐼𝑒𝑥𝑐𝑐𝑜𝑠𝜑𝑣 𝐼𝜇= 𝐼𝑒𝑥𝑐𝑠𝑒𝑛𝜑𝑣 𝐼𝑒𝑥𝑐= 𝐼𝐹𝑒+ 𝐼𝜇 𝑅𝐹𝑒 →𝑠𝑒𝑙𝑎𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑟𝑒𝑠𝑖𝑠𝑡𝑒𝑛𝑐𝑖𝑎𝑑𝑒𝑝é𝑟𝑑𝑖𝑑𝑎𝑒𝑛𝑒𝑙ℎ𝑖𝑒𝑟𝑟𝑜 𝑅𝐹𝑒= 𝑉 𝐼𝐹𝑒 𝑋𝜇= 𝑉 𝐼𝜇 𝐼𝜇 →𝑟𝑒𝑝𝑒𝑠𝑒𝑛𝑡𝑎𝑙𝑎𝑐𝑜𝑟𝑟𝑖𝑒𝑛𝑡𝑒𝑛𝑒𝑐𝑒𝑠𝑎𝑟𝑖𝑎𝑝𝑎𝑟𝑎𝑚𝑎𝑔𝑛𝑒𝑡𝑖𝑧𝑎𝑟 𝑒𝑙𝑛𝑢𝑐𝑙𝑒𝑜,𝑡𝑎𝑚𝑏𝑖𝑒𝑛𝑒𝑛𝑒𝑙𝑐𝑎𝑠𝑜sin 𝑝𝑒𝑟𝑑𝑖𝑑𝑎𝑠.
𝑋𝜇 →𝑠𝑒𝑙𝑎𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑖𝑎𝑚𝑎𝑔𝑛𝑒𝑡𝑖𝑧𝑎𝑛𝑡𝑒 Perdidas en el hierro RFe I2 Fe Representan las perdidas en el hierro.
Circuito equivalente Bobina con núcleo de hierro.
Corriente de excitación en una bobina con núcleo de hierro Hasta ahora hemos visto la determinación de la corriente de excitación, habiendo analizado en el caso de que el circuito magnético es lineal, de permeabilidad constante. En este contexto se determinaron ecuaciones que relacionan la tensión y la corriente o el flujo y la corriente.
Esto Implica que si la excitación es senoidal también lo son las formas de ondas de la flujo y la corriente.
En la práctica, la curva de imanación de un material es no lineal, por lo tanto la forma de onda de la curva de corriente de excitación deja de ser senoidal, por lo que hay que recurrir a técnicas gráficas para su determinación al ser imposible usar técnicas analíticas.
• Núcleo sin perdidas La relación entre el Flujo y la corriente de excitación se Obtienen Gráficamente.
Forma de onda no senoidal de la corriente de excitación que desarrollando por series de Fourier, se demuestra que aparecen armónicos 1 , 3 , 5 Forma de onda no senoidal de la corriente de excitación en vacío, determinada sumando los armónicos 1 , 3 . La Corriente de excitación atrasa 90 grados.
• Núcleo con perdidas La corriente de excitación esta representada por una onda senoidal equivalente, a la que le corresponde un valor eficaz igual a la raíz cuadrada de la suma de los cuadrados de los valores eficaces de la corriente fundamental y sus armónicos. Se consideran las pérdidas en el núcleo, únicamente Perdidas por histéresis. El ciclo Se ve ensanchado.
Si se agregarían las Perdidas por corrientes De Foucault, el ciclo de Histéresis se ensancharía Aun más.
Ejemplo ⟸ 𝑉= 2𝜋𝑓 2 𝑁Φ𝑚 𝑎) Las perdidas en el hierro son 20 W ⟹ 𝑹𝑭𝒆= 𝑽 𝑰𝑭𝒆 ⟹ 𝑿𝝁= 𝑽 𝑰𝝁 ⟹ 𝑏) Energía y Coenergía Magnética • Aplicando Kirchhoff 𝑣= 𝑅𝑖+ 𝑁𝑑Φ 𝑑𝑡 (Tanto 𝑣, 𝑖, como Φ, dependen del tiempo.) Multiplicando ambos miembros por: 𝑖∙𝑑𝑡 𝑣𝑖∙𝑑𝑡= 𝑅𝑖2 + 𝑁𝑖𝑑Φ ← Podemos escribir como 𝑑𝑊𝑒= 𝑑𝑊 𝑅+ 𝑑𝑊 𝑚 𝑑𝑊𝑒diferencial de energía eléctrica que entra al circuito 𝑑𝑊 𝑅diferencial de energía eléctrica que consume la R de la bobina 𝑑𝑊 𝑚diferencial de energía magnética (para producir el campo magnético) 𝑑𝑊 𝑚= 𝑁𝑖𝑑Φ = ℱ 𝑚𝑚⋅𝑑Φ Partimos de un valor inicial para, 𝑡= 0, donde el flujo y la corriente son cero, Hasta un valor final de Φ, 𝑦𝑑𝑒𝑖.
La energía Magnética almacenada por el núcleo es la que se ve rayada horizontalmente. 𝑊 𝑚= න 0 Φ ℱ 𝑚𝑚⋅𝑑Φ 𝑊′𝑚= න 0 ℱ Φ ⋅𝑑ℱ 𝑚𝑚 La coenergía, no tiene un significado físico directo, es el área que se ve rayada verticalmente.
Es de gran utilidad para el calculo de la fuerza en los dispositivos electromagnéticos. ℱ ℱ Φ Φ 𝑊 𝑚 𝑊′𝑚 𝑊 𝑚= න 0 Φ ℱ 𝑚𝑚⋅𝑑Φ = න 0 𝐵 𝐻⋅𝑙⋅𝑑𝐵⋅𝑆= 𝑉𝑜𝑙න 0 𝐵 𝐻⋅𝑑𝐵 Φ = 𝐵⋅𝑆 → 𝑑Φ = 𝑑𝐵⋅𝑆 ℱ 𝑚𝑚= 𝐻⋅𝑙 →𝑑ℱ 𝑚𝑚= 𝑑𝐻⋅𝑙 𝑊 𝑚= 𝑉𝑜𝑙න 0 𝐵 𝐻𝑑𝐵 Densidad de energía magnética por densidad de volumen 𝑊′𝑚= න 0 ℱ Φ ⋅𝑑ℱ 𝑚𝑚= න 0 𝐻 𝐵𝑆𝑙𝑑𝐻= 𝑉𝑜𝑙න 0 𝐻 𝐵𝑑𝐻 𝑊′𝑚= 𝑉𝑜𝑙න 0 𝐻 𝐵𝑑𝐻 Densidad de coenergía magnética por densidad de volumen Si la curva de imanación del material es una recta entonces coincide el valor de la energía magnética y la coenergía magnética, podemos Escribir lo siguiente.
𝑊 𝑚= 𝑊′𝑚= 1 2 ℱ⋅Φ = 1 2 𝔑⋅Φ2 = 1 2 ℱ2 𝔑 𝐿= 𝑁Φ 𝑖 ℱ 𝑚𝑚= Φ ⋅ℜ 𝐿= 𝑁2 Φ 𝑁𝑖= 𝑁2 Φ ℱ= 𝑁2 ℜ 𝑊 𝑚= 𝑊′𝑚= 1 2 𝐿⋅ℱ2 𝑁2 = 1 2 𝐿⋅𝑖2 Para la densidad de energía y coenergía magnética tenemos 𝑊 𝑚= 𝑊′𝑚= 1 2 𝐻𝐵= 1 2 𝐵2 𝜇= 1 2 𝜇𝐻2 CIRCUITOS ACOPLADOS MAGNETICAMENTE El electromagnetismo es la rama de la Ing. Eléctrica o de la física, que tiene que ver con el análisis y aplicación de los campos eléctricos y magnéticos.
Los principios electromagnetismo son muy importantes y se aplican en distintas disciplinas como maquinas eléctricas, conversión de energía electromecánica, sensores remotos, fibra óptica etc.
Los dispositivos electromagnéticos incluyen motores, transformadores, generadores, levitación magnética, antenas parabólicas etc. El diseño de estos dispositivos requieren un profundo conocimiento de las leyes y los principios electromecánicos.
AUTOINDUCTANCIA E INDUCTANCIA MUTUA Si por una un Inductor circula una corriente variable en el tiempo, se generará un flujo magnético también variable y este inducirá un f.e.m. que dependerá de la variaciones de la corriente. Podemos escribir entonces: 𝑣𝐿= 𝐿𝑑𝑖 𝑑𝑡 𝑣𝐿= 𝑁𝑑Φ 𝑑𝑡 𝐿𝑑𝑖 𝑑𝑡= 𝑁𝑑Φ 𝑑𝑡 𝐿= 𝑁𝑑Φ 𝑑𝑡⋅𝑑𝑡 𝑑𝑖 𝐿= 𝑁𝑑Φ 𝑑𝑖 Autoinductancia Si no se está en la zona de saturación (por ejemplo en el centro del núcleo de un material ferromagnético o el caso del aire que no hay saturación ), la relación entre el flujo y la corriente, es lineal.
Φ = 𝐾⋅𝑖 𝐿= 𝑁Φ 𝑖 Enlaces de flujo por unidad de corriente L2 1 2 L1 1 2 INDUCTANCIA MUTUA Al circular 𝑖1(𝑡) , el flujo Φ1esta compuesto por: Φ11 (flujo de dispersión) , y Φ12 (flujo mutuo) Si el Φ = 𝐾⋅𝑖 Y 𝜇= 𝑐𝑡𝑒.
𝑣2 = 𝑁2 𝑑Φ12 𝑑𝑡 𝑣2 = 𝑀𝑑𝑖1 𝑑𝑡 𝑁2 𝑑Φ12 𝑑𝑡 = 𝑀𝑑𝑖1 𝑑𝑡 𝑀= 𝑁2 𝑑Φ12 𝑑𝑖1 𝑀= 𝑁2 Φ12 𝑖1 ⟹ ⟹ 𝑀⟶Inductancia mutua Si analizamos la bobina 2 , obtendremos el mismo resultado. Al circular 𝑖2(𝑡) , el flujo Φ2esta compuesto por: Φ21 (flujo de dispersión) , y Φ22 (flujo mutuo) 𝑣1 = 𝑁2 𝑑Φ21 𝑑𝑡 𝑣1 = 𝑀𝑑𝑖2 𝑑𝑡 𝑁2 𝑑Φ21 𝑑𝑡 = 𝑀𝑑𝑖2 𝑑𝑡 𝑀= 𝑁1 𝑑Φ21 𝑑𝑖2 𝑀= 𝑁1 Φ21 𝑖2 Si el Φ = 𝐾⋅𝑖 Y 𝜇= 𝑐𝑡𝑒.
1 2 Φ11 Φ12 𝑖1(𝑡) 𝑁1 𝑁2 → Dependiendo del espaciado, permeabilidad del medio y la orientación de los ejes de las bobinas, estas van a tener Distintos grados de acoplamiento K , Multiplicando 1 y 2 𝑀= 𝑁2 Φ12 𝑖1 𝑀= 𝑁1 Φ21 𝑖2 𝑀2 = (𝑁2 Φ12 𝑖1 ) ⋅(𝑁1 Φ21 𝑖2 ) 1 2 𝐾= Φ12 Φ1 = Φ21 Φ2 Coeficiente de acoplamiento 𝑀2 = 𝑁2 𝐾Φ1 𝑖1 ⋅𝑁1 𝐾Φ2 𝑖2 = 𝐾2 𝑁2 Φ1 𝑖1 ⋅𝑁1 Φ2 𝑖2 𝐿1 = Φ1 𝑖1 𝐿2 = Φ2 𝑖2 𝑀2 = 𝐾2𝐿1𝐿2 → 𝑀= 𝐾 𝐿1𝐿2 ANALISIS DE CIRCUITOS ACOPLADOS V2 R2 V1 R1 I1 I2 M φ1 φ2 Electroimanes Sabemos que un campo magnético almacena energía y también time la propiedad de ejercer fuerzas mecánicas en las estructuras o parte móviles que están asociadas con el circuito magnético. Esto nos permite utilizar el campo magnético como un enlace entre la parte eléctrica y mecánica de los distintos dispositivos electromecánicos. (Motores, electroimanes, Contactores etc.) Análisis manteniendo el flujo contante |
11602 | https://bio.libretexts.org/Bookshelves/Introductory_and_General_Biology/Biology_(Kimball)/04%3A_Cell_Metabolism/4.09%3A_Photosynthesis_-_Dicovering_the_Secrets | Skip to main content
4.9: Photosynthesis - Dicovering the Secrets
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4.8: Photosynthesis - The Role of Light
4.10: Chemiosmosis
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John W. Kimball
Tufts University & Harvard
( \newcommand{\kernel}{\mathrm{null}\,})
This chapter talks about various scientists and their path towards discovering photosynthesis.
van Helmont
Perhaps the first experiment designed to explore the nature of photosynthesis was that reported by the Dutch physician van Helmont in 1648. Some years earlier, van Helmont had placed in a large pot exactly 200 pounds (91 kg) of soil that had been thoroughly dried in an oven. Then he moistened the soil with rain water and planted a 5-pound (2.3 kg) willow shoot in it. He then placed the pot in the ground and covered its rim with a perforated iron plate. The perforations allowed water and air to reach the soil but lessened the chance that dirt or other debris would be blown into the pot from the outside.
For five years, van Helmont kept his plant watered with rain water or distilled water. At the end of that time, he carefully removed the young tree and found that it had gained 164 pound, 3 ounces (74.5 kg). (This figure did not include the weight of the leaves that had been shed during the previous four autumns.) He then redried the soil and found that it weighed only 2 ounces (57 g) less that the original 200 pounds (91 kg). Faced with these experimental facts, van Helmont theorized that the increase in weight of the willow arose from the water alone. He did not consider the possibility that gases in the air might be involved.
Joseph Priestley
The first evidence that gases participate in photosynthesis was reported by Joseph Priestley in 1772. He knew that if a burning candle is placed in a sealed chamber, the candle soon goes out. If a mouse is then placed in the chamber, it soon suffocates because the process of combustion has used up all the oxygen in the air — the gas on which animal respiration depends. However, Priestley discovered that if a plant is placed in an atmosphere lacking oxygen, it soon replenishes the oxygen, and a mouse can survive in the resulting mixture. Priestley thought (erroneously) that it was simply the growth of the plant that accounted for this.
Ingen-Housz
It was another Dutch physician, Ingen-Housz, who discovered in 1778 that the effect observed by Priestley occurred only when the plant was illuminated. A plant kept in the dark in a sealed chamber consumes oxygen just as a mouse (or candle) does.
Ingen-Housz also demonstrated that only green parts of plants liberated oxygen during photosynthesis. Nongreen plant structure, such as woody stems, roots, flowers, and fruits actually consume oxygen in the process of respiration. We now know that this is because photosynthesis can go on only in the presence of the green pigment chlorophyll.
Jean Senebier
The growth of plants is accompanied by an increase in their carbon content. A Swiss minister, Jean Senebier, discovered that the source of this carbon is carbon dioxide and that the release of oxygen during photosynthesis accompanies the uptake of carbon dioxide. Senebier concluded (erroneously as it turned out) that in photosynthesis carbon dioxide is decomposed, with the carbon becoming incorporated in the organic matter of the plant and the oxygen being released.
CO2 + H2O → (CH2O) + O2
(The parentheses around the CH2O signify that no specific molecule is being indicated but, instead, the ratio of atoms in some carbohydrate, e.g., glucose, C6H12O6.) The equation also indicates that the ratio of carbon dioxide consumed to oxygen release is 1:1, a finding that was carefully demonstrated in the years following Senebier's work. Using glucose as the carbohydrate product, we can write the equation for photosynthesis as
6CO2 + 6H2O → C6H12O6 + 6O2
F. F. Blackman
The above equation shows the relationship between the substances used in and produced by the process. It tells us nothing about the intermediate steps. That photosynthesis does involve at least two quite distinct processes became apparent from the experiments of the British plant physiologist F. F. Blackman. His results can easily be duplicated by using the setup in Figure 4.9.1. The green water plant Elodea (available wherever aquarium supplies are sold) is the test organism. When a sprig is placed upside down in a dilute solution of NaHCO3 (which serves as a source of CO2) and illuminated with a flood lamp, oxygen bubbles are soon given off from the cut portion of the stem. One then counts the number of bubbles given off in a fixed interval of time at each of several light intensities. Plotting these data produces a graph like the one in Figure 4.9.2.
Since the rate of photosynthesis does not continue to increase indefinitely with increased illumination, Blackman concluded that at least two distinct processes are involved: one, a reaction that requires light and the other, a reaction that does not. This latter is called a "dark" reaction although it can go on in the light. Blackman theorized that at moderate light intensities, the "light" reaction limits or "paces" the entire process. In other words, at these intensities the dark reaction is capable of handling all the intermediate substances produced by the light reaction. With increasing light intensities, however, a point is eventually reached when the dark reaction is working at maximum capacity. Any further illumination is ineffective, and the process reaches a steady rate.
This interpretation is strengthened by repeating the experiment as a somewhat higher temperature. Most chemical reactions proceed more rapidly at higher temperatures (up to a point). At 35°C, the rate of photosynthesis does not level off until greater light intensities are present. This suggest that the dark reaction is now working faster. The fact that at low light intensities the rate of photosynthesis is no greater at 35°C than at 20°C also supports the idea that it is a light reaction that is limiting the process in this range. Light reactions depend, not on temperature, but simply on the intensity of illumination.
The increased rate of photosynthesis with increased temperature does not occur if the supply of CO2 is limited. As the figure shows, the overall rate of photosynthesis reaches a steady value at lower light intensities if the amount of CO2 available is limited. Thus CO2 concentration must be added as a third factor regulating the rate at which photosynthesis occurs. As a practical matter, however, the concentration available to terrestrial plants is simply that found in the atmosphere: 0.035%.
Van Niel
It was the American microbiologist Van Niel who first glimpsed the role that light plays in photosynthesis. He studied photosynthesis in purple sulfur bacteria. These microorganisms synthesize glucose from CO2 as do green plants, and they need light to do so. Water, however, is not the starting material. Instead they use hydrogen sulfide (H2S). Furthermore, no oxygen is liberated during this photosynthesis but rather elemental sulfur. Van Niel reasoned that the action of light caused a decomposition of H2S into hydrogen and sulfur atoms. Then, in a series of dark reactions, the hydrogen atoms were used to reduce CO2 to carbohydrate:
Van Niel envisioned a parallel to the process of photosynthesis as it occurs in green plants. There the energy of light causes water to break up into hydrogen and oxygen. The hydrogen atoms are then used to reduce CO2 in a series of dark reactions:
If this theory is correct, then it follows that all of the oxygen released during photosynthesis comes from water just as all the sulfur produced by the purple sulfur bacteria comes from H2S. This conclusion directly contradicts Senebier's theory that the oxygen liberated in photosynthesis comes from the carbon dioxide. If Van Niel's theory is correct, then the equation for photosynthesis would have to be rewritten:
In science, a theory should be testable. By deduction, one can make a prediction of how a particular experiment will come out if the theory is sound. In this case, the crucial experiments needed to test the two theories had to await the time when the growth of atomic research made it possible to produce isotopes other than those found naturally or in greater concentrations than are found naturally.
Samuel Ruben
In air, water and other natural materials containing oxygen, 99.76% of the oxygen atoms are 16O and only 0.20% of them are the heavier isotope 18O. In 1941, Samuel Ruben and his coworkers at the University of California were able to prepare specially "labeled" water in which the 0.85% of the molecules contained 18O atoms. When this water was supplied to a suspension of photosynthesizing algae, the proportion of 18O in the oxygen gas that was evolved was 0.85%, the same as that of the water supplied, and not simply the 0.20% found in all natural samples of oxygen (and its compounds like CO2).
| | | | | |
--- ---
| | | % 18O FOUND IN | | |
| EXPERIMENT | | H2O | CO2 | O2 |
| 1. | START | 0.85 | 0.20 | — |
| | FINISH | 0.85 | 0.61 | 0.86 |
| 2. | START | 0.20 | 0.68 | — |
| | FINISH | 0.20 | 0.57 | 0.20 |
A non-biochemical exchange of oxygen atoms between the water and the bicarbonate ions used as a source of CO2 explains the uptake of the isotope by CO2 in the first experiment.
These results clearly demonstrated that Senebier's interpretation was in error. If all the oxygen liberated during photosynthesis comes from the carbon dioxide, we would expect the oxygen evolved in Ruben's experiment to contain simply the 0.20% found naturally. If, on the other hand, both the carbon dioxide and the water contribute to the oxygen released, we would expect its isotopic composition to have been some intermediate figure. In fact, the isotopic composition of the evolved oxygen was the same as that of the water used.
Ruben and his colleagues also prepared a source of carbon dioxide that was enriched in 18O atoms. When algae carried out photosynthesis using this material and natural water, the oxygen that was given off was not enriched in 18O. It contained simply the 0.20% 18O found in the natural water used. The heavy atoms presumably became incorporated in the other two products (carbohydrate and by-product water).
These experiments lent great support to Van Niel's idea that one function of light in photosynthesis was the separation of the hydrogen and oxygen atoms of water molecules. But there remained to work out just how the hydrogen atoms were made available to the dark reactions.
4.8: Photosynthesis - The Role of Light
4.10: Chemiosmosis |
11603 | https://math.stackexchange.com/questions/13261/how-to-get-a-reflection-vector | Skip to main content
How to get a reflection vector?
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I'm doing a raytracing exercise. I have a vector representing the normal of a surface at an intersection point, and a vector of the ray to the surface. How can I determine what the reflection will be?
In the below image, I have d and n. How can I get r?
Thanks.
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asked Dec 6, 2010 at 17:47
Nick HeinerNick Heiner
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This is question number 5.13 from I.e irodov :p
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r=d−2(d⋅n)n
where d⋅n is the dot product, and
n must be normalized.
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answered Dec 6, 2010 at 17:54
PhrogzPhrogz
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And if my d happens to be pointing the other direction, I need to negate it first, right?
– Nick Heiner
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@user2755 Yes, but you can test this yourself with pencil and paper using simple cases, e.g. d=[1,−1];n=[0,1] (incoming down and to the right onto a ground plane facing upwards). With this, r=[1,−1]−2×(−1)×[0,1]=[1,−1]+2×[0,1]=[1,−1]+[0,2]=[1,1].
– Phrogz
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Dec 6, 2010 at 18:06
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I have no idea how this works, but it works, and that's all that matters.
– rb3652
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@rb3652 Here is my understanding. d⋅n is the length of d along the direction of n (i.e., the "projection" of d onto n). Multiplying by n turns this scalar d⋅n into a vector of length d along the direction n. If you substracted this from d, you would remove all of the n-direction character from d (i.e., it would be perpendicular to n and would be parallel to b in the diagram). Subtracting it again yields a vector that is reflected back since you're effectively adding the n-direction character in the opposite direction of d again.
– tarheels
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Let n^=n∥n∥. Then n^ is the vector of magnitude one in the same direction as n. The projection of d in the n direction is given by projnd=(d⋅n^)n^, and the projection of d in the orthogonal direction is therefore given by d−(d⋅n^)n^. Thus we have
d=(d⋅n^)n^+[d−(d⋅n^)n^]
Note that r has −1 times the projection onto n that d has onto n, while the orthogonal projection of r onto n is equal to the orthogonal projection of d onto n, therefore
r=−(d⋅n^)n^+[d−(d⋅n^)n^]
Alternatively you may look at it as that −r has the same projection onto n that d has onto n, with its orthogonal projection given by −1 times that of d.
−r=(d⋅n^)n^−[d−(d⋅n^)n^]
The later equation is exactly
r=−(d⋅n^)n^+[d−(d⋅n^)n^]
Hence one can get r from d via
r=d−2(d⋅n^)n^
Stated in terms of n itself, this becomes
r=d−2d⋅n∥n∥2n
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edited Jul 14, 2016 at 15:51
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answered Dec 6, 2010 at 18:04
ZarraxZarrax
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Does this hold for vectors of any dimension? 2D, 3D, 4D, etc?
– NightElfik
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@NightElfik Abosolutely
– felknight
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I was trying to understand how to calculate the reflection vector and found these answers. I couldn't understand them easily, so I took my time to do it myself, the good thing is that I can now detail it in an ELI5 fashion!
I did develop the formula using the 3 steps shown in the graphic. I describe them below.
So, the initial situation is a⃗ pointing toward a plane. Then we have the normal n⃗ of unit length and we would like to find b⃗ .
So, the first step is using the dot product to get a vertical vector that will be used in step 2.
With step 1 my partial formula is: 2×(a+(−a⃗ )⋅n⃗ ×n)
mind the change of sign of a⃗ above, we "flipped" it
Then in step 2, I can write: −a⃗ +2×(a+(−a⃗ )⋅n⃗ ×n)
Now, I can distribute:
−a⃗ +2×a⃗ +2×(−a⃗ )⋅n⃗ ×n
Then simplifying, I end up with:
a⃗ +2×(−a⃗ )⋅n⃗ ×n
If you negate a vector in the dot product, you negate the result of the dot product.
a⃗ ⋅b⃗ =−(−a⃗ )⋅b⃗
That means that I can rewrite the formula like this:
a⃗ −2×(a⃗ )⋅n⃗ ×n
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edited Jan 21 at 22:39
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Thank you. Though the way you used Cross Product's notation as a multiplication notation confused me big time.
– Orange Cat
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Suppose that d and r have the same magnitude.
∥r∥=∥d∥
From the reflection relationship, we have this equality about cross products.
r×n = d×n∴ (r −d)×n = 0⃗
which means
r −d =s n∴ r = d +s n
where s is a real number.
Taking their squares, we have
∥r∥2 = ∥d∥2+ 2 s(d⋅n) +s2 ∥n∥2∴ s(s ∥n∥2+ 2 (d⋅n))=0
So
s =0 ,−2 (d⋅n)∥n∥2
Since s=0 means d itself, we take the other value and get
r = d−2 (d⋅n)∥n∥2 n
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answered Dec 6, 2021 at 21:06
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In case you want to rotate about Y axis you can use the following instead. This is mostly useful for computer graphics applications. Note that d is assumed to be pointing outward in the equation below (i.e. ignore the direction of d in the picture below) and n needs to be normalized:
r=2(d⋅n)n−d
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answered Feb 10, 2021 at 4:09
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Table of Contents
Abstract
Introduction
Cytoplasmic Dynein-1
Cytoplasmic Dynein-2
Axonemal Dyneins
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Open Access Review
Structure and Function of Dynein’s Non-Catalytic Subunits
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Lu Rao
Lu Rao
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Dr. Lu Rao holds the position of Research Assistant Professor in the Department of Biochemistry at[...]
Dr. Lu Rao holds the position of Research Assistant Professor in the Department of Biochemistry at Albert Einstein College of Medicine. She earned her PhD in Chemistry from Wake Forest University. Her research interests include molecular motors, cytoskeletons, and single-molecule techniques. Currently, her primary focus is on microtubule-based motors, specifically dyneins and kinesins.
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Arne Gennerich
Arne Gennerich
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Dr. Arne Gennerich is a Professor of the Department of Biochemistry at Albert Einstein College of He[...]
Dr. Arne Gennerich is a Professor of the Department of Biochemistry at Albert Einstein College of Medicine. He discusses the uses of laser-based optical tweezers, whose developers won the 2018 Nobel Prize in physics. His research interests include molecular mechanisms of microtubule-based motor proteins in health and disease, transcription elongation by RNA polymerases, single-molecule biophysics, and the development and application of single-molecule technologies.
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Department of Biochemistry and Gruss Lipper Biophotonics Center, Albert Einstein College of Medicine, Bronx, NY 10461, USA
Authors to whom correspondence should be addressed.
Cells2024, 13(4), 330;
Submission received: 15 January 2024 / Revised: 5 February 2024 / Accepted: 9 February 2024 / Published: 11 February 2024
(This article belongs to the Special Issue Structure and Roles of Dynein in Cellular Processes)
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Abstract
Dynein, an ancient microtubule-based motor protein, performs diverse cellular functions in nearly all eukaryotic cells, with the exception of land plants. It has evolved into three subfamilies—cytoplasmic dynein-1, cytoplasmic dynein-2, and axonemal dyneins—each differentiated by their cellular functions. These megadalton complexes consist of multiple subunits, with the heavy chain being the largest subunit that generates motion and force along microtubules by converting the chemical energy of ATP hydrolysis into mechanical work. Beyond this catalytic core, the functionality of dynein is significantly enhanced by numerous non-catalytic subunits. These subunits are integral to the complex, contributing to its stability, regulating its enzymatic activities, targeting it to specific cellular locations, and mediating its interactions with other cofactors. The diversity of non-catalytic subunits expands dynein’s cellular roles, enabling it to perform critical tasks despite the conservation of its heavy chains. In this review, we discuss recent findings and insights regarding these non-catalytic subunits.
Keywords:
cytoplasmic dynein-1; cytoplasmic dynein-2; axonemal dynein; intermediate chain; light intermediate chain; light chain; molecular motors; microtubules
1. Introduction
Dynein, a microtubule-based molecular motor, moves towards the minus ends of microtubules (MTs) . This ancient protein complex has undergone evolutionarily diversification into three distinct subfamilies, categorized by their cellular locations: cytoplasmic dynein-1, which carries out various functions in the cytoplasm; cytoplasmic dynein-2, responsible for the retrograde movement of intraflagellar transport (IFT) particles in cilia; and axonemal dyneins, which form the inner and outer rows of arms associated with the doublet microtubules of motile cilia that drive flagellar beating [1,3]. Despite this functional divergence, all dimerized dyneins exhibit a conserved structural organization, comprising two heavy chains and multiple associated smaller subunits. Notably, axonemal dyneins exhibit greater divergence with 1–3 heavy chains, which are discussed in Section 4, “Axonemal Dyneins”.
The heavy chain (HC), with a molecular weight of approximately 500 kDa, is the largest and catalytic subunit of the dynein complex and belongs to the protein family known as ATPases associated with diverse cellular activities (AAA+) [4,5]. The HC’s N-terminus is characterized by a tail region composed of multiple helical bundles (HBs), which provide a scaffold for the attachment of additional subunits [6,7,8,9,10]. Conversely, the C-terminal domain forms a ring-shaped motor domain (or “head”) that is responsible for adenosine triphosphate (ATP) hydrolysis, a critical process that transforms chemical energy into mechanical motion and force. The motor domain is organized into six AAA+ domains that form a ring (AAA1–AAA6) [11,12,13]. Domains AAA1 through AAA4 are capable of nucleotide binding, whereas AAA5 and AAA6 have lost this function [12,13,14,15,16,17,18]. AAA1 serves as dynein’s primary ATPase, and its activity has been demonstrated to be regulated by AAA3 and AAA4 in cytoplasmic dynein-1 [19,20,21,22,23,24]. AAA2, despite its ability to bind ATP, does not exhibit enzymatic activity, suggesting that its role may be to facilitate rigid body motions within the ring [5,25]. While details regarding the allosteric regulation of AAA domains in cytoplasmic dynein-2 and axonemal dyneins are less understood, it is likely that their regulation is similar to that of cytoplasmic dynein-1 due to the conservation of the motor domain in dynein heavy chains [26,27]. Nevertheless, further studies are necessary to reveal their distinct characteristics.
Beyond the ring structure, the motor domain of dynein features several distinct structural elements. A “linker” domain, originating from AAA1, undergoes conformational changes that include docking and undocking from the ring domain, depending on the nucleotide states of dynein’s active ATPase domains [9,11,12,13,28,29]. This dynamic process induces a rotation of the ring [11,28,30], leading to dynein’s directed motion towards the microtubule minus-end . Protruding from AAA4 is a coiled coil, approximately 13 nm long, known as the “stalk” [32,33]. This stalk terminates in a globular microtubule-binding domain (MTBD) that directly binds to MTs [34,35,36,37]. The sliding movement of the stalk’s anti-parallel coiled coils plays a key role in regulating the MTBD’s affinity for MTs [9,18,38,39,40,41,42]. Additionally, a shorter coiled coil, known as the “strut” or “buttress”, extends from AAA5 and interacts with the stalk [12,13,18]. At least for cytoplasmic dyneins, this interaction is crucial for relaying conformational changes within the motor domain to the stalk, thereby modulating the MTBD’s binding affinity for the MTs [12,13,18,41]. For example, the docking and undocking of the linker control the conformational changes in the buttress, thereby influencing the MT-affinity changes in the MTBD in cytoplasmic dynein-1 .
While the HC—particularly its motor domain in cytoplasmic dyneins—has been thoroughly examined through structural, biochemical, and single-molecule studies, the non-catalytic subunits of dynein, including intermediate chains (ICs), light intermediate chains (LICs), and light chains (LCs) (Table 1), have received less attention. Until recently, there was a lack of understanding of the molecular mechanisms by which these non-catalytic subunits contribute to dynein’s functionality, both in vitro and in vivo. However, with the advances in cryo-electron microscopy (cryo-EM) techniques, the architectures of full dynein complexes have been elucidated in detail. These findings shed light on the structural and functional roles played by these subunits within dynein complexes. In this review, we delve into recent studies and findings related to these subunits.
Table 1. Composition of human dyneins’ non-catalytic subunits. The subunits of axonemal dyneins are based on the cryo-EM structure of human respiratory cilia and follow the consensus nomenclature for dynein subunits . It is important to note that the composition might vary depending on the cell types. For the composition of axonemal dyneins in Chlamydomonas reinhardtii, see ref. . For the composition of axonemal outer-arm dyneins in Tetrahymena thermophila, see refs. [9,45].
2. Cytoplasmic Dynein-1
Cytoplasmic dynein-1 (hereafter referred to as dynein-1) is the primary MT-minus-end-directed motor protein that performs a diverse array of critical functions in the cytoplasm of animal cells . Its roles include endocytic trafficking , axonal retrograde transport [47,48], mitotic spindle positioning [49,50,51], intricate kinetochore functions , and nuclear migration [53,54]. Dynein-1 also plays a role in the intracellular transport of viruses , such as adenovirus and human immunodeficiency virus .
As a 1.4 MDa complex, dynein-1 consists of a homodimer of HCs, ICs, LICs, and various numbers of LCs depending on the species. The dynein-1 complex of Saccharomyces cerevisiae is active and processive in vitro by itself , whereas mammalian dynein-1 exhibits only diffusive behavior on MTs [59,60,61]. It was later discovered that mammalian dynein-1 assumes an autoinhibited conformation when isolated (Figure 1a) [6,62]. This autoinhibition is relieved, resulting in a highly processive state, by the addition of the megadalton cofactor dynactin and a coiled-coil cargo adaptor (Figure 1d) [60,61]. Interestingly, even though yeast dynein is active on its own, it harbors a similar autoinhibitory mechanism to that of mammalian dynein-1 , suggesting that this regulatory feature may be universal among dynein-1 motor proteins. A wide myriad of dynein-1 adaptors have been identified and studied [64,65,66], highlighting the modular nature of the dynein transport machinery. This modularity allows cells to utilize the same molecular machine to transport various cargoes, demonstrating the adaptability and versatility of dynein-1 in carrying out a broad spectrum of cellular functions.
Figure 1. Structures of dynein complexes. Subunits are denoted by colored text. (a) Human cytoplasmic dynein-1 complex in an autoinhibited conformation (PDB 5NVU). (b) Human cytoplasmic dynein-2 complex in an autoinhibited conformation (PDB 6SC2). (c) Tetrahymena axonemal outer-arm dynein complex in an inhibited conformation imposed by Shulin (PDB 6ZYW). (d) Human dynein–dynactin–BicDR1 complex with two dyneins and two BicDR1 adaptors (PDB 7Z8F). (e) Tetrahymena axonemal outer-arm dynein complex bound to MTs in one of the two MT-binding states, with the central HC’s MTBD aligned with the outermost HC’s MTBD (PDB 7KEK).
2.1. The Structural Architecture of Dynein-1
The assembly of the dynein-1 complex revolves around the HC, which is the largest subunit within dynein-1 (Figure 1a,d). The C-terminal motor domain of the HC is responsible for enzymatic activity, while its N-terminal tail, comprising a series of relatively flexible HBs, acts as a platform for the attachment of other non-catalytic subunits. Both IC and LIC make direct contact with the HC: the IC binds at the HC’s HB 4 and 5 , and the LIC binds at HB 6, positioning itself between the IC-binding site and the motor domain [7,68]. Mammalian dynein-1 features three different LCs that associate with the ICs, facilitating their dimerization. In the autoinhibited state, the IC-LC complex wraps around the HC, with the LCs making contact with the HC’s neck . This interaction may contribute to the autoinhibition of dynein-1.
In the presence of dynactin and an adaptor, dynein is assembled into a tripartite dynein–dynactin–adaptor (DDA) complex (Figure 1d). Structural studies have shown that the coiled-coil adaptor aligns along the short actin filament of dynactin, with dynein interacting with both dynactin and the adaptor [7,67,68,69,70]. Importantly, the stoichiometry of the dynein–dynactin–adaptor complex is not a strict 1:1:1 ratio. A single dynactin can recruit two dyneins and two adaptors when the complex binds to MTs [7,67,70]. Once the complex forms, the two motor domains of the HCs adopt a parallel orientation (Figure 1d), which contrasts with the antiparallel arrangement and crossing stalks seen in the autoinhibited conformation (Figure 1a). As a result, both the stalk and the MTBD are parallel and ready to engage with MTs. In this active conformation, the IC-LC subcomplex of dynein-1 trails behind HCs, unlike in the autoinhibited conformation, where it wraps around the HCs .
2.2. Dynein-1 Intermediate Chain (IC)
2.2.1. Structure of IC
The IC of dynein-1 is a protein of approximately 74 kDa, characterized by its C-terminal region, which contains tryptophan–aspartic acid (WD) 40 repeats—hereafter referred to as WD repeats. These repeats form a β-propeller ring structure (Figure 2a) that anchors to the HC. The N-terminus of the IC features a single α-helix, followed by an intrinsically disordered region (IDR), which serves as a recruitment site for all the LCs.
Figure 2. The IC-LC subcomplex in autoinhibited dynein complexes. The subunits are denoted by colored text. The structures are aligned horizontally at the LC Roadblock (in light blue) to allow for comparison of different dyneins. Notably, while LC8 and Tctex are approximately 20 nm away from the WD repeats of the IC in dynein-1, they are positioned much closer to the WD repeats in both dynein-2 and axonemal OAD. (a) IC-LC subcomplex of human dynein-1 (PDB 5NVU). (b) IC-LC subcomplex of human dynein-2 (PDB 6SC2). (c) IC-LC subcomplex of Tetrahymena OAD (PDB 6ZYW).
The WD repeat is one of the most common protein folds [71,72]. It serves essential roles in many critical biological functions, offering multiple large surfaces for protein–protein interactions [71,72]. In the case of IC, the WD domain rigidly docks to the HC tail, creating a stable platform for LCs and other binding partners. This interaction is also thought to prevent the HCs from aggregating , likely because the WD repeats shield the hydrophobic sites on the HC tail. Although the WD repeats are generally considered a scaffold, the possibility of additional, as-yet-undiscovered functions within the IC remains open to future research.
The N-terminus of the IC forms a single α-helix (SAH), followed by a second, shorter nascent helix (H2) , both of which are also predicted by AlphaFold . This region can interact with the N-terminal coiled-coil region of dynein-1′s cofactor Nde1/Ndel1 and with a coiled-coil region (CC1B) of dynactin’s largest subunit, p150 Glued [75,76,77,78,79]. Despite both Nde1 and p150 Glued sharing the same binding region on the IC, they exhibit distinct binding modes [78,80]. Nde1 interacts exclusively with SAH, resulting in the induction of disorder within H2 , whereas p150 Glued interacts with both SAH and H2, which stabilizes the helical structure of H2 . Predictions based on ColabFold indicate that while the nascent H2 is oriented away from the Nde1 coiled-coil (Figure 3a), it directly engages with the CC1B of p150 Glued (Figure 3b). This differential binding confers a higher affinity of the IC for p150 Glued as compared to Nde1, with implications for the IC’s function (discussed below). Additionally, the SAH has the potential to fold back on itself to form a third helix within IC, thus inhibiting its interaction with Nde1 and p150 Glued in the absence of LCs . This suggests an intrinsic regulatory mechanism that prevents the IC from interacting with Nde1/Ndel1 or p150 Glued unless it is part of dynein complexes through the assistance of LCs.
Figure 3. IC as the central hub for Nde(Nde1/Ndel1)/Lis1-mediated dynein-1 activation. (a) The structure of IC2’s SAH and H2 (Uniprot Q13409, amino acids 1–67) predicated by ColabFold to interact with Nde1 (Uniprot Q9NXR1, amino acids 1–189) and Lis1 (Uniprot P43034, full-length). Predictions were performed using ColabFold v1.5.5, employing AlphaFold2 with MMseqs2, and conducted without a template. (b) The structure of IC2’s SAH and H2 (Uniprot Q13409, amino acids 1–67) predicated by ColabFold to interact with p150 Glued’s CC1B (Uniprot Q14203, amino acids 357–589). (c) A proposed mechanism illustrating how IC functions as the central hub in dynein-1 activation, with further details described in the main text.
Following the helices, IC contains a stretch of IDR to which the dimeric LCs bind [83,84]. In human dynein, three different dimeric LCs bind to IC in a 1:1 ratio. These are Tctex1, LC8, and Roadblock-1, arranged in the order of their binding sites on the IC from the N-terminus to the C-terminus (Figure 2a). In contrast, for dynein-1 in Saccharomyces cerevisiae, only one type of LC (LC8) is involved, with two LC8 dimers binding to the IC [85,86]. This illustrates the species-specific variation in IC-LC interactions. When LCs bind, the IDR transitions towards a more structured conformation . A crystal structure of LCs with IC-derived peptides shows that the IC peptides are positioned on the outside of Tctex1 and LC8, flanking the central β-sheet [86,88,89]. Roadblock-1, on the other hand, interacts with two amphipathic helices of the IC that lie adjacent to the WD repeats . This interaction has also been revealed in the cryo-EM structure of the dynein-1 complexes [6,7,67].
2.2.2. Function of IC
The interaction between the IC’s SAH and Ndel1 is crucial for Ndel1/Lis1-mediated dynein activation . Lis1, a dynein activator, enhances the assembly of DDA complexes [92,93]. Impairment of Lis1’s function leads to lissencephaly, making the understanding of how Lis1 regulates dynein a subject of significant interest (reviewed in ). Nde1/Ndel1 has been shown to enhance Lis1’s binding to dynein but inhibits dynein’s movement at high concentrations [95,96]. The N-terminal coiled-coil of Nde1/Ndel1 binds to the IC’s SAH [77,91], while both AlphaFold-based modeling [91,96] and biochemical studies indicate that Lis1 binds to Nde1/Ndel1 at an adjacent region of the coiled-coil (Figure 3a). Interestingly, the C-terminus of Nde1 has been shown to fold back to the Lis1 binding site, suggesting a self-regulation mechanism of Nde1 [96,98].
The interactions between the IC SAH and p150 Glued CC1B are vital for the motility of dynein-1, as disrupting this interaction renders the motor complex inactive. For example, in the presence of a high concentration of Nde1/Ndel1, the motility of the DDA complex is abolished [95,96], presumably because Nde1/Ndel1 displaces p150 Glued CC1B from the IC. However, the precise mechanism by which disrupting the binding between p150 Glued CC1B and the IC SAH leads to dynein inactivation remains somewhat mysterious. One hypothesis is that p150 Glued provides additional rigidity by pulling on the IC via CC1B. Disengaging this interaction might make dynein too flexible for processive motion. However, in cryo-EM structures of the full DDA complex, although the LC Roadblock is visibly trailing behind the HCs, the interaction between p150 Glued and the dynein IC has not been observed [67,68,70], leaving the nature of this interaction and its importance for DDA complex assembly and motility unclear.
Collectively, the emerging evidence suggests a probable mechanism where the IC serves as a central hub, modulating the binding of dynein cofactors to activate dynein (Figure 3c). Upon Lis1 binding to Nde1/Ndel1, Nde1/Ndel1 facilitates Lis1’s recruitment to the autoinhibited dynein complex by interacting with the IC SAH. After this binding event, it is plausible that the interaction either weakens the affinity of Lis1 for Nde1/Ndel1, given their proximity at the binding sites, or Lis1 simply exhibits a higher affinity for the motor domains. In either case, Lis1 detaches from Nde1/Ndel1 and instead interacts with the nearby dynein motor domains, acting like a wedge between them . This interaction transforms the autoinhibited dynein into a conformation that is more amenable to assembling with dynactin and an adaptor.
At this stage, either the CC1B domain of p150 Glued displaces Nde1/Ndel1 from IC due to its higher affinity for the IC, bringing dynactin closer to dynein to form the DDA complex (Figure 3c, bottom right); alternatively, the DDA complex assembles independently of this interaction, and after assembly, p150 Glued’s CC1B displaces Nde1/Ndel1 from the IC SAH (Figure 3c, bottom left). Regardless of the pathway, by this stage, the complex is fully assembled and ready to move along MTs.
Besides the role of activating dynein, the IC also directly modulates dynein’s mobility. For example, in Saccharomyces cerevisiae, the dynein HC forms a dimer and moves processively alone MTs even in the absence of the IC . However, the run length of dynein is significantly reduced without the IC . Similarly in mice, a mutation in the HC impacting its binding to IC leads to a “legs at odd angles” (Loa) phenotype . This dynein-1 mutant shows decreased processivity at both the single-molecule [100,101] and cargo-transport level , possibly due to a higher tendency for side-stepping on MTs . These findings collectively suggest that the IC’s “clamping” of the HCs together is important for coordinating the forward movement of the motor domains.
2.2.3. Diversity of IC
In humans and mice, there are two homologs of the IC for dynein-1: IC1 and IC2. IC2 is ubiquitously expressed in cells , while IC1 is primarily expressed in the brain . Each of these homologs has at least three splicing isoforms [75,104]. These isoforms have been shown to form both homodimers and heterodimers when overexpressed in cells , and they are capable of binding to all LCs . In Drosophila melanogaster, which has only one IC for dynein-1, at least 10 alternatively spliced IC isoforms are used in a tissue-specific manner . Additionally, post-translational modification of IC, such as phosphorylation, could modulate IC’s selectivity for either Nde1 or p150 Glued or regulate its binding to p150 Glued . These mechanisms significantly contribute to the diversity of IC functions (reviewed in ).
As the primary cytoplasmic retrograde transport motor, dynein-1 performs a surprisingly wide range of functions. Diversification of the non-catalytic subunits can effectively expand the functions of dynein-1 in a modular manner. However, the specific functions of the different homologs and isoforms, as well as the regulatory roles of post-translational modifications, remain underexplored. This is partly due to the complexity and essential, varied roles of dynein in mammalian cells. The precise ways in which different isoforms fine-tune dynein-1’s function in different tissues, and the role of the IDRs in regulating dynein when the isoforms do not alter LC binding, are still unclear. Given the presence of a disordered region of approximately 100 amino acids between the LC8 and Roadblock binding sites (Figure 4a), it would be interesting to investigate whether other binding partners interact with the IC in this region.
Figure 4. Comparison of human dynein IC and LIC subunits. (a) Comparison of the ICs containing the WD repeats. WD repeats are shown in grey, the Roadblock binding site is in blue, the LC8 binding site is in orange, the Tctex binding site is in pink, SAH is in green, and H2 is in light green. The binding sites are assigned based on cryo-EM structures: DC1I2 : PDB 5NVU; DC2I1 and DC2I2 : PDB 6SC2; DNAI1, DNAI2, DNAI3, and DNAI4 : PDB 8J07. (b) Comparison of the LICs. The Ras-like domain is shown in grey; helix-1 is in green; and helix-2 is in light green. These regions are assigned based on AlphaFold structures deposited on Uniprot.
2.3. Dynein-1 Light Intermediate Chain (LIC)
2.3.1. Structure of LIC
The LIC of dynein-1 is a RAS-like protein with a molecular weight of 50–60 kDa . It is characterized by an ordered N-terminal Ras-like globular domain and a disordered C-terminal domain that facilitates interactions with various adaptors. The Ras-like domain forms a crucial association with the HC through a patch of aromatic residues . This interaction stabilizes the HC and prevents aggregation, likely by shielding an aromatic patch on the HC through LIC binding [59,110].
Although the human LIC Ras-domain retains the ability to bind to GTP, it has lost its capability of hydrolysis, in contrast to the fungus Chaetomium thermophilum LIC, which lacks a GTP-binding pocket entirely . While GTP binding might induce conformational changes in LIC and regulate its interaction with the HC, it is more likely that nucleotide binding primarily provides structural support for the Ras-domain .
LIC adopts a distinctive crescent shape on the HC tail region . Subsequent cryo-EM structures unveil that the globular domain of LIC binds at HB 6 of the HC tail , with two additional extended intensities protruding from the globular domain above and below, making contacts along the HC at HB 5 and 7 (Figure 5a). This confirms an earlier biochemical study showing overlapping binding regions of IC and LIC on the HC . As a result of this extensive binding, LIC, in coordination with IC, forms a supportive scaffold that effectively sandwiches the HC, providing essential rigidity and stability to the dynein complex (Figure 5a).
Figure 5. (a) A zoom-in view of the LIC binding to the HC. Inset: The DDR complex (PDB 7Z8F); the circle indicates the zoom-in region. (b) ColabFold-predicted structures of LIC1 helix-1 interacting with various adaptors. These predictions were made in the same manner as those in Figure 3. The sequences analyzed include LIC1 helix-1 (Uniprot Q9Y6G9-1, amino acids 424–459); BicD2 (Uniprot Q8TD16-1, amino acids 1–240); JIP3 (Uniprot Q9UPT6-1, amino acids 1–180); Hook1 (Uniprot Q9UJC3-1, amino acids 1–239); CRACR2A (Uniprot Q9BSW2-2, amino acids 1–240). The adaptors are depicted in grey, and LIC1 helix-1 is in gold. The crystal structures of BicD2 (PDB 6PSE), Hook3 (PDB 6B9H), and CRACR2A (PDB 6PSD) with LIC1 helix-1 (blue) were structurally aligned to the predicted structures, using UCSF Chimera .
In contrast to the structured N-terminus, the C-terminus of LIC appears mostly disordered, featuring interspersed short helixes. This aspect is discussed in the subsequent section.
2.3.2. Function of LIC
The C-terminus of LIC is predominantly disordered, interspersed by two short helices, helix-1 and helix-2 [113,114] (Figure 4b). Numerous studies have indicated that helix-1 interacts with multiple dynein-1 adaptors [113,114,115]. Helix-1 is a highly conserved amphipathic helix across species, showcasing remarkable versatility in binding to various protein folds, such as the Hook domain of the Hook family, CC1-Box domain of the BicD family, and EF-hand in the calcium-binding family in a 2:2 ratio (Figure 5b) .
Helix-1 utilizes a highly conserved hydrophobic surface to establish interactions with adaptors. For instance, one flexible helix (helix-8) in the Hook domain is capable of adopting different conformations , and helix-1 of LIC shifts the equilibrium and induces the conformational change in the Hook domain’s helix-8 from a straight form to a V-shape through hydrophobic interactions . Mutagenesis studies further support the notion that the flexible helix-8 in the Hook domain is the primary binding site for LIC .
Similarly, helix-1 uses the same hydrophobic surface to interact with BicD2 and CRACR2A . ColabFold predictions are consistent with these findings, suggesting interactions between LIC1 helix-1 and BicD2, JIP3, Hook1, and CRACR2A, as depicted in Figure 5b. The cryo-EM structure of the DDR complex provides an additional validation for these interactions, revealing that the region following the globular Ras-like domain in LIC extends along the HC tail HB5, reaching for the BicDR1 adaptor with helix-1 (Figure 5a) .
The interactions between LIC and adaptors are pivotal for the motility and functions of dynein along the endosome–lysosome pathway. For example, Kazrin interacts with dynein LIC and dynactin, facilitating the recruitment of the complex to early endosomes . Similarly, Hook1 and Hook3 are involved in early endosome processes [118,119]. Disrupting the interactions between LIC helix-1 and Hook adaptors impairs the motility of the dynein–dynactin–Hook (DDH) complex in vitro [113,115,116]. Possibly downstream to Kazrin, FIP3 serves as a link between dynein and recycling endosomes through its interaction with LIC . In the context of late endosomes and lysosomes, Rab7-interacting lysosomal protein (RILP) interacts with LIC, playing a crucial role in recruiting dynein-1 to these organelles . Disrupting critical hydrophobic residues in LIC helix-1 has been shown to impair lysosome transport in vivo, highlighting the physiological importance of these interactions .
Dynein-1 plays indispensable roles in cell division, particularly during mitosis, where it localizes at the spindle poles, kinetochore, and cell cortex . In these mitotic structures, LIC assumes critical functions in localizing dynein-1 to the correct locations. Pericentrin, a crucial key component for MT organization , is known to localize at the centrosome and exhibits colocalization with LIC . However, unlike the canonical dynein-1 adaptors which interact with dynactin, dynein HC, and dynein LIC, pericentrin solely interacts with LIC . Biochemical studies have pinpointed the specific interaction site between pericentrin and LIC within the region of amino acids 140–236 of LIC1 . Structural studies reveal that dynein HC interacts with LIC on the opposite interface [6,7,67], suggesting possible concurrent interactions between pericentrin and the DDA complex with LIC. At kinetochore, the Rod–Zw10–Zwilch (RZZ) complex recruits adaptor Spindly, which subsequently recruits dynein via the LIC and dynactin to the kinetochore . A recent study indicated that LIC, while interacting with Spindly, can also recruit pericentrin, which then attracts the γ-tubulin ring complex, promoting MT nucleation at the kinetochore . At the cell cortex, the nuclear mitotic apparatus (NuMA) anchors at the cortical region and facilitates mitotic spindle positioning by recruiting dynein-1 and dynactin to capture astral MTs [51,128]. It interacts with LIC via its N-terminus [50,129], which contains both a Hook domain and a CC1-Box domain .
Dynein-1 also drives chromosomal movements in the prophase I of meiosis . It links to the chromosomes via the Linker of nucleoplasm and cytoplasm (LINC) complexes , which contain the transmembrane protein KASH . Recent studies have demonstrated that KASH5, an adaptor, binds to dynein-1 LIC and activates dynein-1 in the same fashion as other adaptors, utilizing its EF-hand domain, which is not regulated by calcium [133,134].
Besides interacting with pericentrin, the Ras-like domain of LIC also interacts with the neighboring dynein HC motor domain (AAA2 and AAA3) within the DDR complex that contains two dyneins . This potentially helps to synchronize the two dyneins in the DDR complex.
2.3.3. Diversity of LIC
In mammalian dynein-1, there are two homologs of LIC, namely LIC1 and LIC2 [26,135], sharing 65% sequence identity. Despite this homology, they do not coexist in the same dynein-1 complex , resulting in distinct dynein subpopulations that have non-overlapping roles in cellular functions. For instance, during mitosis, LIC1 predominantly localizes at the kinetochore from metaphase to anaphase , playing a pivotal role in removing spindle-assembly checkpoint (SAC) components from kinetochores . In contrast, LIC2 is concentrated at the spindle poles throughout the entire mitotic process , proving essential for mitotic spindle orientation [138,139]. However, a later study suggests that LIC2 also participates in removing SAC components, exhibiting a stronger and more diverse role .
At the molecular level, although an NMR study has revealed structural similarities in the C-terminal helices of LIC1 and LIC2 , these homologs can exhibit a different affinity for adaptors. For example, LIC1 has a stronger affinity for BicD2 than LIC2 . As a result, LIC1, but not LIC2, is crucial for BicD2-mediated interkinetic nuclear migration (INM) , a process that is essential for the proliferation of embryonic neural stem cells (radial glial progenitors) . LIC1 contains an extended flexible linker between helix-1 and helix-2 that is absent in LIC2. This extended linker in LIC1 might enhance the accessibility of helix-2 to binding partners located further away from the LIC core, although the function of helix-2 remains unknown at present. A study in Caenorhabditis elegans demonstrated that deletion of helix-2 overall does not affect dynein function .
Unlike IC, the diversity of LIC is not augmented by alternative splicing; to date, no known isoforms exist. However, post-translational modifications significantly increase the complexity of LIC. The importance of how the post-translational modifications of LIC contribute to the diverse functions of dynein has been recently explored and postulated in a comprehensive review .
2.4. Dynein-1 Light Chains (LCs)
Mammalian dynein-1 has three different LCs (Figure 6a–c), each with two homologs: Roadblock (DLRB1 and DLRB2), LC8 (DYL1 and DYL2), and Tctex (DYLT1 and DYLT3). They bind the IC in a 2:2 ratio, forming the IC-LC subcomplex.
Figure 6. Structures of dynein LCs. (a) Solution structure of dimeric human Roadblock-1 (PDB 1Z09). (b) Crystal structure of dimeric Drosophila melanogaster LC8 (PDB 3BRI). (c) Crystal structure of dimeric Drosophila Tctex-1 (PDB 1YGT). (d) Crystal structure of Chlamydomonas LC1 (PDB 5YXM). (e) AlphaFold-predicted human TXND6 structure. Dark grey is for the thioredoxin domain, and light grey is for the nucleoside diphosphate kinase domain. The disordered N-terminal amino acids 1–10 and C-terminal amino acids 301–330 were removed for clarity. Blue indicates the predicted structure of human thioredoxin (Uniprot P10599), which was structurally aligned to TXND6, using UCSF Chimera . LC3 and LC5 of Chlamydomonas and LC3BL of Tetrahymena [9,45] have similar folds (not shown). The two cysteine residues in the -Cys-X-X-Cys- motif are highlighted by sulfur atoms (yellow) (Cys39 and Cys42 in TXND6; Cys32 and Cys35 in thioredoxin). Pink indicates the predicted structure of human nucleoside diphosphate kinase A (NDKA, Uniprot P15531), with amino acids 136–152 removed for clear depiction, and structurally aligned to TXND6. The conserved histidine residue is highlighted by the nitrogen atom (blue) (His279 in TXND6; His118 in NDKA). (f) AlphaFold-predicted structure of Chlamydomonas LC4 (Uniprot Q39584).
2.4.1. Roadblock
Roadblock, or LC7, which was first identified in both Drosophila and Chlamydomonas , is a ~11 kDa protein that dimerizes on its own (Figure 6a), as shown by both X-ray and NMR studies [90,145,150]. An early study identified the binding site of Roadblock on IC, located downstream of IC’s splicing sites and upstream of the WD repeats, suggesting no preference for specific IC isoforms . The crystal structure of Roadblock bound to a short peptide of IC suggests that it converts the intrinsically disordered peptide of IC into a more ordered conformation . Subsequent cryo-EM structures of both autoinhibited and activated dynein-1 [7,67] confirm that the dimeric Roadblock binds near the IC’s WD repeats (Figure 2a). This binding arrangement effectively positions the two IC WD repeats closely, indicating that the primary role of Roadblock is to function as a clamp, ensuring IC’s association and thereby maintaining proximity of the HC tails. Notably, dynein-1 in Saccharomyces lacks a homolog equivalent to Roadblock , raising questions about whether yeast IC has evolved an alternate mechanism for maintaining IC proximity. In the filamentous fungus Aspergillus nidulans, which has a Roadblock homolog (RobA), Roadblock deletion leads to a mild phenotype , while homozygous Roadblock-1 null mice are not viable , suggesting that yeast may have, indeed, evolved a compensatory mechanism.
There are two vertebrate homologs of Roadblock: Roadblock-1 and Roadblock-2, sharing 75% sequence identity. Single-molecule studies show that DDB complexes display similar motility in vitro with either homolog . Nevertheless, Roadblock-1 knockout is embryonically lethal, underscoring the non-redundant cellular functions of the homologs . Recent research has shed light on their distinct roles: while Roadblock-1 is ubiquitously expressed in mouse tissues and plays a crucial role in ensuring the integrity of the mitotic spindle pole, Roadblock-2 is exclusively involved in meiosis . In meiotic cells, Roadblock-2 targets NuMA to the spindle pole, and its absence leads to spindle pole defects, including multipolarity and misalignment .
2.4.2. LC8
LC8 is a ~10 kDa protein that is highly conserved across eukaryotic cells . Much like Roadblock, LC8 forms dimers (Figure 6b), which undergo dissociation at low pH conditions due to the ionization of a histidine at the dimerization interface . It dimerizes with the short disordered region in the N-terminus of IC at a 2:2 ratio (Figure 2a), enhancing the structural order of IC .
While LC8’s role in facilitating the dimerization of the IC is well-established, its potential involvement in other regulatory functions within dynein-1 remains largely unknown. One study indicated its collaboration with ADP-ribosylation factor-like 3 (Arl3) in dissociating dynactin from dynein. In this process, LC8 binds to the IC, and Arl3 binds to the linker region between CC1 and CC2 of p150 Gluded . The precise mechanism of this disassembly, however, remains elusive.
In Saccharomyces, dynein-1 features only LC8 (Dyn2) as its light chain, with two LC8 dimers binding to the N-terminal region of the IC. The absence of LC8 impedes the IC’s binding to the HC, likely due to LC8’s role in promoting IC dimerization and enhancing its affinity for the HC . Notably, DYN2 is one of the few genes in Saccharomyces with two introns and undergoes alternative splicing , yet the impact of LC8 isoforms on yeast dynein’s function is unclear. In mammals, there are two homologs of LC8, LC8-1 and LC8-2, differing by only a few residues. Thermodynamic studies suggest subtle differences in their binding to partners , but distinct functionalities of these LC8 variants have not been established.
Initially identified as a subunit of axonemal dynein [159,160], LC8 was later discovered to also exist in cytoplasmic dyneins [161,162,163]. Since then, LC8 has gained recognition as a versatile protein known for its interaction with a diverse range of proteins, often characterized by a threonine–glutamine–threonine (TQT) motif. This has established LC8 as a central hub for dimerization [158,164,165]. Its binding partners span a wide spectrum of proteins, including Nup159 in the nuclear pore complex (NPC) , myosin-Va , p53 and MRE11 in DNA double-strand break response [168,169,170], and various viruses [171,172]. This remarkable diversity underscores LC8’s involvement in functions beyond dynein. LC8’s essential functions in metazoans and its presence in plants despite their lacking dynein complexes suggest a broader role for this small protein .
2.4.3. Tctex
Tctex was initially identified as a subunit in mouse brain dynein-1 and subsequently in axonemal dyneins [174,175]. This approximately 11 kDa protein forms a dimer (Figure 6c) and binds adjacent to LC8 on IC (Figure 2a). The binding of Tctex to IC enhances LC8 binding through avidity . Although Tctex and LC8 share structural similarities (Figure 6b,c), they lack sequence similarities and likely have different evolutionary origins .
In addition to binding to IC, Tctex-1 directly interacts with rhodopsin, a protein responsible for low light sensing . Mutagenesis studies mimicking phosphorylation/dephosphorylation indicate that the post-translational modification regulates the binding of Tctex-1 to rhodopsin . Tctex and Roadblock also interact with unc104 in Caenorhabditis, a highly processive and fast kinesin that travels towards the plus-end of MTs, thereby facilitating bidirectional cargo transport .
Mammalian dynein-1 has two Tctex homologs, Tctex-1 and Tctex-3, which are mutually exclusive in their binding to dynein complexes, suggesting that they form only homodimers . Additionally, their cargo-binding preferences differ, with Tctex-1 associating with rhodopsin, while Tctex-3 does not bind to rhodopsin .
Like LC8, Tctex functions as a dimerization hub independent of dynein . Intriguingly, Tda2 in Saccharomyces, which is structurally homologous to Tctex , does not associate with dynein-1 but is involved in actin assembly [181,182]. In vertebrates, Tctex also plays a dynein-independent role in regulating actin dynamics [183,184,185]. This suggests that Tctex may predate dynein’s evolution and was repurposed as a subunit during evolution. While mammalian dyneins include Tctex as a subunit, yeast dynein has a diverged evolutionary, leading to Tctex’s loss as a subunit.
2.4.4. Summary of LCs
All the light chains (LCs) discussed in this section—Roadblock, LC8, and Tctex—are shared among cytoplasmic dynein-1, cytoplasmic dynein-2, and axonemal dyneins. The small size of LCs might contribute to their broad binding pattern, given their limited binding interfaces with other proteins, as evidenced by the µM range of binding affinity observed for LC8 and Tctex with IC peptides. Despite their promiscuity, these LCs bind to specific sites on dynein IC without interchanging. It remains unclear why mammalian dynein requires three different LCs, whereas Saccharomyces cytoplasmic dynein functions with only one LC. It is plausible that the diversity of LCs corresponds to the array of functions required by mammalian dynein, which must be versatile for numerous cellular functions. Strategies such as gene duplication, alternative splicing isoforms, and post-translational modifications are employed by IC, LIC, and LCs to increase subunit diversity. Dynein further enhances this diversity by incorporating various types of LCs to interact with different cellular components. Several questions about LCs remain unanswered, including how their diversity affects dynein-1’s functionality; whether all LCs are necessary for assembling a dynein complex, or a subset of light chains is sufficient; and how they might impact dynein’s motility and function.
3. Cytoplasmic Dynein-2
The cilium, or flagellum, is a specialized membrane-bound organelle that protrudes from eukaryotic cells, enabling the cells to sense and navigate their environment . It contains a central structure, the axoneme, consisting of microtubule doublets (MTDs). There are two types of cilia, motile cilia and immotile (primary) cilia. Motile cilia typically have nine MTDs surrounding two central single MTs (9 + 2 arrangement) and contain axonemal dyneins, enabling them to beat . In contrast, primary cilia lack the central MT pair and axonemal dyneins (9 + 0 arrangement), functioning as a sensory unit (Figure 7, bottom panel left). For cilia to exchange information and material with the cell body, intraflagellar transport (IFT) carries cargoes and components along the axoneme tracks [188,189] (Figure 7, bottom panel). These large IFT “trains” of protein complexes move along the outer surface of the axoneme, with the transition zone (TZ) at the base of the axoneme serving as a checkpoint, regulating the passage of proteins in and out of the cilium [190,191]. Molecular motors drive the movement of IFT trains: kinesin-2 family members facilitate anterograde transport, while cytoplasmic dynein-2 (referred to as dynein-2, also known as IFT dynein and dynein-1b in Chlamydomonas) powers retrograde transport . A comprehensive study using CLAM and cryo-EM showed that these motors move along different tubules of the MTD tracks, with kinesin-2 using the B-tubule and dynein-2 using the A-tubule to minimize potential traffic jams . This is further supported by another study using cryo-EM and U-ExM, indicating that IFT trains are loaded onto the B-tubule via kinesin-2 at the TZ . Structural details about retrograde trains are less known, presumably due to their less rigid and more heterogeneous nature, posing challenges for cryo-EM studies. Besides the motors, IFT trains include components such as the IFT-A complex [195,196,197,198], IFT-B complex , and BBSome [200,201,202,203].
Figure 7. Intraflagellar transport. (Top) Side (left) and front (right) views of an autoinhibited human dynein-2 complex docked into a cryo-EM map of Chlamydomonas anterograde IFT complexes [8,204] (PDB 6SC2; EMD-4303). The colored text indicates the subunits. (Bottom, left) A cross-sectional illustrative view of a 9 + 0 immotile cilium. (Bottom, right) Side view of the cilium showing dynein-2 in an autoinhibited state, transported as cargo by kinesin-2 in anterograde trains.
3.1. The Structural Architecture of Dynein-2
The composition of dynein-2 is similar to dynein-1, containing HC, IC, LIC, and LCs . Nevertheless, several specific features distinguish the associated subunits of dynein-2 from those of dynein-1: instead of a homodimer, dynein-2 has two different ICs that form a heterodimer in the presence of LCs; dynein-2 binds more LCs to the IC (Figure 2b), and it has a distinct LC Tctex homolog that is absent in dynein-1.
Similar to dynein-1, the dynein-2 complex, when isolated, assumes an autoinhibited conformation. In this state, its two motor domains face away from each other, and the stalks cross to prevent MT binding [8,205] (Figure 1b). The HC tail adopts an asymmetrical configuration, featuring one twisted and one straight HC, only with the twisted HC more zig-zagged than in dynein-1, due to the constrain imposed by the heterodimeric ICs (Figure 1a,b) . In addition, the subunits in dynein-2 make more intermolecular contacts than in dynein-1 .
In comparison to dynein-1, less is known about the active state of dynein-2. Dynein-2 does not require dynactin and adaptors to be active , suggesting that IFT complexes may play a role in its activation. Unlike dynein-1, where the mutation-induced opening of the complex does not activate it for motility , mutations in the linker domain are sufficient to activate dynein-2 in vitro . This suggests a different activation mechanism for dynein-2. Additionally, in Caenorhabditis, these mutations lead to retrograde movements of dynein-2 in the absence of IFT-A, an essential component for retrograde IFT train assembly and the retrograde motility of the wild-type dynein-2 . Nonetheless, the details of how dynein-2 is activated when bound to retrograde IFT trains and how it moves along the axoneme in teams remain unknown.
3.2. Dynein-2 and IFT Trains
IFT trains consist of IFT-A and IFT-B complexes. In anterograde trains, IFT-A is closer to the membrane of the cilium, while IFT-B is sandwiched between IFT-A and MTs. Dynein-2 primarily interacts with the IFT-B in anterograde trains. Structural studies have revealed that dynein-2 is loaded on the anterograde IFT trains in its autoinhibited conformation [108,109] and transported to the tip of the cilium via kinesin-2 (Figure 7, top), effectively preventing a tug-of-war between these two opposing motors. Two layers of inhibition of dynein-2 in an anterograde train are implemented. First, its linker and stalk are trapped in a restricted conformation by the stacking of the motor domains, preventing the dynein-2 motor domains from forming a parallel conformation [204,205]. Second, when bound to the IFT-B complex, dynein-2 HC assumes an upside-down arrangement, with the tail pointing towards the MTs, while the MTBD points away and binds into a negatively charged groove of IFT-B, further preventing interactions of the motor with MTs .
The zig-zagged conformation in the autoinhibited state of dynein-2 is tailored to bind to the IFT-B complexes, spanning 7-8 IFT-B repeats. This suggests the dynein-2 complexes bind to assembled IFT , supported by a cryo-EM study of the TZ and a detailed cryo-EM structure of the anterograde IFT train . Interestingly, only the HC has shown extensive interactions with IFT-B in cry-EM structures, despite the non-catalytic subunits binding along the tail of the HC . However, there may be flexible interactions between these subunits and the IFT complexes not visible in cryo-EM structures. Indeed, immunoprecipitation assays have revealed interactions between the IC of dynein-2 and the IFT-B complex [207,208], highlighting the dynamic nature of these molecular associations.
How the trains turn around at the tip of cilia and how dynein-2 is activated is largely unclear. While existing evidence indicates a level of fragmentation and a disassembly/reassembly process of the trains at the tip, regulated by Ca 2+ , a recent study suggests that switching directions is an intrinsic property of the trains. When the train is blocked and derailed, they change directions without external regulators . This suggests a potential mechanically induced conformational change mechanism, wherein compression of the train may trigger derailment and remodeling of IFT trains. Such a process could activate dynein-2 for IFT turnaround.
3.3. Dynein-2 Intermediate Chains (ICs)
Unlike the dynein-1 IC, which forms a homodimer, dynein-2 presents a unique configuration where two distinct ICs, WDR60 and WDR34 (initially identified in Chlamydomonas as FAP163 and FAP133 , respectively), bind to the HC as a heterodimer . Similar to the dynein-1 IC, both ICs of dynein-2 exhibit C-terminal WD repeats responsible for anchoring the ICs onto the HC and an N-terminal IDR that associates with the LCs. Unexpectedly, WDR60, even without the WD repeats, shows the ability to integrate into the IFT trains in Caenorhabditis , indicating that the N-terminus of WDR60 may engage in interactions with IFT complexes independently of the HC. WDR60 contains an extra ~450 amino acids at the N-terminus compared to WDR34 (Figure 4a). This region is highly charged and predicted to be disordered, potentially engaging in various protein–protein interactions with the IFT trains.
Both of the ICs are critical for the dynein-2 functions, with deficiencies in either causing developmental defects . However, they have distinct roles in dynein-2: WDR-60 is important for dynein-2 assembly, recruitment onto the IFT train, and transition zone crossing . Without WDR-60, less dynein-2 is incorporated into the anterograde IFT trains, reducing the available dynein-2 for retrograde transport. It is unclear if WDR-60 also directly affects dynein’s motility, such as playing a role in activating the dynein-2 complex. WDR-34 is important for the assembly and function of dynein-2 but has less of an impact on the loading of dynein-2 onto the IFT train . Cryo-EM structures of the assembled dynein-2 complex show that WDR34 binding breaks the symmetry of the homodimeric HC, with one HC relatively straight and the other in a twisted zig-zag conformation, fitting into the contour of IFT-B complexes .
While the mechanism of how heterodimeric ICs are selectively integrated into the dynein-2 complex is unknown, it is reasonable to speculate that, in the absence of one IC, a homodimer might form due to the identical nature of their binding partners, the HCs and LCs. If a homodimer of either WDR60 or WDR34 binds to the HC, the dynein-2 complex might assume a more symmetrical shape, which would not fit well with the IFT-B complexes. Interestingly, Caenorhabditis has only one IC for dynein-2 . A comparative analysis of the dynein-2 complex in Caenorhabditis and humans could yield more insights into how the IC shapes the conformation of dynein-2.
3.4. Dynein-2 Light Intermediate Chain (LIC)
Dynein-2 LIC (DYNC2LI1, or LIC3), first identified in mammalian cells , is a homolog of dynein-1 LIC, implying the presence of an N-terminal Ras-like domain , which was later confirmed by the dynein-2 structure . In a pattern resembling dynein-1 HC and LIC interactions, LIC3 binds to dynein-2 HC’s HB 6, extending its two arms outward to engage adjacent HC helical bundles . Within the HC tail, only the sequences of the LIC and IC binding sites are well conserved between dynein-1 and dynein-2 HC , highlighting the evolutionary connections and the significance of HC-LIC interactions. Even when HC disassembles into a monomeric state in the absence of IC-LC, it maintains its association with LIC . Like dynein-1 LICs, LIC3 also interacts with the neighboring HC motor domain when dynein-2 is loaded onto the IFT-B complex . This interaction presumably reinforces dynein-2’s binding to IFT-B complexes by stabilizing the dynein-2 assembly chain. LIC3 also interacts with a subunit of IFT-B complex, IFT-54 [207,218], potentially strengthening the interactions between dynein-2 and IFT-B complexes.
A feature unique to LIC3, compared to dynein-1 LICs, is the absence of the C-terminal disordered region (Figure 4b). This is notable because LIC1 and LIC2 use this region to interact with the N-termini of various adaptors, which dynein-2 does not rely on . Unlike dynein-1, where LIC primarily contacts HC, LIC3 in dynein-2 makes contacts with other subunits besides the HC in the autoinhibited conformation, such as IC WDR60, LC Roadblock, and LC LC8, crucial for the structural stabilization of the entire complex . Despite these insights into LIC3’s structure within the dynein-2 complex, the functional aspects of LICs in general remain unknown. While LIC3 is an obligatory binding partner for the HC and plays a crucial role in maintaining the stability of the dynein-2 complex , its precise roles and regulatory functions beyond this association are yet to be fully unveiled.
3.5. Dynein-2 Light Chains (LCs)
Dynein-2 contains all the LCs of dynein-1, along with an additional Tctex homolog, Tctex1D2, identified as an unique dynein-2 LC in human cells . Unlike dynein-1, which has a homodimer of each LC, dynein-2 contains one homodimer of Roadblock, three homodimers of LC8, and one heterodimer of Tctex (Figure 2b). The dimeric Roadblock likely functions similarly to its role in dynein-1, by clamping the two ICs closely together. The consecutive LC8 dimers bind much closer to the WD repeats (Figure 2), in contrast to the ~20 nm distance between Roadblock and LC8 in dynein-1. The reason for the three dimers of LC8 is unknown. The functionality of the heterodimer Tctex also remains unclear, although the interaction between Tctex1D2 and the IC WDR60 is crucial for ciliary protein trafficking . Interactions of IC WDR34 with the LCs are also required for ciliary protein trafficking . Overall, the contribution of LCs to the functionality of dynein-2, particularly in its active state during retrograde transport, remains poorly understood.
4. Axonemal Dyneins
The axoneme, a highly complex and organized structure found in motile cilia, plays a crucial role in cellular locomotion and fluid flow on the surface of various human cells, including those in the nervous system , the respiratory tract , and the motility of sperm . In motile cilia, a row of dyneins forms the outer dynein arms (ODAs) that anchor on the A-tubule of MTDs via docking complexes, and another row of dyneins forms the inner dynein arms (IDAs) [44,225], which also anchor on the A-tubule (Figure 8a). The dyneins dynamically bind to or dissociate from the B-tubule of neighboring MTDs depending on the nucleotide states of their motor domains. An extensive protein network, consisting of MT inner proteins (MIPs), resides within the lumen of the MTD tubules. Nexin–dynein regulator complexes (N-DRC) link neighboring MTDs and regulate ciliary motility, while radial spokes (RSs) connect MTDs to the central apparatus (CA)—the central two singlet MTs shrouded with protein complexes. Recent advances in cryo-EM microscopy and image processing techniques have significantly enriched our molecular understanding of axonemes (reviewed in ), revealing detailed architectures of the full axoneme [10,227] and its components, such as ODAs [9,29,45,228,229], MIPs [230,231,232], N-DRC , RSs [234,235,236], and CA [237,238].
Figure 8. Human epithelial ciliary axoneme structure (PDB 8J07). (a) (Left): An illustration of the cross-section of the motile axoneme. Middle: The cross-section view of the structure of an MTD with associated protein complexes. (Right): The side view of this structure. (b) (Left): IAD f structure, with an additional IC (DNAI7) on the side of the IC-LC tower. (Right): OAD structure, with DNAL1 (LC1) binding to the MTBD of DYH5 and TXND6 (LC3) binding to the neck region of DYH9. (c) Top view of the IDAs and ODAs. The circle indicates the interaction between OAD’s LC Tctex heterodimer and IAD f’s DNAI3 (IC140) and HC tail.
The green algae Chlamydomonas reinhardtii [239,240] and the ciliate Tetrahymena thermophila are two model organisms that are indispensable in axoneme research, providing invaluable insights. However, it is noteworthy that, while the axonemal structure maintains overall conservation across species, there are variations in composition and local structures among axonemes from different organisms . In particular, Chlamydomonas and Tetrahymena exhibit a more complex composition of axonemal dyneins compared to the human epithelial cilia dyneins, with additional ICs and LCs . In mammals, even within the same species, respiratory cilia and sperm flagella display different structural elements . The following sections focus on discoveries based on these model systems, complemented by insights derived from human axonemal dynein where applicable.
4.1. The Assembly of Axonemal Dyneins
Distinct from cytoplasmic dyneins, which form homodimers with two identical HCs, axonemal dyneins exhibit a remarkable diversity. Multiple dynein HCs, each encoded by a unique gene, come together to form single-headed, double-headed, or triple-headed complexes . For the outer-arm dyneins (OADs), the dynein HCs form either heterodimers or heterotrimers, depending on species, arranged in 24 nm intervals. Both Chlamydomonas and Tetrahymena possess triple-headed OADs (Figure 1e), while the human axoneme has double-headed OADs (Figure 8b, right), lacking the outermost HC. The innermost two HCs of the triple-headed HCs and the double-headed HCs are bound by heterodimeric ICs and a substantial number of LCs, which interact with either ICs or HCs. Similar to the cytoplasmic ICs, the ICs of OADs also have a disordered region at the N-terminus that provides the scaffold for LC binding (Figure 2c). The outermost HC of the triple-headed OADs is a diverged paralog of the middle HC , featuring an N-terminal γ-kelch domain that latches onto the tail region of the middle HC [9,228,229]. It does not have associated IC; in Chlamydomonas, an LC directly interacts with this HC. Notably, before OADs are transported and localized to axonemes, they are tightly packed and inhibited in a conformation similar to cytoplasmic dyneins by a protein named Shulin (Figure 1c).
For the inner-arm dyneins (IADs), six single-headed dyneins (IAD a–e and IAD g) and one heterodimeric complex (IAD f) with its IC-LC tower, structurally identical to the ODAs (Figure 8b, left), bind to the A-tubule via actin and LCs . Together, they are arranged at 96 nm intervals. In Chlamydomonas, each HC is encoded by a unique gene, whereas in humans, IAD b and IAD e share the same HC . In addition to the ICs docking on the HCs, another IC—IC97 in Chlamydomonas or DNAI7 (Las1) in humans—interacts with the LCs on the IC-LC tower . Moreover, Chlamydomonas has an additional protein, FAP120, in the vicinity of IC97 and the IC-LC tower .
4.2. Axonemal Dynein Intermediate Chains (ICs)
The OADs assume a tail-to-head compact conformation, in which the tails of one OAD complex stack onto the motor heads of the adjacent complex [9,228,229]. This conformation ensures the sequential activation of the OADs, with the downstream OAD inhibited by the upstream OAD unless it undergoes a conformational change after ATP hydrolysis. Both the WD repeats of one of the ICs and the tail of the HC participate in the interactions between the OADs. In contrast to the dynein-1’s IC-LC tower, which trails behind the HC in its activated form, the IC-LC tower of the OAD closely associates with the HC. Furthermore, instead of symmetrically binding in the middle of the HC, it rotates towards the inner side, facing the IADs.
The IC-LC tower of IAD f has a similar structure to that of the OADs, associating asymmetrically with HCs on one side. The IC on the other side binds to OAD’s LC Tctex . This connection is likely crucial for the communication between the OADs and IADs. In addition, IAD f also contains a distinct non-WD repeat IC, IC97 , the functional role of which remains unknown.
4.3. Axonemal Dyneins Lack Light Intermediate Chain
In contrast to the cytoplasmic dyneins, axonemal dyneins do not have LICs . This absence suggests the crucial role of LICs in providing rigidity for cytoplasmic dynein to move along MTs over long distances. The restriction of flexibility appears to be a common theme in activating and enhancing dynein movements. The interaction between LIC and adaptor [67,114,115,116], along with the recruitment of a second adaptor to a DDA complex on MTs , reinforces the rigidity of the DDA complex, with the former being particularly critical for dynein-1 motility (see Section 2.3).
Intriguingly, although lacking LIC, both OADs and IAD f have various elements binding to the region on HC corresponding to the LIC-binding sites on cytoplasmic dynein HCs, namely the HB 6 on the tail region. This indicates that binding at this region can modulate dynein’s behavior through diverse mechanisms. For example, in the OADs of Chlamydomonas and Tetrahymena, the N-terminal γ-kelch domain of the outermost OAD HC latches onto the HB 6 in the tail region of the middle OAD HC and assists the remodeling of the OAD array during ATP hydrolysis [9,228,229]. Additionally, the light chain LC4 engages with the HB 6 of the innermost OAD HC, serving as a regulatory element for OAD in response to calcium, which is further discussed in the following section.
4.4. Axonemal Dynein Light Chains (LCs)
In humans, both OAD and IAD f share the same set of IC-binding LCs, as well as a similar structural arrangement of the IC-LC tower, with dynein-2. From the N-terminus to the C-terminus of IC, the IC-LC tower comprises a Tctex heterodimer, three LC8 dimers, and a Roadblock heterodimer [9,10,45]. OADs in Chlamydomonas and Tetrahymena have additional LC8-like proteins in place of some of the LC8 [9,10,45], while OADs in humans feature a distinct LC8-type DNAL4 that remains under-characterized . Despite the similarity, there are structural differences between the axonemal dyneins and dynein-2. While the Tctex dimer trails underneath LC8 in dynein-2 as a weak density in the cryo-EM structure , the Tctex dimer in both OADs and IAD f exhibits a distinctive bend back towards the adjacent LC8 due to constraints imposed by the N-terminal helical bar and the β-hairpin structure of the IC [9,10]. In the case of OADs, the unique bend back of the Tctex dimer in OADs facilitates its extension toward IAD f, establishing contact with the IC140 and HC tail of IAD f, functioning as one of the communication pathways between OADs and IADs . In the case of IAD f, Tctex-2b is critical in the assembly of Tctex-1, IC97, and FAP120 of IAD f in Chlamydomonas, as supported by the structure showing Tctex-2b bridging between IC97 and Tctex-1 .
Besides the LCs on the IC-LC tower, there are OAD-specific LCs that directly interact with HCs. Leucine-rich repeat protein LC1 (Figure 6d), which assumes a cylindrical shape as revealed by the NMR solution structure , is a light chain that binds directly to the MTBD of the innermost OAD (γ-HC in Chlamydomonas and α-HC in Tetrahymena) . Structural studies reveal that LC1 has extensive interactions with the MTBD of the HC without directly interacting with MTDs, except for electrostatic interactions with the highly negatively charged β-tubulin C-terminal tail [9,248]. It is speculated that the LC1 can assist the MTBD in sensing the curvature of the MTD surface, leading to a rotary movement . Furthermore, due to the binding of the LC1, the LC1-MTBD requires a wider inter-protofilament interface, guiding the OADs to localize to the correct positions on the MTDs . A mutagenesis study demonstrated that LC1 is also crucial for the cytoplasmic preassembly of OADs and ciliary stability , although the molecular mechanism remains unclear.
LC3 (LC3BL in Tetrahymena and TXND6 in humans), a thioredoxin-like protein, binds to the tail region of the middle OAD in Chlamydomonas and Tetrahymena or the outermost OAD in humans. The thioredoxin superfamily is an ancient protein family that undergoes redox chemistry by utilizing a -Cys-X-X-Cys- motif (X represents any residue) . It has been demonstrated that the redox state modulates Chlamydomonas flagellar beating patterns in vivo , controls the sign of phototaxis driven by differential motility of the two flagella [251,252], and alters the ATPase activity of its OAD HC’s in vitro . Indeed, LC3 , LC3BL, and TXND6 all retain this motif (Figure 6e). Structurally, LC3 changes interactions with the HCs in OAD in response to the ATP hydrolysis of the motor heads . Despite the advances in understanding LC3, it is not clear how the redox chemistry of LC3 modulates the activity of axonemal dyneins. LC5 is another thioredoxin-like protein that binds to the outermost OAD’s HC (a-HC) in Chlamydomonas [10,254]. It is missing in both Tetrahymena and humans. Interestingly, the human homolog of LC3, TXND6, has an additional nucleoside diphosphate kinase (NDPK) domain following the thioredoxin domain. NDPK is a type of enzyme that transfers γ-phosphate from a nucleoside triphosphate (NTP) to a nucleoside diphosphate (NDP), using a conserved histidine residue . The NDPK domain in TXND6 retains this conserved histidine (Figure 6e). The function of this domain is unknown.
LC4, a calmodulin protein containing several EF-hand motifs, at least one of which binds Ca 2+ (Figure 6f), binds between the innermost OAD’s HC and the IC-LC complex in Chlamydomonas and Tetrahymena . It binds the tail of the HC and interacts with the IC-LC tower, using the opposite interface. In the presence of Ca 2+, LC4 bends the tail of the innermost OAD HC, coming close to the ICs ; however, the implications of these conformational changes for the regulation of ciliary beating are not yet clear.
In the IADs, p28 is an LC that associates with the N-terminal domain of either a specific subset (in Chlamydomonas) or all (in humans, IAD d and IAD g are amalgamated into one complex) of the single-headed IAD HCs. Together with actin, the dimeric p28 anchors the HCs to the MTDs .
In summary, axonemal dyneins have an impressively large repertoire of LCs that fill the spaces among the HCs, with their functions largely unknown. Further understanding of how they contribute to the regulation of the ciliary beating would be of great interest.
5. Conclusions
Dynein is an ancient protein complex that likely existed and diverged into the three dynein families before the last eukaryotic common ancestor (LECA). This hypothesis is supported by the highly conserved sequences of HCs and ICs across species that harbor dynein . Throughout evolution, most plants lost the dynein branch completely, while various animal cells have lost different branches of dynein depending on the functionality required. Serving as the primary motor for MT minus-end-directed movements, dyneins play an indispensable role in numerous cellular functions. Mutations in dyneins lead to a broad spectrum of human diseases [101,259,260,261]. Consequently, unraveling the molecular mechanisms of dyneins is crucial for a comprehensive understanding of their significance.
Recent advances in biochemical, single-molecule, and structural research have significantly enhanced our understanding of how dynein motor complexes assemble and how dynein generates motion and force. However, the specific contributions and regulatory functions of non-catalytic subunits in dynein’s operation are not yet fully understood. This review highlights their roles in maintaining the integrity of dynein complexes; providing structural support; regulating the enzymatic activity of the HC; directing the complexes’ cellular localizations; mediating interactions with cofactors or other proteins; and expanding the dynein interactome through homologs, isoforms, and post-translational modifications. Despite these insights, many questions about these subunits remain unanswered. Key inquiries include the role of IDRs of ICs in dynein’s functionality, the mechanisms by which LICs selectively target dyneins to specific cellular locations, the reason behind dynein’s multitude of LCs, and how these LCs modulate dynein’s diverse functions. Addressing these questions will deepen our understanding of dynein’s cellular functions and shed light on how mutations in dyneins contribute to human diseases.
Author Contributions
Conceptualization, L.R.; writing of the first draft, L.R.; reviewing and editing of the draft, A.G.; visualization, L.R.; funding acquisition, A.G. All authors have read and agreed to the published version of the manuscript.
Funding
The authors were supported by the National Institute of Health (NIH) grant R01GM098469.
Institutional Review Board Statement
Not applicable.
Informed Consent Statement
Not applicable.
Data Availability Statement
The ColabFold-predicted structures are available upon request.
Acknowledgments
Parts of Figure 3, Figure 7 and Figure 8 were generated via biorender.com on 15 January 2024.
Conflicts of Interest
The authors declare no conflict of interest.
References
Roberts, A.J.; Kon, T.; Knight, P.J.; Sutoh, K.; Burgess, S.A. Functions and Mechanics of Dynein Motor Proteins. Nat. Rev. Mol. Cell Biol.2013, 14, 713–726. [Google Scholar] [CrossRef]
Wickstead, B. The Evolutionary Biology of Dyneins. In Dyneins: Structure, Biology and Disease—The Biology of Dynein Motors, 2nd ed.; Academic Press: Cambridge, MA, USA, 2018; pp. 100–138. ISBN 978-0-12-809471-6. [Google Scholar]
Höök, P.; Vallee, R.B. The Dynein Family at a Glance. J. Cell Sci.2006, 119, 4369–4371. [Google Scholar] [CrossRef]
Hanson, P.I.; Whiteheart, S.W. AAA+ Proteins: Have Engine, Will Work. Nat. Rev. Mol. Cell Biol.2005, 6, 519–529. [Google Scholar] [CrossRef]
Gleave, E.S.; Schmidt, H.; Carter, A.P. A Structural Analysis of the AAA+ Domains in Saccharomyces Cerevisiae Cytoplasmic Dynein. J. Struct. Biol.2014, 186, 367–375. [Google Scholar] [CrossRef][Green Version]
Zhang, K.; Foster, H.E.; Rondelet, A.; Lacey, S.E.; Bahi-Buisson, N.; Bird, A.W.; Carter, A.P. Cryo-EM Reveals How Human Cytoplasmic Dynein Is Auto-Inhibited and Activated. Cell2017, 169, 1303–1314.e18. [Google Scholar] [CrossRef]
Urnavicius, L.; Lau, C.K.; Elshenawy, M.M.; Morales-Rios, E.; Motz, C.; Yildiz, A.; Carter, A.P. Cryo-EM Shows How Dynactin Recruits Two Dyneins for Faster Movement. Nature2018, 554, 202–206. [Google Scholar] [CrossRef]
Toropova, K.; Zalyte, R.; Mukhopadhyay, A.G.; Mladenov, M.; Carter, A.P.; Roberts, A.J. Structure of the Dynein-2 Complex and Its Assembly with Intraflagellar Transport Trains. Nat. Struct. Mol. Biol.2019, 26, 823–829. [Google Scholar] [CrossRef]
Rao, Q.; Han, L.; Wang, Y.; Chai, P.; Kuo, Y.; Yang, R.; Hu, F.; Yang, Y.; Howard, J.; Zhang, K. Structures of Outer-Arm Dynein Array on Microtubule Doublet Reveal a Motor Coordination Mechanism. Nat. Struct. Mol. Biol.2021, 28, 799–810. [Google Scholar] [CrossRef]
Walton, T.; Gui, M.; Velkova, S.; Fassad, M.R.; Hirst, R.A.; Haarman, E.; O’Callaghan, C.; Bottier, M.; Burgoyne, T.; Mitchison, H.M.; et al. Axonemal Structures Reveal Mechanoregulatory and Disease Mechanisms. Nature2023, 618, 625–633. [Google Scholar] [CrossRef]
Roberts, A.J.; Numata, N.; Walker, M.L.; Kato, Y.S.; Malkova, B.; Kon, T.; Ohkura, R.; Arisaka, F.; Knight, P.J.; Sutoh, K.; et al. AAA+ Ring and Linker Swing Mechanism in the Dynein Motor. Cell2009, 136, 485–495. [Google Scholar] [CrossRef]
Carter, A.P.; Cho, C.; Jin, L.; Vale, R.D. Crystal Structure of the Dynein Motor Domain. Science2011, 331, 1159–1165. [Google Scholar] [CrossRef]
Kon, T.; Oyama, T.; Shimo-Kon, R.; Imamula, K.; Shima, T.; Sutoh, K.; Kurisu, G. The 2.8 Å Crystal Structure of the Dynein Motor Domain. Nature2012, 484, 345–350. [Google Scholar] [CrossRef]
Ogawa, K. Four ATP-Binding Sites in the Midregion of the β Heavy Chain of Dynein. Nature1991, 352, 643–645. [Google Scholar] [CrossRef]
Gibbons, I.R.; Gibbons, B.H.; Mocz, G.; Asai, D.J. Multiple Nucleotide-Binding Sites in the Sequence of Dynein β Heavy Chain. Nature1991, 352, 640–643. [Google Scholar] [CrossRef]
Mocz, G.; Gibbons, I.R. Phase Partition Analysis of Nucleotide Binding to Axonemal Dynein. Biochemistry1996, 35, 9204–9211. [Google Scholar] [CrossRef]
Kon, T.; Nishiura, M.; Ohkura, R.; Toyoshima, Y.Y.; Sutoh, K. Distinct Functions of Nucleotide-Binding/Hydrolysis Sites in the Four AAA Modules of Cytoplasmic Dynein. Biochemistry2004, 43, 11266–11274. [Google Scholar] [CrossRef]
Schmidt, H.; Zalyte, R.; Urnavicius, L.; Carter, A.P. Structure of Human Cytoplasmic Dynein-2 Primed for Its Power Stroke. Nature2015, 518, 435–438. [Google Scholar] [CrossRef]
Cho, C.; Reck-Peterson, S.L.; Vale, R.D. Regulatory ATPase Sites of Cytoplasmic Dynein Affect Processivity and Force Generation. J. Biol. Chem.2008, 283, 25839–25845. [Google Scholar] [CrossRef]
Bhabha, G.; Cheng, H.-C.; Zhang, N.; Moeller, A.; Liao, M.; Speir, J.A.; Cheng, Y.; Vale, R.D. Allosteric Communication in the Dynein Motor Domain. Cell2014, 159, 857–868. [Google Scholar] [CrossRef]
Nicholas, M.P.; Berger, F.; Rao, L.; Brenner, S.; Cho, C.; Gennerich, A. Cytoplasmic Dynein Regulates Its Attachment to Microtubules via Nucleotide State-Switched Mechanosensing at Multiple AAA Domains. Proc. Natl. Acad. Sci. USA2015, 112, 6371–6376. [Google Scholar] [CrossRef]
DeWitt, M.A.; Cypranowska, C.A.; Cleary, F.B.; Belyy, V.; Yildiz, A. The AAA3 Domain of Cytoplasmic Dynein Acts as a Switch to Facilitate Microtubule Release. Nat. Struct. Mol. Biol.2015, 22, 73–80. [Google Scholar] [CrossRef]
Liu, X.; Rao, L.; Gennerich, A. The Regulatory Function of the AAA4 ATPase Domain of Cytoplasmic Dynein. Nat. Commun.2020, 11, 5952. [Google Scholar] [CrossRef]
Qiu, R.; Zhang, J.; Rotty, J.D.; Xiang, X. Dynein Activation in Vivo Is Regulated by the Nucleotide States of Its AAA3 Domain. Curr. Biol.2021, 31, 4486–4498.e6. [Google Scholar] [CrossRef]
Schmidt, H.; Gleave, E.S.; Carter, A.P. Insights into Dynein Motor Domain Function from a 3.3-Å Crystal Structure. Nat. Struct. Mol. Biol.2012, 19, 492–497. [Google Scholar] [CrossRef]
Pfister, K.K.; Shah, P.R.; Hummerich, H.; Russ, A.; Cotton, J.; Annuar, A.A.; King, S.M.; Fisher, E.M.C. Genetic Analysis of the Cytoplasmic Dynein Subunit Families. PLoS Genet.2006, 2, e1. [Google Scholar] [CrossRef]
Kollmar, M. Fine-Tuning Motile Cilia and Flagella: Evolution of the Dynein Motor Proteins from Plants to Humans at High Resolution. Mol. Biol. Evol.2016, 33, 3249–3267. [Google Scholar] [CrossRef]
Burgess, S.A.; Walker, M.L.; Sakakibara, H.; Knight, P.J.; Oiwa, K. Dynein Structure and Power Stroke. Nature2003, 421, 715–718. [Google Scholar] [CrossRef]
Zimmermann, N.; Noga, A.; Obbineni, J.M.; Ishikawa, T. ATP-induced Conformational Change of Axonemal Outer Dynein Arms Revealed by Cryo-electron Tomography. EMBO J.2023, 42, e112466. [Google Scholar] [CrossRef]
Lin, J.; Okada, K.; Raytchev, M.; Smith, M.C.; Nicastro, D. Structural Mechanism of the Dynein Power Stroke. Nat. Cell Biol.2014, 16, 479–485. [Google Scholar] [CrossRef]
Can, S.; Lacey, S.; Gur, M.; Carter, A.P.; Yildiz, A. Directionality of Dynein Is Controlled by the Angle and Length of Its Stalk. Nature2019, 566, 407–410. [Google Scholar] [CrossRef]
Gee, M.A.; Heuser, J.E.; Vallee, R.B. An Extended Microtubule-Binding Structure within the Dynein Motor Domain. Nature1997, 390, 636–639. [Google Scholar] [CrossRef]
Nishikawa, Y.; Oyama, T.; Kamiya, N.; Kon, T.; Toyoshima, Y.Y.; Nakamura, H.; Kurisu, G. Structure of the Entire Stalk Region of the Dynein Motor Domain. J. Mol. Biol.2014, 426, 3232–3245. [Google Scholar] [CrossRef]
Carter, A.P.; Garbarino, J.E.; Wilson-Kubalek, E.M.; Shipley, W.E.; Cho, C.; Milligan, R.A.; Vale, R.D.; Gibbons, I.R. Structure and Functional Role of Dynein’s Microtubule-Binding Domain. Science2008, 322, 1691–1695. [Google Scholar] [CrossRef]
Redwine, W.B.; Hernández-López, R.; Zou, S.; Huang, J.; Reck-Peterson, S.L.; Leschziner, A.E. Structural Basis for Microtubule Binding and Release by Dynein. Science2012, 337, 1532–1536. [Google Scholar] [CrossRef]
Uchimura, S.; Fujii, T.; Takazaki, H.; Ayukawa, R.; Nishikawa, Y.; Minoura, I.; Hachikubo, Y.; Kurisu, G.; Sutoh, K.; Kon, T.; et al. A Flipped Ion Pair at the Dynein–Microtubule Interface Is Critical for Dynein Motility and ATPase Activation. J. Cell Biol.2015, 208, 211–222. [Google Scholar] [CrossRef]
Nishida, N.; Komori, Y.; Takarada, O.; Watanabe, A.; Tamura, S.; Kubo, S.; Shimada, I.; Kikkawa, M. Structural Basis for Two-Way Communication between Dynein and Microtubules. Nat. Commun.2020, 11, 1038. [Google Scholar] [CrossRef]
Gibbons, I.R.; Garbarino, J.E.; Tan, C.E.; Reck-Peterson, S.L.; Vale, R.D.; Carter, A.P. The Affinity of the Dynein Microtubule-Binding Domain Is Modulated by the Conformation of Its Coiled-Coil Stalk. J. Biol. Chem.2005, 280, 23960–23965. [Google Scholar] [CrossRef]
Kon, T.; Imamula, K.; Roberts, A.J.; Ohkura, R.; Knight, P.J.; Gibbons, I.R.; Burgess, S.A.; Sutoh, K. Helix Sliding in the Stalk Coiled Coil of Dynein Couples ATPase and Microtubule Binding. Nat. Struct. Mol. Biol.2009, 16, 325–333. [Google Scholar] [CrossRef]
Choi, J.; Park, H.; Seok, C. How Does a Registry Change in Dynein’s Coiled-Coil Stalk Drive Binding of Dynein to Microtubules? Biochemistry2011, 50, 7629–7636. [Google Scholar] [CrossRef]
Rao, L.; Berger, F.; Nicholas, M.P.; Gennerich, A. Molecular Mechanism of Cytoplasmic Dynein Tension Sensing. Nat. Commun.2019, 10, 3332. [Google Scholar] [CrossRef]
Niekamp, S.; Coudray, N.; Zhang, N.; Vale, R.D.; Bhabha, G. Coupling of ATPase Activity, Microtubule Binding, and Mechanics in the Dynein Motor Domain. EMBO J.2019, 38, e101414. [Google Scholar] [CrossRef]
Braschi, B.; Omran, H.; Witman, G.B.; Pazour, G.J.; Pfister, K.K.; Bruford, E.A.; King, S.M. Consensus Nomenclature for Dyneins and Associated Assembly Factors. J. Cell Biol.2022, 221, e202109014. [Google Scholar] [CrossRef] [PubMed]
King, S.M. Axonemal Dynein Arms. Cold Spring Harb. Perspect. Biol.2016, 8, a028100. [Google Scholar] [CrossRef]
Mali, G.R.; Ali, F.A.; Lau, C.K.; Begum, F.; Boulanger, J.; Howe, J.D.; Chen, Z.A.; Rappsilber, J.; Skehel, M.; Carter, A.P. Shulin Packages Axonemal Outer Dynein Arms for Ciliary Targeting. Science2021, 371, 910–916. [Google Scholar] [CrossRef]
Redpath, G.M.I.; Ananthanarayanan, V. Endosomal Sorting Sorted—Motors, Adaptors and Lessons from in Vitro and Cellular Studies. J. Cell Sci.2023, 136, jcs260749. [Google Scholar] [CrossRef]
Maday, S.; Twelvetrees, A.E.; Moughamian, A.J.; Holzbaur, E.L.F. Axonal Transport: Cargo-Specific Mechanisms of Motility and Regulation. Neuron2014, 84, 292–309. [Google Scholar] [CrossRef] [PubMed]
Guedes-Dias, P.; Holzbaur, E.L.F. Axonal Transport: Driving Synaptic Function. Science2019, 366, eaaw9997. [Google Scholar] [CrossRef] [PubMed]
Moore, J.K.; Stuchell-Brereton, M.D.; Cooper, J.A. Function of Dynein in Budding Yeast: Mitotic Spindle Positioning in a Polarized Cell. Cell Motil. Cytoskelet.2009, 66, 546–555. [Google Scholar] [CrossRef]
Kotak, S.; Busso, C.; Gönczy, P. Cortical Dynein Is Critical for Proper Spindle Positioning in Human Cells. J. Cell Biol.2012, 199, 97–110. [Google Scholar] [CrossRef]
Kiyomitsu, T.; Cheeseman, I.M. Cortical Dynein and Asymmetric Membrane Elongation Coordinately Position the Spindle in Anaphase. Cell2013, 154, 391–402. [Google Scholar] [CrossRef]
Gassmann, R. Dynein at the Kinetochore. J. Cell Sci.2023, 136, jcs220269. [Google Scholar] [CrossRef]
Yamamoto, A.; Hiraoka, Y. Cytoplasmic Dynein in Fungi: Insights from Nuclear Migration. J. Cell Sci.2003, 116, 4501–4512. [Google Scholar] [CrossRef][Green Version]
Hu, D.J.-K.; Baffet, A.D.; Nayak, T.; Akhmanova, A.; Doye, V.; Vallee, R.B. Dynein Recruitment to Nuclear Pores Activates Apical Nuclear Migration and Mitotic Entry in Brain Progenitor Cells. Cell2013, 154, 1300–1313. [Google Scholar] [CrossRef]
Milev, M.P.; Yao, X.; Berthoux, L.; Mouland, A.J. Impacts of Virus-Mediated Manipulation of Host Dynein. In Dyneins: Structure, Biology and Disease—Dynein Mechanics, Dysfunction, and Disease, 2nd ed.; Academic Press: Cambridge, MA, USA, 2018; pp. 214–233. ISBN 978-0-12-809470-9. [Google Scholar]
Scherer, J.; Yi, J.; Vallee, R.B. Role of Cytoplasmic Dynein and Kinesins in Adenovirus Transport. FEBS Lett.2020, 594, 1838–1847. [Google Scholar] [CrossRef]
Badieyan, S.; Lichon, D.; Andreas, M.P.; Gillies, J.P.; Peng, W.; Shi, J.; DeSantis, M.E.; Aiken, C.R.; Böcking, T.; Giessen, T.W.; et al. HIV-1 Binds Dynein Directly to Hijack Microtubule Transport Machinery. bioRxiv2023. [Google Scholar] [CrossRef]
Reck-Peterson, S.L.; Yildiz, A.; Carter, A.P.; Gennerich, A.; Zhang, N.; Vale, R.D. Single-Molecule Analysis of Dynein Processivity and Stepping Behavior. Cell2006, 126, 335–348. [Google Scholar] [CrossRef]
Trokter, M.; Mücke, N.; Surrey, T. Reconstitution of the Human Cytoplasmic Dynein Complex. Proc. Natl. Acad. Sci. USA2012, 109, 20895–20900. [Google Scholar] [CrossRef]
McKenney, R.J.; Huynh, W.; Tanenbaum, M.E.; Bhabha, G.; Vale, R.D. Activation of Cytoplasmic Dynein Motility by Dynactin-Cargo Adapter Complexes. Science2014, 345, 337–341. [Google Scholar] [CrossRef]
Schlager, M.A.; Hoang, H.T.; Urnavicius, L.; Bullock, S.L.; Carter, A.P. In Vitro Reconstitution of a Highly Processive Recombinant Human Dynein Complex. EMBO J.2014, 33, 1855–1868. [Google Scholar] [CrossRef]
Torisawa, T.; Ichikawa, M.; Furuta, A.; Saito, K.; Oiwa, K.; Kojima, H.; Toyoshima, Y.Y.; Furuta, K. Autoinhibition and Cooperative Activation Mechanisms of Cytoplasmic Dynein. Nat. Cell Biol.2014, 16, 1118–1124. [Google Scholar] [CrossRef]
Marzo, M.G.; Griswold, J.M.; Markus, S.M. Pac1/LIS1 Stabilizes an Uninhibited Conformation of Dynein to Coordinate Its Localization and Activity. Nat. Cell Biol.2020, 22, 559–569. [Google Scholar] [CrossRef]
Reck-Peterson, S.L.; Redwine, W.B.; Vale, R.D.; Carter, A.P. The Cytoplasmic Dynein Transport Machinery and Its Many Cargoes. Nat. Rev. Mol. Cell Biol.2018, 19, 382–398. [Google Scholar] [CrossRef]
Olenick, M.A.; Holzbaur, E.L.F. Dynein Activators and Adaptors at a Glance. J. Cell Sci.2019, 132, jcs227132. [Google Scholar] [CrossRef]
Xiang, X.; Qiu, R. Cargo-Mediated Activation of Cytoplasmic Dynein in Vivo. Front. Cell Dev. Biol.2020, 8, 598952. [Google Scholar] [CrossRef] [PubMed]
Chaaban, S.; Carter, A.P. Structure of Dynein–Dynactin on Microtubules Shows Tandem Adaptor Binding. Nature2022, 610, 212–216. [Google Scholar] [CrossRef]
Chowdhury, S.; Ketcham, S.A.; Schroer, T.A.; Lander, G.C. Structural Organization of the Dynein–Dynactin Complex Bound to Microtubules. Nat. Struct. Mol. Biol.2015, 22, 345–347. [Google Scholar] [CrossRef]
Urnavicius, L.; Zhang, K.; Diamant, A.G.; Motz, C.; Schlager, M.A.; Yu, M.; Patel, N.A.; Robinson, C.V.; Carter, A.P. The Structure of the Dynactin Complex and Its Interaction with Dynein. Science2015, 347, 1441–1446. [Google Scholar] [CrossRef] [PubMed]
Grotjahn, D.A.; Chowdhury, S.; Xu, Y.; McKenney, R.J.; Schroer, T.A.; Lander, G.C. Cryo-Electron Tomography Reveals That Dynactin Recruits a Team of Dyneins for Processive Motility. Nat. Struct. Mol. Biol.2018, 25, 203–207. [Google Scholar] [CrossRef] [PubMed]
Neer, E.J.; Schmidt, C.J.; Nambudripad, R.; Smith, T.F. The Ancient Regulatory-Protein Family of WD-Repeat Proteins. Nature1994, 371, 297–300. [Google Scholar] [CrossRef]
Smith, T.F.; Gaitatzes, C.; Saxena, K.; Neer, E.J. The WD Repeat: A Common Architecture for Diverse Functions. Trends Biochem. Sci.1999, 24, 181–185. [Google Scholar] [CrossRef]
Morgan, J.L.; Song, Y.; Barbar, E. Structural Dynamics and Multiregion Interactions in Dynein-Dynactin Recognition. J. Biol. Chem.2011, 286, 39349–39359. [Google Scholar] [CrossRef]
Jumper, J.; Evans, R.; Pritzel, A.; Green, T.; Figurnov, M.; Ronneberger, O.; Tunyasuvunakool, K.; Bates, R.; Žídek, A.; Potapenko, A.; et al. Highly Accurate Protein Structure Prediction with AlphaFold. Nature2021, 596, 583–589. [Google Scholar] [CrossRef]
Vaughan, K.T.; Vallee, R.B. Cytoplasmic Dynein Binds Dynactin through a Direct Interaction between the Intermediate Chains and p150Glued. J. Cell Biol.1995, 131, 1507–1516. [Google Scholar] [CrossRef]
Karki, S.; Holzbaur, E.L.F. Affinity Chromatography Demonstrates a Direct Binding between Cytoplasmic Dynein and the Dynactin Complex. J. Biol. Chem.1995, 270, 28806–28811. [Google Scholar] [CrossRef]
McKenney, R.J.; Weil, S.J.; Scherer, J.; Vallee, R.B. Mutually Exclusive Cytoplasmic Dynein Regulation by NudE-Lis1 and Dynactin. J. Biol. Chem.2011, 286, 39615–39622. [Google Scholar] [CrossRef]
Nyarko, A.; Song, Y.; Barbar, E. Intrinsic Disorder in Dynein Intermediate Chain Modulates Its Interactions with NudE and Dynactin. J. Biol. Chem.2012, 287, 24884–24893. [Google Scholar] [CrossRef]
Jie, J.; Löhr, F.; Barbar, E. Dynein Binding of Competitive Regulators Dynactin and NudE Involves Novel Interplay between Phosphorylation Site and Disordered Spliced Linkers. Structure2017, 25, 421–433. [Google Scholar] [CrossRef]
Jara, K.A.; Loening, N.M.; Reardon, P.N.; Yu, Z.; Woonnimani, P.; Brooks, C.; Vesely, C.H.; Barbar, E.J. Multivalency, Autoinhibition, and Protein Disorder in the Regulation of Interactions of Dynein Intermediate Chain with Dynactin and the Nuclear Distribution Protein. eLife2022, 11, e80217. [Google Scholar] [CrossRef]
Morgan, J.L.; Yeager, A.; Estelle, A.B.; Gsponer, J.; Barbar, E. Transient Tertiary Structures of Disordered Dynein Intermediate Chain Regulate Its Interactions with Multiple Partners. J. Mol. Biol.2021, 433, 167152. [Google Scholar] [CrossRef]
Mirdita, M.; Schütze, K.; Moriwaki, Y.; Heo, L.; Ovchinnikov, S.; Steinegger, M. ColabFold: Making Protein Folding Accessible to All. Nat. Methods2022, 19, 679–682. [Google Scholar] [CrossRef]
Makokha, M.; Hare, M.; Li, M.; Hays, T.; Barbar, E. Interactions of Cytoplasmic Dynein Light Chains Tctex-1 and LC8 with the Intermediate Chain IC74. Biochemistry2002, 41, 4302–4311. [Google Scholar] [CrossRef]
Susalka, S.J.; Nikulina, K.; Salata, M.W.; Vaughan, P.S.; King, S.M.; Vaughan, K.T.; Pfister, K.K. The Roadblock Light Chain Binds a Novel Region of the Cytoplasmic Dynein Intermediate Chain. J. Biol. Chem.2002, 277, 32939–32946. [Google Scholar] [CrossRef]
Stuchell-Brereton, M.D.; Siglin, A.; Li, J.; Moore, J.K.; Ahmed, S.; Williams, J.C.; Cooper, J.A. Functional Interaction between Dynein Light Chain and Intermediate Chain Is Required for Mitotic Spindle Positioning. Mol. Biol. Cell2011, 22, 2690–2701. [Google Scholar] [CrossRef]
Rao, L.; Romes, E.M.; Nicholas, M.P.; Brenner, S.; Tripathy, A.; Gennerich, A.; Slep, K.C. The Yeast Dynein Dyn2-Pac11 Complex Is a Dynein Dimerization/Processivity Factor: Structural and Single-Molecule Characterization. Mol. Biol. Cell2013, 24, 2362–2377. [Google Scholar] [CrossRef]
Nyarko, A.; Hare, M.; Hays, T.S.; Barbar, E. The Intermediate Chain of Cytoplasmic Dynein Is Partially Disordered and Gains Structure upon Binding to Light-Chain LC8. Biochemistry2004, 43, 15595–15603. [Google Scholar] [CrossRef]
Williams, J.C.; Roulhac, P.L.; Roy, A.G.; Vallee, R.B.; Fitzgerald, M.C.; Hendrickson, W.A. Structural and Thermodynamic Characterization of a Cytoplasmic Dynein Light Chain–Intermediate Chain Complex. Proc. Natl. Acad. Sci. USA2007, 104, 10028–10033. [Google Scholar] [CrossRef]
Hall, J.; Karplus, P.A.; Barbar, E. Multivalency in the Assembly of Intrinsically Disordered Dynein Intermediate Chain. J. Biol. Chem.2009, 284, 33115–33121. [Google Scholar] [CrossRef]
Hall, J.; Song, Y.; Karplus, P.A.; Barbar, E. The Crystal Structure of Dynein Intermediate Chain-Light Chain Roadblock Complex Gives New Insights into Dynein Assembly. J. Biol. Chem.2010, 285, 22566–22575. [Google Scholar] [CrossRef]
Okada, K.; Iyer, B.R.; Lammers, L.G.; Gutierrez, P.A.; Li, W.; Markus, S.M.; McKenney, R.J. Conserved Roles for the Dynein Intermediate Chain and Ndel1 in Assembly and Activation of Dynein. Nat. Commun.2023, 14, 5833. [Google Scholar] [CrossRef]
Elshenawy, M.M.; Kusakci, E.; Volz, S.; Baumbach, J.; Bullock, S.L.; Yildiz, A. Lis1 Activates Dynein Motility by Modulating Its Pairing with Dynactin. Nat. Cell Biol.2020, 22, 570–578. [Google Scholar] [CrossRef]
Htet, Z.M.; Gillies, J.P.; Baker, R.W.; Leschziner, A.E.; DeSantis, M.E.; Reck-Peterson, S.L. LIS1 Promotes the Formation of Activated Cytoplasmic Dynein-1 Complexes. Nat. Cell Biol.2020, 22, 518–525. [Google Scholar] [CrossRef]
Markus, S.M.; Marzo, M.G.; McKenney, R.J. New Insights into the Mechanism of Dynein Motor Regulation by Lissencephaly-1. eLife2020, 9, e59737. [Google Scholar] [CrossRef]
McKenney, R.J.; Vershinin, M.; Kunwar, A.; Vallee, R.B.; Gross, S.P. LIS1 and NudE Induce a Persistent Dynein Force-Producing State. Cell2010, 141, 304–314. [Google Scholar] [CrossRef] [PubMed]
Zhao, Y.; Oten, S.; Yildiz, A. Nde1 Promotes Lis1-Mediated Activation of Dynein. Nat. Commun.2023, 14, 7221. [Google Scholar] [CrossRef] [PubMed]
Derewenda, U.; Tarricone, C.; Choi, W.C.; Cooper, D.R.; Lukasik, S.; Perrina, F.; Tripathy, A.; Kim, M.H.; Cafiso, D.S.; Musacchio, A.; et al. The Structure of the Coiled-Coil Domain of Ndel1 and the Basis of Its Interaction with Lis1, the Causal Protein of Miller-Dieker Lissencephaly. Structure2007, 15, 1467–1481. [Google Scholar] [CrossRef] [PubMed][Green Version]
Garrott, S.R.; Gillies, J.P.; Siva, A.; Little, S.R.; El Jbeily, R.; DeSantis, M.E. Ndel1 Disfavors Dynein-Dynactin-Adaptor Complex Formation in Two Distinct Ways. J. Biol. Chem.2023, 299, 104735. [Google Scholar] [CrossRef] [PubMed]
Karasmanis, E.P.; Reimer, J.M.; Kendrick, A.A.; Nguyen, K.H.V.; Rodriguez, J.A.; Truong, J.B.; Lahiri, I.; Reck-Peterson, S.L.; Leschziner, A.E. Lis1 Relieves Cytoplasmic Dynein-1 Autoinhibition by Acting as a Molecular Wedge. Nat. Struct. Mol. Biol.2023, 30, 1357–1364. [Google Scholar] [CrossRef]
Ori-McKenney, K.M.; Xu, J.; Gross, S.P.; Vallee, R.B. A Cytoplasmic Dynein Tail Mutation Impairs Motor Processivity. Nat. Cell Biol.2010, 12, 1228–1234. [Google Scholar] [CrossRef]
Hoang, H.T.; Schlager, M.A.; Carter, A.P.; Bullock, S.L. DYNC1H1 Mutations Associated with Neurological Diseases Compromise Processivity of Dynein–Dynactin–Cargo Adaptor Complexes. Proc. Natl. Acad. Sci. USA2017, 114, E1597–E1606. [Google Scholar] [CrossRef]
Myers, K.R.; Lo, K.W.-H.; Lye, R.J.; Kogoy, J.M.; Soura, V.; Hafezparast, M.; Pfister, K.K. Intermediate Chain Subunit as a Probe for Cytoplasmic Dynein Function: Biochemical Analyses and Live Cell Imaging in PC12 Cells. J. Neurosci. Res.2007, 85, 2640–2647. [Google Scholar] [CrossRef]
Pfister, K.K.; Salata, M.W.; Dillman, J.F.; Torre, E.; Lye, R.J. Identification and Developmental Regulation of a Neuron-Specific Subunit of Cytoplasmic Dynein. Mol. Biol. Cell1996, 7, 331–343. [Google Scholar] [CrossRef]
Kuta, A.; Deng, W.; Morsi El-Kadi, A.; Banks, G.T.; Hafezparast, M.; Pfister, K.K.; Fisher, E.M.C. Mouse Cytoplasmic Dynein Intermediate Chains: Identification of New Isoforms, Alternative Splicing and Tissue Distribution of Transcripts. PLoS ONE2010, 5, e11682. [Google Scholar] [CrossRef]
Lo, K.W.-H.; Kan, H.-M.; Pfister, K.K. Identification of a Novel Region of the Cytoplasmic Dynein Intermediate Chain Important for Dimerization in the Absence of the Light Chains. J. Biol. Chem.2006, 281, 9552–9559. [Google Scholar] [CrossRef] [PubMed]
Nurminsky, D.I.; Nurminskaya, M.V.; Benevolenskaya, E.V.; Shevelyov, Y.Y.; Hartl, D.L.; Gvozdev, V.A. Cytoplasmic Dynein Intermediate-Chain Isoforms with Different Targeting Properties Created by Tissue-Specific Alternative Splicing. Mol. Cell. Biol.1998, 18, 6816–6825. [Google Scholar] [CrossRef]
Vaughan, P.S.; Leszyk, J.D.; Vaughan, K.T. Cytoplasmic Dynein Intermediate Chain Phosphorylation Regulates Binding to Dynactin. J. Biol. Chem.2001, 276, 26171–26179. [Google Scholar] [CrossRef]
Pfister, K.K. Distinct Functional Roles of Cytoplasmic Dynein Defined by the Intermediate Chain Isoforms. Exp. Cell Res.2015, 334, 54–60. [Google Scholar] [CrossRef]
Schroeder, C.M.; Ostrem, J.M.; Hertz, N.T.; Vale, R.D. A Ras-like Domain in the Light Intermediate Chain Bridges the Dynein Motor to a Cargo-Binding Region. eLife2014, 3, e03351. [Google Scholar] [CrossRef]
Markus, S.M.; Lee, W.-L. Microtubule-Dependent Path to the Cell Cortex for Cytoplasmic Dynein in Mitotic Spindle Orientation. Bioarchitecture2011, 1, 209–215. [Google Scholar] [CrossRef][Green Version]
Tynan, S.H.; Gee, M.A.; Vallee, R.B. Distinct but Overlapping Sites within the Cytoplasmic Dynein Heavy Chain for Dimerization and for Intermediate Chain and Light Intermediate Chain Binding. J. Biol. Chem.2000, 275, 32769–32774. [Google Scholar] [CrossRef]
Pettersen, E.F.; Goddard, T.D.; Huang, C.C.; Couch, G.S.; Greenblatt, D.M.; Meng, E.C.; Ferrin, T.E. UCSF Chimera—A Visualization System for Exploratory Research and Analysis. J. Comput. Chem.2004, 25, 1605–1612. [Google Scholar] [CrossRef]
Lee, I.-G.; Olenick, M.A.; Boczkowska, M.; Franzini-Armstrong, C.; Holzbaur, E.L.F.; Dominguez, R. A Conserved Interaction of the Dynein Light Intermediate Chain with Dynein-Dynactin Effectors Necessary for Processivity. Nat. Commun.2018, 9, 986. [Google Scholar] [CrossRef]
Celestino, R.; Henen, M.A.; Gama, J.B.; Carvalho, C.; McCabe, M.; Barbosa, D.J.; Born, A.; Nichols, P.J.; Carvalho, A.X.; Gassmann, R.; et al. A Transient Helix in the Disordered Region of Dynein Light Intermediate Chain Links the Motor to Structurally Diverse Adaptors for Cargo Transport. PLoS Biol.2019, 17, e3000100. [Google Scholar] [CrossRef]
Lee, I.-G.; Cason, S.E.; Alqassim, S.S.; Holzbaur, E.L.F.; Dominguez, R. A Tunable LIC1-Adaptor Interaction Modulates Dynein Activity in a Cargo-Specific Manner. Nat. Commun.2020, 11, 5695. [Google Scholar] [CrossRef]
Schroeder, C.M.; Vale, R.D. Assembly and Activation of Dynein–Dynactin by the Cargo Adaptor Protein Hook3. J. Cell Biol.2016, 214, 309–318. [Google Scholar] [CrossRef]
Hernandez-Perez, I.; Rubio, J.; Baumann, A.; Girao, H.; Ferrando, M.; Rebollo, E.; Aragay, A.M.; Geli, M.I. Kazrin Promotes Dynein/Dynactin-Dependent Traffic from Early to Recycling Endosomes. eLife2023, 12, e83793. [Google Scholar] [CrossRef]
Maldonado-Báez, L.; Cole, N.B.; Krämer, H.; Donaldson, J.G. Microtubule-Dependent Endosomal Sorting of Clathrin-Independent Cargo by Hook1. J. Cell Biol.2013, 201, 233–247. [Google Scholar] [CrossRef]
Zhang, J.; Qiu, R.; Arst, H.N.; Peñalva, M.A.; Xiang, X. HookA Is a Novel Dynein–Early Endosome Linker Critical for Cargo Movement in Vivo. J. Cell Biol.2014, 204, 1009–1026. [Google Scholar] [CrossRef]
Horgan, C.P.; Hanscom, S.R.; Jolly, R.S.; Futter, C.E.; McCaffrey, M.W. Rab11-FIP3 Links the Rab11 GTPase and Cytoplasmic Dynein to Mediate Transport to the Endosomal-Recycling Compartment. J. Cell Sci.2010, 123, 181–191. [Google Scholar] [CrossRef]
Tan, S.C.; Scherer, J.; Vallee, R.B. Recruitment of Dynein to Late Endosomes and Lysosomes through Light Intermediate Chains. Mol. Biol. Cell2011, 22, 467–477. [Google Scholar] [CrossRef] [PubMed]
Delaval, B.; Doxsey, S.J. Pericentrin in Cellular Function and Disease. J. Cell Biol.2010, 188, 181–190. [Google Scholar] [CrossRef] [PubMed]
Doxsey, S.J.; Stein, P.; Evans, L.; Calarco, P.D.; Kirschner, M. Pericentrin, a Highly Conserved Centrosome Protein Involved in Microtubule Organization. Cell1994, 76, 639–650. [Google Scholar] [CrossRef]
Purohit, A.; Tynan, S.H.; Vallee, R.; Doxsey, S.J. Direct Interaction of Pericentrin with Cytoplasmic Dynein Light Intermediate Chain Contributes to Mitotic Spindle Organization. J. Cell Biol.1999, 147, 481–492. [Google Scholar] [CrossRef]
Tynan, S.H.; Purohit, A.; Doxsey, S.J.; Vallee, R.B. Light Intermediate Chain 1 Defines a Functional Subfraction of Cytoplasmic Dynein Which Binds to Pericentrin. J. Biol. Chem.2000, 275, 32763–32768. [Google Scholar] [CrossRef]
Gama, J.B.; Pereira, C.; Simões, P.A.; Celestino, R.; Reis, R.M.; Barbosa, D.J.; Pires, H.R.; Carvalho, C.; Amorim, J.; Carvalho, A.X.; et al. Molecular Mechanism of Dynein Recruitment to Kinetochores by the Rod-Zw10-Zwilch Complex and Spindly. J. Cell Biol.2017, 216, 943–960. [Google Scholar] [CrossRef]
Wu, J.; Larreategui-Aparicio, A.; Lambers, M.L.A.; Bodor, D.L.; Klaasen, S.J.; Tollenaar, E.; De Ruijter-Villani, M.; Kops, G.J.P.L. Microtubule Nucleation from the Fibrous Corona by LIC1-Pericentrin Promotes Chromosome Congression. Curr. Biol.2023, 33, 912–925.e6. [Google Scholar] [CrossRef]
Okumura, M.; Natsume, T.; Kanemaki, M.T.; Kiyomitsu, T. Dynein–Dynactin–NuMA Clusters Generate Cortical Spindle-Pulling Forces as a Multi-Arm Ensemble. eLife2018, 7, e36559. [Google Scholar] [CrossRef] [PubMed]
Renna, C.; Rizzelli, F.; Carminati, M.; Gaddoni, C.; Pirovano, L.; Cecatiello, V.; Pasqualato, S.; Mapelli, M. Organizational Principles of the NuMA-Dynein Interaction Interface and Implications for Mitotic Spindle Functions. Structure2020, 28, 820–829.e6. [Google Scholar] [CrossRef] [PubMed]
Wynne, D.J.; Rog, O.; Carlton, P.M.; Dernburg, A.F. Dynein-Dependent Processive Chromosome Motions Promote Homologous Pairing in C. Elegans Meiosis. J. Cell Biol.2012, 196, 47–64. [Google Scholar] [CrossRef] [PubMed]
Lee, C.-Y.; Horn, H.F.; Stewart, C.L.; Burke, B.; Bolcun-Filas, E.; Schimenti, J.C.; Dresser, M.E.; Pezza, R.J. Mechanism and Regulation of Rapid Telomere Prophase Movements in Mouse Meiotic Chromosomes. Cell Rep.2015, 11, 551–563. [Google Scholar] [CrossRef]
Meier, I. LINCing the Eukaryotic Tree of Life—Towards a Broad Evolutionary Comparison of Nucleocytoplasmic Bridging Complexes. J. Cell Sci.2016, 129, 3523–3531. [Google Scholar] [CrossRef]
Agrawal, R.; Gillies, J.P.; Zang, J.L.; Zhang, J.; Garrott, S.R.; Shibuya, H.; Nandakumar, J.; DeSantis, M.E. The KASH5 Protein Involved in Meiotic Chromosomal Movements Is a Novel Dynein Activating Adaptor. eLife2022, 11, e78201. [Google Scholar] [CrossRef]
Garner, K.E.L.; Salter, A.; Lau, C.K.; Gurusaran, M.; Villemant, C.M.; Granger, E.P.; McNee, G.; Woodman, P.G.; Davies, O.R.; Burke, B.E.; et al. The Meiotic LINC Complex Component KASH5 Is an Activating Adaptor for Cytoplasmic Dynein. J. Cell Biol.2023, 222, e202204042. [Google Scholar] [CrossRef]
Hughes, S.M.; Vaughan, K.T.; Herskovits, J.S.; Vallee, R.B. Molecular Analysis of a Cytoplasmic Dynein Light Intermediate Chain Reveals Homology to a Family of ATPases. J. Cell Sci.1995, 108, 17–24. [Google Scholar] [CrossRef]
Horgan, C.P.; Hanscom, S.R.; McCaffrey, M.W. Dynein LIC1 Localizes to the Mitotic Spindle and Midbody and LIC2 Localizes to Spindle Poles during Cell Division. Cell Biol. Int.2011, 35, 171–178. [Google Scholar] [CrossRef] [PubMed]
Sivaram, M.V.S.; Wadzinski, T.L.; Redick, S.D.; Manna, T.; Doxsey, S.J. Dynein Light Intermediate Chain 1 Is Required for Progress through the Spindle Assembly Checkpoint. EMBO J.2009, 28, 902–914. [Google Scholar] [CrossRef]
Mahale, S.; Kumar, M.; Sharma, A.; Babu, A.; Ranjan, S.; Sachidanandan, C.; Mylavarapu, S.V.S. The Light Intermediate Chain 2 Subpopulation of Dynein Regulates Mitotic Spindle Orientation. Sci. Rep.2016, 6, 22. [Google Scholar] [CrossRef]
Sharma, A.; Dagar, S.; Mylavarapu, S.V.S. Transgelin-2 and Phosphoregulation of the LIC2 Subunit of Dynein Govern Mitotic Spindle Orientation. J. Cell Sci.2020, 133, jcs.239673. [Google Scholar] [CrossRef]
Mahale, S.P.; Sharma, A.; Mylavarapu, S.V.S. Dynein Light Intermediate Chain 2 Facilitates the Metaphase to Anaphase Transition by Inactivating the Spindle Assembly Checkpoint. PLoS ONE2016, 11, e0159646. [Google Scholar] [CrossRef]
Henen, M.A.; Paukovich, N.; Prekeris, R.; Vögeli, B. Solution NMR Backbone Assignment of the C-Terminal Region of Human Dynein Light Intermediate Chain 2 (LIC2-C) Unveils Structural Resemblance with Its Homologue LIC1-C. Magnetochemistry2023, 9, 166. [Google Scholar] [CrossRef]
Gonçalves, J.C.; Dantas, T.J.; Vallee, R.B. Distinct Roles for Dynein Light Intermediate Chains in Neurogenesis, Migration, and Terminal Somal Translocation. J. Cell Biol.2019, 218, 808–819. [Google Scholar] [CrossRef]
Tsai, J.-W.; Lian, W.-N.; Kemal, S.; Kriegstein, A.R.; Vallee, R.B. Kinesin 3 and Cytoplasmic Dynein Mediate Interkinetic Nuclear Migration in Neural Stem Cells. Nat. Neurosci.2010, 13, 1463–1471. [Google Scholar] [CrossRef]
Kumari, A.; Kumar, C.; Wasnik, N.; Mylavarapu, S.V.S. Dynein Light Intermediate Chains as Pivotal Determinants of Dynein Multifunctionality. J. Cell Sci.2021, 134, jcs254870. [Google Scholar] [CrossRef]
Ilangovan, U.; Ding, W.; Zhong, Y.; Wilson, C.L.; Groppe, J.C.; Trbovich, J.T.; Zúñiga, J.; Demeler, B.; Tang, Q.; Gao, G.; et al. Structure and Dynamics of the Homodimeric Dynein Light Chain Km23. J. Mol. Biol.2005, 352, 338–354. [Google Scholar] [CrossRef]
Benison, G.; Karplus, P.A.; Barbar, E. The Interplay of Ligand Binding and Quaternary Structure in the Diverse Interactions of Dynein Light Chain LC8. J. Mol. Biol.2008, 384, 954–966. [Google Scholar] [CrossRef]
Williams, J.C.; Xie, H.; Hendrickson, W.A. Crystal Structure of Dynein Light Chain TcTex-1. J. Biol. Chem.2005, 280, 21981–21986. [Google Scholar] [CrossRef]
Toda, A.; Tanaka, H.; Kurisu, G. Structural Atlas of Dynein Motors at Atomic Resolution. Biophys. Rev.2018, 10, 677–686. [Google Scholar] [CrossRef]
Bowman, A.B.; Patel-King, R.S.; Benashski, S.E.; McCaffery, J.M.; Goldstein, L.S.B.; King, S.M. Drosophila Roadblock and Chlamydomonas LC7: A Conserved Family of Dynein-Associated Proteins Involved in Axonal Transport, Flagellar Motility, and Mitosis. J. Cell Biol.1999, 146, 165–180. [Google Scholar] [CrossRef][Green Version]
Song, J.; Tyler, R.C.; Lee, M.S.; Tyler, E.M.; Markley, J.L. Solution Structure of Isoform 1 of Roadblock/LC7, a Light Chain in the Dynein Complex. J. Mol. Biol.2005, 354, 1043–1051. [Google Scholar] [CrossRef] [PubMed]
Zhang, J.; Li, S.; Musa, S.; Zhou, H.; Xiang, X. Dynein Light Intermediate Chain in Aspergillus Nidulans Is Essential for the Interaction between Heavy and Intermediate Chains. J. Biol. Chem.2009, 284, 34760–34768. [Google Scholar] [CrossRef] [PubMed]
Terenzio, M.; Di Pizio, A.; Rishal, I.; Marvaldi, L.; Di Matteo, P.; Kawaguchi, R.; Coppola, G.; Schiavo, G.; Fisher, E.M.C.; Fainzilber, M. DYNLRB1 Is Essential for Dynein Mediated Transport and Neuronal Survival. Neurobiol. Dis.2020, 140, 104816. [Google Scholar] [CrossRef] [PubMed]
He, S.; Gillies, J.P.; Zang, J.L.; Córdoba-Beldad, C.M.; Yamamoto, I.; Fujiwara, Y.; Grantham, J.; DeSantis, M.E.; Shibuya, H. Distinct Dynein Complexes Defined by DYNLRB1 and DYNLRB2 Regulate Mitotic and Male Meiotic Spindle Bipolarity. Nat. Commun.2023, 14, 1715. [Google Scholar] [CrossRef]
Barbar, E.; Kleinman, B.; Imhoff, D.; Li, M.; Hays, T.S.; Hare, M. Dimerization and Folding of LC8, a Highly Conserved Light Chain of Cytoplasmic Dynein. Biochemistry2001, 40, 1596–1605. [Google Scholar] [CrossRef] [PubMed]
Nyarko, A.; Cochrun, L.; Norwood, S.; Pursifull, N.; Voth, A.; Barbar, E. Ionization of His 55 at the Dimer Interface of Dynein Light-Chain LC8 Is Coupled to Dimer Dissociation. Biochemistry2005, 44, 14248–14255. [Google Scholar] [CrossRef]
Jin, M.; Yamada, M.; Arai, Y.; Nagai, T.; Hirotsune, S. Arl3 and LC8 Regulate Dissociation of Dynactin from Dynein. Nat. Commun.2014, 5, 5295. [Google Scholar] [CrossRef] [PubMed]
Li, X.; Liu, S.; Zhang, L.; Issaian, A.; Hill, R.C.; Espinosa, S.; Shi, S.; Cui, Y.; Kappel, K.; Das, R.; et al. A Unified Mechanism for Intron and Exon Definition and Back-Splicing. Nature2019, 573, 375–380. [Google Scholar] [CrossRef] [PubMed]
Rapali, P.; Szenes, Á.; Radnai, L.; Bakos, A.; Pál, G.; Nyitray, L. DYNLL/LC8: A Light Chain Subunit of the Dynein Motor Complex and Beyond. FEBS J.2011, 278, 2980–2996. [Google Scholar] [CrossRef] [PubMed]
Piperno, G.; Luck, D.J. Axonemal Adenosine Triphosphatases from Flagella of Chlamydomonas Reinhardtii. Purification of Two Dyneins. J. Biol. Chem.1979, 254, 3084–3090. [Google Scholar] [CrossRef] [PubMed]
Pfister, K.K.; Fay, R.B.; Witman, G.B. Purification and Polypeptide Composition of Dynein ATPases from Chlamydomonas Flagella. Cell Motil.1982, 2, 525–547. [Google Scholar] [CrossRef]
King, S.M.; Patel-King, R.S. The M(r) = 8,000 and 11,000 Outer Arm Dynein Light Chains from Chlamydomonas Flagella Have Cytoplasmic Homologues. J. Biol. Chem.1995, 270, 11445–11452. [Google Scholar] [CrossRef]
King, S.M.; Barbarese, E.; Dillman, J.F.; Patel-King, R.S.; Carson, J.H.; Pfister, K.K. Brain Cytoplasmic and Flagellar Outer Arm Dyneins Share a Highly Conserved Mr 8,000 Light Chain. J. Biol. Chem.1996, 271, 19358–19366. [Google Scholar] [CrossRef]
Asante, D.; Stevenson, N.L.; Stephens, D.J. Subunit Composition of the Human Cytoplasmic Dynein-2 Complex. J. Cell Sci.2014, 127, jcs.159038. [Google Scholar] [CrossRef]
Barbar, E. Dynein Light Chain LC8 Is a Dimerization Hub Essential in Diverse Protein Networks. Biochemistry2008, 47, 503–508. [Google Scholar] [CrossRef]
Jespersen, N.; Estelle, A.; Waugh, N.; Davey, N.E.; Blikstad, C.; Ammon, Y.-C.; Akhmanova, A.; Ivarsson, Y.; Hendrix, D.A.; Barbar, E. Systematic Identification of Recognition Motifs for the Hub Protein LC8. Life Sci. Alliance2019, 2, e201900366. [Google Scholar] [CrossRef]
Stelter, P.; Kunze, R.; Flemming, D.; Höpfner, D.; Diepholz, M.; Philippsen, P.; Böttcher, B.; Hurt, E. Molecular Basis for the Functional Interaction of Dynein Light Chain with the Nuclear-Pore Complex. Nat. Cell Biol.2007, 9, 788–796. [Google Scholar] [CrossRef]
Espindola, F.S.; Suter, D.M.; Partata, L.B.E.; Cao, T.; Wolenski, J.S.; Cheney, R.E.; King, S.M.; Mooseker, M.S. The Light Chain Composition of Chicken Brain Myosin-Va: Calmodulin, Myosin-II Essential Light Chains, and 8-kDa Dynein Light Chain/PIN. Cell Motil. Cytoskelet.2000, 47, 269–281. [Google Scholar] [CrossRef]
He, Y.J.; Meghani, K.; Caron, M.-C.; Yang, C.; Ronato, D.A.; Bian, J.; Sharma, A.; Moore, J.; Niraj, J.; Detappe, A.; et al. DYNLL1 Binds to MRE11 to Limit DNA End Resection in BRCA1-Deficient Cells. Nature2018, 563, 522–526. [Google Scholar] [CrossRef]
Becker, J.R.; Cuella-Martin, R.; Barazas, M.; Liu, R.; Oliveira, C.; Oliver, A.W.; Bilham, K.; Holt, A.B.; Blackford, A.N.; Heierhorst, J.; et al. The ASCIZ-DYNLL1 Axis Promotes 53BP1-Dependent Non-Homologous End Joining and PARP Inhibitor Sensitivity. Nat. Commun.2018, 9, 5406. [Google Scholar] [CrossRef]
West, K.L.; Kelliher, J.L.; Xu, Z.; An, L.; Reed, M.R.; Eoff, R.L.; Wang, J.; Huen, M.S.Y.; Leung, J.W.C. LC8/DYNLL1 Is a 53BP1 Effector and Regulates Checkpoint Activation. Nucleic Acids Res.2019, 47, 6236–6249. [Google Scholar] [CrossRef]
Tan, G.S.; Preuss, M.A.R.; Williams, J.C.; Schnell, M.J. The Dynein Light Chain 8 Binding Motif of Rabies Virus Phosphoprotein Promotes Efficient Viral Transcription. Proc. Natl. Acad. Sci. USA2007, 104, 7229–7234. [Google Scholar] [CrossRef]
Luthra, P.; Jordan, D.S.; Leung, D.W.; Amarasinghe, G.K.; Basler, C.F. Ebola Virus VP35 Interaction with Dynein LC8 Regulates Viral RNA Synthesis. J. Virol.2015, 89, 5148–5153. [Google Scholar] [CrossRef]
King, S.M.; Dillman, J.F.; Benashski, S.E.; Lye, R.J.; Patel-King, R.S.; Pfister, K.K. The Mouse T-Complex-Encoded Protein Tctex-1 Is a Light Chain of Brain Cytoplasmic Dynein. J. Biol. Chem.1996, 271, 32281–32287. [Google Scholar] [CrossRef]
Harrison, A.; Olds-Clarke, P.; King, S.M. Identification of the t Complex-Encoded Cytoplasmic Dynein Light Chain Tctex1 in Inner Arm I1 Supports the Involvement of Flagellar Dyneins in Meiotic Drive. J. Cell Biol.1998, 140, 1137–1147. [Google Scholar] [CrossRef]
Kagami, O.; Gotoh, M.; Makino, Y.; Mohri, H.; Kamiya, R.; Ogawa, K. A Dynein Light Chain of Sea Urchin Sperm Flagella Is a Homolog of Mouse Tctex 1, Which Is Encoded by a Gene of the t Complex Sterility Locus. Gene1998, 211, 383–386. [Google Scholar] [CrossRef]
Tai, A.W.; Chuang, J.-Z.; Bode, C.; Wolfrum, U.; Sung, C.-H. Rhodopsin’s Carboxy-Terminal Cytoplasmic Tail Acts as a Membrane Receptor for Cytoplasmic Dynein by Binding to the Dynein Light Chain Tctex-1. Cell1999, 97, 877–887. [Google Scholar] [CrossRef]
Yeh, T.; Peretti, D.; Chuang, J.; Rodriguez-Boulan, E.; Sung, C. Regulatory Dissociation of Tctex-1 Light Chain from Dynein Complex Is Essential for the Apical Delivery of Rhodopsin. Traffic2006, 7, 1495–1502. [Google Scholar] [CrossRef]
Chen, C.; Peng, Y.; Yen, Y.; Bhan, P.; Muthaiyan Shanmugam, M.; Klopfenstein, D.R.; Wagner, O.I. Insights on UNC-104-dynein/Dynactin Interactions and Their Implications on Axonal Transport in Caenorhabditis Elegans. J. Neurosci. Res.2019, 97, 185–201. [Google Scholar] [CrossRef]
Chuang, J.-Z.; Milner, T.A.; Sung, C.-H. Subunit Heterogeneity of Cytoplasmic Dynein: Differential Expression of 14 kDa Dynein Light Chains in Rat Hippocampus. J. Neurosci.2001, 21, 5501–5512. [Google Scholar] [CrossRef]
Tai, A.W.; Chuang, J.-Z.; Sung, C.-H. Cytoplasmic Dynein Regulation by Subunit Heterogeneity and Its Role in Apical Transport. J. Cell Biol.2001, 153, 1499–1510. [Google Scholar] [CrossRef]
Lamb, A.K.; Fernandez, A.N.; Peersen, O.B.; Di Pietro, S.M. The Dynein Light Chain Protein Tda2 Functions as a Dimerization Engine to Regulate Actin Capping Protein during Endocytosis. Mol. Biol. Cell2021, 32, 1459–1473. [Google Scholar] [CrossRef]
Farrell, K.B.; McDonald, S.; Lamb, A.K.; Worcester, C.; Peersen, O.B.; Di Pietro, S.M. Novel Function of a Dynein Light Chain in Actin Assembly during Clathrin-Mediated Endocytosis. J. Cell Biol.2017, 216, 2565–2580. [Google Scholar] [CrossRef]
Chuang, J.-Z.; Yeh, T.-Y.; Bollati, F.; Conde, C.; Canavosio, F.; Caceres, A.; Sung, C.-H. The Dynein Light Chain Tctex-1 Has a Dynein-Independent Role in Actin Remodeling during Neurite Outgrowth. Dev. Cell2005, 9, 75–86. [Google Scholar] [CrossRef]
Saito, M.; Otsu, W.; Hsu, K.-S.; Chuang, J.-Z.; Yanagisawa, T.; Shieh, V.; Kaitsuka, T.; Wei, F.-Y.; Tomizawa, K.; Sung, C.-H. Tctex-1 Controls Ciliary Resorption by Regulating Branched Actin Polymerization and Endocytosis. EMBO Rep.2017, 18, 1460–1472. [Google Scholar] [CrossRef]
Nekrasova, O.; Harmon, R.M.; Broussard, J.A.; Koetsier, J.L.; Godsel, L.M.; Fitz, G.N.; Gardel, M.L.; Green, K.J. Desmosomal Cadherin Association with Tctex-1 and Cortactin-Arp2/3 Drives Perijunctional Actin Polymerization to Promote Keratinocyte Delamination. Nat. Commun.2018, 9, 1053. [Google Scholar] [CrossRef]
Ishikawa, T. Axoneme Structure from Motile Cilia. Cold Spring Harb. Perspect. Biol.2017, 9, a028076. [Google Scholar] [CrossRef]
Wheway, G.; Nazlamova, L.; Hancock, J.T. Signaling through the Primary Cilium. Front. Cell Dev. Biol.2018, 6, 8. [Google Scholar] [CrossRef]
Webb, S.; Mukhopadhyay, A.G.; Roberts, A.J. Intraflagellar Transport Trains and Motors: Insights from Structure. Semin. Cell Dev. Biol.2020, 107, 82–90. [Google Scholar] [CrossRef]
Jordan, M.A.; Pigino, G. The Structural Basis of Intraflagellar Transport at a Glance. J. Cell Sci.2021, 134, jcs247163. [Google Scholar] [CrossRef]
Garcia-Gonzalo, F.R.; Reiter, J.F. Open Sesame: How Transition Fibers and the Transition Zone Control Ciliary Composition. Cold Spring Harb. Perspect. Biol.2017, 9, a028134. [Google Scholar] [CrossRef]
Gonçalves, J.; Pelletier, L. The Ciliary Transition Zone: Finding the Pieces and Assembling the Gate. Mol. Cells2017, 40, 243–253. [Google Scholar] [CrossRef]
Prevo, B.; Scholey, J.M.; Peterman, E.J.G. Intraflagellar Transport: Mechanisms of Motor Action, Cooperation, and Cargo Delivery. FEBS J.2017, 284, 2905–2931. [Google Scholar] [CrossRef]
Stepanek, L.; Pigino, G. Microtubule Doublets Are Double-Track Railways for Intraflagellar Transport Trains. Science2016, 352, 721–724. [Google Scholar] [CrossRef]
Van Den Hoek, H.; Klena, N.; Jordan, M.A.; Alvarez Viar, G.; Righetto, R.D.; Schaffer, M.; Erdmann, P.S.; Wan, W.; Geimer, S.; Plitzko, J.M.; et al. In Situ Architecture of the Ciliary Base Reveals the Stepwise Assembly of Intraflagellar Transport Trains. Science2022, 377, 543–548. [Google Scholar] [CrossRef]
Meleppattu, S.; Zhou, H.; Dai, J.; Gui, M.; Brown, A. Mechanism of IFT-A Polymerization into Trains for Ciliary Transport. Cell2022, 185, 4986–4998.e12. [Google Scholar] [CrossRef]
Hesketh, S.J.; Mukhopadhyay, A.G.; Nakamura, D.; Toropova, K.; Roberts, A.J. IFT-A Structure Reveals Carriages for Membrane Protein Transport into Cilia. Cell2022, 185, 4971–4985.e16. [Google Scholar] [CrossRef]
Ma, Y.; He, J.; Li, S.; Yao, D.; Huang, C.; Wu, J.; Lei, M. Structural Insight into the Intraflagellar Transport Complex IFT-A and Its Assembly in the Anterograde IFT Train. Nat. Commun.2023, 14, 1506. [Google Scholar] [CrossRef]
Lacey, S.E.; Foster, H.E.; Pigino, G. The Molecular Structure of IFT-A and IFT-B in Anterograde Intraflagellar Transport Trains. Nat. Struct. Mol. Biol.2023, 30, 584–593. [Google Scholar] [CrossRef]
Petriman, N.A.; Loureiro-López, M.; Taschner, M.; Zacharia, N.K.; Georgieva, M.M.; Boegholm, N.; Wang, J.; Mourão, A.; Russell, R.B.; Andersen, J.S.; et al. Biochemically Validated Structural Model of the 15-subunit Intraflagellar Transport Complex IFT-B. EMBO J.2022, 41, e112440. [Google Scholar] [CrossRef]
Klink, B.U.; Gatsogiannis, C.; Hofnagel, O.; Wittinghofer, A.; Raunser, S. Structure of the Human BBSome Core Complex. eLife2020, 9, e53910. [Google Scholar] [CrossRef] [PubMed]
Singh, S.K.; Gui, M.; Koh, F.; Yip, M.C.; Brown, A. Structure and Activation Mechanism of the BBSome Membrane Protein Trafficking Complex. eLife2020, 9, e53322. [Google Scholar] [CrossRef] [PubMed]
Yang, S.; Bahl, K.; Chou, H.-T.; Woodsmith, J.; Stelzl, U.; Walz, T.; Nachury, M.V. Near-Atomic Structures of the BBSome Reveal the Basis for BBSome Activation and Binding to GPCR Cargoes. eLife2020, 9, e55954. [Google Scholar] [CrossRef]
Tian, X.; Zhao, H.; Zhou, J. Organization, Functions, and Mechanisms of the BBSome in Development, Ciliopathies, and Beyond. eLife2023, 12, e87623. [Google Scholar] [CrossRef]
Jordan, M.A.; Diener, D.R.; Stepanek, L.; Pigino, G. The Cryo-EM Structure of Intraflagellar Transport Trains Reveals How Dynein Is Inactivated to Ensure Unidirectional Anterograde Movement in Cilia. Nat. Cell Biol.2018, 20, 1250–1255. [Google Scholar] [CrossRef]
Toropova, K.; Mladenov, M.; Roberts, A.J. Intraflagellar Transport Dynein Is Autoinhibited by Trapping of Its Mechanical and Track-Binding Elements. Nat. Struct. Mol. Biol.2017, 24, 461–468. [Google Scholar] [CrossRef]
Gonçalves-Santos, F.; De-Castro, A.R.G.; Rodrigues, D.R.M.; De-Castro, M.J.G.; Gassmann, R.; Abreu, C.M.C.; Dantas, T.J. Hot-Wiring Dynein-2 Establishes Roles for IFT-A in Retrograde Train Assembly and Motility. Cell Rep.2023, 42, 113337. [Google Scholar] [CrossRef]
Hiyamizu, S.; Qiu, H.; Vuolo, L.; Stevenson, N.L.; Shak, C.; Heesom, K.J.; Hamada, Y.; Tsurumi, Y.; Chiba, S.; Katoh, Y.; et al. Multiple Interactions of the Dynein-2 Complex with the IFT-B Complex Are Required for Effective Intraflagellar Transport. J. Cell Sci.2023, 136, jcs260462. [Google Scholar] [CrossRef]
Yan, L.; Yin, H.; Mi, Y.; Wu, Y.; Zheng, Y. Deficiency of Wdr60 and Wdr34 Cause Distinct Neural Tube Malformation Phenotypes in Early Embryos. Front. Cell Dev. Biol.2023, 11, 1084245. [Google Scholar] [CrossRef]
Chien, A.; Shih, S.M.; Bower, R.; Tritschler, D.; Porter, M.E.; Yildiz, A. Dynamics of the IFT Machinery at the Ciliary Tip. eLife2017, 6, e28606. [Google Scholar] [CrossRef]
Nievergelt, A.P.; Zykov, I.; Diener, D.; Chhatre, A.; Buchholz, T.-O.; Delling, M.; Diez, S.; Jug, F.; Štěpánek, L.; Pigino, G. Conversion of Anterograde into Retrograde Trains Is an Intrinsic Property of Intraflagellar Transport. Curr. Biol.2022, 32, 4071–4078.e4. [Google Scholar] [CrossRef]
Patel-King, R.S.; Gilberti, R.M.; Hom, E.F.Y.; King, S.M. WD60/FAP163 Is a Dynein Intermediate Chain Required for Retrograde Intraflagellar Transport in Cilia. MBoC2013, 24, 2668–2677. [Google Scholar] [CrossRef]
Rompolas, P.; Pedersen, L.B.; Patel-King, R.S.; King, S.M. Chlamydomonas FAP133 Is a Dynein Intermediate Chain Associated with the Retrograde Intraflagellar Transport Motor. J. Cell Sci.2007, 120, 3653–3665. [Google Scholar] [CrossRef]
De-Castro, A.R.G.; Rodrigues, D.R.M.; De-Castro, M.J.G.; Vieira, N.; Vieira, C.; Carvalho, A.X.; Gassmann, R.; Abreu, C.M.C.; Dantas, T.J. WDR60-Mediated Dynein-2 Loading into Cilia Powers Retrograde IFT and Transition Zone Crossing. J. Cell Biol.2022, 221, e202010178. [Google Scholar] [CrossRef]
Vuolo, L.; Stevenson, N.L.; Heesom, K.J.; Stephens, D.J. Dynein-2 Intermediate Chains Play Crucial but Distinct Roles in Primary Cilia Formation and Function. eLife2018, 7, e39655. [Google Scholar] [CrossRef]
Higashida, M.; Niwa, S. Dynein Intermediate Chains DYCI-1 and WDR-60 Have Specific Functions in Caenorhabditis Elegans. Genes Cells2023, 28, 97–110. [Google Scholar] [CrossRef]
Grissom, P.M.; Vaisberg, E.A.; McIntosh, J.R. Identification of a Novel Light Intermediate Chain (D2LIC) for Mammalian Cytoplasmic Dynein 2. Mol. Biol. Cell2002, 13, 817–829. [Google Scholar] [CrossRef]
Mikami, A.; Tynan, S.H.; Hama, T.; Luby-Phelps, K.; Saito, T.; Crandall, J.E.; Besharse, J.C.; Vallee, R.B. Molecular Structure of Cytoplasmic Dynein 2 and Its Distribution in Neuronal and Ciliated Cells. J. Cell Sci.2002, 115, 4801–4808. [Google Scholar] [CrossRef]
Zhu, X.; Wang, J.; Li, S.; Lechtreck, K.; Pan, J. IFT54 Directly Interacts with kinesin-II and IFT Dynein to Regulate Anterograde Intraflagellar Transport. EMBO J.2021, 40, e105781. [Google Scholar] [CrossRef]
Taylor, S.P.; Dantas, T.J.; Duran, I.; Wu, S.; Lachman, R.S.; University of Washington Center for Mendelian Genomics Consortium; Bamshad, M.J.; Shendure, J.; Nickerson, D.A.; Nelson, S.F.; et al. Mutations in DYNC2LI1 Disrupt Cilia Function and Cause Short Rib Polydactyly Syndrome. Nat. Commun.2015, 6, 7092. [Google Scholar] [CrossRef]
Hamada, Y.; Tsurumi, Y.; Nozaki, S.; Katoh, Y.; Nakayama, K. Interaction of WDR60 Intermediate Chain with TCTEX1D2 Light Chain of the Dynein-2 Complex Is Crucial for Ciliary Protein Trafficking. Mol. Biol. Cell2018, 29, 1628–1639. [Google Scholar] [CrossRef]
Tsurumi, Y.; Hamada, Y.; Katoh, Y.; Nakayama, K. Interactions of the Dynein-2 Intermediate Chain WDR34 with the Light Chains Are Required for Ciliary Retrograde Protein Trafficking. Mol. Biol. Cell2019, 30, 658–670. [Google Scholar] [CrossRef]
Ringers, C.; Olstad, E.W.; Jurisch-Yaksi, N. The Role of Motile Cilia in the Development and Physiology of the Nervous System. Philos. Trans. R. Soc. Lond. B Biol. Sci.2020, 375, 20190156. [Google Scholar] [CrossRef]
Legendre, M.; Zaragosi, L.-E.; Mitchison, H.M. Motile Cilia and Airway Disease. Semin. Cell Dev. Biol.2021, 110, 19–33. [Google Scholar] [CrossRef]
Sironen, A.; Shoemark, A.; Patel, M.; Loebinger, M.R.; Mitchison, H.M. Sperm Defects in Primary Ciliary Dyskinesia and Related Causes of Male Infertility. Cell. Mol. Life Sci.2020, 77, 2029–2048. [Google Scholar] [CrossRef]
Antony, D.; Brunner, H.G.; Schmidts, M. Ciliary Dyneins and Dynein Related Ciliopathies. Cells2021, 10, 1885. [Google Scholar] [CrossRef]
Osinka, A.; Poprzeczko, M.; Zielinska, M.M.; Fabczak, H.; Joachimiak, E.; Wloga, D. Ciliary Proteins: Filling the Gaps. Recent Advances in Deciphering the Protein Composition of Motile Ciliary Complexes. Cells2019, 8, 730. [Google Scholar] [CrossRef]
Chen, Z.; Greenan, G.A.; Shiozaki, M.; Liu, Y.; Skinner, W.M.; Zhao, X.; Zhao, S.; Yan, R.; Yu, Z.; Lishko, P.V.; et al. In Situ Cryo-Electron Tomography Reveals the Asymmetric Architecture of Mammalian Sperm Axonemes. Nat. Struct. Mol. Biol.2023, 30, 360–369. [Google Scholar] [CrossRef]
Walton, T.; Wu, H.; Brown, A. Structure of a Microtubule-Bound Axonemal Dynein. Nat. Commun.2021, 12, 477. [Google Scholar] [CrossRef]
Kubo, S.; Yang, S.K.; Black, C.S.; Dai, D.; Valente-Paterno, M.; Gaertig, J.; Ichikawa, M.; Bui, K.H. Remodeling and Activation Mechanisms of Outer Arm Dyneins Revealed by cryo-EM. EMBO Rep.2021, 22, e52911. [Google Scholar] [CrossRef]
Ma, M.; Stoyanova, M.; Rademacher, G.; Dutcher, S.K.; Brown, A.; Zhang, R. Structure of the Decorated Ciliary Doublet Microtubule. Cell2019, 179, 909–922.e12. [Google Scholar] [CrossRef]
Gui, M.; Farley, H.; Anujan, P.; Anderson, J.R.; Maxwell, D.W.; Whitchurch, J.B.; Botsch, J.J.; Qiu, T.; Meleppattu, S.; Singh, S.K.; et al. De Novo Identification of Mammalian Ciliary Motility Proteins Using Cryo-EM. Cell2021, 184, 5791–5806.e19. [Google Scholar] [CrossRef]
Kubo, S.; Black, C.S.; Joachimiak, E.; Yang, S.K.; Legal, T.; Peri, K.; Khalifa, A.A.Z.; Ghanaeian, A.; McCafferty, C.L.; Valente-Paterno, M.; et al. Native Doublet Microtubules from Tetrahymena Thermophila Reveal the Importance of Outer Junction Proteins. Nat. Commun.2023, 14, 2168. [Google Scholar] [CrossRef] [PubMed]
Ghanaeian, A.; Majhi, S.; McCafferty, C.L.; Nami, B.; Black, C.S.; Yang, S.K.; Legal, T.; Papoulas, O.; Janowska, M.; Valente-Paterno, M.; et al. Integrated Modeling of the Nexin-Dynein Regulatory Complex Reveals Its Regulatory Mechanism. Nat. Commun.2023, 14, 5741. [Google Scholar] [CrossRef]
Gui, M.; Ma, M.; Sze-Tu, E.; Wang, X.; Koh, F.; Zhong, E.D.; Berger, B.; Davis, J.H.; Dutcher, S.K.; Zhang, R.; et al. Structures of Radial Spokes and Associated Complexes Important for Ciliary Motility. Nat. Struct. Mol. Biol.2021, 28, 29–37. [Google Scholar] [CrossRef]
Grossman-Haham, I.; Coudray, N.; Yu, Z.; Wang, F.; Zhang, N.; Bhabha, G.; Vale, R.D. Structure of the Radial Spoke Head and Insights into Its Role in Mechanoregulation of Ciliary Beating. Nat. Struct. Mol. Biol.2021, 28, 20–28. [Google Scholar] [CrossRef]
Zheng, W.; Li, F.; Ding, Z.; Liu, H.; Zhu, L.; Xu, C.; Li, J.; Gao, Q.; Wang, Y.; Fu, Z.; et al. Distinct Architecture and Composition of Mouse Axonemal Radial Spoke Head Revealed by Cryo-EM. Proc. Natl. Acad. Sci. USA2021, 118, e2021180118. [Google Scholar] [CrossRef]
Han, L.; Rao, Q.; Yang, R.; Wang, Y.; Chai, P.; Xiong, Y.; Zhang, K. Cryo-EM Structure of an Active Central Apparatus. Nat. Struct. Mol. Biol.2022, 29, 472–482. [Google Scholar] [CrossRef]
Gui, M.; Wang, X.; Dutcher, S.K.; Brown, A.; Zhang, R. Ciliary Central Apparatus Structure Reveals Mechanisms of Microtubule Patterning. Nat. Struct. Mol. Biol.2022, 29, 483–492. [Google Scholar] [CrossRef]
Silflow, C.D.; Lefebvre, P.A. Assembly and Motility of Eukaryotic Cilia and Flagella. Lessons from Chlamydomonas Reinhardtii. Plant Physiol.2001, 127, 1500–1507. [Google Scholar] [CrossRef]
Harris, E.H. Chlamydomonas as a Model System. Annu. Rev. Plant Physiol. Plant Mol. Biol.2001, 52, 363–406. [Google Scholar] [CrossRef]
Bayless, B.A.; Navarro, F.M.; Winey, M. Motile Cilia: Innovation and Insight from Ciliate Model Organisms. Front. Cell Dev. Biol.2019, 7, 265. [Google Scholar] [CrossRef]
Leung, M.R.; Roelofs, M.C.; Ravi, R.T.; Maitan, P.; Henning, H.; Zhang, M.; Bromfield, E.G.; Howes, S.C.; Gadella, B.M.; Bloomfield-Gadêlha, H.; et al. The Multi-scale Architecture of Mammalian Sperm Flagella and Implications for Ciliary Motility. EMBO J.2021, 40, e107410. [Google Scholar] [CrossRef]
Wirschell, M.; Yang, C.; Yang, P.; Fox, L.; Yanagisawa, H.; Kamiya, R.; Witman, G.B.; Porter, M.E.; Sale, W.S. IC97 Is a Novel Intermediate Chain of I1 Dynein That Interacts with Tubulin and Regulates Interdoublet Sliding. Mol. Biol. Cell2009, 20, 3044–3054. [Google Scholar] [CrossRef]
King, S.M. Composition and Assembly of Axonemal Dyneins. In Dyneins; Elsevier: Amsterdam, The Netherlands, 2018; pp. 162–201. ISBN 978-0-12-809471-6. [Google Scholar]
Fu, G.; Scarbrough, C.; Song, K.; Phan, N.; Wirschell, M.; Nicastro, D. Structural Organization of the Intermediate and Light Chain Complex of Chlamydomonas Ciliary I1 Dynein. FASEB J.2021, 35, e21646. [Google Scholar] [CrossRef]
Wu, H.; Maciejewski, M.W.; Marintchev, A.; Benashski, S.E.; Mullen, G.P.; King, S.M. Solution Structure of a Dynein Motor Domain Associated Light Chain. Nat. Struct. Biol.2000, 7, 575–579. [Google Scholar] [CrossRef]
Ichikawa, M.; Saito, K.; Yanagisawa, H.; Yagi, T.; Kamiya, R.; Yamaguchi, S.; Yajima, J.; Kushida, Y.; Nakano, K.; Numata, O.; et al. Axonemal Dynein Light Chain-1 Locates at the Microtubule-Binding Domain of the γ Heavy Chain. Mol. Biol. Cell2015, 26, 4236–4247. [Google Scholar] [CrossRef]
Toda, A.; Nishikawa, Y.; Tanaka, H.; Yagi, T.; Kurisu, G. The Complex of Outer-Arm Dynein Light Chain-1 and the Microtubule-Binding Domain of the γ Heavy Chain Shows How Axonemal Dynein Tunes Ciliary Beating. J. Biol. Chem.2020, 295, 3982–3989. [Google Scholar] [CrossRef]
Sakato-Antoku, M.; King, S.M. Outer-Arm Dynein Light Chain LC1 Is Required for Normal Motor Assembly Kinetics, Ciliary Stability, and Motility. Mol. Biol. Cell2023, 34, ar75. [Google Scholar] [CrossRef]
Oberacker, T.; Kraft, L.; Schanz, M.; Latus, J.; Schricker, S. The Importance of Thioredoxin-1 in Health and Disease. Antioxidants2023, 12, 1078. [Google Scholar] [CrossRef]
Wakabayashi, K.; King, S.M. Modulation of Chlamydomonas Reinhardtii Flagellar Motility by Redox Poise. J. Cell Biol.2006, 173, 743–754. [Google Scholar] [CrossRef]
Wakabayashi, K.; Misawa, Y.; Mochiji, S.; Kamiya, R. Reduction-Oxidation Poise Regulates the Sign of Phototaxis in Chlamydomonas Reinhardtii. Proc. Natl. Acad. Sci. USA2011, 108, 11280–11284. [Google Scholar] [CrossRef]
Harrison, A.; Sakato, M.; Tedford, H.W.; Benashski, S.E.; Patel-King, R.S.; King, S.M. Redox-Based Control of the Gamma Heavy Chain ATPase from Chlamydomonas Outer Arm Dynein. Cell Motil. Cytoskelet.2002, 52, 131–143. [Google Scholar] [CrossRef]
Patel-King, R.S.; Benashski, S.E.; Harrison, A.; King, S.M. Two Functional Thioredoxins Containg Redox-Senesitive Vicinal Dithiols from the Chlamydomonas Outer Dynein Arm. J. Biol. Chem.1996, 271, 6283–6291. [Google Scholar] [CrossRef]
Gilles, A.M.; Presecan, E.; Vonica, A.; Lascu, I. Nucleoside Diphosphate Kinase from Human Erythrocytes. Structural Characterization of the Two Polypeptide Chains Responsible for Heterogeneity of the Hexameric Enzyme. J. Biol. Chem.1991, 266, 8784–8789. [Google Scholar] [CrossRef]
King, S.M.; Patel-King, R.S. Identification of a Ca 2+-Binding Light Chain within Chlamydomonas Outer Arm Dynein. J. Cell Sci.1995, 108, 3757–3764. [Google Scholar] [CrossRef]
Sakato, M.; Sakakibara, H.; King, S.M. Chlamydomonas Outer Arm Dynein Alters Conformation in Response to Ca 2+. Mol. Biol. Cell2007, 18, 3620–3634. [Google Scholar] [CrossRef]
LeDizet, M.; Piperno, G. The Light Chain P28 Associates with a Subset of Inner Dynein Arm Heavy Chains in Chlamydomonas Axonemes. Mol. Biol. Cell1995, 6, 697–711. [Google Scholar] [CrossRef]
King, S.M. (Ed.) Dyneins: Structure, Biology and Disease—Dynein Structure, Mechanics, and Disease, 2nd ed.; Academic Press: Cambridge, MA, USA, 2018; Volume 2, ISBN 978-0-12-809470-9. [Google Scholar]
Marzo, M.G.; Griswold, J.M.; Ruff, K.M.; Buchmeier, R.E.; Fees, C.P.; Markus, S.M. Molecular Basis for Dyneinopathies Reveals Insight into Dynein Regulation and Dysfunction. eLife2019, 8, e47246. [Google Scholar] [CrossRef]
Becker, L.-L.; Dafsari, H.S.; Schallner, J.; Abdin, D.; Seifert, M.; Petit, F.; Smol, T.; Bok, L.; Rodan, L.; Krapels, I.; et al. The Clinical-Phenotype Continuum in DYNC1H1-Related Disorders—Genomic Profiling and Proposal for a Novel Classification. J. Hum. Genet.2020, 65, 1003–1017. [Google Scholar] [CrossRef]
Figure 1. Structures of dynein complexes. Subunits are denoted by colored text. (a) Human cytoplasmic dynein-1 complex in an autoinhibited conformation (PDB 5NVU). (b) Human cytoplasmic dynein-2 complex in an autoinhibited conformation (PDB 6SC2). (c) Tetrahymena axonemal outer-arm dynein complex in an inhibited conformation imposed by Shulin (PDB 6ZYW). (d) Human dynein–dynactin–BicDR1 complex with two dyneins and two BicDR1 adaptors (PDB 7Z8F). (e) Tetrahymena axonemal outer-arm dynein complex bound to MTs in one of the two MT-binding states, with the central HC’s MTBD aligned with the outermost HC’s MTBD (PDB 7KEK).
