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10100 | http://www.e-magnetica.pl/doku.php/elementary_charge | Elementary charge [ Encyclopedia Magnetica™ ]
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Elementary charge
Fractional charges
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References
Elementary charge
Stan Zurek, Elementary charge, Encyclopedia Magnetica,
Elementary charge (e) - the smallest amount of electric charge having the value of: e = 1.602 176 634 × 10−19coulomb.1)2). This value is a constant in our universe.3)
A proton has a positive charge of +e, and electron to the exactly opposite, negative value of -e (neutron has zero charge). The matching of the amount of the quantum of positive and negative charges is extremely precise to the highest experimental accuracy that can be attained, at the level of 1 part in 10 20. If this was not the case then matter would violently disintegrate.4)5)
The electric charges in antimatter are reversed, with positron (equivalent of electron) being positive, and antiproton negative. It is possible for positive and negative charges (e.g. electron and positron) to combine and annihilate, converting to other form of energy. It is also possible for two opposing charges to be produced in some sub-atomic interaction. But such interactions always occur in pairs of positive-negative charges, so that the law of charge conservation never violated.6) For example, during radioactive decay it is possible for a positron (e+) to be emitted from a proton (e+), which then becomes a neutron so that the amount of electric charge remains constant.7)
Electrically charged particles also exhibit intrinsic magnetic moment. Electron magnetic moment is especially strong and is responsible for the phenomenon of ferromagnetism.8)
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Fractional charges
Only such sub-atomic particles like quarks are thought to have electric electric charge in non-integer quantities e.g. -1/3 e or +2/3 e, but they only exists in configurations which add up to integer values of charge. For example, proton comprises three quarks (up, up, down), which add up to +1 e. Therefore, in any macroscopic application the charge is always quantised by the elementary amount of 1 e.9)10)
See also
Electric charge
Electron
Proton
References
1)Bureau International des Poids et Mesures, The International System of Units (SI), 9th edition, 2019, {accessed 2021-04-10}
2)P.A. Zyla et al. (Particle Data Group), Prog. Theor. Exp. Phys. 2020, 083C01 (2020)
3)Encyclopædia Britannica, Electric charge, {accessed 2020-10-25}
4), 6), 8), 9)E.M. Purcell, D.J. Morin, Electricity and magnetism, 3rd edition, Cambridge University Press, 2013, ISBN 9781107014022
5)David J. Griffiths, Introduction to electrodynamics, 4th ed., Pearson, Boston, 2013, ISBN 0321856562
7)Charles H. Holbrow, James N. Lloyd, Joseph C. Amato, Enrique Galvez, M. Elizabeth Parks, Modern Introductory Physics, 2nd ed., Springer, New York, ISBN 9780387790794
10)David Tong, Electromagnetism, University of Cambridge Part IB and Part II Mathematical Tripos, Lent Term, 2015, {accessed 2020-10-16}
Electric charge, Elementary electric charge, Elementary charge, Electron, Proton, Antiproton, Positron, Antielectron
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10101 | https://mechanicalc.com/reference/beam-analysis | Stresses & Deflections in Beams
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Beam Calculator
Many structures can be approximated as a straight beam or as a collection of straight beams. For this reason, the analysis of stresses and deflections in a beam is an important and useful topic.
This section covers shear force and bending moment in beams, shear and moment diagrams, stresses in beams, and a table of common beam deflection formulas.
Contents
Related Pages:
• Beam Formula Sheet
• Strength of Materials
• Beam Calculator
Constraints and Boundary Conditions
For a beam to remain in static equilibrium when external loads are applied to it, the beam must be constrained. Constraints are defined at single points along the beam, and the boundary condition at that point determines the nature of the constraint. The boundary condition indicates whether the beam is fixed (restrained from motion) or free to move in each direction. For a 2-dimensional beam, the directions of interest are the x-direction (axial direction), y-direction (transverse direction), and rotation. For a constraint to exist at a point, the boundary condition must indicate that at least one direction is fixed at that point.
Common boundary conditions are shown in the table below. For each boundary condition, the table indicates whether the beam is fixed or free in each direction at the point where the boundary condition is defined.
| Boundary Condition | Direction | | |
--- --- |
| Axial (X) | Transverse (Y) | Rotation |
| Free | Free | Free | Free |
| Fixed | Fixed | Fixed | Fixed |
| Pinned | Fixed | Fixed | Free |
| Guided along X | Free | Fixed | Fixed |
| Guided along Y | Fixed | Free | Fixed |
| Roller along X | Free | Fixed | Free |
| Roller along Y | Fixed | Free | Free |
If the boundary condition indicates that the beam is fixed in a specific direction, then an external reaction in that direction can exist at the location of the boundary condition. For example, if a beam is fixed in the y-direction at a specific point, then a transverse (y) external reaction force may develop at that point. Likewise, if the beam is fixed against rotation at a specific point, then an external reaction moment may develop at that point.
Based on the above discussion, we can see that a fixed boundary condition can develop axial and transverse reaction forces as well as a moment. Likewise, we see that a pinned boundary condition can develop axial and transverse reaction forces, but it cannot develop a reaction moment.
Notice the Free boundary condition in the table above. This boundary condition indicates that the beam is free to move in every direction at that point (i.e., it is not fixed or constrained in any direction). Therefore, a constraint does not exist at this point. This highlights the subtle difference between a constraint and a boundary condition. A boundary condition indicates the fixed/free condition in each direction at a specific point, and a constraint is a boundary condition in which at least one direction is fixed.
Shear Force and Bending Moment
To find the shear force and bending moment over the length of a beam, first solve for the external reactions at each constraint. For example, the cantilever beam below has an applied force shown as a red arrow, and the reactions are shown as blue arrows at the fixed boundary condition.
The external reactions should balance the applied loads such that the beam is in static equilibrium. After the external reactions have been solved for, take section cuts along the length of the beam and solve for the internal reactions at each section cut. (The reaction forces and moments at the section cuts are called internal reactions because they are internal to the beam.) An example section cut is shown in the figure below:
When the beam is cut at the section, either side of the beam may be considered when solving for the internal reactions. The side that is selected does not affect the results, so choose whichever side is easiest. In the figure above, the side of the beam to the right of the section cut was selected. The selected side is shown as the blue section of beam, and section shown in grey is ignored. The internal reactions at the section cut are shown with blue arrows. The reactions are calculated such that the section of beam being considered is in static equilibrium.
Sign Convention
The signs of the shear and moment are important. The sign is determined after a section cut is taken and the reactions are solved for the portion of the beam to one side of the cut. The shear force at the section cut is considered positive if it causes clockwise rotation of the selected beam section, and it is considered negative if it causes counter-clockwise rotation. The bending moment at the section cut is considered positive if it compresses the top of the beam and elongates the bottom of the beam (i.e., if it makes the beam "smile").
Based on this sign convention, the shear force at the section cut for the example cantilever beam in the figure above is positive since it causes clockwise rotation of the selected section. The moment is negative since it compresses the bottom of the beam and elongates the top (i.e., it makes the beam "frown").
The figure below shows the standard sign convention for shear force and bending moment. The forces and moments on the left are positive, and those on the right are negative.
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Shear and Moment Diagrams
The shear force and bending moment throughout a beam are commonly expressed with diagrams. A shear diagram shows the shear force along the length of the beam, and a moment diagram shows the bending moment along the length of the beam. These diagrams are typically shown stacked on top of one another, and the combination of these two diagrams is a shear-moment diagram. Shear-moment diagrams for some common end conditions and loading configurations are shown within the beam deflection tables at the end of this page. An example of a shear-moment diagram is shown in the following figure:
General rules for drawing shear-moment diagrams are given in the table below. All of the rules in this table are demonstrated in the figure above.
| Shear Diagram | Moment Diagram |
--- |
| Point loads cause a vertical jump in the shear diagram. The direction of the jump is the same as the sign of the point load. Uniform distributed loads result in a straight, sloped line on the shear diagram. The slope of the line is equal to the value of the distributed load. The shear diagram is horizontal for distances along the beam with no applied load. The shear at any point along the beam is equal to the slope of the moment at that same point: | The moment diagram is a straight, sloped line for distances along the beam with no applied load. The slope of the line is equal to the value of the shear. Uniform distributed loads result in a parabolic curve on the moment diagram. The maximum/minimum values of moment occur where the shear line crosses zero. The moment at any point along the beam is equal to the area under the shear diagram up to that point: M = ∫ V dx |
Bending Stresses in Beams
The bending moment, M, along the length of the beam can be determined from the moment diagram. The bending moment at any location along the beam can then be used to calculate the bending stress over the beam's cross section at that location. The bending moment varies over the height of the cross section according to the flexure formula below:
where M is the bending moment at the location of interest along the beam's length, Ic is the centroidal moment of inertia of the beam's cross section, and y is the distance from the beam's neutral axis to the point of interest along the height of the cross section. The negative sign indicates that a positive moment will result in a compressive stress above the neutral axis.
The bending stress is zero at the beam's neutral axis, which is coincident with the centroid of the beam's cross section. The bending stress increases linearly away from the neutral axis until the maximum values at the extreme fibers at the top and bottom of the beam.
The maximum bending stress occurs at the extreme fiber of the beam and is calculated as:
where c is the centroidal distance of the cross section (the distance from the centroid to the extreme fiber).
If the beam is asymmetric about the neutral axis such that the distances from the neutral axis to the top and to the bottom of the beam are not equal, the maximum stress will occur at the farthest location from the neutral axis. In the figure below, the tensile stress at the top of the beam is larger than the compressive stress at the bottom.
The section modulus of a cross section combines the centroidal moment of inertia, Ic, and the centroidal distance, c:
The benefit of the section modulus is that it characterizes the bending resistance of a cross section in a single term. The section modulus can be substituted into the flexure formula to calculate the maximum bending stress in a cross section:
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Calculates stresses and deflections in straight beams
Builds shear and moment diagrams
Can specify any configuration of constraints, concentrated forces, and distributed forces
Shear Stresses in Beams
The shear force, V, along the length of the beam can be determined from the shear diagram. The shear force at any location along the beam can then be used to calculate the shear stress over the beam's cross section at that location. The average shear stress over the cross section is given by:
The shear stress varies over the height of the cross section, as shown in the figure below:
The shear stress is zero at the free surfaces (the top and bottom of the beam), and it is maximum at the centroid. The equation for shear stress at any point located a distance y1 from the centroid of the cross section is given by:
where V is the shear force acting at the location of the cross section, Ic is the centroidal moment of inertia of the cross section, and b is the width of the cross section. These terms are all constants. The Q term is the first moment of the area bounded by the point of interest and the extreme fiber of the cross section:
Shear stresses for several common cross sections are discussed in the sections below.
Shear Stresses in Rectangular Sections
The distribution of shear stress along the height of a rectangular cross section is shown in the figure below:
The first moment of area at any given point y1 along the height of the cross section is calculated by:
The maximum value of Q occurs at the neutral axis of the beam (where y1 = 0):
The shear stress at any given point y1 along the height of the cross section is calculated by:
where Ic = b·h3/12 is the centroidal moment of inertia of the cross section. The maximum shear stress occurs at the neutral axis of the beam and is calculated by:
where A = b·h is the area of the cross section.
We can see from the previous equation that the maximum shear stress in the cross section is 50% higher than the average stress V/A.
Shear Stresses in Circular Sections
A circular cross section is shown in the figure below:
The equations for shear stress in a beam were derived using the assumption that the shear stress along the width of the beam is constant. This assumption is valid at the centroid of a circular cross section, although it is not valid anywhere else. Therefore, while the distribution of shear stress along the height of the cross section cannot be readily determined, the maximum shear stress in the section (occurring at the centroid) can still be calculated. The maximum value of first moment, Q, occurring at the centroid, is given by:
The maximum shear stress is then calculated by:
where b = 2r is the diameter (width) of the cross section, Ic = πr4/4 is the centroidal moment of inertia, and A = πr2 is the area of the cross section.
Shear Stresses in Circular Tube Sections
A circular tube cross section is shown in the figure below:
The maximum value of first moment, Q, occurring at the centroid, is given by:
The maximum shear stress is then calculated by:
where b = 2 (ro − ri) is the effective width of the cross section, Ic = π (ro4 − ri4) / 4 is the centroidal moment of inertia, and A = π (ro2 − ri2) is the area of the cross section.
Shear Stresses in I-Beams
The distribution of shear stress along the web of an I-Beam is shown in the figure below:
The equations for shear stress in a beam were derived using the assumption that the shear stress along the width of the beam is constant. This assumption is valid over the web of an I-Beam, but it is invalid for the flanges (specifically where the web intersects the flanges). However, the web of an I-Beam takes the vast majority of the shear force (approximately 90% - 98%, according to Gere), and so it can be conservatively assumed that the web carries all of the shear force.
The first moment of the area of the web of an I-Beam is given by:
The shear stress along the web of the I-Beam is given by:
where tw is the web thickness and Ic is the centroidal moment of inertia of the I-Beam:
The maximum value of shear stress occurs at the neutral axis ( y1 = 0 ), and the minimum value of shear stress in the web occurs at the outer fibers of the web where it intersects the flanges y1 = ±hw/2 ):
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Beam Deflection Tables
The tables below give equations for the deflection, slope, shear, and moment along straight beams for different end conditions and loadings. You can find comprehensive tables in references such as Gere, Lindeburg, and Shigley. However, the tables below cover most of the common cases.
Cantilever Beams
| | | | | | | | | | | | | |
--- --- --- --- --- ---
| Cantilever, End Load | | | | | | | | | --- | | | @ x = L | | | | | | | | --- | | | @ x = L | | | | V = +F | | | | M = −F (L − x) | | | | --- | | Mmax = −FL | @ x = 0 | |
| Cantilever, Intermediate Load | | Deflection: | | | --- | | | ( 0 ≤ x ≤ a ) | | | ( a ≤ x ≤ L ) | | | | --- | | | @ x = L | Slope: | | | --- | | | ( 0 ≤ x ≤ a ) | | | ( a ≤ x ≤ L ) | Shear: | | | --- | | V = +F | ( 0 ≤ x ≤ a ) | | V = 0 | ( a ≤ x ≤ L ) | Moment: | | | --- | | M = −F (a − x) | ( 0 ≤ x ≤ a ) | | M = 0 | ( a ≤ x ≤ L ) | |
| Cantilever, Uniform Distributed Load | | Deflection: | | | | | | | --- | | | @ x = L | Slope: | | | | | | | --- | | | @ x = L | Shear: | | | V = +w (L − x) | | | | --- | | Vmax = +wL | @ x = 0 | Moment: | | | M = −w (L − x)2 / 2 | | | | --- | | Mmax = −wL2 / 2 | @ x = 0 | |
| Cantilever, Triangular Distributed Load | | Deflection: | | | | | | | --- | | | @ x = L | Slope: | | | | | | | --- | | | @ x = L | Shear: | | | --- | | Vmax = +w1L / 2 | @ x = 0 | Moment: | | | --- | | Mmax = −w1L2 / 6 | @ x = 0 | |
| Cantilever, End Moment | | Deflection: | | | | | | | --- | | | @ x = L | Slope: | | | | | | | --- | | | @ x = L | Shear: | | | V = 0 | Moment: | | | M = −M0 | |
Simply Supported Beams
| | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- ---
| Simply Supported, Intermediate Load | | Deflection: | | | --- | | | ( 0 ≤ x ≤ a ) | For a ≥ b: | | | --- | | | | Slope: | | | --- | | | ( 0 ≤ x ≤ a ) | | | @ x = 0 | | | @ x = L | Shear: | | | --- | | V1 = +Fb / L | ( 0 ≤ x ≤ a ) | | V2 = −Fa / L | ( a ≤ x ≤ L ) | Moment: | | | --- | | Mmax = +Fab / L | @ x = a | |
| Simply Supported, Center Load | | Deflection: | | | --- | | | ( 0 ≤ x ≤ L/2 ) | | | @ x = L/2 | Slope: | | | --- | | | ( 0 ≤ x ≤ L/2 ) | | | @ x = 0 | | | @ x = L | Shear: | | | --- | | V1 = +F / 2 | ( 0 ≤ x ≤ L/2 ) | | V2 = −F / 2 | ( L/2 ≤ x ≤ L ) | Moment: | | | --- | | Mmax = FL / 4 | @ x = L/2 | |
| Simply Supported, 2 Loads at Equal Distances from Supports | | Deflection: | | | --- | | | ( 0 ≤ x ≤ a ) | | | ( a ≤ x ≤ L − a ) | | | @ x = L/2 | Slope: | | | --- | | | ( 0 ≤ x ≤ a ) | | | ( a ≤ x ≤ L − a ) | | | @ x = 0 | | | @ x = L | Shear: | | | --- | | V1 = +F | ( 0 ≤ x ≤ a ) | | V2 = −F | ( L − a ≤ x ≤ L ) | Moment: | | | --- | | Mmax = Fa | ( a ≤ x ≤ L − a ) | |
| Simply Supported, Uniform Distributed Load | | Deflection: | | | | | | | --- | | | @ x = L/2 | Slope: | | | | | | | --- | | | @ x = 0 | | | @ x = L | Shear: | | | V = w (L/2 − x) | | | | --- | | V1 = +wL / 2 | @ x = 0 | | V2 = −wL / 2 | @ x = L | Moment: | | | --- | | Mmax = wL2 / 8 | @ x = L/2 | |
| Simply Supported, Moment at Each Support | | Deflection: | | | | | | | --- | | | @ x = L/2 | Slope: | | | | | | | --- | | | @ x = 0 | | | @ x = L | Shear: | | | V = 0 | Moment: | | | M = M0 | |
| Simply Supported, Moment at One Support | | Deflection: | | | | | | | --- | | | @ x = L (1 − √3/3) | Slope: | | | | | | | --- | | | @ x = 0 | | | @ x = L | Shear: | | | V = −M0 / L | Moment: | | | --- | | Mmax = M0 | @ x = 0 | |
| Simply Supported, Center Moment | | Deflection: | | | --- | | | ( 0 ≤ x ≤ L/2 ) | Slope: | | | --- | | | ( 0 ≤ x ≤ L/2 ) | | | | --- | | | @ x = 0 | | | @ x = L | Shear: | | | V = +M0 / L | Moment: | | | --- | | M = M0x / L | ( 0 ≤ x ≤ L/2 ) | | Mmax = M0 / 2 | @ x = L/2 | |
Fixed-Fixed Beams
| | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- ---
| Fixed-Fixed, Center Load | | Deflection: | | | --- | | | ( 0 ≤ x ≤ L/2 ) | | | @ x = L/2 | Shear: | | | --- | | V1 = +F / 2 | ( 0 ≤ x ≤ L/2 ) | | V2 = −F / 2 | ( L/2 ≤ x ≤ L ) | Moment: | | | --- | | M = F (4x − L) / 8 | ( 0 ≤ x ≤ L/2 ) | | | | --- | | M1 = M3 = −FL / 8 | @ x = 0 & x = L | | M2 = +FL / 8 | @ x = L/2 | |
| Fixed-Fixed, Uniform Distributed Load | | Deflection: | | | | | | | --- | | | @ x = L/2 | Shear: | | | V = w (L/2 − x) | | | | --- | | V1 = +wL / 2 | @ x = 0 | | V2 = −wL / 2 | @ x = L | Moment: | | | M = w (6Lx − 6x2 − L2) / 12 | | | | --- | | M1 = M3 = −wL2 / 12 | @ x = 0 & x = L | | M2 = wL2 / 24 | @ x = L/2 | |
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References
Budynas-Nisbett, "Shigley's Mechanical Engineering Design," 8th Ed.
Gere, James M., "Mechanics of Materials," 6th Ed.
Lindeburg, Michael R., "Mechanical Engineering Reference Manual for the PE Exam," 13th Ed.
"Stress Analysis Manual," Air Force Flight Dynamics Laboratory, October 1986.
Shigley, "Mechanical Engineering Design"
Gere, "Mechanics of Materials"
Lindeburg, "Mechanical Engineering Reference Manual" |
10102 | https://www.dadisp.com/webhelp/mergedProjects/refman2/SPLGROUP/ERFINV.htm | ERFINV - DADiSP
ERFINV
Purpose:
Returns the inverse error function.
Syntax:
ERFINV(y)
y-A real or series.
Returns:
A real or series, the inverse error function x where y = erf(x) for -1 <= y<= 1 and -∞<=x<=∞.
Example:
erfinv(.2)
returns 0.17914345, the inverse error function of 0.2.
Example:
erf(erfinv(.2))
returns 0.2, indicating that ERF and ERFINV are inverse functions.
Example:
erfinv(-1..0.01..1);
xlabel("x");ylabel("erfinv(x)");label("Inverse Error Function");
returns 201 samples of erfinv(x).
Remarks:
ERFINV uses the built-in INVPROBN function to find the z value for a given probability value of a normal distribution.
erf(erfinv(x)) == x
erfinv(x) == invprobn((x+1)/2) / sqrt(2)
See Also:
ERF
ERFC
ERFCINV
ERFCX
ERFI
FADDEEVA
DADiSP® Online Reference
Copyright© 1995-2025 DSP Development Corporation All rights reserved. |
10103 | https://kamman-dynamics-control.org/wp-content/uploads/2020/11/09-me471matlabrootlocusanalysis.pdf | Kamman – Introductory Motion and Control – Using MATLAB for a Root Locus Analysis – page: 1/3 Introductory Motion and Control Using MATLAB® for Root Locus Analysis Reference: Dorf & Bishop, Modern Control Systems, Pearson/Prentice-Hall, 11th Ed., 2008. As an example of how to use MATLAB to perform a root locus analysis, consider design problem DP6.4 of Dorf & Bishop. The block diagram of the closed-loop system is shown below. The goal is to use MATLAB to draw a root locus diagram for the parameter K. The characteristic equation of the closed-loop system is 0 ) ( 1 = + s GH or 0 ) ( 1 = + s KP . Substituting the transfer functions from the block diagram into the characteristic equation gives ( )( ) 2 2 3 4 2 6 8 1 1 0 ( 1) s s s s K K s s s s + + + + + = + = − − The MATLAB commands that produce the root locus diagram are: num = [1,6,8]; den = [1, 0, -1, 0]; sys = tf(num,den) rlocus(sys) axis ('equal') title('Root Locus Diagram for K (m=4)') In response to the “tf” command, MATLAB provides the transfer function in the command window. sys = s^2 + 6 s + 8 ------------- s^3 - s Continuous-time transfer function. The three branches of the diagram are shown in green, blue, and red. Note, however, that MATLAB does not show the direction of the movement of the poles as K increases. It is Rocket Dynamics PID Controller Y(s) R(s) + Kamman – Introductory Motion and Control – Using MATLAB for a Root Locus Analysis – page: 2/3 understood the movement is from the poles of ( ) P s to the zeros of ( ) P s . In this case, there are three branches, but only two zeros so one of the branches must go to infinity along an asymptote. As stated in previous notes, if there is only one asymptote, it is at 180 degrees. The “axis” command ensures that the diagram is shown in its true shape. Target Regions for Poles o The damping ratios and settling times of the poles are determined by their location in the s-plane. o To ensure a settling time less than s T , the real parts of all the poles of the system should be to the left of s T 4 . o The damping ratio of each of the complex poles is determined by drawing a vector from the origin to the location of the pole and measuring the angle between this vector and the negative real axis. o The damping ratio is calculated as cos( ) = . For example, poles that lie below the o 45 = line have damping ratios 0.707 . Parameter Values Associated with Poles in the Target Region To find parameter values associated with poles within the target region, use the “rlocfind” command in MATLAB. After executing the “rlocfind” command, click on a desirable pole location on one of the branches of the root locus diagram in the plot window. MATLAB automatically picks the point on the branch that is closest to your selection. X Kamman – Introductory Motion and Control – Using MATLAB for a Root Locus Analysis – page: 3/3 The MATLAB commands are: grid [k, poles] = rlocfind(sys) Note the grid is not the usual rectangular grid but rather a polar grid which aligns with constant values of damping ratio (radial lines) and frequency (concentric circles). In response to the user clicking on the selected point, MATLAB places a “+” at the location of the associated poles, and the following data is provided in the command window. selected point = -3.900473933649288 + 5.249262688362950i k = 9.609310443185597 poles = -3.901567707687948 + 5.228747819083528i -3.901567707687948 - 5.228747819083528i -1.806175027809703 + 0.000000000000000i So, in this case, for a K value of approximately 9.61, the closed-loop system has one real pole at 1.8 − and a pair of complex conjugate poles at 3.90 5.23 j − . The complex poles have a damping ratio of approximately 0.6 = . The frequency associated with the real pole is 1.8 (rad/s), and the frequency associated with the complex poles is 6.52 (rad/s). The settling time associated with the real pole is 4 1.8 2.22 (sec) , and the settling time associated with the complex poles is 4 3.90 1.03 (sec) . |
10104 | https://www.youtube.com/watch?v=OKsIh-pqGaE | Find the Equation of a Circle Given Endpoints of Diameter
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so we're only given these two end points of the diameter of a circle and we want to write the equation in standard form which is this form right here HK is the center of the circle but how do we find the center just given these two end points well we're gonna want to use our midpoint formula which is this one down here and you can see that it's an average of the X's and an average of the Y's so let's go ahead and do that so if we take two plus eight I'm just adding the two x-coordinates in dividing by two and I'm going to take one plus seven adding the two y-coordinates together in dividing by two we'll find that middle point the midpoint so we've got ten divided by two eight divided by two so you can see that our midpoint here is five comma four right five up four somewhere right into there now what we want to do is go ahead and put that in for each and K so that's going to be X minus five the quantity squared plus y minus four the quantity squared equals the radius squared now how do we find the radius well you can do this two different ways one way is you can use the distance formula between the center and either this endpoint or this endpoint or you can do what I like to do which is just to take this equation as we have it in this form now and just pick one of the endpoints that's a point on our circle and put it in for X and Y and what we'll do is we'll solve for R squared so we have 2 minus 5 which is negative 3 squared plus 1 minus 4 which is also negative 3 squared equals our radius squared so this is 9 plus 9 which is 18 which equals R squared now you don't need to take the square root of both sides because we're actually looking for R squared so let's go ahead and just put that in for R squared and you've got the equation of your circle in standard form if you want to see another video that I did talking about finding the equations of circles in standard form by completing the square check out the video I did right there and I'll see over in that video |
10105 | https://www.merckmanuals.com/professional/psychiatric-disorders/anxiety-and-stressor-related-disorders/agoraphobia | honeypot link
IN THIS TOPIC
Symptoms and Signs
Diagnosis
Treatment
OTHER TOPICS IN THIS CHAPTER
Overview of Anxiety Disorders
Specific Phobias
Social Anxiety Disorder
Panic Attacks and Panic Disorder
Agoraphobia
Generalized Anxiety Disorder
Overview of Trauma- and Stressor-Related Disorders
Posttraumatic Stress Disorder (PTSD)
Acute Stress Disorder (ASD)
Adjustment Disorders
Agoraphobia
By
John W. Barnhill, MD, New York-Presbyterian Hospital
Reviewed By
Mark Zimmerman, MD, South County Psychiatry
Reviewed/Revised Modified Aug 2023
v11688162
View Patient Education
Agoraphobia is intense anxiety and/or avoidance of situations (eg, being in crowds or shopping malls, driving) that may be difficult to leave or in which help is not readily available if incapacitating panic-like symptoms were to develop. Diagnosis is based on clinical criteria. Treatment focuses on cognitive-behavioral therapy, specifically, exposure therapy.
Symptoms and Signs
|
Diagnosis
|
Treatment
|
Topic Resources
Overview of Phobias
(See also
Overview of Anxiety Disorders
.)
Agoraphobia is a common consequence of panic disorder, but the 2 disorders can also develop independently.
Agoraphobia affects about 2% of the population in a given year and is more common in women (
1
). Agoraphobia often develops in adolescence and young adulthood, but it can also develop in older adults, especially in the context of fears about safety and their own physical limitations.
General reference
Roest AM, de Vries YA, Lim CCW, et al
: A comparison of DSM-5 and DSM-IV agoraphobia in the World Mental Health Surveys.
Depress Anxiety
36(6):499-510, 2019. doi: 10.1002/da.22885
Overview of Phobias
video
Symptoms and Signs of Agoraphobia
Common examples of situations or places that create fear and anxiety in patients with agoraphobia include leaving home, standing in line, sitting in the middle of a long row in a theater or classroom, and using public transportation, such as a bus or an airplane. Some people develop agoraphobia in response to panic attacks that lead to avoidance of the potential triggers of the panic. Agoraphobia can be relatively mild but can also become so debilitating that the person becomes essentially housebound.
As is true for other anxiety disorders, symptoms of agoraphobia may wax and wane in severity.
Diagnosis of Agoraphobia
Diagnostic and Statistical Manual of Mental Disorders
, 5th edition, Text Revision (DSM-5-TR) criteria
To meet the DSM-5-TR criteria for agoraphobia, patients must have marked, persistent (
≥
6 months) fear of or anxiety about 2 or more of the following situations (
1
):
Using public transportation
Being in open spaces (eg, parking lot, marketplace)
Being in an enclosed place (eg, shop, theater)
Standing in line or being in a crowd
Being alone outside the home
Fear must involve thoughts that escape from the situation might be difficult or that patients would receive no help if they became incapacitated by fear or a panic attack. In addition, all of the following should be present:
The same situations nearly always trigger fear or anxiety.
Patients actively avoid the situation and/or require the presence of a companion.
The fear or anxiety is out of proportion to the actual threat (taking into account sociocultural norms).
The fear, anxiety, and/or avoidance cause significant distress or significantly impair social or occupational functioning.
If another medical condition (eg, inflammatory bowel disease, Parkinson disease) is present, the fear, anxiety, and/or avoidance are clearly excessive.
In addition, the fear and anxiety cannot be better characterized as a different mental disorder (eg,
social anxiety disorder
,
body dysmorphic disorder
).
Diagnosis reference
Diagnostic and Statistical Manual of Mental Disorders
, 5th edition, Text Revision DSM-5-TR. American Psychiatric Association Publishing, Washington, DC, pp 246-250.
Treatment of Agoraphobia
Cognitive-behavioral therapy (CBT)
Exposure therapy
Selective serotonin reuptake inhibitors (SSRI)
The most effective treatment approach, based on the most robust evidence, is exposure therapy that uses CBT principles (
1
). Agoraphobia may resolve without formal treatment, possibly because some affected people conduct their own form of exposure therapy and also because anxiety symptoms (and precipitating stressors) fluctuate with time.
Many patients with agoraphobia also benefit from pharmacotherapy with an SSRI (
2
).
Treatment references
Carpenter JK, Andrews LA, Witcraft SM, et al
: Cognitive behavioral therapy for anxiety and related disorders: A meta-analysis of randomized placebo-controlled trials.
Depress Anxiety
35(6):502-514, 2018. doi: 10.1002/da.22728
Chawla N, Anothaisintawee T, Charoenrungrueangchai K, et al
: Drug treatment for panic disorder with or without agoraphobia: Systematic review and network meta-analysis of randomised controlled trials.
BMJ
376:e066084, 2022. doi: 10.1136/bmj-2021-066084
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Principles of Moving Target Display Radar
Principles of Moving Target Display Radar
Author : Adrian September 12, 2025
Table of Contents
1. The Need for MTD Radar
2. Principle of Operation
3. Key Components of MTD Radar
4. Advantages of MTD Radar
5. Limitations
6. Applications
Conclusion
Principles of Moving Target Display (MTD) Radar
Radar systems are widely used in surveillance, navigation, and defense. One critical function is the ability to distinguish moving targets (such as aircraft, ships, or vehicles) from unwanted stationary echoes (clutter from ground, buildings, or terrain). This is where the Moving Target Display (MTD) Radar concept comes in.
1. The Need for MTD Radar
When a radar transmits a pulse, the receiver picks up reflections from everything in its path:
Ground, sea, and buildings → stationary clutter
Aircraft, ships, vehicles → moving targets
If clutter dominates the radar screen, the true moving objects can be masked. MTD radar solves this problem by filtering out stationary returns and highlighting moving objects.
2. Principle of Operation
The basic idea relies on the Doppler effect:
A stationary object reflects radar waves at the same frequency as transmitted.
A moving object reflects waves with a slightly shifted frequency (Doppler shift), proportional to its velocity.
MTD radar detects this frequency shift to distinguish moving targets from clutter.
3. Key Components of MTD Radar
Pulse Doppler Processing
Transmits a train of pulses and measures phase changes between successive echoes.
Extracts the Doppler frequency of returns.
Clutter Filters (MTI/MTD Filters)
Stationary or slow-moving clutter produces near-zero Doppler shifts.
These are filtered out using high-pass or band-pass filters in the frequency domain.
Fast Fourier Transform (FFT) Analysis
The Doppler spectrum of received signals is computed.
Each Doppler bin corresponds to a velocity range.
Targets with measurable Doppler shifts stand out against clutter.
4. Advantages of MTD Radar
Clear target detection even in heavy clutter environments (e.g., near terrain or sea).
Velocity measurement of targets in addition to range and angle.
Improved reliability for air defense, air traffic control, and maritime surveillance.
5. Limitations
Blind speeds: Certain target velocities may produce zero Doppler shift relative to the radar’s pulse repetition frequency (PRF), making them harder to detect.
Complex signal processing requirements.
Less effective against very slow-moving targets (close to clutter speed).
6. Applications
Air defense radars (detecting aircraft among ground clutter).
Air traffic control systems.
Maritime surveillance (detecting ships in sea clutter).
Weather radar, to separate moving storm systems from static reflections.
Conclusion
The Moving Target Display Radar is built on the principle of exploiting Doppler frequency shifts to distinguish moving objects from stationary clutter. By combining pulse Doppler processing, clutter filters, and FFT analysis, MTD radar provides reliable detection of moving targets, making it a cornerstone of modern surveillance and defense systems.
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10107 | https://www.frontiersin.org/journals/pediatrics/articles/10.3389/fped.2022.911588/full | Your new experience awaits. Try the new design now and help us make it even better
REVIEW article
Front. Pediatr., 30 June 2022
Sec. Neonatology
Volume 10 - 2022 |
This article is part of the Research TopicUnsolved Problems in Congenital Diaphragmatic HerniaView all 17 articles
Use of Prostaglandin E1 in the Management of Congenital Diaphragmatic Hernia–A Review
Srirupa Hari Gopal1Neil Patel2Caraciolo J. Fernandes1
1Section of Neonatology, Department of Pediatrics, Baylor College of Medicine, Houston, TX, United States
2Department of Neonatology, Royal Hospital for Children, Glasgow, United Kingdom
Congenital diaphragmatic hernia (CDH) is a rare congenital anomaly, whose presentation is complicated by pulmonary hypertension (PH), pulmonary hypoplasia, and myocardial dysfunction, each of which have significant impact on short-term clinical management and long-term outcomes. Despite many advances in therapy and surgical technique, optimal CDH management remains a topic of debate, due to the variable presentation, complex pathophysiology, and continued impact on morbidity and mortality. One of the more recent management strategies is the use of prostaglandin E1 (PGE1) infusion in the management of PH associated with CDH. PGE1 is widely used in the NICU in critical congenital cardiac disease to maintain ductal patency and facilitate pulmonary and systemic blood flow. In a related paradigm, PGE1 infusion has been used in situations of supra-systemic right ventricular pressures, including CDH, with the therapeutic intent to maintain ductal patency as a “pressure relief valve” to reduce the effective afterload on the right ventricle (RV), optimize cardiac function and support pulmonary and systemic blood flow. This paper reviews the current evidence for use of PGE1 in the CDH population and the opportunities for future investigations.
Introduction
Congenital diaphragmatic hernia (CDH) is a rare anomaly, characterized by a defect in the diaphragm causing abdominal contents to protrude into the thoracic cavity. The incidence of CDH is 1 in 2,500 to 1 in 3,500 live births (1). It occurs 70%−75% of the time in the posterolateral aspect of the diaphragm, with over 85% occurring on the left side (2). CDH can also be associated with congenital heart defects (25%−40%), urogenital anomalies (18%), musculoskeletal anomalies (16%) and central nervous system anomalies (10%) (3, 4). Despite medical and surgical advances, CDH continues to have high mortality and morbidity rates (5). Pulmonary hypoplasia and pulmonary hypertension (PH) are hallmarks of CDH presentation, resulting from both pulmonary vasculature and respiratory maldevelopment, the severity of which determine outcomes. In addition to these factors, there is also an increasing appreciation of early postnatal cardiac dysfunction as a determinant of outcome in CDH and use of agents to ameliorate cardiac dysfunction (6, 7).
Prostaglandin E1 (PGE1), a prostaglandin analog administered by intravenous infusion, is typically used to maintain ductal patency in newborn infants in the setting of suspected duct-dependent congenital heart disease. In this setting, PGE1 infusion is used to ensure adequate systemic or pulmonary blood flow or mixing until definitive surgical repair is accomplished. In the setting of pulmonary hypertensive disease, including CDH, PGE1 may have other benefits, the most important being reduction of afterload of a failing right ventricle (RV).
In this review, we will discuss the pathophysiology of CDH, the theoretical benefits of PGE1 therapy in the management of CDH, critically review existing evidence of its use, and identify key questions for future areas of research.