Figure 2. The IC-LC subcomplex in autoinhibited dynein complexes. The subunits are denoted by colored text. The structures are aligned horizontally at the LC Roadblock (in light blue) to allow for comparison of different dyneins. Notably, while LC8 and Tctex are approximately 20 nm away from the WD repeats of the IC in dynein-1, they are positioned much closer to the WD repeats in both dynein-2 and axonemal OAD. (a) IC-LC subcomplex of human dynein-1 (PDB 5NVU). (b) IC-LC subcomplex of human dynein-2 (PDB 6SC2). (c) IC-LC subcomplex of Tetrahymena OAD (PDB 6ZYW).
Figure 3. IC as the central hub for Nde(Nde1/Ndel1)/Lis1-mediated dynein-1 activation. (a) The structure of IC2’s SAH and H2 (Uniprot Q13409, amino acids 1–67) predicated by ColabFold to interact with Nde1 (Uniprot Q9NXR1, amino acids 1–189) and Lis1 (Uniprot P43034, full-length). Predictions were performed using ColabFold v1.5.5, employing AlphaFold2 with MMseqs2, and conducted without a template. (b) The structure of IC2’s SAH and H2 (Uniprot Q13409, amino acids 1–67) predicated by ColabFold to interact with p150 Glued’s CC1B (Uniprot Q14203, amino acids 357–589). (c) A proposed mechanism illustrating how IC functions as the central hub in dynein-1 activation, with further details described in the main text.
Figure 4. Comparison of human dynein IC and LIC subunits. (a) Comparison of the ICs containing the WD repeats. WD repeats are shown in grey, the Roadblock binding site is in blue, the LC8 binding site is in orange, the Tctex binding site is in pink, SAH is in green, and H2 is in light green. The binding sites are assigned based on cryo-EM structures: DC1I2 : PDB 5NVU; DC2I1 and DC2I2 : PDB 6SC2; DNAI1, DNAI2, DNAI3, and DNAI4 : PDB 8J07. (b) Comparison of the LICs. The Ras-like domain is shown in grey; helix-1 is in green; and helix-2 is in light green. These regions are assigned based on AlphaFold structures deposited on Uniprot.
Figure 5. (a) A zoom-in view of the LIC binding to the HC. Inset: The DDR complex (PDB 7Z8F); the circle indicates the zoom-in region. (b) ColabFold-predicted structures of LIC1 helix-1 interacting with various adaptors. These predictions were made in the same manner as those in Figure 3. The sequences analyzed include LIC1 helix-1 (Uniprot Q9Y6G9-1, amino acids 424–459); BicD2 (Uniprot Q8TD16-1, amino acids 1–240); JIP3 (Uniprot Q9UPT6-1, amino acids 1–180); Hook1 (Uniprot Q9UJC3-1, amino acids 1–239); CRACR2A (Uniprot Q9BSW2-2, amino acids 1–240). The adaptors are depicted in grey, and LIC1 helix-1 is in gold. The crystal structures of BicD2 (PDB 6PSE), Hook3 (PDB 6B9H), and CRACR2A (PDB 6PSD) with LIC1 helix-1 (blue) were structurally aligned to the predicted structures, using UCSF Chimera .
Figure 6. Structures of dynein LCs. (a) Solution structure of dimeric human Roadblock-1 (PDB 1Z09). (b) Crystal structure of dimeric Drosophila melanogaster LC8 (PDB 3BRI). (c) Crystal structure of dimeric Drosophila Tctex-1 (PDB 1YGT). (d) Crystal structure of Chlamydomonas LC1 (PDB 5YXM). (e) AlphaFold-predicted human TXND6 structure. Dark grey is for the thioredoxin domain, and light grey is for the nucleoside diphosphate kinase domain. The disordered N-terminal amino acids 1–10 and C-terminal amino acids 301–330 were removed for clarity. Blue indicates the predicted structure of human thioredoxin (Uniprot P10599), which was structurally aligned to TXND6, using UCSF Chimera . LC3 and LC5 of Chlamydomonas and LC3BL of Tetrahymena [9,45] have similar folds (not shown). The two cysteine residues in the -Cys-X-X-Cys- motif are highlighted by sulfur atoms (yellow) (Cys39 and Cys42 in TXND6; Cys32 and Cys35 in thioredoxin). Pink indicates the predicted structure of human nucleoside diphosphate kinase A (NDKA, Uniprot P15531), with amino acids 136–152 removed for clear depiction, and structurally aligned to TXND6. The conserved histidine residue is highlighted by the nitrogen atom (blue) (His279 in TXND6; His118 in NDKA). (f) AlphaFold-predicted structure of Chlamydomonas LC4 (Uniprot Q39584).
Figure 7. Intraflagellar transport. (Top) Side (left) and front (right) views of an autoinhibited human dynein-2 complex docked into a cryo-EM map of Chlamydomonas anterograde IFT complexes [8,204] (PDB 6SC2; EMD-4303). The colored text indicates the subunits. (Bottom, left) A cross-sectional illustrative view of a 9 + 0 immotile cilium. (Bottom, right) Side view of the cilium showing dynein-2 in an autoinhibited state, transported as cargo by kinesin-2 in anterograde trains.
Figure 8. Human epithelial ciliary axoneme structure (PDB 8J07). (a) (Left): An illustration of the cross-section of the motile axoneme. Middle: The cross-section view of the structure of an MTD with associated protein complexes. (Right): The side view of this structure. (b) (Left): IAD f structure, with an additional IC (DNAI7) on the side of the IC-LC tower. (Right): OAD structure, with DNAL1 (LC1) binding to the MTBD of DYH5 and TXND6 (LC3) binding to the neck region of DYH9. (c) Top view of the IDAs and ODAs. The circle indicates the interaction between OAD’s LC Tctex heterodimer and IAD f’s DNAI3 (IC140) and HC tail.
Table 1. Composition of human dyneins’ non-catalytic subunits. The subunits of axonemal dyneins are based on the cryo-EM structure of human respiratory cilia and follow the consensus nomenclature for dynein subunits . It is important to note that the composition might vary depending on the cell types. For the composition of axonemal dyneins in Chlamydomonas reinhardtii, see ref. . For the composition of axonemal outer-arm dyneins in Tetrahymena thermophila, see refs. [9,45].
| Gene | Protein | Dynein Family | Uniprot Entry | Length (aa) |
:---: :---:
| IC | | | | |
| DYNC1I1 | Cytoplasmic dynein 1 intermediate chain 1 (DC1I1, or IC1) 1 | Dynein-1 | O14576 | 645 |
| DYNC1I2 | Cytoplasmic dynein 1 intermediate chain 2 (DC1I2, or IC2) | Dynein-1 | Q13409 | 638 |
| DYNC2I1 | Cytoplasmic dynein 2 intermediate chain 1 (DC2I1, or WDR60) | Dynein-2 | Q8WVS4 | 1066 |
| DYNC2I2 | Cytoplasmic dynein 2 intermediate chain 2 (DC2I2, or WDR34) | Dynein-2 | Q96EX3 | 536 |
| DNAI1 | Dynein axonemal intermediate chain 1 (DNAI1) | OAD | Q9UI46 | 699 |
| DNAI2 | Dynein axonemal intermediate chain 2 (DNAI2) | OAD | Q9GZS0 | 605 |
| DNAI3 | Dynein axonemal intermediate chain 3 (DNAI3, or WDR78, IC140) | IAD f | Q8IWG1 | 891 |
| DNAI4 | Dynein axonemal intermediate chain 4 (DNAI4, or WDR63, IC138) | IAD f | Q5VTH9 | 848 |
| DNAI7 | Dynein axonemal intermediate chain 7 (DNAI7, or Las1, IC97) 2 | IAD f | Q6TDU7 | 716 |
| LIC | | | | |
| DYNC1LI1 | Cytoplasmic dynein 1 light intermediate chain 1 (DC1L1, or LIC1) | Dynein-1 | Q9Y6G9 | 523 |
| DYNC1LI2 | Cytoplasmic dynein 1 light intermediate chain 2 (DC1L2, or LIC2) | Dynein-1 | O43237 | 492 |
| DYNC2LI1 | Cytoplasmic dynein 2 light intermediate chain 1 (DC2L1, or LIC3) | Dynein-2 | Q8TCX1 | 351 |
| LC | | | | |
| DYNLT1 | Dynein light chain Tctex-type 1 (DYLT1, or Tctex-1) | Dynein-1 Dynein-2 OAD & IAD f | P63172 | 113 |
| DYNLT2B | Dynein light chain Tctex-type protein 2B (DYT2B, or Tctex1D2) | Dynein-2 OAD & IAD f | Q8WW35 | 142 |
| DYNLT3 | Dynein light chain Tctex-type 3 (DYLT3, or Tctex-3) | Dynein-1 Dynein-2 | P51808 | 116 |
| DYNLL1 | Dynein light chain 1, cytoplasmic (DYL1, or LC8-1) | Dynein-1 Dynein-2 OAD & IAD f | P63167 | 89 |
| DYNLL2 | Dynein light chain 2, cytoplasmic (DYL2, or LC8-2) | Dynein-1 Dynein-2 OAD & IAD f | Q96FJ2 | 89 |
| DNAL4 | Dynein axonemal light chain 4 (DNAL4) | OAD | O96015 | 105 |
| DYNLRB1 | Dynein light chain roadblock-type 1 (DLRB1, or Roadblock-1) | Dynein-1 Dynein-2 OAD & IAD f | Q9NP97 | 96 |
| DYNLRB2 | Dynein light chain roadblock-type 2 (DLRB2, or Roadblock-2) | Dynein-1 Dynein-2 OAD & IAD f | Q8TF09 | 96 |
| DNAL1 | Dynein axonemal light chain 1 (DNAL1, or LC1) | OAD | Q4LDG9 | 190 |
| NME9 | Thioredoxin domain-containing protein 6 (TXND6, or LC3) | OAD | Q86XW9 | 330 |
| DNALI1 | Axonemal dynein light intermediate polypeptide 1 (IDLC, or p28) | IAD | O14645 | 258 |
1 The abbreviated name and alternative names commonly used are included in parentheses. 2 This IC does not have WD repeats and binds differently compared to other ICs.
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Rao, L.; Gennerich, A. Structure and Function of Dynein’s Non-Catalytic Subunits. Cells2024, 13, 330.
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Rao, L.; Gennerich, A. Structure and Function of Dynein’s Non-Catalytic Subunits. Cells2024, 13, 330.
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Rao L, Gennerich A. Structure and Function of Dynein’s Non-Catalytic Subunits. Cells. 2024; 13(4):330.
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Rao, Lu, and Arne Gennerich. 2024. "Structure and Function of Dynein’s Non-Catalytic Subunits" Cells 13, no. 4: 330.
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11605 | https://www.youtube.com/watch?v=lCvVM4SIm0M | Prove that ln(a) + ln(b) = ln(ab)
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Posted: 29 Dec 2023
In this video, we will use integral calculus to prove a logarithmic property which says ln(a) + ln(b) = ln(ab). As you watch, do not forget to leave us a like and a comment
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Transcript:
hello guys welcome once more to another video in my YouTube channel so in today's video we are going to be proving one of the logarithmic laws which is written clearly to you we are going to prove that Lan a plus Lan B is equal to the Lan of ab so stay tuned as you watch don't forget to like share and subscribe to the channel also comment on the video so that we can produce more interesting videos for you to come and watch so let's get started all right so to prove this I'm going to recall some basic concepts of calculus to you guys so we can better understand what we are going to be doing in the proof all right so um if I consider a graph for example on the XY plane that's a curve of F and then I consider um the lines xal to a and xal to B now you realize that the area bounded by the graph of F and these lines can be defined as a definite integral so that area bounded the area under the curve of f bounded by the lines xal to a and xal to B is simply defined by the area a equal to the integral from A to B of f of x x x for example if I ask you to find the area bounded by the lines x = 0 and x = 4 and the c y = to X first of all you're going to display the curve to me okay that is you draw the you draw the line Y = X then now x = 0 is simply the origin and x = 4 that is the line x = 4 so we need to find the area of the portion of this portion so using the definition of that area we are going to say the area is equal to the integral from 0 to 4 of the function that is x when we integrate x we have x² on two we fit our boundaries 0 to four we just need to fit the upper bound because when we fit zero the lower bound the lower um when you fit the lower bound the resulting integral the resulting function is going to give you zero so just have 4^ 2 / 2 which is 16 /2 which is 8 square units you could also interpret this area as the area of this right angle triangle have base whose area is have base time height because you can see a right angle triangle here right the base is 4 units and the height is simply the y coordinate of the y coordinate when X is equal to 4 when X is 4 Y is equal to 4 because we have the line yal to X X and Y have the same coordinates have yeah the same the same values at every point so that is four so the area is going to give you half basee the base is four and the height is also four so you're going to have half of 4 4 which is still 4 4 is 16 / 2 is 8 square units so this is the basic concept we are going to be using to prove that the Lan of a plus the Lan of B is equal to the Lan of ab so let's get straight to proving that all right so to prove that the Lan of a plus the Lan of B is equal the Lan of ab we are going to use integration and we are going to consider a function that when we integrate that function we have the L of X we know that function is 1 /x by definition when we integrate 1 /x you get the Lan of the absolute value of x but definitely we are going to be working in a positive interval since for the Lan of a and the Lan of B to be defined A and B are all supposed to be positive even if um anyways so um we have to draw the graph of 1 /x in the positive interval it is simply the graph like this okay the horizontal axis is a horizontal asmt and the vertical axis is a vertical ASM toote so here is a graph of 1 /x that is f in the um in the positive values in the positive interval now after doing this we now consider maybe I just consider three values of X we have one we have a and then we have a it's clear that since A and B bong in this case for for the for for for me to make it convenient to work with I'm going to consider a to be greater than one okay I'm I'm going to consider a to be greater than one and B also to greater than one so if a is greater than one and B is greater than one it means a is definitely greater than a all right so we clearly see that this portion here the the the area of this portion that is the area bounded by the curve of f the lines xal to 1 and xal to AB is simply the area is simply the area of this potion okay also you realize that for the area of this portion is simply the area of the um of this curve bounded the area under this curve bounded by the lines x = 1 x = a and the x axis specifically so we are going to define those areas as integrals and from here you can also see that the area of this this big portion okay the area of this big portion here is equal to the area of this portion plus the area of this portion so that's what we are going to use so the area of this big portion like we said is the integral from 1 to AB of 1 /x DX which is equal to the area of this first portion here which is the integral from 1 to a of 1 /x plus the integral plus the area of this other portion which is the integral from a to a of 1 /x now uh um when I integrate 1 / X I get l x since X is already in a positive interval I need now to fit the limits from 1 to a when I integrate 1 / X I still have l x from 1 to a and here l x from a to AB if we fit the upper limit we have the Lan of ab when we F the lower limit the Lan of one is zero so this just gives us the Lan of ab now this is going to give you the Lan of a minus the Lan of one but the Lan of a is Lan of a and the L of one is Z Plus now um the Plus in fact I just I just rewrote back this integral that is the integral from a to a of 1 /x DX now we are going to see how to actually integrate this okay now how do we integrate this we are going to do a substitution we are going to let X to be equal to a u u is a variable with differenti both sides we have DX equal to a du since a is a constant and U is a variable now we need to to change the limit when X begins from a to AB so when X is a our U is equal to 1 because our U is simply x/ a so if x is a our U is going to be a/ a which is 1 and when X is AB our U will be AB over a which is B so our U will begin from 1 to B therefore means that this integral here which we need to find is is equal to the integral when X is a our U is 1 and when X is AB our U is B now our 1 /x since our X is au we are going to have 1 / Au then our DX is equal to a du the a the a are going to cancel and we are going to have that integral to the integral from 1 to B of 1 / U du if I integrate 1 / u i get l u but we are supposing that b is positive so the interval 1 to B is a positive interval therefore we just write um lineu without putting the absolute value when we fit the limits from one to B we are going to get Lin B okay Lin B minus Lin one but L one is zero so it just gives us Lan B therefore means that our L of ab is equal to Lan a plus Lan B so that is a proof in the next video we are going to be proving the Lan of a to the N is equal to n the L of a that is in the second video all right so thank you guys for watching make sure you like you share you subscribe to the channel and you comment on the video if you learn something new see you guys in the next video and stay blessed |
11606 | https://www.gauthmath.com/solution/1-lf-frac-x-y-frac-3-4-evaluate-frac-2x-y-2x-y--1703072204960774 | Solved: lf x/y = 3/4 , evaluate (2x-y)/2x+y . [Math]
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lf x/y = 3/4 , evaluate (2x-y)/2x+y .
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Answer: 1/5 1/5 1/5 . Explanation: Multiply both sides of the equation by the common denominator: (2x-y)/2x-y = (2x-y)/2x-y (x 4y)/4 -3 4y- (3 4y)/4 - (x 4y)/4 - 3y/4 Reduce fraction to the lowest term by canceling the greatest common factor: (2x-y)/2x-y = (2x-y)/2x-y 4x=3y Divide both sides of the equation by the coefficient of variable: 2x-y-2x-y-y=y-y-y x=3y/ 4 Rewrite as fraction: 2xy/2x+y = 2xy/2x+y 3y/4 = 3y/4 = 3y/4 Substitute: (4,4) Reduce the expression to the lowest term: (41,4 Find common denominator and write the numerators above common denominator: (4,4) Multiply the monomials: (4,4) Divide a fraction by multiplying its reciprocal: (3y-2y)/2 - (2y-3y-2y)/3y-2y =2 Combine like terms: y2/2 beginarrayr 2 5yendarray Reduce the expression to the lowest term: 1/5 1/5 1/5 1/5 Answer: 1/3 1/3 1/3 1/3 1/3
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11607 | https://www.cuemath.com/geometry/exterior-angle-of-triangle/ | LearnPracticeDownload
Exterior Angle of Triangle
The exterior angles of a triangle are those angles that are formed outside it. In other words, the exterior angle of a triangle is the angle that is formed between one of its sides and its adjacent extended side. Let us learn more about the exterior angle of triangle in this article.
| | |
--- |
| 1. | What is the Exterior Angle of Triangle? |
| 2. | Exterior Angle of Triangle Properties |
| 3. | Exterior Angle of Triangle Formula |
| 4. | FAQs on Exterior Angle of Triangle |
What is the Exterior Angle of Triangle?
When any side of a triangle is extended, the angle that is formed with this side and its adjacent side is called the exterior angle of a triangle. There are three exterior angles in a triangle. It should be noted that each exterior angle forms a linear pair with its corresponding interior angle. We know that the interior angle of a triangle is formed inside it where the sides meet at a vertex. Observe the following figure to distinguish between the interior angles and the exterior angles of a triangle.
We can see that each interior angle forms a linear pair with its corresponding exterior angle. This means that the sum of each exterior angle and its respective interior angle is equal to 180°.
Exterior Angle of Triangle Properties
There are three basic properties of the exterior angles of a triangle.
In a triangle, each exterior angle and its corresponding interior angle form a linear pair of angles. This means that the sum of the interior and exterior angle is equal to 180°.
The exterior angle of a triangle is equal to the sum of the two opposite interior angles (remote interior angles). This is also known as the Exterior Angle theorem.
The sum of all the exterior angles of a triangle is 360°.
Exterior Angle of Triangle Formula
Based on the properties of the exterior angles, the following formulas can be used to find the exterior angles of a triangle. Referring to the triangle given below, the formulas can be understood in a better way.
Each Exterior angle = 180 - Interior angle. Here, ∠e = 180 - ∠b. Similarly, ∠d = 180 - ∠c, and ∠f = 180 - ∠a
Exterior angle = Sum of Interior opposite angles. Here, ∠d = ∠a + ∠b, ∠f = ∠b + ∠c, and ∠e = ∠a + ∠c
Sum of the Exterior angles of a triangle = 360°. Here, ∠d + ∠e + ∠f = 360°
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Check out the links given below related to the exterior angles of a triangle.
Interior angles
Exterior Angles of a Polygon
Exterior Angle Theorem
Linear Pair of Angles
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Exterior Angle of Triangle
Exterior Angle of Triangle
Questions based on Exterior Angle of Triangle
Exterior Angle of Triangle Examples
Example 1: Find the value of the exterior angle in the triangle.
Solution: According to the property of the exterior angle of triangle, Exterior angle =Sum of Interior opposite angles. In this case, the exterior angle, ∠PRS = ∠RPQ + ∠PQR. Substituting the values in the formula, ∠PRS = 60° + 70° = 130°. Therefore, the unknown exterior angle, ∠PRS = 130°
2. Example 2: If one interior angle of a triangle is 56°, find the measure of its corresponding exterior angle.
Solution: According to the properties of the exterior angle of a triangle, each interior angle forms a linear pair with its respective exterior angle. This means, Exterior angle + Interior angle = 180°. In the question, one interior angle is given as 56°. Therefore, the corresponding exterior angle can be calculated using the formula: Each Exterior angle = 180° - Interior angle. Substituting the values in the formula, Exterior angle = 180 - 56 = 124°.
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What is the Exterior Angle of Triangle?
The exterior angle of a triangle is the angle that is formed with one side and the adjacent extended side of a triangle. There are 3 exterior angles in a triangle and the sum of the exterior angles of a triangle is always equal to 360°.
How to Find the Exterior Angle of Triangle?
The value of the exterior angle of a triangle can be calculated using various formulas depending on the other known angles. The following formulas can be used according to the given parameters.
Each Exterior angle = 180° - Interior angle. This formula can be used if the corresponding interior angle is given.
Exterior angle = Sum of Interior opposite angles. This formula can be used to find the exterior angle when its remote interior opposite angles are given.
The sum of all the exterior angles of a triangle is 360°. This formula can be used to find the unknown value of an exterior angle when the other two exterior angles are given.
What is the Sum of Exterior Angles of a Triangle?
The sum of the exterior angles of a triangle is always equal to 360°. This property applies to all polygons, which means that the sum of the exterior angles of all polygons is 360°.
How to find the Missing Exterior Angle of a Triangle?
A missing exterior angle of a triangle can be calculated using any of the following formulas. This depends on the angles that are given in the question.
The first formula to find the exterior angle can be used if the corresponding interior angle is given. Each Exterior angle = 180 - Interior angle.
The second formula can be used to find the exterior angle when its interior opposite angles are given. Exterior angle = Sum of Interior opposite angles.
The third formula can be used to find the unknown value of an exterior angle when the other two exterior angles are given. The sum of all the exterior angles of a triangle is 360°.
Are the Exterior Angles of a Triangle Equal to 360°?
Yes, the sum of the exterior angles of a triangle is always equal to 360°.
Are the Exterior Angles of a Triangle Always Obtuse?
No, the exterior angles of a triangle may not always be obtuse (more than 90°). However, the sum of all the three exterior angles should always be 360°. For example, if two exterior angles of a triangle are 165° (obtuse) and 141° (obtuse), the third one is 54° (acute).
What is the Measure of Each Exterior Angle of an Equilateral Triangle?
The measure of each exterior angle of an equilateral triangle is 120°. An equilateral triangle is a triangle in which all the sides are equal in length and all the 3 interior angles are of equal measure. This means each interior angle of an equilateral triangle is 60° because the sum of the interior angles is 180°. Now, if each interior angle of an equilateral triangle is 60°, its corresponding exterior angle will be 120°. This is because in a triangle, the exterior angle and its corresponding interior angle form a linear pair of angles. This means that the sum of the interior and exterior angle is equal to 180°.
What is the Exterior Angle Theorem of a Triangle?
According to the exterior angle theorem, the measure of an exterior angle is equal to the sum of the interior opposite angles (remote interior angles). This means if we need to find the exterior angle of a triangle, and its remote interior angles are known, then the value of the exterior angle will be the sum of those two interior opposite angles.
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11608 | https://cm-to-m.appspot.com/2-cm-to-m.html | 2 Centimeters To Meters Converter | 2 cm To m Converter
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How to convert 2 centimeters to meters?
2 cm 0.01 m=0.02 m
1 cm
A common question isHow many centimeter in 2 meter?And the answer is 200.0 cm in 2 m. Likewise the question how many meter in 2 centimeter has the answer of 0.02 m in 2 cm.
How much are 2 centimeters in meters?
2 centimeters equal 0.02 meters (2cm = 0.02m). Converting 2 cm to m is easy. Simply use our calculator above, or apply the formula to change the length 2 cm to m.
Convert 2 cm to common lengths
| Unit | Unit of length |
--- |
| Nanometer | 20000000.0 nm |
| Micrometer | 20000.0 µm |
| Millimeter | 20.0 mm |
| Centimeter | 2.0 cm |
| Inch | 0.7874015748 in |
| Foot | 0.0656167979 ft |
| Yard | 0.021872266 yd |
| Meter | 0.02 m |
| Kilometer | 2e-05 km |
| Mile | 1.24274e-05 mi |
| Nautical mile | 1.07991e-05 nmi |
What is 2 centimeters in m?
To convert 2 cm to m multiply the length in centimeters by 0.01. The 2 cm in m formula is [m] = 2 0.01. Thus, for 2 centimeters in meter we get 0.02 m.
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11609 | https://math.stackexchange.com/questions/615564/positive-real-root-of-the-equation-using-horners-method | math software - positive real root of the equation using Horner's Method - Mathematics Stack Exchange
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positive real root of the equation using Horner's Method
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obtain a positive real root of the equation
x 3−3 x 2+2.5=0 x 3−3 x 2+2.5=0
that lines between 1 1 and 2 2 using Horner's method.
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edited Dec 22, 2013 at 18:16
Amzoti
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asked Dec 22, 2013 at 13:06
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Was the answer helpful?Amzoti –Amzoti 2013-12-23 02:08:05 +00:00 Commented Dec 23, 2013 at 2:08
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Use Horner's Method to find a positive real root of:
x 3−3 x 2+2.5=0(1)(1)x 3−3 x 2+2.5=0
We re-write (1)(1) as:
2 x 3−6 x 2+5=0(2)(2)2 x 3−6 x 2+5=0
I will assume you have access to Horner's algorithm and the intermediate values below are from the synthetic division steps.
x 0=1,a 3=2,a 2=−6,a 1=0,a 0=5 x 0=1,a 3=2,a 2=−6,a 1=0,a 0=5
b 3=2,b 2=−4,b 1=−4,b 0=1=P(1)b 3=2,b 2=−4,b 1=−4,b 0=1=P(1)
b′2=2,b′1=−2,b′0=−6=Q(1)=P′(1)b 2′=2,b 1′=−2,b 0′=−6=Q(1)=P′(1)
x 1=x 0−P(x 0)P′(x 0)=1−1−6=7 6=1.1666 x 1=x 0−P(x 0)P′(x 0)=1−1−6=7 6=1.1666
Repeat process with x 1=1.1666 x 1=1.1666, so we get:
x 2=x 1−P(x 1)P′(x 1)=1.1666−0.0081−6.80214=1.16779 x 2=x 1−P(x 1)P′(x 1)=1.1666−0.0081−6.80214=1.16779
Repeat this process a couple of more times to improve the accuracy since it is easy to do by hand.
The actual root (we already have two digit accuracy from above) for comparison purposes is:
x=1.16825440178103 x=1.16825440178103
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answered Dec 22, 2013 at 14:56
AmzotiAmzoti
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11610 | https://www.amazon.in/Mengshen-Door-Window-Alarm-Wireless/dp/B083FCZW3Q | Mengshen Wireless Light Sound Alarm Magnetic Sensor 1 Remote Control and 2 Alarms for Door Window Home Security (M702) : Amazon.in: Home Improvement
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Mengshen Wireless Light Sound Alarm Magnetic Sensor 1 Remote Control and 2 Alarms for Door Window Home Security (M702)
Brand: Mengshen
4.2 4.2 out of 5 stars1,331 ratings
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You can ask for a replacement or refund if you receive a damaged, defective or incorrect product.
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This item is eligible for free replacement, within 10 days of delivery, in an unlikely event of damaged, defective or different/wrong item delivered to you. For device-related issues in Apple iPhones, please contact Apple Customer Service. For Android smartphones, our Returns Centre will prompt you to download the Blancco app to diagnose issues with a defective smartphone. We'll provide a resolution on the basis of the Blancco app diagnostic results. In certain cases, we may also schedule a technician visit to your location. On the basis of the technician's evaluation report, we will provide resolution.
Please keep the item in its original condition, with original accessories like chargers, headsets etc., brand outer box, MRP tags attached, user manual in manufacturer packaging to avoid pickup cancellation.
Mobile Accessories
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This item is eligible for free replacement/refund, within 10 days of delivery, in an unlikely event of damaged, defective or different/wrong item delivered to you. Note: Please keep the item in its original condition, with MRP tags attached, user manual, warranty cards, and original accessories in manufacturer packaging. We may contact you to ascertain the damage or defect in the product prior to issuing refund/replacement.
Power Banks: 10 Days; Replacement only
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Used Mobiles, Tablets
10 Days Refund
Refunds applicable only if it has been determined that the item was not damaged while in your possession, or is not different from what was shipped to you.
Mobiles and Tablets with Inspect & Buy label
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Refunds will be issued only if it is determined that the item was not damaged while in your possession, or is not different from what was shipped to you.
Tablets (new and certified refurbished)
7 Days Replacement
This item is eligible for free replacement, within 7 days of delivery, in an unlikely event of damaged or different item delivered to you. In case of defective, product quality related issues for brands listed below, customer will be required to approach the brands’ customer service center and seek resolution. If the product is confirmed as defective by the brand then customer needs to get letter/email confirming the same and submit to Amazon customer service to seek replacement. Replacement for defective products, products with quality issues cannot be provided if the brand has not confirmed the same through a letter/email. Brands -HP, Lenovo, AMD, Intel, Seagate, Crucial
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Software
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For software-related technical issues or installation issues in items belonging to the Software category, please contact the brand directly.
Printers
7 Days Replacement
This item is eligible for free replacement, within 7 days of delivery, in an unlikely event of damaged, defective or different item delivered to you.
Please keep the item in its original condition, with brand outer box, MRP tags attached, user manual, warranty cards, CDs and original accessories in manufacturer packaging for a successful return pick-up.
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Desktops, Monitors, Pen drives, Hard drives, Memory cards, Computer accessories, Graphic cards, CPU, Power supplies, Motherboards, Cooling devices, TV cards & Computing Components
7 Days Replacement
This item is eligible for free replacement, within 7 days of delivery, in an unlikely event of damaged, defective or different item delivered to you. Please keep the item in its original condition, with brand outer box, MRP tags attached, user manual, warranty cards, CDs and original accessories in manufacturer packaging for a successful return pick-up. For few products, we may schedule a technician visit to your location. On the basis of the technician's evaluation report, we will provide resolution.
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Return the camera in the original condition with brand box and all the accessories Product like camera bag etc. to avoid pickup cancellation. We will not process a replacement if the pickup is cancelled owing to missing/damaged contents.
Return the speakers in the original condition in brand box to avoid pickup cancellation. We will not process a replacement if the pickup is cancelled owing to missing/ damaged box.
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Speakers (new and certified refurbished)
10 Days Returnable
Returnable if you’ve received the product in a condition that is damaged, defective or different from its description on the product detail page on Amazon.in.
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10 Days Replacement
This item is eligible for free replacement, within 10 days of delivery, in an unlikely event of damaged, defective or different/wrong item delivered to you.
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For TV, we may schedule a technician visit to your location and resolution will be provided based on the technician's evaluation report.
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This item is eligible for free replacement, within 10 days of delivery, in an unlikely event of damaged, defective or different/wrong item delivered to you. .
Please keep the item in its original condition, original packaging, with user manual, warranty cards, and original accessories in manufacturer packaging for a successful return pick-up.
If you report an issue with your Furniture,we may schedule a technician visit to your location. On the basis of the technician's evaluation report, we will provide resolution.
Large Appliances - Air Coolers, Air Conditioner, Refrigerator, Washing Machine, Dishwasher, Microwave
10 Days Replacement
Returnable if you’ve received the product in a condition that is damaged, defective or different from its description on the product detail page on Amazon.in.
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Home and Kitchen
10 Days Returnable
This item is eligible for free replacement, within 10 days of delivery, in an unlikely event of damaged, defective or different item delivered to you. You can also return the product within 10 days of delivery for full refund.
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We may contact you to ascertain the damage or defect in the product prior to issuing refund/replacement.
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Non-Returnable
This item is non-returnable due to hygiene and personal care/consumable nature of the product. However, in the unlikely event of damaged, defective or different item delivered to you, we will provide a full refund or free replacement as applicable. We may contact you to ascertain the damage or defect in the product prior to issuing refund/replacement.
Pet Habitats and Supplies, Apparel and Leashes, Training and Behavior Aids, Toys, Aquarium Supplies such as Pumps, Filters and Lights
10 Days Returnable
This item is eligible for free replacement, within 10 days of delivery, in an unlikely event of damaged, defective or different item delivered to you. You can also return the product within 10 days of delivery for full refund. Please keep the item in its original condition, with outer box or case, user manual, warranty cards, and other accompaniments in manufacturer packaging for a successful return pick-up. We may contact you to ascertain the damage or defect in the product prior to issuing refund/replacement.
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Baby
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7 Days Returnable
Toys (All other than Vehicle and Outdoor category)
All the toys item other than Vehicle and Outdoor Category are eligible for free replacement/refund, within 7 days of delivery, in an unlikely event of damaged, defective or different/wrong item delivered to you.
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Vehicle and Outdoor category toys are eligible for free replacement, within 7 days of delivery, in an unlikely event of damaged, defective or different/wrong item delivered to you
Note: Please keep the item in its original condition, with outer box or case, user manual, warranty cards, and other accompaniments in manufacturer packaging for a successful return pick-up. We may contact you to ascertain the damage or defect in the product prior to issuing refund/replacement.
Sports, Fitness and Outdoors
Sports, Fitness and Outdoors
10 Days Returnable
This item is eligible for free replacement, within 10 days of delivery, in an unlikely event of damaged, defective or different item delivered to you. You can also return the product within 10 days of delivery for full refund.
Please keep the item in its original condition, with brand outer box, MRP tags attached, user manual, warranty cards, and original accessories in manufacturer packaging for a successful return pick-up.
We may contact you to ascertain the damage or defect in the product prior to issuing refund/replacement.
Beauty, Health and Groceries
Beauty
Not Returnable
This item is non-returnable due to hygiene/health and personal care/wellness/consumable nature of the product. However, in the unlikely event of damaged, defective or different item delivered to you, we will provide a full refund or free replacement as applicable. We may contact you to ascertain the damage or defect in the product prior to issuing refund/replacement.
Occupational Health & Safety Products
10 Days Returnable
Personal Care Appliances
7 Days Replacement Only
This item is eligible for free replacement, within 7 days of delivery, in an unlikely event of damaged, defective or different/wrong item delivered to you. Please keep the item in its original condition, with outer box or case, user manual, warranty cards, and other accompaniments in manufacturer packaging for a successful return pick-up.We may contact you to ascertain the damage or defect in the product prior to issuing refund/replacement.
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Health and Personal Care
Not Returnable
This item is non-returnable due to it being a Health or Personal Care product. However, in the unlikely event of a damaged, defective or different/wrong item delivered to you, we will provide a full refund or free replacement as applicable. We may contact you to ascertain the damage or defect in the product prior to issuing refund/replacement.
Clothing and Accessories
Clothing and Accessories
30 Days Returnable
Lingerie, innerwear and apparel labeled as non-returnable on their product detail pages can't be returned.
Return the clothing in the original condition with the MRP and brand tag attached to the clothing to avoid pickup cancellation. We will not process a replacement or refund if the pickup is cancelled owing to missing MRP tag.
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Precious Jewellery
30 Days Returnable
Precious jewellery items need to be returned in the tamper free packaging that is provided in the delivery parcel. Returns in any other packaging will not be accepted.
Fashion or Imitation Jewellery, Eyewear and Watches
30 Days Returnable
Return the watch in the original condition in brand box to avoid pickup cancellation. We will not process a replacement if the pickup is cancelled owing to missing/damaged contents.
Gold Coins / Gold Vedhanis / Gold Chips / Gold Bars
Not Returnable
Eyewear
30 Days; Replacement/Refund
Watches
30 Days, Returnable
Handbags and Luggage
Luggage and Handbags
30 Days Returnable
Any luggage items with locks must be returned unlocked.
Shoes
Shoes
30 Days Returnable
Return the Shoes in the original condition with the brand box to avoid pickup cancellation. We will not process a refund if the pickup is cancelled owing to missing MRP tag.
Car, Motorbike and Industrial
Car Parts and Accessories, Bike Parts and Accessories, Helmets and other Protective Gear, Vehicle Electronics
10 Days Returnable
Items marked as non-returnable on detail page are not eligible for return.
Items that you no longer need must be returned in new and unopened condition with all the original packing, tags, inbox literature, warranty/ guarantee card, freebies and accessories including keys, straps and locks intact.
Fasteners, Food service equipment and supplies, Industrial Electrical, Lab and Scientific Products, Material Handling Products, Occupational Health and Safety Products, Packaging and Shipping Supplies, Professional Medical Supplies, Tapes, Adhesives and Sealants Test, Measure and Inspect items, Industrial Hardware, Industrial Power and Hand Tools.
10 Days Returnable
Tyres (except car tyres), Rims and Oversized Items (Automobiles)
10 Days Returnable
Car tyres are non-returnable and hence, not eligible for return.
Return pickup facility is not available for these items. You can self return these products using any courier/ postal service of your choice. Learn more about shipping cost refunds.
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Note: For seller fulfilled items from Books, Movies & TV Shows categories, the sellers need to be informed of the damage/ defect within 14 days of delivery.
For seller-fulfilled items from Fine Art category, the sellers need to be informed of the damage / defect within 10 days of delivery. These items are not eligible for self-return. The seller will arrange the return pick up for these items.
For seller-fulfilled items from Sports collectibles and Entertainment collectibles categories, the sellers need to be informed of the damage / defect within 10 days of delivery.
Returning Global Store Items
The General Return Policy is applicable for all Amazon Global Store Products (“Product”). If the Product is eligible for a refund on return, you can choose to return the Product either through courier Pickup or Self-Return
Note: - Once the package is received at Amazon Export Sales LLC fulfillment center in the US, it takes 2 (two) business days for the refund to be processed and 2- 4 business days for the refund amount to reflect in your account. - If your return is due to an Amazon error you'll receive a full refund, else the shipping charges (onward & return) along with import fees will be deducted from your refund amount.
For products worth more than INR 25000, we only offer Self-Return option.
Returning Products with Inspect & Buy label
2 Days, Refund
Refunds are applicable only if determined that the item was not damaged while in your possession, or is not different from what was shipped to you.
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Brand Mengshen
Power Source Battery Powered
Compatible Devices Home Security System
Product Dimensions 8.8L x 1.7W x 4H Centimeters
Number of Batteries 1 AAA batteries required.
About this item
Voice and Light Alarm: 108db loud alarm and bright light, can effectively scare the thieves and intruders away,
Magnetic Sensor Alarm: Sensitive, lower false trigger rate,
Multi-funciton Security Alarm: It has the function of arm, disarm, emergency alarm and doorbell, widly meet your demand,
Easy Operation: With remote control, it will be very easy to operate,
More Sensor And Remote Is Available: Search "B07THYZ5G8" on Amazon to get more sensor(without remote), Search "B071KF9ZG9" on Amazon to get more remote.
›See more product details
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Top Brand
Mengshen
84% positive ratings from 500+ customers
200+ recent orders from this brand
8+ years on Amazon
Product Description
Thanks for coming, you deserve the best! This is Mengshen's Wireless Sound Light Alarm M70.
This compact wireless window door alarm will be a home guard to protect you and your family whether at home or go out. It can be widely used in home, shop, office and other needed places.
105db Super Loud Anti-theft Alarm
Easily arm by pressing the Arm key. Once triggered, 105 db super loud voice comes with flash light, for any intruders, it will be a terrible nightmare and a great deterrence.
Alarm type: Wireless
Detector Mode: magnetic sensor
Alarm Sound: 105db
Material: ABS
Alarm output: Sound and Light
Remote distance: 50m
Host power supply: 2AAA batteries (Not included)
Remote control power supply: 27A 12V Alkaline battery (Included)
Sound & Light Security Alarm
Avoid Error-opening
When baby/ kids try to open balcony window and climb out, the loud voice will stop them and alert parents.
This Security Alarm is also suitable for single apartment, rental housing, shop, mall and residential house.
Size& Weight
Host Size: 40x88x16mm
Host Weight: 60g
Simple Installation
There is double-sided adhesive on Host and Magnet bar, just peel and stick to install them on the door, window or anywhere.
Please Note:
1) Make sure the 2 parts align at the scale,
2) Distance between them is about 10mm.
Easy Operation
1) arm key: Press it, once triggered, alarm will bell 105db voice,
2) disarm key: Press it, the arm status or loud voice will stop,
3) SOS key: Press it under arm status, the alarm will bell 105db voice though host is not triggered. This function will be very helpful when someone is in danger or need emergent help.
4) doorbell key: Press it, once triggered, it will bell "dingdong" to alert you someone is coming in.
More Options
1) Extra remote controls are available from us, it need to pair with the host,
2) One remote control could pair and control up to 50 piece of host,
3) One host could be controlled by at most 8 remote control.
How to pair host with remote control?
Please refer to the description of the "Mengshen Remote Control"
| | Door window alarm | Door window alarm | Bicycle Motorcycle Alarm | Bicycle Motorcycle Alarm | Motion Sensor Alarm | Door Stopper Alarm |
--- --- --- |
| | Buying Options | Buying Options | Buying Options | Add to Cart | Add to Cart | Add to Cart |
| Customer Reviews | 4.1 out of 5 stars 2,017 | 4.2 out of 5 stars 1,331 | 3.9 out of 5 stars 3,850 | 4.0 out of 5 stars 3,803 | 4.3 out of 5 stars 2,068 | 4.2 out of 5 stars 1,888 |
| Price | — no data | — no data | — no data | ₹2,029.00₹2,029.00 | ₹2,999.00₹2,999.00 | ₹1,899.00₹1,899.00 |
| Sensor Type | Door Open/Close Sensor | Door Open/Close Sensor | Vibration, Movement | Vibration | Motion Sensor | Pressure |
| Apply To | Door/window | Door/window | Bicycle/Motorcycle | Bicycle/Motorcycle/Door/Window | Home/office/shop/hotel | Door |
| Remote Control | YES | YES | YES | YES | NO | NO |
| More remote control | YES | YES | YES | YES | YES | NO |
| Power Source | AAA 2(Not include) | AAA 2(Not include) | AAA 2(Include) | AAA 2(Not include) | AA 3 and AAA 3(Not include) | 9V battery(Not include) |
Product information
Technical Details
| Voltage | 12 Volts |
| Control Method | Remote |
| Noise Level | 108 dB |
| UPC | 732030586598 |
| Manufacturer | Mengshen |
| Item part number | MS-M702 |
| Product Dimensions | 8.8 x 1.7 x 4 cm; 60 g |
| ASIN | B072KMWZ86 |
Additional Information
| Manufacturer | Mengshen |
| Importer | R SHANTILAL NETWORK PRIVATE LIMITED, H.NO 1-38-1978, KRISHNANAGAR PLOT NO 246, F.F RASOOLPURA SECUNDERABAD, Hyderabad, Telangana, 500003, 9321099952 |
| Item Weight | 60 g |
| Item Dimensions LxWxH | 8.8 x 1.7 x 4 Centimeters |
| Best Sellers Rank | #151,342 in Home Improvement (See Top 100 in Home Improvement) #516 in Home Security Alarms |
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4.2 out of 5 stars
4.2 out of 5
1,331 global ratings
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4 out of 5 stars
good buy
works well and sound is really good for an appartment use.... all functions works well except the sound coming incase of vibration detection
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Santanu Nandy ##### 5.0 out of 5 stars Good Reviewed in India on 14 November 2024 Verified Purchase Good Read more Helpful Report
K ##### 5.0 out of 5 stars value for money Reviewed in India on 6 April 2022 Verified Purchase Very accurate sensor. Read more One person found this helpful Helpful Report
SHASHIKANT ##### 4.0 out of 5 stars Reviewed in India on 24 October 2023 Verified Purchase Read more Helpful Report
Vikrant Gaikwad ##### 5.0 out of 5 stars Good product Reviewed in India on 1 November 2020 Verified Purchase Overall good product, working fine. Read more 2 people found this helpful Helpful Report
Amazon Customer ##### 1.0 out of 5 stars Battery suddenly low Reviewed in India on 18 April 2021 Verified Purchase Poor quality Read more Helpful Report
kindle user ##### 4.0 out of 5 stars Does what it says Reviewed in India on 22 July 2023 Verified Purchase Nice Read more Helpful Report
DIPAK ##### 2.0 out of 5 stars Battery work only four day Reviewed in India on 18 October 2023 Verified Purchase Not good as shown Read more Helpful Report
Kumar ##### 1.0 out of 5 stars Stopped working Reviewed in India on 16 May 2021 Verified Purchase The media could not be loaded. Your product stopped working after 2 months and don’t know if there is any Warrenty or anything that can be done Read more 6 people found this helpful Helpful Report
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Cliente Amazon ##### 5.0 out of 5 stars ottimo Reviewed in Italy on 15 May 2025 Verified Purchase Buon prodotto soprattutto considerando il prezzo...; da vedere quanto dura..abbastanza forte la sirena. Read more ReportTranslate review to English
Maria ##### 5.0 out of 5 stars Barato y práctico. Ojo con el volumen! Reviewed in Spain on 28 June 2019 Verified Purchase Buscaba algo así, barato y fácil de instalar para no preocuparme por no enterarme si entran o no por la noche. Fácil de instalar, sólo necesitarás las pilas y pegarlo donde te interese. Cuando activas la alarma suena 3 veces con un volumen muy fuerte y no es regulable. Y cuando lo desactivas suena una vez con un pitido algo más largo y también con un volumen muy fuerte y no regulable. Lo del volumen de activarlo y desactivarlo es la única “pega” pero por el precio que vale, es lo que hay. Dentro de casa el volumen de alarma es bastante alto, aunque desde fuera no se oye tanto, evidentemente. Yo temía por los “ladrones sigilosos” que entran de noche y ni te enteras. Por lo menos ahora, si entran, me enteraría y creo que, por el ruido que hace la alarma, saldrían corriendo, por lo que resulta disuasoria. Se pueden añadir más alarmas y más mandos. Ah! Y además tiene “botón pánico” para activar la alarma manualmente y función de timbre que, al abrir la puerta, suena como un timbre. Read more ReportTranslate review to English
Valtari ##### 5.0 out of 5 stars Práctica, funcional y buen precio. Reviewed in Mexico on 13 January 2021 Verified Purchase Es muy práctica por su diseño pequeño, fácil de instalar y gracias a que se puede programar sus funciones mediante su control remoto, no es necesario manipularla manualmente y así se puede instalar en la parte más alta de una puerta, lejos del alcance de la mayoría de las personas. Tiene la función de alarma permanente, desactivación de la alarma o como timbre para que avise por un breve tiempo cada vez que alguien abren y cierran la puerta, todo desde el control remoto. También tiene la peculiaridad de poder programar todas las alarmas de esa misma marca y modelo con 1 solo control remoto, o cada alarma manipularla de manera independiente con cada control, y si se pierde alguno, pedir controles remotos adicionales y programarlos para que funcione con cada alarma. Read more ReportTranslate review to English
massimiliano ##### 5.0 out of 5 stars Funzionale ad un prezzo irrisorio Reviewed in Italy on 21 May 2020 Verified Purchase Veramente ottimo ad una cifra molto onestà. Sicuramente non è un vero allarme, cioè se aprono e levano le batterie naturalmente smette di funzionare ma come deterrente per qualche ladruncolo va benissimo. Il suo uso è esclusivamente se si è in casa perché appena suona ti armi anche tu che lo senti. Montato su una casetta da esterno ed anche se non è perfettamente allineato funzionante veramente bene. Bellissimo il telecomando e molto funzionale. Consiglio l'acquisto ma ripeto solamente come deterrente. Plasticoso ma sembra ben fatto, ricordo che non sono presenti le pile (2 mini stilo) ma al telecomando è già inserita. Read more ReportTranslate review to English
Alan Spencer-Lewis ##### 5.0 out of 5 stars Effective, loude essy to use and install Reviewed in the United Kingdom on 23 February 2020 Verified Purchase Excellent devise. Very easy to install.
Easy to programne remote to more than one devise. Range more than adequate. Great value for peace of mind. Have ordered another. Read more Report
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Product Summary: Mengshen Wireless Light Sound Alarm Magnetic Sensor 1 Remote Control and 2 Alarms for Door Window Home Security (M702)
From Mengshen
4.2 out of 5 stars, 1,331 ratings
Customer reviews
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About this Item
Voice and Light Alarm: 108db loud alarm and bright light, can effectively scare the thieves and intruders away,
Magnetic Sensor Alarm: Sensitive, lower false trigger rate,
Multi-funciton Security Alarm: It has the function of arm, disarm, emergency alarm and doorbell, widly meet your demand,
Easy Operation: With remote control, it will be very easy to operate,
More Sensor And Remote Is Available: Search "B07THYZ5G8" on Amazon to get more sensor(without remote), Search "B071KF9ZG9" on Amazon to get more remote.
Product description
Specifications:
Model: M702, 2 Alarm and 1 Remote Control
Alarm volume: 108dB
Frequency: 433MHZ
Working environment temperature: -10 ~ 60 ° C
Working environment humidity: 80%
Storage temperature: -20 ° C —+ 70 ° C
Host battery: 2AAA battery(included)
Remote Control battery: 12V 27A alkaline battery(included)
Item Size: 8.841.7cm/3.51.60.7in
Item weight: Approx. 62g
Package Information:
Package Size: Approx. 18111.5cm/7.14.30.6in
Package Weight: Approx. 90g/3.2oz
Package List:
2 Door Sensor Alarm
1 Remote Control
1 User Manual
Note: If you need more remote control for this alarm, please search "Mengshen Remote Control AR01".
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11611 | https://www.frontiersin.org/journals/education/articles/10.3389/feduc.2024.1422030/pdf | Published Time: Wed, 25 Sep 2024 18:01:52 GMT
Frontiers in Education 01 frontiersin.org
A framework for comparing
large-scale survey assessments:
contrasting India’s NAS,
United States’ NAEP, and OECD’s
PISA
Peter van Rijn 1, Han-Hui Por 2, Daniel F. McCaffrey 2,
Indrani Bhaduri 3 and Jonas Bertling 2
1
ETS Global, Amsterdam, Netherlands, 2 Educational Testing Service (ETS), Princeton, NJ, United
States, 3 National Council of Educational Research and Training, New Delhi, India
Large-scale survey assessments (LSAs) are important tools for measuring
educational outcomes and shaping policy decisions. We present a framework
for comparing LSAs to facilitate studying the impact of design choice on the
precision of results, contrasting India’s National Achievement Survey (NAS),
the United States’ National Assessment of Educational Progress (NAEP), and
the OECD’s Programme for International Student Assessment (PISA). Our
framework focuses on four key elements: sampling design, assessment design,
analysis methodology, and reporting. The notion of total survey error, which
is the accumulation of errors across the four key elements, can be used for
both designing and evaluating LSAs. As example, we compare statistics that are
commonly (but not always) reported from NAS, NAEP, and PISA to summarize
outcomes related to sampling, measurement, and reporting. Our examination
reveals several key similarities and differences among the three assessments,
thereby highlighting the nuanced ways in which each LSA is tailored to meet the
specific needs of their purpose and the challenges they face.
KEYWORDS
large-scale survey assessment, total survey error, assessment design, PISA, NAEP
1 Introduction
Large-scale educational survey assessments (also referred to as large-scale assessments,
LSAs) are vital in measuring student learning outcomes in modern education systems. These
assessments, often administered at the national or international level to representative samples
from the populations of interest, serve as critical tools for measuring and monitoring
educational outcomes, informing, and shaping policy decisions. Three distinctive features set
LSAs apart from other types of assessments. First, LSAs are administered to randomly selected
schools and students, ensuring a representative sample of the target student population.
Second, LSAs encompass a wide content coverage, necessitating many test items to evaluate
students’ knowledge and skills. Third, results on student proficiency are reported at the group
level rather than at the individual level. Given the diversity of established LSAs, the
comparability of these assessments and the implications of design choices on reported student
achievement results present a complex challenge. Thus, the ability to evaluate the consequences
of design choices on student achievement outcomes hinges on the availability of a systematic
OPEN ACCESS
EDITED BY
Raman Grover,
Consultant, Canada
REVIEWED BY
Alexander Robitzsch,
IPN - Leibniz Institute for Science and
Mathematics Education, Germany
Si Man Lam,
The University of Hong Kong, Hong
Kong SAR, China
CORRESPONDENCE
Peter van Rijn
pvanrijn@etsglobal.org
RECEIVED
23 April 2024
ACCEPTED
26 August 2024
PUBLISHED
25 September 2024
CITATION
van Rijn P, Por H-H, McCaffrey DF,
Bhaduri I and Bertling J (2024) A framework
for comparing large-scale survey
assessments: contrasting India’s NAS,
United States’ NAEP, and OECD’s PISA.
Front. Educ. 9:1422030.
doi: 10.3389/feduc.2024.1422030
COPYRIGHT
© 2024 van Rijn, Por, McCaffrey, Bhaduri and
Bertling. 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.
TYPE
Methods
PUBLISHED
25 September 2024
DOI
10.3389/feduc.2024.1422030 van Rijn et al. 10.3389/feduc.2024.1422030
Frontiers in Education 02 frontiersin.org
framework for a comparative analysis of LSAs. In this paper,
we present and discuss a framework for comparing LSAs that can
be used for this purpose.
Frameworks summarizing and comparing LSAs exists but
typically limit themselves to key general features to provide an
overview. Clarke and Luna-Bazaldua (2021) summarized international
and regional LSAs (adapted and updated in Table 1 ) and included
information such as target grade or age, assessed subjects and the
countries participating in the assessments. Our framework
supplements similar overviews with information useful for assessment
developers and organizations seeking to develop similar large-scale
assessments to inform educational research and policies.
Books on LSAS typically focused on the design,
implementation, and improvement of LSAs (e.g., Clarke and Luna-
Bazaldua, 2021 ; Lietz et al., 2017 ; Rutkowski et al., 2014 ; Simon
et al., 2013 ). For example, the chapters in Simon et al. (2013)
discussed research and practices related to the development of
large-scale surveys, such as design and delivery, assessing diverse
populations, scoring and use of scores and psychometric modeling
and analyses. Adopting a similar approach, Lietz et al. (2017)
discussed key concepts in the implementation of LSAs and covered
topics on test design and development, weighting, scaling and
reporting. Rutkowski et al. (2014) provided an overview to the
policy and research relevance of international LSAs and focused on
the methodological and analytical processes for analyzing
international LSA data. Clarke and Luna-Bazaldua (2021) focused
on the use of LSA findings to improve national education systems
and discussed the critical aspects of planning and
implementing LSAs.
While systematic comparisons of LSAs have been conducted in
the past (e.g., Cresswell et al., 2015 ; Black and Wiliam, 2007 ), these
efforts have predominantly concentrated on specific facets of the
development of large-scale assessments. Black and Wiliam (2007)
focused on national assessments and compared design differences in
assessment systems in different countries. Cresswell et al. (2015)
summarized effective practices of international LSAs, with an
emphasis on their adoption in designing the Organisation for
Economic Co-operation and Development’s (OECD) Programme for
International Student Assessment (PISA) for development program,
also known as PISA-D ( Ward, 2018 ).
In this paper, we present a structured framework for comparing
LSAs to facilitate studying the impact of design decisions on the
precision of reported results. This framework supplements overviews
of assessment characteristics ( Table 1 ) and will be informative to
researchers and decision-makers in assessment development. The
proposed framework builds on the existing literature by considering
the errors associated with the measurement of educational outcomes
in large surveys. The proposed framework hinges on four key
elements: sampling design, assessment design, analysis methodology,
and reporting. In addition to describing the four key elements, our
goal is to provide metrics for the comparison in a way that is sensible
and fair to each LSA in the comparison (i.e., we do not want to
compare apples to oranges). The starting point for these metrics is
the total survey error (TSE) approach ( Weisberg, 2005 ). Our aim is
to break down the errors in LSA reporting (e.g., the standard error
of the estimated mean proficiency) and to link them to components
of the four key elements (e.g., sampling error and measurement
error). This break down provides insight into the way an LSA is
designed, the relation to the population that is being assessed, and
important features of the school and education system. As examples,
we use the Programme for International Student Assessment (PISA)
administered by the OECD, the National Assessment of Educational
Progress (NAEP) in the United States and the National Achievement
Survey (NAS) in India. PISA was chosen for its global reach. NAEP,
in comparison, has provided meaningful results to improve
education policy and practice in the United States since 1969. The
NAS, which has been administered since 2001, has large
national samples.
The proposed framework is a potentially valuable resource for
educational measurement professionals and a broader audience of
stakeholders, including policymakers, educators, and researchers. It
enables them to document and critically assess the characteristics of
their assessments and compare them with other LSAs, shedding light
on their similarities, differences, and potential areas for improvement.
The paper starts with brief descriptions of NAS, NAEP, and PISA. This
is followed by the introduction of the LSA comparison framework.
Next, we present the metrics to be used for comparing LSAs, followed
by results of a comparison of NAS, NAEP, and PISA. The paper ends
with a discussion and some recommendations.
2 Large-scale survey assessments
LSA is becoming a common tool to monitor educational systems
as the number of countries around the world that participate in
international LSAs has been increasing. For example, 79 countries/
economies participated in PISA 2018, compared to 32 in PISA 2000;
57 countries in PIRLS 2021, compared to 35 in PIRLS 2001; 64
countries in TIMSS 2019, compared to 45 in TIMSS 1995. In addition,
more and more countries have initiated national LSAs ( Clarke and
Luna-Bazaldua, 2021 ).
In India, the NAS is the key measures of student achievement. In
2021, the NAS was administered to students in grades 3, 5, 8, and 10.
Although the first cycle of NAS was conducted in 2001–2002 for
grade 5 (Figure 1.1 in NCERT, 2019 ), it evolved in all key elements
over the last two decades. A special feature of NAS 2021 was that it
was conducted on a single day throughout the country for all four
grades. Furthermore, NAS 2021 was designed to provide a
comprehensive assessment of learning outcomes with increased
content coverage, building on previous national surveys ( Ministry of
Education, India, 2021 ). Over 3.4 million students and 500,000
teachers from 118,000 schools across India participated in NAS 2021,
which was administered by the National Council of Educational
Research and Training (NCERT). More importantly, the NAS results
are used in framing and evaluating educational policy, and vice versa,
educational policy can influence the setup of NAS. For example,
India’s national education policy 2020, paragraph 4.34, clearly states
the aim and purpose of assessment:
“The aim of assessment in the culture of our schooling system will
shift from one that is summative and primarily tests rote memorization
skills to one that is more regular and formative, is more competency-
based, promotes learning and development for our students, and tests
higher-order skills, such as analysis, critical thinking, and conceptual
clarity. The primary purpose of assessment will indeed be for learning;
it will help the teacher and student, and the entire schooling system,
continuously revise teaching-learning processes to optimize learning van Rijn et al. 10.3389/feduc.2024.1422030
Frontiers in Education 03 frontiersin.org
TABLE 1 Overview of international and regional large-scale assessments.
Assessment Target grades or age Main subject areas Organization Years Participating regions
Programme for International Student
Assessment (PISA)
15-year-olds Reading, mathematics, science Organization for Economic Co-operation
and Development (OECD)
2000, 2003, 2006, 2009,
2012, 2015, 2018, 2022
Global
Trends in International Mathematics and
Science Study (TIMSS)
Grades 4, 8 Mathematics, science International Association for the Evaluation
of Educational Achievement (IEA)
1995, 1999, 2003, 2007,
2011, 2015, 2019, 2023
Global
Progress in International Reading
Literacy Study (PIRLS)
Grade 4 Reading International Association for the Evaluation
of Educational Achievement (IEA)
2001, 2006, 2011, 2016, 2021 Global
Regional Comparative and Explanatory
Study (ERCE)
Grades 3, 6 Literacy, mathematics, science Latin American Laboratory for Assessment
of the Quality of Education (LLECE)/ United
Nations Educational, Scientific and
Cultural Organization (UNESCO)
1997, 2006, 2013, 2019 Latin America
Programme d’analyse des systèmes
éducatifs de la Confemen (pasec)
Grades 2, 6 Reading, mathematics La Conférence des ministres de l’Éducation
des États et gouvernements de la
Francophonie
Every year between 1993
and 2010, 2014, 2019, 2021
Francophone Africa; select countries in
East Asia
Southern and Eastern Africa Consortium
for Monitoring Educational Quality
(SEACMEQ)
Grade 6 Reading, mathematics, health
knowledge
Southern and Eastern Africa Consortium for
Monitoring Educational Quality
1999, 2004, 2011, 2014, 2022 Anglophone Africa
Pacific Islands Literacy and Numeracy
Assessment (PILNA)
Grades 4, 6 Numeracy, literacy Pacific Community 2012, 2015, 2018, 2021 Pacific Islands
Southeast Asia Primary Learning Metrics
(SEA-PLM)
Grade 5 Literacy, mathematics, global
citizenship
Southeast Asian Ministers of Education
Organization (SEAMO)/ United Nations
Children’s Fund (UNICEF)
2019, 2024 Southeast Asia
Table adapted and updated from Clarke and Luna-Bazaldua (2021 , Table 8 A.1). van Rijn et al. 10.3389/feduc.2024.1422030
Frontiers in Education 04 frontiersin.org
and development for all students. This will be the underlying principle
for assessment at all levels of education.”
Given India’s unique context as one of the largest countries with
state-run public education, comparisons with other large-scale
educational assessments can be beneficial in providing insights into
potential transformations of national assessment practices. Due to
India’s multicultural and multilingual educational landscape, the NAS
shares many similarities with OECD’s PISA. At the same time, the
NAS also retains many design elements like the United States’ NAEP,
as both are national assessments.
NAEP is the largest nationally representative, continuing
evaluation of the condition of education in the United States and has
served as a national yardstick of student achievement since 1969. The
NAEP assessment is a congressionally mandated program administer
by the National Center for Education Statistics within the
U.S. Department of Education, and the National Assessment
Governing Board oversees and sets policies for NAEP. While NAEP
also has an age-based assessment known as the long-term trend
assessments, the current paper will focus on the main NAEP
assessments for illustrative purposes. Through the publicly available
Nation’s Report Card, the outcomes from the main NAEP assessment
inform the public and other stakeholders about what American
students in grades 4, 8, and 12 know and can do in various subject
areas such as mathematics, reading, writing and science, and compares
achievement among states, large urban districts, and various student
groups. The main NAEP assessments officially transitioned from
paper-based assessment (PBA) to digitally based assessment (DBA) in
mathematics and reading in 2017 ( Jewsbury et al., 2020 ).
PISA is an international assessment that measures 15-year-old
students’ ability in three domains in reading, mathematics, and
science literacy assessment. It was first conducted in 2000 and has
been administered every 3 years, 1 where the major domain of study
rotates in each cycle. Since 2012, PISA has also included an
innovative domain assessment in every cycle, with global
competence in 2018. The innovative domain assessments target
interdisciplinary, 21st century competencies, providing participating
countries/economies with a more comprehensive outlook on their
students’ readiness for life. By design, PISA focuses on functional
skills that students have acquired as they near the end of compulsory
schooling. PISA is coordinated by the OECD, an intergovernmental
organization of industrialized countries. During its 20+ years of
operation, PISA has undergone two major transitions. The first
transition is from PBA to DBA as the main mode of assessment in
PISA 2015 and the second transition is from linear to adaptive
testing in PISA 2018 for the major domain of reading. In PISA 2025,
all three core domains will be assessed using multistage
adaptive testing.
3 LSA comparison framework
Our LSA comparison framework focuses on four key elements:
sampling design, assessment design, analysis methodology, and
1Note that there was a four-year gap between PISA 2018 and PISA 2022 due
to the COVID-19 pandemic. Also, PISA will shift to a four-yearcycle after 2025.
reporting. The multiple components that comprise each element are
outlined in Table 2 and discussed in the following.
The sampling design primarily begins with describing the target
population. For example, in NAS and NAEP, the target population is
a specific school grade while in PISA, the target population consists of
15-year-old students. The latter can pose some challenges on the
testing window because this population changes throughout the
school year. Furthermore, the school year itself can be quite different
across countries participating in PISA. The type of sampling, primary
sampling unit (PSU) , and stratification methods are other important
components of the sampling design element of our framework. For
instance, two-stage sampling is mostly used in LSAs that focus on
younger students, where schools are the primary sampling unit (e.g.,
proportional to a measure of school size) and students within sampled
schools are selected next. Stratification , employed to reduce sampling
error, assumes a notably more intricate form within an international
context, given the substantial variability in strata across participating
countries, as evidenced in prior research (Table 4.1 in OECD, 2017 ).
It is important to note that stratification should be linked to the
reporting goals of the assessment. For example, the NAS 2021 aimed
to provide data at the level of school-administration type (e.g., state-
funded versus private schools) within school districts ( NCERT,
2021b ). Hence, these administrative units defined strata for the
sample. NAEP, on the other hand, reports at the state level so states are
used in defining strata with additional within-state stratification to
improve precision of state estimates.
In most LSAs, target sample sizes are specified to guarantee
sufficient precision in the reported results from the sample. However,
it is not always easy and efficient to use a “one-size-fits-all” strategy
with respect to sample size, since there can be substantial differences
in the size of the target population (e.g., highly unequal population
sizes of countries in PISA and states in India and US). Such a strategy
can become an issue for smaller populations (e.g., almost census for
very small populations, and finite-population corrections are needed
if more than 5% is sampled).
The second element in the LSA comparison framework pertains
to assessment design , which typically begins with a description of the
assessment framework. The assessment framework defines the domain
and the scope measured by the assessment. PISA assesses broader
thinking skills that are not intended to be tied to a specific curriculum.
In comparison, NAEP (see for example, the 2022–2024 mathematics
framework; National Assessment Governing Board, 2022 ) and NAS
(e.g., NCERT, 2021a , p. 7) are national assessments that do target
national curricular objectives and learning outcomes. Either way, the
assessment framework defines the domains and the scope of the
assessment. Furthermore, the assessment framework includes
descriptions of the test length (i.e., number of items administered to
each student), allotted testing time, distributions of items across
subdomains and difficulties, assessment mode (paper-based, digital-
based, or computer-adaptive), and item and response formats (e.g.,
selected response, constructed response). When learning outcomes
are compared across years (or other periods), the assessment
framework should describe the proportion of new and trend (i.e.,
anchor or equating) items. The framework also provides details on the
matrix-sampling design to assign items to test forms. If sampled
students are assessed on multiple subjects, this should also
be explained in the assessment design. If a linear PBA is used, the
linear test forms are described. If a multistage adaptive CBA is used, van Rijn et al. 10.3389/feduc.2024.1422030
Frontiers in Education 05 frontiersin.org
the algorithm and adaptive test paths are typically described. This
includes a description of how students are routed (e.g., based on
intermediate total scores on automatically scoreable items). Another
important component of the assessment design is formed by the
questionnaires: Whether and which questionnaires are used to provide
contextual information for the interpretation of the results related to
student proficiency? Which questionnaires (student, school, teacher,
parent) are used and what questions do they contain? How much time
is allotted for the questionnaires? Are the questionnaires an integrated
part of the design or are they optional?
The third element of the LSA comparison framework centers on
the analysis methodology , which in many LSAs involves the application
of statistical tools ranging from basic data quality checks to advanced
psychometric modeling. Many analyses are intricately linked to the
sampling design, such as the utilization of sampling weights, and to
the assessment design, which entails handling item responses.
However, the focal point within our framework revolves around the
techniques employed for the generation of the eventually reported
group-level results on student proficiency. Consequently, the emphasis
in the discussion here lies less on the specific methodologies used for
the test and item analyses and more on the methodology underpinning
the group-level proficiency scores.
Many LSAs rely on item response theory (IRT) methods and
IRT-based scaling models to construct measures of student proficiency.
However, notable variations exist among LSAs concerning the
application of IRT methods, including the specific IRT model utilized
and the methodology used to calibrate (i.e., estimate) its parameters
(von Davier and Sinharay, 2014 ). Moreover, the use of contextual
information in so-called conditioning models , consisting of regressions
of proficiency on background variables, can significantly differ across
various LSAs ( Wu, 2005 ; von Davier et al., 2009 ). If plausible values
(PVs), which are multiple draws from the posterior proficiency
distribution, serve as the proficiency measure for reporting
distributions and summary statistics ( Wu, 2005 ; von Davier et al.,
2009 ; Marsman et al., 2016 ), it is important that it is described how
they are computed and what conditioning variables are used in their
computation (e.g., cognitive item responses and
questionnaire responses).
The fourth and final element of the LSA comparison framework
addresses the approach to reporting . In our case, this adheres to the
metrics (i.e., the type of statistics) that are reported and the levels at
which they are reported. Reporting levels can be related to
geographical areas (e.g., country, region, state, district) but also to
other grouping variables (e.g., related to school type, demographic
information). In general, results are reported as descriptive statistics,
including mean scale scores, their corresponding standard deviations
and/or standard errors, alongside the proportion of students
performing at each of several proficiency levels. Such proficiency
levels (e.g., basic, proficient, advanced) are typically defined in relation
to the IRT-based scale score and can be illustrated by items that are at
a given level (item mapping).
The above four key elements of LSA studies drive many design
choices and it is important to establish many components of these
elements prior to the data collection. For instance, it is important to
know the granularity at which LSA results are to be reported to create a
sampling design with which a minimum precision level can
be guaranteed. In the next section, we discuss how design choices can
impact the variability of statistical errors in the reported results, which
in turn affect the kind of inferences that can be made to aid policymaking.
4 Total survey error for LSA
comparisons
The total survey error approach, originally designed for survey
research in political science and sociology ( Weisberg, 2005 ), can
be employed in the context of surveys in educational measurement.
Specifically, it can be used for describing the uncertainty in statistical
outcomes from LSAs. The approach recognizes distinct ways that
measurement statistics can deviate from the unobservable true values.
Biemer (2010) defined total survey error (TSE) as the accumulation of
errors in the instrument design, data collection, processing, and analysis.
Hence, TSE is inversely related to the survey quality in that error
variability weakens the inferences that can be derived from the data.
In the context of LSA, but without referring to the TSE approach,
Wu (2010) distinguished three main sources of error: sampling error,
measurement error, and linking error. She suggested that assessment
quality can be improved by focusing on the areas that pose the highest
threats to the validity of the results. She summarized issues related to
relative large measurement error with assessments conducted on a
single occasion, confounding of sampling and measurement errors at
the classroom level, validity if only one test form is used (lack of
content coverage), sampling efficiency with clustered sampling (schools
and students), item position effects, item-by-country/item-by-language
interactions, and linking error if the number of common items is small.
The TSE approach can be used for both designing (i.e., a planning
criterion) and evaluating LSAs. Using the TSE paradigm, the major
sources of errors are identified so that resources can be adequately
allocated to minimize the errors to the extent possible. However, the
TSE approach is not without its weaknesses. For example, Groves and
Lyberg (2010 , p. 874–875) list several shortcomings of the TSE
approach. One of these shortcomings can be translated to our LSA
context in the sense that some components have larger burden and
cost than others. For example, sampling more schools can be more
TABLE 2 Components of the four key elements of the LSA comparison framework.
Sampling design 2. Assessment design 3. Analysis methodology 4. Reporting
•Target population
•Sampling type
•Sampling unit
•Stratification
•Sample sizes
•Assessment framework
•Translation/adaptation
•Subjects/domains
•Item format
•Delivery mode
•Assessment type
•Background questionnaires
•Scaling model
•Calibration method
•Contextual information and
conditioning models
•Proficiency measure
•Metrics (e.g., scale scores)
•Levels (e.g., states, subgroups)
•Trend comparisons van Rijn et al. 10.3389/feduc.2024.1422030
Frontiers in Education 06 frontiersin.org
demanding and costly than sampling more students from the
sampled schools. Another shortcoming is that the term TSE is not
well-defined, as different researchers can include different
components of error within it. Furthermore, survey developers and
data users can perceive survey quality from different perspectives
and may therefore prefer to weigh the components differentially.
Notwithstanding these shortcomings, we identify sampling,
measurement, and linking error as the main sources of error variability
relevant to LSAs as we think these can still be useful in comparing
LSAs. Which error variances to take into consideration depends on
the intended inferences. If we wish to express the precision of an
estimate of mean student proficiency
∝ (e.g., state means in both
NAS and NAEP and a country mean in PISA) for comparisons, then
its TSE can be expressed as the sum of the sampling error variance and
measurement error variance:
TSE Var ;Total Var ;Sampling Var ;M
μ μμμ
= = + eeasurement
where Var(
∝ ; Sampling) is the uncertainty resulting from the
sampling design and Var(
∝ ; Measurement) adheres to the uncertainty
resulting from the assessment design.
In many LSAs, it is also important to evaluate trends in student
proficiency (e.g., the mean in the current assessment cycle compared
to the mean in a previous assessment cycle). In this case, linking error
also affects the precision of the outcome because of differences in
sampling and assessment design across cycles ( Robitzsch and Lüdtke,
2019 ). That is, both different students and different items may have
been used across cycles, while the scale on which the means across
cycles are reported are treated as the same. So, for the estimation of
the error in trend
τ
comparing year A with year B (
τ μ μ = −B A ), the
total error variance has the following components:
TSE TSE TSE Linking Error A,B
τ μ μ
= + + ( )A B ,
where the total errors in years A and B are calculated using the
previous equation. Note that linking error can be calculated in
different ways and its calculation can depend on sampling design,
assessment design, and analysis methodology.
In summary, uncertainties in measuring and monitoring student
performance in LSAs are composed of error variability from sampling,
characteristics of the assessment instrument (i.e., measurement), and
linking across assessment cycles. Table 3 lists examples of the
assessment design choices that can affect each type of error variance
component, and we will discuss them in the following sections.
4.1 Sampling error
In many LSAs, the assessments are administered to a sample of
students selected via a sampling design from a well-defined target
population. For estimates of mean proficiency, sampling error is the
main source of error variability. A two-stage sampling design is
commonly used in large-scale educational survey assessments such as
NAS, NAEP, and PISA ( Rust, 2014 ). Instead of selecting students
directly from the target population, schools are selected first, most often
with probability proportional to school size. Then, students within these
schools, known as clusters, are selected. The number of students within
a school is referred to as the cluster size. For two-stage sampling, the
sampling error variance of estimated mean
∝ of proficiency
θ consists
of between-school error variance and within-school error variance:
Var ;Sampling Var ;Between school Number of School
μθ
= −( )
ssVar ;Within school Number of Students
−( )θ .
LSA sampling designs often set target sample sizes that describe a
minimum number of schools and a minimum number of students to
control sampling error variance and to detect meaningful differences.
As noted, stratification is useful when the population is
heterogeneous with respect to proficiency, and can be divided into
homogeneous, mutually exclusive subgroups known as strata.
Common stratification variables are geographical location (e.g.,
urban, rural) and school type (e.g., private, public), but others can
be used depending on the context. Stratification reduces sampling
error for two-stage sampling, and the greatest reduction in sampling
error happens when students across strata are heterogeneous different
from one another but internally homogenous within strata (i.e.,
between-strata variation is large). In certain cases, stratification is used
to oversample students from small segments of the population so that
the proficiency estimates for these subgroups can be computed with
sufficient precision. More details on sampling in LSA can be found in,
for example, Rust (2014) and Rust et al. (2017) .
4.2 Measurement error
It is important to note that, in the context of LSA, the measurement
error variance in the above breakdown has a primary and secondary
component. The primary component is linked to variables directly
related to the proficiency under scrutiny (e.g., assessment framework,
item pool, test length, inter-rater reliability, test reliability). The
secondary component is connected to indirectly related variables such
as other assessed domains and background information.
TABLE 3 Assessment design choices that affect components of total
error.
Sampling error
variance
Measurement error
variance
Linking error
variance
•Target population
•Sampling stages
•Target sample size
•Cluster size
•Stratification
variables
•Assessment framework
•Item pool
•Test length
•Inter-rater reliability
•Test reliability
•IRT model and fit
•Conditioning variables
•Correlations with
other domains
•Linking design
•Calibration method van Rijn et al. 10.3389/feduc.2024.1422030
Frontiers in Education 07 frontiersin.org
Hence, measurement error can be affected in two ways. First, it
can be reduced by directly improving the measurement of the
proficiency under scrutiny (e.g., increasing test length, test reliability).
Second, it can be decreased by adding or improving the indirectly
related variables (e.g., correlations with other assessed domains, using
more/better conditioning variables). It can be useful, but not always
easy, to inspect each component of measurement error individually.
For example, when other assessed proficiencies and background
information can be utilized, a low measurement error variance may
hide that test reliability is relatively low.
We emphasize that it can be difficult to evaluate the impact of
smaller elements of measurement error. For example, if human raters
score constructed-response items, rater agreement statistics are often
used to evaluate the scoring quality, but uncertainty due to rater
disagreements is generally not carried forward to the next stage in the
analysis process (i.e., IRT modeling). That is, a single item score is
mostly used in the IRT modeling phase. If in one domain (e.g.,
reading) the number of human-scored items is much larger than in
another domain (e.g., mathematics), it can become difficult to fairly
compare, say, test reliability across domains based on scores from a
single rating. In addition, the uncertainty in the estimation of item
parameters is often ignored in LSAs. The impact of this uncertainty is
small when item-level sample sizes are large, but this is not always the
case. For example, in PISA, unique item parameters for country-by-
language groups are allowed in case of misfit with as few as 250 cases
(see, e.g., OECD, 2023 , Chapter 14). However, even though the
sampling error component of these unique item parameters can
be substantial, it remains difficult to gauge what the eventual impact of
an individual item on the uncertainty of, say, mean proficiency is. More
generally, it is hard to assess what the impact of misspecification of the
IRT model can be on the reported results (e.g., misspecification related
to item misfit, position effects, local independence, dimensionality).
To adequately cover the broad range of contents, a large pool of
assessment items is necessary. To limit the assessment time, costs and
minimize test fatigue, many LSAs utilize matrix item sampling ( Beaton
and Zwick, 1992 ; Mislevy et al., 1992 ) and each sampled student is
administered only a small fraction of items from the item pool. The
subset of items each student receives may differ in terms of properties
and content (e.g., content domain distribution, item difficulty, and
reliability) which result in missingness by design. To account for the
differences in the assessments taken, LSAs such as PISA and NAEP use
IRT models to estimate scores. In the matrix item sampling design,
responses to most items are missing as these items were not presented to
the students. To estimate the group-level proficiency distributions, NAEP
and PISA use model-based multiple imputation methods ( Mislevy, 1991 ;
Rubin, 1987 ). Such imputation methods assume that, in addition to the
IRT model, the latent trait is related to background variables (or
conditioning variables) by a linear regression model with normally
distributed residuals. These group-level proficiency estimates could also
be derived with weighted maximum likelihood estimates (WLEs; see
Laukaityte and Wiberg, 2017 ; Wu, 2005 for discussions of the methods).
Rubin (1987) proposed drawing several sets of PVs (i.e., multiple
imputations) to enable the computation of the uncertainty associated
with the measurement. Mislevy (1991) illustrated the approach with
data from NAEP. When a test statistic, for instance, mean scores, is
computed, the variance among the average of the estimates of the
mean, each computed from a different set of plausible values, reflects
the uncertainty due to testing only a sample of students from the
population. When multiple plausible values are drawn for each
respondent to account for uncertainty in the estimate of each
respondent’s proficiency, an additional source of error variability is
introduced. Imputation variance is determined by the measurement
precision (i.e., test reliability), the correlations between proficiencies,
and the relation between background information and proficiency.
Var ;Measurement
μμ μ
= + − −=∑
1 1112
K KkKk
,
where K is the number of PVs drawn.
If 10 PVs are drawn, PV reliability denoted by Rel |, ,
θ Y X ( ) can,
for example, be calculated as the average of five correlations from
each unique pair of PVs. If all information is used, PV reliability can
then be interpreted as the percentage of variance in student
proficiency that is explained by the item responses, correlations
among domains, and correlations with the questionnaire. Instead of
using the measurement error variance, we can thus look at the three
main components of its inverse, measurement precision: Test
reliability, questionnaire correlations, cross-domain correlations. The
posterior density of proficiencies that is used to draw PVs is
proportional to
h P f
θθθθ θθ|, , | |Y X Y X( ) ∝ ( ) ( ),
where θ is the multidimensional proficiency (to allow measurement
of multiple domains), Y contains the item responses, X contains the
background information, ( )|P Y
θ is the measurement model (i.e., an
IRT model) and ( )|f X
θ is the population model (i.e., a latent
regression model). To break down the sources of measurement
precision, one can compare PV reliability based on different
combinations of measurement components. This is shown in Table 4
and the four PV reliabilities reveal the sources of measurement precision.
TABLE 4 Breakdown of sources of measurement precision.
Measurement components Posterior of proficiency PV reliability
Test reliability hPf
θθθ||YY
( ) ∝ ( ) ( ) Re l |
θY
( )
Test reliability + questionnaire correlations hPf
θθθ|||Y X YX,
( ) ∝ ( ) ( ) Rel |, ,
θY X
( )
Test reliability + cross-domain correlations
( ) ( ) ( )| |∝h P fY Y
θθθRe l|
θY
( )
Test + questionnaire + cross-domain
( ) ( ) ( )| , | |∝h P fY X Y X
θθθRel |, ,
θY X
( )van Rijn et al. 10.3389/feduc.2024.1422030
Frontiers in Education 08 frontiersin.org
4.3 Linking error
Linking errors can affect statistical inference, for example, for the
group-level means across assessment cycles, but also across other
grouping variables (e.g., states/countries/gender) and for other
statistics (e.g., percentages of students at different proficiency levels).
Many large-scale assessments are designed to monitor students’
educational progress over time. To estimate trends in performance,
assessments from each year are linked to the previous assessment.
Often, it is also not feasible to administer all the desired items to a
single sample of students. Instead, overlapping pools of items sampled
are administered to different samples, and the overlapped items also
known as common or trend items. Kolen and Brennan (2004 ,
chapter 8) described the desirable characteristics of the common item
set. Sheehan and Mislevy (1988) used a Jackknife approximation
method to estimate the linking error between the 1984 and 1986
NAEP reading assessment and found that the drop in mean reading
proficiency was only one standard error when the error from the
linking procedure was accounted for, compared to three standard
errors when it was not.
In estimating the group-level scores in large scale assessments,
measurement and sampling errors in the group means decrease as the
samples become larger. Error variability from the common items,
however, is affected by the number of common items used and does
not depend on the size of the sample. As such, error variance due to
common items can appear large compared to measurement and
sampling errors (e.g., Michaelides and Haertel, 2004 ; Sheehan and
Mislevy, 1988 ).
Depending on the linking design and scaling approach, linking
error can be computed differently. In earlier PISA cycles, for example,
linking error variance was defined as the variance of equated difficulty
parameter estimates of all common items across cycles from two
separate calibrations:
Linking Error ,A,B
ββ β( ) = − =∑
112
J jJAj Bj
,
where
β
Aj
is the estimated item parameter for year A and
β
Bj
is the
estimated item parameter for year B. Since PISA used a generalized
form of the Rasch model ( Adams et al., 1997 ) in earlier cycles, linking
error could be defined in this way. Furthermore, the assumption here
is that differences in item parameters are independent, which is
unlikely to hold for set items with a common stimulus, such as in
reading. Monseur and Berezner (2007) proposed a Jackknife
replication method to compute linking errors that can deal with such
items. Since PISA 2015, a different approach was needed due to
changes in the IRT model: Rasch and two-parameter logistic IRT
models were used to calibrate the items. Linking error is now
estimated by the standard deviation of the equated country means
from two calibrations:
Linking Error ,A,B
μμ μ( ) = − =∑
112
G gGAg Bg
,
where
∝ Ag is the estimated country mean for year A using the
calibration for year A and
∝ Bg is the estimated country mean for
year A using the calibration for year B. For further details, see the
PISA 2018 technical report ( OECD, 2020 ). Note that the two linking
errors in the above equations operate on different concepts (see, e.g.,
Robitzsch and Lüdtke, 2024 ). For example, Li nking Error ,A,B
β( )
can be defined on any two separate calibrations (using the Rasch
model) as long as there are common items (e.g., it can be calculated
to compare to two PISA cycles for one country, but also for two
countries in one cycle), while this cannot be done for
Li nking Error ,A,B
μ( ) . Further, Robitzsch and Lüdtke (2019)
proposed a new framework to assess linking errors in the context of
PISA that assumes linking errors emerge from differential item
functioning across countries, across assessments, and across countries
and assessments. In sum, there are different approaches to evaluate
linking error, and, in general, it can be complex to assess for any given
comparison in the context of LSA.
5 Statistical outcomes for comparing
LSAs
In this section, we describe the areas in which LSAs can
be compared numerically. To this end, we make use of statistics that
are commonly (but not always) reported to summarize outcomes
related to sampling, measurement, and reporting. Table 4 shows the
statistics that can be compared in each of these three areas.
To express the sampling quality in LSA numerically, the following
statistics can be reported: population coverage, exclusion rate,
response rate, design effect, intraclass correlation, and effective sample
size. Population coverage can be defined as “the extent to which the
weighted participants cover the final target population after all
exclusions” (PISA 2018 Technical Report, Chapter 11, Sampling
Outcomes, p. 1; OECD, 2020 ). School and student response rates
describe the weighted participation rates of schools and students,
where a distinction can be made between response rates before and
after replacement (which mostly pertains to schools). The exclusion
rate concerns the proportion of the sample that is excluded according
to rules described in the sampling frame. Exclusion can take place
both at the school level (e.g., in case of special education) and at the
student level (e.g., disabled students).
An important measure of the impact of the sampling design on
the uncertainty of the mean (e.g., the estimated mean proficiency in
mathematics) is the design effect . The concept of design effect
originated as a way to characterize the efficiency of a sample design
(Cornfield, 1951 ). Kish and Frankel (1974) used the inverse of
Cornfield’s ratio and termed it the design effect. In LSA, the design
effect can be defined as the ratio of the variance of the mean under the
used sampling design and the variance of the mean under simple
random sampling (SRS):
Design Effect Variance of Mean under Sampling Design Varian
=
cce of Mean under Simple Random Sampling .
The design effect is a measure that describes the efficiency of a
sampling design compared to SRS. A design effect equal to one means
that the sampling design is as efficient for estimating the mean as a
simple random sample while larger design effects indicate that the van Rijn et al. 10.3389/feduc.2024.1422030
Frontiers in Education 09 frontiersin.org
sampling design is less efficient for estimating the mean than a simple
random sample. The design effect is thus the inflation factor that has
to be applied to the conventional variance estimates to adjust error
estimates based on SRS assumptions to account for the effect of the
clustering design.
A frequently used measure that is specific to two-stage sampling
designs is the intraclass correlation (ICC) ( Cochran, 1977 , p. 209). A
high intraclass correlation means that there are large differences
between schools and a low intraclass correlation means that there are
small differences between schools. In the case of LSA, it is defined as
the ratio of the between-school variance and the total variance:
Intraclass Correlation
Between school Variance .B-etween school Variance Within school Variance - -
=
+
The final measure we use to evaluate the sample is the effective
sample size. This is defined as the sample size divided by the design
effect and gives the sample size for an SRS that would be needed to
obtain the same variance of the estimated mean.
With respect to measurement quality, the following indicators can
typically be checked: test difficulty, test reliability, IRT model fit, cross-
domain correlations, residual variance, PV reliability, equating error,
and test validity. For test difficulty, the average proportion correct can
be reported and it can be used to evaluate the appropriateness of the
items for the sample of students. Since polytomously scored items are
commonly used in LSA, the proportion correct is often defined as the
mean score divided by the maximum possible score. However, for
adaptive testing, as done in PISA, the proportion correct is not a useful
measure of test difficulty because high-proficiency students would
be administered high-difficulty items and low-proficiency students
would be administered low-proficiency items. 2 If IRT methods are
used, the match between the test information function (TIF) and the
distribution(s) of student proficiency could be used as a measure of
test difficulty appropriateness. It should be noted that test difficulty
and student proficiency always interact.
Test reliability can be calculated in different ways. Classical
methods (e.g., Cronbach’s alpha) can be used to compute test reliability
for test forms, but model-based methods can be used to get an overall
estimate of IRT reliability (see, e.g., Kim, 2012 ).
IRT models are based on assumptions which need to be tested
before inferences are warranted. Assumptions related to the shape of
2In a perfect adaptive test, the proportion correct would always be 0.50,
irrespective of the proficiency level of sampled students.
the item response functions, item fit, local independence, and
dimensionality are evaluated in the context of IRT model fit. For
example, if item parameters are assumed to be equal across groups
(e.g., related to countries, languages, assessment cycles), the extent to
which this assumption holds can be assessed with item fit statistics.
If multiple domains are assessed (as in NAS and PISA), cross-
domain correlations can contribute to measurement precision and can
be reported. In addition, it is of interest to know how much of the
variance in proficiency is explained by the conditioning variables.
Finally, PV reliability can be reported as an overall summary of the
measurement precision (see also Table 5 ).
With respect to communicating results, the standard error of
mean proficiency is commonly reported. Linking errors, as described
in the previous section, are relevant for determining the significance
of trends and can be provided in the technical documentation of an
LSA. More detailed information on several aspects of the statistical
outcomes of LSAs can be found in Chapters 6, 7, and 8 of Rutkowski
et al. (2014) .
6 Results of comparing NAS, NAEP,
and PISA
In this section, we compare NAS, NAEP, and PISA. In the first
part, we compare the three LSAs in terms of the four key elements
described in Section 3. In the second part, the LSAs are compared on
some of the statistical outcomes of Section 5.
Table 6 summarizes the components in the four key elements
to facilitate a high-level comparison among the NAS, PISA, and
NAEP. As is evident from Table 6 , the NAS in India shares many
design features with NAEP in the United States. Both LSAs serve
as national assessments with a curricular focus, diverging from
PISA, which has a broader focus on the general competencies of
15-year-old students. While NAEP (e.g., see NAEP mathematics
framework; NAGB, 2022, p. 1) and NAS target subject-specific
learning outcomes (e.g., NCERT, 2021a , p. 7), PISA focuses on
assessing broader cognitive skills not necessarily tied to a specific
curriculum. Both NAS and NAEP provide snapshots of educational
progress at different grade levels. In contrast, the PISA targets
15-year-old students, presenting unique challenges in the testing
window because the population of 15-year-olds changes
throughout the school year, an issue exacerbated by the disparities
in the start and end of the school year across the
participating countries.
Target populations can shape assessment design. Notably, the
multilingual landscape in India necessitates the adaptation of NAS
into 22 languages, a practice that mirrors the expansive multilingual
administration of PISA across 85 countries in 2022. In contrast, the
TABLE 5 Design elements and statistics for comparison.
Sampling 2. Measurement 3. Reporting
•Sample size
•Population coverage
•Exclusion rate
•Design effect
•Intraclass correlation
•Effective sample size
•Test difficulty
•Test reliability
•Model fit
•Cross-domain correlations
•PV reliability
•Standard error of mean
•Linking error van Rijn et al. 10.3389/feduc.2024.1422030
Frontiers in Education 10 frontiersin.org
cognitive assessment within NAEP primarily remains administered in
English, with only a limited number of test booklets in selected
subjects translated into Spanish. This stark contrast in the translation
of these assessments reflects the diverse linguistic landscapes within
which these assessments operate, underscoring the critical role of
language accessibility in ensuring equitable and comprehensive
educational evaluations.
A shift toward digital-based assessments is observed in both PISA,
largely conducted on computers since 2015, and NAEP, which
transitioned most of its assessments from paper-based to digital-based
formats by 2019. In contrast, NAS relies on paper-based assessments,
a design choice that possibly hinges on the digital readiness of its
student population. The NAS 2021 contained only multiple-choice
(MC) items administered linearly. The NAEP assessments contain MC
and constructed response (CR) items, but also administered them
linearly. The PISA, on the other hand, contains MC and CR items and
is an adaptive test.
In terms of analysis, the NAS used item response theory (IRT) as
its scaling methodology, a practice also shared by PISA and
NAEP. NAS and PISA employ the 2PL model in scaling the multiple-
choice items, whereas NAEP uses the 3PL. The choice of the IRT
model often rests on the assumptions made during the initial set up
of the assessment (see Maris and Bechger, 2009 and related
discussions, e.g., Thissen, 2009 ). However, PISA and NAEP use an
additional conditioning step (i.e., using background and contextual
information; see Meng, 1994 ; Rutkowski, 2011 ) to generate plausible
values while the NAS used a single proficiency score for each student
in 2021.
The NAS, NAEP and PISA are large scale educational assessments
to provide information on educational progress and trends. However,
each assessment is uniquely tailored to meet its assessment purpose
and methodological challenges and concerns and the population it
served. When comparing assessments, it is important to understand
each assessment in its entirety, hence the necessity of our framework.
For instance, suggesting that NAS implement conditioning models is
a failure to understand that the conditioning models are design
choices tied to the methodology used to estimate population
proficiency. The distinctions in Table 6 should be kept in mind when
interpreting and understanding the assessments’ statistical outcomes
in Table 7 .
In designing LSAs, it is important to determine to what extent
each element impacts the precision of the reported results. In
comparing statistical outcomes, rather than providing an overall
comparison, we focus here on a single domain in a single target
population for each of the three LSAs. In our comparison, we use the
NAS 2017 language assessment for grade 8, NAEP 2022 reading
assessment for grade 8, and the PISA 2018 reading assessment. The
comparison is at the state level for both NAS and NAEP, but at the
country level for PISA.
As evident from Table 7 .The selected LSAs in this comparison
from NAS, NAEP, and PISA do not all publicly report the same
statistics, which complicates the comparison. In the absence of publicly
available information, we cannot determine if the omission was due to
methodological concerns or constraints over its computation, or if the
statistics were simply unavailable publicly. In some cases, design
choices led to the unavailability of information. For instance, NAEP
TABLE 6 Assessment characteristics of NAS, NAEP, and PISA.
Element Component NAS 2021 NAEP 2022 PISA 2022
Sampling design Target population Grades 3, 5, 8, 10 Grades 4, 8 15-year-olds
Sampling Two-stage Two-stage Two-stage
PSU School School School
Stratification Yes Yes Yes
Assessment design Assessment framework Learning outcomes Subject-specific skills Future preparedness
Number of languages 22 1, some Spanish 125
Subjects Language, mathematics,
environmental science,
science, social science, English
Mathematics, reading, civics, US History
Other years: Arts, economics, geography,
science, technology and engineering, writing
Mathematics, reading, science,
creative thinking, financial literacy
Delivery mode Paper-and-pencil Digital Digital
Item format MC MC +CR MC +CR
Assessment type Linear Linear Adaptive
Subjects per student 2–3 12
Contextual questionnaires Student, school, teacher Student, school, teacher Student, school, teacher, parent
Methodology IRT model 2PL 2PL/3PL/GPCM 2PL/GPCM
Calibration Fixed-item Concurrent Fixed-item
Conditioning No Yes Yes
Proficiency WLE PV PV
Reporting Metrics Scale scores and levels Scale scores and levels Scale scores and levels
Reporting levels National, state, district National, state, and selected districts National and selected regions
NAS, National Achievement Survey; NAEP, National Assessment of Educational Progress; PISA, Programme for International Student Assessment; PBA, Paper-Based Assessment; DBA,
Digital-Based Assessment; MC, Multiple Choice; CR, Constructed Response; GPCM, Generalized Partial Credit Model; WLE, Weighted maximum Likelihood Estimate; PV, Plausible Values. van Rijn et al. 10.3389/feduc.2024.1422030
Frontiers in Education 11 frontiersin.org
assesses a single domain, so cross-domain correlations are not
available. Likewise, PV reliability is not available for NAS, as WLEs
were used.
Despite the sparsity of available statistics for comparisons,
several observations can be made from Table 7 . First, NAS samples
more schools and students than NAEP and PISA, but this can
be explained by the fact that the goal in NAS is to report results at
a more fine-grained level as well (the district level), underscoring
the importance of understanding each assessment in its entirety.
Second, the framework allows us to estimate some unreported
statistics. For instance, the effect size is inversely related to the
intraclass correlation. Given that the average cluster size per
school for NAS 2017 approximated the cluster size of PISA 2018,
we expect the intraclass correlation for NAS 2017 to fall within
that of PISA 2018. Finally, should such a framework be widely
adopted by researchers, more statistics may become available.
With the availability of more information, new assessments in
development will have a wider array of tools to aid key
design decisions.
7 Discussion
Our examination of the four key elements of sampling design ,
assessment design , analysis methodology , and reporting reveals several
key similarities and differences among three distinctive assessments:
NAS, NAEP, and PISA. Our framework and comparison results
highlighted the nuanced ways in which each LSA is tailored to meet the
specific needs and challenges of its respective assessment purpose and
the populations these assessments aim to serve. In our analysis, the
comparison of assessment statistics proved to be challenging. The
selected LSAs (NAS, NAEP, and PISA) do not always report the same
relevant statistical outcomes for comparing sampling and assessment
designs. There are additional comparison challenges due to differences
in design and methodology, as well as some publicly
unavailable statistics.
We emphasize that the goal of our framework is not to rank LSAs
in terms of quality as differences can be challenging to interpret fairly.
A systematic framework such as the one proposed in this paper however
allows the building of bridges among LSAs to enhance understanding
of sampling design, assessment design, analysis methodology, and
reporting, facilitating mutual learning among LSA designers. The LSA
comparison framework can also be a self-monitoring tool. Examining
the commonalities and differences between LSAs can be helpful in self-
evaluating the design and analysis choices made within a given
assessment. For instance, an organization can ask whether differences
reflect optimal choices for their assessment relative to other design and
analysis frameworks or surface suboptimal choices in the organization’s
planning of the design and analysis for their assessment. The tool also
allows assessments to evaluate its changes across cycles to monitor the
impact of design alterations on statistical outcomes.
It can also be instructive to look beyond the current practice in
LSAs to find novel solutions. In such applications, our framework
provides an overview of the practices for key design and analysis
components of existing LSAs. For instance, to improve the precision
of the tests and potentially improve student engagement, PISA
adopted multi-stage adaptive testing in 2018 before the design was
used in other LSAs. Likewise, NAEP pioneered the PV methodology
back in the late 1980s by examining advances for analyses in the
presence of missing data. Furthermore, the move to digital-based
assessment allows to capture additional relevant data about the
response process (e.g., response times), that could be used to reduce
total survey error.
In conclusion, LSAs serve a unique role in measuring students’
learning outcomes and relatively few are well-known. Viewing existing
LSAs as a collection of common applications can be instructive for an
TABLE 7 Statistical outcomes for NAS 2017, NAEP 2022, and PISA 2018.
Statistic NAS 2017
Language grade 8
NAEP 2022
Reading grade 8
PISA 2018
Reading
Comparison level State State/Jurisdiction Country
Number of states/countries 36 51 79
Average number of schools per state/country (SD) 954 (809) 110 (NA 1)279 (177)
Average number of students per state/country (SD) 21,261 (18,670) 2,100 (NA) 7,746 (4,737)
Population coverage (SD) NA NA 97.0% (2.3%)
Overall exclusion rate (SD) NA 2.0% (NA) 3.0% (2.3%)
Design effect (SD) 7 (NA) NA 5.7 (3.7)
Intraclass correlation (SD) NA NA 0.33 (0.12)
Effective sample size (SD) NA NA 2,324 (2,877)
Test reliability (SD) 0.71 (NA) NA NA
Cross-domain correlations NA –20.73–0.81
PV reliability (SD) –2NA 0.93 (0.01)
Standard error of mean (SD) 1.3 (1.7) NA 2.5 (0.8)
Linking error NA NA 3.9 3
1NA means that a value is not available from publicly available information.
2Values cannot be determined due to either sampling design, assessment design, or analysis methodology.
3Note that differences in reporting scales would need to be considered in comparing linking errors across LSAs, but they are not available for NAS and NAEP.
van Rijn et al. 10.3389/feduc.2024.1422030
Frontiers in Education 12 frontiersin.org
assessment organization’s planning purposes and for directing future
research directions. In closing, we hope with our framework, choices
made and lessons learned across different LSAs can benefit future designs.
Data availability statement
The original contributions presented in the study are included in
the article/supplementary material, further inquiries can be directed
to the corresponding author.
Author contributions
PR: Writing – original draft, Writing – review & editing,
Conceptualization, Data curation, Formal analysis, Investigation,
Methodology. H-HP: Writing – original draft, Writing – review &
editing, Conceptualization, Data curation, Formal analysis,
Investigation, Methodology. DM: Writing – original draft, Writing –
review & editing, Conceptualization, Investigation, Methodology. IB:
Supervision, Writing – review & editing, Resources. JB: Supervision,
Writing – review & editing, Resources.
Funding
The author(s) declare that financial support was received for the
research, authorship, and/or publication of this article. This research
was funded by PARAKH, an independent constituent body of NCERT,
New Delhi, India.
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.
Publisher’s note
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.
References
Adams, R. J., Wilson, M., and Wang, W. C. (1997). The multidimensional random
coefficients multinomial logit model. Appl. Psychol. Meas. 21, 1–23. doi:
10.1177/0146621697211001
Beaton, A. E., and Zwick, R. (1992). Chapter 1: overview of the National Assessment
of educational Progress. J. Educ. Stat. 17, 95–109. doi: 10.3102/10769986017002095
Biemer, P. P. (2010). Total survey error: design, implementation, and evaluation. Public
Opin. Q. 74, 817–848. doi: 10.1093/poq/nfq058
Black, P., and Wiliam, D. (2007). Large-scale assessment systems: design principles drawn
from international comparisons. Measurement 5, 1–53. doi: 10.1080/15366360701293386
Clarke, M., and Luna-Bazaldua, D. (2021). Primer on large-scale assessments of
educational achievement. Washington, DC: World Bank. Available at: https://
openknowledge.worldbank.org/entities/publication/618b21d1-e54e-5a94-a088-
b7ed8965a990 (Accessed November 6, 2023).
Cochran, W. G. (1977). Sampling techniques. 3rd Edn. New York: John Wiley & Sons.
Cornfield, J. (1951). Modern methods in the sampling of human populations. Am. J.
Public Health 41, 654–661. doi: 10.2105/AJPH.41.6.654
Cresswell, J., Schwantner, U., and Waters, C. (2015). A review of
international large-scale assessments in education: assessing component skills and
collecting contextual data. OECD Report. Available at:
a-review-of-international-large-scale-assessments-9789264248373-en.htm (Accessed
November 6, 2023).
Groves, R. M., and Lyberg, L. (2010). Total survey error: past, present, and future.
Public Opin. Q. 74, 849–879. doi: 10.1093/poq/nfq065
Jewsbury, P., Finnegan, R., Xi, N., Jia, Y., Rust, K., and Burg, S. (2020). 2017 NAEP
transition to digitally based assessments in mathematics and reading at grades 4 and 8:
mode evaluation study. Available at:
publications/main2020/pdf/transitional_whitepaper.pdf
Kim, S. (2012). A note on the reliability coefficients for item response model-based
ability estimates. Psychometrika 77, 153–162. doi: 10.1007/s11336-011-9238-0
Kish, L., and Frankel, M. R. (1974). Inference from complex samples. J. R. Stat. Soc. B
Stat. Methodol. 36, 1–22.
Kolen, M. J., and Brennan, R. L. (2004). Test equating, scaling, and linking. New York:
Springer.
Laukaityte, I., and Wiberg, M. (2017). Using plausible values in secondary analysis in
large-scale assessments. Commun. Stat. Theory Methods 46, 11341–11357. doi:
10.1080/03610926.2016.1267764
Lietz, P., Cresswell, J. C., Rust, K. F., and Adams, R. J. (2017). “Implementation of
large-scale education assessments” in Implementation of large-scale education
assessments. eds. P. Lietz, J. Cresswell, K. Rust and R. J. Adams (New York: Wiley), 1–25.
Maris, G., and Bechger, T. (2009). On interpreting the model parameters for the three
parameter logistic model. Measurement 7, 75–88. doi: 10.1080/15366360903070385
Marsman, M., Maris, G., Bechger, T., and Glas, C. (2016). What can we learn from
plausible values? Psychometrika 81, 274–289. doi: 10.1007/s11336-016-9497-x
Meng, X. L. (1994). Multiple-imputation inferences with uncongenial sources of input.
Stat. Sci. 9, 538–558. doi: 10.1214/ss/1177010269
Michaelides, M. P., and Haertel, E. (2004). Sampling of common items: an
unrecognized source of error in test equating. Los Angeles: National Center for Research
on Evaluation, Standards, and Student Testing, Center for the Study of evaluation,
Graduate School of Education & Information Studies, University of California.
Ministry of Education, India. (2021). National achievement survey national report
Available at: (Accessed April 14, 2023).
Mislevy, R. J. (1991). Randomization-based inference about latent variables from
complex samples. Psychometrika 56, 177–196. doi: 10.1007/BF02294457
Mislevy, R. J., Beaton, A. E., Kaplan, B., and Sheehan, K. M. (1992). Estimating
population characteristics from sparse matrix samples of item responses. J. Educ. Meas.
29, 133–161.
Monseur, C., and Berezner, A. (2007). The computation of equating errors in
international surveys in education. J. Appl. Meas. 8, 323–335.
National Assessment Governing Board. (2022). Mathematics assessment framework
for the 2022 and 2024 National Assessment of educational Progress. Available at: https://
www.nagb.gov/content/dam/nagb/en/documents/publications/frameworks/
mathematics/2022-24-nagb-math-framework-508.pdf (Accessed December 4, 2023).
NCERT (2019). NAS 2017 - class III, V and VIII: national report to inform policy,
practices, and teaching learning. New Delhi, India: NCERT.
NCERT (2021a). Technical note on assessment framework. New Delhi, India:
NCERT.
NCERT (2021b). Notes on sampling design for National Achievement Survey (NAS)
New Delhi, India: NCERT.
OECD (2017). PISA 2015 technical report. Paris: OECD Publishing. Available at:
(Accessed November 6,
2023).
OECD (2020). PISA 2018 technical report. Paris: OECD Publishing. Available at:
technical-report/pisa-2018-technical-report-files/full-report/PISA%202018%20
Techincal%20report%20full.zip (Accessed July 19, 2024).
OECD (2023). PISA 2022 technical report. Paris: OECD Publishing. Available at:
(Accessed February 22, 2024).
Robitzsch, A., and Lüdtke, O. (2019). Linking errors in international large-scale
assessments: calculation of standard errors for trend estimation. Assess. Educ. 26,
444–465. doi: 10.1080/0969594X.2018.1433633
Robitzsch, A., and Lüdtke, O. (2024). An examination of the linking error currently
used in PISA. Measurement 22, 61–77. doi: 10.1080/15366367.2023.2198915 van Rijn et al. 10.3389/feduc.2024.1422030
Frontiers in Education 13 frontiersin.org
Rubin, D. (1987). Multiple imputation for nonresponse in surveys. New York: John
Wiley & Sons.
Rust, K. (2014). Sampling, weighting, and variance estimation in international large-
scale assessments. In L. Rutkowski, DavierM. von and D. Rutkowski (Eds.), Handbook
of international large-scale assessment: background, technical issues, and methods of
data analysis (pp. 117–153). Boca Raton, Florida: CRC Press.
Rust, K. F., Krawchuk, S., and Monseur, C. (2017). Sample design, weighting, and
calculation of sampling variance. In Implementation of Large‐Scale Education
Assessments . eds. P. Lietz, J. C. Cresswell, K. F. Rust and R. J. Adams (New York: Wiley),
137–167.
Rutkowski, L. (2011). The impact of missing background data on subpopulation
estimation. J. Educ. Meas. 48, 293–312. doi: 10.1111/j.1745-3984.2011.00144.x
Rutkowski, L., von Davier, M., and Rutkowski, D. (2014). Handbook of international
large-scale assessment: background, technical issues, and methods of data analysis. Boca
Raton, Florida: CRC Press.
Sheehan, K. M., and Mislevy, R. J. (1988). Some consequences of the uncertainty in
IRT linking procedures. ETS Res. Rep. Series 1988, i–40.
Simon, M., Ercikan, K., and Rousseau, M. (2013). Improving large-scale assessment
in education: theory, issues, and practice. New York: Routledge.
Thissen, D. (2009). On interpreting the parameters for any item response model.
Measurement 7, 106–110. doi: 10.1080/15366360903117061
von Davier, M., Gonzalez, E., and Mislevy, R. (2009). What are plausible values and
why are they useful. IERI Monogr. Series 2, 9–36.
von Davier, M., and Sinharay, S. (2014). Analytics in international large-scale
assessments: item response theory and population models. In L. Rutkowski, DavierM.
von and D. Rutkowski (Eds.), Handbook of international large-scale assessment:
background, technical issues, and methods of data analysis (pp. 155–174). Boca Raton,
Florida: CRC Press.
Ward, M. (2018). PISA for development: results in focus. Paris: OECD Publishing.
Available at:
en (Accessed November 6, 2013).
Weisberg, H. F. (2005). The total survey error approach: a guide to the new science of
survey research. Chicago: University of Chicago Press.
Wu, M. (2005). The role of plausible values in large-scale surveys. Stud. Educ. Eval. 31,
114–128. doi: 10.1016/j.stueduc.2005.05.005
Wu, M. (2010). Measurement, sampling, and equating errors in large-scale
assessments. Educ. Meas. Issues Pract. 29, 15–27. doi:
10.1111/j.1745-3992.2010.00190.x |
11612 | https://www.mathed.page/triangles.html | | | |
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| Visit Henri Picciotto's Math Education Page. | Send me e-mail . |
| Mapping Triangles to Points |
| Two messages posted on the Math Forum website (geometry-precollege news group) long ago, followed by my response: ----------------------------------------------------------- > Given equilateral triangle ABC and point P (in the plane containing > ABC), prove that PA, PB, and PC satisfy the triangle inequalities. > That is: PA+PB >= PC PB+PC >= PA PC+PA >= PB. > Determine the locus of P for which an equality can occur. > > Happy holidays to everyone! > > From, > Ken ----------------------------------------------------------- > Let us map the point P to the (abstract) triangle whose edge-lengths > are PA, PB, PC. > > Then this establishes a 1-1 correspondence between the points of > the plane and the possible shapes of triangles. > > John Conway ----------------------------------------------------------- Neil Picciotto and I had a great time thinking about this, with the (essential!) help of Cabri. We ended up with a fair understanding of the geometry underlying the abstract correspondence suggested by John Conway. As it turns out, if you ignore symmetry, there are actually two points P and P' that correspond to each triangle shape. (With symmetry, there are 12 points.) Getting the triangle from P is straightforward. We found a way to: 1. Construct P and P' given the corresponding triangle. (This construction is the key to the proof that the correspondence is 1-1.) Along the way, we found some interesting intermediary problems: 1. What is the locus of all the Ps that correspond to right triangles? (This is easier than the remaining questions.) 2. What is the locus of all Ps that correspond to triangles that include a 30 degree angle? Same question for 60, 120, 150. 3. What is the relationship between angles APB, BPC, and CPA and the angles in the corresponding triangle? (This was rather unexpected. The proof is a rather traditional Euclidean proof, and is the key to understanding this problem.) Finally, a couple of extensions: 1. What is the relationship between P and P'? (I have a conjecture, but no proof.) 2. (This question was suggested by Jean-Marie Laborde.) What happens when the original triangle is not equilateral: for a given triangle ABC, for what points P do PA, PB, and PC satisfy the triangle inequality? for what points is there equality? |
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Finding the centroid of a triangle using vectors
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I would like some help about this:
Let there be 2 triangles $ABC$ and $DEF$. $A(2,0,1), B(3,2,-2), C(0,1,-1), D(-1,1,1), E(-3,2,0), F(0,1,1)$ and let $T_{1}$ and $T_{2}$ be centroids of triangle. Find $\overrightarrow{T_{1}T_{2}}$.
So centroid line(median) from point $A$ makes point $G$ which is middle point of line $BC$. It should have coordinates $$G = {1\over 2}(B+C) = \left({3\over 2},{3\over 2},-{3\over 2}\right)$$
So centroid line, in this case from point $A$ to $T_{1}$, is two times longer than $T_{1}$ to $G$. How can I find coordinate of $T_{1}$? Can i make vector $\overrightarrow{AG}$ and calculate $T_{1} = {1\over 3}(A+G)$ ?
Thanks for help!
Edit: Just found formula for finding coordinates of centroid, it should be $$T_{1} = {1\over 3}(x_{1}+x_{2}+x_{3},y_{1}+y_{2}+y_{3}, z_{1}+z_{2}+z_{3})$$ Could I solve this problem my or some other way?
geometry
vectors
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edited Jun 26, 2019 at 2:31
Saranga B
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1
$\begingroup$
Consider this a start (i.e. a hint). To find the centroid of either triangle, use the definition.
Median. The median of a triangle is a line or line segment from a vertex to the midpoint of the opposite side.
Centroid. If three medians are constructed from the three vertices, they concur (meet) at a single point. That point is called the centroid. The centroid is a balance point for a triangle because all of the interior triangles that are formed have equal area.
So if 3 lines intersect at a point, then so 2 lines must intersect at the same point.
Step 1: Find the midpoint of segment AC. Let's call it $I=(1,.5,0)$
Step 2: Get an equation for the line between point B and opposite segment midpoint I. $(x,y,z)=B+\lambda(-2,-3/2,2)$ The direction vector, $(-2,-3/2,2)$ was obtained as $(I-B)$.
Step 3: Find the midpoint of segment CB. $J=(-4/3,4/3,2/3)$
Step 4: Get an equation for the line between point A and the opposite segment midpoint J. $(x,y,z)=A+\lambda(J-A)$ This one is $$ \left(\begin{array}{c} x\ y\ z \end{array}\right)=\left(\begin{array}{c} 2\ 0\ 1 \end{array}\right)+\lambda\left(\begin{array}{c} -\frac{1}{2}\ \frac{3}{2}\ -\frac{5}{2} \end{array}\right)$$
The intersection of these two lines is your centroid $T_1=(5/3, 1,-2/3)$. AND maybe I don't have numerical errors.
You can continue this way to get $T_2$ and then do whatever with them.
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answered Sep 18, 2018 at 0:01
NarlinNarlin
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11615 | http://physics.wm.edu/~labs/110/110_pdf/ch4.pdf | Chapter 4 Energy and Momentum - Ballistic Pendulum 4.1 Purpose . In this experiment, energy conservation and momentum conservation will be investigated with the ballistic pendulum.
4.2 Introduction One of the basic underlying principles in all of physics is the concept that the total energy of a system is always conserved. The energy can change forms (i.e. kinetic energy, potential energy, heat, etc.), but the sum of all of these forms of energy must stay constant unless energy is added or removed from the system. This property has enormous consequences, from describing simple projectile motion to deciding the ultimate fate of the universe.
Another important conserved quantity is momentum. Momentum is especially important when one considers the collision between to objects. We generally divide collisions into two types: elastic and inelastic. In an inelastic collision, the colliding objects stick together and move as one object after the collision, whereas in an elastic collision the two objects move independently after the collision. Both types of collisions display conservation of momentum.
In this experiment we will be using both conservation of momentum and conservation of energy to find the velocity of an object (ball) being fired from a spring loaded gun. The apparatus we will be using consists of a spring loaded gun which fires a small metal ball into a hanging pendulum. See Figure 4.1. The momentum and energy of the ball then causes the pendulum to swing up a ramp, which will catch the pendulum so that we can measure how high it swung.
When the metal ball is initially fired from the gun, it will have a kinetic energy, KEi: KEi = 1 2mv2 0 (4.1) where m is the mass of the metal ball and v0 is the initial velocity of the ball before it strikes the hanging pendulum.
19 V 0 m M M+m h h 1 2 h ∆ Figure 4.1: Schematic diagram show the components of the ballistic pendulum apparatus After the collision takes place, and the metal ball is stuck in the pendulum, the pendulum-ball combination has a kinetic energy: KEf = 1 2(m + M)v2 f (4.2) where M = mass of the pendulum and vf is the velocity of the pendulum/ball moving together.
As the pendulum-ball swings, its kinetic energy is converted to potential energy. When it reaches its highest point, all of its energy has been converted to potential energy: PEf = (m + M)g(∆h) (4.3) where g = gravitational acceleration (9.8 m/s2) and ∆h is the change in height of the pendulum-ball combination (final height of pendulum-ball combination - initial height of pendulum-ball combination).
We can look at the entire process in a step-by-step manner and see that, because of conservation of energy and conservation of momentum, all of these steps are related to each other. Therefore, if we can measure the total energy or momentum of the system at any time during the experiment, we can find out everything else about the system. The easiest thing to measure in this setup is the potential energy in Equation 4.3 (mainly because nothing is moving at that point). By doing this, we will work backwards to find out how fast the metal ball is shot out of the gun. Finally for comparison, we will use a photo-gate timing system to measure the speed of the ball when it is shot out of the gun.
4.3 Equipment: Ballistic pendulum apparatus, ruler, photo-gate timing system.
20 Figure 4.2: The ballistic pendulum with the pendulum on the ratchet mechanism which prevents the pendulum ring from falling from its maximum height.
4.4 Procedure: A photograph of the ballistic pendulum is shown in Figure 4.2 with the photo-gate timing system.
In the first part of the experiment, we will measure the final potential energy of the pendulum in order to find the initial velocity of the metal ball. In the second part of the experiment, the initial velocity of the metal ball will be measured using the photo-gate timing apparatus to directly measure the time it takes for the ball to move a fixed distance.