Pathophysiology of CDH
Congenital diaphragmatic hernia is associated with pulmonary hypoplasia of varying extent typically affecting both the ipsilateral and contralateral lungs. This abnormal pathophysiology has been hypothesized to begin at ~8th to 10th-week of gestation, after failure of the normal physiological closure of the diaphragm and the establishment of the separation between abdominal and thoracic organs (8). A 2-hit hypothesis has been proposed to explain this spectrum of pulmonary hypoplasia. Following the initial “hit”, possibly genetic or environmental, during the early stages of organ development, bilateral lung hypoplasia occurs. This is followed by the second “hit” - compression of the ipsilateral lung by the hernia itself (9). Accompanying these changes in lung architecture are changes in the pulmonary vasculature, specifically early maturation, underdevelopment and increased muscularization of the pulmonary arterial vessels leading to altered vessel tone and reduced vessel caliber (10). Molecular pathways that are implicated in pulmonary vascular remodeling in CDH, which have been studied in humans and nitrofen rat models (11), include the retinol pathway (12), vascular endothelial growth factor (VEGF) (13), endothelin (14), Bone Morphogenic Protein (BMP) and Apelin (15). Alternations in these pathways may affect endothelial cell function, molecular signaling to the pulmonary arterial smooth muscle cells contributing to pulmonary arterial smooth muscle cell proliferation, and the characteristically hypertrophic pulmonary arterioles found in CDH-associated PH (CDH-PH) (6, 16).
The pulmonary alveolar and vascular maldevelopment results in increased pulmonary vascular resistance and associated PH. Studies have shown that over 70% of CDH infants exhibit CDH-PH (17), which is independently associated with increased mortality risk, oxygen support at 30 days, and utilization of extracorporeal life support (ECLS) (18). PH manifests as hypoxia due to right-to-left shunting across the atria, patent ductus arteriosus and any ventricular septal defect, if present, as well as increased afterload on the RV. In response to the increased afterload, the RV exhibits an initial adaptive dilatory response that may be followed by maladaptive hypertrophy and subsequent failure, which may be exacerbated by a restrictive ductus arteriosus. This RV failure may in turn result in impairment of diastolic filling of the left ventricle (LV) and reduced systemic blood flow. Myocardial ischemia of the RV plays an important role in the pathophysiology of cardiac failure in PH; specifically, through compromised right coronary blood flow. In PH, the right coronary perfusion gradient may be reduced due to the sustained increase in RV pressures and decrease in the aortic pressures (due to reduced LV preload and cardiac output) (19). Decreased RV coronary perfusion in the context of increasing myocardial oxygen consumption may predispose the RV to ischemia and dysfunction (20).
In addition, CDH has been described to be associated with both structural and functional left ventricular abnormalities (7, 21, 22). Fetal LV hypoplasia is well-described and possibly occurs due to mechanical compression, reduced fetal LV blood flow from reduced pulmonary venous return and altered streaming of venous return due to mediastinal shift (23–25). In a LV that may already be relatively hypoplastic due to the aforementioned reasons, the increase in afterload during the transition at birth, combined with the interdependent impacts of RV dilatation and dysfunction, can lead to significant LV dysfunction with adverse cardiopulmonary and hemodynamic outcomes (26). In a recent study, Patel et al. reported that early LV systolic function correlated with prenatal and postnatal markers of clinical disease severity (27). This observation underscores the importance of appropriate management of early PH in order to prevent biventricular dysfunction and associated impairment of systemic blood flow and oxygen delivery.
The combination and spectrum of pulmonary hypoplasia, pulmonary hypertension, and ventricular dysfunction makes CDH a unique clinical management challenge. Historically, PH therapeutic strategies in CDH have focused mainly on pulmonary vasodilation. The main therapeutic targets are cytokine pathways regulating pulmonary artery smooth muscle tone, specifically such the nitric oxide (NO) pathway, prostacyclin pathway and endothelin pathways (28).
Inhaled nitric oxide (iNO), a potent pulmonary vasodilator, acts by stimulating guanylyl cyclase in the vascular smooth muscle cells to produce cyclic guanosine monophosphate (cGMP). Elevated intracellular concentrations of cGMP activate cGMP-dependent protein kinases and lower cytosolic calcium concentrations, which in turn promote vascular smooth muscle cell relaxation (29). Phosphodiesterases (PDEs) are a large family of enzymes that hydrolyze cyclic nucleotides (cGMP and cAMP). Inhibition of PDEs leads to vasodilator effects. PDE5 (a cGMP-specific PDE), PDE3 and 4 (which hydrolyze cAMP) are expressed in the lung (28). Sildenafil, a PDE5 inhibitor that acts via the NO pathway, has been widely used in the management of CDH-PH (30). Milrinone, a PDE3 inhibitor, has also been studied in the CDH population (28, 31). Another important pulmonary vasodilator is prostacyclin; agents targeting this pathway include epoprostenol and inhaled iloprost (28, 32, 33). Inhibition of PDE3 causes lower pulmonary arterial pressures by acting via the PGI2 pathway (28). Endothelin (ET)-1 is a potent vasoconstrictor, and, hence, a target for modulating pulmonary vascular resistance. In a randomized control trial comparing the use of Bosentan, a drug which acts on the ETA and ETB receptors, with placebo as treatment for neonates with persistent pulmonary hypertension (PPHN), Mohamed et al. reported that Bosentan was superior to placebo for the treatment of PPHN (34).
The RV in PH Disease: Animal Models
Among studies in adults with PH, the major cause of mortality of patients with PH was RV failure (35). Experimental animal models to investigate the effect of pressure overload on the RV include the monocrotaline (MCT) and chronic hypoxic mouse models (36) and the pulmonary artery banding (PAB) mouse model (37) which was developed to study the RV-specific effects independent of the pulmonary circulation. RV failure molecular mechanisms involve abnormal metabolism, impaired angiogenesis, mitochondrial dysfunction and increased oxidative stress (38–40). Further, the “sick lung circulation” hypothesis postulates that altered lung vascular cells from the “sick lung”, such as those containing cell fragments, free DNA and microRNA, can be cytotoxic to the RV and can re-program endothelial cell genes, thus contributing to the RV failure (40).
Several important concepts regarding ventricular response to elevated pulmonary pressures investigated in animal models can be translated to clinical medicine. Firstly, as demonstrated by Urashima et al. (41), the RV and LV do not respond identically to pressure overload; thus, treatment strategies that focus individually and specifically to each ventricle are important. Additionally, it has been shown that acute RV pressure overload impairs LV function by altering septal strain and apical rotation (42). The RV's molecular adaptation varies based on the degree of pressure overload as well as the type of pressure overload (proximal type as seen in the PAB model vs peripheral type seen in PH models vs combined pressure and volume overload as in the presence of a shunt) (43, 44). Severe PH can result not only in systolic, but also diastolic dysfunction (45). Hemodynamic measurements of the RV in response to PH have been shown to correlate and predict biomechanical changes in the myocardium (46). Pressure overload on the RV significantly alters the pressure-volume relationship, leading to greater end-diastolic pressures, and concurrently increasing the longitudinal elastic modulus [Elastic modulus (E) or the amount of force required to deform a tissue] in the PAB rat model (46).
Existing studies of RV function in CDH, though limited, indicate similar morphological and functional changes. Echocardiographic studies of early RV function have demonstrated RV systolic and diastolic dysfunction, and evidence of interdependent impairment of LV function (47, 48). Furthermore, early RV dysfunction pre- and post CDH repair have been shown to be associated with adverse outcomes, including increased mortality, ECLS use and length of hospitalization in survivors, in single center cohorts as well as large registry-based analyses (49–51).
An important conclusion that can be drawn from these animal model studies and clinical studies in CDH is the importance of unloading the RV in the setting of elevated pulmonary pressures, and tailoring PH therapy to target both biomechanical and hemodynamic function with the aim of optimizing RV function and improving outcomes.
Role OF PGE1 in CDH
Prostaglandin E1 is a potent dilator of the ductus arteriosus in human neonates (52). The first studies ex-vivo in fetal lambs in 1973 by Coceani and Olley (53) led to clinical trials (54, 55) and approval for use by Food and Drug Administration (FDA) in 1981 (56).
The therapeutic benefits that PGE 1 offers in the setting of elevated pulmonary vascular resistance in CDH are theoretically three-fold. This has been summarized in Figure 1.
1) By acting as pressure “blow-off” valve, reducing the effective afterload on the pressure loaded RV, alleviating RV dilatation and myocardial dysfunction. LV function in turn may also improve by mechanisms of ventricular interdependence. A similar strategy of having a “pop off” conduit in supra-systemic pulmonary pressures has been demonstrated by the use of the Pott's shunt (anastomosis between left pulmonary artery to descending aorta) in pediatric hypertension and in patients with Eisenmenger syndrome (57). Evidence from pediatric patients with pulmonary hypertension have shown that a Pott's shunt improves RV-systolic function and RV-PA coupling, resulting in overall improved functional status and transplant-free survival (58).
2) By augmenting systemic blood flow in the setting of LV failure, by facilitating right-to-left shunting via the ductus arteriosus. The evidence of the benefits of using PGE1 to augment systemic blood flow is best noted in single ventricle pathologies such as hypoplastic left heart syndrome, where there exists an uncertain balance among systemic, pulmonary and coronary blood flows, with the systemic and pulmonary circulations in parallel rather than in series. The use of PGE1 in this situation ensures systemic blood flow to the vital organs, and also balances the systemic and pulmonary cardiac output (59).
3) By its direct pulmonary vasodilating action in pulmonary artery smooth muscle. PGE1 increases intracellular cyclic AMP leading to decreased pulmonary vascular resistance (60), reducing RV afterload and potentially improving coronary perfusion to the RV (20). The pulmonary vasodilator benefits of PGE1 in primary pediatric pulmonary hypertension (61) and in neonatal PPHN have been demonstrated previously (62, 63).
FIGURE 1
Figure 1. Effect of PGE1 in PH-RV afterload reduction, direct pulmonary vasodilation and augmenting systemic blood flow. RV, right ventricle; LV, left ventricle; PH, pulmonary hypertension.
Animal studies support these potential benefits. Sakuma et al. (64), demonstrated in a monocrotaline rat PH model that PGE1 administration significantly reduced the production of cytokines IL-1, IL-6 and TNF, previously implicated in pulmonary hypertension. In another study by Ono et al. (65), PGE1 had a dose-dependent suppression of RV hypertrophy and pulmonary hypertension in a MCT rat model.
Though the potential benefits of PGE1 use in the cardiopulmonary physiology of CDH appear compelling, there are potential adverse effects. In the short-term, PGE1 may induce apnea, peripheral vasodilation, fever and hypotension (66). With long term use (>5 days), cortical hyperostosis, brown fat necrosis, gastric outlet obstruction and intimal mucosal damage have been reported (66). Worsening hypoxia due to right-to left ductal shunting should also be considered (67).
Review of Clinical Studies
To date, the investigation of PGE1 use in CDH has been limited to case reports and retrospective chart reviews. We performed a review of literature using the electronic bibliographic databases PubMed and Embase, and of ongoing trials in www.clinicaltrials.gov. Additionally, we also used PubMed's related citations feature to identify relevant studies. We included chart reviews, case control studies, case series and case reports. Once a list of studies was obtained, we analyzed the studies for methodology and outcomes measures as described below.
A summary of these studies is provided in Table 1.
TABLE 1
Table 1. Summary of clinical studies on PGE1 use in CDH.
Methodology and Indications of PGE1 Use
All but one of the studies of PGE1 use in CDH-PH have been retrospective chart reviews, comparing patients who received PGE1 with those who did not (26, 68–70), one retrospective study compared the combined therapy of PGE+iNO with those only receiving iNO (23). Three studies reported initiating PGE1 based only on echocardiographic parameters. Inamura et al. (68, 69) reported initiating PGE1 infusion when the duration of right-to-left shunting via the DA was longer than that of left-to-right shunting, whereas Shiyanagi et al. (23) reported using PGE1 for PH based on echocardiographic signs: dominant right-to-left shunt through a PDA, decrease in pulmonary arterial blood flow on the affected side, and tricuspid regurgitation velocity (TRV) more than 2.5 m/s. Two of the studies reported using PGE1 based on a specific criterion; Lawrence et al. delineated specific indications for PGE1 initiation based on institutional CDH guidelines, which included (1) echocardiographic findings of PH with a restrictive PDA, (2) persistent metabolic acidosis (pH <7.25 with base deficit or elevation of lactate) without left heart dysfunction on echocardiography, or (3) persistent post-ductal arterial oxygen content <30 mmHg (70). In the study by Le Duc et al. (26), PGE1 was initiated when the maximal right-to-left blood flow velocities were >1.5m/s in CDH infants with acute worsening of the cardiorespiratory status.
As noted, most of the cited studies specified a right-to-left ductal shunting pattern as an indication for initiation of PGE1. One study mentioned a “restrictive PDA” as a criterion, but no specific duct size measurement was reported (70). None of the studies report echocardiographic evidence of abnormal RV size or function as a marker for PGE1 initiation, although clinical signs of RV/LV failure were used as criteria in two of the cited studies (26, 70). One study used plasma BNP before and after PGE1 initiation as a measure of PH and RV strain, and demonstrated decline in BNP measurements and improvement in echocardiographic measures PH (70). Plasma BNP peptides are secreted in response to wall stress by both the ventricles. However, a study by Koch et al. demonstrated the rapid decrease in plasma BNP levels during the first week of life, and the use of plasma BNP as a marker of clinical improvement may not necessarily reflect the effect of PGE1 use (71).
Outcome Measures to Assess Response to PGE1
Echocardiographic markers have been used to assess the effect of PGE1: two studies used LV size and function (measuring LV diastolic diameter, total pulmonary artery index (TPAI), left ventricular end-diastolic diameter and LV-Tei index (a composite measure of LV function based on systolic and diastolic time intervals) (68, 69). One study used echocardiographic markers of PH (estimated RV systolic pressure using tricuspid regurgitation jet velocity, direction of flow across a patent ductus arteriosus, and ventricular septal position) (70), and one study reported ductal flow direction and velocities (26).
In the studies by Inamura et al. (68, 69), the authors concluded that in instances of severe PH keeping the ductus open plays an important role in the circulatory management of these patients by improving LV function (as indicated by a higher LV Tei index in infants receiving PGE1). Lawrence et al. (70) observed that use of PGE1 significantly reduced B-Natriuretic peptide levels (BNP, a plasma biomarker of pulmonary hypertension and associated cardiac strain) and echocardiographic indices of PH, as assessed by tricuspid regurgitation jet velocity, ductus arteriosus direction, and ventricular septum position. Le Duc et al. (26) concluded that use of PGE1 in CDH decreased FiO2 requirements (median FiO2 decreased from 80% to 34% to target preductal SpO2 between 88 and 96%), and improved circulatory function, thus preventing cardiorespiratory failure in this population. Echocardiographic markers for PH used in this study include ductal flow velocities and flow patterns and mean pulmonary arterial pressures in relation to systemic blood pressures and classified as suprasystemic when mean PAP > systemic blood pressure +10 mm Hg (26). An observational study conducted by Hofmann et al. (72) reported that use of PGE1 in addition to circulatory management with catecholamines in two of their patients with CDH-PH relieved and stabilized right ventricular function. Although these studies reported on common echocardiographic markers of PH and LV dysfunction parameters, it is notable that none of the cited studies assessed RV function.
Most of the studies report improvement in cardiopulmonary outcomes with the use of PGE1 in CDH. However, a retrospective study by Shiyanagi et al. (23) demonstrated no significant clinical effects with the use of PGE1 combined with iNO, and concluded that use of iNO alone would simplify the management of PH due to CDH. The study reported no significant difference in survival to discharge, however, a shorter duration of hospitalization and earlier dates of repair were observed for those receiving iNO alone (23). Interestingly, this was also the only study to report PDA diameter, timing of spontaneous PDA closure and other long-term outcomes such as length of stay and survival to discharge (23).
In terms of adverse effects of PGE1, Shiyanagi et al. (23) reported lower systemic BP in the group that received PGE1 in comparison to those that did not. Lawrence et al. (70) reported seven patients with side-effects due to PGE1 which included pulmonary overcirculation due to L–R shunting (2%), cortical proliferation of their long bones (5%), temperature elevation (1.8%) and GI bleed (1.8%). However, none of the studies reported any life-threatening adverse effects or mortality attributed to the use of the medication.
In addition to the above studies, case reports describing improvement of cardio-respiratory function following use of PGE1 infusion are summarized in Table 2 (73–76).
TABLE 2
Table 2. Summary of case reports on PGE1 use in CDH.
The above-described retrospective clinical studies and case reports indicate a potential cardio-respiratory benefit in using PGE1 in the managing PH in the CDH population. Why then is PGE1 not a routine component of the management of patients with CDH? (77, 78). Likely factors include uncertainty in identifying the appropriate subset of patients with CDH-PH, who might potentially benefit from this approach, and possibly the concerns relating to the short-term adverse effects and/or long-term adverse effects of having the ductus open. Possible ways to address these relevant concerns are two-fold:
1) Advocating for a more pronounced pathophysiology-based approach using serial echocardiograms.
2) Promoting further research on the use of PGE1 in this patient population to address these clinical concerns.
To our knowledge, there have not been any prospective studies investigating the effect of PGE1 on cardiorespiratory outcomes in CDH. Studies on significant and/or long-term outcomes, such as need for ECMO, duration of ventilation, length of stay, need for oxygen at discharge, need for additional PH medications and neurodevelopmental outcomes, are lacking. Use of PGE1 infusion either alone or in combination with other PH management strategies is a potential area for future research, which may further open the doors to other aspects of research such as the long-term effects of the presence of a ductus in patients with congenital diaphragmatic hernia.
Future Investigations of Pge1 Infusion in CDH
Ongoing areas of uncertainty about the use of PGE1 in CDH which require further investigation include:
1) Appropriate timing of the PGE1 infusion in CDH (e.g., earlier prophylactic administration vs. later after echocardiographic evidence of PH).
2) Dosing regimens of PGE1 for PH (e.g., fixed dosing or dose titration based on clinical and echocardiographic response). The studies described above have used variable dosing patterns for PGE1 use in CDH. Although this area may need to be further investigated, higher doses of PGE1 may have utility in situations of acute severe PH with RV failure with duct closure, analogous to those used in resuscitation of infants presenting with duct closure in critical congenital heart disease.
3) Potential side effects of its use in this population (both short- and long-term).
4) Duration of therapy (e.g., fixed or based on clinical and echocardiographic response).
5) Use of PGE1 in relation to ECLS.
6) Impact of concomitant use of other pulmonary vasodilators, such as iNO, Sildenafil and/or Bosentan.
7) Impact on LV/RV performance.
8) The specific subset of CDH that might best benefit from it use, which may require a pathophysiology-based, targeted approach to PH management in CDH.
In terms of “long-term” effects, areas of uncertainty ripe for further investigation include:
1) Impact on short- and long-term outcomes, including survival, need for ECLS, duration of ventilation, duration of hospital stay, need for oxygen at discharge, and neurodevelopmental outcome.
2) Timing and need for additional PH medications, such as Sildenafil and/or Bosentan at discharge.
3) Impact on intervention for ductal management due to the potential effect from L–R shunting.
With the evidence from a recent study that the severity of early postnatal PH has a significant impact on long-term outcome, an important question to be answered is the timing of the first echocardiogram in this population (18). This study stressed the importance of an early echocardiogram as a valuable prognostic tool that could potentially provide information that can impact the clinical course and management of PH.
Theoretically a well-designed clinical trial of PGE1 in CDH may help to address the current evidence gap. Ideally this would be a randomized double-blinded placebo trial of PGE1 in a priori risk-stratified subgroups of CDH patients with echocardiographic and clinical evidence of elevated PAP, biventricular dysfunction and a restrictive ductus and with outcome measures that include cardiopulmonary outcomes such as RV and LV performance, need for ECLS, effect of ventilatory needs, vasopressor needs, survival at discharge and mortality, managed using standardized management guidelines. Risk stratification should include prenatal imaging markers of severity, such as percent liver herniation (%LH), observed to expected Total Fetal Lung Volume (O/E TFLV), observed to expected Lung-Head Ratio (O/E LHR) (79), location of birth (in-born vs. out-born patients), etc.
Previous milestone trials in CDH include Ventilation in Infants with Congenital diaphragmatic hernia (VICI), the TOTAL trial of fetal tracheal occlusion, and the Neonatal Inhaled Nitric Oxide Study (NINOS) (80–82). Others are in progress include Congenital Diaphragmatic hernia Nitric Oxide vs. Sildenafil (CoDiNOS) trial and a randomized pilot trial of milrinone in congenital diaphragmatic hernia (31, 83). However, study recruitment for well-powered trials is a common challenge, which unfortunately can lead to smaller pilot studies without adequate power (84). Research in the CDH population is challenging due to small number of patients with isolated CDH and lack of evidence-based treatment strategies. Registry-based studies may be useful, but the wide variability in CDH management amongst institutions within the registry limit researchers' ability to draw meaningful conclusions or extrapolate the results to clinical practice (84). Investigating a therapeutic strategy of the use of PGE1 in patients with CDH, based on their clinical markers and echocardiographic indices of RV/LV dysfunction, could be thought of as a “pathophysiology-based approach” toward promoting precision medicine in this population. Such trials are sorely needed and will likely require multi-center collaboration to be completed in a timely fashion.
Conclusion
In conclusion, with continuing research to improve cardio-pulmonary and long-term outcomes in CDH, new management strategies are being proposed and studied. Supra-systemic RV pressures are associated with poor clinical outcomes in this population. There is a pathophysiological rationale for the use of PGE1 in CDH to maintain ductal patency and promote right-to-left shunting, thereby reducing effective RV afterload and supporting systemic blood flow. In addition, PGE1 may have direct pulmonary vasodilating actions. Although existing, single-center retrospective studies and case reports suggest benefit from the use of PGE1 in terms of reducing severity of PH and improving short-term cardiopulmonary stability, uncertainties remain around its optimal pragmatic clinical use in CDH, and current evidence from these studies may not strongly support clinical recommendations. Conducting pharmacological trials in neonates can be challenging due to physiological changes, variable pharmacokinetics in the early newborn period and the ethical considerations involved. However, a well-designed a-priori prospective study as outlined above should be considered to definitively understand the implications of the use of PGE1 in CDH and its impact on meaningful outcomes.
Author Contributions
SH: conceptualization, design, methodology, and drafting and revising. NP: design, methodology, and reviewing and revising. CF: conceptualization, design, reviewing and revising, and supervision. All authors contributed to manuscript revision, read, approved the submitted version and agreed to be accountable for all aspects of the work.
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.
Acknowledgments
We thank Amaryllis Fernandes for assistance with editing Figure 1 (Effect of PGE1 in PH-RV afterload reduction, direct pulmonary vasodilation and augmenting systemic blood flow) included in this paper.
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Keywords: Congenital Diaphragmatic Hernia (CDH), pulmonary hypertension, prostagladin E1, Patent Ductus Arteriosus (PDA), ventricular dysfunction
Citation: Hari Gopal S, Patel N and Fernandes CJ (2022) Use of Prostaglandin E1 in the Management of Congenital Diaphragmatic Hernia–A Review. Front. Pediatr. 10:911588. doi: 10.3389/fped.2022.911588
Received: 02 April 2022; Accepted: 10 June 2022;
Published: 01 July 2022.
Edited by:
Fiammetta Piersigilli, Cliniques Universitaires Saint-Luc, Belgium
Reviewed by:
Thomas Schaible, University of Heidelberg, Germany
Alba Perez Ortiz, University of Heidelberg, Germany
Florian Kipfmueller, University Hospital Bonn, Germany
Copyright © 2022 Hari Gopal, Patel and Fernandes. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms.
Correspondence: Srirupa Hari Gopal, srirupa.harigopal@bcm.edu
Disclaimer: All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article or claim that may be made by its manufacturer is not guaranteed or endorsed by the publisher. |
10108 | https://pubmed.ncbi.nlm.nih.gov/28397377/ | Effectiveness of 12-13-week scan for early diagnosis of fetal congenital anomalies in the cell-free DNA era - PubMed
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. 2018 Apr;51(4):463-469.
doi: 10.1002/uog.17487. Epub 2018 Mar 4.
Effectiveness of 12-13-week scan for early diagnosis of fetal congenital anomalies in the cell-free DNA era
M J A Kenkhuis1,M Bakker1,F Bardi1,F Fontanella1,M K Bakker12,J H Fleurke-Rozema1,C M Bilardo1
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1 Fetal Medicine Unit, Department of Obstetrics & Gynecology, University Medical Centre Groningen, University of Groningen, Groningen, The Netherlands.
2 Eurocat Northern Netherlands, Department of Genetics, University Medical Center Groningen, University of Groningen, Groningen, The Netherlands.
PMID: 28397377
DOI: 10.1002/uog.17487
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Comparative Study
Effectiveness of 12-13-week scan for early diagnosis of fetal congenital anomalies in the cell-free DNA era
M J A Kenkhuis et al. Ultrasound Obstet Gynecol.2018 Apr.
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. 2018 Apr;51(4):463-469.
doi: 10.1002/uog.17487. Epub 2018 Mar 4.
Authors
M J A Kenkhuis1,M Bakker1,F Bardi1,F Fontanella1,M K Bakker12,J H Fleurke-Rozema1,C M Bilardo1
Affiliations
1 Fetal Medicine Unit, Department of Obstetrics & Gynecology, University Medical Centre Groningen, University of Groningen, Groningen, The Netherlands.
2 Eurocat Northern Netherlands, Department of Genetics, University Medical Center Groningen, University of Groningen, Groningen, The Netherlands.
PMID: 28397377
DOI: 10.1002/uog.17487
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Objectives: The main aim of this study was to assess the proportion and type of congenital anomalies, both structural and chromosomal, that can be detected at an early scan performed at 12-13 weeks' gestation, compared with at the 20-week structural anomaly scan offered under the present screening policy. Secondary aims were to evaluate the incidence of false-positive findings and ultrasound markers at both scans, and parental choice regarding termination of pregnancy (TOP).
Methods: Sonographers accredited for nuchal translucency (NT) measurement were asked to participate in the study after undergoing additional training to improve their skills in late first-trimester fetal anatomy examination. The early scans were performed according to a structured protocol, in six ultrasound practices and two referral centers in the north-east of The Netherlands. All women opting for the combined test (CT) or with an increased a-priori risk of fetal anomalies were offered a scan at 12-13 weeks' gestation (study group). All women with a continuing pregnancy were offered, as part of the 'usual care', a 20-week anomaly scan.
Results: The study group consisted of 5237 women opting for the CT and 297 women with an increased a-priori risk of anomalies (total, 5534). In total, 51 structural and 34 chromosomal anomalies were detected prenatally in the study population, and 18 additional structural anomalies were detected after birth. Overall, 54/85 (63.5%) anomalies were detected at the early scan (23/51 (45.1%) structural and all chromosomal anomalies presenting with either an increased risk at first-trimester screening or structural anomalies (31/34)). All particularly severe anomalies were detected at the early scan (all cases of neural tube defect, omphalocele, megacystis, and multiple severe congenital and severe skeletal anomalies). NT was increased in 12/23 (52.2%) cases of structural anomaly detected at the early scan. Of the 12 cases of heart defects, four (33.3%) were detected at the early scan, five (41.7%) at the 20-week scan and three (25.0%) after birth. False-positive diagnoses at the early scan and at the 20-week scan occurred in 0.1% and 0.6% of cases, respectively, whereas ultrasound markers were detected in 1.4% and 3.0% of cases, respectively. After first- or second-trimester diagnosis of an anomaly, parents elected TOP in 83.3% and 25.8% of cases, respectively.
Conclusions: An early scan performed at 12-13 weeks' gestation by a competent sonographer can detect about half of the prenatally detectable structural anomalies and 100% of those expected to be detected at this stage. Particularly severe anomalies, often causing parents to choose TOP, are amenable to early diagnosis. The early scan is an essential part of modern pregnancy care. Copyright © 2017 ISUOG. Published by John Wiley & Sons Ltd.
Keywords: chromosomal anomalies; early diagnosis; first-trimester ultrasound; nuchal translucency; screening for congenital anomalies; structural anomalies.
Copyright © 2017 ISUOG. Published by John Wiley & Sons Ltd.
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Reply.Bilardo CM.Bilardo CM.Ultrasound Obstet Gynecol. 2018 Oct;52(4):550-551. doi: 10.1002/uog.20091.Ultrasound Obstet Gynecol. 2018.PMID: 30284364 No abstract available.
Re: Effectiveness of 12-13-week scan for early diagnosis of fetal congenital anomalies in the cell-free DNA era.Yagel S.Yagel S.Ultrasound Obstet Gynecol. 2018 Oct;52(4):550. doi: 10.1002/uog.20090.Ultrasound Obstet Gynecol. 2018.PMID: 30284366 No abstract available.
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10109 | https://math.stackexchange.com/questions/5090934/an-olympiad-like-inequality-problem-with-square-roots-sum-c-y-c-a-sqrt3-a | contest math - An olympiad-like inequality problem with square roots: $\sum_{c y c} a \sqrt{3 a^2+5(a b+b c+c a)} \geq \sqrt{2}(a+b+c)^2$ - Mathematics Stack Exchange
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An olympiad-like inequality problem with square roots: \sum_{c y c} a \sqrt{3 a^2+5(a b+b c+c a)} \geq \sqrt{2}(a+b+c)^2
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\begingroup
The problem seeks to prove \sum_{\text{cyc}} a \sqrt{3 a^2+5(a b+b c+c a)} \geq \sqrt{2}(a+b+c)^2 when a, b, c are nonnegative reals.
I exerted all my effort, only found that the equal conditions are a=b=c and a=b, c=0.
I really cannot find any way to deal with it. Any direct application of AM-GM, Cauchy, Chebyshev ... makes the inequal sign unhold or reversed. Even Schur's inequality cannot be applied, because of the existence of the square root. How should I deal with this type of problem?
This problem comes from the 6 th problem in the problem set Problems Proposed by Vasc and Arqady - Amir Hossein Parvardi.
My attempt:
I tried partial inequality a\sqrt{3 a^2 + 5 (ab+bc+ca)} \geq \sqrt{2} (a^2 + \frac{2}{3} (a b + b c + c a)), which, by squaring both sides, gives \frac{1}{9} (3 a^2 - ab-bc-ca) (3a^2+8ab+8bc+8ca) \geq 0, which is obviously wrong.
Squaring both sides, and applying Cauchy's inequality \sqrt{3b^2+5(ab+bc+ca)}\sqrt{3c^2+5(ab+bc+ca)} \geq 5ab+8bc+5ca etc. to remove the square root, but this doe not align with the equal conditio a=b and c=0, and therefore this does not hold.
Chebyshev's inequality: \text{LHS} \geq \frac{1}{3} (a+b+c) \sum_{\text{cyc}} \sqrt{3 a^2+5(a b+b c+c a)}. However, the inequality \sum_{\text{cyc}} \sqrt{3 a^2+5(a b+b c+c a)} \geq 3\sqrt{2}(a+b+c) does not hold.
Hölder's inequality \text{LHS}^2 \cdot \sum_{\text{cyc}} \frac{a}{3a^2+5(ab+bc+ca)} \geq (a+b+c)^3 but still, the equal sign does not hold for the a=b, c=0 case.
inequality
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cauchy-schwarz-inequality
symmetric-polynomials
jensen-inequality
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edited Aug 19 at 17:30
Michael Rozenberg
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asked Aug 19 at 5:03
Mr. EggMr. Egg
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\begingroup
We will use the Equal Variable Method.
In your inequality, let a+b+c and a^2+b^2+c^2 be fixed, then ab+bc+ca is also fixed. We need to minimize f(a)+f(b)+f(c) where f(x)=x\sqrt{3x^2+5\sum ab}. Since \frac{\operatorname d^2}{\operatorname dx^2}f'(x)=f'''(x)=\frac{225\big(\sum ab\big)^2}{\big(3x^2+5\sum ab\big)^{\frac52}}>0, then f'(x) is strictly convex. Clearly f(x) is continuous at x=0. Thus, by Corollary 1.4 in the linked article, it suffices to prove the inequality for a=0 or b=c.
If a=0, we need to prove that b\sqrt{3b^2+5bc}+c\sqrt{3c^2+5bc}\ge\sqrt2(b+c)^2. We have \begin{align}\text{LHS}^2&=3b^4+3c^4+5b^3c+5bc^3+2bc\sqrt{3b^2+5bc}\sqrt{3c^2+5bc}\&\ge3b^4+3c^4+5b^3c+5bc^3+2bc\cdot8bc\ge2(b+c)^4,\end{align} where we used AM-GM and then (b-c)^2\left(b^2-bc+c^2\right)\ge0.
If b=c>0, let \frac ab=u, then we need to prove that u\sqrt{3u^2+10u+5}+2\sqrt{10u+8}\ge\sqrt2(u+2)^2,\tag1 or after squaring both sides, 4u\sqrt{3u^2+10u+5}\sqrt{10u+8}\ge24u+43u^2+6u^3-u^4. If the right side is negative, then we are done. If it is positive, squaring both sides again gives u^2(u-1)^2\left(64+144u+71u^2+10u^3-u^4\right)\ge0, which is true since 64+144u+71u^2+10u^3\ge24u+43u^2+6u^3\ge u^4.
I am not sure whether (1) can be proven more easily.
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edited Aug 19 at 10:51
answered Aug 19 at 10:38
youthdooyouthdoo
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\begingroup You need to prove that g(x)=f'\left(\frac{1}{x}\right) is a convex function for x>0, which is true. I think, you need to fix your post.\endgroup Michael Rozenberg –Michael Rozenberg 2025-08-19 19:43:37 +00:00 Commented Aug 19 at 19:43
\begingroup@MichaelRozenberg See this (i.sstatic.net/FyribcDV.jpg). According to the article, if we fix \sum a and \color{red}\prod a, then we need f'\left(\frac1x\right) convex.\endgroup youthdoo –youthdoo 2025-08-19 23:49:16 +00:00 Commented Aug 19 at 23:49
\begingroup Now, I see. You are right. Thank you! Nice solution.+1\endgroup Michael Rozenberg –Michael Rozenberg 2025-08-20 04:33:35 +00:00 Commented Aug 20 at 4:33
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\begingroup
Another solution.
By AM-GM \sum_{cyc}a\sqrt{3a^2+5(ab+ac+bc)}=\sum_{cyc}\frac{2\sqrt2(a+b+c)a(3a^2+5(ab+ac+bc))}{2\sqrt2(a+b+c)\sqrt{3a^2+5(ab+ac+bc)}}\geq\geq\sum_{cyc}\tfrac{2\sqrt2(a+b+c)a(3a^2+5(ab+ac+bc))}{2(a+b+c)^2+3a^2+5(ab+ac+bc)}=\sum_{cyc}\tfrac{2\sqrt2(a+b+c)a(3a^2+5(ab+ac+bc))}{5a^2+2b^2+2c^2+9(ab+ac+bc)} and it's enough to prove: \sum_{cyc}\frac{a(3a^2+5(ab+ac+bc))}{5a^2+2b^2+2c^2+9(ab+ac+bc)}\geq\frac{a+b+c}{2} or \sum_{cyc}\left(\frac{a(3a^2+5(ab+ac+bc))}{5a^2+2b^2+2c^2+9(ab+ac+bc)}-a\right)\geq-\frac{a+b+c}{2} or -\sum_{cyc}\frac{2a(a+b+c)^2}{5a^2+2b^2+2c^2+9(ab+ac+bc)}\geq-\frac{a+b+c}{2} or 4(a+b+c)\sum_{cyc}\frac{a}{5a^2+2b^2+2c^2+9(ab+ac+bc)}\leq1, which we can prove by SOS again, but I think it's better a full expanding: \sum_{sym}(2a^6+16a^5b-9a^4b^2-11a^3b^3+5a^4bc-10a^3b^2c+7a^2b^2c^2)\geq0, which is true by Schur twice and Muirhead: \sum_{sym}(2a^6+16a^5b-9a^4b^2-11a^3b^3+5a^4bc-10a^3b^2c+7a^2b^2c^2)==\sum_{sym}(2a^6-4a^4b^2+2a^2b^2c^2+5a^4bc-10a^3b^2c+5a^2b^2c^2)++\sum_{sym}(16a^5b-5a^4b^2-11a^3b^3)\geq0.
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edited Aug 19 at 17:32
answered Aug 19 at 17:26
Michael RozenbergMichael Rozenberg
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Let us assume that (b+c,a+c,a+b) are the side lengths of a triangle.
The inequality can be stated as
\sum_{cyc} \cot\left(\frac{A}{2}\right)\sqrt{3\cot^2\left(\frac{A}{2}\right)+5\left(1+4\frac{R}{r}\right)}\geq 4\sqrt{2}\left(\frac{s}{r}\right)^2
where the LHS is \geq \sum_{cyc} f(A)=\sum_{cyc}\cot\left(\frac{A}{2}\right)\sqrt{3\cot^2\left(\frac{A}{2}\right)+45} via R\geq 2r.
f is a convex function over (0,\pi), hence f(B)+f(C)\geq 2 f\left(\frac{B+C}{2}\right)=2f\left(\frac{\pi}{2}-\frac{A}{2}\right) and it is enough to show that 2f\left(\frac{\pi}{2}-\frac{A}{2}\right)+f(A)\geq 4\sqrt{2}\left(\frac{s}{r}\right)^2. By letting z=\cot\frac{A}{4} we have to show that
\inf_{z > 1} \frac{z^2-1}{2z}\sqrt{3\left(\frac{z^2-1}{2z}\right)^2+45}+2\frac{z+1}{z-1}\sqrt{3\left(\frac{z+1}{z-1}\right)^2+45}\geq 4\sqrt{2}\left(\frac{s}{r}\right)^2
or, since z\mapsto\frac{z+1}{z-1} is an involution,
\inf_{z > 1} \frac{2z}{z^2-1}\sqrt{3\left(\frac{2z}{z^2-1}\right)^2+45}+2z\sqrt{3z^2+45}\geq 4\sqrt{2}\left(\frac{s}{r}\right)^2.
Now it is tedious (but doable) to check that such \inf is attained at z=\sqrt{3} only by factoring a derivative, and this finishes the proof.
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answered Aug 19 at 10:13
Jack D'AurizioJack D'Aurizio
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\begingroup What happens if it's not a triangle?\endgroup Michael Rozenberg –Michael Rozenberg 2025-08-19 10:36:35 +00:00 Commented Aug 19 at 10:36
1 \begingroup@MichaelRozenberg It is (a+b,\dots) what's assumed to be triangle side lengths.\endgroup youthdoo –youthdoo 2025-08-19 10:41:23 +00:00 Commented Aug 19 at 10:41
\begingroup I don't understand how you obtained\sum_{cyc} \cot\left(\frac{A}{2}\right)\sqrt{3\cot^2\left(\frac{A}{2}\right)+5\left(1+4\frac{R}{r}\right)}\geq 4\sqrt{2}\left(\frac{s}{r}\right)^2.Also I don't know how to check the last inequality. There is \frac sr on the right. Do we need to express this in terms of z?\endgroup youthdoo –youthdoo 2025-08-19 10:47:54 +00:00 Commented Aug 19 at 10:47
\begingroup a+b = r\cot\frac{C}{2} and en.wikipedia.org/wiki/Incircle_and_excircles#Other_properties_3\endgroup Jack D'Aurizio –Jack D'Aurizio 2025-08-19 12:28:13 +00:00 Commented Aug 19 at 12:28
1 \begingroup@MichaelRozenberg: if a,b,c are positive then a+b,b+c,c+a surely are the sides of a triangle by Ravi's substitution.\endgroup Jack D'Aurizio –Jack D'Aurizio 2025-08-19 14:03:13 +00:00 Commented Aug 19 at 14:03
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SOS helps.
We need to prove: \sum_{cyc}\left(3a^4+5a^3b+5a^3c+5a^2bc+2ab\sqrt{(3a^2+5(ab+ac+bc))(3a^2+5(ab+ac+bc))}\right)\geq\geq2(a+b+c)^4 or \sum_{cyc}(3a^4+5a^3b+5a^3c+5a^2bc+2+ab(3a^2+3b^2+10(ab+ac+bc))\geq\geq2(a+b+c)^4+\sum_{cyc}ab\left(\sqrt{3a^2+5(ab+ac+bc)}-\sqrt{3b^2+5(ab+ac+bc)}\right)^2 or \sum_{cyc}(a^4-2a^2b^2+a^2bc)\geq\sum_{cyc}\frac{9ab(a-b)^2(a+b)^2}{\left(\sqrt{3a^2+5(ab+ac+bc)}+\sqrt{3b^2+5(ab+ac+bc)}\right)^2} or \sum_{cyc}(2a^4-2a^2b^2-2a^2b^2+2a^2bc)\geq\geq\sum_{cyc}\frac{18ab(a-b)^2(a+b)^2}{\left(\sqrt{3a^2+5(ab+ac+bc)}+\sqrt{3b^2+5(ab+ac+bc)}\right)^2} or \sum_{cyc}(a-b)^2\left((a+b)^2-c^2-\frac{18ab(a+b)^2}{\left(\sqrt{3a^2+5(ab+ac+bc)}+\sqrt{3b^2+5(ab+ac+bc)}\right)^2}\right)\geq0 and the rest is smooth:
By Minkowski we obtain: \sum_{cyc}(a-b)^2\left((a+b)^2-c^2-\tfrac{18ab(a+b)^2}{\left(\sqrt{3a^2+5(ab+ac+bc)}+\sqrt{3b^2+5(ab+ac+bc)}\right)^2}\right)\geq\geq \sum_{cyc}(a-b)^2\left((a+b)^2-c^2-\frac{18ab(a+b)^2}{3(a+b)^2+20(ab+ac+bc)}\right) and since ((a+b)^2-c^2)(3(a+b)^2+20(ab+ac+bc))-18ab(a+b)^2==3a^4+14a^3b+22a^2b^2+14ab^3+3b^4++20(a+b)^3c-(3a^2+26ab+3b^2)c^2-20(a+b)c^3\geq\geq3(a^2+b^2)(a^2+b^2-c^2)+20(a+b)((a+b)^2-c^2)+26abc(a+b-c), it's enough to prove three following inequalities.
\sum_{cyc}\frac{(a-b)^2(a^2+b^2)(a^2+b^2-c^2)}{3(a+b)^2+20(ab+ac+bc)}\geq0,
\sum_{cyc}\frac{(a-b)^2(a+b)(a+b-c)}{3(a+b)^2+20(ab+ac+bc)}\geq0 and
\sum_{cyc}\frac{(a-b)^2(a+b-c)}{3(a+b)^2+20(ab+ac+bc)}\geq0. Indeed, let a\geq b\geq c.
The first inequality: \sum_{cyc}\frac{(a-b)^2(a^2+b^2)(a^2+b^2-c^2)}{3(a+b)^2+20(ab+ac+bc)}\geq\geq\frac{(a-c)^2(a^2+c^2)(a^2+c^2-b^2)}{3(a+c)^2+20(ab+ac+bc)}+\frac{(b-c)^2(b^2+c^2)(b^2+c^2-a^2)}{3(b+c)^2+20(ab+ac+bc)}\geq\geq\frac{(b-c)^2(a^2+c^2)(a^2-b^2)}{3(a+c)^2+20(ab+ac+bc)}+\frac{(b-c)^2(b^2+c^2)(b^2-a^2)}{3(b+c)^2+20(ab+ac+bc)}==\frac{2(b-c)^2(a-b)^2(a+b)(10ab(a+b)+(10a^2+23ab+10b^2)c-3c^3)}{(3(a+c)^2+20(ab+ac+bc))(3(b+c)^2+20(ab+ac+bc))}\geq0. The second inequality: \sum_{cyc}\frac{(a-b)^2(a+b)(a+b-c)}{3(a+b)^2+20(ab+ac+bc)}\geq\geq\frac{(a-c)^2(a+c)(a+c-b)}{3(a+c)^2+20(ab+ac+bc)}+\frac{(b-c)^2(b+c)(b+c-a)}{3(b+c)^2+20(ab+ac+bc)}\geq\geq\frac{(b-c)^2(a+c)(a-b)}{3(a+c)^2+20(ab+ac+bc)}+\frac{(b-c)^2(b+c)(b-a)}{3(b+c)^2+20(ab+ac+bc)}==\frac{(b-c)^2(a-b)^2(17(ab+ac+bc)-3c^2)}{(3(a+c)^2+20(ab+ac+bc))(3(b+c)^2+20(ab+ac+bc))}\geq0. Third inequality: b^2\sum_{cyc}\frac{(a-b)^2(a+b-c)}{3(a+b)^2+20(ab+ac+bc)}\geq\geq\frac{b^2(a-c)^2(a+c-b)}{3(a+c)^2+20(ab+ac+bc)}+\frac{b^2(b-c)^2(b+c-a)}{3(b+c)^2+20(ab+ac+bc)}\geq\geq\frac{a^2(b-c)^2(a-b)}{3(a+c)^2+20(ab+ac+bc)}+\frac{b^2(b-c)^2(b-a)}{3(b+c)^2+20(ab+ac+bc)}==\frac{(b-c)^2(a-b)^2(20ab(a+b)+(20a^2+46ab+20b^2)c+3(a+b)c^2)}{(3(a+c)^2+20(ab+ac+bc))(3(b+c)^2+20(ab+ac+bc))}\geq0, which finishes the solution.
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edited Aug 19 at 13:58
answered Aug 19 at 10:04
Michael RozenbergMichael Rozenberg
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Alternative solution.
WLOG, assume that a + b + c = 3. We split into two cases.
Case 1. ab + bc + ca \le \frac94
By Minkowski inequality, we have \begin{align} &\sum_{\mathrm{cyc}} a \sqrt{3 a^2+5(a b+b c+c a)} \ ={}& \sum_{\mathrm{cyc}} \sqrt{3a^4 + 5(ab + bc + ca)a^2}\ \ge{}& \sqrt{3(a^2 + b^2 + c^2)^2 + 5(ab + bc + ca)(a + b + c)^2}. \end{align}
It suffices to prove that 3(a^2 + b^2 + c^2)^2 + 5(ab + bc + ca)(a + b + c)^2 \ge 162, or 12(9/4 - ab - bc - ca)(3 - ab - bc - ca) \ge 0 which is true.
Case 2. ab + bc + ca > \frac94
The desired inequality is written as \sum_{\mathrm{cyc}} a \sqrt{\frac{3a^2 + 5(ab + bc + ca)}{18}} \ge 3.
Let x_1 := \frac{3a^2 + 5(ab + bc + ca)}{18}, x_2 := \frac{3b^2 + 5(ab + bc + ca)}{18}, and x_3 := \frac{3c^2 + 5(ab + bc + ca)}{18}. Then x_1, x_2, x_3 \ge \frac58.
Using \sqrt{x} \ge 1 + \frac12(x-1) - \frac38(x-1)^2 for all x \ge 1/40, letting f(x) := 1 + \frac12(x-1) - \frac38(x-1)^2, it suffices to prove that a f(x_1) + bf(x_2) + cf(x_3) \ge 3. \tag{1}
We use the pqr method. Let p := a + b + c = 3, q := ab + bc + ca, r := abc. (1) is 15(3-q)r+20q^2-105q+135 \ge 0.
Using degree three Schur inequality r \ge \frac{4pq - p^3}{9}, we have 15(3-q)r+20q^2-105q+135 \ge 15(3-q)\cdot \frac{12q - 27}{9}+20q^2-105q+135 = 0.
We are done.
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answered Aug 20 at 4:50
River LiRiver Li
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10110 | https://en.wikipedia.org/wiki/Glycogen_storage_disease_type_III | Jump to content
Glycogen storage disease type III
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From Wikipedia, the free encyclopedia
Not to be confused with Cori cycle.
Medical condition
| Glycogen storage disease type III | |
--- |
| Other names | Cori Disease, Debrancher Deficiency, Forbes Disease |
| | |
| Micrograph of glycogen storage disease with histologic features consistent with Cori disease. Liver biopsy. H&E stain. | |
| Specialty | Endocrinology |
| Symptoms | Hypotonia |
| Causes | AGL gene mutation |
| Diagnostic method | Biopsy, Elevated transaminases |
| Treatment | Currently no cure, Diet regime |
Glycogen storage disease type III (GSD III) is an autosomal recessive metabolic disorder and inborn error of metabolism (specifically of carbohydrates) characterized by a deficiency in glycogen debranching enzymes.
It is also known as Cori's disease in honor of the 1947 Nobel laureates Carl Cori and Gerty Cori. Other names include Forbes disease in honor of clinician Gilbert Burnett Forbes (1915–2003), an American physician who further described the features of the disorder, or limit dextrinosis, due to the limit dextrin-like structures in cytosol. Limit dextrin is the remaining polymer produced after hydrolysis of glycogen. Without glycogen debranching enzymes to further convert these branched glycogen polymers to glucose, limit dextrinosis abnormally accumulates in the cytoplasm.
Glycogen is a molecule the body uses to store carbohydrate energy. Symptoms of GSD-III are caused by a deficiency of the enzyme amylo-1,6 glucosidase, or debrancher enzyme. This causes excess amounts of abnormal glycogen to be deposited in the liver, muscles, and, in some cases, the heart.[medical citation needed]
Signs and symptoms
[edit]
Glycogen storage disease type III presents during infancy with hypoglycemia and failure to thrive. Clinical examination usually reveals hepatomegaly. Muscular disease, including hypotonia and cardiomyopathy, usually occurs later. The liver pathology typically regresses as the individual enters adolescence, as does splenomegaly, should the individual develop it.
Genetics
[edit]
In regards to genetics glycogen storage disease type III is inherited in an autosomal recessive pattern (which means both parents need to be a carrier), and occurs in about 1 of every 100,000 live births. The highest incidence of glycogen storage disease type III is in the Faroe Islands where it occurs in 1 out of every 3,600 births, probably due to a founder effect.
There seem to be two mutations in exon 3 (c.17_18delAG) being one of them, which are linked to the subtype IIIb.
The amylo-alpha-1, 6-glucosidase, 4-alpha-glucanotransferase gene and its mutations, are at the root of this condition. The gene is responsible for creating glycogen debranching enzyme, which in turn helps in glycogen decomposition.
Diagnosis
[edit]
In terms of the diagnosis of glycogen storage disease type III, the following tests/exams are carried out to determine if the individual has the condition:
Biopsy (muscle or liver)
CBC
Ultrasound
DNA mutation analysis (helps ascertain GSD III subtype)
Differential diagnosis
[edit]
The differential diagnosis of glycogen storage disease type III includes GSD I, GSD IX and GSD VI. This however does not mean other glycogen storage diseases should not be distinguished as well.
Classification
[edit]
Clinical manifestations of glycogen storage disease type III are divided into four classes:
GSD IIIa, is the most common, (along with GSD IIIb) and clinically includes muscle and liver involvement
GSD IIIb, which clinically has liver involvement but no muscle involvement
GSD IIIc which clinically affects liver and muscle.
GSD IV affects the liver only (not muscle)
Treatment
[edit]
Treatment for glycogen storage disease type III may involve a high-protein diet, to facilitate gluconeogenesis. Additionally the individual may need:
IV glucose (if oral route is inadvisable)
Nutritional specialist
Vitamin D (for osteoporosis/secondary complication)
Hepatic transplant (if a complication occurs)
References
[edit]
^ Jump up to: a b c d Dagli, Aditi; Sentner, Christiaan P.; Weinstein, David A. (1 January 1993). "Glycogen Storage Disease Type III". GeneReviews. PMID 20301788. Archived from the original on 29 May 2023. Retrieved 11 August 2016.update 2012
^ Jump up to: a b c d "Genetics of Glycogen-Storage Disease Type III Clinical Presentation: History, Physical, Causes". emedicine.medscape.com. Archived from the original on 2017-02-06. Retrieved 2016-08-11.
^ Jump up to: a b c d Reference, Genetics Home. "glycogen storage disease type III". Genetics Home Reference. Archived from the original on 2019-04-23. Retrieved 2016-08-07.
^ Jump up to: a b "Glycogen storage disease type 3 | Genetic and Rare Diseases Information Center (GARD) – an NCATS Program". rarediseases.info.nih.gov. Archived from the original on 18 June 2020. Retrieved 2 January 2018.
^ J. G. Salway (2012). Medical Biochemistry at a Glance. John Wiley & Sons. p. 60. ISBN 9780470654514. Archived from the original on 2023-10-29. Retrieved 2020-11-11.
^ Santer, René; Kinner, Martina; Steuerwald, Ulrike; Kjærgaard, Susanne; Skovby, Flemming; Simonsen, Henrik; Shaiu, Wen-Ling; Chen, Yuan-Tsong; Schneppenheim, Reinhard; Schaub, Jürgen (May 2001). "Molecular genetic basis and prevalence of glycogen storage disease type IIIA in the Faroe Islands". European Journal of Human Genetics. 9 (5): 388–391. doi:10.1038/sj.ejhg.5200632. ISSN 1476-5438. PMID 11378828. S2CID 448760.
^ "OMIM Entry - # 232400 - Glycogen Storage Disease III; GSD3". www.omim.org. Archived from the original on 2017-03-29. Retrieved 2016-08-11.
^ Reference, Genetics Home. "AGL". Genetics Home Reference. Archived from the original on 2016-08-25. Retrieved 2016-08-11.
^ "Glycogen Storage Disorders. Inborn errors of metabolism | Patient". Patient. Archived from the original on 2017-12-06. Retrieved 2016-08-11.
^ Jump up to: a b Kishnani, Priya S.; Austin, Stephanie L.; Arn, Pamela; Bali, Deeksha S.; Boney, Anne; Case, Laura E.; Chung, Wendy K.; Desai, Dev M.; El-Gharbawy, Areeg; Haller, Ronald; Smit, G. Peter A.; Smith, Alastair D.; Hobson-Webb, Lisa D.; Wechsler, Stephanie Burns; Weinstein, David A.; Watson, Michael S. (1 July 2010). "Glycogen Storage Disease Type III diagnosis and management guidelines". Genetics in Medicine. 12 (7): 446–463. doi:10.1097/GIM.0b013e3181e655b6. ISSN 1098-3600. PMID 20631546.
Further reading
[edit]
Mayorandan, Sebene; Meyer, Uta; Hartmann, Hans; Das, Anibh Martin (1 January 2014). "Glycogen storage disease type III: modified Atkins diet improves myopathy". Orphanet Journal of Rare Diseases. 9 196. doi:10.1186/s13023-014-0196-3. ISSN 1750-1172. PMC 4302571. PMID 25431232.
Sentner, Christiaan P.; Hoogeveen, Irene J.; Weinstein, David A.; Santer, René; Murphy, Elaine; McKiernan, Patrick J.; Steuerwald, Ulrike; Beauchamp, Nicholas J.; Taybert, Joanna; Laforêt, Pascal; Petit, François M.; Hubert, Aurélie; Labrune, Philippe; Smit, G. Peter A.; Derks, Terry G. J. (22 April 2016). "Glycogen storage disease type III: diagnosis, genotype, management, clinical course and outcome". Journal of Inherited Metabolic Disease. 39 (5): 697–704. doi:10.1007/s10545-016-9932-2. ISSN 0141-8955. PMC 4987401. PMID 27106217.
External links
[edit]
Scholia has a topic profile for Glycogen storage disease type III.
Media related to Glycogen storage disease type III at Wikimedia Commons
| | |
--- |
| Classification | D ICD-10: E74.0 ICD-9-CM: 271.0 OMIM: 232400 610860 MeSH: D006010 DiseasesDB: 5302 |
| External resources | eMedicine: med/909 ped/479 GeneReviews: Glycogen Storage Disease Type III |
| v t e Inborn error of carbohydrate metabolism: monosaccharide metabolism disorders Including glycogen storage diseases (GSD) | |
--- |
| Sucrose, transport (extracellular) | | | | --- | | Disaccharide catabolism | Congenital alactasia Sucrose intolerance | | Monosaccharide transport | Glucose-galactose malabsorption Inborn errors of renal tubular transport (Renal glycosuria) Fructose malabsorption De Vivo Disease (GLUT1 deficiency) Fanconi-Bickel syndrome (GLUT2 deficiency) | |
| Hexose → glucose | | | | | | | | --- --- --- | | Monosaccharide catabolism | | | | --- | | Fructose: | Essential fructosuria Fructose intolerance | | Galactose / galactosemia: | GALK deficiency GALT deficiency/GALE deficiency | | |
| Glucose ⇄ glycogen | | | | --- | | Glycogenesis | GSD type 0 (glycogen synthase deficiency) GSD type IV (Andersen's disease, branching enzyme deficiency) Adult polyglucosan body disease (APBD) Lafora disease GSD type XV (glycogenin deficiency) | | Glycogenolysis | | | | --- | | Extralysosomal: | GSD type III (Cori's disease, debranching enzyme deficiency) GSD type VI (Hers' disease, liver glycogen phosphorylase deficiency) GSD type V (McArdle's disease, myophosphorylase deficiency) GSD type IX (phosphorylase kinase deficiency) Phosphoglucomutase deficiency (PGM1-CDG, CDG1T, formerly GSD-XIV) | | Lysosomal (LSD): | Glycogen storage disease type II (Pompe's disease, glucosidase deficiency, formerly GSD-IIa) Danon disease (LAMP2 deficiency, formerly GSD-IIb) | | |
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| Pentose phosphate pathway | Glucose-6-phosphate dehydrogenase deficiency Transaldolase deficiency SDDHD (Transketolase deficiency) 6-phosphogluconate dehydrogenase deficiency |
| Other | Hyperoxaluria + Primary hyperoxaluria Pentosuria Fatal congenital nonlysosomal cardiac glycogenosis (AMP-activated protein kinase deficiency, PRKAG2) |
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10111 | https://www.geeksforgeeks.org/maths/continuity-at-a-point/ | Tutorials
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Continuity at a Point
Continuity at a point means that a function doesn’t have any sudden jumps, breaks, or holes at a particular spot. Imagine you are drawing a graph of a function on a piece of paper. If you can draw it without lifting your pencil from the paper, the function is continuous at that point.
For example, think of driving a car along a road. If the road is smooth and you don’t have to stop or make sharp turns, that’s like a function being continuous. But if there’s a gap in the road or a sudden bump, it’s like a break in the function, meaning it's not continuous at that point.
In this article, we will discuss Continuity of a function at a Point in detail.
Table of Content
Continuity of Function at a Point
Continuity of a function at a point refers to the behavior of the function at that point and how its value behaves in the neighbourhood of that point. A function is considered continuous at a point if it satisfies the following conditions at that point.
Conditions for Continuity at a Point
Mathematically, this can be written as:
\lim_{x \to c} f(x) = f(c)
If any of the above conditions fail, the function is said to be discontinuous at c.
Example of Continuity
Consider the function f(x) = x2. To check the continuity at x = 2:
Since all three conditions are satisfied, f(x) = x2 is continuous at x = 2.
How to Determine Continuity at a Point?
Here are the steps to determine the continuity of a function at a point x = c:
Step 1: Check if the function is defined at x = c.
Step 2: Check if the limit exists as x approaches c.
Step 3: Verify if the limit equals the function value at x = c.
If all three steps are satisfied, the function is continuous at x = c.
Solved Examples
Example 1: Determine if the function f(x) = 2x + 3 is continuous at x = 1.
Solution:
For a function to be continuous at x = 1, we need to check three conditions:
Check if f(1) exists:
f(1) = 2(1) + 3 = 5.
Find \lim_{{x \to 1}} f(x):
\lim_{{x \to 1}} (2x + 3) = 2(1) + 3 = 5.
Compare the limit and the function value:
Since \lim_{{x \to 1}} f(x) = f(1) = 5, the function is continuous at x = 1.
Thus, f(x) = 2x + 3 is continuous at x = 1.
Example 2: Check if the function g(x) = \frac{x^2 - 1}{x - 1} is continuous at x = 1.
Solution:
Check if g(1) exists:
g(1) = \frac{1^2 - 1}{1 - 1} = \frac{0}{0}, which is undefined.
Find \lim_{{x \to 1}} g(x):
g(x) = \frac{(x - 1)(x + 1)}{x - 1} = x + 1 for x ≠ 1.
Now, \lim_{{x \to 1}} g(x) = \lim_{{x \to 1}} (x + 1) = 2.
Since g(1) is undefined but \lim_{{x \to 1}} g(x) = 2, the function is not continuous at x = 1.
Example 3: Is the function h(x) = |x| continuous at x = 0?
Solution:
Check if h(0) exists:
h(0) = |0| = 0.
Find \lim_{{x \to 0}} h(x):
For x > 0, h(x) = x, and for x < 0, h(x) = -x.
So, \lim_{{x \to 0^+}} h(x) = \lim_{{x \to 0^+}} x = 0, and \lim_{{x \to 0^-}} h(x) = \lim_{{x \to 0^-}} (-x) = 0.
Since both the left-hand and right-hand limits are equal, \lim_{{x \to 0}} h(x) = 0.
\lim_{{x \to 0}} h(x) = h(0) = 0, so h(x) = |x| is continuous at x = 0.
Example 4: Determine if the function f(x) = \begin{cases}
x^2 & \text{if} \ x < 2 \
4 & \text{if} \ x = 2 \
x + 2 & \text{if} \ x > 2
\end{cases} is continuous at x = 2.
Solution:
Check if f(2) exists: f(2) = 4.
Find \lim_{{x \to 2}} f(x):
\lim_{{x \to 2^-}} f(x) = \lim_{{x \to 2^-}} x^2 = 4, and
\lim_{{x \to 2^+}} f(x) = \lim_{{x \to 2^+}} (x + 2) = 4.
Since both the left-hand and right-hand limits are equal, \lim_{{x \to 2}} f(x) = 4.
\lim_{{x \to 2}} f(x) = f(2) = 4, so the function is continuous at x = 2.
Practice Questions
Question 1: Check whether the function f(x) = \frac{2x + 1}{x - 3} is continuous at x = 3.
Question 2: Determine if the following piecewise function is continuous at x = 1:
f(x) = \begin{cases}
x^2 + 2x & \text{if} \ x < 1 \
3 & \text{if} \ x = 1 \
x + 2 & \text{if} \ x > 1
\end{cases}
Question 3: Is the function g(x) = sin(x) continuous at x = π/2?
Question 4: Check the continuity of f(x) = \begin{cases}
2x - 1 & \text{if} \ x \leq 2 \
x^2 & \text{if} \ x > 2
\end{cases} at x = 2.
Question 5: Determine if the function h(x) = \frac{x^2 - 4}{x - 2} is continuous at x = 2.
Question 6: Is the absolute value function f(x) = |x - 3| continuous at x = 3?
Question 7: Determine if the function f(x) = ex is continuous at x = 0.
Question 8: Check the continuity of the following piecewise function at x = 0:
f(x) = \begin{cases}
x^2 & \text{if} \ x < 0 \
0 & \text{if} \ x = 0 \
x & \text{if} \ x > 0
\end{cases}
Answer Key
Conclusion
In simple terms, continuity at a point means that a function behaves smoothly without any breaks, jumps, or holes at that specific point.
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How do I explain to students that $0 \bmod n$ equals $0$?
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$\begingroup$
I am teaching a beginning programming class in Visual Basic (for non-CS majors). I told my students that the mod operator basically gives the remainder of the division. So, when seeing $0 \bmod 10$, some students (apparently) reasoned that, "$10$ goes into $0$ zero times and there are $10$ leftover."
What would be the best way to explain this to (non-math and non-CS major) students?
I would rather counter the "$10$ goes into $0$ zero times with $10$ leftover" reasoning than change my definition, if possible. I checked the documentation for the language, and it gave the same "remainder" definition that I gave.
intuition
modular-arithmetic
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edited Sep 2, 2019 at 4:07
Lee David Chung Lin
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asked Feb 20, 2012 at 17:33
jtpereydajtpereyda
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$\begingroup$ The notion you have considered is fine, Modulus operator just gives you the remainder. So $0$ is the only number that fits the division giving out another $0$ after multiplication and gives $0$ finally. I think 'chessmath' even posted the same thing. $\endgroup$
IDOK
– IDOK
2012-02-20 17:48:32 +00:00
Commented Feb 20, 2012 at 17:48
20
$\begingroup$ "10 goes into 0 zero times with 10 leftover" means $\:0 - 10\cdot 0 = 10$. This is the error you need to address. $\endgroup$
Math Gems
– Math Gems
2012-02-20 18:09:19 +00:00
Commented Feb 20, 2012 at 18:09
1
$\begingroup$ I usually think of this relation, (x + y)mod(y) = (x)mod(y) so using your example, (0 + 10)mod(10) = 0 $\endgroup$
CleoR
– CleoR
2015-11-05 13:28:36 +00:00
Commented Nov 5, 2015 at 13:28
3
$\begingroup$ Using the example below from @will, I guess you could say "0 pizzas divided by 10 students results in each student getting zero pizzas allocated to them, with zero unallocated pizzas leftover." (Thus 0/10 === 0 and also 0%10 === 0) $\endgroup$
mikermcneil
– mikermcneil
2019-03-19 23:41:30 +00:00
Commented Mar 19, 2019 at 23:41
$\begingroup$ I simply don't understand why your students think $10$ is left over. How could a nonzero number be left over from dividing nothing? $\endgroup$
Allawonder
– Allawonder
2019-09-02 04:10:00 +00:00
Commented Sep 2, 2019 at 4:10
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$\begingroup$
$10$ goes into $0$ zero times and there are $0$ left over.
Go back to long division the way you learned it in third or fourth grade: $$ \begin{array}{ccccccc} & & 0 \ \ 10 & ) & 0 \ & & 0 \ \hline & & 0 \end{array} $$ The remainder is $0$.
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edited Jun 9, 2016 at 19:05
answered Feb 20, 2012 at 17:47
Michael HardyMichael Hardy
1
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11
2
$\begingroup$ This will probably work best for my situation. Most of the students may be confused if I bring in the division theorem, although that would likely be best for math students. This idea goes back to what they know. $\endgroup$
jtpereyda
– jtpereyda
2012-02-20 18:45:16 +00:00
Commented Feb 20, 2012 at 18:45
2
$\begingroup$ How are there 0 left over if 10 didn't go anywhere any times? $\endgroup$
Deji
– Deji
2016-06-30 14:02:42 +00:00
Commented Jun 30, 2016 at 14:02
$\begingroup$ @Deji : Zero minus zero is zero. $\endgroup$
Michael Hardy
– Michael Hardy
2021-05-23 15:42:26 +00:00
Commented May 23, 2021 at 15:42
1
$\begingroup$ @Deji : $$ 10\times0=0. $$ So you have an issue with that? $\qquad$ $\endgroup$
Michael Hardy
– Michael Hardy
2021-05-30 20:08:58 +00:00
Commented May 30, 2021 at 20:08
1
$\begingroup$ @Deji : There are good reasons why it's done the way it is, but unfortunately mathematicians, who are fully non-dogmatic about proofs, are dogmatic about definitions, so they don't explain things like this explicitly. $\endgroup$
Michael Hardy
– Michael Hardy
2021-05-30 20:35:15 +00:00
Commented May 30, 2021 at 20:35
| Show 6 more comments
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$\begingroup$
If you want intuitive explanation say to them that leftover must not be greater that a size of cup (divisor) with which you take water from container (dividend) because if there was more water in container than size of the cup you could always take one full cup more.
I also think that showing them division algorithm equation would be also good: $$a = bq + r$$ than substitute variables $$0 = 10 q + r$$ and show them that only valid substitution for $q$ and $r$ is $0$ and $0$ because $0$ and $10$ would produce false equation $0 = 10$.
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edited Jan 22, 2020 at 18:17
vijayant
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answered Feb 20, 2012 at 17:46
TrismegistosTrismegistos
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1
$\begingroup$ Nicely explained through the division algorithm theorem. $\endgroup$
dacabdi
– dacabdi
2017-09-02 18:27:22 +00:00
Commented Sep 2, 2017 at 18:27
Add a comment |
5
$\begingroup$
Well it is actually more easy to explain than most try to do. Example:
4mod2 = the 2 is inside the 4 two times and rest is 0 So if you do "if(4%2 == 0)" this would be true. in a case of "if(i%2 == 0)" this would also be true for i=0. Why is that so? Because the rest of the modulo does not depend on the denominator. if you have 0 as a numerator , the rest will never be more than the numerater itself. It doesnt matter how big your denominator grows. eg x%y => the value will never be more than x. It is simply not possible. Everytime y is bigger than x, the rest will become 0 automaticaly. 0%0+n = 0 1%1+n = 0 2%2+n = 0 x%x+n = 0
Tell me, if you have zero pizzas and you want to give 10 students pizzas, how many pizzas will be left? Excatly, zero ones. Your students reasoned 10 students wants to eat 0 zero pizzas and there are 10 pizzas left. So actually you have to show them, there are 10 students left with empty stomachs but still zero pizzas on the table. (students are always hungry, this will work)
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answered Jan 24, 2016 at 16:33
WillWill
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If you're teaching future programmers, you will need to bring up the division theorem at some point. While Hardy's answer is surely best for the specific question you asked, your students will also need some guidance when negative numbers are involved.
For example, what do you expect from -22 Mod 3 or 7 Mod -2? The results will seem arcane at first, but are the clear consequences of the relationship to integer division, which I believe always rounds towards 0 in Visual Basic.
Other languages may handle the rounding differently, but the equation a = b (a div b) + (a mod b) seems to be universal.
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answered Feb 20, 2012 at 19:44
user14972user14972
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Well there is no problem $0$ is equal $0\times10+0$ what means that the rest of the Euclidean division is zero. You can do that for every positive integer.
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answered Feb 20, 2012 at 17:45
checkmathcheckmath
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I also have a hard time putting together the question statement. But if someone were to ask me to explain $0 \mod 10 \equiv 0$ to them, I might say one of the following three things.
Everything is congruent to itself. That is, if I were to say $x \mod 10 \equiv x$, this is true for all $x$. But perhaps that's uninspired.
$10$ divides $0-0$. This is the standard (at least, the one I consider standard) definition of mod, and so $0 \equiv 0$.
By the division algorithm, we see that $0 = 0 \cdot 10 + 0$, so that (reading the two outside numbers) $0 \equiv 0$.
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answered Feb 20, 2012 at 17:46
davidlowryduda♦davidlowryduda
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This helped me, I'm a programming student.
0%2 = 0 (0/2 = 0 remainder 0)
1%2 = 1 (1/2 = 0 remainder 1)
2%2 = 0 (2/2 = 1 remainder 0)
4%2 = 0 (4/2 = 2 remainder 0)
5%2 = 1 (5/2 = 2 remainder 1)
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answered Oct 16, 2014 at 22:38
VictorVictor
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The remainder at the division with $k>0$ must be equal to some number between $0$ and $k-1$. The division remainder theorem should handle the rest.
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answered Feb 20, 2012 at 17:47
Beni BogoselBeni Bogosel
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Think of it this way.
x mod n where n > 0 and x > 0.
For example x = 49 and n = 10. As we no there are 9 remaining.
However, how does x have anything remaining if there was nothing there to start with.I.E x = 0.
0 mod n == 0 as there was nothing to divide into to begin with so it can't have a remainder.
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answered May 23, 2013 at 23:07
SpooSpoo
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You said intuition?
The modulo operation finds the remainder after division of one number by another.
A mod B
But when it comes to intuition and understanding the logic behind the scene, modulo operation is the distance from ZERO by moving to direction of A in range B.