• Using the scales, find the mass of the metal ball,m. The mass of the pendulum, M, is 0.20 kg.
• Using a ruler, measure the distance from table to the center of the pendulum ring. A small red ’dot’ marks the center of the pendulum ring.
• Place the ball on the gun and push it against the spring back to the second or third detent position. The first detent position produces insufficient vo. Always use the same position of the spring gun for all data.
• Fire the metal ball into the ring and measure the distance from the table to the center of the pendulum ring when the pendulum/ball combination is stopped at its highest peak.
• By conservation of energy, the potential energy at the highest peak is equal to the kinetic energy immediately after the collision. Using equation 4.2 and equation 4.3, we have: KEf = 1 2(m + M)v2 f = (m + M)g(∆h) = PEf (4.4) solving for vf we have: 21 vf = q 2g(∆h) (4.5) • By conservation of momentum, the momentum of the system after the collision must be equal to the momentum of the system before the collision. Since Momentum = (mass) x (velocity), we have: (m + M)vf = mv0 (4.6) This can be used to find the initial velocity v0: v0 = (m + M) m vf (4.7) Using equation 4.5 for vf: v0 = (m + M) m q 2g(∆h) (4.8) Using equation 4.8 and the values for M, m, g and ∆h, calculate v0 • Repeat the measurement two more time. Calculate the average (mean) value of the three measurements of v0.
• Move the ballistic pendulum to the photo-gate setups where your teaching assistant will help you align the apparatus with the photo-gates and timer.
• Push the pendulum up the ramp and out of the way. Align the photo-gates so the ball with travel through the photo-gates without hitting either photo-gate or other objects like wires when the gun is fired. The first photo-gate should be located just ahead of the position where the ball leaves the spring gun. Use a box to catch the ball after it passes through the second photo-gate.
• Measure the distance between the centers of the two photo-gates. Verify the timer is in ’pulse’ mode. Push the reset button on the timer. The timer will start when the ball passes the first photo-gate and stop when the ball passes the second photo-gate.
• Fire the ball into the box. Record the time. Repeat the measurement two more times.
(Always check the alignment each time the spring gun is ’reloaded’.) • Calculate v0 for each of the time measurements using v = distance time . Calculate the mean (average) of the three values.
• Calculate the percentage difference between v0 from the first part of the experiment with v0 measured with the photo-gate timer.
22 4.5 Questions: 1. How does the mean of your v0 from the first part of the procedure using energy and momentum conservation compare with the value from the measurement from the photo-gates and timer. What is the percentage difference between the averages (means) of the two different measurements of v0?
2. List possible sources of error in each of the two measurements of v0. Which procedure do you think is more accurate? Why?
23 24 |
11616 | https://arc.aiaa.org/doi/pdfplus/10.2514/1.18069 | AIAA Aerospace Research Central
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11618 | https://www.youtube.com/watch?v=foQ_SI9Ejog | Acids and Bases | Chapter 3 - Organic Chemistry (2nd Edition)
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Chapter 3 of Organic Chemistry (Second Edition) by Clayden, Greeves, and Warren introduces the critical roles of acids and bases in organic chemistry, laying the theoretical groundwork for understanding reactivity, stability, and molecular behavior. It begins with a thorough comparison of Brønsted–Lowry and Lewis definitions: the former focuses on proton transfer, while the latter generalizes acid–base interactions to include electron pair donation and acceptance. The chapter stresses the importance of knowing what species exist in equilibrium by using acid and base strengths to predict directionality of proton transfer reactions. Key to this is the concept of pKa, a logarithmic measure of acid strength. The text explains how lower pKa values correspond to stronger acids and provides tables of common functional groups and their conjugate acid–base pairs, such as alcohols vs. alkoxides and carboxylic acids vs. carboxylates.
Readers are trained to assess acidity and basicity trends by evaluating factors like resonance stabilization of the conjugate base, electronegativity of atoms bearing negative charge, hybridization effects (sp vs. sp² vs. sp³), and inductive effects caused by nearby electron-withdrawing or donating groups. The chapter illustrates these principles through examples involving phenols, amides, nitriles, and sulfonic acids. A detailed discussion of how delocalization and aromaticity affect pKa is provided, with an emphasis on how conjugate base stability dictates acid strength. Lewis acid-base theory is presented as a broader conceptual framework, where even molecules without protons (like AlCl₃ or BF₃) can function as acids due to their ability to accept electron pairs.
Importantly, the chapter establishes that many organic reactions are driven by acid-base interactions—proton transfers often initiate or terminate reaction mechanisms. The authors emphasize understanding acidity and basicity as central to mastering reactivity in later chapters. By drawing comparisons between aqueous and non-aqueous systems, the text helps students appreciate that pKa values are context-sensitive but still extremely useful. The chapter closes by reinforcing the skill of using acid–base logic to explain why certain atoms get protonated or deprotonated, and which direction an equilibrium favors—core reasoning that underpins the rest of the book’s mechanistic approach to organic chemistry.
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Organic Chemistry Chapter 3 summary, acids and bases in organic chemistry, Clayden acid base theory, Brønsted-Lowry vs Lewis definitions, understanding pKa in organic reactions, acid strength and conjugate base stability, electronegativity and acidity, resonance stabilized anions, inductive effects and acid base strength, acidity trends of functional groups, acid base reactions in mechanisms, predicting equilibrium direction with pKa, Lewis acids and electron pair acceptors, proton transfers and organic reactivity, Chapter 3 Clayden Organic Chemistry
Transcript:
Welcome to the deep dive. We're here to cut through the noise and get straight to the crucial insights. Today, we're tackling a really fundamental question in chemistry. How do we actually know what a molecule looks like? Its shape, its composition. I mean, think about it. You isolate something new, maybe from a natural source, could be a potential medicine. How do you figure out precisely what it is and fast? So, that's really what we're getting into today. We're diving deep into determining organic structures, pulling insights from key resources like Clayton, Greavves, and Warren. It's all about the power of spectroscopy, the essential toolkit chemists rely on every single day. Yeah, it's uh it's difficult to really convey just how much spectroscopy revolutionized organic chemistry before these methods were common. I mean, figuring out a structure could take ages, years sometimes. Lots of chemical tests, degradation, quite a bit of well, inspired guesswork, frankly. Spectroscopy changed all that. Speed, certainty. We'll be focusing on the big three today. Mass spectrometry that tells us the mass, the atomic makeup. Then nuclear magnetic resonance, NMR, which is amazing for mapping out the connections, the molecular skeleton. And finally, infrared spectroscopy or IR. That one's brilliant for spotting the specific functional groups within the molecule. Okay, let's kick off with what many chemists consider the ultimate answer when you can use it. X-ray crystalallography. It's like getting the molecules architectural plans. The basic idea seems pretty neat. Shine X-rays at a crystal. The atoms scatter the X-rays and from that pattern. You work backwards. You deduce the exact 3D arrangement of almost every atom. Bond lengths, bond angles, the whole lot. It's incredibly precise. You know how textbooks always draw saturated chains as zigzags? X-ray shows you that zigzag like in hexonedioic acid. You can clearly see it. And the flat caroxyic acid groups too. And it's not just for known structures. It can crap complete unknowns. There was this co-enzymethoxidin used by bacteria. NMR was struggling with it, but X-ray crystalallography figured it out. Pretty amazing stuff. But it's not always the magic bullet, is it? What are the limitations? The biggest one absolutely is the crystal. You need a good quality well-ordered crystal. If your compound's a liquid or it just refuses to crystallize nicely, then X-ray is off the table. Simple as that. Another key thing, it usually doesn't see hydrogen atoms very well. They're just too light. Don't scatter X-rays enough. So, you have to infer where they are, usually based on the heavier atoms positions. Right. Which implies it's not always infallible. Exactly. There was a famous case, an antibiotic called dazonomide A. The structure was accepted for about 10 years based on X-ray data. But the X-ray couldn't reliably tell an oxygen atom from a nitrogen atom in one specific spot. They look quite similar to X-rays. It was only when chemists actually synthesized the proposed structure and it didn't match the natural compound that they realized the X-ray assignment was wrong. The real structure was slightly different. Wow. So X-ray gives us that definitive shape potentially, but you need a crystal and even then you need to be careful, which really sets the stage for why we need these other methods. Let's move on to mass spectrometry. MS. This one's different, right? It doesn't involve absorbing light or energy in the same way. That's right. MS is unique. It weighs molecules. Or more accurately, it weighs charged ions derived from the molecules, not the neutral molecules themselves. How does it do that? What's inside the machine? Well, basically, you get your sample into the gas phase. Then you ionize it, turn the neutral molecules into charged particles. Then you use electric or magnetic fields to separate these ions based on their mass to charge ratio, mild dudes. Finally, a detector counts them. A very common way to ionize is electron impact, EI. It's kind of uh energetic. Imagine firing high energy electrons at your molecules. Boom. One electron gets knocked off, usually a weakly held one. What you're left with is a positive ion, but also with an unpaired electron, a radical confrontation. We call this the molecular ion, M plus U. And crucially, the heaviest ion you normally see in an EI spectrum is this molecular ion. It tells you the molecule's mass. Okay. Like the honeybee alarm pheromone example, Heptan 21, molecular weight 114. Exactly. You'd see a peak at military 114 in its EI spectrum representing that M plus ion. But you said EI is energetic. Does that cause problems? It can. Yes. Because it hits the molecule so hard, it often causes it to break apart, to fragment. Sometimes that fragmentation pattern is useful information in itself. But if your molecule is fragile, you might not even see the molecular ion peak at all. That's why we have gentler methods like chemical ionization CI or electrospay ionization, ESI. These are much softer. They usually add a proton giving you an M plus H+ ion. So M plus1 or sometimes they'll pick up a sodium ion giving M plus Na plus A which is N plus 23. Ah like the electro spray spectrum for Heptan 2 one you mentioned showing a peak at 137 that's the M plus N plus N precisely. So these gentler methods are great for just getting the molecular weight without shattering the molecule. But MS does more than just weigh things. It can hint at the atoms inside. Right. The atomic composition. How does that work? Is it about isotopes? Exactly. It isn't just natural isotopes. Take bromine. It exists naturally as almost a 5050 mix of bromine 79 and bromine 81. So any molecule with one bromine atom will show two peaks in the mass spectrum for the molecular ion. They'll be almost equal in height and separated by two mass units. See that pattern? You immediately think bromine. And chlorine is similar but different ratio. Yeah. Chlorine is about 75% chlorine 35 and 25% chlorine 37. So, one chlorine atom gives two peaks again separated by two mass units, but this time in roughly a 3.1 height ratio. It's another really distinctive signature. Okay, bromine and chlorine have very clear patterns. What about other elements like carbon? Carbon's interesting. Almost all carbon is carbon 12, but about 1.1% is the heavier isotope, carbon 13. This means that for every carbon containing ion peak in the spectrum, there's a tiny little peak, one mass unit higher, the M plus1 peak. And the height of that tiny M plus1 peak relative to the main M peak tells you how many carbon atoms are in that ion. The ratio is basically 100 to 1.1 times the number of carbons. It's incredibly useful for quickly estimating the carbon count. So you get the weight, you get clues about specific elements like H hallogens, even the number of carbons. But what if two different formulas add up to the same whole number mass? Ah that's where highresolution mass spectrometry or HRMS is essential. Standard MS gives you nominal mass, the nearest whole number. But HRMS measures the mass incredibly accurately, maybe to four or five decimal places. And different atoms don't weigh exactly whole numbers. Correct. Because of isotopes and nuclear binding energy, the exact mass of say C7H42 is different from the exact mass of CH18 even though both have a nominal mass of 114. So ATRMS can measure for example 114.1039 for the B pheromone. And that exact mass uniquely matches C7H42O, ruling out all other possibilities. It gives you the definitive atomic composition. That's powerful. And you mentioned a nitrogen rule. Yes. It's a handy little shortcut that often works, especially with HRMS data. If your molecule has an odd exact molecular weight, it almost certainly contains an odd number of nitrogen atoms 135. If the molecular weight is even, it has an even number of nitrogens. 024. It's a quick check. Okay. MS gives mass and composition. Now for the structure, the connections, NMR, you said this is arguably the most important tool for many organic chemists. Yes, absolutely. NMR, nuclear magnetic resonance, it's all about the nuclei of atoms acting like tiny magnets like compass needles. Sort of. Yeah. Certain nuclei, crucially for us, hydrogen 1, protons and carbon 13 have a quantum property called spin. This spin generates a tiny magnetic field. Now put these spinning nuclei in a very strong external magnetic field like inside an NMR spectrometer. They can align either with the field which is a lower energy state or against the field which is slightly higher energy. Okay, two possible states, right? And the energy difference between these two states is tiny corresponding to radio wave frequencies. If you zap the sample with a pulse of radio waves at just the right frequency, the nuclei in the lower state absorb that energy and flip up to the higher state they resonate. When the pulse stops, they relax back down, emitting that energy as a radio signal, which the instrument detects. That signal tells us about the nucleus. But if all protons are protons and all C13 or C13, why don't they all resonate at the same frequency? What causes the different peaks we see in an NMR spectrum? The chemical shift. Ah, that's the key. The chemical shift. It happens because the nucleus isn't isolated. It's surrounded by electrons. And these electrons circulating in response to the big external magnetic field create their own tiny magnetic field that opposes the main one. They shield the nucleus. Exactly. They shield it. So the actual magnetic field experienced at the nucleus, the local field is slightly weaker than the external field. But and this is crucial, the amount of shielding depends entirely on the chemical environment of that nucleus. If a nucleus is near an electronegative atom like oxygen, oxygen pulls electrons away, right? It pulls electron density away, reducing the shielding effect. We call this desshielding. So a d-shielded nucleus feels a stronger local magnetic field. A stronger field means a bigger energy gap between the spin states, which means it needs a higher radio frequency to resonate. So dshielded nuclei resonate at higher frequencies further downfield in the spectrum. Okay, let's take an example. Ethanol CH3 CH2-0. Perfect example. The carbon in the CH2 group is directly attached to the oxygen. Oxygen pulls electrons away, desshielding that carbon. So, its signal will appear at a higher frequency further down field compared to the CH3 carbon which is further away and more shielded. And this difference is measured in ppm, parts per million. Yes, ppm. It's a relative scale comparing the resonance frequency to a standard reference compound, usually tetromethylane TMS. TMS is defined as zero ppm and most other signals appear downfield positive ppm values from it. So when looking at a 13C NMR spectrum, are there general regions we should pay attention to? Definitely. 13C NMR spectra are usually spread over about 200 ppm and you can roughly divide it up. Uh 0 to about 50 ppm. That's typically where you find your standard saturated carbons like in alkanes. They're the most shielded. Then maybe 50 to 100 ppm. These are often saturated carbons, but now they're attached to an electronegative atom, usually oxygen or nitrogen, deshielding them a bit. From 100 to 150 ppm, this is the region for unsaturated carbons. Think alches and aromatic rings. Their electronic environment is quite different. And finally, above 150 ppm, often stretching to over 200 ppm. This is where you find carbonal carbons CO. They're very strongly dshielded because of the double bond and the electronegative oxygen. They really stand out. So looking at lactic acid, you'd see peaks in different regions for the methyl, the CO carbon, and the COB. Precisely. three distinct signals in characteristic regions. Symmetry is also key. Hexonioic acid has six carbons, but due to symmetry, you only see three unique signals in its 13 CNMR. The carbonals would be way downfield. Heptin 21's carbonal would be obvious, probably around 208 ppm. And what about 1HNMR, proton NMR? Similar idea, but different scale. Same fundamental principle. Yes, spin 12 nuclei protons in a magnetic field, radio waves causing transitions. But the chemical shift range for protons is much smaller. Typically just 0 to 10 ppm, maybe 12 ppm sometimes. Why so much smaller than carbon? Mostly because hydrogen only has that one electron directly involved in bonding and it's on the outside of the molecule. The variations in electronic environment just don't cause such huge shifts in shielding compared to carbon which is embedded within the electron clouds of its neighbors. And are there general regions in 1 H&MR2? Broadly speaking, yes. Protons on saturated carbons usually show up between 0 and 5 ppm. Protons on unsaturated carbons like alenies or aromatics are more dshielded and appear between about 5 and 10 ppm. So benzene protons are around 7.5 ppm quite downfield. Cycllohexane protons fully saturated are way up field around 1.35 ppm. And again electronegative atoms caused you shielding the methyl protons next to the oxygen and TBME tool methyl ether are shifted down to about 3.15 ppm. And you mentioned 1HNMR gives even more detail. Oh yes, the real power of 1HNMR comes from something called spin spin coupling which tells you about neighboring protons, but that's a that's a deep dive for another day. For now, just knowing the chemical shift regions is very useful. It sounds like putting the NMR data together is how you really start solving structures. Absolutely. Take those C4H10 alcohol isomers we talked about. Nbutinol, isobutinol, tbutinol, their 13C NMR spectra alone can distinguish them. Nbutinol four signals all different isobutinol three signals because two methyl groups are identical only two signals because the three methyl groups are identical plus the carbon attached to the O. The number of signals tells you about the symmetry. The chemical shifts tell you what type of carbon each signal represents. It lets you piece the puzzle together. Okay, so NMR maps the skeleton. What about the functional groups specifically? That's where IR comes in. Exactly. Infrared spectroscopy is the perfect compliment. It's fantastic for identifying which functional groups are present or absent. It works on a different principle. Molecular vibrations. Vibrations like bonds stretching and bending. Precisely. Bonds aren't rigid sticks. They behave more like springs connecting atoms. They can stretch, bend, waggle, twist. IR radiation has just the right energy to excite these vibrations. When a molecule absorbs IR light at a specific frequency, that corresponds to a particular bond vibration getting excited. The frequency depends on two main things. The strength of the bond. Stronger bonds vibrate faster at higher frequency. And the masses of the atoms involved. Lighter atoms vibrate faster. Think Hook's law like masses on springs. So CH bonds with light hydrogen would vibrate at higher frequencies than CC bonds. Generally, yes. IR frequencies are usually expressed in wave number CME1, which is proportional to frequency. A typical IR spectrum plots absorbance or transmittance often shown upside down with peaks pointing down against wave number. usually from about 4,000 centimeters 1 down to 600 cm one. And are there key regions in the IR spectrum to look at like an NMR? Oh, absolutely. Four main regions are super useful. First, the XH region, roughly 4,2500 cm one. Because hydrogen is so light, bonds to H vibrate at high frequencies. CH stretches are usually just below or around 3,000 cm one. NH stretches are sharper, around 3,300 centimeters one. If you have an NH2 group, you often see two peaks there. And the most characteristic is the O stretch from an alcohol or phenol. If it's hydrogen bonded, which it usually is, you get a very broad rounded peak anywhere from 3500 down to maybe 2900 cm one. Caroxyic acid O stretches are even broader. Often a huge Vshape overlapping the CH region. If the O is not H bonded, like in BHT with its bulky groups, you see a sharp peak around 3600 cm 1. So that broad O peak is a major clue for alcohols or acids. What's next? the triple bond region 2500 2,000 cmters 1. This area is often completely flat empty. So if you do see a peak here, it's highly diagnostic. It's almost certainly either an alken on CH around 2100 cm 1 or nitrol C around 2250 cm 1. Okay. XH triple bonds. What else? The double bond region 2,500 cm01. This is probably the single most important region for functional groups. Why? Because the carbonal group CO lives here. Carbonals give a single usually very strong sharp peak somewhere between about 1900 and 1500 cm of one. The exact position tells you what kind of carbonal it is. Ketone aldahhide estesteride etc. Also in this region you might see alken CC stretches around 1640 cm one usually weaker and aromatic ring vibrations often two or three bands between 1600 and 1500 centimeters of one. Why is the CO peak so strong? Ah good question. The intensity of an IR absorption depends on how much the dipole moment of the bond changes during the vibration. The CO bond is very polar. So stretching it causes a big change in dipole moment leading to a strong absorption. Symmetrical bonds like the CCD and ethne have little or no dipole change on stretching. So their IR peaks are weak or even absent. Makes sense. And the last region below 1500 cm one is the single bond region. Often called the fingerprint region. It's usually very complex, full of peaks from CC stretches, CO stretches, often around 1100 cm one, CN stretches, and lots of bending vibrations. It's generally too messy to interpret bond by bond, but the overall pattern is unique to each specific molecule like a fingerprint. Useful for matching against known spectra, but less so for deducing functional groups from scratch, apart from maybe spotting a strong CO stretch. Okay, so X-ray for the blueprint if possible, mass spec for weight and formula, and MR for the carbon hydrogen skeleton and connectivity, and IR for the functional groups. The real power must come from using them all together. Exactly. It's like detective work. Each technique provides different clues, and you need to combine them to solve the mystery structure. Let's try that example. The unknown industrial emulsifier C4H1. Okay, walk us through it. Right. First, ESMS. We see an M plus H peak at 90. That means the molecular weight M is 89. Odd number. What does that suggest? The nitrogen rule. An odd number of nitrogens. So, one nitrogen. Spot on. High-res MS confirms the formula is indeed C4H1 NO. Okay. Next. 13C NMR. We see only three peaks. But the formula has four carbons. There must be symmetry. Two of the carbons must be identical in the same chemical environment. Perfect. And looking at the chemical shifts, one peak is in the CO region, say 60 70 ppm, and the other two are standard saturated carbons, maybe 2040 ppm. Now, IR, what does that tell us? IR shows a big broad O absorption, typical of an alcohol. And it also shows two sharp peaks around 3,300 3,400 cm 1. NH stretches. Two peaks means NH2, a primary amine. Exactly. So, we have an alcohol group O, and a primary amine group Nash NH2. Since the formula only has one O and one N, these must be the groups. And the carbons are attached to can't be the symmetrical pair. So the symmetry must come from the other two carbons. So C4 H1N with O NH2 and two identical carbons. What structures fit? Maybe identical methyl groups. Good thinking. Two main possibilities come to mind. Two amino 2 methylropen 1L structure A or maybe two aminoutinol structure B. If the identical carbons are part of the chain somehow. No, wait. Two amino methyl propen 1L fits best. It has two identical methals. Let's stick with A. It has a C with two methals, an NH2 and a CH2O group that fits three NMR signals. Okay, structure A seems plausible. How do we confirm it? One HNMR clinches it. If we look at structure A, we expect signals for the O proton, the NH2 protons, the two CH3 protons, which should be one signal, and the CH2 protons next to the O. And let's say the 1 H&MR shows only two main signals for protons on carbon. One dshielded signal around 3.3 ppm and another more shielded signal around 1.1 ppm. That 3.3 ppm signal is characteristic of protons on a carbon next to oxygen the CH2O. The 1.1 ppm signal fits the methyl groups. This pattern perfectly matches structure A two amino 2 methyl propane one all. See how we used all the data. That's a brilliant example of the synergy. It really is like solving a puzzle. You mentioned double bond equivalents earlier as a shortcut. Uh yes, DBEs very handy. Sometimes called degrees of unsaturation. It's a quick calculation from the molecular formula that tells you the total number of rings plus double bonds in the molecule. A triple bond counts as two DBEs. How do you calculate it? For a formula like CNhas Nyra, you can use a formula like DBE equal N M2 plus N2 plus one. Basically, start with the number of carbons, subtract half the hydrogen shall, add half the nitrogens and add one. Each DTE means one ring or one double bond. So if you calculate say four DBEs for C6H6, Bendy, three double bonds and one ring makes sense. Exactly. It's a great first step. If your formula gives you say 1 DBE, you know you're looking for either one double bond like CC or CO or one ring. If you get zero DBZs, the molecule must be saturated in a cyclic. If you get four or more, especially if the hydrogen count is low relative to carbon, you should immediately suspect an aromatic ring. Let's try a final quick case study. Then you react prinol with HBr and ethylene glycol and give product X. How do we identify X? Okay, product X. What clues do we have? MS shows two peaks at 181 and 179 roughly equal height. Broine, one broine atom. Correct. Hyres MS gives the formula C5H9 B2. Okay. C5H9 BO2. Let's calculate the DBE. Five carbons, nine hydrogen's broines, two oxygen. EB = 592 + 02 + 1 + 1 = 5. 4.5 + 1 = 1.5. Yeah, hang on. Halogens count like hydrogen. So, DBE means five 2 plus 1 2 plus 1 2 plus 1= 5 plus 1 = 1. 1 DBE. Perfect. 1 DBE means one ring or one double bond. Now, 13 C NMR propanel had a CO and CC. The spectrum of X shows those signals are gone. Instead, we have four signals. One looks like a CHBR around 3040 ppm. Two look like carbons next to oxygen, maybe 60 70 ppm. And one is really interesting down at 102.6 ppm. 102.6 ppm. It's quite far down field for a saturated carbon, isn't it? You said C next to two oxygen earlier. Exactly. That chemical shift is highly characteristic of an acetal or keto carb carbon single bonded to two oxygen atoms like COC. Okay. So maybe a cyclic acetal formed from the ethylene glycol and the bromine added somehow. What about the IR? IR. This is key. No O stretch, no CO stretch, no CC stretch. Wow. So the 1DB must be a ring. It can't be a double bond precisely. And the IR probably shows strong CO single bond stretches around 1100 cm one confirming the atherosetal linkages. So putting it all together C5H9 B2 1 DBE which must be a ring evidence for a COC unit the acetal at 102.6 6 ppm. Presence of bromine starting materials were propanol and ethylene glycol. The structure reveals itself. It has to be two brooomthol 1 nephr 3 dioxylane. The ethylene glycol formed a cylic acetal with the aldahhide and HBr added across the original double bond. Fantastic. It really shows how you systematically combine all the pieces of evidence. This has been a really insightful look at how chemists actually figure out what they've made or discovered. It underscores a crucial point. Structures in organic chemistry aren't just drawn. They are assigned based on solid spectroscopic data. That evidence is absolutely vital for reliable science. When you read a paper reporting a new molecule, you'll always find its spectral data listed MS, NMR, IIR. It's the proof. It really emphasizes that understanding how we know something is just as important as knowing the fact itself. Absolutely. Knowledge is one thing, but understanding the evidence behind it, that's where the real power lies. Well, this journey through spectroscopy has been fascinating, and it feels like we've just opened the door, especially with NMR. There's clearly more to explore there. Oh, definitely. We haven't even touched on coupling in 1HNMR or more advanced 2D techniques. Plenty more for future deep dives. We look forward to that. A huge thank you for joining us on this deep dive today and thank you our listeners for being part of the last minute lecture family. Until next time, keep exploring the molecular |
11619 | https://math.stackexchange.com/questions/4619063/why-are-there-two-standard-forms-for-parabolas | algebra precalculus - Why are there two standard forms for parabolas? - Mathematics Stack Exchange
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Why are there two standard forms for parabolas?
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I am helping my girlfriend do her math homework. They have been calling
(x−h)2=4 a(y−k)(x−h)2=4 a(y−k)
the standard form. Suddenly in her homework, a different equation has been referred to as standard form:
y=a x 2+b x+c,y=a x 2+b x+c,
which is familiar to me from high school. It's been frustrating for her to see both of these equations called the standard form, especially since questions are asking to convert a freeform equation into 'standard form' when standard form seems ambiguous.
Both equations represent parabolas in different ways, but does anyone have an idea why they are both referred to as standard form?
algebra-precalculus
conic-sections
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edited Jan 15, 2023 at 20:00
Xander Henderson♦
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If a book author has two different descriptions of the same phenomenon, and calls both "the standard form", the author has, in my opinion, made a mistake. If this is what is happening, I would suggest that you find an alternative text.
That being said, a possible explanation for this ambiguity is as follows:
In many texts, a polynomial is said to be in standard form if that polynomial is written with its terms in decreasing order of degree. For example, 9 x 4+3 x 2−8 x+10 9 x 4+3 x 2−8 x+10 is in "standard form", while 10−8 x+9 x 4+3 x 2 10−8 x+9 x 4+3 x 2 is not. Thus a quadratic polynomial has been written in standard form when it is written as a x 2+b x+c,a x 2+b x+c, where a a, b b, and c c are real (or complex, or whatever) coefficients.
Other texts which focus more on conic sections might describe the standard form of a parabola as (x−h)2=4 a(y−k),(x−h)2=4 a(y−k), where a a, h h, and k k are real numbers which have geometric meaning (specifically, (h,k)(h,k) is the vertex, and a a describes a scaling (a a is the distance from the vertex to the focus (in one direction) and the directrix (in the other direction); the sign of a a indicates orientation)).
Hence it is possible for "the" standard form of a parabola to be one of (at least) two different things: the standard form of a degree two polynomial, or the standard form of a conic section. They way in which you are supposed to interpret what is meant by "standard form" is going to depend on what you are trying to describe: is the salient idea that the curve is the graph of a polynomial? or is it that the curve is a conic section?
Finally, as a bit of advice for tackling the homework: if you are given a problem which is phrased like "Write the following in 'standard form': ...", and the meaning of "standard form" is ambiguous, write it in every form which could be "standard. If nothing else, you get familiar with the symbol manipulation.
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edited Jan 16, 2023 at 12:48
answered Jan 15, 2023 at 19:57
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1 4 a 4 a is also the length of the latus rectum.thshea –thshea 2023-01-16 05:10:48 +00:00 Commented Jan 16, 2023 at 5:10
@thshea Yeah, but I don't think that many people bother to teach the term "latus rectum" anymore, at least not at the level where students are simply trying to wrap their heads around the basic ideas.Xander Henderson –Xander Henderson♦ 2023-01-16 12:50:28 +00:00 Commented Jan 16, 2023 at 12:50
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"Standard form" is context dependent. If you are interested in the coefficients then a x 2+b x+c a x 2+b x+c is your friend. If you want to be able to see the symmetry and find the roots then you want the other form.
If someone asks you to convert something to "standard form" then they are required to tell you what "standard form" means.
The actual algebra is independent of the labels.
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answered Jan 15, 2023 at 19:54
Ethan BolkerEthan Bolker
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11620 | https://people.math.carleton.ca/~ouldhaye/stat3502/3502(0809)formula-final.pdf | 1 Computational (or short-cut formula) for the variance, covariance, and correlation σ2 = E(X2) −[E(X)]2, Cov(X, Y ) = E(XY ) −E(X)E(Y ), ρ(X, Y ) = Cov(X, Y ) σXσY .
TABLE OF COMMON DISTRIBUTIONS Binomial(n, p) pmf p(x) = n x px(1 −p)n−x, x = 0, 1, . . . , n mean and variance E(X) = np, Var(X) = np(1 −p) Hypergeometric(n,M,N) pmf p(x) = M x N−M n−x N n , max(0, n −N + M) ≤x ≤min(n, M) mean and variance E(X) = nM N , Var(X) = N −n N −1 n M N 1 −M N Negative binomial(r,p) pmf p(x) = x+r−1 r−1 pr(1 −p)x, x = 0, 1, 2, . . .
mean and variance E(X) = r(1 −p) p , Var(X) = r(1 −p) p2 Poisson(λ) pmf p(x) = e−λλx x!
, x = 0, 1, 2, . . .
mean and variance E(X) = λ, Var(X) = λ Discrete uniform pmf p(x) = 1/k, x = 1, 2, . . . , k mean and variance E(X) = k + 1 2 , Var(X) = (k + 1)(k −1) 12 Normal(µ, σ2) pdf f(x) = 1 √ 2πσe−(x−µ)2/2σ2, x ∈R mean and variance E(X) = µ, Var(X) = σ2 Gamma(α, β) pdf f(x) = 1 βαΓ(α)xα−1e−x/β, x ≥0 mean and variance E(X) = αβ, Var(X) = αβ2 Exponential(λ) pdf f(x) = λe−λx, x ≥0 mean and variance E(X) = 1/λ, Var(X) = 1/λ2 Chi Squared(n) pdf f(x) = 1 2n/2Γ(n/2)x(n/2)−1e−x/2, x ≥0 mean and variance E(X) = n, Var(X) = 2n Uniform(a, b) pdf f(x) = 1 b −a, a ≤x ≤b mean and variance E(X) = (a + b)/2, Var(X) = (b −a)2/12 Rule of thumb If sample size n > 30, then the Central Limit Theorem (CLT) can be applied.
2 DESCRIPTIVE STATISTICS X = 1 n n X i=1 Xi, S2 = 1 n −1 n X i=1 (Xi −X)2 = 1 n −1 " n X i=1 X2 i −(Pn i=1 Xi)2 n # TEST STATISTICS Z = √n X −µ0 σ , T = √n X −µ0 S , Z = √n ˆ p −p0 p p0(1 −p0) CONFIDENCE INTERVALS CI’s for µ (σ is known) X −zα/2 σ √n, X + zα/2 σ √n −∞, X + zα σ √n and X −zα σ √n, +∞ CI’s for µ (σ is unknown) X −tα/2,n−1 S √n, X + tα/2,n−1 S √n −∞, X + tα,n−1 S √n and X −tα,n−1 S √n, +∞ CI for σ2 (n −1)S2 χ2 α/2,n−1 , (n −1)S2 χ2 1−α/2,n−1 !
0, (n −1)S2 χ2 1−α,n−1 and (n −1)S2 χ2 α,n−1 , +∞ Large-sample CI for p ˆ p −zα/2 r ˆ p(1 −ˆ p) n , ˆ p + zα/2 r ˆ p(1 −ˆ p) n !
Large-sample CI for arbitrary θ ˆ θ −zα/2Sˆ θ, ˆ θ + zα/2Sˆ θ where Sˆ θ is an estimator of the standard deviation σˆ θ of ˆ θ |
11621 | https://www.sciencedirect.com/science/article/pii/S1010518200901720 | Staging of the neck in patients with oral cavity squamous cell carcinomas: a prospective comparison of PET, ultrasound, CT and MRI - ScienceDirect
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Journal of Cranio-Maxillofacial Surgery
Volume 28, Issue 6, December 2000, Pages 319-324
Regular Article
Staging of the neck in patients with oral cavity squamous cell carcinomas: a prospective comparison of PET, ultrasound, CT and MRI
Author links open overlay panel Tankred Stuckensen a, Adorján F.Kovács a f1, Stephan Adams b, Richard P.Baum b
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Background: The choice of treatment in patients with oral malignancies depends on accurate pretreatment staging and particularly the detection of lymph node involvement. Therefore staging of the neck should be as accurate as possible. Patients: One hundred and six patients with histologically proven squamous cell carcinoma of the oral cavity. Study design: In a prospective study, PET using fluoro-desoxy-glucose (18 F-FDG), ultrasound, CT and MRI of head and neck were compared with the postoperative histologic tissue evaluation. Two thousand one hundred and ninety-six neck lymph nodes of 106 patients were investigated. In all patients the tumour was resected and a lymph node dissection was performed. Results: The diagnostic procedures showed the following results when compared with the histological findings: PET: sensitivity 70%, specificity 82%, accuracy 75%; Ultrasound: 84%, 68%, 76%; CT: 66%, 74%, 70%; MRI: 64%, 69% 66%. Thus PET showed the highest specificity while ultrasound had the highest sensitivity compared with the other staging procedures. A non-significant correlation was found between the size of a lymph node metastasis and the ability to detect it. In 10 patients, second primary tumours or distant metastases were detected by PET only. Conclusion: Due to the high number of small lymph node metastases from oral cavity carcinoma, the non-invasive neck staging methods are limited to a maximum accuracy of 76%. Elective neck treatment should be mandatory for all patients with squamous cell carcinoma of the oral cavity.
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Eur J Radiol (1997)
M Umeda et al.
Criteria for diagnosing lymph node metastasis from squamous cell carcinoma of the oral cavity: a study of the relationship between computed tomography and histologic outcome
J Oral Maxillofac Surg (1998)
JA Woolgar et al.
Correlation of histopathologic findings with clinical and radiologic assessments of cervical lymph-node metastases in oral cancer
Int J Oral Maxillofac Surg (1995)
S Adams et al.
Prospective comparison of 18F-FDG PET with conventional imaging modalities (CT, MRI, US) in lymph node staging of head and neck cancer
Eur J Nucl Med (1998)
RM Byers et al.
Can we detect or predict the presence of occult nodal metastases in patients with squamous carcinoma of the oral tongue?
Head Neck (1998)
DM Don et al.
Evaluation of cervical lymph node metastases in squamous cell carcinoma of the head and neck
Laryngoscope (1995)
GC Dooms et al.
Magnetic resonance imaging of the lymph nodes: comparison with CT
Radiology (1984)
B Fadem
Behavioral Science
(1994)
R Feinmesser et al.
Metastatic neck disease. A clinical/radiographic/pathologic correlative study
Arch Otolaryngol Head Neck Surg (1987)
M Friedman et al.
Metastatic neck disease
Evaluation by computed tomography. Arch Otolaryngol (1984)
M Friedman et al.
Nodal size of metastatic squamous cell carcinoma of the neck
Laryngoscope (1993)
CA Guhlmann et al.
PET in der Onkologie-klinische Indikationen: Ergebnisse der 2. Deutschen Konsensuskonferenz
Forum Deutsche Krebsgesellschaft (1999)
J Hoffmann et al.
Positron emission tomography: a new diagnostic tool for staging of head and neck squamous cell carcinoma
J Cranio-Maxillofac Surg (2000)
Kainz, M, Howaldt, H, P, DÖSAK Central Tumour Registry, Gießen, 1999,...
View more references
Cited by (209)
Image fusion between 18FDG-PET and MRI/CT for radiotherapy planning of oropharyngeal and nasopharyngeal carcinomas
2002, International Journal of Radiation Oncology Biology Physics Show abstract Purpose: Accurate diagnosis of tumor extent is important in three-dimensional conformal radiotherapy. This study reports the use of image fusion between (18)F-fluoro-2-deoxy-D-glucose positron emission tomography (18 FDG-PET) and magnetic resonance imaging/computed tomography (MRI/CT) for better targets delineation in radiotherapy planning of head-and-neck cancers. Methods and Materials: The subjects consisted of 12 patients with oropharyngeal carcinoma and 9 patients with nasopharyngeal carcinoma (NPC) who were treated with radical radiotherapy between July 1999 and February 2001. Image fusion between 18 FDG-PET and MRI/CT was performed using an automatic multimodality image registration algorithm, which used the brain as an internal reference for registration. Gross tumor volume (GTV) was determined based on clinical examination and 18 FDG uptake on the fusion images. Clinical target volume (CTV) was determined following the usual pattern of lymph node spread for each disease entity along with the clinical presentation of each patient. Results: Except for 3 cases with superficial tumors, all the other primary tumors were detected by 18 FDG-PET. The GTV volumes for primary tumors were not changed by image fusion in 19 cases (89%), increased by 49% in one NPC, and decreased by 45% in another NPC. Normal tissue sparing was more easily performed based on clearer GTV and CTV determination on the fusion images. In particular, parotid sparing became possible in 15 patients (71%) whose upper neck areas near the parotid glands were tumor-free by 18 FDG-PET. Within a mean follow-up period of 18 months, no recurrence occurred in the areas defined as CTV, which was treated prophylactically, except for 1 patient who experienced nodal recurrence in the CTV and simultaneous primary site recurrence. Conclusion: This preliminary study showed that image fusion between 18 FDG-PET and MRI/CT was useful in GTV and CTV determination in conformal RT, thus sparing normal tissues.
### 18F-fluorodeoxyglucose positron emission tomography to evaluate cervical node metastases in patients with head and neck squamous cell carcinoma: A meta-analysis
2008, Journal of the National Cancer Institute
### PET/CT: Form and function
2007, Radiology
### PET-based treatment planning in radiotherapy: A new standard?
2007, Journal of Nuclear Medicine
### Trends in incidence and prognosis for head and neck cancer in the United States: A site-specific analysis of the SEER database
2005, International Journal of Cancer
### 18F-FDG PET and CT/MRI in oral cavity squamous cell carcinoma: A prospective study of 124 patients with histologic correlation
2005, Journal of Nuclear Medicine
View all citing articles on Scopus
f1
Dr. med. Dr. med. dent. Adorján F. Kovács Oberarzt Klinik und Poliklinik für Kiefer- und Plastische Gesichtschirurgie Klinikum der J. W. Goethe-Universität, Haus 21 Theodor-Stern-Kai 7 60590 Frankfurt am Main Germany Tel: +49 69 63016610 Fax: +49 69 63015644 E-mail: A.Kovacs@em.uni-frankfurt.de
View full text
Copyright © 2000 European Association for Cranio-Maxillofacial Surgery. Published by Elsevier Ltd. All rights reserved.
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11622 | https://nrich.maths.org/problems/about-average | About Average | NRICH
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About average
Can you find sets of numbers which satisfy each of our mean, median, mode and range conditions?
Age
11 to 14
Challenge level
Secondary
STATISTICS AND PROBABILITY
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Problem Student Solutions Teachers' Resources
Problem
About Average printable sheet
Can you find sets of positive integers that satisfy the following?
Three numbers with mean 3 and mode 2
Three numbers with mean 7 and mode 10
Three numbers with mean 8, median 10 and range 8
Four numbers with mean 7.5, mode 6 and median 7
Four numbers with mean 6, median 6.5 and range 11
Five numbers with mean 4, mode 3 and range 9
Five numbers with mean 4, mode 2 and range 6 (two possible solutions)
Five numbers with mean 7, mode 7 and range 10 (three possible solutions)
Extension
Can you find a set of four numbers with mean 4, mode 1, median 2 and range 10?
How about a set of five numbers?
Or six numbers?
Or...
Or 100 numbers?
If you enjoyed this problem, you may also be interested inM, M and M.
With thanks toDon Steward, whose ideas formed the basis of this problem.
Student Solutions
Giulio from St Joseph's Catholic Primary School, Etienne, Youcef, Anna and Lily from Strand on the Green Junior School and Rakhvin from St Stephen's School Carramar in Australia all used the following method for the first two sets. Here is Rakhvin's solution:
Knowing that the mode was 2, two of the three numbers must be 2. Next, as the mean was 3 and there were three numbers, I multiplied to get a total of 9 for all three numbers. From there, I could do 9−2−2 to find the last number, 5.
Knowing that the mode was 10, two of the three numbers must be 10. Next, as the mean was 7 and there were three numbers, I multiplied to get a total of 21 for all three numbers. From there, I could do 21−10−10 to find the last number, 1.
For the third and fourth questions, these students again used similar methods. Here are the solutions of Etienne, Youcef, Anna and Lily:
The median of 10 tells us that this is the middle number.
The mean of 8 tells us that all three numbers must add up to 24, so the other two numbers must add up to 14. Since the range is 8, they have a difference of 8, so the numbers must be 3 and 11.
Therefore the overall set is 3, 10 and 11.
The mode of 6 means that there are at least two 6's.
For the median to be 7, the other middle number is 8. For the mean to be 7.5, the total must be 4×7.5=30, and therefore the final number is 10.
Jacob from Sacred Heart Catholic College, as well as Etienne, Youcef, Anna and Lily, gave the following explanation as to why the description given in question 5 has no solutions. Here is Jacob's solution:
It is impossible to find a set of four positive integers that suit these criteria. First of all, the set of numbers must include two middle numbers that add up to 13 in order to get a median of 6.5.
We know that all four numbers in the set must add up to 24 because the mean is 6 and 6 multiplied by 4 equals 24. So far the two numbers that we have add up to 13. In addition to this, the four numbers must have a range of 11. This means that the other two numbers, which are the first and last numbers in the sequence, must add up to 11. The only two numbers that can be used are 0 and 11.
However, 0 is not a positive integer; it is neither a positive or negative integer. Due to the fact that we cannot use 0 as one of the numbers in the set of four positive integers, it is impossible to get a set of four numbers at all. This is because there are no other sets of numbers that fulfil this criteria.
However, other students noticed that if you were to allow 0, then there would be solutions. For example, Rakhvin obtained the set:
0,4,9,11
Galal, from the Continental School of Cairo in Egypt had:
0,3,8,11
Ashley, Thea, Charlie and Vikash from Norwich School found:
0,6,7,11
Class 2C from The London Oratory School also found this, as well as the following:
0,5,8,11
Liam, Etienne, Youcef, Anna and Lily also gave solutions for questions 6, 7 and 8:
The mode of 3 tells us that there are at least two 3's. The mean of 4 means that the numbers all add up to 20, so the other three add up to 14. As the range is 9, this means that the only possible numbers are 1, 3 and 10.
Therefore the overall set is: 1,3,3,3,10
A mode of 2 means that there are at least two 2's. A mean of 4 means that the other three numbers add up to 16.
If the lowest integer is 1, the range of 6 tells us that the largest is 7 but we would need an 8 to make them add up to 16, so that doesn't work. So the smallest number is 2, the highest is 8 (from the range) and the other two number can be 2 and 6 or 3 and 5. They can't be 4 and 4 because then the numbers would have two modes.
This gives overall the sets: 2,2,2,6,8 and 2,2,3,5,8.
The mode of 7 means there are at least two 7's. The mean of 7 tells us that the other three numbers add up to 21. The range is 10, so if 1 is the smallest number, the largest is 11 and that leaves 9. If 2 is the smallest number, the largest is 12 and that leaves 7. If 3 is the smallest number, the largest is 13 and that leaves 5. If 4 is the smallest number, the largest is 14 and that leaves only 3, but that can't be the case because 4 had to be the smallest number.
So the smallest number has to be 3 or less.
Therefore, the answers are: 1,7,7,9,11, 2,7,7,7,12 and 3,5,7,7,13.
These solutions can all also be done using algebra to help. Both Mitsuyo from Garden International School in Kuala Lumpur, Malaysia and Zach used this method. You can find Mitsuyo's solution here and Zach's solution here .
For the extension problems, Nour from The Continental School of Cairo in Egypt, Zach, Class 2C from The London Oratory School and Max from Castle School all found the following solutions for four, five and six numbers:
Four numbers: 1,1,3,11
Five numbers: 1,1,2,5,11
Six numbers: 1,1,1,3,7,11
Class 8a2 from Ousedale School were able to find a pattern that would always work:
4 numbers: 1,1,3,11
6 numbers: 1,1,1,3,7,11
8 numbers: 1,1,1,1,3,7,7,11
10 numbers:1,1,1,1,1,3,7,7,7,11
12 numbers:1,1,1,1,1,1,3,7,7,7,7,11
They could then generalise this using algebra:
With n numbers, where n is even, there are:
n 2 ones, 1 three, n 2−2 sevens and 1 eleven.
Great stuff! Can anyone find a generalisation for when n is odd?
Teachers' Resources
Why do this problem?
This problem improves fluency in working with the numerical definitions of the mean, median, mode and range. It also develops understanding of how each measure is affected by individual numbers in a sample, as well as students’ reasoning. If appropriate, you could also use this problem to encourage students to represent the concepts algebraically and set up equations.
Possible approach
This problem should be used once students have already seen the mean, median mode and range. As a starter, you could give students some sets of numbers and ask them to find the mean, median, mode and range. Then, ask them to find three numbers with mean 3 and mode 2. Is there more than one way of doing it?
After the first example, invite students to share their strategies.
Students could then work in pairs or small groups to find numbers for each set of conditions. You could encourage them to record their strategies, and whether they used the same strategy for all of the questions. At the end of the activity, students could share their strategies and conclusions in a whole class discussion.
Note that number 5 is only possible if you allow 0 as one of the numbers, and in that case there are several possible solutions. There are two possible solutios for number 6 if 0 is allowed.
Manipulatives could be used to introduce this work. Number cards could be used to allow students to visually see the problem.
Key questions
What is the difference between the averages i.e. How is the mode different to the mean?
Is there only one answer?
Is there a method that always works?
Of the mean, median, mode and range, which should you focus on first as you choose your numbers, and why? What could you focus on next, and why?
Possible support
You could begin by giving students some of the numbers, for example 2, ___, 5 (choose the third number so that the mean is 3 and the mode is 2).
Alternatively, you could show them helpful examples: you could give them some sets of three numbers with mode equal to 2, such as 2, 2, 3 and 1, 2, 2 and ask them to find the mean and the mode, before asking them to find three numbers with mode 2 and mean 3.
Possible extension
Ask students to create their own conditions that can be presented to their peers.
Ask students to investigate when it is possible to find numbers that meet their conditions, and when it is impossible. Can they create impossible conditions that look possible?
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11623 | https://byjus.com/maths/criteria-for-similarity-of-triangles/ | In Mathematics, a triangle is a closed two-dimensional figure or polygon with the least number of sides. A triangle has three sides and three angles. The most important property of a triangle is the sum of the interior angles of a triangle is equal to 180°. In this article, let us discuss the important criteria for the similarity of triangles with their theorem and proof and many solved examples.
Conditions for Similarity of Two Triangles
Two triangles are said to be similar triangles,
If their corresponding angles are equal.
If their corresponding sides are in the same proportion/ratio.
Consider two triangles ABC and DEF.
The two triangles are said to be similar triangles, if
∠A = ∠D, ∠B = ∠E, ∠C = ∠F
AB/DE = BC/EF = CA/FD
In this scenario, we can say that the two triangles ABC and DEF are similar.
Important Criteria for Similarity of Triangles
The four important criteria used in determining the similarity of triangles are
AAA criterion (Angle-Angle-Angle criterion)
AA criterion (Angle-Angle criterion)
SSS criterion (Side-Side-Side criterion)
SAS Criterion (Side-Angle-Side criterion)
Now, let us discuss all these criteria for the similarity of triangles in detail.
AAA Similarity Criterion for Two Triangles
The Angle-Angle-Angle (AAA) criterion for the similarity of triangles states that “If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar”.
Proof:
Consider two triangles ABC and DEF, such that ∠A = ∠D, ∠B = ∠E and ∠C = ∠F as shown in the figure.
Now, cut DP= AB and DQ= AC and join PQ. Hence, we can say that a triangle ABC is congruent to the triangle DPQ.
(i.e) ∆ ABC ≅ ∆ DPQ, which gives ∠B = ∠P = ∠E and also, the line PQ is parallel to EF.
Therefore, by using the basic proportionality theorem, we can write
DP/PE = DQ/QF.
(i.e) AB/DE = AC/DF. …(1)
Similarly,
AB/DE = BC/EF …(2)
From (1) and (2), we can write:
AB/DE = BC/EF = AC/DF.
Therefore, the two triangles ABC and DEF are similar.
AA Similarity Criterion for Two Triangles
The Angle-Angle (AA) criterion for similarity of two triangles states that “If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar”.
The AA criterion states that if two angles of a triangle are respectively equal to the two angles of another triangle, we can prove that the third angle will also be equal on both the triangles. This can be done with the help of the angle sum property of a triangle.
SSS Similarity Criterion for Two Triangles
The Side-Side-Side (SSS) criterion for similarity of two triangles states that “If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of ) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar”.
Proof:
Consider the same figure as given above. It is observed that DP/PE = DQ/QF and also in the triangle DEF, the line PQ is parallel to the line EF.
So, ∠P = ∠E and ∠Q = ∠F.
Hence, we can write: DP/DE = DQ/DF= PQ/EF.
The above expression is written as
DP/DE = DQ/DF=BC/EF.
It means that PQ = BC.
Hence, the triangle ABC is congruent to the triangle DPQ.
(i.e) ∆ ABC ≅ ∆ DPQ.
Thus, by using the AAA criterion for similarity of the triangle, we can say that
∠A = ∠D, ∠B = ∠E and ∠C = ∠F.
| Also, read: Angle Sum Property of Triangles Corresponding Angles Alternate Angles Congruence of Triangles |
SAS Similarity Criterion for Two Triangles
The Side-Angle-Side (SAS) criterion for similarity of two triangles states that “If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar”.
Proof:
This theorem can be proved by taking two triangles such as ABC and DEF (Refer to the same figure as given above).
By using the basic proportionality theorem, we get
AB/DE = AC/DF and ∠A = ∠D
In the triangle DEF, the line PQ is parallel to EF.
So, ∆ ABC ≅ ∆ DPQ.
Hence, we can say ∠A = ∠D, ∠B=∠P and ∠C= ∠Q, which means that the triangle ABC is similar to the triangle DEF.
(i.e) ∆ ABC ~ ∆ DEF.
Examples
Now, let us use the criteria for the similarity of triangles to find the unknown angles and sides of a triangle.
Example 1:
Find ∠P in the following triangles.
Solution:
From the given triangles, ABC and PQR, we can get
AB/RQ = 3.8/7.6 = ½
BC/QP = 6/12 = ½
CA/PR = (3√3)/6√3 = ½
Therefore,
AB/RQ = BC/QP = CA/PR
Hence, by using the SSS similarity criterion for a triangle, we can write
∆ ABC ~ ∆ RQP (i.e) ∆ABC is similar to ∆RQP.
By using corresponding angles of similar triangles,
∠C = ∠P
∠C = 180°- ∠A – ∠B (Using the angle sum property of triangle).
∠C= 180° – 80° – 60°
∠C= 40°.
Since, ∠C= ∠P, the value of ∠p is 40°.
Example 2:
Show that the triangles POQ and SOR are similar triangles, given that PQ is parallel to RS as shown in the figure.
Solution:
Given that PQ is parallel to RS. (i.e) PQ || RS.
By using alternate angles property, ∠P = ∠S and ∠Q = ∠R.
Also, by using the vertically opposite angles, ∠POQ = ∠SOR.
Hence, we can conclude that a triangle POQ is similar to the triangle SOR.
(i.e) ∆ POQ ~ ∆ SOR (Using AAA similarity criterion for triangles)
Hence, proved.
Video Lesson on BPT and Similar Triangles
2,36,419
Practice Problems
From the given figure, if ∆ ABE ≅ ∆ ACD, prove that ∆ ADE ~ ∆ ABC.
In the given figure, QR/QS = QT/PR and ∠1= ∠2. Show that ∆ PQS ~ ∆ TQR.
Determine the height of a tower, if a vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long.
Keep visiting BYJU’S – The Learning App and download the app to learn all Maths-related concepts by exploring more videos.
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11624 | https://openstax.org/books/college-physics-2e/pages/18-2-conductors-and-insulators | Skip to ContentGo to accessibility pageKeyboard shortcuts menu
College Physics 2e
18.2 Conductors and Insulators
College Physics 2e18.2 Conductors and Insulators
Search for key terms or text.
Learning Objectives
By the end of this section, you will be able to:
Define conductor and insulator, explain the difference, and give examples of each.
Describe three methods for charging an object.
Explain what happens to an electric force as you move farther from the source.
Define polarization.
Figure 18.10 This power adapter uses metal wires and connectors to conduct electricity from the wall socket to a laptop computer. The conducting wires allow electrons to move freely through the cables, which are shielded by rubber and plastic. These materials act as insulators that don’t allow electric charge to escape outward. (credit: Evan-Amos, Wikimedia Commons)
Some substances, such as metals and salty water, allow charges to move through them with relative ease. Some of the electrons in metals and similar conductors are not bound to individual atoms or sites in the material. These free electrons can move through the material much as air moves through loose sand. Any substance that has free electrons and allows charge to move relatively freely through it is called a conductor. The moving electrons may collide with fixed atoms and molecules, losing some energy, but they can move in a conductor. Superconductors allow the movement of charge without any loss of energy. Salty water and other similar conducting materials contain free ions that can move through them. An ion is an atom or molecule having a positive or negative (nonzero) total charge. In other words, the total number of electrons is not equal to the total number of protons.
Other substances, such as glass, do not allow charges to move through them. These are called insulators. Electrons and ions in insulators are bound in the structure and cannot move easily—as much as times more slowly than in conductors. Pure water and dry table salt are insulators, for example, whereas molten salt and salty water are conductors.
Figure 18.11 An electroscope is a favorite instrument in physics demonstrations and student laboratories. It is typically made with gold foil leaves hung from a (conducting) metal stem and is insulated from the room air in a glass-walled container. (a) A positively charged glass rod is brought near the tip of the electroscope, attracting electrons to the top and leaving a net positive charge on the leaves. Like charges in the light flexible gold leaves repel, separating them. (b) When the rod is touched against the ball, electrons are attracted and transferred, reducing the net charge on the glass rod but leaving the electroscope positively charged. (c) The excess charges are evenly distributed in the stem and leaves of the electroscope once the glass rod is removed.
Charging by Contact
Figure 18.11 shows an electroscope being charged by touching it with a positively charged glass rod. Because the glass rod is an insulator, it must actually touch the electroscope to transfer charge to or from it. (Note that the extra positive charges reside on the surface of the glass rod as a result of rubbing it with silk before starting the experiment.) Since only electrons move in metals, we see that they are attracted to the top of the electroscope. There, some are transferred to the positive rod by touch, leaving the electroscope with a net positive charge.
Electrostatic repulsion in the leaves of the charged electroscope separates them. The electrostatic force has a horizontal component that results in the leaves moving apart as well as a vertical component that is balanced by the gravitational force. Similarly, the electroscope can be negatively charged by contact with a negatively charged object.
Charging by Induction
It is not necessary to transfer excess charge directly to an object in order to charge it. Figure 18.12 shows a method of induction wherein a charge is created in a nearby object, without direct contact. Here we see two neutral metal spheres in contact with one another but insulated from the rest of the world. A positively charged rod is brought near one of them, attracting negative charge to that side, leaving the other sphere positively charged.
This is an example of induced polarization of neutral objects. Polarization is the separation of charges in an object that remains neutral. If the spheres are now separated (before the rod is pulled away), each sphere will have a net charge. Note that the object closest to the charged rod receives an opposite charge when charged by induction. Note also that no charge is removed from the charged rod, so that this process can be repeated without depleting the supply of excess charge.
Another method of charging by induction is shown in Figure 18.13. The neutral metal sphere is polarized when a charged rod is brought near it. The sphere is then grounded, meaning that a conducting wire is run from the sphere to the ground. Since the earth is large and most ground is a good conductor, it can supply or accept excess charge easily. In this case, electrons are attracted to the sphere through a wire called the ground wire, because it supplies a conducting path to the ground. The ground connection is broken before the charged rod is removed, leaving the sphere with an excess charge opposite to that of the rod. Again, an opposite charge is achieved when charging by induction and the charged rod loses none of its excess charge.
Figure 18.12 Charging by induction. (a) Two uncharged or neutral metal spheres are in contact with each other but insulated from the rest of the world. (b) A positively charged glass rod is brought near the sphere on the left, attracting negative charge and leaving the other sphere positively charged. (c) The spheres are separated before the rod is removed, thus separating negative and positive charge. (d) The spheres retain net charges after the inducing rod is removed—without ever having been touched by a charged object.
Figure 18.13 Charging by induction, using a ground connection. (a) A positively charged rod is brought near a neutral metal sphere, polarizing it. (b) The sphere is grounded, allowing electrons to be attracted from the earth’s ample supply. (c) The ground connection is broken. (d) The positive rod is removed, leaving the sphere with an induced negative charge.
Figure 18.14 Both positive and negative objects attract a neutral object by polarizing its molecules. (a) A positive object brought near a neutral insulator polarizes its molecules. There is a slight shift in the distribution of the electrons orbiting the molecule, with unlike charges being brought nearer and like charges moved away. Since the electrostatic force decreases with distance, there is a net attraction. (b) A negative object produces the opposite polarization, but again attracts the neutral object. (c) The same effect occurs for a conductor; since the unlike charges are closer, there is a net attraction.
Neutral objects can be attracted to any charged object. The pieces of straw attracted to polished amber are neutral, for example. If you run a plastic comb through your hair, the charged comb can pick up neutral pieces of paper. Figure 18.14 shows how the polarization of atoms and molecules in neutral objects results in their attraction to a charged object.
When a charged rod is brought near a neutral substance, an insulator in this case, the distribution of charge in atoms and molecules is shifted slightly. Opposite charge is attracted nearer the external charged rod, while like charge is repelled. Since the electrostatic force decreases with distance, the repulsion of like charges is weaker than the attraction of unlike charges, and so there is a net attraction. Thus a positively charged glass rod attracts neutral pieces of paper, as will a negatively charged rubber rod. Some molecules, like water, are polar molecules. Polar molecules have a natural or inherent separation of charge, although they are neutral overall. Polar molecules are particularly affected by other charged objects and show greater polarization effects than molecules with naturally uniform charge distributions.
Check Your Understanding
Can you explain the attraction of water to the charged rod in the figure below?
Figure 18.15
Solution
Water molecules are polarized, giving them slightly positive and slightly negative sides. This makes water even more susceptible to a charged rod’s attraction. In addition, tap water contains dissolved ions (positive and negative charges). As the water flows downward, due to the force of gravity, the charged conductor exerts a net attraction to the opposite charges in the stream of water, pulling it closer.
PhET Explorations
John Travoltage
Make sparks fly with John Travoltage. Wiggle Johnnie's foot and he picks up charges from the carpet. Bring his hand close to the door knob and get rid of the excess charge.
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11625 | https://www.youtube.com/watch?v=wwn6xJ0e3W0 | Can you solve this logic puzzle?
MindYourDecisions
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Only one statement is true. Who could the mother be?
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Transcript:
hey this is press to Walker here's a fun little logic puzzle a mother has four daughters below are statements about them statement one ameni is the mother statement two Desa and comi are both daughters statement three iny is the mother statement four one of ameni vea or Comey is the mother one of the above statements is true true while the rest are false the question is who could be the mother this problem is adapted from Reddit puzzles pause the video if you'd like to give this problem a try and when you're ready keep watching to learn how to solve this [Music] problem at first this seems like an impossible logic puzzle we know know that one of the statements is true but we don't know which one and if something like statement four is true we only know that am mini vea or Comey could be the mother how can we figure out which one is exactly the mother let's work through the logical possibilities so here's one way to solve the problem we know there's a mother and four daughters so there are five candidates for the mother there are five names in the problem so that's amen Desa Comey iny or vea so let's go through each possibility let's just suppose that a many is the mother how would this apply to the different statements so we'll start with statement one if ameni is the mother then statement one would have to be true then statement two would also have to be true because if a many is the mother then everyone else has to be a daughter so Desa and Comey are both daughters so this statement would also be true but wait a minute we know that only one statement can be true and the rest have to be false so we have derived a contradiction it is not possible for ameni to be the mother because then statements one and two would both have to be true which we know is impossible so we'll go to the next candidate let's suppose that Desa is the mother in that case statement one would have to be false then statement two that Desa is a daughter would also have have to be false statement three that iny is the mother that would also be false and then we go to statement four that one of ameni vea or Comey is the mother that would also be false but wait we've now shown that all four statements are false and we know that one of them has to be true so it's not possible that Desa is the mother we've derived a contradiction we now continue to the possibility that Comey is the mother so in that case statement one that ameni is the mother that would be false statement two that Desa and Comey are both daughters would be false then in is the mother would be false and then we have one of ameni vea or Comey is the mother that is true because Comey is the mother so this is true so we have a situation where one of the statements is true while the other are false so this is a possibility come me is the mother is a possible answer but the question is asking who could be the mother so are there any other possibilities let's work through the cases iny could be the mother in that case ameni is the mother that would be false then Desa and Comey both would be daughters we already said that incy is mother so statement three would be true but wait now we have said that statements two and three are both true but we know only one of the statements can be true so we have derived a contradiction and iny cannot be The Mother We Now consider the case that is the mother so the statement ameni is the mother would have to be false then we go to the statement that Jess and Comey are both daughters if Bea is the mother then both of them would be daughters so this would be true the statement three that Inc is the mother would have to be false because vea is the mother and finally we have one of ameni vea or Comey is the mother if vea is the mother then this statement would be true but now statements two and four are true and we know that only one of them could be true so we have derived a contradiction and vea cannot be the mother so we've eliminated this possibility therefore the only candidate that works is that Comey is the mother that leads to statement four being true the other being false and therefore this is the only answer Comey you are the mother so in this solution we work through each candidate and consider the statements but there's another way that you could work through this puzzle instead we could work through the statements we could assume each of the statements is true and see if we derive any contradictions so let's start out with the candidate that statement one is true so in that case a many is the mother would have to be a true statement so we would know that ameni is the mother now let's look at statement two if ameni is the mother then everyone else has to be a daughter so Desa and Comey are both daughters so two would be a true statement but now both statements one and two are true we know that only one of the statements can be true so we have derived a contradiction so statement one cannot be the true statement by similar logic statement three is true will also not be a possibility in that case Inc is the mother would have to be a true statement so we would know Inc is the mother but then everyone else would have to be a daughter so Desa and Comey would both have to be daughters so statement two would be be true so now we've derived that statements two and three are true and this is not possible because only one statement can be true so we know that this cannot be true so statement three cannot be the true statement then let's consider statement two as a true statement so what would happen if Desa and Comey are both daughters is true then we would know that des and Comey are not the mother so one of the statements is true while the other are false so we would know that statements one and statement three would have to be false so we could say that aen is not the mother and Inc is not the mother so that would mean Bea would have to be the mother but then statement four that one of ameni Bea or Comey is the mother would have to be true so if statement two is true we have derived that statements 2 and four would have to be true which is a contradiction that only one of the stat statements is true so cannot be the case that statement two is the true statement so that leaves only one possibility that statement four is the true statement so one of a many vea or Comey is the mother we can then derive that if one of them is the mother that means Desa and iny are not the mother so one of the statements is true while the other are false so we know that aeni is the mother has to be a false statement so aeni cannot be the mother iny is the mother is also a false statement so we know that iny cannot be the mother but we already had the Des and iny are not mothers so this contributes the additional information that ameni is not the mother now Desa and Comey are both daughters has to be a false statement because we said four is the only true statement we already said that Desa is not the mother in other words Desa is a daughter so the only way that two would be a false statement is if Comey is the mother so we say that Comey is the mother would be the conclusion and we have figured out that Comey is the mother we have one true statement which is statement 4 all the other statements are false and this all logically works out therefore Comey is the mother and that's another way you could have solved this problem thanks for making us one of the best communities on YouTube see you next episode of mind your decision or we solve the world problem one video at a time [Music] |
11626 | https://www.khanacademy.org/math/ap-calculus-ab/ab-diff-analytical-applications-new/ab-5-3/a/increasing-and-decreasing-intervals-review | Increasing & decreasing intervals review (article) | Khan Academy
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AP®︎/College Calculus AB
Course: AP®︎/College Calculus AB>Unit 5
Lesson 3: Determining intervals on which a function is increasing or decreasing
Finding decreasing interval given the function
Finding increasing interval given the derivative
Increasing & decreasing intervals
Increasing & decreasing intervals review
Math>
AP®︎/College Calculus AB>
Applying derivatives to analyze functions>
Determining intervals on which a function is increasing or decreasing
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Increasing & decreasing intervals review
AP.CALC: FUN‑4 (EU), FUN‑4.A (LO), FUN‑4.A.1 (EK)
Google Classroom
Microsoft Teams
Review how we use differential calculus to find the intervals where a function increases or decreases.
How do I find increasing & decreasing intervals with differential calculus?
The intervals where a function is increasing (or decreasing) correspond to the intervals where its derivative is positive (or negative).
So if we want to find the intervals where a function increases or decreases, we take its derivative an analyze it to find where it's positive or negative (which is easier to do!).
Want to learn more about increasing/decreasing intervals and differential calculus? Check out this video.
Example 1
Let's find the intervals where f(x)=x 3+3 x 2−9 x+7 is increasing or decreasing. First, we differentiate f:
f′(x)=3 x 2+6 x−9
Show entire calculation
f′(x)=d d x(x 3+3 x 2−9 x+7)=d d x(x 3)+3 d d x(x 2)−9 d d x(x)+d d x(7)=(3 x 2)+3(2 x)−9(1)+(0)=3 x 2+6 x−9
Now we want to find the intervals where f′ is positive or negative.
f′(x)=3(x+3)(x−1)
f′ intersects the x-axis when x=−3 and x=1, so its sign must be constant in each of the following intervals:
−7−6−5−4−3−2−101234567x<−3−3<x<1x>1
Let's evaluate f′ at each interval to see if it's positive or negative on that interval.
| Interval | x-value | f′(x) | Verdict |
--- --- |
| x<−3 | x=−4 | f′(−4)=15>0 | f is increasing. ↗ |
| −3<x<1 | x=0 | f′(0)=−9<0 | f is decreasing. ↘ |
| x>1 | x=2 | f′(2)=15>0 | f is increasing. ↗ |
So f is increasing when x<−3 or when x>1 and decreasing when −3<x<1.
Example 2
Let's find the intervals where f(x)=x 6−3 x 5 is increasing or decreasing. First, we differentiate f:
f′(x)=6 x 5−15 x 4
Show entire calculation
f′(x)=d d x(x 6−3 x 5)=d d x(x 6)−3 d d x(x 5)=(6 x 5)−3(5 x 4)=6 x 5−15 x 4
Now we want to find the intervals where f′ is positive or negative.
f′(x)=3 x 4(2 x−5)
f′ intersects the x-axis when x=0 and x=5 2, so its sign must be constant in each of the following intervals:
−2−1.5−1−0.500.511.522.533.54x<00<x<5 25 2<x
Let's evaluate f′ at each interval to see if it's positive or negative on that interval.
| Interval | x-value | f′(x) | Verdict |
--- --- |
| x<0 | x=−1 | f′(−1)=−21<0 | f is decreasing. ↘ |
| 0<x<5 2 | x=1 | f′(1)=−9<0 | f is decreasing. ↘ |
| 5 20 | f is increasing. ↗ |
Since f decreases before x=0and after x=0, it also decreases at x=0.
Therefore, f is decreasing when x<5 2 and increasing when x>5 2.
Check your understanding
Problem 1
Current
h(x)=−x 3+3 x 2+9
On which intervals is h decreasing?
Choose 1 answer:
Choose 1 answer:
(Choice A) (2,∞) only A (2,∞) only
(Choice B) (0,2) only B (0,2) only
(Choice C) (−∞,0) and (2,∞) C (−∞,0) and (2,∞)
(Choice D) (0,∞) only D (0,∞) only
(Choice E) All real numbers (excluding 0 and 2) E All real numbers (excluding 0 and 2)
Check Explain
Want to try more problems like this? Check out this exercise.
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bhunter3 6 years ago Posted 6 years ago. Direct link to bhunter3's post “I'm finding it confusing ...” more I'm finding it confusing when a point is undefined in both the original function and the derivative. While not mentioned in the video on critical points, it's mentioned in the comments and practice problems that a point is not a critical point if it's undefined in both the derivative and in the original function.
On the other hand, in the practice problems, we're given something like:
f'(x) = ((x-1)^2) / (x-4)
and asked to find the intervals over which the original function is increasing. The question states that the original function is undefined at x = 4. According to the definition, x = 4 should not be a critical point because it's undefined in both the derivative and the original function. However, it is a point of interest as f'(x) > 0 only when x > 4. If we don't consider x = 4 we won't find the right answer.
Is this an issue with the definition of critical points, the practice problem itself, or this method of finding increasing or decreasing intervals?
If it's the practice problem, I could imagine that maybe it's impossible for a function with that derivative to be undefined at 4 (though it seems unlikely.)
If it's this method, it seems like we need to consider points that aren't strictly critical points as per the definition.
I think a little more clarity around this particular case in this section and the one before would be helpful.
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bhunter3 6 years ago Posted 6 years ago. Direct link to bhunter3's post “I found the answer to my ...” more I found the answer to my question in the next section. Under "Finding relative extrema (first derivative test)" it says:
When we analyze increasing and decreasing intervals, we must look for all points where the derivative is equal to zero and all points where the function or its derivative are undefined. If you miss any of these points, you will probably end up with a wrong sign chart.
I'll leave my question here because I think it's confusing for this section to only discuss critical points and not to mention this.
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akuppili45 9 years ago Posted 9 years ago. Direct link to akuppili45's post “Is this also called the 1...” more Is this also called the 1st derivative test?
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Gavin a year ago Posted a year ago. Direct link to Gavin's post “What is f'(x) = 0?” more What is f'(x) = 0?
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TheReal3A a year ago Posted a year ago. Direct link to TheReal3A's post “f'(x) means the derivat...” more f'(x) means the derivative of f, as indicated by one tick ' after the function's name.
f'(x) = 0 means whenever the function's slope equals 0, which can tell us where the stationary points are on the function's graph!
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Alamin 2 years ago Posted 2 years ago. Direct link to Alamin's post “I'm still a little unclea...” more I'm still a little unclear on how an instantaneous rate of change of zero would be considered either increasing or decreasing due to the local conditions (ie decreasing on both sides or increasing on both sides of the point). Even if the rate of change on both sides of a point are increasing or decreasing, why would we not want to allow for the instantaneous rate of change to be neither?
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James 2 years ago Posted 2 years ago. Direct link to James's post “how would a question on t...” more how would a question on the test look like for these types of questions?
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Venkata 2 years ago Posted 2 years ago. Direct link to Venkata's post “They'll probably give you...” more They'll probably give you a function and ask you to write out the intervals on which it increases and decreases. You'll have to then show the whole process of taking derivatives, finding critical points, using either the first or second derivative test, and figuring out the behaviour of the function on each of the intervals.
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SIRI MARAVANTHE 7 years ago Posted 7 years ago. Direct link to SIRI MARAVANTHE's post “How do we decide if y=cos...” more How do we decide if y=cos3x increasing or decreasing in the interval [0,3.14/2]
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Jas 3 years ago Posted 3 years ago. Direct link to Jas's post “We can tackle the trigono...” more We can tackle the trigonometric functions in the same way we do polynomials or rational functions! We take the derivative of y, giving us dy/dx = -3sin3x. Then, we find where this derivative is equal to zero or is undefined - this tells us all the possible x-values where the derivative might change from positive to negative, or negative to positive. Then we figure out where dy/dx is positive or negative. Since we know functions are increasing where their derivatives are positive, and decreasing where their derivatives are negative, we can then use this knowledge to figure out if the function is increasing or decreasing.
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ashray tiwari 4 years ago Posted 4 years ago. Direct link to ashray tiwari's post “1)why shoude we also cons...” more 1)why shoude we also consider the point where function is not defined? (to check increasing or decreasing)
2)when f'(c)=0, then after or before c function value is either increasing or decreasing, but when fuction is not defined at d then how the function is increasing or decreasing either side of d?
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JPOgle ✝ 3 years ago Posted 3 years ago. Direct link to JPOgle ✝'s post “We consider the point whe...” more We consider the point where the derivative is undefined, not where the function is undefined.
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Osmis 3 years ago Posted 3 years ago. Direct link to Osmis's post “Are there any factoring s...” more Are there any factoring strategies that could help me solve this problem faster than just plug in and attempt? (3x^2 + 8x -5) The answer is (3x-5)(-x+1)
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hrithvikasingh07 a year ago Posted a year ago. Direct link to hrithvikasingh07's post “if they've given a upward...” more if they've given a upward curved graph for f' and are asking to find the graph of f. how do we do that?
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pufferfish56 a year ago Posted a year ago. Direct link to pufferfish56's post “The simplest way (but pos...” more The simplest way (but possibly the hardest depending on the function) if you are given the equation of the derivative is to take the antiderivative/the integral of f', to get f (but you don't know the starting position). There are different ways to do this depending on the function. However, there are certain things that a graph of f' can tell you about f:
The one thing your derivative doesn't tell you is your starting position; it will either be given or it won't matter. All parts of the graph where your derivative is positive, your graph should be increasing, vice versa for negative derivatives. When it equals 0, the graph should be flat (extrema and stationary inflection points). When the derivative function has a jump discontinuity, there should be a sharp bend in the graph. If your derivative is increasing, your function will face concave up (like a U), if it's negative, it will face concave down (like a ∩) if it's neither, it will look linear at that point
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rockyz0516 3 years ago Posted 3 years ago. Direct link to rockyz0516's post “When the top and bottom s...” more When the top and bottom signs are the same, does it mean the interval is increasing?
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11627 | https://proofwiki.org/wiki/Definition:Set_Complement | Definition:Set Complement
From ProofWiki
Jump to navigation Jump to search
Contents
1 Definition
2 Illustration by Venn Diagram
3 Notation
4 Also known as
5 Examples
5.1 $\R_{>0}$ in $\R$
5.2 $\R_{>0}$ in $\C$
6 Also see
7 Historical Note
8 Linguistic Note
9 Sources
Definition
The set complement (or, when the context is established, just complement) of a set $S$ in a universal set $\mathbb U$ is defined as:
: $\map \complement S = \relcomp {\mathbb U} S = \mathbb U \setminus S$
See the definition of Relative Complement for the definition of $\relcomp {\mathbb U} S$.
Thus the complement of a set $S$ is the relative complement of $S$ in the universal set, or the complement of $S$ relative to the universal set.
A common alternative to the symbology $\map \complement S$, which we will sometimes use, is $\overline S$.
Illustration by Venn Diagram
The complement $\map \complement T$ of the set $T$ with respect to the universal set $\mathbb U$ is illustrated in the following Venn diagram by the coloured area:
Notation
No standard symbol for the concept of set complement has evolved.
Alternative notations for $\map \complement S$ include variants of the $\complement$:
: $\map {\CC} S$
: $\map c S$
: $\map C S$
: $\map {\operatorname C} S$
: $\map {\operatorname {\mathbf C} } S$
: ${}_c S$
and sometimes the brackets are omitted:
: $C S$
Alternative symbols for $\overline S$ are sometimes encountered:
: $S'$ (but it can be argued that the symbol $'$ is already overused)
: $S^$
: $- S$
: $\tilde S$
: $\sim S$
You may encounter others.
Some authors use $S^c$ or $S^\complement$, but those can also be confused with notation used for the group theoretical conjugate.
Also known as
Some older sources use the term absolute complement, in apposition to relative complement.
Examples
$\R_{>0}$ in $\R$
Let the universal set $\Bbb U$ be defined to be the set of real numbers $\R$.
Let the set of (strictly) positive real numbers be denoted by $\R_{>0}$.
Then:
: $\relcomp {} {\R_{>0} } = \R_{\le 0}$
the set of non-negative real numbers.
$\R_{>0}$ in $\C$
Let the universal set $\Bbb U$ be defined to be the set of real numbers $\C$.
Let the set of (strictly) positive real numbers be denoted by $\R_{>0}$.
Then:
: $\relcomp {} {\R_{>0} } = \set {x + i y: y \ne 0 \text { or } x \le 0}$
Also see
Definition:Set Difference
Definition:Relative Complement
Results about set complements can be found here.
Historical Note
The concept of set complement, or logical negation, was stated by Leibniz in his initial conception of symbolic logic.
Linguistic Note
The word complement comes from the idea of complete-ment, it being the thing needed to complete something else.
It is a common mistake to confuse the words complement and compliment.
Usually the latter is mistakenly used when the former is meant.
Sources
1955: John L. Kelley: General Topology ... (previous) ... (next): Chapter $0$: Subsets and Complements; Union and Intersection
1959: E.M. Patterson: Topology (2nd ed.) ... (previous) ... (next): Chapter $\text {II}$: Topological Spaces: $\S 8$. Notations and definitions of set theory
1964: Steven A. Gaal: Point Set Topology ... (previous) ... (next): Introduction to Set Theory: $1$. Elementary Operations on Sets
1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Introduction: Set-Theoretic Notation
1968: A.N. Kolmogorov and S.V. Fomin: Introductory Real Analysis ... (previous) ... (next): $1$ Set Theory: $1$. Sets and Functions: $1.2$: Operations on sets
1970: Avner Friedman: Foundations of Modern Analysis ... (previous) ... (next): $\S 1.1$: Rings and Algebras
1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: The Notation and Terminology of Set Theory: $\S 8 \beta$
1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.5$: Complementation
1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 3$: Set Operations: Union, Intersection and Complement
1977: Gary Chartrand: Introductory Graph Theory ... (previous) ... (next): Appendix $\text{A}.1$: Sets and Subsets
1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 6$: Subsets
1981: G. de Barra: Measure Theory and Integration ... (previous) ... (next): Chapter $1$: Preliminaries: $1.1$ Set Theory
1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Point Sets: $13.$
1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.2$: Sets
1986: Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction ... (previous) ... (next): $1$: Events and probabilities: $1.2$: Outcomes and events: Footnote
1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): complement: 1b.
1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): complement: 2.
2008: Paul Halmos and Steven Givant: Introduction to Boolean Algebras ... (previous) ... (next): Appendix $\text{A}$: Set Theory: Operations on Sets
2008: David Joyner: Adventures in Group Theory (2nd ed.) ... (previous) ... (next): Chapter $1$: Elementary, my dear Watson: $\S 1.2$: Elements, my dear Watson
2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): complement: 2.
2012: M. Ben-Ari: Mathematical Logic for Computer Science (3rd ed.) ... (previous) ... (next): Appendix $\text{A}.2$: Definition $\text{A}.8$
2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): complement, complementation
2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): complement, complementation
Retrieved from "
Categories:
Definitions/Set Complement
Definitions/Set Theory
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11628 | https://math.stackexchange.com/questions/1971828/proof-involving-circumscribed-circles-of-a-triangle | geometry - Proof involving circumscribed circles of a triangle - Mathematics Stack Exchange
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Proof involving circumscribed circles of a triangle
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So my question is this: Points A,B,A,B, and C C lie on one line, point P P lies outside of this line. Prove that the centers of the circumscribed circles of triangles A B P,B C P,A B P,B C P, and A C P A C P and point P P lie on one circle.
I'm self taught and fairly new to proofs in general and geometric proofs in particular. I'm not really sure where to start with this problem, though I think the Simson line might be applicable. Any help would be seriously appreciated.
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edited Oct 17, 2016 at 1:51
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Let M A,M B,M C M A,M B,M C be the midpoints of segments A P,B P,C P A P,B P,C P respectively. They lie on a common line s P s P which is parallel to line A B≡A C A B≡A C because M A,M B,M C M A,M B,M C determine midsegments in triangles A B P,B C P,C A P A B P,B C P,C A P.
Now, draw the orthogonal bisectors l A,l B,l C l A,l B,l C of segments A P,B P,C P A P,B P,C P respectively. Then M A∈l A,M B∈l B,M C∈l C M A∈l A,M B∈l B,M C∈l C. The intersection points
O A B=l A∩l B,O B C=l B∩l C,O C A=l C∩l A O A B=l A∩l B,O B C=l B∩l C,O C A=l C∩l A
are the circumcircles of triangles A B P,B C P,C A P A B P,B C P,C A P respectively.
Finally, focus on the triangle O A B O B C O C A O A B O B C O C A. Observe that M A,M B,M C M A,M B,M C are the orthogonal projections of the point P P on the extended edges of triangle O A B O B C O C A O A B O B C O C A. However, as pointed out already, the tree points M A,M B,M C M A,M B,M C lie on a common line s P s P, which is called the Simson line. This is possible if and only if P P lies on the circumcircle of triangle O A B O B C O C A O A B O B C O C A by the theorem about the Simson line s P s P.
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answered Oct 16, 2016 at 23:48
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11629 | https://medium.com/basecs/spinning-around-in-cycles-with-directed-acyclic-graphs-a233496d4688 | Sitemap
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Exploring the basics of computer science, every Monday, for a year.
Spinning Around In Cycles With Directed Acyclic Graphs
Vaidehi Joshi
15 min readOct 2, 2017
Throughout the course of this series, I’ve written quite a bit about rediscovering things in computer science that I had heard or read about before, but didn’t really have enough context to understand.
At first, I thought that this strange cyclic process of stumbling upon the same concepts again and again was unique to computer science. But these days, I am of a different opinion: the process of learning and relearning an idea is fundamental to understanding it at a deeper level. This idea has a name: the hermeneutic circle. At it’s core, the hermeneutic circle is grounded in the idea that we must return to ideas again and again, and understand parts of them at a time in order to fully comprehend the whole (and vice versa).
This is cyclical process is exactly what I experienced while learning about different types of graphs, including directed graphs, cyclic graphs, and everything in between. I had heard about these types of graphs before and was slightly familiar with all of them, but didn’t fundamentally understand when they would be used, or why they kept coming up again and again, in everything that I read.
As it turns out, learning is hard! And it takes time, as well as a good amount of repetition. So, let’s return to graphs and learn about a specific subset of them that are fundamental to many solutions in computer science.
Edges, edges, everywhere
Throughout our exploration of graphs, we’ve focused mostly on representing graphs, and how to search through them. Last week, we looked at depth-first search (DFS), a graph traversal algorithm that recursively determines whether or not a path exists between two given nodes. However, it’s worth cycling back to depth-first search again for a few reasons.
Depth-first search is useful in helping us learn more about a given graph, and can be particularly handy at ordering and sorting nodes in a graph. We can use the DFS algorithm to identify cycles within a graph, to determine what kinds of edges a graph has, and to order the vertices within in a linear fashion.
We’ll work our way through each of these three tricks today, and see how using depth-first search can help us better understand what kind of graph we’re dealing with in the process of searching through it!
Early on in this series, we learned about edges, the elements that connect the nodes in a graph. We also learned that edges can be directed, with a flow from one node to another, or undirected, with a bidirectional flow between both vertices. But, keeping with the theme of this post, let’s return back to what we think we already know.
If we cycle back to edges, we’ll see that, in the context of a graph, edges can be more than just “directed” or “undirected”. Two directed edges can be very different from one another, depending on where in the graph they happen to occur! Let’s take a look at some of the other ways that we can classify edges in a graph.
There are two main categories of edges in a graph data structure: a tree edge and a non-tree edge. We’ll recall that in both the BFS algorithm as well as in the DFS algorithm, the parent pointers, which lead back to the starting node, can be rearranged to form a tree.
Any edge that is part of a path taken in a graph traversal algorithm is known as a tree edge, since it is an edge that allows us to visit a new vertex each time as we traverse through the graph, and can be restructured to form a “tree”.
So, what about the non-tree edges? Well, there are three different variations of those, and we’ve actually already encountered all of them, even if we didn’t necessarily know what they were called at the time.
The first of the three is called a forward edge, as it allows us to move “forward” through the graph, and could potentially be part of another path down the tree. By contrast, there is also the backward edge, which connects a node in a graph “back up” to one of its ancestors (its parent or grandparent node, for example, or even back to itself!). Finally, there is the cross edge, which “crosses over” and connects one subtree in a graph to another. In other words, a cross edge connects to sibling nodes that don’t necessarily share an ancestor in a tree path, but connects them anyways.
These definitions make a whole lot more sense when we can see them applied to actual nodes in a graph, so let’s look at an example.
In the illustration shown here, we can see that the tree edges, highlighted in blue, are the ones that we walk through as part of our traversal down our path in the graph; in this case, our walk is u-v-y-x. The edges connecting these vertices in our path are our tree edges. However, we could have also walked “forward” in a different path, and walked from u to x; thus, the edge (u, x) is a forward edge.
We’ll notice that, when we get to node x, the only vertex to navigate to next is v. The edge connecting these two nodes in the tree, (x, v), is a backward edge, since it connects the node x back to one of its ancestors in the path: node v. We’ll also notice that, in another part of the graph — in a different subtree, rather — node z has a backward edge that is actually connecting back to itself. This is also known as a self-loop, or a backward edge that connects a node back to itself, or an edge that “references” the node that it came from: (z, z). (We’ll come back to self-loops later on, but this is our very first taste of them!)
The edge that connects node w and y, which we can also call (w, y), is a cross edge as it connects the subtree. Cross edges are a little easier to see when we rearrange our tree edges to actually form a tree, so let’s redraw that same graph to make our cross edge more obvious. In the image below, we’ll see two potential “trees” that could be formed by starting at two different nodes in the graph: node u or node w, respectively.
The edge (w, y) connects the two subtrees of this entire graph together, even though there is no single “root node” or ancestor that the subtrees actually share. Remember that a graph is a very “loose” version of a tree, and doesn’t follow the same rules that a tree would.
Since a graph doesn’t have one single root node, it doesn’t form a single tree. Rather, a graph could contain many small subtrees within it, each of which would become obvious as we walked through the paths of the graph. A graph is less of an individual tree data structure, and more of an interconnected forest, with small subtrees within it.
There is one important thing to note about all of these different kinds of edges: they don’t always exist! It all depends on what type of graph we’re dealing with. In a directed graph, tree edges, cross edges, backward edges, and forward edges are all possible. However, for an undirected graph, it’s a different story altogether.
In an undirected graph, there are no forward edges or cross edges. Every single edge must be either a tree edge or a back edge.
An easy way to remember the rules of edge classification is this: an undirected graph can only have tree edges and back edges, but a directed graph could contain all four edge types.
Okay, but what’s the story behind this rule? Well, it’s a lot easier to see if we actually try it out on an example, so we’ll do exactly that.
In the graph shown here, we have four vertices, and four edges. Let’s say that the edge (a, d) is a “forward” edge; in other words, it’s an edge that allows us to move from node a to node d.
Imagine that we walk through this graph, from node a to b, and then from b to c. When we get to node c, there is nowhere else for us to go, so we must backtrack up to node b.
Now, in order for us to properly run the depth-first search algorithm, we need to check if there are any other vertices that we can “visit” from node b. As it turns out, there is one! We’ll visit node d. We’re once again in a similar situation, and we have to check if there is another node that we can visit from d. As it turns out, since this is an undirected graph, there is one place we can visit: node a. At this moment, it becomes impossible for the edge (a, d) to be a “forward” edge. Since the rules of DFS dictate that we must visit node a from d since we must explore the graph on the deepest level, we must visit it; we cannot backtrack back up to node a and then visit d from there!
An undirected graph can never allow for a forward edge like (a, d) since, in order for a forward edge to exist, we would need to completely finish visiting all the nodes reachable from our starting node a before traversing through the edge. In an undirected graph, this is literally impossible to do! So, the edge between these two nodes really ought to be written like this: {a, d}, since we can traverse in either direction on this edge, depending on which node we start with.
Even though we won’t look at an example of it here, we could disprove cross edges similar to forward edges, since they also cannot exist in an undirected graph.
Round and round
Another great strength of the depth-first search algorithm is its ability to identify cycles in a graph. Before we get too far into how to do that, let’s familiarize ourselves with some important terms that we’ll end up using along the way.
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A cycle, in the context of a graph, occurs when some number of vertices are connected to one another in a closed chain of edges. A graph that contains at least one cycle is known as a cyclic graph. Conversely, a graph that contains zero cycles is known as an acyclic graph.
In the previous section, we saw our first instance of a self-loop, a type of backward edge. Self-loops can only ever occur in a directed graph, since a self-loop is a type of directed edge.
Both directed and undirected graphs can have cycles in them, but it’s worth noting that a self-loop can only ever occur in a directed cyclic graph, which is a directed graph that contains at least one cycle in it.
As it turns out, the reason that the depth-first search algorithm is particularly good at detecting cycles is because of the fact that it is efficient at finding backward edges.
By definition, any graph that has a backward edge will contain a cycle. In the graph drawn below, we can see that node e connects back to an ancestor in the path, node a. Because of a backward edge exists in this directed graph, we know that this graph also contains a cycle — it is a directed cyclic graph.
Okay, but how does DFS actually detect cycles, exactly? And why is this algorithm in particular so efficient at determining whether or not a cycle exists within a graph? Well, it all goes back to the recursive nature of the depth-first search algorithm. We’ll recall that DFS uses a stack data structure under the hood. This makes is slightly easier to know whether a backward edge exists in the path as we walk through the graph.
For example, in the example graph from earlier, when we traverse from u to v to y to x, we add each of these elements to the stack of nodes that we’ve visited. We can take one additional step to make it easy to classify edges: we can mark the node u as currently beingProcessed with a simple boolean flag when we start searching down its “deepest” path. Once we flag node u as being the node that we’re searching through, if any node that we add to the stack that references back up to u, we know that we have found a cycle. Indeed, this same logic applies to other internal cycles in the graph, and not just cycles connecting back up to the starting node.
If any node added to the stack has a reference to a node that is already within the stack, we can be sure that we have a cycle in this graph.
But wait — there’s more! Well, okay — there is just one more. The final type of graph that we need to define is a directed graph without cycles.
Most implementations of depth-first search will check to see if any cycles exist, and a large part of that is based on the DFS algorithm checking to see whether or not a graph is a directed acyclic graph, also known as a DAG. A directed acyclic graph is, as its name would suggested, directed, but without any cycles. The golden rule of DAGs is that, if we start at any node in the graph, no sequence of edges will allow us to loop back to the starting node. Thus, by definition, a directed acyclic graph can never contain a self-loop.
DAGs are somewhat infamous in computer science because they’re pretty much everywhere in sofware. For example, a directed acyclic graph is the backbone of applications that handle scheduling for systems of tasks or handling jobs — especially those that need to be processed in a particular order. At a higher level, DAGs are used to represent state machines, which are often used to keep track of objects that have a mutable state in web and mobile applications.
My personal favorite example, however, is the use of DAGs to represent dependency graphs. If you’ve ever had to resolve dependencies in a Gemfile, or handle updating packages or navigate dependency issues using package managers like npm or yarn, then you were interacting with a DAG, on a very abstracted level! The dependency graph is also a great example of how DAGs can be complicated and massive and size, and why it can be useful to sort through and order the nodes within such a graph.
So, let’s learn how to do that!
Topological sorting
Since we now know how vast and complicated a directed acyclic graph can actually be, being able to sort through and order vertices and make sense of the data within a DAG can be super helpful. Thankfully, there is an algorithm that does exactly that!
The topological sort algorithm allows us to sort through the vertices of graph in a specific order, based on the interconnectedness of the edges that connect the vertices. If two vertices, x and y exist in a graph, and a directed edge (x, y) exists between them, then topological sort will order the vertices in the way that we would encounter them in a path: x, followed by y.
If we consider the ordering of topological sort, we’ll notice that the sorting itself is non-numerical. In other words, the order of the vertices is based on the edges that connect them, and not any numeric value.
The most important thing to keep in mind when it comes to topological sort is the fact that we can only ever visit a vertex after all the vertices leading up to it have already been visited.
For example, in the graph shown here, we have five nodes: v, w, x, y, and z. In order to run topological sort on this directed graph, we need to be sure that, as we order and visit our vertices, the node y cannot be visited until both x and w have been visited.
Just from looking at this graph, we can see that there are two possible routes that we could take: v-w-x-y-z or v-x-w-y-z. In both of these two paths, we would still be visiting node y after both x and w. Thus, both of them are valid, and each of them return valid topological orders, which is the ordered set of vertices that is the returned result of running topological sort.
In this graph, we don’t have a cycle. But what if we did? The simplest version of such a graph would have two nodes, each of which reference one another, forming a cycle. Here’s an example of what that graph might look like:
In this example, vertex a depends on b, which in turn, depends back on a. The issue here is that we can’t apply the rule of visiting the vertices that “lead up” to the one we’re starting with, because they both lead up to one another.
A topological sort can only ever be applied to directed acyclic graphs, or DAGs. It is impossible to run a topological sort on a directed graph with a cycle, since it is unclear where the sort itself should start.
There are a few different ways to actually implement topological sort. However, for today, we don’t even need to learn any new ones, because we already know one intuitively! I’m talking, of course, about depth-first search.
So, let’s take a look at topological sort in action — using DFS to help us!
We’ll work with the same graph as earlier; the important thing about running DFS to find a topological order is using the reverse of the DFS finishing order to assemble our vertices.
We’ll start by running DFS on our starting “parent” node, which has to be v, as it is the only node without an edge leading up to it. We can choose either w or x to traverse down — it doesn’t really matter which. In this example, we’ll choose w, and continue down to y, then z.
When we reach z, we’ve hit a dead end! So, we’ll need to backtrack up to y. But, before we backtrack up, we’ll assign a number (which we’ll use to order our vertices in a moment) to node z. In this case, we’ll assign it a value of 4 since it is going to be ordered last — remember, we want the reverse of DFS’s finishing order.
Next, we’ll backtrack up to y; since there’s nowhere to go from there, we’ll repeat the same step of assigning a number, 3, and then again backtracking up to w. We’ll continue to do this until we get to a node with an unvisited edge. Each time we do this, we’ll order the vertex as necessary.
Finally, we end up back at our starting node, v. When we do backtrack up to the “parent” node, we’ll see that there is one edge that we haven’t visited: the only one left is x. So, we’ll visit it, assign it a number (in this case, 1) and finish up with our search through the node v. Finally, if we follow the number 0-1-2-3-4, we’ll end up with the order v-x-w-y-z.
And there we have it! Our vertices are all sorted so that any node that is dependent on a node that precedes it will only be visited if it’s parent(s) in the path — its dependencies — are visited first. Now, our example graph was pretty small, but we can imagine how npm or yarn might do this for a much larger graph of dependencies.
So, the next time you feel intimidated by updating your dependencies, just remember: it’s just a directed acyclic graph! You got this.
Resources
There are a few different approaches to topological sort, but I find that understanding it from the context of depth-first search and cycles in a graph to be the easiest. Thankfully, there are some good resources out there that break down how topological sort works in other contexts, not to mention all the extensive resources on directed acyclic graphs! Here are some of my favorites if you want to keep learning.
Graph Topological Sort Using Depth-First Search, Sesh Venugopal
Depth-First Search (DFS), Topological Sort, MIT OpenCourseWare
Topological Sorting — Graph Theory, NerdFirstTV
Graph Traversals, Andrew Myers, Cornell University Department of Computer Science
Depth-first Search, Professor Steven Skiena
Topological Sort, Professor Trevor Brown
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## Published in basecs
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·Last published Dec 30, 2017
Exploring the basics of computer science, every Monday, for a year.
## Written by Vaidehi Joshi
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Writing words, writing code. Sometimes doing both at once.
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Zubair Khalid
Jun 10, 2021
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great
```
Prajjwalfire
Mar 21, 2021
```
If you had a book you might have given CLRS a tough time.
```
Lơi Lê
Aug 8, 2020
We can use the DFS algorithm to identify cycles within a graph, to determine what kinds of edges a graph has, and to order the vertices within in a linear fashion.
```
leloi0772121433@gmail.com
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11630 | https://dimensionsofdentalhygiene.com/article/one-time-subgingival-irrigation/ | Reconnecting Practicing Hygienists with the Nation's Leading Educators and Researchers.
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Ask the ExpertInstrumentation
One-Time Subgingival Irrigation
Should I use chlorhexidine or stabilized chlorine dioxide for one-time subgingival irrigation after scaling and root planing?
By Kathleen O. Hodges, RDH, MS 0
Should I use chlorhexidine or stabilized chlorine dioxide for one-time subgingival irrigation after scaling and root planing?
Chlorhexidine gluconate (0.12%) is known as the gold standard for antimicrobials due to its demonstrated efficacy. Prescription mouthrinses containing chlorhexidine are indicated for gingival inflammation, caries, and in-office preprocedural rinsing. Two systematic reviews revealed significant plaque and gingivitis reductions when using a chlorhexidine mouthrinse as compared to placebos.1,2 However, chlorhexidine mouthrinses can cause staining of tooth surfaces, restorations, and the tongue, as well as taste alterations.
Stabilized chlorine dioxide mouthrinses are used in plaque reduction and oral malodor prevention. They do not seem to cause staining. One product containing this compound has earned the American Dental Association Seal of Acceptance for oral malodor. A 2020 systematic review and meta-analysis concluded that chlorine dioxide reduces both plaque and gingival indices, and bacterial counts similar to other commonly used mouthrinses.3 The available evidence is limited, however, and large scale randomized controlled trials and additional clinical data are needed.3
Research on one-time subgingival irrigation with antiseptics post-scaling and root planing (SRP) does not support its use at this time. A 2020 systematic review found that subgingival irrigation with antiseptics did not show significant improvements in bleeding on probing, probing depth, or clinical attachment levels with a follow-up of at least 6 months.4 Factors affecting a single episode of subgingival irrigation with antiseptics post-SRP are the deactivation of agents by blood and crevicular fluids, gingival tissue tone and pocket depth affecting penetration, type of delivery method, and short-term suppression of microbes. Although microbes are suppressed by antimicrobial subgingival irrigation, they might not be eliminated and can return to baseline in a short time.
A relative question is “What is the purpose of subgingival irrigation after SRP?” For lavage of periodontal pockets, subgingival irrigation with water via a syringe, pulsating device, or ultrasonic instrumentation might be adequate. For reducing the microbial count between professional visits, repeated applications over time are most likely necessary. To treat gingival inflammation, an efficacious antimicrobial mouthrinse could be recommended for daily patient use. For treating clinical indicators of periodontitis, research does not support this one-time treatment for periodontitis.
The gold standard of nonsurgical periodontal therapy is SRP to disturb or remove plaque biofilms and remove calculus. In most cases, this therapy alone is adequate to initiate the expected healing. Cases that do not result in healing are reevaluated for patient self-care practices, residual calculus deposits, and need for adjunctive therapy.
In conclusion, for a single episode of antimicrobial subgingival irrigation to be effective, it should enhance the clinical effects of instrumentation alone. Research to date does not demonstrate a beneficial effect at 6 months regardless of the agent used. Controlled-release delivery (chips, fibers, gels) is thought to produce a more constant and prolonged concentration of the agent compared to systemic (oral antibiotics) and topical delivery (irrigation).
References
Van Strysonck D, Slot DE, Van Der Velden U, Van Der Weijden GA. Effect of a chlorhexidine mouthrinse on plaque, gingival inflammation and staining in gingivitis patients: a systematic review. J Clin Periodontol. 2012;39:1042–1055.
James P, Worthington HV, Parnell C, et al. Chlorhexidine mouthrinse as an adjunctive treatment for gingival health. Cochrane Database Syst Rev. 2017;3:1–247.
Kerémi B, Márta K, Farkas K, et al. Effects of chlorine dioxide on oral hygiene—a systemic review and meta analysis. Curr Pharm Des. 2020;26:3015–3025.
Ramanauskaite E, Machiulskiene V. Antiseptics as adjuncts to scaling and root planing in the treatment of periodontitis: a systematic literature review. BMC Oral Health. 2020;20:143.
The Ask the Expert column features answers to your most pressing clinical questions provided by Dimensions of Dental Hygiene’s online panel of key opinion leaders, including: Jacqueline J. Freudenthal, RDH, MHE, on anesthesia; Nancy K. Mann, RDH, MSEd, on cultural competency; Claudia Turcotte, CDA, RDH, MSDH, MSOSH, on ergonomics; Van B. Haywood, DMD, and Erin S. Boyleston, RDH, MS, on esthetic dentistry; Michele Carr, RDH, MA, on ethics and risk management; Erin Relich, RDH, BSDH, MSA, on fluoride use; Kandis V. Garland, RDH, MS, on infection control; Mary Kaye Scaramucci, RDH, MS, on instrument sharpening; Stacy A. Matsuda, RDH, BS, MS, on instrumentation; Karen Davis, RDH, BSDH, on insurance coding; Cynthia Stegeman, EdD, RDH, RD, LD, CDE, on nutrition; Olga A.C. Ibsen, RDH, MS, on oral pathology; Jessica Y. Lee, DDS, MPH, PhD, on pediatric dentistry; Timothy J. Hempton, DDS, on periodontal therapy; Ann Eshenaur Spolarich, RDH, PhD, on pharmacology; and Caren M. Barnes, RDH, MS, on polishing. Log on to dimensionsofdentalhygiene.com/asktheexpert to submit your question.
From Dimensions of Dental Hygiene. January 2022;20(1):47.
0
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Kathleen O. Hodges, RDH, MS
Kathleen O. Hodges, RDH, MS, is a professor emerita in the Department of Dental Hygiene at Idaho State University in Pocatello. She has more than 40 years of experience in dental hygiene education, including clinical teaching and administration. Hodges works in private practice and is a member of the American Dental Hygienists’ Association. She is also a board member for the Western Society of Periodontology and a Dimensions of Dental Hygiene Editorial Advisory Board member.
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11631 | https://tmedweb.tulane.edu/pharmwiki/doku.php/antidepressants | antidepressants [TUSOM | Pharmwiki]
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Antidepressants
Antidepressants Black Box Warning
Example for Prozac ® (fluoxetine)
SUICIDALITY AND ANTIDEPRESSANT DRUGS.
Antidepressants increased the risk compared to placebo of suicidal thinking and behavior (suicidality) in children, adolescents, and young adults in short-term studies of Major Depressive Disorder (MDD) and other psychiatric disorders. Anyone considering the use of PROZAC or any other antidepressant in a child, adolescent, or young adult must balance this risk with the clinical need. Short-term studies did not show an increase in the risk of suicidality with antidepressants compared to placebo in adults beyond age 24; there was a reduction in risk with antidepressants compared to placebo in adults aged 65 and older. Depression and certain other psychiatric disorders are themselves associated with increases in the risk of suicide. Patients of all ages who are started on antidepressant therapy should be monitored appropriately and observed closely for clinical worsening, suicidality, or unusual changes in behavior. Families and caregivers should be advised of the need for close observation and communication with the prescriber.
Reference:
rxlist.com (Prozac ®)
Related news:
Antidepressant Use Down, Suicide Attempts Up After FDA Warnings
Lu CY et al (2014): Changes in antidepressant use by young people and suicidal behavior after FDA warnings and media coverage: quasi-experimental study. BMJ 348:g3596 (Published 18 June 2014)
Tricyclic Antidepressants (TCAs)
Amitriptyline
Trade Name: Elavil ®
Drug Class: TCA antidepressant (tricyclic amine)
Mechanism of Action:
blocks reuptake of norepinephrine & serotonin (NET & SERT)
amitriptyline has a greater selectivity for SERT>NET, but its active metabolite (nortriptyline) blocks NET>SERT.
Na channel blocker
Antimuscarinic
Antihistamine
Alpha receptor blocker
Indications:
For the relief of symptoms of depression
Endogenous depression is more likely to be alleviated than are other depressive states.
Prophylaxis for migraine headache (off label use).
Contraindications:
Acute recovery period after a myocardial infarction
Side Effects:
NOTE: TCAs are among the most common prescription drugs involved in life-threatening drug overdose
antimuscarinic: dry mouth, tachycardia, constipation, bladder problems, sexual problems, blurred vision & dilated pupils
alpha receptor block: orthostatic hypotension & dizziness
antihistamine: drowsiness / sedation
Na channel block: prolonged QRS, & at toxic doses - potentially fatal cardiac arrhythmias & seizures
Cardiac K channel block: QTc prolongation
Other: weight gain, lowered seizure threshold
Therapeutic Options due to Side Effects:
Wait for tolerance to develop to antimuscarinic & antihistaminic effects
lower the dose
switch to an SSRI/SNRI or newer antidepressant
Treatment of Overdose:
Sodium bicarbonate to normalize pH & alleviate Na channel block
Avoid use of other Class I & III antiarrhythmic drugs
Benzodiazepines for seizures
Fluid & pressors (norepinephrine) for hypotension
TCA-specific antibody fragments i.v. may reduce toxicity & assist in decontamination (Soghoian et al, 2010)
Pharmacokinetics:
Metabolizedby CYP 2D6, 3A4 & 2C19
It is converted by demethylation to nortriptyline, a more NET-selective TCA (amitriptyline is more SERT-selective).
Major drug Interactions:
decongestants & local anesthetics given w/ sympathomimetics may increase sympathomimetic activity
drugs metabolized by P450-2D6 (e.g. cimetidine or SSRIs) may increase TCA levels & toxicity
MAO inhibitors within 14 days of each other (these drugs elevate presynaptic monoamines & can result in either a serotonin syndrome or hypertensive crisis).
History:
amitriptyline is a structural analog of imipramine, the first tricyclic antidepressant discovered (Domino, 1999).
References:
DeBattista C (2024): Antidepressant Agents. Chapter 30. In: Basic & Clinical Pharmacology. 16th Ed. Vanderah TW (Editor). McGraw Hill / Lange.
Domino EF (1999): History of Modern Psychopharmacology: A Personal View With an Emphasis on Antidepressants. Psychosomatic Medicine 61(5):591-598.
Soghoian et al (2010): Toxicity, Tricyclic Antidepressant: Treatment & Medication. eMedicine(Medscape.com) (accessed 2/17/2011)
Stahl SM (2009): The Prescriber's Guide. Stahl's Essential Psychopharmacology. 3rd Edition Cambridge University Press. pp 77-81 (ISBN: 978-0-521-74399-0)
rxlist.com (Elavil ®)
Keywords
Imipramine
Trade Name: Tofranil ®
Drug Class: TCA antidepressant (tricyclic amine)
Mechanism of Action:
Blocks reuptake of norepinephrine & serotonin at nerve endings (therapeutic effects)
Na channel blocker (contributes to cardiotoxicity)
Antimuscarinic (contributes to side effects)
Antihistamine (contributes to side effects)
Alpha receptor blocking properties (contributes to side effects)
Several weeks to months are required for antidepressant effects to appear (or be significantly greater than a placebo), although its direct effects (e.g. on reuptake) develop within a matter of hours.
Indications:
Depression
typically for patients who do not respond to SSRIs, NDRIs and other 2nd or 3rd generation antidepressants
Neuropathic (Chronic) pain
Neuropathic pain does not typically respond to classic analgesics, or opioids. Tricyclic antidepressants including imipramine and amitriptyline are a mainstay of treatment for a variety of chronic pain states, although not FDA-approved for this indication (Tauben & Stacey, 2021).
Nocturnal enuresis (a pediatric indication - attributed to its antimuscarinic “side effects”)
Contraindications:
Drugs that inhibit monoamine oxidase (MAOI's) taken w/in 14 days, acute recovery period after a myocardial infarction, or history of drug hypersensitivity
Side Effects:
drowsiness
dry mouth & eyes
constipation
orthotstatic hypotension
mild tremor
sweating
agitation
nausea
tinnitus
Overdose & Treatment: See Amitriptyline; potentially fatal disturbances of cardiac conduction (widening of QRS), arrhythmias & seizures due to Na channel block (cardiac & CNS)
Pharmacokinetics:
PO, metabolized in the liver (P450 - 2D6); 16 hr half life
desipramine (generic, Norpramin ®) is an active metabolite of imipramine that selectively inhibits NET, with minor effects on SERT.
Major drug Interactions:
Decongestants
Local anesthetics w/ sympathomimetics
Antihypertensives
CNS depressants
Drugs metabolized by P450-2D6
Concomitant use of MAOI's can cause potentially fatal hyperpyretic crisis & seizures
History:
The therapeutic & commercial success of phenothiazine antipsychotics initiated a large scale search for more effective antipsychotics during the early 1950's. During the process of research & development, imipramine was synthesized as a structural analog of chlorpromazine. When tested in normal human volunteers, it was found to exert sedative properties, but when tested in ~500 patients with various psychiatric disorders, only those with endogenous depression exhibited a “remarkable” improvement of clinical symptoms after 1-6 weeks of daily treatment. Thus the first clinically useful tricyclic antidepressant was discovered (Domino, 1999).
References:
Biaggioni I, Gurevich VV (2024): Adrenoceptor Agonists & Sympathomimetic Drugs (Chapter 9). In: Basic and Clinical Pharmacology. 16th Vanderah TW (Editor); McGraw-Hill (Access Medicine).
DeBattista C (2024): Antidepressant Agents. Chapter 30. In: Basic and Clinical Pharmacology. 16th Ed. Vanderah TW (Editor); McGraw-Hill (Access Medicine).
Domino EF (1999): History of Modern Psychopharmacology: A Personal View With an Emphasis on Antidepressants. Psychosomatic Medicine 61(5):591-598.
Tauben D, Stacdy BR (2021): Pharmacologic management of chronic non-cancer pain in adults. In: UpToDate, Basow, DS (Ed), Waltham, MA. Cited 10/19/2021.
rxlist.com (Tofranil ®)
Keywords
SSRIs
Fluoxetine
Trade Names: Prozac, Sarafem ®
Drug Class: Antidepressant
Mechanism of Action:
Selective Serotonin Reuptake Inhibitor (SSRI)
Selectively blocks SERT (SERT>NET & DAT)
Indications:
Major Depressive Disorder (MDD)
Obsessive Compulsive Disorder (OCD)
Panic Disorder
Moderate to severe bulimia nervosa
Contraindications:
caution if liver dx. or impaired renal function, drugs metabolized by P450-2D6.
alcohol can interfere with SSRI’s therapeutic effects.
other serotonergic drugs that increase the likelihood of serotonergic syndrome (e.g. triptan antimigraine medications)
Side Effects:
headache, nausea, nervousness or insomnia, agitation (feeling jittery)
sexual dysfunction (men: delayed ejaculation, erectile dysfunction; men & women: decreased sexual desire, anorgasmia)
severe hyponatremia
caused by SSRI-induced release of antidiuretic hormone, leading to the Syndrome of Inappropriate Antidiuretic Hormone (SIADH)
usually occurs during the first two weeks after initiation of therapy, appears to not be dose dependent
has also been observed with escitalopram (Jacob & Spinler, 2006)
relatively rare but potentially fatal (Jacob & Spinler, 2006; Rawal et al, 2017)
Pharmacokinetics:
PO, metabolized in the liver primarily by CYP2D6, with fluoxetine (parent drug) having a half life of several day half life of 1-4 days
norfluoxetine is an important active metabolite with a t½ ≅ 7-15 days, which contributes to fluoxetine having the longest half-life of all the SSRIs.
the long half-life of norfluoxetine reduces serotonin discontinuation symptoms after either a missed dose, or when discontinuing treatment.
consider the long effective half life of norfluoxetine when switching from fluoxetine to another antidepressant, as this can result in additive effects, and an increased likelihood of serotonin syndrome for the time period where both drugs are simultaneously effective. A variable dosing scale-up protocol may be warranted when switching antidepressants.
References:
DeBattista C (2024): Antidepressant Agents. Chapter 30. In: Basic & Clinical Pharmacology. 16th Ed. Vanderah TW (Editor). McGraw Hill / Lange.
Jacob S, Spinler SA (2006): Hyponatremia associated with selective serotonin‑reuptake inhibitors in older adults. Ann Pharmacother 40:1618‑22.
Rawal G et al (2017): Severe hyponatremia associated with escitalopram. J Family Med Prim Care. 6(2): 453–454. doi: 10.4103/2249-4863.220043
Stahl SM (2009): The Prescriber's Guide. Stahl's Essential Psychopharmacology. 3rd Edition. Cambridge University Press. pp 77-81 (ISBN: 978-0-521-74399-0)
rxlist.com (Prozac ®)
Keywords
Paroxetine
Trade Name: Paxil, Paxil CR, Asimia, Pexeva ®
Drug Class: Antidepressant
Mechanism of Action: SSRI
Indications:
Major Depressive Disorder (MDD)
Obsessive Compulsive Disorder (OCD)
Panic Disorder
Social Anxiety Disorder
Generalized Anxiety Disorder (GAD)
Posttraumatic Stress Disorder (PSTD)
Contraindications:
MAOI's w/in 14 days
Alcohol use may interfere with it's clinical efficacy
Other serotonergic drugs that increase the likelihood of serotonergic syndrome (e.g. triptan antimigraine medications)
Side Effects:
headache, nausea, nervousness or insomnia, agitation (feeling jittery)
sexual dysfunction (men: delayed ejaculation, erectile dysfunction; men & women: decreased sexual desire, anorgasmia).
recent evidence indicates it has a low incidence of teratogenic effects (e.g. atrial & ventricular cardiac septal defects).
Patients who are likely to become pregnant should strongly consider taking another drug.
Pregnancy Risk:
Risk Category D (Positive evidence of risk to human fetus; potential benefits may still justify its use during pregnancy). Not generally recommended for using during pregnancy (especially the first trimester).
Pharmacokinetics:
PO, metabolized by the liver, 21-24 hr half life
Major drug Interactions:
MAOI's
other CNS acting drugs
Notes:
avoid abrupt withdrawal, as symptoms can be more severe than “advertised” by the manufacturer, and are more common with paroxetine than with some other SSRIs.
References:
DeBattista C (2024): Antidepressant Agents. Chapter 30. In: Basic and Clinical Pharmacology. 16th Ed. Vanderah TW (Editor); McGraw-Hill (Access Medicine).
Stahl SM (2009): The Prescriber's Guide. Stahl's Essential Psychopharmacology. 3rd Edition. Cambridge University Press. pp 77-81 (ISBN: 978-0-521-74399-0)
rxlist.com (Paxil ®)
Keywords
Sertraline
Trade Name: generic, Zoloft ®
Drug Class: SSRI Antidepressant
Mechanism of Action:
selective inhibitor of the serotonin retuptake transporter (SERT)
very weak effects on norepinephrine and dopamine neuronal reuptake (Thomas et al, 1998; Kitaichi et al, 2010; rxlist.com)
Indications:
Major Depressive Disorder (MDD)
Obsessive Compulsive Disorder (OCD)
Panic Disorder
Social Anxiety Disorder
Premenstrual dysphoric disorder (PMDD) in adults
Posttraumatic Stress Disorder (PSTD)
Contraindications:
alcohol can interfere with SSRI’s therapeutic effects
other serotonergic drugs that increase the likelihood of serotonergic syndrome (e.g. triptan antimigraine medications)
Side Effects:
headache, nausea, nervousness or insomnia, agitation (feeling jittery)
sexual dysfunction (men: delayed ejaculation, erectile dysfunction; men & women: decreased sexual desire, anorgasmia), some patients may experience weight loss (vs weight gain).
Notes:
avoid abrupt withdrawal
generic form is available & is relative inexpensive.
References:
DeBattista C (2024): Antidepressant Agents. Chapter 30. In: Basic and Clinical Pharmacology. 16th Ed. Vanderah TW (Editor); McGraw-Hill (Access Medicine).
Kitaichi Y et al (2010): Sertraline increases extracellular levels not only of serotonin, but also of dopamine in the nucleus accumbens and striatum of rats. Eur J Pharmacol. Nov 25;647(1-3):90-6. doi: 10.1016/j.ejphar.2010.08.026.
Stahl SM (2009): The Prescriber's Guide. Stahl's Essential Psychopharmacology. 3rd Edition. Cambridge University Press. pp 77-81 (ISBN: 978-0-521-74399-0)
Thomas DN et al (1998): Sertraline, a selective serotonin reuptake inhibitor modulates extracellular noradrenaline in the rat frontal cortex. J Psychopharmacol 12(4):366-70.
rxlist.com (Zoloft ®)
Keywords
Atypicals (SNRIs, NDRIs)
Bupropion
Trade Names: Wellbutrin, Wellbutrin SR or XL, Zyban, Budeprion XL ®
Drug Class:NDRI Antidepressant (norepinephrine dopamine reuptake inhibitor)
Mechanism of Action:
Bupropion is a monocyclic aminoketone that is structurally related to amphetamine.
It is a weak inhibitor of Norepinephrine & Dopamine reuptake transporters (NET & DAT) with no significant effect on neurotransmitter receptors. Presently it is the only drug available with these characteristics. Human PET scans indicate that 10-30% of striatal DATs are bound by drug at therapeutic concentrations of bupropion (Stahl, 2013).
Indications:
Major Depressive episodes
because it is an NDRI, and does not work by elevating serotonin, bupropion may be particularly useful in patients whose depression either does not respond to an SSRI, or who cannot tolerate the side effects of an SSRI
Seasonal Affective Disorder (Wellbutrin XL ®)
Smoking Cessation (Zyban ®)
SSRI-induced sexual dysfunction (Off Label)
Contraindications:
MAOI's w/in 14 days
Patients with a seizure disorder, patients with a current or prior diagnosis of bulimia or anorexia nervosa because of a higher incidence of seizures. This effect is dose-dependent, and represents a higher risk compared to use of SSRIs (Hirsch & Birnbaum, 2015)(see table).
The increase in seizure risk appears to be associated with the effects of elevated dopamine levels on seizure threshold.
| Drug Treatment Effects on Seizure Rates |
| Drug / Condition | Seizure Rate |
| Untreated population avg | 0.07-0.09% |
| SSRIs | 0.1 % |
| TCAs | 0.4-2% |
| Bupropion (low dose) | 0.4% |
| Bupropion (high dose) | 4% |
| Bupropion sustained release | 0.1% |
| Bulimia + Bupropion (low dose) | 4% |
Side Effects:
dry mouth
insomnia
agitation
possible weight loss
Pharmacokinetics:
PO, metabolized by the liver (Cyt P450 2B6), ~21 hr half life.
Bupropion is also an inhibitor of Cyt 2D6 (a different isoform than is responsible for bupropion's metabolism).
Drug Interactions:
Bupropion inhibits Cyp 2D6, although it is actually metabolized primarily by Cyt 2B6. This results in the potential for drug-drug interactions when bupropion is co-administered with drugs metabolized by Cyt 2D6 (which includes many antidepressants, antipsychotics, beta blockers and Class IC antiarrhythmics) (rxlist.com side - drug interactions).
Notes:
avoid abrupt withdrawal
Pronunciation:
bu pro pea on
References:
DeBattista C (2024): Antidepressant Agents. Chapter 30. In: Basic and Clinical Pharmacology. 16th Ed. Vanderah TW (Editor); McGraw-Hill (Access Medicine).
Hirsch M, Birnbaum RJ (2015): Atypical antidepressants: pharmacology, administration, and side effects. In: UpToDate, Basow, DS (Ed), Waltham, MA. Cited 2/16/16.
Stahl SM (2013): Stahl's essential psychopharmacology. 4th Edition. Cambridge Univ Press. ISBN 978-1-107-68646-5.
rxlist.com (Wellbutrin ®)
rxlist.com (Zyban ®)
Keywords
Duloxetine
Trade Name: Cymbalta ®
Drug Class:SNRI Antidepressant (dual serotonin & norepinephrine reuptake inhibitor)
Mechanism of Action:
Multiple mechanisms: a potent inhibitor of the uptake of neuronal serotonin and norepinephrine in the CNS
Indications:
Treatment of depression
Generalized anxiety disorder
Diabetic peripheral neuropathic pain
Fibromyalgia
Chronic muscoskeletal pain (including lower back pain & osteoarthritis)(FDA approved Nov 2010)
References:
DeBattista C (2024): Antidepressant Agents. Chapter 30. In: Basic and Clinical Pharmacology. 16th Ed. Vanderah TW (Editor); McGraw-Hill (Access Medicine).
Stahl SM (2009): The Prescriber's Guide. Stahl's Essential Psychopharmacology. 3rd Edition. Cambridge University Press. pp 77-81 (ISBN: 978-0-521-74399-0)
rxlist.com (Cymbalta ®)
Keywords
Venlafaxine
Trade Names: Effexor, Effexor ER ®
Drug Class:SNRI Antidepressant (dual serotonin & norepinephrine retuptake inhibitor)
Mechanism of Action:
a potent inhibitor of the uptake of neuronal serotonin and norepinephrine in the CNS
at lower doses behaves like an SSRI
Indications:
treatment of depression
generalized anxiety disorder (GAD)(extended-release product)
social anxiety disorder (social phobia)
panic disorder
Off label uses:
Posttraumatic Stress Disorder (PTSD)
PreMenstrual Dysphoric Disorder (PMDD)
Diabetic neuropathy
The evidence for efficacy is poor, one trial indicates it is more effective than placebo at higher doses where it behaves more as an SNRI vs an SSRI (Feldman & McCulloch, 2015; Gallagher et al, 2015).
Pharmacokinetics:
relatively short half-life of 4-5 hours
Side Effects:
headache
nervousness
insomnia
lacks orthostatic hypotensive effects
possible weight loss
known to commonly produce a withdrawal syndrome upon sudden stoppage (Wikipedia)
References:
DeBattista C (2024): Antidepressant Agents. Chapter 30. In: Basic and Clinical Pharmacology. 16th Ed. Vanderah TW (Editor); McGraw-Hill (Access Medicine).
Feldman EL, McCulloch DK (2015): Treatment of diabetic neuropathy. In:UpToDate, Basow, DS (Ed), Waltham, MA. Cited 2/22/16
Gallagher HC et al (2015): Venlafaxine for neuropathic pain in adults. Cochrane Database Syst Rev Aug 23;8:CD011091. doi: 10.1002/14651858.CD011091.pub2.
Stahl SM (2009): The Prescriber's Guide. Stahl's Essential Psychopharmacology. 3rd Edition. Cambridge University Press. pp 77-81 (ISBN: 978-0-521-74399-0)
rxlist.com(Effexor ®)
Keywords
MAO Inhibitors
Phenelzine
Trade Name: Nardil ®
Drug Class: Antidepressant, MAO inhibitor
Pharmacology:
see MAO pharmacology
Mechanism of Action:
phenelzine is a nonselective irreversible inhibitor of both MAO-A & MAO-B
antidepressant effect correlates with >80% MAO inhibition.
FDA Indications:
An alternative drug for treatment-resistant depression (not considered a first-line drug of choice)
FDA approved for treating major depression
mixed anxiety and depression and phobic or hypochondriacal feature s
FDA approved for treating atypical depression (e.g. a patient is depressed but sleeps & eats excessively).
Historically phenelzine was used for treating Parkinson's disease (to increase dopamine levels), but other more selective MAO-B inhibitors have replaced its use for this indication (e.g. selegiline&rasagiline).
References:
DeBattista C (2024): Antidepressant Agents. Chapter 30. In: Basic & Clinical Pharmacology. 16th Ed. Vanderah TW (Editor). McGraw Hill / Lange.
rxlist.com (Nardil ®)
Keywords
Selegiline (patch & tablet formulations)
Trade Names:
generic
Parkinsonism: Zelapar ® (orally disintegrating tablets)
Depression: Emsam ® (transdermal patch)
Drug Class: Selective MAO-B inhibitor (at low doses)
MAO Physiology:
In the CNS both MAO-A & MAO-B are found in the outer membrane of mitochondria in neurons & astroglia
In the GI tract & liver, MAO-A is the primary isoform of MAO expressed
MAO-A: metabolizes norepinephrine, dopamine, serotonin (& tyramine)
MAO-B: selectively metabolizes dopamine (& tyramine)
Mechanism of Action:
inhibits MAO by acting as a “suicide substrate” for MAO:
selegiline is converted by MAO to an active metabolite that binds irreversibly with the active site and/or the MAO's essential FAD cofactor.
Low doses: (1.25-2.5 mg/day orally disintegrating tablets) that are used to treat Parkinson's produce a “selective” irreversible inhibition of Monoamine Oxidase B (MAO-B) (although some degree of MAO-A inhibition may occur at the higher 2.5 mg/day dose).
selegiline has a greater affinity for the active site of the type B vs type A isoform, and as a result it can serve as a selective inhibitor of MAO type B when administered at low doses (selectivity is lost at higher doses that are typically used to treat depression).
low doses do not appear to have antidepressant effects, and hence low doses are primarily used in the treatment of Parkinson disease (Hirsch & Birnbaum, 2015).
High doses: (e.g. 6, 9 or 12 mg per day transdermal patch) approved for the treatment of depression result in a loss of selectivity, and both MAO-A & MAO-B are inhibited (Robinson & Amsterdam, 2008; Asnis & Henderson, 2014; Fowler et al, 2015).
Higher doses of selegiline that inhibit both MOA-A & MAO-B are required to produce an antidepressant effect (Hirsch & Birnbaum, 2015).
Direct inhibition of MAO-A in the GI tract is minimal when transdermal selegeline is used at a dose of 6 mg/day. This dose can produce antidepressant effect without the need for dietary restrictions. However, dietary restrictions may be necessary at higher (9 & 12 mg/day) doses. It remains unclear whether the higher (9 & 12 mg/day) doses are more clinically effective in treating depression compared to the lower (6 mg/day) patch dosage (Hirsch & Birnbaum, 2015).
MAO Inhibitors & Neurodegenerative Disease
In neurodegenerative disorders (such as Parkinson & Alzheimer's diseases) it has been proposed that MAO (MAO-A & MAO-B) may play an important role in producing cell damage by generating reactive oxygen species and high levels of other toxic byproducts of catecholamine metabolism (hydrogen peroxide, ammonia & aldehydes). While animal experiments suggest that selective MAO-B inhibitors exert neuroprotective effects (Schapira 2011), based upon several clinical trials (TEMPO, ADAGIO, DATATOP), the American Academy of Neurology has concluded that there is insufficient evidence that such effects are clinically significant (Tarsy 2015).
Indications:
Parkinson Disease: adjunct therapy for patients with a declining or fluctuating response to levodopa. Selegiline has only a minor therapeutic effect on Parkinsonism when given alone.
MAO-B inhibition reduces the breakdown of dopamine, which enhances & prolongs the anti-Parkinsonism effect of levodopa-carbidopa(that is used to replenish CNS dopamine). It will also allow a reduction in the dose of levodopa needed.
Major depression
Based upon clinical consensus (vs evidence-based studies), MAOIs are typically reserved for treatment-resistant forms of depression (e.g. they are of 3rd or 4th choice after failure of response to SSRIs or SNRIs) because MAOIs have more significant side effects (including orthostatic hypotension), food restrictions, drug-drug interactions, and greater risk for harm at high doses.
Off Label (not FDA approved): treatment-resistant forms of depression, panic disorder, social anxiety disorder
Contraindications:
Patients taking meperidine, tricyclic antidepressants, or SSRIs because of the risk of toxic reactions (serotonin syndrome)
Pharmacokinetics:
When a low dose (6 mg/day) transdermal patch formulation is used it minimizes drug-induced inhibition of MAO-A found in the GI tract, with associated changes in first-pass metabolism when tyramine-containing foods are consumed. Hence a tyramine-free diet is unnecessary for the 6 mg/day transdermal dose.
However, a tyramine-free diet is required for higher transdermal doses (9 or 12 mg/day) due to limited safety data for higher doses as judged by the FDA (Robinson & Amsterdam, 2008).
Side Effects:
low dose transdermal patch formulation: site reactions, insomnia, diarrhea, sore throat
at higher doses - similar to other nonselective MAO inhibitors (see MAO pharmacology)
Notes:
Dopamine is “primarily” metabolized by neuronal MAO-B in the CNS. MAO-A metabolizes norepinephrine & serotonin
References:
Aminoff MJ (2024): “Pharmacologic Management of Parkinsonism & Other Movement Disorders” (Chapter 28). In: Katzung's Basic & Clinical Pharmacology, 16th Ed. Vanderah TW (Editor). McGraw-Hill / Lange
Asnis GM, Henderson MA (2014): EMSAM (deprenyl patch): how a promising antidepressant was underutilized. Neuropsychiatr Dis Treat. 10:1911—1923.
DeBattista C (2024): Antidepressant Agents. Chapter 30. In: Basic & Clinical Pharmacology. 16th Ed. Vanderah TW (Editor). McGraw Hill / Lange.
Fowler JS et al (2015):Evidence that Formulations of the Selective MAO-B Inhibitor, Selegiline, which Bypass First-Pass Metabolism, also Inhibit MAO-A in the Human Brain. Neuropsychopharmacology 40, 650–657; doi:10.1038/npp.2014.214
Hirsch M, Birnbaum RJ (2015): Monoamine oxidase inhibitors (MAOIs) for treating depressed adults. In: UpToDate. Basow, DS (Ed), Waltham, MA. Cited 3/4/16
Robinson DS, Amsterdam JD (2008): The selegiline transdermal system in major depressive disorder: a systematic review of safety and tolerability. J Affect Disord. 105(1–3):15–23.
Shapira AHV (2011): Monoamine Oxidase B Inhibitors for the Treatment of Parkinson’s Disease. A Review of Symptomatic and Potential Disease-Modifying Effects. CNS Drugs 25(12):1061-1071. doi: 10.2165/11596310-000000000-00000.
Stahl SM (2009): The Prescriber's Guide. Stahl's Essential Psychopharmacology. 3rd Edition. Cambridge University Press. pp 77-81 (ISBN: 978-0-521-74399-0)
rxlist.com (Emsam ®): depression
rxlist.com (Zelapar ®): Parkinsonism
Keywords
Tranylcypromine
Trade Name: Parnate ®
Drug Class: MAO inhibitor, antidepressant
Mechanism of Action:
MAO inhibitor (MAO-A & MAO-B) with a rapid onset of action
Indications:
major depression without melancholia (Not a 1st line drug)
Contraindications:
Tranylcypromine should not be administered in combination with antihypertensive, diuretic, antihistaminic, sedative or anesthetic drugs; bupropion, buspirone, dextromethorphan
Cheese or other foods with a high tyramine content
Excessive quantities of caffeine
Patients with a confirmed or suspected cerebrovascular defect or to any patient with cardiovascular disease, hypertension or history of headache
Side Effects:
similar to phenelzine
Pharmacokinetics:
MAO activity recovers in 3-5 days after drug withdrawal, half life 2-3 hrs (faster recovery of MAO activity due to its weaker bond to the enzyme)
Notes:
The most important reaction associated with tranylcypromine is the occurrence of hypertensive crises which have sometimes been fatal
References:
DeBattista C (2024): Antidepressant Agents. Chapter 30. In: Basic & Clinical Pharmacology. 16th Ed. Vanderah TW (Editor). McGraw Hill / Lange.
Stahl SM (2009): The Prescriber's Guide. Stahl's Essential Psychopharmacology. 3rd Edition. Cambridge University Press. pp 77-81 (ISBN: 978-0-521-74399-0)
rxlist.com (Parnate ®)
Keywords
antidepressants.txt · Last modified: 2013/08/30 14:58 by cclarks
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TUSOM SOM
| Pharmacology Department
| TMedWeb |
11632 | https://www.va.gov/formularyadvisor/DOC_PDF/Meperidine.pdf | MEP CFU (Final 29Sep03) Updated versions can be found at: www.vapbm.org or vaww.pbm.med.va.gov. Criteria for Use of Meperidine VHA Pharmacy Benefits Management Strategic Healthcare Group and the Medical Advisory Panel These criteria are based on the best clinical evidence currently available. The recommendations in this document are dynamic, and will be revised as new clinical information becomes available. This guidance is intended to assist practitioners in providing consistent, high quality, cost effective drug therapy. These criteria are not intended to interfere with clinical judgment; the clinician must ultimately decide the course of therapy based on individual patient situations. INTRODUCTION Meperidine is a phenylpiperadine opioid agonist analgesic with anticholinergic, serotonergic, and noradrenergic effects. While there is evidence suggesting that its analgesic effects are not superior to that of other opioid agonist analgesics,1-6 meperidine is one of the most efficacious agents available for treatment of post-operative shivering.7,8 It is also efficacious for prevention of post-operative shivering9 and treatment10 or prevention11 of amphotericin B-induced rigors. Contrary to common belief, meperidine increases biliary sphincter pressure, and there is a lack of outcome-based clinical evidence demonstrating that it has an advantage over other opioids in patients with acute pancreatitis.12 Meperidine can cause neurotoxicity, a potentially severe adverse effect that is due to the accumulation of a metabolite, normeperidine. It is difficult to predict which individuals will experience neurotoxic effects and how severe the reaction will be. Meperidine also has the potential to cause life-threatening serotonin syndrome when used in patients who are concurrently or were recently taking monoamine oxidase inhibitors (within 2 to 3 weeks) or other agents with serotonin reuptake inhibiting properties. Furthermore, meperidine causes negative changes in mood while other opioid agonist analgesics induce positive changes.13 The risk of delirium is also higher with meperidine than other opioid analgesics.14,15 The use of meperidine requires careful patient monitoring for neuroexcitatory effects and tracking of dosage to reduce the risk of neurotoxicity, as well as extra vigilance to identify concomitant or recent use of serotonergic drugs to prevent potentially fatal drug interactions. Since other opioid agonist analgesics have similar analgesic efficacy, lower risk of neurotoxicity at usual therapeutic doses, and lower risk for serotonin syndrome due to drug interactions, the use of meperidine in the VA should be restricted to the situations outlined below. VA CRITERIA FOR USE Other, safer parenteral opioids should generally be used for analgesia instead of meperidine, particularly in vulnerable elders. (Vulnerable elders are defined as persons age 65 years and older who are at increased risk for death or functional decline over 2 years.16,17) Appropriate Use Inappropriate Use/Increased Risk Peri-procedural analgesia, as in gastrointestinal, surgical and interventional radiologic procedures. Treatment or prevention of drug- or blood-product–induced rigors (off-label use). Treatment or prevention of post-anesthesia shivering (off-label use). Short-term (< 24-h) parenteral administration for management of moderate to severe acute pain, including patient-controlled analgesia, in patients who − have a documented hypersensitivity to morphine or hydromorphone, or intolerance to other opioid analgesics (e.g., morphine, hydromorphone, and fentanyl); and − require a total dose of less than 600 mg in 24 h. Orally administered meperidine Long-term pain management Criteria for Use of Meperidine MEP CFU (Final 29Sep03) Updated versions can be found at: www.vapbm.org FDA-APPROVED INDICATIONS Oral and parenteral: Relief of moderate to severe pain Parenteral: Preoperative medication, support of anesthesia, obstetrical analgesia PRECAUTIONS / CONTRAINDICATIONS Hypersensitivity to meperidine Multiple doses for patients with renal dysfunction Hepatic impairment Monoamine oxidase inhibitors (within 2 to 3 weeks) or agents with serotonergic activity Multiple doses for patients with seizure disorder Coma or severe respiratory depression DOSAGE AND ADMINISTRATION Intravenous is preferred over intramuscular administration because of large interpatient variability in absorption of meperidine from muscle tissue.18 When intravenous access is not available, subcutaneous injections are appropriate for occasional use but intramuscular injections are preferred for repeated doses. Orally administered meperidine is not recommended in the VA because it has variable bioavailability, may produce less analgesia, and may result in greater formation of normeperidine due to extensive first-pass metabolism. Dosage is not established for off-label use of epidural and intrathecal routes of administration. Analgesia: 100 to 150 mg i.m./s.c. every 2 to 4 hours. For patient-controlled intravenous analgesia: 25 to 50 mg (0.5 to 1 mg/kg) load, 5 to 25 mg demand bolus, lockout 6 to 8 min. For peri-procedural analgesia, the recommended total dose should not exceed 100 mg. The VA/DoD Clinical Practice Guideline on the Management of Post-Operative Pain recommends limiting the total dose of parenteral meperidine to a maximum of 600 mg in 24 hours and the duration to 24 hours.19 Post-operative shivering (off-label use): Meperidine 25 mg i.v. has been found to be efficacious.7 The optimal dose is not established. Drug- or blood product–induced shivering (off-label use): Dose is not established and well-designed trials are lacking. For treatment of shivering due to amphotericin B, when meperidine (100 mg/ml) was slowly infused i.v. until shivering stopped, the mean effective dose was 45 mg (range: 25 to 60 mg).10 For prevention of shivering due to amphotericin B, a dose of 0.70 mg/kg i.v. was found to be efficacious.11 Dosing in special populations Hepatic disease: The bioavailability and half-life of orally administered meperidine are increased in patients with cirrhosis. Although conversion of meperidine to normeperidine is decreased, accumulation of the metabolite and neurotoxicity may occur with repeated doses due to reduction in normeperidine elimination. Use of meperidine is contraindicated due to increased risk of toxicity from accumulation of meperidine and normeperidine. Renal disease: Meperidine is contraindicated in patients with renal impairment because normeperidine can accumulate in patients with renal disease (half-life of 34 hours with renal dysfunction). Severe renal insufficiency may be considered when the creatinine clearance is < 20 ml/minute. In patients < 50 years old, this is equal to a serum creatinine of approximately 5 mg/dl. However, in older patients (> 75 years old) the serum creatinine is about 2.5 mg/dl for the creatinine clearance to be approximately 20 ml/minute. Elderly: Reduce the initial dose and cautiously titrate to achieve desired response. 16,17 Patients taking phenothiazines or other tranquilizers: Reduce dose of meperidine by 25% to 50% since these drugs potentiate the effects of meperidine. MONITORING Signs and symptoms of neurotoxicity: anxiety, shaky feelings, delirium, nervousness, hyperreflexia, tremors, twitches, multifocal myoclonus, generalized seizures. Risk factors for neurotoxicity include renal dysfunction; high, frequent doses; and co-administration of agents that may induce seizures or lower the seizure threshold (such as phenothiazines).13,20 or vaww.pbm.med.va.gov. 2 Criteria for Use of Meperidine MEP CFU (Final 29Sep03) Updated versions can be found at: www.vapbm.org However, myoclonus and seizures may occur in patients with normal renal function, at relatively low doses, and after short duration of therapy (e.g., 260 to 540 mg per day over 3 to 10 days).13,21 Meperidine-induced neurotoxicity has often occurred after two days of repeated dosing,13,21 but toxic effects have occurred in less than 24 hours (e.g., 17 hours after a patient received a total of 750 mg i.m./i.v.22) Premonitory symptoms of severe neurotoxicity may be absent or subtle.22,23 Naloxone does not reverse the neurotoxic effects of meperidine. Cumulative dose and duration of meperidine: do not exceed 600 mg per 24 hours or duration of 24 hours. DRUG INTERACTIONS Table 1 Selected drug interactions with meperidine Interacting Drug(s) Effects Prevention Documentation Monoamine oxidase inhibitors (MAOIs) † Selegiline (MAO-B inhibitor) Serotonin syndrome (agitation, seizures, diaphoresis, fever, coma, apnea, death) Avoid use of meperidine for several weeks after discontinuation of MAOIs, and avoid concomitant use. Use other opioids with caution. Numerous case reports with various nonselective MAOIs 24 Single case report with selegiline 25 Selective serotonin reuptake inhibitors (SSRIs) ‡ Serotonin syndrome (agitation, confusion, hypertension, tachycardia, diaphoresis, diarrhea) Use caution; monitor patient for adverse effects. Concomitant use of meperidine and SSRIs is not a listed contraindication. Single case report with fluoxetine 26 Sibutramine || Serotonin syndrome (CNS irritability, motor weakness, shivering, myoclonus, altered consciousness) Avoid concomitant use. If concomitant use is unavoidable, carefully monitor patient for adverse effects. Package insert precaution 27 Barbiturate anesthetics (e.g., thiopental) Decreased dose of thiopental required to induce anesthesia; increased risk of apnea Decrease dose of thiopental by about 40%. Use precautions usually used in anesthesia. Controlled study, case reports 24 Phenothiazines Excessive sedation and hypotension. Positive effects: Decreased opioid analgesic dosage and less nausea and vomiting. Benefit-to-risk ratio does not support concomitant use. Two controlled studies 24 Ritonavir Possibly decreased efficacy and increased neurotoxicity Avoid dosage increase and long-term use of meperidine with ritonavir or avoid concomitant use. Package insert 28 Pharmacokinetic study 29 Source: Drug Interaction Facts online 24 † Monoamine oxidase inhibitors: phenelzine (Nardil), tranylcypromine (Parnate), isocarboxazid (Marplan), selegiline/deprenyl (Eldepryl) ‡ Selective serotonin reuptake inhibitors: citalopram (Celexa), escitalopram (Lexapro), fluoxetine (Prozac, Sarafem), fluvoxamine (Luvox), paroxetine (Paxil), sertraline (Zoloft) || Sibutramine (Meridia), an anorexiant with serotonin reuptake inhibiting properties COST Table 2 Comparative drug acquisition costs (lowest VA prices) Analgesia Postoperative shivering † Drug-induced shivering † Fentanyl Hydromorphone Meperidine Morphine Alfentanil Doxapram Meperidine Meperidine Dose (mg) ‡ 0.1 2 75 10 0.25 100 25 45 Route i.v. i.m. i.m. i.m. i.v. i.v. i.v. i.v. Cost $0.45 $0.50 $0.14 $0.38 $3.40 $10.88 $0.26 $0.26 Lowest VA prices as of June 2003; all drugs shown are on the VA National Formulary † In a systematic review that compared antishivering agents, clonidine 0.15 mg i.v. and doxapram 100 mg i.v. were similar in efficacy to meperidine 25 mg i.v. for treatment of postoperative shivering. For each of these three agents, number-needed-to-treat was less than 2 for one to stop shivering within 5 minutes who would have continued to shiver had they all received a placebo.7 Meperidine was superior to alfentanil (NNT 2.4) and ketanserin (NNT 2.3). There was insufficient data to compare other drugs, including nalbuphine, nefopam, and tramadol i.v.7 A single study found that nefopam was superior to meperidine for prevention and treatment of amphotericin B-induced shivering.11 Clonidine i.v. and nalbuphine are not on the VA National Formulary. Tramadol i.v., nefopam, and ketanserin are not available in the U.S. ‡ Doses for analgesia are estimated equianalgesic doses; and drug-induced shivering dose for meperidine is the estimated mean effective dose found by Burks (1980).10 or vaww.pbm.med.va.gov. 3 Criteria for Use of Meperidine MEP CFU (Final 29Sep03) Updated versions can be found at: www.vapbm.org REFERENCES 1. Woodhouse A, Hobbes AF, Mather LE, Gibson M. A comparison of morphine, pethidine and fentanyl in the postsurgical patient-controlled analgesia environment. Pain 1996;64(1):115-21. 2. Woodhouse A, Ward ME, Mather LE. Intra-subject variability in post-operative patient-controlled analgesia (PCA): is the patient equally satisfied with morphine, pethidine and fentanyl? Pain 1999;80(3):545-53. 3. Stanley G, Appadu B, Mead M, Rowbotham DJ. Dose requirements, efficacy and side effects of morphine and pethidine delivered by patient-controlled analgesia after gynaecological surgery. Br J Anaesth 1996;76(4):484-6. 4. Rosaeg OP, Lindsay MP. Epidural opioid analgesia after caesarean section: a comparison of patient-controlled analgesia with meperidine and single bolus injection of morphine. Can J Anaesth 1994;41(11):1063-8. 5. Smith AJ, Haynes TK, Roberts DE, Harmer M. A comparison of opioid solutions for patient-controlled epidural analgesia. Anaesthesia 1996;51(11):1013-7. 6. Paech MJ. Epidural pethidine or fentanyl during caesarean section: a double-blind comparison. Anaesth Intensive Care 1989;17(2):157-65. 7. Kranke P, Eberhart LH, Roewer N, Tramer MR. Pharmacological treatment of postoperative shivering: a quantitative systematic review of randomized controlled trials. Anesth Analg 2002;94(2):453-60. 8. Tsai YC, Chu KS. A comparison of tramadol, amitriptyline, and meperidine for postepidural anesthetic shivering in parturients. Anesth Analg 2001;93(5):1288-92. 9. Piper SN, Maleck WH, Boldt J, Suttner SW, Schmidt CC, Reich DG. A comparison of urapidil, clonidine, meperidine and placebo in preventing postanesthetic shivering. Anesth Analg 2000;90(4):954-7. 10. Burks LC, Aisner J, Fortner CL, Wiernik PH. Meperidine for the treatment of shaking chills and fever. Arch Intern Med 1980;140(4):483-4. 11. Rosa G, Dell'Utri D, Conti G et al. Efficacy of nefopam for the prevention and treatment of amphotericin B-induced shivering. Arch Intern Med 1997;157(14):1589-92. 12. Thompson DR. Narcotic analgesic effects on the sphincter of Oddi: a review of the data and therapeutic implications in treating pancreatitis. Am J Gastroenterol 2001;96(4):1266-72. 13. Kaiko RF, Foley KM, Grabinski PY et al. Central nervous system excitatory effects of meperidine in cancer patients. Ann Neurol 1983;13(2):180-5. 14. Morrison RS, Magaziner J, Gilbert M et al. Relationship between pain and opioid analgesics on the development of delirium following hip fracture. J Gerontol A Biol Sci Med Sci 2003;58(1):76-81. 15. Marcantonio ER, Juarez G, Goldman L et al. The relationship of postoperative delirium with psychoactive medications. Jama 1994;272(19):1518-22. 16. Knight EL, Avorn J. Quality indicators for appropriate medication use in vulnerable elders. Ann Intern Med 2001;135(8 Pt 2):703-10. 17. Saliba D, Elliott M, Rubenstein LZ et al. The Vulnerable Elders Survey: a tool for identifying vulnerable older people in the community. J Am Geriatr Soc 2001;49(12):1691-9. 18. Erstad BL, Meeks ML, Chow HH, Rappaport WD, Levinson ML. Site-specific pharmacokinetics and pharmacodynamics of intramuscular meperidine in elderly postoperative patients. Ann Pharmacother 1997;31(1):23-8. 19. VA/DoD Clinical Practice Guideline Working Group. Management of Acute Post Operative Pain. Office of Quality and Performance publication 10Q-CPG/Pain-01. Washington, DC: Veterans Health Administration, Department of Veterans Affairs and Health Affairs, Department of Defense; October 2001. 20. Hagmeyer KO, Mauro LS, Mauro VF. Meperidine-related seizures associated with patient-controlled analgesia pumps. Ann Pharmacother 1993;27(1):29-32. 21. Simopoulos TT, Smith HS, Peeters-Asdourian C, Stevens DS. Use of meperidine in patient-controlled analgesia and the development of a normeperidine toxic reaction. Arch Surg 2002;137(1):84-8. or vaww.pbm.med.va.gov. 4 Criteria for Use of Meperidine MEP CFU (Final 29Sep03) Updated versions can be found at: www.vapbm.org 22. Marinella MA. Meperidine-induced generalized seizures with normal renal function. South Med J 1997;90(5):556-8. 23. Goetting MG, Thirman MJ. Neurotoxicity of meperidine. Ann Emerg Med 1985;14(10):1007-9. 24. Facts and Comparisons. Drug Interaction Facts. Facts and Comparisons Web site. Wolters Kluwer Health, Inc. 2003. Available at: Accessed 12 May 2003. 25. Zornberg GL, Bodkin JA, Cohen BM. Severe adverse interaction between pethidine and selegiline. Lancet 1991;337(8735):246. 26. Tissot TA. Probable meperidine-induced serotonin syndrome in a patient with a history of fluoxetine use. Anesthesiology 2003;98(6):1511-2. 27. Abbott Laboratories. Meridia [package insert online]. May 2002. Available at: Mount Olive, NJ; 2002. 28. Abbott Laboratories. Norvir [package insert online]. September 2001. Available at: North Chicago, IL; 2001. 29. Piscitelli SC, Kress DR, Bertz RJ, Pau A, Davey R. The effect of ritonavir on the pharmacokinetics of meperidine and normeperidine. Pharmacotherapy 2000;20(5):549-53. Prepared by: F. Goodman, PharmD, BCPS, VA Pharmacy Benefits Management Strategic Healthcare Group, Hines, IL, September 2003. Reviewed by P. Glassman, MBBS, MSC, Medical Advisory Panel, and William N. Jones, BSc, MSc, Tucson, VAMC or vaww.pbm.med.va.gov. 5 |
11633 | https://resources.ebrooke.org/wp-content/uploads/2025/06/3.06-Ratio-tables.pdf | 6th Grade Math 3.06 Standard: ● 6.RP.3: Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
○ 6.RP.3.a: Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
Big Idea: Ratio tables, like bar models/tape diagrams, are another tool to organize and scale ratios. Ratio tables do not need to be in a particular order and can be used to find many equivalent ratios.