The distance colored in orange:
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edited Jan 22, 2017 at 12:04
Glorfindel
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answered Jan 22, 2017 at 11:39
Stav BodikStav Bodik
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1.Group 17
1.1 Physical Properties of the Group 17 Elements
1.1.1 Trends in Bond Strength of Halogen Molecules
1.1.2 Volatility Explained by Intermolecular Forces
1.1.3 Colours and Volatility of Chlorine, Bromine and Iodine
1.2 The Chemical Properties of the Halogen Elements and the Hydrogen Halides
1.2.1 Relative Reactivity of Halogen Elements as Oxidising Agents
1.2.2 Reactions of Halogens with Hydrogen
1.2.3 Thermal Stability of Hydrogen Halides
1.3 Some Reactions of the Halide Ions
1.3.1 Relative Reactivity of Halide Ions as Reducing Agents
1.3.2 Reactions of Halide Ions with Aqueous Silver Ions and Ammonia
1.3.3 Reactions of Halide Ions with Concentrated Sulfuric Acid
1.4 The Reactions of Chlorine
1.4.1 Reaction of Chlorine with Aqueous Sodium Hydroxide (Cold and Hot)
1.4.2 Use of Chlorine in Water Purification
2.Electrochemistry
2.1 Redox Processes: Electron Transfer and Changes in Oxidation Number
2.1.1 Calculating Oxidation Numbers
2.1.2 Using Oxidation Numbers to Balance Equations
2.1.3 Redox, Oxidation, Reduction, Disproportionation: Definitions
2.1.4 Oxidising and Reducing Agents: Definitions
2.1.5 Indicating Oxidation Numbers Using Roman Numerals
3.Chemical Energetics
3.1 Enthalpies of Solution and Hydration
3.1.1 Calculations Involving Solution and Hydration Energies
3.1.2 Effect of Ionic Charge and Radius on Hydration Enthalpy
3.1.3 Definition and Use of Enthalpy Change of Solution and Hydration
3.1.4 Construction and Use of Energy Cycles for Solution and Hydration
3.2 Entropy Change ΔS
3.2.1 Definition of Entropy
3.2.2 Prediction and Explanation of Entropy Changes
3.2.3 Calculation of Entropy Change for Reactions
3.3 Gibbs Free Energy Change ΔG
3.3.1 Gibbs Free Energy Equation
3.3.2 Calculations Using ΔG = ΔH – TΔS
3.3.3 Prediction of Reaction Feasibility
3.3.4 Effect of Temperature on Reaction Feasibility
3.4 Lattice Energy and Born–Haber Cycles
3.4.1 Definition and Use of Enthalpy Change of Atomisation and Lattice Energy
3.4.2 First Electron Affinity and Trends in Electron Affinity
3.4.3 Construction and Use of Born–Haber Cycles
3.4.4 Calculations Involving Born–Haber Cycles
3.4.5 Effect of Ionic Charge and Radius on Lattice Energy
4.Group 2
4.1 Magnesium to Barium, and Their Compounds
4.1.1 Variation in Solubility and Enthalpy Change of Solution of Hydroxides and Sulfates
4.1.2 Trend in Thermal Stability of Group 2 Nitrates and Carbonates
4.1.3 Effect of Ionic Radius on Polarisation of Large Anions
5.Atoms, Molecules and Stoichiometry
5.1 Formulas
5.1.1 Using Appropriate State Symbols
5.1.2 Empirical and Molecular Formulas: Definitions
5.1.3 Anhydrous, Hydrated and Water of Crystallisation
5.1.4 Calculation of Empirical and Molecular Formulas
5.1.5 Writing Ionic Compound Formulas from Ionic Charges and Oxidation Numbers
5.1.6 Predicting Ionic Charges from Periodic Table Positions
5.1.7 Common Ions: Names and Formulas
5.1.8 Writing and Balancing Chemical and Ionic Equations
5.2 Reacting Masses and Volumes (of Solutions and Gases)
5.2.1 Calculations Involving Reacting Masses and Percentage Yield
5.2.2 Calculations Involving Volumes of Gases
5.2.3 Calculations Involving Volumes and Concentrations of Solutions
5.2.4 Limiting Reagent and Excess Reagent Calculations
5.2.5 Deducing Stoichiometric Relationships from Calculations
5.3 Relative Masses of Atoms and Molecules
5.3.1 Unified Atomic Mass Unit
5.3.2 Relative Atomic Mass (Ar)
5.3.3 Relative Isotopic Mass
5.3.4 Relative Molecular Mass (Mr)
5.3.5 Relative Formula Mass
5.4 The Mole and the Avogadro Constant
5.4.1 Definition and Use of the Mole
5.4.2 Understanding the Avogadro Constant
6.Nitrogen Compounds
6.1 Primary and Secondary Amines
6.1.1 Production of Primary and Secondary Amines from Halogenoalkanes
6.1.2 Reduction of Amides and Nitriles to Amines
6.1.3 Reaction of Amines with Acyl Chlorides
6.1.4 Basicity of Aqueous Solutions of Amines
6.2 Phenylamine and Azo Compounds
6.2.1 Preparation of Phenylamine from Nitrobenzene
6.2.2 Reactions of Phenylamine with Bromine and Diazotisation
6.2.3 Relative Basicities of Ammonia, Ethylamine and Phenylamine
6.2.4 Formation and Use of Azo Compounds as Dyes
6.3 Amides
6.3.1 Production of Amides from Acyl Chlorides and Ammonia/Amines
6.3.2 Hydrolysis of Amides with Acids or Alkalis
6.3.3 Reduction of Amides to Amines
6.3.4 Weak Basic Nature of Amides
6.4 Amino Acids
6.4.1 Acid–Base Properties and Zwitterions
6.4.2 Formation of Peptide Bonds Between Amino Acids
6.4.3 Electrophoresis of Amino Acids and Peptides
7.Halogen Compounds
7.1 Halogenoalkanes
7.1.1 Elimination Reactions of Halogenoalkanes
7.1.2 SN1 and SN2 Mechanisms of Nucleophilic Substitution
7.1.3 Reactivity of Halogenoalkanes and Bond Strength
7.1.4 Production of Halogenoalkanes: Free-Radical Substitution and Electrophilic Addition
7.1.5 Classification of Halogenoalkanes: Primary, Secondary, Tertiary
7.1.6 Nucleophilic Substitution Reactions of Halogenoalkanes
7.1.7 Identification of Halogens by Reaction with Silver Nitrate
8.Chemistry of Transition Elements
8.1 General Physical and Chemical Properties of the First Row of Transition Elements
8.1.1 Titanium to Copper: Definition of Transition Elements
8.1.2 Titanium to Copper: Sketching 3dxy and 3dz² Orbitals
8.1.3 Titanium to Copper: General Properties: Variable Oxidation States, Catalytic Behavior, Formation of
8.1.4 Titanium to Copper: Explanation of Variable Oxidation States and Catalysis
8.2 General Characteristic Chemical Properties of the First Set of Transition Elements
8.2.1 Formation of Complexes and Ligand Types
8.2.2 Shapes and Coordination Numbers of Complexes
8.2.3 Ligand Exchange Reactions
8.2.4 Redox Reactions Involving Transition Elements
8.2.5 Redox Calculations Using E° Values
8.3 Colour of Complexes
8.3.1 Degenerate and Non-Degenerate d Orbitals
8.3.2 Splitting of d Orbitals in Octahedral and Tetrahedral Complexes
8.3.3 Explanation of Colour Based on Light Absorption
8.3.4 Effect of Ligand Types on Colour
8.4 Stereoisomerism in Transition Element Complexes
8.4.1 Geometrical (cis-trans) and Optical Isomerism in Complexes
8.4.2 Deduction of Overall Polarity of Complexes
8.5 Stability Constants
8.5.1 Definition and Expression of Stability Constants (Kstab)
8.5.2 Ligand Exchange and Kstab Values
9.Hydroxy Compounds
9.1 Alcohols
9.1.1 Production of Alcohols: Steam Addition, Oxidation, Reduction and Hydrolysis
9.1.2 Reactions of Alcohols: Combustion, Substitution, Oxidation, Dehydration, Esterification
9.1.3 Classification of Alcohols: Primary, Secondary, Tertiary
9.1.4 Tri-iodomethane Test for CH₃CH(OH)– Group
9.1.5 Acidity of Alcohols Compared to Water
10.Equilibria
10.1 Chemical Equilibria: Reversible Reactions and Dynamic Equilibrium
10.1.1 Reversible Reactions and Dynamic Equilibrium
10.1.2 Le Chatelier’s Principle
10.1.3 Effect of Temperature, Concentration, Pressure and Catalysts on Equilibrium
10.1.4 Equilibrium Constant Expressions (Kc)
10.1.5 Mole Fraction and Partial Pressure
10.1.6 Equilibrium Constant Expressions (Kp)
10.1.7 Calculations Using Kc and Kp
10.1.8 Changes Affecting the Value of Equilibrium Constants
10.1.9 Industrial Applications: Haber Process and Contact Process
10.2 Brønsted–Lowry Theory of Acids and Bases
10.2.1 Common Acids and Alkalis: Names and Formulas
10.2.2 Brønsted–Lowry Theory: Acids and Bases
10.2.3 Strong and Weak Acids and Bases
10.2.4 pH Scale and Differences Between Strong and Weak Acids
10.2.5 Neutralisation and Salt Formation
10.2.6 Acid–Base Titration Curves
10.2.7 Choosing Suitable Indicators for Titrations
11.Nitrogen and Sulfur
11.1 Nitrogen and Sulfur
11.1.1 Lack of Reactivity of Nitrogen Explained by Bond Strength and Polarity
11.1.2 Basicity of Ammonia and Structure of the Ammonium Ion
11.1.3 Displacement of Ammonia from Ammonium Salts
11.1.4 Natural and Man-made Sources of Nitrogen Oxides
11.1.5 Role of Nitrogen Oxides in Photochemical Smog Formation
11.1.6 Role of Nitrogen Oxides in Acid Rain Formation
12.Carbonyl Compounds
12.1 Aldehydes and Ketones
12.1.1 Production of Aldehydes and Ketones by Oxidation of Alcohols
12.1.2 Reduction of Aldehydes and Ketones
12.1.3 Reaction of Aldehydes and Ketones with Hydrogen Cyanide
12.1.4 Mechanism of Nucleophilic Addition Reactions
12.1.5 Detection of Carbonyl Compounds Using 2, 4-DNPH
12.1.6 Distinguishing Aldehydes from Ketones Using Fehling’s and Tollens’ Tests
12.1.7 Tri-iodomethane Test for CH₃CO– Group
13.Chemical Bonding
13.1 Electronegativity and Bonding
13.1.1 Definition of Electronegativity
13.1.2 Factors Affecting Electronegativity
13.1.3 Trends in Electronegativity Across a Period and Down a Group
13.1.4 Predicting Bond Type Using Electronegativity Differences
13.2 Ionic Bonding
13.2.1 Definition of Ionic Bonding
13.2.2 Ionic Bonding Examples: Sodium Chloride, Magnesium Oxide, Calcium Fluoride
13.3 Metallic Bonding
13.3.1 Definition of Metallic Bonding
13.4 Covalent Bonding and Coordinate (Dative Covalent) Bonding
13.4.1 Definition of Covalent Bonding
13.4.2 Covalent Bonding in Common Molecules (H₂, O₂, N₂, Cl₂, HCl, CO₂, NH₃, CH₄, C₂H₆, C₂H₄)
13.4.3 Expanded Octet in Period 3 Elements (SO₂, PCl₅, SF₆)
13.4.4 Definition of Coordinate (Dative Covalent) Bonding
13.4.5 Coordinate Bonding in NH₄⁺ and Al₂Cl₆
13.4.6 Sigma (σ) and Pi (π) Bonds: Orbital Overlap
13.4.7 Hybridisation: sp, sp² and sp³ Orbitals
13.4.8 Bond Energy and Bond Length Concepts
13.5 Shapes of Molecules
13.5.1 Shapes and Bond Angles Using VSEPR Theory (Examples: BF₃, CO₂, CH₄, NH₃, H₂O, SF₆, PF₅)
13.5.2 Predicting Shapes and Bond Angles of Similar Molecules and Ions
13.6 Intermolecular Forces, Electronegativity and Bond Properties
13.6.1 Hydrogen Bonding: Examples in Water and Ammonia
13.6.2 Anomalous Properties of Water Due to Hydrogen Bonding
13.6.3 Bond Polarity and Dipole Moments
13.6.4 Van der Waals' Forces: Instantaneous Dipole–Induced Dipole and Permanent Dipole–Dipole Forces
13.6.5 Comparison of Bond Strengths: Ionic, Covalent, Metallic vs. Intermolecular Forces
13.7 Dot-and-Cross Diagrams
13.7.1 Using Dot-and-Cross Diagrams to Represent Ionic, Covalent and Coordinate Bonding
14.Electrochemistry
14.1 Electrolysis
14.1.1 Prediction of Products During Electrolysis
14.1.2 Faraday Constant and Relationship with Avogadro Constant and Electron Charge
14.1.3 Calculations Involving Charge, Mass and Volume in Electrolysis
14.1.4 Determination of Avogadro Constant by Electrolysis
14.2 Standard Electrode Potentials and the Nernst Equation
14.2.1 Definition of Standard Electrode Potential and Standard Cell Potential
14.2.2 Description of the Standard Hydrogen Electrode
14.2.3 Measurement of Standard Electrode Potentials
14.2.4 Calculation of Standard Cell Potentials
14.2.5 Deduction of Electron Flow and Reaction Feasibility Using Electrode Potentials
14.2.6 Reactivity Predictions Using E° Values
14.2.7 Construction of Redox Equations from Half-Equations
14.2.8 Variation of Electrode Potential with Concentration
14.2.9 Using the Nernst Equation for Electrode Potentials
14.2.10 Relationship Between ΔG° and E°cell
15.Introduction to Organic Chemistry
15.1 Formulas, Functional Groups and the Naming of Organic Compounds
15.1.1 Definition of Hydrocarbons
15.2 Formulas
15.2.1 Functional Groups and Physical/Chemical Properties
15.2.2 Interpretation of General, Structural, Displayed and Skeletal Formulas
15.2.3 Systematic Nomenclature of Simple Aliphatic Organic Compounds
15.2.4 Deducing Molecular and Empirical Formulas
15.3 Characteristic Organic Reactions
15.3.1 Terminology of Organic Reactions: Homologous Series, Saturated/Unsaturated, Fission, Radicals, Nucle
15.3.2 Types of Organic Reactions: Addition, Substitution, Elimination, Hydrolysis, Condensation, Oxidation
15.3.3 Mechanisms of Organic Reactions: Free Radical Substitution, Electrophilic Addition, Nucleophilic Sub
15.4 Shapes of Organic Molecules; σ and π Bonds
15.4.1 Shapes and Bond Angles in Organic Molecules
15.4.2 Sigma (σ) and Pi (π) Bond Arrangements
15.4.3 Planarity in Organic Molecules
15.5 Isomerism: Structural Isomerism and Stereoisomerism
15.5.1 Types of Structural Isomerism: Chain, Positional, Functional Group
15.5.2 Types of Stereoisomerism: Geometrical (cis-trans) and Optical
15.5.3 Geometrical Isomerism in Alkenes
15.5.4 Chirality and Optical Isomerism
15.5.5 Identifying Chiral Centres and Isomer Types
16.Genetic technology
16.1 Genetically modified organisms in agriculture
16.1.1 GM crops and animals and their implications
16.2 Primary Amines
16.2.1 Production of Primary Amines by Reaction of Halogenoalkanes with Ammonia
16.3 Nitriles and Hydroxynitriles
16.3.1 Production of Nitriles by Reaction of Halogenoalkanes with KCN
16.3.2 Production of Hydroxynitriles by Reaction of Aldehydes and Ketones with HCN
16.3.3 Hydrolysis of Nitriles to Produce Carboxylic Acids
17.Atomic Structure
17.1 Particles in the Atom and Atomic Radius
17.1.1 Structure of the Atom: Nucleus and Electron Shells
17.1.2 Subatomic Particles: Protons, Neutrons, Electrons
17.1.3 Atomic Number, Proton Number, Mass Number, Nucleon Number
17.1.4 Mass and Charge Distribution in Atoms
17.1.5 Deflection of Charged Particles in Electric Fields
17.1.6 Calculations Involving Protons, Neutrons, Electrons
17.1.7 Trends in Atomic and Ionic Radii Across Periods and Down Groups
17.2 Isotopes
17.2.1 Definition and Notation of Isotopes
17.2.2 Chemical Properties of Isotopes
17.2.3 Physical Properties of Isotopes: Mass and Density Differences
17.3 Electrons
17.3.1 Energy Levels, and Atomic Orbitals: Shells, Sub-shells and Orbitals
17.3.2 Energy Levels,and Atomic Orbitals: Principal Quantum Number (n) and Ground State Configurations
17.3.3 Energy Levels and Atomic Orbitals: Order of Sub-shell Energy Levels (s, p, d)
17.3.4 Energy Levels and Atomic Orbitals: Full and Shorthand Electronic Configurations
17.3.5 Energy Levels and Atomic Orbitals: Electron Box Diagrams
17.3.6 Energy Levels and Atomic Orbitals: Shapes of s and p Orbitals
17.3.7 Energy Levels and Atomic Orbitals: Free Radicals and Unpaired Electrons
17.4 Ionisation Energy
17.4.1 First Ionisation Energy: Definition and Equations
17.4.2 Trends in Ionisation Energy Across Periods and Down Groups
17.4.3 Successive Ionisation Energies and Electron Shells
17.4.4 Factors Affecting Ionisation Energy: Nuclear Charge, Radius, Shielding
17.4.5 Deducing Electronic Configurations Using Ionisation Energy Data
17.4.6 Position of Elements Using Ionisation Energy Patterns
18.Polymerisation (Condensation Polymers)
18.1 Condensation Polymerisation
18.1.1 Formation of Polyesters from Diols and Dicarboxylic Acids or Dioyl Chlorides
18.1.2 Formation of Polyamides from Diamines and Dicarboxylic Acids or Dioyl Chlorides
18.1.3 Formation of Polymers from Amino Acids
18.2 Predicting the Type of Polymerisation
18.2.1 Identifying the Type of Polymerisation from Monomers or Polymer Structure
18.3 Degradable Polymers
18.3.1 Chemical Inertness and Disposal of Poly(alkenes)
18.3.2 Biodegradability of Polyesters and Polyamides
18.3.3 Degradation by Light
19.Carboxylic Acids and Derivatives
19.1 Carboxylic Acids
19.1.1 Production of Carboxylic Acids: Oxidation, Hydrolysis of Nitriles and Esters
19.1.2 Reactions of Carboxylic Acids with Metals, Alkalis and Carbonates
19.1.3 Esterification Reactions with Alcohols
19.1.4 Reduction of Carboxylic Acids
19.1.5 Relative Acidity of Carboxylic Acids, Phenols and Alcohols
19.1.6 Acidity of Chlorine-Substituted Carboxylic Acids
19.2 Esters
19.2.1 Production of Esters by Condensation of Carboxylic Acids and Alcohols
19.2.2 Hydrolysis of Esters with Acid or Alkali
20.Introduction to A Level Organic Chemistry
20.1 Formulas, Functional Groups and the Naming of Organic Compounds
20.1.1 Functional Groups in Aromatic and Extended Organic Molecules
20.1.2 Use of Structural, Displayed and Skeletal Formulas
20.1.3 Systematic Nomenclature for Aliphatic and Aromatic Molecules
20.2 Characteristic Organic Reactions
20.2.1 Electrophilic Substitution and Addition–Elimination Mechanisms
20.3 Shapes of Aromatic Organic Molecules; σ and π Bonds
20.3.1 Shape and Bonding in Benzene: sp² Hybridisation and Delocalised π System
20.4 Isomerism: Optical
20.4.1 Physical and Biological Properties of Optical Isomers
20.4.2 Concepts: Optically Active Compounds, Racemic Mixtures, Chiral Catalysts
21.Reaction Kinetics
21.1 Rate of Reaction
21.1.1 Definition of Rate of Reaction
21.1.2 Collision Theory: Effective and Non-Effective Collisions
21.1.3 Effect of Concentration and Pressure on Reaction Rate
21.1.4 Calculating Reaction Rate from Experimental Data
21.2 Effect of Temperature on Reaction Rates and the Concept of Activation Energy
21.2.1 Definition of Activation Energy
21.2.2 Boltzmann Distribution and Activation Energy
21.2.3 Effect of Temperature on Reaction Rate and Collision Frequency
21.3 Homogeneous and Heterogeneous Catalysts
21.3.1 Definition and Mechanism of Catalysis
21.3.2 Catalysts Lowering Activation Energy
21.3.3 Reaction Pathway Diagrams With and Without Catalysts
22.Organic Synthesis
22.1 Organic Synthesis
22.1.1 Identification of Functional Groups in Organic Molecules
22.1.2 Prediction of Properties and Reactions of Organic Molecules
22.1.3 Designing Multi-Step Synthetic Routes
22.1.4 Analysis of Synthetic Routes and Possible By-products
23.Hydrocarbons (Arenes)
23.1 Arenes
23.1.1 Substitution Reactions of Benzene and Methylbenzene
23.1.2 Mechanism of Electrophilic Substitution in Arenes
23.1.3 Predicting Halogenation: Side-Chain vs. Aromatic Ring
23.1.4 Directing Effects of Substituents in Electrophilic Substitution
24.Equilibria
24.1 Acids and Bases
24.1.1 Conjugate Acids and Bases
24.1.2 pH, Ka, pKa and Kw: Definitions and Calculations
24.1.3 Calculations of pH for Strong Acids, Strong Alkalis and Weak Acids
24.1.4 Definition and Formation of Buffer Solutions
24.1.5 Buffer Solutions in pH Control and Blood Chemistry
24.1.6 Calculations Involving Buffer Solutions
24.1.7 Solubility Product (Ksp) and Calculations
24.1.8 Common Ion Effect and Solubility Calculations
24.2 Partition Coefficients
24.2.1 Definition and Calculation of Partition Coefficient (Kpc)
24.2.2 Factors Affecting Partition Coefficients Based on Solute and Solvent Polarity
25.Analytical Techniques
25.1 Thin-Layer Chromatography
25.1.1 Terms and Interpretation: Stationary Phase, Mobile Phase, Rf Value, Solvent Front, Baseline
25.1.2 Explanation of Differences in Rf Values
25.2 Gas / Liquid Chromatography
25.2.1 Terms and Interpretation: Stationary Phase, Mobile Phase, Retention Time
25.2.2 Composition Analysis and Retention Time Explanation
25.3 Carbon-13 NMR Spectroscopy
25.3.1 Interpretation of C-13 NMR Spectra
25.3.2 Prediction of Number and Type of C Environments
25.4 Proton (¹H) NMR Spectroscopy
25.4.1 Interpretation of ¹H NMR Spectra: Chemical Shifts, Peak Areas, Splitting Patterns (n+1 Rule)
25.4.2 Prediction of Spectral Features
25.4.3 Use of TMS and Deuterated Solvents
25.4.4 Identification of O–H and N–H Protons by D₂O Exchange
26.Halogen Compounds (Arenes)
26.1 Halogen Compounds
26.1.1 Production of Halogenoarenes by Substitution
26.1.2 Reactivity Comparison: Halogenoalkanes vs. Halogenoarenes
27.Hydroxy Compounds (Phenol)
27.1 Alcohols
27.1.1 Reaction of Alcohols with Acyl Chlorides to Form Esters
27.2 Phenol
27.2.1 Directing Effects of Hydroxyl Group in Phenol Reactions
27.2.2 Reactions of Phenol with Bases, Sodium and Diazonium Salts
27.2.3 Nitration and Bromination of Phenol
27.2.4 Acidity of Phenol Compared to Water and Ethanol
27.2.5 Different Reaction Conditions for Phenol vs. Benzene
27.2.6 Production of Phenol from Diazonium Salts
28.Polymerisation
28.1 Addition Polymerisation
28.1.1 Description of Addition Polymerisation
28.1.2 Deduction of Repeat Units from Monomers
28.1.3 Identification of Monomers from Polymer Structures
28.1.4 Challenges of Disposal: Non-biodegradability and Harmful Combustion Products
29.Hydrocarbons
29.1 Alkanes
29.1.1 Production of Alkanes: Hydrogenation and Cracking
29.1.2 Combustion of Alkanes: Complete and Incomplete
29.1.3 Free-Radical Substitution of Alkanes
29.1.4 Mechanism of Free-Radical Substitution: Initiation, Propagation, Termination
29.1.5 Cracking for Useful Alkanes and Alkenes
29.1.6 Environmental Impact of Alkane Combustion
29.2 Alkenes
29.2.1 Production of Alkenes: Elimination and Dehydration Reactions
29.2.2 Electrophilic Addition Reactions of Alkenes
29.2.3 Oxidation of Alkenes: Cold Dilute and Hot Concentrated KMnO₄
29.2.4 Test for C=C Bond Using Aqueous Bromine
29.2.5 Mechanism of Electrophilic Addition in Alkenes
29.2.6 Stability of Carbocations and Markovnikov’s Rule
30.States of Matter
30.1 The Gaseous State: Ideal and Real Gases and pV = nRT
30.1.1 Origin of Pressure in Gases
30.1.2 Ideal Gas Assumptions
30.1.3 Ideal Gas Equation: pV = nRT
30.1.4 Applications of the Ideal Gas Equation
30.2 Bonding and Structure
30.2.1 Lattice Structures of Crystalline Solids
30.2.2 Structures and Bonding: Effect on Physical Properties
30.2.3 Deducing Structure and Bonding from Data
31.The Periodic Table: Chemical Periodicity
31.1 Periodicity of Physical Properties of the Elements in Period 3
31.1.1 Trends in Atomic Radius, Ionic Radius, Melting Point and Electrical Conductivity
31.1.2 Explanation of Melting Point and Conductivity Based on Structure and Bonding
31.2 Periodicity of Chemical Properties of the Elements in Period 3
31.2.1 Reactions of Elements with Oxygen, Chlorine and Water
31.2.2 Oxidation Numbers of Oxides and Chlorides
31.2.3 Reactions of Oxides with Water and pH of Solutions
31.2.4 Acid-Base Behaviour of Oxides and Hydroxides
31.2.5 Reactions of Chlorides with Water and pH of Solutions
31.2.6 Trends Explained by Bonding and Electronegativity
31.2.7 Bonding Types from Chemical and Physical Properties
31.3 Chemical Periodicity of Other Elements
31.3.1 Predicting Properties Based on Group Trends
31.3.2 Deducing Element Identity from Physical and Chemical Data
32.Organic Synthesis
32.1 Organic Synthesis
32.1.1 Identification of Organic Functional Groups
32.1.2 Predicting Properties and Reactions
32.1.3 Designing Multi-Step Synthetic Routes
32.1.4 Analyzing Synthetic Routes and Identifying By-products
33.Reaction Kinetics
33.1 Simple Rate Equations, Orders of Reaction and Rate Constants
33.1.1 Terms: Rate Equation, Order of Reaction, Overall Order, Rate Constant, Half-life, Rate-determining S
33.1.2 Deducing Orders from Graphs or Experimental Data
33.1.3 Interpretation of Concentration–Time and Rate–Concentration Graphs
33.1.4 Calculating Initial Rate from Data
33.1.5 Half-life of First-Order Reactions and Calculations
33.1.6 Calculating Rate Constants Using Different Methods
33.1.7 Predicting Reaction Mechanisms and Rate-Determining Steps
33.1.8 Identifying Catalysts and Intermediates from Mechanisms
33.1.9 Effect of Temperature on Rate Constants
33.2 Homogeneous and Heterogeneous Catalysts
33.2.1 Mode of Action of Heterogeneous Catalysts (Adsorption and Desorption)
33.2.2 Examples of Industrial Heterogeneous Catalysis
33.2.3 Mode of Action of Homogeneous Catalysts
33.2.4 Examples of Homogeneous Catalysis in Atmospheric Chemistry
34.Analytical Techniques
34.1 Infrared Spectroscopy
34.1.1 Analysis of Infrared Spectra to Identify Functional Groups
34.2 Mass Spectrometry
34.2.1 Interpretation of Mass Spectra: m/e Values and Isotopic Abundances
34.2.2 Calculation of Relative Atomic Mass and Molecular Mass
34.2.3 Identification of Molecules by Fragmentation Patterns
34.2.4 Determination of Carbon Atom Number Using [M+1]+ Peak
34.2.5 Detection of Bromine and Chlorine Atoms Using [M+2]+ Peak
35.Chemical Energetics
35.1 Enthalpy Change ΔH
35.1.1 Exothermic and Endothermic Reactions
35.1.2 Reaction Pathway Diagrams: Enthalpy Change and Activation Energy
35.1.3 Standard Conditions and Standard Enthalpy Changes
35.1.4 Enthalpy Changes: Reaction, Formation, Combustion, Neutralisation
35.1.5 Energy Transfer in Breaking and Making Bonds
35.1.6 Calculations Using Bond Energies
35.1.7 Calculations Using Experimental Data: q = mcΔT and ΔH = –mcΔT/n
35.2 Hess’s Law
35.2.1 Application of Hess’s Law to Construct Energy Cycles
35.2.2 Calculations Using Hess’s Law and Bond Energy Data
36.Group 2
36.1 Magnesium to Barium, and their Compounds
36.1.1 Reactions of Group 2 Elements with Oxygen, Water and Acids
36.1.2 Reactions of Group 2 Oxides, Hydroxides and Carbonates with Water and Acids
36.1.3 Thermal Decomposition of Group 2 Nitrates and Carbonates
36.1.4 Trends in Physical and Chemical Properties
36.1.5 Variation in Solubility of Hydroxides and Sulfates
37.Carboxylic Acids and Derivatives (Aromatic)
37.1 Esters
37.1.1 Production of Esters from Alcohols and Acyl Chlorides
37.2 Acyl Chlorides
37.2.1 Comparison of Hydrolysis of Acyl Chlorides, Alkyl Chlorides and Aryl Chlorides
37.2.2 Production of Acyl Chlorides from Carboxylic Acids
37.2.3 Addition–Elimination Mechanism of Acyl Chlorides
37.2.4 Reactions of Acyl Chlorides with Water, Alcohols, Phenols, Ammonia and Amines
37.3 Carboxylic Acids
37.3.1 Further Oxidation of Methanoic Acid and Ethanedioic Acid
37.3.2 Reactions of Carboxylic Acids with PCl₃, PCl₅ and SOCl₂
37.3.3 Effect of Chlorine Substitution on Acid Strength
37.3.4 Relative Acidities of Carboxylic Acids, Phenols and Alcohols
37.3.5 Production of Benzoic Acid from Alkylbenzene
Show
AS & A Level
Chemistry
Chemistry - 9701
Chemical Bonding
Covalent Bonding and Coordinate (Dative Covalent) Bonding
Sigma (σ) and Pi (π) Bonds: Orbital Overlap
Revision Notes
Sigma (σ) and Pi (π) Bonds: Orbital Overlap
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Sigma (σ) and Pi (π) Bonds: Orbital Overlap
Introduction
Understanding sigma (σ) and pi (π) bonds is fundamental in the study of chemical bonding, particularly within the framework of covalent bonding and coordinate (dative covalent) bonding. This topic is essential for students in the AS & A Level Chemistry curriculum (9701), as it provides insight into the nature of bond formation and molecular structure, which are critical for predicting the behavior of compounds in various chemical reactions.
Key Concepts
Definition of Sigma (σ) and Pi (π) Bonds
Sigma (σ) and pi (π) bonds are two primary types of covalent bonds that differ in their formation and properties. A sigma bond is the first bond formed between two atoms and is characterized by head-on orbital overlap, providing a strong and stable connection. In contrast, a pi bond results from the side-to-side overlap of p-orbitals and typically forms after a sigma bond has been established between the same two atoms.
Orbital Overlap in Sigma Bonds
Sigma bonds involve the overlap of atomic orbitals along the internuclear axis (the line connecting the nuclei of the bonding atoms). This overlap can occur between various types of orbitals, such as:
s-s Overlap: Occurs when two s-orbitals from each atom overlap directly, forming a sigma bond. An example is the H₂ molecule.
s-p Overlap: Involves the overlap of an s-orbital from one atom with a p-orbital from another atom, seen in molecules like HCl.
p-p Overlap: Happens when two p-orbitals overlap along the internuclear axis, as in F₂.
The effective overlap in sigma bonds leads to a strong bonding interaction, resulting in a stable single bond between atoms.
Orbital Overlap in Pi Bonds
Pi bonds are formed by the lateral overlap of parallel p-orbitals. Unlike sigma bonds, pi bonds do not involve the direct overlap along the internuclear axis. This type of bonding typically occurs in double and triple bonds:
Double Bonds: Consist of one sigma bond and one pi bond, as seen in ethylene (C₂H₄).
Triple Bonds: Comprise one sigma bond and two pi bonds, exemplified by acetylene (C₂H₂).
The presence of pi bonds introduces areas of electron density above and below the plane of the nuclei, which affects the molecular geometry and reactivity.
Bond Strength and Stability
Sigma bonds are generally stronger than pi bonds due to the greater extent of orbital overlap. The direct overlap in sigma bonds allows for more effective electron sharing, resulting in a lower bond energy. Pi bonds, with their side-to-side overlap, have less effective orbital overlap and therefore lower bond strength. This difference in bond strength influences the properties of molecules, such as bond lengths and rotational freedom:
Single bonds (sigma only) allow free rotation around the bond axis.
Double and triple bonds restrict rotation due to the presence of pi bonds.
Molecular Geometry and Hybridization
The formation of sigma and pi bonds is closely related to the concept of hybridization, which describes the mixing of atomic orbitals to form new hybrid orbitals:
sp³ Hybridization: Involves one s and three p-orbitals, forming four sigma bonds as seen in methane (CH₄).
sp² Hybridization: Combines one s and two p-orbitals, resulting in three sigma bonds and facilitating the formation of one pi bond, as in ethylene.
sp Hybridization: Merges one s and one p-orbital, allowing for the formation of two sigma bonds and two pi bonds, characteristic of acetylene.
Hybridization affects the geometry of molecules, influencing angles between bonds and overall molecular shape.
Bond Formation and Energy
Bond formation involves the release of energy as atoms achieve a more stable electronic configuration. Sigma bonds release more energy compared to pi bonds due to their stronger bonding interaction. The bond energy is a measure of the strength of a bond and is defined as the energy required to break one mole of bonds in gaseous molecules:
Single bonds typically have lower bond energies than double or triple bonds.
Breaking a double bond requires more energy than breaking a single bond due to the additional pi bond.
Understanding bond energies is crucial for predicting reaction energetics and the stability of compounds.
Examples of Sigma and Pi Bonds in Compounds
Numerous compounds exhibit sigma and pi bonding, illustrating their roles in molecular structure:
Methane (CH₄): Contains four sigma bonds formed by sp³ hybridization.
Ethylene (C₂H₄): Features a double bond consisting of one sigma and one pi bond.
Acetylene (C₂H₂): Comprises a triple bond with one sigma and two pi bonds.
Oxygen (O₂): Exhibits a double bond between oxygen atoms, involving sigma and pi bonding.
These examples demonstrate how sigma and pi bonds contribute to the diversity of molecular structures and properties.
Advanced Concepts
Bond Order and Multiple Bonding
Bond order refers to the number of shared electron pairs between two atoms. It is a key concept in understanding the strength and stability of bonds: Bond Order=Number of Bonding Electrons−Number of Antibonding Electrons 2 \text{Bond Order} = \frac{\text{Number of Bonding Electrons} - \text{Number of Antibonding Electrons}}{2} Bond Order=2 Number of Bonding Electrons−Number of Antibonding Electrons Higher bond order indicates stronger and shorter bonds. For example, a double bond (bond order of 2) is stronger and shorter than a single bond (bond order of 1), while a triple bond (bond order of 3) is even stronger and shorter.
Resonance and Delocalized Pi Bonds
In molecules with resonance, pi electrons are delocalized over multiple atoms, leading to resonance structures. This delocalization stabilizes the molecule by distributing electron density more evenly:
Benzene (C₆H₆): Features a ring structure with alternating single and double bonds, where pi electrons are delocalized above and below the plane of the ring.
Nitrate Ion (NO₃⁻): Exhibits resonance among three equivalent structures, with delocalized pi electrons contributing to its stability.
Resonance affects molecular geometry, bond lengths, and overall stability, making it a critical concept in advanced chemistry studies.
Orbital Hybridization and Molecular Orbitals
Hybridization not only explains sigma bond formation but also integrates with molecular orbital theory to describe the distribution of electrons in molecules:
Molecular Orbitals: Formed by the combination of atomic orbitals from interacting atoms, resulting in bonding and antibonding molecular orbitals.
Bonding Molecular Orbitals: Lower in energy and contribute to bond formation.
Antibonding Molecular Orbitals: Higher in energy and can weaken bonds if populated.
The interplay between hybridization and molecular orbitals provides a comprehensive understanding of bond formation, bond strength, and molecular stability.
Stereochemistry and Pi Bond Restrictions
The presence of pi bonds imposes restrictions on the rotation around bond axes, leading to distinct stereochemical outcomes:
E/Z Isomerism: Observed in alkenes with restricted rotation, resulting in geometric isomers based on the relative positions of substituents.
Planarity: Molecules with pi bonds, such as carbonyl compounds, tend to be planar to maximize pi orbital overlap.
These stereochemical implications are essential for understanding the reactivity and physical properties of compounds containing pi bonds.
Interdisciplinary Connections
Sigma and pi bonds are not only fundamental in chemistry but also intersect with other scientific disciplines:
Materials Science: The strength and flexibility of polymers are influenced by the types of bonds between monomer units, involving sigma and pi bonding.
Biochemistry: The structure and function of biomolecules, such as DNA and proteins, are stabilized by sigma and pi interactions.
Physics: The principles of orbital overlap and bonding relate to quantum mechanics and the behavior of electrons in atoms and molecules.
These connections highlight the relevance of sigma and pi bonds across various fields, demonstrating their integral role in the broader scientific landscape.
Advanced Bonding Theories
Beyond basic orbital overlap, advanced theories provide deeper insights into bond formation and behavior:
Valence Bond Theory: Emphasizes the role of orbital hybridization and overlap in bond formation.
Molecular Orbital Theory: Describes bonding as the creation of molecular orbitals that extend over the entire molecule, accommodating delocalized electrons.
Natural Bond Orbital (NBO) Analysis: Offers a detailed view of electron distribution and bonding interactions within molecules.
These theories enhance the understanding of chemical bonding, offering frameworks to predict and explain complex molecular phenomena.
Comparison Table
| Aspect | Sigma (σ) Bonds | Pi (π) Bonds |
---
| Orbital Overlap | Head-on overlap along the internuclear axis | Side-to-side overlap above and below the internuclear axis |
| Bond Strength | Stronger due to greater orbital overlap | Weaker compared to sigma bonds |
| Formation Order | First bond formed between two atoms | Subsequent bonds after a sigma bond is established |
| Rotation | Allows free rotation around the bond axis | Restricts rotation due to overlapping orbitals |
| Presence in Multiples Bonds | Present in all single, double, and triple bonds | Present in double and triple bonds only |
| Molecular Geometry Impact | Determines the basic bonding framework | Influences bond angles and molecular shape |
Summary and Key Takeaways
Sigma (σ) bonds involve head-on orbital overlap, forming the primary connection between atoms.
Pi (π) bonds result from side-to-side overlap of p-orbitals, supplementing sigma bonds in multiple bond scenarios.
Understanding sigma and pi bonds is crucial for comprehending molecular geometry, bond strength, and chemical reactivity.
Advanced concepts like hybridization and molecular orbital theory provide deeper insights into bond formation and stability.
The comparison between sigma and pi bonds highlights their distinct roles and properties in chemical bonding.
Coming Soon!
Examiner Tip
Tips
Remember the Order: Sigma bonds form before pi bonds—think "Sigma starts the show."
Use Mnemonics: "SP3 suits Methane" helps recall sp³ hybridization in methane molecules.
Visual Learning: Practice drawing molecular structures to differentiate between sigma and pi bonds effectively.
Understand Hybridization: Relate the type of hybridization to the number of sigma bonds and lone pairs for accurate predictions on exams.
Did You Know
Did You Know
Sigma (σ) bonds are the strongest type of covalent bonds and play a crucial role in the stability of essential biological molecules like DNA and proteins. Interestingly, pi (π) bonds are responsible for the distinctive colors of many organic compounds, as they allow the absorption of specific wavelengths of light. Additionally, the unique electrical properties of graphene, a single layer of carbon atoms, are largely due to the extensive network of pi bonds that enable electron mobility across the material.
Common Mistakes
Common Mistakes
Mistake 1: Believing that pi bonds form before sigma bonds.
Incorrect: Pi bonds are the first to form between two atoms.
Correct: Sigma bonds always form first, followed by pi bonds.
Mistake 2: Thinking that pi bonds are stronger than sigma bonds.
Incorrect: Pi bonds hold atoms together more strongly.
Correct: Sigma bonds are generally stronger due to better orbital overlap.
Mistake 3: Misassigning hybridization without considering lone pairs.
Incorrect: Assigning sp³ hybridization solely based on four bonds.
Correct: Consider both bonding and lone pairs when determining hybridization.
FAQ
What is the main difference between sigma and pi bonds?
Sigma bonds result from head-on orbital overlap along the internuclear axis, while pi bonds are formed by side-to-side overlap of p-orbitals above and below this axis.
Which bond type is stronger, sigma or pi?
Sigma bonds are generally stronger than pi bonds due to more effective orbital overlap.
Can molecules rotate freely around pi bonds?
No, the presence of pi bonds restricts rotation around the bond axis, leading to fixed geometries in double and triple bonds.
How does hybridization affect sigma and pi bond formation?
Hybridization determines the arrangement of sigma bonds by mixing atomic orbitals, which in turn influences the formation and presence of pi bonds in multiple bonding scenarios.
Why are pi bonds important in organic chemistry?
Pi bonds contribute to the reactivity, color, and structural properties of organic molecules, playing a key role in reactions like addition and polymerization.
What is bond order and how is it calculated?
Bond order indicates the number of shared electron pairs between two atoms. It is calculated using the formula: (Number of Bonding Electrons - Number of Antibonding Electrons) / 2.
1.Group 17
1.1 Physical Properties of the Group 17 Elements
1.1.1 Trends in Bond Strength of Halogen Molecules
1.1.2 Volatility Explained by Intermolecular Forces
1.1.3 Colours and Volatility of Chlorine, Bromine and Iodine
1.2 The Chemical Properties of the Halogen Elements and the Hydrogen Halides
1.2.1 Relative Reactivity of Halogen Elements as Oxidising Agents
1.2.2 Reactions of Halogens with Hydrogen
1.2.3 Thermal Stability of Hydrogen Halides
1.3 Some Reactions of the Halide Ions
1.3.1 Relative Reactivity of Halide Ions as Reducing Agents
1.3.2 Reactions of Halide Ions with Aqueous Silver Ions and Ammonia
1.3.3 Reactions of Halide Ions with Concentrated Sulfuric Acid
1.4 The Reactions of Chlorine
1.4.1 Reaction of Chlorine with Aqueous Sodium Hydroxide (Cold and Hot)
1.4.2 Use of Chlorine in Water Purification
2.Electrochemistry
2.1 Redox Processes: Electron Transfer and Changes in Oxidation Number
2.1.1 Calculating Oxidation Numbers
2.1.2 Using Oxidation Numbers to Balance Equations
2.1.3 Redox, Oxidation, Reduction, Disproportionation: Definitions
2.1.4 Oxidising and Reducing Agents: Definitions
2.1.5 Indicating Oxidation Numbers Using Roman Numerals
3.Chemical Energetics
3.1 Enthalpies of Solution and Hydration
3.1.1 Calculations Involving Solution and Hydration Energies
3.1.2 Effect of Ionic Charge and Radius on Hydration Enthalpy
3.1.3 Definition and Use of Enthalpy Change of Solution and Hydration
3.1.4 Construction and Use of Energy Cycles for Solution and Hydration
3.2 Entropy Change ΔS
3.2.1 Definition of Entropy
3.2.2 Prediction and Explanation of Entropy Changes
3.2.3 Calculation of Entropy Change for Reactions
3.3 Gibbs Free Energy Change ΔG
3.3.1 Gibbs Free Energy Equation
3.3.2 Calculations Using ΔG = ΔH – TΔS
3.3.3 Prediction of Reaction Feasibility
3.3.4 Effect of Temperature on Reaction Feasibility
3.4 Lattice Energy and Born–Haber Cycles
3.4.1 Definition and Use of Enthalpy Change of Atomisation and Lattice Energy
3.4.2 First Electron Affinity and Trends in Electron Affinity
3.4.3 Construction and Use of Born–Haber Cycles
3.4.4 Calculations Involving Born–Haber Cycles
3.4.5 Effect of Ionic Charge and Radius on Lattice Energy
4.Group 2
4.1 Magnesium to Barium, and Their Compounds
4.1.1 Variation in Solubility and Enthalpy Change of Solution of Hydroxides and Sulfates
4.1.2 Trend in Thermal Stability of Group 2 Nitrates and Carbonates
4.1.3 Effect of Ionic Radius on Polarisation of Large Anions
5.Atoms, Molecules and Stoichiometry
5.1 Formulas
5.1.1 Using Appropriate State Symbols
5.1.2 Empirical and Molecular Formulas: Definitions
5.1.3 Anhydrous, Hydrated and Water of Crystallisation
5.1.4 Calculation of Empirical and Molecular Formulas
5.1.5 Writing Ionic Compound Formulas from Ionic Charges and Oxidation Numbers
5.1.6 Predicting Ionic Charges from Periodic Table Positions
5.1.7 Common Ions: Names and Formulas
5.1.8 Writing and Balancing Chemical and Ionic Equations
5.2 Reacting Masses and Volumes (of Solutions and Gases)
5.2.1 Calculations Involving Reacting Masses and Percentage Yield
5.2.2 Calculations Involving Volumes of Gases
5.2.3 Calculations Involving Volumes and Concentrations of Solutions
5.2.4 Limiting Reagent and Excess Reagent Calculations
5.2.5 Deducing Stoichiometric Relationships from Calculations
5.3 Relative Masses of Atoms and Molecules
5.3.1 Unified Atomic Mass Unit
5.3.2 Relative Atomic Mass (Ar)
5.3.3 Relative Isotopic Mass
5.3.4 Relative Molecular Mass (Mr)
5.3.5 Relative Formula Mass
5.4 The Mole and the Avogadro Constant
5.4.1 Definition and Use of the Mole
5.4.2 Understanding the Avogadro Constant
6.Nitrogen Compounds
6.1 Primary and Secondary Amines
6.1.1 Production of Primary and Secondary Amines from Halogenoalkanes
6.1.2 Reduction of Amides and Nitriles to Amines
6.1.3 Reaction of Amines with Acyl Chlorides
6.1.4 Basicity of Aqueous Solutions of Amines
6.2 Phenylamine and Azo Compounds
6.2.1 Preparation of Phenylamine from Nitrobenzene
6.2.2 Reactions of Phenylamine with Bromine and Diazotisation
6.2.3 Relative Basicities of Ammonia, Ethylamine and Phenylamine
6.2.4 Formation and Use of Azo Compounds as Dyes
6.3 Amides
6.3.1 Production of Amides from Acyl Chlorides and Ammonia/Amines
6.3.2 Hydrolysis of Amides with Acids or Alkalis
6.3.3 Reduction of Amides to Amines
6.3.4 Weak Basic Nature of Amides
6.4 Amino Acids
6.4.1 Acid–Base Properties and Zwitterions
6.4.2 Formation of Peptide Bonds Between Amino Acids
6.4.3 Electrophoresis of Amino Acids and Peptides
7.Halogen Compounds
7.1 Halogenoalkanes
7.1.1 Elimination Reactions of Halogenoalkanes
7.1.2 SN1 and SN2 Mechanisms of Nucleophilic Substitution
7.1.3 Reactivity of Halogenoalkanes and Bond Strength
7.1.4 Production of Halogenoalkanes: Free-Radical Substitution and Electrophilic Addition
7.1.5 Classification of Halogenoalkanes: Primary, Secondary, Tertiary
7.1.6 Nucleophilic Substitution Reactions of Halogenoalkanes
7.1.7 Identification of Halogens by Reaction with Silver Nitrate
8.Chemistry of Transition Elements
8.1 General Physical and Chemical Properties of the First Row of Transition Elements
8.1.1 Titanium to Copper: Definition of Transition Elements
8.1.2 Titanium to Copper: Sketching 3dxy and 3dz² Orbitals
8.1.3 Titanium to Copper: General Properties: Variable Oxidation States, Catalytic Behavior, Formation of
8.1.4 Titanium to Copper: Explanation of Variable Oxidation States and Catalysis
8.2 General Characteristic Chemical Properties of the First Set of Transition Elements
8.2.1 Formation of Complexes and Ligand Types
8.2.2 Shapes and Coordination Numbers of Complexes
8.2.3 Ligand Exchange Reactions
8.2.4 Redox Reactions Involving Transition Elements
8.2.5 Redox Calculations Using E° Values
8.3 Colour of Complexes
8.3.1 Degenerate and Non-Degenerate d Orbitals
8.3.2 Splitting of d Orbitals in Octahedral and Tetrahedral Complexes
8.3.3 Explanation of Colour Based on Light Absorption
8.3.4 Effect of Ligand Types on Colour
8.4 Stereoisomerism in Transition Element Complexes
8.4.1 Geometrical (cis-trans) and Optical Isomerism in Complexes
8.4.2 Deduction of Overall Polarity of Complexes
8.5 Stability Constants
8.5.1 Definition and Expression of Stability Constants (Kstab)
8.5.2 Ligand Exchange and Kstab Values
9.Hydroxy Compounds
9.1 Alcohols
9.1.1 Production of Alcohols: Steam Addition, Oxidation, Reduction and Hydrolysis
9.1.2 Reactions of Alcohols: Combustion, Substitution, Oxidation, Dehydration, Esterification
9.1.3 Classification of Alcohols: Primary, Secondary, Tertiary
9.1.4 Tri-iodomethane Test for CH₃CH(OH)– Group
9.1.5 Acidity of Alcohols Compared to Water
10.Equilibria
10.1 Chemical Equilibria: Reversible Reactions and Dynamic Equilibrium
10.1.1 Reversible Reactions and Dynamic Equilibrium
10.1.2 Le Chatelier’s Principle
10.1.3 Effect of Temperature, Concentration, Pressure and Catalysts on Equilibrium
10.1.4 Equilibrium Constant Expressions (Kc)
10.1.5 Mole Fraction and Partial Pressure
10.1.6 Equilibrium Constant Expressions (Kp)
10.1.7 Calculations Using Kc and Kp
10.1.8 Changes Affecting the Value of Equilibrium Constants
10.1.9 Industrial Applications: Haber Process and Contact Process
10.2 Brønsted–Lowry Theory of Acids and Bases
10.2.1 Common Acids and Alkalis: Names and Formulas
10.2.2 Brønsted–Lowry Theory: Acids and Bases
10.2.3 Strong and Weak Acids and Bases
10.2.4 pH Scale and Differences Between Strong and Weak Acids
10.2.5 Neutralisation and Salt Formation
10.2.6 Acid–Base Titration Curves
10.2.7 Choosing Suitable Indicators for Titrations
11.Nitrogen and Sulfur
11.1 Nitrogen and Sulfur
11.1.1 Lack of Reactivity of Nitrogen Explained by Bond Strength and Polarity
11.1.2 Basicity of Ammonia and Structure of the Ammonium Ion
11.1.3 Displacement of Ammonia from Ammonium Salts
11.1.4 Natural and Man-made Sources of Nitrogen Oxides
11.1.5 Role of Nitrogen Oxides in Photochemical Smog Formation
11.1.6 Role of Nitrogen Oxides in Acid Rain Formation
12.Carbonyl Compounds
12.1 Aldehydes and Ketones
12.1.1 Production of Aldehydes and Ketones by Oxidation of Alcohols
12.1.2 Reduction of Aldehydes and Ketones
12.1.3 Reaction of Aldehydes and Ketones with Hydrogen Cyanide
12.1.4 Mechanism of Nucleophilic Addition Reactions
12.1.5 Detection of Carbonyl Compounds Using 2, 4-DNPH
12.1.6 Distinguishing Aldehydes from Ketones Using Fehling’s and Tollens’ Tests
12.1.7 Tri-iodomethane Test for CH₃CO– Group
13.Chemical Bonding
13.1 Electronegativity and Bonding
13.1.1 Definition of Electronegativity
13.1.2 Factors Affecting Electronegativity
13.1.3 Trends in Electronegativity Across a Period and Down a Group
13.1.4 Predicting Bond Type Using Electronegativity Differences
13.2 Ionic Bonding
13.2.1 Definition of Ionic Bonding
13.2.2 Ionic Bonding Examples: Sodium Chloride, Magnesium Oxide, Calcium Fluoride
13.3 Metallic Bonding
13.3.1 Definition of Metallic Bonding
13.4 Covalent Bonding and Coordinate (Dative Covalent) Bonding
13.4.1 Definition of Covalent Bonding
13.4.2 Covalent Bonding in Common Molecules (H₂, O₂, N₂, Cl₂, HCl, CO₂, NH₃, CH₄, C₂H₆, C₂H₄)
13.4.3 Expanded Octet in Period 3 Elements (SO₂, PCl₅, SF₆)
13.4.4 Definition of Coordinate (Dative Covalent) Bonding
13.4.5 Coordinate Bonding in NH₄⁺ and Al₂Cl₆
13.4.6 Sigma (σ) and Pi (π) Bonds: Orbital Overlap
13.4.7 Hybridisation: sp, sp² and sp³ Orbitals
13.4.8 Bond Energy and Bond Length Concepts
13.5 Shapes of Molecules
13.5.1 Shapes and Bond Angles Using VSEPR Theory (Examples: BF₃, CO₂, CH₄, NH₃, H₂O, SF₆, PF₅)
13.5.2 Predicting Shapes and Bond Angles of Similar Molecules and Ions
13.6 Intermolecular Forces, Electronegativity and Bond Properties
13.6.1 Hydrogen Bonding: Examples in Water and Ammonia
13.6.2 Anomalous Properties of Water Due to Hydrogen Bonding
13.6.3 Bond Polarity and Dipole Moments
13.6.4 Van der Waals' Forces: Instantaneous Dipole–Induced Dipole and Permanent Dipole–Dipole Forces
13.6.5 Comparison of Bond Strengths: Ionic, Covalent, Metallic vs. Intermolecular Forces
13.7 Dot-and-Cross Diagrams
13.7.1 Using Dot-and-Cross Diagrams to Represent Ionic, Covalent and Coordinate Bonding
14.Electrochemistry
14.1 Electrolysis
14.1.1 Prediction of Products During Electrolysis
14.1.2 Faraday Constant and Relationship with Avogadro Constant and Electron Charge
14.1.3 Calculations Involving Charge, Mass and Volume in Electrolysis
14.1.4 Determination of Avogadro Constant by Electrolysis
14.2 Standard Electrode Potentials and the Nernst Equation
14.2.1 Definition of Standard Electrode Potential and Standard Cell Potential
14.2.2 Description of the Standard Hydrogen Electrode
14.2.3 Measurement of Standard Electrode Potentials
14.2.4 Calculation of Standard Cell Potentials
14.2.5 Deduction of Electron Flow and Reaction Feasibility Using Electrode Potentials
14.2.6 Reactivity Predictions Using E° Values
14.2.7 Construction of Redox Equations from Half-Equations
14.2.8 Variation of Electrode Potential with Concentration
14.2.9 Using the Nernst Equation for Electrode Potentials
14.2.10 Relationship Between ΔG° and E°cell
15.Introduction to Organic Chemistry
15.1 Formulas, Functional Groups and the Naming of Organic Compounds
15.1.1 Definition of Hydrocarbons
15.2 Formulas
15.2.1 Functional Groups and Physical/Chemical Properties
15.2.2 Interpretation of General, Structural, Displayed and Skeletal Formulas
15.2.3 Systematic Nomenclature of Simple Aliphatic Organic Compounds
15.2.4 Deducing Molecular and Empirical Formulas
15.3 Characteristic Organic Reactions
15.3.1 Terminology of Organic Reactions: Homologous Series, Saturated/Unsaturated, Fission, Radicals, Nucle
15.3.2 Types of Organic Reactions: Addition, Substitution, Elimination, Hydrolysis, Condensation, Oxidation
15.3.3 Mechanisms of Organic Reactions: Free Radical Substitution, Electrophilic Addition, Nucleophilic Sub
15.4 Shapes of Organic Molecules; σ and π Bonds
15.4.1 Shapes and Bond Angles in Organic Molecules
15.4.2 Sigma (σ) and Pi (π) Bond Arrangements
15.4.3 Planarity in Organic Molecules
15.5 Isomerism: Structural Isomerism and Stereoisomerism
15.5.1 Types of Structural Isomerism: Chain, Positional, Functional Group
15.5.2 Types of Stereoisomerism: Geometrical (cis-trans) and Optical
15.5.3 Geometrical Isomerism in Alkenes
15.5.4 Chirality and Optical Isomerism
15.5.5 Identifying Chiral Centres and Isomer Types
16.Genetic technology
16.1 Genetically modified organisms in agriculture
16.1.1 GM crops and animals and their implications
16.2 Primary Amines
16.2.1 Production of Primary Amines by Reaction of Halogenoalkanes with Ammonia
16.3 Nitriles and Hydroxynitriles
16.3.1 Production of Nitriles by Reaction of Halogenoalkanes with KCN
16.3.2 Production of Hydroxynitriles by Reaction of Aldehydes and Ketones with HCN
16.3.3 Hydrolysis of Nitriles to Produce Carboxylic Acids
17.Atomic Structure
17.1 Particles in the Atom and Atomic Radius
17.1.1 Structure of the Atom: Nucleus and Electron Shells
17.1.2 Subatomic Particles: Protons, Neutrons, Electrons
17.1.3 Atomic Number, Proton Number, Mass Number, Nucleon Number
17.1.4 Mass and Charge Distribution in Atoms
17.1.5 Deflection of Charged Particles in Electric Fields
17.1.6 Calculations Involving Protons, Neutrons, Electrons
17.1.7 Trends in Atomic and Ionic Radii Across Periods and Down Groups
17.2 Isotopes
17.2.1 Definition and Notation of Isotopes
17.2.2 Chemical Properties of Isotopes
17.2.3 Physical Properties of Isotopes: Mass and Density Differences
17.3 Electrons
17.3.1 Energy Levels, and Atomic Orbitals: Shells, Sub-shells and Orbitals
17.3.2 Energy Levels,and Atomic Orbitals: Principal Quantum Number (n) and Ground State Configurations
17.3.3 Energy Levels and Atomic Orbitals: Order of Sub-shell Energy Levels (s, p, d)
17.3.4 Energy Levels and Atomic Orbitals: Full and Shorthand Electronic Configurations
17.3.5 Energy Levels and Atomic Orbitals: Electron Box Diagrams
17.3.6 Energy Levels and Atomic Orbitals: Shapes of s and p Orbitals
17.3.7 Energy Levels and Atomic Orbitals: Free Radicals and Unpaired Electrons
17.4 Ionisation Energy
17.4.1 First Ionisation Energy: Definition and Equations
17.4.2 Trends in Ionisation Energy Across Periods and Down Groups
17.4.3 Successive Ionisation Energies and Electron Shells
17.4.4 Factors Affecting Ionisation Energy: Nuclear Charge, Radius, Shielding
17.4.5 Deducing Electronic Configurations Using Ionisation Energy Data
17.4.6 Position of Elements Using Ionisation Energy Patterns
18.Polymerisation (Condensation Polymers)
18.1 Condensation Polymerisation
18.1.1 Formation of Polyesters from Diols and Dicarboxylic Acids or Dioyl Chlorides
18.1.2 Formation of Polyamides from Diamines and Dicarboxylic Acids or Dioyl Chlorides
18.1.3 Formation of Polymers from Amino Acids
18.2 Predicting the Type of Polymerisation
18.2.1 Identifying the Type of Polymerisation from Monomers or Polymer Structure
18.3 Degradable Polymers
18.3.1 Chemical Inertness and Disposal of Poly(alkenes)
18.3.2 Biodegradability of Polyesters and Polyamides
18.3.3 Degradation by Light
19.Carboxylic Acids and Derivatives
19.1 Carboxylic Acids
19.1.1 Production of Carboxylic Acids: Oxidation, Hydrolysis of Nitriles and Esters
19.1.2 Reactions of Carboxylic Acids with Metals, Alkalis and Carbonates
19.1.3 Esterification Reactions with Alcohols
19.1.4 Reduction of Carboxylic Acids
19.1.5 Relative Acidity of Carboxylic Acids, Phenols and Alcohols
19.1.6 Acidity of Chlorine-Substituted Carboxylic Acids
19.2 Esters
19.2.1 Production of Esters by Condensation of Carboxylic Acids and Alcohols
19.2.2 Hydrolysis of Esters with Acid or Alkali
20.Introduction to A Level Organic Chemistry
20.1 Formulas, Functional Groups and the Naming of Organic Compounds
20.1.1 Functional Groups in Aromatic and Extended Organic Molecules
20.1.2 Use of Structural, Displayed and Skeletal Formulas
20.1.3 Systematic Nomenclature for Aliphatic and Aromatic Molecules
20.2 Characteristic Organic Reactions
20.2.1 Electrophilic Substitution and Addition–Elimination Mechanisms
20.3 Shapes of Aromatic Organic Molecules; σ and π Bonds
20.3.1 Shape and Bonding in Benzene: sp² Hybridisation and Delocalised π System
20.4 Isomerism: Optical
20.4.1 Physical and Biological Properties of Optical Isomers
20.4.2 Concepts: Optically Active Compounds, Racemic Mixtures, Chiral Catalysts
21.Reaction Kinetics
21.1 Rate of Reaction
21.1.1 Definition of Rate of Reaction
21.1.2 Collision Theory: Effective and Non-Effective Collisions
21.1.3 Effect of Concentration and Pressure on Reaction Rate
21.1.4 Calculating Reaction Rate from Experimental Data
21.2 Effect of Temperature on Reaction Rates and the Concept of Activation Energy
21.2.1 Definition of Activation Energy
21.2.2 Boltzmann Distribution and Activation Energy
21.2.3 Effect of Temperature on Reaction Rate and Collision Frequency
21.3 Homogeneous and Heterogeneous Catalysts
21.3.1 Definition and Mechanism of Catalysis
21.3.2 Catalysts Lowering Activation Energy
21.3.3 Reaction Pathway Diagrams With and Without Catalysts
22.Organic Synthesis
22.1 Organic Synthesis
22.1.1 Identification of Functional Groups in Organic Molecules
22.1.2 Prediction of Properties and Reactions of Organic Molecules
22.1.3 Designing Multi-Step Synthetic Routes
22.1.4 Analysis of Synthetic Routes and Possible By-products
23.Hydrocarbons (Arenes)
23.1 Arenes
23.1.1 Substitution Reactions of Benzene and Methylbenzene
23.1.2 Mechanism of Electrophilic Substitution in Arenes
23.1.3 Predicting Halogenation: Side-Chain vs. Aromatic Ring
23.1.4 Directing Effects of Substituents in Electrophilic Substitution
24.Equilibria
24.1 Acids and Bases
24.1.1 Conjugate Acids and Bases
24.1.2 pH, Ka, pKa and Kw: Definitions and Calculations
24.1.3 Calculations of pH for Strong Acids, Strong Alkalis and Weak Acids
24.1.4 Definition and Formation of Buffer Solutions
24.1.5 Buffer Solutions in pH Control and Blood Chemistry
24.1.6 Calculations Involving Buffer Solutions
24.1.7 Solubility Product (Ksp) and Calculations
24.1.8 Common Ion Effect and Solubility Calculations
24.2 Partition Coefficients
24.2.1 Definition and Calculation of Partition Coefficient (Kpc)
24.2.2 Factors Affecting Partition Coefficients Based on Solute and Solvent Polarity
25.Analytical Techniques
25.1 Thin-Layer Chromatography
25.1.1 Terms and Interpretation: Stationary Phase, Mobile Phase, Rf Value, Solvent Front, Baseline
25.1.2 Explanation of Differences in Rf Values
25.2 Gas / Liquid Chromatography
25.2.1 Terms and Interpretation: Stationary Phase, Mobile Phase, Retention Time
25.2.2 Composition Analysis and Retention Time Explanation
25.3 Carbon-13 NMR Spectroscopy
25.3.1 Interpretation of C-13 NMR Spectra
25.3.2 Prediction of Number and Type of C Environments
25.4 Proton (¹H) NMR Spectroscopy
25.4.1 Interpretation of ¹H NMR Spectra: Chemical Shifts, Peak Areas, Splitting Patterns (n+1 Rule)
25.4.2 Prediction of Spectral Features
25.4.3 Use of TMS and Deuterated Solvents
25.4.4 Identification of O–H and N–H Protons by D₂O Exchange
26.Halogen Compounds (Arenes)
26.1 Halogen Compounds
26.1.1 Production of Halogenoarenes by Substitution
26.1.2 Reactivity Comparison: Halogenoalkanes vs. Halogenoarenes
27.Hydroxy Compounds (Phenol)
27.1 Alcohols
27.1.1 Reaction of Alcohols with Acyl Chlorides to Form Esters
27.2 Phenol
27.2.1 Directing Effects of Hydroxyl Group in Phenol Reactions
27.2.2 Reactions of Phenol with Bases, Sodium and Diazonium Salts
27.2.3 Nitration and Bromination of Phenol
27.2.4 Acidity of Phenol Compared to Water and Ethanol
27.2.5 Different Reaction Conditions for Phenol vs. Benzene
27.2.6 Production of Phenol from Diazonium Salts
28.Polymerisation
28.1 Addition Polymerisation
28.1.1 Description of Addition Polymerisation
28.1.2 Deduction of Repeat Units from Monomers
28.1.3 Identification of Monomers from Polymer Structures
28.1.4 Challenges of Disposal: Non-biodegradability and Harmful Combustion Products
29.Hydrocarbons
29.1 Alkanes
29.1.1 Production of Alkanes: Hydrogenation and Cracking
29.1.2 Combustion of Alkanes: Complete and Incomplete
29.1.3 Free-Radical Substitution of Alkanes
29.1.4 Mechanism of Free-Radical Substitution: Initiation, Propagation, Termination
29.1.5 Cracking for Useful Alkanes and Alkenes
29.1.6 Environmental Impact of Alkane Combustion
29.2 Alkenes
29.2.1 Production of Alkenes: Elimination and Dehydration Reactions
29.2.2 Electrophilic Addition Reactions of Alkenes
29.2.3 Oxidation of Alkenes: Cold Dilute and Hot Concentrated KMnO₄
29.2.4 Test for C=C Bond Using Aqueous Bromine
29.2.5 Mechanism of Electrophilic Addition in Alkenes
29.2.6 Stability of Carbocations and Markovnikov’s Rule
30.States of Matter
30.1 The Gaseous State: Ideal and Real Gases and pV = nRT
30.1.1 Origin of Pressure in Gases
30.1.2 Ideal Gas Assumptions
30.1.3 Ideal Gas Equation: pV = nRT
30.1.4 Applications of the Ideal Gas Equation
30.2 Bonding and Structure
30.2.1 Lattice Structures of Crystalline Solids
30.2.2 Structures and Bonding: Effect on Physical Properties
30.2.3 Deducing Structure and Bonding from Data
31.The Periodic Table: Chemical Periodicity
31.1 Periodicity of Physical Properties of the Elements in Period 3
31.1.1 Trends in Atomic Radius, Ionic Radius, Melting Point and Electrical Conductivity
31.1.2 Explanation of Melting Point and Conductivity Based on Structure and Bonding
31.2 Periodicity of Chemical Properties of the Elements in Period 3
31.2.1 Reactions of Elements with Oxygen, Chlorine and Water
31.2.2 Oxidation Numbers of Oxides and Chlorides
31.2.3 Reactions of Oxides with Water and pH of Solutions
31.2.4 Acid-Base Behaviour of Oxides and Hydroxides
31.2.5 Reactions of Chlorides with Water and pH of Solutions
31.2.6 Trends Explained by Bonding and Electronegativity
31.2.7 Bonding Types from Chemical and Physical Properties
31.3 Chemical Periodicity of Other Elements
31.3.1 Predicting Properties Based on Group Trends
31.3.2 Deducing Element Identity from Physical and Chemical Data
32.Organic Synthesis
32.1 Organic Synthesis
32.1.1 Identification of Organic Functional Groups
32.1.2 Predicting Properties and Reactions
32.1.3 Designing Multi-Step Synthetic Routes
32.1.4 Analyzing Synthetic Routes and Identifying By-products
33.Reaction Kinetics
33.1 Simple Rate Equations, Orders of Reaction and Rate Constants
33.1.1 Terms: Rate Equation, Order of Reaction, Overall Order, Rate Constant, Half-life, Rate-determining S
33.1.2 Deducing Orders from Graphs or Experimental Data
33.1.3 Interpretation of Concentration–Time and Rate–Concentration Graphs
33.1.4 Calculating Initial Rate from Data
33.1.5 Half-life of First-Order Reactions and Calculations
33.1.6 Calculating Rate Constants Using Different Methods
33.1.7 Predicting Reaction Mechanisms and Rate-Determining Steps
33.1.8 Identifying Catalysts and Intermediates from Mechanisms
33.1.9 Effect of Temperature on Rate Constants
33.2 Homogeneous and Heterogeneous Catalysts
33.2.1 Mode of Action of Heterogeneous Catalysts (Adsorption and Desorption)
33.2.2 Examples of Industrial Heterogeneous Catalysis
33.2.3 Mode of Action of Homogeneous Catalysts
33.2.4 Examples of Homogeneous Catalysis in Atmospheric Chemistry
34.Analytical Techniques
34.1 Infrared Spectroscopy
34.1.1 Analysis of Infrared Spectra to Identify Functional Groups
34.2 Mass Spectrometry
34.2.1 Interpretation of Mass Spectra: m/e Values and Isotopic Abundances
34.2.2 Calculation of Relative Atomic Mass and Molecular Mass
34.2.3 Identification of Molecules by Fragmentation Patterns
34.2.4 Determination of Carbon Atom Number Using [M+1]+ Peak
34.2.5 Detection of Bromine and Chlorine Atoms Using [M+2]+ Peak
35.Chemical Energetics
35.1 Enthalpy Change ΔH
35.1.1 Exothermic and Endothermic Reactions
35.1.2 Reaction Pathway Diagrams: Enthalpy Change and Activation Energy
35.1.3 Standard Conditions and Standard Enthalpy Changes
35.1.4 Enthalpy Changes: Reaction, Formation, Combustion, Neutralisation
35.1.5 Energy Transfer in Breaking and Making Bonds
35.1.6 Calculations Using Bond Energies
35.1.7 Calculations Using Experimental Data: q = mcΔT and ΔH = –mcΔT/n
35.2 Hess’s Law
35.2.1 Application of Hess’s Law to Construct Energy Cycles
35.2.2 Calculations Using Hess’s Law and Bond Energy Data
36.Group 2
36.1 Magnesium to Barium, and their Compounds
36.1.1 Reactions of Group 2 Elements with Oxygen, Water and Acids
36.1.2 Reactions of Group 2 Oxides, Hydroxides and Carbonates with Water and Acids
36.1.3 Thermal Decomposition of Group 2 Nitrates and Carbonates
36.1.4 Trends in Physical and Chemical Properties
36.1.5 Variation in Solubility of Hydroxides and Sulfates
37.Carboxylic Acids and Derivatives (Aromatic)
37.1 Esters
37.1.1 Production of Esters from Alcohols and Acyl Chlorides
37.2 Acyl Chlorides
37.2.1 Comparison of Hydrolysis of Acyl Chlorides, Alkyl Chlorides and Aryl Chlorides
37.2.2 Production of Acyl Chlorides from Carboxylic Acids
37.2.3 Addition–Elimination Mechanism of Acyl Chlorides
37.2.4 Reactions of Acyl Chlorides with Water, Alcohols, Phenols, Ammonia and Amines
37.3 Carboxylic Acids
37.3.1 Further Oxidation of Methanoic Acid and Ethanedioic Acid
37.3.2 Reactions of Carboxylic Acids with PCl₃, PCl₅ and SOCl₂
37.3.3 Effect of Chlorine Substitution on Acid Strength
37.3.4 Relative Acidities of Carboxylic Acids, Phenols and Alcohols
37.3.5 Production of Benzoic Acid from Alkylbenzene
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10114 | https://dlmf.nist.gov/8.21 | About the Project
8 Incomplete Gamma and Related FunctionsRelated Functions8.20 Asymptotic Expansions of 8.22 Mathematical Applications
§8.21 Generalized Sine and Cosine Integrals
ⓘ
Keywords:
: cosine integrals, generalized, generalized sine and cosine integrals, sine integrals
Permalink:
:
See also:
: Annotations for Ch.8
Contents
§8.21(i) Definitions: General Values
§8.21(ii) Definitions: Principal Values
§8.21(iii) Integral Representations
§8.21(iv) Interrelations
§8.21(v) Special Values
§8.21(vi) Series Expansions
§8.21(vii) Auxiliary Functions
§8.21(viii) Asymptotic Expansions
§8.21(i) Definitions: General Values
ⓘ
Keywords:
: analytic properties, definitions, general values, generalized sine and cosine integrals
Referenced by:
: §8.21(iv)
Permalink:
:
See also:
: Annotations for §8.21 and Ch.8
With and denoting here the general values of the incomplete gamma functions (§8.2(i)), we define
| | | |
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| 8.21.1 | | |
| ⓘ Defines: : generalized cosine integral and : generalized sine integral Symbols: : the ratio of the circumference of a circle to its diameter, : base of natural logarithm, : imaginary unit, : incomplete gamma function, : complex variable and : parameter Referenced by: §8.21(ii) Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(i), §8.21 and Ch.8 |
| 8.21.2 | | |
| ⓘ Defines: : generalized cosine integral and : generalized sine integral Symbols: : the ratio of the circumference of a circle to its diameter, : base of natural logarithm, : imaginary unit, : incomplete gamma function, : complex variable and : parameter Referenced by: §8.21(ii) Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(i), §8.21 and Ch.8 |
From §§8.2(i) and 8.2(ii) it follows that each of the four functions , , , and is a multivalued function of with branch point at . Furthermore, and are entire functions of , and and are meromorphic functions of with simple poles at and , respectively.
§8.21(ii) Definitions: Principal Values
ⓘ
Keywords:
: definitions, generalized sine and cosine integrals, principal values
Referenced by:
: §8.21(iii)
Permalink:
:
See also:
: Annotations for §8.21 and Ch.8
When (and when , in the case of , or , in the case of ) the principal values of , , , and are defined by (8.21.1) and (8.21.2) with the incomplete gamma functions assuming their principal values (§8.2(i)). Elsewhere in the sector the principal values are defined by analytic continuation from ; compare §4.2(i).
From here on it is assumed that unless indicated otherwise the functions , , , and have their principal values.
Properties of the four functions that are stated below in §§8.21(iii) and 8.21(iv) follow directly from the definitions given above, together with properties of the incomplete gamma functions given earlier in this chapter. In the case of §8.21(iv) the equation
| | |
--- |
| 8.21.3 | |
| , |
| ⓘ Symbols: : gamma function, : the ratio of the circumference of a circle to its diameter, : differential of , : base of natural logarithm, : imaginary unit, : integral, : real part and : parameter Referenced by: §8.21(iv) Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(ii), §8.21 and Ch.8 |
(obtained from (5.2.1) by rotation of the integration path) is also needed.
§8.21(iii) Integral Representations
ⓘ
Keywords:
: gamma function, generalized sine and cosine integrals, integral representations
Notes:
: Follow the prescription given in the first paragraph of §8.21(ii). For (8.21.4) and (8.21.5) replace by with in (8.2.2), deform the path of integration to run along the positive imaginary axis, and replace by . Then extend to the sector by analytic continuation. Similarly for (8.21.6) and (8.21.7).
Referenced by:
: §8.21(ii)
Permalink:
:
See also:
: Annotations for §8.21 and Ch.8
| | | |
---
| 8.21.4 | | |
| , |
| ⓘ Symbols: : differential of , : generalized sine integral, : integral, : real part, : sine function, : complex variable and : parameter A&S Ref: 6.5.8 (This function is called in AMS 55.) Referenced by: §8.21(iii), §8.21(iii), §8.21(v), §8.21(vii) Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(iii), §8.21 and Ch.8 |
| 8.21.5 | | |
| , |
| ⓘ Symbols: : cosine function, : differential of , : generalized cosine integral, : integral, : real part, : complex variable and : parameter A&S Ref: 6.5.7 (This function is called in AMS 55.) Referenced by: §8.21(iii), §8.21(iii), §8.21(v), §8.21(vii) Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(iii), §8.21 and Ch.8 |
| 8.21.6 | | |
| , |
| ⓘ Symbols: : differential of , : generalized sine integral, : integral, : real part, : sine function, : complex variable and : parameter Referenced by: §8.21(iii), §8.21(vi) Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(iii), §8.21 and Ch.8 |
| 8.21.7 | | |
| . |
| ⓘ Symbols: : cosine function, : differential of , : generalized cosine integral, : integral, : real part, : complex variable and : parameter Referenced by: §8.21(iii), §8.21(iv), §8.21(vi) Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(iii), §8.21 and Ch.8 |
In these representations the integration paths do not cross the negative real axis, and in the case of (8.21.4) and (8.21.5) the paths also exclude the origin.