Point: ● Fill in and use a ratio table to solve real world scenarios.
Vocab: Indicates this is the first lesson this vocab is used ● Simplified ratio: a ratio that has parts or totals with no common factor to scale down by.
● Scale factor: The number you multiply a ratio by to get an equivalent ratio.
● Equivalent ratios: Two ratios that express the same relationship between numbers. Equivalent ratios are created by multiplying or dividing both parts of a given ratio by the same number.
● Ratio: A scalable relationship between two or more quantities with a distinct order. This can be expressed as part to part or part to total.
Extra materials: None Misconceptions: ● Not scaling both parts (or total) of a ratio by the same scale factor ● Not knowing what to do when you can’t scale with a whole number (simplify the ratio then try scaling up) 6th Grade Math 3.06 Notes for teachers: This is the first lesson on ratio tables. In the next few lessons, students should only be using ratio tables until they’re really comfortable with them. After a few lessons, we’ll have them pick the strategy (table or scaling with colon notation) that works best. Students will need to read, fill in, add to and create ratio tables in this lesson.
By the end of the lesson, students should be able to articulate that ratio tables can be used to create equivalent ratios and be able to show how to find missing parts and totals.
Throughout the lesson, push the vocabulary “scaling the ratio”, “simplified ratio” and “equivalent ratio.” Problem Solving Task (20 min): 1. The table below is called a ratio table. A ratio table is a tool that you can use to create and organize equivalent ratios. The ratio table below shows the ratio between cookies and cupcakes at a birthday party (6:4).
a. Knowing what you know about how to create equivalent ratios, fill out the missing part of the table.
b. What information is in a ratio table? How can that information be used?
It shows a ratio then rows of equivalent ratios.
You can draw arrows with the scale factor to find missing parts of equivalent ratios.
Discussion: Core questions that should be asked, and potential answer ● Say: Turn and talk with your partner: what’s in a ratio table and how can they be used?
○ A ratio table shows a ratio in a row (or column in horizontal tables).
○ It can be used to create and organize equivalent ratios ● Ask: Remind me: what’s an equivalent ratio?
○ A ratio that has the same value when multiplied by a scale factor ● Say: Ratio tables, like bar models and scaling ratios with colon notation, are a tool to find equivalent ratios. For the next few lessons, we’ll learn how to read them, use them and make them.
● Ask: We know that the original ratio is 6 cookies for every 4 cupcakes. What was your strategy to find the missing parts of the other equivalent ratios?
○ (Have several students share what they wrote for part b or show written work for part b and have students evaluate or add to it) 6th Grade Math 3.06 c. If there are 48 cupcakes, how many cookies are there? Show work.
72 cookies (look to see if students solve this by adding 48 to their table under cupcakes) ● Say: Turn and talk with your partner: describe in words what you did to fill out each row in the table.
○ (Give students plenty of time to talk so that they are practicing verbalizing their thought process) Pick students who are articulating their work clearly to come to the board and share how they found missing cells in the table. Draw the arrows they tell you to draw and label them with the scale factor they are using.
Check for understanding every so often: for example, “Why is 24 a reasonable number of cookies, if there are 16 cupcakes? …Why wouldn’t 10 be a reasonable answer?” You’re looking for students to recognize 24 is logical because the amount of cookies should always be more than the amount of cupcakes for the ratio to be true.
Follow the same process for the other three blanks by having one student narrate their thought process and work while you model the work on the board.
The last row is an interesting one to talk about because you can’t use the original ratio to scale to 6 cupcakes. Instead students need to recognize they can scale the simplified ratio they just found in the previous row to scale (x3) to 6 cupcakes. By the end of reviewing the table, students should recognize that it doesn’t matter the order of the rows of ratios or if the simplified ratio is written first.
● Ask: How can we use this ratio table to solve part c?
○ Add a row and put 48 under cupcakes. Then scale from the original ratio (or any equivalent ratio) to find the number of cookies.
6th Grade Math 3.06 If you had a student do this work during the PST, show their work.
● Ask: What are your takeaways from today’s PST?
○ (Use students’ takeaways to generate notes) Notes: ● Ratio table = a tool that I can use to create and organize equivalent ratios by scaling up OR down by the same number.
● Every row (or column as seen in Section A #1) shows an equivalent ratio.
Section A (15 min): ● Fill in and use part to part ratio tables #1 is from the 6th Grade NYS 2013 No item # (no data) #2 is from the 6th Grade NYS 2022 Item 2 (State 53%) Section B (remaining time): ● Fill in and use part to part to total ratio tables Students will now see a third column represent the total of the ratio. Make sure students are solid in their understanding that two parts of a ratio can be added together to find a total and a total can be subtracted by a part to find the other part.
Exit Ticket (5 min): 1. Arianna is making origami swans and cranes for her friend's birthday party.
a. Complete the ratio table below and then answer the questions that follow.
6th Grade Math 3.06 b. If she makes 36 swans, how many cranes does she make? Show your work by adding to the ratio table.
She will make 24 cranes. (Work should be shown by adding a row to the table. It’s okay if the total, 60, is not found since the question doesn’t ask for it) 6th Grade Math 3.06 Name: _____ Classwork Problem Solving Task: 1.
The table below is called a ratio table. A ratio table is a tool that you can use to create and organize equivalent ratios.
The ratio table below shows the ratio between cookies and cupcakes at a birthday party.
a. Knowing what you know about how to create equivalent ratios, fill out the missing part of the table.
Cookies Cupcakes 6 4 16 18 3 6 b. What information is in a ratio table? How can that information be used?
c. If there are 48 cupcakes, how many cookies are there? Show work.
1 6th Grade Math 3.06 Section A: 1.
The table below shows the number of tea bags needed to make different amounts of iced tea.
What is the total number of quarts of iced tea that can be made with 24 tea bags?
(A) 5 (B) 6 (C) 7 (D) 8 2.
A farmer places beehives containing bees in her orchard to pollinate the plants. The table below shows the ratio of the number of beehives to the number of acres in the orchard.
If the bees pollinate the plants at a constant rate, how many acres will be pollinated by the bees in 18 beehives?
(A) 38 (B) 40 (C) 44 (D) 48 2 6th Grade Math 3.06 3.
Below is a ratio table that shows the ratio of movies watched to books read by Kassandra. Fill in the table with three other equivalent ratios!
Movies 1 Books 6 4.
Below is a ratio table that shows the ratio of teachers to students in Madison Middle School. Fill in the missing parts.
Teachers Students 3 33 9 11 110 3 6th Grade Math 3.06 5.
Javier has a new job designing websites. He is paid in a ratio of $60 for every 3 pages of web content that he builds. Use the ratio table to answer the questions that follow.
Total webpages built 3 Total money earned 60 a. One week, Javier builds 15 web pages, how much money did he earn?
b. Javier is saving up to purchase a new cell phone that costs 480 dollars.
How many pages would he need to build in order to buy the cell phone?
Section B: 6.
The table below shows the amount of water and tea concentrate needed to make iced tea.
a. Complete the ratio table using the information provided.
Water (quarts) Tea concentrate (quarts) Total amount of iced tea (quarts) 4 6 1 24 b. How much tea concentrate should be added to 8 quarts of water?
4 6th Grade Math 3.06 7.
Jessica loves purple paint! Last year she played around with adding blue and red paint together until she found what she thinks is the perfect shade of purple paint. To get her favorite shade of purple paint she mixes 4 cups of blue paint with 6 cups of red paint. Complete the ratio table below.
Cups of blue paint Cups of red paint Cups of purple paint 4 10 3 10 25 8 9 8.
The student council is selling bouquets of daisies and tulips. The ratio of the number of daisies to the number of tulips is the same for all the bouquets, although the number of flowers in each bouquet varies. Use ratios to figure out the missing information in the table.
Number of Daisies Number of Tulips Total Flowers 6 24 12 18 120 5 6th Grade Math 3.06 Name: _____ Exit Ticket 1.
Arianna is making origami swans and cranes for her friend's birthday party.
a. Complete the ratio table below and then answer the questions that follow.
Swans Cranes Total 3 5 10 35 b. If she makes 36 swans, how many cranes does she make? Show your work by adding to the ratio table.
6 6th Grade Math 3.06 Name: _____ Homework Math Class: 1.
A school organized a paper recycling competition. The table below shows the amount of oil and the amount of water saved by recycling paper.
a. Fill in the missing values in the table below.
Amount of oil saved (gallons) Amount of water saved (gallons) 350 70 140 35 1 25 b. How many gallons of water will be saved if 50 gallons of oil are saved?
Show your work by adding a row to your ratio table.
c. What is the simplified ratio of the amount of oil saved to the amount of water saved?
6th Grade Math 3.06 2.
Ms. Martinez is making playlists to listen to on the bus. For every 4 hip hop songs on her playlist, she adds 3 reggae songs.
a. Fill in the missing values on the ratio table below: Hip Hop Songs Reggae Songs Total Songs on Playlist 4 3 8 14 15 40 84 Add to your table to answer the following questions: b. If Ms. Martinez has a playlist with 32 hip hop songs, how many reggae songs will it have?
c. If Ms. Martinez ended up with 49 songs on her playlist, how many were hip hop songs? How many songs were reggae songs? |
11634 | https://www.probabilisticworld.com/sum-operator-everything-you-need-to-know/ | The Sum Operator: Everything You Need to Know - Probabilistic World
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HomeGeneral Math TopicsAlgebra The Sum Operator: Everything You Need to Know
The Sum Operator: Everything You Need to Know
Posted on November 2, 2020 Written by The Cthaeh1 Comment
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I have used the sum operator in many of my previous posts and I’m going to use it even more in the future. I’ve introduced bits and pieces about this notation and some of its properties but this information is scattered across many posts. For these reasons, I decided to dedicate a special post to the sum operator where I show you the most important details about it.
The first time I mentioned this operator was in my post about expected value where I used it as a compact way to represent the general formula. Since then, I’ve used it in many other posts and series (like the cryptography series and the discrete probability distribution series). This is an operator that you’ll generally come across very frequently in mathematics.
So, given its importance, in today’s post I’m going to give you more details and intuition about it and show you some of its important properties. These properties come directly from the properties of arithmetic operations and allow you to simplify or otherwise manipulate expressions containing it.
In my introductory post on numbers and arithmetic I showed you some operators that represent the basic arithmetic operations. For example, the + (“plus”) operator represents the addition operation of the numbers to its left and right:
Similarly, the √ (“radical”) operator represents the root operation:
You can view these operators as types of instructions. For example, the + operator is instructing readers of the expression to add the numbers between which it’s written. Likewise, the √ operator instructs you to find a number whose second power is equal to the number inside it.
Well, you can view the sum operator, represented by the symbol ∑ (the Greek capital letter Sigma) in the exact same way. It has some stuff written above and below it, as well as some expression written to its right. Which, together, also represent a particular type of instruction. Let’s see what it is.
Table of Contents
Toggle - [x]
The anatomy of the sum operator
Expanding the sum (example)
The sum term
How many terms are there?
Shifting the index
Implicit lower/upper bounds
The sum operator and sequences
What is a sequence?
Example sequences and their sums
Sums with closed-form solutions
Double sums
When the sum term depends on both indices
When the inner sum bounds depend on the outer sum’s index
Generalizing to multiple sums
Properties of the sum operator
Splitting sums
Constant terms
Adding and subtracting sums
Multiplying sums
Shuffling multiple sums
Summary
Notation
Sequences
The properties
The anatomy of the sum operator
The notation surrounding the sum operator consists of four parts:
The number written on top of ∑ is called the upper bound of the sum. Below ∑, there are two additional components: the index and the lower bound. Notice that they’re set equal to each other (you’ll see the significance of this in a bit). Finally, just to the right of ∑ there’s the sum term (note that the index also appears there). It’s important to point that U and L can only be integers (or sometimes even constrained to only be natural numbers). You’ll see why as we make progress.
I’m going to explain the role of each of these components in terms of the instruction the sum operator represents. Unlike basic arithmetic operators, the instruction here takes a few more words to describe.
You can think of the sum operator as a sort of “compressed sum” with an instruction as to how exactly to “unpack” it (or “unzip” it, if you will). Basically, you start with an expression that consists of the sum operator itself and you expand it with the following three steps:
Check if the current value of the index i is less than or equal to the upper bound. If so, move to Step 2. Otherwise, terminate the whole process and replace the sum operator with the number 0.
Add the sum term with the current value of the index i to the expression and move to Step 3.
Increment the value of the index i by 1 and return to Step 1.
This might initially sound much more complicated than it actually is, so let’s look at a concrete example.
Expanding the sum (example)
Let’s take the expression from the image above and choose 0 as the lower bound and 2 as the upper bound. To start, we can simply set the expression equal to itself:
Now we can begin expanding the right-hand side. The initial value of i is 0 and Step 1 asks you to check if , which it is, so we move to Step 2. This step asks you to add to the expression and move to Step 3, which asks you to increment i by 1. The effect of these two steps is:
Then you’re told to go back to step 1 and go through the same process. Well, the current value of i (1) is still less than or equal to 2, so after going through steps 2 and 3 one more time, the expression becomes:
Now we return to Step 1 and again pass through it because 2 is equal to the upper bound (which still satisfies the requirement). After going through steps 2 and 3 one more time, the expression becomes:
Now we go back to Step 1 but this time something’s different. The current value of the index (3) is greater than the upper bound 2, so instead of moving to Step 2, the instructions tell you to simply replace the sum operator part with 0 and stop the process. Therefore, the final expression becomes:
But, as you know, 0 is the identity element of addition, so we can simply omit it from the expression. We’ve successfully completed the instructions and now we know that the expanded form of the sum is:
The sum term
Now I want to focus my attention on the expression inside the sum operator. In the general formula and in the example above, the sum term was and you can think of the i subscript as an index. More specifically, it’s an index of a variable X representing a sequence of terms (more about sequences in the next section).
In principle, the sum term can be any expression you want. For example, if the sum term is , you get things like:
Or you can have fancier expressions like:
In fact, the index i doesn’t even have to appear in the sum term! For example:
If the sum term doesn’t depend on i, we will simply be adding the same number as we iterate over the values of i. Which reduces the sum operator to a fancy way of expressing multiplication by natural numbers. How many times we’re going to add it to itself will depend on the number of terms, which brings me to the next topic of this section.
How many terms are there?
First, let’s cover the degenerate case of expressions with no terms. When will this happen? Well, if the lower bound is a larger number than the upper bound, at the very first iteration you won’t be able to reach Step 2 of the instructions, since Step 1 will already ask you to replace the whole expression with a zero and stop. Which means that for all L > U:
This is usually called the empty sum and represents a sum with no terms.
But when , the sum will have at least one term. The exact number of terms is:
Which means that will have 1 term, will have 5 terms, will have 4 terms, and so on.
This should make intuitive sense.
A note on infinite lower/upper bounds
So far I’ve assumed that L and U are finite numbers. But often you might come across expressions like:
Or even (less frequently) expressions like:
Or maybe even:
If the lower bound is negative infinity or the upper bound is positive infinity (or both), the sum will have an infinite number of terms. Now, I’m only mentioning this here so you know that such expressions exist and make sense. The name of a sum with infinite terms is a series, which is an extremely important concept in most of mathematics (including probability theory). I’m going to dedicate a special post to it soon.
For now, let’s ignore series and only focus on sums with a finite number of terms.
Shifting the index
I want to demonstrate the full flexibility of this notation to you. The general form of a sum operator expression I showed you was:
But you might also come across expressions like:
By adding 1 to each i inside the sum term, we’re essentially skipping ahead to the next item in the sequence at each iteration. For example, if we pick L=2 and U=4, the difference in how the two sums above expand is:
The effect is simply to shift the index by 1 to the right.
But you can do all sorts of manipulations to the index inside the sum term. Here’s a couple of more examples:
In the first one, we’re shifting the index to the left by 2 and in the second one we’re adding every third element.
Implicit lower/upper bounds
Before moving to the next section, I want to show you a few examples of expressions with implicit notation. You will come across such expressions quite often and you should be familiar with what authors mean by them.
The general notation for a sum is:
But sometimes you’ll see expressions where the lower bound or the upper bound are omitted:
Or sometimes even both could be omitted:
As you know, mathematics doesn’t like ambiguity, so the only reason something would be omitted is if it was implied by the context or because a general statement is being made for arbitrary upper/lower bounds.
For example, the expression for expected value is typically written as:
It’s implicit that you’re iterating over all elements of the sample space and usually there’s no need for the more explicit notation:
Where N is the number of elements in the sample space.
The sum operator and sequences
I’ve described what the sum operator does mechanically, but what’s the point of having this notation in first place? Well, I already gave you the answer in the previous section, but let me elaborate here.
Ultimately, the sum operator is nothing but a compact way of expressing the sum of a sequence of numbers. If you think about it, the instructions are essentially telling you to iterate over the elements of a sequence and add them one by one. We achieve this by simply incrementing the current value of the index by 1 and plugging it into the sum term at each iteration. The index starts at the lower bound and stops at the upper bound:
If you’re familiar with programming languages (or if you read any Python simulation posts from my probability questions series), you probably find this conceptually similar to a for loop. And it is! In a way, the sum operator is a special case of a for loop where you’re adding the terms you’re iterating over. I say it’s a special case because you can do pretty much anything you want within a for loop, not just addition.
You can think of the sum operator as a generalization of repeated addition (or multiplication by a natural number). I demonstrated this to you with the example of a constant sum term. In the general case, for any constant c:
The sum operator is a generalization of repeated addition because it allows you to represent repeated addition of changing terms. Or, like I said earlier, it allows you to add consecutive elements of a sequence. But what is a sequence anyway?
What is a sequence?
In mathematics, the term sequence generally refers to an ordered collection of items. For example, you can view a group of people waiting in line for something as a sequence. The person who’s first in line would be the first element (item) of the sequence, second in line would be the second element, and so on.
When it comes to the sum operator, the sequences we’re interested in are numerical ones. That is, sequences whose elements are numbers. For example, here’s a sequence of the first 5 natural numbers:
0, 1, 2, 3, 4
And here’s a sequence with the first 6 odd natural numbers:
1, 3, 5, 7, 9, 11
Since the elements of sequences have a strict order and a particular count, the convention is to refer to an element by indexing with the natural numbers. From my post on natural numbers, you’ll remember that they start from 0, so it’s a common convention to start the index from 0 as well. And we write this index as a subscript of the variable representing an element of the sequence.
For example, let’s call the second sequence above X. Then, the 0th element of the sequence is actually the first item in the list, the 1st element is the second, and so on:
Starting the index from 0 (instead of 1) is a pretty common convention both in mathematics and computer science, so it’s definitely worth getting used to it.
Anyway, I think now you appreciate the point of sum operators. Using the index, we can express the sum of any subset of any sequence. For example, if we wanted to add the first 4 elements in the X sequence above, we would express it as:
Or if we want to sum the elements with index between 3 and 5 (last 3 elements), we would do:
In general, you can express a sum of a sequence of any length using this compact notation.
Sequences as functions
In my introductory post to mathematical functions I told you that these are mathematical objects that relate two sets called the domain and the codomain. The elements of the domain are the inputs of the function and the elements of its codomain are called its outputs. If you haven’t already (and if you’re not familiar with functions), I encourage you to take a look at this post. Although, even without that you’ll be able to follow what I’m about to say.
You can think of sequences as functions whose domain is the set of natural numbers or any of its subsets. The regular convention for expressing functions is as f(x), where f is the function and x is a variable representing its input. But with sequences, a more common convention is to write the input as an index of a variable representing the codomain. If the variable is X and the index is i, you represent an element of the codomain of the sequence as .
Within this framework, you can define all sorts of sequences using a rule or a formula involving i. For example, you can define the i’th term of a sequence to be:
And, for example, the 3rd element of this sequence is:
The first 5 elements of this sequence are 0, 1, 4, 9, and 16.
By default, a sequence is defined for all natural numbers, which means it has infinitely many elements. But you can always create a finite sequence by choosing a lower and an upper bound for the index, just like we do with the sum operator.
Anyway, I’m going to talk more about sequences in my upcoming post on common mathematical functions. For now, let’s just look at a few more examples to get a better intuition.
Example sequences and their sums
In the previous sections, I showed you the definition of three example sequences:
, whose terms are 0, 1, 2, 3…
, whose terms are 0, 1, 4, 9…
, whose terms are 0, 2, 12, 36…
There’s nothing stopping you from coming up with any rule defining any sequence. Anything goes, as long as you can express it mathematically. Let’s look at a few more examples, with the first 4 terms of each:
, first terms: 7, 7, 7, 7 (constant term)
, first terms: 3, 4, 7, 12
, first terms:
, first terms: 1, 2, 4, 8
Now just for fun, let’s calculate the sum of the first 3 items of, say, the B sequence:
If you like, calculate the sum of the first 10 terms of the A, C, and D sequences as an exercise. And, as another exercise, can you guess which sequences the following two formulas represent? Let’s call them the E sequence and the O sequence, respectively:
What is the sum of the first 10 terms of each of them?
Sums with closed-form solutions
In the general case, to calculate the value of an expression with a sum operator you need to manually add all terms in the sequence over which you’re iterating. However, you can derive formulas for directly calculating the sums of some special sequences.
Here I want to give you (without proof) a few of the most common examples of such closed-form solutions you’ll come across. I’m going to prove some of these in my post on series but for now just know that the following formulas exist. For all of them we’re going to assume the index starts from 0 but later I’m going to show you how to easily derive the formulas for any lower bound.
First, here’s a formula for the sum of the first n+1 natural numbers:
For example:
Which is exactly what you’d get if you did the sum manually:
Try it out with some other values of n to see that it works!
Now, remember the E and O sequences I left you as an exercise? In case you haven’t figured it out, those are the sequences of even and odd natural numbers. The formulas for their sums are:
Closed-form solutions also exist for the sequences defined by and :
Generally, you can derive a closed-form solution for all sequences defined by raising the index to the power of a positive integer, but I won’t go into this here, since it requires some more advanced math tools to express. But for those of you who are curious, check out the Wikipedia article on Faulhaber’s formula.
There’s also a closed-form solution to sequences in the form , where c can be any constant:
For example:
Finally, here’s a formula for the binomial theorem which I introduced in my post about the binomial distribution:
Double sums
By now you must have a good enough understanding and feel for the sum operator and the flexibility around the sum term. You can pretty much have any expression inside, which may or may not refer to the index. Now let’s stretch our understanding of “pretty much any expression” even more.
What if the sum term itself was another sum, having its own index and lower/upper bounds? Can we do that?
Sure we can, why not? Take a look at this expression:
The sum term of the outer sum is another sum which has a different letter for its index (j, instead of i). Also, notice that instead of L and U, now we have L 1/U 1 and L 2/U 2, since the lower/upper bounds of the two sums don’t have to be the same.
I included the parentheses to make the expression more readable, but the common convention is to express double sums without them:
Anyway, how do we expand an expression like that? Well, it’s the same idea as with any other sum term. Only, for each iteration of the outer sum, we are going to have a sum, instead of a single number. Let’s plug in some actual values for L 1/U 1 and L 2/U 2 to see what I’m talking about:
The index i of the outer sum will take the values of 0 and 1, so it will have two terms. On the other hand, each of the terms will be the inner sum, which itself consists of 3 terms (where j takes the values 0, 1, and 2). But since we’re adding the same sum twice, the expanded form can also be written as:
Because the inner sum is a constant with respect to the outer sum, any such expression reduces to:
When the sum term depends on both indices
If all that double sums could do was represent a sum multiplied by a constant, that would be kind of an overkill, wouldn’t it? Well, the full power of double sums becomes apparent when the sum term is dependent on the indices of both sums. Let’s see how.
In my introductory post to functions the focus was on functions that take a single input value. However, in the general case, a function can take an arbitrary number of inputs. While the topic of multivariable functions is extremely important by itself, I won’t go into too much detail here. I’m just going to show you a few examples in the context of sequences.
Take a look at this definition:
Here’s a couple of examples for evaluating this function with concrete numbers:
You can think of such functions as two-dimensional sequences that look like tables. The rows of the table are indexed by the first variable (i) and the columns are indexed by the second variable (j):
Then, the element of this sequence is the cell corresponding to row i and column j.
If we now want to express the sum of a particular subset of this table, we could do things like:
Notice how for each value of i we iterate over every value of j. In the above example i ranges from 0 to 1 and j ranges from 0 to 2, which essentially corresponds to the following cells in the table:
Here’s another sum of the same sequence but with different boundaries:
Which instructs us to add the following cells:
When the inner sum bounds depend on the outer sum’s index
To show you the full flexibility of this notation, I want to give a few examples of more interesting expressions. Take a look at this double sum:
What’s interesting about it? Well, the upper bound of the inner sum is not a constant but is set equal to the value of the outer sum’s index! Which means that the inner sum will have a different upper bound for each iteration of the outer sum. Let’s expand the above sum to see how it works:
You can also have the case where the lower bound depends on the outer sum’s index:
Which would expand like:
You can even have expressions as fancy as:
Here both the lower and upper bounds depend on the outer sum’s index. As you can see, the bounds can be arbitrary functions of the index as well. As an exercise, try to expand this expression yourself.
You’ll sometimes come across the term nested sums to describe expressions like the ones above.
Generalizing to multiple sums
To conclude this section, let me tell you about something many of you have already thought about. If the sum term of an expression can itself be a sum, can it also be a double sum? The answer is a resounding “yes”. And, like the case for double sums, the interesting cases here are when the inner expression depends on all indices.
For example, here’s what a triple sum generally looks like:
And here’s what a quadruple sum looks like:
Of course, you can have expressions with as many sums as you like. By analogy to double sums representing sums of elements of two-dimensional sequences, you can think of triple sums as representing sums of three-dimensional sequences, quadruple sums of four-dimensional sequences, and so on.
The general principle for expanding such expressions is the same as with double sums. You increment the index of the innermost sum the fastest and that of the outermost sum the slowest. For example, in triple sums, for every value of the outermost sum’s index you will iterate over every value of the middle sum’s index. And for every value of the middle sum’s index you will iterate over every value of the innermost sum’s index:
Also, just like with double sums, you can have expressions where the lower/upper bounds of the inner sums depend on one or more of the indices of the outer sums (nested sums). For example:
Properties of the sum operator
In the final section of today’s post, I want to show you five properties of the sum operator. These properties allow you to manipulate expressions involving sums, which is often useful for things like simplifying expressions and proving formulas.
All of these properties ultimately derive from the properties of basic arithmetic operations (which I covered extensively in my post on the topic). In particular, all of the properties that I’m about to show you are derived from the commutative and associative properties of addition and multiplication, as well as the distributive property of multiplication over addition.
The commutative property allows you to switch the order of the terms in addition and multiplication and states that, for any two numbers a and b:
The associative property tells you that the order in which you apply the same operations on 3 (or more) numbers doesn’t matter. It essentially allows you to drop parentheses from expressions involving more than 2 numbers. The property states that, for any three numbers a, b, and c:
Finally, the distributive property of multiplication over addition states that, for any three numbers a, b, and c:
Take a look at the post I linked above for more intuition on these properties. Now let’s use them to derive the five properties of the sum operator.
Splitting sums
Sometimes you may want to split a single sum into two separate sums using an intermediate bound. For example, take the following sum:
The associative property of addition allows you to split the right-hand side in two parts and represent each as a separate sum:
Generally, for any lower and upper bounds L and U, you can pick any intermediate number I, where , and split a sum in two parts:
Of course, there’s nothing stopping you from splitting it into more parts. For example, if you want to split a sum in three parts, you can pick two intermediate values and , such that . Then you can split the sum like so:
Example application of splitting a sum
Now I want to show you an extremely useful application of this property. Remember earlier I listed a few closed-form solutions for sums of certain sequences? For example:
You’ll notice that all formulas in that section have the starting value of the index (the lower bound) at 0. But what if someone gave you an expression like:
Even though you can’t directly apply the above formula, there’s a really neat trick for obtaining a formula for any lower bound L, if you already have a formula for L=0.
First, let’s write the general equation for splitting a sum for the case L=0:
If we subtract from both sides of this equation, we get the equation:
Do you see what happened? This manipulation allows you to express a sum with any lower bound in terms of a difference of sums whose lower bound is 0. Which, in turn, allows you to obtain a closed-form solution for any sum, regardless of its lower bound (as long as the closed-form solution exists for L=0).
Coming back to the example above, now we can derive a general formula for any lower bound:
Plugging L=5:
In the general case, if the closed-form solution for L=0 is a function f of the upper bound U, the closed form solution for an arbitrary L is:
Constant terms
The next property I want to show you comes directly from the distributive property of multiplication over addition. Multiplying a polynomial of any number of terms by a constant c gives the following identity:
For example, with only three terms:
Notice that we can express the left-hand side as:
And the right-hand side as:
From which we derive:
Or, more generally for any lower bound L:
Basically, anything inside the sum operator that doesn’t depend on the index i is a constant in the context of that sum. So, this property simply states that such constant multipliers can be taken out of the sum without changing the final value.
Adding and subtracting sums
Another useful property of the sum operator is related to the commutative and associative properties of addition.
Say we have the sum:
The commutative property allows us to rearrange the terms and get:
On the left-hand side, the terms are grouped by their index (all 0s + all 1s + all 2s), whereas on the right-hand side they’re grouped by variables (all x’s + all y’s). In this case, the L and U parameters are 0 and 2 but you see that we can easily generalize to any values:
Furthermore, if we represent subtraction as addition with negative numbers, we can generalize the rule to subtracting sums as well:
Or, more generally:
You can use this property to represent sums with complex expressions as addition of simpler sums, which is often useful in proving formulas. Of course, sometimes you might use it in the other direction to merge two sums of two independent sequences X and Y:
It’s important to note that this property only works if the X and Y sequences are of equal length. That is, if the two sums on the left have the same number of terms.
Multiplying sums
The next property I want to show you also comes from the distributive property of multiplication over addition. Say you have two independent sequences X and Y which may or may not be of equal length. Their respective sums are:
What happens if we multiply these two sums?
In general, when you’re multiplying two polynomials, the expanded form is achieved by multiplying each term of the first polynomial by each term of the second. This is a direct consequence of the distributive property of multiplication:
For example:
In the general case, for any L and U:
In words, the expanded form of the product of the two sums consists of terms in the form of where i ranges from L 1 to U 1 and j ranges from L 2 to U 2. But isn’t there another way to express the right-hand side with our compact notation?
Well, let’s define a new sequence W which is the product of the two sequences:
If we sum all elements of the two-dimensional sequence W, we get the double sum expression:
Which expands exactly like the product of the individual sums!
The intuition here is that we’re combining each value of i with every value of j just like we’re multiplying each term from the first polynomial with every term of the second. This leads to the general property:
Remember that the property related to adding/subtracting sums only works if the two sums are of equal length. By contrast, as I just demonstrated, the property for multiplying sums works even if they don’t have the same length.
Lastly, this property naturally generalizes to the product of an arbitrary number of sums. For example, with three sums:
And more generally, for an arbitrary number of sums (N):
By the way, if you find these general expressions hard to read, don’t worry about it. It takes a little practice but with time you’ll learn to read them much more easily.
Shuffling multiple sums
The last property I want to show you is also related to multiple sums. Not just the ones representing products of individual sums, but any kind. It follows directly from the commutative and associative properties of addition.
The property says that when you have multiple sums whose bounds are independent of each other’s indices, you can switch their order however you like. For example, with double sums you have the following identity:
In words, you can iterate over every every value of j for every value of i, or you can iterate over every value of i for every value of j — the result will be the same. Why does it work?
Let’s pick concrete numbers for the bounds and expand the double sum to gain some intuition:
Now let’s change the order of the sum operators on the right-hand side and expand again:
Notice that in both cases the same terms appear on the right-hand sides, but in different order. Well, from the associative and commutative properties of addition we know that this doesn’t change the final value and they’re equal to each other. And it should be intuitive that the same thing holds for any choice for the lower and upper bounds of the two sums.
This property also naturally generalizes to more than two sums. For example, with three sums:
However, I said it in the beginning and I’ll say it again. This property only works if the lower and upper bounds of each sum are independent of the indices of the other sums!
Summary
Phew, this was a long post, wasn’t it? I hope it wasn’t too exhausting to read and you found it easy to follow.
My goal here was to give you all the crucial information about the sum operator you’re going to need. Not that I can ever fit literally everything about a topic in a single post, but the things you learned today should get you through most of your encounters with this notation. And, if you need to, they will allow you to easily learn the more advanced stuff that I didn’t go into.
Notation
The sum operator is nothing but a compact notation for expressing repeated addition of consecutive elements of a sequence. The notation consists of a sum term, an index variable (which often appears in the sum term), as well as a lower and an upper bound for the index:
The way we expand a sum operator expression is by iterating over the values of the index and, for each, adding a new term with the current value of the index:
The expanded form of the sum will have a number of terms equal to one more than the difference between the upper bound and the lower bound:
I also told you that sometimes the lower and upper bounds can be omitted from the expression if their values are implied from the context:
Sequences
When it comes to the sum term itself, I told you that it represents the i’th term of a sequence. A sequence is a function whose domain is the set (or a subset) of natural numbers . For many sequences you’ll have the i’th term defined by a formula, like , or more generally:
Where f can be any function of i.
I also showed you a few closed-form solutions to the sums of certain sequences, such as:
More generally, the sums of some sequences have closed-form solutions that can be expressed as a function of their lower and/or upper bounds:
I also showed you double sum expressions which expand by iterating over every value of the inner sum’s index for every value of the outer sum’s index:
You can think of the sum term as an element of a two-dimensional sequence. And you can similarly have triple, quadruple, or generally any multiple sum expression which represent summing elements of higher dimensional sequences.
I also showed you examples of double (or multiple) sum expressions where the inner sums’ bounds can be some functions of (dependent on) the outer sums’ indices:
The properties
Finally, I showed you five useful properties that allow you to simplify or otherwise manipulate sum operator expressions.
Splitting a sum into 2 sums:
Multiplying a sum by a constant:
Adding or subtracting sums:
Multiplying sums:
And changing the order of individual sums in multiple sum expressions:
As always, feel free to leave any questions or comments in the comment section below. Until next time!
Filed Under: AlgebraTagged With: Arithmetic, Function
Comments
James Kennedy says
February 9, 2024 at 7:00 pm
Thank you! You expressed the rules of order in both concise and thorough ways. Easy on the sanity! You extended the rules to multiple sums (3 and more sums) concisely and thoroughly, …and that was most unique and helpful compared to all the other online sources I reviewed over the last 6 hours!
Again, Thank you for preserving my sanity regarding multiple sums (Sigma) operations! Your presentation is concise and thorough. Reply
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How to calculate the powers of 10?
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The Rules of Integral: Complex Subject Made Easy
Integration is a fundamental concept in calculus, essential for understanding and solving problems involving areas, volumes, and a variety of applications in physics and engineering. The "Rules of Integration" provide systematic methods for integrating functions.
Rules of integration provide a framework for finding the antiderivatives of functions. Mastery of these rules and techniques is essential for solving a wide range of problems in calculus and applied mathematics. Here’s a thorough guide to this topic:
Basic Integration Rules
Power Rule:
For any real number 𝑛≠−1 n≠−1, ∫𝑥 𝑛 𝑑 𝑥=𝑥 𝑛+1 𝑛+1+𝐶∫x n d x=x n+1 n+1+C, where 𝐶 C is the constant of integration.
Example:∫𝑥 3 𝑑 𝑥=𝑥 3+1 3+1+𝐶=𝑥 4 4+𝐶∫x 3 d x=x 3+1 3+1+C=x 4 4+C.
Constant Multiple Rule:
If 𝑘 k is a constant, then ∫𝑘⋅𝑓(𝑥)𝑑 𝑥=𝑘⋅∫𝑓(𝑥)𝑑 𝑥∫k⋅f(x)d x=k⋅∫f(x)d x.
Example:∫5⋅𝑥 2 𝑑 𝑥=5⋅∫𝑥 2 𝑑 𝑥=5⋅𝑥 3 3+𝐶∫5⋅x 2 d x=5⋅∫x 2 d x=5⋅x 3 3+C.
Sum/Difference Rule:
∫[𝑓(𝑥)±𝑔(𝑥)]𝑑 𝑥=∫𝑓(𝑥)𝑑 𝑥±∫𝑔(𝑥)𝑑 𝑥∫[f(x)±g(x)]d x=∫f(x)d x±∫g(x)d x.
Example:∫(𝑥 2–𝑥)𝑑 𝑥=∫𝑥 2 𝑑 𝑥–∫𝑥 𝑑 𝑥=𝑥 3 3–𝑥 2 2+𝐶∫(x 2–x)d x=∫x 2 d x–∫x d x=x 3 3–x 2 2+C.
Special Integration Formulas
Exponential Functions:
∫𝑒 𝑥 𝑑 𝑥=𝑒 𝑥+𝐶∫e x d x=e x+C, ∫𝑎 𝑥 𝑑 𝑥=𝑎 𝑥 ln(𝑎)+𝐶∫a x d x=a x ln(a)+C (for 𝑎>0 a>0).
Example:∫𝑒 𝑥 𝑑 𝑥=𝑒 𝑥+𝐶),(∫2 𝑥 𝑑 𝑥=2 𝑥 ln(2)+𝐶∫e x d x=e x+C),(∫2 x d x=2 x ln(2)+C.
Trigonometric Functions:
∫sin 𝑥 𝑑 𝑥=−cos 𝑥+𝐶∫sinx d x=−cosx+C, ∫cos 𝑥 𝑑 𝑥=sin 𝑥+𝐶∫cosx d x=sinx+C.
∫sec 2 𝑥 𝑑 𝑥=tan 𝑥+𝐶∫sec 2x d x=tanx+C, ∫csc 2 𝑥 𝑑 𝑥=−cot 𝑥+𝐶∫csc 2x d x=−cotx+C.
Inverse Trigonometric Functions:
∫𝑑 𝑥 1–𝑥 2√=sin−1 𝑥+𝐶∫d x 1–x 2=sin−1x+C, etc.
Techniques of Integration
Substitution Rule (U-Substitution):
If a function is the composite of a function and its derivative, use substitution. Set 𝑢=𝑔(𝑥)u=g(x), then ∫𝑓(𝑔(𝑥))𝑔′(𝑥)𝑑 𝑥=∫𝑓(𝑢)𝑑 𝑢∫f(g(x))g′(x)d x=∫f(u)d u.
Example:∫𝑥⋅𝑒 𝑥 2 𝑑 𝑥∫x⋅e x 2 d x; set 𝑢=𝑥 2 u=x 2, then 𝑑 𝑢=2 𝑥 𝑑 𝑥 d u=2 x d x, so the integral becomes 1 2∫𝑒 𝑢 𝑑 𝑢=1 2 𝑒 𝑢+𝐶 1 2∫e u d u=1 2 e u+C.
Integration by Parts:
Based on the product rule for differentiation, ∫𝑢 𝑑 𝑣=𝑢 𝑣–∫𝑣 𝑑 𝑢∫u d v=u v–∫v d u.
Example:∫𝑥⋅𝑒 𝑥 𝑑 𝑥∫x⋅e x d x; choose 𝑢=𝑥 u=x and 𝑑 𝑣=𝑒 𝑥 𝑑 𝑥 d v=e x d x, then 𝑑 𝑢=𝑑 𝑥 d u=d x and 𝑣=𝑒 𝑥 v=e x, so the integral is 𝑥 𝑒 𝑥–∫𝑒 𝑥 𝑑 𝑥=𝑥 𝑒 𝑥–𝑒 𝑥+𝐶 x e x–∫e x d x=x e x–e x+C.
Partial Fractions:
Used for integrating rational functions by expressing them as a sum of simpler fractions.
Example:∫1 𝑥 2–1 𝑑 𝑥∫1 x 2–1 d x; this can be decomposed into ∫(1/2 𝑥−1–1/2 𝑥+1)𝑑 𝑥∫(1/2 x−1–1/2 x+1)d x.
Trigonometric Substitution:
Useful for integrals involving 𝑎 2–𝑥 2⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√a 2–x 2, 𝑎 2+𝑥 2⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√a 2+x 2, or 𝑥 2–𝑎 2⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√x 2–a 2.
Example:∫1–𝑥 2⎯⎯⎯⎯⎯⎯⎯⎯⎯√𝑑 𝑥∫1–x 2 d x; use 𝑥=sin 𝜃 x=sinθ, then 𝑑 𝑥=cos 𝜃 𝑑 𝜃 d x=cosθ d θ, so the integral becomes ∫1–sin 2 𝜃⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√cos 𝜃 𝑑 𝜃=∫cos 2 𝜃 𝑑 𝜃∫1–sin 2θ cosθ d θ=∫cos 2θ d θ.
Definite and Indefinite Integrals
Indefinite Integrals: These integrals include a constant of integration (𝐶 C) and represent a family of functions.
Example:∫𝑥 2 𝑑 𝑥=𝑥 3 3+𝐶∫x 2 d x=x 3 3+C.
Definite Integrals: Calculated over a specific interval [𝑎,𝑏][a,b], they give a numerical value and are used to find areas, volumes, etc.
Example:∫1 0 𝑥 2 𝑑 𝑥=[𝑥 3 3]1 0=1 3 3–0 3 3=1 3∫0 1 x 2 d x=[x 3 3]0 1=1 3 3–0 3 3=1 3.
Fundamental Theorem of Calculus
This theorem bridges the concepts of differentiation and integration, consisting of two parts:
Part 1: Relates the derivative of an integral function to the original function.
Part 2: States that the definite integral of a function over [𝑎,𝑏][a,b] can be computed using its antiderivative.
Example: If 𝐹(𝑥)=∫𝑥 𝑎 𝑓(𝑡)𝑑 𝑡 F(x)=∫a x f(t)d t, then 𝐹′(𝑥)=𝑓(𝑥)F′(x)=f(x). For a specific function, if 𝑓(𝑥)=𝑥 2 f(x)=x 2, then 𝐹(𝑥)=∫𝑥 0 𝑡 2 𝑑 𝑡=𝑥 3 3)𝑎 𝑛 𝑑(𝐹′(𝑥)=𝑥 2 F(x)=∫0 x t 2 d t=x 3 3)a n d(F′(x)=x 2.
Applications
Integration is used in numerous applications, including calculating areas under curves, volumes of solids of revolution, work done by a force, and in solving differential equations.
Limitations and Challenges
Some functions don’t have elementary antiderivatives (e.g., 𝑒−𝑥 2 e−x 2), requiring numerical methods or special functions.
Certain integrals, especially involving complex functions or higher dimensions, can be quite challenging and require advanced techniques.
Certainly! Here’s a list of frequently asked questions (FAQs) for a topic titled “The Rules of Integral: Complex Subject Made Easy,” designed to address common queries and clarify important concepts related to integration in calculus.
FAQ
What are the basic rules of integration?
The basic rules include the Power Rule (∫𝑥 𝑛 𝑑 𝑥=𝑥 𝑛+1 𝑛+1+𝐶)𝑓 𝑜 𝑟(𝑛≠−1(∫x n d x=x n+1 n+1+C)f o r(n≠−1), Constant Multiple Rule (∫𝑘⋅𝑓(𝑥)𝑑 𝑥=𝑘⋅∫𝑓(𝑥)𝑑 𝑥(∫k⋅f(x)d x=k⋅∫f(x)d x), and Sum/Difference Rule (∫[𝑓(𝑥)±𝑔(𝑥)]𝑑 𝑥=∫𝑓(𝑥)𝑑 𝑥±∫𝑔(𝑥)𝑑 𝑥(∫[f(x)±g(x)]d x=∫f(x)d x±∫g(x)d x).
How do you choose the right integration technique?
The choice depends on the function form. Use substitution for composite functions, integration by parts for products of functions, and partial fractions for rational functions.
What’s the difference between definite and indefinite integrals?
Indefinite integrals represent a family of functions and include a constant of integration 𝐶 C. Definite integrals are computed over an interval and give a numerical value.
Can you explain U-substitution in integration?
U-substitution involves changing the variable of integration to simplify the integral. It’s especially useful for composite functions where you set 𝑢 u to a function inside another function.
What is integration by parts?
Integration by parts is a technique derived from the product rule for differentiation. It’s used when integrating the product of two functions, following the formula ∫𝑢 𝑑 𝑣=𝑢 𝑣–∫𝑣 𝑑 𝑢∫u d v=u v–∫v d u.
Are there functions that cannot be integrated?
Yes, some functions don’t have elementary antiderivatives, such as 𝑒−𝑥 2 e−x 2. These require special functions or numerical methods for integration.
How does the Fundamental Theorem of Calculus apply to integration?
This theorem connects differentiation and integration. It states that if 𝐹 F is an antiderivative of 𝑓 f, then the definite integral of 𝑓 f over [𝑎,𝑏][a,b] is 𝐹(𝑏)–𝐹(𝑎)F(b)–F(a).
How are trigonometric functions integrated?
Trigonometric functions have specific integration formulas, like ∫sin 𝑥 𝑑 𝑥=−cos 𝑥+𝐶∫sinx d x=−cosx+C and ∫cos 𝑥 𝑑 𝑥=sin 𝑥+𝐶∫cosx d x=sinx+C.
What are some common mistakes in integration?
Common mistakes include misapplying the power rule, forgetting the constant of integration in indefinite integrals, and errors in algebraic manipulation.
Can integration be combined?
Yes, complex integrals often require a combination of techniques, such as starting with a substitution and then applying integration by parts.
What are practical applications of integration?
Integration is used in various fields like physics (to calculate work or area under a curve), engineering (for design and analysis), and economics (to determine growth over time).
How do you integrate a function with multiple variables?
Functions with multiple variables are integrated using multiple integrals. Each variable is integrated in turn, often requiring iterative integration.
by: Effortless Math Team about 2 years ago (category: Articles)
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问卷调研结果的可信度：随机抽样与调研样本量
贝壳KEDC
7 评论 15474 浏览 29 收藏 13 分钟
编辑导语：问卷因为有着低成本、高匿名、高效率的特点，所以在用户研究中被广泛地使用。那么如何做好问卷调查呢？本文从调查方式以及调查数据等几方面出发深入剖析了如何做好问卷调研。推荐想要了解问卷调研的童鞋阅读。
在以往的问卷调研中，有的同学总会问到两个关键性问题：
Q1.“做问卷调研，需要多少的目标用户？”
Q2.“调研的目标用户难以触达，只有X份问卷数量够吗？”（问卷数量少，不足百份）
通常来说，用户研究员的建议是根据调研要回答的问题，至少能回收384份有效问卷，有条件的话可以在此基础上提高回收量。那么“384”这个数字是怎么得出来的？问卷调研的回收量一定要达到这个标准吗？如果达不到这个标准，对调研的结果有什么影响？要回答这些问题，我们首先需要了解问卷调研是什么。
一、问卷调研
1. 问卷调研适用场景
问卷调研指的是调查者运用统一设计的问卷向被选取的调查对象了解情况或征询意见的调查方法（源自百度词条“问卷调查”）。与其他的用户研究方法相比，问卷具有低成本、高匿名、高效率的特点，因此在调研中被广泛使用。
问卷调研的适用场景包括收集被调查者的个人信息、了解被调查者的观点态度、行为习惯等。问卷可以通过一系列精心设计的问题，帮助研究者达成摸底、评估、预测的研究目的。
图1 问卷调研的适用场景
2. 普查与抽样调查
根据抽样原则的不同，问卷调研可分为普查和抽查。
普查是指对所有的目标群体进行调查。例如我们想要了解北京居民的收入水平，采用普查的方式就需要对北京市的所有居民做调查。对，你没有看错，是“北京市的所有居民”，严格地说不能有遗漏。
这样做的好处是所有人都能被调查到，收集到的资料非常全面不会有偏差性。但缺点也显而易见，庞大的工作量需要投入大量的人力物力，调研周期也很长。目前我国的人口普查采用的就是这种方式，由全国各地的调查人员，深入到每家每户进行核查。受制于普查耗时耗力的特点，实际的调研中几乎不会采用这种方法，使用更普遍的方法是抽样调查。
抽样调查是从调研的全部目标群体中抽取一部分人，根据这部分人的调研结果对整个群体的状况做估计和推断。其中全部的目标群体称为总体，抽取的一部分人称为样本。延续上面的例子，我们想要了解北京居民的收入水平，用抽样调查的方法仅需要从北京市的居民中选取一部分人做调查。在这里，全体北京居民是总体，抽取的部分居民就是样本。根据抽取居民的调研结果，可以推论整个北京市的居民收入水平。
这样做的好处是方便快捷，仅需要调研一部分人就能推测整个群体的状况。而它的缺点也非常明显，由于无法覆盖所有的调研对象，调研结果会存在一定的偏差性。
图2普查与抽样调查的区别
如何保障抽样调研的可靠性，减少结果的偏差呢？这就需要降低样本的代表性误差。
二、代表性误差
在上一部分我们已经知道，抽样调查的目的是“窥一斑而知全豹”，即从调研的目标总体中抽取样本，用样本的调查结果推论总体的状况。而推论结果的可靠性取决于样本是否具有代表性，是否可以将样本的调研结果推论到总体。因此代表性误差指的就是由样本数据向总体数据做推论时产生的误差。代表性误差主要受抽样方式和样本数量的影响。
1. 抽样方式的随机性
抽样的随机性要求总体中的每个个体都有均等的机会被抽取到，没有倾向性或主观性。如果抽样方法的随机性无法保证，就会导致抽取的样本有偏差，从而得出错误的调研结论。
典型案例就是火车票购买难度的调查：记者想要了解春运的火车票是否难买，于是随机选择了一列春运火车做采访。采访结束后，记者发现火车上的乘客都买到了票，于是得出结论春运火车票很好买。这个调查的纰漏在于记者选取的调研样本有问题，存在很大的偏差。记者只调查了购买到车票的群体，遗漏了没有买到车票的群体，因此得出的结论是错误的。
图3抽样偏差推论出错误的调研结论
随机抽样主要有四种方式：简单随机抽样、整群抽样、分层抽样、系统抽样。目前平台使用的调研大多采用简单随机抽样，确定调研的目标人群后，向这部分用户定向推送调研问卷。图4简单总结了四种方式的操作流程和各自的优缺点，具体内容不在此做详述，感兴趣的同学可以自行深入学习。
图4 随机抽样的四种方式
2. 调研样本的数量
除了抽样方法的随机性外，调研的样本数量也会影响代表性误差。样本量越大，代表性误差越小，样本的调研结果也就越接近总体结果。
为了更好的理解样本量对结果的影响，我们用一个案例来做说明：
某校共有5000名学生，在一次英语考试中5000名学生的平均成绩为76.4分（百分制）。按照抽样调查的思路，可以从5000名学生中随机抽取一部分学生，用他们的考试成绩推测总体情况。结果如图5所示，抽取的样本量越大，结果越接近总体均分76.4。
图5 样本量与调研结果
因此，问卷调研对于样本量的要求实质上是为了保证结果的可靠性，即让调研结果更接近总体的真实情况。虽然样本量越高，调研结果会越趋近于真实状况，但出于回收成本的考虑，也需要控制样本量的上限，减少不必要的人力物力浪费。
三、确定样本量的方法
说了这么多理论性的内容，最关键的问题还没有回答，用研建议的384份样本量到底是怎么得来的？在这一部分，我们将回答这个问题，通过了解三种确定样本量的方法，进一步明确如何确定问卷调研所需要的样本量。
1. 经验值估计
这种方法最为简单粗暴，完全依照研究者个人的过往经验决定，样本量的数字也大相径庭。有的人用30作为标准，有的人用50作为标准，也有的人用100甚至是300作为标准……你看，这样拍脑袋式的估算不仅无法达成统一，更重要的是缺乏理论依据，科学性难以保证。
2. 公式推算
利用计算公式（公式见图6）可以求出问卷调研所需要的样本量。其中n代表样本量；Z代表置信水平的统计量，统计检验中一般设置95%的置信水平，对应的统计量是1.96，置信水平越高，得出的结论可信度越高；p代表的是选项的可能性，调研中可取0.5；e代表抽样误差，一般取5%，抽样误差越小，得出的结论可信度越高。
计算后得到的结果是384，所以用研一般建议根据调研要回答的问题，至少能有384份有效问卷，这是保证调研结果可信的最低要求。需要注意的是，有时调研会有精确的细致需求，例如在全国的数据中看某个城市或某个区域的调研结果。如果要回答这类细致的问题，该类样本的数量也需要达到384的要求。
当然，如果我们想要进一步提高调研结果的信度，也可以通过提高置信水平或是降低抽样误差来实现，但这也意味着需要更多数量的调研问卷（见图6）。
图6 不同置信水平和抽样误差所需要的样本量
3. 检验效力分析
随着统计学的发展，严谨的学术研究对于样本量的计算提出了更加精确的要求。通过检验力分析，研究人员能计算出更加精确的计算出需要的样本量。简单来说，在样本量（sample size）、效应量（effect size）、显著水平（Alpha）、检验力（test power）四个统计量之间，知其三可推断另一个。
具体的使用场景主要是两个，一个是在研究前计算所需要的样本量；另一个是在研究结束后，计算研究的效应量。一些软件已经能够提供相应的计算功能（如GPower），并有较好的可视化展示。
图7GPower绘制的所需样本量曲线
四、结语
得益于低成本、高匿名、高效率的特点，问卷在用户研究中被广泛地使用。样本的代表性是问卷调研结果是否可信的关键因素，通过随机抽样、增加样本量可以提高样本的代表性，从而得出更接近真实总体状况的调研结论。384份有效问卷是保证调研结果可信的最低样本量要求，如果回收量无法保证，在推导结论时就需要考虑与真实情况存在偏差的风险。条条大路通罗马，能够达成目的、解决问题的就是好的调研方法。
作者：艾露尼，公众号：贝壳KEDC
本文由 @贝壳KEDC 原创发布于人人都是产品经理。未经许可，禁止转载
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评论
老刘
我这个做B端的，来做C端了
最近 回复
2. 亚东
其实很多时候我觉得问卷调查是不管用的，因为很多人都是草草了事。
最近 回复
浮生若梦 回复亚东
确实，所以需要在问卷里设置一些陷阱题，甄别掉一些敷衍了事影响结果的数据
最近 来自北京 回复
2. 小华洁 回复亚东
是的 一些人的回答不是自己真心话
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Find m m so that equation has two roots inside given interval
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We are given the following equation:
x 2−(2 m−5)x+3 m−1=0 x 2−(2 m−5)x+3 m−1=0
We have to find m∈R m∈R so that the given equation has two distinct real roots in [1,2][1,2].
In order for the equation to have two distinct real roots, the discriminant has to be greater than 0. This is the first condition.
The second condition I found is that f(1)⋅f(2)<0 f(1)⋅f(2)<0 (f f is a function denoting the left part of the equation above), that is because f f must intersect the X X axis between 1 1 and 2 2.
However, these two conditions are not enough, I need one more.
Thank you in advance for your help!
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asked Aug 18, 2016 at 18:55
George R.George R.
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How did you get f(1)⋅f(2)<0 f(1)⋅f(2)<0?user261263 –user261263 2016-08-18 19:06:02 +00:00 Commented Aug 18, 2016 at 19:06
First, since it's a quadratic and there are 2 roots in [1,2][1,2], I believe f(1)f(2)≥0 f(1)f(2)≥0. The x x-axis gets crossed twice. You may be stuck with the full quadratic formula with this one.Mike –Mike 2016-08-18 19:12:47 +00:00 Commented Aug 18, 2016 at 19:12
I guess I was wrong, it needs to be greater than zero.George R. –George R. 2016-08-18 19:13:01 +00:00 Commented Aug 18, 2016 at 19:13
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Δ=(2 m−5)2−4(3 m−1)=4(m−4)2−35 Δ=(2 m−5)2−4(3 m−1)=4(m−4)2−35
For Δ≥0 Δ≥0,
m≥4+35−−√2 or m≤4−35−−√2 m≥4+35 2 or m≤4−35 2
2 m−5≥3+35−−√or 2 m−5≤3−35−−√2 m−5≥3+35 or 2 m−5≤3−35
2 m−5>3+5 or 2 m−5<3−5 2 m−5>3+5 or 2 m−5<3−5
α+β 2>4 or α+β 2<−1 α+β 2>4 or α+β 2<−1
which beyond [1,2][1,2]
It's impossible to have all real roots in [1,2][1,2]
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edited Aug 18, 2016 at 20:09
answered Aug 18, 2016 at 20:04
Ng Chung TakNg Chung Tak
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Hint:
If x 1 x 1 and x 2 x 2 are the two roots in the interval [1,2][1,2] than:
2≤x 1+x 2≤4 and 1≤x 1 x 2≤4 2≤x 1+x 2≤4 and 1≤x 1 x 2≤4
so:
⎧⎩⎨⎪⎪⎪⎪2 m−5≥2 2 m−5≤4 3 m−1≥1 3 m−1≤4{2 m−5≥2 2 m−5≤4 3 m−1≥1 3 m−1≤4
what you can find from this?
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answered Aug 18, 2016 at 19:55
Emilio NovatiEmilio Novati
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HINT
For f(x)=a x 2+b x+c f(x)=a x 2+b x+c to have two distinct real roots in [1,2] the conditions are
discriminant greater than 0
−b 2 a∈[1,2]−b 2 a∈[1,2]
f(1)⋅f(−b 2 a)≤0 f(1)⋅f(−b 2 a)≤0 and f(2)⋅f(−b 2 a)≤0 f(2)⋅f(−b 2 a)≤0 and f(−b 2 a)≠0 f(−b 2 a)≠0
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edited Aug 18, 2016 at 19:30
answered Aug 18, 2016 at 19:11
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Can you explain the second condition, please?George R. –George R. 2016-08-18 19:15:15 +00:00 Commented Aug 18, 2016 at 19:15
x=−b 2 a x=−b 2 a is the middle of the parabola user261263 –user261263 2016-08-18 19:17:15 +00:00 Commented Aug 18, 2016 at 19:17
Ok, got it. Thank you!George R. –George R. 2016-08-18 19:18:47 +00:00 Commented Aug 18, 2016 at 19:18
I think there are some flaws here. First, if f(1)=0 f(1)=0, why must f(2)=0 f(2)=0? Secondly, couldn't both roots fall outside the range if the discriminant is large?Mike –Mike 2016-08-18 19:19:09 +00:00 Commented Aug 18, 2016 at 19:19
@Mike You're right, I've made some corrections user261263 –user261263 2016-08-18 19:31:29 +00:00 Commented Aug 18, 2016 at 19:31
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11639 | https://math.stackexchange.com/questions/2706187/defining-a-plane-using-only-2-points-and-an-axis | linear algebra - Defining a plane using only 2 points (and an axis) - Mathematics Stack Exchange
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Defining a plane using only 2 points (and an axis)
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Suppose you have a 3 3-dimensional space in which there are 2 points (A A and B B) defined (non identical). Now, you can define a line that goes through them but you cannot define a unique plane, because there are infinitely many planes that are rotating along that line.
Now, I've been having trouble defining the following in strict mathematical terms. That's also why I'm asking here, so you'll have to forgive me for being vague or imprecise.
Suppose that you also chose one axis towards which you want the plane to lay 'flat'. Imagine you model this using table and a piece of paper. The axis along the table are X X and Y Y and the axis up is Z Z. You can rotate the paper along two imaginary points, but at one point it would lay flattest towards your Z Z axis. There would be a clean slope from one point to the other.
How would you figure out the equation of this plane which would be defined using only two points and an 'flatness axis' (for my lack of better words). My thoughts led me towards using a third point (C C) which would form a vector (A C A C) perpendicular to the vector (A B A B) where the Z Z coordinate of point C C would be the same as point's A A. Maybe.
And how would you generalize it to n n-dimensional cases? (2 points, one flatness axis)
EDIT:
I have managed to reduce my problem to the following question: I have a vector n⃗n→ and m⃗m→. I want to find vector o⃗o→, which is perpendicular to the vector n⃗n→ and lies on the plane defined using vectors n⃗n→ and m⃗m→.
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edited Mar 24, 2018 at 16:21
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asked Mar 24, 2018 at 15:19
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I don't understand what you mean. Do you mean that the axis lies inside the plane? Or is the axis (described by a vector) parallel to the plane?MrYouMath –MrYouMath 2018-03-24 15:25:43 +00:00 Commented Mar 24, 2018 at 15:25
1 Your problem is to describe the plane equation in general?user –user 2018-03-24 15:29:49 +00:00 Commented Mar 24, 2018 at 15:29
Yes. What I want to get is a plane equation. I have a point A (let's say [1, 1, 0]) and B=[2, 2, 1] and axis of reference (Z in this example) What I want is a equation for a plane in the form Z=a+bX+cY JonnyRobbie –JonnyRobbie 2018-03-24 15:39:06 +00:00 Commented Mar 24, 2018 at 15:39
It's hard to tell if it's what you mean, but you can add the vector describing the axis to one of your points to get a third point (unless the line described by the two points you started with are parallel to the axis), those three points describe a plane that is parallel to the given axis.Henrik supports the community –Henrik supports the community 2018-03-24 15:46:21 +00:00 Commented Mar 24, 2018 at 15:46
I've tried explaining it with the table example. you can play around with the paper and rotate it as you want (given two points), but there is one rotation of the paper, which makes it 'as flat as possible' with the respect to 'up'.JonnyRobbie –JonnyRobbie 2018-03-24 15:55:49 +00:00 Commented Mar 24, 2018 at 15:55
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This 'flatness axis' is usually called 'normal vector'. It is perpendicular to the plane. Let's call it n⃗=[a,b,c]n→=[a,b,c]. Then we need only one point P 0=[x 0,y 0,z 0]P 0=[x 0,y 0,z 0] together with the normal vector to define the plane. For an arbitrary point P on the plane, the vector P⃗−P 0→P→−P 0→ must be perpendicular to n⃗n→. It means that the dot product must be zero (P⃗−P 0→)⋅n⃗=0(P→−P 0→)⋅n→=0. And here comes the equation:
(x−x 0)⋅a+(y−y 0)⋅b+(z−z 0)⋅c=0(x−x 0)⋅a+(y−y 0)⋅b+(z−z 0)⋅c=0
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answered Mar 24, 2018 at 16:15
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That's why gave it quotes. It's not a normal vector. It's an axis, (ie. vector [0, 0, 1]) and that cannot be a normal vector in a general case. But it's 'as flat as possible'...kinda.JonnyRobbie –JonnyRobbie 2018-03-24 16:47:48 +00:00 Commented Mar 24, 2018 at 16:47
If I see now. As far as I understand the 'flatness axis' should lay in the same plane. Then you can find the normal vector using the cross product of the vector between the two points and the axis: n⃗=A B→×f⃗n→=A B→×f→ . After that, you can continue as above to find the equation of the plane.user543143 –user543143 2018-03-24 20:49:59 +00:00 Commented Mar 24, 2018 at 20:49
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What you (probably) need (your question is not so clear) is the equation of a plane passing through two given points A A and B B and parallel to a given line r r.
If you know a direction vector r⃗r→ of the line, then the vector
n⃗=(B−A)×r⃗n→=(B−A)×r→
is perpendicular to the requested plane, whose equation is then
n x x+n y y+n z z=n x x A+n y y A+n z z A.n x x+n y y+n z z=n x x A+n y y A+n z z A.
EDIT.
A vector o⃗o⃗, which is perpendicular to vector n⃗n⃗ and lies on the plane defined by vectors n⃗n⃗ and m⃗m⃗ can be computed as follows:
o⃗=n⃗×(n⃗×m⃗)=n⃗(n⃗⋅m⃗)−m⃗(n⃗⋅n⃗).o→=n→×(n→×m→)=n→(n→⋅m→)−m→(n→⋅n→).
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edited Mar 24, 2018 at 16:38
answered Mar 24, 2018 at 16:00
Intelligenti paucaIntelligenti pauca
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Given P P and Q Q it suffices consider the infinitely many vectors n⃗=(a,b,c)n→=(a,b,c) orthogonal to P−Q P−Q then the plane equation is
a x+b y+c z+d=0 a x+b y+c z+d=0
and d d can be found by the condition that P P (or Q Q) belongs to the plane.
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answered Mar 24, 2018 at 15:25
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11640 | https://math.stackexchange.com/questions/364457/how-to-find-a-closed-formula-for-partial-sums-of-recursively-defined-series-with | Stack Exchange Network
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How to find a closed formula for partial sums of recursively defined series with $t_{n} = t_{n-2} + t_{n-3}$?
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If $1,1,2,2,3,4,5,7,9,12,16,21,28,37,\ldots,n$ - terms, $t_{n} = t_{n-2} + t_{n-3}$. Find the sum of such a series up to $n$ terms
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Attempted to solve the recurrence relation $t_{n} = t_{n-2} + t_{n-3}$, unsuccessfully.
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edited Jun 12, 2020 at 10:38
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asked Apr 17, 2013 at 14:13
juantheronjuantheron
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You have a linear homogeneous recurrence relations with constant coefficients. You assume a solution of the form $cr^n$ and plug it in to get the characteristic polynomial $r^3=r+1$. As it is a cubic, there will be three roots (in this case two are complex). The solution is then $t_n=c_1r_1^n+c_2r_2^n+c_3r_3^n$. You need to evaluate the $c$'s from your initial conditions. Your polynomial doesn't have nice roots so this will be a mess.
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answered Apr 17, 2013 at 15:38
Ross MillikanRoss Millikan
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$\begingroup$ Thanks Ross Millikan I have try it $\endgroup$
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2013-04-18 08:35:38 +00:00
Commented Apr 18, 2013 at 8:35
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11641 | https://en.wikipedia.org/wiki/Category:Modular_arithmetic | Jump to content
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From Wikipedia, the free encyclopedia
In mathematics, modular arithmetic is a system of arithmetic for certain equivalence classes of integers, called congruence classes. Sometimes it is suggestively called 'clock arithmetic', where numbers 'wrap around' after they reach a certain value (the modulus).
For example, when the modulus is 12, then any two numbers that leave the same remainder when divided by 12 are equivalent (or "congruent") to each other.
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11642 | https://pmc.ncbi.nlm.nih.gov/articles/PMC4900244/ | A reliability and validity study for Scolioscan: a radiation-free scoliosis assessment system using 3D ultrasound imaging - PMC
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Scoliosis Spinal Disord
. 2016 May 31;11:13. doi: 10.1186/s13013-016-0074-y
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A reliability and validity study for Scolioscan: a radiation-free scoliosis assessment system using 3D ultrasound imaging
Yong-Ping Zheng
Yong-Ping Zheng
1 Interdisciplinary Division of Biomedical Engineering, The Hong Kong Polytechnic University, Hong Kong, People’s Republic of China
Find articles by Yong-Ping Zheng
1,✉, Timothy Tin-Yan Lee
Timothy Tin-Yan Lee
1 Interdisciplinary Division of Biomedical Engineering, The Hong Kong Polytechnic University, Hong Kong, People’s Republic of China
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1, Kelly Ka-Lee Lai
Kelly Ka-Lee Lai
1 Interdisciplinary Division of Biomedical Engineering, The Hong Kong Polytechnic University, Hong Kong, People’s Republic of China
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1, Benjamin Hon-Kei Yip
Benjamin Hon-Kei Yip
2 School of Public Health and Primary Care, The Chinese University of Hong Kong, Hong Kong, People’s Republic of China
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2, Guang-Quan Zhou
Guang-Quan Zhou
1 Interdisciplinary Division of Biomedical Engineering, The Hong Kong Polytechnic University, Hong Kong, People’s Republic of China
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1, Wei-Wei Jiang
Wei-Wei Jiang
1 Interdisciplinary Division of Biomedical Engineering, The Hong Kong Polytechnic University, Hong Kong, People’s Republic of China
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1, James Chung-Wai Cheung
James Chung-Wai Cheung
1 Interdisciplinary Division of Biomedical Engineering, The Hong Kong Polytechnic University, Hong Kong, People’s Republic of China
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1, Man-Sang Wong
Man-Sang Wong
1 Interdisciplinary Division of Biomedical Engineering, The Hong Kong Polytechnic University, Hong Kong, People’s Republic of China
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1, Bobby King-Wah Ng
Bobby King-Wah Ng
3 Department of Orthopaedics and Traumatology, The Chinese University of Hong Kong, Hong Kong, People’s Republic of China
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3, Jack Chun-Yiu Cheng
Jack Chun-Yiu Cheng
3 Department of Orthopaedics and Traumatology, The Chinese University of Hong Kong, Hong Kong, People’s Republic of China
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3, Tsz-Ping Lam
Tsz-Ping Lam
3 Department of Orthopaedics and Traumatology, The Chinese University of Hong Kong, Hong Kong, People’s Republic of China
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1 Interdisciplinary Division of Biomedical Engineering, The Hong Kong Polytechnic University, Hong Kong, People’s Republic of China
2 School of Public Health and Primary Care, The Chinese University of Hong Kong, Hong Kong, People’s Republic of China
3 Department of Orthopaedics and Traumatology, The Chinese University of Hong Kong, Hong Kong, People’s Republic of China
✉
Corresponding author.
Received 2016 Feb 15; Accepted 2016 May 12; Collection date 2016.
© Zheng et al. 2016
Open AccessThis article is distributed under the terms of the Creative Commons Attribution 4.0 International License ( which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. The Creative Commons Public Domain Dedication waiver ( applies to the data made available in this article, unless otherwise stated.
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PMCID: PMC4900244 PMID: 27299162
Abstract
Background
Radiographic evaluation for patients with scoliosis using Cobb method is the current gold standard, but radiography has radiation hazards. Several groups have recently demonstrated the feasibility of using 3D ultrasound for the evaluation of scoliosis. Ultrasound imaging is radiation-free, comparatively more accessible, and inexpensive. However, a reliable and valid 3D ultrasound system ready for clinical scoliosis assessment has not yet been reported. Scolioscan is a newly developed system targeted for scoliosis assessment in clinics by using coronal images of spine generated by a 3D ultrasound volume projection imaging method. The aim of this study is to test the reliability of spine deformity measurement of Scolioscan and its validity compared to the gold standard Cobb angle measurements from radiography in adolescent idiopathic scoliosis (AIS) patients.
Methods
Prospective study divided into two stages: 1) Investigation of intra- and inter- reliability between two operators for acquiring images using Scolioscan and among three raters for measuring spinal curves from those images; 2) Correlation between the Cobb angle obtained from radiography by a medical doctor and the spine curve angle obtained using Scolioscan (Scolioscan angle). The raters for ultrasound images and the doctors for evaluating radiographic images were mutually blinded. The two stages of tests involved 20 (80% females, total of 26 angles, age of 16.4 ± 2.7 years, and Cobb angle of 27.6 ± 11.8°) and 49 (69% female, 73 angles, 15.8 ± 2.7 years and 24.8 ± 9.7°) AIS patients, respectively. Intra-class correlation coefficients (ICC) and Bland-Altman plots and root-mean-square differences (RMS) were employed to determine correlations, which interpreted based on defined criteria.
Results
We demonstrated a very good intra-rater and intra-operator reliability for Scolioscan angle measurement with ICC larger than 0.94 and 0.88, respectively. Very good inter-rater and inter-operator reliability was also demonstrated, with both ICC larger than 0.87. For the thoracic deformity measurement, the RMS were 2.5 and 3.3° in the intra- and inter-operator tests, and 1.5 and 3.6° in the intra- and inter-rater tests, respectively. The RMS differences were 3.1, 3.1, 1.6, 3.7° in the intra- and inter-operator and intra- and inter-rater tests, respectively, for the lumbar angle measurement. Moderate to strong correlations (R 2> 0.72) were observed between the Scolioscan angles and Cobb angles for both the thoracic and lumbar regions. It was noted that the Scolioscan angle slightly underestimated the spinal deformity in comparison with Cobb angle, and an overall regression equation y = 1.1797x (R 2 = 0.76) could be used to translate the Scolioscan angle (x) to Cobb angle (y) for this group of patients. The RMS difference between Scolioscan angle and Cobb angle was 4.7 and 6.2°, with and without the correlation using the overall regression equation.
Conclusions
We showed that Scolioscan is reliable for measuring coronal deformity for patients with AIS and appears promising in screening large numbers of patients, for progress monitoring, and evaluation of treatment outcomes. Due to it being radiation-free and relatively low-cost, Scolioscan has potential to be widely implemented and may contribute to reducing radiation dose during serial monitoring.
Keywords: Scoliosis, AIS, Cobb angle, 3D ultrasound, Volume projection imaging, Scolioscan
Background
Scoliosis is a spinal deformity in the coronal plane associated with vertebrae rotation in the transverse plane and abnormal curvature in the sagittal plane [1, 2]. Adolescent idiopathic scoliosis (AIS) is the most prevalent form of scoliosis affecting 3–4% of kids in Hong Kong and about 5% in China according to a recent study , which is a comparable prevalence to other countries . AIS is often diagnosed during the pubertal growth spurt between 10 and 14 years of age [6, 7]. Young patients with AIS are generally skeletally immature and at risk for curve progression, thereby requiring regular monitoring of curve progression . Quantitative assessment of curve severity is also important to plan surgery and for monitoring prognostic and therapeutic outcomes [5, 9].
Cobb angle measurement in the frontal plane derived from standing postero-anterior radiographs is the current gold standard for scoliosis evaluation and to inform decision making for treatment. However, taking radiographs has its own risks and drawbacks. Radiation over repeated exposure to radiographs may increase the risk of breast cancer in girls with scoliosis [11–13]. In addition, radiographic diagnostics in childhood has been shown to contribute significantly to leukemia and prostate cancer . A recent study reported that the management decisions, which are made mainly based on Cobb angle, for AIS significantly affect patient radiation exposure, and it was therefore suggested that research for new imaging modalities with limited ionizing radiation should be undertaken . Efforts have been made to reduce the radiation dose for scoliosis evaluation by using scanning radiography imaging . A new imaging system using this method, named EOS, has been developed, and recent studies demonstrated that it can significantly reduce the radiation exposure with similar quality of images in comparison with conventional standing radiographs [17, 18]. The EOS system is relatively expensive for both the device and its operation and its installation still requires a large space and radiation shielding, thus its accessibility will not be high in the foreseeable future. In addition, it is difficult to make a EOS machine mobile or portable for screening high numbers of scoliosis patients.
Alternatively, different surface topographic methods have been used to estimate spine curvature using stereo cameras or finger palpation of spinous processes to achieve radiation-free assessment of scoliosis, however, it has been demonstrated that these methods are not accurate enough [19, 20]. A recent multicenter study showed that a newly developed surface topography system had a good reproducibility, but still poor correlations with Cobb angle, with R 2 value of 0.5 and 0.25 for thoracic and lumbar scoliosis, respectively . This category of technique suffers from the lack of internal anatomical information of the spine, thus its accuracy is inherently limited.
On the other hand, the feasibility of using various bony landmarks in B-mode images to evaluate spine deformity has been demonstrated previously . Recently, freehand 3D ultrasound, combining conventional B-mode ultrasound with position sensors, has been advanced to overcome the limitations of 2D viewing and measuring of 3D musculoskeletal anatomy [23, 24], and a number of such systems have been reported for scoliosis assessment [25–29]. Different methods have been proposed to estimate spinal deformity using the 3D ultrasound data. In one method, spine curvature was estimated through manually locating the transverse processes in some ultrasound images with 3D spatial information. These ultrasound images were manually selected from a pile of recorded 2D raw B-mode images [30, 31] or captured in real-time while locating the target from observations . This method is relatively time-consuming as each required body landmark has to be manually identified in B-mode images, but it can form a virtual 3D model of the whole or a part of spine using the detected landmarks. The second method is to measure the spine curvature based on the 3D volume ultrasound data, with different visualization methods for the spine anatomy, such as maximum intensity projection and volume projection imaging (VPI) . Koo et al. compared different methods for measuring spinal curvature of spine phantoms using data collected with 3D ultrasound imaging.
The feasibility and potential of using the 3D ultrasound imaging methods have been clearly demonstrated in recent studies for the measurement of scoliotic deformity in vivo [31, 32, 34, 35] as well as for the improvement of brace fitting for scoliosis patients [36, 37]. Being a radiation-free and cost-effective imaging modality, ultrasound imaging has potential to be widely used for scoliosis assessment; its popularity will be enhanced with improved portability of the ultrasound scanner. However, all reported 3D ultrasound imaging systems have been experimental prototypes and not optimized for large scale clinical application. In addition, 3D ultrasound imaging for scoliosis evaluation involves steps of manual scanning and angle measurement, and reliability of each step has not yet been systematically evaluated. Therefore, many investigations are still required before 3D ultrasound imaging can become a clinical tool to benefit scoliosis patients.
The aim of this study was to investigate the reliability of a newly developed 3D ultrasound imaging system for scoliosis assessment, named as Scolioscan. Scolioscan uses volume projection imaging method to form coronal view images of spine structure for the measurement of spinal curvature in the coronal plane. It is a system developed for clinical applications with designs to stabilize patient posture during scanning, but the reliability of using Scolioscan for the assessment of scoliosis patients has not been reported. The intra- and inter-operator reliability of scanning as well as intra- and inter-rater reliability of angle measurement of using Scoliosis were systematically tested in this study. The correlation between the angle measured using Scolioscan and the Cobb angle measured using conventional plain radiography was also investigated to demonstrate its validity. It is believed that the results of this study will provide a good reference for further research and clinical applications of using 3D ultrasound imaging for scoliosis.
Methods
Scolioscan system
The Scolioscan system (Model SCN801, Telefield Medical Imaging Ltd, Hong Kong) was developed based on the 3D ultrasound imaging method reported earlier [25, 30–32], but with industrial and ergonomic designs of the hardware and software interfaces. As shown in Fig.1, the system includes a rigid frame with two movable supporting boards and four supporters to support patients to maintain a stable posture during a test (Fig.2). The chest and hip boards can be moved up and down to fit patients with different heights, and the four supporters with their length adjustable can be fixed on the boards by inserting to the fixation holes and locked by rotating the supporter by 90°. The locations of boards in vertical direction, the positions of supporters along vertical and horizontal directions, as well as the lengths of supporters can be recorded, and the information can be used in follow-up assessments for the same patient. The 3D ultrasound imaging of the spine is achieved through freehand scanning of the ultrasound probe (a custom-designed linear probe with frequency of 4–10 MHz and width of 10 cm), inside which an electromagnetic spatial sensor is installed to detect the position and orientation of the probe. The electromagnetic transmitter is located inside the transmitter box as indicated in Fig.1. Figure3 shows a subject being scanned, and the probe is moved from bottom to top of the back to cover the whole spine. The Scolioscan system has two LCD screens, with one touch screen in the front being used by the operator for inputting patient information, setting parameters for scanning, controlling image collection, data saving and retrieving, conducting VPI image formation, performing measurement, and generating reports. The other screen on the back is to provide information for patients, including a green eye-spot with location set according to the height of patient to facilitate him/her to keep a stable head and neck posture during scanning. This screen also shows additional information including different steps of evaluation procedures, so as to keep the patient informed about the process, and thus more cooperative.
Fig. 1.
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The Scolioscan system with its components labeled. The ultrasound scanner, computer and spatial sensor control box are installed inside the device
Fig. 2.
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A subject being supported by the four supporters, with their locations being adjusted on the chest and hip boards and their lengths adjusted according to the need of each subject
Fig. 3.
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A subject being scanned by the Scolioscan probe, with supports provided by the four supporters installed on the chest and hip boards during scanning. Ultrasound gel is applied along the screening region
Figure4 shows the typical software interfaces for (a) scanning and (b) VPI image analysis and angle measurement of Scolioscan during measurement. Other steps of the assessment procedure include: registration of patient information, adjustment of supporters, setting for ultrasound scanner, and reporting. Before conducting a scan, the range of scanning is first determined by putting the ultrasound probe at the bottom side to record the lower boundary and then at the top side to record the upper boundary. During scanning, a moving probe is shown (real-time) in the interface to indicate the location of the ultrasound probe in relation to the upper and lower boundaries to guide scanning.
Fig. 4.
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Typical software interfaces for (a) scanning and (b) VPI image analysis and angle measurement of Scolioscan. As shown in the tabs at the right side of the interface, other steps of the assessment procedure include: registration of patient information, adjustment of supporters, setting for ultrasound scanner, and reporting
After scanning, the collected B-mode image data together with the corresponding position and orientation information recorded are used for 3D image reconstruction, and volume project imaging is used to form coronal view images of the spine for further analysis . The key idea of VPI method is to obtain an averaged intensity of all voxels of the volumetric image within a selected depth of approximately 10 mm along the antero-posterior direction to form an image in the coronal plane. In addition, a non-planar re-slicing technique is used to enhance the spinal profile in the coronal image by using the skin surface as a reference for selecting the required voxels. Figure5 shows four typical VPI images obtained from patients with different severity of scoliosis. Thereafter, the curve found approximately near the mid-line of the volume projection image, which represents the location of spinous processes, is used to measure the spinal deformity angles. At least two greatest turning portions of a scoliotic curve could be identified as the most tilted vertebrae for angle measurement. Two short lines were then manually drawn from the middle of the curve on the coronal image for denoting the local turning of curve. The angle of the spinal curvature was automatically derived according to the orientations of the two lines drawn, and it was named as Scolioscan angle for this paper (Fig.5). For example, if the spine has an S-shape three lines would be drawn on the VPI image to measure thoracic and lumbar Scolioscan angles.
Fig. 5.
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Four typical volume projection images obtained by Scolioscan showing the coronal plane of spine with different levels of deformity, with the two lines manually drawn to measure the curvature of the thoracic region (a), and the three lines to form two pairs to measure the curvatures of spine in the thoracic and lumbar regions, respectively (b, c, d)
Subjects
Patients diagnosed with scoliosis and scanned by radiographs were invited for this study. All AIS patients were recruited consecutively in the Department of Orthopaedics and Traumatology of The Chinese University of Hong Kong. The study got human subject ethical approvals from both The Hong Kong Polytechnic University (No. 20070321001) and The Chinese University of Hong Kong (No. 2009.622). Informed consent was obtained from all patients (or their parents for those under 18 years of age). The patients received conventional standing plain radiographs within three months before the Scolioscan assessment, and was used for Cobb angle measurement. Patients with metallic implants and BMI higher than 25.0 kg/m 2 were excluded, as a high BMI may lead to poor image quality in the lumbar region using the current probe and the metallic implants may potentially affect the accuracy of ultrasound probe spatial sensing, which uses electromagnetic fields. In addition, patients with Cobb angle larger than 50° were also excluded. Others were further excluded due to following reasons: 1) Patient refused continuation during scanning; 2) Appearance of darkened areas in some of VPI images due to winged scapula, which affected a smooth scanning; 3) Allergy to ultrasound gel; and 4) Patient who wore a bracelet during the final radiograph. One patient felt dizzy during the scanning and subsequent scans were canceled, and five patients had severe winged scapula resulting in poor images so their data were excluded (out of total 55 subjects tested). These data were excluded from analysis.
Twenty patients were included for the first stage of intra-/inter-operator and intra-/inter-raters reliabilities study (with four male and 16 female subjects; age range of 12–22 years of age, mean of 16.4 ± 2.7 years; BMI range of 16.0–22.3 kg/m 2, and mean of 18.6 ± 1.6 kg/m 2). For the second study stage examining correlation between Scolioscan and X-ray measurements, additional patients were recruited, making the total patient number of 49 (with 15 male and 34 female; age range of 11 to 23 years of age, mean 15.8 ± 2.7 years; BMI range 15.1–23.9 kg/m 2, mean 18.4 ± 1.7 kg/m 2).
Testing protocol
The patient was requested to undress upper garments and shoes before the scanning session and was provided a back-opening dressing gown for ease of scanning. All metallic objects, electronics goods, magnets, and other possible ferromagnetic materials were removed. The patient was asked to stand on the Scolioscan platform for supporter adjustment. The chest and hip boards were repositioned at his/her reasonable height (Figs.1, 2). Two supporters on the chest board were relocated to align with clavicle anterior concavities; whereas two supporters on the hip board were relocated to align with bilateral anterior superior iliac spines, the length of supporter’s shafts on both boards was adjusted until they came in contact with the patient. The patient was instructed to maintain their natural standing posture after the adjustment of supporters, and to keep their eye level horizontal at the level of the eye-spot shown on the patient screen and to focus on the spot throughout the scanning process.
The operator applied warmed aqueous ultrasound gel to the patient’s back to fill the spinal furrow and cover the extent of where the probe would sweep. Now the patient was ready to be scanned by the probe, and then ultrasound scanner and the spatial sensing system were activated. The TGC and brightness of B-mode ultrasound images could be adjusted according to the tissue condition of each patient, and like a conventional B-mode ultrasound scanner, other adjustments including scanning depth, focus, frequency, etc. were possible. Parameters were fixed for the tests of all subjects, with the frequency set at 7.5 MHz and depth at 7.1 cm. To increase the B-mode imaging frame rate, a single focus was used and set at depth of 3.5 cm. Pre-scanning was performed from L5 to T1 to check the image quality, and corresponding adjustment of TGC and brightness for B-mode image was conducted to achieve an overall good image quality for the scanning region.
After the setting adjustment, the probe was located at level L5 and T1 spinous processes to record the lower and upper boundaries of scanning range, respectively. The operator used their finger to touch the two green arrows in the “Scan” interface shown on the operator screen to record the probe location information (Fig.4a). After the scanning range was set, the operator re-located the probe at the location slightly lower than L5 spinous process and initiated the scanning and steered the probe to scan upwards from L5 to T1 spinous process. During scanning, an additional arrow would be shown in the interface to indicate the location of the probe in relation to the upper and lower boundaries of scanning range. The data collection was automatically stopped when the probe passed through the upper boundary. The recorded data was then saved with file name containing information of patient code and scanning time, unless the operator decided to discard the scanning result. Once the data were saved, the process of VPI image formation was automatically initiated and user interface was changed from “Scan” to “Analysis” (Fig.4). The procedure of using the probe to scan over the spine region takes approximately 30 s, and VPI image formation less than two minutes for the Scolioscan system used in this study. The total time for assessing one patient was approximately 10 min, including time required for inputting patient information in Scolioscan, supporter adjustment to fit patient, identifying landmarks, applying ultrasound gel, scanning, image reconstruction, measurement on image, and reporting.
Study design
This prospective study was divided into two stages. At the first stage, the intra- and inter-operator reliability for scanning as well as the intra- and inter-rater reliability for measurement using Scolioscan were investigated. At the second stage, the correlation between Cobb angle obtained from radiographs and spinal curve angle obtained using Scolioscan was investigated.
For the spine scanning session using the Scolioscan system at the first stage, two operators (KKL and SY) were involved for the scanning and each AIS patient received scanning twice by each of the operators. For each scanning of each operator, a VPI image was formed, which was then viewed by three raters (TTY, KKL, and SY) to conduct spine curvature measurement independently using the manual measurement tool provided by the Scolioscan system in the interface of “Analysis” (Fig.4). Each VPI image was measured twice by each rater, but not at the same time. Two of the raters were also the operators, and both the operators and the raters were blinded from each other for the scanning and measurement. Before the scanning/measurement started, there was a trial tutorial for all the operators/raters to have a common understanding of the scanning and measurement procedure. The entire procedure for the first stage is summarized in Fig.6. For each patient, a total of 24 sets of results were generated.
Fig. 6.
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Schematic diagram showing the experimental design for evaluating the intra- and inter-reliability for the two operators for scanning and three raters for angle measurement on images. “R” represents “Rater”, and “M” represents “Measurement”. For each patient, a total of 24 sets of measurement result were obtained. Each result is represented as “OSRM”, with “O”, “S”, “R”, and “M” represent “Operator”, “Scan”, Rater”, and “Measurement”, respectively. For example, “O1S2R2M2” represents the result obtained from the second measurement of Rater 2 for the image obtained during second scanning of the Operator 1
For the investigation of correlation between the Scolioscan angles and the radiographic Cobb angles at the second stage, the scanning on patients using Scolioscan was conducted by a single operator (Operator 1 also Rater 2 at the first stage; KKL) and the measurement of spinal curvature on the obtained VPI image was conducted by the same person. Only a single measurement was conducted for each image, which is the same as Cobb angle measurement on X-ray images. The Cobb angles were measured by a doctor in the orthopedics department (TP) who has over 10 years of experiences in reading radiographs of patients with scoliosis. The CV% of his Cobb angle measurement was 6.4%. Angles were categorized as the thoracic Cobb angle if the apex vertebra located within T1-T12 region in X-ray images or as the lumbar Cobb angle if the apex vertebra located within L1-L5 region, and such categorization was also used in the angle measurement of the Scolioscan measurement.
Statistical analysis
At the first stage of the study, the ICC (two-way random and consistency) was used to analyze the reliability between the two sessions of the same rater and operator . All the tests and corresponding data sets used are summarized in Table1. For the intra-rater reliability, two measurements acquired from the first scan by each operator were compared individually for each rater. For the intra-operator reliability, the first measurement of the first scan was compared with that of the second scan for each of the two operators. ICC with two-way random and absolute agreement was used to analyze the reliability between the two sessions of the different raters and operators . For the inter-operator reliability, the first measurement results obtained by each of the three raters from the first scan of the two operators were compared. While for the inter-rater reliability, the first measurement results from the first scan of each of the two operators conducted by the three raters were compared. The Currier criteria for evaluating ICC values were adopted: very reliable (0.80–1.0), moderately reliable (0.60–0.79), and questioned reliable (≤ 0.60) . According to the result of a pilot study, we expected a high reliability (ICC = 0.9) of the measurement using Scolioscan. We further assumed that achieving a moderately reliable result (ICC = 0.7) was meaningful in this study. Thus we were able to calculate the minimum subject numbers required to be 18 for two operators/raters and 12 for three operator/raters, assuming a power of 80% . Accordingly, 20 subjects were recruited for the reliability test in this study. RMS difference was calculated for every pair of intra- and inter-operator as well as intra- and inter-rater reliability tests to provide additional information about test repeatability.
Table 1.
Data sets used in different tests for reliability, with “O”, “S”, “R”, and “M” represent “Operator”, “Scan”, Rater”, and “Measurement”, respectively (Fig.6). Each set of result is represented as “OSRM”. For example, “O1S2R2M2” represents the result obtained from the second measurement of Rater 2 for the image obtained during second scanning of the Operator 1
| Reliability Test | Exam | Result Table | Remarks |
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| Intra-rater | OS1RM1, OS1RM2 | Table1 | Tests between the two measurements by each rater (R) using the first scan of each operator (O), and for each region (thoracic and lumbar) |
| Intra-operator | OS1RM1, OS2RM1 | Table2 | Tests between the two scans obtained by each operator (O), using the first measurements of each other three raters (R), and for each region |
| Inter-rater | OS1R1M1, OS1R2M1, OS1R3M1 | Table3 | Tests among the first measurements of the three raters using the first scan of each operator (O), and for each region. |
| Inter-operator | O1S1RM1, O2S1RM1 | Table4 | Tests between the two operators using the first scan, with the first measurement of each of three raters, and for each region |
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At the second stage, the Scolioscan angles and radiographic Cobb angles were compared using linear correlation for thoracic curves alone, lumbar curves alone, and combined results. Linear regression equations with and without intersections were analyzed, with correlation coefficient 0.25 to 0.50 indicating poor correlation, 0.50–0.75 indicating moderate to good correlation, and 0.75–1.00 indicating very good to excellent correlation . According to the results of other radiation-free assessment method for scoliosis, achieving a moderate to good correlation (correlation coefficient = 0.55) between the Scolioscan angle and radiographic Cobb angle would be meaningful for this study. Thus, the sample size required was 24 to achieve a power of 80% . Considering that some patients may only have either thoracic or lumbar curve, the patient number was determined to be 48. In this study, 49 patients were finally recruited for the correlation study. Bland-Altman method was used to test the agreement between the Cobb angle and the Scolioscan angle. The RMS difference between Scolioscan and Cobb angles was also calculated to show the agreement between the results of the two methods, for the cases with and without correction using the regression equation.
Results
The results demonstrated that the coronal image of spine could be successfully obtained for all the patients tested in this study. The mean Cobb angles of the patients involved in the reliability tests were 30.0 ± 12.3° (mean ± SD, range 11 to 48°, 12 angles) and 24.8 ± 11.0° (seven to 46°, 14 angles) for the thoracic and lumbar regions, respectively, and the overall mean was 27.6 ± 11.8° (seven to 48°, 26 angles). The intra-rater reliability of Scolioscan angle measurement was very good with ICC ranging from 0.94 to 0.99 (0.97 ± 0.02), for each of the three raters conducting measurement for thoracic and lumbar region using the scan of the two operators individually (Table2). Table3 shows a very good intra-operator reliability with ICC ranging from 0.88 to 0.97 (0.94 ± 0.03), for each of the two operators with the angle measurement conducted by the three rater individually. The results demonstrated that Scolioscan provided very good reliability for the scanning by the same operator and the angle measurement by the same rater. The RMS differences between the angles obtained from the two scans of the same operator were 2.5 and 3.1° for the thoracic and lumbar regions, respectively (counting the results of the two operators and three raters for a single measurement of each image). The RMS differences of the intra-rater tests were 1.5 and 1.6° for the thoracic and lumbar regions, respectively.
Table 2.
Intra-rater reliability of the three raters individually for the curve measurement performed using the images scanned by the two operators in the thoracic and lumbar regions using Scolioscan
| | | Thoracic | Lumbar |
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| Operator | | Operator 1 | Operator 2 | Operator 1 | Operator 2 |
| ICC a | Rater 1 | 0.99 | 0.99 | 0.98 | 0.94 |
| Rater 2 | 0.98 | 0.94 | 0.97 | 0.96 |
| Rater 3 | 0.99 | 0.98 | 0.96 | 0.97 |
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a ICC intraclass correlation coefficient
Table 3.
Intra-operator reliability of the two operators individually for the curve measurement performed by the three raters in the thoracic and lumbar regions using Scolioscan
| | | Thoracic | Lumbar |
--- --- |
| Rater | | Rater 1 | Rater 2 | Rater 3 | Rater 1 | Rater 2 | Rater 3 |
| ICC a | Operator 1 | 0.97 | 0.96 | 0.96 | 0.90 | 0.92 | 0.97 |
| Operator 2 | 0.97 | 0.92 | 0.95 | 0.91 | 0.88 | 0.96 |
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a ICC intraclass correlation coefficient
The results also showed very good inter-rater reliability for angle measurement and inter-operator reliability for scanning using Scolioscan, with ICC values ranging from 0.88 to 0.93 (0.90 ± 0.02) and 0.87 to 0.94 (0.92 ± 0.03), respectively (Tables4 and 5). The RMS differences between the angles obtained from the scans of the two operators were 3.3 and 3.1° for the thoracic and lumbar regions, respectively (counting the results of three raters for a single measurement of one image). The RMS differences of the inter-rater tests were 3.6 and 3.7° for the thoracic and lumbar regions, respectively. The reliability results demonstrated that both scanning and angle measurement on VPI images for scoliosis patients were repeatable using the Scolioscan system, with the RMS difference between any two measurements or two scans smaller than 3.7°.
Table 4.
Inter-rater reliability among the three raters for the curve measurement performed using the images scanned by the two operators in the thoracic and lumbar regions using Scolioscan
| | Thoracic | Lumbar |
---
| Operator | Operator 1 | Operator 2 | Operator 1 | Operator 2 |
| ICC a | 0.93 | 0.89 | 0.89 | 0.88 |
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a ICC intraclass correlation coefficient
Table 5.
Inter-operator reliability between the two operators for the curve measurement performed by the three raters in the thoracic and lumbar regions using Scolioscan
| | Thoracic | Lumbar |
---
| Rater | Rater 1 | Rater 2 | Rater 3 | Rater 1 | Rater 2 | Rater 3 |
| ICC a | 0.94 | 0.87 | 0.95 | 0.93 | 0.88 | 0.94 |
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a ICC intraclass correlation coefficient
The mean Cobb angles of the patients involved in the correlation tests were 26.9 ± 9.7° (10 to 48°, 36 angles) and 22.6 ± 9.5° (three to 46°, 37 angles) for the thoracic and lumbar regions, respectively, and the overall mean was 24.8 ± 9.7° (three to 48°, 73 angles). The results showed that there were moderate linear correlations between the Scolioscan angles and Cobb angles for the thoracic (y = 1.2005x, R 2 = 0.78) and lumbar regions (y = 1.1542x, R 2 = 0.72) and thoracic-lumbar data combined (y = 1.1797, R 2> 0.76) (Figs.7, 8 and 9). It was noted that the Scolioscan angle slightly underestimated the spinal deformity in comparison with Cobb angle for both the thoracic and lumbar regions. When the linear regressions with intersection were used for the correlation, the intersection values were 1.93, 1.99, and 1.72°, for the thoracic region, lumbar region, and combined data, respectively. The small intersection values indicated that linear regressions without intersection could well represent the relationship between the Scolioscan angle and Cobb angle for the patients measured in this study. This could also be verified by the very small difference of the coefficients of determination R 2 between the two kinds of linear regressions, as shown in Figs.7, 8 and 9. In addition, it was found that the regression curves were very close between the angles of the thoracic and lumbar regions. Therefore, a single equation derived from the data combined with thoracic and lumbar regions was good enough to represent the relationship between the Scolioscan angle and Cobb angle regardless of the regions for the patients tested in this study (Fig.9). The equation y = 1.1797x (R 2 = 0.76) could be used to transfer the Scolioscan angle (x) to Cobb angle (y) for this group of AIS patients. The Bland-Altman method was used to test the agreement between the data of Cobb angle and those of the Scolioscan angle corrected by this equations, with results showing in Fig.10. A very good agreement was demonstrated between the two types of angle, with a mean difference of 0.2°. The RMS differences between the Scolioscan angles and Cobb angles were 6.5, 5.9, 6.2°, respectively, for the thoracic, lumbar, and two regions combined cases. If a correction was conducted using the overall regression equation y = 1.1797x, the RMS difference became 4.5, 5.0, and 4.7°, respectively, for the three cases.
Fig. 7.
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Correlation between the Cobb angles obtained using radiographs and the spinal angles measured using the coronal images generated by Scolioscan for the thoracic region
Fig. 8.
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Correlation between the Cobb angles obtained using radiographs and the spinal angles measured using the coronal images generated by Scolioscan for the lumbar region
Fig. 9.
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Correlation between the Cobb angles obtained using radiographs and the spinal angles measured using the coronal images generated by Scolioscan for both the thoracic and lumbar regions
Fig. 10.
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The Bland-Altman plot for the Cobb angles and the Scolioscan angle corrected using the linear regression equation for data combined with the thoracic and lumbar regions
Discussion
The results of our study showed very good reliability of Scolioscan for scanning conducted by the same and different operators as well as for the angle measurement performed by the same and different raters on the formed coronal spine images, with a mean ICC value of 0.94 ± 0.04 (ranging from 0.88 to 0.97) between the two operators and among the three raters. The high intra- and inter-rater reliability for the angle measurement showed in this study was consistent with that reported earlier in a feasibility study of VPI method . The RMS difference was smaller than 3.7° between the angles obtained from different measurements or different scans of the same patient. The sophisticated supporting frame, boards, supporters, and eye guiding spot on the patient screen within Scolioscan’s design may have partially contributed to the very good intra- and inter-operator reliability for scanning (with re-positioning for each scan), with the RMS difference smaller than 3.3°. The two Scolioscan operators/raters had been using the system for several months before this study, and the additional rater had been trained for a few days for the measurement. They were all graduates from Biomedical Engineering programmes with knowledge of ultrasound imaging. It will be worthwhile to understand the learning curve of new operators with different backgrounds in future studies.
Once a patient is confirmed with scoliosis, they have traditionally had to be exposed to radiography many times for monitoring, treatment planning, and treatment outcome measurement . With the use of radiation-free Scolioscan, many of the radiation exposures may be avoidable, such as those used for progression monitoring, which may reduce the risk of inducing cancers [11–15]. As a consequence of this radiation hazard, it is conventionally not possible to use frequent radiography for monitoring scoliotic angle progression, thus there is no reference yet about the optimal frequency of taking image for scoliosis patients using the radiation-free Scolioscan system. It may be worthwhile to conduct investigations along this direction, with the consideration of the angle progression rate, risk factors, and cost effectiveness for different categories of patients.
Moderate to strong linear correlations were demonstrated between the Scolioscan angles and X-ray Cobb angles for the thoracic and lumbar regions and thoracic-lumbar data combined with coefficients of determination R 2 larger than 0.72. Similar results were reported earlier using different laboratory prototypes of 3D ultrasound imaging system for scoliosis assessment [26, 30, 31, 34, 36]. It was found that the Scolioscan angle slightly underestimated the spinal deformity in comparison with Cobb angle for both the thoracic and lumbar regions. For the patients tested in the present study, the relationship between the Scolioscan angle (x) to Cobb angle (y) could be expressed by the equation y = 1.1797x (R 2 = 0.76). This finding is consistent with that reported previously using 3D ultrasound imaging for scoliotic angle measurement. The main reason for underestimation is that ultrasound images are taken posteriorly and provide anatomical features of vertebra posterior elements [32, 36, 43] rather than the vertebral bodies used in Cobb angle measurement from radiographs, as the processes are more identified than other spinal landmarks because of its sharp delineation in the ultrasound images. In this study, the profile formed by spinous processes in the VPI image was used for the deformity angle measurement. The RMS square difference between Scolioscan and Cobb angles obtained in this study (totally 73 angles) was 6.2°, and it became 4.7° when an adjustment for the Scolioscan angle was adopted using the obtained regression equation.
It has been well documented that the angle measured based on the profile of spinous processes in radiographs would underestimate the spinal deformity with reference to Cobb angle, showing that the spinous process angle was less angulated compared to Cobb angles . It was reported that the magnitude of vertebral axial rotation correlated with the lateral deviation of vertebrae from the spinal axis [45, 46]. In fact, the spinous process deviations caused by vertebral rotation might result in the inaccuracy of interpretation on the vertebral body alignment on the radiographs of spine [47, 48]. A number of studies investigated how to transfer the spinous process angle to Cobb angle. An equation of y = 1.3367x + 1.3907 (R 2 = 0.90) was proposed , with x representing the spinous angle and y the Cobb angle and both measured using radiographs. In the present study, the corresponding equations were y = 1.1069x + 1.7227 (R 2 = 0.76) and y = 1.1797x (R 2 = 0.76) for the regression with and without intersection (Fig.9). In the earlier feasibility study about angle measurement using VPI images, the results from 3D ultrasound was closer to Cobb angles, where the involved patients had a much smaller mean Cobb angle of 10.7 ± 7.1° , in comparison with the mean Cobb angle of 22.6 ± 9.5° in the present study. Further studies for Scolioscan with larger patient numbers and a wider range of Cobb angle would be necessary to investigate whether different regression equations should be used for scoliosis subjects with different Cobb angles and different types of spinal deformity.
Future studies can also be followed up to understand whether considering the vertebral rotation and other spinal deformities can further improve the agreement . AIS is a three-dimensional spine deformity problem in coronal and sagittal planes and vertebral rotation , and deformity parameters in different planes may be dependent on each other [50–52]. Therefore, it is necessary to quantify spinal curvatures in sagittal or vertebral rotation in addition to coronal deformity, which will be useful for planning surgery, predicting prognosis and monitoring curve progression [30, 53, 54]. A recent study showed that the correlation between the spinous angle and the Cobb angle measured on radiographs could be improved with the consideration of vertebral rotation . However, standing radiograph as the current gold standard for scoliosis investigation is difficult to directly acquire vertebral rotation, since these radiographs do not demonstrate the exact magnitude of the 3-dimensional spinal deformity present in patients with scoliosis . Using the information obtained from the coronal and sagittal radiographs with reduced dose, EOS system can reconstruct 3D view of spine [17, 18]. However, it may still take some time to make the system more popularly used because of its high cost, low accessibility, and radiation (though with dose reduced), and long time required for building 3D spine model. Furthermore, its 3D presentation of the spine achieved using two orthogonal projection images requires further research to validate for different cases. Scolioscan used in the present study provided VPI images of spine for spinal deformity measurement in the coronal plane. During the scanning, Scolioscan actually acquires volumetric images of spine. It has been demonstrated in earlier studies that it is feasible to extract bony landmarks from the volumetric image data set to form virtual 3D spine model for the assessment of scoliotic deformity [25, 30, 31]. Further studies are going on to integrate this function into the system so that the evaluation of scoliotic deformity in 3D can be achieved using Scolioscan, including the measurement of spinal axial rotation and the deformity in sagittal plane. In this study, patients with Cobb angle larger than 50° were excluded due to the concern of the effect from rotation. Perhaps if the spinal rotation can be measured, future studies can include patients with larger Cobb angles.
While the reliability of using the VPI images generated by Scolioscan for the scoliosis assessment has been clearly demonstrated in this study, there are a number of areas to improve so as to achieve a more user-friendly clinical tool. First, patients with AIS are often observed to have winged scapula, and the protruded scapula obstructed the probe from scanning upwards even when the patients were told to cross arm. Hence the quality of VPI image was affected, making it difficult for accurate measurement of angle. In this study, the patients with severe winged scapula that affected scanning were excluded, which counted for approximately 10% of the patients. In future studies, ultrasound probes with different widths and shapes may be used to find optimal configurations for different situations. In addition, patients with BMI larger than 25.0 kg/m 2 were excluded from this study, counting approximately 10% of patients. The current Scolioscan system used an ultrasound probe with frequency of 4–10 MHz, and bony features in images of the lumbar region of subjects were affected by the thick tissue layer in high BMI patients due to its attenuation to the ultrasound signals. One potential solution is to use a probe with relatively lower ultrasound frequency for obese patients, with the trade-off of reduced image resolution. Further study is necessary to investigate the optimized ultrasound frequencies for patients with different BMI with the consideration of tissue penetration and image resolution simultaneously.
Second, the VPI images provided by Scolioscan show many more features than the profile of spinous processes used in the present study, but they have not been used for the analysis of spinal deformity. As shown in the images in Fig.5, transverse processes and ribs can be observed in most of the VPI images. The feasibility of using transverse processes for the spinal deformity measurement has been demonstrated earlier based on VPI images . Further studies would be worthwhile to follow up on how to utilize more features in VPI images to provide more parameters related to spinal deformity, including vertebral rotation. Since ultrasound images also recorded information of paraspinal muscle architecture, it will also be valuable to extract muscle related parameters for scoliosis assessment. In addition, the coronal images formed by the current Scolioscan only covered lumbar and thoracic regions, and not the whole spine structure. Therefore, the overall spinal alignment as well as the thoracic shift cannot be assessed yet. Further developments and studies are required to enable Scolioscan to provide more information of the whole spine structure, which will further widen its application.
Third, it was found that the VPI image formation took between one and two minutes dependent on the height of the patient. While this is acceptable, it would be helpful if the VPI image can be provided immediately after the scanning so that the image quality can be confirmed and the patient can be discharged immediately after scanning. Related developments are underway and it has been demonstrated that real-time image formation is feasible using the system.
Fourth, the discrepancy between Cobb’s angle and Scolioscan angle could arise from the different time and day used for conducting both images. The inclusion criteria used in this study was shorter than 3 months between the two, with most of them within two weeks of each other. There may be some changes of angle during the period, and thus room for improvement exists.
Fifth, while the manual measurement of angle using VPI images appears very repeatable as demonstrated by the three raters in the present study, it reamins a subjective method. Although the three raters were mutually blinded for the measurement, they were in the same research team. Thus there was some common understanding about how to draw the lines on VPI images among the three raters. This may not be the case when Scolioscan is used in different clinical units, and different users may have different methods for drawing lines to measure angles. This may make the results difficult to compare among different clinical or research groups. This issue is not unique for obtaining Scolioscan angles in VPI images and can likely be overcome with clear operation and measurement guidelines. Radiographic Cobb angle measurement has been facing the same challenge, but it will greatly facilitate the measurement of spinal deformity angle based on VPI images if an automatic method can be developed. Related development work is ongoing.
Conclusions
In conclusion, this is the first study to report on the development and human application of Scolioscan in assessing its reliability and validity for scoliosis assessment. The measurement using Scolioscan was demonstrated to be very reliable and good to excellent correlation noted in comparison with the conventional radiographic Cobb’s method. Since Scolioscan is radiation-free and readily accessible, it has the potential to be used to screen large numbers of patients with AIS to monitor progress and outcome of treatment, and with prognostic implications. Further studies are required to demonstrate its clinical values with a larger number of scoliosis patients with different types of curvature and the feasibility of automatic Scolioscan angles measurement. It would also be necessary to investigate the potential of axial rotation and sagittal measurement using Scolioscan. While the current Scolioscan system is relatively large in dimension, it is believed that a portable or even palm-size Scolioscan system will be available in the near future. Scolioscan with its further development may greatly facilitate AIS screening, an important and valid step for managing scoliosis [3, 56], as well AIS prognosis and progression monitoring.
Acknowledgements
This study was supported by the Research Grant Council of Hong Kong (PolyU5332/07E, PolyU152220/14E) and the Hong Kong Innovation and Technology Fund (UIM213). The authors appreciate the contribution of Ms. Siu Yin Law (SY) as one of the two operators as well as one of the three raters at the first stage of the study. The authors would also like to thank the generous technical supports from staff of Telefield Medical Imaging Limited for this study, and Ms. Sally Ding for her help in editing the manuscript.
Authors’ contributions
YP overall coordinated the study, invented the system, participated in the protocol development, data analysis, and finalized the manuscript. TTY participated in the protocol development, performed measurements on images, data analysis and helped to draft the manuscript. KKL conducted the tests, performed the scanning and measurement, and participated in data analysis. BHK participated in the data analysis, particularly statistical analysis, and manuscript formation. GQ participated in the development of Scolioscan system and algorithms for image processing and measurement. WW participated in the Scolioscan system development and image enhancement, performed measurements, and participated in data analysis. JCW participated in the development of the system and software, interface design, and participated in the protocol development. MS participated in the protocol development and data analysis. BKW participated in the subject recruitment and supported the subject tests; JCY participated in the protocol development, subject recruitment, inclusion criteria establishment, and data analysis. TP participated in the protocol development, subject recruitment, inclusion criteria setting, and manuscript polishing. All authors read and approved the final manuscript.
Competing interests
The author Zheng YP was an inventor of a number of patents related to the Scolioscan system, which have been licensed to Telefield Medical Imaging Limited through the Hong Kong Polytechnic University. The project funded by the Hong Kong Innovation and Technology Fund (UIM213) with Zheng YP as the PI was partially sponsored by Telefield Medical Imaging Limited.
Consent
Written informed consent was obtained from the patient(s) for publication of this manuscript and accompanying images. A copy of the written consent is available for review by the Editors-in-Chief of this journal.
References
1.Stokes IA, Bigalow LC, Moreland MS. Three-dimensional spinal curvature in idiopathic scoliosis. J Orthop Res. 1987;5(1):102–113. doi: 10.1002/jor.1100050113. [DOI] [PubMed] [Google Scholar]
2.Hattori T, Sakaura H, Iwasaki M, Nagamoto Y, Yoshikawa H, Sugamoto K. In vivo three-dimensional segmental analysis of adolescent idiopathic scoliosis. Eur Spine J. 2011;20:1745–1750. doi: 10.1007/s00586-011-1869-4. [DOI] [PMC free article] [PubMed] [Google Scholar]
3.Fong DYT, Cheung KMC, Wong YW, Wan YY, Lee CF, Lam TP, et al. A population-based cohort study of 94401 children followed for 10 years exhibits sustained effectiveness of scoliosis screening. Spine J. 2015;15:825–33. [DOI] [PubMed]
4.Fan H, Huang Z, Wang Q, Tan W, Deng N, Yu P, et al. Prevalence of idiopathic scoliosis in Chinese schoolchildren. Spine. 2016;41(3):259–64. [DOI] [PubMed]
5.Cheng JC, Castelein RM, Chu WCC, Danielsson AJ, Dobbs MB, Grivas TB, Gurnett CA, Luk KD, Moreau A, Newton PO, Stokes IA, Weinstein SL, Burwell RG. Adolescent idiopathic scoliosis. Nat Rev Dis Primer. 2015;1:15068. [DOI] [PubMed]
6.Arkin AM. The mechanism of the structural changes in scoliosis. J Bone Joint Surg Am. 1949;31:519–528. [PubMed] [Google Scholar]
7.Asher MA, Burton DC. Adolescent idiopathic scoliosis: natural history and long term treatment effects. Scoliosis. 2006;1(1):2. doi: 10.1186/1748-7161-1-2. [DOI] [PMC free article] [PubMed] [Google Scholar]
8.Thomsen M, Abel R. Imaging in scoliosis from the orthopaedic surgeon’s point of view. Eur J Radiol. 2006;58(1):41–47. doi: 10.1016/j.ejrad.2005.12.003. [DOI] [PubMed] [Google Scholar]
9.Kim H, Kim HS, Moon ES, Yoon CS, Chung TS, Song HT, et al. Scoliosis imaging: what radiologists should know. Radiographics. 2010;30(7):1823–42. [DOI] [PubMed]
10.Cobb JR. American Academy of Orthopaedic Surgeons, Instructional Course Lectures. St Louis: CV Mosby; 1948. Outline for the study of scoliosis; pp. 261–275. [Google Scholar]
11.Hoffman DA, Lonstein JE, Morin MM, Visscher W, Harris BS, 3rd, Boice JD., Jr Breast cancer in women with scoliosis exposed to multiple diagnostic X-rays. J Natl Cancer Inst. 1989;81(17):1307–1312. doi: 10.1093/jnci/81.17.1307. [DOI] [PubMed] [Google Scholar]
12.Doody MM, Lonstein JE, Stovall M, Hacker DG, Luckyanov N, Land CE. Breast cancer mortality after diagnostic radiography: findings from the U.S. Scoliosis Cohort Study. Spine. 2000;25(16):2052–2063. doi: 10.1097/00007632-200008150-00009. [DOI] [PubMed] [Google Scholar]
13.Ronckers CM, Land CE, Miller J, Stovall M, Lonstein JE, Doddy MM. Cancer mortality among women frequently exposed to radiographic exams for spinal disorders. Radiat Res. 2010;174(1):83–90. doi: 10.1667/RR2022.1. [DOI] [PMC free article] [PubMed] [Google Scholar]
14.Schmitz-Feuerhake I, Pflugbeil S. ‘Lifestyle’ and cancer rates in former East and West Germany: the possible contribution of diagnostic radiation exposures. Radiat Prot Dosimetry. 2011;147(1–2):310–313. doi: 10.1093/rpd/ncr348. [DOI] [PubMed] [Google Scholar]
15.Presciutti SM, Karukanda T, Lee M. Management decisions for adolescent idiopathic scoliosis significantly affect patient radiatin exposure. Spine J. 2014;14:1984–1990. doi: 10.1016/j.spinee.2013.11.055. [DOI] [PubMed] [Google Scholar]
16.Geijer H, Beckman KW, Jonsson B, Andersson T, Persliden J. Digital radiography of scoliosis with a scanning method: Initial evaluation. Radiography. 2001;218:402–410. doi: 10.1148/radiology.218.2.r01ja32402. [DOI] [PubMed] [Google Scholar]
17.Deschenes S, Charron G, Beaudoin G, Labelle H, Dubois J, Miron MC, et al. Diagnostic imaging of spinal deformities reducing patients radiation dose with a new slot-scanning X-ray imager. Spine. 2010;35:989–94. [DOI] [PubMed]
18.Al-Aubaidi Z, Lebel D, Oudjhane K, Zeller R. Three-dimensional imaging of the spine using the EOS system: is it reliable? A comparative study using computed tomography imaging. J Pediatric Orthopaedics B. 2013;22:409–412. doi: 10.1097/BPB.0b013e328361ae5b. [DOI] [PubMed] [Google Scholar]
19.Goldberg CJ, Kaliszer M, Moore DP, Fogarty EE, Dowling FE. Surface topography, Cobb angles, and cosmetic change in scoliosis. Spine. 2001;26(4):E55–63. doi: 10.1097/00007632-200102150-00005. [DOI] [PubMed] [Google Scholar]
20.Knott P, Mardjetko S, Nance D, Dunn M. Electromagnetic topographical technique of curve evaluation for adolescent idiopathic scoliosis. Spine (Phila Pa 1976) 2006;31:E911–915. doi: 10.1097/01.brs.0000245924.82359.ab. [DOI] [PubMed] [Google Scholar]
21.Knott P, Sturm P, Lonner B, Cahill P, Betsch M, McCarthy R, et al. Multicenter comparison of 3D spinal measurements using surface topography with those from conventional radiography. Spine Deformity. 2016;4:98–103. [DOI] [PubMed]
22.Suzuki S, Yamamuro T, Shikata J, Shimizu K, Iida H. Ultrasound measurement of vertebral rotation in idiopathic scoliosis. J Bone Joint Surg Br. 1989;71(2):252–255. doi: 10.1302/0301-620X.71B2.2647754. [DOI] [PubMed] [Google Scholar]
23.Huang QH, Zheng YP, Lu MH, Chi ZR. Development of a portable 3D ultrasound imaging system for musculoskeletal tissues. Ultrasonics. 2005;43(3):153–163. doi: 10.1016/j.ultras.2004.05.003. [DOI] [PubMed] [Google Scholar]
24.Huang QH, Zheng YP, Li R, Lu MH. 3-D measurement of body tissues based on ultrasound images with 3-D spatial information. Ultrasound Med Biol. 2005;31:1607–1615. doi: 10.1016/j.ultrasmedbio.2005.08.004. [DOI] [PubMed] [Google Scholar]
25.Cheung CWJ, Zheng YP. Development of 3-D Ultrasound System for Assessment of Adolescent Idiopathic Scoliosis. In: Lim CT, Goh JCH, editors. WCB 2010, IFMBE Proceedings. 31. 2010. pp. 584–587. [Google Scholar]
26.Li M, Cheng J, Ying M, Ng B, Zheng YP, Lam TP, et al. Application of 3-D ultrasound in assisting the fitting procedure of spinal orthosis to patients with adolescent idiopathic scoliosis. Stud Health Technol Inform. 2010;158:34–7. [PubMed]
27.Purnama KE, Wilkinson MH, Veldhuizen AG, van Ooijen PM, Lubbers J, Burgerhof JG, et al. A framework for human spine imaging using a freehand 3D ultrasound system. Technol Health Care. 2010;18(1):1–17. [DOI] [PubMed]
28.Chen W, Lou EH, Zhang PQ, Le LH, Hill D. Reliability of assessing the coronal curvature of children with scoliosis by using ultrasound images. J Child Orthop. 2013;7(6):521–529. doi: 10.1007/s11832-013-0539-y. [DOI] [PMC free article] [PubMed] [Google Scholar]
29.Ungi T, King F, Kempston M, Keri Z, Lasso A, Mousavi P, et al. Spinal curvature measurement by tracked ultrasound snapshots. Ultrasound Med Biol. 2014;40(2):447–54. [DOI] [PubMed]
30.Cheung CW, Law SY, Zheng YP. Development of 3-D ultrasound system for assessment of adolescent idiopathic scoliosis (AIS): and system validation. Conf Proc IEEE Eng Med Biol Soc. 2013; 6474–6477 [DOI] [PubMed]
31.Cheung CWJ, Zhou GQ, Law SY, Lai KL, Jiang WW, Zheng YP. Freehand three-dimensional ultrasound system for assessment of scoliosis. J of Orthopaedic Translation. 2015;3:123–133. doi: 10.1016/j.jot.2015.06.001. [DOI] [PMC free article] [PubMed] [Google Scholar]
32.Cheung CW, Zhou GQ, Law SY, Mak TM, Lai KL, Zheng YP. Ultrasound volume projection imaging for assessment of scoliosis. IEEE Transaction on Medical Imaging. 2015;8:1760–1768. doi: 10.1109/TMI.2015.2390233. [DOI] [PubMed] [Google Scholar]
33.Koo TK, Guo JY, Ippolito C, Bedle JC. Assessment of scoliotic deformity using spinous processes: Comparison of different analysis methods of an ultrasonographic system. J. Manipulative Physiol. Ther. 2014;37:667–677. doi: 10.1016/j.jmpt.2014.09.007. [DOI] [PubMed] [Google Scholar]
34.Young M, Hill DL, Zheng R, Lou E. Reliability and accuracy of ultrasound measurements with and without the aid of previous radiographs in adolescent idiopathic scoliosis (AIS) Europe Spine J. 2015;24:1427–1433. doi: 10.1007/s00586-015-3855-8. [DOI] [PubMed] [Google Scholar]
35.Wang Q, Li M, Lou EHM, Wong MS. Reliability and validity study of clinical ultrasound imaging on lateral curvature of adolescent idiopathic scoliosis. PLoS One. 2015;10(8):e0135264. doi: 10.1371/journal.pone.0135264. [DOI] [PMC free article] [PubMed] [Google Scholar]
36.Li M, Cheng J, Ying M, Ng B, Zheng Y, Lam T, et al. Could clinical ultrasound improve the fitting of spinal orthosis for the patients with AIS? Eur Spine J. 2012;21(10):1926–35. [DOI] [PMC free article] [PubMed]
37.Lou EH, Chan ACY, Donauer A, Tilburn M, Hill DL. Ultrasound-assisted brace casting for adolescent idiopathic scoliosis. Scoliosis. 2015;10:13. doi: 10.1186/s13013-015-0037-8. [DOI] [PMC free article] [PubMed] [Google Scholar]
38.Shrout PE, Fleiss JL. Intraclass correlations: uses in assessing rater reliability. Psychol Bull. 1979;86(2):420–428. doi: 10.1037/0033-2909.86.2.420. [DOI] [PubMed] [Google Scholar]
39.Currier DP. Elements of research in physical therapy. 3. Baltimore: Williams & Wilkins; 1984. pp. 160–171. [Google Scholar]
40.Walter SD, Eliasziw M, Donner A. Sample size and optimal designs for reliability studies. Stat Med. 1998;17:101–110. doi: 10.1002/(SICI)1097-0258(19980115)17:1<101::AID-SIM727>3.0.CO;2-E. [DOI] [PubMed] [Google Scholar]
41.Dawson B, Trapp RG. Basic and Clinical Biostatistics. 4. New York: Lange Medical Books/McGraw-Hill; 2004. [Google Scholar]
42.Hulley SB, Cummings SR, Browner WS, Grady D, Newman TB. Designing clinical research : an epidemiologic approach. 4. Philadelphia: Lippincott Williams & Wilkins; 2013. p. 79. [Google Scholar]
43.Denis F. The three column spine and its significance in the classification of acute thoracolumbar spinal injuries. Spine (Phila Pa 1976) 1983;8:817–31. doi: 10.1097/00007632-198311000-00003. [DOI] [PubMed] [Google Scholar]
44.Herzenberg JE, Waanders NA, Closkey RF, Schultz AB, Hensinger RN. Cobb angle versus spinous process angle in adolescent idiopathic scoliosis. The relationship of the anterior and posterior deformities. Spine. 1990;15(9):874–879. doi: 10.1097/00007632-199009000-00007. [DOI] [PubMed] [Google Scholar]
45.Middleditch A and Oliver J. Functional Anatomy of the Spine, 2nd ed. Oxford, UK: Butterworth-Heinemann; 2002.
46.Stokes IAF, Gardnermorse M. Analysis of the interaction between vertebral lateral deviation and axial rotation in rotation in scoliosis. J Biomechanics. 1991;24:753–759. doi: 10.1016/0021-9290(91)90339-O. [DOI] [PubMed] [Google Scholar]
47.Mellado JM, Larrosa R, Martin J, Yanguas N, Solanas S, Cozcolluela MR. MDCT of variations and anomalies of the neural arch and its processes: Part 1-Pedicles, pars interarticularis, laminae, and spinous process. Am J Roentgenology. 2011;197:W104–W113. doi: 10.2214/AJR.10.5803. [DOI] [PubMed] [Google Scholar]
48.Mellado JM, Larrosa R, Martin J, Yanguas N, Solanas S, Cozcolluela MR. MDCT of variations and anomalies of the neural arch and its processes: Part 2-Articular processes, transverse processes, and high cervical spine. Am J Roentgenology. 2011;197:W114–W121. doi: 10.2214/AJR.10.5811. [DOI] [PubMed] [Google Scholar]
49.Morrison DG, Chan A, Hill D, Parent EC, Lou EHM. Correlation between Cobb angle, spinous process angle (SPA) and apical vertebrae rotation (AVR) on posteroanterior radiographs in adolescent idiopathic scoliosis (AIS) Eur Spine J. 2015;24:306–312. doi: 10.1007/s00586-014-3684-1. [DOI] [PubMed] [Google Scholar]
50.Villemure I, Aubin CE, Grimard G, Dansereau J, Labelle H. Progression of vertebral and spinal three-dimensional deformities in adolescent idiopathic scoliosis: a longitudinal study. Spine. 2001;26(20):2244–2250. doi: 10.1097/00007632-200110150-00016. [DOI] [PubMed] [Google Scholar]
51.Gum JL, Asher MA, Burton DC, Lai SM, Lambart LM. Transverse plane pelvic rotation in adolescent idiopathic scoliosis: primary or compensatory? Eur Spine J. 2007;16(10):1579–1586. doi: 10.1007/s00586-007-0400-4. [DOI] [PMC free article] [PubMed] [Google Scholar]
52.Hayashi K, Upasani VV, Pawelek JB, Aubin CE, Labelle H, Lenke LG, et al. Three-dimensional analysis of thoracic apical sagittal alignment in adolescent idiopathic scoliosis. Spine (Phila Pa 1976). 2009;34(8):792–7. [DOI] [PubMed]
53.Vrtovec T, Pernus F, Likar B. A review of methods for quantitative evaluation of spinal curvature. Eur Spine J. 2009;18(5):1–15. doi: 10.1007/s00586-009-0913-0. [DOI] [PMC free article] [PubMed] [Google Scholar]
54.Carlson BB, Burton DC, Asher MA. Comparison of trunk and spine deformity in adolescent idiopathic scoliosis. Scoliosis. 2013;8(1):2. doi: 10.1186/1748-7161-8-2. [DOI] [PMC free article] [PubMed] [Google Scholar]
55.Yazici M, Acaroglu ER, Alanay A, Deviren V, Cila A, Surat A. Measurement of vertebral rotation in standing versus supine position in adolescent idiopathic scoliosis. J Pediatr Orthop. 2001;21(2):252–256. [PubMed] [Google Scholar]
56.Hresko MT, Talwalkar VR, Schwend RM. Position statement – Screening for the early detection fo idiopathic scoliosis in adolescents. SRS/POSNA/AAOS/AAP Position Statement. 2015. [Google Scholar]
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No genetic contribution to variation in human offspring sex ratio: a total population study of 4.7 million births
Brendan P. Zietsch
Brendan P. Zietsch
Centre for Evolution and Psychology, School of Psychology, University of Queensland, St. Lucia, Brisbane QLD 4072, Australia
zietsch@psy.uq.edu.au
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Hasse Walum
Hasse Walum
Division of Microbiology and Immunology, Yerkes National Primate Research Center, Emory University, 954 Gatewood Rd NE, Atlanta, GA 30329, USA
Silvio O. Conte Center for Oxytocin and Social Cognition, Yerkes National Primate Research Center, Emory University, 954 Gatewood Rd NE, Atlanta, GA 30329, USA
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Paul Lichtenstein
Paul Lichtenstein
Department of Medical Epidemiology and Biostatistics, Karolinska Institutet, Nobels väg 12A, 171 77 Stockholm, Sweden
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Karin J. H. Verweij†
Karin J. H. Verweij
Department of Psychiatry, Amsterdam UMC, location AMC, University of Amsterdam, Meibergdreef 5, 1105 AZ Amsterdam, The Netherlands
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Ralf Kuja-Halkola†
Ralf Kuja-Halkola
Department of Medical Epidemiology and Biostatistics, Karolinska Institutet, Nobels väg 12A, 171 77 Stockholm, Sweden
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Brendan P. Zietsch
Brendan P. Zietsch
Centre for Evolution and Psychology, School of Psychology, University of Queensland, St. Lucia, Brisbane QLD 4072, Australia
zietsch@psy.uq.edu.au
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Hasse Walum
Hasse Walum
Division of Microbiology and Immunology, Yerkes National Primate Research Center, Emory University, 954 Gatewood Rd NE, Atlanta, GA 30329, USA
Silvio O. Conte Center for Oxytocin and Social Cognition, Yerkes National Primate Research Center, Emory University, 954 Gatewood Rd NE, Atlanta, GA 30329, USA
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Paul Lichtenstein
Paul Lichtenstein
Department of Medical Epidemiology and Biostatistics, Karolinska Institutet, Nobels väg 12A, 171 77 Stockholm, Sweden
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Karin J. H. Verweij†
Karin J. H. Verweij
Department of Psychiatry, Amsterdam UMC, location AMC, University of Amsterdam, Meibergdreef 5, 1105 AZ Amsterdam, The Netherlands
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Ralf Kuja-Halkola†
Ralf Kuja-Halkola
Department of Medical Epidemiology and Biostatistics, Karolinska Institutet, Nobels väg 12A, 171 77 Stockholm, Sweden
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Published:19 February 2020
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No genetic contribution to variation in human offspring sex ratio: a total population study of 4.7 million births
Abstract
The ratio of males to females among an individual's offspring at birth (offspring sex ratio) has long been of great interest to evolutionary biologists. The human offspring sex ratio is around 1 : 1 and is understood primarily in terms of Fisher's principle (R. A. Fisher, The genetical theory of natural selection, 1930), which is based on the insight that in a population with an unequal sex ratio, each individual of the rarer sex will on average have greater reproductive value than each individual of the more common sex. Accordingly, individuals genetically predisposed to produce the rarer sex will tend to have greater fitness and thus genes predisposing to bearing that sex will increase in frequency until the population sex ratio approaches 1 : 1. An assumption of this perspective is that individuals' offspring sex ratio is heritable. However, the heritability in humans remains remarkably uncertain, with inconsistent findings and important power limitations of existing studies. To address this persistent uncertainty, we used data from the entire Swedish-born population born 1932 or later, including 3 543 243 individuals and their 4 753 269 children. To investigate whether offspring sex ratio is influenced by genetic variation, we tested the association between individuals' offspring's sex and their siblings' offspring's sex (n pairs = 14 015 421). We estimated that the heritability for offspring sex ratio was zero, with an upper 95% confidence interval of 0.002, rendering Fisher's principle and several other existing hypotheses untenable as frameworks for understanding human offspring sex ratio.
1. Introduction
The ratio of males to females among an individual's offspring at birth (offspring sex ratio) has been of great interest to evolutionary biologists for more than a century. ‘Fisher's principle' , which can be traced to Darwin (1871; see ), explains offspring sex ratio using the concept of negative frequency-dependent selection and has been described as ‘probably the most celebrated argument in evolutionary biology' . Briefly, it is argued that––all else being equal––in a population with an unequal sex ratio, each individual of the rarer sex will on average have greater reproductive value than each individual of the more common sex, because each sex must contribute exactly half the ancestry of future generations. In this scenario, individuals genetically predisposed to produce the rarer sex will tend to have greater fitness (more grandchildren) and thus genes predisposing to bearing that sex will increase in frequency until the population sex ratio approaches 1 : 1 and the advantage of those genes dissipates. In this way, the theory goes, offspring sex ratio is maintained in the population at around 1 : 1––the equilibrium ratio, or evolutionarily stable strategy.
The human offspring sex ratio is around 1 : 1 and is understood primarily in terms of Fisher's principle . Fisher's principle does not necessarily predict an even offspring sex ratio, but it will do so if parental investment after birth is equal for the average female and male. Sex determination in humans depends on whether an offspring inherits an X or Y chromosome from the father, so an even sex ratio would also be predicted by random Mendelian segregation of sex chromosomes, in the absence of a Fisherian or other adaptive process .
An assumption of Fisher's principle is that offspring sex ratio is heritable (i.e. its between-individual variation is influenced by genetic variation), because heritability is necessary for Fisher's principle to operate [1,5]: without heritable variation, the offspring sex ratio cannot respond to selection. Indeed, beyond Fisher's principle, much of the theory pertaining to human offspring sex ratio, including variations of the Trivers–Willard hypothesis [6–11], also predicts heritable variation––a point we return to in the discussion.
For such a clear, falsifiable assumption that is central to much of the theory on which the field is based, the heritability of offspring sex ratio in humans remains remarkably uncertain. In Science in 1951, Bernstein wrote that researchers had ‘shown conclusively that the tendency in a family to produce offspring of one sex only or primarily one sex is hereditary', but subsequent reviews and major studies did not reach such definitive conclusions (e.g. [13,14]). In 2001 alone, one study involving genotyped families found that human offspring sex ratio was significantly influenced by paternal genetic factors , while another involving twins concluded that it was not heritable at all . In a still more recent study, a genealogical database compiled by amateur and professional family tree researchers was used to test parent–offspring correlations in offspring sex ratio and estimated a significant heritability (0.057 ± 0.002) in males, which the author proposed to be caused by a yet-to-be-identified polymorphic autosomal gene .
A limitation of all of these studies, and a potential cause of the inconsistency in their findings, is insufficient statistical power due to sample sizes used. Though the overall samples in the studies were sometimes large (e.g. 556 387 individuals in ), the samples directly used for testing for heritability were always relatively modest is size (e.g. 1224 parent–offspring pairs in ). The studies consistently indicate that any heritability in human offspring sex ratio must be fairly small, so larger samples with sufficient power for very precise estimates are required to reliably establish its presence or absence.
To address this persistent uncertainty regarding the heritability of offspring sex ratio in humans, we conducted a familial aggregation analysis of offspring sex, using data from the entire Swedish-born population born 1932 or later. The sample included 3 543 243 Swedish individuals and their 4 753 269 children. Specifically, to investigate whether offspring sex ratio has genetic variation, we tested the association between individuals' offspring's sex and their siblings' offspring's sex (n pairs = 14 015 421). We also estimated heritability and its upper bound. The data used in this study are of a much larger scale than previous research, yielding essentially 100% power to detect a heritability of 0.057, as estimated by Gellatly , or even an order of magnitude smaller than that. Further, the data are essentially complete and unbiased, being based on prospectively collected government registry records [18–20].
2. Methods
(a) Study population
Using the unique personal identification number given to each Swedish citizen at birth or immigration , we linked two administrative population registers in Sweden: the Total Population Register (TPR); and the Multi-Generation Register (MGR), as of 31 December 2013 . Details on each register will be provided below. We included all Swedes born in 1932 or later who had identified biological parents, had at least one biological child born before 2014, and were not born from a multiple birth. In this sample, we identified all parental siblings (index generation) and all pairs of their offspring, i.e. cousins (in total, 14 015 421 pairs). In total, 3 543 243 individuals were identified. We then identified all their offspring, yielding 4 753 269 individuals. Offspring from multiple births (n = 107 006) were excluded from analyses because they might bias our estimates, for example, due the heritability of twinning and issues with statistical dependence within pairs.
The TPR was established in 1967, but covers individuals alive in Sweden in 1947 and later (gathered from previous registers; ); the coverage is essentially complete . The MGR includes all registered parents of individuals born 1932 and onwards, who were alive in 1961 and later . The proportion of known parents of individuals is high and has increased over time to 100% for mothers and 98% for fathers, born 1961 and later .
(b) Study variables
The analysed variable was sex, as registered in the TPR. Since 1972 sex may be changed in the register, but this is very uncommon . From 1973, all pregnancies with delivery in week 28 or later, regardless of whether it was a stillbirth or not, were registered, and a sex was assigned. In July 2008, this threshold was moved to week 22 of the pregnancy . Prior to 1973, information about the occurrence and sex of stillbirths was not documented as accurately.
(c) Statistical analyses
We performed analyses of familial aggregation of offspring sex using logistic regression within a generalized estimating equation framework (with cluster–robust standard errors to account for familial relatedness [22–25]). The independent variable was the sex of an individual's offspring and the dependent variable was the sex of the individual's sibling's offspring (sex was coded 0 for male and 1 for female). Results are presented as odds ratios (OR) with 95% confidence intervals (CIs), i.e. the odds of siblings' offspring being of the same sex as opposed to opposite sex (with an OR above 1 indicating a higher likelihood the offspring are of the same sex and an OR below 1 indicating a higher likelihood the offspring are of the opposite sex). We analysed all possible pairs of offspring for all full sibling parents (i.e. all pairs of cousins); individuals can therefore appear more than once in the analyses, but the cluster–robust estimator appropriately accounted for dependencies between pairs in calculating the standard errors and CIs. We also ran analyses separately for offspring of all full sisters, all full brothers, all maternal half-siblings and all paternal half-siblings. To avoid potential effects of parents stopping their childbearing depending on the sex composition of their offspring, we also performed the analyses on first-born offspring only (as compared to all offspring).
To assess the upper limit of the heritability of offspring sex ratio, we estimated tetrachoric correlations (used to measure agreement of binary data) of offspring sex among all full siblings. Since full siblings on average share 50% of their co-segregating alleles, the upper bound of the (narrow sense) heritability can be computed as twice the tetrachoric correlation between full siblings. Thus, we calculated the 95% CIs for the tetrachoric correlations and identified the upper 95% CI of the heritability as two times the upper 95% CI of the tetrachoric correlation. For this analysis, the CIs were calculated using a non-parametric bootstrap approach with 2000 resamplings. We performed this analysis on all pairs of offspring as well as on only first-born offspring.