§8.21(iv) Interrelations
ⓘ
Keywords:
: generalized sine and cosine integrals, interrelations
Notes:
: Temporarily restrict . Then (8.21.8) and (8.21.9) follow immediately from (8.21.3)–(8.21.7). Subsequently, ease the restrictions on by analytic continuation with respect to ; compare §8.21(i).
Referenced by:
: §8.21(ii)
Permalink:
:
See also:
: Annotations for §8.21 and Ch.8
| | |
--- |
| 8.21.8 | |
| , |
| ⓘ Symbols: : gamma function, : the ratio of the circumference of a circle to its diameter, : generalized sine integral, : generalized sine integral, : sine function, : complex variable and : parameter Referenced by: §8.21(iv), §8.21(v) Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(iv), §8.21 and Ch.8 |
| | |
--- |
| 8.21.9 | |
| . |
| ⓘ Symbols: : gamma function, : the ratio of the circumference of a circle to its diameter, : cosine function, : generalized cosine integral, : generalized cosine integral, : complex variable and : parameter Referenced by: §8.21(iv), §8.21(v) Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(iv), §8.21 and Ch.8 |
§8.21(v) Special Values
ⓘ
Keywords:
: generalized sine and cosine integrals, relation to sine and cosine integrals, special values
Notes:
: For (8.21.12) and (8.21.13) use (8.21.8) and (8.21.9), and also (8.21.4) and (8.21.5).
Permalink:
:
See also:
: Annotations for §8.21 and Ch.8
| | | |
---
| 8.21.10 | | |
| | |
| ⓘ Symbols: : cosine integral, : generalized cosine integral, : generalized sine integral, : sine integral and : complex variable Referenced by: §8.21(v) Permalink: Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for §8.21(v), §8.21 and Ch.8 |
| | |
--- |
| 8.21.11 | |
| ⓘ Symbols: : generalized sine integral, : sine integral and : complex variable Referenced by: §8.21(v) Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(v), §8.21 and Ch.8 |
For the functions on the right-hand sides of (8.21.10) and (8.21.11) see §6.2(ii).
| | | |
---
| 8.21.12 | | |
| , |
| ⓘ Symbols: : gamma function, : the ratio of the circumference of a circle to its diameter, : generalized sine integral, : sine function and : parameter Referenced by: §8.21(v) Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(v), §8.21 and Ch.8 |
| 8.21.13 | | |
| . |
| ⓘ Symbols: : gamma function, : the ratio of the circumference of a circle to its diameter, : cosine function, : generalized cosine integral and : parameter Referenced by: §8.21(v) Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(v), §8.21 and Ch.8 |
§8.21(vi) Series Expansions
ⓘ
Notes:
: (8.21.14) and (8.21.15) are obtained by expansion of the trigonometric functions in (8.21.6), (8.21.7), and termwise integration. See also Luke (1975, p. 115).
Referenced by:
: §8.25(i)
Permalink:
:
See also:
: Annotations for §8.21 and Ch.8
Power-Series Expansions
ⓘ
Keywords:
: generalized sine and cosine integrals, power-series expansions
See also:
: Annotations for §8.21(vi), §8.21 and Ch.8
| | |
--- |
| 8.21.14 | |
| , |
| ⓘ Symbols: : factorial (as in ), : generalized sine integral, : complex variable, : parameter and : nonnegative integer Referenced by: §8.21(vi) Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(vi), §8.21(vi), §8.21 and Ch.8 |
| | |
--- |
| 8.21.15 | |
| . |
| ⓘ Symbols: : factorial (as in ), : generalized cosine integral, : complex variable, : parameter and : nonnegative integer Referenced by: §8.21(vi) Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(vi), §8.21(vi), §8.21 and Ch.8 |
Spherical-Bessel-Function Expansions
ⓘ
Keywords:
: expansions in series of spherical Bessel functions, generalized sine and cosine integrals
See also:
: Annotations for §8.21(vi), §8.21 and Ch.8
| | | |
---
| 8.21.16 | | |
| , |
| ⓘ Symbols: : Pochhammer’s symbol (or shifted factorial), : generalized sine integral, : spherical Bessel function of the first kind, : complex variable, : parameter and : nonnegative integer Referenced by: §8.21(vi) Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(vi), §8.21(vi), §8.21 and Ch.8 |
| 8.21.17 | | |
| . |
| ⓘ Symbols: : Pochhammer’s symbol (or shifted factorial), : generalized cosine integral, : spherical Bessel function of the first kind, : complex variable, : parameter and : nonnegative integer Referenced by: §8.21(vi) Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(vi), §8.21(vi), §8.21 and Ch.8 |
For see §10.47(ii). For (8.21.16), (8.21.17), and further expansions in series of Bessel functions see Luke (1969b, pp. 56–57).
§8.21(vii) Auxiliary Functions
ⓘ
Keywords:
: auxiliary functions, generalized sine and cosine integrals, integral representations
Notes:
: (8.21.22) and (8.21.23) follow from (8.21.4), (8.21.5), (8.21.18), and (8.21.19). For (8.21.24) and (8.21.25) assume , and in the integrals for obtained from (8.21.4) and (8.21.5) set , rotate the integration paths in the -plane through , and apply (8.21.18) and (8.21.19). The restriction is eased to by analytic continuation.
Permalink:
:
See also:
: Annotations for §8.21 and Ch.8
| | | |
---
| 8.21.18 | | |
| ⓘ Defines: : auxiliary function (locally) Symbols: : cosine function, : generalized cosine integral, : generalized sine integral, : sine function, : complex variable and : parameter A&S Ref: 5.2.6 (This generalizes the form in AMS 55.) Referenced by: §8.21(vii) Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(vii), §8.21 and Ch.8 |
| 8.21.19 | | |
| ⓘ Defines: : auxiliary function (locally) Symbols: : cosine function, : generalized cosine integral, : generalized sine integral, : sine function, : complex variable and : parameter A&S Ref: 5.2.7 (This generalizes the form in AMS 55.) Referenced by: §8.21(vii) Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(vii), §8.21 and Ch.8 |
| 8.21.20 | | |
| ⓘ Symbols: : cosine function, : generalized sine integral, : sine function, : complex variable, : parameter, : auxiliary function and : auxiliary function Referenced by: §8.21(viii) Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(vii), §8.21 and Ch.8 |
| 8.21.21 | | |
| ⓘ Symbols: : cosine function, : generalized cosine integral, : sine function, : complex variable, : parameter, : auxiliary function and : auxiliary function Referenced by: §8.21(viii) Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(vii), §8.21 and Ch.8 |
When and ,
| | |
--- |
| 8.21.22 | |
| ⓘ Symbols: : differential of , : integral, : sine function, : complex variable, : parameter and : auxiliary function A&S Ref: 5.2.12 (This generalizes the form in AMS 55.) Referenced by: §8.21(vii) Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(vii), §8.21 and Ch.8 |
| | |
--- |
| 8.21.23 | |
| ⓘ Symbols: : cosine function, : differential of , : integral, : complex variable, : parameter and : auxiliary function A&S Ref: 5.2.13 (This generalizes the form in AMS 55.) Referenced by: §8.21(vii) Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(vii), §8.21 and Ch.8 |
When ,
| | |
--- |
| 8.21.24 | |
| ⓘ Symbols: : differential of , : base of natural logarithm, : imaginary unit, : integral, : complex variable, : parameter and : auxiliary function Referenced by: §8.21(vii), §8.21(viii) Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(vii), §8.21 and Ch.8 |
| | |
--- |
| 8.21.25 | |
| ⓘ Symbols: : differential of , : base of natural logarithm, : imaginary unit, : integral, : complex variable, : parameter and : auxiliary function Referenced by: §8.21(vii), §8.21(viii) Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(vii), §8.21 and Ch.8 |
§8.21(viii) Asymptotic Expansions
ⓘ
Keywords:
: asymptotic expansions for large variable, auxiliary functions, cosine integrals, generalized, generalized sine and cosine integrals, sine integrals
Notes:
: Apply Watson’s lemma to (8.21.24) and (8.21.25), and then extend the sector of validity from to ; see §2.4(i).
Permalink:
:
See also:
: Annotations for §8.21 and Ch.8
When with (),
| | | |
---
| 8.21.26 | | |
| ⓘ Symbols: : Pochhammer’s symbol (or shifted factorial), : Poincaré asymptotic expansion, : complex variable, : parameter, : nonnegative integer and : auxiliary function Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(viii), §8.21 and Ch.8 |
| 8.21.27 | | |
| ⓘ Symbols: : Pochhammer’s symbol (or shifted factorial), : Poincaré asymptotic expansion, : complex variable, : parameter, : nonnegative integer and : auxiliary function Permalink: Encodings: TeX, pMML, png See also: Annotations for §8.21(viii), §8.21 and Ch.8 |
For the corresponding expansions for and apply (8.21.20) and (8.21.21).
8.20 Asymptotic Expansions of 8.22 Mathematical Applications
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10115 | http://sigmaa.maa.org/mcst/SANTAFENM.pdf | Pairing Math Competitions with Math Wrangles throughout a School District Joint Mathematics Meetings 2024 4 January 2024 James C Taylor, MathAmigos Geoffrey Moon, Santa Fe Public Schools 1/9/24 9:00:50 PM (mathamigos.org) In spring of 2017 a small group of math education-enthusiastic Santa Feans started meeting in order to raise awareness and levels of mathematics understanding in Santa Fe, New Mexico public schools (SFPS). We initially focused on teacher professional development.
Since then, we have entered the classrooms, teaching over two hundred students each year and modeling our methods for the teachers in those classrooms.
Two years ago, we began greatly expanding the use of mathematics competitions—and adding math wrangles.
School District Context v Total SFPS students: 12,875 in 30 schools v District math proficiency average: 24% (grades 3-8, MathAmigos‘ core grades) v The district’s highest-rated schools (small schools in affluent neighborhoods) demonstrate proficiency in the 30%-45% range, with one at 58%.
v Working within primarily Title I schools (25 of 30 schools in Santa Fe, NM), most of which test around 12% math proficiency, we have been able help more than 200 upper elementary and middle school students a year develop positive math identities through district-sponsored math competitions and wrangles (math debates). We tripled participation in one competition in one year, and have increased participation again this year. v More than half of the students come from underserved communities. v Many of our students are English language learners Elements of the New Program v Weekly sessions in 9 schools v Alternating weeks in 6 additional schools v Grades 2 through 8 v SA2GE classes (Services for Advanced Academics and Gifted Education) v Note: Gifted does not necessarily mean strong in math or reading skills v Most of the SA2GE teachers do not have much background in mathematics v Nearly all are Title I schools v Program: v Begin with math circles (grades 4-8) for quarter 1, then sporadically throughout the year v Introduce problem-solving techniques through Olympiad (MOEMS ) problems to entire class v Ongoing student participation in MOEMS competition v Class-level problem solving morphs into two teams solving math wrangle sets in separate rooms v Wrangle practice v Wrangle discussion, debrief, exploration of alternate approaches v Interschool wrangles Math Olympiads for Elementary & Middle School Students Some Math Circle Problems 1.
(Find the) Polite Numbers to 20 v A polite number can be written as the sum of two or more consecutive positive integers v Only addition is needed, so accessible to all grades and skills v Is about noticing patterns: v “Rude” or impolite numbers v Especially polite numbers (can be expressed as a sum of different numbers v Useful for introducing algebra at early (as early as 3rd) grades for generating the polite numbers that are the sum of 2, 3, or 4 (or more) successive numbers 2.
The Game of Set v A game, so no overt math v Introducing dimensions: 4D, 3D, 2D using Zometools. And to understand 4D Set better, “make it small” by going to 2D v How many sets in 2D? Students notice that playing 2D is is like Tic-Tac-Toe, until someone finds a set that is not in a row v 2D Set is Tic-Tac-Toe (students notice this right away), but on a torus v Counting sets and a taste of combinatorics School Program Growth 2021-2024 v School, teacher, and student numbers have been increasing steadily over 3 years v Not all of these students are also participating in the MOEMS competition and wrangles MOEMS ® Competitions Overview v Mathematical Olympiads for Elementary and Middle Schools is an international math competition program that provides 5 monthly problem-solving contests for elementary and middle school students v There are two divisions: v The elementary division is for grades 4, 5 and 6 v The middle school division for grades 6, 7 and 8 v 5 monthly contests, 5 questions each, of increasing difficulty v Problem sets administered online each month from November through March. Online administration option became available during SY 20-21.
v Floor to ceiling range is quite high. Most students can get one correct answer, 2-3 correct with reasoning, and 4-5 with well-developed skills.
v Fully worked solutions and extension problems are available post-test A Type of MOEMS Competition Problem 1. Cryptarithms: Each letter represents a different digit. What 4-digit number does TOUR represent? v A puzzle that help develop number sense v AI apps such as ChatGPT fail to solve these, despite “recognizing” them or solve using brute force substitutions for all possible letters TWO +TWO FOUR MOEMS Competitions Growth 2021-2024 v School, teacher, and student numbers have been steadily over 3 years v Not all of these students are also participating in the circles and wrangles v Mixture of students identified for gifted education and interested/motivated others.
MOEMS Elementary (Grades 4-6) 23-24 reports average from first two contests MOEMS Middle (Grades 7-8 – 6th omitted) 23-24 reports average from first two contests Newest Effort: Competitions & Wrangles v In partnership with the district’s SA2GE program, which also runs the math competitions for the district, since 2021 we have been dramatically increasing the numbers of grades 3-8 students taking the MOEMS tests. We tripled the numbers in 2022-23—to 160—from the previous year, and now have 21 schools participating, most of them being Title I schools (most students qualifying for free or reduced lunch).
v Several students ranked nationally or internationally in year 2 (2022-23) v The number of students performing above average has increased, but the more important change may have to do with their attitude towards math. A majority of the students surveyed after the 2022-23 school year expressed positive feelings about “hard problems,” and about two-thirds reported positive changes to the ways they felt about and approached difficult math.
v In May of 2023 the SFPS hosted its first district-wide math wrangle, one elementary level and one middle school level wrangle. v We believe this is the first such district-wide wrangle in the US.
SA2GE = Services for Advanced Academics and Gifted Education Math Olympiads for Elementary & Middle School Students Benefits of the Competitions to Wrangles Project v For the students v Competition Prep: Teamwork, strategic thinking, conversing about math v Competitions: Problem solving skills, resilience, math, confidence v Wrangles: Presenting mathematical reasoning, conversations, confidence, public speaking, debating skills and strategies, logic, teamwork, use of clear and proper language, respectful critiquing of presentations, etc.
v For the teachers v Confidence with hard math v Independence in training competition and wrangle teams v For the schools: v Visibility and pride How do you feel about hard math problems?
v 1st year Students v I feel pretty good about them when I'm solving them, happy or sad when I know they're right or wrong, and eager to know what I scored on them.
v Hard math problems are fun (and brain frying).
v MAD AND CONFUSED v I like to experiment with hard math problems over time.
v Multi-Year Students v I have mixed feelings about hard to harder math problems. One side of me thinks hard math problems are enjoyable while the other thinks they are frustrating.
v I can sometimes feel stressed or confused but I do love challenges and I think it can be really fun.
v I LOVE them, they are so much fun.
v I honestly feel stressed and a little bit rushed, when I'm stuck on a problem and everyone is done.
How have your feelings and approaches to hard problems changed over the course of this Math Olympiad season?
v 1st Year Students v "My feelings have changed because originally, I thought Math Problems were either too hard or too easy, and I like the Math Olympiad problems because they feel right.” v "I went from scared to fine with the math.” v "At first I would just try one way to solve it and if the solution seemed correct, but now I try a lot of things to make it work.” v "I've felt more frustrated.” v Multi-Year Students v "I've learned that rather than using my energy to be angry at a problem, I can use it to solve the problem. Sometimes.” v "My feelings and approaches to hard problems have changed because instead of just knowing all of the answers like I normally do, I really have to think more about what the question is asking me so I can know what my answer will be.” v "I do it in more creative ways” v "I have found out more ways to use the strategies I have been taught but I still am not that good at complex problems.” Competitions & Resilience v Late in the 2022-23 school year I asked a teacher in one of my weekly classes whether regular practice with competition problems and wrangles had improved his students’ math performance in their regular classes.
v He said that these (accelerated) students normally did well in those classes and so he saw no significant improvement there, but… v …when they had encountered hard problems, tears were often the result.
v But now, no tears. Now they were accustomed to hard problems.
v We hope to measure resilience in the coming school year District Math Wrangle Overview v A math wrangle is a mathematics debate, where the two teams go to separate rooms and try to solve a set of hard problems. Solutions are written on large pads.
v Pairing: Our wrangle used MOEMS-style problems of the type that the students had seen during the year for wrangle prep, critical especially when they’d had little wrangle practice.
v The Math Wrangle: v A coin toss, and one team challenges the other to solve one of the problems v The challenged team presents and explains a complete solution, not just an answer v The other team may challenge (rebut) the result, and judges assign points. And so on.
v No student or student pair may present or rebut more than once v May 2023 district-wide wrangle v 58 students from 12 district schools participated in wrangles or small math circles v Students not on the 4 wrangle teams formed small math circles around one of our school-year mentored teachers and worked on the wrangle problems so that they would be ready to understand the wrangle session v About 20 teachers worked as wrangle proctors or math circle leaders v About 20 parents attended and helped transport the students that school day v We surveyed all participants Wrangles: Responses & Surveys v When the wrangle was done, I asked everyone if they might want to do this again, and there was a resounding “YES!” from the participants v Students reported that they: v felt challenged, had fun and cared more about the hard math problems v grown in confidence, comfort, concentration and determination v had learned to use better strategies, find patterns and think creatively rather than assume they should already have answers v found the problems “are very challenging, but once you find the pattern or the secret to the problem it becomes really easy, which makes me happy.” v felt the event helped them gain confidence, with one writing the wrangle, “only solidified that I truly love math.” v And one middle school student said it was the best thing he’d done all year v Parents noted students’ enjoyment at working together, writing, “the students looked so happy throughout.” v In 2023-24 we are planning wrangles between pairs of school throughout the year Impact on Mathematical Motivation v 5th and 6th grader SA2GE students begged to be permitted to bring their classmates to their math circles and competition class--and then fetched them. (Dual-Bilingual School) v When told at the end of a math circle session in a general education class that the teacher had some bad news, students shouted “oh no! you’re not coming back”. But no, the bad news was that they were only almost correct about their solutions. They showed visible relief. (Average Elementary) Impact on Number Sense v General education fourth grade students at one school have been participating in Math Circles and their teachers have commented on some of the positive impacts they have observed to their SA2GE teacher partner. v One teacher saw that the students were able to grasp the concept of rounding numbers more quickly and easily after those students had explored patterns/sequences in a Math Circle activity. This impact was more noticeable in the group of students who had lower scores on standardized tests. Providing students the opportunity to explore the numbers “in-between” the tens gave them insights into the process of rounding and the general properties of counting numbers.
Impact on Playful Mathematical Affect v Another teacher noted that these students do not often have the chance to ’play’ with numbers, and this exploration through simple counting and addition was beneficial to their number sense. Most of the math experiences for these students is rigid and formulaic, when they work with numbers it’s much more a tedious task coupled with fear and shame of getting the wrong answer. On the other hand, Math Circles provide an opportunity to freely play and explore with numbers without as many constraints or consequences. The exploration and “playing” with numbers is an important process in the development of mathematical thinking.
v SA2GE teacher connecting with partner 4th Grade General Ed Teacher Impact on one SA2GE Teacher “I never saw myself as a strong math student. After a certain point, it just didn't make sense to me. I didn't understand fractions until I started baking in high school. I paid my sister to do my geometry work in high school. My brother (who was in 4th grade at the time) helped me with pre algebra in 8th grade...I felt stupid anytime I couldn't understand what I was supposed to do.” “I never wanted to teach math past the 3rd grade level-I didn't know how I could teach it and explain it when I wasn't understanding it myself. When I started teaching gifted ed, I realized I had to teach math. I avoided it as much as possible, never really getting into complicated math concepts-just having the kids do number and critical thinking puzzles.” Impact on one SA2GE Teacher “Math Circles, Math Amigos, MOEMs changed everything. I enjoy teaching math/creative problem solving now. I prefer it over teaching ELA. I learned how playful and creative math can be. I realized I can do the work right alongside my students and share my problem solving strategies and not worry if I didn't get the correct answer. I love modeling problem solving for them, talking them through challenges they may face with a MOEMs problem.” “I love learning from my students-often times they are able to solve problems in completely different ways than what I thought of. I am able to see patterns, find shortcuts and make some pretty cool connections in math now. I have enjoyed watching my youngest students work on problem solving along with the older students. My students see my SA2GE math time as a time to play, ask questions, discover new things and to learn from their mistakes.” Program Leaders: Lessons Learned (James) v Grade levels & ages: v Early work more grades 6-9. Shifted to more earlier grades, 2-6, with some 7 & 8 v Cultivating a culture of mathematical inquiry, weeks 2 to 8: respectful conversation (with the teacher, classmates, teacher of record), challenging the teacher, “leaping to the board” v Coping with wide range of grade levels/skills in math v Need to focus on patterns and early grade level math circles v Experience with the above giving insight and motivation to… v …branch out year 3 into general education classes at two schools v Teaching an entire grade level v 3-4 grade levels of math skill in a single class v Requires low threshold/high ceiling play and pattern exploration benefitting all students v Teachers have requested materials to follow up on our weekly sessions v Seeing the uses and value of competition problems, especially with group solving leading into wrangles v Gratifying levels of teacher growth as mathematical thinkers and doers, leading to independence with the materials we use. Many have been studying and using AMS Math Circle Library books, Art of Problem Solving materials, and more, on their own.
Program Leaders: Lessons Learned (Geoffrey) v Growth across a horizon is not linear.
v Easily available data such as grades, teacher observations, and grade-level achievement can lead to narrowing of focus in math content instruction. That may boomerang.
v Low mathematical self-efficacy is culturally self-reinforcing. Changing that requires work with teachers, students, and the community.
v More reasoning-focused problems support more heterogeneous student grouping.
v Domain-specific identification of mathematical talent can lead to too little focus on language development.
Retention Questions Years In MOEMS N ≥3 28 2 83 1 140 v Who stays?
v Who leaves?
v Why?
v What impacts retention?
v What is the year one experience like?
More Questions v How does student mathematical identity and sense of belonging change over time? v To what extent do circle-competition-wrangles change mathematical problem solving abilities and temperament?
v How does the programming affect other teachers?
v What skills are/can be developed to improve contest performance?
v Does the program have effects on mathematical creativity?
v Does the program have effects on self-management?
v How far and in what ways can we expand the program to more broadly affect the school district and broader community?
v How can this program be used by the schools to trumpet some successes in math?
v What should we be studying further? Resilience? Detailed pre- and post-measures of attitudes? Thanks!
Geoffrey Moon, Santa Fe Public Schools gmoon@sfps.k12.nm.us James C Taylor, MathAmigos.org jtaylor505@gmail.com |
10116 | https://www.merriam-webster.com/thesaurus/censures | Est. 1828
Synonyms of censures
noun
as in reprimands
verb
as in condemns
as in denounces
as in criticizes
as in reprimands
as in condemns
as in denounces
as in criticizes
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censures 1 of 2
Definition of censures
plural of censure
as in reprimands
an often public or formal expression of disapproval a rare censure of a senator by the full United States Senate for misconduct
Synonyms & Similar Words
riot acts
tongue-lashings
dressing-downs
businesses
talking-tos
belittlements
Antonyms & Near Antonyms
citations
commendations
indorsements
endorsements
honors
sanctions
tributes
praises
acclamations
plaudits
eulogies
encomiums
approvals
blessings
panegyrics
encomia
censures
2 of 2
verb
present tense third-person singular of censure
1
as in condemns
to express public or formal disapproval of a vote to censure the President for conduct that was unbecoming to his office
Synonyms & Similar Words
reprimands
condemns
denounces
punishes
criticizes
scolds
rebukes
reproaches
reproves
admonishes
calls to account
chastises
objurgates
brings to account
upbraids
lambasts
castigates
berates
disparages
belittles
lectures
flays
deprecates
lambastes
bawls out
rails (at or against)
jaws
cuts up
rags
tells off
scores
gibbets
depreciates
dresses down
chews out
keelhauls
rates
Antonyms & Near Antonyms
commends
cites
endorses
indorses
applauds
honors
hails
praises
approves
lauds
sanctions
acclaims
blesses
eulogizes
2
as in denounces
to declare to be morally wrong or evil our society generally censures the taking of another person's life
Synonyms & Similar Words
denounces
condemns
criticizes
blames
attacks
decries
damns
anathematizes
faults
sentences
execrates
disparages
belittles
reprehends
deprecates
reprobates
excommunicates
rebukes
convicts
upbraids
reproaches
scolds
reprimands
chides
pans
chastises
hates
slams
admonishes
dooms
reproves
knocks
berates
abominates
castigates
blasts
lambasts
reviles
detests
ostracizes
dispraises
abhors
loathes
rakes
lambastes
disses
blacklists
curses
vituperates
imprecates
Antonyms & Near Antonyms
approves
sanctions
endorses
blesses
indorses
praises
commends
lauds
exalts
applauds
extols
hails
salutes
glorifies
eulogizes
touts
honors
sanctifies
consecrates
extolls
acclaims
reveres
hallows
venerates
3
as in criticizes
to express one's unfavorable opinion of the worth or quality of critics have striven to outdo each other in censuring that pop novelist's latest work
Synonyms & Similar Words
criticizes
blames
condemns
faults
denounces
knocks
attacks
slams
scolds
tweaks
pans
finds fault (with)
complains
comes down hard (on)
slags
reprehends
takes to task
slashes
dispraises
kicks
disses
lashes
blasts
reprimands
assails
admonishes
rebukes
reproaches
upbraids
moans
chides
reproves
whines
murmurs
lambasts
grumbles
mutters
growls
berates
disparages
clobbers
excoriates
cavils
derides
crucifies
skewers
beefs
belittles
gripes
pillories
grouses
decries
castigates
quibbles
flays
bitches
nicks (at)
lambastes
bellyaches
kvetches
carps
bad-mouths
crabs
niggles
snipes (at)
croaks
gibbets
hammers
drubs
fusses
discommends
keelhauls
puts down
Antonyms & Near Antonyms
praises
lauds
extols
approves
commends
recommends
sanctions
endorses
indorses
extolls
Example Sentences
Recent Examples of Synonyms for censures
reprimands
condemns
denounces
criticizes
condemnations
punishes
blames
faults
Noun
Most job losses start at the district level, where teachers can face discipline that ranges from reprimands to outright termination.
— Tiffani Jackson, Fort Worth Star-Telegram, 17 Sep. 2025
Definition of reprimands
Verb
Latino civil rights organization condemns attack The League of United Latin American Citizens (LULAC), a national civil rights organization for Latinos, condemned the attack at the ICE facility.
— Jeanine Santucci, USA Today, 24 Sep. 2025
The measure, introduced by House Speaker Mike Johnson, a Louisiana Republican, passed on a vote of 310 to 58 and condemns the killing of Kirk.
— Kacen Bayless, Kansas City Star, 20 Sep. 2025
Definition of condemns
Verb
Qatar denounces cowardly Israeli strike targeting Hamas leaders Israel launched an aerial strike against senior Hamas leaders in Doha on Tuesday, in its first direct attack on Qatari soil, marking a significant escalation in regional tensions.
— Jeronimo Gonzalez, semafor.com, 10 Sep. 2025
For instance, ProPublica cited his tax experience as exceptionally limited, pointing out that his only qualification centers around being a Certified Tax & Business Advisor, which ProPublica denounces as a dubious and frivolous designation.
— Nathan Goldman, Forbes.com, 11 Aug. 2025
Definition of denounces
Verb
Former Vice President Kamala Harris criticizes President Biden's decision to run for re-election.
— NBC news, NBC news, 14 Sep. 2025
The judges new ruling at turns criticizes the Ivy League school for its response to antisemitism, but the judge noted the school was now taking steps to address it.
— Lexi Lonas Cochran, The Hill, 3 Sep. 2025
Definition of criticizes
Noun
Emirati leaders response, marked by swift condemnations and the prompt visit of a historically high-ranking delegation to Doha, indicate that Israels government seriously misjudged the situation.
— Monica Marks, Time, 19 Sep. 2025
Mangione allegedly kept journals that described his plans for the attack, his intent to send a message and condemnations of the health insurance industry.
— Michael Ruiz, FOXNews.com, 18 Sep. 2025
Definition of condemnations
Verb
This regressive move punishes our own workers and residents to try to plug budget holes.
— Phillip Molnar, San Diego Union-Tribune, 5 Sep. 2025
Continue reading ¦ WOKE OVERREACH Parents outraged as school punishes boys over trans locker room confrontation.
— , FOXNews.com, 20 Aug. 2025
Definition of punishes
Verb
The trouble started with the SAVE Plan The Education Department largely blames these delays in debt cancellation on the Biden administration and the federal courts.
— Cory Turner, NPR, 19 Sep. 2025
But more often than not Harris blames somebody else for campaign difficulties.
— Francesca Chambers, USA Today, 19 Sep. 2025
Definition of blames
Verb
Cindy also faults her decision to bring Stephenie with her on the reward.
— EW.com, EW.com, 13 Aug. 2025
Definition of faults
Examples are automatically compiled from online sources to show current usage. Read More Opinions expressed in the examples do not represent those of Merriam-Webster or its editors. Send us feedback.
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censuring
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[### Censor vs. Censure
What the [bleep] is the difference?](/grammar/differences-between-censor-and-censure)
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10117 | https://www.vedantu.com/maths/lines-of-symmetry-in-a-parallelogram | Lines of Symmetry in a Parallelogram Explained Simply
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Lines of Symmetry in a Parallelogram: Complete Guide
Lines of Symmetry in a Parallelogram: Complete Guide
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How to Identify Symmetry in Parallelograms with Examples
Before we begin with the lines of symmetry of a parallelogram, we need to understand the concept of a parallelogram, its properties, its sides, angles and the corresponding relationships. A parallelogram can be defined as a special or unique kind of quadrilateral which is a closed four-sided figure with each of the opposite sides that are parallel to each other and have equal length. The interior opposite angles in any given parallelogram have equal value and any pair of adjacent interior angles present can be said as supplementary angles that is they have a sum of 180°.
What is the Symmetry of a Parallelogram?
The parallelogram has no lines of symmetry and, as with the rectangle, students should experiment with folding a copy to see what happens with the lines through the diagonals as well as horizontal and vertical lines.
For understanding the line of symmetry we need to analyse what exactly a line of symmetry is. We can say that a line of symmetry is an axis or imaginary line that can pass through the centre of a shape, facing in any direction, in such a manner that it represents mirror images of each other when cut into two equal halves for example if we cut a square or rectangle, it will have a line of symmetry because at least one imaginary line can be drawn through the centre of the shape that cuts it into two equal halves in such a manner that mirror images of each other are provided. A shape can have multiple lines of symmetry given its properties etc.
After looking at the key characteristics and other observations, it turns out that a parallelogram does not have any line of symmetry. It is a very curious question if we ask that why doesn't the parallelogram not have lines of symmetry, will the simplest answer to this question can be that it is impossible to construct a line of symmetry, an axis or an imaginary line that passes through the centre cutting its image in half where each side would represent a mirror image of the other, in order to test this you can simply try and construct a line of symmetry on any parallelogram and figure out that it is almost impossible. In addition, parallelograms do have rotational symmetry, when we see from geometry perspective of rotational symmetry refers to when a shape or figure is exactly the same as its preimage after it has been rotated a number of degrees, a parallelogram will give the same result as the original or pre-image has been rotated 180°.
After reviewing the characteristics properly and analyzing a parallelogram from all the sides we can conclude that parallelograms do not have any lines of symmetry in turn after reviewing the properties of parallelograms namely that they are quadrilaterals, we can conclude that shapes like squares and rectangles do have lines of symmetry. In the case of a parallelogram, even if they have rotational symmetry, it will result in the same exact image as we started with after a rotation of 180°.
Parallelogram and Lines of Symmetry
Have you ever wondered how many lines of symmetry a parallelogram has? A Parallelogram in general has no lines of symmetry. However, a parallelogram does have a decisive (rotational) symmetry - the half-turn around the median point at which the two diagonals intersect. As shown in the image below, because triangles WXO and YZO are congruent (have the same size and shape), thus
WO= YO
XO = ZO
(Image will be uploaded soon)
In all Parallelograms’ Angles that are not opposite of each other will supplement up to 180 degrees. That being said, there are some specific kinds of parallelograms that have lines of symmetry. They are listed below.
Number of Symmetry Lines in Different Parallelogram
Parallelogram Number of Symmetry Lines
Rhombus 2
Rectangle 2
Square 4
Symmetry Lines in Different Parallelogram
Lines of Symmetry in a Rhombus-
Rhombus is a unique kind of parallelogram and it has 2 lines of symmetry - its diagonals. It means that a rhombus has reflection symmetry over either of its diagonals. Same as a parallelogram, it also has rotational symmetry of 180º about its midpoint.
(image will be uploaded soon)
Lines of Symmetry in a Rectangle -
Rectangle, which is a quadrilateral with four right angles, has 2 lines of symmetry - two lines moving through the central points of opposite sides. A rectangle has reflection symmetry when reflected over the line across the central point of its opposite sides. Same as the parallelogram, it also has rotational symmetry of 180º about its midpoint.
(image will be uploaded soon)
Lines of Symmetry in a Square
Square is a unique kind of parallelogram, with four equal sides and four equal angles. It has 4 lines of symmetry - two diagonals and two lines running through the central points of opposite sides. A square has reflection symmetry when reflected over the line across the central point of its opposite sides as well as over its diagonals. Same as the parallelogram, it also has rotational symmetry of 90º about its midpoint.
(image will be uploaded soon)
Types of Symmetry
While you now know about how many symmetries a parallelogram has, we must also know what exactly symmetry is. Symmetry is the characteristic of being composed or created of exactly equivalent parts facing each other or around an axis. Keeping the definition of symmetry in mind, know that there are various types of symmetry in geometry. However, different types of symmetrical shapes may or may not have all or a particular type of symmetry. Thus, it is important to learn about different types of symmetrical shapes which possess or do not possess a specific type of symmetry. There are 3 types of symmetry which are as follows:-
Linear Symmetry: It has1 a line of symmetry i.e. perpendicular bisector of AB
Point Symmetry: It has point symmetry centre point Z of line segment AB
Rotational Symmetry: It has rotational symmetry of order 2 about Z.
(image will be uploaded soon)
Fun Facts
The term "parallelogram" is derived from the Greek word "parallelogramma" (which means fenced by parallel lines).
Parallelograms are quadrilaterals with four sides.
The area is bisected by Any of the lines passing through the centre of a parallelogram.
A parallelogram has to its name 2 sets of parallel sides (which never meet) and four edges.
Opposite sides of a parallelogram are equally long (they are the same in length) and are parallel to each other.
Squares, Rectangles, and rhombuses are all parallelograms.
A trapezoid is not a parallelogram. A trapezoid has exactly two parallel sides whereas a parallelogram has two pairs of parallel sides.
A trapezoid is a special kind of parallelogram with at least one pair of parallel sides but neither has reflectional symmetry nor a rotational symmetry. Therefore you cannot make observations based upon symmetry.
An isosceles trapezoid is a parallelogram for which all four sides are the same (are equally long).
An isosceles trapezoid that has only one pair of parallel sides has reflectional symmetry but no rotational symmetry.
FAQs on Lines of Symmetry in a Parallelogram: Complete Guide
What is the definition of a line of symmetry in geometry?
In geometry, a line of symmetry is a line that divides a figure into two parts that are perfect mirror images of each other. If you were to fold the shape along this line, the two halves would coincide exactly. This property is also known as reflectional symmetry or an axis of symmetry.
How many lines of symmetry does a general parallelogram have?
A general parallelogram has zero lines of symmetry. While its opposite sides and angles are equal, no single line can be drawn to divide it into two identical, mirror-image halves. It does, however, possess rotational symmetry.
Why does a parallelogram have no lines of reflectional symmetry?
A parallelogram lacks lines of reflectional symmetry because its non-opposite vertices and sides are not equidistant from any potential symmetry line. If you attempt to fold a parallelogram along its diagonals or any other line connecting midpoints, the two resulting halves will not perfectly overlap, proving that they are not mirror images.
What is the difference between line symmetry and rotational symmetry in a parallelogram?
Line symmetry requires a figure to be split into two mirror images by a line, which a general parallelogram does not have. In contrast, rotational symmetry occurs when a figure looks identical after being rotated less than 360°. A parallelogram has rotational symmetry of order 2, as it appears unchanged after a 180° rotation around the point where its diagonals intersect.
Can a parallelogram ever have lines of symmetry? Explain with examples.