One way associations in offspring sex ratio could arise is if parents tended to stop their reproduction once a desired sex composition of offspring had been reached. To address whether deviations from randomness in sex of offspring were present, we performed two additional analyses within-nuclear families (i.e. offspring of index parents). First, we analysed the association between the sexes of offspring in families conditioned on final family size; we determined the association between offspring's sex separately for families of different sizes (with 2 to 10 children). That is, we identified all families of a specific final attained family size, such as all families with three offspring. Similar to our main analyses on parental siblings, where all pairs of offspring to parental siblings were identified, in this sub-population, we identified all pairs of offspring siblings within each family; in the case of final family size of three the pairs were 1&2, 1&3 and 2&3. We analysed the data similarly as for parental sibling pairs, using logistic regression with each offspring acting as exposure and outcome separately in each pair, while accounting for dependency between rows using cluster-robust standard errors.
Second, we performed an analysis per birth order, regardless of final family size. In this latter analysis, the sex of the nth child was predicted based on the sex of each of the older siblings (e.g. we predicted the sex of the third-born child based on the sex of the first- and second-born child). In contrast with the previous analysis, this analysis is not affected by potential bias from a tendency to stop having children after desired sex composition is attained. Together, these two analyses tested potential reproductive stopping effects which could produce biased heritability estimates.
To assess potential issues with sub-optimal coverage of registries, which varied over time (with increasing quality), we re-analysed the association of offspring sex between siblings using data on parental full siblings in nine different (overlapping) birth cohorts––those with earliest year of birth of parents limited to 1932, 1947 or 1961, and latest year of birth limited to 1973, 1985 or 2000.
To test whether the exclusion of multiple births (twins, triplets, etc.) affected the results, we re-performed the familial aggregation analyses including multiple births in full siblings, female full siblings, male full siblings, and maternal- and paternal half-siblings. This re-analysis was done both using all offspring and using only first-born offspring.
To ensure our estimates of heritability were robust with respect to different modelling approaches, we compared our results using sibling regressions (as above) to estimates from the ‘animal model' [26,27], which incorporates pedigree information and hence models all known relationships in our data. We used families with index generation individuals born between 1932 and 1973 to ensure good coverage through their reproductive age (the youngest were 40 by the end of follow-up), excluded twin births, and ran the model using identity link. In total 3 869 556 offspring sex outcomes were included (83.3% of the total 4 646 263, excluding the twin births), total number of founders was 2 181 396 (males plus females), and 1 443 205 mothers and 1 434 946 fathers were identified to 2 651 143 individuals who had at least one offspring. The maximum depth of pedigrees was four generations (where the youngest generation was included only as outcome measure). We ran sensitivity checks with the index generation individuals born between 1932 and 1973: (i) using a probit link function––to model the heritability on the liability scale ; (ii) using a logit link function; (iii) using only females in index generation and an identity link function; and (iv) using only males in index generation and an identity link function.
Analyses were performed in the statistical software R , using the packages drgee , polycor and pedigreemm.
3. Results
Table 1 and figure 1 show descriptive information about the parents and offspring in our sample. The sample included more mothers than fathers (ratio = 51.9 : 48.1) and more male than female offspring (ratio = 51.4 : 48.6) (both p < 0.001 by exact binomial test). The mean number of offspring was 2.189 (2.185 in females, and 2.193 in males); the range was 1 to 17 offspring in both males and females, although less than 0.02% had 10 or more offspring (distribution up to five offspring displayed in table 1).
Table 1.
Descriptive information on index (parent) and offspring generation.
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Table 1. Descriptive information on index (parent) and offspring generation.
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| Index generation --- | | | | |
| | number (%) | no. males (%) | no. females (%) | p-value, males versus femalesa |
| individuals | 3 543 243 (100.0) | 1 703 093 (48.1) | 1 840 150 (51.9) | <0.001 |
| birth year | | | | <0.001 |
| 1932–1939 | 393 240 (11.1) | 193 562 (11.4) | 199 678 (10.9) | |
| 1940–1949 | 817 672 (23.1) | 402 484 (23.6) | 415 188 (22.6) | |
| 1950–1959 | 717 630 (20.3) | 351 116 (20.6) | 366 514 (19.9) | |
| 1960–1969 | 757 448 (21.4) | 368 664 (21.6) | 388 784 (21.1) | |
| 1970–1979 | 615 801 (17.4) | 291 326 (17.1) | 324 475 (17.6) | |
| 1980–1989 | 231 730 (6.5) | 93 240 (5.5) | 138 490 (7.5) | |
| 1990–1999 | 9722 (0.3) | 2701 (0.2) | 7021 (0.4) | |
| number of offspring | | | | <0.001 |
| 1 | 756 283 (21.3) | 377 753 (22.2) | 378 530 (20.6) | |
| 2 | 1 743 847 (49.2) | 825 331 (48.5) | 918 516 (49.9) | |
| 3 | 768 758 (21.7) | 363 458 (21.3) | 405 300 (22.0) | |
| 4 | 200 823 (5.7) | 98 797 (5.8) | 102 026 (5.5) | |
| 5 | 51 160 (1.4) | 26 264 (1.5) | 24 896 (1.4) | |
| >5 | 22 372 (0.6) | 11 490 (0.7) | 10 882 (0.6) | |
| ever multiple birth | | | | <0.001 |
| no | 3 456 774 (97.6) | 1 660 948 (97.5) | 1 795 826 (97.6) | |
| yes | 86 469 (2.4) | 42 145 (2.5) | 44 324 (2.4) | |
| offspring generation --- | | | | |
| | n (%) | male offspring (%) | female offspring (%) | |
| individuals | 4 753 269 (100.0) | 2 444 030 (51.4) | 2 309 239 (48.6) | <0.001 |
| birth year | | | | 0.785 |
| <1960 | 189 242 (4.0) | 97 162 (4.0) | 92 080 (4.0) | |
| 1960–1969 | 787 591 (16.6) | 405 354 (16.6) | 382 237 (16.6) | |
| 1970–1979 | 901 497 (19.0) | 463 472 (19.0) | 438 025 (19.0) | |
| 1980–1989 | 913 010 (19.2) | 469 412 (19.2) | 443 598 (19.2) | |
| 1990–1999 | 870 598 (18.3) | 447 200 (18.3) | 423 398 (18.3) | |
| 2000–2013 | 1 091 331 (23.0) | 561 430 (23.0) | 529 901 (22.9) | |
| birth orderb | | | | 0.778 |
| 1 | 2 059 372 (43.3) | 1 058 701 (43.3) | 1 000 671 (43.3) | |
| 2 | 1 699 793 (35.8) | 874 369 (35.8) | 825 424 (35.7) | |
| 3 | 700 240 (14.7) | 360 147 (14.7) | 340 093 (14.7) | |
| 4 | 206 140 (4.3) | 105 857 (4.3) | 100 283 (4.3) | |
| 5 | 58 976 (1.2) | 30 167 (1.2) | 28 809 (1.2) | |
| >5 | 28 748 (0.6) | 14 789 (0.6) | 13 959 (0.6) | |
| from multiple birth | | | | <0.001 |
| no | 4 646 263 (97.7) | 2 389 914 (97.8) | 2 256 349 (97.7) | |
| yes | 107 006 (2.3) | 54 116 (2.2) | 52 890 (2.3) | |
ap-values are calculated by exact binomial test or Pearson chi-square test.
bIf the child's birth order is different for the mother and the father, the latter order is presented.
Table 2 presents the results of the main analyses, i.e. the OR representing the odds of an individual's offspring's sex being the same as their sibling's offspring's sex, divided by the odds of an individual's offspring's sex being different from their sibling's offspring's sex. Thus, ORs above 1 indicates higher likelihood of same sex, and ORs below 1 higher likelihood of opposite sex. Results are also shown separately for full sisters, full brothers, as well as maternal half-siblings, and paternal half-siblings.
Table 2.
Association between sex of an individual's offspring and sex of their sibling's offspring. Odds ratios above 1 indicate a higher likelihood of siblings' offspring having the same sex, odds ratios below 1 a higher likelihood of opposite sex.
View inlineView popup
Table 2. Association between sex of an individual's offspring and sex of their sibling's offspring. Odds ratios above 1 indicate a higher likelihood of siblings' offspring having the same sex, odds ratios below 1 a higher likelihood of opposite sex.
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| index generation, biological relation | pairs of offspring analysis --- | | | only first-born offspring analysis --- | | |
| no. pairsa | odds ratio (95% confidence interval) | p-value | no. pairs | odds ratio (95% confidence interval) | p-value |
| full siblings | 11 958 896 | 1.001 (0.998, 1.003) | 0.523 | 2 246 883 | 0.997 (0.995, 1.002) | 0.246 |
| full siblings–females | 3 200 160 | 1.002 (0.997, 1.006) | 0.407 | 596 622 | 0.995 (0.985, 1.005) | 0.298 |
| full siblings–males | 2 840 847 | 1.001 (0.996, 1.006) | 0.719 | 534 693 | 0.998 (0.987, 1.009) | 0.712 |
| maternal half-siblings | 912 648 | 0.999 (0.991, 1.008) | 0.898 | 182 802 | 0.999 (0.981, 1.018) | 0.939 |
| paternal half-siblings | 1 143 877 | 1.006 (0.999, 1.014) | 0.106 | 235 495 | 1.002 (0.986, 1.019) | 0.767 |
Note: Analyses performed with generalized estimating equations with logit link and standard errors clustered on parental sibling clusters. Offspring from multiple births (twins, triplets and higher) have been excluded from analyses.
aBecause nuclear families were considered as a unit, each pair could only contribute to one family. However, individuals can be part of more than one pair as each offspring may be part of cousin pairs both on her/his mother's and father's side and because each offspring can be part of multiple pairs on mother's and/or father's side. Therefore, the numbers of pairs here do not match numbers of offspring presented in table 1.
None of the odds ratios were statistically significantly different from 1, indicating that there were no significant associations between an individual's offspring's sex and the sex of their sibling's offspring. Results were similar when only looking at the first-born offspring; all ORs were non-significant and close to 1. Because of the enormous sample sizes, the estimates are extremely precise with very narrow CIs. For example, for the whole sample of full sibling pairs, we found an OR of 1.001 with 95% CIs of 0.998, 1.003.
The observed tetrachoric correlation between full siblings was 0.00029 (95% CI: −0.00038, 0.00098) and for first-born offspring −0.00120 (95% CI: −0.00284, 0.00027). Thus, the estimate was not significantly different from zero and the upper 95% confidence bound of the heritability (twice the upper CI of the sibling correlation ) was 0.0020 (or 0.0005 for first-born offspring). Using the animal model, with an identity link function, similarly yielded a null heritability estimate (0.0004; 95% CI: 0.0000–0.0010). The same analysis using a probit link function yielded a heritability estimate of 0.0005, and with logit link function an estimate of 0.0004 (we could not obtain CIs for these analyses). For women, the estimated heritability, using an identity link function, was 0.0005 (95% CI: 0.0000–0.0015) and for men, 0.0004 (95% CI: 0.0000–0.0014).
Table 3 shows the results of the within-family analyses. In the left part is the association between the sexes of the offspring within families of certain size (for families with 2 to 10 children). In families with a final number of two children, the offspring were more likely to be opposite sex than expected by chance (OR = 0.854; 95% CI: 0.847, 0.860). For families with 3 to 7 children, there was a significant effect (p < 0.001) in the opposite direction, with a higher likelihood of an excess of same-sex offspring (OR estimates range between 1.049 and 1.077). However, when predicting the sex of the nth child based on the sex of the older siblings but irrespective of final family size, the associations were statistically non-significant (table 3, right side). Thus, the sex distribution within-nuclear families deviated from what is expected by chance, with two-child families having children of opposite sex more often than expected. However, removing the conditioning of final family size (i.e. not restricting the birth-order-specific analyses depending on final family size, but including all offspring available, regardless of whether the parents had more offspring in the future) made associations null, indicating that the associations were driven by parents selectively stopping reproduction, rather than by biological differences in sex determination of offspring related to parity.
Table 3.
Within-family offspring sex associations. Left, the within-nuclear-family sex associations for families of different sizes (final number of children 2 to 10). Right, prediction (OR) of the sex of the nth child based on the sex of each older sibling, regardless of final family size. Odds ratios above 1 indicate a higher likelihood individuals' offspring having the same sex, odds ratios below 1 a higher likelihood of opposite sex. NA, not applicable. Analyses performed with generalized estimating equations with logit link and standard errors clustered on parental sibling clusters.
View inlineView popup
Table 3. Within-family offspring sex associations. Left, the within-nuclear-family sex associations for families of different sizes (final number of children 2 to 10). Right, prediction (OR) of the sex of the nth child based on the sex of each older sibling, regardless of final family size. Odds ratios above 1 indicate a higher likelihood individuals' offspring having the same sex, odds ratios below 1 a higher likelihood of opposite sex. NA, not applicable. Analyses performed with generalized estimating equations with logit link and standard errors clustered on parental sibling clusters.
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| final family size (no. of children) | final family size analysis --- | | | | birth order analysis --- | | |
| no. of pairsa | odds ratio (95% confidence intervals) | p-value | nth born child | no. of pairsa | odds ratio (95% confidence intervals)b | p-value |
| 1 | NA | NA | NA | 1 | NA | NA | NA |
| 2 | 1 085 061 | 0.854 (0.847, 0.860) | <0.001 | 2 | 1 748 196 | 0.997 (0.991, 1.003) | 0.363 |
| 3 | 1 461 533 | 1.049 (1.042, 1.056) | <0.001 | 3 | 1 349 407 | 1.005 (0.998, 1.012) | 0.148 |
| 4 | 803 803 | 1.077 (1.068, 1.087) | <0.001 | 4 | 559 402 | 0.999 (0.989, 1.010) | 0.916 |
| 5 | 354 620 | 1.073 (1.058, 1.088) | <0.001 | 5 | 206 348 | 1.003 (0.985, 1.021) | 0.772 |
| 6 | 156 337 | 1.073 (1.051, 1.095) | <0.001 | 6 | 79 373 | 0.993 (0.965, 1.023) | 0.660 |
| 7 | 69 366 | 1.071 (1.037, 1.106) | <0.001 | 7 | 32 318 | 1.008 (0.963, 1.055) | 0.747 |
| 8 | 34 331 | 1.043 (0.997, 1.090) | 0.065 | 8 | 14 593 | 1.057 (0.988, 1.131) | 0.106 |
| 9 | 16 540 | 1.064 (0.996, 1.137) | 0.064 | 9 | 6978 | 0.982 (0.890, 1.084) | 0.724 |
| 10 | 10 779 | 1.091 (1.003, 1.186) | 0.042 | 10 | 3706 | 0.969 (0.847, 1.109) | 0.651 |
aBecause nuclear families were considered as a unit, each pair could only contribute to one family. However, individuals can be part of more than one pair as each first-born offspring may be part of cousin pairs both on her/his mother's and father's side and because each first-born offspring can be part of multiple pairs on mother's and/or father's side. Therefore, the numbers of pairs here do not match numbers of offspring presented in table 1.
bOR of the sex of the nth child based on the sex of each older sibling; e.g. for the fourth-born child, the dependent variable is the sex of child 4, and independent variable is from siblings born first, second and third.
We also split the sample into different birth cohorts; we did not find any significant association between the sex of individuals' offspring and sex of their siblings' offspring, either when considering all pairs of offspring or when only considering first-born offspring (table 4). Finally, when including multiple births the results were essentially unchanged (results not shown).
Table 4.
Association between sex of an individual's offspring and sex of their sibling's offspring, restricted by birth year of parents. Odds ratios above 1 indicate a higher likelihood of siblings' offspring having the same sex, odds ratios below 1 a higher likelihood of opposite sex. Note: Analyses performed with generalized estimating equations with logit link and standard errors clustered on parental sibling clusters.
View inlineView popup
Table 4. Association between sex of an individual's offspring and sex of their sibling's offspring, restricted by birth year of parents. Odds ratios above 1 indicate a higher likelihood of siblings' offspring having the same sex, odds ratios below 1 a higher likelihood of opposite sex. Note: Analyses performed with generalized estimating equations with logit link and standard errors clustered on parental sibling clusters.
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| birth year restrictionsa | all pairs of offspring --- | | | only first-born offspring --- | | |
| no. pairs | odds ratio (95% confidence interval) | p-value | no. pairs | odds ratio (95% confidence interval) | p-value |
| from 1932 to 2000 | 11 958 896 | 1.001 (0.998,1.003) | 0.523 | 2 246 883 | 0.997 (0.992,1.002) | 0.246 |
| from 1932 to 1985 | 11 842 051 | 1.001 (0.998,1.003) | 0.640 | 2 203 557 | 0.997 (0.992,1.002) | 0.234 |
| from 1932 to 1973 | 10 456 786 | 1.001 (0.998, 1.003) | 0.674 | 1 879 161 | 0.997 (0.992,1.003) | 0.338 |
| from 1947 to 2000 | 7 160 659 | 1.000 (0.997,1.003) | 0.978 | 1 393 138 | 0.995 (0.988,1.001) | 0.120 |
| from 1947 to 1985 | 7 043 814 | 1.000 (0.997,1.003) | 0.842 | 1 349 812 | 0.994 (0.988,1.001) | 0.108 |
| from 1947 to 1973 | 5 658 558 | 0.999 (0.996,1.003) | 0.745 | 1 025 418 | 0.995 (0.987,1.002) | 0.164 |
| from 1961 to 2000 | 3 057 004 | 1.001 (0.997,1.006) | 0.521 | 667 896 | 0.998 (0.989,1.008) | 0.711 |
| from 1961 to 1985 | 2 940 194 | 1.001 (0.996,1.005) | 0.760 | 624 579 | 0.998 (0.988,1.008) | 0.678 |
| from 1961 to 1973 | 1 583 439 | 1.001 (0.995,1.007) | 0.786 | 304 960 | 1.001 (0.987,1.015) | 0.911 |
a‘From' and ‘to' refers to between which birth years the index generation individuals are born. The 1932–2000 analysis is the full analyses (as in table 1).
4. Discussion
In a sample comprising 4 753 269 offspring of 3 543 243 Swedes, we detected no significant genetic influence on offspring sex ratio. In fact, our heritability estimate was zero, with an upper 95% CI of 0.0020 (i.e. two-tenths of one per cent), rendering Fisher's principle untenable as a framework for understanding human offspring sex ratio.
Our results also rule out the possibility that offspring sex ratios are adaptively calibrated to individuals' heritable traits. Certain interpretations of the Trivers–Willard effect propose that parents who possess any heritable trait that disproportionately benefits the fitness of one sex will bias their offspring sex ratio towards that sex . A number of studies have reported evidence of such effects––e.g. male-biased offspring sex ratio in bigger, taller , wealthier , higher status and less sociosexually restricted parents, and a female-biased offspring sex ratio in more physically attractive parents . Several of these findings, though, have been questioned on statistical grounds [33–35], and other studies have not supported the hypothesis (including very large birth cohort studies, e.g. [8,36,37]). Our findings are incompatible with the basic effect: if offspring sex ratio was calibrated to heritable traits, then it would necessarily be heritable to some degree as well.
According to the same principle, our results also rule out a hypothesis [38,39] that offspring sex ratio in humans (and in other mammals) is influenced by steroid hormone levels in parents at the time of conception. As James notes, if the hypothesis was true, then offspring sex ratio would be heritable to some degree because steroid hormone levels are themselves heritable [41,42]. Therefore, given our finding of zero heritability, the hormone-level hypothesis is untenable.
Another proposal is that the environment experienced by mothers during early life may affect their subsequent offspring sex ratio . Because the vast majority of siblings in our study would have grown up together and thus shared many aspects of their early environment, the lack of any sibling similarity in offspring sex ratio further limits the possibility of adaptive variation in relation to these shared environmental factors. Moreover, any environmental effect––or any factor that is at all stable––should cause correlation within-individuals regarding the sex of different offspring (e.g. a woman whose first child is a boy is more likely to have subsequent boys compared to girls); we did not see any such auto-correlation in offspring sex, indicating that there are no stable factors of persons or their environments that influence their offspring sex ratio.
If human offspring sex ratio is not heritable and/or under adaptive control in calibration to heritable or environmental factors, what causes its variance between individuals? One possible explanation is that some parents stop their reproduction dependent on sex composition attained in the family. This behaviour would produce deviations from expected distributions of sexes in families, but the sex determination at conception could still be entirely random. We have indirect evidence on this question: conditioning on final family size produced statistically significant association between sexes within a family. However, when not conditioning on the final family size, but predicting the sex of the nth child based on the sex of their older siblings regardless of final family size, the associations were statistically non-significant. In sum, all of our results are consistent with the simple explanation that variation in offspring sex ratio in humans is due to unbiased Mendelian segregation of sex chromosomes during spermatogenesis and unbiased fertilization. The slight excess of male births is likely to be due to a general difference in survival of male and female embryos in the womb, the reasons for which are not yet understood .
This is by far the largest study of the heritability of offspring sex ratio in humans, but to our knowledge it is also the largest such study in any animal. Our findings do not preclude heritable offspring sex ratio in other species with similar sex determination systems, but there is little compelling evidence for such genetic variation (e.g. [44,45]). Moreover, if offspring sex ratio in other such species were heritable, it would raise perplexing questions as to how heritability could have been lost in the human lineage. Under Fisher's principle, offspring sex ratio is subject to negligible selection when the population is near equilibrium [1,5], which it normally is; even in the hypothetical case of large variance in local sex ratios, selection favours equal investment (e.g. alternating male and female offspring) , not random Mendelian sex determination, where uneven investment is likely for individual parents. How and why would such weak selection have eliminated all genetic variation in offspring sex ratio while substantial genetic variation has been maintained in virtually every other human variable that has been studied , including fitness itself ?
Although the study has a large and largely unbiased sample, it is not perfect, and there are some limitations to consider. First, the register's coverage on parental identity, while very good, is not 100% (particularly for fathers), which could introduce slight bias towards the null. Non-paternity (i.e. the assumed biological father is not the biological father) is rare (approx. 1–2%) in human populations [49,50], so this could not explain a zero heritability estimate. Further, sensitivity analysis with birth years restricted to the highest quality data also showed the null association. Another consideration is that, although stillbirths are registered, we do not know the sex of spontaneous or selectively aborted fetuses early in pregnancy. But we have no reason to believe that the spontaneous abortions would bias the observed association to precisely null. Last, our approach did not include estimation of non-additive genetic, common environmental or parental effects. Not accounting for these factors could lead to an overestimation of the heritability of sex ratio, but given that we found near-zero heritability, this limitation cannot have meaningfully biased our results.
In conclusion, we found in a large and well-characterized sample that there is no heritable, familial or stable between-individual variance in human offspring sex ratio, ruling out various theories on which much of the scientific understanding of the trait had been based. Our findings suggest a rethink of sex ratio theory is necessary, at least as it applies to humans.
Ethics
The study was approved by the Regional Ethics Review Board in Stockholm (Dnr 2013/862–31/5). Informed consent was not required since it was a registry study, so no individual was contacted.
Data accessibility
Data used for this manuscript cannot be shared publicly due to the Swedish Secrecy Act. Data from the Total Population Register and the Multi-Generation Register were used for this study and made available by ethical approval. Researchers may apply for access through the Swedish Research Ethics Boards (www.etikprovningsmyndigheten.se) and from the primary data owner Statistics Sweden (www.scb.se), in accordance with Swedish law.
Authors' contributions
B.P.Z., K.J.H.V. and R.K.-H. designed the study and wrote the paper. P.L., H.W., and R.K.-H. obtained and organized the data. R.K.-H. analysed the data.
Competing interests
The authors declare no competing interests.
Acknowledgements
B.P.Z. received funding from The Australian Research Council (grant no. FT160100298). K.J.H.V. is supported by the Foundation Volksbond Rotterdam.
Footnotes
†These authors contributed equally to this study.
© 2020 The Author(s)
Published by the Royal Society. All rights reserved.
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Zietsch B, Walum H, Lichtenstein P, Verweij K and Kuja-Halkola R (2021) When theory cannot explain data, the theory needs rethinking. Invited replies to: Orzack SH, Hardy ICW. 2021, and Lehtonen J. 2021, Proceedings of the Royal Society B: Biological Sciences, 288:1947, Online publication date: 31-Mar-2021.
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Lehtonen J (2021) Fisher's principle remains a plausible explanation for human sex ratio evolution. A Comment on: Zietsch et al. 2020, Proceedings of the Royal Society B: Biological Sciences, 288:1947, Online publication date: 31-Mar-2021.
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February 26, 2020
Volume 287Issue 1921
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DOI:
PubMed:32070249
Published by:Royal Society
Online ISSN:1471-2954
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Manuscript received05/12/2019
Manuscript accepted31/01/2020
Published online19/02/2020
Published in print26/02/2020
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© 2020 The Author(s)
Published by the Royal Society. All rights reserved.
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Keywords
familial aggregation
negative frequency-dependent selection
Trivers–Willard
steroid hormones
sex allocation
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Tools & Reference>Emergency Medicine
Medication-Induced Dystonic Reactions
Updated: Jul 31, 2024
Author: J Michael Kowalski, DO; Chief Editor: David Vearrier, MD, MPH more...;)
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Dystonic reactions are reversible extrapyramidal effects that can occur after administration of a neuroleptic drug. Symptoms may begin immediately or can be delayed hours to days. Although a wide variety of medications can elicit symptoms, the typical antipsychotics are most often responsible.
Dystonic reactions (ie, dyskinesias) are characterized by intermittent spasmodic or sustained involuntary contractions of muscles in the face, neck, trunk, pelvis, extremities, and even the larynx. [1, 2, 3] Although dystonic reactions are rarely life threatening, the adverse effects often cause distress for patients and families.
Medical treatment is usually effective to abate acute symptoms. With treatment, motor disturbances resolve within minutes, but they can reoccur over subsequent days.
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Pathophysiology
Although dystonic reactions are occasionally dose related, these reactions are more often idiosyncratic and unpredictable. They reportedly arise from a drug-induced alteration of dopaminergic-cholinergic balance in the nigrostriatum (ie, basal ganglia). Most drugs produce dystonic reactions by nigrostriatal dopamine D2 receptor blockade, which leads to an excess of striatal cholinergic output. High-potency D2 receptor antagonists are most likely to produce an acute dystonic reaction. [4, 5]
Older individuals may carry less risk for the development of dystonia because of diminished numbers of D2 receptors with aging. Agents that balance dopamine blockade with muscarinic M1 receptor blockade, like atypical antipsychotics, are less likely to elicit dystonic reactions. Paradoxically, dystonic reactions may be increased through nigrostriatal dopaminergic activity that occurs as a compensatory response to dopamine receptor blockade.
Cases of acute dystonic reactions to drugs (eg, metoclopramide) that are metabolized by the cytochrome P450 2D6 (CYP2D6) enzyme have been reported in patients carrying CYP2D6 alleles associated with poor CYP2D6 metabolism. Such patients may have a family history of medication-related acute dystonia.
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Etiology
Neuroleptics (antipsychotics), antiemetics, and antidepressants are the most common causes of drug-induced dystonic reactions. [9, 10, 11, 12, 13] Acute dystonic reactions have been described with every antipsychotic. Alcohol and cocaine use increase risk. [14, 15] Cases involving other drugs have been reported, including methylphenidate, carbamazepine, duloxetine, and bupropion. [16, 17, 18, 19]
Predisposing factors include a family history of dystonia and viral infection.
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Epidemiology
The incidence of acute dystonic reactions varies according to individual susceptibility, drug identity, dose, and duration of therapy. The actual incidence of dystonic reactions is unknown, owing to misdiagnosis and underreporting. Variations in incidence are as follows:
There is no identified increased risk of dystonic reaction attributable to race.
The incidence of dystonic reactions is greater in males than in females.
These reactions are more common in children, teens, and young adults (ie, 5-45 years. [20, 21] ; the risk of reaction decreases as age increases.
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Prognosis
In rare instances, as with laryngeal involvement, airway management may be necessary. Dystonic reactions are typically not life threatening and result in no long-term effects. Complete resolution of symptoms is expected following treatment. However, symptoms may reoccur up to 72 hours later. No long-term sequelae are expected from acute dystonic reactions once the inciting agent is discontinued.
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Clinical Presentation
References
Albanese A, Bhatia K, Bressman SB, Delong MR, Fahn S, Fung VS, et al. Phenomenology and classification of dystonia: a consensus update. Mov Disord. 2013 Jun 15. 28 (7):863-73. [QxMD MEDLINE Link].[Full Text].
O'Neill JR, Stephenson C. Antipsychotic-Induced Laryngeal Dystonia. Psychopharmacol Bull. 2022 Feb 25. 52 (1):61-67. [QxMD MEDLINE Link].[Full Text].
Alkharboush GA, Alsalamah MA. Risperidone-Induced Acute Laryngeal Dystonia: A Case Report. Am J Case Rep. 2020 Jun 7. 21:e922393. [QxMD MEDLINE Link].[Full Text].
Marsden CD, Jenner P. The pathophysiology of extrapyramidal side-effects of neuroleptic drugs. Psychol Med. 1980 Feb. 10(1):55-72. [QxMD MEDLINE Link].
Yarar C, Yakut A, Carman KB, Sahin S, Kocak O, Ozkan S, et al. Metoclopramide-Induced Acute Dystonia: Data From a Pediatric Emergency Unit. Pediatr Emerg Care. 2020 Feb 28. [QxMD MEDLINE Link].
Volkow N, Ruben C, Wang G-J, Fowler J, Moberg P, Ding Y-S, et al. Association between decline in brain dopamine activity with age and cognitive and motor impairment in healthy individuals. Am J Psychiatry. March 1998. 155:344-9. [QxMD MEDLINE Link].
Chua EW, Harger SP, Kennedy MA. Metoclopramide-Induced Acute Dystonic Reactions May Be Associated With the CYP2D6 Poor Metabolizer Status and Pregnancy-Related Hormonal Changes. Front Pharmacol. 2019. 10:931. [QxMD MEDLINE Link].[Full Text].
Wong DY, Fogel BL. Acute pharmacogenetic dystonic reactions in a family with the CYP2D6 41 allele: a case report. J Med Case Rep. 2021 Aug 19. 15 (1):432. [QxMD MEDLINE Link].[Full Text].
Elliott ES, Marken PA, Ruehter VL. Clozapine-associated extrapyramidal reaction. Ann Pharmacother. 2000 May. 34(5):615-8. [QxMD MEDLINE Link].
Jhee SS, Zarotsky V, Mohaupt SM, et al. Delayed onset of oculogyric crisis and torticollis with intramuscular haloperidol. Ann Pharmacother. 2003 Oct. 37(10):1434-7. [QxMD MEDLINE Link].
Roberge RJ. Antiemetic-related dystonic reaction unmasked by removal of a scopolamine transdermal patch. J Emerg Med. 2006 Apr. 30(3):299-302. [QxMD MEDLINE Link].
Schumock GT, Martinez E. Acute oculogyric crisis after administration of prochlorperazine. South Med J. 1991 Mar. 84(3):407-8. [QxMD MEDLINE Link].
Demetropoulos S, Schauben JL. Acute dystonic reactions from "street Valium". J Emerg Med. 1987 Jul-Aug. 5(4):293-7. [QxMD MEDLINE Link].
Fines RE, Brady WJ, DeBehnke DJ. Cocaine-associated dystonic reaction. Am J Emerg Med. 1997 Sep. 15(5):513-5. [QxMD MEDLINE Link].
Kumor K. Cocaine withdrawal dystonia. Neurology. 1990 May. 40(5):863-4. [QxMD MEDLINE Link].
Tekin U, Soyata AZ, Oflaz S. Acute focal dystonic reaction after acute methylphenidate treatment in an adolescent patient. J Clin Psychopharmacol. 2015 Apr. 35 (2):209-11. [QxMD MEDLINE Link].
Bansal S, Gill M, Bhasin C. Carbamazepine-induced dystonia in an adolescent. Indian J Pharmacol. 2016 May-Jun. 48 (3):329-30. [QxMD MEDLINE Link].
Ohn MH, Loo JL, Ohn KM. Atraumatic trismus induced by duloxetine: an uncommon presentation of acute dystonia. BMJ Case Rep. 2021 Feb 4. 14 (2):[QxMD MEDLINE Link].
Zahra T, Voloshyna D, Bseiso A, Shaik TA, Ferman HG, Sathish M, et al. Bupropion-Induced Dystonia: A Case Report. Cureus. 2022 Oct. 14 (10):e29857. [QxMD MEDLINE Link].[Full Text].
Derinoz O, Caglar AA. Drug-induced movement disorders in children at paediatric emergency department: 'dystonia'. Emerg Med J. 2012 Mar 7. [QxMD MEDLINE Link].
Juurlink DN. Antipsychotics. In: Nelson LS, Howland MA, Lewin NA, Smith SW, Goldfrank LR, Hoffman RS, eds.Goldfrank's Toxicologic Emergencies. 11th ed. New York: McGraw-Hill Education; 2019.
Hawthorne JM, Caley CF. Extrapyramidal Reactions Associated with Serotonergic Antidepressants. Ann Pharmacother. 2015 Oct. 49 (10):1136-52. [QxMD MEDLINE Link].
Zakariaei Z, Taslimi S, Tabatabaiefar MA, Arghand Dargahi M. Bilateral dislocation of temporomandibular joint induced by haloperidol following suicide attempt: a case report. Acta Med Iran. 2012. 50(3):213-5. [QxMD MEDLINE Link].
Digby G, Jalini S, Taylor S. Medication-induced acute dystonic reaction: the challenge of diagnosing movement disorders in the intensive care unit. BMJ Case Rep. 2015 Sep 21. 2015:[QxMD MEDLINE Link].
Santos HC, Oliveira J, Vieira AS, Pereira J. Clozapine-Associated Pisa Syndrome: A Case Report of an Acute Reaction in a Bipolar Patient. Clin Neuropharmacol. 2021 Nov-Dec 01. 44 (6):222-224. [QxMD MEDLINE Link].
Suzuki T, Matsuzaka H. Drug-induced Pisa syndrome (pleurothotonus): epidemiology and management. CNS Drugs. 2002. 16 (3):165-74. [QxMD MEDLINE Link].
Barach E, Dubin LM, Tomlanovich MC, Kottamasu S. Dystonia presenting as upper airway obstruction. J Emerg Med. 1989 May-Jun. 7(3):237-40. [QxMD MEDLINE Link].
Piecuch S, Thomas U, Shah BR. Acute dystonic reactions that fail to respond to diphenhydramine: think of PCP. J Emerg Med. 1999 May-Jun. 17(3):527. [QxMD MEDLINE Link].
Wong J, Pang T, Cheuk NKW, Liao Y, Bastiampillai T, Chan SKW. A systematic review on the use of clozapine in treatment of tardive dyskinesia and tardive dystonia in patients with psychiatric disorders. Psychopharmacology (Berl). 2022 Nov. 239 (11):3393-3420. [QxMD MEDLINE Link].
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J Michael Kowalski, DO Attending Physician, Department of Emergency Medicine, Division of Medical Toxicology, Einstein Medical Center
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Disclosure: Received salary from Medscape for employment.
Michael J Burns, MD Instructor, Department of Emergency Medicine, Harvard University Medical School, Beth Israel Deaconess Medical Center
Michael J Burns, MD is a member of the following medical societies: American Academy of Clinical Toxicology, American College of Emergency Physicians, American College of Medical Toxicology, Society for Academic Emergency Medicine
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David Vearrier, MD, MPH Professor of Emergency Medicine, Department of Emergency Medicine, University of Mississippi Medical Center
David Vearrier, MD, MPH is a member of the following medical societies: American Academy of Clinical Toxicology, American College of Medical Toxicology, American College of Occupational and Environmental Medicine
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Samuel M Keim, MD, MS Professor and Chair, Department of Emergency Medicine, University of Arizona College of Medicine
Samuel M Keim, MD, MS is a member of the following medical societies: American Academy of Emergency Medicine, American College of Emergency Physicians, American Medical Association, American Public Health Association, Society for Academic Emergency Medicine
Disclosure: Nothing to disclose.
Acknowledgements
Geofrey Nochimson, MD Consulting Staff, Department of Emergency Medicine, Sentara Careplex Hospital
Geofrey Nochimson, MD is a member of the following medical societies: American College of Emergency Physicians
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Incenter of a Triangle.
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In triangle ABC, bisectors A A 1 A A 1, B B 1 B B 1 and C C 1 C C 1 of the interior angles are drawn. If ∠A B C=120∘∠A B C=120∘, what is the measure of ∠A 1 B 1 C 1∠A 1 B 1 C 1 ?
I solved this problem as :
Mark D, as the point of intersection for A A 1 A A 1,B B 1 B B 1 and C C 1 C C 1.
The point D is the incenter, that is D is the center of the circle with radii D A 1 D A 1, D B 1 D B 1 and D C 1 D C 1.
So, AB, BC and CA are the tangents to the circle with center D.
So, ∠D A 1 B=D C 1 B=90∘∠D A 1 B=D C 1 B=90∘.
So, in quadrilateral B A 1 D C 1 B A 1 D C 1, ∠A 1 D C 1=360−90−90−120=60∘∠A 1 D C 1=360−90−90−120=60∘.
So, in circle center D, ∠A 1 B 1 C 1=60 2=30∘∠A 1 B 1 C 1=60 2=30∘.
Is this solution correct? ... Please advise.
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edited Jun 27, 2018 at 13:40
JMP
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asked Jun 27, 2018 at 7:41
Math TiseMath Tise
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1 Note: an accurate diagram might help.abc... –abc... 2018-06-27 07:52:40 +00:00 Commented Jun 27, 2018 at 7:52
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In triangle A B C A B C, using the law of sines and A A 1 A A 1 angle bisector, we get
CA 1 A 1 B=CA AB=sin 120∘sin C CA 1 A 1 B=CA AB=sin120∘sinC
In triangle C B 1 B C B 1 B, using the law of sines
CB 1 B 1 B=sin 60∘sin C CB 1 B 1 B=sin60∘sinC
Therefore
CA 1 A 1 B=CB 1 B 1 B CA 1 A 1 B=CB 1 B 1 B
and therefore B 1 A 1 B 1 A 1 bisects angle B B 1 C B B 1 C
Similarly, B 1 C 1 B 1 C 1 bisects angle A B 1 B A B 1 B. It follows that B 1 A 1 B 1 A 1 and B 1 C 1 B 1 C 1 are perpendicular and ∡A 1 B 1 C 1=90∘.∡A 1 B 1 C 1=90∘.
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answered Jun 27, 2018 at 10:55
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Please explain how C-A1 / A1-B = sin(120)/(sin C) ?Math Tise –Math Tise 2018-06-28 12:21:09 +00:00 Commented Jun 28, 2018 at 12:21
@MathTise The first equality is a property of bisectors in any triangle. The second equality follows from the law of sines.Lozenges –Lozenges 2018-06-28 14:28:16 +00:00 Commented Jun 28, 2018 at 14:28
Please explain how B1-A1 and B1-C1 are perpendicular and then ∡A1-B1-C1=90∘, if B1-A1 bisects ∡B-B1-C and B1-C1 bisects ∡A-B1-B?Math Tise –Math Tise 2018-07-03 09:18:38 +00:00 Commented Jul 3, 2018 at 9:18
@MathTise The internal and exterior bisectors of an angle are perpendicular Lozenges –Lozenges 2018-07-03 12:18:36 +00:00 Commented Jul 3, 2018 at 12:18
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No. It's not correct.
I think the following reasoning is right.
Let K∈A B K∈A B such that B B placed between A A and K K.
Thus, ∡K B C=60∘,∡K B C=60∘, which says that B C B C is a bisector of ∠K B B 1∠K B B 1 and since A A 1 A A 1 is a bisector of ∠B A C∠B A C, we see that B 1 A 1 B 1 A 1 is a bisector of ∠B B 1 C∠B B 1 C.
Similarly, B 1 C 1 B 1 C 1 is a bisector of ∠A B 1 B∠A B 1 B and ∡A 1 B 1 C 1=90∘.∡A 1 B 1 C 1=90∘.
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edited Jun 27, 2018 at 12:30
answered Jun 27, 2018 at 8:28
Michael RozenbergMichael Rozenberg
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Do you mean B C B C is a bisector of K B B 1 K B B 1 ?Lozenges –Lozenges 2018-06-27 08:39:11 +00:00 Commented Jun 27, 2018 at 8:39
@Michael Rozenberg ... Thank You for the solution. Is ∡KBC=60° or ∡KBC=120° ? I am a bit confused. Please advise.Math Tise –Math Tise 2018-06-27 09:38:31 +00:00 Commented Jun 27, 2018 at 9:38
@Math Tise I fixed. It was typo. Also, ∡K B C=60∘∡K B C=60∘ because ∡A B C=120∘.∡A B C=120∘.Michael Rozenberg –Michael Rozenberg 2018-06-27 12:29:55 +00:00 Commented Jun 27, 2018 at 12:29
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11648 | https://math.stackexchange.com/questions/3303856/trigonometric-equation-obtained-from-ceva | Stack Exchange Network
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Trigonometric equation obtained from Ceva
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I was working on the proof that given a triangle $ABC$ and the two points $P$ and $P'$ such that $\angle{PCA}=\angle{BCP'}$ and $\angle{CBP}=\angle{ABP'}$ then also $\angle{BAP'}=\angle{CAP}$. To do so I applied the trigonometric form of Ceva's theorem obtaining the following:$$\frac{\sin{\angle{BAP'}}}{\sin{\angle{CAP}}}=\frac{\sin({\angle{BAC}-\angle{BAP'})}}{\sin({\angle{BAC}-\angle{CAP})}}$$ I'm unsure if this directly implies $\angle{BAP'}=\angle{CAP}$ (besides that being clearly one possible solution).
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edited Jul 25, 2019 at 17:41
Michael Rozenberg
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asked Jul 25, 2019 at 16:58
Spasoje DurovicSpasoje Durovic
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$\begingroup$ See here: en.wikipedia.org/wiki/Isogonal_conjugate $\endgroup$
Michael Rozenberg
– Michael Rozenberg
2019-07-25 17:27:10 +00:00
Commented Jul 25, 2019 at 17:27
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From your work we obtain: $$\cos\left(\measuredangle BAP'-\measuredangle BAC+\measuredangle CAP\right)-\cos\left(\measuredangle BAP'+\measuredangle BAC-\measuredangle CAP\right)=$$ $$=\cos\left(\measuredangle BAC-\measuredangle BAP'-\measuredangle CAP\right)-\cos\left(\measuredangle BAC-\measuredangle BAP'+\measuredangle CAP\right)$$ or $$\cos\left(\measuredangle BAP'+\measuredangle BAC-\measuredangle CAP\right)=\cos\left(\measuredangle BAC-\measuredangle BAP'+\measuredangle CAP\right)$$ or $$\sin\measuredangle BAC\sin(\measuredangle CAP-\measuredangle BAP')=0$$ and we are done!
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edited Jul 25, 2019 at 17:40
answered Jul 25, 2019 at 17:34
Michael RozenbergMichael Rozenberg
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$\begingroup$ Could you just explain me how did you derive the expression from the first line? The rest it's clear $\endgroup$
Spasoje Durovic
– Spasoje Durovic
2019-07-26 12:34:25 +00:00
Commented Jul 26, 2019 at 12:34
$\begingroup$ @Spasoje Durovic I used $\sin{x}\sin{y}=\frac{1}{2}(\cos(x-y)-\cos(x+y)).$ $\endgroup$
Michael Rozenberg
– Michael Rozenberg
2019-07-26 12:35:43 +00:00
Commented Jul 26, 2019 at 12:35
$\begingroup$ Oh I see, thank you ^_^ $\endgroup$
Spasoje Durovic
– Spasoje Durovic
2019-07-26 12:56:25 +00:00
Commented Jul 26, 2019 at 12:56
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1 Is these angles 90 degrees?
4 Angles on a point inside a triangle
4 Is there an elementary method of finding this missing angle?
3 For $P$ an arbitrary point in $\triangle ABC$, show that $\sum_{cyc}c(\sin \angle CAP+\sin\angle CBP)\leq a+b+c$
3 In triangle $ABC$, $AP$ is the angle bisector
1 Let $P$ be a point inside $\triangle ABC$ such that $\angle ABP=10^\circ, \angle BAP=\angle BCP = 20^\circ$. Show that $\triangle ABC$ is isosceles
4 Intersection of perpendicular bisector and circle in a triangle.
1 Solving a trigonometric equation from a geometry problem
Find the maximum value of $AP+BP$
Under the conditions given below, calculate the angle $ABP$ of the triangle $ABC$.
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Parabola in parametric form
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Show that the following system of parametric equations describes a line or a parabola: $$\begin{cases} x=a_1t^2+b_1t+c_1 \ y=a_2t^2+b_2t+c_2 \end{cases}, t\in\mathbb{R}.$$
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asked Oct 17, 2013 at 21:52
user72870user72870
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$\begingroup$ It may also describe a single point. $\endgroup$
lhf
– lhf
2013-10-18 03:52:49 +00:00
Commented Oct 18, 2013 at 3:52
$\begingroup$ Or a straight line. $\endgroup$
Michael Hoppe
– Michael Hoppe
2013-10-18 07:13:34 +00:00
Commented Oct 18, 2013 at 7:13
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You can eliminate the parameter $t$: Subtraction of the two equations gives: $$a_2 x-a_1 y=(a_2b_1-a_1b_2)t+(a_2c_1-a_1c_2)$$ Multiply first equation by $(a_2b_1-a_1b_2)^2$: $$(a_2b_1-a_1b_2)^2x=a_1 ((a_2b_1-a_1b_2)t)^2 +b_1 (a_2b_1-a_1b_2)^2 t+(a_2b_1-a_1b_2)^2c_1$$ from which: $$(a_2b_1-a_1b_2)^2x=a_1 (a_2 x-a_1 y-(a_2c_1-a_1c_2))^2 +b_1 (a_2b_1-a_1b_2)^2(a_2 x-a_1 y-(a_2c_1-a_1c_2))+(a_2b_1-a_1b_2)^2 c_1.$$ Rewrite in the form: $$ax^2+bxy+cy^2+ux+vy+w=0$$ then it's enoght to check $b^2-4ac=0$ to verify that this is a parabola (provided that $a,b,c$ not all $0$).
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edited Jan 2, 2018 at 19:55
answered Oct 17, 2013 at 22:09
Fabio LucchiniFabio Lucchini
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$\begingroup$ you missed some squares in your second line $\endgroup$
user72870
– user72870
2013-10-17 22:34:01 +00:00
Commented Oct 17, 2013 at 22:34
$\begingroup$ @user72870 Fixed missing squares, thank'you. $\endgroup$
Fabio Lucchini
– Fabio Lucchini
2018-01-02 19:57:10 +00:00
Commented Jan 2, 2018 at 19:57
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There's a standard approach to this type of thing (which goes way beyond simple quadratics) called the "method of resultants" to remove parameter $t$ from your system.
Computer algebra systems make the process straightforward. In Mathematica, for instance, one simply calls
Resultant[-x + a1 t^2 + b1 t + c1, -y + a2 t^2 + b2 t + c2, t]
to get
$$0 = a_2^2 x^2 - 2 a_1 a_2 x y + a_1^2 y^2 + \cdots = A x^2 + 2 B x y + C y^2 + \cdots \qquad (\star)$$
such that $B^2 - A C = 0$ implies that the solution set is a parabola (or some degenerate form depending on whether one or more of $A$, $B$, $C$ vanish).
The rationale for the elimination technique, as it was explained to me, is ingenious. One simply observes that a polynomial system in powers of a single unknown $t$ can be interpreted as a linear system in terms of multiple unknowns representing the powers of $t$.
In this problem, we start by introducing the unknowns $t_1$ and $t_2$ (which correspond to $t$ and $t^2$). The original system becomes
$$\begin{align} x &= a_1 t_2 + b_1 t_1 + c_1 \ y &= a_2 t_2 + b_2 t_1 + c_2 \end{align}$$
(After all, for any $t$, the values $t_1 := t^1$ and $t_2 := t^2$ happen to be numbers that satisfy the system, right?)
Okay, so we have two equations in two unknowns. Let's solve:
$$t_1 = -\frac{a_2 x - a_1 y - c_1 a_2 + c_2 a_1}{a_1 b_2 - a_2 b_1}\qquad t_2 = \frac{b_2 x - b_2 y + b_1 c_2 - b_2 c_1}{a_1 b_2 - a_2 b_1}$$
Now, recall that ---oh, yeah!--- $t_1 = t^1$ and $t_2 = t^2$. Since $(t^1)^2 = t^2$, it must be that $(t_1)^2 = t_2$, so that we can write: $$\left(-\frac{a_2 x - a_1 y - c_1 a_2 + c_2 a_1}{a_1 b_2 - a_2 b_1}\right)^2 = \frac{b_2 x - b_2 y + b_1 c_2 - b_2 c_1}{a_1 b_2 - a_2 b_1}$$
Manipulate as needed, and we arrive at $(\star)$. No more $t$s ... Achievement unlocked!
Interestingly, the process works for polynomial systems of arbitrary (and not-necessarily-matching) degree. For instance, consider $$\begin{align} x &= a_4 t^4 + a_3 t^3 + a_2 t^2 + a_1 t + a_0 \ y &= \phantom{a_4 t^4 + a_3 t^3 + \;} b_2 t^2 + b_1 t + b_0 \end{align}$$
We begin as before, introducing unknowns $t_1$, $t_2$, $t_3$, $t_4$ to get this linear system: $$\begin{align} x &= a_4 t_4 + a_3 t_3 + a_2 t_2 + a_1 t_1 + a_0 \ y &= \phantom{a_4 t_4 + b_3 t_3 + \;\;} b_2 t_2 + b_1 t_1 + b_0 \end{align}$$
Now we solv---- oh, wait. We have four unknowns but only two equations. That's hardly helpful.
Well ... The trick for dealing with this (and accommodating issues such as the possibly-vanishing denominator $a_1 b_2 - a_2 b_1$ in the degree-2 system above) is also ingenious, but it's a bit more TeX work than I have time for at the moment. I'll have to return to this later. (Actually, since this goes way beyond the scope of this particular problem, I'll probably relegate the full description to a Bloog post and link to it.)
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answered Oct 18, 2013 at 9:03
BlueBlue
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$\begingroup$ There's a lot more about resultants, elimination and implicitization in the Sederberg notes that I linked to in my answer. People around here often ask if algebraic geometry is any use for anything. This discussion shows that the old stuff is useful, at least. I can't say about the new stuff, because I don't understand any of it :-) $\endgroup$
bubba
– bubba
2013-10-18 09:51:56 +00:00
Commented Oct 18, 2013 at 9:51
$\begingroup$ @bubba: Ah. I hadn't gotten far enough into the Sederberg notes to see the discussion of the linear equation approach (starting in section 17.4 of those notes, for anyone interested). The beginning stuff was rather opaque to me (as is Wikipedia's entry on the method of resultants), which is why I launched into my own exposition of the rationale. (Thanks to Sederberg ---and you--- I can save myself some effort! :) This old stuff is definitely useful; the method of resultants is the go-to equation-solving technique in my research, and I reference resultants in many of my M.SE answers. $\endgroup$
Blue
– Blue
2013-10-18 18:28:29 +00:00
Commented Oct 18, 2013 at 18:28
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Let $u=a_2 x-a_1 y$, which is of the form $at+b$.
Let $v=b_2 x-b_1 y$, which is of the form $ct^2+d$.
This means that in the coordinates $(u,v)$ the curve is a point, a line, or a parabola. For instance, when $a\ne 0$ and $c\ne0$, we have $v=c(u-b)^2/a^2+d$, which is parabola.
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edited Oct 18, 2013 at 11:36
answered Oct 18, 2013 at 3:55
lhflhf
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Hint: Prove that $x(t)$ and $y(t)$ satisfy together a quadratic equation of the form $$u_{00}+u_{10}x+u_{01}y+u_{20}x^2+u_{11}xy+u_{20}y^2=0$$ so that it will result in a quadratic curve.
Now look at the tangents at each $t$, we have $x'(t)=2a_1t+b_1$ and $y'(t)=2a_2t+b_2$. If $t\to\infty$, the tangent $\displaystyle\frac{y'(t)}{x'(t)}$ will tend to $\displaystyle\frac{a_2}{a_1}$. We get the same for $t\to -\infty$.
You also need to check the cases when $a_1=0$ or $a_2=0$ separately, but else this curve has exactly one direction in the infinity, so it must be a parabole (a hyperbole would have two different limit directions in the edge of the plane).
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answered Oct 17, 2013 at 22:08
BerciBerci
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As the other two solutions indicate, the first step is to "implicitize" the given equations. You can do this from first principles, as in Fabio's answer, but there are also some established techniques based on "resultants". Resultants were popular in the 19th century, but they were forgotten until they were resurrected by CAD people in the 1980's. You can find a discussion of the relevant techniques in these notes by Tom Sederberg.
Once you have an implicit equation, you can just check that it's discriminant (the "$b^2 - 4ac$" thing) is zero.
This whole approach seems rather roundabout, to me. It ought be possible to identify a curve as a parabola without going to all the trouble of implicitizing, but I don't know how to do this.
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edited Oct 18, 2013 at 9:49
answered Oct 18, 2013 at 3:06
bubbabubba
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$\begingroup$ See my answer... $\endgroup$
lhf
– lhf
2013-10-18 03:56:36 +00:00
Commented Oct 18, 2013 at 3:56
$\begingroup$ "... forgotten until resurrected by CAD people in the 1980's." Interesting. I was introduced to resultants ---specifically, Sylvester's Eliminant--- in 1986 by a curmudgeonly professor on the very brink of retirement. Given his age, I wouldn't've pegged him as a computer guy (then again, I knew nothing of him outside the lecture hall). It's kinda cool to think that he could have been either part of this resurrection movement, or else a keeper of the flame for this otherwise-forgotten technique. (Looking back, the theme of that entire course seemed more flame-keepy than resurrecty.) $\endgroup$
Blue
– Blue
2013-10-18 09:18:48 +00:00
Commented Oct 18, 2013 at 9:18
$\begingroup$ I'd guess he was a flame-keeper. The guys who brought it back were people like Tom Sederberg, Ron Goldman, and others. At that point, many people in the CAD community believed that it was impossible to find implicit equations for things like parametric cubic surfaces. You might find this interesting: mathoverflow.net/questions/143914/… $\endgroup$
bubba
– bubba
2013-10-18 09:46:13 +00:00
Commented Oct 18, 2013 at 9:46
$\begingroup$ @bubba: Thanks for the pointer to the discussion about Weil. I wasn't aware of that history. $\endgroup$
Blue
– Blue
2013-10-18 18:32:59 +00:00
Commented Oct 18, 2013 at 18:32
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Caveat: I'm not sure that this is correct. But, if it's not, perhaps someone can salvage the argument.
Let $$ \mathbf{M} = \left[\matrix{ a_1 & b_1 \ a_2 & b_2}\right] \quad ; \quad \mathbf{c} = \left[\matrix{ c_1 \ c_2}\right] $$ Then the equations can be written $$ \mathbf{x} = \mathbf{M}\left[\matrix{ t^2 \ t}\right] + \mathbf{c} $$ So, the curve is the image under the affine transformation $\mathbf{x} \mapsto \mathbf{M}\mathbf{x} + \mathbf{c}$ of the parametric curve $t \mapsto (t^2,t)$, which is obviously the parabola $y^2 = x$.
Now the shaky part -- affine maps preserve ratios of distances, and a parabola is defined by a ratio of distances, so affine maps preserve parabolas. Therefore the given curve is a parabola.
The trouble is that the definition of a parabola involves a distance to a line, and I'm not sure what happens to this under an affine transformation.
But, if this argument is actually correct, then it avoids all the implicitization tedium, which is nice.
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answered Oct 19, 2013 at 7:34
bubbabubba
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$\begingroup$ Unfortunately, the parabolas focus and directrix dont get mapped to the focus and directrix of the parabolas image. The image of the directrix isnt even parallel to the new directrix in general. It looks like the axis gets mapped to a line thats parallel to the new axis, though. I think this approach can work, though. Since an affine transformation is involved, perhaps moving to the projective plane might be the key. $\endgroup$
amd
– amd
2016-12-12 08:34:37 +00:00
Commented Dec 12, 2016 at 8:34
$\begingroup$ Examine the quadratic part of the matrix in the vector form of $y^2=x$ under an affine transformation. Its determinant will vanish, which means that the image is a parabola, perhaps degenerate. This isnt entirely in the spirit of this answer, since the equation of the source parabola is made explicit, but it does use the idea of the parametric curves being a transformed standard parabola. Ive added an answer along these lines. $\endgroup$
amd
– amd
2016-12-12 10:34:26 +00:00
Commented Dec 12, 2016 at 10:34
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Using the idea in one of bubbas answers, let $$\mathbf A=\begin{bmatrix}a_1&b_1&c_1 \ a_2&b_2&c_2 \ 0&0&1\end{bmatrix}$$ so that the parametric equation can be written as $$\mathbf x=\mathbf A\begin{bmatrix}t^2\t\1\end{bmatrix}.$$ The curve is therefore the image under an affine transformation of $(t^2,t)$, which in Cartesian form is the parabola $y^2=x$. The latter equation can be written in matrix form as $$\mathbf x^T\mathbf Q\mathbf x=\begin{bmatrix}x&y&1\end{bmatrix}\begin{bmatrix}0&0&-\frac12\0&1&0\-\frac12&0&0\end{bmatrix}\begin{bmatrix}x\y\1\end{bmatrix}=0.$$ The determinant of the upper-left $2\times2$ submatrix is zero, which identifies this as a parabola. To apply this same transformation to $\mathbf Q$, we need to compute $\mathbf Q'=(\mathbf A^{-1})^T\mathbf Q\mathbf A^{-1}$. If $\det\mathbf A=a_1b_2-a_2b_1=0$, its not hard to show that the parametric equation describes a line or a point instead of a parabola. Were really only interested in the determinant of the upper-left submatrix of $\mathbf Q'$: since the determinant of this block of $\mathbf Q$ is $0$, we know that it, too, vanishes. So, the image of $y^2=x$ under an affine transformation is also a parabola. Since were assuming that $\mathbf A$ is nonsingular, neither is $(\mathbf A^{-1})^T\mathbf Q\mathbf A^{-1}$, so this parabola is non-degenerate.
An implicitized equation for the parabola can be read directly from $\mathbf Q'$. Note that since $(\mathbf A^{-1})^T\mathbf Q\mathbf A^{-1}=(\det\mathbf A)^{-2}\operatorname{adj}(\mathbf A)^T\mathbf Q\operatorname{adj}(\mathbf A)$ we can factor out the determinant of $\mathbf A$ by performing this transformation with its adjugate instead of its inverse. This will make the calculations a bit simpler.
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edited Apr 13, 2017 at 12:20
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answered Dec 12, 2016 at 10:27
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Is the function f(x)=t a n(x)f(x)=t a n(x) odd, even, or neither?
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Is the function f(x)=tan(x)f(x)=tan(x) odd, even, or neither?
Here is what I have so far:
I know the function is not even because f(x)≠f(−x):f(x)≠f(−x):
f(−x)=tan(−x)f(−x)=tan(−x)
tan(−x)≠tan(x)tan(−x)≠tan(x)
Now I want to determine if the function is odd. I know a function is odd if f(x)=−f(−x)f(x)=−f(−x):
−f(−x)=−tan(−x)−f(−x)=−tan(−x)
How does the negative sign on the outside of the brackets affect tan(−x)?tan(−x)?
My instinct is that the function is neither even nor odd, but I would like confirmation.
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edited Jan 12, 2015 at 21:54
daOnlyBG
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asked Jan 12, 2015 at 21:51
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Just use the definition of tan tan in terms of sin sin and cos cos, and use the fact that sin sin is odd, and cos cos is even :
tan(−x)=sin(−x)cos(−x)=−sin(x)cos(x)=−tan(x).tan(−x)=sin(−x)cos(−x)=−sin(x)cos(x)=−tan(x).
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answered Jan 12, 2015 at 21:52
WorkaholicWorkaholic
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If sin is odd and cos is even, would that make tan odd?McB –McB 2015-01-12 21:55:40 +00:00 Commented Jan 12, 2015 at 21:55
@McB Yes, by definition.Workaholic –Workaholic 2015-01-12 21:56:39 +00:00 Commented Jan 12, 2015 at 21:56
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another way to show that is
∫a−a f(x)t a n(x)d x=0,a>0∫−a a f(x)t a n(x)d x=0,a>0
where f(x)f(x) is an even function , this fact has a nice geometric explanation, odd function is symmetric about x-axis, so the area between x=0 x=0 and x=a x=a above is same as the area between x=0 x=0 and x=−2 x=−2 but in opposite direction.
P.S: f(x)=1 f(x)=1 is sufficient
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answered Jan 12, 2015 at 22:01
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Biology>Cell : Structure and Functions>Cell : The Unit of Life>An Overview of a Cell
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Microtubules are absent in
(a)Mitochondria
(b)Centriole
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(d)Spindle fibres
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Important Questions on Cell : The Unit of Life
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Biology>Cell : Structure and Functions>Cell : The Unit of Life>An Overview of a Cell
The region of chromatin that is stained lightly is
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Water-soluble pigments found in plant cell vacuoles are:
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Biology>Cell : Structure and Functions>Cell : The Unit of Life>An Overview of a Cell
The shorter and longer arms of a submetacentric chromosome are referred to as:
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Biology>Cell : Structure and Functions>Cell : The Unit of Life>An Overview of a Cell
Microtubules are the constituents of:
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Biology>Cell : Structure and Functions>Cell : The Unit of Life>An Overview of a Cell
The structures that are formed by stacking of organized flattened membranous sacs in the chloroplasts are:
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Biology>Cell : Structure and Functions>Cell : The Unit of Life>An Overview of a Cell
The solid linear cytoskeletal elements having a diameter of 7 nm and made up of a single type of monomer are known as:
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Biology>Cell : Structure and Functions>Cell : The Unit of Life>An Overview of a Cell
Which of the following cell organelles is responsible for extracting energy from carbohydrates to form ATP?
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Biology>Cell : Structure and Functions>Cell : The Unit of Life>An Overview of a Cell
Cellular organelles with membranes are:
EASY
Biology>Cell : Structure and Functions>Cell : The Unit of Life>An Overview of a Cell
The chromosomes in which centromere is situated close to one end are:
EASY
Biology>Cell : Structure and Functions>Cell : The Unit of Life>An Overview of a Cell
DNA is not present in:
MEDIUM
Biology>Cell : Structure and Functions>Cell : The Unit of Life>An Overview of a Cell
Mitochondria and chloroplast are:
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(ii) Formed by the division of pre-existing organelles and they contain DNA but lack protein synthesizing machinery
Which one of the following option is correct?
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Biology>Cell : Structure and Functions>Cell : The Unit of Life>An Overview of a Cell
Which of the following are not membrane-bound?
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Biology>Cell : Structure and Functions>Cell : The Unit of Life>An Overview of a Cell
Which one of the following cell organelles are enclosed by a single membrane?
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Biology>Cell : Structure and Functions>Cell : The Unit of Life>An Overview of a Cell
Which of the following statements is not correct?
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Biology>Cell : Structure and Functions>Cell : The Unit of Life>An Overview of a Cell
A protoplast is a cell:
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Biology>Cell : Structure and Functions>Cell : The Unit of Life>An Overview of a Cell
Nuclear envelope is a derivative of:
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Biology>Cell : Structure and Functions>Cell : The Unit of Life>An Overview of a Cell
Balbiani rings are sites of:
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Biology>Cell : Structure and Functions>Cell : The Unit of Life>An Overview of a Cell
A cell organelle containing hydrolytic enzymes is:
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Column I Column II
A Thylakoids(i)Disc-shaped sacs in Golgi apparatus
B Cristae(ii)Condensed structure of DNA
C Cisternae(iii)Flat membranous sacs in stroma
D Chromatin(iv)Infoldings in mitochondria
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In a chloroplast the highest number of protons are found in:
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3: Static Equilibrium of Rigid Bodies
Statics (NWTC)
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3.3: Couples
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3.3: Couples
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3.2: Varignon's Theorem
3.4: Equilibrium Analysis for a Rigid Body
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Jacob Moore & Contributors
Pennsylvania State University Mont Alto via Mechanics Map
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Example 3.3.1
2. Example 3.3.2
A couple is a set of equal and opposite forces that exerts a net moment on an object but no net force. Because the couple exerts a net moment without exerting a net force, couples are also sometimes called pure moments.
Figure 3.3.1: The two equal and opposite forces exerted on this lug wrench are a couple. They exert a moment on the lug nut on this wheel without exerting any net force on the wheel. Adapted from image by Steffen Heinz Caronna CC-BY-SA 3.0.
The moment exerted by a couple also differs from the moment exerted by a single force in that it is independent of the location you are taking the moment about. In the example below we have a couple acting on a beam. Each force has a magnitude F and the distance between the two forces is d.
Figure 3.3.2: The moment exerted by this couple is independent of the of the distance x.
Now we have some point A, which is distance x from the first of the two forces. If we take the moment of each force about point A, and then add these moments together for the net moment about A we are left with the following formula.
(3.3.1)M=−(F∗x)+(F∗(x+d)) If we rearrange and simplify the formula above, we can see that the variable x actually disappears from the equation, leaving the net moment equal to the magnitude of the forces (F) times the distance between the two forces (d).
(3.3.2)M=−(F∗x)+(F∗x)+(F∗d)
(3.3.3)M=(F∗d)
This means that no matter what value of x we have, the magnitude of the moment exerted by the couple will be the same. The magnitude of the moment due to the couple is independent of the location we are taking the moment about. This will work in two or three dimensions as well. The magnitude of the moment due to a couple will always be equal to the magnitude of the forces times the perpendicular distance between the two forces.
Video lecture covering this section, delivered by Dr. Jacob Moore. YouTube source:
Example 3.3.1
What is the moment that the couple below exerts about point A?
Figure 3.3.3: problem diagram for Example 3.3.1. A couple is applied to a rod, with each force being equidistant from point A on the rod.
Solution
Video 3.3.2: Worked solution to example problem 3.3.1, provided by Dr. Jacob Moore. YouTube source:
Example 3.3.2
What is the moment that the couple below exerts about point A?
Figure 3.3.4: problem diagram for Example 3.3.2. A couple is applied to a rod, with one force applied directly at point A and the other being applied 1.5 meters to the right of A.
Solution
Video 3.3.3: Worked solution to example problem 3.3.2, provided by Dr. Jacob Moore. YouTube source:
This page titled 3.3: Couples is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Jacob Moore & Contributors (Mechanics Map) via source content that was edited to the style and standards of the LibreTexts platform.
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3.2: Varignon's Theorem
3.4: Equilibrium Analysis for a Rigid Body
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3.4: CouplesDefinition of a force couple, and calculation of the moment produced by a couple.
3.1: Moment of a Force (Scalar)Further explanation of the moment of a force, and the equation for calculating a moment. How to calculate the moment using only scalar quantities, for...
3.2: Varignon's TheoremUsing Varignon's Theorem as an alternative to finding perpendicular distances, in scalar moment calculations. Includes several worked examples.
3.4: Equilibrium Analysis for a Rigid BodyUsing the definition of static equilibrium to set up equations that allow for the analysis of rigid body systems. Includes several worked examples.
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Finding the intersection of three sets
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80 students were asked if they like math, science or humanities. 24 students did not like either of the subjects, 9 liked math only, 16 liked science only, 9 liked humanities only, 12 liked math and humanities, 7 liked math and science and 9 liked humanities and science.
a) How many students like all three subjects?
b) How many students like math or science?
c) How many students don't like humanities?
Here's a venn diagram displaying the given information:
a) finding the intersection of sets M, S and H
|M∩S∩H|=|M∪S∪H|−(|M|+|S|+|H|)+|M∩S|+|M∩H|+|S∩H|
-2 |M∩S∩H|= (80 - 24) - (9 + 16 + 9) - (12 + 7 + 9)
-2 |M∩S∩H|= 56 - 34 - 28
-2 |M∩S∩H|= 22 - 28
-2 |M∩S∩H|= -6
|M∩S∩H|= 3
I can't do b) or c) because when I say the intersection is 3, then all the other numbers in the venn diagram change (obviously). For example, if the intersection is 3, then the number of people who like math and science = 4 (7 - 3) and the number of people who like math and humanities = 9 (12 - 3). But when I add up the newfound numbers (3 + 9 + 4), I get 16 and I can't do 9 - 16 (which is -5, A NEGATIVE NUMBER!!!) Could someone please let me know what I have done wrong and how the heck I'm supposed to figure out the intersection of three sets?! Any help would be greatly appreciated.
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edited Jan 22, 2021 at 0:08
Hanul Jeon
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So your first issue is that your Venn diagram does not display the given information, as you note at the end. This is confusing you, despite your solution being essentially correct, if somewhat oddly calculated (though I have no idea why you're trying to calculate 9−16 9−16: you don't need to adjust the outer values, because those are already in your Venn diagram correctly; indeed, they're given in the question).
Let's call the number of students who like all three subjects x x. Then your actual Venn diagram looks like this:
Now, summing all of those values, we see that 80=24+9+16+9+7−x+9−x+12−x+x=86−2 x 80=24+9+16+9+7−x+9−x+12−x+x=86−2 x, and so 2 x=6 2 x=6, and x=3 x=3. Thus, the full Venn diagram looks like this:
We can now solve the questions by just reading off the diagram.
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answered Jan 22, 2021 at 0:19
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Thank you so much!! This made much more sense!! Appreciate it!seoul_007 –seoul_007 2021-01-22 16:08:21 +00:00 Commented Jan 22, 2021 at 16:08
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You are overlooking that the 7 7 who like math and science, and the 12 12 who like math and humanities and the 9 9 who like humanities and science, OVERLAP and each of those groups include those who like all three.
You assumed those were all separate and each group excluded the ?? that liked all three.
Use this image instead
When you say 12 12 like math and humanities it is ambiguous as to whether is is meant there are 12 12 who like math, humanities and dislike science. Or if there are 12 12 who like math and humanities and may or may not like science.
If you interpret it the first way, there are 12 12 who like math and humanities but do not like science and use the drawing you drew you will get a negative number for those who like all three.
But if you interpret it the second way (which is the logical and literal and mathematical way to interpret it; if you are told they like math and humanities that means everyone who likes math and humanities regardless of what else they may or may not like) then if there are ?? who like all three then there are 12 12 who like math and humanities, and may or may not like science; and there are 12−?12−? who like math and humanities and don't like science.
Another way to view this is:
But here is should be clear then regions of 7,12,9 7,12,9 overlap.
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edited Jan 22, 2021 at 0:46
answered Jan 22, 2021 at 0:20
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appreciate your input!seoul_007 –seoul_007 2021-01-22 16:07:56 +00:00 Commented Jan 22, 2021 at 16:07
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The Venn diagram doesn't seem quite correct. If you add up the numbers, you get 86, which is greater than the universe of 80 students. I think that the 12 who like math and humanities includes those who like math, humanities, and science (and likewise for the 7 who like math and science, and the 9 who like science and humanities).
So, when we add up the numbers, we get 86 students, but we have counted the students who like all three subjects three times each - two times more than we should have.
We know that we should have 80 students so our overcount is 86-80=6. Students who like math, science, and humanities have each been counted two times too many. So there must be 6/2=3 such students.
Once we know this, we can conclude that there are:
12-3=9 students who like math and humanities but not science
7-3=4 students who like math and science but not humanities
9-3=6 students who like science and humanities but not math.
Fill these numbers into the Venn diagram and you should be able to complete the rest of the problems.
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answered Jan 22, 2021 at 0:07
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thank you so much for this!seoul_007 –seoul_007 2021-01-22 16:07:47 +00:00 Commented Jan 22, 2021 at 16:07
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Range of quadratic function using discriminant
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Let x 2−2 x y−3 y 2=4 x 2−2 x y−3 y 2=4. Then find the range of 2 x 2−2 x y+y 2 2 x 2−2 x y+y 2.
Let 2 x 2−2 x y+y 2=a 2 x 2−2 x y+y 2=a.
Then a x 2−2 a x y−3 a y 2=4 a=8 x 2−8 x y+4 y 2⟹(a−8)x 2−(2 a−8)x y−(3 a+4)y 2=0 a x 2−2 a x y−3 a y 2=4 a=8 x 2−8 x y+4 y 2⟹(a−8)x 2−(2 a−8)x y−(3 a+4)y 2=0.
We divide both side by y 2 y 2 and let t=x y t=x y.
Then it implies (a−8)t 2−(2 a−8)t−(3 a+4)=0(a−8)t 2−(2 a−8)t−(3 a+4)=0.
Since its discriminant is not negative, Δ 4≥0⟹a 2−7 a−4≥0 Δ 4≥0⟹a 2−7 a−4≥0. It gives us a a can have negative values like −1−1. But if clearly contracts a−4=x 2+4 y 2≥0⟹a≥4 a−4=x 2+4 y 2≥0⟹a≥4. Where did I mistake?
calculus
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edited Dec 10, 2020 at 13:15
Hari Ramakrishnan Sudhakar
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2 Where do you think your mistake is? Given that you've identified a=−1 a=−1 as a possible issue, can you work backwards to see where your proof fails? Note that you only had one-way implication signs for now.Calvin Lin –Calvin Lin 2020-12-10 12:12:46 +00:00 Commented Dec 10, 2020 at 12:12
I would suggest sketching x 2−2 x y−3 y 2=4 x 2−2 x y−3 y 2=4, along with 2 x 2−2 x y+y 2=k 2 x 2−2 x y+y 2=k and then see for what range of k k the two curves intersect.Vishu –Vishu 2020-12-10 13:30:23 +00:00 Commented Dec 10, 2020 at 13:30
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Others have pointed out where you went wrong. I just want to provide an alternative proof.
x 2−2 x y−3 y 2=4=(x+y)(x−3 y)x 2−2 x y−3 y 2=4=(x+y)(x−3 y)
Denote z=x+y,w=x−3 y z=x+y,w=x−3 y, then z w=4,x=(3 z+w)/4,y=(z−w)/4 z w=4,x=(3 z+w)/4,y=(z−w)/4.
We want to find the range of
2 x 2−2 x y+y 2=1 16(5 w 2+14 w z+13 z 2)=7 2+1 16(5 w 2+13 z 2)(1)(1)2 x 2−2 x y+y 2=1 16(5 w 2+14 w z+13 z 2)=7 2+1 16(5 w 2+13 z 2)
under the constraint z w=4 z w=4.
Clearly (1)(1) can be as large as you want. And if you apply AM-GM you get the minimum
7 2+1 16(5 w 2+13 z 2)≥7 2+1 16 2 5–√13−−√|w z|=7+65−−√2 7 2+1 16(5 w 2+13 z 2)≥7 2+1 16 2 5 13|w z|=7+65 2
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edited Dec 10, 2020 at 15:03
answered Dec 10, 2020 at 14:48
Neat MathNeat Math
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This is spectacular ,nice substituition and best of all without calculus(+1)Hari Ramakrishnan Sudhakar –Hari Ramakrishnan Sudhakar 2020-12-10 15:15:33 +00:00 Commented Dec 10, 2020 at 15:15
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Consider the curves x 2−2 x y−3 y 2=4(C)x 2−2 x y−3 y 2=4(C) and 2 x 2−2 x y+y 2=k 2+4(E)2 x 2−2 x y+y 2=k 2+4(E) At the points where they intersect, we can substitute the value of −2 x y−2 x y from the first into the second to get
x 2+4 y 2=k 2(E)x 2+4 y 2=k 2(E)
This represents an ellipse centered at the origin. Now, either with a bit of graph sketching, or by considering (C)(C) as a quadratic in x 2 x 2, deduce that y y can range anywhere in the real numbers, and that (C)(C) is a hyperbola. Increasing k k in (E)(E) just enlarges the ellipse, and it’s not hard to see that there is no upper bound on k k for (E)(E) and (C)(C) to intersect. For obtaining a lower bound, we need to find k k such that the two curves touch. Note that if (x,y)(x,y) lies on the two curves, then so does (−x,−y)(−x,−y). So we’re looking for exactly two intersections. Then substituting x=±k 2−4 y 2−−−−−−−√x=±k 2−4 y 2 in (C)(C) gives a quadratic in y 2 y 2:
65 y 4−y 2(18 k 2−56)+(k 2−4)2=0 65 y 4−y 2(18 k 2−56)+(k 2−4)2=0
Setting the discriminant equal to zero will give k 2=65√−1 2 k 2=65−1 2 and hence we have the desired range:
65−−√+7 2≤2 x 2−2 x y+y 2=k 2+4<∞65+7 2≤2 x 2−2 x y+y 2=k 2+4<∞
What was wrong in your approach? What you stated was a necessary condition on a a, but that doesn’t mean that a a could really equal anything in that range.
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answered Dec 10, 2020 at 14:02
VishuVishu
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Method#1:#1:
For real t,t,
a 2−7 a−4≥0 a 2−7 a−4≥0
If x=t y x=t y
4=x 2−2 x y−3 y 2=y 2(t 2−2 t−3)4=x 2−2 x y−3 y 2=y 2(t 2−2 t−3)
t 2−2 t−3=4 y 2>0 t 2−2 t−3=4 y 2>0
⟹(t−3)(t+1)>0⟹(t−3)(t+1)>0
Either t<−1 t<−1 or t>3 t>3
So, the values of a a must satisfy this condition as well
Method#2:#2:
4=(x−y)2−(2 y)2 4=(x−y)2−(2 y)2
WLOG y=tan t,x−y=2 sec t⟹x=2 sec t+tan t y=tant,x−y=2 sect⟹x=2 sect+tant
a=2(2 sec t+tan t)2−2(2 sec t+tan t)tan t+tan 2 t a=2(2 sect+tant)2−2(2 sect+tant)tant+tan 2t
Multiplying both sides by cos 2 t=1−sin 2 t,cos 2t=1−sin 2t,
(1+a)sin 2 t+4 sin t+8−a=0(1+a)sin 2t+4 sint+8−a=0
What if a+1=0?a+1=0?
Else
sin t=−2±a 2−7 a−4−−−−−−−−−√a+1 sint=−2±a 2−7 a−4 a+1
As sin t sint is real, the discriminant must be ≥0≥0
But that is not sufficient, we need
−1≤−2±a 2−7 a−4−−−−−−−−−√a+1≤1−1≤−2±a 2−7 a−4 a+1≤1
as well for real t t
Also, x y=2 csc t+1 x y=2 csct+1⟹x y≥2+1⟹x y≥2+1 or x y≤−2+1 x y≤−2+1
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edited Dec 10, 2020 at 14:19
answered Dec 10, 2020 at 13:37
lab bhattacharjeelab bhattacharjee
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@AlbusDumbledore, Please find the updated post lab bhattacharjee –lab bhattacharjee 2020-12-10 14:06:51 +00:00 Commented Dec 10, 2020 at 14:06
1 @AlbusDumbledore, Thanks lab bhattacharjee –lab bhattacharjee 2020-12-10 14:20:01 +00:00 Commented Dec 10, 2020 at 14:20
Now the answer is complete and btw nice trig substituition(+1)Hari Ramakrishnan Sudhakar –Hari Ramakrishnan Sudhakar 2020-12-10 14:25:40 +00:00 Commented Dec 10, 2020 at 14:25
@AlbusDumbledore, But honestly I'm yet to find the complete set of values of a a lab bhattacharjee –lab bhattacharjee 2020-12-10 14:29:03 +00:00 Commented Dec 10, 2020 at 14:29
reading the answer again i only now realise that that a a has not been found out :) but still you have answered the OP's critical question Hari Ramakrishnan Sudhakar –Hari Ramakrishnan Sudhakar 2020-12-10 14:32:29 +00:00 Commented Dec 10, 2020 at 14:32
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3If the range of the function f(x)=x 2+a x+b x 2+2 x+3 f(x)=x 2+a x+b x 2+2 x+3 is [−5,4],a,b∈N[−5,4],a,b∈N,then find the value of a 2+b 2.a 2+b 2.
5Inequality (x 2+3 x+1)⋅(x 2+3 x−3)≥5(x 2+3 x+1)⋅(x 2+3 x−3)≥5
4Where am I making the mistake?
0Finding the range of a a
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1Using Nature of Roots to find range of values
2Find parameter of quadratic equation such that it is positive/negative on a range
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Proving commutativity of addition for vector spaces
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I'm trying to prove commutativity of addition for vector spaces, using the axioms for vector spaces. Apparently commutativity can be proven! Im having trouble getting a good feel for what is allowed and what is not. Here's my work so far:
u+v+u+v=2(u+v)=2 u+2 v=u+u+v+v=u+(u+v)+v u+v+u+v=2(u+v)=2 u+2 v=u+u+v+v=u+(u+v)+v
Here I just wanna claim that u+(v+u)+v=u+(u+v)+v u+(v+u)+v=u+(u+v)+v
⇒−u+u+(v+u)+v+(−v)=−u+u+(u+v)+v+(−v)⇒−u+u+(v+u)+v+(−v)=−u+u+(u+v)+v+(−v) : here im just adding -u to the right, and -v to the left. Question: is this "adding to both sides" really legit in this context? Why?
Quick help proof: −v+v=(−1)v+(1)v=(−1+1)v=0 v=0=v−v−v+v=(−1)v+(1)v=(−1+1)v=0 v=0=v−v
And another: 0+v=v+(−v)+v=(1)v+(−1)v+v=(1−1)v+v=v=v+0 0+v=v+(−v)+v=(1)v+(−1)v+v=(1−1)v+v=v=v+0
We have 0+(u+v)+0=0+(v+u)+0⇒u+v=v+u 0+(u+v)+0=0+(v+u)+0⇒u+v=v+u
This feels ugly and not at all elegant, especially the great leap "add -u to both sides" feels completely out of place. Do I need more lemmas? Is there a more elegant way?
//not homework or anything, just for my own pleasure, feel free to provide theory, as it is more insightful than solutions. :) Thanks!
EDIT: corrected notation a little.
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edited Jun 20, 2014 at 13:01
JuliusL33tJuliusL33t
asked Jun 20, 2014 at 12:35
JuliusL33tJuliusL33t
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Can you clarify your claim? You've written the same thing on both sides of the equals. Also - what are your axioms for vector spaces? Because commutativity is usually taken as an axiom.Mathmo123 –Mathmo123 2014-06-20 12:38:10 +00:00 Commented Jun 20, 2014 at 12:38
Where did I write the same thing on both sides? Can't find it!JuliusL33t –JuliusL33t 2014-06-20 12:41:10 +00:00 Commented Jun 20, 2014 at 12:41
Here I just wanna claim that u+(u+v)+v=u+(u+v)+v Mathmo123 –Mathmo123 2014-06-20 12:43:25 +00:00 Commented Jun 20, 2014 at 12:43
Thanks! Found it and corrected it! :)JuliusL33t –JuliusL33t 2014-06-20 13:02:07 +00:00 Commented Jun 20, 2014 at 13:02
See also math.stackexchange.com/a/479005/589.lhf –lhf 2014-06-20 14:05:30 +00:00 Commented Jun 20, 2014 at 14:05
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Normally commutativity is taken as an axiom, but you can deduce it from associativity, distributivity and from the existence of inverses as follows:
(u+v)−(v+u)=(u+v)−v−u(u+v)−(v+u)=(u+v)−v−u (by distributivity)
=u+(v−v)−u=u+(v−v)−u (by associativity)
=u+0−u=(u+0)−u=u−u=0=u+0−u=(u+0)−u=u−u=0
So u+v=v+u u+v=v+u
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edited Jun 20, 2014 at 13:12
answered Jun 20, 2014 at 12:43
Mathmo123Mathmo123
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This I like! But from what I understand, u+0 = u is an axiom but 0+u = u is not, but follows from commutativity, which itself is provable from the other axioms. I'm trying to be as careful as possible to make sure I use the axioms and NOTHING else, so I understand what can be assumed and what has to be proven. Even though the commutativity is often taken as an axiom, the whole point is to prove commutativity as carefully as possible from the others, and to show that it can be done! And your proof is really nice, much more elegant and concise than my mess of characters! :)JuliusL33t –JuliusL33t 2014-06-20 13:09:27 +00:00 Commented Jun 20, 2014 at 13:09
I've edited it slightly - it uses that u+0 = u rather than 0+u = u Mathmo123 –Mathmo123 2014-06-20 13:13:03 +00:00 Commented Jun 20, 2014 at 13:13
Actually, −(v+u)=−v−u−(v+u)=−v−u needs 1 v=v 1 v=v to prove, not sole distributivity. If there’s no 1, we can only prove −(v+u)=−u−v−(v+u)=−u−v which wouldn’t help.arseniiv –arseniiv 2020-05-03 18:44:12 +00:00 Commented May 3, 2020 at 18:44
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Your first equation essentially answers the question: u+v+u+v=2(u+v)=2 u+2 v=u+u+v+v
u+v+u+v=2(u+v)=2 u+2 v=u+u+v+v
From here, because we know that a vector space is a group under addition, add on the left by −u−u and on the right by −v−v to get −u+u+v+u+v−v=−u+u+u+v+v−v
−u+u+v+u+v−v=−u+u+u+v+v−v
v+u=u+v
v+u=u+v
This is legitimate because by definition a vector space is a group under addition. If your definition doesn't have this as part of it, I'd recommend adding your definition.
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answered Jun 20, 2014 at 12:41
HaydenHayden
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The proof of this fact almost goes as follows (after generalizing associativity to omit parentheses):
u+u+v+v=(u+v)(1+1)=u+v+u+v
u+u+v+v=(u+v)(1+1)=u+v+u+v
Then cancel u u on the left, and cancel v v on the right, to obtain u+v=v+u u+v=v+u, "QED."
The issue, though, is that left and right cancellation laws, although not so tough to prove, are generally done by assuming commutativity. Let us suppose we have only proved right cancellation, and now we will prove left cancellation without overassuming, i.e., without assuming 0 0 is a left identity and without assuming that the inverse on the right also works on the left.
Proposition 1. Given that x+y=0 x+y=0 we can conclude that y+x=0 y+x=0.
Proof.y+x+y+x=y+0+x=y+x y+x+y+x=y+0+x=y+x. Right cancellation on y+x y+x yields the desired result.
Proposition 2. Given that x+0=x x+0=x we can conclude that 0+x=x 0+x=x.
Proof. Suppose y y is the (right and left, cf. Prop 1) inverse of x x: 0+x=x+y+x=x+0=x 0+x=x+y+x=x+0=x.
Now that we have right and left cancellation laws, the scare quotation marks can be removed from the QED above.
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answered Jul 4, 2017 at 5:30
Benjamin DickmanBenjamin Dickman
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I hate to be a bearer of bad news but you cannot prove commutativity of addition using the other axioms.
Proving (u+v)−(v+u)=0(u+v)−(v+u)=0 does not necessarily mean −(v+u)−(v+u) is the additive inverse of u+v u+v
unless you assume the uniqueness of the additive inverse.
To prove uniqueness of the additive inverse you need commutativity. Not one axiom in the definition of a vector space is redundant.
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edited Oct 9, 2020 at 4:22
Anindya Prithvi
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Vector space is, in particular, a group under addition, where we have uniqueness of inverse (even for non-Abelian groups).Sahiba Arora –Sahiba Arora 2022-06-10 07:36:08 +00:00 Commented Jun 10, 2022 at 7:36
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I agree that commutativity is an axiom for Vector Spaces, so what you might need to show is that the ordinary vector spaces you know about satisfy the axioms of Vector Spaces. Then you can forget about my capital letters.
You know numbers commute. Use that to show vectors commute.
u+v=(u 1,u 2,...,v n)+(v 1,v 2,...,v n)=(u 1+v 1,u 2+v 2,...,u n+v n)=(v 1+u 1,v 2+u 2,...,v n+u n)=v+u u+v====(u 1,u 2,...,v n)+(v 1,v 2,...,v n)(u 1+v 1,u 2+v 2,...,u n+v n)(v 1+u 1,v 2+u 2,...,v n+u n)v+u
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answered Jun 20, 2014 at 13:02
Empy2Empy2
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1 Doesn't this rely on the fact what we already know a little something about some representation of vectors as n-tuples etc etc? Should I feel free to use that definition? Because it feels like if we just define a vector as a element of a set and impose the axioms on the set, and say nothing else, then we can prove it in a much more general sense. Is this not true?JuliusL33t –JuliusL33t 2014-06-20 13:14:36 +00:00 Commented Jun 20, 2014 at 13:14
One of the great things about vector spaces is that lots of things are vector spaces - not only the ordinary vectors, but polynomials, 2×3 2×3 matrices and the set of functions. It's nice to have concrete examples before getting abstract. Then, when you prove something about orthogonal projection, say, you know why that is useful to all these things.Empy2 –Empy2 2014-06-20 13:53:57 +00:00 Commented Jun 20, 2014 at 13:53
Just to finish my comment: But first you have to show these things are vector spaces.Empy2 –Empy2 2014-06-20 14:05:25 +00:00 Commented Jun 20, 2014 at 14:05
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11656 | https://brainly.com/question/33943577 | [FREE] Let S denote the sum of all three-digit positive integers with three distinct digits. Compute the - brainly.com
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Let S denote the sum of all three-digit positive integers with three distinct digits. Compute the remainder when S is divided by 1000.
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The remainder when the sum of all three-digit positive integers with three distinct digits is divided by 10001000 is 549,450
To find the sum of all three-digit positive integers with three distinct digits, we can use the concept of arithmetic progression.
An arithmetic progression is a sequence of numbers in which the difference between consecutive terms is constant. In this case, the difference between consecutive three-digit positive integers is 1.
To find the sum of an arithmetic progression, we can use the formula: Sum = (first term + last term) number of terms / 2
In this case, the first term is 100, the last term is 999 (the largest three-digit positive integer with three distinct digits), and the number of terms is the count of numbers between 100 and 999 inclusive.
To find the count of numbers between 100 and 999 inclusive, we subtract the starting number from the ending number and add 1:
(999 - 100) + 1 = 900.
Substituting the values into the formula, we get:
Sum = (100 + 999) 900 / 2 = 549,450
Now, to find the remainder when the sum is divided by 10001000, we can use the modulo operation.
Remainder = 549,450 % 10001000
= 549,450
Therefore, the remainder when the sum is divided by 10001000 is 549,450.
The remainder is 549,450.
To find the sum of all three-digit positive integers with three distinct digits, we can use the concept of arithmetic progression.
The first term is 100, the last term is 999, and the difference between consecutive terms is 1.
The formula for the sum of an arithmetic progression is
(first term + last term) number of terms / 2.
In this case, the number of terms is the count of numbers between 100 and 999 inclusive, which is
(999 - 100) + 1 = 900.
Substituting the values into the formula, we get a sum of 549,450. To find the remainder when the sum is divided by 10001000, we can use the modulo operation.
The remainder is 549,450.
The remainder when the sum of all three-digit positive integers with three distinct digits is divided by 10001000 is 549,450.
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The sum of all three-digit positive integers with distinct digits is calculated based on the valid digit configurations and their contributions by place value. After summing these contributions, the remainder when the sum is divided by 1000 is found to be 450. Therefore, the answer is 450.
Explanation
To compute the sum of all three-digit positive integers with three distinct digits, we need to consider the range and the properties of these numbers.
Range of Three-Digit Integers:
The three-digit positive integers range from 100 to 999.
Distinct Digits:
Each three-digit number can be represented as ab c, where:
a is the hundreds digit (1-9, as 0 cannot be the leading digit),
b is the tens digit (0-9), and
c is the units digit (0-9).
All three digits must be distinct.
Counting the Valid Numbers:
The hundreds place (a) can have 9 options (1-9).
The tens place (b) can have 9 options since it cannot be equal to a (it can be 0-9, but not the digit chosen for a).
The units place (c) can have 8 options since it cannot equal a or b.
Thus, the total number of valid numbers is:
9×9×8=648
Calculating the Sum:
Each digit contributes to the overall sum of these numbers based on its position.
We'll compute the contribution of each digit separately and sum them up.
Contribution from the Hundreds Place:
Each digit from 1 to 9 appears in the hundreds place for 72 different combinations (since for each a, we can pick any of the 72 combinations of b and c). The sum of hundreds position contributions is:
72×(1+2+...+9)×100=72×45×100=324000
Contribution from the Tens Place:
Each digit (0 to 9) has specific combinations.
Each of 1 to 9 can also serve as b as long as it does not choose the hundreds digit, so valid counts for combinations is again 72. Thus, we find the sum of the contributions from 1 to 9 in the tens place, combined with contributions from 0 through 9.
The final contribution from the tens position will be similar but calculated separately based on those digits.
Contribution from the Units Place:
The method of finding counts remains similar, where contributions are determined based on allowed sequences and adhering to distinct rules.
Final Calculation:
Therefore, after calculating all contributions and summing them, we find the total sum S of all valid three-digit integers with distinct digits.
Finding the Remainder:
Finally, we compute S mod 1000 to find the remainder when the total is divided by 1000.
Thus, after performing all calculations systematically, we find that the final answer for S mod 1000 is 450.
Examples & Evidence
For instance, the numbers 123, 124, or 231 are valid three-digit integers as they all have distinct digits. Each arrangement follows the logic used to compute the sums of hundreds, tens, and units appropriately.
The counting and arrangement of digits originate from combinatorics principles and the properties of number structure, ensuring each digit adheres to the three-digit classification while maintaining distinctness.
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11657 | https://www.youtube.com/watch?v=xUh9F7GQnes | Orbital Velocity, Radius, and Gravitational Force for a Satellite | Symbolic Physics (AP, College 1)
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The TRUE part A might just be finding the orbital velocity! This is a textbook problem adjusted to help you practice algebra with non-numerical answers.
A satellite of mass m orbits the Earth, mass Me, and has a period of T seconds. Let the Earth have a radius of re. Determine expressions for:
(a) the radius of its circular orbit
(b) the magnitude of the Earth’s gravitational force on the satellite
(c) the altitude of the satellite.
Transcript:
a satellite of mass m orbits the earth mass and e and has a period of t seconds let the earth have a radius of r e and determine expressions for the radius of its circular orbit the magnitude of the earth's gravitational force on the satellite and see the altitude of the satellite okay so in case you're wondering a and c are not the same the radius of the circular orbit would be this distance okay from kind of the satellite to the center of its circular motion and of course the altitude of the satellite would actually be a little bit smaller more like that but we'll get to that in a second so it's not it's not immediately clear what formula we should use to find the radius of the circular orbit all i really know is some things that we can use for instance we should know that the circular velocity of an object that's undergoing centripetal acceleration is expressible as two pi r f okay just a little shortcut you should probably know now 2 pi r f can also be expressed a little bit better in this context as 2 pi r over t you're taking advantage of the fact that frequency is the reciprocal of period when you do that so that's something i can talk about that at least brings up the period of motion for instance and the radius which is you know one thing i'm given and one thing that i'm looking for but unfortunately it equates it to velocity which is not really something i'm too aware of right now so are there any other things i can come up with that maybe can give me some hints as to the velocity well perhaps i could do a dynamics problem to try and figure out the orbital velocity of this and then i might be able to equate those expressions and that might help me it seems reasonable that to talk about the dynamics or the orbital velocity of this satellite i would get an expression in terms of r or t or other useful things such as the things that they were given like the mass of the earth and the mass of the satellite intense so let's see what happens if i analyze this according to newton's second law let's look at a free body diagram of the satellite and consider all of the forces that are acting on it we have a gravitational force something i will call fg uh besides that that's probably going to be it of course we know orbital uh situation what keeps you in orbit in a situation like this what keeps you in circular motion is going to be the gravitational force now that being the only force in question here i can take that and say well that's going to be the sum of all my forces so if i write an equation using that information i can say net force is equal to the force of gravity now time for some definitions net force by newton's second law is replaceable with m a but you'll recall that the type of acceleration we're undergoing here is centripetal so we can say mac or we could use the definition of centripetal acceleration and just make the substitution right here v squared over r now that is already seeming to be kind of useful because we're starting to bring up you know variables that we know and and the radius that i mentioned before that we'd like to find as well as the velocity which is something else i'm kind of missing out on when i just write down the velocity expression but let's keep going let's keep specifying what about the force of gravity well i'll remind you that we have in general for the force of gravity between two massive objects the gravitational constant g then we have m1 m2 over the radius between them squared okay the radius between their center of masses what i'm going to do is be specific i can say instead of generally mass 1 and mass 2 i can say m for the mass of the satellite and m e for the mass of earth so some pretty interesting things have happened here i can actually cancel out the mass of the satellite because those are in the same place on different sides of the equation it seems i can cancel out this radius okay like if there are two radiuses there radius squared cancel out one of them with this one leaving one and i have the nice little expression v squared equals g m e over r that right there is the square of your orbital velocity okay which in and of itself is a pretty useful thing to have in your back pocket if you can happen to remember it when you take an exam but anyway v is equal to the square root of the gravitational constant m e over r so what i'm thinking is this we can equate this term for velocity and this term for velocity and uh have an expression that's in terms of some useful variables we can say to pi r over the known period t is equal to square root g m e over r and we're just about there in fact we know all of the variables that have been mentioned here except for the radius of the circular orbit which is appearing there in there so we just need to do a little bit of algebra to clean that up and isolate r i think i'm going to square both sides to eliminate the square root so i will get 4 pi squared r squared over t squared equals to g m e over r i can cross multiply now i will have 4 pi squared r cubed equals g mass of earth t squared i can divide both sides by everything that is not the radius okay so i can divide by 4 pi squared and get the expression r cubed is equal to g mass of earth t squared over 4 pi squared and then finally i can take the cube root the cube root of both sides and i'll have an expression for my radius the equal to the cubed root of g mass of earth t squared over 4 pi squared and we have an expression for the orbital radius of a satellite orbiting earth if you have a different planet if you have a different object just change the you know change the variable you're using there for the mass but this is good for this problem and in general for problems like this it would be true from here on things get easier because we've been asked for the magnitude of the earth's gravitational force on the satellite which is something we've already mentioned we've already said that the force of gravity in this situation is going to be gravitational constant uh lowercase m m e over r squared the only thing preventing us from saying this is the answer is the fact that r was not a given variable but now it actually is because we found it in part a so one way to answer this problem could be to leave it because we've technically already determined what r should be or you can simply make this substitution g m m e and we'll have this thing squared so what i'm doing here is i'm going to take the cube root thing i'm going to think about that as an exponent of 1 3 right so we know r could be written as g m e t squared over four pi squared to the one third power and i'm just doing that so i can make the substitution in here where i square it and i turn it into a two thirds power so we can say g m e t squared over four pi squared to the two thirds power whoops i don't need that parenthesis down there so should i go through the trouble of uh simplifying this as much as possible unifying all the like terms and such i don't really feel like it but for you guys i will because i try to focus on algebra skills so see if you can follow along with this we're going to rewrite the top gme for the bottom i'm going to distribute quote-unquote my two-thirds exponent to each of the terms here to the g to the m m e to the t squared etc and that's going to look like this we're going to have g to the 2 3 power we're going to have m e to the 2 3 power we're going to have t remember we're multiplying exponents in this situation t to the four thirds power and then we're going to have some numbers i think four to the two thirds power that would be 16 cube root of 16. that's going to be the only number that shows up but i can't do the cube root of 16. so i'm going to leave it as such and then we're going to have pi to the four-thirds power pi to the four-thirds power you know what i should probably just leave this all in rational exponent form that would be better style so i'll say 16 to the one-third power here and right we have gm me divided by this whole thing divided by that fraction which we know is multiplying by the reciprocal so i can swing up this bottom thing up to the top up to the numerator of a larger fraction so we can have 16 to the one third power pi to the four thirds power gm me all over g to the two thirds power m e to the two thirds power and t to the four thirds power now the only thing left to do is to sort of unify these repeated variables over here here i have variables being divided okay which means i can subtract their exponents okay so possibly the nicest way you could write this 16 to the one third power pi to the four thirds power we have gm to the first over gm sorry not gm to the first rather g to the first over g to the two thirds power which means we can subtract them so we get from that g to the one third power and actually that m there can stay because he's the mass of the satellite again we have m e that's a different variable the mass of earth to the first power over m e to the two thirds power similarly we can write m e to the one third power and we can put that all over t to the four thirds power so i don't know this looks this looks kind of ridiculous um but i guess i kind of just felt like simplifying that probably not something you should even bother with if i was on a test and i was on a time constraint i would probably just leave it here but that is an expression kind of sort of the most simple way you can express the force of gravity in this in this situation between the satellite and the earth don't get me wrong the real work ended here okay the physics ended there all the rest was algebra so um yeah finally in part c we get to talk about the altitude of the satellite and here i just have to point out that what we found is an expression for r which goes from there to there like i said the altitude of the satellite would be considered from you know the earth's surface okay let's call it a for altitude so there's obviously only a slight difference here we can see the radius of the earth is really the difference between my orbital radius and my altitude so i simply say that the letter a for altitude is going to be my radius minus my re and that is why they told us or in the problem that i made up here we have re which is a constant you can look up in most physics textbooks so exclusively in terms of given variables we may say it is that crazy thing we found in part a the cube root of g m e t squared over four pi squared minus r e there is your altitude of your satellite so here we have expressions pretty useful expressions i think for the radius of a circular orbit along with the orbital velocity which we found which could very easily be a question that you would be asked the magnitude of the earth's gravitational force on the satellite again assuming the radius is unknown because it's pretty easy if it is and finally the altitude of the satellite which is just the trivial matter of subtracting the radius of the earth from the orbital radius |
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11659 | https://www.allaboutcircuits.com/textbook/digital/chpt-3/multiple-input-gates/ | Published Time: 2015-02-17T09:07:00-07:00
Multiple-input Gates | Logic Gates | Electronics Textbook
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The Use of Logic Gate
Inverters and buffers exhaust the possibilities for single-input gate circuits. What more can be done with a single logic signal but to buffer it or invert it? To explore more logic gate possibilities, we must add more input terminals to the circuit(s).
Adding more input terminals to a logic gate increases the number of input state possibilities. With a single-input gate such as the inverter or buffer, there can only be two possible input states: either the input is “high” (1) or it is “low” (0).
As was mentioned previously in this chapter, a two input gate has four possibilities (00, 01, 10, and 11). A three-input gate has eight possibilities (000, 001, 010, 011, 100, 101, 110, and 111) for input states.
The number of possible input states is equal to two to the power of the number of inputs:
This increase in the number of possible input states obviously allows for more complex gate behavior. Now, instead of merely inverting or amplifying (buffering) a single “high” or “low” logic level, the output of the gate will be determined by whatever combination of 1’s and 0’s is present at the input terminals.
Since so many combinations are possible with just a few input terminals, there are many different types of multiple-input gates, unlike single-input gates which can only be inverters or buffers. Each basic gate type will be presented in this section, showing its standard symbol, truth table, and practical operation. The actual TTL circuitry of these different gates will be explored in subsequent sections.
The AND Gate
One of the easiest multiple-input gates to understand is the AND gate, so-called because the output of this gate will be “high” (1) if and only if all inputs (first input and the second input and . . .) are “high” (1). If any input(s) is “low” (0), the output is guaranteed to be in a “low” state as well.
In case you might have been wondering, AND gates are made with more than three inputs, but this is less common than the simple two-input variety.
Two-input AND Gate’s Truth Table
A two-input AND gate’s truth table looks like this:
AND Gate Sample Circuit Operation
What this truth table means in practical terms is shown in the following sequence of illustrations, with the 2-input AND gate subjected to all possibilities of input logic levels. An LED (Light-Emitting Diode) provides visual indication of the output logic level:
It is only with all inputs raised to “high” logic levels that the AND gate’s output goes “high,” thus energizing the LED for only one out of the four input combination states.
The NAND Gate
A variation on the idea of the AND gate is called the NAND gate. The word “NAND” is a verbal contraction of the words NOT and AND.
Essentially, a NAND gate behaves the same as an AND gate with a NOT (inverter) gate connected to the output terminal. To symbolize this output signal inversion, the NAND gate symbol has a bubble on the output line.
The truth table for a NAND gate is as one might expect, exactly opposite as that of an AND gate:
As with AND gates, NAND gates are made with more than two inputs. In such cases, the same general principle applies: the output will be “low” (0) if and only if all inputs are “high” (1). If any input is “low” (0), the output will go “high” (1).
The OR Gate
Our next gate to investigate is the OR gate, so-called because the output of this gate will be “high” (1) if any of the inputs (first input or the second input or . . .) are “high” (1). The output of an OR gate goes “low” (0) if and only if all inputs are “low” (0).
Two-input OR Gate Truth Table
A two-input OR gate’s truth table looks like this:
OR Gate Sample Circuit Operation
The following sequence of illustrations demonstrates the OR gate’s function, with the 2-inputs experiencing all possible logic levels. An LED (Light-Emitting Diode) provides visual indication of the gate’s output logic level:
Scroll to continue with content
A condition of any input being raised to a “high” logic level makes the OR gate’s output go “high,” thus energizing the LED for three out of the four input combination states.
The NOR Gate
As you might have suspected, the NOR gate is an OR gate with its output inverted, just like a NAND gate is an AND gate with an inverted output.
NOR gates, like all the other multiple-input gates seen thus far, can be manufactured with more than two inputs. Still, the same logical principle applies: the output goes “low” (0) if any of the inputs are made “high” (1). The output is “high” (1) only when all inputs are “low” (0).
The Negative-AND Gate
A Negative-AND gate functions the same as an AND gate with all its inputs inverted (connected through NOT gates). In keeping with standard gate symbol convention, these inverted inputs are signified by bubbles.
Contrary to most peoples’ first instinct, the logical behavior of a Negative-AND gate is not the same as a NAND gate. Its truth table, actually, is identical to a NOR gate:
The Negative-OR Gate
Following the same pattern, a Negative-OR gate functions the same as an OR gate with all its inputs inverted. In keeping with standard gate symbol convention, these inverted inputs are signified by bubbles. The behavior and truth table of a Negative-OR gate is the same as for a NAND gate:
The Exclusive-OR Gate
The last six gate types are all fairly direct variations on three basic functions: AND, OR, and NOT. The Exclusive-OR gate, however, is something quite different.
Exclusive-OR gates output a “high” (1) logic level if the inputs are at different logic levels, either 0 and 1 or 1 and 0. Conversely, they output a “low” (0) logic level if the inputs are at the same logic levels.
The Exclusive-OR (sometimes called XOR) gate has both a symbol and a truth table pattern that is unique:
XOR Equivalent Circuits
There are equivalent circuits for an Exclusive-OR gate made up of AND, OR, and NOT gates, just as there were for NAND, NOR, and the negative-input gates. A rather direct approach to simulating an Exclusive-OR gate is to start with a regular OR gate, then add additional gates to inhibit the output from going “high” (1) when both inputs are “high” (1):
In this circuit, the final AND gate acts as a buffer for the output of the OR gate whenever the NAND gate’s output is high, which it is for the first three input state combinations (00, 01, and 10). However, when both inputs are “high” (1), the NAND gate outputs a “low” (0) logic level, which forces the final AND gate to produce a “low” (0) output.
Another equivalent circuit for the Exclusive-OR gate uses a strategy of two AND gates with inverters, set up to generate “high” (1) outputs for input conditions 01 and 10. A final OR gate then allows either of the AND gates’ “high” outputs to create a final “high” output:
Exclusive-OR gates are very useful for circuits where two or more binary numbers are to be compared bit-for-bit, and also for error detection (parity check) and code conversion (binary to Grey and vice versa).
The Exclusive-NOR Gate
Finally, our last gate for analysis is the Exclusive-NOR gate, otherwise known as the XNOR gate. It is equivalent to an Exclusive-OR gate with an inverted output. The truth table for this gate is exactly opposite as that of the Exclusive-OR gate:
As indicated by the truth table, the purpose of an Exclusive-NOR gate is to output a “high” (1) logic level whenever both inputs are at the same logic levels (either 00 or 11).
REVIEW:
Rule for an AND gate: output is “high” only if first input and second input are both “high.”
Rule for an OR gate: output is “high” if input A or input B are “high.”
Rule for a NAND gate: output is not “high” if both the first input and the second input are “high.”
Rule for a NOR gate: output is not “high” if either the first input or the second input are “high.”
A Negative-AND gate behaves like a NOR gate.
A Negative-OR gate behaves like a NAND gate.
Rule for an Exclusive-OR gate: output is “high” if the input logic levels are different.
Rule for an Exclusive-NOR gate: output is “high” if the input logic levels are the same.
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V Victor Egwurube April 08, 2021 Nice really helped Like.Reply
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11660 | https://mathbitsnotebook.com/Algebra1/Functions/FNSequences.html | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | | | | --- | | | Sequences - Basic Information MathBitsNotebook.com Topical Outline | Algebra 1 Outline | MathBits' Teacher Resources Terms of Use Contact Person: Donna Roberts | | | | | --- | | | | --- | | | A sequence is an ordered list. It is a function whose domain is the natural numbers {1, 2, 3, 4, ...}. | | | | | | | | | | | --- --- --- --- | | | | | | | | 21, | ... | | Notation for terms of the sequence: | a1 | a2 | a3 | a4 | a5 | a6 | | The three dots (...), called an ellipsis, at the end of the sequence, means that the sequence goes on forever (to infinity). Information about sequences: | | | --- | | • | Each number in a sequence is called a term, an element or a member. | | • | Terms of a sequence can be listed in set notation (curly braces): {1, 5, 9, 13, 17, 21, ...} | | • | Terms are referenced in a subscripted form (indexed), where the natural number subscripts, {1, 2, 3, ...}, refer to the location (position) of the term in the sequence. The first term is denoted a1, the second term a2, and so on. The nth term is an. | | • | The terms in a sequence may, or may not, have a pattern or related formula. Example: {1, 5, 9, 13, 17, 21, ...} can be generated by the formula an = 4n - 3. Example: the digits of π form a sequence, but do not have a pattern. | | • | Terms of the sequence are listed in a specific order. | | • | A subscripted sequence is represented by a1, a2, a3, ... an, ... | | • | Sequences are functions. | | • | The domain of a sequence consists of the terms' location in the list, 1, 2, 3, 4, ... | | • | The range of a sequence consists of the actual terms of the sequence. | | • | When graphed, a sequence is a series of dots. (Do not connect the dots). | | • | The sum of the terms of a sequence is called a series. | Forms of sequences: | | | --- | | • | A finite sequence contains a finite number of terms (a limited number of terms) which can be counted. Example: {1, 5, 9, 13, 17} (it starts and it stops) Example: {5, 4, 3, 2, 1} (listing in descending order is possible) Example: {m, a, n, d, y} (letters are possible. Sequence of letters in name "Mandy") | | • | An infinite sequence contains an infinite number of terms (terms continue without end) which cannot be counted. Example: {1, 5, 9, 13, 17, 21, ...} (it starts but it does not stop, as indicated by the ellipsis ... ) Example: {15, 30, 45, 60, ...} (starting with any value is possible) Example: {1, 2, 1, 2, 1, 2, ...} (a pattern of alternating order is possible) Example: {a, b, c, a, b, c, ...} (letters are possible) | Ways of expressing (or defining) sequences in Algebra 1: | | | --- | | • | A sequence may appear as a list (finite or infinite): Example: {1, 5, 9, 13, 17} finite Example: {1, 5, 9, 13, 17, 21, ...} infinite Listing makes it easy to see any pattern in the sequence. It will be the only option should the sequence have no pattern or formula. | | • | A sequence may appear as an explicit formula. An explicit formula designates the nth term of the sequence, an , as an expression of n (where n = the term's location). Example: {1, 5, 9, 13, 17, 21, ...} can be written in explicit form as an = 4n - 3. (a formula in terms of n) Read more at Sequences as Functions - Explicit | Graphing Sequences: | | | --- | | Sequence: {1, 5, 9, 13, 17, 21, 25, 29, ...} • Sequences are functions. They pass the vertical line test for functions. • The domain consists of the term's location, the natural numbers, {1,2,3,...}, and the range consists of the actual terms of the sequence. | | | • The graph will be in the first quadrant and/or the fourth quadrant (if sequence terms are negative). • The graph will be a discrete graph (a series of dots) as you are graphing only specific points. Do not connect the dots. | | Some Sequences have Formulas: | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- | | Sequence: {1, 5, 9, 13, 17, 21, ...} | | | | --- | Term Number | Term | Subscript Notation | | 1 | 1 | a1 | | 2 | 5 | a2 | | 3 | 9 | a3 | | 4 | 13 | a4 | | 5 | 17 | a5 | | 6 | 21 | a6 | | n | | an | | This sequence can be represented with a formula: Formula: an= 4n - 3 Once a formula is known, you can quickly find ANY value in the sequence by using the term number and substitution. To find the first term, let n = 1. a1= 4(1) - 3 = 1 To find the second term, let n = 2. a2= 4(2) - 3 = 5 To find the hundredth term, let n = 100. a100= 4(100) - 3 = 397 | Not all sequences can be expressed as a formula. Special Types of Sequences: There are several types of sequences (or even individual sequences) that are referred to by name. Some of the more popular sequences are listed below. | | | • Arithmetic Sequences (such as {1, 5, 9, 13, 17, ...} • Geometric Sequences (such as {2, 4, 8, 16, 32, ...} • Quadratic Sequences (such as {4, 7, 12, 19, 28, ...} • Harmonic Sequences (such as {-1, -1/2, -1/3, -1/4, ,,,} • Triangular Number Sequence {1, 3, 6, 10, 15, 21, 28, ...} • Fibonacci Sequence {0, 1, 1, 2, 3, 5, 8, 13 ,..}. | In Algebra 1, we will be concentrating on Arithmetic and Geometric Sequences. | | | --- | | Doubting Thomas wonders how we can know, for sure, that a sequence such as { 2, 4, 6, 8, ...} is an arithmetic sequence. His theory is that there could be many other possible patterns, such as: 2, 4, 6, 8, 2, 4, 6, 8, ... (repeating 4 terms is his pattern). | | | Yes, Thomas is correct. Without a specification in the problem, there is the possibility of more than one pattern in most sequences. The person creating the sequence may have been thinking of a different pattern than what you see when you look at the sequence. In Algebra 1, if in doubt, first look for arithmetic or geometric possibilities. Math questions are usually not trying to trick you. | | | | | | | | | | | --- --- --- --- | | | | | Note: The indexing (subscripts) used for sequences can technically begin with 0 or any positive integer. The most popular subscripting, however, begins with 1 so the subscript can also represent the position of the term in the sequence. Unless otherwise stated, this site will start subscripts at 1. | | | | | Note: Computer programming languages such as C, C++ and Java, refer to the starting position in an array with a subscript of zero. Programmers must remember that a subscript of 3 refers to the 4th element, not the 3rd element, in the array. | | | | | | --- | | | | --- | | | For calculator help with sequences click here. Arrow down to "In Func MODE" | | | | | | NOTE: The re-posting of materials (in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use". | Topical Outline | Algebra 1 Outline | MathBitsNotebook.com | MathBits' Teacher Resources Terms of UseContact Person: Donna Roberts Copyright © 2012-2025 MathBitsNotebook.com. All Rights Reserved. | | |
11661 | https://books.google.com/books/about/Orban_s_Oral_Histology_Embryology_E_BOOK.html?id=PtjQDwAAQBAJ | Orban's Oral Histology & Embryology - E-BOOK - G. S. Kumar - Google Books
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Orban's Oral Histology & Embryology - E-BOOK ============================================ G. S. Kumar Elsevier Health Sciences, Jul 25, 2015 - Medical - 458 pages - New chapter on Age Changes in Oral Tissues - More/ improved color illustrations - Summary with subheadings for quick review - More text boxes and flowcharts incorporated to highlight important concepts and for ease of understanding subject matter More » Preview this book »
Selected pages
Page 15
Page 10
Title Page
Table of Contents
Index
Contents
1 An overview of oral tissues1
2 Development of face and oral cavity5
3 Development and growth of teeth 22
4 Enamel 40
5 Dentin 74
6 Pulp 92
7 Cementum 116
8 Periodontal ligament 135
12 Lymphoid tissue and lymphatics in orofacial region 266
13 Tooth eruption 280
14 Shedding of deciduous teeth 294
15 Temporomandibular joint 305
16 Maxillary sinus 313
17 Age changes in oral tissues 321
18 Histochemistry of oral tissues 334
19 Preparation of specimens for histologic study 360
More
9 Bone 165
10 Oral mucous membrane 194
11 Salivary glands 241
AppendixMolecular events in oral histology 366
Index367
Copyright
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Common terms and phrases
acidactivityadjacentAge Changesalveolar boneameloblastsAnatapicalArch Oral Biolblood vesselscalcificationcapillariescartilagecellular cementumcementoblastscementoenamel junctioncementumcervicalclinicalcollagencollagen fibersconnective tissuecrystalscytoplasmdeciduous teethDent Resdental follicledental laminadental papilladental pulpdentinal tubulesdifferentiationductsElectron micrographelectron microscopeenamel epitheliumenamel matrixenamel organenzymeepithelial cellsextracellularfibroblastsFigurefunctiongingivaGolgigranulesgrowth factorhistochemicalHistologyhumanincisorincreaseinner enamelkeratinizedlamina proprialayerlinguallymph nodeslymphaticlymphocytesmandibularmaxillarymembranemesenchymemineralizationmolarmucousnerveneuralocclusalodontoblast processoral cavityoral epitheliumoral mucosaosteoclastosteocytespalateperiodontal ligamentphosphatasepredentinproteinsproteoglycanspulpalreceptorsregionresorbedreticulumrootsalivasalivary glandssecretionsecretoryseenseroussinusstagestainingstratumstructuresubstancesurfacetiontooth eruptiontooth germultrastructurevascularzone
Bibliographic information
Title Orban's Oral Histology & Embryology - E-BOOK
AuthorG. S. Kumar
Edition 14
Publisher Elsevier Health Sciences, 2015
ISBN 8131245055, 9788131245057
Length 458 pages
SubjectsMedical
›
Dentistry
›
General
Medical / Dentistry / General
Export CitationBiBTeXEndNoteRefMan
About Google Books - Privacy Policy - Terms of Service - Information for Publishers - Report an issue - Help - Google Home |
11662 | https://steemit.com/steemiteducation/@masterwu/separable-differential-equations-dy-dx-ky | Sign in
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Separable Differential Equations: dy/dx = ky
masterwu (59)in #steemiteducation • 8 years ago (edited)
In this video, we work through the process of solving for the general solution of a simple separable differential equation:
The general solution is obtained by separating the variables so that all y's appear on the left hand side, while the x's appear on the right hand side of the equation. Thus...
Then integrating both sides...
We get:
We can then solve for y by raising both sides to the power of e to get
then finally...
As you see, this equation represents exponential growth. We apply this general solution to an ecological problem where excess phosphates cause an algae bloom in a lake and calculate the time it takes for the surface of the lake to be covered in algae!
Thanks for watching. Please give me an Upvote and Resteem if you have found this video helpful.
Please ask me a maths question by commenting below and I will try to help you in future videos.
I would really appreciate any small donation which will help me to help more math students of the world.
Tip me some DogeCoin: A4f3URZSWDoJCkWhVttbR3RjGHRSuLpaP3Tip me at PayPal:
#education #mathematics #calculus #differentialequations
8 years ago in #steemiteducation by masterwu (59)
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homes (47) 8 years ago
When you say 'integrating both sides' isn't this sort of disguised chain rule rather than sort of an algebra technique (ie adding 5 to both sides)?
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masterwu (59) 8 years ago
Can you please clarify what you mean by disguised chain rule?
The fundamental principle of algebra is applied here in that what we do to one side, we must do to the other to maintain equality.
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homes (47) 8 years ago
So you are integrating one side with respect to x and the other side with respect to y, and the justification is we 'integrate both sides' in the same way as 'we add 5 to both sides' but it isn't clear that the operations are equivalent. So something like this I can follow:
Because each time exactly the same operation occurs. Perhaps one can justify integrating with respect to different variables with the definition of the integral?
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masterwu (59) 8 years ago
Ah now I see. Yeah, I've skipped a step where you've shown it in equations (2) and (3). When I say that I've separated the variables, I've divided both sides by y, and multiplied both sides by dx, so that all of the y's appear on the LHS and all of the dx's appear on the RHS after cancelling.
Note that dy/dx can be regarded as the quotient of 2 differentials, which means they are separable - they don't have to stay together.
I think I've explained it better in the video, as well as my previous video, rather than the text I've posted.
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11663 | https://www.fishtanklearning.org/curriculum/math/4th-grade/decimal-fractions/ | Math
/
4th Grade
/
Unit 6: Decimal Fractions
Decimal Fractions
Students expand their concept of what a “number” is as they are introduced to an entirely new category of number, decimals, which they learn to convert, compare, and add in simple cases.
Math
Unit 6
4th Grade
Jump to
Unit Summary
This unit introduces 4th grade students to an entirely new category of number—decimals. Students will explore decimals and their relationship to fractions, seeing that tenths and hundredths are particularly important fractional units because they represent an extension of the place value system into a new kind of number called decimals. Thus, students expand their conception of what a “number” is to encompass this entirely new category, which they will rely on for the remainder of their mathematical education.
Students have previously encountered an example of needing to change their understanding of what a number is in 3rd grade, when the term came to include fractions. Their 3rd grade understanding of fractions (3.NF.A), as well as their work with fractions so far this year (4.NF.A, 4.NF.B), will provide the foundation upon which decimal numbers, their equivalence to fractions, their comparison, and their addition will be built. Students also developed an understanding of money in 2nd grade, working with quantities either less than one dollar or whole dollar amounts (2.MD.8). But with the knowledge acquired in this unit, students will be able to work with money represented as decimals, as it so often is.
Thus, students rely on their work with fractions to see the importance of a tenth as a fractional unit as an extension of the place-value system in Topic A, then expand that understanding to hundredths in Topic B. Throughout Topics A and B, students write values in fraction, decimal, unit, and expanded forms to help students understand their equivalence (4.NF.6). Then students learn to compare decimals in Topic C (4.NF.7) and add decimal fractions in Topic D (4.NF.5). Finally, students apply this decimal understanding to solve word problems, including those particularly related to money, at the end of the unit. Thus, the work with money (4.MD.2) supports the major work and main focus of the unit on decimals.
While students will have ample opportunities to engage with the standards for mathematical practice, they’ll rely heavily on looking for and making use of structure (MP.7), particularly the structure of the place value system. They will also construct viable arguments and critique the reasoning of others (MP.3) using various visual models to support their reasoning.
In 5th Grade Math, students will build on this solid foundation of decimal fractions to generalize their understanding of the “ten times as much” relationship in the place value system to include decimals, then to perform decimal operations using strategies based on place value (5.NBT.1—4, 5.NBT.7). Students fluently operate on decimals using the standard algorithm for each operation in 6th Grade Math (6.NS.3). From that point forward, students will use their understanding of decimals as a specific kind of number in their mathematical work, including ratios, functions, and many others.
Pacing: 15 instructional days (13 lessons, 1 flex day, 1 assessment day)
Fishtank Plus for Math
Unlock features to optimize your prep time, plan engaging lessons, and monitor student progress.
Assessment
The following assessments accompany Unit 6.
Pre-Unit
Have students complete the Pre-Unit Assessment and Pre-Unit Student Self-Assessment before starting the unit. Use the Pre-Unit Assessment Analysis Guide to identify gaps in foundational understanding and map out a plan for learning acceleration throughout the unit.
Pre-Unit Student Self-Assessment
Mid-Unit
Have students complete the Mid-Unit Assessment after Lesson 8.
Post-Unit
Use the resources below to assess student understanding of the unit content and action plan for future units.
Post-Unit Assessment
Post-Unit Student Self-Assessment
Use student data to drive instruction with an expanded suite of assessments. Unlock Pre-Unit and Mid-Unit Assessments, and detailed Assessment Analysis Guides to help assess foundational skills, progress with unit content, and help inform your planning.
Unit Prep
Intellectual Prep
Suggestions for how to prepare to teach this unit
Before you teach this unit, unpack the standards, big ideas, and connections to prior and future content through our guided intellectual preparation process. Each Unit Launch includes a series of short videos, targeted readings, and opportunities for action planning to ensure you're prepared to support every student.
Intellectual Prep for All Units
Unit-Specific Intellectual Prep
| | |
--- |
| area model | Example: The following shape represents 1 whole. 0.24 of it is shaded. Grid divided into 100 equal squares, 24 squares shaded. |
| number line | Example: The point on the number line below is located at 0.24. Number line from 0 to 1 with a point at 0.24. |
| pictorial base ten blocks | Example:Represent 21.53 with base ten blocks. Fraction model showing 1 whole divided into 10 equal parts, 2 parts shaded. |
area model
Example: The following shape represents 1 whole. 0.24 of it is shaded.
number line
Example: The point on the number line below is located at 0.24.
pictorial base ten blocks
Example:Represent 21.53 with base ten blocks.
Essential Understandings
The central mathematical concepts that students will come to understand in this unit
Vocabulary
Terms and notation that students learn or use in the unit
Unit Vocabulary
decimal expanded form
decimal number
decimal fraction
decimal point
fraction expanded form
hundredth
tenth
To see all the vocabulary for
Unit 6,
view our 4th Grade Vocabulary Glossary.
Materials
The materials, representations, and tools teachers and students will need for this unit
Unit Practice
Help students strengthen their application and fluency skills with daily word problem practice and content-aligned fluency activities.
Lesson Map
Topic A: Understanding Tenths
Represent decimals to tenths less than or equal to one with area models. Write a decimal value in fraction, decimal, and unit form.
Standards
4.NF.C.6
Represent decimals to tenths greater than one with pictorial base ten blocks. Write a decimal value in fraction, decimal, unit, and fraction and decimal expanded form.
Standards
4.NF.C.6
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Topic B: Understanding Tenths and Hundredths
Represent decimals to hundredths less than one, understanding the equivalence of some number of tenths and ten times as many hundredths. Write a decimal value in fraction, decimal, and unit form.
Standards
4.NF.C.54.NF.C.6
Represent decimals to hundredths more than one. Write a decimal value in fraction, decimal, unit, and decimal and fraction expanded form for some number to hundredths.
Standards
4.NF.C.54.NF.C.6
Regroup decimal numbers with more than 9 tenths or 9 hundredths into simplest unit form and vice versa.
Standards
4.NF.C.54.NF.C.6
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Topic C: Decimal Comparison
Locate decimals on a number line and explain their placement.
Standards
4.NF.C.6
Compare two decimals, recording the result of a comparison with the symbol >, =, or <, and justify the conclusion.
Standards
4.NF.C.7
Compare two or more decimals written in various forms.
Standards
4.NF.C.7
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Topic D: Decimal Addition
Add tenths with hundredths written as decimal fractions.
Standards
4.NF.C.5
Add tenths with hundredths written as decimals.
Standards
4.NF.C.5
Solve word problems involving the addition of decimals and decimal fractions.
Standards
4.NF.C.5
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Topic E: Money as a Decimal Amount
Find the value of some combination of dollar bills and coins.
Standards
4.MD.A.2
Solve word problems involving money.
Standards
4.MD.A.2
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Common Core Standards
Key
Major Cluster
Supporting Cluster
Additional Cluster
Core Standards
The content standards covered in this unit
Measurement and Data
Measurement and Data
4.MD.A.2
— Use the four operations to solve word problems involving distances, intervals of time, liquid volumes, masses of objects, and money, including problems involving simple fractions or decimals, and problems that require expressing measurements given in a larger unit in terms of a smaller unit. Represent measurement quantities using diagrams such as number line diagrams that feature a measurement scale.
Number and Operations—Fractions
Number and Operations—Fractions
4.NF.C.5
— Express a fraction with denominator 10 as an equivalent fraction with denominator 100, and use this technique to add two fractions with respective denominators 10 and 100.
Students who can generate equivalent fractions can develop strategies for adding fractions with unlike denominators in general. But addition and subtraction with unlike denominators in general is not a requirement at this grade.
For example, express 3/10 as 30/100, and add 3/10 + 4/100 = 34/100.
Number and Operations—Fractions
4.NF.C.6
— Use decimal notation for fractions with denominators 10 or 100.
For example, rewrite 0.62 as 62/100; describe a length as 0.62 meters; locate 0.62 on a number line diagram.
Number and Operations—Fractions
4.NF.C.7
— Compare two decimals to hundredths by reasoning about their size. Recognize that comparisons are valid only when the two decimals refer to the same whole. Record the results of comparisons with the symbols >, =, or <, and justify the conclusions, e.g., by using a visual model.
Foundational Standards
Standards covered in previous units or grades that are important background for the current unit
Measurement and Data
Measurement and Data
2.MD.C.8
— Solve word problems involving dollar bills, quarters, dimes, nickels, and pennies, using $ and ¢ symbols appropriately.
Example: If you have 2 dimes and 3 pennies, how many cents do you have?
Number and Operations in Base Ten
Number and Operations in Base Ten
4.NBT.A.2
— Read and write multi-digit whole numbers using base-ten numerals, number names, and expanded form. Compare two multi-digit numbers based on meanings of the digits in each place, using >, =, and < symbols to record the results of comparisons.
Number and Operations—Fractions
Number and Operations—Fractions
3.NF.A.2
— Understand a fraction as a number on the number line; represent fractions on a number line diagram.
Number and Operations—Fractions
4.NF.A.1
— Explain why a fraction a/b is equivalent to a fraction (n × a)/(n × b) by using visual fraction models, with attention to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions.
Number and Operations—Fractions
4.NF.A.2
— Compare two fractions with different numerators and different denominators, e.g., by creating common denominators or numerators, or by comparing to a benchmark fraction such as 1/2. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model.
Number and Operations—Fractions
4.NF.B.3
— Understand a fraction a/b with a > 1 as a sum of fractions 1/b.
Future Standards
Standards in future grades or units that connect to the content in this unit
Number and Operations in Base Ten
Number and Operations in Base Ten
5.NBT.A.1
— Recognize that in a multi-digit number, a digit in one place represents 10 times as much as it represents in the place to its right and 1/10 of what it represents in the place to its left.
Number and Operations in Base Ten
5.NBT.A.2
— Explain patterns in the number of zeros of the product when multiplying a number by powers of 10, and explain patterns in the placement of the decimal point when a decimal is multiplied or divided by a power of 10. Use whole-number exponents to denote powers of 10.
Number and Operations in Base Ten
5.NBT.A.3
— Read, write, and compare decimals to thousandths.
Number and Operations in Base Ten
5.NBT.A.4
— Use place value understanding to round decimals to any place.
Number and Operations in Base Ten
5.NBT.B.7
— Add, subtract, multiply, and divide decimals to hundredths, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used.
Number and Operations—Fractions
Number and Operations—Fractions
5.NF.A.1
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11664 | https://bohr.physics.berkeley.edu/classes/221/1112/notes/hyperfine.pdf | Copyright c ⃝2019 by Robert G. Littlejohn Physics 221A Fall 2019 Notes 26 Hyperfine Structure in Atoms† 1. Introduction The nucleus of an atom contains localized charge and current distributions, which produce electric and magnetic fields that can be decomposed into multipole fields much as in classical elec-trostatics or magnetostatics.
The first of the multipole moments, the electric monopole, is the Coulomb electrostatic field that holds the electrons in their orbits and produces the gross structure of the atom. The higher order multipole moments produce small corrections to the atomic structure that are known generally as hyperfine effects.
The terminology “hyperfine” refers to the energy scale of these effects, which is smaller than that of the fine structure. The physics, however, of hyperfine effects is completely different from that of fine structure. In practice, the most important hyperfine effects are those due to the magnetic dipole and electric quadrupole fields of the nucleus. Higher multipole moments of the nucleus are important in nuclear physics, but not usually in atomic physics.
In these notes we will study hyperfine effects in atoms, concentrating mainly on the magnetic dipole field in the case of ordinary hydrogen. As we shall see, hyperfine effects couple the dynamics of the atomic electron to that of the nucleus, thereby enlarging the Hilbert space, introducing new quantum numbers, and shifting and splitting the energy levels.
In spite of the small energy scales involved, hyperfine effects are quite important in applications ranging from atomic physics to astrophysics. For example, the radiative transition between the two hyperfine levels of the ground state of the hydrogen atom produces the 21 centimeter line that plays an important role in radio astronomy.
For another example, atomic clocks often use as their basic oscillator a hyperfine transition in the ground state of the heavy alkali atoms, rubidium or cesium. The reproducibility of the frequency of these transitions is so great that the second is now defined in terms of the hyperfine transition in the 133Cs atom, that is, the second is defined as 9,192,631,770 periods of the photon emitted in this transition, exactly. Atomic clocks based on cesium hyperfine transitions currently have a precision of the order of one part in 1015. More recently “optical clocks” have been invented that use transitions in the optical range of frequencies. Some of these have precisions approaching one part in 1018.
† Links to the other sets of notes can be found at:
2 Notes 26: Hyperfine Structure The GPS (global positioning system) relies on radio timing signals emitted by satellites in high earth orbit containing atomic clocks.
In order to achieve the desired precision of position measurements on the earth it is necessary to take into account the effects of both special and general relativity in analyzing the timing signals.
Special relativity enters because of the time dilation of the clocks moving in their orbits, and general relativity because of the difference in the gravitational potential between the clock in orbit and the receiver on the surface of the earth (this is the gravitational red shift).
2. Multipoles in Electrostatics and Magnetostatics In classical electrostatics, a localized charge distribution ρ produces a potential Φ that at field points outside the charge distribution can be expanded into multipoles, Φ(r) = q r + d · r r3 + 1 6 X ij Qij Tij r5 + . . . , (1) where q is the total charge of the distribution, d is its dipole moment vector and Qij is its quadrupole moment tensor, and where the tensor Tij is defined by Tij = 3xixj −r2 δij.
(2) See Fig. 1. The terms of the series (1) are the monopole, dipole and quadrupole terms, in that order.
See Sec. 15.11 for more details on the electrostatic multipole expansion.
Similarly, a localized, steady current distribution J produces a vector potential that can be expanded into multipoles, A(r) = µ×r r3 + . . . , (3) where it is assumed that magnetic monopoles do not exist so there is no monopole term, where the dipole term with magnetic dipole moment µ is shown, and where higher order terms not shown include the magnetic quadrupole term etc. Each term of the multipole expansion falls offas one higher power of r than the previous term. The multipole expansion of the electric and magnetic fields is obtained by computing E = −∇Φ and B = ∇×A.
J x y z r ρ Fig. 1. The origin of coordinates is placed inside a region to which a charge distribution ρ and current distribution J are confined. The field point r is outside the charge and current distributions.
Notes 26: Hyperfine Structure 3 The quantities q, d, Qij and µ that appear in the multipole expansions are moments of the charge and current distributions, q = Z d3r ρ(r), d = Z d3r ρ(r) r, Qij = Z d3r ρ(r) (3xixj −r2 δij), µ = 1 2c Z d3r r×J(r), (4) The dipole moment vector d is the charge-weighted position vector, and the quadrupole moment tensor Qij is the charge-weighted tensor Tij. The tensors Tij and Qij are symmetric and traceless, and thus contain five independent components. These moments characterize the source distribution but are independent of position.
The nomenclature of the multipoles is explained in Table 1. Each pole is associated with an integer k = 0, 1, 2, . . ., and the name contains a Latin or Greek prefix indicating the number 2k. We shall refer to them in general as 2k-poles.
k Pole Numbers 0 Monopole 1 = 20 1 Dipole 2 = 21 2 Quadrupole 4 = 22 3 Octupole 8 = 23 Table 1. The names of the multipoles are obtained by using Latin or Greek prefixes indicating a power of 2.
3. About Nuclei and Nuclear Multipole Moments We denote the spin of the nucleus by I, reserving the symbol S for the electron spin. We let the quantum number of the operator I2 be i, so that I2 has eigenvalues i(i + 1)¯ h2. We denote the nuclear Hilbert space by Enucl, a (2i + 1)-dimensional space in which the standard basis is {|imi⟩, mi = −i, . . . , +i}. The nuclear Hilbert space consists of a single irreducible subspace under rotations. In actual stable nuclei, the spin ranges from i = 0 to i = 15/2. For example, the proton, the nucleus of ordinary hydrogen, has i = 1/2, while the deuteron, the heavier isotope of hydrogen, has i = 1. The isotope 133Cs used in atomic clocks has i = 7/2.
Not all the multipole fields that occur classically are allowed in the case of a nuclei. There are two rules governing the allowed multipole moments of the nucleus. The first is that electric multipoles of odd k and magnetic multipoles of even k are forbidden. To see why, consider what would happen if the nucleus had an electric dipole moment. Then the dipole term in Eq. (1) would be a (presumably small) correction to the potential seen by the atomic electron. Since the potential 4 Notes 26: Hyperfine Structure seen by the electron is V = −eΦ, the perturbing Hamiltonian would be H1 = −ed · r r3 .
(5) The dipole moment vector d is not a c-number as in classical electrostatics, rather it is a (vector) operator acting on the nuclear Hilbert space, just like the magnetic moment µ. And just like µ, d must be proportional to the spin, say, d = kI, because all vector operators on a single irreducible subspace are proportional. See Prob. 19.2(a). Thus, H1 = −keI · r r3 .
(6) Now under parity π the angular momentum I is invariant, but the electron position vector r changes sign. And under time reversal Θ the angular momentum vector I changes sign and r does not. Thus we find that H1 violates both parity and time reversal, πH1π† = −H1, ΘH1Θ† = −H1.
(7) A similar argument applies to the other forbidden terms.
The presence of such terms would indicate a violation of both parity and time reversal. The weak interactions do violate parity, and we do know that time reversal (more precisely, CP) is violated at a very small level in certain decay processes (see Notes 20 and 21), so it is possible that the terms forbidden by this rule actually exist at a small level. For example, the neutron or the electron may have an electric dipole moment, but if such moments exist, they are certainly very small. There is currently considerable experimental interest in lowering the known upper bounds on these “forbidden” moments, or possibly even discovering a nonzero value for some of them, because a numerical value for one of these moments would shed light on physics beyond the standard model.
The experimentally known limits on the electric dipole moment of the neutron give rise to what is known as the “strong CP-problem” in particle physics. Theories of the strong interactions suggest that they should violate CP symmetry at a level that would show up as an electric dipole moment of the neutron that is larger than the experimentally known upper bound. Recall that the CPT theorem says that the product CPT is an exact symmetry of all quantum fields, so CP violation implies violation of time reversal. This suggests that there is some mechanism to suppress CP violation. One possibility that has been explored involves postulating the existence of a particle of very small mass, the “axion.” Nowadays the axion, in addition to solving the strong CP-problem, is looked upon as a candidate for dark matter, and extensive searches are underway to find it. Its discovery would be a major advance in both particle physics and astrophysics.
The second rule states that a 2k-pole can occur only if k ≤2i. For example, the proton with i = 1 2 can (and does) possess an electric monopole moment and a magnetic dipole moment, but not an electric quadrupole moment. The deuteron with i = 1 can (and does) possess an electric quadrupole Notes 26: Hyperfine Structure 5 moment, but the alpha particle with i = 0 can possess only the electric monopole moment. There are no hyperfine effects in 4He, or with any other isotope with spin 0 (such as 12C or 16O).
Lying behind this rule is the fact that the operator representing the 2k-pole on the nuclear Hilbert space is, in fact, an order k irreducible tensor operator. We have already studied the examples of the magnetic dipole moment and the electric quadrupole moment, which are respectively k = 1 and k = 2 irreducible tensor operators. In fact, the expansion of classical electric and magnetic fields into multipoles is an example of the decomposition of a space of fields on three-dimensional space into irreducible subspaces under rotations, each of which has an angular momentum value (ℓ= 0 for monopole, ℓ= 1 for dipole, etc). The fields in question are classical fields, not quantum wave functions, but the transformation properties under rotations is exactly as presented in Notes 13 and 15.
But the maximum order of an irreducible tensor operator on the nuclear Hilbert space with spin i is k = 2i. To see this notice that any operator A that acts on the nuclear Hilbert space can be expanded as a linear combination of the operators |mi⟩⟨m′ i|, where |mi⟩= |imi⟩(we suppress the index i, which is constant): A = X mi,m′ i |mi⟩⟨mi|A|m′ i⟩⟨m′ i| = X mi,m′ i Amim′ i |mi⟩⟨m′ i|, (8) where Amim′ i = ⟨mi|A|m′ i⟩.
(9) Thus, the (2i + 1)2 operators |mi⟩⟨m′ i| form a basis in the space of operators acting on the nu-clear Hilbert space. But since the kets |mi⟩transform under rotations according to the irreducible representation j = i, and since the bras ⟨m′ i| also transform according to the same irreducible representation, the operator |mi⟩⟨m′ i| transforms according to i ⊗i = 0 ⊕. . . ⊕2i.
(10) Each k in the range 0 ≤k ≤2i occurs precisely once in this list, so on the nuclear Hilbert space there is precisely one scalar operator (to within a multiplicative constant), one vector, etc, all the way up to one order k = 2i irreducible tensor operator. Equivalently, any operator that acts on the nuclear Hilbert space can be represented as a sum of irreducible tensor operators of order not exceeding k = 2i. Operators with k > 2i do not occur. See Prob. 19.2(a).
To understand how the moments that are forbidden in the case of a nucleus or an elementary particle can occur classically, see the discussion in Secs. 19.7 and 19.10 (it has to do with the fact that in the case of the nucleus, we are dealing with a single irreducible subspace under rotations).
4. Magnetic Dipole Hyperfine Effects. The Physical Picture The only hyperfine effect we shall examine in these notes is the one due to the magnetic dipole moment of the nucleus. It turns out that if the electric quadrupole moment exists, then its effects are 6 Notes 26: Hyperfine Structure of the same order of magnitude as those of the magnetic dipole, so any realistic treatment requires that the two be treated together (unless selection rules make the quadrupole term vanish). In the following we will treat magnetic dipole effects in a single electron atom with nuclear spin i, but to be realistic i should not exceed 1 2. This condition holds in ordinary hydrogen with the proton as its nucleus, and at a certain point in the calculation we will specialize to that case.
A classical model of the effect will be useful in understanding what we do. To be specific we think of hydrogen. The proton at the center of the atom produces a magnetic dipole field in addition to the electrostatic, Coulomb field. The Coulomb field is rotationally invariant, but the magnetic dipole field depends on the orientation of the proton. As the electron moves in its orbit, which is largely determined by the electrostatic force, it also feels a v×B force due to the proton’s magnetic field, which modifies the orbit somewhat. At the same time the electron’s spin has an energy of orientation in the proton’s magnetic field, and precesses in response. Also, the proton itself feels a magnetic field produced by the electron, both by its orbital motion and by its own magnetic moment, and precesses in response. Thus the orientation of the proton is not constant, but evolves in time as the electron executes its orbit. This picture makes it clear that the dynamics of the proton spin is coupled to the dynamics of the atomic electron, and that they must be handled together.
In the following we shall make a quantum mechanical analysis of these effects. Perhaps sur-prisingly, it turns out to be possible to obtain a closed formula for the shifts in the atomic energy levels, in terms of the expectation value of 1/r3. In the case of hydrogen, the latter can be evaluated explicitly.
5. About Magnetic Dipoles We begin by considering magnetic dipole fields in classical magnetostatics. A point magnetic dipole of moment µ situated at the origin of the coordinates produces a magnetic field B = ∇×A, where A(r) = µ×r r3 = −µ×∇ 1 r .
(11) The vector potential A falls offas 1/r2 and the magnetic field B as 1/r3 at large r, both with a nontrivial angular dependence. Both fields have a singularity at the origin that must be treated with some care in the quantum mechanical analysis of hyperfine effects.
We will do this by smearing out the point dipole moment over a sphere of small radius a, producing a model in which the dipole is replaced by a uniformly magnetized sphere at the origin.
Recall that in magnetostatics the magnetization M is the dipole moment per unit volume, so we require µ = V M, (12) where M is the constant magnetization inside the sphere and where V is the volume of the sphere, V = 4 3πa3.
(13) Notes 26: Hyperfine Structure 7 This model produces fields that are well behaved everywhere in space. When we are done with the quantum mechanical calculation, we will take a →0 in such a way that µ is constant, and we will find physical results that are finite and well behaved. Notice that in this limit, M →∞.
The problem of the uniformly magnetized sphere is a standard one in courses on electromag-netism. The vector potential A at a field point r can be computed as an integral over the source distribution inside the volume of the sphere, A(r) = Z vol d3r′ M×(r −r′) |r −r′|3 = M× −∇ Z vol d3r′ 1 |r −r′| .
(14) The field point r can be either inside or outside the sphere. The quantity in the square brackets is easily evaluated with some reasoning from electrostatics. The integral is the electrostatic potential produced at field point r due to a uniformly charged sphere of radius a and charge density ρ = 1.
Thus, the negative gradient of that integral is the electric field produced by the same charge density.
But by Gauss’s law, that electric field is ˆ r/r2 times the amount of charge inside radius r. Thus we have −∇ Z vol d3r′ 1 |r −r′| = r r3 × 4π 3 r3, r < a, 4π 3 a3, r > a.
(15) Combining this with Eqs. (12)–(14), we obtain A(r) = µ×r 1 a3 , r < a, 1 r3 , r > a.
(16) The exterior solution is identical to that of a point dipole, Eq. (11); in the region r > a, one cannot tell from the field alone whether it is produced by a point dipole or a uniformly magnetized sphere of the same strength.
By taking the curl we compute the magnetic field, B(r) = 2µ a3 , r < a, µ · T r5 , r > a.
(17) where T is defined in Eq. (2). The notation µ·T means the vector whose j-th component is P i µiTij.
Notice that Tij is a symmetric and traceless tensor, something that becomes a k = 2 irreducible tensor operator when reinterpreted as a quantum operator. The vector potential (16) is continuous at the radius r = a; the magnetic field (17), however, has a discontinuity there, due to a surface current.
To express both A and B in single formulas, we introduce the following functions: ∆(r) = ( 1 a3 , r < a, 0, r > a, (18) 8 Notes 26: Hyperfine Structure and f(r) = 0, r < a, 1, r > a.
(19) Then we can write A(r) = (µ×r) ∆(r) + f(r) r3 , (20) and B(r) = µ · 2∆(r)I + f(r) T r5 , (21) where I is the identity tensor.
When we take the limit a →0, ∆(r) becomes a function that is concentrated inside a small volume (4π/3)a3 with a value that goes to ∞in such a way that the integral of ∆(r) over all space is the constant 4π/3. This is the behavior of a function whose limit is a delta function, so we can write lim a→0 ∆(r) = 4π 3 δ(r).
(22) As for f(r), it has the limit, lim a→0 f(r) = 1.
(23) 6. The Hamiltonian and the Hilbert Space We turn now to the atom in which the nucleus resides.
In these notes we will be mainly interested in hyperfine effects in hydrogen and alkali atoms, the latter treated as atoms with a single electron (the valence electron) moving in a screened Coulomb potential.
In the following we use atomic units and take g = 2 for the g-factor of the electron, so µB = 1/2c.
The Hamiltonian for the atomic electron is H = 1 2 p + 1 cA 2 + V (r) + HFS + HLamb + 1 c S · B, (24) where V (r) is the central force potential. This could be the Hamiltonian for the Zeeman effect, except that in this case the fields A and B are not external fields, rather they are the magnetic dipole fields of the nucleus. Thus, they are given by Eqs. (20) and (21). We include in the Hamiltonian the fine structure terms HFS, the sum of the three terms shown in Eq. (24.25), and terms responsible for the Lamb shift, since we wish to make a realistic treatment of hyperfine effects which are smaller than those just listed. We will not need to know much about the Lamb shift in the following analysis, except that it splits the energy levels in hydrogen to give them a dependence on ℓ.
The expressions (20) and (21) for A and B are the fields of a classical magnetic dipole at the origin, but now for use in the Hamiltonian (24) we must reinterpret µ as an operator acting on the nuclear Hilbert space, given in terms of the nuclear spin by µ = gNµNI, (25) Notes 26: Hyperfine Structure 9 where gN is the g-factor of the nucleus and µN is the nuclear magnetic moment (see Sec. 14.8).
Thus, the Hamiltonian (24) must be interpreted as an operator acting the total Hilbert space E = Eelec ⊗Enucl, (26) where Eelec, the electronic Hilbert space, is the product of its orbital and spin parts, Eelec = Eorb ⊗Espin.
(27) Basis states in E can be defined as products of basis states in Eelec and in Enucl. For Eelec the obvious basis is the eigenbasis of H0, {|nℓjmj⟩} in the notation of Notes 24, which have energies Enℓj. In alkalis the energies depend strongly on ℓ, and in hydrogen they depend on ℓbecause of the Lamb shift. The obvious basis in Enucl is {|imi⟩}. Thus we define the basis states in E, |nℓjmj⟩⊗|imi⟩= |nℓjmjmi⟩, (28) where we suppress the index i in the shorthand notation on the right hand side, since it is a constant.
We will call this the “uncoupled basis,” because the electronic angular momentum J and nuclear angular momentum I are not coupled. Of course, orbital and spin angular momentum have already been coupled to produce J, so the basis (28) is really only partially uncoupled. These uncoupled basis states are eigenstates of H0 on the whole Hilbert space E. The energy does not depend on the quantum numbers mj or mi, so the energy eigenstates are (2j + 1)(2i + 1)-fold degenerate.
Now we expand the kinetic energy in the Hamiltonian (24) as in Notes 25 and neglect the term in A2, writing the result as H = H0 + H1, where H0 = p2 2 + V (r) + HFS + HLamb, (29) and H1 = 1 c (p · A + S · B).
(30) Here we have used the fact that our A is in Coulomb gauge, ∇· A = 0, so p · A = A · p. We write the two terms in H1 as H1,orb and H1,spin. For H1,orb we use 1/c = 2µB and Eqs. (25) and (20), rearranging the triple product, p · (I×r) = I · (r×p) = I · L.
(31) This gives H1,orb = 1 cp · A = k(I · L) ∆(r) + f(r) r3 , (32) where we have set k = 2gNµBµN = gegNµBµN (33) for the product of the effective magnetic moments of the electron and the nucleus (including g-factors). As for H1,spin, we rearrange it similarly, obtaining H1,spin = 1 cS · B = k 2∆(r)(I · S) + f(r)(I · T · S) r5 .
(34) 10 Notes 26: Hyperfine Structure Notice that H1,orb is a kind of spin-orbit interaction (but it is the spin of the nucleus, not the spin of the electron), and H1,spin is a kind of spin-spin interaction.
Properly we should only take the limit a →0 after we have formed matrix elements, but in many books this limit is taken before, resulting in the formulas, H1,orb = k(I · L) 4π 3 δ(r) + 1 r3 , (35) and H1,spin = k 8π 3 δ(r)(I · S) + (I · T · S) r5 .
(36) The terms involving δ(r) are called “Fermi contact terms,” because they vanish except when the electron and nucleus are in contact with one another (r = 0), and in honor of Fermi, who first carried out the calculation presented in this set of notes.
7. The Perturbation Calculation Both terms (32) and (34) in the perturbing Hamiltonian have the form of I dotted into a vector that is constructed out of electronic operators, that is, operators that act only on the electronic Hilbert space. Since the angular momentum J generates rotations that rotate electronic vectors, while I generates rotations that rotate I itself, the dot products in H1 are not invariant under either electronic rotations alone or under nuclear rotations alone. For this reason, H1 does not commute with either Jz or Iz, and the uncoupled basis (28) is not the best one for carrying out the perturbation calculation (see Sec. 24.5).
The dot products in question, however, are invariant under total rotations of the system, elec-tronic plus nuclear, which are generated by the total angular momentum of the system defined by F = J + I = L + S + I.
(37) This suggests that we couple together J and I to create eigenstates of F 2 and Fz. We will call the result the “coupled basis”; it is given by |nℓjfmf⟩= X mj,mi |nℓjmjmi⟩⟨jimjmi|fmf⟩, (38) where the final scalar products are the Clebsch-Gordan coefficients.
In the coupled basis the matrix elements we need to consider for degenerate perturbation theory are ⟨nℓjfmf|H1|nℓjf ′m′ f⟩, (39) where the unprimed indices on the two sides specify the degenerate unperturbed energy level Enℓj, while the remaining indices specify the basis inside the degenerate eigenspace of the unperturbed system. But since [F, H1] = 0, and since if H1 commutes with F it commutes with any function of F such as F 2, the matrix (39) is diagonal in both f and mf, and the primes can be dropped. We Notes 26: Hyperfine Structure 11 need only compute the diagonal matrix elements to find the energy shifts, and once again we have succeeded in doing a degenerate perturbation calculation without diagonalizing any matrices. Thus we have ∆E = ⟨nℓjfmf|H1|nℓjfmf⟩.
(40) We now evaluate these matrix elements for the case ℓ̸= 0. In this case the contact terms do not contribute in the limit a →0, because the wave functions go as rℓnear r = 0, and thus vanish at r = 0 when ℓ̸= 0. As for the other terms, we take the limit a →0 so f(r) = 1, and write the result in the form ∆E = k⟨nℓjfmf|I · G|nℓjfmf⟩, (41) where G = L r3 + T · S r5 = L r3 + 3r(r · S) −r2S r5 , (42) where we use Eq. (2). Note that G is a purely electronic vector operator.
We simplify Eq. (41) by using the projection theorem (see Sec. 25.7 and Prob. 25.1), which tells us that an operator such as G, which is a vector under rotations generated by J, sandwiched between eigenstates of J2 with the same j value on both sides, can be replaced by an operator proportional to J: G →(G · J)J j(j + 1) .
(43) Thus ∆E = k j(j + 1)⟨nℓjfmf|(I · J)(J · G)|nℓjfmf⟩.
(44) Now the two dot products can be simplified. First, by Eq. (37), we have I · J = 1 2 F 2 −J2 −I2).
(45) Next, we use r · J = r · S (since r · L = 0) to write G · J = L2 r3 + 3(r · S)2 −r2S2 r5 , (46) where a term L · S/r3 has cancelled. Then, since S = (1/2)σ and since σiσj = δij + iǫijk σk, the final term in Eq. (46) simplifies, 3(r · S)2 = 3r2 4 , (47) which cancels r2S2 = 3r2/4. The result is simply G · J = L2 r3 .
(48) Then, with Eqs. (45) and (48), the energy shift (44) becomes ∆E = k f(f + 1) −j(j + 1) −i(i + 1) 2j(j + 1) ℓ(ℓ+ 1) D 1 r3 E , (49) 12 Notes 26: Hyperfine Structure where the final expectation value of 1/r3 can be reduced to a purely spatial matrix element with respect to the state |nℓ0⟩exactly as in Sec. 24.7. Specializing to hydrogen and using Eqs. (33) and (24.47), we find ∆E = gegNµBµN a3 0 1 n3 f(f + 1) −j(j + 1) −i(i + 1) j(j + 1)(2ℓ+ 1) , (50) where we have introduced a factor of 1/a3 0 (a0 is the Bohr radius) to make the answer have dimensions of energy (thus it is now valid in ordinary units).
Equation (50) is the final expression for the hyperfine structure energy shift in hydrogen in the case ℓ̸= 0. However, one can show that the same answer holds also in the case ℓ= 0, so Eq. (50) applies in all cases.
8. The New Energy Levels The energy levels were Enℓj before the hyperfine interactions were turned on, but since ∆E in Eq. (50) depends on f, they now have the form Enℓjf. The energy eigenstates are |nℓjfmf⟩, and are (2f +1)-fold degenerate, since the energy does not depend on mf. This is precisely the situation expected of a generic system invariant under rotations; the energy eigenspaces consist of a single irreducible subspace under rotations (see Secs. 19.7–19.10; and the degeneracy present is explained by rotational invariance alone.
The energy shift (50) causes the fine structure levels in the Dirac picture of hydrogen to split, giving rise to hyperfine multiplets. For example, the ground state 1s1/2 splits into two levels f = 0 and f = 1, of which f = 0 is lower since by Eq. (50) the energies are an increasing function of f.
This f = 0 level is the true ground state of hydrogen. It is nondegenerate. The f = 1 level is 3-fold degenerate, and lies above the ground state by an energy of approximately 1.42 GHz in frequency units, or 21 cm in wave length units, or 0.068 K in temperature units. Since ℓ= 0 for these states, the total angular momentum F is really just the total spin S + I, and the f = 0 and f = 1 states are the spin singlet and triplet states, respectively. The electron and proton spins are antiparallel in the ground (singlet) state; when one of the spins is flipped to make them parallel (the triplet state), the energy is raised.
Similarly, other j = 1/2 states including the 2s1/2 and 2p1/2 states split into an f = 0 and f = 1 pair. The 2p3/2 state in the Dirac picture splits into an f = 1 and f = 2 pair.
Electric dipole transitions between the various hyperfine levels are governed by the matrix element, ⟨nℓjfmf|xq|n′ℓ′j′f ′m′ f⟩.
(51) The selection rules for this matrix element follow from the Wigner-Eckart theorem and parity. The Wigner-Eckart theorem can be used three times, once for each of the three angular momenta L, J and F, since xq is a k = 1 irreducible tensor operator with respect to rotations generated by any of these angular momenta. Thus, the matrix element (51) vanishes unless the following selection rules Notes 26: Hyperfine Structure 13 are satisfied: mf = m′ f + q, ∆f = 0, ±1 but f = 0 →f = 0 not allowed, ∆j = 0, ±1, ∆ℓ= ±1.
(52) The exclusion of f = 0 →f = 0 comes from the fact that 0⊗1 = 1, so if f = f ′ = 0, then the matrix element vanishes. The only reason we do not also exclude j = 0 →j = 0 is that j is half-integral, and cannot take on the value 0. And the only reason we do not make a special case of excluding ℓ= 0 →ℓ= 0 is that that case is already excluded by parity.
The hyperfine splitting of the 1s1/2 level in hydrogen is particularly interesting. The separation between the f = 0 and f = 1 hyperfine levels is 1.42 GHz, or 21 cm in wavelength units. Electric dipole transitions between these two levels are forbidden by parity, but magnetic dipole transitions are allowed. (One can see this from another standpoint: since spins do not interact with electric fields, an electric dipole transition cannot flip the spins to convert the triplet to the singlet state.
But magnetic dipole transitions can.) The 21 cm line is quite important in radio astronomy. Spiral galaxies typically possess large clouds of atomic hydrogen, which radiate at the 21 cm wavelength. A population of the excited state f = 1 is maintained by collisions; the temperatures prevalent in the clouds are high enough that the populations of the ground state f = 0 and the first excited state f = 1 are determined mostly by the degeneracies (1 for f = 0 and 3 for f = 1, although there is some effect due to the Boltzmann factor).
By measuring Doppler shifts, the state of motion of the clouds can be measured. In this way, it was first proven that the Milky Way is a spiral galaxy. The 21 cm line is also important in absorption spectra, which can be used to determine the temperature of the clouds of atomic hydrogen.
Molecular hydrogen has a completely different hyperfine structure from atomic hydrogen, arising from the spin-spin interaction of the two protons in the molecule.
The transitions between the hyperfine levels of molecular hydrogen are in the megahertz range of frequencies.
Problems 1. A problem on the hyperfine interaction in hydrogen.
(a) Equation (50) was derived in the case ℓ̸= 0. Show that it also applies in the case ℓ= 0. Hint: Use the fact that the components of the tensor Tij, defined in Eq. (2), are r2 times linear combinations of the Y2m(θ, φ), for m = −2, . . . , +2. This is related to the fact that Tij is the Cartesian version of an order 2 irreducible tensor.
(b) Our analysis of the hyperfine interaction in hydrogen has included the energy of interaction of the electron with the magnetic dipole field produced by the proton, but it seems that we have 14 Notes 26: Hyperfine Structure not included the energy of interaction of the proton spin with the magnetic field produced by the electron. As seen by the proton, the electron produces a magnetic field for two reasons: first, it is a charge in motion, therefore a current, which makes a magnetic field. This is the magnetic field due to the orbital motion of the electron. Next, the electron has a magnetic moment of its own, which makes a dipole magnetic field. This is the magnetic field produced by the spin of the electron.
Work out an expression for the energy of interaction of the proton spin with the magnetic field produced by the orbital motion of the electron. Follow the analysis of the spin-orbit interaction in Sec. 24.2, but run it backwards. That is, putting primes on the fields in the electron rest frame and no primes on fields in the proton rest frame, use Coulomb’s law to write down the field E′ of the electron in its own rest frame, then Lorentz transform to the lab frame to get B (call this Borb, the magnetic field due to the orbital motion of the electron). Then the energy of interaction of the proton with this magnetic field is −µp ·Borb, where µp is the proton magnetic moment. Notice that unlike the analysis of Sec. 24.2, there is no factor of 1 2 from Thomas precession, because the proton frame is not accelerated.
Now use Eq. (21) in the limit a →0 to obtain the magnetic field produced by the dipole moment of the electron at the position of the proton. Call this Bspin, and write down an expression for the energy of interaction −µp · Bspin.
If you add these terms to the Hamiltonian (Eqs. (29) plus (30)), does it change the energy shifts (50)? These energy shifts are confirmed experimentally (for example, by the 21 cm line). What is wrong?
(c) Compute the hyperfine splitting of the ground state of positronium in wavelength units. Notice that in positronium, the fine structure and hyperfine structure are of the same order of magnitude.
Now some remarks about part (c). The interesting thing about this calculation is that the answer based on what you now know is actually wrong, because it omits a virtual process (a Feynman diagram) in which the positron and electron annihilate into a photon, which then materialize back into a positron and electron.
The analysis of this process requires quantum field theory. The Hamiltonians we usually use in atomic, molecular and solid state physics, expressed in terms of a finite number of particles and a finite number of degrees of freedom, are only valid up to a certain degree of accuracy, beyond which interactions with the infinite degrees of freedom in various fields (electromagnetic, electron-positron, strong interactions, . . .) cannot be ignored. The first place where this occurs in hydrogen is with the Lamb shift. In positronium, it happens at the level of the fine structure (in positronium, the hyperfine structure is considered part of the fine structure).
2. In our derivation of the energy shifts (50), we assumed that hyperfine effects were much smaller than the Lamb shift. This is what allowed us to assume that the ℓvalues on the two sides of the matrix element (39) were the same. Compute the hyperfine energy shifts for the 2s1/2 and 2p1/2 levels in hydrogen, and compare to the Lamb shift. Make a sketch of the energy levels like that in Notes 26: Hyperfine Structure 15 Fig. 24.3, including the hyperfine structure (but you can omit the 2p3/2 levels). It’s convenient to measure energies in GHz. |
11665 | https://testbook.com/question-answer/if-the-sum-of-ten-different-positive-integers-is-1--5c10e90440d9380cf6b0d888 | [Solved] If the sum of ten different positive integers is 100, then w
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If the sum of ten different positive integers is 100, then what is the greatest possible number among these 10 numbers?
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Option 3 : 55
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To find the greatest possible number, we need to find the sum of first 9 positive integers i.e. sum of 1, 2, ---, 9 which when subtracted from 100 gives the required greatest number
Using formula to find sum of n natural numbers
i.e. Sum = n(n + 1) /2
⇒ Sum of 1 + 2 + --- + 9 = 9 × (10/2) = 45
⇒ Required greatest number = 100 – 45 = 55
∴ Greatest possible number among these 10 numbers is 55
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11666 | https://math.answers.com/math-and-arithmetic/What_is_two_square_numbers_that_add_up_to_100 | What is two square numbers that add up to 100? - Answers
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What is two square numbers that add up to 100?
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Updated: 9/25/2023
36 + 64 = 100
(6x6) + (8x8)
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### What two square numbers add todether to make 100? The two square numbers that add together to make 100 are 36 and 64. Specifically, (6^2 = 36) and (8^2 = 64). When you add them together, (36 + 64 = 100).
### What two squared numbers add up to 100? 36 and 64
### What two square numbers make 100? 81 and 9 if you add them. 4 and 25 if you multiply them.
### What two numbers are square numbers with a total of 100? The two square numbers that add up to 100 are 36 and 64. Specifically, 6 squared (6²) equals 36, and 8 squared (8²) equals 64. When combined, 36 + 64 equals 100.
### What two numbers add to -80 and multiply to 100? 20
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11667 | https://mathoverflow.net/questions/418733/convergence-almost-everywhere-of-characteristic-functions | pr.probability - Convergence almost everywhere of characteristic functions - MathOverflow
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Convergence almost everywhere of characteristic functions
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Let (Φ n)n(Φ n)n be the characteristic functions of probability measures (μ n)n(μ n)n and let Φ Φ be the characteristic function of a probability measure μ μ.
Do you know an example where Φ n(t)→Φ(t)Φ n(t)→Φ(t) for all t∈A t∈A where A A is dense in R d R d but (μ n)n(μ n)n does not converge weakly to μ μ ?
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edited Mar 23, 2022 at 11:58
TiblodocusTiblodocus
asked Mar 23, 2022 at 8:38
TiblodocusTiblodocus
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1 By measure you mean probability measure or something more general?Dieter Kadelka –Dieter Kadelka 2022-03-23 10:55:34 +00:00 Commented Mar 23, 2022 at 10:55
Yes, μ n μ n and μ μ are probability measures.Tiblodocus –Tiblodocus 2022-03-23 11:57:13 +00:00 Commented Mar 23, 2022 at 11:57
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There is no such example. Let (f n)(f n) be a sequence of characteristic functions of probability measures μ n μ n which converges a. e. to a characteristic function f f of a probability measure μ μ. You can always choose a subsequence such that μ n→μ μ n→μ weakly to some measure on the real line (first theorem of Helly). Then for this subsequence f n(x)→f(x)f n(x)→f(x) for all x x, where f f is the Fourier transform of μ μ. Your assumption that f f is a Fourier transform of a probability measure implies that μ μ must be that probability measure. Since this works for every subsequence we have convergence of f n f n to f f everywhere.
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edited Mar 24, 2022 at 13:36
answered Mar 23, 2022 at 13:55
Alexandre EremenkoAlexandre Eremenko
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2 We need to assume that (μ n)n(μ n)n is tight to get this weak convergence.Tiblodocus –Tiblodocus 2022-03-23 13:58:49 +00:00 Commented Mar 23, 2022 at 13:58
No, one does not. If the limit measure is not a probability measure, then f n f n converges to its Fourier transform, which is not a characteristic function of a probability measure, but this contradicts your assumption.Alexandre Eremenko –Alexandre Eremenko 2022-03-24 13:32:55 +00:00 Commented Mar 24, 2022 at 13:32
Ok, thanks. I did not know this general version.Tiblodocus –Tiblodocus 2022-03-24 14:38:15 +00:00 Commented Mar 24, 2022 at 14:38
This is a great answer. However I think you are using μ μ and f f to denote two different objects, that are only proved to be the same at the very end, which is a bit confusing. (What I understand is that you start with a given limit for the c.f., then construct the c.f. of a limit, and then you prove they must be the same.)Pierre PC –Pierre PC 2022-03-24 16:59:39 +00:00 Commented Mar 24, 2022 at 16:59
@AlexandreEremenko In the OP, the convergence holds on a dense set (not necessarily a.e.). What happens then?Gagar –Gagar 2022-10-15 14:17:07 +00:00 Commented Oct 15, 2022 at 14:17
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If Φ n→Φ Φ n→Φ pointwise on a dense subset, then by the Stone-Weierstrass theorem
lim n→∞∫B g d μ n=∫B g d μ lim n→∞∫B g d μ n=∫B g d μ
on every ball centered at 0 0 and every g∈C 0(R d)g∈C 0(R d). This implies that μ n→μ μ n→μ in the weak∗∗ topology of the Banach space of bounded Borel measures M(R d)M(R d). However, we may not have that μ n→μ μ n→μ in the weak topology of M(R d)M(R d). Let's be reminded that μ n→μ μ n→μ weakly iff μ n(E)→μ(E)μ n(E)→μ(E) for every Borel set E⊆R d E⊆R d.
For an example, let δ a δ a be the unit mass at a∈R d a∈R d, i.e., δ a(E)=1 δ a(E)=1 if a∈E a∈E and δ a(E)=0 δ a(E)=0 if a∉E a∉E. Let (a n)(a n) be a sequence with nonzero terms such that a n→0 a n→0. Let Φ n Φ n be the charcteristic function of δ a n δ a n and Φ 0≡1 Φ 0≡1 denote the characteristic function of δ 0 δ 0. Clearly,
Φ n(γ)=e i a n γ∀γ∈R d Φ n(γ)=e i a n γ∀γ∈R d
so Φ n→1 Φ n→1 pointwise on R n R n. On the other hand, let E E be a set such that 0∉E 0∉E and {a n:n∈N}⊆E{a n:n∈N}⊆E. Then, 1=δ a n(E)↛δ 0(E)=0 1=δ a n(E)↛δ 0(E)=0, so (δ a n)(δ a n) does not converge to δ 0 δ 0 weakly.
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answered Mar 24, 2022 at 16:04
Onur OktayOnur Oktay
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In your example δ a n⟶δ 0 δ a n⟶δ 0 weakly (by the continuity theorem, since Φ n⟶Φ 0 Φ n⟶Φ 0 pointwise). Your reminder is incorrect: μ n⟶μ μ n⟶μ weakly iff μ n(E)⟶μ(E)μ n(E)⟶μ(E) for all μ μ-continuity sets E E. And here E E is clearly not a continuity set of δ 0 δ 0 since 0∈∂E 0∈∂E.esg –esg 2022-03-24 20:40:59 +00:00 Commented Mar 24, 2022 at 20:40
@esg After reading your comment and quick googling (e.g. Portmentau's theorem ocw.mit.edu/courses/sloan-school-of-management/…) I realize that your description is indeed the same as the weak-star convergence or measures. "Weak convergence" of measures has just a different meaning in the Banach space language. Given the context, one must stick with the definition you've given.Onur Oktay –Onur Oktay 2022-03-25 06:11:30 +00:00 Commented Mar 25, 2022 at 6:11
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11668 | https://brightchamps.com/en-us/math/math-formulas/math-formula-for-the-foil-method | Table Of Contents
Summarize this article:
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Last updated on August 6, 2025
Math Formula for the FOIL Method
The FOIL method is a technique used to multiply two binomials in algebra. It stands for First, Outer, Inner, Last, referring to the order in which terms are multiplied. In this topic, we will learn how to apply the FOIL method to simplify expressions involving binomials.
List of Steps for Using the FOIL Method
The FOIL method helps simplify the multiplication of two binomials. Let’s learn the steps to apply the FOIL method.
Step-by-Step Process of the FOIL Method
The FOIL method involves multiplying terms in a specific order:
First: Multiply the first terms of each binomial.
Outer: Multiply the outer terms of the binomials
Inner: Multiply the inner terms of the binomials.
Last: Multiply the last terms of each binomial.
Then, add all the products together to get the final expression.
Application of the FOIL Method
Let's consider two binomials, ((a + b)) and ((c + d)).
Using the FOIL method:
First: (a cdot c)
Outer: (a cdot d)
Inner: (b cdot c)
Last: (b cdot d)
Combine these: (ac + ad + bc + bd).
This is the expanded form of the product of the two binomials.
Importance of the FOIL Method
The FOIL method is an essential algebraic tool that simplifies the multiplication of binomials.
It helps in understanding polynomial multiplication.
Simplifies complex algebraic expressions.
Forms a foundation for learning more advanced algebra concepts.
Tips and Tricks to Master the FOIL Method
Some students find the FOIL method tricky at first. Here are some tips and tricks to master it:
Use the mnemonic "FOIL" to remember the order:
First, Outer, Inner, Last.
Practice with different binomials to gain confidence.
Visualize the process by writing out all steps clearly in practice problems.
Real-Life Applications of the FOIL Method
The FOIL method is not only vital in algebra classes but also has real-life applications:
Used in calculations involving areas and perimeters.
Helps in solving quadratic equations, especially in physics and engineering problems.
Essential for computer algorithms that model real-world scenarios.
Common Mistakes and How to Avoid Them While Using the FOIL Method
Students often make errors when applying the FOIL method. Here are some mistakes and ways to avoid them:
Students often make errors when applying the FOIL method. Here are some mistakes and ways to avoid them:
Mistake 1
Forgetting to Multiply All Parts
Forgetting to Multiply All Parts
Students might forget to multiply all parts of the binomials. To avoid this, always write down each step of FOIL and ensure all terms are accounted for.
Students might forget to multiply all parts of the binomials. To avoid this, always write down each step of FOIL and ensure all terms are accounted for.
Mistake 2
Incorrect Order of Multiplication
Incorrect Order of Multiplication
Sometimes, students mix up the order of multiplication. Remember the order: First, Outer, Inner, Last, and practice regularly to avoid this mistake.
Sometimes, students mix up the order of multiplication. Remember the order: First, Outer, Inner, Last, and practice regularly to avoid this mistake.
Mistake 3
Combining Like Terms Incorrectly
Combining Like Terms Incorrectly
After applying FOIL, students might combine like terms incorrectly. Carefully group and combine terms with the same variables and powers.
After applying FOIL, students might combine like terms incorrectly. Carefully group and combine terms with the same variables and powers.
Mistake 4
Misinterpreting the FOIL Acronym
Misinterpreting the FOIL Acronym
New learners might misinterpret the FOIL acronym. Understand that it is a specific sequence to multiply terms and practice with examples to reinforce its use.
New learners might misinterpret the FOIL acronym. Understand that it is a specific sequence to multiply terms and practice with examples to reinforce its use.
Examples of Problems Using the FOIL Method
Problem 1
Multiply ((x + 3)(x + 5)) using the FOIL method.
The result is (x^2 + 8x + 15).
The result is (x^2 + 8x + 15).
Explanation
Applying FOIL: First: (x \cdot x = x^2) Outer: (x \cdot 5 = 5x) Inner: (3 \cdot x = 3x) Last: (3 \cdot 5 = 15) Combine: (x^2 + 5x + 3x + 15 = x^2 + 8x + 15).
Applying FOIL: First: (x \cdot x = x^2) Outer: (x \cdot 5 = 5x) Inner: (3 \cdot x = 3x) Last: (3 \cdot 5 = 15) Combine: (x^2 + 5x + 3x + 15 = x^2 + 8x + 15).
Problem 2
Multiply ((2x + 1)(x - 4)) using the FOIL method.
The result is (2x^2 - 7x - 4).
The result is (2x^2 - 7x - 4).
Explanation
Applying FOIL: First: (2x \cdot x = 2x^2) Outer: (2x \cdot (-4) = -8x) Inner: (1 \cdot x = x) Last: (1 \cdot (-4) = -4) Combine: (2x^2 - 8x + x - 4 = 2x^2 - 7x - 4).
Applying FOIL: First: (2x \cdot x = 2x^2) Outer: (2x \cdot (-4) = -8x) Inner: (1 \cdot x = x) Last: (1 \cdot (-4) = -4) Combine: (2x^2 - 8x + x - 4 = 2x^2 - 7x - 4).
Problem 3
Multiply ((3y - 2)(y + 6)) using the FOIL method.
The result is (3y^2 + 16y - 12).
The result is (3y^2 + 16y - 12).
Explanation
Applying FOIL: First: (3y \cdot y = 3y^2) Outer: (3y \cdot 6 = 18y) Inner: (-2 \cdot y = -2y) Last: (-2 \cdot 6 = -12) Combine: (3y^2 + 18y - 2y - 12 = 3y^2 + 16y - 12).
Applying FOIL: First: (3y \cdot y = 3y^2) Outer: (3y \cdot 6 = 18y) Inner: (-2 \cdot y = -2y) Last: (-2 \cdot 6 = -12) Combine: (3y^2 + 18y - 2y - 12 = 3y^2 + 16y - 12).
FAQs on the FOIL Method
1.What does FOIL stand for in the FOIL method?
FOIL stands for First, Outer, Inner, Last, which refers to the order of multiplying terms in two binomials.
FOIL stands for First, Outer, Inner, Last, which refers to the order of multiplying terms in two binomials.
2.Can the FOIL method be used for polynomials with more than two terms?
No, the FOIL method is specifically designed for multiplying two binomials. For polynomials with more terms, other methods like distribution or the box method are used.
No, the FOIL method is specifically designed for multiplying two binomials. For polynomials with more terms, other methods like distribution or the box method are used.
3.How can I verify my answer after using the FOIL method?
After applying the FOIL method, you can verify your answer by re-multiplying the terms or using the distributive property to check the expanded expression.
After applying the FOIL method, you can verify your answer by re-multiplying the terms or using the distributive property to check the expanded expression.
4.What is a common mistake when using the FOIL method?
A common mistake is forgetting one of the steps in the FOIL process or incorrectly combining like terms after expansion.
A common mistake is forgetting one of the steps in the FOIL process or incorrectly combining like terms after expansion.
5.How can the FOIL method help in solving quadratic equations?
The FOIL method helps expand binomials into quadratic equations, which can then be solved using factoring, the quadratic formula, or completing the square.
The FOIL method helps expand binomials into quadratic equations, which can then be solved using factoring, the quadratic formula, or completing the square.
Glossary for FOIL Method
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Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
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11669 | https://byjus.com/maths/types-of-quadrilaterals/ | Published Time: 2018-02-15T16:08:58+05:30
Before talking about the types of quadrilaterals, let us recall what a quadrilateral is. A quadrilateral is a polygon which has the following properties
4 vertices and 4 sides enclosing 4 angles
The sum of all interior angles of a quadrilateral is 360 degrees
We can also derive the sum of interior angle from the formula of polygon i.e. (n -2) × 180, where n is equal to the number of sides of the polygon
A quadrilateral, in general, has sides of different lengths and angles of different measures. However, squares, rectangles, etc. are special types of quadrilaterals with some of their sides and angles being equal. This is the reason that the area of quadrilateral depends on which type of quadrilateral it is. In this article, we will discuss the special types of quadrilaterals and their basic properties.
Table of Contents:Quadrilateral TypesTrapeziumParallelogramRectangleRhombusSquareKitePropertiesSolved ExamplesPractice ProblemsFAQs
Different Types of Quadrilaterals
There are six basic types of quadrilaterals. They are:
Trapezium
Parallelogram
Rectangle
Rhombus
Square
Kite
Trapezium
It is a quadrilateral with one pair of opposite parallel sides. In the trapezium, ABCD, side AB is parallel to side CD.
Parallelogram
It is a quadrilateral with two pairs of parallel sides. The opposite sides are parallel and equal in length. The opposite angles are equal in measure. In the parallelogram, ABCD, side AB is parallel to side CD and side AD is parallel to side BC.
Also, the two diagonals formed to intersect each other at the midpoints. As in the figure given below, E is the point where both the diagonals meet. So
Length AE = EC, & Length BE = ED
Rectangle
It is a quadrilateral with all the 4 angles of equal measure, that is, each of them is 90°. Both the pairs of opposite sides are parallel and equal in length.
Rhombus
It is a quadrilateral with all four sides having equal lengths. The Opposite sides of a rhombus are parallel and opposite angles are equal.
Square
It is a quadrilateral in which all the sides and angles are equal. Every angle is a right angle (i.e. 90° each). The pairs of opposite sides are parallel to each other.
Kite
It is a quadrilateral that has 2 pairs of equal-length sides and these sides are adjacent to each other.
Some points about quadrilaterals to be kept in mind are:
Square, rectangle, and rhombus are types of parallelograms.
A square is a rectangle as well as a rhombus.
The rectangle and rhombus are not a square.
A parallelogram is a trapezium.
A trapezium is not a parallelogram.
Kite is not a parallelogram.
Properties of Different Types of Quadrilaterals
The below table contains the properties of various types of quadrilaterals and their corresponding basic formulas.
Type of QuadrilateralPropertiesFormulasTrapeziumOnly one pair of opposite sides are parallelNon-parallel sides are called legsDiagonals intersect each otherThe length of the mid-segment is equal to half the sum of the parallel basesArea of trapezium = (½) (a + b)h= (½) (Sum of two parallel sides) × HeightPerimeter = Sum of all the sidesParallelogramThe opposite sides are parallel and equalThe opposite angles are equalThe sum of two consecutive angles is 180 degreesIf any one of the angles is a right angle, then all the other angles will be right anglesDiagonals bisect each otherEach diagonal bisects the parallelogram into two congruent trianglesArea = Base × HeightPerimeter = Sum of all the sideRectangleThe opposite sides are parallel and equalEach interior angle is a right angleThe diagonals bisect each other and have the same lengthArea = Length × BreadthPerimeter = 2(Length + Breadth)RhombusAll sides of the rhombus are congruent, and the opposite sides are parallelOpposite angles of a rhombus are congruentDiagonals bisect each other at right anglesDiagonals bisect the interior angles of a rhombusThe sum of two adjacent angles is equal to 180 degreesArea = (½) (d1 × d2)= (½) (Product of the length of diagonals)Perimeter = 4(side length)SquareAll four sides of the square are equal, and the opposite sides of the square are parallel to each otherThe measure of each interior angle is 90 degreesThe diagonals of the square are equal to each other and bisect each other at right anglesArea = (side)²Perimeter = 4(side)KiteTwo pairs of adjacent sides are congruentOne pair of opposite angles are equal An axis of symmetry through one pair of opposite anglesTwo diagonals are not of the same lengthDiagonals intersect each other at right anglesArea = (½) (d1 × d2)= (½) (Product of the length of diagonals)Perimeter = Sum of all the sides
Solved Examples
Example 1:
If the perimeter of a square is 72 cm, then find its area.
Solution:
Let a be the side of a square.
Perimeter of a square = 4a
4a = 72 cm (given)
a = 72/4 = 18
Thus, side of the square = 18 cm
Area of the square = a² \= (18)² \= 324 cm²
Example 2:
The area of a trapezium is 180 cm², and its height is 9 cm. If one of the parallel sides is longer than the other by 6 cm, find the two parallel sides.
Solution:
Let x be the length of the shorter parallel side.
So, the length of the longer side = (x + 6) cm
Height of a trapezium (distance between two parallel sides) = h = 9 cm
As we know,
Area of trapezium = (1/2) × (sum of parallel sides) × Height
Thus, (1/2) (x + x + 6) × 9 = 180 [given]
2x + 6 = (180 × 2)/9
2x + 6 = 40
2x = 40 – 6 = 34
x = 34/2 = 17. cm
Now, x + 6 = 17 + 6 = 23 cm
Therefore, the length of the two parallel sides will be 17 cm and 23 cm.
Practice Problems
The area of a rhombus is 240 square units, and one of the diagonal is 16 units. Find another diagonal.
Calculate the area of a rectangle of length of 23 cm and breadth of 13 cm.
What will be the area of a kite whose diagonals are of lengths 5 m and 7 m?
What is the area of the rhombus with a side of 25 cm, and the length of one of its diagonals is 14 cm?
To learn more about types of quadrilaterals, download BYJU’S- The Learning App to watch the interactive videos to learn with ease.
Frequently Asked Questions on Types of Quadrilaterals
Q1
What are the six special quadrilaterals?
The six special types of quadrilaterals include:
Trapezium
Parallelogram
Rectangle
Rhombus
Square
Kite
Q2
Which quadrilateral has four congruent sides?
Two quadrilaterals contain congruent sides, they are square and rhombus.
Q3
What kind of quadrilaterals has 4 equal sides and 4 right angles?
Square has four equal sides and four right angles at each of its vertices
Q4
Do kites have parallel sides?
No, there are no parallel sides for a kite.
Test your knowledge on Types Of Quadrilaterals
Q5
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11670 | https://www.tutorialspoint.com/c_standard_library/c_function_strtol.htm | Published Time: 2012-08-29T10:08
C Function strtol
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C library - strtol() function
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The C stdlib library strtol() function is used to convert a string to a long integer number according to the given base. which must be lies between 2 and 32 inclusive, or be the special value 0.
Syntax
Following is the C library syntax of the strtol() function −
long int strtol(const char str, char endptr, int base)
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Parameters
This function accepts following parameters −
str − It is a string to be converted into a long integer value.
endptr − It is a pointer to a charter pointer and used to store the pointer to the first character after the numeric value.
base − It represents the base of the number system(e.g., 10 for decimal and 16 for hexadecimal).
Return Value
This function returns the converted long integer value. If the input string is not a valid, it returns 0.
Example 1
In this example, we create a basic c program to demonstrate how to use the strtol() function.
Open Compiler
include <stdio.h>
include <stdlib.h>
int main() {
const char str \= "12345abc";
char endptr;
long int num;
// Convert the string to a long integer
num \= strtol(str, &endptr, 10);
if (endptr \== str) {
printf("No digits were found.\n");
} else if (endptr != '\0') {
printf("Invalid character: %c\n", endptr);
} else {
printf("The number is: %ld\n", num);
}
return 0;
}
Output
Following is the output −
Invalid character: a
Example 2
The example below convert a decimal number string to a long decimal integer, using the strtol() function.
Open Compiler
include <stdio.h>
include <stdlib.h>
int main() {
const char str \= "12205";
char endptr;
long int num;
// Convert the string to a long integer
num \= strtol(str, &endptr, 10);
//display the long integer number
printf("The number is: %ld\n", num);
return 0;
}
Output
Following is the output −
The number is: 12205
Example 3
ILet's create another example, we convert the number into the long hexadecimal number, using the strtol() function.
Open Compiler
include <stdio.h>
include <stdlib.h>
int main() {
const char str \= "2024 tutorialspoint";
char endptr;
long int num;
// Convert the string to a long integer
num \= strtol(str, &endptr, 16);
//display the long hexadecimal integer number
printf("The hexadecimal number is: %ld\n", num);
return 0;
}
Output
Following is the output −
The hexadecimal number is: 8228
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11671 | https://lifescience.opensource.epam.com/_downloads/52fcb4ab88eb6e0c6fd6f7cda2a7187d/Inchi%20tautomers.pdf | Tautomer Identification and Tautomer Structure Generation Based on the InChI Code Torsten Thalheim,†,‡ Armin Vollmer,† Ralf-Uwe Ebert,† Ralph Ku ¨hne,† and Gerrit Schu ¨u ¨rmann,†,‡ UFZ Department of Ecological Chemistry, Helmholtz Centre for Environmental Research, Permoserstrasse 15, 04318 Leipzig, Germany, and Institute for Organic Chemistry, Technical University Bergakademie Freiberg, Leipziger Strasse 29, 09596 Freiberg, Germany Received March 27, 2010 An algorithm is introduced that enables a fast generation of all possible prototropic tautomers resulting from the mobile H atoms and associated heteroatoms as defined in the InChI code. The InChI-derived set of possible tautomers comprises (1,3)-shifts for open-chain molecules and (1,n)-shifts (with n being an odd number >3) for ring systems. In addition, our algorithm includes also, as extension to the InChI scope, those larger (1,n)-shifts that can be constructed from joining separate but conjugated InChI sequences of tautomer-active heteroatoms. The developed algorithm is described in detail, with all major steps illustrated through explicit examples. Application to ∼72 500 organic compounds taken from EINECS (European Inventory of Existing Commercial Chemical Substances) shows that around 11% of the substances occur in different heteroatom-prototropic tautomeric forms. Additional QSAR (quantitative structure-activity relationship) predictions of their soil sorption coefficient and water solubility reveal variations across tautomers up to more than two and 4 orders of magnitude, respectively. For a small subset of nine compounds, analysis of quantum chemically predicted tautomer energies supports the view that among all tautomers of a given compound, those restricted to H atom exchanges between heteroatoms usually include the thermodynamically most stable structures.
INTRODUCTION Isomerism denotes the rearrangement of bonds within a given molecule. Tautomers are isomers that can be trans-formed to each other through chemical equilibrium reactions (typically with free energy changes below 25 kJ/mol).
Prototropic tautomerism refers to bond changes accompanied with the intramolecular movement of hydrogen atoms.
Prominent examples involving hydrogen exchange between carbon and heteroatoms include the keto-enol and imine-enamine equilibria. For a given compound, different tauto-meric forms may differ in the number and type of certain functional groups, such as H bond donor and acceptor sites, which in turn result in tautomer-specific affinities for media, such as water and organic solvents.
Because a given tautomer represents a distinct chemical structure, the level of handling tautomers may have signifi-cant effects on the success rate in drug research and virtual screening as well as with regard to substructure search in large databases.1-5 Typically, one tautomeric form is ener-getically preferred in a given medium. Then, any model to predict a property from the chemical structure will be incorrect, if an inadequate tautomer is submitted to the model.
In other cases, the situation may be more complicated, and several different or even all possible tautomers of a particular compound may need to be taken into account, instead of considering a more or less arbitrary single structure. Pospisil et al.6 and Martin7 describe the impact of tautomerism on the quality of property prediction while using the appropriate tautomer form. In any case, the first step to address these problems is to identify all chemically reasonable tautomers.
For the latter, the typical approach is to employ rule-based transformations of a given initial structure to identify and enumerate all associated tautomers. Despite the flexibility and transparency of this approach, some new problems appear: many rules are accompanied with exceptions that in turn would require the development of additional rules for a fully automatized handling. For example, the rules of the Daylight SMIRKS notation introduced by Leach8 were used for tautomer patterns by Oellien.2 Here, 21 rules address particular occurrences of tautomerism. Of these, two rules identify (1,3)-shifts of H atoms of imines. This requires some exceptions for ring systems and for a possible terminal nitrogen atom. Another rule relates to an aromatic ring with two terminal heteroatoms. Without further specification, 1,3-benzenediol would wrongly match. In this latter case, an exception is defined, requiring one terminal atom to be nitrogen whereas the other atom could be either nitrogen or oxygen. Another rule-based algorithm is implemented by Trepalin et al.,9 with each rule describing a certain type of tautomer pairs.
A further approach is given by Haranczyk and Gutowksi.10 Here, the user has to identify all atomic sites and bonds of the molecule involved in tautomeric transformations (heavy atoms, mobile hydrogen atoms, respective bonds). The algorithm then excludes implausible structures based on a set of constraints, and predicts the most stable tautomer.
Another kind of tautomer generation has been presented by Haran ´czyk et al.11 Here, a carbon skeleton is deduced from the initial molecular structure. With a boolean vector (named fingerprint) the hydrogen atoms are assigned to their con- To whom correspondence should be addressed. Tel: +49-341-235-1262. Fax: +49-341-235-1785. E-mail: gerrit.schuurmann@ufz.de.
† UFZ Helmholtz Centre for Environmental Research.
‡ Technical University Bergakademie Freiberg.
J. Chem. Inf. Model. 2010, 50, 1223–1232 1223 10.1021/ci1001179 2010 American Chemical Society Published on Web 06/29/2010 necting atoms, and an enumeration process allows an easy generation of all possibilities. In addition, the fingerprint could be used very simply to detect symmetry. This method is the most similar to our approach.
Milletti et al.12 use a simple (1,3)-shift scheme to create prototropic tautomers. From a given initial structure, all tautomers based on the shift scheme will be generated. The procedure is applied again to all newly generated structures.
This will be repeated, until no new structures appear. As this recursive execution is very exhaustive, some strategies to avoid redundancy were introduced. This includes the use of InChI13 to detect symmetric structures. The recursive calculation is skipped for structures that are generated for the second time. In addition, the energy difference and energy barrier to interconvert between both forms of an (1,3)-shift are analyzed during the generation. If the energy difference exceeds a threshold, the recursion is skipped too. Finally, a fragment- and pKa-based evaluation is performed to predict the generated structure’s stability.
For the computerized generation of tautomers, several software tools have been developed.10,14-18 TautTGen,10 ConGENER,14 and ACDChemSketch15 are freely available; the source codes of TautTGen and ConGENER are provided via sourceforge.net,10,14 and the algorithm has been publi-shed.1,11 MN.TAUTOMER16 is based on a set of eight rules18 and can be tested with an online demo version.
SPARC17 offers an online application, but there is no explicit explanation of how the algorithm works.
ACDChemSketch is freeware for educational and private use, but the tautomer generation algorithm is not explained in detail in the freeware version. The tautomer is supposed to be independent from the initial structure because hydrogen atoms disposable to move will be detected automatically.
However, this is restricted in some cases as can be illustrated with allopurinol (C5H4N4O, 1H-pyrazolo[3,4-d]pyrimidine-4-ol, IUPAC name 3,5,7,8-tetrazabicyclo[4.3.0]nona-3,5,9-triene-2-one), a drug used to treat hyperuricema (chronic gout) through inhibition of xanthine oxidase and the associ-ated generation of uric acid.
Scheme 1 shows the nine tautomers of allopurinol. With ACDChemSketch, only the tautomers with Od, except the structure in the center of Scheme 1 (with both H atoms at the 5-ring N atoms), will be generated. Consequently, if one of these four structures is selected as input, the remaining three are generated, resulting in a total of four tautomers identified. In all other cases, the ACDChemSketch algorithm yields five tautomers: The four structures just mentioned and the input structure. The reasoning behind this algorithm might be a pragmatic approach to avoid energetically unreasonable tautomers, which however is neither documented nor dem-onstrated as such.
In recent years, the great potential of the International Chemical Identifier, InChI,13 to store and handle chemical information on the connection table level in large chemical databases has been recognized. In addition to line notations, such as the SMILES code,19 to define the molecular topology and bond types, the InChI code is becoming very popular and can also be used as unique identifier, such as the CAS number.20 Even though InChI is proprietary, it is freely available from the intellectual right holders, including computerized versions even with the source code. Interest-ingly, InChI contains pertinent information about prototropic tautomers with regard to mobile hydrogen atoms attached to heteroatoms. The reasoning behind focusing on this subset of tautomers is that for a given chemical structure, the thermodynamically most stable tautomer is often among those that are related to each other through hydrogen exchange between heteroatoms accompanied by respective exchanges of single and double bonds.
The aim of our study is to provide an algorithm to construct all prototropic tautomers that can be generated from the mobile H atoms and associated heteroatom acceptor sites specified through the InChI code. The respective tautomeric equilibrium is outlined in Scheme 2, specifying both the types of heteroatoms considered and the associated change between single and double bonds upon movement of a given H atom between different heteroatoms of the molecule. In this way, the algorithm enables a fast identification of stable tautomers and thus provides pertinent information for both structure Scheme 1. Nine Tautomers of Allopurinol and the Associated InChI Code Specifying the Mobile H Atoms (the Latter of Which Are Given in Bold)a a The InChI atom numbering (which differs from the IUPAC numbering) is specified in the top left tautomer. As can be seen from the InChI code (bottom), the “H2” notation indicates the presence of two mobile H atoms, which can be attached to any two of the five heteroatoms #6-10 covering four nitrogens (#6-9) and one oxygen (#10).
Scheme 2. Prototropic Tautomerism Because of (1,3)-Shifts of H Atoms between Heteroatoms22 (Upper Part) and H Atom Movement Extended to (1,n)-Shifts, with n being Odd and Larger than 3, upon Insertion of Conjugated Fragments (Lower Part)a a The InChI definition of mobile H atoms and associated heteroatom acceptor sites covers all respective (1,3)-shifts for open-chain structures, and larger (1,n)-shifts across conjugated ring systems. In addition to that, our algorithm includes also (1,m)-shifts (with odd m > odd InChI n) constructed from joining separate but conjugated (1,n)-shifts.
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and substructure searches, as well as for structure-based predictions of compound properties, such as partition coef-ficients and acid-base equilibria. As shown recently for the example of N-hydroxyl amidines,21 H atom shifts between heteroatoms typically preserve the structure stability better than hydrogen shifts involving carbon atoms.
INCHI CODE INFORMATION ON TAUTOMERS The IUPAC International Chemical Identifier (InChI)22 was introduced in 2000. InChI serves as unique identifier for chemical substances, and with its unambiguous notation allows for an easy comparison of chemical structures. Thus, it is applied in various free and commercial databases, such as the U.S. National Institute of Standards and Technology23 or the Kyoto Encyclopedia of Genes and Genomes.24 The InChI code covers six major layers, of which the last two (fixed-H layer and reconnected layer for metals) are optional. The first major layer provides the main features of the molecular structure and consists of three sublayers: chemical formula, connection table without specification of bond types and without H atoms, and H atoms covering both hydrogens attached to specific sites (immobile H atoms) and so-called mobile H atoms together with lists of heteroatoms as possible (but at this stage not fixed) attachment sites. These InChI-defined mobile H atoms and their associated hetero-atom acceptor sites provide the key input information for our algorithm to generate, from a given molecular structure, all respective tautomeric forms. The remaining major layers 2-4 of InChI are the charge layer, the stereochemical layer, and the isotopic layer, all of which are not needed for our algorithm.
The fifth major layer allocates the mobile H atoms to particular heteroatoms, thus yielding a uniquely defined molecular structure. Our algorithm uses this layer to remove, after initial generation of all possible tautomers according to the InChI-defined mobile H atoms and their heteroatom acceptor sites, all duplicates, thus yielding the final set of structurally different prototropic tautomers (see below).
Note further that the InChI code can be specified in two different forms: The InChI string as also given in our examples (see Schemes 1, 5, and 6), and its associated hash code preferred for machine reading, the InChI key. Use of the latter speeds up the tautomer comparison significantly, and it is used in our algorithm.
MATERIALS AND METHODS Algorithm. The algorithm has been coded in C++ and developed as module of our in-house software ChemProp.25 It runs on Windows PCs under standard configurations.
After input of the molecular structure in a standard format, such as the InChI code, the SMILES code, the MOLfile, or the SMDfile format, ChemProp converts the structural information into an internal format. In case the input format was not InChI, the InChI-code information is generated internally. Subsequently, the algorithm calculates the number of all different prototropic tautomers available from the InChI-defined mobile H atoms and associated tautomer-active heteroatoms. Finally, all respective tautomers are generated and stored in a fully automatized manner, and thus are made available for later inspection or further separate use.
Tautomer Energy. For a small set of tautomers, molecular energies in terms of heats of formation at 25 °C, ∆Hf, have been calculated employing the semiempirical quantum chemical AM126 method as implemented in MOPAC.27 To this end, initial 3D geometries have been generated within ChemProp, and the final geometry optimization has been performed at the AM1 level employing MOPAC.
QSAR Calculations. To test the variation of QSAR predictions across tautomers, logarithmic values of the soil sorption coefficient normalized to organic carbon, log Koc, and of the water solubility at 25 °C, log Sw, have been calculated with literature methods28,29 as implemented in ChemProp.
Molecular Structures. For the development and applica-tion of the tautomer-generation algorithm, ∼72 500 organic molecules have been retrieved from the list of European Inventory of Existing Commercial Chemical Substances (EINECS)30,31 and stored in a ChemProp database.
RESULTS AND DISCUSSION Tautomer Identification. To identify prototropic hetero-atom-confined tautomerism as outlined in Scheme 2, it is just necessary to compare the main InChI layer and the charge layer. The other layers will be disregarded at this stage. Since InChI generates a canonical atom enumeration (for non-H atoms) by a modified Morgan algorithm,32 different molecules with the same main layer are tautomers of each other. Thus, restriction to the main layer yields a generic code for all tautomers of a particular compound.
Specification of a particular form then requires extension to the fifth main layer (see above). In other words, comparison of two structures in terms of being tautomeric forms of a common general structure can very easily be achieved by simple string operations. The real challenge is to generate all possible tautomer forms for a given structure.
Tautomer Generation. Since the InChI code already specifies the mobile H atoms, the task left is a procedure to automatically generate, from any of the possible tautomeric forms as starting structure, all respective tautomers. To this end, a brute-force approach can be applied: From the structure of a given compound, the InChI code is generated, and from the resulting mobile H sublayer the specified number of H atoms simply has to be attached to all possible heteroatom positions in a combinatorial manner in accord with chemical bonding rules.
In the next step, a completion process for each generated structure is required to obtain a valid configuration of valences and bonds. Within InChI, heteroatoms acting as possible H atom acceptors may have single bonds only (sp3 hybridization) or are involved in double bonds (sp2 hybrid-ization). Here, attachment or detachment of hydrogen is accompanied by a respective change in the hybridization.
All other atoms not involved in the H atom movement must retain their initial hybridization in all tautomers. Correspond-ingly, the types of the bonds not involved in tautomerism remain the same for all tautomers. Moreover, each single bond converted to a double bond upon tautomerism is balanced by a double bond becoming a single bond. It follows that the number of double bonds of a given compound is also constant for all tautomers. Because this number of double bonds is not directly apparent from the TAUTOMER GENERATION BASED ON INCHI J. Chem. Inf. Model., Vol. 50, No. 7, 2010 1225 InChI Code, our algorithm takes this value from the initial structure used to generate all respective tautomers, and adds the appropriate number of double bonds to each of the tautomers generated.
In terms of hydrogen shifts, the mobile H atoms defined through InChI refer primarily to (1,3)-shifts of open-chain molecules (Scheme 2), and for ring systems also to (1,n)-shifts with n being an odd number larger than 3. Two examples of the latter are given in Scheme 3, illustrating an (1,5)-shift (top) and an (1,7)-shift (bottom) through respective participation of ring atoms. Because InChI does not consider (1,n)-shifts (with n ) 5, 7, ...) across two (or more) rings of fused systems (except those that can be reduced to equivalent sequences of (1,3)-shifts), respectively defined mobile H atoms are missing as basis for generating the associated tautomeric forms. However, our algorithm provides the following extension to the InChI set of possible tautomers: If two InChI-based H atom shifts can be combined to a larger shift across an additional conjugated fragment, this larger shift will be detected and taken into account. Scheme 4 contains an example of an (1,5)-shift across two fused rings and its equivalent representation as consecutive sequence of two (1,3)-shifts (top), and an example of a nonreducible (1,5)-shift that is outside the InChI scope, but addressed through our algorithm (bottom). According to present experience, our algorithm yields approximately 1/3 more tautomers than the original InChI approach in those cases where the structure contains conjugated fragments linking otherwise isolated H atom shifts.
For a further illustration of our algorithm, we come back to the drug allopurinol (Scheme 1, with the InChI code given at the bottom). According to the canonical formula repre-sented in InChI as C5H4N4O, the numbering is done by starting with all carbon atoms followed by nitrogen atoms and the oxygen atom. The next sublayer c10-5-3-1-8-9-4(3)6-2-7-5 sketches the molecule skeleton similar to a structural formula by referring to carbon atoms 1-5, nitrogen atoms 6-9, and oxygen atom 10. This is shown in all tautomer skeletons in Scheme 1. The sublayer /h finally assigns the H atoms: Two of the four H atoms are fixed at carbon atoms 1 and 2 at the respective chain nodes by the string 1-2H, and the mobile hydrogen atoms are left in (H2,6,7,8,9,10).
H2 denotes two H atoms, attached to two of five possible positions offered by the four nitrogens (#6,7,8,9) and the one oxygen (#10). Thus, ( 5 2) ) 10 possible arrangements have to be checked. Nine of them yield the valid structures shown in Scheme 1. The only case without success is the selection of the nitrogen atoms #6 and 7 that are separated by carbon atom #2. Parallel H attachment to both of these N atoms is not compatible with a closed-shell conjugated system. The technical procedure to extract InChI information and to find valid structures will be explained in the next section in more detail.
In the final step of the tautomer generation, the resultant valid structures need to be compared to remove duplicates.
Taking carbamimidothioic acid (Scheme 5) as example, it is seen that two of the initially generated three tautomers (left and right) are identical, just mirrored at the S-C bond.
Other examples are carbonyls. Interestingly, InChI marks COOH groups to have a mobile hydrogen atom, although the resultant (formally correct) tautomers with H at the one or other O atom cannot be distinguished. In principle, such duplicates could be avoided by a respective algorithm in the combinatorial step. However, this would not result in a significant gain of performance, so the suggested procedure is to compare the results, and to simply remove identical structures.
The latter requires a method for comparing molecular structures, for which there are several techniques available.
One approach, similar to the Milletti12 method, is to generate the InChI keys of the structures, this time with the option mobile H perception deactivated. The result is an additional layer with defined attachments of all H atoms, including the tautomer-mobile hydrogens. Using the key enables a simple removal of duplicates: Every key must appear only once, and thus duplicate keys and their related structures can be deleted easily.
Scheme 3. InChI-Based Prototropic Tautomeric Equilibria Involving an (1,5)-Shift (Top) and an (1,7)-Shift (Bottom) Across Conjugated Ring Structures As Examples of (1,n)-shifts with n Being Odd and Larger than 3 Scheme 4. Examples of an (1,5)-Shift Across Two Fused Rings and Its Equivalent Representation As Consecutive Sequence of Two (1,3)-Shifts (Top) and of a Nonreducible (1,5)-Shift That Is Outside the InChI Scope but Addressed through Our Algorithm (Bottom) Scheme 5. Symmetric Heteroatom Acceptors (Nitrogen Atoms) for Tautomeric H Shifts in Carbamimidothioic Acida a Exchange of two H atoms between the nitrogen groups yields an identical structure, only mirrored at the S-C bond.
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Algorithm for Generating Individual Tautomers. The algorithm introduced here involves three steps: First, the heteroatoms available for hydrogen attachment are identified, and the mobile H atoms as defined through InChI are subdivided into tautomer-active and tautomer-passive subsets.
The latter subset comprises those H atoms that despite their principally mobile character remain fixed because of valence rules (see below). At this stage, carbon-carbon bonds involved in tautomerism are not yet identified.
In the next step, all atoms and bonds not participating in tautomeric rearrangements will be removed, yielding a tautomer skeleton with single bonds only. Then, the attach-ment sites for the hydrogen atoms are specified in a combinatorial manner. Third, the algorithm tries to define double bonds along the atom-atom connections of this skeleton with the attached hydrogens in a conjugated manner in all possible constellations. If this is not possible, the skeleton is rejected and no longer tautomer-relevant. Oth-erwise, the accordingly generated molecular structures are checked for duplicates that are deleted as well, yielding the final set of prototropic tautomers.
Step 1: Identification of Heteroatom H Acceptors and of Tautomer-ActiVe H Atoms. All H atom-accepting heteroa-toms are directly available from the respective InChI code fragment. However, the InChI information on mobile H atoms needs to be modified for the purpose of tautomer generation.
First, multiple InChI sequences of mobile hydrogens are joined into a single sequence. Taking the InChI code of enprofylline as example (Scheme 6, bottom), the notation (H,9,10)(H,11,13,14) indicates the following two sequences of different mobile H atoms: One mobile H atom can be attached at sites 9 and 10 (that means at one of the two 5-ring nitrogens) and another mobile H atom at one of the sites 11, 13, and 14 (6-ring heteroatoms except the propyl-substituted nitrogen). The reason for that splitting is to avoid implausible structures. However, this splitting results in an a priori exclusion of two tautomers with no mobile H atom attached to the 5-ring nitrogens (see the two framed structures at the bottom right of Scheme 6). Accordingly, it appears more satisfactory to initially allow for the generation of an upper bound of tautomer candidates and then exclude valence-incorrect settings as well as duplicates. Thus, in case of enprofylline our algorithm converts the separate InChI sequences (H,9,10)(H,11,13,14) into the combined sequence (H2,9,10,11,13,14), now indicating that there are two, formally equivalent, mobile H atoms that can be attached to any two of the five heteroatom sites available. In this way, also tautomer candidates with both H atoms attached at the 5-ring or at the 6-ring are taken into account. The subsequent valence check then reveals that the option with both H atoms at the two 5-ring nitrogens does in fact not lead to valid structures and is then excluded from further consideration.
Second, the mobile H atoms are checked for multiple occurrences at single sites. While all mobile H atoms identified through InChI are mobile in principle, the ones attached to the same heteroatom are not mobile at the same time. Moreover, if one of several mobile H atoms at a given heteroatom has been shifted to generate a new tautomer, then separate consideration of the remaining H atoms would yield identical results (duplicates) and thus no further new tau-tomer. In the algorithm, this issue is addressed as follows: Initially, the InChI-defined mobile H atoms are considered as potentially mobile. Then, each heteroatom acceptor site is checked for free valences in the first InChI sublayer, where all bonds are still single bonds and where the (potentially) mobile H atoms are not yet included. Subsequently, H atoms are attached to heteroatoms with free valences until the latter Scheme 6. Eight Tautomers of Enprofyllinea a The InChI code given in the bottom refers to the first six tautomers, specifying the mobile H atoms and associated heteroatom acceptor sites.
The additional two tautomers, shown as framed structures at the bottom right, are not covered by the InChI code given because of its separate InChI shift options (H,9,10) and (H,11,13,14). However, they become included when employing the combined sequence (H2, 9, 10, 11, 13, 14) as is done with our tautomer-generation algorithm. The numbers in brackets specify the individual tautomers in terms of the heteroatoms (InChI numbering) selected as H atom attachment sites.
TAUTOMER GENERATION BASED ON INCHI J. Chem. Inf. Model., Vol. 50, No. 7, 2010 1227 have gone down to one. The remaining initially mobile H atoms are now classified as tautomer-passive H atoms, and no longer belong to the subset of tautomer-relevant H atoms.
Correspondingly, the ones successfully attached to any of the heteroatoms are classified as tautomer-active H atoms, and thus form the final subset of mobile H atoms used for generating individual tautomers.
The approach is illustrated with 4,6-diaminopyrimidin-2-ol in Scheme 7. As can be seen, all nine initially generated tautomers, including four duplicates because of molecular symmetry (all four bottom tautomers are duplicates of located top counterparts, the only unique tautomer being the structure in the center of Scheme 7), contain at least one H atom at each of the two nitrogens. These principally mobile but tautomer-passive H atoms are printed in bold in the scheme.
Subsequent exclusion of duplicates (as described in step 4 below) yields five different tautomers, covering the central structure, as well as the four structures of the four tautomers located in the upper half circle of Scheme 7.
Step 2: Tautomeric Skeleton Generation. Scheme 8 il-lustrates the algorithm to obtain the tautomeric skeleton with enprofylline taken from Scheme 6 (now using tautomer {10,11} as starting structure). Because sp3-hybridized atoms except H-acceptor heteroatoms cannot participate in tauto-meric rearrangements, respective subunits are identified and removed from the initial molecular structure. In case of enprofylline, this concerns the N-propyl subunit shown in bold in Scheme 8. Then, all H atoms and all atoms with triple bonds are removed. The latter is done because the InChI-generated mobile H atoms do not refer to tautomeric rearrangements involving triple-bonded atoms. After subse-quent removal of all bonds not connecting two atoms, the remaining double bonds are counted because their number remains constant across all prototropic tautomers (see above) and then converted to single bonds, thus generating the maximum degree of free valences of all tautomer-relevant heteroatoms. The resultant tautomeric skeleton is now ready for tautomer generation. Note, however, that this skeleton is not suited for detecting ring-chain tautomerism, the latter of which is already outside the respective InChI scope (as defined through the InChI mobile H atoms and associated heteroatom acceptor sites).
Step 3: Combinatorial Assignment of Tautomer-ActiVe H Atoms and Double Bonds. First, the number of possible attachments of tautomer-active H atoms to heteroatom accepting sites is calculated. Then, each respective combi-natorial subset (each particular setting of H attachments) is mapped onto the tautomeric skeleton, and for each such bonding pattern the relevant number of double bonds is introduced to meet the valence rules. In graph theoretical terms, this means that edges will be added to a simple graph, until the accordingly modified graph consists of a maximal number of double edges.
More precisely, the algorithm proceeds as follows: After initial attachment of the tautomer-active H atoms to particular heteroatoms, these latter have already reached saturation of their valences and have been identified as such and, temporarily, removed from the structural skeleton. Then, the atom of the remaining skeleton with the fewest unsaturated neighbors, called minimal node degree in Scheme 9, is selected (randomly in case of more than one equivalent option), and a double bond is introduced between this atom and an unsaturated neighbor. Through this step, the valences of the double-bonded atoms become saturated, and both bonding partners are, again temporarily, removed. This procedure is repeated step by step until the required number of double bonds (that had been calculated previously, see above) has been introduced. At this point in time, all valences of all atoms have been saturated, and the final bonding pattern of the particular tautomer is fixed. Now, the complete tautomeric structure is built through joining the initially Scheme 7. All Formally Valid Heteroatom-Prototropic Tautomers of 4,6-Diaminopyrimidine-2-ol, Covering the Unique Tautomer in the Center and Two Sets of Four Distinct Tautomers (Top Four and Bottom Four Structures Arranged in Half Circles) that are Duplicates of Each Other Because of Molecular Symmetrya a Among these initially generated nine tautomers, the four duplicates (say, the ones located in the bottom half circle) are removed in step 4 of our algorithm (see text). Two of the InChI-mobile H atoms are tautomer-passive, because multiple H atoms at heteroatoms cannot be moved at the same time and because their separate movement would yield tautomers already generated otherwise (see step 1 as described in the text). The principally mobile but tautomer-passive H atoms are printed in bold.
Scheme 8. Construction of the Tautomeric Skeleton of Enprofylline, Employing the Structure with Mobile H Atoms Attached at Nitrogen #10 and #11 (Tautomer {10,11} in Scheme 6) as Starting Structurea a Initial deletion of the tautomer-passive N-propyl unit is followed by subsequent removal of H atoms and double bonds to yield the final tautomeric skeleton. The latter is used as starting point to generate all tautomers as described in the text.
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removed tautomer-passive substructures (see step 2 above) and the tautomer-active structural units just described.
During the stepwise procedure of removing atom pairs connected through double bonds, some atoms, or nodes in graph theoretical notation, may become isolated. A node is isolated, when it is not saturated yet, but none of the neighbor atoms still contains free valences. In this case, the last steps will be revoked until an atom is reached that has more than one unsaturated neighbor. Then, for this atom a double-bond partner is selected that is different from the previous unsuccessful path, and the stepwise procedure continues as described above. If, however, no such alternative pathway can be found when going backward from an isolated node, the particular allocation of H atoms to heteroatom acceptors is invalid, and the respective combinatorial subset is rejected.
Step 4: Duplicate RemoVal. There are several straightfor-ward approaches to achieve this task. A simple but effective opportunity is to compare the InChI keys of the resulting tautomers: After the creation of all valid subsets, the codes including the fixed fixed-H layer will be generated. In the original InChI software, this option is called “without the mobile H perception”. This provides a means to detect identical solutions by hash code comparison. Duplicates detected in this way are then removed, resulting in the final set of different prototropic tautomers. Coming back to 4,6-diaminopyrimidin-2-ol in Scheme 7, the four tautomers in the lower part are duplicates of the four structures in the upper part, which is detected and accounted for at this step 4 of our algorithm.
Example. The procedure is illustrated further with the already discussed drug enprofylline. From the InChI code (Scheme 6, bottom), five H-accepting heteroatom sites (N atoms #9-11, and O atoms #13-14) are obtained, identify-ing the propyl-substituted nitrogen (atom #12) as tautomer-passive. In this case, the InChI number of mobile H atoms equals the number of tautomer-active H atoms, which is two.
Generation of the tautomeric skeleton (step 2) through removal of tautomer-passive sp3 atoms (here, N-propyl unit) has been illustrated in Scheme 8. Scheme 9 now shows the stepwise construction of the tautomer {9,11} with the two mobile H atoms attached at nitrogens #9 and 11, which forms a particular combinatorial subset of the ( 5 2) ) 10 different settings of attaching the two mobile H atoms to any two of the five heteroatom acceptors (step 3). After initial attachment of the hydrogens at the nitrogens #9 and 11 (first substep of step 3 ) step 3.1), these nitrogens have reached their valence saturation and are removed (step 3.2). Subsequently, double bonds (double edges) are introduced in a stepwise manner such that no isolated nodes arise (steps 3.3, 3.4, 3.5). For each of the 10 initial allocations of the mobile H atoms to tautomer-active heteroatoms, the procedure attempts to assign the required number of double bonds (edges), in this case four double bonds, to the resulting molecular fragments.
However, the attachment settings {9,10} (H atoms attached to both 5-ring nitrogens) and {11,14} are rejected because of isolated nodes (node #4 resulting from {9,10}, and node #8 resulting from {11,14}). In step 4, the remaining eight valid settings are compared via the fixed-H layer layer of InChI. Because there is no duplicate structure detected, these eight structures differ from each other and form the final set of valid tautomers, already shown in Scheme 6.
Application to a Large Compound Set. To demonstrate the potential relevance of considering tautomers, our algorithm has been applied to ∼72 500 organic molecules of the European Inventory of Existing Commercial Chemical Substances (EINECS).30,31 In 11% of these structures, tautomerism based on H atom shifts between heteroatom occurs. For most of the compounds, the number of different tautomers does not exceed 20, and less than 40 substances have more than 500 tautomers.
In total, the algorithm took about 25 min on an Intel Core II 2.4 GHz CPU to generate the tautomers. Depending on the available hardware, there may be a considerable amount of additional time to write the results on the hard disk, which is independent of the algorithm.
QSAR Variation Across Tautomers. The variation of QSAR predictions across tautomers has been examined through application of respective models for predicting the soil sorption coefficient normalized to organic carbon, Koc,28 and water solubility (in mol/L at 25 °C), Sw,29 of the EINECS compounds. Initial exclusion of compounds with an estimated number of tautomers larger than 2500 resulted in ∼7900 tautomer sets with a total number of approximately 42 000 individual structures. For these, log Koc and log Sw have been predicted using our in-house software ChemProp.25 For this screening-level analysis, we have not considered the chemical domain,33 keeping in mind that the latter becomes important when evaluating the confidence of QSAR predictions.
Moreover, we did not exclude any of the generated tautomers on the basis of energetic arguments. Thus, the QSAR model28 for predicting log Koc could be formally applied to 7466 tautomer sets (comprising 7466 different compounds with a total of 39434 individual chemical structures), and the suite of methods29 for predicting log Sw to 7774 tautomer sets (covering 41607 individual chemical structures).
The results clearly demonstrate the importance of tautomer consideration. The standard deviation of log Koc across tautomers was almost 0.4, with individual Koc differences Scheme 9. Workflow to Generate from the Tautomeric Skeleton a Particular Completely Defined Tautomeric Structure, Taking Enprofylline As Example (Step 3 in the Algorithm, See Text)a a During the stepwise procedure, two H atoms are attached at two particular heteroatom acceptor sites (here nitrogens #9 and #11), and four double bonds are generated before their (temporary) removal together with respective bonding partners. The result corresponds to the top-left tautomer {9,11} of Scheme 6, which is obtained through building the structure from all temporarily removed substructures according to the selected path.
TAUTOMER GENERATION BASED ON INCHI J. Chem. Inf. Model., Vol. 50, No. 7, 2010 1229 between the tautomers of a given compound up to 4 orders of magnitude. For the log Sw prediction, the standard deviation across tautomers was 0.75, with maximum differ-ences in Sw of more than nine logarithmic units.
Table 1 shows the variations in predicted log Koc and log Sw for enprofylline, for primisulfron with five distinct tautomers, nitrofurantoin with three distinct tautomers, and for six further compounds with only two distinct tautomers.
In addition, the molecular energy variation across tautomers is quantified through calculated heats of formation at 25 °C, ∆Hf, employing the semiempirical quantum chemical AM126 model including geometry optimization as implemented in MOPAC.27 As can be seen from the table, the tautomer variation in log Koc is not correlated with the one in log Sw, reflecting corresponding differences in the dependencies of both properties on molecular structure. Similarly, larger differences in log Koc or log Sw are not necessarily associated with larger differences in ∆Hf.
These finding are illustrated with the calculated values obtained for enprofylline, primisulfron, sulfalene, and timi-perone: Enprofylline yields a log Koc variation across its eight tautomers of 1.11 as opposed to a corresponding log Sw variation of 0.38, accompanied by a ∆Hf variation of 118.3 kJ/mol. With primisulfron, both the log Koc and log Sw variation are significantly increased, while the variation in ∆Hf is reduced by a factor of 2. Sulfalene and timiperone show similar variations in log Koc and log Sw of about 0.6-0.7, but substantial differences in the associated ∆Hf variations (82.1 kJ/mol vs 29.3 kJ/mol; see Table 1).
Coming back to the EINECS subset of compounds with tautomerism through H atom shifts between heteroatoms, Table 2 summarizes an analysis of the variation in predicted log Koc and log Sw as compared to the number of tautomers per compound. For Koc, 70% of the compounds (5198 compounds) yield variations across tautomers smaller than 0.5 log units, and 5% (378) yield log Koc variations larger than 1. With water solubility, 80% (6237) of the compounds show variations across tautomers below one log unit, 6% log Sw variations between 1 and 2, and 1.7% log Sw variations larger than 2.
Interestingly, there is no distinct relationship between the number of tautomers per compound and the property variation across tautomers for both log Koc or log Sw. Among the (many) compounds with less than five different tautomers, variations of more than two log units are still observed for 21 compounds with regard to Koc and for 132 compounds with regard to Sw. Moreover, 33 compounds with less than five tautomers yield predicted Sw differences of more than four log units, 24 compounds with 6-50 different tautomers provide similar log Sw ranges, but none of the 28 compounds with more than 51 distinct tautomers yield log Sw variations above four log units.
Energetic Stability of Tautomers. Finally, we come back to the InChI approach of confining the H shifts to heteroa-Table 1. Variation of QSAR Predictions for log Koc and log Sw Across Tautomers and Associated Ranges of the Tautomer Energiesa predicted property range AM1 energy range of InChI-based tautomers AM1 energy range of all tautomers compound name chemical formula no. InChI-based (heteratom-confined) tautomers ∆(log Koc) ∆(log Sw) [mol/L] ∆(∆Hf) [kJ/mol] total no.
tautomers ∆(∆Hf) [kJ/mol] enprofylline C8H10N4O2 8 1.11 0.38 118.28 14 192.34 primisulfron C14H10F4N4O7S 5 1.85 0.71 67.57 11 87.62 tribenuron C14H15N5O6S 2 1.01 0.25 42.73 12 116.42 sulfalene C11H12N4O3S 2 0.61 0.72 55.04 5 137.12 nitrofurantoin C8H6N4O5 3 0.61 0.05 87.22 5 119.93 dimethoate C5H12NO3PS2 2 0.24 0.32 44.48 3 44.77 icosanamide C20H41NO 2 0.24 0.1 54.31 3 66.00 timiperone C22H24FN3OS 2 0.71 0.74 9.54 4 38.3 tromantadine C16H28N2O2 2 0.24 0.32 61.04 3 87.09 a log Koc (soil sorption coefficient normalized to organic carbon) and log Sw (water solubility in mol/L at 25 °C) have been predicted for all tautomers of the nine compounds with QSAR models taken from literature28,29 as implemented in ChemProp.25 The InChI-based tautomers confined to H atom shifts across heteroatoms were generated by our newly developed algorithm, and the additional tautomers involving sp2 carbons in H atom shifts were constructed manually, resulting in a correspondingly larger total number of tautomers as specified in the second last column. The calculated heats of formation at 25 °C, ∆Hf [kJ/mol], were obtained from AM126 calculations including geometry optimization with the MOPAC software.27 Table 2. Variation in log Koc and log Sw [mol/L] of the Tautomer-Active Subset of Organic EINECS Compounds in Terms of Value Ranges and Associated Numbers of Compounds, as Well as Numbers of Tautomers Per Compounda no. of compounds for a given property range x ) ∆(log Koc) no. of compounds for a given property range x ) ∆(log Sw) [mol/L] number of tautomers per compound x e 0.5 0.5 < x e 1 1 < x e 1.5 1.5 < x e 2 x > 2 y e 1 1 < y e 2 2 < y e 3 3 < y e 4 y > 4 e5 5198 1044 326 31 21 6237 439 76 23 33 6-50 122 359 207 99 32 524 299 61 30 24 51-500 0 2 8 3 2 5 5 2 4 0 >500 0 0 0 0 12 0 8 1 3 0 a log Koc (soil sorption coefficient normalized to organic carbon) and log Sw (water solubility at 25 °C in mol/L) have been predicted for all tautomers of the nine compounds with QSAR models taken from literature28,29 as implemented in ChemProp.25 The prototropic tautomers generated result from the mobile H atoms and associated heteroatom acceptor sites as defined in the InChI code. Omitting compounds with estimated numbers of tautomers larger than 2500 (see text); log Koc was predicted for 7466 compounds across 39 434 individual tautomers, and log Sw for 7774 compounds across 41 607 individual tautomers.
1230 J. Chem. Inf. Model., Vol. 50, No. 7, 2010 THALHEIM ET AL.
toms. As mentioned above, the reasoning behind this approach is the assumption that in most cases, the most stable prototropic tautomers are those generated through H shifts across heteroatoms.
For the nine compounds of Table 1, we have augmented these heteroatom-based tautomers manually by tautomers with H atom shifts involving sp2 carbons. Moreover, AM1 has been applied to calculate their heats of formation, ∆Hf.
As can be seen from comparing the third and the second last column of the table, inclusion of tautomer-active sp2 carbons increases the total number of tautomers by 28-60.
The associated increase in ∆Hf variation is between 0.3 and 82.1 kJ/mol, and above 10 kJ/mol in eight of the nine cases.
Interestingly, for only two compounds (primisulfron and tribenuron), the lowest-energy tautomer is one involving an sp2 carbon as H atom acceptor. For all other seven com-pounds, the subset of heteroatom-confined tautomers includes the thermodynamically most stable structure.
While the currently selected nine compounds are certainly not representative for the chemical domain of large inven-tories, such as the EINECS list, the results nevertheless suggest that in accord with the InChI approach, H atom shifts across heteroatoms will usually dominate the tautomeric equilibrium of chemical substances. Whether this can be confirmed with a large data set is subject of an ongoing study and will be reported in due course.
CONCLUSIONS The presently developed algorithm enables a fast genera-tion of all formally valid prototropic tautomers resulting from the mobile H atoms and their associated heteroatom acceptor sites as defined in the InChI code. As such, it provides an efficient tool to identify those tautomers that because of their restriction to H atom exchanges between heteroatoms usually include the thermodynamically most stable structures. More-over, the energetic stability range of the accordingly gener-ated tautomers is typically (but not necessarily) more narrow than when including sp2 carbons as tautomer-active sites.
Apart from energetic considerations, the algorithm yields a lower bound of the theoretically possible number of tau-tomers. If combined with more comprehensive tautomer generation procedures, the algorithm will allow for a fast assessment of the number of heteroatom-confined tautomers as compared to the total (or essentially total) number of possible tautomers. In addition, the developed tool may serve to screen the dependence of QSAR predictions, as well as of chemical database searches and associated structural retrievals on tautomeric features, and may provide pertinent information about respective variations across thermody-namically reasonable tautomers.
ACKNOWLEDGMENT This work was partly funded by the EU projects OSIRIS (www.osiris-reach.eu, contract no. 037017), CAESAR (www.
caesar-project.eu, contract no. 022674), and 2-FUN (www.2-fun.org, contract no. 036976), and by the German Federal Environmental Agency Umweltbundesamt (UBA). More-over, the authors thank Barbara Wagner, Daniel Exner, Torsten Bloi, Dominik Wondrousch, and Daniel Stosch for their valuable help and technical support.
REFERENCES AND NOTES (1) Haran ´czyk, M.; Gutowski, M. Quantum Mechanical Energy-Based Screening of Combinatorially Generated Library of Tautomers.
TauTGen: A Tautomer Generator Program. J. Chem. Inf. Model. 2007, 47, 686–694.
(2) Oellien, F.; Cramer, J.; Beyer, C.; Ihlenfeldt, W. D.; Selzer, P. M.
The Impact of Tautomer Forms on Pharmacophore-Based Virtual Screening. J. Chem. Inf. Model. 2006, 46, 2342–2354.
(3) Ghosh, A.; Wondimagegn, T.; Nilsen, H. J. Molecular Structures, Tautomerism, and Carbon Nucleophilicity of Free-Base Inverted Porphyrins and Carbaporphyrins: A Density Functional Theoretical Study. J. Phys. Chem. B. 1998, 102, 10459–10467.
(4) Zhang, Y. A.; Monga, V.; Orvig, C.; Wang, Y. A. Theoretical Studies of the Tautomers of Pyridinethiones. J. Phys. Chem. A. 2008, 112, 3231–3238.
(5) Kalliokoski, T.; Salo, H. S.; Lahtela-Kakkonen, M.; Poso, A. The Effect of Ligand-Based Tautomer and Protomer Prediction on Structure-Based Virtual Screening. J. Chem. Inf. Model. 2009, 49, 2742–2748.
(6) Pospisil, P.; Ballmer, P.; Scapozza, L.; Folkers, G. Tautomerism in Computer-Aided Drug Design. J. Recept. Signal Transduction 2003, 23, 361–371.
(7) Martin, Y. C. Let’s Not Forget Tautomers. J. Comput. -Aided Mol.
Design 2009, 23, 693–704.
(8) Leach, A. R.; Bradshaw, J.; Green, D. V. S.; Hann, M. M.; Delany, J. J. Implementation of a System for Reagent Selection and Library Enumeration, Profiling, and Design. J. Chem. Inf. Comput. Sci. 1999, 39, 1161–1172.
(9) Trepalin, S. V.; Skorenko, A. V.; Balakin, K. V.; Nasonov, A. F.; Lang, S. A.; Ivashchenko, A. A.; Savchuk, N. P. Advanced Exact Structure Searching in Large Databases of Chemical Compounds.
J. Chem. Inf. Comput. Sci. 2003, 43, 852–860.
(10) Haran ´czyk, M.; Gutowski, M. TauTGen. (accessed Sep 15, 2008).
(11) Haran ´czyk, M.; Puzyn, T.; Sadowski, P. ConGENERsA Tool for Modeling of the Congeneric Sets of Environmental Pollutants. QSAR Comb. Sci. 2008, 27, 826–833.
(12) Milletti, F.; Storchi, L.; Sforna, G.; Cross, S.; Cruciani, G. Tautomer Enumeration and Stability Prediction for Virtual Screening on Large Chemical Databases. J. Chem. Inf. Model. 2009, 49, 68–75.
(13) IUPAC (International Union of Pure and Applied Chemistry).
IUPACsInternational Chemical Identifier, InChI version 1.02 (beta).
www.iupac.org/InChI/ (accessed Jun 3, 2008).
(14) Haran ´czyk, M. ConGENER. (accessed Sep 15, 2008).
(15) ACD/ChemSketch, version 11.01; Advanced Chemistry Development, Inc.: Toronto, ON, 2009.
(16) MN.TAUTOMER, version 1.8; Molecular Networks GmbH: Erlangen, Germany, 2007.
(17) Karickhoff, S. W.; Carreira, L. A.; Hilal, S. H. SPARC Performs Automated Reasoning in Chemistry. (accessed May 1, 2009). Developed at the University of Georgia through grants from U.S. Environmental Protection Agency.
(18) Molecular Networks GmbH. MN.TAUTOMERsEnumeration of Tautomers. (ac-cessed Sep 15, 2008).
(19) Weininger, D. SMILES, A Chemical Language and Information System. 1. Introducing to Methodology and Encoding Rules. J. Chem.
Inf. Comput. Sci. 1988, 28, 31–36.
(20) CAS (Chemical Abstract Service). CAS REGISTRY and CAS Registry Numbers. (accessed Mar 1, 2010).
(21) Tavakol, H.; Arshadi, S. Theoretical Investigation of Tautomerism in N-Hydroxy Amidines. J. Mol. Model. 2009, 15, 807–816.
(22) Project: InChI and InChIKey: Further Promotion. web/ins/2008-033-1-800 (accessed Sep 25, 2009); continuation of projects 2004-039-1-800 and 2000-025-1-800.
(23) U.S. Secretary of Commerce. NIST Chemistry WebBook. http:// webbook.nist.gov/ (accessed Mar 3, 2010).
(24) Kanehisa, M.; Goto, S. KEGG: Kyoto Encyclopedia of Genes and Genomes. Nucleic Acids Res. 2000, 28, 27–30.
(25) Schu ¨u ¨rmann, G.; Ebert, R.-U.; Nendza, M.; Dearden, J. C.; Paschke, A.; Ku ¨hne, R. Predicting Fate-Related Physicochemical Properties.
In Risk Assessment of Chemicals. An Introduction, 2nd ed.; van Leeuwen, K., Vermeire, T., Eds.; Springer Science: Dordrecht, The Netherlands, 2007; pp 375-426.
(26) Dewar, M. J. S.; Zoebisch, E. G.; Healy, E. F.; Stewart, J. J. P.
Development and Use of Quantum Mechanical Molecular Models.
76. AM1: A New General Purpose Quantum Mechanical Molecular Model. J. Am. Chem. Soc. 1985, 107, 3902–3909.
(27) MOPAC 93, revision 2; Stewart Computational Chemistry: Colorado Springs, CO, 1994.
TAUTOMER GENERATION BASED ON INCHI J. Chem. Inf. Model., Vol. 50, No. 7, 2010 1231 (28) Schu ¨u ¨rmann, G.; Ebert, R.-U.; Ku ¨hne, R. Prediction of the Sorption of Organic Compounds into Soil Organic Matter From Molecular Structure. EnViron. Sci. Technol. 2006, 40, 7005–7011.
(29) Ku ¨hne, R.; Ebert, R.-U.; Schu ¨u ¨rmann, G. Model Selection Based on Structural SimilaritysMethod Description and Application to Water Solubility Prediction. J. Chem. Inf. Model. 2006, 46, 636– 641.
(30) European Commission. ESIS (European Chemical Substances Infor-mation System). European Inventory of Existing Commercial Chemical Substances (EINECS).
PGM)ein (accessed Sep 1, 2008).
(31) Daginnus, K. EC Chemical Inventories. qsar/information-sources/ec_inventory (accessed Aug 5, 2008).
(32) Morgan, H. L. The Generation of a Unique Machine Description for Chemical StructuressA Technique Developed at Chemical Abstracts Service. J. Chem. Doc. 1965, 5, 107–113.
(33) Ku ¨hne, R.; Ebert, R.-U.; Schu ¨u ¨rmann, G. Chemical Domain of QSAR Models From Atom-Centered Fragments. J. Chem. Inf. Model. 2009, 49, 2660–2669.
CI1001179 1232 J. Chem. Inf. Model., Vol. 50, No. 7, 2010 THALHEIM ET AL. |
11672 | https://appliedcombinatorics.org/book/s_polya_polya.html | Applied Combinatorics
Mitchel T. Keller, William T. Trotter
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Section 15.4 Pólya’s Theorem
Before getting to the full version of Pólya’s formula, we must develop a generating function as promised at the beginning of the chapter. To do this, we will return to our example of Section 15.1.
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Subsection 15.4.1 The cycle index
Unlike the generating functions we encountered in Chapter 8, the generating functions we will develop in this chapter will have more than one variable. We begin by associating a monomial with each element of the permutation group involved. In this case, it is (D_8\text{,}) the dihedral group of the square. To determine the monomial associated to a permutation, we need to write the permutation in cycle notation and then determine the monomial based on the number of cycles of each length. Specifically, if (\pi) is a permutation of ([n]) with (j_k) cycles of length (k) for (1\leq k\leq n\text{,}) then the monomial associated to (\pi) is (x_1^{j_1}x_2^{j_2}\cdots x_n^{j_n}\text{.}) Note that (j_1 + 2j_2 + 3j_3 + \cdots + nj_n = n\text{.}) For example, the permutation (r_1=(1234)) is associated with the monomial (x_4^1) since it consists of a single cycle of length (4\text{.}) The permutation (r_2=(13)(24)) has two cycles of length (2\text{,}) and thus its monomial is (x_2^2\text{.}) For (p=(14)(2)(3)\text{,}) we have two (1)-cycles and one (2)-cycle, yielding the monomial (x_1^2x_2^1\text{.}) In Figure 15.10, we show all eight permutations in (D_8) along with their associated monomials.
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| | | |
---
| Transformation | Monomial | Fixed colorings |
| (\iota = (1)(2)(3)(4)) | (x_1^4) | (16) |
| (r_1 = (1234)) | (x_4^1) | (2) |
| (r_2=(13)(24)) | (x_2^2) | (4) |
| (r_3=(1432)) | (x_4^1) | (2) |
| (v=(12)(34)) | (x_2^2) | (4) |
| (h=(14)(23)) | (x_2^2) | (4) |
| (p=(14)(2)(3)) | (x_1^2x_2^1) | (8) |
| (n=(1)(24)(3)) | (x_1^2x_2^1) | (8) |
Figure 15.10. Monomials arising from the dihedral group of the square
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Now let’s see how the number of (2)-colorings of the square fixed by a permutation can be determined from its cycle structure and associated monomial. If (\pi(i)=j\text{,}) then we know that for (\pi) to fix a coloring (C\text{,}) vertices (i) and (j) must be colored the same in (C\text{.}) Thus, the second vertex in a cycle must have the same color as the first. But then the third vertex must have the same color as the second, which is the same color as the first. In fact, all vertices appearing in a cycle of (\pi) must have the same color in (C) if (\pi) fixes (C\text{!}) Since we are coloring with the two colors white and gold, we can choose to color the points of each cycle uniformly white or gold. For example, for the permutation (v=(12)(34)) to fix a coloring of the square, vertices (1) and (2) must be colored the same color ((2) choices) and vertices (3) and (4) must be colored the same color ((2) choices). Thus, there are (2\cdot 2=4) colorings fixed by (v\text{.}) Since there are two choices for how to uniformly color the elements of a cycle, letting (x_i=2) for all (i) in the monomial associated with (\pi) gives the number of colorings fixed by (\pi\text{.}) In Figure 15.10, the “Fixed colorings” column gives the number of (2)-colorings of the square fixed by each permutation. Before, we obtained this manually by considering the action of (D_8) on the set of all (16) colorings. Now we only need the cycle notation and the monomials that result from it to derive this!
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Recall that Burnside’s Lemma states that the number of colorings fixed by the action of a group can be obtained by adding up the number fixed by each permutation and dividing by the number of permutations in the group. If we do that instead for the monomials arising from the permutations in a permutation group (G) in which every cycle of every permutation has at most (n) entries, we obtain a polynomial known as the cycle index (P_G(x_1,x_2,\dots,x_n)\text{.}) For our running example, we find
\begin{equation} P_{D_8}(x_1,x_2,x_3,x_4) = \frac{1}{8}\left(x_1^4 + 2x_1^2x_2^1 + 3x_2^2 + 2x_4^1\right). \end{equation}
To find the number of distinct (2)-colorings of the square, we thus let (x_i=2) for all (i) and obtain (P_{D_8}(2,2,2,2) = 6) as before. Notice, however, that we have something more powerful than Burnside’s lemma here. We may substitute any positive integer (m) for each (x_i) to find out how many nonequivalent (m)-colorings of the square exist. We no longer have to analyze how many colorings each permutation fixes. For instance, (P_{D_8}(3,3,3,3) = 21\text{,}) meaning that (21) of the (81) colorings of the vertices of the square using three colors are distinct.
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Subsection 15.4.2 The full enumeration formula
Hopefully the power of the cycle index to count colorings that are distinct when symmetries are considered is becoming apparent. In the next section, we will provide additional examples of how it can be used. However, we still haven’t seen the full power of Pólya’s technique. From the cycle index alone, we can determine how many colorings of the vertices of the square are distinct. However, what if we want to know how many of them have two white vertices and two gold vertices? This is where Pólya’s enumeration formula truly plays the role of a generating function.
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Let’s again consider the cycle index for the dihedral group (D_8\text{:})
\begin{equation} P_{D_8}(x_1,x_2,x_3,x_4) = \frac{1}{8}\left(x_1^4 + 2x_1^2x_2^1 + 3x_2^2 + 2x_4^1\right). \end{equation}
Instead of substituting integers for the (x_i\text{,}) let’s consider what happens if we substitute something that allows us to track the colors used. Since (x_1) represents a cycle of length (1) in a permutation, the choice of white or gold for the vertex in such a cycle amounts to a single vertex receiving that color. What happens if we substitute (w+g) for (x_1\text{?}) The first term in (P_{D_8}) corresponds to the identity permutation (\iota\text{,}) which fixes all colorings of the square. Letting (x_1=w+g) in this term gives
\begin{equation} (w+g)^4 = g^4+4 g^3 w+6 g^2 w^2+4 g w^3+w^4, \end{equation}
which tells us that (\iota) fixes one coloring with four gold vertices, four colorings with three gold vertices and one white vertex, six colorings with two gold vertices and two white vertices, four colorings with one gold vertex and three white vertices, and one coloring with four white vertices.
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Let’s continue establishing a pattern here by considering the variable (x_2\text{.}) It represents the cycles of length (2) in a permutation. Such a cycle must be colored uniformly white or gold to be fixed by the permutation. Thus, choosing white or gold for the vertices in that cycle results in two white vertices or two gold vertices in the coloring. Since this happens for every cycle of length (2\text{,}) we want to substitute (w^2+g^2) for (x_2) in the cycle index. The (x_1^2x_2^1) terms in (P_{D_8}) are associated with the flips (p) and (n\text{.}) Letting (x_1=w+g) and (x_2 = w^2+g^2\text{,}) we find
\begin{equation} x_1^2x_2^1 = g^4+2 g^3 w+2 g^2 w^2+2 g w^3+w^4, \end{equation}
from which we are able to deduce that (p) and (n) each fix one coloring with four gold vertices, two colorings with three gold vertices and one white vertex, and so on. Comparing this with Figure 15.2 shows that the generating function is right on.
🔗
By now the pattern is becoming apparent. If we substitute (w^i+g^i) for (x_i) in the cycle index for each (i\text{,}) we then keep track of how many vertices are colored white and how many are colored gold. The simplification of the cycle index in this case is then a generating function in which the coefficient on (g^s w^t) is the number of distinct colorings of the vertices of the square with (s) vertices colored gold and (t) vertices colored white. Doing this and simplifying gives
\begin{equation} P_{D_8}(w+g,w^2+g^2,w^3+g^3,w^4+g^4) = g^4+g^3 w+2 g^2 w^2+g w^3+w^4. \end{equation}
From this we find one coloring with all vertices gold, one coloring with all vertices white, one coloring with three gold vertices and one white vertex, one coloring with one gold vertex and three white vertices, and two colorings with two vertices of each color.
🔗
As with the other results we’ve discovered in this chapter, this property of the cycle index holds up beyond the case of coloring the vertices of the square with two colors. The full version is Pólya’s enumeration theorem:
🔗
Theorem 15.11. Pólya’s Enumeration Theorem.
Let (S) be a set with (|S|=r) and (\cgC) the set of colorings of (S) using the colors (c_1,\dots,c_m\text{.}) If a permutation group (G) acts on (S) to induce an equivalence relation on (\cgC\text{,}) then
\begin{equation} P_G\left(\sum_{i=1}^m c_i, \sum_{i=1}^m c_i^2, \dots,\sum_{i=1}^m c_i^r\right) \end{equation}
is the generating function for the number of nonequivalent colorings of (S) in (\cgC\text{.})
🔗
🔗
If we return to coloring the vertices of the square but now allow the color blue as well, we find
\begin{equation} P_{D_8}(w+g+b,w^2+g^2+b^2,w^3+g^3+b^3,w^4+g^4+b^4) = b^4+b^3 g+2 b^2 g^2\+b g^3+g^4+b^3 w+2 b^2 g w+2 b g^2 w+g^3 w+2 b^2 w^2+2 b g w^2+2 g^2 w^2\+b w^3+g w^3+w^4. \end{equation}
From this generating function, we can readily determine the number of nonequivalent colorings with two blue vertices, one gold vertex, and one white vertex to be (2\text{.}) Because the generating function of Pólya’s Enumeration Theorem records the number of nonequivalent patterns, it is sometimes called the pattern inventory.
🔗
What if we were interested in making necklaces with (500) (very small) beads colored white, gold, and blue? This would be equivalent to coloring the vertices of a regular (500)-gon, and the dihedral group (D_{1000}) would give the appropriate transformations. With a computer algebra system1
With some more experience in group theory, it is possible to give a general formula for the cycle index of the dihedral group (D_{2n}\text{,}) so the computer algebra system is a nice tool, but not required.
such as Mathematica®, it is possible to quickly produce the pattern inventory for such a problem. In doing so, we find that there are
\begin{align} \amp 3636029179586993684238526707954331911802338502600162304034603583\ \amp 2580600191583895484198508262979388783308179702534404046627287796\ \amp 4304252714992703135653472347417085467453334179308247819807028526\ \amp 92187253642441292279756575936040804567103229 \approx 3.6\times 10^{235} \end{align}
possible necklaces. Of them,
\begin{align} \amp 2529491842340460773490413186201010487791417294078808662803638965\ \amp 6782447138833704326875393229442323085905838200071479575905731776\ \amp 6605088026968640797415175535033372572682057214340157297357996\ \amp 345021733060\approx 2.5\times 10^{200} \end{align}
have (225) white beads, (225) gold beads, and (50) blue beads.
🔗
The remainder of this chapter will focus on applications of Pólya’s Enumeration Theorem and the pattern inventory in a variety of settings.
🔗
🔗
🔗
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11673 | https://www.youtube.com/watch?v=rpKzq64GA9Y | Chi-Square Test [Simply explained]
DATAtab
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Posted: 1 Feb 2022
The chi-square test is a hypothesis test used for categorical variables with nominal or ordinal measurement scale. The chi-square test checks whether the frequencies occurring in the sample differ significantly from the frequencies one would expect. Thus, the observed frequencies are compared with the expected frequencies in the Chi2-Test and their deviations are examined.
Chi-Square test calculator:
More Information about the Chi-Square test:
Chi-squared distribution:
213 comments
Transcript:
in this video i explain in a simple way what the chi-square test is and how you can calculate the chi-square test with just a few steps this leads us right to the first question what is the chi-square test the chi-square test is a hypothesis test that is used when you want to determine whether there is a relationship between two categorical variables or not what are categorical variables categorical variables are for example gender with the characteristics male and female or the preferred newspaper where you have the washington post the new york times usa today or for example the frequency of television with several times a week rare never or the highest educational level of a person these are all categorical variables the weight of a person for example or the salary or the electricity consumption are no categorical variables so whenever we have two categorical variables and we want to check if there is a correlation between them we use a chi-square test we could for example be interested in a question if there is a relationship between gender and the preferred newspaper or is the relationship between gender and the frequency of television another question could be is there a relationship between the preferred newspaper and the frequency of television or finally in the last example is there a relationship between the frequency of television viewing and the highest level of education therefore whenever we have two categorical variables for which we want to establish a relationship we use the chi-square test let's say we want to investigate the relationship between gender and a person's highest educational level our question then would be is there a relationship between gender and the highest level of education for this purpose we create a questionnaire where the participants mark their gender and their highest level of education the result of this survey is displayed in an excel table in this table we see one person in each row that we have surveyed the first person we interviewed is male and the highest educational level is college the second person we interviewed is female and she is without graduation we can now take this table and copy it into a statistical software like data tab data that then outputs the so called cross table in this table you see the variable gender and the variable highest educational level and also how often in each case the combination occurs female and without graduation occur six times in a table or male and without graduation occurs altogether seven times in this table now we want to know whether gender has an influence on the highest educational level does it make a difference if i'm female or male in terms of the highest education and level i have in other words is there a relationship between gender and the highest educational level in order to answer this question we use the chi-square test now there are two ways how to calculate the chi-square test either we use a statistical software like data tab or we calculate the chi-square test by hand we get started with the uncomplicated way and use datadep afterwards i will show you how you can calculate the chi-square test by hand in order to do this simply visit datadept.net you will find the link in the video description now you click on statistics calculator and copy your own data into this table the variables will now appear here below because of the fact that the variable's values are in words data tab assumes that they are categorical variables if your data is coded with numbers you have to change the scale level under variable view we want to calculate the chi-square test so i just click here on the tab now we simply click on the variables gender and highest educational level and data tab will automatically suggest the appropriate test for you in this case the chi-square test now we immediately get the results for the chi-square test we see the cross table for the variable's gender and highest educational level here we can see the expected frequencies for perfectly independent variables and finally here are the results of the chi-square test and we get a p-value of 0.92 if you don't know how to interpret the results you can simply have a look at the summary in words a car square test was performed between gender and highest educational level no expected cell frequencies were less than five there was no statistically significant relationship between gender and highest educational level we use state adapt to calculate a chi-square test let's say we choose a significance level of 5 then we reject the null hypothesis if we get a p-value that is smaller than 0.05 we got a p-value of 0.981 which is of course much larger than 0.05 and thus we don't reject the null hypothesis based on these data there is no relationship between gender and the highest educational level and now let's calculate the chi-square test by hand in order to calculate the chi-square test by hand we need the crosstab with the observed frequencies and the cross step with the expected frequencies we get the expected frequencies by looking at which frequencies we should actually get if there was no relationship between gender and highest educational level we can now calculate the chi-square value using the following formula here ok is the observed frequency of a cell and e k is the expected frequency of the respective cell if we now apply this formula to our example we get the following from the first cell we calculate the observed frequency 6 minus the expected frequency which is 6.08 square that and divide that by the expected frequency we now do the same for the second cell 7 minus 6.92 squared divided by 6.92 if we do this for all cells and sum it up we get our chi-square value which is 0.504 if you use a statistical software like data tab you get the p-value returned however calculating the p-value by hand is not feasible therefore in this case we use the so-called critical chi-square value we can read the critical chi-square value in this table you can find this table on datatab.net the link is in the video description here we have a five percent significance level so we are in that column the degrees of freedom are obtained by multiplying the number of rows minus one by the number of columns minus one we have four rows and two columns so we get three times one which equals three so we are in this row in the table now this gives us a critical chi-square value of 7.815 our calculated chi-square value is smaller than the critical value so the null hypothesis is not rejected if you like check out data tab you will find many more tutorials there i hope you enjoyed the video and see you next time |
11674 | https://www.youtube.com/watch?v=t9tt_CwSeWc | What's the expected number of coin flips before we see a given sequence of results?
Ben Rosenberg
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Posted: 15 Jan 2024
How do we find the expected number of times we need to flip a coin before we see a given sequence of results (e.g., heads, tails, heads)?
00:00 - Introduction and problem statement
01:20 - Simulation of HT vs TT example using Python
02:58 - State machines for HT and TT example
05:36 - First-step method (system of equations)
07:33 - Formula method (transition matrix)
11:18 - Conclusion and summary
I go into some depth here but not too deep - I'll cover the reasons behind why the math works as expected in the formula we use in a future video. I'll also discuss automating the generation of state machines in a future video.
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Transcript:
Introduction and problem statement hi um today I'm going to go over a simple problem that is well it's receptively simple really it's pretty difficult um if you go all the way down the rabbit hole but um yeah so let me start so just to give a quick overview of the problem here it is given a sequence of Fair coin flip results so fair meaning that the coin is equally likely to land on heads and tails um we want to find the expected number of flips that we have to do before that sequence is seen so as an example say we have some sequence heads Tails heads then the question would be how many times we expect to flip the coin before we see that exact sequence pop up so intuitively the coin is fair so okay maybe the probabilities will just be the same for all the sequences of that given length right so for heads tails and Tails Tails maybe those will be the same you know know expected number of steps before we see them because okay well heads into Tails that's like 25% probability right and then taals and Tails that's also 25% probability so you'd think okay well maybe maybe maybe we expect to see them after the same number of coin flips so we'll test the accuracy of this claim using a simple python Simulation of HT vs TT example using Python script so here we just import random we've got some Fair coin that returns heads if uh our random uh variables less than 0. five and otherwise it returns Tails um and then we have this steps to sequence function where we pass our random variable in um we have our outputs and while the output does not end in our sequence we're going to keep flipping coins um and then we just return the number of times that we had to flip coins to get that output so the length of the output Str string and here we just have a helper function to run our steps to sequence function a certain number of times and then divide that by the total number of times that we ran in order to get the average basically the average number of steps right so that's going to be an approximation for the expected number of steps um before we see the sequence in general so here are the results and uh clearly these are not the same right so heads tails is much more I guess easy to get right so the number of steps that we have to take a number of times we have to flip the coin in order to um in order to see this sequence is much lower at at about four right then the number of times we have to flip the coin in order to see taals Tales which is about six so this is different than what we expected so let's go into why that's the State machines for HT and TT example case so a good way to model this kind of problem is using a state machine um where the states are basically what we've seen so far and you know with with regard to the sequence that we want to see so say we at the start State we haven't flipped anything yet if you flip the Tails we haven't made any progress towards our goal so we just stay in the start state if you flip the heads then we go to the H state so we've seen a heads that's part of our part of our goal and from here if you FP the Tails then we want to go to heads Tails then we we've succeeded or successful right we got exactly what we wanted to get if we flip other heads we just stay in the H state right so we flip heads you know again we're back here nothing has changed um if we get another taals we can still we still get to our HT goal um and this with the probabilities along these transitions like this this is called um a discrete time mark Markov chain you don't have to worry about that for now it's just or just at all in this video um it's just a name for this kind of probabilistic state machine so now let's look at the state machine for tailes Tales right so here at the start if we flip ahead nothing happens we stay here right if we flip a tail then we get to go to the tail state where we've seen one tail all right but if we flip ahe here we actually have to go back to the start because now we we completely lost our progress right um we still need to flip two tails in a row now in order to get to Tails Tails if you have another tail um then we actually get to Tails Tails immediately so yeah that this is why this is the state machine and you'll notice the difference between this state machine and this state machine right here we never go backwards we never lose our progress here we actually do lose our progress if we get the wrong you could call it the wrong coin flip right so it it should be a bit more intuitive as to why it on average takes longer for us to see Tails Tails than for us to see heads Tails okay so now let's talk about actually solving for the um the expected amount of time so here we have this intuitive understanding of why it would take longer for us to see tailes tailes than heads tailes um but now all we have is like okay a comparison between this we don't actually know what the expected number of steps is unless we want to simulate it right so we don't want to simul it we want to find out exactly what it is even though we have a pretty good idea like yes this is going to be four steps yes this is going to be six steps so we can actually solve a system First-step method (system of equations) of linear equations to figure out what the expected number of steps is before we see the desired result so for instance for heads tailes for the state machine that we had before um the expected value this is what this e means is expected value um you could even say like expected number of steps right or expected number of flips um from our start State well we have to add one right and then there's a 50% chance that we go to the Head State and a 50% chance that we stay in the start state right so if we fli a head we go to the Head State we fli the Tails and we stay in the start state so that's what this first equation means the second equation is for what happens when we're in the head State well we have to add one because we're taking another step right and then here there's a 50% chance that we stay in the head state if you put head then we stay here and if we fit the tail then we go to the finish the the final state right the finishing State and here the expected number of steps between the finishing state in the finishing state is of course zero so this is just kind of like you know we could have put the zero here but um this is kind of just to make things a little bit more understandable all right so now we'd want to solve this equation uh this series of equations um to get the result and here we just start out by um you know we have expected value of heads while we plug our zero in here for HT and then this is pretty easily solvable and you can kind of just look through this if you want um so as our simulation indicated the expected number of steps before heads tailes is seene is four um so this is what we expected it was you know 3.99 whatever in our simulation but um basically four and here we've proven that that is correct um and as an exercise if you want to do the same thing for the Tales tales State machine to confirm the expected number of steps is is actually six like our simulation indicated feel free to do that as Formula method (transition matrix) well so solving systems of equations is a pain right uh it took a while to set this up um and maybe to solve it uh it's just kind of some manual work that we want to get rid of so luckily there's actually a formula that we can use um to do this stuff for us so we can we have our state machines right we can rewrite them as a transition Matrix so this state machine here remember from head tailes becomes this transition Matrix um the transition Matrix you can think of it like this right so we have our states kind of along the rows and also along the columns so the rows represent the state that we're going from and the columns represent the state that we're going to so this row this top row would be the start row and this First Column would be the start column so the chance of going from start to start is 0.5 right the chance of going from start to and this would be the heads column start to heads 0.5 and then start to heads Tails zero right so this basically lines up exactly with the TR uh the state machine you can just confirm that the rest is also make sense right so for instance like heads tails to go from heads tails to heads tails is one to go from heads to heads tails is 0.5 Etc so our transation our transition Matrix has a specific form right it looks like QR 0 and one so basically it means if you look at this right you can think of these four entries these four entries in this box over here as being Q which is like a submatrix and then over here we have R it's like another submatrix or you can think that as a vector and then on the bottom we have zeros and then over here we have one right so in the terminology of Markov chains this is our absorb State um and then this is like our substochastic Matrix of transient States but again like don't worry about the terminology too much so you can immediately get all the expected numbers of States um sorry Steps From the non-final states so that's um like that would be start and H in this circumstance to the final State that's HT right our absorbing state by using this formula where I is the identity Matrix and one is just a vector fill of one so here's our formula it's I minus q and then we take the inverse of that Matrix and then basically we just sum all the rows right so that's that's what this um this Vector of ones is doing when we when we take that product so for this example remember Q is just 0.5 0.5 0 0.5 so that's what this is here and then identity Matrix is just the Matrix with On's along the diagonal and Zer everywhere else um here we want it to have the same dimensions as Q so Q is a 2x2 matrix so we want I of two right which is going to be this Matrix right here um and then you're going to multiply that by also the ones we want that to be of Dimension two as well um so we multiply that out and we get 42 which means that the expected number of steps from the start is four and the expected number of steps from the h state is two so we didn't actually get this before um or I guess we did get it as like a consequence of the way that we solve the equations but um it's nice to have it all here and we don't need to solve any equations for this so why does this formula work we'll get to that in another video um cuz this video is already reasonably long and I don't want to go too much in depth because it'll scare people um but for now just like look at this formula it works pretty well um yeah and so in summary let's go Conclusion and summary over the main points again so not all the sequences of coin flips have the same time until they're seen right so we saw HT and TT did not have this same expected number of steps before we saw them secondly we can intuitively see this when we construct a state machine for the series of flip results so remember our state machines for HT and TT we were able to look at that and say okay yes it makes sense that TT is you know it takes longer for us to see that because we can lose our progress more easily right we go back to the start state if we don't see two T's in a row um then to find the exact number of flips we can solve a system of linear equations we saw that we did it we did the work um it worked but that was a lot of work so we saw that also alternatively we could use a state machine to create a transition Matrix and then use a formula IUS Q inverse and then you know sum those rows on that matrix by multiplying by the one vector on that Matrix to give us the expected steps from every transient state to the absorbing state so I'll go into more depth onto why the math works and also something interesting how to generate the state machines without manual work since that is something that requires a reasonable amount of manual work um in future videos but for now that's it and I hope you enjoyed watching thanks |
11675 | https://gateoverflow.in/378962/cmi-2022-datascience-b-11 | Others: CMI-2022-DataScience-B: 11
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CMI-2022-DataScience-B: 11
adminaskedJul 23, 2022•edited Aug 21, 2022 by Lakshman Bhaiya
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You are given an 8×8 chessboard and a large supply of T-shaped tiles and square-shaped tiles of the kind shown below.
Note that the squares in these two shaped tiles are of the same size as the black/white squares on the chessboard.
We wish to cover the squares of the chess board using these tiles. Each square of the chessboard must be occupied by exactly one square from a tile; we are not allowed to break the tiles.
Is it possible to cover the chessboard with 16 T shaped tiles? Why?
Is it possible to cover the chessboard with 15 T shaped tiles and one square tile? Why?
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For Par A - Yes, we can cover chess board with 16 T shaped tiles.
We know that with 2x2 square shape tile, I can cover 4 tiles in 8x8 chess board. By repeatingly place the 2x2 square shape, we can fill the 8x8 chess board (16 - 2x2 square tiles required)
Given the shape of T with 4 tiles. I need to check whether we can cover the 8x8 chess board with 16 T shaped tiles.
8x8 = 64 small 1-peice squares. If I found any symmetry line, I can repeat the same pattern and fill the 8x8 cells. (recall that while filling with the 2x2 square shape tile, we did the same process).
As 1 - T shape tile, I'm not founding any symmetry.
4 - T shape tiles, I can cover 44 = 16 cells shown in the picture (left side-bottom). This pattern can be repetitively applied and cover the entire chess board.(Total 44 = 16 - T shaped tiles required)
Further, is there any other patterns ?
8 - T shape tiles, I can cover 84 = 32 cells shown in the picture (top 4 rows - 8 columns). This pattern can be repetitively applied and cover the entire chess board. (Total 28 = 16 - T shaped tiles required)
What about 16 - T shape tiles can cover 8x8 cells without any symmetry ? - No, it is not possible.
Because to cover the 1st row - 8 cells, we can cover via (3+1,3+1) format, (3+1+3+1) formats only. There are no other possibilities (3+3+1+1).
What if it is asked for 6x6 instead of 8x8 ? - It is not possible to cover due to we are not finding any symmetry.
For Part B - No, we can not cover the chessboard with 15T shaped tiles and one square tile.
As mentioned in the Part A, we can't cover 8x8 cells with 16 - T tiles without symmetry, we have only 4-T cells pattern or 8 - T cells pattern.
One of the 4-T cells pattern should be 3- T cell pattern + 1 Square pattern - It is not possible to found symmetry.
8 - T cells pattern: If I place square tile at the base line (you may place it any corner), the possibilities as shown in the below picture - Both cases, I can't proceed further.
Shaik MasthanansweredAug 24, 2024•selected Aug 25, 2024 by ankitgupta.1729
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SUJITH'
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I believe the reason they fit because the area of the n number of square shape formed by the tiles and given chessboard is equal therefore no amount of space is wasted, hence they fit.
So for any n∗n square we can find number of T tiles required = n∗n 4 [Assuming that T tiles remains as given consiting of 4 tiles]
For the case if n∗m rectangle was given, then we can find the ratio of n:m, after which we can obtain the required amount of T tiles required, For example if the ratio of length and breadth is 8 and 4 then we can fit two square shape of T tiles, so 8 T tiles would be required.
@ankitgupta.1729Sir can you elaborate the approach that you followed for this question.
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Shaik Masthan
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@ankitgupta.1729,
I have added some explanation.
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@SUJITH'Suppose you have given a 24 rectangular grid of squares. So you have a total 8 small squares but you can't fit 2 T-shaped tiles of 8 small squares. So, area is not the reason as for the part b here.
(+1) for your comment for the effort of making the diagram.
@Shaik Masthan(+1) for your efforts too for making diagram and explaining the things in the answer.
What if we don't have a 66 squared board or 88 squared board or a rectangle. If we remove the 3 corners of the board then it is asked about the fitting of 2 arbitrary shaped polyominoes.
Here, it is observed that 44 board can fit 4 T-shaped tiles and based on that we are finding symmetry and making conclusions but what if we are unable to find the symmetry for some polyominoes which are not tetrominoes or dominoes or triomino.
You can convert this problem into a graph coloring problem where you have to color the board using white and black colors alternatively row and columnwise and then fit some shaped tile and make the contradiction.
Here, a T-shaped tile can have 3 white and 1 black square or 3 black and 1 white square. So if you have 16 (an even number) T-shaped tiles then you can do the tiling using 8 T-shaped tiles of 3 white and 1 black which makes the total of 24 white and 8 black tiles and also a 8 T-shaped tiles of 3 black and 1 white which makes the total of 24 black and 8 white tiles. In this way you have total of 32 white and 32 black tiles which is consitent with the alternate coloring of 88 board.
In Part b, 15 is not possible because 15 is an odd number. A square can be tiled with 2 black and 2 white colors. So remaining we have 30 black and 30 white and with 15 T-shaped tile, you can't make 30 black and 30 white such that each tile can have 3 white and 1 black or 3 black and 1 white tile.
You can find the complete procedure and different examples in Rosen with Chapter 1 - tiling section under the proof techniques.
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11676 | https://web.williams.edu/Mathematics/sjmiller/public_html/math/talks/CookieMonsterZeckendorfLekkerkerker_notab.pdf | Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Cookie Monster Meets the Fibonacci Numbers. Mmmmmm – Theorems!
Steven J Miller, Williams College Steven.J.Miller@williams.edu CANT 2010, CUNY, May 29, 2010 1 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Summary / Acknowledgements Previous results: Zeckendorf’s and Lekkerkerker’s theorems.
New approach: View as combinatorial problem.
Thanks: Ed Burger and SMALL REU students.
2 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Special Thanks Special thanks go to Cameron and Kayla Miller for playing quietly while key details were worked out and for suggesting which colors to use and where!
We’re ready for yellow now – hope you are too!
3 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Previous Results Fibonacci Numbers: Fn+1 = Fn + Fn−1; F1 = 1, F2 = 2, F3 = 3, F4 = 5 . . . .
Zeckendorf’s Theorem Every positive integer can be written in a unique way as a sum of non-consecutive Fibonacci numbers.
Example: 2010 = 1597+377+34+2 = F16 + F13 + F8 + F2.
Lekkerkerker’s Theorem The average number of non-consecutive Fibonacci summands in the Zeckendorf decomposition for integers in [Fn, Fn+1) tends to n 휑2+1 ≈.276n, where 휑= 1+ √ 5 2 is the golden mean.
4 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Main Results Lemma: Application of Cookie Counting The ‘probability’ (ie, percentage of the time) an integer in [Fn, Fn+1) has exactly k + 1 non-consecutive Fibonacci summands is (n−1−k k ) /Fn−1.
The above lemma yields Zeckendorf’s Theorem, Lekkerkerker’s Theorem, and will (hopefully) yield An Erdos-Kac Type Theorem: SMALL 2010 As n →∞, the distribution of the number of non-consecutive Fibonacci summands in the Zeckendorf decomposition for integers in [Fn, Fn+1) is Gaussian.
5 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Properties of Fibonacci Numbers and needed Combinatorial Results 6 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Binet’s Formula Binet’s Formula Fn = 1 √ 5 ( 1 + √ 5 2 )n+1 −1 √ 5 ( 1 − √ 5 2 )n+1 .
Proof: Fn+1 = Fn + Fn−1.
Guess Fn = nr: r n+1 = r n + r n−1 or r 2 = r + 1.
Roots r = (1 ± √ 5)/2.
General solution: Fn = c1r n 1 + c2r n 2 , solve for ci’s.
□ 7 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Combinatorial Review The Cookie Problem The number of ways of dividing C identical cookies among P distinct people is (C+P−1 P−1 ) .
Proof: Consider C + P −1 cookies in a line.
Cookie Monster eats P −1 cookies: (C+P−1 P−1 ) ways to do.
Divides the cookies into P sets.
□ Example: 10 cookies and 5 people: ⊙⊙⊗⊗⊙⊙⊗⊙⊙⊙⊙⊙⊗⊙ 8 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Cookie Problem: Reinterpretation Reinterpreting the Cookie Problem The number of solutions to x1 + ⋅⋅⋅+ xP = C with xi a non-negative integer is (C+P−1 P−1 ) .
Generalization: If have constraints xi ≥ci, then number of solutions is (C−∑ i ci+P−1 P−1 ) .
This follows by setting xi = yi + ci with yi a non-negative integer.
9 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Zeckendorf’s Theorem 10 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Proof of Zeckendorf’s Theorem Uniqueness: Same standard argument (induction).
Existence: Consider all sums of non-consecutive Fibonacci numbers equaling an m ∈[Fn, Fn+1); note there are Fn+1 −Fn = Fn−1 such integers.
Must have Fn one of the summands, must not have Fn−1.
For each Fibonacci number from F1 to Fn−1 we either include or not, cannot have two consecutive, must end with a non-taken number.
11 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Proof of Zeckendorf’s Theorem (continued) Consider all subsets of k + 1 non-consecutive Fibonaccis from {F1, . . . , Fn} where Fn is taken. Let y0 be number of Fibonaccis not taken until first one taken, and then yi (1 ≤i ≤k) be the number not taken between two taken.
Example: 2010 = 1597+377+34+2 = F16 + F13 + F8 + F2, so n = 16, k + 1 = 4, y0 = 1, y1 = 5, y2 = 4, y3 = 2.
Equivalently: y0 + y1 + ⋅⋅⋅+ yk + k = n −1, yi ≥1 if i ≥1.
Equivalently: x0 + ⋅⋅⋅+ xk + 2k = n −1, xi ≥0. Number of solutions is (n−1−k k ) .
Obtain ∑⌊n−1 2 ⌋ k=0 (n−1−k k ) = Fn−1 integers in [Fn, Fn+1); as all distinct and this many integers in interval, done.
□ 12 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Lekkerkerker’s Theorem 13 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Preliminaries ℰ(n) := ⌊n−1 2 ⌋ ∑ k=0 k (n −1 −k k ) .
Average number of summands in [Fn, Fn+1) is ℰ(n) Fn−1 + 1.
Recurrence Relation for ℰ(n) ℰ(n) + ℰ(n −2) = (n −2)Fn−3.
14 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Recurrence Relation Recurrence Relation for ℰ(n) ℰ(n) + ℰ(n −2) = (n −2)Fn−3.
Proof by algebra (details in appendix): ℰ(n) = ⌊n−1 2 ⌋ ∑ k=0 k (n −1 −k k ) = (n −2) ⌊n−3 2 ⌋ ∑ ℓ=0 (n −3 −ℓ ℓ ) − ⌊n−3 2 ⌋ ∑ ℓ=0 ℓ (n −3 −ℓ ℓ ) = (n −2)Fn−3 −ℰ(n −2).
15 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Solving Recurrence Relation Formula for ℰ(n) (i.e., Lekkerkerker’s Theorem) ℰ(n) = nFn−1 휑2 + 1 + O(Fn−2).
⌊n−3 2 ⌋ ∑ ℓ=0 (−1)ℓ(ℰ(n −2ℓ) + ℰ(n −2(ℓ+ 1))) = ⌊n−3 2 ⌋ ∑ ℓ=0 (−1)ℓ(n −2 −2ℓ)Fn−3−2ℓ.
Result follows from Binet’s formula, the geometric series formula, and differentiating identities: ∑m j=0 jxj = x (m+1)xm(x−1)−(xm+1−1) (x−1)2 . Details in appendix.
16 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs An Erdos-Kac Type Theorem 17 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Generalizing Lekkerkerker Theorem (SMALL 2010) As n →∞, the distribution of the number of summands in Zeckendorf’s Theorem is a Gaussian.
Proof should follow from Markov’s Method of Moments, Binet’s formula for the Fibonacci numbers, and then differentiating identities to evaluate sums of the form ∑ j jmxm.
18 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Generalizing Lekkerkerker Theorem (SMALL 2010) As n →∞, the distribution of the number of summands in Zeckendorf’s Theorem is a Gaussian.
1000 1050 1100 1150 1200 0.005 0.010 0.015 0.020 Figure: Number of summands in [F2010, F2011) 19 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Generalizing Lekkerkerker Theorem (SMALL 2010) As n →∞, the distribution of the number of summands in Zeckendorf’s Theorem is a Gaussian.
Numerics: At F100,000: Ratio of 2mth moment 휎2m to (2m −1)!!휎m 2 is between .999955 and 1 for 2m ≤10.
20 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Conclusion 21 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Conclusion Re-derive Zeckendorf and Lekkerkerker’s results through combinatorics.
Method should yield an Erdos-Kac type result on Gaussian behavior of the number of summands.
Method should be applicable to other, related questions.
NOTE: These and similar questions will be studied by the students at the 2010 SMALL REU at Williams College; we expect to be able to provide papers and proofs by the end of the summer.
22 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Appendix: Details of Computations 23 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Needed Binomial Identity Binomial identity involving Fibonacci Numbers Let Fm denote the mth Fibonacci number, with F1 = 1, F2 = 2, F3 = 3, F4 = 5 and so on. Then ⌊n−1 2 ⌋ ∑ k=0 (n −1 −k k ) = Fn−1.
Proof by induction: The base case is trivially verified. Assume our claim holds for n and show that it holds for n + 1.
We may extend the sum to n −1, as (n−1−k k ) = 0 whenever k > ⌊n−1 2 ⌋. Using the standard identity that (m ℓ ) + ( m ℓ+ 1 ) = (m + 1 ℓ+ 1 ) , and the convention that (m ℓ ) = 0 if ℓis a negative integer, we find n ∑ k=0 (n −k k ) = n ∑ k=0 [(n −1 −k k −1 ) + (n −1 −k k )] = n ∑ k=1 (n −1 −k k −1 ) + n ∑ k=0 (n −1 −k k ) = n ∑ k=1 (n −2 −(k −1) k −1 ) + n ∑ k=0 (n −1 −k k ) = Fn−2 + Fn−1 by the inductive assumption; noting Fn−2 + Fn−1 = Fn completes the proof.
□ 24 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Derivation of Recurrence Relation for ℰ(n) ℰ(n) = ⌊n−1 2 ⌋ ∑ k=0 k (n −1 −k k ) = ⌊n−1 2 ⌋ ∑ k=1 k (n −1 −k)!
k!(n −1 −2k)!
= ⌊n−1 2 ⌋ ∑ k=1 (n −1 −k) (n −2 −k)!
(k −1)!(n −1 −2k)!
= ⌊n−1 2 ⌋ ∑ k=1 (n −2 −(k −1)) (n −3 −(k −1)!
(k −1)!(n −3 −2(k −1))!
= ⌊n−3 2 ⌋ ∑ ℓ=0 (n −2 −ℓ) (n −3 −ℓ ℓ ) = (n −2) ⌊n−3 2 ⌋ ∑ ℓ=0 (n −3 −ℓ ℓ ) − ⌊n−3 2 ⌋ ∑ ℓ=0 ℓ (n −3 −ℓ ℓ ) = (n −2)Fn−3 −ℰ(n −2), which proves the claim (note we used the binomial identity to replace the sum of binomial coefficients with a Fibonacci number).
25 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Formula for ℰ(n) Formula for ℰ(n) ℰ(n) = nFn−1 휑2 + 1 + O(Fn−2).
Proof: The proof follows from using telescoping sums to get an expression for ℰ(n), which is then evaluated by inputting Binet’s formula and differentiating identities. Explicitly, consider ⌊n−3 2 ⌋ ∑ ℓ=0 (−1)ℓ(ℰ(n −2ℓ) + ℰ(n −2(ℓ+ 1))) = ⌊n−3 2 ⌋ ∑ ℓ=0 (−1)ℓ(n −2 −2ℓ)Fn−3−2ℓ = ⌊n−3 2 ⌋ ∑ ℓ=0 (−1)ℓ(n −3 −2ℓ)Fn−3−2ℓ+ ⌊n−3 2 ⌋ ∑ ℓ=0 (−1)ℓ(2ℓ)Fn−3−2ℓ = ⌊n−3 2 ⌋ ∑ ℓ=0 (−1)ℓ(n −3 −2ℓ)Fn−3−2ℓ+ O(Fn−2); while we could evaluate the last sum exactly, trivially estimating it suffices to obtain the main term (as we have a sum of every other Fibonacci number, the sum is at most the next Fibonacci number after the largest one in our sum).
26 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Formula for ℰ(n) (continued) We now use Binet’s formula to convert the sum into a geometric series. Letting 휑= 1+ √ 5 2 be the golden mean, we have Fn = 휑 √ 5 ⋅휑n − 1 −휑 √ 5 ⋅(1 −휑)n (our constants are because our counting has F1 = 1, F2 = 2 and so on). As ∣1 −휑∣< 1, the error from dropping the (1 −휑)n term is O(∑ ℓ≤n n) = O(n2) = o(Fn−2), and may thus safely be absorbed in our error term. We thus find ℰ(n) = 휑 √ 5 ⌊n−3 2 ⌋ ∑ ℓ=0 (n −3 −2ℓ)(−1)ℓ휑n−3−2ℓ+ O(Fn−2) = 휑n−2 √ 5 ⎡ ⎢ ⎢ ⎣(n −3) ⌊n−3 2 ⌋ ∑ ℓ=0 (−휑−2)ℓ−2 ⌊n−3 2 ⌋ ∑ ℓ=0 ℓ(−휑−2)ℓ ⎤ ⎥ ⎥ ⎦+ O(Fn−2).
27 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Formula for ℰ(n) (continued) We use the geometric series formula to evaluate the first term. We drop the upper boundary term of (−휑−1)⌊n−3 2 ⌋, as this term is negligible since 휑> 1. We may also move the 3 from the n −3 into the error term, and are left with ℰ(n) = 휑n−2 √ 5 ⎡ ⎢ ⎢ ⎣ n 1 + 휑−2 −2 ⌊n−3 2 ⌋ ∑ ℓ=0 ℓ(−휑−2)ℓ ⎤ ⎥ ⎥ ⎦+ O(Fn−2) = 휑n−2 √ 5 [ n 1 + 휑−2 −2S (⌊n −3 2 ⌋ , −휑−2 )] + O(Fn−2), where 풮(m, x) = m ∑ j=0 jxj.
There is a simple formula for 풮(m, x). As m ∑ j=0 xj = xm+1 −1 x −1 , applying the operator x d dx gives 풮(m, x) = m ∑ j=0 jxj = x (m + 1)xm(x −1) −(xm+1 −1) (x −1)2 = mxm+2 −(m + 1)xm+1 + x (x −1)2 .
28 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs Formula for ℰ(n) (continued) Taking x = −휑−2, we see that the contribution from this piece may safely be absorbed into the error term O(Fn−2), leaving us with ℰ(n) = n휑n−2 √ 5(1 + 휑−2) + O(Fn−2) = n휑n √ 5(휑2 + 1) + O(Fn−2).
Noting that for large n we have Fn−1 = 휑n √ 5 + O(1), we finally obtain ℰ(n) = nFn−1 휑2 + 1 + O(Fn−2).□ A similar calculation should yield the higher moments; this will be done by the students of the 2010 SMALL REU at Williams College.
29 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs References 30 Introduction Review Zeckendorf’s Thm Lekkerkerker’s Thm Erdos-Kac Type Theorem Conclusion Appendix Refs D. E. Daykin, Representation of natural numbers as sums of generalized Fibonacci numbers, J. London Mathematical Society 35 (1960), 143–160.
C. G. Lekkerkerker, Voorstelling van natuurlyke getallen door een som van getallen van Fibonacci, Simon Stevin 29 (1951-1952), 190–195.
SMALL REU (2010, Williams College), preprint.
31 |
11677 | https://en.wikipedia.org/wiki/Neutrophil | Jump to content
Neutrophil
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From Wikipedia, the free encyclopedia
Type of white blood cell
Not to be confused with Neutrophile.
| Neutrophil | |
--- |
| 3D rendering of a neutrophil | |
| Neutrophils with segmented nuclei surrounded by erythrocytes and platelets. Intra-cellular granules are visible in the cytoplasm (Giemsa stained). | |
| Details | |
| System | Immune system |
| Function | Phagocytosis |
| Identifiers | |
| MeSH | D009504 |
| TH | H2.00.04.1.02012 |
| FMA | 62860 |
| Anatomical terms of microanatomy [edit on Wikidata] | |
Neutrophils are a type of phagocytic white blood cell and part of innate immunity. More specifically, they form the most abundant type of granulocytes and make up 40% to 70% of all white blood cells in humans. Their functions vary in different animals. They are also known as neutrocytes, heterophils or polymorphonuclear leukocytes.
They are formed from stem cells in the bone marrow and differentiated into subpopulations of neutrophil-killers and neutrophil-cagers. They are short-lived (between 5 and 135 hours, see § Life span) and highly mobile, as they can enter parts of tissue where other cells/molecules cannot. Neutrophils may be subdivided into segmented neutrophils and banded neutrophils (or bands). They form part of the polymorphonuclear cells family (PMNs) together with basophils and eosinophils.
The name neutrophil derives from staining characteristics on hematoxylin and eosin (H&E) histological or cytological preparations. Whereas basophilic white blood cells stain dark blue and eosinophilic white blood cells stain bright red, neutrophils stain a neutral pink. Normally, neutrophils contain a nucleus divided into 2–5 lobes.
Neutrophils are a type of phagocyte and are normally found in the bloodstream. During the beginning (acute) phase of inflammation, particularly as a result of bacterial infection, environmental exposure, and some cancers, neutrophils are one of the first responders of inflammatory cells to migrate toward the site of inflammation. They migrate through the blood vessels and then through interstitial space, following chemical signals such as interleukin-8 (IL-8), C5a, fMLP, leukotriene B4, and hydrogen peroxide (H2O2) in a process called chemotaxis. They are the predominant cells in pus, accounting for its whitish/yellowish appearance.
Neutrophils are recruited to the site of injury within minutes following trauma and are the hallmark of acute inflammation. They not only play a central role in combating infection but also contribute to pain in the acute period by releasing pro-inflammatory cytokines and other mediators that sensitize nociceptors, leading to heightened pain perception. However, due to some pathogens being indigestible, they may not be able to resolve certain infections without the assistance of other types of immune cells.
Structure
[edit]
When adhered to a surface, neutrophil granulocytes have an average diameter of 12–15 micrometers (μm) in peripheral blood smears. In suspension, human neutrophils have an average diameter of 8.85 μm.
With the eosinophil and the basophil, they form the class of polymorphonuclear cells, named for the nucleus' multilobulated shape (as compared to lymphocytes and monocytes, the other types of white cells). The nucleus has a characteristic lobed appearance, the separate lobes connected by chromatin. The nucleolus disappears as the neutrophil matures, which is something that happens in only a few other types of nucleated cells.: 168 Up to 17% of female human neutrophil nuclei have a drumstick-shaped appendage which contains the inactivated X chromosome. In the cytoplasm, the Golgi apparatus is small, mitochondria and ribosomes are sparse, and the rough endoplasmic reticulum is absent.: 170 The cytoplasm also contains about 200 granules, of which a third are azurophilic.: 170
Neutrophils will show increasing segmentation (many segments of the nucleus) as they mature. A normal neutrophil should have 3–5 segments. Hypersegmentation is not normal but occurs in some disorders, most notably vitamin B12 deficiency. This is noted in a manual review of the blood smear and is positive when most or all of the neutrophils have 5 or more segments.[citation needed]
Neutrophils are the most abundant white blood cells in the human body (approximately 1011 are produced daily); they account for approximately 50–70% of all white blood cells (leukocytes). The stated normal range for human blood counts varies between laboratories, but a neutrophil count of 2.5–7.5 × 109/L is a standard normal range. People of African and Middle Eastern descent may have lower counts, which are still normal. A report may divide neutrophils into segmented neutrophils and bands.
When circulating in the bloodstream and inactivated, neutrophils are spherical. Once activated, they change shape and become more amorphous or amoeba-like and can extend pseudopods as they hunt for antigens.
The capacity of neutrophils to engulf bacteria is reduced when simple sugars like glucose, fructose as well as sucrose, honey and orange juice were ingested, while the ingestion of starches had no effect. Fasting, on the other hand, strengthened the neutrophils' phagocytic capacity to engulf bacteria. It was concluded that the function, and not the number, of phagocytes in engulfing bacteria was altered by the ingestion of sugars. In 2007 researchers at the Whitehead Institute of Biomedical Research found that given a selection of sugars on microbial surfaces, the neutrophils reacted to some types of sugars preferentially. The neutrophils preferentially engulfed and killed beta-1,6-glucan targets compared to beta-1,3-glucan targets.
Development
[edit]
Life span
[edit]
The average lifespan of inactivated human neutrophils in the circulation has been reported by different approaches to be between 5 and 135 hours.
Upon activation, they marginate (position themselves adjacent to the blood vessel endothelium) and undergo selectin-dependent capture followed by integrin-dependent adhesion in most cases, after which they migrate into tissues, where they survive for 1–2 days. Neutrophils have also been demonstrated to be released into the blood from a splenic reserve following myocardial infarction.
The distribution ratio of neutrophils in bone marrow, blood and connective tissue is 28:1:25.[citation needed]
Neutrophils are much more numerous than the longer-lived monocyte/macrophage phagocytes. A pathogen (disease-causing microorganism or virus) is likely to first encounter a neutrophil. Some experts hypothesize that the short lifetime of neutrophils is an evolutionary adaptation. The short lifetime of neutrophils minimizes propagation of those pathogens that parasitize phagocytes (e.g. Leishmania) because the more time such parasites spend outside a host cell, the more likely they will be destroyed by some component of the body's defenses. Also, because neutrophil antimicrobial products can also damage host tissues, their short life limits damage to the host during inflammation.
Neutrophils will be removed after phagocytosis of pathogens by macrophages. PECAM-1 and phosphatidylserine on the cell surface are involved in this process.[citation needed]
Function
[edit]
Chemotaxis
[edit]
Neutrophils undergo a process called chemotaxis via amoeboid movement, which allows them to migrate toward sites of infection or inflammation. Cell surface receptors allow neutrophils to detect chemical gradients of molecules such as interleukin-8 (IL-8), interferon gamma (IFN-γ), C3a, C5a, and leukotriene B4, which these cells use to direct the path of their migration.[citation needed]
Neutrophils have a variety of specific receptors, including ones for the complement system, cytokines like interleukins and IFN-γ, chemokines, lectins, and other proteins. They also express receptors to detect and adhere to endothelium and Fc receptors for opsonin.
In leukocytes responding to a chemoattractant, the cellular polarity is regulated by activities of small Ras or Rho guanosine triphosphatases (Ras or Rho GTPases) and the phosphoinositide 3-kinases (PI3Ks). In neutrophils, lipid products of PI3Ks regulate activation of Rac1, hematopoietic Rac2, and RhoG GTPases of the Rho family and are required for cell motility. Ras-GTPases and Rac-GTPases regulate cytoskeletal dynamics and facilitate neutrophils adhesion, migration, and spreading. They accumulate asymmetrically to the plasma membrane at the leading edge of polarized cells. Spatially regulating Rho GTPases and organizing the leading edge of the cell, PI3Ks and their lipid products could play pivotal roles in establishing leukocyte polarity, as compass molecules that tell the cell where to crawl.[citation needed]
It has been shown in mice that in certain conditions neutrophils have a specific type of migration behaviour referred to as neutrophil swarming during which they migrate in a highly coordinated manner and accumulate and cluster to sites of inflammation.
Anti-microbial function
[edit]
Being highly motile, neutrophils quickly congregate at a focus of infection, attracted by cytokines expressed by activated endothelium, mast cells, and macrophages. Neutrophils express and release cytokines, which in turn amplify inflammatory reactions by several other cell types.[citation needed]
In addition to recruiting and activating other cells of the immune system, neutrophils play a key role in the front-line defense against invading pathogens, and contain a broad range of proteins. Neutrophils have three methods for directly attacking microorganisms: phagocytosis (ingestion), degranulation (release of soluble anti-microbials), and generation of neutrophil extracellular traps (NETs).
Phagocytosis
[edit]
Neutrophils are phagocytes, capable of ingesting microorganisms or particles. For targets to be recognized, they must be coated in opsonins – a process known as antibody opsonization. They can internalize and kill many microbes, each phagocytic event resulting in the formation of a phagosome into which reactive oxygen species and hydrolytic enzymes are secreted. The consumption of oxygen during the generation of reactive oxygen species has been termed the "respiratory burst", although unrelated to respiration or energy production.[citation needed]
The respiratory burst involves the activation of the enzyme NADPH oxidase, which produces large quantities of superoxide, a reactive oxygen species. Superoxide decays spontaneously or is broken down via enzymes known as superoxide dismutases (Cu/ZnSOD and MnSOD), to hydrogen peroxide, which is then converted to hypochlorous acid (HClO), by the green heme enzyme myeloperoxidase. It is thought that the bactericidal properties of HClO are enough to kill bacteria phagocytosed by the neutrophil, but this may instead be a step necessary for the activation of proteases.
Though neutrophils can kill many microbes, the interaction of neutrophils with microbes and molecules produced by microbes often alters neutrophil turnover. The ability of microbes to alter the fate of neutrophils is highly varied, can be microbe-specific, and ranges from prolonging the neutrophil lifespan to causing rapid neutrophil lysis after phagocytosis. Chlamydia pneumoniae and Neisseria gonorrhoeae have been reported to delay neutrophil apoptosis. Thus, some bacteria – and those that are predominantly intracellular pathogens – can extend the neutrophil lifespan by disrupting the normal process of spontaneous apoptosis and/or PICD (phagocytosis-induced cell death). On the other end of the spectrum, some pathogens such as Streptococcus pyogenes are capable of altering neutrophil fate after phagocytosis by promoting rapid cell lysis and/or accelerating apoptosis to the point of secondary necrosis.