Yes, a parallelogram has lines of symmetry only when it is a special type of quadrilateral. The key examples are:
A rectangle, a type of parallelogram, has two lines of symmetry that connect the midpoints of its opposite sides.
A rhombus, another parallelogram, has two lines of symmetry along its diagonals.
A square, which is both a rectangle and a rhombus, has four lines of symmetry (two diagonals and two lines connecting opposite midpoints).
A general parallelogram, with no right angles and unequal adjacent sides, has none.
What is a common misconception about the diagonals of a parallelogram and symmetry?
A frequent misconception is that the diagonals of a parallelogram act as its lines of symmetry. While diagonals are key properties (they bisect each other), they only function as lines of symmetry in special cases like a rhombus or a square. For a general parallelogram, folding the shape along a diagonal will not produce two identical halves.
How do the lines of symmetry in a parallelogram compare to other quadrilaterals?
A general parallelogram has zero lines of symmetry, which differs from other quadrilaterals. Here's a comparison:
Square: 4 lines of symmetry.
Rectangle: 2 lines of symmetry.
Rhombus: 2 lines of symmetry.
Isosceles Trapezoid: 1 line of symmetry.
Kite: 1 line of symmetry.
This highlights that while a parallelogram is a fundamental quadrilateral, it is one of the least symmetrical in terms of reflection.
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10118 | http://chethermo.net/screencasts/section-4.3 | 04.03 The Macroscopic View of Entropy | Introductory Chemical Engineering Thermodynamics, 2nd ed.
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ScreenCast Supplements:
Chapter 4 - Entropy
04.03 The Macroscopic View of Entropy
Elliott
11 years 4 months ago
Entropy of the Universe in a Glass of Water
Heat and entropy in a glass of water(uakron, 9min) Taking a glass from the refrigerator causes heat to flow from the room to the water. The temperature of the water slowly rises while the temperature of the (relatively large) room remains fairly constant. Applying the macroscopic definition of entropy makes it easy to compute the entropy changes, but is one larger than the other? Are all entropy changes greater than zero? What does the second law mean exactly?
Comprehension Questions:
Describe your own example of a process with an entropy decrease and explain why it doesn't violate the second law.
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Elliott
10 years 4 months ago
Molecular Nature of S: Micro to Macro in the Einstein Solid
Molecular Nature of S: Micro to Macro of Thermal Entropy(uakron.edu, 20min) We can explain configurational entropy by studying particles in boxes, but only at constant temperature. How does the entropy change if we change the temperature? Why should it change if we change the temperature? The key is to recognize that energy is quantized, as best exemplified in the Einstein Solid model. We learned in Chapter 1 that energy increases when temperature increases. If we have a constant number of particles confined to lattice locations, then the only way for the energy to increase is if some of the molecules are in higher energy states. These "higher energy states" correspond to faster (higher frequency) vibrations that stretch the bonds (Hookean springs) to larger amplitudes. We can count the number of molecules in each energy state similar to the way we counted the number of molecules in boxes. Then we supplement the formula for configurational entropy changes to arrive at the following simple relation for all changes in entropy for ideal gases: Δ S = Cv ln(T 2/T 1) + R ln(V 2/V 1). Note that we have related the entropy to changes in state variables. This observation has two significant implications: (1) entropy must also be a state function (2) we can characterize the entropy by specifying any two variables. For example, substituting V = RT/P into the above equation leads to: Δ S = Cp ln(T 2/T 1) - R ln(P 2/P 1).
Comprehension Questions:
Show the steps required to derive Δ S = Cp ln(T 2/T 1) - R ln(P 2/P 1) from Δ S = Cv ln(T 2/T 1) + R ln(V 2/V 1).
We derived a memorable equation for adiabatic, reversible, ideal gases in Chapter 2. Hopefully, you have memorized it by now! Apply this formula to compute the change in entropy for adiabatic, reversible, ideal gases as they go through any change in temperature and pressure.
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Elliott
10 years 4 months ago
Getting to Know Entropy: Micro to Macro to Micro
You might better understand the macroscopic definition of entropy(uakron, 9min) if you consider isothermal reversible expansion of an ideal gas. Note the word "isothermal" is different from "adiabatic." If the expansion was an adiabatic and reversible expansion of an ideal gas, then we know from Chapter 2 that the temperature would go down, ie. T2/T1=(P2/P1)^(R/Cp)=(V1/V2)^(R/Cv). Therefore, holding the temperature constant must require the addition of heat. We can calculate the change in entropy for this isothermal process from the microscopic balance, then show that the amount of heat added is exactly equal to the change in entropy (of this reversible process) times the (isothermal) temperature. Studying the energy and entropy balance for the irreversible process helps us to appreciate how entropy is a state function. As suggested by the hint at the end of this video, you can turn this perspective around and infer the relation of entropy to volume by starting with the macroscopic definition and calculating exactly how much heat must be added after adiabatic, reversible expansion in order to recover the original (isothermal) temperature. Through this thought process, you should start to appreciate that the micro and macro definitions are really interchangeable expressions of the same quantity.
Comprehension Questions: (Hint: entropy is a state function.)
Use the macroscopic definition of entropy to compute the change in entropy (J/mol-K) of N2 in a piston/cylinder from 450K and 1cm3/mol to 450K and 4cm3/mol.
Use the macroscopic definition of entropy to compute the change in entropy (J/mol-K) of N2 in a piston/cylinder from 450K and 4cm3/mol to 258.46K and 4cm3/mol.
Use the macroscopic definition of entropy to compute the change in entropy (J/mol-K) of N2 in a piston/cylinder from 450K and 1cm3/mol to 258.46K and 4cm3/mol.
Use the macroscopic definition of entropy to compute the change in entropy (J/mol-K) of N2 in a piston/cylinder from 450K and 1cm3/mol to 300K and 3 cm3/mol.
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Elliott
7 years 4 months ago
Entropy Changes for Ideal Gases
Once we establish equations relating macroscopic properties to entropy changes, it becomes straightforward to compute entropy changes for all sorts of situations. To begin, we can compute entropy changes of ideal gases (learncheme, 3 min). Entropy change calculations may also take a more subtle form in evaluating reversibility (learncheme, 3min).
Comprehension Questions:
Nitrogen at 298K and 2 bars is adiabatically compressed to 375K and 5 bars in a continuous process. (a) Compute the entropy change. (b) Is this process reversible, irreversible, or impossible?
Nitrogen at 350K and 2 bars is adiabatically compressed to 575K and 15 bars in a piston/cylinder. (a) Compute the entropy change. (b) Is this process reversible, irreversible, or impossible?
Steam at 450K and 2 bars is adiabatically compressed to 575K and 15 bars in a continuous process. (a) Compute the entropy change. (b) Is this process reversible, irreversible, or impossible?
Steam at 450K and 2 bars is isothermally compressed to 8 bars in a continuous process. (a) Compute the entropy change. (b) Is this process reversible, irreversible, or impossible?
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Book navigation
Chapter 1 - Basic concepts
Chapter 2 - The energy balance
Chapter 3 - Energy balances for composite systems.
Chapter 4 - Entropy
04.02 The Microscopic View of Entropy
04.03 The Macroscopic View of Entropy
04.04 The Entropy Balance
04.09 Turbine calculations
04.10 Pumps and Compressors
Chapter 5 - Thermodynamics of Processes
Chapter 6 - Classical Thermodynamics - Generalization to any Fluid
Chapter 7 - Engineering Equations of State for PVT Properties
Chapter 8 - Departure functions
Chapter 9 - Phase Equlibrium in a Pure Fluid
Chapter 10 - Introduction to Multicomponent Systems
Chapter 11 - An Introduction to Activity Models
Chapter 12 - Van der Waals Activity Models
Chapter 13 - Local Composition Activity Models
Chapter 14 - Liquid-liquid and solid-liquid equilibria
Chapter 16 - Advanced Phase Diagrams
Chapter 15 - Phase Equilibria in Mixtures by an Equation of State
Chapter 17 - Reaction Equilibria
Chapter 18 - Electrolyte Solutions
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Prentice-Hall web site. |
10119 | https://www.slideshare.net/slideshow/casio-fx991ex-classwiz-user-manual-and-commands/242261442 | Uploaded byFarhanAhmade
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Casio fx-991EX classWiz user manual and commands.
The document is a user's guide for Casio's fx-570es plus and fx-991es plus calculators, detailing essential information, safety precautions, operation instructions, and configuration settings. It covers topics such as initializing the calculator, performing calculations, using menus, and setting up the calculator for various modes and displays. The manual includes specifications, troubleshooting information, and frequent questions for user reference.
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Casio fx-991EX classWiz user manual and commands.
E fx-570ES PLUS fx-991ES PLUS User’sGuide CASIO Worldwide Education Website CASIO EDUCATIONAL FORUM EEE Yönetmeliğine Uygundur
2.
Contents Important Information .............................................................2 Sample Operations.................................................................. 2 Initializing the Calculator........................................................ 2 Safety Precautions .................................................................. 2 Handling Precautions.............................................................. 2 Removing the Hard Case ........................................................ 3 Turning Power On and Off ...................................................... 3 Adjusting Display Contrast .................................................... 3 Key Markings ........................................................................... 3 Reading the Display ................................................................ 4 Using Menus ............................................................................ 5 Specifying the Calculation Mode ........................................... 5 Configuring the Calculator Setup .......................................... 5 Inputting Expressions and Values ......................................... 7 Toggling Calculation Results ................................................. 9 Basic Calculations ................................................................ 10 Function Calculations........................................................... 13 Complex Number Calculations (CMPLX) ............................ 18 Using CALC............................................................................ 19 Using SOLVE.......................................................................... 20 Statistical Calculations (STAT)............................................. 22 Base-n Calculations (BASE-N)............................................. 26 Equation Calculations (EQN) ............................................... 28 Matrix Calculations (MATRIX)............................................... 29 Creating a Number Table from a Function (TABLE) ........... 32 Vector Calculations (VECTOR)............................................. 33 Scientific Constants.............................................................. 35 Metric Conversion ................................................................. 37 Calculation Ranges, Number of Digits, and Precision......................................................................... 38 Errors...................................................................................... 40 Before Assuming Malfunction of the Calculator... ............. 41 Replacing the Battery............................................................ 42 Specifications ........................................................................ 42 Frequently Asked Questions................................................ 43 E-1
3.
E-2 Important Information • Thedisplays and illustrations (such as key markings) shown in this User’s Guide are for illustrative purposes only, and may differ somewhat from the actual items they represent. • The contents of this manual are subject to change without notice. • In no event shall CASIO Computer Co., Ltd. be liable to anyone for special, collateral, incidental, or consequential damages in connection with or arising out of the purchase or use of this product and items that come with it. Moreover, CASIO Computer Co., Ltd. shall not be liable for any claim of any kind whatsoever by any other party arising out of the use of this product and the items that come with it. • Be sure to keep all user documentation handy for future reference. Sample Operations Sample operations in this manual are indicated by a icon. Unless specifically stated, all sample operations assume that the calculator is in its initial default setup. Use the procedure under “Initializing the Calculator” to return the calculator to its initial default setup. For information about the B, b, v, and V marks that are shown in the sample operations, see “Configuring the Calculator Setup”. Initializing the Calculator Perform the following procedure when you want to initialize the calculator and return the calculation mode and setup to their initial default settings. Note that this operation also clears all data currently in calculator memory. !9(CLR)3(All)=(Yes) Safety Precautions Battery • Keep batteries out of the reach of small children. • Use only the type of battery specified for this calculator in this manual. Handling Precautions • Even if the calculator is operating normally, replace the battery at least once every three years (LR44 (GPA76)) or two years (R03 (UM-4)). A dead battery can leak, causing damage to and malfunction of the calculator. Never leave a dead battery in the calculator. Do not try using the calculator while the battery is completely dead (fx-991ES PLUS). • The battery that comes with the calculator discharges slightly during shipment and storage. Because of this, it may require replacement sooner than the normal expected battery life. • Do not use an oxyride battery or any other type of nickel-based primary battery with this product. Incompatibility between such batteries and product specifications can result in shorter battery life and product malfunction. • Avoid use and storage of the calculator in areas subjected to temperature extremes, and large amounts of humidity and dust. • Do not subject the calculator to excessive impact, pressure, or bending.
4.
E-3 • Never tryto take the calculator apart. • Use a soft, dry cloth to clean the exterior of the calculator. • Whenever discarding the calculator or batteries, be sure to do so in accordance with the laws and regulations in your particular area. Company and product names used in this manual may be registered trademarks or trademarks of their respective owners. Removing the Hard Case Before using the calculator, slide its hard case downwards to remove it, and then affix the hard case to the back of the calculator as shown in the illustration nearby. Turning Power On and Off Press O to turn on the calculator. Press 1A(OFF) to turn off the calculator. Auto Power Off Your calculator will turn off automatically if you do not perform any operation for about 10 minutes. If this happens, press the O key to turn the calculator back on. Adjusting Display Contrast Display the CONTRAST screen by performing the following key operation: 1N(SETUP)c6(]CONT'). Next, use d and e to adjust contrast. After the setting is the way you want, press A. Important: If adjusting display contrast does not improve display readability, it probably means that battery power is low. Replace the battery. Key Markings Pressing the 1 or S key followed by a second key performs the alternate function of the second key. The alternate function is indicated by the text printed above the key. The following shows what the different colors of the alternate function key text mean. If key marking text is this color: It means this: Yellow Press 1 and then the key to access the applicable function. Red Press S and then the key to input the applicable variable, constant, or symbol. Purple (or enclosed in purple brackets) Enter the CMPLX Mode to access the function. Green (or enclosed in green brackets) Enter the BASE-N Mode to access the function. sin–1 D s Alternate function Keycap function sin–1 D s Alternate function Keycap function
5.
E-4 Reading the Display Thedisplay of the calculator shows expressions you input, calculation results, and various indicators. Input expression Indicators Calculation result • If a ' indicator appears on the right side of the calculation result, it means the displayed calculation result continues to the right. Use e and d to scroll the calculation result display. • If a g indicator appears on the right side of the input expression, it means the displayed calculation continues to the right. Use e and d to scroll the input expression display. Note that if you want to scroll the input expression while both the ' and g indicators are displayed, you will need to press A first and then use e and d to scroll. Display indicators This indicator: Means this: The keypad has been shifted by pressing the 1 key. The keypad will unshift and this indicator will disappear when you press a key. The alpha input mode has been entered by pressing the S key. The alpha input mode will be exited and this indicator will disappear when you press a key. M There is a value stored in independent memory. STO The calculator is standing by for input of a variable name to assign a value to the variable. This indicator appears after you press 1t(STO). RCL The calculator is standing by for input of a variable name to recall the variable’s value. This indicator appears after you press t. STAT The calculator is in the STAT Mode. CMPLX The calculator is in the CMPLX Mode. MAT The calculator is in the MATRIX Mode. VCT The calculator is in the VECTOR Mode. 7 The default angle unit is degrees. 8 The default angle unit is radians. 9 The default angle unit is grads. FIX A fixed number of decimal places is in effect. SCI A fixed number of significant digits is in effect. Math Natural Display is selected as the display format. $` Calculation history memory data is available and can be replayed, or there is more data above/below the current screen. Math Math Math Math
6.
E-5 Disp The display currentlyshows an intermediate result of a multi-statement calculation. Important: For some type of calculation that takes a long time to execute, the display may show only the above indicators (without any value) while it performs the calculation internally. Using Menus Some of the calculator’s operations are performed using menus. Pressing N or w, for example, will display a menu of applicable functions. The following are the operations you should use to navigate between menus. • You can select a menu item by pressing the number key that corresponds to the number to its left on the menu screen. • The $ indicator in the upper right corner of a menu means there is another menu below the current one. The ` indicator means another menu above. Use c and f to switch between menus. • To close a menu without selecting anything, press A. Specifying the Calculation Mode When you want to perform this type of operation: Perform this key operation: General calculations N1(COMP) Complex number calculations N2(CMPLX) Statistical and regression calculations N3(STAT) Calculations involving specific number systems (binary, octal, decimal, hexadecimal) N4(BASE-N) Equation solution N5(EQN) Matrix calculations N6(MATRIX) Generation of a number table based on an expression N7(TABLE) Vector calculations N8(VECTOR) Note: The initial default calculation mode is the COMP Mode. Configuring the Calculator Setup First perform the following key operation to display the setup menu: 1N(SETUP). Next, use c and f and the number keys to configure the settings you want. Underlined (___) settings are initial defaults. 1MthIO 2LineIO Specifies the display format. Natural Display (MthIO) causes fractions, irrational numbers, and other expressions to be displayed as they are written on paper. Math Math
7.
E-6 MthIO: Selects MathOor LineO. MathO displays input and calculation results using the same format as they are written on paper. LineO displays input the same way as MathO, but calculation results are displayed in linear format. Linear Display (LineIO) causes fractions and other expressions to be displayed in a single line. Note: • The calculator switches to Linear Display automatically whenever you enter the STAT, BASE-N, MATRIX, or VECTOR Mode. • In this manual, the B symbol next to a sample operation indicates Natural Display (MathO), while the b symbol indicates Linear Display. 3Deg 4Rad 5Gra Specifies degrees, radians or grads as the angle unit for value input and calculation result display. Note: In this manual, the v symbol next to a sample operation indicates degrees, while the V symbol indicates radians. 6Fix 7Sci 8Norm Specifies the number of digits for display of a calculation result. Fix: The value you specify (from 0 to 9) controls the number of decimal places for displayed calculation results. Calculation results are rounded off to the specified digit before being displayed. Example: b 100 ÷ 7 = 14.286 (Fix 3) 14.29 (Fix 2) Sci: The value you specify (from 1 to 10) controls the number of significant digits for displayed calculation results. Calculation results are rounded off to the specified digit before being displayed. Example: b 1 ÷ 7 = 1.4286 × 10–1 (Sci 5) 1.429 × 10–1 (Sci 4) Norm: Selecting one of the two available settings (Norm 1, Norm 2) determines the range in which results will be displayed in non-exponential format. Outside the specified range, results are displayed using exponential format. Norm 1: 10–2 |x|, |x| 1010 Norm 2: 10–9 |x|, |x| 1010 Example: b 1 ÷ 200 = 5 × 10–3 (Norm 1) 0.005 (Norm 2) c1ab/c c2 d/c Specifies either mixed fraction (ab/c) or improper fraction (d/c) for display of fractions in calculation results. c3CMPLX 1a+bi ; 2r∠ Specifies either rectangular coordinates (a+bi) or polar coordinates (r∠) for EQN Mode solutions. c4STAT 1ON ; 2OFF Specifies whether or not to display a FREQ (frequency) column in the STAT Mode Stat Editor. c5Disp 1Dot ; 2Comma Specifies whether to display a dot or a comma for the calculation result decimal point. A dot is always displayed during input. Note: When dot is selected as the decimal point, the separator for multiple results is a comma (,). When comma is selected, the separator is a semicolon (;). c6]CONT' Adjusts display contrast. See “Adjusting Display Contrast” for details.
8.
E-7 Initializing Calculator Settings Performthe following procedure to initialize the calculator, which returns the calculation mode to COMP and returns all other settings, including setup menu settings, to their initial defaults. 19(CLR)1(Setup)=(Yes) Inputting Expressions and Values Basic Input Rules Calculations can be input in the same form as they are written. When you press = the priority sequence of the input calculation will be evaluated automatically and the result will appear on the display. 4 × sin30 × (30 + 10 × 3) = 120 4s30)(30+103)= 1 2 3 1 Input of the closing parenthesis is required for sin, sinh, and other functions that include parentheses. 2 These multiplication symbols (×) can be omitted. A multiplication symbol can be omitted when it occurs immediately before an opening parenthesis, immediately before sin or other function that includes parentheses, immediately before the Ran# (random number) function, or immediately before a variable (A, B, C, D, E, F, M, X, Y), scientific constants, π or e. 3 The closing parenthesis immediately before the = operation can be omitted. Input example omitting 2 and )3 operations in the above example. 4s30)(30+103= Note: • If the calculation becomes longer than the screen width during input, the screen will scroll automatically to the right and the ] indicator will appear on the display. When this happens, you can scroll back to the left by using d and e to move the cursor. • When Linear Display is selected, pressing f will cause the cursor to jump to the beginning of the calculation, while c will jump to the end. • When Natural Display is selected, pressing e while the cursor is at the end of the input calculation will cause it to jump to the beginning, while pressing d while the cursor is at the beginning will cause it to jump to the end. • You can input up to 99 bytes for a calculation. Each numeral, symbol, or function normally uses one byte. Some functions require three to 13 bytes. • The cursor will change shape to k when there are 10 bytes or less of allowed input remaining. If this happens, end calculation input and then press =. Calculation Priority Sequence The priority sequence of input calculations is evaluated in accordance with the rules below. When the priority of two expressions is the same, the calculation is performed from left to right. Math Math Math Math
9.
E-8 1st Parenthetical expressions 2ndFunctions that require an argument to the right and a closing parenthesis “)” following the argument. 3rd Functions that come after the input value (x2 , x3 , x–1 , x!, °’ ”, °, r , g , %, 't), powers (x^), roots () 4th Fractions 5th Negative sign (–), base-n symbols (d, h, b, o) Note: When squaring a negative value (such as –2), the value being squared must be enclosed in parentheses ((-2)w =). Since x2 has a higher priority than the negative sign, inputting -2w= would result in the squaring of 2 and then appending a negative sign to the result. Always keep the priority sequence in mind, and enclose negative values in parentheses when required. 6th Metric conversion commands (cm'in, etc.), STAT Mode estimated values (m, n, m1, m2) 7th Multiplication where the multiplication sign is omitted 8th Permutation (nPr), combination (nCr), complex number polar coordinate symbol (∠) 9th Dot product (·) 10th Multiplication, division (×, ÷) 11th Addition, subtraction (+, –) 12th Logical AND (and) 13th Logical OR, XOR, XNOR (or, xor, xnor) Inputting with Natural Display Selecting Natural Display makes it possible to input and display fractions and certain functions (log, x2 , x3 , x^, ), #, , x−1 , 10^, e^, ∫ , d/dx, Σ, Abs) just as they are written in your textbook. 2 + ' 2 1 + ' 2 B '2+!2ee1+!2= Important: • Certain types of expressions can cause the height of a calculation formula to be greater than one display line. The maximum allowable height of a calculation formula is two display screens (31 dots × 2). Further input will become impossible if the height of the calculation you are inputting exceeds the allowable limit. • Nesting of functions and parentheses is allowed. Further input will become impossible if you nest too many functions and/or parentheses. If this happens, divide the calculation into multiple parts and calculate each part separately. Note: When you press = and obtain a calculation result using Natural Display, part of the expression you input may be cut off. If you need to view the entire input expression again, press A and then use d and e to scroll the input expression. Math Math
10.
E-9 Using Values andExpressions as Arguments (Natural Display only) A value or an expression that you have already input can be used as the argument of a function. After you have input 7 6 , for example, you can make it the argument of ', resulting in 7 6 '. To input 1 + 7 6 and then change it to 1 + 7 6 ' B 1+7'6 dddd1Y(INS) ! As shown above, the value or expression to the right of the cursor after 1Y(INS) are pressed becomes the argument of the function that is specified next. The range encompassed as the argument is everything up to the first open parenthesis to the right, if there is one, or everything up to the first function to the right (sin(30), log2(4), etc.) This capability can be used with the following functions: ', , 7, 17(F), 1(8), 16(), 1l($), 1i(%), !, 6, 1!(#), 1w(Abs). Overwrite Input Mode (Linear Display only) You can select either insert or overwrite as the input mode, but only while Linear Display is selected. In the overwrite mode, text you input replaces the text at the current cursor location. You can toggle between the insert and overwrite modes by performing the operations: 1Y(INS). The cursor appears as “I” in the insert mode and as “ ” in the overwrite mode. Note: Natural Display always uses the insert mode, so changing display format from Linear Display to Natural Display will automatically switch to the insert mode. Correcting and Clearing an Expression To delete a single character or function: Move the cursor so it is directly to the right of the character or function you want to delete, and then press Y. In the overwrite mode, move the cursor so it is directly under the character or function you want to delete, and then press Y. To insert a character or function into a calculation: Use d and e to move the cursor to the location where you want to insert the character or function and then input it. Be sure always to use the insert mode if Linear Display is selected. To clear all of the calculation you are inputting: Press A. Toggling Calculation Results While Natural Display is selected, each press of f will toggle the currently displayed calculation result between its fraction form and decimal form, its ' form and decimal form, or its π form and decimal form. Math Math Math Math Math Math
11.
E-10 π ÷ 6= 1 6 π = 0.5235987756 B 15(π)/6= 1 6 π f 0.5235987756 (' 2 + 2) × ' 3 = ' 6 + 2' 3 = 5.913591358 B (!2e+2)!3= ' ' 6 + 2' 3 f 5.913591358 While Linear Display is selected, each press of f will toggle the currently displayed calculation result between its decimal form and fraction form. 1 ÷ 5 = 0.2 = 1 5 b 1/5= 0.2 f 1{5 1 – 4 5 = 1 5 = 0.2 b 1-4'5= 1{5 f 0.2 Important: • Depending on the type of calculation result that is on the display when you press the f key, the conversion process may take some time to perform. • With certain calculation results, pressing the f key will not convert the displayed value. • You cannot switch from decimal form to mixed fraction form if the total number of digits used in the mixed fraction (including integer, numerator, denominator, and separator symbols) is greater than 10. Note: With Natural Display (MathO), pressing 1= instead of = after inputting a calculation will display the calculation result in decimal form. Pressing f after that will switch to the fraction form or π form of the calculation result. The ' form of the result will not appear in this case. Basic Calculations Fraction Calculations Note that the input method for fractions is different, depending upon whether you are using Natural Display or Linear Display. 2 + 1 = 7 3 2 6 B 2 ' 3 e+ 1 ' 2 = 7 6 or ' 2c3e+'1c2 = 7 6 b 2 ' 3 + 1 ' 2 = 7{6 1 = 1 2 2 4 − 3 B 4-1'(()3e1c2= 1 2 b 4-3'1'2= 1{2 Note: • Mixing fractions and decimal values in a calculation while Linear Display is selected will cause the result to be displayed as a decimal value. • Fractions in calculation results are displayed after being reduced to their lowest terms. To switch a calculation result between improper fraction and mixed fraction form: Perform the following key operation: 1f()
12.
E-11 To switch acalculation result between fraction and decimal form: Press f. Percent Calculations Inputting a value and pressing 1((%) causes the input value to become a percent. 150 × 20% = 30 150201((%)= 30 Calculate what percentage of 880 is 660. (75%) 660/8801((%)= 75 Increase 2500 by 15%. (2875) 2500+2500151((%)= 2875 Discount 3500 by 25%. (2625) 3500-3500251((%)= 2625 Degree, Minute, Second (Sexagesimal) Calculations Performing an addition or subtraction operation between sexagesimal values, or a multiplication or division operation between a sexagesimal value and a decimal value will cause the result to be displayed as a sexagesimal value. You also can convert between sexagesimal and decimal. The following is the input format for a sexagesimal value: {degrees} $ {minutes} $ {seconds} $. Note: You must always input something for the degrees and minutes, even if they are zero. 2°20´30˝ + 39´30˝ = 3°00´00˝ 2$20$30$+0$39$30$= 3°0´0˝ Convert 2°15´18˝ to its decimal equivalent. 2$15$18$= 2°15´18˝ (Converts sexagesimal to decimal.) $ 2.255 (Converts decimal to sexagesimal.) $ 2°15´18˝ Multi-Statements You can use the colon character (:) to connect two or more expressions and execute them in sequence from left to right when you press =. 3 + 3 : 3 × 3 3+3S7(:)33= 6 = 9 Using Engineering Notation A simple key operation transforms a displayed value to engineering notation. Transform the value 1234 to engineering notation, shifting the decimal point to the right. 1234= 1234 W 1.234×103
13.
E-12 W 1234×100 Transform thevalue 123 to engineering notation, shifting the decimal point to the left. 123= 123 1W(←) 0.123×103 1W(←) 0.000123×106 Calculation History In the COMP, CMPLX, or BASE-N Mode, the calculator remembers up to approximately 200 bytes of data for the newest calculation. You can scroll through calculation history contents using f and c. 1 + 1 = 2 1+1= 2 2 + 2 = 4 2+2= 4 3 + 3 = 6 3+3= 6 (Scrolls back.) f 4 (Scrolls back again.) f 2 Note: Calculation history data is all cleared whenever you press O, when you change to a different calculation mode, when you change the display format, or whenever you perform any reset operation. Replay While a calculation result is on the display, you can press d or e to edit the expression you used for the previous calculation. 4 × 3 + 2.5 = 14.5 b 43+2.5= 14.5 4 × 3 − 7.1 = 4.9 (Continuing) dYYYY-7.1= 4.9 Note: If you want to edit a calculation when the ' indicator is on the right side of a calculation result display (see “Reading the Display”), press A and then use d and e to scroll the calculation. Answer Memory (Ans) The last calculation result obtained is stored in Ans (answer) memory. Ans memory contents are updated whenever a new calculation result is displayed. To divide the result of 3 × 4 by 30 b 34= (Continuing) /30= 123 + 456 = 579 B 123+456= 789 – 579 = 210 (Continuing) 789-G= Math Math
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E-13 Variables (A, B,C, D, E, F, X,Y) Your calculator has eight preset variables named A, B, C, D, E, F, X, and Y. You can assign values to variables and also use the variables in calculations. To assign the result of 3 + 5 to variable A 3+51t(STO)y(A) 8 To multiply the contents of variable A by 10 (Continuing) Sy(A)10= 80 To recall the contents of variable A (Continuing) ty(A) 8 To clear the contents of variable A 01t(STO)y(A) 0 Independent Memory (M) You can add calculation results to or subtract results from independent memory. The “M” appears on the display when there is any value other than zero stored in independent memory. To clear the contents of M 01t(STO)l(M) 0 To add the result of 10 × 5 to M (Continuing) 105l 50 To subtract the result of 10 + 5 from M (Continuing) 10+51l(M–) 15 To recall the contents of M (Continuing) tl(M) 35 Note: Variable M is used for independent memory. Clearing the Contents of All Memories Ans memory, independent memory, and variable contents are retained even if you press A, change the calculation mode, or turn off the calculator. Perform the following procedure when you want to clear the contents of all memories. !9(CLR)2(Memory)=(Yes) Function Calculations For actual operations using each function, see the “Examples” section following the list below. π π: π is displayed as 3.141592654, but π = 3.14159265358980 is used for internal calculations. e: e is displayed as 2.718281828, but e = 2.71828182845904 is used for internal calculations. sin, cos, tan, sin−1 , cos−1 , tan−1 : Trigonometric functions. Specify the angle unit before performing calculations. See 1. sinh, cosh, tanh, sinh−1 , cosh−1 , tanh−1 : Hyperbolic functions. Input a function from the menu that appears when you press w. The angle unit setting does not affect calculations. See 2.
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E-14 °, r , g :These functions specify the angle unit. ° specifies degrees, r radians, and g grads. Input a function from the menu that appears when you perform the following key operation: 1G(DRG'). See 3. $, %: Exponential functions. Note that the input method is different depending upon whether you are using Natural Display or Linear Display. See 4. log: Logarithmic function. Use the l key to input logab as log (a, b). Base 10 is the default setting if you do not input anything for a. The key also can be used for input, but only while Natural Display is selected. In this case, you must input a value for the base. See 5. ln: Natural logarithm to base e. See 6. x2 , x3 , x^, ), #, , x−1 : Powers, power roots, and reciprocals. Note that the input methods for x^, ), #, and are different depending upon whether you are using Natural Display or Linear Display. See 7. Note: • The following functions cannot be input in consecutive sequence: x2 , x3 , x^, x−1 . If you input 2ww, for example, the final w will be ignored. To input 2 22 , input 2w, press the d key, and then press w(B). • x2 , x3 , x−1 can be used in complex number calculations. : Function for performing numerical integration using the Gauss-Kronrod method. Natural Display input syntax is ∫a b f(x), while Linear Display input syntax is ∫( f(x), a, b, tol). tol specifies tolerance, which becomes 1 × 10–5 when nothing is input for tol. Also see “Integration and Differential Calculation Precautions” and “Tips for Successful Integration Calculations” for more information. See 8 . F: Function for approximation of the derivative based on the central difference method. Natural Display input syntax is dx d ( f(x))x=a , while Linear Display input syntax is dx d ( f(x), a, tol). tol specifies tolerance, which becomes 1 × 10–10 when nothing is input for tol. Also see “Integration and Differential Calculation Precautions” for more information. See 9. 8: Function that, for a specified range of f(x), determines sum Σ ( f(x)) x=a b = f(a) + f(a+1) + f(a+2) + ...+ f(b). Natural Display input syntax is Σ ( f(x)) x=a b , while Linear Display input syntax is Σ(f(x), a, b). a and b are integers that can be specified within the range of –1 × 1010 a b 1 × 1010 . See 10. Note: The following cannot be used in f(x), a, or b: Pol, Rec, ∫, d/dx, Σ. Pol, Rec: Pol converts rectangular coordinates to polar coordinates, while Rec converts polar coordinates to rectangular coordinates. See 11 . Pol(x, y) = (r, ) Rec(r, ) = (x, y) Rectangular Polar Coordinates (Rec) Coordinates (Pol) Specify the angle unit before performing calculations. The calculation result for r and and for x and y are each assigned respectively to variables X and Y. Calculation result θ is displayed in the range of −180° θ 180°.
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E-15 x!: Factorial function.See 12. Abs: Absolute value function. Note that the input method is different depending upon whether you are using Natural Display or Linear Display. See 13 . Ran#: Generates a 3-digit pseudo random number that is less than 1. The result is displayed as a fraction when Natural Display is selected. See 14 . RanInt#: For input of the function of the form RanInt#(a, b), which generates a random integer within the range of a to b. See 15 . nPr, nCr: Permutation (nPr) and combination (nCr) functions. See 16 . Rnd:Theargumentofthisfunctionismadeadecimalvalueandthenrounded in accordance with the current number of display digits setting (Norm, Fix, or Sci). With Norm 1 or Norm 2, the argument is rounded off to 10 digits. With Fix and Sci, the argument is rounded off to the specified digit. When Fix 3 is the display digits setting, for example, the result of 10 ÷ 3 is displayed as 3.333, while the calculator maintains a value of 3.33333333333333 (15 digits) internally for calculation. In the case of Rnd(10÷3) = 3.333 (with Fix 3), both the displayed value and the calculator’s internal value become 3.333. Because of this a series of calculations will produce different results depending on whether Rnd is used (Rnd(10÷3) × 3 = 9.999) or not used (10 ÷ 3 × 3 = 10.000). See 17 . Note: Using functions can slow down a calculation, which may delay display of the result. Do not perform any subsequent operation while waiting for the calculation result to appear. To interrupt an ongoing calculation before its result appears, press A. Integration and Differential Calculation Precautions • Integration and differential calculations can be performed in the COMP Mode (,1) only. • The following cannot be used in f(x), a, b, or tol: Pol, Rec, ∫, d/dx, Σ. • When using a trigonometric function in f(x), specify Rad as the angle unit. • A smaller tol value increases precision, but it also increases calculation time. When specifying tol, use value that is 1 × 10–14 or greater. Precautions for Integration Calculation Only • Integration normally requires considerable time to perform. • For f(x) 0 where a x b (as in the case of ∫0 1 3x2 – 2 = –1), calculation will produce a negative result. • Depending on the content of f(x) and the region of integration, calculation error that exceeds the tolerance may be generated, causing the calculator to display an error message. Precautions for Differential Calculation Only • If convergence to a solution cannot be found when tol input is omitted, the tol value will be adjusted automatically to determine the solution. • Non-consecutive points, abrupt fluctuation, extremely large or small points, inflection points, and the inclusion of points that cannot be differentiated, or a differential point or differential calculation result that approaches zero can cause poor precision or error.