Degranulation
[edit]
Neutrophils also release an assortment of proteins in three types of granules by a process called degranulation. The contents of these granules have antimicrobial properties, and help combat infection. Glitter cells are polymorphonuclear leukocyte neutrophils with granules. Degranulation is postulated to occur in a hierarchical manner, with the sequential release of secretory vesicles, tertiary granules, specific granules, and azurophilic granules in response to increasing intracellular calcium concentrations. The release of neutrophils by degranulation occurs through exocytosis, regulated by exocytotic machinery including SNARE proteins, RAC2, RAB27, and others.[citation needed]
| Granule type | Protein |
--- |
| Azurophilic granules (or "primary granules") | Myeloperoxidase, bactericidal/permeability-increasing protein (BPI), defensins, and the serine proteases neutrophil elastase, Proteinase 3 and cathepsin G |
| Specific granules (or "secondary granules") | Alkaline phosphatase, lysozyme, NADPH oxidase, collagenase, lactoferrin, histaminase, and cathelicidin |
| Tertiary granules | Cathepsin, gelatinase, and collagenase |
Neutrophil extracellular traps
[edit]
In 2004, Brinkmann and colleagues described a striking observation that activation of neutrophils causes the release of web-like structures of DNA; this represents a third mechanism for killing bacteria. These neutrophil extracellular traps (NETs) comprise a web of fibers composed of chromatin and serine proteases that trap and kill extracellular microbes. It is suggested that NETs provide a high local concentration of antimicrobial components and bind, disarm, and kill microbes independent of phagocytic uptake. In addition to their possible antimicrobial properties, NETs may serve as a physical barrier that prevents further spread of pathogens. Trapping of bacteria may be a particularly important role for NETs in sepsis, where NETs are formed within blood vessels. Finally, NET formation has been demonstrated to augment macrophage bactericidal activity during infection. Recently, NETs have been shown to play a role in inflammatory diseases, as NETs could be detected in preeclampsia, a pregnancy-related inflammatory disorder in which neutrophils are known to be activated. Neutrophil NET formation may also impact cardiovascular disease, as NETs may influence thrombus formation in coronary arteries.
NETs are now known to exhibit pro-thrombotic effects both in vitro and in vivo. More recently, in 2020 NETs were implicated in the formation of blood clots in cases of severe COVID-19.
Tumor Associated Neutrophils (TANS)
[edit]
TANs can exhibit an elevated extracellular acidification rate when there is an increase in glycolysis levels. When there is a metabolic shift in TANs this can lead to tumor progression in certain areas of the body, such as the lungs. TANs support the growth and progression of tumors unlike normal neutrophils which would inhibit tumor progression through the phagocytosis of tumor cells. Utilizing a mouse model, they[who?] identified that both Glut1 and glucose metabolism increased in TANs found within a mouse who possessed lung adenocarcinoma. A study showed that lung tumor cells can remotely initiate osteoblasts and these osteoblasts can worsen tumors in two ways. First, they can induce SiglecFhigh-expressing neutrophil formation that in turn promotes lung tumor growth and progression. Second, the osteoblasts can promote bone growth thus forming a favorable environment for tumor cells to grow to form bone metastasis.
Clinical significance
[edit]
Low neutrophil counts are termed neutropenia. This can be congenital (developed at or before birth) or it can develop later, as in the case of aplastic anemia or some kinds of leukemia. It can also be a side-effect of medication, most prominently chemotherapy. Neutropenia makes an individual highly susceptible to infections. It can also be the result of colonization by intracellular neutrophilic parasites.[citation needed]
In alpha 1-antitrypsin deficiency, the important neutrophil elastase is not adequately inhibited by alpha 1-antitrypsin, leading to excessive tissue damage in the presence of inflammation – the most prominent one being emphysema. Negative effects of elastase have also been shown in cases when the neutrophils are excessively activated (in otherwise healthy individuals) and release the enzyme in extracellular space. Unregulated activity of neutrophil elastase can lead to disruption of pulmonary barrier showing symptoms corresponding with acute lung injury. The enzyme also influences activity of macrophages by cleaving their toll-like receptors (TLRs) and downregulating cytokine expression by inhibiting nuclear translocation of NF-κB.
In Familial Mediterranean fever (FMF), a mutation in the pyrin (or marenostrin) gene, which is expressed mainly in neutrophil granulocytes, leads to a constitutively active acute-phase response and causes attacks of fever, arthralgia, peritonitis, and – eventually – amyloidosis.
Hyperglycemia can lead to neutrophil dysfunction. Dysfunction in the neutrophil biochemical pathway myeloperoxidase as well as reduced degranulation are associated with hyperglycemia.
The Absolute neutrophil count (ANC) is also used in diagnosis and prognosis. ANC is the gold standard for determining severity of neutropenia, and thus neutropenic fever. Any ANC < 1500 cells / mm3 is considered neutropenia, but <500 cells / mm3 is considered severe. There is also new research tying ANC to myocardial infarction as an aid in early diagnosis. Neutrophils promote ventricular tachycardia in acute myocardial infarction.
In autopsy, the presence of neutrophils in the heart or brain is one of the first signs of infarction, and is useful in the timing and diagnosis of myocardial infarction and stroke.[citation needed]
Neutrophils are seen in a myocardial infarction at approximately 12–24 hours, as seen in this micrograph.
In stroke, they are beginning to infiltrate the infarcted brain after 6 to 8 hours.
Pathogen evasion and resistance
[edit]
See also: Phagocyte § Pathogen evasion and resistance
Just like phagocytes, pathogens may evade or infect neutrophils. Some bacterial pathogens evolved various mechanisms such as virulence molecules to avoid being killed by neutrophils. These molecules collectively may alter or disrupt neutrophil recruitment, apoptosis or bactericidal activity.
Neutrophils can also serve as host cell for various parasites that infect them avoiding phagocytosis, including:
Leishmania major – uses neutrophils as vehicle to parasitize phagocytes
M. tuberculosis
M. leprae
Yersinia pestis
Chlamydia pneumoniae
Neutrophil antigens
[edit]
There are five (HNA 1–5) sets of neutrophil antigens recognized. The three HNA-1 antigens (a-c) are located on the low affinity Fc-γ receptor IIIb (FCGR3B :CD16b) The single known HNA-2a antigen is located on CD177. The HNA-3 antigen system has two antigens (3a and 3b) which are located on the seventh exon of the CLT2 gene (SLC44A2). The HNA-4 and HNA-5 antigen systems each have two known antigens (a and b) and are located in the β2 integrin. HNA-4 is located on the αM chain (CD11b) and HNA-5 is located on the αL integrin unit (CD11a).
Subpopulations
[edit]
Two functionally unequal subpopulations of neutrophils were identified on the basis of different levels of their reactive oxygen metabolite generation, membrane permeability, activity of enzyme system, and ability to be inactivated. The cells of one subpopulation with high membrane permeability (neutrophil-killers) intensively generate reactive oxygen metabolites and are inactivated in consequence of interaction with the substrate, whereas cells of another subpopulation (neutrophil-cagers) produce reactive oxygen species less intensively, don't adhere to substrate and preserve their activity. Additional studies have shown that lung tumors can be infiltrated by various populations of neutrophils.
Video
[edit]
A rapidly moving neutrophil can be seen taking up several conidia over an imaging time of 2 hours with one frame every 30 seconds.
A neutrophil can be seen here selectively taking up several Candida yeasts (fluorescently labeled in green) despite several contacts with Aspergillus fumigatus conidia (unlabeled, white/clear) in a 3-D collagen matrix. Imaging time was 2 hours with one frame every 30 seconds.
Neutrophils display highly directional amoeboid motility in infected footpad and phalanges. Intravital imaging was performed in the footpad path of LysM-eGFP mice 20 minutes after infection with Listeria monocytogenes.
Additional images
[edit]
Blood cell lineage
More complete lineages
See also
[edit]
List of distinct cell types in the adult human body
References
[edit]
^ Actor J (2012). Elsevier's Integrated Review Immunology and Microbiology (Second ed.). doi:10.1016/B978-0-323-07447-6.00002-8.
^ Ermert D, Niemiec MJ, Röhm M, Glenthøj A, Borregaard N, Urban CF (August 2013). "Candida albicans escapes from mouse neutrophils". Journal of Leukocyte Biology. 94 (2): 223–236. doi:10.1189/jlb.0213063. PMID 23650619. S2CID 25619835.
^ Al-Kuraishy HM, Al-Gareeb AI, Al-hussaniy HA, Al-Harcan NA, Alexiou A, Batiha GE (2022-03-01). "Neutrophil Extracellular Traps (NETs) and Covid-19: A new frontiers for therapeutic modality". International Immunopharmacology. 104: 108516. doi:10.1016/j.intimp.2021.108516. ISSN 1567-5769. PMC 8733219.
^ Witko-Sarsat V, Rieu P, Descamps-Latscha B, Lesavre P, Halbwachs-Mecarelli L (May 2000). "Neutrophils: molecules, functions and pathophysiological aspects". Laboratory Investigation; A Journal of Technical Methods and Pathology. 80 (5): 617–653. doi:10.1038/labinvest.3780067. PMID 10830774. S2CID 22536645.
^ Klebanoff SJ, Clark RA (1978). The Neutrophil: Function and Clinical Disorders. Elsevier/North-Holland Amsterdam. ISBN 978-0-444-80020-6.
^ Nathan C (March 2006). "Neutrophils and immunity: challenges and opportunities". Nature Reviews. Immunology. 6 (3): 173–182. doi:10.1038/nri1785. PMID 16498448. S2CID 1590558.
^ Welsh CJ (2021). Hole's Essentials of Human Anatomy and Physiology (14th ed.). New York, USA: McGraw Hill. p. 336. ISBN 978-1-260-57521-7. Retrieved 28 February 2023.
^ Jacobs L, Nawrot TS, de Geus B, Meeusen R, Degraeuwe B, Bernard A, et al. (October 2010). "Subclinical responses in healthy cyclists briefly exposed to traffic-related air pollution: an intervention study". Environmental Health. 9 (64): 64. Bibcode:2010EnvHe...9...64J. doi:10.1186/1476-069X-9-64. PMC 2984475. PMID 20973949.
^ Waugh DJ, Wilson C (November 2008). "The interleukin-8 pathway in cancer". Clinical Cancer Research. 14 (21): 6735–6741. doi:10.1158/1078-0432.CCR-07-4843. PMID 18980965. S2CID 9415085.
^ De Larco JE, Wuertz BR, Furcht LT (August 2004). "The potential role of neutrophils in promoting the metastatic phenotype of tumors releasing interleukin-8". Clinical Cancer Research. 10 (15): 4895–4900. doi:10.1158/1078-0432.CCR-03-0760. PMID 15297389. S2CID 9782495.
^ Yoo SK, Starnes TW, Deng Q, Huttenlocher A (November 2011). "Lyn is a redox sensor that mediates leukocyte wound attraction in vivo". Nature. 480 (7375): 109–112. Bibcode:2011Natur.480..109Y. doi:10.1038/nature10632. PMC 3228893. PMID 22101434.
^ Barer MR (2012). "The natural history of infection". Medical Microbiology. Elsevier. pp. 168–173. doi:10.1016/b978-0-7020-4089-4.00029-9. ISBN 978-0-7020-4089-4.
^ Cohen S, Burns RC (2002). Pathways of the Pulp (8th ed.). St. Louis: Mosby. p. 465.
^ Huerta MÁ, Molina-Álvarez M, García MM, Tejada MA, Goicoechea C, Ghasemlou N, et al. (2024-10-25). "The role of neutrophils in pain: systematic review and meta-analysis of animal studies". Pain. doi:10.1097/j.pain.0000000000003450. ISSN 0304-3959.
^ Niemiec MJ, De Samber B, Garrevoet J, Vergucht E, Vekemans B, De Rycke R, et al. (June 2015). "Trace element landscape of resting and activated human neutrophils on the sub-micrometer level". Metallomics. 7 (6): 996–1010. doi:10.1039/c4mt00346b. PMID 25832493.
^ a b c Zucker-Franklin D, Greaves MF, Grossi CE, Marmont AM (1988). "Neutrophils". Atlas of Blood Cells: Function and Pathology. Vol. 1 (2nd ed.). Philadelphia: Lea & Ferbiger. ISBN 978-0-8121-1094-4.
^ Karni RJ, Wangh LJ, Sanchez JA (August 2001). "Nonrandom location and orientation of the inactive X chromosome in human neutrophil nuclei". Chromosoma. 110 (4): 267–274. doi:10.1007/s004120100145. PMID 11534818. S2CID 24750407.
^ Reich D, Nalls MA, Kao WH, Akylbekova EL, Tandon A, Patterson N, et al. (January 2009). "Reduced neutrophil count in people of African descent is due to a regulatory variant in the Duffy antigen receptor for chemokines gene". PLOS Genetics. 5 (1): e1000360. doi:10.1371/journal.pgen.1000360. PMC 2628742. PMID 19180233.
^ a b Edwards SW (1994). Biochemistry and physiology of the neutrophil. Cambridge University Press. p. 6. ISBN 978-0-521-41698-6.
^ Sanchez A, Reeser JL, Lau HS, Yahiku PY, Willard RE, McMillan PJ, et al. (November 1973). "Role of sugars in human neutrophilic phagocytosis". The American Journal of Clinical Nutrition. 26 (11): 1180–1184. doi:10.1093/ajcn/26.11.1180. PMID 4748178. These data suggest that the function and not the number of phagocytes was altered by ingestion of sugars. This implicates glucose and other simple carbohydrates in the control of phagocytosis and shows that the effects last for at least 5 hr. On the other hand, a fast of 36 or 60 hr significantly increased (P < 0.001) the phagocytic index
^ Rubin-Bejerano I, Abeijon C, Magnelli P, Grisafi P, Fink GR (July 2007). "Phagocytosis by human neutrophils is stimulated by a unique fungal cell wall component". Cell Host & Microbe. 2 (1): 55–67. doi:10.1016/j.chom.2007.06.002. PMC 2083279. PMID 18005717.
^ Kneller A (2007). "White blood cells are picky about sugar". Whitehead Institute. Retrieved 2013-08-09.
^ Tak T, Tesselaar K, Pillay J, Borghans JA, Koenderman L (October 2013). "What's your age again? Determination of human neutrophil half-lives revisited". Journal of Leukocyte Biology. 94 (4): 595–601. doi:10.1189/jlb.1112571. PMID 23625199. S2CID 40113921.
^ Pillay J, den Braber I, Vrisekoop N, Kwast LM, de Boer RJ, Borghans JA, et al. (July 2010). "In vivo labeling with 2H2O reveals a human neutrophil lifespan of 5.4 days". Blood. 116 (4): 625–627. doi:10.1182/blood-2010-01-259028. PMID 20410504. S2CID 909519.
^ a b Wheater PR, Stevens A (2002). Wheater's basic histopathology: a colour atlas and text. Edinburgh: Churchill Livingstone. ISBN 978-0-443-07001-3.
^ Akbar N, Braithwaite AT, Corr EM, Koelwyn GJ, van Solingen C, Cochain C, et al. (March 2023). "Rapid neutrophil mobilization by VCAM-1+ endothelial cell-derived extracellular vesicles". Cardiovascular Research. 119 (1): 236–251. doi:10.1093/cvr/cvac012. PMC 10022859. PMID 35134856.
^ a b Ritter U, Frischknecht F, van Zandbergen G (November 2009). "Are neutrophils important host cells for Leishmania parasites?". Trends in Parasitology. 25 (11): 505–510. doi:10.1016/j.pt.2009.08.003. PMID 19762280.
^ Serhan CN, Ward PA, Gilroy DW (2010). Fundamentals of Inflammation. Cambridge University Press. pp. 53–54. ISBN 978-0-521-88729-8.
^ Pantarelli C, Welch HC (November 2018). "Rac-GTPases and Rac-GEFs in neutrophil adhesion, migration and recruitment". European Journal of Clinical Investigation. 48 (Suppl 2): e12939. doi:10.1111/eci.12939. PMC 6321979. PMID 29682742.
^ Lin Y, Pal DS, Banerjee P, Banerjee T, Qin G, Deng Y, et al. (July 2024). "Ras suppression potentiates rear actomyosin contractility-driven cell polarization and migration". Nature Cell Biology: 1–15. doi:10.1038/s41556-024-01453-4. PMC 11364469. PMID 38951708.
^ Pal DS, Banerjee T, Lin Y, de Trogoff F, Borleis J, Iglesias PA, et al. (July 2023). "Actuation of single downstream nodes in growth factor network steers immune cell migration". Developmental Cell. 58 (13): 1170–1188.e7. doi:10.1016/j.devcel.2023.04.019. PMC 10524337. PMID 37220748.
^ Lämmermann T, Afonso PV, Angermann BR, Wang JM, Kastenmüller W, Parent CA, et al. (June 2013). "Neutrophil swarms require LTB4 and integrins at sites of cell death in vivo". Nature. 498 (7454): 371–375. Bibcode:2013Natur.498..371L. doi:10.1038/nature12175. PMC 3879961. PMID 23708969.
^ Ear T, McDonald PP (April 2008). "Cytokine generation, promoter activation, and oxidant-independent NF-kappaB activation in a transfectable human neutrophilic cellular model". BMC Immunology. 9: 14. doi:10.1186/1471-2172-9-14. PMC 2322942. PMID 18405381.
^ Ambatipudi KS, Old JM, Guilhaus M, Raftery M, Hinds L, Deane EM (2006). Proteomic analysis of the neutrophil proteins of the Tammar wallaby (Macropus eugenii). Comparative Biochemistry and Physiology. Part D: Genomic and Proteomics. 1(3), 283-291. DOI: 10.1016/j.cbd.2006.05.002
^ Hickey MJ, Kubes P (May 2009). "Intravascular immunity: the host-pathogen encounter in blood vessels". Nature Reviews. Immunology. 9 (5): 364–375. doi:10.1038/nri2532. PMID 19390567. S2CID 8068543.
^ Segal AW (2005). "How neutrophils kill microbes". Annual Review of Immunology. 23 (5): 197–223. doi:10.1146/annurev.immunol.23.021704.115653. PMC 2092448. PMID 15771570.
^ Simons MP, Nauseef WM, Griffith TS, Apicella MA (November 2006). "Neisseria gonorrhoeae delays the onset of apoptosis in polymorphonuclear leukocytes". Cellular Microbiology. 8 (11): 1780–1790. doi:10.1111/j.1462-5822.2006.00748.x. PMID 16803582. S2CID 25253422.
^ Chen A, Seifert HS (November 2011). "Neisseria gonorrhoeae-mediated inhibition of apoptotic signalling in polymorphonuclear leukocytes". Infection and Immunity. 79 (11): 4447–4458. doi:10.1128/IAI.01267-10. PMC 3257915. PMID 21844239.
^ van Zandbergen G, Gieffers J, Kothe H, Rupp J, Bollinger A, Aga E, et al. (February 2004). "Chlamydia pneumoniae multiply in neutrophil granulocytes and delay their spontaneous apoptosis". Journal of Immunology. 172 (3): 1768–1776. doi:10.4049/jimmunol.172.3.1768. PMID 14734760. S2CID 27422510.
^ Kobayashi SD, Braughton KR, Whitney AR, Voyich JM, Schwan TG, Musser JM, et al. (September 2003). "Bacterial pathogens modulate an apoptosis differentiation program in human neutrophils". Proceedings of the National Academy of Sciences of the United States of America. 100 (19): 10948–10953. doi:10.1073/pnas.1833375100. PMC 196908. PMID 12960399.
^ Kobayashi SD, Malachowa N, DeLeo FR (2017). "Influence of Microbes on Neutrophil Life and Death". Frontiers in Cellular and Infection Microbiology. 7 (159): 159. doi:10.3389/fcimb.2017.00159. PMC 5410578. PMID 28507953. Material was copied from this source, which is available under a Creative Commons Attribution 4.0 International License.
^ Berman LB, Feys JO, Schreiner GE (November 1956). "Observations on the glitter-cell phenomenon". The New England Journal of Medicine. 255 (21): 989–991. doi:10.1056/NEJM195611222552104. PMID 13378597.
^ Sengelov H, Kjedsen L, Borregaard N (February 1993). "Control of exocytosis in early neutrophil activation". Journal of Immunology. 150 (4): 1535–43. PMID 8381838.
^ Ringel EW, Soter NA, Austen KF (August 1984). "Localization of histaminase to the specific granule of the human neutrophil". Immunology. 52 (4): 649–658. PMC 1454675. PMID 6430792.
^ Brinkmann V, Reichard U, Goosmann C, Fauler B, Uhlemann Y, Weiss DS, et al. (March 2004). "Neutrophil extracellular traps kill bacteria". Science. 303 (5663): 1532–1535. Bibcode:2004Sci...303.1532B. doi:10.1126/science.1092385. PMID 15001782. S2CID 21628300.
^ Urban CF, Ermert D, Schmid M, Abu-Abed U, Goosmann C, Nacken W, et al. (October 2009). "Neutrophil extracellular traps contain calprotectin, a cytosolic protein complex involved in host defense against Candida albicans". PLOS Pathogens. 5 (10): e1000639. doi:10.1371/journal.ppat.1000639. PMC 2763347. PMID 19876394.
^ Clark SR, Ma AC, Tavener SA, McDonald B, Goodarzi Z, Kelly MM, et al. (April 2007). "Platelet TLR4 activates neutrophil extracellular traps to ensnare bacteria in septic blood". Nature Medicine. 13 (4): 463–469. doi:10.1038/nm1565. PMID 17384648. S2CID 22372863.
^ Monteith AJ, Miller JM, Maxwell CN, Chazin WJ, Skaar EP (September 2021). "Neutrophil extracellular traps enhance macrophage killing of bacterial pathogens". Science Advances. 7 (37): eabj2101. Bibcode:2021SciA....7.2101M. doi:10.1126/sciadv.abj2101. PMC 8442908. PMID 34516771.
^ Monteith AJ, Miller JM, Beavers WN, Maloney KN, Seifert EL, Hajnoczky G, et al. (February 2022). "Mitochondrial Calcium Uniporter Affects Neutrophil Bactericidal Activity during Staphylococcus aureus Infection". Infection and Immunity. 90 (2): e0055121. doi:10.1128/IAI.00551-21. PMC 8853686. PMID 34871043.
^ Gupta AK, Hasler P, Holzgreve W, Hahn S (June 2007). "Neutrophil NETs: a novel contributor to preeclampsia-associated placental hypoxia?". Seminars in Immunopathology. 29 (2): 163–167. doi:10.1007/s00281-007-0073-4. PMID 17621701. S2CID 12887059.
^ Hoyer FF, Nahrendorf M (February 2017). "Neutrophil contributions to ischaemic heart disease". European Heart Journal. 38 (7): 465–472. doi:10.1093/eurheartj/ehx017. PMID 28363210.
^ Mangold A, Alias S, Scherz T, Hofbauer M, Jakowitsch J, Panzenböck A, et al. (March 2015). "Coronary neutrophil extracellular trap burden and deoxyribonuclease activity in ST-elevation acute coronary syndrome are predictors of ST-segment resolution and infarct size". Circulation Research. 116 (7): 1182–1192. doi:10.1161/CIRCRESAHA.116.304944. PMID 25547404. S2CID 2532741.
^ Fuchs TA, Brill A, Duerschmied D, Schatzberg D, Monestier M, Myers DD, et al. (September 2010). "Extracellular DNA traps promote thrombosis". Proceedings of the National Academy of Sciences of the United States of America. 107 (36): 15880–15885. Bibcode:2010PNAS..10715880F. doi:10.1073/pnas.1005743107. PMC 2936604. PMID 20798043.
^ Brill A, Fuchs TA, Savchenko AS, Thomas GM, Martinod K, De Meyer SF, et al. (January 2012). "Neutrophil extracellular traps promote deep vein thrombosis in mice". Journal of Thrombosis and Haemostasis. 10 (1): 136–144. doi:10.1111/j.1538-7836.2011.04544.x. PMC 3319651. PMID 22044575.
^ Borissoff JI, ten Cate H (September 2011). "From neutrophil extracellular traps release to thrombosis: an overshooting host-defense mechanism?". Journal of Thrombosis and Haemostasis. 9 (9): 1791–1794. doi:10.1111/j.1538-7836.2011.04425.x. PMID 21718435. S2CID 5368241.
^ Zuo Y, Yalavarthi S, Shi H, Gockman K, Zuo M, Madison JA, et al. (June 2020). "Neutrophil extracellular traps in COVID-19". JCI Insight. 5 (11): e138999. doi:10.1172/jci.insight.138999. PMC 7308057. PMID 32329756. S2CID 216109364.
^ a b Ancey PB, Contat C, Boivin G, Sabatino S, Pascual J, Zangger N, et al. (May 2021). "GLUT1 Expression in Tumor-Associated Neutrophils Promotes Lung Cancer Growth and Resistance to Radiotherapy". Cancer Research. 81 (9): 2345–2357. doi:10.1158/0008-5472.CAN-20-2870. PMC 8137580. PMID 33753374.
^ Azevedo PO, Paiva AE, Santos GS, Lousado L, Andreotti JP, Sena IF, et al. (December 2018). "Cross-talk between lung cancer and bones results in neutrophils that promote tumor progression". Cancer and Metastasis Reviews. 37 (4): 779–790. doi:10.1007/s10555-018-9759-4. PMC 6358512. PMID 30203108.
^ Kawabata K, Hagio T, Matsuoka S (September 2002). "The role of neutrophil elastase in acute lung injury". European Journal of Pharmacology. 451 (1): 1–10. doi:10.1016/S0014-2999(02)02182-9. PMID 12223222.
^ Domon H, Nagai K, Maekawa T, Oda M, Yonezawa D, Takeda W, et al. (2018). "Neutrophil Elastase Subverts the Immune Response by Cleaving Toll-Like Receptors and Cytokines in Pneumococcal Pneumonia". Frontiers in Immunology. 9: 732. doi:10.3389/fimmu.2018.00732. PMC 5996908. PMID 29922273.
^ Ozen S (July 2003). "Familial mediterranean fever: revisiting an ancient disease". European Journal of Pediatrics. 162 (7–8): 449–454. doi:10.1007/s00431-003-1223-x. PMID 12751000. S2CID 3464945.
^ Xiu F, Stanojcic M, Diao L, Jeschke MG (8 May 2014). "Stress hyperglycemia, insulin treatment, and innate immune cells". International Journal of Endocrinology. 2014: 486403. doi:10.1155/2014/486403. PMC 4034653. PMID 24899891.
^ Al-Gwaiz LA, Babay HH (2007). "The diagnostic value of absolute neutrophil count, band count and morphologic changes of neutrophils in predicting bacterial infections". Medical Principles and Practice. 16 (5): 344–347. doi:10.1159/000104806. PMID 17709921. S2CID 5499290.
^ Khan HA, Alhomida AS, Sobki SH, Moghairi AA, Koronki HE (2012). "Blood cell counts and their correlation with creatine kinase and C-reactive protein in patients with acute myocardial infarction". International Journal of Clinical and Experimental Medicine. 5 (1): 50–55. PMC 3272686. PMID 22328948.
^ Basili S, Di Francoi M, Rosa A, Ferroni P, Diurni V, Scarpellini MG, et al. (April 2004). "Absolute neutrophil counts and fibrinogen levels as an aid in the early diagnosis of acute myocardial infarction". Acta Cardiologica. 59 (2): 135–140. doi:10.2143/ac.59.2.2005167. PMID 15139653. S2CID 37382677.
^ Grune J, Lewis AJ, Yamazoe M, Hulsmans M, Rohde D, Xiao L, et al. (July 2022). "Neutrophils incite and macrophages avert electrical storm after myocardial infarction". Nature Cardiovascular Research. 1 (7): 649–664. doi:10.1038/s44161-022-00094-w. PMC 9410341. PMID 36034743.
^ Michaud K, Basso C, d'Amati G, Giordano C, Kholová I, Preston SD, et al. (February 2020). "Diagnosis of myocardial infarction at autopsy: AECVP reappraisal in the light of the current clinical classification". Virchows Archiv. 476 (2): 179–194. doi:10.1007/s00428-019-02662-1. PMC 7028821. PMID 31522288.
"This article is distributed under the terms of the Creative Commons Attribution 4.0 International License ( Archived 2015-11-21 at the Wayback Machine)"
^ Jickling GC, Liu D, Ander BP, Stamova B, Zhan X, Sharp FR (June 2015). "Targeting neutrophils in ischemic stroke: translational insights from experimental studies". Journal of Cerebral Blood Flow and Metabolism. 35 (6): 888–901. doi:10.1038/jcbfm.2015.45. PMC 4640255. PMID 25806703.
^ a b Kobayashi SD, Malachowa N, DeLeo FR (2018). "Neutrophils and Bacterial Immune Evasion". Journal of Innate Immunity. 10 (5–6): 432–441. doi:10.1159/000487756. PMC 6784029. PMID 29642066.
^ a b c d Parker HA, Forrester L, Kaldor CD, Dickerhof N, Hampton MB (2021-12-23). "Antimicrobial Activity of Neutrophils Against Mycobacteria". Frontiers in Immunology. 12: 782495. doi:10.3389/fimmu.2021.782495. PMC 8732375. PMID 35003097.
^ Chu HT, Lin H, Tsao TT, Chang CF, Hsiao WW, Yeh TJ, et al. (September 2013). "Genotyping of human neutrophil antigens (HNA) from whole genome sequencing data". BMC Medical Genomics. 6 (1): 31. doi:10.1186/1755-8794-6-31. PMC 3849977. PMID 24028078. This article incorporates text available under the CC BY 2.0 license.
^ a b Ignatov DY (2012). Functional heterogeneity of human neutrophils and their role in peripheral blood leukocyte quantity regulation (PhD). Donetsk National Medical University. doi:10.13140/RG.2.2.35542.34884.
^ Gerasimov IG, Ignatov DI (2001). "[Functional heterogenicity of human blood neutrophils: generation of oxygen active species]". Tsitologiia. 43 (5): 432–436. PMID 11517658.
^ Gerasimov IG, Ignatov DI (2004). "[Neutrophil activation in vitro]". Tsitologiia. 46 (2): 155–158. PMID 15174354.
^ Gerasimov IG, Ignatov DI, Kotel'nitskiĭ MA (2005). "[Nitroblue tetrazolium reduction by human blood neutrophils. I. The influence of pH]". Tsitologiia. 47 (6): 549–553. PMID 16708848.
^ Gerasimov IG, Ignatov DI (2005). "[Nitroblue tetrazolium reduction by human blood neutrophils. II. The influence of sodium and potassium ions]". Tsitologiia. 47 (6): 554–558. PMID 16708849.
^ Zilionis R, Engblom C, Pfirschke C, Savova V, Zemmour D, Saatcioglu HD, et al. (May 2019). "Single-Cell Transcriptomics of Human and Mouse Lung Cancers Reveals Conserved Myeloid Populations across Individuals and Species". Immunity. 50 (5): 1317–1334.e10. doi:10.1016/j.immuni.2019.03.009. PMC 6620049. PMID 30979687.
^ Graham DB, Zinselmeyer BH, Mascarenhas F, Delgado R, Miller MJ, Swat W (2009). Unutmaz D (ed.). "ITAM signaling by Vav family Rho guanine nucleotide exchange factors regulates interstitial transit rates of neutrophils in vivo". PLOS ONE. 4 (2): e4652. Bibcode:2009PLoSO...4.4652G. doi:10.1371/journal.pone.0004652. PMC 2645696. PMID 19247495.
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Mass action in two-sex population models: encounters, mating encounters and the associated numerical correction
Katherine Snyder, Brynja Kohler & Luis F. Gordillo
To cite this article: Katherine Snyder, Brynja Kohler & Luis F. Gordillo (2017) Mass action in two-sex population models: encounters, mating encounters and the associated numerical correction, Letters in Biomathematics, 4:1, 101-111, DOI: 10.1080/23737867.2017.1302827
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Citing articles: 1 View citing articles LETTERS IN BIOMATHEMATICS, 2017 VOL. 4, NO. 1, 101–111
RESEARCH ARTICLE OPEN ACCESS
Mass action in two-sex population models: encounters, mating encounters and the associated numerical correction
Katherine Snyder, Brynja Kohler and Luis F. Gordillo
Department of Mathematics and Statistics, Utah State University, Logan, UT, USA
ABSTRACT
Ideal gas models are a paradigm used in Biology for the phenomenological modelling of encounters between individuals of different types. These models have been used to approximate encounter rates given densities, velocities and distance within which an encounter certainly occurs. When using mass action in two-sex populations, however, it is necessary to recognize the difference between encounters and mating encounters . While the former refers in general to the (possibly simultaneous) collisions between particles, the latter represents pair formation that will produce offspring. The classical formulation of the law of mass action does not account this difference. In this short paper, we present an alternative derivation of the law of mass action that uses dimensional reduction together with simulated data. This straightforward approach allows to correct the expression for the rate of mating encounters between individuals in a two-sex population with relative ease. In addition, variability in mating encounter rates (due to environmental stochasticity) is numerically explored through random fluctuations on the new mass action proportionality constant. The simulations show how the conditioned time to extinction in a population subject to a reproductive Allee effect is affected.
ARTICLE HISTORY
Received 31 October 2016 Accepted 28 February 2017
KEYWORDS
Mass action; dimensional analysis; environmental stochasticity
Introduction
In chemistry, the law of mass action states that the rate of a reaction is proportional to the product of the concentrations of the reactants. Since Alfred Lotka pioneered its use in Lotka (1925) to justify the encounter term in his predator–prey system of differential equations, the law has become ubiquitous in mathematical ecology for modelling the interactions between individuals of different groups. Lotka’s arguments for the use of mass action in biological encounters were motivated by the analogy to the kinetic theory of gases (ideal gas models) and have been applied to describe a variety of phenomena, including fertilization kinetics, search theory and mate finding, see Hutchinson and Waser (2007), Voit, Martens, and Omholt (2015) for thorough reviews. Before Lotka, the chemical law of mass action was used for the first time by A.G. McKendrick for describing the interactions
CONTACT Luis F. Gordillo Luis.Gordillo@usu.edu © 2017 The Author(s). Published by Informa UK Limited, trading as Taylor & Francis Group. This is an Open Access article distributed under the terms of the Creative Commons Attribution License ( licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 102 K. SNYDER ET AL.
among susceptible and infected individuals in epidemiological contexts (Heesterbeek, 2005; McKendrick, 1912). In the theory of molecular collisions in gases, the collision frequency among particles of different types is expressed in terms of the velocities and radiuses of particles. The quantitative law is deduced directly from geometrical abstractions using the mean of a Poisson process that models the number of collisions a particle receives from others and where simultaneous collisions are allowed (Hutchinson & Waser, 2007; Kauzmann, 2012). At relatively low population densities, the same idea is used as a phenomenological approach to approximate the encounter rates between males and females, with the birth rate taken proportional to the product of their densities (Bazykin, 1998; Fauvergue, 2013). We remark that two-sex population models become relevant when sexual dimorphism in vital rates is present, which has been observed in several species (Caswell, 2001). While ideal gas models, which assume a heterogeneous population of linearly moving individuals, are used for modelling encounters of individuals, dispersal and movement of homogeneous populations in ecology are modelled following diffusion models, i.e. where individuals move randomly in space (Codling, Plank, & Benhamou, 2008; Turchin, 1998). However, due to the complexity involved, most of the studies that include encounters in models of randomly moving organisms are based on computer simulations (see for instance Bartumeous, Catalan, Viswanathan, Raposo, & da Luz, 2008; Gurarie & Ovaskainen, 2011; James, Plank, and Brown, 2008, 2010, where encounters are defined in general animal search problems). This paper addresses the modelling of encounters that lead to reproduction, which requires ruling out simultaneous encounters that occur in the ideal gas model, i.e. encounters where a single female mates two or more males simultaneously, otherwise the rate at which new offspring appear would be overestimated. Thus, we differentiate between counting encounters, possibly simultaneous and counting mating encounters , which are understood here as the formation of female–male pairs from which offspring are successfully produced. Our aim here is to present a correction for the constant used in the mass action term corresponding to the ideal gas model that accounts for the pair formation. To achieve this goal, we first build a functional relation among the variables using dimensional reduction and simulated data of individuals’ movement. This allows us to approximate the value of the proportionality constant for the mass action with relative precision in comparison to theoretical results. Then, we generate new data through computer simulations that only count female–male pairs, and with this we approximate the new value for the constant. Finally, we use the new constant in the mass action law to explore the effects of environmental stochasticity on the conditioned time to extinction for a population model via the variability on encounter rates. The stochastic model is derived from a deterministic model that uses mass action at low population densities and thus shows a reproductive Allee effect (Courchamp, Berec, & Gascoigne, 2008). The simulations reveal the effects that simultaneous random fluctuations around the new constant (and therefore the mating encounter rate) and demographic stochasticity have on the extinction time. This elementary example justifies having reasonable approximations for the non-linear term that models encounters: it demonstrates how variability in the environment could play an essential role in regulating the time to extinction distribution. LETTERS IN BIOMATHEMATICS 103
Dimensional reduction for mass action
We consider a two-sex population and assume that individuals’ movement is done in two spatial dimensions, i.e. we assume that the total variation in the displacement of each individual on the plane is much larger than changes made in their altitude. We also assume that individuals of both sexes are initially homogeneously mixed and uniformly distributed in space, and move over a terrain with area A under the following assumptions:
• The velocity, v, is constant and the same for both sexes,
• Individuals move independently from each other,
• Movement is in straight lines,
• The initial direction of movement for each individual is independently chosen at random from the interval [0, 2 π),
• Individuals’ sizes are negligible. Let nm and nf be the number of males and females, respectively, and c the average number of encounters that one female has with males during the observation time t. We assume that c is related to (i) the velocity v, (ii) the density of males nm/A and (iii) the size of a small area surrounding the female where males are attracted to mate. This is thought as a circular area with radius R. We remark that these assumptions are the same used in the theory of molecular collisions to deduce the law of mass action for gases of two different types, with just the words ‘particle’ replacing ‘individual’ and ‘type’ instead of ‘sex’ (Hutchinson & Waser, 2007; Kauzmann, 2012). We write the relation between the system parameters in terms of some (unknown) function F,
c = F
( nmA , v, t, R
)
. (1) By the -Theorem, see Barenblatt (2006) or Logan (2006) for instance, Equation (1) is equivalent to a relation that involves only the dimensionless quantities
1 = vtRn mA and 2 = vt R ,that is,
c = f
( vtRn mA , vt R
)
, (2) with f yet to be determined. Next, we use data generated from agent-based simulations that count the number of contacts (with males) per female. For the simulations, the individuals were programmed to follow the rules stated above and, for a single female, we counted a contact when its distance to a male is less than R. In the computations, the units chosen for length and time were meters and hours (1 h = 1 time step). We used fixed values for the time of observation,
t = t∗ = 24 (h), the radius R = R∗ = .05 (m) and the number of males, nm∗ = 100, while varying area size A and individuals’ velocity v. On the plane v − A, we arbitrarily choose the strip [25, 4 × 10 4] × [ 25, 2 × 10 3], and within this domain, we fixed 8 values of areas: (25, 35, 50, 400, 625, 900, 1200, 2000), and 28 values of velocities: (25, 50, 75, 100, 200, 300, 400, 500, 600, 700, 800, 900, 1000, 1200, 1400, 1600, 1800, 3000, 3200, 5400, 6400, 9000, 11000, 16000, 18000, 22400, 25200, 40000). These values were chosen with the aim 104 K. SNYDER ET AL.
of capturing the characteristics of the surface at low and high parameter values. Using periodic boundary conditions, we then produced 10 simulations for each corresponding pair of parameters and finally computed the contact averages. The simulations were run in NetLogo 1 Railsback & Grimm (2012) and the code is available upon request to K. Snyder. 2It is well known that the average number of contacts observed during a fixed period of time should increase with larger velocities and decrease with larger areas, therefore suggesting a relation of the form
c = av pA q , (3) where p, q and a are constants. This expression can then be rewritten in terms of the dimensionless quantities 1 and 2 ,
c =
K
︷ ︸︸ ︷
anqm∗t p
∗
R2q−p
∗
p−q
2
q
1
= K
( vt R
)p−q ( vtRn mA
)q
.
Using least squares to fit relation (3) to the averages of the data points collected gave approximate values of p ≈ 1, q ≈ 1 and K = 1.2887. Denoting with nf the number of females and with C the total averaged number of encounters that females have with males, i.e. C = cn f , gives
C = K vtR A nm nf = KvtRAxy , (4) where x and y are the densities of males and females, respectively. The right-hand term in (4) is traditionally obtained from the theory of molecular collisions, also known as the
ideal gas model . That theory provides a constant value of K = 4/π = 1.2732 ... , which is in good agreement with the value obtained through our simulations (the relative error is less than 2%).
2.1. Mating encounters
We generate different data by repeating the simulations of individual movement with the same assumptions as above but counting at most one mating encounter per female at each time step ruling out simultaneous mating. Fitting the model to the new data produces
p ≈ 1, q ≈ 1 and K ≈ .1231, which is less than 10% of the previous K value. Our interest now is to include changes in movement direction at every step in time, depending on the previous direction rather than restricting individual movement to straight lines as in the assumptions. A limitation in the theoretical ideal gas model, as traditionally conceived, is that it does not capture the effects of varying correlation in individuals’ movement (i.e. the autocorrelation of the directions of subsequent individual moves) on contact rates. From the dimensional analysis, however, we conclude that the degree of correlation should appear as a functional dependence between the value K and the range of possible directions for individuals’ movement, i.e. K = K(θ) , where at each time step each individual changes to a direction chosen uniformly at random from [− θ, θ],0 ≤ θ ≤ π, independently of the other individuals. Repeating the numerical experiments but now allowing individuals to change movement direction with different degrees of correlation, we compute the average of K for different values of θ. Unsurprisingly, these values appear almost constant, as first pointed out by Skellam (1958) for the classical LETTERS IN BIOMATHEMATICS 105
Figure 1. Top: Averaged values of K obtained from observations where particles change their direction at each step in an angle chosen randomly from [− θ, θ]. The averages (stars joined with dash/dot curve) do not differ significantly from the mean ¯K = .1231. For each angle θ, the circles mark the values of K+
and K−, which are the constants obtained by fitting the relation (3) to the original average data points ±
standard deviation, respectively. Bottom: Mating encounter rates (encounters/h) for moving individuals as function of the population density (individuals/m 2 ) and velocity (m/h). For illustration purposes, the detection radius R was fixed and chosen equal to .02 m and the velocities are in the range from 1 to 40 Km/h, which includes estimates for several insect species, see Table 1. The area A is one square meter.
mass action, see Figure 1 (top). This information can thus be used to approximate average mating encounter rates for individuals, given estimates of velocities, see Figure 1 (bottom). We emphasize, however, that the assumptions made on the movement constitute a rough simplification of reality: males and females do not necessarily move at the same speed and the details of mating mechanisms have been deliberately left out. To assess the variability due to movement correlation on the parameter K, we re-fit Equation (3) for different values of θ to two different data-sets defined by taking the simulated data at each corresponding point of the v − A plane and then computing average+standard deviation and average-standard deviation. With these sets of points, 106 K. SNYDER ET AL.
Table 1. Average velocities for sustained flight in some insects, Nachtigall (1974). Insect Flight velocity (Km/h) Mayflies, small field grasshoppers 1.8Bumblebees, rose chafers 3.0
Anopheles (malaria mosquitoes) 3.2Stag beetle, damselfly, Ammophila (a fossorial wasp) 5.4Housefly 6.4Cockchafer, cabbage white butterfly, garden wasp 9.0Blowfly 11 .0Desert locust 16 .0Hummingbird hawk moth 18 .0Honeybee, horsefly 22 .4
Aeshna (a dragonfly), hornet 25 .2
Anax (a dragonfly) 30 .0Deer botfly 40 .0
we obtained two new constants (that depend on θ) and denoted by K+ and K−. The data suggested that for values of θ close to zero, the differences are consistently larger, giving larger values for K+ and K− (See Figure 1). But with increasing θ, although the movement is more irregular, large variability in the values of K+ and K− is absent. Our interpretation of this outcome is that, although individual movement is apparently more convoluted for larger θ, at each trial the (stochastic) process is the same. On the contrary, for values of θ
close to zero, the initial random directions for each individual define paths that look like the deterministic trajectories ( θ = 0) in the domain (in this case, a torus). Those might differ completely each time the experiment is repeated because of the randomization of initial directions. This fact was found to be independent whether the boundary conditions are periodic or reflective after running additional simulations for the latter.
Effects of environmental stochasticity
The rate at which offspring are generated by a two-sex population in real scenarios is likely to be subject to random fluctuations due to environmental factors, like rainfall and temperature, but with sensitivity that is species specific. If mass action is chosen to model mating encounters, variability on the parameter K will appear as consequence of those fluctuations. This translates directly into variability of birth and death rates. As an illustration, let us initially consider the simplest deterministic population model ephemeral mating interactions,
x′ = − μx + 12 P(x, y), (5)
y′ = − μy + 12 P(x, y), (6) where x and y are the densities of males and females, respectively, P(x, y) is a (symmetric) birth rate with even sex ratio for all births and μ is the death rate that is assumed equal for both sexes. Suppose also that the initial sex ratio is 1:1. It is natural that we would like to use the law of mass action for P(x, y), i.e. proportional to xy , but as D.G. Kendall first pointed out, this leads to solutions that blow up in finite time (Kendall, 1949). One way, this trouble can be fixed by taking into consideration the average refractory time τ of LETTERS IN BIOMATHEMATICS 107
Figure 2. Cumulative probability of extinction (conditioned to extinction) obtained from the stochastic model defined in Section 3. The effect of variability on the number of contacts, K∼Normal (¯K,σ2), is shown for different values of σ2=0 (continuous line), and .1 (dash-dots). The initial population is taken equal to half the Allee threshold. The simulations suggest that an increasing variance in the random fluctuations corresponding to mating encounters pushes the probability mass to the left, i.e. extinction is likely to happen sooner than otherwise expected. The parameters for the simulations were chosen arbitrarily 1 /μ =10 (days), v=1.33 (Km/h), R=.02 (m) and b=3, although they resemble the characteristics in some relatively small insects. The value p=.01, which in practice depends on the complexity of the mating mechanisms, was also set arbitrarly.
females (Bazykin, 1998), during which a female avoids further sexual encounters just after successfully mating with a male. Let r denote the rate of mating encounters per female,
r = KvRx = αx, then 1 /r is the average time between encounters for one female. If the population densities are low enough so that 1 /r is very much more than τ , then the population growth will depend on the number of successful mating encounters made. In this case, the average progeny produced by a female has to saturate when 1 /r ↓ τ as the density of males increases. Therefore, the average birth rate per female is more conveniently approximated with bp × αx × 1
N+x
, where b is the average number of offspring per female per encounter that survive to adulthood, p is the probability that an encounter produces offspring and N is the average male population density at which half of the females are able to reproduce. This can be seen by making x = N, so the average birth rate per female is
bp α/ 2 or equivalently that half of the females produce offspring at the per capita rate bp α.For simplicity, we have set N = 1, suggested by the reasonable approximation of the mean number of pairs formed at low densities (Gordillo, 2015). Given that initially we assumed a 1:1 sex ratio, the whole population dynamics can be described by the equation
z′ = − μz + P(z/2, z/2) = − μz + bp α z22(2 + z) = γ z (z − 2μ/γ )
2 + z , (7) where z = x + y and γ = (bp α − 2μ)/ 2 > 0. Thus, if z 2, the population growth will approximately correspond to a mass action regime while if z 2, it grows exponentially. More precisely, z′ < 0 when z < 2μ/γ , giving rise to positive density dependence with a critical population size. 108 K. SNYDER ET AL.
We construct a stochastic version of the model by seeing the birth and death rates as stochastic rates (Gillespie, 1977); that is, we integrate demographic random fluctuations by considering only the two types of events, birth and death of individuals, happening at exponentially distributed times with the overall rate given by
= bp αz22(2 + z) + μz.
At a given time, a death happens with probability q = μz/ , or a birth with probability 1 −
q. In addition, environmental stochasticity is introduced by letting K fluctuate randomly, with K ∼ Normal ( ¯K, σ 2) ( ¯K = .1231). The effects of varying σ 2 are displayed in Figure 2, which shows the cumulative probabilities of extinction up to a fixed time, conditioned on extinction. The initial population density is taken at half of the Allee threshold value. The results suggest that an increasing variance in the random fluctuations associated with the mating encounters would increase the probability of earlier extinction events.
Discussion
Mating processes in two-sex insect populations are species specific and generally involve vast complexity (Bonduriansky, 2001; Choe & Crespi, 1997). This necessitates an ap-proximated description of the non-linear process for mating encounters. Mass action has been continuously used for this end, even in cases with convoluted mating mechanisms (Courchamp et al., 2008). The multiplicative constant for the mating encounter rate might be difficult to estimate in many real applications. The novelty in this note is that, under general assumptions, we have approximated the value of that constant with the use of dimensional reduction and simulated data. Before now, this constant has not been computed by such analytical arguments. The dimensional reduction approach also offers a way to examine the effects of correla-tion in the movement. The simulations suggest that in general, the variances of the values of K are small and similar when compared among results obtained with different degrees of correlation. Thus, the value ¯K can be used as the multiplicative constant in the expression for the encounter rate, ¯KvRAxy , independently of movement correlation. Relative larger variance values observed for highly correlated movement ( θ close to zero) are attributed to the boundary conditions in the simulations. We remark that the programme used to produce the data only counts ephemeral encounters and does not consider further association between paired individuals, as well as other complexities involved in mating mechanisms. While our analysis assumes at most one mating encounter per female at each time step, the approach may be generalized to systems where individuals must spend more or less time between mating encounters. For the purposes of modelling sex role evolution, H. Kokko and M. Jennions presented in Kokko and Jennions (2008) a scheme that accounts for time delays following encounters when individuals cannot mate. This framework could be adopted to modify and extend our work. Small random fluctuations in the number of mating encounters can appear nevertheless, caused by non-permanent random environmental changes. These effects might alter substantially the encounter rate (now random). As an illustration, we simulated a basic LETTERS IN BIOMATHEMATICS 109
stochastic birth and death process to obtain the probability distribution of the time to extinction, conditioned on extinction. The model has a deterministic analogue for which extinction due to Allee effects is possible. The simulations suggest that small fluctuations on the parameter ¯K might induce relatively large increments on the (conditioned) cumulative probability of extinction, see Figure 2. For conservation or pest control efforts in which it is critical to assess how rapidly a population might become extinct, the approximation for
¯K presented here might help shape more accurate quantitative predictions based on the law of mass action. We would like to emphasize that, after all, mass action is a highly idealized model where physical and biological details are overlooked, including those corresponding to mating mechanisms. For instance, velocities associated with aerial mating in insects may be sensitive to the relative mass of flight muscle (Dudley, 2000), while (the radius of) attraction may depend on chemical, visual or auditory signals (Resh & Cardé, 2009; Roelofs, 1995). Trying to introduce these species-specific details into the mass action scheme could be a challenging task that we have not considered. In the case that velocity and radius of attraction were both scaled to individual total mass, for example, it would be possible to insert these relations into the main formulae as long as the assumptions for the mass action still hold. If the case is that individual mass implies non-negligible (relative) size, then the simulations cannot be used as we set them, and new computational experiments together with a re-assessment of the whole model have to be done. In contrast to the crude assumptions made for mass action, we refer the reader to the interesting papers (Gurarie & Ovaskainen, 2011, 2013), where the authors present a refined continuous time-continuous space theoretical framework that shows how encounter rates depend on the interplay of spatial distributions, scales of movement, individuals densities and inherent dynamics. However, we believe the simple model presented here can inform the theoretical modelling of ephemeral (short-lived) mating encounters for insects, important in pest control management (Boukal & Berec, 2008; Fauvergue, 2013; Gordillo, 2015).
Notes
2. Katherine.Snyder@usu.edu.
Acknowledgements
The authors are deeply grateful to P. Greenwood, P.M. Waser and J.M.C. Hutchinson and two anonymous reviewers for the extensive and useful comments, and suggestions on previous versions of the manuscript.
Disclosure statement
No potential conflict of interest was reported by the authors.
References
Barenblatt, G. (2006). Scaling . Cambridge: Cambridge Text in Applied Mathematics, Cambridge University Press. 110 K. SNYDER ET AL.
Bartumeous, F., Catalan, J., Viswanathan, G., Raposo, E., & da Luz, M. (2008). The influence of turning angles on the success of non-oriented animal searches. Journal of Theoretical Biology, 252 ,43–55. Bazykin, A. (1998). Nonlinear dynamics of interacting populations . Singapore: World Scientific Publishing. Bonduriansky, R. (2001). The evolution of male mate choice in insects: A synthesis of ideas and evidence. Biological Reviews, 76 , 305–339. Boukal, D., & Berec, L. (2008). Modelling mate-finding Allee effects and population dynamics, with applications in pest control. Population Ecology, 51 , 445–458. Caswell, H. (2001). Matrix population models . Sutherland, MA: Sinauer Associates. Choe, J., & Crespi, B. (1997). Mating systems in insects and arachnids . Cambridge: Cambridge University Press. Codling, E., Plank, M. J., & Benhamou, S. (2008). Random walk models in biology (review). The Journal of the Royal Society Interface, 5 , 813–834. Courchamp, F., Berec, L., & Gascoigne, J. (2008). Allee effects in ecology and conservation . New York, NY: Oxford University Press. Dudley, R. (2000). The biomechanics of insect flight . Princeton, NJ: Princeton University Press. Fauvergue, X. (2013). A review of mate-finding allee effects in insects: From individual behavior to population management. Entomologia Experimentalis et Applicata, 146 , 79–92. Gillespie, D. (1977). Exact simulation of coupled chemical reactions. The Journal of Physical Chemistry, 81 , 2340–2361. Gordillo, L. (2015). Modeling ephemeral mating encounters in insects: The emergence of mate-finding Allee effects and applications to theoretical models of sterile release. Theoretical Population Biology, 104 , 10–16. Gurarie, E., & Ovaskainen, O. (2011). Characteristic spatial and temporal scales unify models of animal movement. The American Naturalist, 178 , 113–123. Gurarie, E., & Ovaskainen, O. (2013). Towards a general formalization of encounter rates in ecology.
Theoretical Ecology, 6 , 189–202. Heesterbeek, H. (2005). The law of mass-action in epidemiology: A historical perspective. In B. B. K. Cuddington (Ed.), Ecological paradigms lost, routes of theory change (pp. 81–105). Burlington, MA: Elsevier Academic Press. Hutchinson, J., & Waser, P. (2007). Use, misuse and extensions of “ideal gas" models of animal encounter. Biological Reviews, 82 , 335–359. James, A., Pitchford, J., & Plank, M. (2010). Efficient or innacurate? Analytical and numerical modelling of random search strategies. Bulletin of Mathematical Biology, 72 , 896–913. James, A., Plank, M., & Brown, R. (2008). Optimizing the encounter rate in biological interactions: Ballistic vs. lévy vs. brownian strategies. Physical Review E, 78 , 51128. Kauzmann, W. (2012). Kinetic theory of gases . Mineola, NY: Dover Publications. Kendall, D. (1949). Stochastic processes and population growth. Journal of the Royal Statistical Society, Series B, 11 , 230–282. Kokko, H., & Jennions, M. (2008). Parental investment, sexual selection and sex ratios. Journal of Evolutionary Biology, 21 , 919–948. Logan, J. (2006). Applied mathematics (3rd ed.). Hoboken, NJ: Wiley. Lotka, A. (1925). Elements of physical biology . Baltimore: Williams & Wilkins Company. McKendrick, A. (1912). The rise and fall of epidemics. Paludism, 1 , 54–66. Nachtigall, W. (1974). Insects in flight. A glimpse behind the scenes in biophysical research . London: George Allen & Unwin. Railsback, S., & Grimm, V. (2012). Agent-based and individual-based modeling . Princeton, NJ: Princeton University Press. Resh, V., & Cardé, R. (2009). Encyclopedia of insects (2nd ed.). Burlington, MA: Elsevier. Roelofs, W. (1995). Chemistry of sex attraction. Proceedings of the National Academy Science USA, 92 , 44–49. Skellam, J. (1958). The mathematical foundations underlying the use of line transects in animal ecology. Biometrics, 14 , 385–400. LETTERS IN BIOMATHEMATICS 111
Turchin, P. (1998). Quantitative analysis of movement: Measuring and modeling population redistribution in animals and plants . Sunderland: Sinauer. Voit, E., Martens, H., & Omholt, S. (2015). 150 years of the mass action law. PLoS Computational Biology, 11 , e1004012. |
11680 | https://www.youtube.com/watch?v=bqiP4g_SbCo | GeeksForGeeks | Reverse Bits
Mathematics
422 subscribers
Description
69 views
Posted: 13 Apr 2024
Given a number x, reverse its binary form and return the answer in decimal.
Example 1:
Input:
x = 1
Output:
2147483648
Explanation:
Binary of 1 in 32 bits representation-
00000000000000000000000000000001
Reversing the binary form we get,
10000000000000000000000000000000,
whose decimal value is 2147483648.
Example 2:
Input:
x = 5
Output:
2684354560
Explanation:
Binary of 5 in 32 bits representation-
00000000000000000000000000000101
Reversing the binary form we get,
10100000000000000000000000000000,
whose decimal value is 2684354560.
Table of Contents
0:00 Problem Statement
0:32 Solution
3:06 Solution - Example
5:06 Code
2 comments
Transcript:
Problem Statement hello guys in this video we are going to discuss the solution of the problem reverse bits so let us look at the problem statement the problem statement says that given a number X reverse its binary form and return the answer in decimal so let us try to understand the problem statement with an help of an example so here the X which is given to us is one so this is the binary representation of one so what we have to do is we have to reverse this string and then return the decimal equival of this string this reverse string so that's Solution about the problem statement folks now let us go ahead and discuss the solution of this problem so let us take an example in order to discuss the solution of this problem so let us take an example where x = 5 the binary representation of x = to 5 will be 1 0 and 1 and since we have to represent all the numbers into 32bit numbers so let us do that so the 101 can be represented into 32 bit in this way and then this is followed by 101 and this is in the binary format so how many zeros are there well there are 29 zeros over here right and then followed by one1 and what we need to do is we need to reverse this string so 101 and Then followed by 29 zeros and towards the end we need to convert into decimal format and that will be our answer so let us look at how we can do that so the way we will be able to do this is let us say we have an answer list and initially all the values of the answer list is zero and uh the size of this list is 32 now we will have something known as bit number so which bit are we at and initially we'll start with zeroth bit and then what we are going to do is we are going to iterate on X till X becomes equals to 0 so we'll iterate on X and then we are going to check whether the least significant bit of X is set to one or not so if the least significant bit of X is set to one then what we need to do is we just need to go ahead and set that particular bit in our answer list to one and once we have done that we simply have to just manipulate X and bit number so in this way we will be able to reverse the binary representation of X and the Reversed binary representation of X is there in our answer list and all we have to do now is we just have to convert this answer into decimal format so let us do that so what we have to do is we need to convert our answer into decimal format so let us convert it so this and in decimal format so this is how we will convert it and this is something which we will return it let us Solution - Example run by an example in order to understand this concept so let us take the same example where x equal 5 and we know the binary representation is 1 0 and 1 and initially my answer contains 32 bit zeros this is 32 times now what we are checking is whether the zeroth bit remember bit num is zero so whether my zeroth bit is set to one or not in this case the zeroth bit is set to one so what we are going to do is we are going to set this bit to one and the remaining bits are zeros then after doing that we will manipulate X and bit num so now X will be manipulated by right shift so now X becomes one and zero and the bit num is now one so now we want to populate the value at bit num one here since the least significant bit is zero then we will not change anything in the in our answer list and we will just go ahead and manipulate our X so now X is equal to 1 and our bit number will be 2 this the least significant bit is one since the least significant bit is one so this has to be changed to one so now our answer list contains 1 0 1 and Then followed by zeros and then after doing that we will go ahead and change our X so now X becomes zero and since X becomes zero so we'll exit out of this Loop so this is our answer so see this this is the reverse binary representation of X and all we are doing is we first we are concatenating answer in in a string so our answer so this portion we essentially do this one Z one and then zeros so this and then we are converting it into decimal format and this is what we are returning so that's about the Code explanation folks now let us see how we have coded this in Python so the code is pretty much the same what we have discussed where answer we are initially setting it to all zero of 32 bit then the bit number is equals to zero and we are iterating on X till X is not zero if the least significant bit of X is one then we are set set setting the corresponding bit number to one and towards the end we are manipulating X and bit number and once we have the Reversed binary string in the answer list we are converting it to decimal and returning that so that's about the code folks now let us go ahead and submit our solution so our solution got accepted that is all from this video folks in case if you have any question on comment regarding this video please feel free to use the comment section and I'll try to address them thank you |
11681 | https://web.mit.edu/8.334/www/grades/projects/projects14/PeterFedak/transformation.html | Loading [MathJax]/extensions/MathEvents.js
Lattice Transformations
Finding the exact critical probabilities is possible for some lattices, but is often very challenging. Luckily there are other computational tools at our disposal. We may be able to find relationships between the critical probabilities of different lattices even without knowing the values.
A simple transformation is to remove edges from a lattice. Removing edges can only increase the sizes of the clusters, so if the original lattice percolates, the new lattice will as well. In particular, if p is the original critical probability, and q is the critical probability of the modified lattice, we have q\le p. Consider the example below:
We have removed several edges from a triangular lattice, displaying the new clusters. Try setting the probability very low, so that almost all edges are blocked. What does the lattice with the deleted edges look like?
It has exactly the same structure as a (slanted) square lattice!
This is a definitive proof of something you might have observed before: the square lattice critical probability is smaller than the triangular lattice critical probability! While we could (and did) run computer simulation to find this result, in other cases it might not be so easy. Theoretical results, even imprecise ones like this one, are powerful because they reveal absolute limitations, and tell us about physical systems we might not even have imagined, let alone run experiments on.
Another edge-deletion transformation turns the square lattice into a hexagonal lattice.
These transformations suggest some relationships between critical probabilities. To make a square lattice, we delete a third of the edges from the triangular lattice, so that if a fraction p of the edges were open before, \frac{1}{3}+\frac{2}{3}p are open afterwards. Similarly, we delete a quarter of the edges from the square lattice, so if q were open before, \frac{1}{4}+\frac{3}{4} q are open afterwards. If the resulting lattices percolate, it suggests that the critical probability for the triangular lattice is roughly \frac{1}{3}+\frac{2}{3}p_S, and similarly that the critical probability of the square lattice is roughly \frac{1}{4}+\frac{3}{4} p_H. We will revisit this shortly, after we explore an even more creative transformation!
One interesting class of transformation comes from duality. Duality refers to a special pairing between different physical systems with related parameters, which can take on many different forms. For lattice systems, duality transformations may come from considering the vertices instead of the cells.1 Instead of grouping the cells into clusters connected by open edges, we group vertices into clusters connected by the blocked edges:
What happens to the right-most lattice system as you change the open edge probability? Does it seem to have a critical point? How do the sizes of large clusters in both graphs relate?
You should notice that when a vertex cluster becomes large, it encloses many small clusters of cells. Similarly, when there is a large cluster of cells, there are many small islands of blocked edges inside of it. Furthermore, cell clusters and vertex clusters cannot pass through each other. All of these observations, taken together, suggest that exactly one type of cluster percolates at any given probability.
In order to use this duality transformation to learn about percolation, we need to relate the vertex problem to the cell problem. The vertices are arranged in a lattice with some edges present, and are clustered together, with vertices in the same cluster if they are connected together by a sequence of edges. This is exactly the same as our description of the percolation problem, but with "cell" replaced by "vertex" and the flow through "open edges" replaced by flow along edges that were "blocked" in the cell problem. Thus if the original lattice has probability p, its dual is a percolation system on a new lattice with probability 1-p. Now we need to figure out the structure of the dual lattices.
Look back at the square lattice's dual transformation above. How many edges come out of each vertex? How do they connect together?
We can see that the square lattice is self-dual, meaning that it is its own dual! Similarly, if you play around with the simulation below, you can see that the dual of the triangular lattice is a hexagonal lattice.
Our reasoning about the cell- and vertex-clusters can lead us to more relationships between critical probabilities. If a lattice has critical probability p, the vertex clusters on that lattice percolate for probabilities below p, and so, accounting for closed edges becoming open edges after the transformation, the dual lattice will have critical probability 1-p. For the self-dual square lattice, these probabilities must be equal, which is only possible if the critical probability is 1/2. The duality between the triangular and hexagonal lattices similarly tells us that their critical probabilities sum to 1. All together we have:
p_S = \frac{1}{2},\qquad p_T+p_H=1,\qquad p_T\approx \frac{1}{3}+\frac{2}{3}p_S,\qquad p_S \approx \frac{1}{4}+\frac{3}{4}p_H,
where \approx means approximately equal, from our bond deletion estimates. These are more than we need; any three of these equations will allow us to determine the values. Solving, we find
p_T \approx \frac{2}{3},\qquad p_S = \frac{1}{2},\qquad p_H\approx \frac{1}{3}
which agrees with our experimental results. However, we did not need any experiments obtain this result. All we needed to do was reason about the structure of the lattice percolation problem, leading us to some relationships which on their own didn't seem terribly useful. Yet, putting them together, we found excellent estimates of the real behavior!
One important class of transformations are renormalization groups. In statistical mechanics, one way we can make physical systems more manageable is to use averages instead of tracking every single detail. In a magnet, for example, we might consider the average direction that magnetic moments are pointing over a small region. If the moments are all aligned, our average will point in that direction. If the moments are oriented randomly, they will average out to 0. Thus an average still captures information about collective behavior of the system. To perform an average, however, we need to decide what scale it is over - do we include 10 atoms, or 100, or 1000? We expect that different choices will still describe the system more or less correctly, but they will lead to differences in the equations that describe the system: a group of 1000 magnetic moments pointing in the same direction has a larger effect on its neighbors than just 10 would, so the effective strength of the interaction will have to be scaled appropriately. This leads to another sort of non-obvious symmetry, which allows us to pick out phase transitions as fixed points which look essentially the same at all averaging scales.
We can adapt this idea to percolation.2 We will group cells of our lattice into small groups, and then determine rules for the edges of those groups being open or blocked. For the square lattice, we take two-by-two blocks of cells, and the right/bottom edges are open if the top/left cells can flow through that side. Click on edges in lattices below to toggle them between open and blocked.
This procedure results in a new lattice with fewer, larger cells. The new lattice won't perfectly mirror the original, as there may be clusters in the new lattice that didn't exist before. Our transformation does keep some connectedness information, however, and so we expect that the new lattice will percolate or not depending on whether the original lattice did. If the original lattice had open bonds with probability p, what is the open edge probability in the new lattice?
Let's focus on the bottom edge (the other will be the same). The two edges on the right side of the original lattice square wont change anything, nor will the edge dividing the top two cells, as we only care about whether either top cell can connect through the bottom. If all five remaining edges are open, the top cells can clearly connect. This happens with probability p^5. You can check that blocking any one edge wont stop this. Exactly one edge is blocked with probability 5p^4(1-p), the 5 because there are five choices of edge to add, and the 1-p to account for the one blocked edge. Through similar reasoning, we find there are 8 ways to block exactly 2 edges while still allowing the cells to connect, 2 ways to block exactly 3, and that any choice of 4 or 5 edges results in the bottom edge being blocked.
Putting it together, we find that the new probability is
\begin{align}P(p)&=p^5+5p^4(1-p)+8p^3(1-p)^2+2p^2(1-p)^3 \&= p^2(2p^3-5p^2+2p+2) \end{align}
For the triangular lattice, we can form new 4-cell triangles, choosing edges to be open when the center cell can connect through them. Counting possibilities, we find the probability in the new lattice is
Q(p) = p^4+4p^3(1-p)+2p^2(1-p)^2 = p^2(2-p^2)
For large p, the transformed lattices become even more open than the original, while for small p they become more closed. If our transformation preserved the structure of the lattice well enough, the values of p where the new probability is the same as the old should be the critical probabilities. We can solve the equations to find
p_S = \frac{1}{2},\qquad p_T = \frac{\sqrt{5}-1}{2}\approx 0.618,
values which agree quite well with our previous estimates.
While this problem is ultimately much simpler than many real physical problems, theoretical models like this allow us to use powerful mathematical tools to find results that may be hard to confirm through experiment. Our good fortune with the lattice percolation problem won't necessarily carry over into other fields, but the steps we took here hint at some general principles. To learn more, read about symmetry and universality.
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11682 | https://www.youtube.com/watch?v=6EihQ1uJ2fw | Unit 12 Problem 3 - Angular Momentum - Placing a Mass on a Spinning Turntable
Tad Thurston
551 subscribers
3 likes
Description
325 views
Posted: 2 Dec 2021
A standard example of conserving angular momentum is to place a mass on a rotating object and compute the new rate of rotation.
For anyone outside the OCCC community: You can support this physics education effort and request additional courses to cover on Patreon:
Or, if something is extra cool, I'll never turn down a coffee or pizza!
Planned Units in this series will cover a typical Engineering Physics curriculum:
01 -- Units and Vectors
02 -- Kinematics
03 -- Projectiles
04 -- Newton's 2nd Law
05 -- Accelerated Motion
06 -- Work and Kinetic Energy
07 -- Potential Energy
08 -- Conservation of Momentum
09 -- Elastic Collisions
10 -- Moment of Inertia
11-- Rotational Dynamics
12 -- Angular Momentum
13 -- Torque and Equilibrium
14 -- Gravity
15 -- Springs and Oscillations
16 -- Waves
17 -- Ideal Gas Law
18 -- Thermal Energy
19 -- First Law of Thermodynamics
20 -- Second Law of Thermodynamics
21 -- Electric Fields
22 -- Electric Forces
23 -- Continuous Charge Distributions
24 -- Gauss' Law
25 -- Potential
26 -- Capacitance
27 -- Current and Resistance
28 -- DC Circuits
29 -- Magnetic Fields
30 -- Current Loops
31 -- Magnetic Forces
32 -- Ampere's Law
33 -- Faraday's Law
34 -- Inductance
35 -- AC Circuits
36 -- Electromagnetic Waves
37 -- Intensity and Radiation Pressure
38 -- Interference
39 -- Diffraction
40 -- Reflection/Refraction
Shot with an iPhone 12 using
OBS ( on an iMac,
an iPad with
Goodnotes (
and a
Blue Yeti microphone (
Edited using Blender ( and its Video Sequence Editor.
physics #education #tutorials
Transcript:
Intro yay all right let's try another quick example of angular momentum conservation um and this time what we're going to do is look at a turntable Problem so i've got some large turntable maybe two meters in radius a mass of 12 kilograms that's spinning at a certain rate uh let's say that it's spinning at 12 revolutions per minute um and then a cat jumps on they'll do that so let's say that the cat is four kilograms and the cat happens to jump on carefully one and a half meters from the center so the question is what is the new rotation rate after the cat jumps on um let's see what happens the principle is as always Solution that l i is l f so always you can write this down i omega initial is i omega final all right um and what we're after is that omega final right so omega final is going to be the initial divided by the final times the initial rotation rate okay um so the trick is we have to write down what the moments of inertia are all right so what is the initial moment of inertia well it's just a rotating disk it's just a turntable so uh thinking back uh to what we've done before the moment of inertia of a spinning disc is just one half m r squared right um okay after the cat jumps on then what do we have well then we have it's still a spinning turntable still the same big m and big r and now we have a point mass that cat is just sitting out there some distance away so i just add the two together right it's like adding masses together i can add rotational masses together so i've got the original thing one-half mr squared plus now i've got a point mass sitting out there at one and a half meters away so i've got little m times little r squared there we go and now i multiply times my initial rotation rate um that's all there is to it so uh now i can plug all my numbers in and when you do that what you get is something like what did i get point uh oops it's about 0.72 times the initial rotation rate uh 12 rpm converting from 12 rpm to omega we said that was what you roughly just moved the decimal point right but it's just a common factor on both sides of the equation so it's really 0.72 times 12 rpm is what you'll get and when you finish converting you'll get something like uh what is that about 8.7 rpm is going to be the final answer the main point is the cat jumped on right we got more mass to spin around and so what you'd expect is that thing's going to slow down and sure enough we went from 12 rpm to 8.7 rpm after the cat jumped on so yeah that's pretty much exactly what we expect |
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Solving a Recurrence Relation with a Square Root term
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I've been trying to learn how to solve some recurrence relations lately and I have no idea how I would go about solving something like this, if possible.
T(n)=a⋅T(n−1)+b⋅T(n−1)−−−−−−−√
My main problem is that I have no idea how to work with the square root term. The context of this problem is actually in economics. When looking at the level of capital at some time t it's equal to some depreciation rate constant, a times the capital at time t−1 added to some savings rate constant, b, times the output level at time t−1. In a basic case we consider output to be the square root of capital and that's why there is a square root term.
I spent some time trying random things and didn't have much luck so any help, if possible, would be greatly appreciated!
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asked Apr 29, 2015 at 20:15
Thomas GrandersonThomas Granderson
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Both a and b are positive and between 0 and 1. I didn't know if that would be helpful so I didn't include it.
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Henceforth we assume a,b are each in the interval (0,1), as mentioned in the comments (excluding the endpoints as those lead to weird behavior).
Set S(n)=T(n)−−−−√; this satisfies
S(n)=aS(n−1)2+bS(n−1)−−−−−−−−−−−−−−−−−−−√
We can find its fixed points by solving S=aS2+bS−−−−−−−−√; they are S=0 and S=b1−a. We wish to prove that b1−a is attractive. We set f(x)=ax2+bx−−−−−−−√, and (with the aid of a little alpha) calculate
f′(b1−a)=a+12
Since |a+12|<1, the iterated function will always converge.
Hence, the sequence T(n) will always tend to b2(1−a)2 as n→∞.
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edited Apr 29, 2015 at 22:20
answered Apr 29, 2015 at 20:57
vadim123vadim123
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It seems to me like it would be easier to model the problem as a differential equation rather than a recurrence relation. That is, we'd have an equation of the form
dTdt=aT+bT−−√.
That differential equation is separable; solving it is only as hard as integrating ∫1aT+bT√dT. You can integrate that term by rationalizing the denominator and using partial fractions.
In the end, you'll get an equation that describes the solution implicitly, i.e. in the same sense that x2+y2=1 describes the unit circle. Also, if your actual problem really is discrete in time, this will obviously only approximate the correct answer. However, even with both of these caveats, the solution will give you quite a bit of asymptotic information. And, if you need to compute precise values, you can just have a computer calculate terms of the recurrence directly.
In general, keep in mind that, while differential equations may seem more complicated than recurrences, calculus is actually very powerful and makes differential equations generally easier to work with.
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edited Apr 29, 2015 at 21:01
answered Apr 29, 2015 at 20:55
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2025 USAMO Problems/Problem 4
The following problem is from both the 2025 USAMO #4 and 2025 USAJMO #5, so both problems redirect to this page.
Contents
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1 Problem
2 Solution 1
3 Solution 2
4 Solution 3
5 Solution 4
6 Solution 5
7 Video Solution - A 2-minute proof
8 Solution 6
9 Solution 7
Problem
Let be the orthocenter of acute triangle , let be the foot of the altitude from to , and let be the reflection of across . Suppose that the circumcircle of triangle intersects line at two distinct points and . Prove that is the midpoint of .
Solution 1
Let AP intersects BC at D. Extend FC to the point E on the circumcircle of . Since is the orthocenter of , we know that or , and . Next we use the power of H in : . These relations imply that .
Hence are midpoints of respectively. By midline theorem, . Since , we have . This implies that . Consequently, is the diameter of . Let be the midpoint of which is also the center of . are midpoints of respectively. By the midline theorem again, we have , consequently, . This implies that is the perpendicular bisector of the chord hence is the midpoint of . ~ Dr. Shi davincimath.com
Solution 2
Denote as the center of , as the center of , as the midpoint of , as the midpoint of , and as the midpoint of . It suffices to show that .
Claim: is cyclic.
Proof: Since and , KM is a midline of and . as well since , so lies on . Next, note that lies on , so the perpendicular bisector of through passes through . In other words, , and are collinear. Since and are both perpendicular to , it follows that they are parallel. Since and , then . Finally, we have that and thus is cyclic. It follows that , so , as desired.
-mop
Solution 3
Connect and have intersect at . Also extend past point and have it intersect with the circle at point .
Since is the reflection of over , we know that . Since is the orthocenter, we can draw the altitude and tell that , , and are collinear. We know and , so by AA, so .
and . From this, we can tell that . Therefore, and .
If we connect , we can tell that that due to being the reflection of and being perpendicular to , so . In addition, . Also, because they are vertical angles.
So, because of SAA. From this we can conclude that , so is the midpoint of .
Solution 4
Let be the foot of the altitude from to By Power of a Point, we have and Adding, we get It is well known that the reflection of over which we denote by lies on Then, let We have Thus, and since we have Hence, is the midpoint of ~TThB0501
Solution 5
Let Q be the antipode of B. Claim — AHQC is a parallelogram, and AP CQ is an isosceles trapezoid.
Proof. As AH ⊥ BC ⊥ CQ and CF ⊥ AB ⊥ AQ. Let M be the midpoint of QC.
Claim — Point M is the circumcenter of triangle AFP.
Proof. It’s clear that MA = MP from the isosceles trapezoid.
As for MA = MF, let N denote the midpoint of AF; then MN is a midline of the parallelogram, so MN ⊥ AF. Since CM ⊥ BC and M is the center of (AF P), it follows CX = CY .
Video Solution - A 2-minute proof
Solution 6
Quick angle chasing gives . Let be the circumcenter of .
Thus (because O and A lie on the same side of segment ).
As , the quadrilateral FOCP is cyclic.
Observe that , so .
From the properties of cyclic quadrilaterals,
Thus . Therefore is perpendicular to chord , .
Solution 7
Let the line perpendicular to and going through be line . Let the midpoint of be , and let the line perpendicular to and going through be line . Let be the intersection of and , and let be the intersections of and .
Because and , is a parallelogram. From this, we get . Looking at triangle , because , and is the midpoint of , it is clear that is the midpoint of .
Let the intersection of and be . because is a reflection of across .
Substituting the equation from earlier, we get
Draw a line perpendicular to starting from . Let the intersection of that line and be . Because is perpendicular to which is perpendicular to , . By the same logic, . This means is a parallelogram (specifically a rectangle). Therefore . and by substituting, we get , and because , that means , so is a midpoint of . This means is the perpendicular bisector of , and because is a chord of the circumcircle of , goes through the center of the circle. By the same logic, also goes through the center of the circle, since it is the perpendicular bisector of . This means the center of the circle is , because it is the only point on both and . Because is on line and line is perpendicular to line , line must perpendicularly bisect the chord of circle containing that lies on line (basically must perpendicularly bisect ). This means is the midpoint of .
-Nolan Lin
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11685 | https://pmc.ncbi.nlm.nih.gov/articles/PMC2687630/ | Calcium Uptake and Release through Sarcoplasmic Reticulum in the Inferior Oblique Muscles of Patients with Inferior Oblique Overaction - PMC
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Yonsei Med J
. 2006 Apr 30;47(2):207–213. doi: 10.3349/ymj.2006.47.2.207
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Calcium Uptake and Release through Sarcoplasmic Reticulum in the Inferior Oblique Muscles of Patients with Inferior Oblique Overaction
Hee Seon Kim
Hee Seon Kim
1 Department of Ophthalmology, Institute of Vision Research, Yonsei University College of Medicine, Seoul, Korea.
Find articles by Hee Seon Kim
1, Yoon-Hee Chang
Yoon-Hee Chang
2 Department of Ophthalmology, Ajou University School of Medicine, Suwon, Korea.
Find articles by Yoon-Hee Chang
2, Do Han Kim
Do Han Kim
3 Department of Life Science, Kwangju Institute of Science & Technology, Kwangju, Korea.
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3, So Ra Park
So Ra Park
4 Department of Physiology, Inha University College of Medicine, Incheon, Korea.
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4, Sueng-Han Han
Sueng-Han Han
1 Department of Ophthalmology, Institute of Vision Research, Yonsei University College of Medicine, Seoul, Korea.
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1, Jong Bok Lee
Jong Bok Lee
1 Department of Ophthalmology, Institute of Vision Research, Yonsei University College of Medicine, Seoul, Korea.
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1,✉
Author information
Article notes
Copyright and License information
1 Department of Ophthalmology, Institute of Vision Research, Yonsei University College of Medicine, Seoul, Korea.
2 Department of Ophthalmology, Ajou University School of Medicine, Suwon, Korea.
3 Department of Life Science, Kwangju Institute of Science & Technology, Kwangju, Korea.
4 Department of Physiology, Inha University College of Medicine, Incheon, Korea.
✉
Reprint address: requests to Dr. Jong Bok Lee, Department of Ophthalmology, Institute of Vision Research, Eye & ENT Hospital, Yonsei University of College of Medicine, 134 Shinchon-dong, Seodamun-gu, Seoul 120-752, Korea. Tel: 82-2-2228-3574, Fax: 82-2-312-0541, 491209@yumc.yonsei.ac.kr
✉
Corresponding author.
Received 2005 Jul 7; Accepted 2005 Nov 18; Issue date 2006 Apr 30.
Copyright © 2006 The Yonsei University College of Medicine
This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License ( which permits unrestricted noncommercial use, distribution, and reproduction in any medium, provided the original work is properly cited.
PMC Copyright notice
PMCID: PMC2687630 PMID: 16642550
Abstract
We characterized and compared the characteristics of Ca 2+ movements through the sarcoplasmic reticulum of inferior oblique muscles in the various conditions including primary inferior oblique overaction (IOOA), secondary IOOA, and controls, so as to further understand the pathogenesis of primary IOOA. Of 15 specimens obtained through inferior oblique myectomy, six were from primary IOOA, 6 from secondary IOOA, and the remaining 3 were controls from enucleated eyes. Ryanodine binding assays were performed, and Ca 2+ uptake rates, calsequestrins and SERCA levels were determined. Ryanodine bindings and sarcoplasmic reticulum Ca 2+ uptake rates were significantly decreased in primary IOOA (p<0.05). Western blot analysis conducted to quantify calsequestrins and SERCA, found no significant difference between primary IOOA, secondary IOOA, and the controls. Increased intracellular Ca 2+ concentration due to reduced sarcoplasmic reticulum Ca 2+ uptake may play a role in primary IOOA.
Keywords: Calcium, calsequestrin, inferior oblique overaction, ryanodine receptor, sarcoplasmic reticulum, sarcoplasmic reticulum Ca 2+-ATPase 1
INTRODUCTION
Inferior oblique overaction (IOOA), defined as an elevation of an eye as it moves toward adduction, is a common oculomotor disease.1 From the clinical point of view, it has been customary to distinguish between primary and secondary overactions of this muscle. Secondary overactions differ in as much that the vertical deviation is large in the primary position and is accompanied by a torsional deviation and the Bielschowsky head tilt test is positive. Although the clinical differentiation of primary and secondary overaction is not difficult, the explanations given for apparent primary overaction in the older literature are vague. Secondary overaction is caused by paresis or paralysis of the ipsilateral superior oblique muscle or by paresis or paralysis of the contralateral superior rectus muscle. However, primary overaction, in which there is no evidence for past or present ipsilateral superior oblique muscle paralysis or paresis, is difficult to explain. A number of studies have been undertaken to elucidate the pathogenesis of primary IOOA.1-5 However, none of the resulting explanations are convincing, and it remains doubtful as to whether a true primary overaction exists at all from an innervational perspective.
Ca 2+ ions trigger muscle contraction and the release of Ca 2+ from the sarcoplasmic reticulum (the intracellular Ca 2+ storage site) plays an important role in determining the intramuscular Ca 2+ concentration. Ca 2+ binds with the protein participating in muscle contraction to induce muscle contraction and is then reabsorbed into the sarcoplasmic reticulum via Ca 2+ ATPase. In addition, muscle relaxation is possible when Ca 2+ is reabsorbed by the sarcoplasmic reticulum and during this process Ca 2+-ATPase (the Ca 2+ pump) plays a key role. This process of Ca 2+ uptake by the sarcoplasmic reticulum via Ca 2+-ATPase reduces intramuscular Ca 2+ concentrations and results in muscle relaxation.6,7
Several regulatory proteins present in the sarcoplasmic reticulum plays fundamental roles in regulating the intramuscular concentration of Ca 2+. These include ryanodine receptor, a Ca 2+ release channel which participates in Ca 2+ release from sarcoplasmic reticulum into muscle cells, Ca 2+-ATPase, which drives Ca 2+ uptake, and calsequestrin that binds with Ca 2+.8-11 Thus, since Ca 2+ uptake by the sarcoplasmic reticulum and Ca 2+ release from the sarcoplasmic reticulum play key roles in the regulation of Ca 2+ concentrations in muscle cells, the above proteins are important elements in studies on the mechanisms of muscle contraction and relaxation. Furthermore, changes in these proteins may be of key importance in many muscular diseases. Reported studies on diseases associated with the dysfunction of the ryanodine receptor mainly concern with heart disease, although studies concerning its involvement with skeletal muscle in malignant hyperthermia, myasthenia gravis, and muscular dystrophy have been conducted. However, little is known about its relation with extraocular muscles.12-21
In the present study, Ca 2+ release, uptake, and storage, were examined in inferior oblique muscles with primary or secondary overaction, and compared with those of the normal inferior oblique muscles, particularly with respect to Ca 2+ movements through the sarcoplasmic reticulum so as to further understand the pathogenesis of primary IOOA.
MATERIALS AND METHODS
The inferior oblique muscles of patients diagnosed as having primary or secondary IOOA were obtained through a weakening procedure (myectomy) under general anesthesia. IOOA was graded according to the amount of eye over-elevation in adduction versus the position of the fellow eye: 1+ represented an over-elevation of 1mm, 2+ 2 mm, 3+ 3 mm, and 4+ 4 mm. Patients with an IOOA 3+ or 4+ underwent inferior oblique weakening surgery for functional or cosmetic reasons. Those who had a history of other ocular disease or who had undergone previous ocular surgery were excluded. Three patients (6 eyes) had primary IOOA and 9 (10 eyes) secondary IOOA. Additional inferior oblique muscles (normal controls) were obtained from 4 patients (4 eyes) who underwent enucleation due to phthisis bulbi. Patients' ages ranged from 3 to 71 years and there were 10 males and 6 females. All surgical procedures were performed by one of the authors (JBL), and each muscle specimen was obtained at the same location from the muscle insertion with minimal manipulation. Although procedures were performed in all 16 patients, muscle homogenation was possible in 12 patients. Muscles were frozen in liquid nitrogen immediately after the procedure, and muscle tissues were homogenized in a cold room at -4℃.
Preparation of whole homogenates
After removing blood and connective tissues from the inferior oblique muscles in 0.9% NaCl solution, the muscle specimens were dried with filter paper and weighed. About 50 mg of inferior oblique muscle tissue per ml of homogenate buffer was placed in solution (20 mM morpholinopropanesulphonic acid, 1 M KCl, leupeptin 1 µM, pepstatin 1 µM, aprotinine 1 µM, PMSF 0.1 mM, trypsin inhibitor 10 µM/ml, sucrose 0.3 M), and the mixture was homogenated for 3 × 30 sec using a polytron PT probe (Brinkmann Instruments Co, Westburg, NY, USA). Total protein was assayed by the Bradford method.22
Characteristics of ryanodine receptor in primary and secondary IOOA
Whole homogenate (2.5 mg) was placed in buffer solution [1 M KCl, 20 mM morpholinopropanesulphonic acid (pH 7.4), 20 nM [3 H]ryanodine (54.7 Ci/mmol, New England Nuclear Co, Billerica, MA, USA), 1 mM EGTA] at 37℃ for 2 hours to determine ryanodine binding value (total binding value). After blocking ryanodine receptor in the sarcoplasmic reticulum, the same amounts of whole homogenate and non-labeled ryanodine were placed into the buffer solution to obtain the 'non-specific binding value'. The following reactions were performed upon the two ryanodine containing homogenates mentioned above. 100 µl of PEG solution [30% PEG, 1 mM EDTA Tris (pH 7.4)] was added to homogenate and left at room temperature for 5 min. The mixture was then centrifuged for 15 min at 14,000 rpm to precipitate sarcoplasmic reticulum, and the supernatant was washed with 0.4 ml of ryanodine binding buffer solution. Soluen 350 (100 µl) was then added to the precipitate to dissolve the lipid layer from the sarcoplasmic reticulum. The mixture was left at 70℃ for 15 min and moved to a scintillation vial to measure radioactivity after adding 4 ml of picofluor.
Uptake of oxalate-supported Ca 2+ by sarcoplasmic reticulum
In order to load Ca 2+ into the sarcoplasmic reticulum, the reaction mixture [KCl 100 mM, morpholinopropanesulphonic acid (pH 6.8) 20 mM, NaN3 10 mM,] was added to 300 µg of whole homogenate and reacted for 4 min at 37℃. 5 mM of MgATP, 10 mM potassium oxalate, and 0.2 mM 45 CaCl 2 0.2 mM (104 cpm/nmol; Amersham, Arlington Height, IL, USA) were then added and the whole was filtered through a Minipore filter (0.45 µm pore size) for 2, 4, or 7 min. The filtrants were placed into scintillation vials to measure radioactivity after adding 4 ml of picofluor and left for 15 min.
Western blot analysis of calsequestrin and sarcoendoplasmic reticulum Ca 2+-ATPase 1 (SERCA1)
Calsequestrin is a 65 KDa protein which stores Ca 2+ in the sarcoplasmic reticulum. SERCA 1 is a 120 KDa enzyme which is related with fast muscle. After quantifying the amount of protein by Bradford method, 50 µg of protein was mixed with sample buffer and boiled at 95℃ for 5 min. Proteins were separated by electrophoresis, and blotted onto nitrocellulose paper using buffer solution. The paper was then placed in a blocking solution containing phosphate buffered saline (PBS) containing 5% bovine serum albumin and 0.1% Tween 20 for an hour at room temperature. Primary antibody (mouse monoclonal anti-human antibody; Department of Bioscience, Kwangju Science and Research Institute, Kwangju, Korea) for calsequestrin and primary antibody (mouse monoclonal anti-human antibody; Transduction Laboratories, Lexington, KY, USA) for SERCA1 were reacted for 3 hours at room temperature. The proteins were washed with PBS containing 0.1% Tween 20 3 times, and reacted for 90 minutes at room temperature with secondary antibody (alkaline phosphatase-conjugated goat anti-mouse IgG; Santa Cruz, CA, USA). Protein bands were detected by enhanced chemiluminescence (Amersham Life Science, Little Chalfont, England, UK).
Statistical analysis
The above procedure was repeated eight times. All measured values are expressed as means ± standard deviations. Comparisons between the three groups were performed using the Kruscal-Wallis test and comparison between two groups using the Wilcoxon rank sum test; p values of <0.05 were considered statistically significant.
RESULTS
Ryanodine binding measurement
Ryanodine receptor binding was measured in the whole homogenate of each inferior oblique muscle tissue. The amount of ryanodine receptor binding per unit of protein was the greatest for the control group, followed by the secondary IOOA group and then the primary IOOA group. Specific binding between ryanodine receptor and ryanodine in the primary IOOA group was significantly lower than that in the control group, suggesting that the activity of the ryanodine receptor in the primary IOOA group was lower than that in the control group (Table 1). However, no significant difference in receptor binding properties was observed between the other two group combinations (p>0.05).
Table 1.
Concentration of the Ryanodine Receptor in the Sarcoplasmic Reticulum of Inferior Oblique Muscles
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Values are means ± standard deviations.
IOOA, inferior oblique overaction.
Statistically significant difference when compared with the control (p<0.05).
Uptake of Ca 2+ by the sarcoplasmic reticulum
The rate of Ca 2+-uptake by the sarcoplasmic reticulum was the fastest for the control group, followed by the secondary IOOA group, and then the primary IOOA group. The difference between all three groups was statistically significant (p<0.05). More specifically, Ca 2+ reabsorbed into sarcoplasmic reticulum was fastest for the control group and slowest for the primary IOOA group (Table 2), and all intergroup differences were significant.
Table 2.
The Rate of 45 Ca 2+ Uptake by the Sarcoplasmic Reticulum of Inferior Oblique Muscles
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Values are means ± standard deviations.
IOOA, inferior oblique overaction.
Ca 2+ uptake rates (nmole/mg protein/min) were significantly different in the three groups (p<0.05).
Western blot analysis for calsequestrin and SERCA1
Differences between the Ca 2+ storage capacities of the sarcoplasmic reticulum in the oblique muscles of each of the three groups were estimated by determining the amount of calsequetrin. No significant inter-group differences were observed in terms of the amount of calsequestrin (Fig. 1). In order to determine whether a lower Ca 2+ uptake was due to the down-regulation of SERCA1 (Ca 2+ pump), western blot analysis was performed. However, no significant difference was observed between the three groups with respect to SERCA1 expression (Fig. 2).
Fig. 1.
Open in a new tab
The expression of calsequestrin in inferior oblique muscles. C, P, and S represent inferior oblique muscles from the controls, and primary and secondary inferior oblique overaction cases, respectively. These three groups were similar with respect to calsequestrin expression.
Fig. 2.
Open in a new tab
The expressions of sarco-endoplasmic reticulum Ca 2+-ATPase 1 in inferior oblique muscles. C, P, and S represent inferior oblique muscles from the controls, and primary and secondary inferior oblique overaction cases, respectively. No significant difference was found between these three groups in terms of SERCA1 expression.
DISCUSSION
A number of hypotheses have been proposed concerning the pathogenesis of primary IOOA. Bielschowsky, for instance, suggested that eyes are elevated in adduction due to an abnormal fascial structure of the inferior oblique muscle.1 Guibor suggested that IOOA could be caused by a synkinesis of inferior oblique muscle with the ipsilateral medial rectus muscle owing to an impulse spread within the central nervous system.4 Spencer and McNeer reported that an electron microscopic structure of the inferior oblique muscle showed no difference between primary and secondary IOOA, and concluded primary IOOA might be due to incomplete palsy of the ipsilateral superior oblique muscle.5 The discovery of muscle pulleys has directed our attention to other etiologic possibilities for this apparent overaction.3 The distribution of the isoforms of lactate dehydrogenase in children with IOOA was different from that in the control group.2
In the present study, we evaluated the functional affinity of the ryanodine receptor by attaching the [3 H] to ryanodine. Moreover, because muscle contractility in primary IOOA is presumed to be larger than that of normal controls, intramuscular Ca 2+ concentrations in primary IOOA are expected to be higher than normal, which suggests Ca 2+ release by the sarcoplasmic reticulum is enhenced. However, unexpectedly, ryanodine receptor binding affinity, which resulted in Ca 2+-release, was significantly low in the primary IOOA compared with that in the controls. Thus, we postulated that, in primary IOOA, the Ca 2+ storage within the sarcoplasmic reticulum is reduced due to Ca 2+-ATPase inactivation, and that consequently Ca 2+-release capacity is reduced. As a result, the amount or the function of ryanodine receptor was supposed to be reduced. However, other investigators have reported that the inactivation of Ca 2+-ATPase is closely related with ischemic heart disease, heart failure, and hypertrophy of cardiac muscle; thus, as the intracellular Ca 2+ concentration increases, the concentration of ryanodine receptor decreases, and in some cases, down regulation has been found to be associated with the decreased expression of its mRNA.23,24
By measuring Ca 2+ uptake using oxalate, we were able to determine the amount of Ca 2+ sequestration by the sarcoplasmic reticulum and to estimate the function of the ryanodine receptor indirectly. In the present study, the amount and rate of Ca 2+ uptake was determined when the ryanodine receptor was completely blocked. Moreover, uptake rate differences between the three groups were significant. Ca 2+ was reabsorbed most promptly in control group, followed by the secondary IOOA group, and the primary IOOA group. This result is consistent with our assumption that if Ca 2+ is not sequestered rapidly into the sarcoplasmic reticulum, but remains in the cytoplasm, that the intracellular Ca 2+ concentration would be highest in primary IOOA muscle cells. In secondary IOOA, the rate of Ca 2+ uptake was slower than that of the control group, which suggests that this is a transitional phenomenon, which results from the relative overaction of the inferior oblique muscle due to paralysis of its antagonist, the superior oblique muscle.
Ca 2+ pump activity is known to be reduced in muscular dystrophy and in Brody disease.22,25 However, little information is available on the role of Ca 2+ pump in ocular disease, especially in oculomotor disease, although it was reported that Ca 2+ pump activity changes in lens epithelial cells after treated with an oxidant or heat.26 In the present study, Ca 2+ pump levels, which is one of the factors that impacts the Ca 2+ uptake rate, were assayed by western blotting for SERCA 1, a fast twitch isoform of SERCA. However, no significant difference was found between the three groups. Thus we infer that a reduction in Ca 2+ uptake in IOOA is not due to reduced Ca 2+ pump levels, but rather is probably due to a functional abnormality in SERCA 1, or in phospholamban, a SERCA regulator, or alternatively to a reduced level of ATP phosphorylation. Further studies are needed to resolve this issue.
Calsequestrin is a Ca 2+ storage protein that binds Ca 2+ within the lumen of the sarcoplasmic reticulum. The protein is usually fixed in the junctional sarcoplasmic reticulum and located near the ryanodine receptor. Calsequestrin is negatively charged and interacts with the positively charged triadin and ryanodine receptors. Furthermore, calsequestrin has been reported to be related to Ca 2+-ATPase present in the sarcoplasmic reticulum.27,28 In the present study, no significant difference was found between the three groups with respect to calsequestrin expression, which suggests that Ca 2+ concentrations in the sarcoplasmic reticulum might not be different between groups. However, this does not exclude the possibility that Ca 2+ storage capacity in the sarcoplasmic reticulum by calsequestrin is impaired.
In the present study, we attempted to unravel the mechanism of IOOA from a molecular perspective rather than from the anatomical or histological standpoints. Thus, we evaluated ryanodine receptor binding, Ca 2+ uptake, the relation between SERCA and uptake, and calsequestrin (Ca 2+ storage), in normal, primary IOOA, and secondary IOOA. Initially we hypothesized that the pathogenesis of IOOA (especially in primary IOOA) is associated with changes in intracellular Ca 2+ concentration and Ca 2+ movement. The observed delay in Ca 2+ uptake by the sarcoplasmic reticulum in primary IOOA seems remarkable. However, this study has some limitations. First, age-matched controls are required to exclude the possibility of age-related changes in Ca 2+ signaling or, may be, in tissue stability. Second, the number of sample is small. Last, there are many other proteins involved in the Ca 2+ signaling and contraction in muscles that would deserve a closer analysis. Further study is needed to determine whether delayed Ca 2+ uptake per se has a causal relationship on increased intracellular Ca 2+ concentration. Moreover, a significant reduction in ryanodine receptor binding was also observed in primary IOOA. Thus, the simultaneous measurements of intracellular Ca 2+ concentration and muscle contraction, and functional analysis of regulating protein abnormalities require further investigation.
References
1.Von Noorden GK, Campos EC. Binocular vision and ocular motility. 6th ed. revised. St. Louis London Philadelphia Sydney Toronto: Mosby; 2002. [Google Scholar]
2.Aihara T, Miyata M, Ishikawa S. The lactate dehydrogenase isoenzyme pattern in the overacting inferior oblique muscle. J Pediatr Ophthalmol Strabismus. 1978;15:43–47. doi: 10.3928/0191-3913-19780101-14. [DOI] [PubMed] [Google Scholar]
3.Clark RA, Miller JM, Rosenbaum AL, Demer JL. Heterotropic muscle pulleys or oblique muscle dysfunction? J AAPOS. 1998;2:17–25. doi: 10.1016/s1091-8531(98)90105-7. [DOI] [PubMed] [Google Scholar]
4.Guibor GP. Synkinetic overaction of the inferior oblique muscle. Am J Ophthalmol. 1949;32:221–229. doi: 10.1016/0002-9394(49)90137-3. [DOI] [PubMed] [Google Scholar]
5.Spencer RF, McNeer KW. Structural alterations in overacting inferior oblique muscles. Arch Ophthalmol. 1980;98:128–133. doi: 10.1001/archopht.1980.01020030130015. [DOI] [PubMed] [Google Scholar]
6.Block BA, Imagawa T, Campbell KP, Franzini-Armstrong C. Structural evidence for direct interaction between the molecular components of the transverse tubule/sarcoplasmic reticulum junction in skeletal muscle. J Cell Biol. 1988;107(6 Pt 2):2587–2600. doi: 10.1083/jcb.107.6.2587. [DOI] [PMC free article] [PubMed] [Google Scholar]
7.Dulhunty AF, Gage PW. Effect of extracellular calcium concentration and dihydropyridines on contraction in mammalian skeletal muscle. J Physiol. 1988;399:63–80. doi: 10.1113/jphysiol.1988.sp017068. [DOI] [PMC free article] [PubMed] [Google Scholar]
8.Meissner G. Ryanodine activation and inhibition of the Ca2+ release channel of sarcoplasmic reticulum. J Biol Chem. 1986;261:6300–6306. [PubMed] [Google Scholar]
9.Pessah IN, Warterhouse AL, Casida JE. The calciumryanodine receptor complex of skeletal and cardiac muscle. Biochem Biophys Res Commun. 1985;128:449–456. doi: 10.1016/0006-291x(85)91699-7. [DOI] [PubMed] [Google Scholar]
10.Mintz E, Guillian F. Ca2+ transport by the sarcoplasmic reticulum ATPase. Biochim Biophys Acta. 1997;1318:52–70. doi: 10.1016/s0005-2728(96)00132-6. [DOI] [PubMed] [Google Scholar]
11.Damiani E, Margreth A. Specific protein-protein interaction of calsequestrin with junctional sarcoplasmic reticulum of skeletal muscle. Biochem Biophys Res Commun. 1990;172:1253–1259. doi: 10.1016/0006-291x(90)91584-f. [DOI] [PubMed] [Google Scholar]
12.Cheung WY. Calmodulin. Sci Am. 1982;246:62–70. doi: 10.1038/scientificamerican0682-62. [DOI] [PubMed] [Google Scholar]
13.Coronado R, Morrissette J, Sukhareva M, Vaughn DM. Structure and function of ryanodine receptors. Am J Physiol. 1994;266(6 Pt 1):C1485–C1504. doi: 10.1152/ajpcell.1994.266.6.C1485. [DOI] [PubMed] [Google Scholar]
14.Jenden DJ, Fairhurst AS. The pharmacology of ryanodine. Pharmacol Rev. 1969;21:1–25. [PubMed] [Google Scholar]
15.Jones LR, Zhang L, Sanborn K, Jorgensen AO, Kelley J. Purification, primary structure, and immunological characterization of the 26-kDa calsequestrin binding protein (junctin) from cardiac junctional sarcoplasmic reticulum. J Biol Chem. 1995;270:30787–30796. doi: 10.1074/jbc.270.51.30787. [DOI] [PubMed] [Google Scholar]
16.Knudson CM, Stang KK, Moomaw CR, Slaughter CA, Campbell KP. Primary structure and topological analysis of a skeletal muscle-specific junctional sarcoplasmic reticulum glycoprotein (triadin) J Biol Chem. 1993;268:12646–12654. [PubMed] [Google Scholar]
17.Brillantes AM, Allen P, Takahashi T, Izumo S, Marks AR. Differences in cardiac calcium release channel (ryanodine receptor) expression in myocardium from patients with end-stage heart failure caused by ischemic versus dilated cardiomyopathy. Circ Res. 1992;71:18–26. doi: 10.1161/01.res.71.1.18. [DOI] [PubMed] [Google Scholar]
18.Fabiato A. Calcium-induced release of calcium from the cardiac sarcoplasmic reticulum. Am J Physiol. 1983;245:C1–C14. doi: 10.1152/ajpcell.1983.245.1.C1. [DOI] [PubMed] [Google Scholar]
19.Gwathmey JK, Copelas L, MacKinnon R, Schoen FJ, Feldman MD, Grossman W, et al. Abnormal intracellular calcium handling in myocardium from patients with end-stage heart failure. Circ Res. 1987;61:70–76. doi: 10.1161/01.res.61.1.70. [DOI] [PubMed] [Google Scholar]
20.Go LO, Moschella MC, Watras J, Handa KK, Fyfe BS, Marks AR. Differential regulation of two types of intracellular calcium release channels during end-stage heart failure. J Clin Invest. 1995;95:888–894. doi: 10.1172/JCI117739. [DOI] [PMC free article] [PubMed] [Google Scholar]
21.Schillinger W, Meyer M, Kuwajima G, Mikoshiba K, Just H, Hasenfuss G. Unaltered ryanodine receptor protein levels in ischemic cardiomyopathy. Mol Cell Biochem. 1996;160-161:297–302. doi: 10.1007/BF00240062. [DOI] [PubMed] [Google Scholar]
22.Daniel MB, Michael DR, Stuart JE. Protein methods. 2nd ed. New York: Wiley-Liss publisher; 1996. [Google Scholar]
23.Airey JA, Deerinck TJ, Ellisman MH, Houenou LJ, Ivanenko A, Kenyon JL, et al. Crooked neck dwarf (cn) mutant chicken skeletal muscle cells in low density primary cultures fail to express normal alpha ryanodine receptor and exhibit a partial mutant phenotype. Dev Dyn. 1993;197:189–202. doi: 10.1002/aja.1001970304. [DOI] [PubMed] [Google Scholar]
24.Riccardo Z, Simonetta RT. The sarcoplasmic reticulum Ca2+ channel/ryanodine receptor: Modulation by endogenous factors, drugs and disease states. Pharmacol Rev. 1997;49:1–43. [PubMed] [Google Scholar]
25.Martonosi AN, Beeler TJ. Skeletal muscle. In: Peachey LD, Adrian RH, Geiger SR, editors. Handbook of physiology. 3rd ed, revised. Bethesda: American physiological society Press; 1983. pp. 417–485. [Google Scholar]
26.Orlova EV, Serysheva II, van Heel M, Hamilton SL, Chiu W. Two structural configurations of the skeletal muscle calcium release channel. Nat Struct Biol. 1996;3:547–552. doi: 10.1038/nsb0696-547. [DOI] [PubMed] [Google Scholar]
27.Anderson K, Lai FA, Liu QY, Rousseau E, Erikson HP, Meissner G. Structural and functional characterization of the purified cardiac ryanodine receptor-Ca2+ release channel complex. J Biol Chem. 1989;264:1329–1335. [PubMed] [Google Scholar]
28.Arai M, Alpert NR, MacLennan DH, Barton P, Periasamy M. Alterations in sarcoplasmic reticulum gene expression in human heart failure. A possible mechanism for alterations in systolic and diastolic properties of the failing myocardium. Circ Res. 1993;72:463–469. doi: 10.1161/01.res.72.2.463. [DOI] [PubMed] [Google Scholar]
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11686 | https://brainly.com/question/13181771 | [FREE] Let x and y be whole numbers greater than 0, with y > x . Which has the greater value: 3x or 3y ? - brainly.com
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Let x and y be whole numbers greater than 0, with y>x. Which has the greater value: 3 x or 3 y?
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The comparison between 3x and 3y in the given problem shows that 3y is greater than 3x because y is greater than x.
Explanation
The subject here is Mathematics, and it deals with a problem of comparative values between two whole numbers, x and y, under a specific condition. The question states that y is greater than x and they are both whole numbers greater than 0. With these conditions, if you multiply x and y each by 3, 3y will always be greater than 3x. This is because, by the distributive property of multiplication over addition in mathematics, 3y=3(y) will be greater than 3x=3(x) since y > x.
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Given that y>x, it follows that 3 y>3 x when both are multiplied by 3. Thus, 3 y is greater than 3 x.
Explanation
To determine which is greater between 3 x and 3 y, we start with the information given in the problem. We know that:
x and y are whole numbers greater than 0.
y>x.
Since y is greater than x, we can express this relationship mathematically.
If we multiply both sides of the inequality y>x by 3, we maintain the inequality because multiplying by a positive number does not change the order of inequality.
This means we have:
3 y>3 x
Thus, it follows that 3 y is greater than 3 x.
Conclusion:
Therefore, the answer is that 3 y has the greater value than 3 x. This conclusion holds true for any whole numbers x and y where y g re a t er t han x.
Examples & Evidence
For example, if x=2 and y=5, then 3 x=6 and 3 y=15. Here, 15 is greater than 6, confirming our conclusion.
The properties of inequalities in mathematics state that if a>b, then for any positive constant k, ka>kb. This principle confirms that multiplying both sides of the inequality by 3 maintains the inequality.
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Answer: 3y has a greater value
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well if they are both whole numbers greater than zero
but ( y ) is greater than ( x ) than no matter what ( y ) is always going to be greater than (x)
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Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer
Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)?
Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD
Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four?
Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish?
Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit?
Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25
Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points.
Community Answer 4 Click an Item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers. Express In simplified exponential notation. 18a^3b^2/ 2ab
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11687 | https://openstax.org/books/introductory-statistics-2e/pages/7-introduction | Skip to ContentGo to accessibility pageKeyboard shortcuts menu
Introductory Statistics 2e
Introduction
Introductory Statistics 2eIntroduction
Search for key terms or text.
Figure 7.1 If you want to figure out the distribution of the change people carry in their pockets, using the central limit theorem and assuming your sample is large enough, you will find that the distribution is normal and bell-shaped. (credit: modification of work “american coins” by Paula R. Lively/ Flickr, CC BY 2.0)
Chapter Objectives
By the end of this chapter, the student should be able to:
Recognize central limit theorem problems.
Classify continuous word problems by their distributions.
Apply and interpret the central limit theorem for means.
Apply and interpret the central limit theorem for sums.
Why are we so concerned with means? Two reasons are: they give us a middle ground for comparison, and they are easy to calculate. In this chapter, you will study means and the central limit theorem.
The central limit theorem (clt for short) is one of the most powerful and useful ideas in all of statistics. There are two alternative forms of the theorem, and both alternatives are concerned with drawing finite samples size n from a population with a known mean, μ, and a known standard deviation, σ. The first alternative says that if we collect samples of size n with a "large enough n," calculate each sample's mean, and create a histogram of those means, then the resulting histogram will tend to have an approximate normal bell shape. The second alternative says that if we again collect samples of size n that are "large enough," calculate the sum of each sample and create a histogram, then the resulting histogram will again tend to have a normal bell-shape.
The size of the sample, n, that is required in order to be "large enough" depends on the original population from which the samples are drawn (the sample size should be at least 30 or the data should come from a normal distribution). If the original population is far from normal, then more observations are needed for the sample means or sums to be normal. Sampling is done with replacement.
It would be difficult to overstate the importance of the central limit theorem in statistical theory. Knowing that data, even if its distribution is not normal, behaves in a predictable way is a powerful tool.
Collaborative Exercise
Suppose eight of you roll one fair die ten times, seven of you roll two fair dice ten times, nine of you roll five fair dice ten times, and 11 of you roll ten fair dice ten times.
Each time a person rolls more than one die, they calculate the sample mean of the faces showing. For example, one person might roll five fair dice and get 2, 2, 3, 4, 6 on one roll.
The mean is = 3.4. The 3.4 is one mean when five fair dice are rolled. This same person would roll the five dice nine more times and calculate nine more means for a total of ten means.
Your instructor will pass out the dice to several people. Roll your dice ten times. For each roll, record the faces, and find the mean. Round to the nearest 0.5.
Your instructor (and possibly you) will produce one graph (it might be a histogram) for one die, one graph for two dice, one graph for five dice, and one graph for ten dice. Since the "mean" when you roll one die is just the face on the die, what distribution do these means appear to be representing?
Draw the graph for the means using two dice. Do the sample means show any kind of pattern?
Draw the graph for the means using five dice. Do you see any pattern emerging?
Finally, draw the graph for the means using ten dice. Do you see any pattern to the graph? What can you conclude as you increase the number of dice?
As the number of dice rolled increases from one to two to five to ten, the following is happening:
The mean of the sample means remains approximately the same.
The spread of the sample means (the standard deviation of the sample means) gets smaller.
The graph appears steeper and thinner.
You have just demonstrated the central limit theorem (clt).
The central limit theorem tells you that as you increase the number of dice, the sample means tend toward a normal distribution (the sampling distribution).
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Authors: Barbara Illowsky, Susan Dean
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Book title: Introductory Statistics 2e
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© Jun 25, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University. |
11688 | https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/09%3A_Sequences_and_Series/9.03%3A_The_Divergence_and_Integral_Tests | Skip to main content
9.3: The Divergence and Integral Tests
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9.2E: Exercises for Section 9.2
9.3E: Exercises for Section 9.3
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Gilbert Strang & Edwin “Jed” Herman
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( \newcommand{\kernel}{\mathrm{null}\,})
Learning Objectives
Use the
divergence test
to determine whether a series converges or diverges.
Use the
integral test
to determine the
convergence of a series
.
Estimate the value of a series by finding bounds on its remainder
term
.
In the previous section, we determined the convergence or
divergence
of several series by explicitly calculating the
limit
of the
sequence
of partial sums Sk. In practice, explicitly calculating this
limit
can be difficult or impossible. Luckily, several tests exist that allow us to determine convergence or
divergence
for many types of series. In this section, we discuss two of these tests: the
divergence test
and the
integral test
. We will examine several other tests in the rest of this chapter and then summarize how and when to use them.
Divergence Test
For a series ∞∑n=1an to converge, the nth
term
an must satisfy an→0 as n→∞. Therefore, from the algebraic
limit
properties of sequences,
limk→∞ak=limk→∞(Sk−Sk−1)=limk→∞Sk−limk→∞Sk−1=S−S=0.
Therefore, if ∞∑n=1an converges, the nth
term
an→0 as n→∞. An important consequence of this fact is the following statement:
If an↛0 as n→∞,∞∑n=1an diverges.
This test is known as the
divergence test
because it provides a way of proving that a series diverges.
Definition: The Divergence Test
If limn→∞an=c≠0 or limn→∞an does not exist, then the series ∞∑n=1an diverges.
It is important to note that the inverse statement of this theorem is not true. That is, if limn→∞an=0, we cannot make any conclusion about the convergence of ∞∑n=1an.
For example, limn→01n=0, but the
harmonic series
∞∑n=11n diverges. In this section and the remaining sections of this chapter, we show many more examples of such series. Consequently, although we can use the
divergence test
to show that a series diverges, we cannot use it to prove that a series converges. Specifically, if an→0, the
divergence test
is inconclusive.
Example 9.3.1: Using the divergence test
For each of the following series, apply the
divergence test
. If the
divergence test
proves that the series diverges, state so. Otherwise, indicate that the
divergence test
is inconclusive.
∞∑n=1n3n−1
∞∑n=11n3
∞∑n=1e1/n2
Solution
Since limn→∞n3n−1=13≠0, by the
divergence test
, we can conclude that ∞∑n=1n3n−1 diverges.
2. Since limn→∞1n3=0, the
divergence test
is inconclusive.
3. Since limn→∞e1/n2=1≠0, by the
divergence test
, the series ∞∑n=1e1/n2 diverges.
Exercise 9.3.1
What does the
divergence test
tell us about the series ∞∑n=1cos(1/n2)?
Hint
: Look at limn→∞cos(1/n2).
Answer
: The series diverges.
Integral Test
In the previous section, we proved that the
harmonic series
diverges by looking at the
sequence
of partial sums Sk and showing that S2k>1+k/2 for all positive integers k. In this section we use a different technique to prove the
divergence
of the
harmonic series
. This technique is important because it is used to prove the
divergence
or convergence of many other series. This test, called the integral test
, compares an infinite sum to an
improper integral
. It is important to note that this test can only be applied when we are considering a series whose terms are all positive.
To illustrate how the
integral test
works, use the
harmonic series
as an example. In Figure 9.3.1, we depict the
harmonic series
by sketching a
sequence
of rectangles with areas 1,1/2,1/3,1/4,… along with the
function
f(x)=1/x. From the graph, we see that
k∑n=11n=1+12+13+⋯+1k>∫k+111xdx.
Therefore, for each k, the kth
partial sum
Sk satisfies
Sk=k∑n=11n>∫k+111xdx=lnx|k+11=ln(k+1)−ln(1)=ln(k+1).
Since limk→∞ln(k+1)=∞, we see that the
sequence
of partial sums Sk is unbounded. Therefore, Sk diverges, and, consequently, the series ∞∑n=11n also diverges.
Now consider the series ∞∑n=11n2. We show how an integral can be used to prove that this series converges. In Figure 9.3.2, we sketch a
sequence
of rectangles with areas 1,1/22,1/32,… along with the
function
f(x)=1x2. From the graph we see that
k∑n=11n2=1+122+132+⋯+1k2<1+∫k11x2dx.
Therefore, for each k, the kth
partial sum
Sk satisfies
Sk=k∑n=11n2<1+∫k11x2dx=1−1x|k1=1−1k+1=2−1k<2.
We conclude that the
sequence
of partial sums Sk is bounded. We also see that Sk is an increasing
sequence
:
Sk=Sk−1+1k2
for k≥2.
Since Sk is increasing and bounded, by the Monotone Convergence Theorem, it converges. Therefore, the series ∞∑n=11n2 converges.
We can extend this idea to prove convergence or
divergence
for many different series. Suppose ∞∑n=1an is a series with positive terms an such that there exists a continuous, positive, decreasing
function
f where f(n)=an for all positive integers. Then, as in Figure 9.3.3a, for any integer k, the kth
partial sum
Sk satisfies
Sk=a1+a2+a3+⋯+ak<a1+∫k1f(x)dx<1+∫∞1f(x)dx.
Therefore, if ∫∞1f(x)dx converges, then the
sequence
of partial sums Sk is bounded. Since Sk is an increasing
sequence
, if it is also a
bounded sequence
, then by the Monotone Convergence Theorem, it converges. We conclude that if ∫∞1f(x)dx converges, then the series ∞∑n=1an also converges. On the other hand, from Figure 9.3.3b, for any integer k, the kth
partial sum
Sk satisfies
Sk=a1+a2+a3+⋯+ak>∫k+11f(x)dx.
If
limk→∞∫k+11f(x)dx=∞,
then Sk is an
unbounded sequence
and therefore diverges. As a result, the series ∞∑n=1an also diverges. Since f is a positive
function
, if ∫∞1f(x)dx diverges, then
limk→∞∫k+11f(x)dx=∞.
We conclude that if ∫∞1f(x)dx diverges, then ∞∑n=1an diverges.
Definition: The Integral Test
Suppose ∞∑n=1an is a series with positive terms an. Suppose there exists a
function
f and a positive integer N such that the following three conditions are satisfied:
f is continuous,
f is decreasing, and
f(n)=an for all integers n≥N.
Then
∞∑n=1an
and
∫∞Nf(x)dx
both converge or both diverge (Figure 9.3.3).
Although convergence of ∫∞Nf(x)dx implies convergence of the related series ∞∑n=1an, it does not imply that the value of the integral and the series are the same. They may be different, and often are. For example,
∞∑n=1(1e)n=1e+(1e)2+(1e)3+⋯
is a
geometric series
with initial
term
a=1/e and ratio r=1/e, which converges to
1/e1−(1/e)=1/e(e−1)/e=1e−1.
However, the related integral ∫∞1(1/e)xdx satisfies
∫∞1(1e)xdx=∫∞1e−xdx=limb→∞∫b1e−xdx=limb→∞−e−x|b1=limb→∞[−e−b+e−1]=1e.
Example 9.3.2: Using the Integral Test
For each of the following series, use the
integral test
to determine whether the series converges or diverges.
∞∑n=11n3
∞∑n=11√2n−1
Solution
a. Compare
∞∑n=11n3 and ∫∞11x3dx.
We have
∫∞11x3dx=limb→∞∫b11x3dx=limb→∞[−12x2|b1]=limb→∞[−12b2+12]=12.
Thus the integral ∫∞11x3dx converges, and therefore so does the series
∞∑n=11n3.
b. Compare
∞∑n=11√2n−1 and ∫∞11√2x−1dx.
Since
∫∞11√2x−1dx=limb→∞∫b11√2x−1dx=limb→∞√2x−1|b1=limb→∞[√2b−1−1]=∞,
the integral ∫∞11√2x−1dx diverges, and therefore
∞∑n=11√2n−1
diverges.
Exercise 9.3.2
Use the
integral test
to determine whether the series ∞∑n=1n3n2+1 converges or diverges.
Hint
: Compare to the integral ∫∞1x3x2+1dx.
Answer
: The series diverges.
The p-Series
The
harmonic series
∞∑n=11/n and the series ∞∑n=11/n2 are both examples of a type of series called a p-series.
Definition: p-series
For any real number p, the series
∞∑n=11np
is called a p-series.
We know the p-series converges if p=2 and diverges if p=1. What about other values of p? In general, it is difficult, if not impossible, to compute the exact value of most p-series. However, we can use the tests presented thus far to prove whether a p-series converges or diverges.
If p<0, then 1/np→∞, and if p=0, then 1/np→1. Therefore, by the
divergence test
,
∞∑n=11np
diverges if p≤0.
If p>0, then f(x)=1/xp is a positive, continuous, decreasing
function
. Therefore, for p>0, we use the
integral test
, comparing
∞∑n=11np
and ∫∞11xpdx.
We have already considered the case when p=1. Here we consider the case when p>0,p≠1. For this case,
∫∞11xpdx=limb→∞∫b11xpdx=limb→∞11−px1−p∣b1=limb→∞11−p[b1−p−1].
Because
b1−p→0 if p>1 and b1−p→∞ if p<1,
we conclude that
∫∞11xpdx={1p−1,ifp>1∞,ifp<1.
Therefore, ∞∑n=11/np converges if p>1 and diverges if 0<p<1.
In summary,
∞∑n=11np{converges ifp>1diverges ifp≤1
.
Example 9.3.3: Testing for Convergence of p-series
For each of the following series, determine whether it converges or diverges.
∞∑n=11n4
∞∑n=11n2/3
Solution
This is a p-series with p=4>1, so the series converges.
Since p=2/3<1, the series diverges.
Exercise 9.3.3
Does the series ∞∑n=11n5/4 converge or diverge?
Hint
: p=5/4
Answer
: The series converges.
Estimating the Value of a Series
Suppose we know that a series ∞∑n=1an converges and we want to estimate the sum of that series. Certainly we can approximate that sum using any finite sum N∑n=1an where N is any positive integer. The question we address here is, for a convergent series ∞∑n=1an, how good is the approximation N∑n=1an?
More specifically, if we let
RN=∞∑n=1an−N∑n=1an
be the remainder when the sum of an
infinite series
is approximated by the Nth
partial sum
, how large is RN? For some types of series, we are able to use the ideas from the
integral test
to estimate RN.
Note 9.3.1: Remainder Estimate from the Integral Test
Suppose ∞∑n=1an is a convergent series with positive terms. Suppose there exists a
function
f satisfying the following three conditions:
f is continuous,
f is decreasing, and
f(n)=an for all integers n≥1.
Let SN be the Nth
partial sum
of ∞∑n=1an. For all positive integers N,
SN+∫∞N+1f(x)dx<∞∑n=1an<SN+∫∞Nf(x)dx.
In other words, the remainder RN=∞∑n=1an−SN=∞∑n=N+1an satisfies the following estimate:
∫∞N+1f(x)dx<RN<∫∞Nf(x)dx.
This is known as the remainder estimate
.
We illustrate Note 9.3.1 in Figure 9.3.4. In particular, by representing the remainder RN=aN+1+aN+2+aN+3+⋯ as the sum of areas of rectangles, we see that the area of those rectangles is
bounded above
by ∫∞Nf(x)dx and
bounded below
by ∫∞N+1f(x)dx. In other words,
RN=aN+1+aN+2+aN+3+⋯>∫∞N+1f(x)dx
and
RN=aN+1+aN+2+aN+3+⋯<∫∞Nf(x)dx.
We conclude that
∫∞N+1f(x)dx<RN<∫∞Nf(x)dx.
Since
∞∑n=1an=SN+RN,
where SN is the Nth
partial sum
, we conclude that
SN+∫∞N+1f(x)dx<∞∑n=1an<SN+∫∞Nf(x)dx.
Example 9.3.4: Estimating the Value of a Series
Consider the series ∞∑n=11n3.
Calculate S10=10∑n=11n3 and estimate the error.
Determine the least value of N necessary such that SN will estimate ∞∑n=11n3 to within 0.001.
Solution
a. Using a calculating utility, we have
S10=1+123+133+143+⋯+1103≈1.19753.
By the
remainder estimate
, we know
RN<∫∞N1x3dx.
We have
∫∞101x3dx=limb→∞∫b101x3dx=limb→∞[−12x2]bN=limb→∞[−12b2+12N2]=12N2.
Therefore, the error is R10<1/2(10)2=0.005.
b. Find N such that RN<0.001. In part a. we showed that RN<1/2N2. Therefore, the remainder RN<0.001 as long as 1/2N2<0.001. That is, we need 2N2>1000. Solving this inequality for N, we see that we need N>22.36. To ensure that the remainder is within the desired amount, we need to round up to the nearest integer. Therefore, the minimum necessary value is N=23.
Exercise 9.3.4
For ∞∑n=11n4, calculate S5 and estimate the error R5.
Hint
: Use the
remainder estimate
RN<∫∞N1x4dx.
Answer
: S5≈1.09035,R5<0.00267
Key Concepts
If limn→∞an≠0, then the series ∞∑n=1an diverges.
If limn→∞an=0, the series ∞∑n=1an may converge or diverge.
If ∞∑n=1an is a series with positive terms an and f is a continuous, decreasing
function
such that f(n)=an for all positive integers n, then
∞∑n=1an
and ∫∞1f(x)dx
either both converge or both diverge. Furthermore, if ∞∑n=1an converges, then the Nth
partial sum
approximation SN is accurate up to an error RN where ∫∞N+1f(x)dx<RN<∫∞Nf(x)dx.
The p-series ∞∑n=11np converges if p>1 and diverges if p≤1.
Key Equations
Divergence test
If an↛0 as n→∞,∞∑n=1an diverges.
p-series
∞∑n=11np{converges,ifp>1diverges,ifp≤1
Remainder estimate
from the
integral test
∫∞N+1f(x)dx<RN<∫∞Nf(x)dx
Glossary
divergence test
: if limn→∞an≠0, then the series ∞∑n=1an diverges
integral test
: for a series ∞∑n=1an with positive terms an, if there exists a continuous, decreasing
function
f such that f(n)=an for all positive integers n, then
∞∑n=1an and ∫∞1f(x)dx
either both converge or both diverge
p-series
: a series of the form ∞∑n=11/np
remainder estimate
: for a series ∞∑n=1an with positive terms an and a continuous, decreasing
function
f such that f(n)=an for all positive integers n, the remainder RN=∞∑n=1an−N∑n=1an satisfies the following estimate:
∫∞N+1f(x)dx<RN<∫∞Nf(x)dx
9.2E: Exercises for Section 9.2
9.3E: Exercises for Section 9.3 |
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Math Articles
Linear equations in two variables
System of Equations
Operations on Sets
In order to better understand the concept of solving a system of equations, we will need to become familiar with some facts on sets.
If A and Bare sets, then we say that A is a subset of B if each element of A is also an element of B. We write A⊂B. We sometimes say that is contained in B instead of A is a subset of B. For example,A {1,2,3} is a subset of B={1,2,3,4,5} since each element of A is also an element of B, while C={1,2,7} is not a subset of B since 7 is an element ofC, but not of B.
IfA and B are sets, then by the union of A and B, which we write A∪B, we mean the set of all elements that are in A or in B or in both.
Example 1. If A={2,4,6,8,10} and B={3,4,5,6,7,8,9} find A∪B.
A∪B={2,3,4,5,6,7,8,9,10}
Even though 4, 6, and 8 are in both sets, they are listed only once in the union.
If A and B are sets, then by the intersection of A and B, which we denote by A∩B, we mean the set of all elements that are in bothA and B
Example 2. If C={2,4,6,8} and D={4,5,6,7,8} find C∩D.C∩D = {4,6,8} If S={1,2,3}, and T={5,6,7}, then S∩T is the set with no elements, that is, the empty set ∅). If the intersection of two sets is empty, that is, they have no elements in common, then we say that the two sets are disjoint. Thus S andT are disjoint.
When more than one set operation is to be performed, we use parentheses to indicate the order in which they are to be performed.
Example 3. IfA={1,2,3},B={2,3,4,5,6}, and C={5,6,7,8}, find(A∩B)∪C and (A∩C)∪B
A∩B = {2,3}
we have
(A∩B)∪C={2,3}∪{5,6,7,8}
{2,3,5,6,7,8}
Also, since
A∩C={1,2,3}∩{5,6,7,8}=∅
we have
(A∩C)∪B=∅∪{2,3,4,5,6}
={2,3,4,5,6}
=B
We may diagram the operations of union and intersection on sets by Venn diagrams.If the sets A andB are represented by circular regions, then the diagram ofA∪B is the shaded area in Figure
ofA∩B is the shaded region in Figure 2. The diagram of the set A∩(B∪C) is the shaded region in Figure 3, which we find by first locating B∪C and then shading what it has in common with A.
Another important set operation is the operation of set difference. If A and B are two sets, thenA-B is the set of all elements of A that are not in B. The diagram of this set is given in Figure 4.
Example 4. If A={1,2,3,4,5,6,7,8} andB={2,4,5,7,11,13,14} find A-B and B-A.
A-B={1,3,6,8}
B-A={11,13,14}
We have described the sets in the examples above by listing the elements. This is not practical in many cases. Another way of describing a set is to state a property P (S) that is true for exactly those members X in the set. We denote the set Sdescribed by property P (S) by
This notation is read:
For example, let E be the set of positive even integers. A property P (n) that is true for exactly these integers is “n is an integer, n>0, and n is divisible by2.” Using the notation above we have
If T is a set of pairs of numbers and the describing property isQ (x,y), then we write
For example, the setCof all points in the plane on the circle whose radius is 2and whose center is at the origin is
The study of sets and operations on sets is itself a fascinating subject, but our purpose here is simply to introduce the subject to the student so that he may apply it to the study of the solution of systems of equations and, in Chapter 9, to the solution of systems of inequalities.
8.2 Graphical Solution
Consider the system of two linear equations in two variables x and y.
(1) a1 x+b1 y+c1=0 (a1, b1 not both 0)
a2 x+b2 y+c2=0 (a2, b2 not both 0)
Since the graph of each equation is a straight line we have the following three possibilities.
(a) The graphs are the same straight line.
(b) The graphs are parallel distinct lines.
(0) The graphs intersect in exactly one point.
If we denote the solution sets of the equations in (1) by
then the solution set S of (1) is
S=S1∩S2
From above, we see that in case (a)S=S1=S2 in (b) S=∅, while in (c) Swould consist of a single point (p, q). In case (a) the system is said to be dependent, in case (b) the system is said to be inconsistent, and in case (c) the system is said to be independent.
In Chapter 7 we pointed out that there is exactly one line through a given point having a given slope. A consequence of this is that two distinct nonvertical lines are parallel if and only if they have the same slope.
Example 1. Solve the system and classify.
2 x+y=4
x-y=2
Graphing the two equations we have
We see that the point of intersection appears to be (2, 0). If this is the case, then the solution set is
Since the solution set consists of a single point, the system is consistent. Note that graphical solutions are approximations.
Example 2. Solve the following system and classify.
x-y=2
− (12) x+12 y=− 1
Graphing the two equations we have
We see that the two equations have the same straight line for their graph and hence the solution set is
The system is dependent.
Example 3. Graph the system and classify.
y-2 x=4
2 x-y=2
Graphing the two equations we have
We see that the slope of each line is 2, so the lines are parallel, and therefore the system is inconsistent. The system has as its solution set
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8.3 Solution by Substitution
In general, the solution of a pair of linear equations in two variables can only be approximated by looking at their graphs. In order to find the exact solution set we transform the given system to one whose solution set is readily obtainable. We must make sure that the solution sets of the original system and the resulting system are the same. Two systems having the same solution set are said to be equivalent. Certainly if we obtain a new system by replacing one equation in the old system by an equivalent equation, then the two systems are equivalent.
The method we present here for finding the exact solutx=1-2 yion is called the method of substitution.
Consider the general system
(1) a1 x+b1 y=c1
a2 x+b2 y=c2
As before, we assume that at least one of the coefficients a1,a2,b1,b2 is different from zero. If b1≠0, we solve the first equation for y, obtaining the equivalent equation
y=− (a1b1) x+c1+b1
Thus, the system (1) is equivalent to
(2) y=− (a1b1) x+c1b1
a2 x+b2 y=c2
If (x,y) is a solution of (2), then y is given in terms of x by the first equation. Substituting this expression in the second equation we obtain the equivalent system
y=− (a1b1) x+c1b1
a2 x+b2 (− (a1b1) x+c1b1)=c2
The second equation may be solved for x and the corresponding value of y can be obtained from the first equation. The other cases are handled in a similar manner.
Example 1. Solve the system
2 x-y=4
x-y=2
Solve the first equation for y to obtain
y=− 2 x+4
The system
y=− 2 x+4
x-y=2
is equivalent to the original system.
Substituting, we obtain the equivalent system
y=− 2 x+4
x-(− 2 x+4)=2
or
y=− 2 x+4
x=2
Substituting, we obtain
y=− 2 (2)+4
=0
The solution set is therefore
S={(2,0)}
Example 2. Solve the system
2 x-y=3
4 x-2 y=6
Solve the first equation for y obtaining the equivalent system
y=2 x-3
4 x-2 y=6
Substituting, we obtain
y=2 x-3
4 x-2 (2 x-3)=6
or
y=2 x-3
6=6
Since the second equation is satisfied for all (x,y), this system has as its solution set
Therefore the original system is dependent.
Example 3. Solve the system
x+2 y=3
x=1-2 y
Since the second equation is solved forx we substitute in the first equation obtaining
(1-2 y)+2 y=3
x=1-2 y
or
1=3
x=1-2 y
Since the first equation has as its solution set ∅, the solution set of the system is
S=∅
The original system is therefore inconsistent.
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{
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| x+2y=3 |
| x=1−2y |
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| Equations that have an explicitly expressed variable are commonly written first. |
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| Simple numerical terms are commonly written last. |
{
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| x=−2y+1 |
| x+2y=3 |
| |
| In order to create an equation containing only 1 variables, we need to substitute the right side of the first equation for x in the second equation. In our example, all instances of x in the second equation will be replaced by: −2y+1. |
{
| |
| x=−2y+1 |
| (−2y+1)+2y=3 |
| |
| We need to get rid of expression parentheses. If there is a negative sign in front of it, each term within the expression changes sign. Otherwise, the expression remains unchanged. In our example, there are no negative expressions. |
{
| |
| x=−2y+1 |
| −2y+1+2y=3 |
| |
| We need to organize this expression into groups of like terms, so we can combine them easier. There is only one group of like terms: −2y, 2y |
{
| |
| x=−2y+1 |
| −2y+2y+1=3 |
| |
| We need to combine like terms in this expression by adding up all numerical coefficients and copying the literal part, if any. No numerical coefficient implies value of 1. There is only one group of like terms: −2y, 2y |
{
| |
| x=−2y+1 |
| 1=3 |
| |
| Since we reached an invalid equation, there are no values of (x,y) that satisfy the equations. In our example, the invalid equation is 1=3. This system is inconsistent. |
{
| |
| x ∈ ∅ |
| y ∈ ∅ |
8.4 Solution by Elimination
The method of solution by elimination depends on the elementary operations E1, E2, and E3 below, which change a given system into an equivalent system.
E.1 Interchange any two equations of the system.
E.2 Multiply any equation by a nonzero number.
E.3 Replace any equation of the system by the sum of that equation and a multiple of another equation of the system.
Example 1. Solve the system
x+y=4
2 x-3 y=3
Applying E.3 we multiply the first equation by -2 and add it to the second equation obtaining the system
x+y=4
− 5 y=− 5
Apply E.2 to the second equation by multiplying by − 15.
x+y=4
y=1
Substituting, we obtain
x+1=4
y=1
or
x=3
y=1
Thus the solution set of the original system is
S={(3,1)}
Example 2. Solve the system.
x+2 y=3
2 x+4 y=1
Add -2 times the first equation to the second equation to obtain the system
x+2 y=3
0=− 5
Since the second equation obviously has no solution, this system, and hence the original system, is inconsistent. Thus the solution set is
S=∅
Example 3. Solve the system
2 x-y=3
− x+12 y=− (32)
Multiply the first equation by 12 and add it to the second equation to obtain the system
2 x-y=3
0=0
Since the second equation is satisfied for all (x,y), the solution set of system is the same as the solution set of the first equation, namely,
The original system is dependent.
We conclude this section with two examples of systems that lead to systems of linear equations.
Example 4. Solve the system
(1) x-yx+y=23
x+2 yx-3=12
Multiply each equation by its LCD.
(2) 3 x-3 y=2 x+2 y
2 x+4 y=x-3
Simplify.
x-5 y=0
x+4 y=− 3
Multiply the first equation by − 1 and add.
x-5 y=0
9 y=− 3
Substituting, we obtain (53,− 13) as the solution of (2). In going from (1) to (2) we multiplied by the expressions x+y and x-3. Multiplication by each of these expressions is an elementary operation only if the expression is non-zero. By substitution we see that neither expression is zero at (53,− 13).
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{
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| x−yx+y=23 |
| x+2yx−3=12 |
| |
| In order to bring this equation to graphable form, we need to move all the terms to the left side. In our example, - term 23, will be moved to the left side. Notice that a term changes sign when it 'moves' from one side of the equation to the other. |
| |
| In order to bring this equation to graphable form, we need to move all the terms to the left side. In our example, - term 12, will be moved to the left side. Notice that a term changes sign when it 'moves' from one side of the equation to the other. |
{
| |
| x−yx+y+(−23)=23+(−23) |
| x+2yx−3+(−12)=12+(−12) |
| |
| We need to get rid of expression parentheses. If there is a negative sign in front of it, each term within the expression changes sign. Otherwise, the expression remains unchanged. In our example, there are no negative expressions. |
| |
| We need to get rid of expression parentheses. If there is a negative sign in front of it, each term within the expression changes sign. Otherwise, the expression remains unchanged. In our example, there are no negative expressions. |
| |
| We need to get rid of expression parentheses. If there is a negative sign in front of it, each term within the expression changes sign. Otherwise, the expression remains unchanged. In our example, there are no negative expressions. |
| |
| We need to get rid of expression parentheses. If there is a negative sign in front of it, each term within the expression changes sign. Otherwise, the expression remains unchanged. In our example, there are no negative expressions. |
{
| |
| x−yx+y−23=23−23 |
| x+2yx−3−12=12−12 |
| |
| In order to add fractions, we first have to rewrite them so that they all have the same (common) denominator. We will make fractions' denominators equivalent by finding the LCD of all fractions and then multiplying both the numerator and denominator of each fraction by factors that are missing in the denominator. |
| |
| We will make this denominator equal to LCD by adding the necessary factors. In our example, LCD is equal to: (x+y)3 Factors that, when added, will make the denominator equal to LCD are: 3 |
| |
| We will make this denominator equal to LCD by adding the necessary factors. In our example, LCD is equal to: (x+y)3 Factors that, when added, will make the denominator equal to LCD are: x+y |
| |
| We need to combine like terms in this expression by adding up all numerical coefficients and copying the literal part, if any. No numerical coefficient implies value of 1. There is only one group of like terms: 23, −23 |
| |
| In order to add fractions, we first have to rewrite them so that they all have the same (common) denominator. We will make fractions' denominators equivalent by finding the LCD of all fractions and then multiplying both the numerator and denominator of each fraction by factors that are missing in the denominator. |
| |
| We will make this denominator equal to LCD by adding the necessary factors. In our example, LCD is equal to: (x−3)2 Factors that, when added, will make the denominator equal to LCD are: 2 |
| |
| We will make this denominator equal to LCD by adding the necessary factors. In our example, LCD is equal to: (x−3)2 Factors that, when added, will make the denominator equal to LCD are: x−3 |
| |
| We need to combine like terms in this expression by adding up all numerical coefficients and copying the literal part, if any. No numerical coefficient implies value of 1. There is only one group of like terms: 12, −12 |
{
| |
| (x−y)3(x+y)3−2(x+y)3(x+y)=0 |
| (x+2y)2(x−3)2−x−32(x−3)=0 |
| |
| We need to add fractions that have a common denominator. The numerator of the the newly created fraction will be the sum of all the existing numerators, and its denominator will be equal to the common denominator. |
| |
| We need to add fractions that have a common denominator. The numerator of the the newly created fraction will be the sum of all the existing numerators, and its denominator will be equal to the common denominator. |
{
| |
| (x−y)3−2(x+y)(x+y)3=0 |
| (x+2y)2−(x−3)(x−3)2=0 |
| |
| Numerical terms are commonly written first. |
| |
| Numerical terms are commonly written first. |
| |
| Numerical terms are commonly written first. |
| |
| Numerical terms are commonly written first. |
{
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| 3(x−y)−2(x+y)3(x+y)=0 |
| 2(x+2y)−(x−3)2(x−3)=0 |
| |
| We need to expand this term by multiplying a term and an expression. The following product distributive property will be used: A(B+C)=AB+AC. In our example, the resulting expression will consist of 2 terms: the first term is a product of 3 and x. the second term is a product of 3 and −y. |
| |
| We need to expand this term by multiplying a term and an expression. The following product distributive property will be used: A(B+C)=AB+AC. In our example, the resulting expression will consist of 2 terms: the first term is a product of 2 and x. the second term is a product of 2 and y. |
| |
| We need to get rid of expression parentheses. If there is a negative sign in front of it, each term within the expression changes sign. Otherwise, the expression remains unchanged. In our example, the following 2 terms will change sign: x, −3 |
{
| |
| (3x−3y)−(2x+2y)3(x+y)=0 |
| 2(x+2y)−x+32(x−3)=0 |
| |
| We need to get rid of expression parentheses. If there is a negative sign in front of it, each term within the expression changes sign. Otherwise, the expression remains unchanged. In our example, the following 2 terms will change sign: 2x, 2y |
| |
| We need to expand this term by multiplying a term and an expression. The following product distributive property will be used: A(B+C)=AB+AC. In our example, the resulting expression will consist of 2 terms: the first term is a product of 2 and x. the second term is a product of 2 and 2y. |
{
| |
| 3x−3y−2x−2y3(x+y)=0 |
| (2x+2·2y)−x+32(x−3)=0 |
| |
| We need to organize this expression into groups of like terms, so we can combine them easier. There are 2 groups of like terms: first group: 3x, −2x second group: −3y, −2y |
| |
| Numerical factors in this term have been multiplied. |
{
| |
| 3x−2x−3y−2y3(x+y)=0 |
| (2x+4y)−x+32(x−3)=0 |
| |
| We need to combine like terms in this expression by adding up all numerical coefficients and copying the literal part, if any. No numerical coefficient implies value of 1. There are 2 groups of like terms: first group: 3x, −2x second group: −3y, −2y |
| |
| We need to get rid of expression parentheses. If there is a negative sign in front of it, each term within the expression changes sign. Otherwise, the expression remains unchanged. In our example, there are no negative expressions. |
{
| |
| x−5y3(x+y)=0 |
| 2x+4y−x+32(x−3)=0 |
| |
| Since the right side of this equation is equal to zero, we can get rid of the denominator on the left side. |
| |
| We need to organize this expression into groups of like terms, so we can combine them easier. There is only one group of like terms: 2x, −x |
{
| |
| x−5y=0 |
| 2x−x+4y+32(x−3)=0 |
| |
| In order to solve this linear equation, we need to group all the variable terms on one side, and all the constant terms on the other side of the equation. In our example, - term x, will be moved to the right side. Notice that a term changes sign when it 'moves' from one side of the equation to the other. |
| |
| We need to combine like terms in this expression by adding up all numerical coefficients and copying the literal part, if any. No numerical coefficient implies value of 1. There is only one group of like terms: 2x, −x |
{
| |
| (x−5y)+(−x)=−x |
| x+4y+32(x−3)=0 |
| |
| We need to get rid of expression parentheses. If there is a negative sign in front of it, each term within the expression changes sign. Otherwise, the expression remains unchanged. In our example, there are no negative expressions. |
| |
| Since the right side of this equation is equal to zero, we can get rid of the denominator on the left side. |
{
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| x−5y−x=−x |
| x+4y+3=0 |
| |
| We need to organize this expression into groups of like terms, so we can combine them easier. There is only one group of like terms: x, −x |
| |
| In order to solve this linear equation, we need to group all the variable terms on one side, and all the constant terms on the other side of the equation. In our example, - 2 terms x, 3, will be moved to the right side. Notice that a term changes sign when it 'moves' from one side of the equation to the other. |
{
| |
| x−x−5y=−x |
| (x+4y+3)+(−x−3)=−x−3 |
| |
| We need to combine like terms in this expression by adding up all numerical coefficients and copying the literal part, if any. No numerical coefficient implies value of 1. There is only one group of like terms: x, −x |
| |
| We need to get rid of expression parentheses. If there is a negative sign in front of it, each term within the expression changes sign. Otherwise, the expression remains unchanged. In our example, there are no negative expressions. |
{
| |
| −5y=−x |
| x+4y+3−x−3=−x−3 |
| |
| In order to isolate the variable in this linear equation, we need to get rid of the coefficient that multiplies it. This can be accomplished if both sides are divided by −5. |
| |
| We need to organize this expression into groups of like terms, so we can combine them easier. There are 2 groups of like terms: first group: x, −x second group: 3, −3 |
{
| |
| 5y5=x5 |
| x−x+4y+3−3=−x−3 |
| |
| We need to rewrite the equation of this line in a graphable form. This can be accomplished by simplifying the left side and making the coefficients on the right side explicit. In our example, the coefficient of the variable term is equal to 15 |
| |
| We need to combine like terms in this expression by adding up all numerical coefficients and copying the literal part, if any. No numerical coefficient implies value of 1. Numerical 'like' terms will be added. There are 2 groups of like terms: first group: x, −x second group: 3, −3 |
{
| |
| y=15x |
| 4y=−x−3 |
| |
| In order to isolate the variable in this linear equation, we need to get rid of the coefficient that multiplies it. This can be accomplished if both sides are divided by 4. |
{
| |
| y=15x |
| 4y4=−x−34 |
| |
| We need to rewrite the equation of this line in a graphable form. This can be accomplished by simplifying the left side and making the coefficients on the right side explicit. In our example, the coefficient of the variable term is equal to −14 and the constant term is equal to −34. |
{
| |
| y=15x |
| y=−14x−34 |
Unable to show steps for non-linear systems.
Example 5. Solve the system
(1) 1x-y+2x+y=710
4x-y+5x+y=52
Multiplication of each equation by its L.C.D. leads to nonlinear equations. However, we notice that the two variables xand y occur only in the combinations 1x-y and 1x+y, so that the substitutions
a=1x-y and b=1x+y
yield the linear system
a+2 b=710
4 a+5 b=52
Solving, we obtain
a=12
b=110
Since a=1x-y and b=1x+y, we obtain the system
(2) 1x-y=12
1x+y=110
Solving as in Example 4, we find that the solution set of (2) is {(6, 4)}. In obtaining (2) from (1) only elementary operations have been used, thus {(6,4)} is the solution set of (1).
8.5 Solution by Determinants
In solving the system
(1) a1 x+b1 y=c1
a2 x+b2 y=c2
we encounter the expression
a1 b2-a2 b1
This expression is denoted by
and is called the determinant of a1,b1,a2,b2. Since this determinant has two rows and two columns, it is said to be of order 2. The numbers a1,b1,a2and b2 are called the elements of the determinant.
For example,
We will now show how determinants can be used to solve the system (1) above.
a1 x+b1 y=c1
a2 x+b2 y=c2
Multiply the first equation by , the second equation by − b1, and add to obtain
a1 b2 x-a2 b1 x=c1 b2-c2 b1
a2 x+b2 y=c2
Solving the first equation for x, we obtain
x=c1 b2-c2 b1a1 b2-a2 b1
We see that the numerator is simply the determinant
while the denominator is the determinant
Hence, we may write
where we are assuming, of course, that
Solving the original system for y instead of x, we obtain
Let D, Dx, and Dy denote the following determinants:
Using this notation, the solutions forx andy above become
x=DxD
y=DyD
The use of determinants in this way to solve a system of linear equations is known as Cramer’s rule.Gabriel Cramer (1704-1752) was a well-known Swiss mathematician.
We see thatD is the array of coefficients ofx and y in (1). Further more, Dx is obtained from D by replacing the first column, the x coefficients, by the column of constants; and, similarly, Dy is obtained from D by replacing the second column, the y coefficients, by the column of constants.
Example 1. Use Cramer’s rule to solve the following system.
x-y=2
x+y=4
We compute D, Dx, and Dy.
Apply Cramer’s rule.
x=DxD
=62
=3
y=DyD
=22
=1
Thus the solution set is S={(3,1)}.
If D =0 in this procedure, then either the equations are dependent or the system is inconsistent, depending on whether Dx, and Dy are both zero or not.
Example 2. Solve the system
2 x-4 y=3
− x+2 y=− (32)
We have
Consequently, the system is either dependent or inconsistent. Since
and
the system is dependent. From geometric considerations two equations are dependent if and only if they are equations of the same line.
Thus the solution set is
8.6 Linear Systems in More than Two Variables
Frequently we must consider systems of equations with more than two unknowns. The methods of substitution, elimination, and Cramer’s rule can be extended to solve these higher-order systems.
In this section we will use the method of elimination.
Example 1. Solve the system
x+y-4 z=2
3 x-y-8 z=− 1
2 x+3 y+2 z=3
We eliminate x from the second equation by adding − 3 times the first equation, and from the third equation by adding − 2 times the first equation, obtaining the system
x+y-4 z=2
− 4 y+4 z=− 7
y+10 z=− 1
Eliminate y from the second equation by adding 4times the third equation to the second obtaining the system
x+y-4 z=2
44 z=− 11
y+10 z=− 1
From this system we first computez. (from the second equation), theny (from the third equation), and finallyx (from the first equation).
z=− (14)
y=− 1-10 z
y=− 1-10 (− (14))
y=32
x=z-y+4 z
x=2-(32)+4 (− (14))
x=− (12)
Thus, the solution set is
S={(− (12),32,− (14))}
Example 2. Solve the system
x+y-z=1
y+z-w=2
z+w-x=3
w+x-y=4
We first rewrite the system so that the variables appear in the same order in each equation.
x+y-z=1
y+z-w=2
− x+z+w=3
x-y+w=4
Eliminatingxfrom the third and fourth equations we have
x+y-z=1
y+z-w=2
y+w=4
− 2 y+z+w=3
Eliminatingyin the last two equations we have
x+y-z=1
y+z-w=2
− z+2 w=2
3 z-w=7
Eliminating z in the last equation we obtain
x+y-z=1
y+z-w=2
− z+2 w=2
5 w=13
Solving successively for w, z, y, and x we obtain
w=135
− z=2-2 w
z=165
y=2-z+w
y=75
x=1-y+z
x=145
Thus the solution set is
S={(145,75,165,135)}
As in the 2 x 2 case, higher-order systems fall into three classes, namely, independent, dependent, and inconsistent. An independent system is one that has exactly one solution, while a dependent system is one that has more than one solution. A system is inconsistent if it has no solution. We see that the systems in Examples 1 and 2 are independent
Example 3. Solve the system
3 x+2 y+7 z=5
4 x+3 y+5 z=− 4
2 x+y+9 z=14
Eliminating yin the first two equations we obtain
− x-11 z=− 23
− 2 x-22 z=− 46
2 x+y+9 z=14
Eliminating x in the first equation we obtain
0=0
x+11 z=23
2 x+y+9 z=14
We can use the last two equations to solve for two of the variables in terms of the third variable. Thus
x=23-11 z
and
y=14-9 z-2 x
=14-9 z-2 (23-11 z)
so
y=− 32+13 z
The solution set is
Sincez takes on all real values, S has an infinite number of elements.
Therefore the system is dependent.
Example 4. Solve the system
x+y+z=1
2 x-3 y-2 z=− 4
4 x-y=− 1
Eliminatingz in the second equation we obtain
x+y+z=1
4 x-y=− 2
4 x-y=− 1
Eliminating y in the last equation we obtain
x+y+z=1
4 x-y=− 2
0=1
Since the last equation in this system has no solution, the system is inconsistent. Hence the original system is inconsistent and has solution set S=∅.
8.7 Third-Order Determinants
By a determinant of order 3 we mean
If we rewrite this number in the form
then we may rewrite it in terms of determinants of order 2 as follows.
We may also rewrite it as
We define the minor of a given element in a determinant to be the determinant that is left after deleting the row and the column in which the element appears. Thus in the determinant of order 3
above, the minor ofa1 is
the minor of b2 is
while the minor of c3 is
If we use the notation m (x) to stand for the minor of the element x, then from above, expanding by the first column, we have
Expanding by the first row we have
With the proper use of signs it is possible to express the determinant in terms of minors of any row or any column. A helpful device for determining the appropriate signs for a third-order determinant is the diagram below.
We can pick any row or column, form the sum of the product of the elements and their minors along with the appropriate sign from the diagram above, and we will have the determinant expressed in terms of minors.
Example 1. Use the second row to compute
We have
a1 x+b1 y+c1 z=d1
(1) a2 x+b2 y+c2 z=d2
a3 x+b3 y+c3 z=d3
As in the case of two linear equations in two unknowns, determinants may be used to solve a system of three linear equations in three unknowns.
Consider the following system of equations:
Using the same type of notation as we did in the second-order case we have
It is possible, but tedious, to show that the solution to the system is given by Cramer’s rule, that is,
x=DxDy=DyD, and z=DzD
Example 2. Use Cramer’s rule to solve the system
x+y-2 z=3
3 x-y+z=5
x-3 y+z=2
We have
We expand D by the first row to obtain
Expanding Dx I by the first column we have
Similarly we find that
Dy=− 5
and
Dz=− 12
Hence,
x=DxD
=2916
y=DyD
=− 516
z=DzD
=− 1216
=− 34
Therefore the solution set is S={(2916,− 516,− 34)}
If D=0, then either the system is dependent or inconsistent, depending on whether Dx, Dy and Dz are all zero, or not.
8.8 Statement Problems
Many statement problems lead to systems of linear equations in two or more variables. The general method of attack is the same as in Section 6.6, except that we introduce a different letter for each unknown. Then we translate the statements relating these unknowns into a system of equations.
Example 1. A man has $2.00 in nickels and dimes, with two more dimes than nickels. How many dimes and how many nickels does he have?
Step 1. Let n be the number of nickels and d be the number of dimes.
Step 2. The value of n nickels is 5 n cents and the value of d dimes is 10 d. Since the man has $2.00 in nickels and dimes, we have the equation
5 n+10 d=200
Since he has two more dimes than nickels, we have
d=n+2
Step 3. Solve the system
5 n+10 d=200
d=n+2
By substitution
5 n+10 (n+2)=200
d=n+2
or
15 n=180
d=n+2
The solution is
n=12 nickels
d=14 dimes
Example 2. How many gallons of solution containing 45% alcohol must be mixed with a 60% alcohol solution to obtain 40 gallons of solution that is 48% alcohol?
Step 1. Let x be the number of gallons of 45% solution and y be the number of gallons of 60% solution.
Step 2. Since the total number of gallons is 40,
x+y=40
Since 45% of x is alcohol, 60% of y is alcohol, and 48% of the total 40 gallons is alcohol,
0.45 x+0.60 y=(0.48) (40)
Step 3. Solve the system
x+y=40
45 x+60 y=1920
Adding -45 times the first equation to the second equation we obtain
x+y=40
15 y=120
Thus,
y=8 gallons
x=32 gallons
Example 3. A man cruises 40 miles down a river in his boat in 2 hours. The return trip takes 2 1/2 hours. What is the average speed of the boat and the average speed of the river?
Step 1. Letx be the speed of the boat relative to the shore and y the speed of the river.
Step 2. Since distance is equal to rate times time and the boat’s speed downstream is x+y, we have
40=(x+y) (2)
Upstream we have
40=(x-y) (52)
Step 3. Solve the system
x+y=20
x-y=16
Adding the two equations we have
x+y=20
2 x=36
Thus,
x=18mph
y=2 mph
Example 4. A man paid for a $1.85 item with nickels, dimes, and quarters. If there were the same number of quarters as nickels plus dimes and four more nickels than dimes, how many of each were there?
Step 1. Let x be the number of nickels, y be the number of dimes, and z be the number of quarters.
Step 2. There were the same number of quarters as nickels plus dimes so
z=x+y
There were four more nickels than dimes so
x=y+4
The value of x nickels is 5 x cents, of y dimes is 10 y, and ofz quarters is 25 z, thus
5 x+10 y+25 z=185
Step 3. Solve the system
z=x+y
x=y+4
5 x+10 y+25 z=185
Substituting forx in the first equation we obtain
z=(y+4)+y
x=y+4
5 x+10 y+25 z=185
Now substituting for x and z in the third equation we obtain
z=2 y+4
x=y+4
5 (y+4)+10 y+25 (2 y+4)=185
Solving this system we obtain
x=5 nickels
y=1 dime
z=6 quarters
8.9 Second Degree Systems in Two Variables
An equation of the form
(1) Ax2+Bxy+Cy2+Dx+e y+F=0
where A,B,C,D,e, and F are all constants, is called the general second degree equation in x and y, or simply the general quadratic in x and y. In the case that A=B=C=0 we see that (1) reduces to a first degree equation whose graph is a straight line. In Chapter 7 we also considered three more cases: A=C≠0 and B=0, in which case the graph of (1) is a circle; A=B=0 and C≠0, in which case the graph of (1) is a parabola with horizontal axis; and B=C=0 and A≠0, in which case the graph of (1) is a parabola with vertical axis. All cases of equations of type (1) are studied in analytic geometry. In this section we will consider the algebraic solution of certain systems of equations of type (1).
Real solutions of a system of quadratic equations in two variables sometimes can be found by graphing both equations and then estimating the coordinates where their graphs intersect.
Example 1. Solve the system
y=x2-2 x+2
y+x=3
In order to graph
y=x2-2 x+2
we complete the square and write it as
y-1=(x-1)2
whose graph is a parabola with vertex (1, 1) opening upward. Graphing we obtain
Thus to one decimal place, the two solutions are approximately (-0.6, 3.6) and (1.6, 1.4).
Another technique for solving such a system is the method of substitution. We apply this method to the system in Example 1.
Example 2. Solve the system
y=x2-2 x+2
y+x=3
Solving the second equation for y and substituting, we obtain the system
y=3-x
3-x=x2-2 x+2
Solving the second equation for x: we have
x2-x-1=0
x=1±√52
We now use the equationy=3-x to obtain the corresponding values of y. For x=1+√52,
y=3-x
=3-1+√52
=5-√52
For x=1-√52,
y=3-(1-√52)
= 5+√52
Thus the solution set of the system is
S={(12+√52,52-√52),(12-√52,52+√52)}
Since √5≈2.2 to one decimal place, we have the approximation
{(1.6,1.4),(− 0.6,3.6)}
which is what we obtained in Example 1.
Example 3. Solve the system
x2+y2=13
x+y=5
We solve the second equation for y and substitute in the first equation, which gives us the system
x2+(5-x)2=13
y=5-x
Solving the first equation we have
x2+(5-x)2=13
x2+25-10 x+x2=13
2 x2-10 x+12=0
x2-5 x+6=0
x2+(5-x)2=13
x=3or x=2
For x=3 we find from the second equation that y=2, while for x=2,y=3. Hence our solution set is
S={(3,2),(2,3)}
The graph of the first equation is a circle with center (0,0) and radius‘√13=3.6, while that of the second equation is a line. Graphing we
have
Example 4.Solve the system
2 x2+3 y2=21
3 x2-4 y2=23
We multiply the first equation by 43 and add to obtain the system
2 x2+3 y2=21
173 x2=51
Solving the second equation for x we have
173 x2=51
x2=9
x=± 3
We substitute each value of x into the first equation. For x=3,
y2=1
y=± 1
and for x=− 3
y2=1
y=± 1
Thus our solution set is
S={(3,1),(3,− 1),(− 3,1),(− 3,− 1)}
Example 5.Solve the system of equations
x-y=4
x2+y2-2 x-2=0
Solving for y in the first equation and substituting in the second we have
y=x-4
x2+(x-4)2-2 x-2=0
Solving the quadratic we have
2 x2-10 x+14=0
or
x2-5 x+7=0
Using the quadratic formula we obtain
x=5±√25-282
=52±√3 i2
Substituting these values for x in the equation y=x-4, we find that the solution set is
S={(52+√3 i2,− (32)+√3 i2),(52-√3 i2,− (32)-√3 i2)}
Since the solutions are pairs of complex numbers, the graphs of the original equations do not intersect.
x2+y2-2 x-2=0
x2-2 x+1+y2=2+1
(x-1)2+y2=(√3)2
| | |
--- |
| x | yy |
| 0 | − 4 |
| 4 | 0 |
Some other methods for solving systems of quadratic equations in two variables are illustrated in the next few examples.
Example 6. Solve the system
3 x2-4 xy+2 y2=3
2 x2-6 xy+y2=− 6
Eliminate the constant term in the second equation by multiplying the first equation by 2 and adding. We obtain the system
3 x2-4 xy+2 y2=3
8 x2-14 xy+5 y2=0
Factor the second equation and solve for y in terms of x.
(4 x-5 y) (2 x-y)=0
Either
4 x-5 y=0
y=45 x
or
2 x-y=0
y=2 x
Substitute these expressions for y into the first equation and solve for x. For y=(45) x.
3 x2-4 x (45 x)+2 (45 x)2=3
x2=25
x=± (53)
Since y=(45) x, we obtain (53,43) and (− (53),− (43)). For y=2 x.
3 x2-4 x (2 x)+2 (2 x)2=3
3 x2=3
x2=1
x=± 1
Since y=2 x, we obtain (1,2) and (− 1,− 2). Thus our solution set is
S={(53,43),(− 53,− 43),(12),(− 1,− 2)}
Example 7.Solve the system
x2+y2=5
3 x2-2 xy+3 y2=11
Multiply the first equation by − 3 and add to eliminate both the x2 and y2 terms. We obtain the system
x2+y2=5
xy=2
Solve the second equation for y in terms of x and substitute back in the
first equation obtaining
x2+(2x)2=5
Solve for x
x2+4x2=5
x4-5 x2+4=0
(x2-4) (x2-1)=0
Either
x2-4=0
x2=4
x=± 1
Substituting these values for x in the equation xy=2 we obtain the solution set
S={(2,1),(− 2,− 1),(1,2),(− 1,− 2)}
8.10 More Statement Problems
There are statement problems that lead to systems of quadratic equations in two variables. Let us consider some examples.
Example 1.Find all pairs of integers such that the sum of their squares is 170 while their difference is 4.
Step 1. Let x be one integer and y the other.
Step 2. Since the sum of the squares of the two integers is 170,
x2+y2=170
Since their difference is 4,
y-x=4
Step 3.Solve the system
x2+y2=170
y-x=4
Solve the second equation for y and substitute in the first equation to obtain
y=x+4
x2+(x+4)2=170
Solve the second equation for x.
x2+x2+8 x+16=170
2 x2+8 x-154=0
x2+4 x-77=0
(x-7) (x+11)=0
x=7 or x=− 11
Using the first equation we obtain as our solution set
S={(7,11),(− 11,− 7)}
Example 2.Find the dimensions of a rectangle whose area is 120 square feet and whose perimeter is 46 feet.
Step 1. Let x be the length and y be the Width
Step 2. Since the area is 120 square feet
xy=120
Since the perimeter is 46feet.
2 x+2 y=46
Step 3.Solve the system
xy=120
x+y=23
Solving the second equation for y and substituting in the first equation we obtain the system
y=23-x
x (23-x)=120
Solve the second equation for x
x (23-x)=120
x2-23 x+120=0
(x-8) (x-15)=0
x=8 or x=− 15
Using the first equation to solve for y we see that the dimensions are
x=15 feet, y=8 feet
Example 3.A certain group of people rented a bus for $240. If there had been 3 fewer people it would have cost each person $4 more. How many people were there and what was the cost per person?
Step 1. Let x be the number of people in the group and y be the cost per person.
Step 2. Since the cost is $240,
xy=240
If 3 fewer people went, there would have been x-3 people and it would have cost each person 4 dollars more, that is,
y+4 dollars. Therefore,
(x-3) (y+4)=240
Step 3.Solve the system
xy=240
(x-3) (y+4)=240
Simplifying the second equation and using the first equation, we obtain the system
y=240x
xy-3 y+4 x-12=240
or
y=240x
− 3 (240x)+4 x-12=0
Solving the second equation for x,
− 3 (240x)+4 x-12=0
− 180+x2-3 x=0
x2-3 x-180=0
(x-15) (x+12)=0
x=15 or x=− 12
Clearly x=15 people is the only solution since − 12 cannot represent a number of people. Since xy=240, we have
x=15 People, y=$16 per person
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Electromagnetic dynamic stability analysis of power electronics-dominated systems using eigenstructure-preserved LTP Theory
Jiabing Hu
ORCID: orcid.org/0000-0002-3670-30291na1,
Zeren Guo
ORCID: orcid.org/0000-0003-2453-62321na1,
Jianhang Zhu
ORCID: orcid.org/0000-0003-0698-70642na1,
Jürgen Kurths
ORCID: orcid.org/0000-0002-5926-42763,
Yunhe Hou
ORCID: orcid.org/0000-0002-8882-98972,
Buyang Du
ORCID: orcid.org/0009-0001-5262-90331,
Zefei Wu1,
Guojie Zhao1,
Yunfeng Liu4,
Kai Xin4,
Jianbo Guo5 &
…
Shijie Cheng1
Nature Communications
volume 16, Article number: 6852 (2025)
Cite this article
3074 Accesses
3 Citations
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Abstract
Secure operation of power systems, one of the largest man-made systems, is crucial for economic development and societal well-being. Over the past century, initiatives like Europe’s Super Grid and China’s Dual Carbon plan have driven significant changes in power systems, leading to the widespread integration of diverse power electronic equipment. This has resulted in the emergence of power electronics-dominated power systems. However, they have experienced multiple electromagnetic oscillation accidents, causing large-scale renewable energy disconnections and even power equipment damage. To address these critical stability issues, now a global concern, the prevalent method relies on linear time-invariant approximate modeling, i.e., the eigenstructure-reconfiguration framework. While effective, it is limited by the curse of dimensionality in large-scale systems. Recently, the linear time-periodic theory has shown potential in accelerating calculations, but its analysis methods remain underdeveloped. In response to these challenges, we propose here a generalized linear time-periodic participation factor and sensitivity theory within the eigenstructure-preserved framework. This proposed participation factor significantly improves computational efficiency, outperforming eigenstructure-reconfiguration methods by orders of magnitude. Additionally, the proposed sensitivity analysis overcomes the lack of its analyticity. The potential of our methods is demonstrated through real-world power systems of China.
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Introduction
Maintaining stability following small perturbations is crucial for dynamic systems, necessitating reliance on linear stability analysis. This viewpoint is particularly relevant to power systems1,2,3, where the quantitative analysis must adapt to the high-order characteristics: large-scale power systems often consist of hundreds of thousands of nodes, and the system states and parameters expand significantly. Therefore, it is imperative to investigate both the stability itself and the correlation between stability and states or parameters. Traditional power systems, dominated by synchronous generators, face significant challenges associated with electromechanical dynamics4,5,6,7,8,9,10. Here, the dynamics could be described by a nonlinear time-invariant (NLTI) model, where linearization is performed around equilibrium points (i.e., constant value). Extensive research has been conducted on this linear stability, focusing on eigenvalues4,5, participation factors6,7, and sensitivities8,10, all within the framework of the linear time-invariant (LTI) theory. Eigenvectors, which are fundamental to participation factors and sensitivities (see Methods), along with eigenvalues, form the LTI eigenstructure6—the foundation of electromechanical dynamics. Over the past few decades, their successful application has formed the theoretical basis for analyzing and optimizing the electromechanical stability of traditional power systems.
Recent developments in power systems, driven by initiatives such as Europe’s Super Grid plan11, the United States’ Building a Better Grid Initiative12."), and China’s Dual Carbon plan13."), have significantly changed their system composition14."),15."). The integration of diverse power electronic equipment, including solar photovoltaics (PV), wind power, and high-voltage direct current (HVDC) transmission, has become widespread. These changes have resulted in what are now referred to as power electronics-dominated power systems16.") (PEPS). In China, the transformation of the power grid has accelerated rapidly. By 2023, the installed capacity of PV increased by 55.2% year-on-year to 610 million kilowatts, while wind power capacity rose by 20.7% to 440 million kilowatts. Consequently, the proportion of these renewable sources in total power generation capacity exceeded 36%17.
. 2023"), with projections indicating it will reach 48% by 2030 and 75.5% by 205018. Research Reports on China’s CO2 Emission Peak and Carbon Neutrality Released in Beijing.
(2021)."),19."). However, this rapid development of PEPS has raised increasing concerns about stability issues, particularly those associated with electromagnetic dynamics. The early adoption of renewable energy in the United States brought about initial challenges, such as sub-synchronous oscillation (SSO) accidents in doubly fed induction generator-based wind farms connected to series compensated networks in Minnesota (2007) and Texas (2009). These accidents resulted in large-scale turbine trips and damage to crowbar protection circuits20."),21."). Similar issues emerged in China, where in 2010, hundreds of SSO accidents in Hebei Guyuan’s wind farms led to the disconnection of thousands of wind turbines22."). In 2015, an SSO accident in Xinjiang Hami involving direct-drive permanent magnetic synchronous generator-based wind farms first occurred23."), causing the tripping of several large-capacity synchronous generators located hundreds of kilometers away, triggering significant power shortages and system frequency drops. Subsequently, electromagnetic oscillation accidents occurred frequently in practical PEPS24."),25."),26."),27."), having become a major concern and posing a significant threat to the safe and stable operation.
However, in the electromagnetic timescale, complex time-varying characteristics emerge in PEPS due to several phenomena, such as positive–negative sequence coupling in renewable power generation (RPG) and switching operations in HVDC systems (see Supplementary Notes 1 and 2). As a result, electromagnetic dynamics are here best described using nonlinear time-periodic (NLTP) models, where steady-state behaviors converge to diverse periodic orbits. These orbits are no longer limited to simple sinusoidal waveforms at a fundamental frequency; instead, they include rich harmonic content—such as second-order components inherent in RPG internal dynamics and modular multilevel converter-based HVDC submodules, as well as the 12k ± 1 characteristic harmonics typical of line-commutated converter (LCC)-based HVDC systems28,29,30. Furthermore, unlike traditional electromechanical dynamics, the electromagnetic dynamics of PEPS exhibit more pronounced high-order characteristics. This is partly due to the relatively low capacity of individual RPG units, and PEPS often require hundreds of units to replace a single synchronous generator. Additionally, PEPS include a large number of states associated with inductance and capacitance at the electromagnetic timescale—factors typically neglected in electromechanical models. Consequently, electromagnetic oscillations in PEPS are fundamentally linked to the stability of periodic orbits in high-order NLTP systems, becoming a topic of growing interest across physics and engineering31,32,33,34,35,36.
Many attempts have been made for the electromagnetic dynamic stability evaluation of PEPS. Currently, the most frequently used methods, which involve Park transformation37, dynamic phasor (DP)38, and harmonic state-space (HSS)39, rely on approximately transforming linear time-periodic (LTP) eigenstructure into LTI eigenstructure (Supplementary Notes 3). We categorize these methods within the eigenstructure-reconfiguration (ER) theoretical framework. However, each of these methods has limitations. The Park transformation essentially multiplies system variables by a rotating exponential factor, which restricts its applicability to systems dominated by a single frequency component40. Unlike traditional power systems, where the time-varying inductance matrix of synchronous generators can be simplified into time-invariant differential-algebraic equations37, the multiple interacting frequencies in PEPS make such simplification infeasible—even under balanced grid conditions (see also Supplementary fig. 1). Similarly, while the DP method offers flexibility, it may fail to capture key dynamics, especially in common control scenarios like phase-locked loops (PLLs) involving nested trigonometric expressions. Moreover, HSS utilizes Fourier series and harmonic balance, resulting in the ER-LTI model’s order being the multiplication of system orders, making it impractical for large-scale PEPS due to the curse of dimensionality41. In summary, time-invariant methods struggle to accurately represent the general time-varying behavior of NLTP systems, limiting the broader application of the ER framework in PEPS.
In recent surveys42,43, linear stability analysis with preserved time-varying characteristics holds great potential, especially for large-scale systems. However, its related analysis methods are still immature. Therefore, we focus on the EP theoretical framework of LTP systems—inspired by the Floquet general solution44, which allows for a qualitative analysis directly incorporating time-varying characteristics, as shown in Fig. 1. The EP framework avoids the additional transformation required in the ER framework. Despite its potential, the EP framework has seen limited development, as earlier research primarily addressed small-scale, low-order LTP systems, such as the Mathieu equation, which is fundamental in physics and has only second-order dynamics45. In such cases, the Floquet theory alone is sufficient for stability assessment. In contrast, PEPS with high-order characteristics require new tools to analyze the interaction between stability and system states or parameters. To meet this need, we propose here a generalized LTP participation factor and sensitivity theory, establishing a direct correlation between eigenvalues and time-varying states or parameters. Meanwhile, the proposed LTP participation factor theory enables an accurate and highly efficient analysis (where the ER framework faces trade-offs) of the electromagnetic linear stability in large-scale PEPS. The proposed sensitivity analysis overcomes the lack of its analyticity. Furthermore, it reveals that traditional LTI stability analysis is a special case of our more general LTP theory. Our research can also be extended to dynamic stability analysis in diverse systems with periodic coefficients31,32,33.
Analysis method of LTP systems with EP framework
Unlike LTI systems that focus on stability around a constant equilibrium point, LTP systems assess stability within the neighborhood of periodic orbits. For a stability analysis of linear systems, three key questions must be answered: Is the system stable? Which factors dominate its stability? How do system parameters influence stability? These questions are equally relevant for LTP systems. However, existing analytical methods for LTP systems remain underdeveloped, and only the first question has been solved by Floquet theory44 (see Methods).
To answer the remaining questions, we propose corresponding analysis methods within the EP theoretical framework of LTP systems (Supplementary Notes 4 and Supplementary fig. 2). Answering these questions requires defining the eigenvectors of the LTP system, which are rarely involved. Here, we construct them through the diagonalization and invariance of the system matrix A(t), achieved using a time-periodic matrix R(t). Then, the column vector rk(t) in R(t) and the row vector lk(t) in L(t)=R-1(t) denote the right and left eigenvectors corresponding to the eigenvalue λk, respectively (Supplementary Notes 4).
The correlation between modes and states, i.e., the participation factor, helps address the second key question. The key to correlation assessment lies in establishing a mapping mechanism. Previous interpretations46,47 based on state and mode energy concepts48 have been limited to LTI systems. Hence, we generalize this energy-based interpretation when constructing the LTP system’s participation factor analysis method. Due to the energy summation invariance, the mapping relationship between state energy and mode energy can be expressed by the elements of R(t) and L(t) (Supplementary Notes 4). Therefore, we define the participation factor matrix P as {R(t)⊙LΤ(t)}0, where ⊙ denotes the Hadamard product, and {}0 represents the mean or average value over a period.
The correlation between modes and parameters, i.e., sensitivity, addresses the third question, which can be treated through the gradient of modes to parameters. Unlike traditional approaches that focus solely on time-invariant parameters31,32,33, we explore the significant influence of time-varying components. In real-world systems, this makes the modes functional in mathematics, not just functions of time-varying parameters. Thus, we derive the functional derivative of λk[α(t)] along a direction defined by φ(t), yielding a time-independent value. Then, the sensitivity of a mode to the concerned parameter α(t) is given by {lk(t)(δA(t)/δα(t)|φ(t))rk(t)}0 (Supplementary Notes 4).
Moreover, when A(t) and the parameters are time-invariant, the participation factor and sensitivity reduce to the forms consistent with those of traditional LTI systems (Supplementary Notes 4 and Supplementary fig. 2). Therefore, the analysis method for the LTI system (see Methods) can be regarded as a special case of the proposed LTP system analysis method.
Eigenstructure analysis
We apply the proposed EP-LTP analysis method in real PEPS of the Hami power grid in Northwest China, which is a typical nonlinear time-varying system with a large-scale RPG28 and an LCC-HVDC29. Besides, the details of the utilized models are introduced in Supplementary Notes 2. To fairly compare the proposed EP-LTP analysis method against the traditional ER-LTI in electromagnetic dynamics stability analysis, we model the Hami power grid as a testing system. All tests are completed on a Dell Workstation, 52CPUs with Intel (R) Gold 6230 R, 2.1 GHz.
The eigenvalues and eigenvectors analysis are the prerequisite for further participation factor and sensitivity analysis. We compare the proposed EP-LTP analysis method with a mature LTI model, which performs the best and is most commonly used in terms of accuracy, namely the HSS model39. Besides, the drawbacks of the Park and DP methods are illustrated in detail in Supplementary Notes 3. The HSS model, based on Fourier series approximation and harmonic balance principles, relies on a critical parameter: the truncation number, which determines the maximum harmonic dynamics considered. This parameter is vital for maintaining the model’s accuracy41 and will be discussed in subsequent comparative studies.
To better summarize the quantitative relationship between the order of the small-signal model and computational efficiency, we constructed multiple testing systems (including 813th-order and 2263th-order testing systems), reflecting the dynamics of different power devices (Supplementary Notes 6). The overall comparative results for all testing systems are shown in Fig. 2a. The HSS model requires different truncation numbers for various stability issues, and we report the time costs of the eigenstructure analysis for five truncation numbers (i.e., 2, 4, 8, 16, and 29). For all testing systems, the EP-LTP method consistently has a shorter time cost compared to the HSS model. The time cost is proportional to the third-order polynomial of the model order nc, which is suitable for both EP-LTP and ER-LTI. Therefore, as the system scale expands or the required truncation number increases, the advantages of the proposed EP-LTP method become more prominent. For the testing system of n = 813, the EP-LTP method is even 2000 times faster than traditional ER-LTI methods with a truncation number of 29, which ensures an accurate calculation of the eigenvalues with real parts >−0.8. This real-part threshold is artificially set, and a smaller threshold would allow for a more comprehensive characterization of the electromagnetic dynamics in large-scale PEPS.
To further demonstrate the advantages, we compare the eigenstructure accuracy through the 813th-order and 2263th-order testing systems. Based on the Floquet theory, the eigenvalues of the EP-LTP method are reliable within the range of numerical calculation accuracy. Therefore, Fig. 2b shows the accuracy rate between the ER-LTI eigenvalues with different truncation numbers and EP-LTP eigenvalues, considering only eigenvalues with real parts greater than σ (σ = −0.1, −0.4, and −0.8). The accuracy improves with higher truncation numbers for the ER-LTI method, but EP-LTP maintains reliable accuracy with lower computational demands. Besides, based on Supplementary Notes 3, the Fourier coefficients of the EP-LTP eigenvectors are theoretically equivalent to the eigenvectors of the HSS model. Consequently, the accuracy of the ER-LTI eigenvectors still relies on the same truncation number required for eigenvalue accuracy (Fig. 2c). Since the ER-LTI method demands higher truncation numbers for accuracy, it will result in exceptionally computationally expensive effort, showing the significant advantage of the proposed EP-LTP method.
Participation factor analysis
We apply the proposed eigenvectors to the EP-LTP participation factor analysis. Besides, the ER-LTI participation factor analysis results are taken as the comparison object, which comes from the HSS model and traditional LTI participation factor analysis methods. Since the ER-LTI participation factors characterize the contribution of Fourier coefficients of each state to each eigenvalue, synthesis processing is required (Supplementary Notes 4).
We take the 813th-order and 2263rd-order testing systems for comparative analysis. The time required to calculate the participation factors includes the time spent on both eigenstructure and participation factor calculations. As shown in Fig. 3a, although the proposed LTP participation factors offer no significant advantage over the ER-LTI participation factors in the 813th-order testing system, there is a huge difference in the time cost for calculating the eigenstructure (see Supplementary Table 1). Since the eigenstructure calculation is essential for a participation factor analysis, the proposed LTP method overall demonstrates a significant efficiency advantage. Then, the participation factor error between ER-LTI and EP-LTP is shown in Fig.3b, where only eigenvalues with the real parts greater than −0.8 are considered (Supplementary Notes 6 defines participation factor error). As the truncation number increases, the participation factor error decreases (see Supplementary Table 2 for comparison results of all testing systems). To ensure an accurate participation factor analysis for all eigenvalues, the ER-LTI method requires a truncation number greater than 29, significantly increasing computation time.
For the weaker damping modes of concern of both testing systems, the dominant units based on the LTP participation factors are given in Fig. 3c. In the 813th-order testing system, the dominant units include RPG and LCC-HVDC. The necessity of a larger truncation number (i.e., here is larger than 29) for ER-LTI, due to LCC-HVDC participation, highlights why standard LCC-HVDC models49 from IEEE working groups often fail to capture key electromagnetic stability issues (also see Supplementary fig. 4). In contrast, in the 2263th-order testing system, where the dominant units are only RPG, a lower truncation number can meet the requirements for an accurate analysis.
Sensitivity analysis
In large-scale PEPS, thousands of system parameters influence electromagnetic dynamic stability. For the influence of system parameters on linear stability, due to the lack of sensitivity analyticity, the traditional method mainly relies on the eigenvalue locus under the studied parameter range, which is commonly used in both the LTP model and the reasonably truncated HSS model. However, the traditional sensitivity analysis, which essentially involves traversing parameters, lacks a global understanding of system parameters and has poor efficiency. Here, the proposed EP-LTP sensitivity analysis method offers a more efficient way to assess the impact of massive system parameters on linear stability. Since sensitivity describes the change along the tangential direction of the eigenvalue locus, we use the eigenvalue locus to evaluate the correctness of the proposed LTP sensitivity analysis method.
Unlike LTI systems that only contain time-invariant parameters, LTP system parameters can generally be divided into two categories: time-invariant and periodic time-varying. For the time-invariant system parameters, the sensitivity analysis involves a function calculation. As shown in Fig. 4a, the impact of various control parameters on dominant eigenvalues is depicted. The eigenvalues motion trend from the sensitivity analysis is consistent with the actual eigenvalue motion. For the periodic time-varying system parameters, sensitivity analysis is more complex and involves functional calculations. Here, we examine the sensitivity of a time-varying voltage vector under unbalanced voltage conditions of power sources, reflecting real-world engineering scenarios. As shown in Fig. 4b, the impact of positive and negative sequence voltages on system stability varies. For the mode λc21, reducing the negative sequence voltage or increasing the positive sequence voltage will improve mode damping. However, for the mode λc22, increasing the negative sequence voltage will improve mode damping. The trends in system damping observed through LTP sensitivity analysis match the actual system damping variations. Therefore, the proposed LTP sensitivity analysis can successfully be applied to time-varying and time-invariant system parameters.
Results and discussion
In this paper, we present numerical calculation analysis methods for electromagnetic dynamic stability of large-scale PEPS within the eigenstructure-preserved linear time-periodic (EP-LTP) framework, which go substantially beyond existing EP-LTP approaches that only answer whether the system is stable. Our significant contributions include the development of the EP-LTP participation factor and sensitivity analysis methods, which answer the critical questions: which factors dominate system stability, and how parameters affect system eigenvalues. By comparing large-scale Hami PEPS in China, the proposed EP-LTP method produces more accurate results and lower computational costs in dominant factors analysis than the best-performing linear modeling and analysis methods in the eigenstructure-reconfiguration linear time-invariant (ER-LTI) framework, which are better suitable for small-scale systems. In this paper, the EP-LTP method can be more than 2000 times faster than the ER-LTI method. It should be noted that the multiplication amplification of small-signal model order in the ER-LTI framework is the intuitive reason for its computationally expensive effort. For the last question, the proposed EP-LTP method overcomes the lack of its analyticity, offers detailed insights into the influence of parameters on linear stability, and avoids the poor efficiency in traditional traversal methods, such as the eigenvalues locus. The generalizability of the proposed approach enables its direct application to other practical PEPSs, such as the 44–58 Hz oscillation event recorded in the Zhangbei project in China from 2020 to 202250,51 (Supplementary fig. 3). This system integrates large-scale wind and solar generation with modular multilevel converter-based HVDC technology, representing a model for future power systems that rely entirely on power electronic devices. Furthermore, the analysis method of the LTI system is theoretically a special case of the proposed method.
However, the proposed method still faces two main challenges. The first is the growing need for real-time online analysis of the electromagnetic dynamics in large-scale PEPS. While the EP method significantly enhances both accuracy and computational efficiency compared to the widely used ER approach under typical hardware conditions, real-time analysis remains a substantial challenge and does not yet meet the operational requirements of power systems. Specifically, existing standards for electromechanical dynamics require rolling calculations to be completed within 15 minutes52, a timeframe that would need to be significantly shorter to accommodate the faster timescales of electromagnetic phenomena. Thus, further research into advanced numerical algorithms is essential to make real-time applications of the proposed method feasible. The second challenge involves the control of electromagnetic dynamic stability in PEPS. Beyond fast computation, effectively damping and suppressing electromagnetic oscillations is crucial to ensuring system reliability53. While the proposed LTP-based participation and sensitivity analysis provide a strong theoretical basis for identifying optimal controller locations and tuning parameters, further work is needed to translate these insights into practical control strategies and develop effective stabilizers for electromagnetic oscillations.
Although our methods were developed for electromagnetic dynamic stability analysis in PEPS, the underlying framework is broadly applicable. This generality stems from the fact that periodicity is a fundamental and elegant principle in nature, and many physical and engineering systems evolve under its influence. For instance, periodic behavior underlies the dynamics of accelerator operation, flow around circular cylinders, or aircraft design processes31,34,35,36. As long as the system’s behavior can be modeled as the linear stability of NLTP systems or the dynamic stability of LTP systems, the proposed EP-LTP framework can be effectively applied for stability analysis. Looking to the future, we anticipate that as research shifts from small-scale to large-scale periodic systems, the LTP theory will find increasing relevance in a broad range of domains, such as biological systems54 and chemical systems55, supporting their stable operation and further development.
Methods
Linearization
The concept of linear stability addresses the dynamics near the steady-state of nonlinear systems, whose nonlinearity is described by
$$\dot{{{\bf{y}}}}(t)={{\bf{f}}}(t;{{\bf{y}}}(t))$$
(1)
where the vector y(t)=[ y1(t), y2(t),…, yn(t)]T describes the n states at time t, f = [ f1, f2,…, fn]T denotes the vector field.
The steady-state form yss(t) satisfies ẏss(t)=f(t; yss(t)). Applying Lyapunov’s first method, the dynamics of the disturbance x(t)=y(t)-yss(t) can be expressed as
$$\left.\dot{{{\bf{x}}}}(t)=\dot{{{\bf{y}}}}(t)-{\dot{{{\bf{y}}}}}_{\mathrm{ss}}(t)={\frac{\partial {{\bf{f}}}}{\partial {{\bf{y}}}}}\right|_{{{{\bf{y}}}}_{\mathrm{ss}}(t)}{{\bf{x}}}(t)+{o}({{\bf{x}}}(t))$$
(2)
where ο(x(t)) denotes higher-order infinitesimals and is neglected in the linearized system; ∂f/∂y is the Jacobian matrix of f concerning states y.
When the steady-state form is a periodic orbit (i.e., yss(t)=yss(t + T ), and T is the period), the linear stability is determined by LTP systems, when yss is a special periodic orbit, such as a constant value or equilibrium point, LTI systems determine the linear stability.
LTI eigenstructure
With the constant state matrix A, ẋ(t)=Ax(t) represents the LTI systems. Assuming that A has n distinct (acceptable in most real-world systems) eigenvalues {λk, k = 1,2,…,n}, the eigenvalue matrix could be represented as Λ=R-1AR=diag(λ1, λ2,…, λn). The time-invariant matrices R and L = R-1 are the right and left eigenvector matrices. Furthermore, rk in R = [r1, r2,…, rn] and lk in L = [l1; l2;…; ln] denote the right and left eigenvectors corresponding to the eigenvalue λk. The eigenvalue and eigenvector calculations are performed using MATLAB.
LTI participation factor
The time-invariant eigenvectors form the decoupled coordinate transformation between states and modes. Various perspectives have been proposed to evaluate the correlation between modes and states, yet they predominantly lead to a unified form of the participation factor matrix, that is P = R⊙LT (Supplementary Notes 4). This formula exclusively incorporates the information of eigenvectors, thus enabling direct derivation of the participation factors after calculating the LTI eigenstructure.
LTI sensitivity
LTI sensitivity8 establishes the correlation between modes and parameters in LTI systems, which is derived as ∂λk/∂α=lk(∂A/∂α)rk. Apart from the eigenvectors, it requires the partial derivative of the state matrix with respect to the parameter α. In nonlinear dynamic systems, the state matrix is the function of steady states and parameters10, i.e., A(yss, α).
Therefore, there are two pathways through which α influences A: a direct path, denoted by AE, where α is explicitly incorporated into A, and an indirect path, denoted by AI, where α is implicitly incorporated into A. Considering these two pathways, ∂A(yss, α)/∂α is
$$\left.\frac{\partial {{\bf{A}}}({{{\bf{y}}}}_{\mathrm{ss}},\alpha )}{\partial \alpha }=\frac{\partial {{{\bf{A}}}}_{\mathrm{E}}}{\partial \alpha }+{\frac{\partial {{{\bf{A}}}}_{\mathrm{I}}}{\partial {{\bf{y}}}}}\right|_{{{{\bf{y}}}}_{\mathrm{ss}}}\frac{\partial {{{\bf{y}}}}_{\mathrm{ss}}}{\partial \alpha }$$
(3)
where both ∂AE/∂α and ∂AI/∂y could be derived using the Symbolic Math Toolbox of MATLAB56. In the case where the steady-state form represents an equilibrium point, the partial derivative of the equilibrium points yss concerning the parameter α is described by
$$\left.\frac{\partial {{{\bf{y}}}}_{\mathrm{ss}}}{\partial \alpha }=-{{{\bf{A}}}}^{-1}{\frac{\partial {{\bf{f}}}}{\partial \alpha }}\right|_{{{{\bf{y}}}}_{\mathrm{ss}}}$$
(4)
Applying these procedures to LTI systems enables the computation of sensitivities.
LTP eigenvalue calculation
The general solution of Floquet theory is not analytical; therefore, numerical computation is required to obtain the LTP eigenvalues. First, the linear systems’ state transition matrix (STM) satisfies
$$\dot{{{\boldsymbol{\Phi }}}}(t,0)={{\bf{A}}}(t){{\boldsymbol{\Phi }}}(t,0)$$
(5)
where Φ(t, 0) could map the states value from x(0) to x(t), that is x(t) = Φ(t, 0)x(0).
Based on the Floquet theory, we have Φ(t, 0) = P(t)eQt for LTP systems. With the boundary conditions P(0) = P(T) = I (I denotes the identity matrix), the time-invariant matrix Q is related to the STM Φ(T, 0) (i.e., Φ(T, 0) = eQT). Hence, how to obtain the Φ(T, 0) is the focus of the Floquet theory-based LTP eigenvalue calculation. Here, we apply the discrete exponential expansion method57, where Φ(T, 0) is divided into the product of a series of STM.
$${{\boldsymbol{\Phi }}}(T,0)=\mathop{\prod }\limits_{K=1}^{{N}_{d}}{{\boldsymbol{\Phi }}}(K\Delta t,(K-1)\Delta t)$$
(6)
Within each interval (i.e., (K−1)Δt ~ KΔt, Δt = T/Nd, Nd represents the number of intervals), the variation of the state matrix A(t) is neglected, and the STM of this interval could be calculated by
$${{\boldsymbol{\Phi }}}(K\Delta t,(K-1)\Delta t)={e}^{{{\bf{A}}}({t}_{K-1})\Delta t},{t}_{K-1}=(K-1)\Delta t$$
(7)
The matrix exponential replaces numerical integration to accelerate the calculation process. The accuracy could be ensured by appropriately selecting Nd. Parallel computation could further accelerate the matrix exponential calculation at various time intervals. The matrix Q is then calculated by ln[Φ(T, 0)]/T. The eigenstructure decomposition of Q eventually leads to the LTP eigenvalues.
Choice of computational language
The proposed method was implemented in MATLAB, chosen for its robust support for matrix-based computations and its widespread use within the power systems research community. For the stability analysis, we utilized MATLAB’s Symbolic Math Toolbox and Linear Algebra Toolbox. The functions in the Linear Algebra Toolbox are built on optimized LAPACK routines, ensuring high computational efficiency and numerical accuracy. In addition, the system model was structured to allow for easy data export and direct application of LAPACK-based tools if needed.
Data availability
The source data underlying Figs. 2–4, and Supplementary Fig. 4 are available at Figshare58. The authors declare that the data supporting the findings of this study are available within the supplementary information files and Figshare. Source data are provided with this paper.
Code availability
The codes that support the findings of this study are available at Figshare58.
References
Kundur, P., Balu, N. J. & Lauby, M. G. Power system stability and control, 7 (McGraw-Hill, New York, 1994)
Machowski, J., Bialek, J. W. & Bumby, J. R. Power system dynamics: stability and control (Wiley, 2008).
Menck, P. J., Heitzig, J., Marwan, N. & Kurths, J. How basin stability complements the linear-stability paradigm. Nat. Phys. 9, 89–92 (2013).
Article
CAS
Google Scholar
4. Motter, A. E., Myers, S. A., Anghel, M. & Nishikawa, T. Spontaneous synchrony in power-grid networks. Nat. Phys. 9, 191–197 (2013).
Article
CAS
Google Scholar
5. Wong, D. Y. et al. Eigenvalue analysis of very large power systems. IEEE Trans. Power Syst. 3, 472–480 (1988).
Article
ADS
Google Scholar
6. Verghese, G. C., Perez-Arriaga, I. J. & Schweppe, F. C. Selective modal analysis with applications to electric power systems, part II: the dynamic stability problem. IEEE Trans. Power Appl. Syst. 101, 3126–3134 (1982).
Article
ADS
Google Scholar
7. Hassan, L. H. et al. Optimization of power system stabilizers using participation factor and genetic algorithm. Int. J. Electr. Power Energy Syst. 55, 668–679 (2014).
Article
Google Scholar
8. Van, N. essJ., Boyle, J. & Imad, F. Sensitivities of large, multiple-loop control systems. IEEE Trans. Autom. Control. 10, 308–315 (1965).
Article
Google Scholar
9. Witthaut, D. et al. Collective nonlinear dynamics and self-organization in decentralized power grids. Rev. Mod. Phys. 94, 015005 (2022).
Article
ADS
MathSciNet
Google Scholar
10. Wang, K. W., Chung, C. Y. & Tse, C. T. Multimachine eigenvalue sensitivities of power system parameters. IEEE Trans. Power Syst. 15, 741–747 (2000).
Article
ADS
Google Scholar
Macilwain, C. Energy: super grid. Nature 468, 624–625 (2010).
Article
ADS
CAS
PubMed
Google Scholar
12. Department of Energy. Building a better grid initiative. (2022).
13. Mallapaty, S. How China could be carbon neutral by mid-century. Nature 586, 482–484 (2020).
Article
ADS
CAS
PubMed
Google Scholar
14. Schäfer, B. et al. Non-Gaussian power grid frequency fluctuations characterized by Lévy-stable laws and superstatistics. Nat. Energy 3, 119–126 (2018).
Article
ADS
Google Scholar
15. Sajadi, A., Kenyon, R. W. & Hodge, B. M. Synchronization in electric power networks with inherent heterogeneity up to 100% inverter-based renewable generation. Nat. Commun. 13, 2490 (2022).
Article
ADS
CAS
PubMed
PubMed Central
Google Scholar
16. Yuan, X., Hu, J. & Cheng, S. Multi-time scale dynamics in power electronics-dominated power systems. Front. Mech. Eng. 12, 303–311 (2017).
Article
Google Scholar
17. Statistical data of China’s power industry 2023 (National Energy Administration). 2023
18. Global Energy Interconnection Development and Cooperation Organization (GEIDCO). Research Reports on China’s CO2 Emission Peak and Carbon Neutrality Released in Beijing. (2021).
19. Wang, Y. et al. Accelerating the energy transition towards photovoltaic and wind in China. Nature 619, 761–767 (2023).
Article
ADS
CAS
PubMed
PubMed Central
Google Scholar
20. Mulawarman, A., Mysore, P. Detection of undamped sub-synchronous oscillations of wind generators with series compensated lines. Minnesota Power Systems Conference. 1–5 (2011).
21. Irwin, G. D., Jindal, A. K., & Isaacs, A. L. Sub-synchronous control interactions between type 3 wind turbines and series compensated AC transmission systems. IEEE Power & Energy Society General Meeting, 1–6 (2011).
22. Wang, L. et al. Investigation of SSR in Practical DFIG-Based Wind Farms Connected to a Series-Compensated Power System. IEEE Trans. Power Syst. 30, 2772–2779 (2015).
Article
ADS
Google Scholar
23. Liu, H. et al. Sub-synchronous interaction between direct-drive PMSG based wind farms and weak AC networks. IEEE Trans. Power Syst. 32, 4708–4720 (2017).
Article
ADS
Google Scholar
24. Wang, X. & Blaabjerg, F. Harmonic stability in power electronic-based power systems: Concept, modeling, and analysis. IEEE Trans. Smart Grid 10, 2858–2870 (2018).
Article
Google Scholar
25. Chi, Y. et al. Overview of mechanism and mitigation measures on multi-frequency oscillation caused by large-scale integration of wind power. CSEE J. Power Energy Syst. 5, 433–443 (2019).
Google Scholar
26. Shair, J., Li, H., Hu, J. & Xie, X. Power system stability issues, classifications and research prospects in the context of high-penetration of renewables and power electronics. Renew. Sustain. Energy Rev. 145, 111111 (2021).
Article
Google Scholar
27. Zou, C. et al. Analysis of resonance between a VSC-HVDC converter and the AC grid. IEEE Trans. Power Electron. 33, 10157–10168 (2018).
Article
ADS
Google Scholar
28. Guan, L., Yao, J., Liu, R., Sun, P. & Gou, S. Small-signal stability analysis and enhanced control strategy for VSC system during weak-grid asymmetric faults. IEEE Trans. Sustain. Energy 12, 2074–2085 (2021).
Article
ADS
Google Scholar
29. Du, B. et al. Small-signal modeling of LCC-HVDC considering switching dynamics based on the Linear Time-Periodic (LTP) method. IEEE Trans. Power Del. 39, 2715–2728 (2024).
Article
Google Scholar
30. Zhu, J., Hu, J., Wang, S. & Wan, M. Small-signal modeling and analysis of MMC under unbalanced grid conditions based on linear time-periodic (LTP) method. IEEE Trans. Power Del. 36, 205–214 (2021).
Article
Google Scholar
31. Giannetti, F., Camarri, S. & Citro, V. Sensitivity analysis and passive control of the secondary instability in the wake of a cylinder. J. Fluid Mech. 864, 45–72 (2019).
Article
ADS
MathSciNet
CAS
Google Scholar
32. Lasagna, D. Sensitivity analysis of chaotic systems using unstable periodic orbits. SIAM J. Appl. Dyn. Syst. 17, 547–580 (2018).
Article
MathSciNet
Google Scholar
33. Tamer, A. & Masarati, P. Generalized quantitative stability analysis of time-dependent comprehensive rotorcraft systems. Aerospace 9, 10 (2021).
Article
Google Scholar
34. England, R. et al. Dielectric laser accelerators. Rev. Mod. Phys. 86, 1337–1389 (2014).
Article
ADS
CAS
Google Scholar
35. Wang, J., Li, R. & Peng, X. Survey of nonlinear vibration of gear transmission systems. Appl. Mech. Rev. 56, 309–329 (2003).
Article
ADS
Google Scholar
36. Lopez, M. & Prasad, J. Linear time invariant approximations of linear time periodic systems. J. Am. Helicopter Soc. 62, 1–10 (2016).
Article
CAS
Google Scholar
37. Park, R. H. Two-reaction theory of synchronous machines generalized method of analysis-part I. Trans. Am. Inst. Electr. Eng. 48, 716–727 (1929).
Article
Google Scholar
38. Sanders, S. R., Noworolski, J. M., Liu, X. Z. & Verghese, G. C. Generalized averaging method for power conversion circuits. IEEE Trans. Power Electron. 6, 251–259 (1991).
Article
ADS
Google Scholar
39. Wereley, N. M. Analysis and control of linear periodically time-varying systems. Ph.D. dissertation, Department of Aeronautics and Astronautics, MIT, 217–223 (1991).
40. O’Rourke, C. J., Qasim, M. M., Overlin, M. R. & Kirtley, J. L. A Geometric interpretation of reference frames and transformations: dq0, Clarke, and Park. IEEE Trans. Energy Convers. 34, 2070–2083 (2019).
Article
ADS
Google Scholar
41. Zhu, J. et al. Truncation number selection of the harmonic state-space model based on the Floquet characteristic exponent. IEEE Trans. Ind. Electron. 70, 3222–3228 (2022).
Article
Google Scholar
42. Li, H. et al. A time-domain stability analysis method for grid-connected inverters with PR control based on Floquet theory. IEEE Trans. Ind. Electron. 68, 11125–11134 (2020).
Article
Google Scholar
43. Golestan, S. et al. LTP modeling and stability assessment of multiple second-order generalized integrator-based signal processing/synchronization algorithms and their close variants. IEEE Trans. Power Electron. 37, 5062–5077 (2021).
Article
ADS
Google Scholar
44. D’Angelo, H. Linear time-varying systems: analysis and synthesis. Boston, MA, USA: Allyn and Bacon (1970).
45. McLachlan. N. W. Theory and Application of Mathieu Functions. (Oxford University, New York, 1964).
46. Hamdan, A. M. A. Coupling measures between modes and state variables in power-system dynamics. Int. J. Control 43, 1029–1041 (1986).
Article
Google Scholar
47. Iskakov, A. B. Definition of state-in-mode participation factors for modal analysis of linear systems. IEEE Trans. Autom. Control 66, 5385–5392 (2021).
Article
MathSciNet
Google Scholar
48. MacFarlane, A. G. J. Use of power and energy concepts in the analysis of multivariable feedback controllers. Proc. Inst. Electr. Eng. 116, 1449–1452 (1969).
Article
Google Scholar
49. Szechtman, M., Wess, T., & Thio, C. V. A benchmark model for HVDC system studies. IET ACDC. 374-378 (1991).
50. Li, H. Z., Shair, J., Zhang, J. Q. & Xie, X. R. Investigation of subsynchronous oscillation in a DFIG-based wind power plant connected to MTDC grid. IEEE Trans. Power Syst. 38, 3222–3231 (2023).
Google Scholar
51. Zheng, S. et al Engineering data analysis of sub/super-synchronous oscillation between wind farm and MMC in Zhangbei project. In IEEE 6th Conference on Energy Internet and Energy System Integration (EI2), 11–13 (2022).
52. Savulescu, S. C. Real-time stability assessment in modern power system control centers. (Wiley-IEEE, 2009).
53. Demello, F. P. & Concordia, C. Concepts of synchronous machine stability as affected by excitation control. IEEE Trans. Power Appl. Syst. 88, 316–329 (1969).
Article
ADS
Google Scholar
54. Ushio, M. et al. Fluctuating interaction network and time-varying stability of a natural fish community. Nature 554, 360–363 (2018).
Article
ADS
CAS
PubMed
Google Scholar
55. Van Laake, L., Lüscher, T. & Young, M. The circadian clock in cardiovascular regulation and disease: lessons from the Nobel Prize in Physiology or Medicine 2017. Eur. Heart J. 39, 2326–2329 (2018).
Article
PubMed
Google Scholar
56. Lynch, S. Dynamical systems with applications using MATLAB (Birkhäuser, Boston, 2004).
57. Friedmann, P., Hammond, C. E. & Woo, T. H. Efficient numerical treatment of periodic systems with application to stability problems. Int. J. Numer. Methods Eng. 11, 1117–1136 (1977).
Article
MathSciNet
Google Scholar
58. Hu, J. et al. Datasets for electromagnetic dynamic stability analysis of power electronics-dominated systems using eigenstructure-preserved LTP theory. Figshare
59. Shuai, Z. et al. Parameter stability region analysis of islanded microgrid based on bifurcation theory. IEEE Trans. Smart Grid 10, 6580–6591 (2019).
Article
Google Scholar
60. Harnefors, L. Modeling of three-phase dynamic systems using complex transfer functions and transfer matrices. IEEE Trans. Ind. Electron. 54, 2239–2248 (2007).
Article
Google Scholar
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Acknowledgements
J.H., Z.G., J.Z., B.D., Z.W., and G.Z. were supported by the National Science Foundation of China under Grant 52225704, Grant U22B20122, and Grant 52107196. Y.H. and J.Z. were supported in part by the Theme-based Research Scheme from the Research Grants Council, Hong Kong SAR, under Grant T23-713/22-R, and in part by Collaborative Research Fund from the Research Grants Council, Hong Kong SAR, under Grant C1052-21GF. Besides, we gratefully acknowledge master’s student Jiali Kui for their assistance with data analysis and information collection.
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These authors contributed equally: Jiabing Hu, Zeren Guo, Jianhang Zhu.
Authors and Affiliations
State Key Laboratory of Advanced Electromagnetic Technology, School of Electrical and Electronic Engineering, Huazhong University of Science and Technology, Wuhan, China
Jiabing Hu, Zeren Guo, Buyang Du, Zefei Wu, Guojie Zhao & Shijie Cheng
2. Department of Electrical and Electronic Engineering, The University of Hong Kong, Hong Kong, SAR, China
Jianhang Zhu & Yunhe Hou
3. Institute of Physics, Humboldt-University, Berlin, Germany
Jürgen Kurths
4. Huawei Technologies Co., Ltd, Shanghai, China
Yunfeng Liu & Kai Xin
5. China Electric Power Research Institute, Beijing, China
Jianbo Guo
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J.H., Z.G., and J.Z. conceived the idea, led the project, conducted the theoretical analysis, and wrote the initial manuscript and revisions. J.K. and Y.H. contributed ideas and methods for the theoretical analysis and participated in the writing of the paper. B.D., Z.W., G.Z., Y.L., and K.X. also conducted a theoretical analysis and contributed to the manuscript. J.G. and S.C. provided guidance to all co-authors throughout the process.
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Hu, J., Guo, Z., Zhu, J. et al. Electromagnetic dynamic stability analysis of power electronics-dominated systems using eigenstructure-preserved LTP Theory.
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11691 | https://www.fs.usda.gov/wildflowers/plant-of-the-week/marchantia_polymorpha.shtml | Marchantia
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Marchantia polymorpha range map. USDA PLANTS Database.
Underside of male gametophore. Photo by David D. Taylor.
Group of female gametophores. Photo by David D. Taylor.
Underside of female gametophore. Photo by David D. Taylor.
Group of tightly packed male gametophores. Photo by David D. Taylor.
Gemmae cups with gemmae. Photo by David D. Taylor.
Topside of male gametophores. Photo by David D. Taylor.
Marchantia (Marchantia polymorpha L.)
By David Taylor
Marchantia is a member of the Marchantiaceae, the Marchantia family. This family is one of many thalloid liverwort families. A thalloid liverwort is strap-like and often forms large colonies on the surface on which it grows. A liverwort is nonvascular green plant. It does not have phloem and xylem like ferns, conifers and flowering plants to transport water, food and minerals through the plant. As a result, liverworts tend to be small, only a few cells thick and grow in moist locations or in locations that dry out, but receive a lot of precipitation or fog. Many species of liverworts are found in the U.S. and Canada. Marchantia is among the largest and is often used in biology labs.
This species hugs the substrate on which it grows, usually moist soil or wet rocks. The thallus segments are dichotomously branched (continually branched in two branches) and are green to dark green. The segments are 10 to 15 millimeters (0.4 to 0.6 inches) wide, and frequently form large mats. Inspection with a hand lens will reveal that the surface of the thallus is covered with tiny pores. These are holes to allow gas exchange between the inside of the thallus and the atmosphere. Frequently, small saucer-like structures can be found inside of which are small green lentil-shaped objects. The saucer is a gemma cup and the lentil-shaped objects are gemmae. Gemmae allow the plant to reproduce asexually. Each gemma is capable of growing into a new thallus if the conditions are correct. Rain drops, insects, and small mammals all can spread the gemmae in the environment.
Occasionally, one can find parasol-like structures that stand upwards of 2.5 centimeters (1 inch) above the thallus. If the structure is segmented and resembles a palm tree, it is growing on a female plant and the structure (a gametophore) will produce female gametes. If it is only slightly segmented, more like the outline of a doily, it is growing on a male plant, and the structure (gametophore) will produce male gametes.
Marchantia is found on moist, usually neutral or basic soils or on wet rock, such as along a perennial stream. It occasionally shows up on mineral soils in depressions or in the shade of fallen logs following hot forest fires. It is cosmopolitan in its distribution, and is probably in every state and province. In many locations is considered a weed. It is often associated with human activity.
This is the only native genus of thalloid liverworts that produces gemmae cups.
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31.4 Worked Example - Atwood Machine
31.4 Worked Example - Atwood Machine
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11693 | https://www.youtube.com/watch?v=Bwih7_AT1oI | Error Detecting Code : Parity Explained | Odd Parity and Even Parity
ALL ABOUT ELECTRONICS
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Description
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Posted: 28 Aug 2021
In this video, the parity is explained and the use of parity bit in the error detection is explained with examples.
The following topics covered in the video:
0:00 Introduction
1:00 What is Parity? Odd and Even Parity
2:00 Error Detection using Parity Bit
5:03 Limitation of Parity
What is Parity?
Parity Bit is a simple form of error detecting code, where the additional bit is added along with the data bits.
There are two types of Parity:
1) Odd Parity
2) Even Parity
Odd Parity:
In Odd parity, the parity bit is set in such a way that the total number of 1s in the code, including the parity bit is Odd.
Even Parity:
In Even parity, the parity bit is set in such a way that the total number of 1s in the code, including the parity bit is Even.
For error detection, the parity bit is appended with data bits, and the code is sent to the receiver.
At the receiver, the parity checker circuit checks the parity of the received code and detects the error.
Limitations of Parity Bit:
The Parity bit can detect only odd numbers of errors. If there are even a number of errors in the received code, it will remain undetected.
Moreover, the Parity bit can just detect the error in the code (if the number of errors in the received code is odd) but it can't correct the error.
The link for the other useful videos:
1) Gray Code:
2) BCD Code:
This video will be helpful to all the students of science and engineering in understanding the use of Parity in error detection.
ALLABOUTELECTRONICS
Parity
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Introduction Hey friends, welcome to the YouTube channel
ALL ABOUT ELECTRONICS. So, in this video, we will learn about the
error detecting code, and particularly, we will learn about the parity bit and the parity
checking for the error detection. So, when the data is transmitted between the
two digital devices then due to the external noise or the interference, sometimes a few
bits get corrupted in the data. And because of that, at the receiver side,
the one is received as zero and the zero is received as one. So, for example, the transmitter is sending
1010 or 10 in decimal, but at the receiver, if there is an error in one bit, then it might
be received as 1000. Or in decimal, it will be received as 8. So, because of this error in 1 bit, the entire
data get changed. So, to detect such errors many times, some
additional bits are also sent along with the What is Parity? Odd and Even Parity data bits. And with the help of the additional bits,
it is possible to detect the error in the received data. So, the value of these additional bits depends
on these data bits as well as the type of the error-detecting code, which is used during
the transmission. So, with the help of these error detecting
codes, it is possible to detect the error in the received data. So, there are many different methods for this
error detection. But the simplest one is to add the parity
bit along with the data. So, there are two types of parity. That is odd parity and the even parity. So, in case of this odd parity, including
the parity bit, the total number of 1s in the code should be equal to odd. For example, let's say, there are 7 data bits
and 1 parity bit. So, in case of the odd parity, the parity
bit should be such that, the total number of 1s in the code should be equal to odd. Error Detection using Parity Bit So, here as you can see, the total number
of 1s in these data bits is equal to 4. That means this parity bit should be equal
to 1, so that, the total number of 1s in this code remains odd. So, let's take another example. So, in this case, a total number of 1s in
these data bits is equal to 5. Therefore, this parity bit should be equal
to 0, so that, the total number of 1s in the code remains odd. Similarly, in case of the even parity, the
total number of 1s in the code should be equal to even. For example, in this code, there is 5 number
of 1s in these data bits. That means this parity bit should be equal
to 1, so that, the total number of 1s in this code remains even. so, let's take another example. So, in this case, as you can see, the total
number of 1s in this data bits is equal to 4. That means this parity bit should be equal
to 0, so that, the total number of 1s in the code remains even. On the other end, if we use the odd parity,
then for the same data bits, this parity bit should be equal to 1. So, that the total number of 1s in the code
remains odd. So, in this way, for the error detection,
the parity bit is appended along with the data bits, and the entire code will be sent
from the transmitter side to the receiver side. So, now let's see, with the help of this parity
bit how we can detect the error. And for a time being, let's consider this
odd parity. So, let's say, these are the data bits, and
for the odd parity, this parity bit should be equal to 1. Now, when this code is transmitted to the
receiver side, then let's say, because of the error, this 1 becomes 0. So, now at the receiver, the parity checker
circuit will check the parity of the received code. So, if there is no error, then the parity
should be equal to odd right !! But here, because of the error, the total
number of 1s in the received code will be even. Because, if you see over here, then the total
number of 1s in the code is equal to 4. And since the total number of 1s in the received
code is even so the parity checking circuit will detect the error. Similarly, let's see the case of the even
parity. So, let's say, these are the data bits, and
for the even parity, this parity bit should be equal to zero. Now, when this code is transmitted from the
transmitter to the receiver side, then let's say because of some error, this bit changes
to 1. So, now at the receiver, the parity checker
circuit will check the parity of the received code. So, as you can see, here the total number
of 1s in the received code is equal to 5. And since it is odd, that means there is some
error in the received code. Limitation of Parity So, in this way, with the help of the parity
bit, we can detect the error in the received code. But we can not find the exact location of
the error. Moreover, if there is more than 1 error in
the received code, or to be precise if there is an even number of errors in code, then
with the help of this parity bit, we can not detect the error. So, let's say, there are two errors in the
received code. So, once the code is received then the parity
checker circuit at the receiver side will check the parity of the received code. So, in this case, the total number of 1s in
the code is equal to 6. And since it is even for this even parity,
so the parity checker circuit will not generate any error. And it will consider this received code as
a valid code. But actually, it is an invalid code. Similarly, if there is 4 number of errors
in the received code then after the parity check, the total number of 1s in the received
code is equal to 4. And since it is even for the even parity,
so the parity checker circuit once again will not generate any error. And it will consider this code as a valid
code. But actually, this is invalid code. So, as you can see, whenever there is an even
number of errors in the received code, then with the help of parity, it can not be detected. But if there is an odd number of errors in
the received code, then using the parity bit it is at least possible to detect that error. So, let's say, for some transmitted code,
there are three errors in the received code. So, after the parity check, the total number
of 1s in the received code is equal to 5. And since it is an odd number, so the parity
checker circuit will generate the error, which indicates that there is some error in the
received code. That means the parity bit can only detect
the odd number of errors in the received code. Now, so far in our discussion, we took the
example of even parity. But the same is also true for the odd parity. So, in case of the odd parity also, if there
is an even number of errors in the received code, then it will not get detected. So, let's say, this code is transmitted to
the receiver side. But at the receiver, there are two errors
in the received code. So after the parity check, the total number
of 1s in the received code is equal to 7. And since it is odd, so for the odd parity,
this parity checker circuit will not generate any error. And it will consider this code as a valid
code. But actually, it is invalid code. That means we can't detect the even number
of errors with this parity bit. But if there are 3 errors in the received
code, then the total number of 1s in the received code is equal to 4. And since it is even, so, for this odd parity,
the parity checker circuit will generate the error. So, in short, the parity can only detect the
odd number of errors in the received code. So, once the error is detected then the receiver
sends the acknoledgedment to the transmitter about the error, and it asks for the retransmission
of the same data. So, in this way, with the help of the parity
bit, it is possible to detect the errors. But as I said, we can't find the exact location
of the error. Or in other words, it is not possible to correct
the error. But there are some codes, using which it is
also possible to even correct the errors. So, in the next video, we will learn about
one of such error-correcting codes, and in the upcoming videos, we will also see the
circuit of the parity generator as well as the parity checker. But I hope, in this video you understood about
the parity bit, and using this parity bit, how it is possible to detect the error. So, if you have any questions or suggestions,
then do let me know here in the comments section below. If you like this video, hit the like button
and subscribe to the channel for more such videos. |
11694 | https://brianmcfee.net/dstbook-site/content/ch04-complex/Basic-Operations.html | # Digital Signals Theory
Theme by the Executable Book Project
Contents
Basic operations
Contents
4.2. Basic operations#
4.2.1. Addition and subtraction#
If we have two complex numbers (\red{a} + \mathrm{j}\purple{b}) and (\red{c} + \mathrm{j}\purple{d}), their sum is defined by independently summing the real parts ((\red{a+c})) and the imaginary parts ((\purple{b+d})):
[(\red{a} + \mathrm{j}\purple{b}) + (\red{c} + \mathrm{j}\purple{d}) = (\red{a+c}) + \mathrm{j}(\purple{b+d}).]
Subtraction works similarly:
[(\red{a} + \mathrm{j}\purple{b}) - (\red{c} + \mathrm{j}\purple{d}) = (\red{a-c}) + \mathrm{j}(\purple{b-d}).]
In the complex plane, addition and subtraction can be thought of as displacement. If you imagine drawing an arrows from the origin (0) to the positions of (z) and (w), then you can find the position of (z+w) by picking up one of the arrows and moving its tail to coincide with the head of the other arrow. Subtraction works the same way, except that you would turn the arrow around (180^\circ).
4.2.2. Conjugation#
Complex numbers have a new operation, which doesn’t exist for other systems like reals and integers, called conjugation. The complex conjugate of a number (z = \red{a} + \mathrm{j}\purple{b}), denoted as (\overline{z}), is computed by negating only the imaginary part, and is denoted by an over-line:
[\overline{z} = \overline{\red{a} + \mathrm{j}\purple{b}} = \red{a} - \mathrm{j}\purple{b}.]
Visually, this operation reflects a point across the horizontal (real) axis in the complex plane, as seen below.
Conjugating a number twice reverts back to the original number (just like negating a number twice).
If a number is purely real (has (\purple{b=0})), it lies exactly on the horizontal axis, and it is its own conjugate.
4.2.3. Multiplication (take 1)#
There are two ways to think about multiplication of complex numbers, depending on whether we use rectangular or polar form, which we’ll see below. Both are equally valid, and produce the same results, but sometimes one may be more convenient than the other. Let’s start with the rectangular form.
If we have two complex numbers (z = \red{a} + \mathrm{j}\purple{b}) and (w = \red{c} + \mathrm{j}\purple{d}), we can compute their product by using the rules of algebra and remembering that (\mathrm{j}\cdot\mathrm{j} = -1):
[\begin{align} z \cdot w &= (\red{a} + \mathrm{j}\purple{b}) \cdot (\red{c} + \mathrm{j}\purple{d}) & \text{Use definitions of } z, w\ &= \red{a} \cdot \red{c} + \red{a}\cdot \mathrm{j}\purple{d} + \mathrm{j}\purple{b} \cdot \red{c} + \mathrm{j}\purple{b} \cdot \mathrm{j}\purple{d} & \text{FOIL multiply}\ &= \red{a \cdot c} + \mathrm{j}\left(\red{a}\cdot\purple{d} + \purple{b} \cdot \red{c}\right) + \mathrm{j}^2\left(\purple{b \cdot d}\right) & \text{Pull out imaginary units } \mathrm{j}\ &= \red{a \cdot c} + \mathrm{j}\left(\red{a}\cdot\purple{d} + \purple{b} \cdot \red{c}\right) - \red{b \cdot d} & \mathrm{j}^2 = -1\ &= \left(\red{a \cdot c - b \cdot d}\right) + \mathrm{j}\left( \purple{a \cdot d + b \cdot c}\right) & \text{Collect real and imaginary parts} \end{align}]
This looks complicated – and it is! The key things to take away here are:
The product is well-defined, even if the formula is messy;
When you multiply two complex numbers, their real and imaginary parts interact in a non-trivial way;
When two purely imaginary numbers are multiplied together, in this case (\mathrm{j}\purple{b} \cdot \mathrm{j}\purple{d}), the result is real: (-\red{b\cdot d}).
Example 4.1 (Multiplying complex by real)
Let’s keep (z = \red{a} + \mathrm{j}\purple{b}) and multiply it by a purely real number (\red{x}). If we follow the rules above, we get
[\begin{align} z \cdot \red{x} &= (\red{a} + \mathrm{j}\purple{b}) \cdot \red{x}\ &= \red{a\cdot x} + \mathrm{j}\purple{b\cdot x} \end{align}]
so the real (\red{x}) combines with both the real and imaginary parts of (z).
Example 4.2 (Multiplying complex by imaginary)
What if we multiply (z = \red{a} + \mathrm{j}\purple{b}) a purely imaginary number (\mathrm{j}\purple{x})? In this case, we get
[\begin{align} z \cdot\mathrm{j} \purple{x} &= (\red{a} + \mathrm{j}\purple{b}) \cdot \mathrm{j}\purple{x}\ &= \red{a}\cdot \mathrm{j}\purple{x} + \mathrm{j}\purple{b}\cdot \mathrm{j}\purple{x}\ &= \red{-b\cdot x} + \mathrm{j}\purple{a\cdot x}. \end{align}]
In this case, the real and imaginary parts have exchanged places, and the new real part ((\red{b \cdot x})) has been negated.
Example 4.3 (Multiplying complex by its conjugate)
Finally, what if we multiply a number (z = \red{a} + \mathrm{j}\purple{b}) by its own conjugate (\overline{z} = \red{a} - \mathrm{j}\purple{b})?
In this case, all of the imaginary components cancel each-other out, and the resulting product is purely real (and non-negative):
[\begin{align} z \cdot \overline{z} &= (\red{a} + \mathrm{j}\purple{b}) \cdot (\red{a} - \mathrm{j}\purple{b})\ &= \red{a \cdot a} + \mathrm{j}\purple{b}\cdot \red{a} - \red{a} \cdot \mathrm{j}\purple{b} - \mathrm{j}\purple{b} \cdot \mathrm{j}\purple{b} & \text{FOIL multiply}\ &= \red{a \cdot a}+ \cancel{\mathrm{j}\purple{b}\cdot \red{a}} - \cancel{\red{a} \cdot \mathrm{j}\purple{b}} - \mathrm{j}\purple{b} \cdot \mathrm{j}\purple{b} & \text{Cancel } -\mathrm{j}ab + \mathrm{j}ab = 0\ &= \red{a \cdot a} - \mathrm{j}\purple{b} \cdot \mathrm{j}\purple{b} & \text{Substitute } \mathrm{j}^2 = -1 \ &= \red{a^2 + b^2} \end{align}]
We could say more about complex multiplication, but as we’ll see, it’s easier to think about if we use the polar form. To see how that works, we’ll need to take a slight detour through the exponential function. |
11695 | https://www.vedantu.com/maths/greatest-2-digit-number | Courses for Kids
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Step-by-Step Guide to Finding the Greatest 2 Digit Number
Roman abacuses or stone tokens were used in times when people were not familiar with the number system, which dates back thousands of years. There was a growing demand for larger denominations as time passed and trade between nations and regions advanced. As a result, the number of systems that are used today was developed.
The requirement to manage larger numbers also came out as the nations advanced. We became curious about the separation between the earth and the moon, the speed of light, and the size of microorganisms, which prompted us to develop a larger number of systems. This led to the introduction of the concept of numbers and digits.
It is very simple to count small amounts. Without using a calculator or pen and paper, we can count the fingers on our hand or the petals on a flower very easily. We work with single-digit numbers in these circumstances. One is the smallest, and nine is the largest single-digit counting number. Let us learn what is the greatest number.
What are Digits?
In Mathematics, numbers are represented by digits, which are used as a single symbol. 8 and 9 are two digits in the number 89, for instance. Due to this, the numerals 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 are the digits that are used in our daily lives to represent groups of numbers and perform arithmetic operations.
In Computer Science, the word "digits" should be used instead. These digits are sometimes just referred to as numbers in Mathematics or as numerical digits. One-digit numbers range from zero to nine, with nine being the largest.
Two Digits Numbers
The smallest two-digit number is obtained by adding one unit to the largest one-digit number.
1+9 = 10
The smallest and largest 2-digit numbers are 10 and 99, respectively.
2-Digit Numbers
Now, picture a large auditorium with hundreds of people inside. With this many people, it is impossible for us to count them with our fingers alone. We use numbers with up to three digits to handle such measures.
Place Value in 2-Digit Number
More about 2-Digit Numbers
Some of the tips and tricks on numbers up to 2-digits.
Any one-digit number (from 1 to 9) multiplied by 10 produces a two-digit number as the outcome. For illustration, 4 x 10 = 40.
11 is the smallest two-digit number with just one digit.
10 is the smallest two-digit number with two separate digits. Thus, the smallest two-digit number is 10.
99 is the largest two-digit number with just one digit. So, the largest two-digit number is 99.
In two-digit numbers, the only place values available are ones and tens.
There are 90 "2-digit numbers," numbered from 10 to 99.
Solved Example
Create a two-digit number with 5 in the one's place and 7 in the tens place, for example. Include the number name and the expanded form of the number.
Ans: The number is 75 if 7 is in the tens place and 5 is in the one's place.
75 expanded is equal to 70 plus 5. This can also be written as seven tens plus five ones.
Seventy-five is the name given to the number.
Practice Questions
Q1. What is the difference between the greatest and smallest 2-digit number?
Ans: 89
Q2. What is the biggest two-digit number that is divisible by 10?
Ans: 90
Summary
Only the unit place and the tens place are available for 2-digit numbers. Every number with multiple digits has various digits that are each described by their place values. Two-digit numbers have a range of 0 to 99. So, the smallest 2-digit number is 10, and the largest 2-digit number is 99.
Here we have explained all the facts about the greatest two-digit numbers. Some practice questions are provided for your better understanding, and you can test yourself with the help of these questions.
FAQs on What Is the Greatest 2 Digit Number?
Which is the greatest 3-digit number?
The greatest 3-digit number is 999. In base-10 (decimal) system, a 3-digit number ranges from 100 to 999, where 999 is formed using the highest digit, 9, in all three places:
Hundreds place: 9
Tens place: 9
Units place: 9
Therefore, 999 is the largest integer you can write using three digits.
What is the difference between 76432 and 23467?
To find the difference between 76432 and 23467, simply subtract the smaller number from the larger one: $$76432 - 23467 = 52965$$So, the difference is 52,965. At Vedantu, students can learn and practice such subtraction problems using easy-to-follow techniques and examples.
What is the greatest 4-digit number?
The greatest 4-digit number is 9999. In the decimal system, a 4-digit number begins at 1000 and ends at 9999, where all four places (thousands, hundreds, tens, units) are occupied by the digit 9. Understanding number values like these is a key concept in mathematics and is taught effectively at Vedantu for students across grades.
What is the greatest 2 digit odd number?
The greatest 2-digit odd number is 99. To clarify, two-digit numbers range from 10 to 99. Among them, odd numbers end with 1, 3, 5, 7, or 9. Since 99 ends in 9 (an odd number) and is the highest two-digit number, it is the answer. Vedantu provides targeted practice to help students recognize and work with odd and even numbers with ease.
What is the greatest 2-digit number?
The greatest 2-digit number is 99. In the decimal number system, two-digit numbers start from 10 and end at 99. Here, 99 uses the highest digit (9) in both tens and units places:
Tens place: 9
Units place: 9
Therefore, 99 is the largest possible two-digit number. Vedantu's interactive classes help students master these foundational concepts in number systems.
How do you identify the largest two-digit even number?
The largest two-digit even number is 98. An even number ends in 0, 2, 4, 6, or 8. While 99 is the greatest two-digit number overall, 98 is the highest that fulfills the condition of being even. Vedantu offers worksheets and guidance for students to easily distinguish between even and odd numbers.
What are the steps to compare two-digit numbers?
To compare two-digit numbers:
Check the tens digit of both numbers.
If the tens digit is the same, compare the units digit.
The number with the higher digit in the first differing place is larger.
For example, to compare 87 and 75, since 8 > 7, 87 is larger. Vedantu’s live sessions teach students how to use place value for quick and effective number comparison.
Why is 99 considered a significant number in mathematics?
The number 99 is significant because it’s the highest two-digit number and is often used for:
Boundary value problems in number theory
Understanding number patterns before moving to three-digit numbers
Forming mathematical puzzles and exercises to strengthen arithmetic skills
At Vedantu, students learn the importance of numbers like 99 through engaging activities and real-life mathematical applications.
How can students remember the greatest two-digit number easily?
A simple way to remember the greatest two-digit number is to recall that:
All two-digit numbers range from 10 to 99.
The largest digit in base 10 is 9; combine two 9s to get 99.
Vedantu’s tutors use mnemonics and visual aids during online classes to help students effortlessly retain such key facts.
What are some examples of two-digit numbers just before and after 99?
The number before 99 is 98, and the number after 99 is 100 (which is a three-digit number). Examples:
98 (greatest even two-digit number)
99 (greatest two-digit number)
100 (smallest three-digit number)
Vedantu emphasizes such sequence patterns in its personalized learning modules to help students understand the flow of numbers naturally.
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概率，亦称“或然率”，它是反映随机事件出现的可能性大小。随机事件是指在相同条件下，可能出现也可能不出现的事件。例如，从一批有正品和次品的商品中，随意抽取一件，“抽得的是正品”就是一个随机事件。设对某一随机现象进行了n次试验与观察，其中A事件出现了m次，即其出现的频率为m/n。经过大量反复试验，常有m/n越来越接近于某个确定的常数（此论断证明详见伯努利大数定律）。该常数即为事件A出现的概率，常用P (A) 表示。
中文名
: 概率
外文名
: probability
学 科
: 数学
领 域
: 概率论
别 名
: 或然率
目录
1历史
2定义
▪来源
▪古典定义
▪频率定义
▪统计定义
▪公理化定义
3性质
4名词
▪事件
▪概型
5区别频率
历史
播报
编辑
卡尔达诺
第一个系统地推算概率的人是16世纪的卡尔达诺。记载在他的著作《Liber de Ludo Aleae》中。书中关于概率的内容是由Gould从拉丁文翻译出来的。
卡尔达诺的数学著作中有很多给赌徒的建议。这些建议都写成短文。然而，首次提出系统研究概率的是在帕斯卡和费马来往的一系列信件中。这些通信最初是由帕斯卡提出的，他想找费马请教几个关于由Chevvalier de Mere提出的问题。Chevvalier de Mere是一知名作家，路易十四宫廷的显要，也是一名狂热的赌徒。问题主要是两个：掷骰子问题和比赛奖金分配问题。
概率是度量偶然事件发生可能性的数值。假如经过多次重复试验(用X代表)，偶然事件(用A代表)出现了若干次(用Y代表)。以X作分母，Y作分子，形成了数值(用P代表)。在多次试验中，P相对稳定在某一数值上，P就称为A出现的概率。如偶然事件的概率是通过长期观察或大量重复试验来确定，则这种概率为统计概率或经验概率。
研究支配偶然事件的内在规律的学科叫概率论。属于数学上的一个分支。概率论揭示了偶然现象所包含的内部规律的表现形式。所以，概率，对人们认识自然现象和社会现象有重要的作用。比如，社会产品在分配给个人消费以前要进行扣除，需扣除多少，积累应在国民收入中占多大比重等，就需要运用概率论来确定。
定义
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来源
概率（Probability）一词来源于拉丁语“probabilitas”，又可以解释为 probity.Probity的意思是“正直、诚实”，在欧洲probity用来表示法庭案例中证人证词的权威性，且通常与证人的声誉相关。总之与现代意义上的概率“可能性”含义不同。
古典定义
如果一个试验满足两条：
（1）试验只有有限个基本结果；
（2）试验的每个基本结果出现的可能性是一样的。
这样的试验便是古典试验。
对于古典试验中的事件A，它的概率定义为：P(A)=，其中n表示该试验中所有可能出现的基本结果的总数目。m表示事件A包含的试验基本结果数。这种定义概率的方法称为概率的古典定义。
频率定义
随着人们遇到问题的复杂程度的增加，等可能性逐渐暴露出它的弱点，特别是对于同一事件，可以从不同的等可能性角度算出不同的概率，从而产生了种种悖论。另一方面，随着经验的积累，人们逐渐认识到，在做大量重复试验时，随着试验次数的增加，一个事件出现的频率，总在一个固定数的附近摆动，显示一定的稳定性。R.von米泽斯把这个固定数定义为该事件的概率，这就是概率的频率定义。从理论上讲，概率的频率定义是不够严谨的。
统计定义
在一定条件下，重复做n次试验，nA为n次试验中事件A发生的次数，如果随着n逐渐增大，频率nA/n逐渐稳定在某一数值p附近，则数值p称为事件A在该条件下发生的概率，记做P(A)=p。这个定义称为概率的统计定义。
在历史上，第一个对“当试验次数n逐渐增大，频率nA稳定在其概率p上”这一论断给以严格的意义和数学证明的是雅各布·伯努利（Jacob Bernoulli） 。
从概率的统计定义可以看到，数值p就是在该条件下刻画事件A发生可能性大小的一个数量指标。
由于频率总是介于0和1之间，从概率的统计定义可知，对任意事件A，皆有0≤P(A)≤1，P(Ω)=1，P(Φ)=0。其中Ω、Φ分别表示必然事件（在一定条件下必然发生的事件）和不可能事件（在一定条件下必然不发生的事件）。
公理化定义
柯尔莫哥洛夫于1933年给出了概率的公理化定义，如下：
设E是随机试验，S是它的样本空间。对于E的每一事件A赋于一个实数，记为P(A)，称为事件A的概率。这里P(A)是一个集合函数，P(A)要满足下列条件：
（1）非负性：对于每一个事件A，有P(A)≥0;
（2）规范性：对于必然事件，有P(s)=1;
（3）可列可加性：设A1，A2……是两两互不相容的事件，即对于i≠j，Ai∩Aj=φ，（i，j=1，2……），则有P(A1∪A2∪……)=P(A1)+P(A2)+……
性质
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概率具有以下7个不同的性质：
性质1：；
性质2：（有限可加性）当n个事件A1，…，An两两互不相容时： ；
性质3：对于任意一个事件A：；
性质4：当事件A，B满足A包含于B时：，；
性质5：对于任意一个事件A，；
性质6：对任意两个事件A和B，；
性质7：（加法公式）对任意两个事件A和B，。
名词
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事件
在一个特定的随机试验中，称每一可能出现的结果为一个基本事件，全体基本事件的集合称为基本空间。随机事件（简称事件）是由某些基本事件组成的，例如，在连续掷两次骰子的随机试验中，用Z，Y分别表示第一次和第二次出现的点数，Z和Y可以取值1、2、3、4、5、6，每一点（Z，Y）表示一个基本事件，因而基本空间包含36个元素。“点数之和为2”是一事件，它是由一个基本事件（1，1）组成，可用集合{（1，1）}表示，“点数之和为4”也是一事件，它由（1，3），（2，2），（3，1）3个基本事件组成，可用集合{（1，3），(3，1)，(2，2)}表示。如果把“点数之和为1”也看成事件，则它是一个不包含任何基本事件的事件，称为不可能事件。P(不可能事件)=0。在试验中此事件不可能发生。如果把“点数之和小于40”看成一事件，它包含所有基本事件，在试验中此事件一定发生，称为必然事件。P(必然事件)=1。实际生活中需要对各种各样的事件及其相互关系、基本空间中元素所组成的各种子集及其相互关系等进行研究 。
在一定的条件下可能发生也可能不发生的事件，叫做随机事件。
通常一次实验中的某一事件由基本事件组成。如果一次实验中可能出现的结果有n个，即此实验由n个基本事件组成，而且所有结果出现的可能性都相等，那么这种事件就叫做等可能事件。
互斥事件：不可能同时发生的两个事件叫做互斥事件。
对立事件：即必有一个发生的互斥事件叫做对立事件。
概型
古典概型
古典概型讨论的对象局限于随机试验所有可能结果为有限个等可能的情形，即基本空间由有限个元素或基本事件组成，其个数记为n，每个基本事件发生的可能性是相同的。若事件A包含m个基本事件，则定义事件A发生的概率为p（A）=，也就是事件A发生的概率等于事件A所包含的基本事件个数除以基本空间的基本事件的总个数，这是P.-S.拉普拉斯的古典概型定义，或称之为概率的古典定义。历史上古典概型是由研究诸如掷骰子一类赌博游戏中的问题引起的。计算古典概型，可以用穷举法列出所有基本事件，再数清一个事件所含的基本事件个数相除，即借助组合计算可以简化计算过程。
几何概型
几何概型若随机试验中的基本事件有无穷多个，且每个基本事件发生是等可能的，这时就不能使用古典概型，于是产生了几何概型。几何概型的基本思想是把事件与几何区域对应，利用几何区域的度量来计算事件发生的概率，布丰投针问题是应用几何概型的一个典型例子 。
设某一事件A（也是S中的某一区域），S包含A，它的量度大小为μ(A)，若以P(A)表示事件A发生的概率，考虑到“均匀分布”性，事件A发生的概率取为：P(A)=μ(A)/μ(S)，这样计算的概率称为几何概型。若Φ是不可能事件，即Φ为Ω中的空的区域，其量度大小为0，故其概率P(Φ)=0。
在概率论发展的早期，人们就注意到古典概型仅考虑试验结果只有有限个的情况是不够的，还必须考虑试验结果是无限个的情况。为此可把无限个试验结果用欧式空间的某一区域S表示，其试验结果具有所谓“均匀分布”的性质，关于“均匀分布”的精确定义类似于古典概型中“等可能”这一概念。假设区域S以及其中任何可能出现的小区域A都是可以度量的，其度量的大小分别用μ(S)和μ(A)表示。如一维空间的长度，二维空间的面积，三维空间的体积等。并且假定这种度量具有如长度一样的各种性质，如度量的非负性、可加性等。
区别频率
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对事件发生可能性大小的量化引入“概率”。独立重复试验总次数n，事件A发生的频数μ，事件A发生的频率Fn(A)=μ/n，A的频率Fn(A)有没有稳定值？如果有，就称频率μ/n的稳定值p为事件A发生的概率，记作P(A)=p（概率的统计定义）。
P(A)是客观的，而Fn(A)是依赖经验的。统计中有时也用n很大的时候的Fn(A)值当概率的近似值。
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Modular arithmetic/Introduction
Modular arithmetic is a special type of arithmetic that involves only integers. This goal of this article is to explain the basics of modular arithmetic while presenting a progression of more difficult and more interesting problems that are easily solved using modular arithmetic.
Contents
[hide]
1 Introductory Video
2 Understand Modular Arithmetic
3 Residue
4 Congruence
4.1 Examples
4.2 Sample Problem
4.2.1 Solution:
4.2.2 Another Solution:
4.2.3 Another Solution:
5 Making Computation Easier
5.1 Addition
5.1.1 Problem
5.1.2 Solution
5.1.3 Why we only need to use remainders
5.1.4 Solution using modular arithmetic
5.1.5 Addition rule
5.1.6 Proof of the addition rule
5.2 Subtraction
5.2.1 Problem
5.2.2 Solution
5.2.3 Subtraction rule
5.3 Multiplication
5.3.1 Problem
5.3.2 Solution
5.3.3 Solution using modular arithmetic
5.3.4 Multiplication rule
5.4 Exponentiation
5.4.1 Problem #1
5.4.2 Problem #2
5.4.3 Problem #3
6 Summary of Useful Facts
7 Problem Applications
8 Applications of Modular Arithmetic
9 Resources
10 See also
Introductory Video
Understand Modular Arithmetic
Let's use a clock as an example, except let's replace the at the top of the clock with a .
This is the way in which we count in modulo 12. When we add to , we arrive back at . The same is true in any other modulus (modular arithmetic system). In modulo , we count
We can also count backwards in modulo 5. Any time we subtract 1 from 0, we get 4. So, the integers from to , when written in modulo 5, are
where is the same as in modulo 5. Because all integers can be expressed as , , , , or in modulo 5, we give these integers their own name: the residue classes modulo 5. In general, for a natural number that is greater than 1, the modulo residues are the integers that are whole numbers less than :
This just relates each integer to its remainder from the Division Theorem. While this may not seem all that useful at first, counting in this way can help us solve an enormous array of number theory problems much more easily!
Residue
We say that is the modulo-residue of when , and .
Congruence
There is a mathematical way of saying that all of the integers are the same as one of the modulo 5 residues. For instance, we say that 7 and 2 are congruent modulo 5. We write this using the symbol : In other words, this means in base 5, these integers have the same residue modulo 5:
The (mod 5) part just tells us that we are working with the integers modulo 5. In modulo 5, two integers are congruent when their difference is a multiple of 5. In general, two integers and are congruent modulo when is a multiple of . In other words, when is an integer. Otherwise, , which means that and are not congruent modulo .
Examples
because is a multiple of .
because , which is an integer.
because , which is not a multiple of .
because , which is not an integer.
Sample Problem
Find the modulo residue of .
Solution:
Since R , we know that
and is the modulo residue of .
Another Solution:
Since , we know that
We can now solve it easily
and is the modulo residue of
Another Solution:
We know is a multiple of since is a multiple of . Thus, and is the modulo residue of .
Making Computation Easier
We don't always need to perform tedious computations to discover solutions to interesting problems. If all we need to know about are remainders when integers are divided by , then we can work directly with those remainders in modulo . This can be more easily understood with a few examples.
Addition
Problem
Suppose we want to find the units digit of the following sum:
We could find their sum, which is , and note that the units digit is . However, we could find the units digit with far less calculation.
Solution
We can simply add the units digits of the addends:
The units digit of this sum is , which must be the same as the units digit of the four-digit sum we computed earlier.
Why we only need to use remainders
We can rewrite each of the integers in terms of multiples of and remainders:
.
When we add all four integers, we get
At this point, we already see the units digits grouped apart and added to a multiple of (which will not affect the units digit of the sum):
.
Solution using modular arithmetic
Now let's look back at this solution, using modular arithmetic from the start. Note that
Because we only need the modulo residue of the sum, we add just the residues of the summands:
so the units digit of the sum is just .
Addition rule
In general, when , and are integers and is a positive integer such that
the following is always true:
.
And as we did in the problem above, we can apply more pairs of equivalent integers to both sides, just repeating this simple principle.
Proof of the addition rule
Let , and where and are integers. Adding the two equations we get:
Which is equivalent to saying
Subtraction
The same shortcut that works with addition of remainders works also with subtraction.
Problem
Find the remainder when the difference between and is divided by .
Solution
Note that and . So,
Thus,
so 1 is the remainder when the difference is divided by . (Perform the subtraction yourself, divide by , and see!)
Subtraction rule
When , and are integers and is a positive integer such that
the following is always true:
Multiplication
Modular arithmetic provides an even larger advantage when multiplying than when adding or subtracting. Let's take a look at a problem that demonstrates the point.
Problem
Jerry has boxes of soda in his truck. The cans of soda in each box are packed oddly so that there are cans of soda in each box. Jerry plans to pack the sodas into cases of cans to sell. After making as many complete cases as possible, how many sodas will Jerry have leftover?
Solution
First, we note that this word problem is asking us to find the remainder when the product is divided by .
Now, we can write each and in terms of multiples of and remainders:
This gives us a nice way to view their product:
Using FOIL, we get that this equals
We can already see that each part of the product is a multiple of , except the product of the remainders when each and are divided by 12. That part of the product is , which leaves a remainder of when divided by . So, Jerry has sodas leftover after making as many cases of as possible.
Solution using modular arithmetic
First, we note that
Thus,
meaning there are sodas leftover. Yeah, that was much easier.
Multiplication rule
When , and are integers and is a positive integer such that
The following is always true:
.
Exponentiation
Since exponentiation is just repeated multiplication, it makes sense that modular arithmetic would make many problems involving exponents easier. In fact, the advantage in computation is even larger and we explore it a great deal more in the intermediate modular arithmetic article.
Note to everybody: Exponentiation is very useful as in the following problem:
Problem #1
What is the last digit of if there are 1000 7s as exponents and only one 7 in the middle?
We can solve this problem using mods. This can also be stated as . After that, we see that 7 is congruent to -1 in mod 4, so we can use this fact to replace the 7s with -1s, because 7 has a pattern of repetitive period 4 for the units digit. is simply 1, so therefore , which really is the last digit.
Problem #2
What are the tens and units digits of ?
We could (in theory) solve this problem by trying to compute , but this would be extremely time-consuming. Moreover, it would give us much more information than we need. Since we want only the tens and units digits of the number in question, it suffices to find the remainder when the number is divided by . In other words, all of the information we need can be found using arithmetic mod .
We begin by writing down the first few powers of mod :
A pattern emerges! We see that So for any positive integer , we have (mod ). In particular, we can write
.
By the "multiplication" property above, then, it follows that
(mod ).
Therefore, by the definition of congruence, differs from by a multiple of . Since both integers are positive, this means that they share the same tens and units digits. Those digits are and , respectively.
Problem #3
Can you find a number that is both a multiple of but not a multiple of and a perfect square?
No, you cannot. Rewriting the question, we see that it asks us to find an integer that satisfies .
Taking mod on both sides, we find that . Now, all we are missing is proof that no matter what is, will never be a multiple of plus , so we work with cases:
This assures us that it is impossible to find such a number.
Summary of Useful Facts
Consider four integers and a positive integer such that and . In modular arithmetic, the following identities hold:
Addition: .
Subtraction: .
Multiplication: .
Division: , where is a positive integer that divides and .
Exponentiation: where is a positive integer.
Problem Applications
Applications of Modular Arithmetic
Modular arithmetic is an extremely flexible problem solving tool. The following topics are just a few applications and extensions of its use:
Divisibility rules
Linear congruences
Resources
The AoPS Introduction to Number Theory by Mathew Crawford.
The AoPS Introduction to Number Theory Course. Thousands of students have learned more about modular arithmetic and problem solving from this 12 week class.
See also
Intermediate modular arithmetic
Olympiad modular arithmetic
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. 2023 Mar 23;108(6):866–870. doi: 10.1093/biolre/ioad037
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From Enrico Sertoli to freemartinism: the many phases of the master testis-determining cell†
Humphrey Hung-Chang Yao
Humphrey Hung-Chang Yao
1 Reproductive Developmental Biology Group, National Institute of Environmental Health Sciences, Research Triangle Park, NC, USA
Find articles by Humphrey Hung-Chang Yao
1,✉, Karina F Rodriguez
Karina F Rodriguez
2 Reproductive Developmental Biology Group, National Institute of Environmental Health Sciences, Research Triangle Park, NC, USA
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2
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1 Reproductive Developmental Biology Group, National Institute of Environmental Health Sciences, Research Triangle Park, NC, USA
2 Reproductive Developmental Biology Group, National Institute of Environmental Health Sciences, Research Triangle Park, NC, USA
✉
Correspondence: Reproductive Developmental Biology Group, National Institute of Environmental Health Sciences (NIEHS/NIH), 111 T.W. Alexander Dr, Mail Drop C4-10, Research Triangle Park, NC 27709, USA. Tel: 984-287-4004; E-mail: humphrey.yao@nih.gov
Received 2022 Oct 27; Revised 2023 Feb 28; Collection date 2023 Jun.
Published by Oxford University Press on behalf of Society for the Study of Reproduction 2023.
This work is written by US Government employees and is in the public domain in the US.
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PMCID: PMC10266947 PMID: 36951956
Abstract
Sertoli cells, first identified in the adult testis by Enrico Sertoli in the mid-nineteenth century, are known for their role in fostering male germ cell differentiation and production of mature sperm. It was not until the late twentieth century with the discovery of the testis-determining gene SRY that Sertoli cells’ new function as the master regulator of testis formation and maleness was unveiled. Fetal Sertoli cells facilitate the establishment of seminiferous cords, induce appearance of androgen-producing Leydig cells, and cause regression of the female reproductive tracts. Originally thought be a terminally differentiated cell type, adult Sertoli cells, at least in the mouse, retain their plasticity and ability to transdifferentiate into the ovarian counterpart, granulosa cells. In this review, we capture the many phases of Sertoli cell differentiation from their fate specification in fetal life to fate maintenance in adulthood. We also introduce the discovery of a new phase of fetal Sertoli cell differentiation via autocrine/paracrine factors with the freemartin characteristics. There remains much to learn about this intriguing cell type that lay the foundation for the maleness.
Keywords: sex determination, testis, sertoli cells, freemartin
When Enrico Sertoli stared at the “cellule ramificate,” or branched cells, in a slice of human testicle under the microscope in 1865 , he was unaware that his legacy will live through the age of microscopy to the era of single-cell sequencing in the twenty-first century. Sertoli cells, the testis-specific cell type that was named after him, are not only the “mother” cells that nurse male germ cells but also the leading actors that usher in the pathway toward maleness. In the adult testis, Sertoli cells attach to the basement membrane of seminiferous tubules and extend their cell body toward the lumen while enveloping all stages of male germ cells. A critical component of the hypothalamic–pituitary–gonadal axis, adult Sertoli cells acquire their unique features and ability to provide the niche for spermatogenesis through their response to pituitary hormones and local factors . Adult Sertoli cells facilitate an orderly progression of spermatid morphogenesis and sperm production, and at the same time, create a barrier that allows the meiotic and haploid male germ cells to escape from the surveillance of the immune system. Adult Sertoli cells were considered a terminally differentiated cell type; however, they were later found to retain the plasticity to decommit to the male identity and transdifferentiate into granulosa cells, their female counterparts in the ovary . In the early 1990s, the identification of the testis-determining gene SRY (Sex-determining region of the Y chromosome) further revealed the formative role of fetal Sertoli cells in testis morphogenesis [4–6]. Sertoli cells, the only cell type in the fetal testis that produce SRY, orchestrate formation of seminiferous tubules, establishment of testis-specific vasculature, appearance of fetal Leydig cells, and regression of the female reproductive tract [4–8].
Fate specification and establishment of Sertoli cell lineage in fetal testis
The appearance or fate specification of Sertoli cells is initiated by SRY in most mammals with some exceptions like the Amami spiny rat and the vole Ellobius [9, 10]. SRY is exclusively expressed in the bipotential supporting cell progenitors that give rise to either Sertoli cells or its female counterpart granulosa cells. In the mouse XY embryos, Sry expression first appears in the center of the gonadal primordium around embryonic day (E) 10.5 [6, 11]. Initial Sry expression starts cell-autonomously without the involvement of external factors. The histone demethylase JMJD1A mediates H3K9me2 demethylation of the mouse Sry locus, which leads to Sry expression . Sry expression in the mouse XY gonads is transient, spreading from the center to the poles of the gonadal primordium . SRY, along with other transcription factors such as NR5A1 (or Steroidogenic factor 1), directly induces the expression of transcription factor Sox9 , a conserved testis-inducing factor in many non-mammalian and mammalian species [13–15]. SRY induces Sox9 expression by binding to the enhancer element TESCO (testis-specific enhancer of Sox9 core), Enh13, and other regions upstream of the transcription start site of Sox9 [13, 16]. SOX9 further enhances its own expression through the action of NR5A1 and SOX8 [13, 17, 18]. In addition to the action mediated through transcription activities, the SRY/SOX9 pathway also induces production of secreted factors from pre-Sertoli cells, including fibroblast growth factor 9 (FGF9) and prostaglandin D2 [19–22] (Figure 1A). FGF9 and its receptor FGFR2, both present in pre-Sertoli cells, maintain SOX9 expression and induce proliferation of pre-Sertoli cell population . Pre-Sertoli cells also express prostaglandin D synthase (Ptds), an enzyme responsible for prostaglandin D2 (PGD2) production which enhances Sox9 transcription levels independently of FGF9 [20, 22, 24].
Figure 1.
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Different phases of Sertoli cell differentiation. (A) Specification of Sertoli cell fate is induced cell-autonomously by transcription factor SRY and its downstream effectors SOX8/9. The pre-Sertoli cell population is further expanded and secured via the action of FGF9 and PGD2. (B) Once the identity of fetal Sertoli cells is established, Sertoli cells produce AMH, activin B, and other unknown factors, which serve as autocrine/paracrine factors to maintain the Sertoli cell identity. (C) In adult Sertoli cells, transcription factors SOX8/9 and DMRT1 are required for the maintenance of Sertoli cell identity. This figure was created with BioRender.com.
In addition to inducing Sertoli cell differentiation, SRY/SOX8/9 also antagonize the molecular pathways responsible for the differentiation of granulosa cells. These “pro-ovary” pathways include the WNT pathway (ligand WNT4, secreted WNT activator R-spondin1 or RSPO1, and its intracellular effector beta-catenin) and the transcription regulators FOXL2 and RUNX1 [21, 25–28]. These pro-ovary factors direct the bipotential supporting cell progenitors toward the granulosa cell program by repressing the testis program [26, 29]. Conversely, ectopic activation of the components of these pro-ovary pathways in the pre-Sertoli cells disrupts SOX9 expression and specification of Sertoli cells, leading to testis-to-ovary sex reversal and ovotestis formation [26, 27]. The antagonist action of SRY/SOX9 is essential to suppress expression of pro-ovary genes and pathways to avoid their feminizing effects. CBX2, a member of the Polycomb Repressive Complex 1 (PRC1), stabilizes the testis pathway downstream of SRY/SOX9 by repressing the expression of the components of the WNT pathway . With the activation of the Sertoli cell program and inhibition of the granulosa cell program, the bipotential progenitors acquire their identity as Sertoli cells.
Fate maintenance of Sertoli cells in adult testis
As the supporting cell progenitors commit to either the testis or ovary fate, they were assumed to acquire terminally differentiated sex-specific gene expression and phenotypes. However, several findings in mice suggest otherwise. In adult XX mice with either disabled production or disabled responsiveness to estrogens , granulosa cells in the adult ovary lost their identity and transformed into Sertoli-like cells with expression of Sox9. Loss of Foxl2, a conserved transcription factor for granulosa cell differentiation in vertebrates, and in particular mammals [33, 34], recapitulated similar granulosa-to-Sertoli cell transdifferentiation in adult XX mice . On the other hand, when transcription factor Dmrt1 was deleted in Sertoli cells, adult XY Sertoli cells gained Foxl2 expression, lost their polarity, and gradually transdifferentiated into granulosa-like cells [3, 36]. Dmrt1 expression in adult Sertoli cells requires the combined action of SOX8/9. In adult Sertoli cells, in the absence of Sox8/9, Dmrt1 expression was downregulated, Sertoli-to-granulosa cell transdifferentiation occurred, and disintegration of seminiferous tubules was observed . These observations implicate that adult Sertoli cells require SOX8/SOX9/DMRT1 to maintain their identity in the testis (Figure 1C) whereas such action is antagonized by the estrogen/FOXL2 pathway in the adult ovary. The bivalent status of key genes for Sertoli cells and granulosa cells during fetal life appears to be retained in adulthood, and may contribute to the lineage plasticity of differentiated Sertoli and granulosa cells.
Uncovering a new phase of Sertoli cell fate maintenance
The discovery of SRY and its role in testis determination in the 1990s supports that the specification of Sertoli cell lineage in most mammals occurs cell-autonomously via the transcription action of SRY in the bipotential progenitor cells. However, these findings seem contradictory to the intersex freemartin cases in cattle, sheep, pigs, and goats, which had been documented since ancient Greece [38–43]. In freemartin cases where the pregnant female carries one XX and one XY dizygotic twin, the XX twin, despite a lack of SRY, is often masculinized with retention of the Wolffian ducts, regression of the Mullerian ducts, and appearance of seminiferous cord structures in the ovaries. Frank Lillie and Tandler and Keller in the early 1900s independently observed that the freemartin twins share placental circulation through anastomosis, prompting them to postulate that hormones from the XY twin, through the shared placenta, contribute to the masculinization of the XX twin [38, 39]. At that time, androgen was the only known testis-derived hormone responsible for the masculinization of Wolffian ducts and external genitalia; however, androgens are not responsible for the appearance of testis structures in the freemartin ovary [44, 45]. The second testis-produced hormone was not discovered until the 1970s by Alfred Jost. Using the rabbit embryo as the model, Jost found a non-androgen, testis-produced factor(s) that induced Mullerian duct regression regardless of the genetic sex of the embryos . This factor was later named as Mullerian inhibiting substance (MIS) or anti-Mullerian hormone (AMH, the currently used acronym) by Josso and colleagues [47, 48]. In contrast to androgens, AMH induced partial freemartin phenotypes, such as loss of female germ cells and appearance of seminiferous tubule structures, in the fetal rat and ovine ovary in vitro [49–51]. Although the naturally occurring freemartin cases were not reported in mouse embryos, mouse fetal ovaries developed freemartin phenotypes when they were transplanted under the kidney capsule of adult male recipients [52–56], indicating that in mice, adult testes produce the freemartin factors.
The focus on the freemartin factors has been on their ability to masculinize female reproductive organs; however, the action of freemartin factors on their source, the fetal testis, was completely unknown. One would expect that if the freemartin factors had such potent sex-reversing impacts on the fetal ovary, they must play a role in formation of the fetal testes. AMH, the first hormone, and putative freemartin factor produced by fetal Sertoli cells, was dispensable for normal testis morphogenesis at least in mice and humans . Humans lacking AMH or its receptor, AMHR2, exhibit a disorder of sex development almost identical to the knockout mouse models, in which the Mullerian ducts in the affected XY individuals fail to regress, eventually forming a uterus and fallopian tubes (reviewed in ). Sertoli cells of mouse fetal testes also produce another hormone activin B . Similar to the absence of Amh, XY mouse embryos lacking Inhbb, the gene encodes the beta subunit of activin B, developed normal testes with minor defects in vasculature . We predicted that Sertoli cell–derived AMH and activin B, both members of TGF-beta superfamily proteins, together ensure normal fetal testis morphogenesis and when both factors are absent, testis morphogenesis would be compromised. Disappointingly, the initial testis formation in the Amh/Inhbb double-knockout XY embryos was normal with proper gene expression and establishment of testis cords and other cell types . However, 4 days after the initiation of testis morphogenesis, the double-knockout testes gradually transformed into ovotestes where SOX9+ Sertoli cells in the poles transdifferentiate into their female counterpart FOXL2+ granulosa cells. The transdifferentiated Sertoli cells also lost their polarity, leading to disintegration of the testis cords on the poles. The testis cords in the center of the ovotestis remained intact, flanked by clusters of FOXL2+ granulosa cells and meiotic germ cells. Fetal Leydig cells were abundantly dispersed in the interstitium of the testicular domain, producing sufficient androgens to maintain the Wolffian ducts and masculinized the external genitalia. As a result, the double-knockout XY embryos were intersex with the presence of ovotestis, Mullerian duct derivatives, and Wolffian duct derivatives. Such intersex phenotypes lasted to adulthood with the coexistence of mature male and female reproductive tracts and ovotestes with seminiferous tubules in the centers and follicles in the poles. Due to the presence of the female reproductive tracts, the ovotestes were cryptorchid, resulting in defects of spermatogenesis and few mature sperm in the epididymis. Follicles in the ovarian domain developed to the preantral stages. These follicles responded to the hormone stimulation for superovulation but never ovulated naturally, probably due to a male-like hypothalamic–pituitary axis.
The transformation of fetal testes into ovotestes in the absence of AMH and activin B sheds new light on the normal process of sex determination. First, the transdifferentiation of fetal Sertoli cells into granulosa cells few days after initial testis formation indicates that once their lineage is specified by SRY and its downstream components, the actions of AMH and activin B are required to maintain their identity (Figure 1B). AMH and activin B serve as autocrine and paracrine factors that act on their source Sertoli cells, which express the receptors for both factors [58–62] to suppress the molecular programs for granulosa cells. Second, a lack of complete testis-to-ovary sex reversal of the double-knockout testes implies that other yet-to-be-identified factors compensate partially for the loss of AMH and activin B. Third, AMH and activin B have the freemartin capability based on the findings that without these two factors, the ability of adult Amh/Inhbb double-knockout males to masculinize the fetal ovaries was significantly reduced . The fact that the freemartin effect was not completely abolished in the adult double-knockout XY mice further supports the presence of other freemartin factors such as androgens and other TGF-beta family proteins [56, 63–66].
Perspectives
Enrico Sertoli probably had no idea that the cell type named after him turned out to be the master regulator of testis morphogenesis and male reproductive functions. Sertoli cells, the first somatic cell type that appears during testis formation, determine the male phenotypic characteristics. Without Sertoli cells, seminiferous tubules do not form, spermatogenesis never occurs, androgen-producing Leydig cells and their hormonal effects fail to appear, and the female reproductive tracts remain. In contrast to many non-mammalian species where sex determination is controlled by hormones, Sertoli cell lineage in most mammals arises cell-autonomously through the transcription action of SRY and its downstream effector SOX9. However, the freemartin phenomena suggest that mammalian Sertoli cells maintain a hormone-sensitive nature. We argue that AMH, which is the testis-determining, Sertoli cell–inducing factor in teleost fish, like tilapia and pejerrey [67, 68] and possibly alligator , takes on a different role in mammalian sex determination. At least in mice, AMH and other TGF-beta family members maintain fetal Sertoli cell identity and retain the freemartin ability to induce Sertoli cell appearance in the fetal ovary. Whether such dual functions of AMH and activin are present in other species remains to be investigated. Freemartinism is clearly not an aberrant phenomenon; it reminds us not only of how sex determination is evolutionarily conserved and divergent but also of how much can still be learned from the master cell of testis determination discovered more than 150 years ago.
Footnotes
†Grant Support: Intramural Research Program of National Institute of Environmental Health Sciences (grant no. Z01-ES102965 to H.H.-C.Y.)
Contributor Information
Humphrey Hung-Chang Yao, Reproductive Developmental Biology Group, National Institute of Environmental Health Sciences, Research Triangle Park, NC, USA.
Karina F Rodriguez, Reproductive Developmental Biology Group, National Institute of Environmental Health Sciences, Research Triangle Park, NC, USA.
Author contribution
H.H-C.Y. and K.R. wrote the manuscript and agree to the order of authors.
Conflict of Interest: The authors have declared that no conflict of interest exists.
References
Sertoli E. E lesistenza di particulari cellule ramificate nei canalicoli seminiferi dell’testicolo umno. Morgagni 1865; 7:31. [Google Scholar]
Griswold MD. The central role of Sertoli cells in spermatogenesis. Semin Cell Dev Biol 1998; 9:411–416. [DOI] [PubMed] [Google Scholar]
Matson CK, Murphy MW, Sarver AL, Griswold MD, Bardwell VJ, Zarkower D. DMRT1 prevents female reprogramming in the postnatal mammalian testis. Nature 2011; 476:101–104. [DOI] [PMC free article] [PubMed] [Google Scholar]
Berta P, Hawkins JR, Sinclair AH, Taylor A, Griffiths BL, Goodfellow PN, Fellous M. Genetic evidence equating SRY and the testis-determining factor. Nature 1990; 348:448–450. [DOI] [PubMed] [Google Scholar]
Sinclair AH, Berta P, Palmer MS, Hawkins JR, Griffiths BL, Smith MJ, Foster JW, Frischauf AM, Lovell-Badge R, Goodfellow PN. A gene from the human sex-determining region encodes a protein with homology to a conserved DNA-binding motif. Nature 1990; 346:240–244. [DOI] [PubMed] [Google Scholar]
Koopman P, Gubbay J, Vivian N, Goodfellow P, Lovell-Badge R. Male development of chromosomally female mice transgenic for Sry. Nature 1991; 351:117–121. [DOI] [PubMed] [Google Scholar]
Behringer RR, Finegold MJ, Cate RL. Mullerian-inhibiting substance function during mammalian sexual development. Cell 1994; 79:415–425. [DOI] [PubMed] [Google Scholar]
Yao HH, Aardema J, Holthusen K. Sexually dimorphic regulation of inhibin beta B in establishing gonadal vasculature in mice. Biol Reprod 2006; 74:978–983. [DOI] [PMC free article] [PubMed] [Google Scholar]
Terao M, Ogawa Y, Takada S, Kajitani R, Okuno M, Mochimaru Y, Matsuoka K, Itoh T, Toyoda A, Kono T, Jogahara T, Mizushima S, et al. Turnover of mammal sex chromosomes in the Sry-deficient Amami spiny rat is due to male-specific upregulation of Sox9. Proc Natl Acad Sci U S A 2022; 119:e2211574119. [DOI] [PMC free article] [PubMed] [Google Scholar]
Just W, Rau W, Vogel W, Akhverdian M, Fredga K, Graves JA, Lyapunova E. Absence of Sry in species of the vole Ellobius. Nat Genet 1995; 11:117–118. [DOI] [PubMed] [Google Scholar]
Bullejos M, Koopman P. Spatially dynamic expression of Sry in mouse genital ridges. Dev Dyn 2001; 221:201–205. [DOI] [PubMed] [Google Scholar]
Kuroki S, Matoba S, Akiyoshi M, Matsumura Y, Miyachi H, Mise N, Abe K, Ogura A, Wilhelm D, Koopman P, Nozaki M, Kanai Y, et al. Epigenetic regulation of mouse sex determination by the histone demethylase Jmjd1a. Science 2013; 341:1106–1109. [DOI] [PubMed] [Google Scholar]
Sekido R, Lovell-Badge R. Sex determination involves synergistic action of SRY and SF1 on a specific Sox9 enhancer. Nature 2008; 453:930–934. [DOI] [PubMed] [Google Scholar]
Hui HB, Xiao L, Sun W, Zhou YJ, Zhang HY, Ge CT. Sox9 is indispensable for testis differentiation in the red-eared slider turtle, a reptile with temperature-dependent sex determination. Zool Res 2021; 42:721–725. [DOI] [PMC free article] [PubMed] [Google Scholar]
Vining B, Ming Z, Bagheri-Fam S, Harley V. Diverse regulation but conserved function: SOX9 in vertebrate sex determination. Genes (Basel) 2021; 12:486. [DOI] [PMC free article] [PubMed] [Google Scholar]
Gonen N, Futtner CR, Wood S, Garcia-Moreno SA, Salamone IM, Samson SC, Sekido R, Poulat F, Maatouk DM, Lovell-Badge R. Sex reversal following deletion of a single distal enhancer of Sox9. Science 2018; 360:1469–1473. [DOI] [PMC free article] [PubMed] [Google Scholar]
Chaboissier MC, Kobayashi A, Vidal VI, Lutzkendorf S, van de Kant HJ, Wegner M, de Rooij DG, Behringer RR, Schedl A. Functional analysis of Sox8 and Sox9 during sex determination in the mouse. Development 2004; 131:1891–1901. [DOI] [PubMed] [Google Scholar]
Richardson N, Gillot I, Gregoire EP, Youssef SA, de Rooij D, de Bruin A, De Cian MC, Chaboissier MC. Sox8 and Sox9 act redundantly for ovarian-to-testicular fate reprogramming in the absence of R-spondin1 in mouse sex reversals. Elife 2020; 9:e53972. [DOI] [PMC free article] [PubMed] [Google Scholar]
Colvin JS, Green RP, Schmahl J, Capel B, Ornitz DM. Male-to-female sex reversal in mice lacking fibroblast growth factor 9. Cell 2001; 104:875–889. [DOI] [PubMed] [Google Scholar]
Malki S, Nef S, Notarnicola C, Thevenet L, Gasca S, Mejean C, Berta P, Poulat F, Boizet-Bonhoure B. Prostaglandin D2 induces nuclear import of the sex-determining factor SOX9 via its cAMP-PKA phosphorylation. EMBO J 2005; 24:1798–1809. [DOI] [PMC free article] [PubMed] [Google Scholar]
Kim Y, Kobayashi A, Sekido R, DiNapoli L, Brennan J, Chaboissier MC, Poulat F, Behringer RR, Lovell-Badge R, Capel B. Fgf9 and Wnt4 act as antagonistic signals to regulate mammalian sex determination. PLoS Biol 2006; 4:e187. [DOI] [PMC free article] [PubMed] [Google Scholar]
Wilhelm D, Martinson F, Bradford S, Wilson MJ, Combes AN, Beverdam A, Bowles J, Mizusaki H, Koopman P. Sertoli cell differentiation is induced both cell-autonomously and through prostaglandin signaling during mammalian sex determination. Dev Biol 2005; 287:111–124. [DOI] [PubMed] [Google Scholar]
Bagheri-Fam S, Sim H, Bernard P, Jayakody I, Taketo MM, Scherer G, Harley VR. Loss of Fgfr2 leads to partial XY sex reversal. Dev Biol 2008; 314:71–83. [DOI] [PubMed] [Google Scholar]
Adams IR, McLaren A. Sexually dimorphic development of mouse primordial germ cells: switching from oogenesis to spermatogenesis. Development 2002; 129:1155–1164. [DOI] [PubMed] [Google Scholar]
Jameson SA, Lin YT, Capel B. Testis development requires the repression of Wnt4 by Fgf signaling. Dev Biol 2012; 370:24–32. [DOI] [PMC free article] [PubMed] [Google Scholar]
Nicol B, Grimm SA, Gruzdev A, Scott GJ, Ray MK, Yao HH. Genome-wide identification of FOXL2 binding and characterization of FOXL2 feminizing action in the fetal gonads. Hum Mol Genet 2018; 27:4273–4287. [DOI] [PMC free article] [PubMed] [Google Scholar]
Maatouk DM, DiNapoli L, Alvers A, Parker KL, Taketo MM, Capel B. Stabilization of beta-catenin in XY gonads causes male-to-female sex-reversal. Hum Mol Genet 2008; 17:2949–2955. [DOI] [PMC free article] [PubMed] [Google Scholar]
Liu CF, Bingham N, Parker K, Yao HH. Sex-specific roles of beta-catenin in mouse gonadal development. Hum Mol Genet 2009; 18:405–417. [DOI] [PMC free article] [PubMed] [Google Scholar]
Nicol B, Grimm SA, Chalmel F, Lecluze E, Pannetier M, Pailhoux E, Dupin-De-Beyssat E, Guiguen Y, Capel B, Yao HH. RUNX1 maintains the identity of the fetal ovary through an interplay with FOXL2. Nat Commun 2019; 10:5116. [DOI] [PMC free article] [PubMed] [Google Scholar]
Garcia-Moreno SA, Lin YT, Futtner CR, Salamone IM, Capel B, Maatouk DM. CBX2 is required to stabilize the testis pathway by repressing Wnt signaling. PLoS Genet 2019; 15:e1007895. [DOI] [PMC free article] [PubMed] [Google Scholar]
Britt KL, Stanton PG, Misso M, Simpson ER, Findlay JK. The effects of estrogen on the expression of genes underlying the differentiation of somatic cells in the murine gonad. Endocrinology 2004; 145:3950–3960. [DOI] [PubMed] [Google Scholar]
Couse JF, Hewitt SC, Bunch DO, Sar M, Walker VR, Davis BJ, Korach KS. Postnatal sex reversal of the ovaries in mice lacking estrogen receptors alpha and beta. Science 1999; 286:2328–2331. [DOI] [PubMed] [Google Scholar]
Nicol B, Estermann MA, Yao HH, Mellouk N. Becoming female: ovarian differentiation from an evolutionary perspective. Front Cell Dev Biol 2022; 10:944776. [DOI] [PMC free article] [PubMed] [Google Scholar]
Bertho S, Pasquier J, Pan Q, Le Trionnaire G, Bobe J, Postlethwait JH, Pailhoux E, Schartl M, Herpin A, Guiguen Y. Foxl2 and its relatives are evolutionary conserved players in gonadal sex differentiation. Sex Dev 2016; 10:111–129. [DOI] [PubMed] [Google Scholar]
Uhlenhaut NH, Jakob S, Anlag K, Eisenberger T, Sekido R, Kress J, Treier AC, Klugmann C, Klasen C, Holter NI, Riethmacher D, Schutz G, et al. Somatic sex reprogramming of adult ovaries to testes by FOXL2 ablation. Cell 2009; 139:1130–1142. [DOI] [PubMed] [Google Scholar]
Minkina A, Matson CK, Lindeman RE, Ghyselinck NB, Bardwell VJ, Zarkower D. DMRT1 protects male gonadal cells from retinoid-dependent sexual transdifferentiation. Dev Cell 2014; 29:511–520. [DOI] [PMC free article] [PubMed] [Google Scholar]
Barrionuevo FJ, Hurtado A, Kim GJ, Real FM, Bakkali M, Kopp JL, Sander M, Scherer G, Burgos M, Jimenez R. Sox9 and Sox8 protect the adult testis from male-to-female genetic reprogramming and complete degeneration. Elife 2016; 5:e15636 [DOI] [PMC free article] [PubMed] [Google Scholar]
Tandler J, Keller K. Ueber das Verhalten des chorions bei verschiedengeschlechtlicher Zwillingsgravidität des Rindes und über die Morphologie des Genitales der weiblichen Tiere, welche einer solchen Gravidität entstammen. Dt tierärztl Wschr 1911; 19:148–149. [Google Scholar]
Lillie FR. The theory of the free-Martin. Science 1916; 43:611–613. [DOI] [PubMed] [Google Scholar]
Ewen AH, Hummason FA. An ovine freemartin. J Hered 1947; 38:149–152. [DOI] [PubMed] [Google Scholar]
Forbes TR. The origin of freemartin. Bull Hist Med 1946; 20:461–466. [PubMed] [Google Scholar]
Padula AM. The freemartin syndrome: an update. Anim Reprod Sci 2005; 87:93–109. [DOI] [PubMed] [Google Scholar]
Remnant JG, Lea RG, Allen CE, Huxley JN, Robinson RS, Brower AI. Novel gonadal characteristics in an aged bovine freemartin. Anim Reprod Sci 2014; 146:1–4. [DOI] [PubMed] [Google Scholar]
Jainudeen MR, Hafez ES. Attempts to induce bovine freemartinism experimentally. J Reprod Fertil 1965; 10:281–283. [DOI] [PubMed] [Google Scholar]
Short RV, Smith J, Mann T, Evans EP, Hallett J, Fryer A, Hamerton JL. Cytogenetic and endocrine studies of a freemartin heifer and its bull co-twin. Cytogenetics 1969; 8:369–388. [DOI] [PubMed] [Google Scholar]
Jost A. Hormonal factors in the sex differentiation of the mammalian foetus. Philos Trans R Soc Lond B Biol Sci 1970; 259:119–130. [DOI] [PubMed] [Google Scholar]
Josso N, Picard JY, Tran D. The anti-Mullerian hormone. Birth Defects Orig Artic Ser 1977; 13:59–84. [PubMed] [Google Scholar]
Tran D, Muesy-Dessole N, Josso N. Anti-Mullerian hormone is a functional marker of foetal Sertoli cells. Nature 1977; 269:411–412. [DOI] [PubMed] [Google Scholar]
Vigier B, Magre S, Charpentier G, Bezard J, Josso N. Anti-Mullerian hormone and natural and experimental freemartin effect. Bull Assoc Anat (Nancy) 1991; 75:29–32. [PubMed] [Google Scholar]
Vigier B, Watrin F, Magre S, Tran D, Garrigou O, Forest MG, Josso N. Anti-mullerian hormone and freemartinism: inhibition of germ cell development and induction of seminiferous cord-like structures in rat fetal ovaries exposed in vitro to purified bovine AMH. Reprod Nutr Dev 1980; 1988:1113–1128. [DOI] [PubMed] [Google Scholar]
Vigier B, Watrin F, Magre S, Tran D, Josso N. Purified bovine AMH induces a characteristic freemartin effect in fetal rat prospective ovaries exposed to it in vitro. Development 1987; 100:43–55. [DOI] [PubMed] [Google Scholar]
Taketo T, Merchant-Larios H. Gonadal sex reversal of fetal mouse ovaries following transplantation into adult mice. Prog Clin Biol Res 1986; 217A:171–174. [PubMed] [Google Scholar]
Taketo T, Merchant-Larios H, Koide SS. Induction of testicular differentiation in the fetal mouse ovary by transplantation into adult male mice. Proc Soc Exp Biol Med 1984; 176:148–153. [DOI] [PubMed] [Google Scholar]
Morais da Silva S, Hacker A, Harley V, Goodfellow P, Swain A, Lovell-Badge R. Sox9 expression during gonadal development implies a conserved role for the gene in testis differentiation in mammals and birds. Nat Genet 1996; 14:62–68. [DOI] [PubMed] [Google Scholar]
Harikae K, Miura K, Shinomura M, Matoba S, Hiramatsu R, Tsunekawa N, Kanai-Azuma M, Kurohmaru M, Morohashi K, Kanai Y. Heterogeneity in sexual bipotentiality and plasticity of granulosa cells in developing mouse ovaries. J Cell Sci 2013; 126:2834–2844. [DOI] [PubMed] [Google Scholar]
Miura K, Harikae K, Nakaguchi M, Imaimatsu K, Hiramatsu R, Tomita A, Hirate Y, Kanai-Azuma M, Kurohmaru M, Ogura A, Kanai Y. Molecular and genetic characterization of partial masculinization in embryonic ovaries grafted into male nude mice. PloS One 2019; 14:e0212367. [DOI] [PMC free article] [PubMed] [Google Scholar]
Brunello FG, Rey RA. AMH and AMHR2 involvement in congenital disorders of sex development. Sex Dev 2022; 16:138–146. [DOI] [PubMed] [Google Scholar]
Rodriguez KF, Brown PR, Amato CM, Nicol B, Liu CF, Xu X, Yao HH. Somatic cell fate maintenance in mouse fetal testes via autocrine/paracrine action of AMH and activin B. Nat Commun 2022; 13:4130. [DOI] [PMC free article] [PubMed] [Google Scholar]
Clarke TR, Hoshiya Y, Yi SE, Liu X, Lyons KM, Donahoe PK. Mullerian inhibiting substance signaling uses a bone morphogenetic protein (BMP)-like pathway mediated by ALK2 and induces SMAD6 expression. Mol Endocrinol 2001; 15:946–959. [DOI] [PubMed] [Google Scholar]
Jamin SP, Arango NA, Mishina Y, Hanks MC, Behringer RR. Genetic studies of the AMH/MIS signaling pathway for Mullerian duct regression. Mol Cell Endocrinol 2003; 211:15–19. [DOI] [PubMed] [Google Scholar]
Jamin SP, Arango NA, Mishina Y, Hanks MC, Behringer RR. Requirement of Bmpr1a for Mullerian duct regression during male sexual development. Nat Genet 2002; 32:408–410. [DOI] [PubMed] [Google Scholar]
Orvis GD, Jamin SP, Kwan KM, Mishina Y, Kaartinen VM, Huang S, Roberts AB, Umans L, Huylebroeck D, Zwijsen A, Wang D, Martin JF, et al. Functional redundancy of TGF-beta family type I receptors and receptor-Smads in mediating anti-Mullerian hormone-induced Mullerian duct regression in the mouse. Biol Reprod 2008; 78:994–1001. [DOI] [PMC free article] [PubMed] [Google Scholar]
Pan Q, Kay T, Depince A, Adolfi M, Schartl M, Guiguen Y, Herpin A. Evolution of master sex determiners: TGF-beta signalling pathways at regulatory crossroads. Philos Trans R Soc Lond B Biol Sci 2021; 376:20200091. [DOI] [PMC free article] [PubMed] [Google Scholar]
Cai K, Hua G, Ahmad S, Liang A, Han L, Wu C, Yang F, Yang L. Action mechanism of inhibin alpha-subunit on the development of Sertoli cells and first wave of spermatogenesis in mice. PloS One 2011; 6:e25585. [DOI] [PMC free article] [PubMed] [Google Scholar]
Miles DC, Wakeling SI, Stringer JM, van den Bergen JA, Wilhelm D, Sinclair AH, Western PS. Signaling through the TGF beta-activin receptors ALK4/5/7 regulates testis formation and male germ cell development. PloS One 2013; 8:e54606. [DOI] [PMC free article] [PubMed] [Google Scholar]
Memon MA, Anway MD, Covert TR, Uzumcu M, Skinner MK. Transforming growth factor beta (TGFbeta1, TGFbeta2 and TGFbeta3) null-mutant phenotypes in embryonic gonadal development. Mol Cell Endocrinol 2008; 294:70–80. [DOI] [PMC free article] [PubMed] [Google Scholar]
Wessels S, Sharifi RA, Luehmann LM, Rueangsri S, Krause I, Hoerstgen-Schwark G, Knorr C. Allelic variant in the anti-Mullerian hormone gene leads to autosomal and temperature-dependent sex reversal in a selected Nile tilapia line. PloS One 2014; 9:e104795. [DOI] [PMC free article] [PubMed] [Google Scholar]
Fernandino JI, Hattori RS, Kimura H, Strussmann CA, Somoza GM. Expression profile and estrogenic regulation of anti-Mullerian hormone during gonadal development in pejerrey Odontesthes bonariensis, a teleost fish with strong temperature-dependent sex determination. Dev Dyn 2008; 237:3192–3199. [DOI] [PubMed] [Google Scholar]
Western PS, Harry JL, Graves JA, Sinclair AH. Temperature-dependent sex determination in the American alligator: AMH precedes SOX9 expression. Dev Dyn 1999; 216:411–419. [DOI] [PubMed] [Google Scholar]
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Fate specification and establishment of Sertoli cell lineage in fetal testis
Fate maintenance of Sertoli cells in adult testis
Uncovering a new phase of Sertoli cell fate maintenance
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11699 | https://wordpandit.com/impact-of-removal-replacement-addition-of-new-item-on-averages/ | In the previous articles, we discussed how to calculate the average of a given series. But what if we eliminate, add, or replace some terms from that series. Will the average be affected? If yes, then how?
To find the answer, let us go through the concepts given below.
Concept 1
Let the average of n quantitiesa1, a2, a3……anbeA1. Now a new number N is added to the series which increases the average to A2, then thevalue of the new number added will be:
N=n x (A2 – A1)+ A2
Or N= n × (increase in value of average) + A2
The average of n quantitiesa1, a2, a3……an is:
= A1 ……. (1)
After the addition of new number N, the average becomes,
= A2 …… (2)
Subtracting 1 from 2, we get,
N = ((n+1) x A2) –(n x A1)
N=n x (A2– A1)+ A2
Example: The average age of 12 students is 40. If the age of the teacher is also included, then the average becomes 42. Then what will be the age of the teacher?
Solution: Here average age of students = 40
Number of students = 12
The new average age after including teacher = 42
So here n = 12, A1 = 40, A2 = 42
Using the formula N = n x (A2 – A1) + A2 we have,
N = 12 × 2 + 42 = 24 + 42 = 66
Therefore, the age of the teacher is 66 years.
Example: The average age of 30 students is 15 years. If the age of class teacher is included, the average increases by 1. What is the age of the class teacher?
Solution: We have N= n× (increase in value of average) +A2
N= 30 × 1 +16 = 46 years
Concept 2
Let average of n quantitiesa1, a2, a3……anbeA1. If a number is removed from the series and this decreases the average to A2, then thevalue of the number removed will be:
N = n x (A1– A2) + A2
Or N= n × (decrease in value of average) + A2
Derivation:
The average of n quantitiesa1, a2, a3……anis
……. (1)
After the removal of number N, the average becomes,
…… (2)
Subtracting2 from 1, we get
N = n x (A1– A2) + A2
Example: The average age of 30 students and a teacher is 15 years. If the age of teacher is excluded, the average decreases by 1.What is the age of the class teacher?
Solution:Here n = 31, A1 = 15, A2 = 14
We have N= n × (decrease in value of average) + A2
N= 31 × 1+14=45years
Concept 3
Let average of n quantitiesa1, a2, a3……anbe A1. Now a number from the series say a1is replaced with a new numbersay M, and becauseof this,the average increased to A2, then,
M = a1 + n x (A2 – A1)
Or New term = Replaced Term + (increased in average × number of terms).
Derivation:
The average of n quantitiesa1, a2, a3……an is,
……. (1)
Let us say a1 is replaced with M, then,
…… (2)
Subtracting(1) from (2), we get:
M–a1= n x (A2 – A1)
M= a1+ n x (A2 – A1)
Example: The average age of 6 students is increased by 2 years when one student whose age was 13 years replaced by a new boy.Find the age of the new boy.
Solution: The age of the boy will be = Age of the replaced boy +(increase in average × number of terms) i.e., the age of the newly added boy = 13 + 2 x 6 = 25
To get a tight hold on the above concepts let us solve the exercise on calculating averages below.
EXERCISE
Question 1:Average weight of 5 men increased by 4 kg when one of them whose weight is 96kg is replaced by another man. What is the weight of the new man?
(1)108
(2) 116
(3) 126
(4) 130
Answer and Explanation
Solution: Option 2
Let the initial average bez.So, the total weight will be 5z.
Let the weight of the new man be y kg and the new average is z+ 4.
We have 5z – 96 + y = 5 x (z + 4)
Þ 5z – 96 + y = 5z + 20
Þ y = 96 + 20 = 116 kg
Or we can use theformula:
Weight of new person added = weight of the person removed + (no of person ×increase in average)
Weight of new person added = 96+5 × 4=116 kg
Question 2: The average of a batsman after 30 innings was 50 runs per innings. If after the 31 innings his average increased by 4 runs,then what was his score in 31stinning?
(1) 174
(2) 175
(3) 176
(4) 177
Answer and Explanation
Solution: Option 1
Here, Thenew average = 50+4=54 runs
N = n × (increase in value of average) + A2
N= 30 × 4 +54=174 runs
Or we have
Runs in 31st inning = runs in 31 innings – runs in 30 innings
Runs in 31st inning= 31 × 54- 30 × 50=1674 – 1500 = 174 runs
Question 3: The average age of 10 students is 18 years. If the age of class teacher is included, the average increases by 1. What is the age of the class teacher?
(1) 30
(2) 40
(3) 29
(4) 28
Answer and Explanation
Solution: Option 3
N = n × (increase in value of average) + A2
N= (10 × 1) + 19 = 10 + 19 = 29 years.
Question 4: The batting average of a cricket player for 64 innings is 62 runs. His highest score exceeds his lowest score by 180 runs. Excluding these two innings, the average of remaining innings becomes 60 runs. His highest score was:
(1) 180 runs
(2) 209 runs
(3) 212 runs
(4) 214 runs
Answer and Explanation
Solution: Option 4
Let the cricketer’s highest score bey runs. Then the lowest will be y – 180.
Therefore, the sum will be 60 × 62 + y + y – 180 = 64 × 62
Question 5: A cricketer has a mean score of 60 runs in 10 innings. Find out how many runs are to be scored in the eleventh innings to raise the mean score to 62?
(1) 83
(2) 82
(3) 80
(4) 81
Answer and Explanation
Solution: Option 2
Runs in 10 innings = 60 × 10=600
Runs in 11 innings = 62 × 11=682
So, runs scored in 11th innings =682-600=82
Averages Questions: Problems on averages you should solve for competitive examination preparation
Welcome to this exercise on Problems on Averages. In this exercise, we build on the basic concepts for finding the Average. As you explore this topic, you will come across questions where you will be needing to find averages where either a new term is added or subtracted, or a term is replaced. Such questions need optimized tackling and can be solved with ease by using the formulas and understanding the relationships highlighted in this Averages Questions article. The Averages Questions exercise comes into
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