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E-16 Tips for SuccessfulIntegration Calculations When a periodic function or integration interval results in positive and negative f(x) function values Perform separate integrations for each cycle, or for the positive part and the negative part, and then combine the results. When integration values fluctuate widely due to minute shifts in the integration interval Divide the integration interval into multiple parts (in a way that breaks areas of wide fluctuation into small parts), perform integration on each part, and then combine the results. Examples sin 30°= 0.5 bv s30)= 0.5 sin−1 0.5 = 30° bv 1s(sin−1 )0.5)= 30 sinh 1 = 1.175201194 wb(sinh)1)= 1.175201194 cosh–1 1 = 0 wf(cosh−1 )1)= 0 π/2 radians = 90°, 50 grads = 45° v (15(π)/2)1G(DRG')c(r )=90 501G(DRG')d(g )= 45 To calculate e5 × 2 to three significant digits (Sci 3) 1N(SETUP)7(Sci)3 B 1i(%)5e2= 2.97×102 b 1i(%)5)2= 2.97×102 log10 1000 = log 1000 = 3 l1000)= 3 log2 16 = 4 l21)(,)16)= 4 B 2e16= 4 To calculate ln 90 (= loge 90) to three significant digits (Sci 3) 1N(SETUP)7(Sci)3 i90)= 4.50×100 1.2 × 103 = 1200 B 1.21063= 1200 (1+1)2+2 = 16 B (1+1)62+2= 16 (52 )3 = 15625 (5x)1w(x3 )= 15625 32 5 = 2 B 16() 5e32= 2 b 516()32)= 2 S Positive S Negative S Positive S Negative ∫ ∫ ∫ a b f(x)dx = a c f(x)dx + (– c b f(x)dx) Positive Part (S Positive) Negative Part (S Negative) ∫ ∫ ∫ a b f(x)dx = a c f(x)dx + (– c b f(x)dx) Positive Part (S Positive) Negative Part (S Negative) b a x1 x2 x3 x4 x 0 f (x) b a x1 x2 x3 x4 x 0 f (x) a b f(x)dx = a x1 f(x)dx + x1 x2 f(x)dx + ..... ∫ ∫ ∫ x4 b f(x)dx ∫ + a b f(x)dx = a x1 f(x)dx + x1 x2 f(x)dx + ..... ∫ ∫ ∫ x4 b f(x)dx ∫ + 1 1 2 2 3 3 4 4 5 5 6 6 7 7
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E-17 To calculate ' 2× 3(= 3' 2 = 4.242640687...) to three decimal places (Fix 3) 1N(SETUP)6(Fix)3 B !2e3= 3' 2 1= 4.243 b !2)3= 4.243 ∫1 e ln(x) = 1 B 7iS)(X))e1eS5(e)= 1 b 7iS)(X))1)(,) 11)(,)S5(e))= 1 To obtain the derivative at point x = π/2 for the function y = sin(x) V B 17(F)sS)(X)) e'15(π)e2= 0 b 17(F)sS)(X)) 1)(,)15(π)'2)= 0 Σ x =1 5 (x + 1) = 20 B 1(8)S)(X)+1e1e5= 20 b 1(8)S)(X)+11)(,)1 1)(,)5)= 20 To convert rectangular coordinates (' 2 , ' 2 ) to polar coordinates v B 1+(Pol)!2e1)(,)!2e)= r=2,=45 b 1+(Pol)!2)1)(,)!2))= r= 2 = 45 To convert polar coordinates (' 2 , 45°) to rectangular coordinates v B 1-(Rec)!2e1)(,)45)= X=1,Y=1 (5 + 3) ! = 40320 (5+3)1E(x!)= 40320 |2 – 7| × 2 = 10 B 1w(Abs)2-7e2= 10 b 1w(Abs)2-7)2= 10 To obtain three random three-digit integers 10001.(Ran#)= 459 = 48 = 117 (Results shown here are for illustrative purposes only. Actual results will differ.) 8 8 9 9 10 10 11 11 12 12 13 13 14 14
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E-18 To generate randomintegers in the range of 1 to 6 S.(RanInt)11)(,)6)= 2 = 6 = 1 (Results shown here are for illustrative purposes only. Actual results will differ.) To determine the number of permutations and combinations possible when selecting four people from a group of 10 Permutations: 101(nPr)4= 5040 Combinations: 101/(nCr)4= 210 To perform the following calculations when Fix 3 is selected for the number of display digits: 10 ÷ 3 × 3 and Rnd(10 ÷ 3) × 3 b 1N(SETUP)6(Fix)3 10/33= 10.000 10(Rnd)10/3)3= 9.999 Complex Number Calculations (CMPLX) To perform complex number calculations, first press N2(CMPLX) to enter the CMPLX Mode. You can use either rectangular coordinates (a+bi) or polar coordinates (r∠) to input complex numbers. Complex number calculation results are displayed in accordance with the complex number format setting on the setup menu. (2 + 6i) ÷ (2i) = 3 – i (Complex number format: a + bi) (2+6W(i))/(2W(i))= 3–i 2 ∠ 45 = ' 2 + ' 2 i Bv (Complex number format: a + bi) 21-(∠)45= ' 2 +' 2 i ' 2 + ' 2 i = 2 ∠ 45 Bv (Complex number format: r∠) !2e+!2eW(i)= 2∠45 Note: • If you are planning to perform input and display of the calculation result in polar coordinate format, specify the angle unit before starting the calculation. • The value of the calculation result is displayed in the range of –180° 180°. • Display of the calculation result while Linear Display is selected will show a and bi (or r and ) on separate lines. CMPLX Mode Calculation Examples (1 – i)–1 = 1 2 1 2 + i B (Complex number format: a + bi) (1-W(i))E= 1 2 1 2 + i (1 + i)2 + (1 – i)2 = 0 B (1+W(i))w+(1-W(i))w= 0 15 15 16 16 17 17
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E-19 To obtain theconjugate complex number of 2 + 3i (Complex number format: a + bi) 12(CMPLX)2(Conjg)2+3W(i))= 2–3i To obtain the absolute value and argument of 1 + i Bv Absolute Value: 1w(Abs)1+W(i)= ' 2 Argument:12(CMPLX)1(arg)1+W(i))= 45 Using a Command to Specify the Calculation Result Format Either of two special commands ('r∠ or 'a+bi) can be input at the end of a calculation to specify the display format of the calculation results. The command overrides the calculator’s complex number format setting. ' 2 + ' 2 i = 2 ∠ 45, 2 ∠ 45 = ' 2 + ' 2 i Bv !2e+!2eW(i)12(CMPLX)3('r∠)= 2∠45 21-(∠)4512(CMPLX)4('a+bi)= ' 2 +' 2 i Using CALC CALC lets you save calculation expressions that contain variables, which you can then recall and execute in the COMP Mode (N1) and the CMPLX Mode (N2). The following describes the types of expressions you can save with CALC. • Expressions: 2X + 3Y, 2AX + 3BY + C, A + Bi • Multi-statements: X + Y : X (X + Y) • Equalities with a single variable on the left and an expression including variables on the right: A = B + C, Y = X2 + X + 3 (Use Ss(=) to input the equals sign of the equality.) To store 3A + B and then substitute the following values to perform the calculation: (A, B) = (5, 10), (7, 20) 3S-(A)+Se(B) s Prompts for input of a value for A Current value of A 5=10= s (or =) Math Math Math Math Math Math Math Math
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E-20 7=20= To exit CALC:A To store A + Bi and then determine ' 3 + i, 1 + ' 3 i using polar coordinates (r∠) v N2(CMPLX) S-(A)+Se(B)W(i) 12(CMPLX)3('r∠) s!3)=1= s (or =)1=!3)= To exit CALC: A Note: During the time from when you press s until you exit CALC by pressing A, you should use Linear Display input procedures for input. Using SOLVE SOLVE uses Newton’s Law to approximate the solution of equations. Note that SOLVE can be used in the COMP Mode (N1) only. The following describes the types of equations whose solutions can be obtained using SOLVE. • Equations that include variable X: X2 + 2X – 2,Y = X + 5, X = sin(M), X + 3 = B + C SOLVE solves for X. An expression like X2 + 2X – 2 is treated as X2 + 2X – 2 = 0. • Equations input using the following syntax: {equation}, {solution variable} SOLVE solves for Y, for example, when an equation is input as: Y = X + 5, Y Important: • If an equation contains input functions that include an open parenthesis (such as sin and log), do not omit the closing parenthesis. • The following functions are not allowed inside of an equation: ∫, d/dx, Σ, Pol, Rec. To solve y = ax2 + b for x when y = 0, a = 1, and b = –2 Sf(Y)Ss(=)S-(A) S)(X)w+Se(B) 1s(SOLVE) Prompts for input of a value for Y Current value of Y Math Math Math CMPLX Math CMPLX Math Math Math Math
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E-21 0=1=-2= Current value ofX Input an initial value for X (Here, input 1): 1= To exit SOLVE: A Solution screen Note: During the time from when you press 1s(SOLVE) until you exit SOLVE by pressing A, you should use Linear Display input procedures for input. Important: • Depending on what you input for the initial value for X (solution variable), SOLVE may not be able to obtain solutions. If this happens, try changing the initial value so they are closer to the solution. • SOLVE may not be able to determine the correct solution, even when one exists. • SOLVE uses Newton’s Law, so even if there are multiple solutions, only one of them will be returned. • Due to limitations in Newton’s Law, solutions tend to be difficult to obtain for equations like the following: y = sin(x), y = ex , y = ' x . Solution Screen Contents Solutions are always displayed in decimal form. Equation (The equation you input.) Variable solved for Solution (Left Side) – (Right Side) result “(Left Side) – (Right Side) result” shows the result when the right side of the equation is subtracted from the left side, after assigning the obtained value to the variable being solved for. The closer this result is to zero, the higher the accuracy of the solution. Continue Screen SOLVE performs convergence a preset number of times. If it cannot find a solution, it displays a confirmation screen that shows “Continue: [=]”, asking if you want to continue. Press = to continue or A to cancel the SOLVE operation. To solve y = x2 – x + 1 for x when y = 3, 7, and 13 Sf(Y)Ss(=) S)(X)w-S)(X)+1 1s(SOLVE) Math Math Math Math Math Math Math Math Math Math
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E-22 3= Input an initialvalue for X (Here, input 1): 1= =7== =13== Statistical Calculations (STAT) To start a statistical calculation, perform the key operation N3(STAT) to enter the STAT Mode and then use the screen that appears to select the type of calculation you want to perform. To select this type of statistical calculation: (Regression formula shown in parentheses) Press this key: Single-variable (X) 1(1-VAR) Paired-variable (X, Y), linear regression (y = A + Bx) 2(A+BX) Paired-variable (X, Y), quadratic regression (y = A + Bx + Cx2 ) 3(_+CX2 ) Paired-variable (X, Y), logarithmic regression (y = A + Blnx) 4(ln X) Paired-variable (X, Y), e exponential regression (y = AeBx ) 5(e^X) Paired-variable (X, Y), ab exponential regression (y = ABx ) 6(A•B^X) Paired-variable (X, Y), power regression (y = AxB ) 7(A•X^B) Paired-variable (X, Y), inverse regression (y = A + B/x) 8(1/X) Pressing any of the above keys (1 to 8) displays the Stat Editor. Note: When you want to change the calculation type after entering the STAT Mode, perform the key operation 11(STAT)1(Type) to display the calculation type selection screen. Inputting Data Use the Stat Editor to input data. Perform the following key operation to display the Stat Editor: 11(STAT)2(Data). The Stat Editor provides 80 rows for data input when there is an X column only, 40 rows when there are X and FREQ columns or X and Y columns, or 26 rows when there are X, Y, and FREQ columns. Math Math Math Math Math Math Math Math
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E-23 Note: Use theFREQ (frequency) column to input the quantity (frequency) of identical data items. Display of the FREQ column can be turned on (displayed) or off (not displayed) using the Stat Format setting on the setup menu. To select linear regression and input the following data: (170, 66), (173, 68), (179, 75) N3(STAT)2(A+BX) 170=173=179=ce 66=68=75= Important: • All data currently input in the Stat Editor is deleted whenever you exit the STAT Mode, switch between the single-variable and a paired- variable statistical calculation type, or change the Stat Format setting on the setup menu. • The following operations are not supported by the Stat Editor: m, 1m(M–), 1t(STO). Pol, Rec, and multi-statements also cannot be input with the Stat Editor. To change the data in a cell: In the Stat Editor, move the cursor to the cell that contains the data you want to change, input the new data, and then press =. To delete a line: In the Stat Editor, move the cursor to the line that you want to delete and then press Y. To insert a line: In the Stat Editor, move the cursor to the location where you want to insert the line and then perform the following key operation: 11(STAT)3(Edit)1(Ins). To delete all Stat Editor contents: In the Stat Editor, perform the following key operation: 11(STAT)3(Edit)2(Del-A). Obtaining Statistical Values from Input Data To obtain statistical values, press A while in the Stat Editor and then recall the statistical variable (σx, Σx2 , etc.) you want. Supported statistical variables and the keys you should press to recall them are shown below. For single-variable statistical calculations, the variables marked with an asterisk () are available. Sum: Σx2 , Σx, Σy2 , Σy, Σxy, Σx3 , Σx2 y, Σx4 11(STAT) 3(Sum) 1 to 8 Number of Items: n, Mean: o, p, Population Standard Deviation: σx, σy, Sample Standard Deviation: sx, sy 11(STAT) 4(Var) 1 to 7 Regression Coefficients: A, B, Correlation Coefficient: r, Estimated Values: m, n 11(STAT) 5(Reg) 1 to 5 Regression Coefficients for Quadratic Regression: A, B, C, Estimated Values: m1, m2, n 11(STAT) 5(Reg) 1 to 6 1 1 STAT STAT STAT STAT STAT STAT
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E-24 • See thetable at the beginning of this section of the manual for the regression formulas. • m, m1, m2 and n are not variables. They are commands of the type that take an argument immediately before them. See “Calculating Estimated Values” for more information. Minimum Value: minX, minY, Maximum Value: maxX, maxY 11(STAT) 6(MinMax) 1 to 4 Note: While single-variable statistical calculation is selected, you can input the functions and commands for performing normal distribution calculation from the menu that appears when you perform the following key operation: 11(STAT) 5 (Distr). See “Performing Normal Distribution Calculations” for details. To input the single-variable data x = {1, 2, 2, 3, 3, 3, 4, 4, 5}, using the FREQ column to specify the number of repeats for each items ({xn; freqn} = {1;1, 2;2, 3;3, 4;2, 5;1}), and calculate the mean and population standard deviation. 1N(SETUP)c4(STAT)1(ON) N3(STAT)1(1-VAR) 1=2=3=4=5=ce 1=2=3=2= A11(STAT)4(Var)2(o)= A11(STAT)4(Var)3(σx)= Results: Mean: 3 Population Standard Deviation: 1.154700538 To calculate the linear regression and logarithmic regression correlation coefficients for the following paired-variable data and determine the regression formula for the strongest correlation: (x, y) = (20, 3150), (110, 7310), (200, 8800), (290, 9310). Specify Fix 3 (three decimal places) for results. 1N(SETUP)c4(STAT)2(OFF) 1N(SETUP)6(Fix)3 N3(STAT)2(A+BX) 20=110=200=290=ce 3150=7310=8800=9310= A11(STAT)5(Reg)3(r)= A11(STAT)1(Type)4(In X) A11(STAT)5(Reg)3(r)= A11(STAT)5(Reg)1(A)= A11(STAT)5(Reg)2(B)= Results: Linear Regression Correlation Coefficient: 0.923 Logarithmic Regression Correlation Coefficient: 0.998 Logarithmic Regression Formula: y = –3857.984 + 2357.532lnx 2 2 STAT STAT 3 3 STAT FIX STAT FIX
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E-25 Calculating Estimated Values Basedon the regression formula obtained by paired-variable statistical calculation, the estimated value of y can be calculated for a given x-value. The corresponding x-value (two values, x1 and x2, in the case of quadratic regression) also can be calculated for a value of y in the regression formula. To determine the estimate value for y when x = 160 in the regression formula produced by logarithmic regression of the data in 3. Specify Fix 3 for the result. (Perform the following operation after completing the operations in 3.) A16011(STAT)5(Reg)5(n)= Result: 8106.898 Important: Regression coefficient, correlation coefficient, and estimated value calculations can take considerable time when there are a large number of data items. Performing Normal Distribution Calculations While single-variable statistical calculation is selected, you can perform normal distribution calculation using the functions shown below from the menu that appears when you perform the following key operation: 11(STAT)5(Distr). P, Q, R: These functions take the argument t and determine a probability of standard normal distribution as illustrated below. 't: This function is preceded by the argument X, and determines the normalized variate . For the single variable data {xn ; freqn} = {0;1, 1;2, 2;1, 3;2, 4;2, 5;2, 6;3, 7;4, 9;2, 10;1}, to determine the normalized variate ('t) when x = 3, and P(t) at that point up to three decimal places (Fix 3). 1N(SETUP)c4(STAT)1(ON) 1N(SETUP)6(Fix)3N3(STAT)1(1-VAR) 0=1=2=3=4=5=6=7=9= 10=ce1=2=1=2=2=2=3= 4=2=1= A311(STAT)5(Distr)4('t)= 4 4 P(t) Q(t) R(t) 0 t 0 t 0 t P(t) Q(t) R(t) 0 t 0 t 0 t 5 5 STAT FIX STAT FIX STAT FIX STAT FIX
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E-26 11(STAT)5(Distr)1(P()G)= Results: Normalized variate('t): –0.762 P(t): 0.223 Base-n Calculations (BASE-N) Press N4(BASE-N) to enter the BASE-N Mode when you want to perform calculations using decimal, hexadecimal, binary, and/or octal values. The initial default number mode when you enter the BASE-N Mode is decimal, which means input and calculation results use the decimal number format. Press one of the following keys to switch number modes: w(DEC) for decimal, 6(HEX) for hexadecimal, l(BIN) for binary, or i(OCT) for octal. To enter the BASE-N Mode, switch to the binary mode, and calculate 112 + 12 N4(BASE-N) l(BIN) 11+1= Continuing from above, switch to the hexadecimal mode and calculate 1F16 + 116 A6(HEX)1t(F)+1= Continuing from above, switch to the octal mode and calculate 78 + 18 Ai(OCT)7+1= Note: • Use the following keys to input the letters A through F for hexadecimal values: -(A), $(B), w(C), s(D), c(E), t(F). • In the BASE-N Mode, input of fractional (decimal) values and exponents is not supported. If a calculation result has a fractional part, it is cut off. • The input and output ranges is 16 bits for binary values, and 32 bits for other types of values. The following shows details about input and output ranges. Base-n Mode Input/Output Ranges Binary Positive: 0000000000000000 x 0111111111111111 Negative: 1000000000000000 x 1111111111111111 Octal Positive: 00000000000 x 17777777777 Negative: 20000000000 x 37777777777 Decimal –2147483648 x 2147483647 STAT FIX STAT FIX
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E-27 Hexadecimal Positive: 00000000 x 7FFFFFFF Negative: 80000000 x FFFFFFFF Specifying the Number Mode of a Particular Input Value You can input a special command immediately following a value to specify the number mode of that value. The special commands are: d (decimal), h (hexadecimal), b (binary), and o (octal). To calculate 1010 + 1016 + 102 + 108 and display the result as a decimal value Aw(DEC) 13(BASE)c1(d)10+ 13(BASE)c2(h)10+ 13(BASE)c3(b)10+ 13(BASE)c4(o)10= 36 Converting a Calculation Result to another Type of Value You can use any one of the following key operations to convert the currently displayed calculation result to another type of value: x(DEC) (decimal), 6(HEX) (hexadecimal), l(BIN) (binary), i(OCT)(octal). To calculate 1510 × 3710 in the decimal mode, and then convert the result to hexadecimal, binary, and octal Ax(DEC)1537= 555 6(HEX) 0000022B l(BIN) 0000001000101011 i(OCT) 00000001053 Logical and Negation Operations Your calculator provides you with logical operators (and, or, xor, xnor) and functions (Not, Neg) for logical and negation operations on binary values. Use the menu that appears when you press 13(BASE) to input these logical operators and functions. All of the following examples are performed in the binary mode (l(BIN)). To determine the logical AND of 10102 and 11002 (10102 and 11002) A101013(BASE)1(and)1100= 0000000000001000 To determine the logical OR of 10112 and 110102 (10112 or 110102) A101113(BASE)2(or)11010= 0000000000011011 To determine the logical XOR of 10102 and 11002 (10102 xor 11002) A101013(BASE)3(xor)1100= 0000000000000110 To determine the logical XNOR of 11112 and 1012 (11112 xnor 1012) A111113(BASE)4(xnor)101= 1111111111110101
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E-28 To determine thebitwise complement of 10102 (Not(10102)) A13(BASE)5(Not)1010)= 1111111111110101 To negate (take the two’s complement) of 1011012 (Neg(1011012)) A13(BASE)6(Neg)101101)= 1111111111010011 Note: In the case of a negative binary, octal or hexadecimal value, the calculator converts the value to binary, takes the two’s complement, and then converts back to the original number base. For decimal (base-10) values, the calculator merely adds a minus sign. Equation Calculations (EQN) You can use the following procedure in the EQN Mode to solve simultaneous linear equations with two or three unknowns, quadratic equations, and cubic equations. 1. Press N5(EQN) to enter the EQN Mode. 2. On the menu that appears, select an equation type. To select this calculation type: Press this key: Simultaneous linear equations with two unknowns 1(anX + bnY = cn) Simultaneous linear equations with three unknowns 2(anX + bnY + cnZ = dn) Quadratic equation 3(aX2 + bX + c = 0) Cubic equation 4(aX3 + bX2 + cX + d = 0) 3. Use the Coefficient Editor that appears to input coefficient values. • To solve 2x2 + x – 3 = 0, for example, press 3 in step 2, and then input the following for the coefficients (a = 2, b = 1, c = –3): 2=1=- 3=. • To change a coefficient value you already have input, move the cursor to the appropriate cell, input the new value, and then press =. • Pressing A will clear all of the coefficients to zero. Important: The following operations are not supported by the Coefficient Editor: m, 1m(M–), 1t(STO). Pol, Rec, and multi-statements also cannot be input with the Coefficient Editor. 4. After all the values are the way you want, press =. • This will display a solution. Each press of = will display another solution. Pressing = while the final solution is displayed will return to the Coefficient Editor. • You can scroll between the solutions using the c and f keys. • To return to the Coefficient Editor while any solution is displayed, press A. Note: • Even if Natural Display is selected, the solutions of simultaneous linear equations are not displayed using any form that includes '. • Values cannot be converted to engineering notation on the solution screen. Changing the Current Equation Type Setting Press N5(EQN) and then select an equation type from the menu that appears. Changing the equation type causes the values of all Coefficient Editor coefficients to change to zero.
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E-29 EQN Mode CalculationExamples x + 2y = 3, 2x + 3y = 4 N5(EQN)1(anX + bnY = cn) 1=2=3= 2=3=4= = (X=) –1 c (Y=) 2 x – y + z = 2, x + y – z = 0, –x + y + z = 4 N5(EQN)2(anX + bnY + cnZ = dn) 1=-1=1=2= 1=1=-1=0= -1=1=1= 4= = (X=) 1 c (Y=) 2 c (Z=) 3 x2 + x + 4 3 = 0 B N5(EQN)3(aX2 + bX + c = 0) 1=1=3'4== (X1=) 1 2 + – ' 2 2 i c (X2=) 1 2 – – ' 2 2 i x2 – 2' 2x + 2 = 0 B N5(EQN)3(aX2 + bX + c = 0) 1=-2!2)=2== (X=) ' 2 x3 – 2x2 – x + 2 = 0 N5(EQN)4(aX3 + bX2 + cX + d = 0) 1=-2=-1=2== (X1=) –1 c (X2=) 2 c (X3=) 1 Matrix Calculations (MATRIX) Use the MATRIX Mode to perform calculations involving matrices of up to 3 rows by 3 columns. To perform a matrix calculation, you first assign data to special matrix variables (MatA, MatB, MatC), and then use the variables in the calculation as shown in the example below. To assign 2 1 1 1 to MatA and 2 –1 –1 2 to MatB, and then perform the following calculations: × 2 1 1 1 2 –1 –1 2 (MatA×MatB), + 2 1 1 1 2 –1 –1 2 (MatA+MatB) Math Math Math Math 1 1
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E-30 1. Press N6(MATRIX)to enter the MATRIX Mode. 2. Press 1(MatA)5(2×2). • This will display the Matrix Editor for input of the elements of the 2 × 2 matrix you specified for MatA. “A” stands for “MatA”. 3. Input the elements of MatA: 2=1=1=1=. 4. Perform the following key operation: 14(MATRIX)2(Data) 2(MatB)5(2×2). • This will display the Matrix Editor for input of the elements of the 2 × 2 matrix you specified for MatB. 5. Input the elements of MatB: 2=-1=-1=2=. 6. Press A to advance to the calculation screen, and perform the first calculation (MatA×MatB): 14(MATRIX)3(MatA)14(MATRIX) 4(MatB)=. • This will display the MatAns screen with the calculation results. “Ans” stands for “MatAns”. Note: “MatAns” stands for “Matrix Answer Memory”. See “Matrix Answer Memory” for more information. 7. Perform the next calculation (MatA+MatB): A14(MATRIX) 3(MatA)+14(MATRIX)4(MatB)=. Matrix Answer Memory Whenever the result of a calculation executed in the MATRIX Mode is a matrix, the MatAns screen will appear with the result. The result also will be assigned to a variable named “MatAns”. The MatAns variable can be used in calculations as described below. • To insert the MatAns variable into a calculation, perform the following key operation: 14(MATRIX)6(MatAns). • Pressing any one of the following keys while the MatAns screen is displayed will switch automatically to the calculation screen: +, -, , /, E, w, 1w(x3 ). The calculation screen will show the MatAns variable followed by the operator or function for the key you pressed. Assigning and Editing Matrix Variable Data Important: The following operations are not supported by the Matrix Editor: m, 1m(M–), 1t(STO). Pol, Rec, and multi-statements also cannot be input with the Matrix Editor. To assign new data to a matrix variable: 1. Press 14(MATRIX)1(Dim), and then, on the menu that appears, select the matrix variable to which you want to assign data. 2. On the next menu that appears, select dimension (m×n). MAT MAT → MAT MAT → MAT MAT → MAT MAT → MAT MAT
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E-31 3. Use theMatrix Editor that appears to input the elements of the matrix. To assign 1 0 –1 0 –1 1 to MatC 14(MATRIX) 1(Dim)3(MatC)4(2×3) 1=0=-1=0=-1=1= To edit the elements of a matrix variable: 1. Press 14(MATRIX)2(Data), and then, on the menu that appears, select the matrix variable you want to edit. 2. Use the Matrix Editor that appears to edit the elements of the matrix. • Move the cursor to the cell that contains the element you want to change, input the new value, and then press =. To copy matrix variable (or MatAns) contents: 1. Use the Matrix Editor to display the matrix you want to copy. • If you want to copy MatA, for example, perform the following key operation: 14(MATRIX)2(Data)1(MatA). • If you want to copy MatAns contents, perform the following to display the MatAns screen: A14(MATRIX)6(MatAns)=. 2. Press 1t(STO), and then perform one of the following key operations to specify the copy destination: -(MatA), $(MatB), or w(MatC). • This will display the Matrix Editor with the contents of the copy destination. Matrix Calculation Examples The following examples use MatA = 2 1 1 1 and MatB = 2 –1 –1 2 from 1, and MatC = 1 0 –1 0 –1 1 from 2. You can input a matrix variable into a key operation by pressing 14(MATRIX) and then pressing one of the following number keys: 3(MatA), 4(MatB), 5(MatC). 3 × MatA (Matrix scalar multiplication). A3MatA= Obtain the determinant of MatA (det(MatA)). A14(MATRIX)7(det) MatA)= 1 Obtain the transposition of MatC (Trn(MatC)). A14(MATRIX)8(Trn) MatC)= Obtain the inverse matrix of MatA (MatA–1 ). Note: You cannot use 6 for this input. Use the E key to input “–1 ”. AMatAE= 2 2 MAT MAT 3 3 4 4 5 5 6 6
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E-32 Obtain the absolutevalue of each element of MatB (Abs(MatB)). A1w(Abs) MatB)= Determine the square and cube of MatA (MatA2 , MatA3 ). Note: You cannot use 6 for this input. Use w to specify squaring, and 1w(x3 ) to specify cubing. AMatAw= AMatA1w(x3 )= Creating a Number Table from a Function (TABLE) TABLE generates a number table for x and f(x) using an input f(x) function. Perform the following steps to generate a number table. 1. Press N7 (TABLE) to enter the TABLE Mode. 2. Input a function in the format f(x), using the X variable. • Be sure to input the X variable (S)(X)) when generating a number table. Any variable other than X is handled as a constant. • The following cannot be used in the function: Pol, Rec, ∫, d/dx, Σ. 3. In response to the prompts that appear, input the values you want to use, pressing = after each one. For this prompt: Input this: Start? Input the lower limit of X (Default = 1). End? Input the upper limit of X (Default = 5). Note: Make sure that the End value is always greater than the Start value. Step? Input the increment step (Default = 1). Note: The Step specifies by how much the Start value should be sequentially incremented as the number table is generated. If you specify Start = 1 and Step = 1, X sequentially will be assigned the values 1, 2, 3, 4, and so on to generate the number table until the End value is reached. • Inputting the Step value and pressing = generates and displays the number table in accordance with the parameters you specified. • Pressing A while the number table screen is displayed will return to the function input screen in step 2. To generate a number table for the function f(x) = x2 + 2 1 for the range –1 x 1, incremented in steps of 0.5 B N7(TABLE) 7 7 8 8 Math Math
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E-33 S)(X)x+1'2 =-1=1=0.5= Note: • Youcan use the number table screen for viewing values only. Table contents cannot be edited. • The number table generation operation causes the contents of variable X to be changed. Important: The function you input for number table generation is deleted whenever you display the setup menu in the TABLE Mode and switch between Natural Display and Linear Display. Vector Calculations (VECTOR) Use the VECTOR Mode to perform 2-dimensional and 3-dimensional vector calculations. To perform a vector calculation, you first assign data to special vector variables (VctA, VctB, VctC), and then use the variables in the calculation as shown in the example below. To assign (1, 2) to VctA and (3, 4) to VctB, and then perform the following calculation: (1, 2) + (3, 4) 1. Press N8(VECTOR) to enter the VECTOR Mode. 2. Press 1(VctA)2(2). • This will display the Vector Editor for input of the 2-dimensional vector for VctA. “A” stands for “VctA”. 3. Input the elements of VctA: 1=2=. 4. Perform the following key operation: 15(VECTOR)2(Data) 2(VctB)2(2). • This will display the Vector Editor for input of the 2-dimensional vector for VctB. 5. Input the elements of VctB: 3=4=. 6. Press A to advance to the calculation screen, and perform the calculation (VctA + VctB): 1 5 (VECTOR)3 (VctA)+ 1 5 (VECTOR) 4(VctB)=. • This will display the VctAns screen with the calculation results. “Ans” stands for “VctAns”. Note: “VctAns” stands for “Vector Answer Memory”. See “Vector Answer Memory” for more information. Math Math Math Math 1 1 VCT VCT → VCT VCT → VCT VCT
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E-34 Vector Answer Memory Wheneverthe result of a calculation executed in the VECTOR Mode is a vector, the VctAns screen will appear with the result. The result also will be assigned to a variable named “VctAns”. The VctAns variable can be used in calculations as described below. • To insert the VctAns variable into a calculation, perform the following key operation: 15(VECTOR)6(VctAns). • Pressing any one of the following keys while the VctAns screen is displayed will switch automatically to the calculation screen: +, -, , /. The calculation screen will show the VctAns variable followed by the operator for the key you pressed. Assigning and Editing Vector Variable Data Important: The following operations are not supported by the Vector Editor: m, 1m(M–), 1t(STO). Pol, Rec, and multi-statements also cannot be input with the Vector Editor. To assign new data to a vector variable: 1. Press 15(VECTOR)1(Dim), and then, on the menu that appears, select the vector variable to which you want to assign data. 2. On the next menu that appears, select dimension (m). 3. Use the Vector Editor that appears to input the elements of the vector. To assign (2, –1, 2) to VctC 15(VECTOR)1(Dim)3(VctC)1(3) 2=-1=2= To edit the elements of a vector variable: 1. Press 15(VECTOR)2(Data), and then, on the menu that appears, select the vector variable you want to edit. 2. Use the Vector Editor that appears to edit the elements of the vector. • Move the cursor to the cell that contains the element you want to change, input the new value, and then press =. To copy vector variable (or VctAns) contents: 1. Use the Vector Editor to display the vector you want to copy. • If you want to copy VctA, for example, perform the following key operation: 15(VECTOR)2(Data)1(VctA). • If you want to copy VctAns contents, perform the following to display the VctAns screen: A15(VECTOR)6(VctAns)=. 2. Press 1t(STO), and then perform one of the following key operations to specify the copy destination: -(VctA), $(VctB), or w(VctC). • This will display the Vector Editor with the contents of the copy destination. Vector Calculation Examples The following examples use VctA = (1, 2) and VctB = (3, 4) from 1, and VctC = (2, –1, 2) from 2. You can input a vector variable into a key operation by pressing 15(VECTOR) and then pressing one of the following number keys: 3(VctA), 4(VctB), 5(VctC). 2 2 VCT VCT
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E-35 3 × VctA(Vector scalar multiplication), 3 × VctA – VctB (Calculation example using VctAns) A3VctA= -VctB= VctA • VctB (Vector dot product) AVctA15(VECTOR)7(Dot)VctB= VctA × VctB (Vector cross product) AVctAVctB= Obtain the absolute values of VctC. A1w(Abs)VctC)= Determine the angle formed by VctA and VctB to three decimal places (Fix 3). v (cos = (A•B) AB , which becomes = cos–1 (A•B) AB ) 1N(SETUP)6(Fix)3 A(VctA15(VECTOR)7(Dot)VctB)/ (1w(Abs)VctA)1w(Abs) VctB))= 1c(cos–1 )G)= Scientific Constants Your calculator comes with 40 built-in scientific constants that can be used in any mode besides BASE-N. Each scientific constant is displayed as a unique symbol (such as π), which can be used inside of calculations. To input a scientific constant into a calculation, press 17(CONST) and then input the two-digit number that corresponds to the constant you want. 3 3 VCT VCT VCT VCT 4 4 VCT VCT 5 5 VCT VCT 6 6 VCT VCT 7 7 VCT FIX VCT FIX VCT FIX VCT FIX
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E-36 To input thescientific constant C0 (speed of light in a vacuum), and display its value A17(CONST) 28(C0)= To calculate C0 = 1 ε0μ0 B A'1c!17(CONST)32(ε0) 17(CONST)33(0)= The following shows the two-digit numbers for each of the scientific constants. 01: (mp) proton mass 02: (mn) neutron mass 03: (me) electron mass 04: (m) muon mass 05: (a0) Bohr radius 06: (h) Planck constant 07: (N) nuclear magneton 08: (B) Bohr magneton 09: (h) Planck constant, rationalized 10: (α) fine-structure constant 11: (re) classical electron radius 12: (λc) Compton wavelength 13: (γp) proton gyromagnetic ratio 14: (λcp) proton Compton wavelength 15: (λcn) neutron Compton wavelength 16: (R∞) Rydberg constant 17: (u) atomic mass constant 18: (p) proton magnetic moment 19: (e) electron magnetic moment 20: (n) neutron magnetic moment 21: () muon magnetic moment 22: (F) Faraday constant 23: (e) elementary charge 24: (NA) Avogadro constant 25: (k) Boltzmann constant 26: (Vm) molar volume of ideal gas 27: (R) molar gas constant 28: (C0) speed of light in vacuum 29: (C1) first radiation constant 30: (C2) second radiation constant 31: (σ) Stefan-Boltzmann constant 32: (ε0) electric constant Math Math Math Math
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E-37 33: (0) magneticconstant 34: (φ0) magnetic flux quantum 35: (g) standard acceleration of gravity 36: (G0) conductance quantum 37: (Z0) characteristic impedance of vacuum 38: (t) Celsius temperature 39: (G) Newtonian constant of gravitation 40: (atm) standard atmosphere The values are based on CODATA recommended values (March 2007). Metric Conversion The calculator’s built-in metric conversion commands make it simple to convert values from one unit to another. You can use the metric conversion commands in any calculation mode except for BASE-N and TABLE. To input a metric conversion command into a calculation, press 18(CONV) and then input the two-digit number that corresponds to the command you want. To convert 5 cm into inches b A518(CONV) 02(cm'in)= To convert 100 g into ounces b A10018(CONV)22(g'oz)= To convert –31°C into Fahrenheit b A-3118(CONV)38(°C'°F)= The following shows the two-digit numbers for each of the metric conversion commands. 01: in'cm 02: cm'in 03: ft'm 04: m'ft 05: yd'm 06: m'yd 07: mile'km 08: km'mile 09: n mile'm 10: m'n mile 11: acre'm2 12: m2 'acre 13: gal (US)'R 14: R'gal (US) 15: gal (UK)'R 16: R'gal (UK) 17: pc'km 18: km'pc 19: km/h'm/s 20: m/s'km/h 21: oz'g 22: g'oz 23: lb'kg 24: kg'lb
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E-38 25: atm'Pa 26:Pa'atm 27: mmHg'Pa 28: Pa'mmHg 29: hp'kW 30: kW'hp 31: kgf/cm2 'Pa 32: Pa'kgf/cm2 33: kgf • m'J 34: J'kgf • m 35: lbf/in2 'kPa 36: kPa'lbf/in2 37: °F'°C 38: °C'°F 39: J'cal 40: cal'J Conversion formula data is based on the “NIST Special Publication 811 (1995)”. Note: The J'cal command performs conversion for values at a temperature of 15°C. Calculation Ranges, Number of Digits, and Precision The calculation range, number of digits used for internal calculation, and calculation precision depend on the type of calculation you are performing. Calculation Range and Precision Calculation Range ±1 × 10–99 to ±9.999999999 × 1099 or 0 Number of Digits for Internal Calculation 15 digits Precision In general, ±1 at the 10th digit for a single calculation. Precision for exponential display is ±1 at the least significant digit. Errors are cumulative in the case of consecutive calculations. Function Calculation Input Ranges and Precision Functions Input Range sinx DEG 0 |x| 9 × 109 RAD 0 |x| 157079632.7 GRA 0 |x| 1 × 1010 cosx DEG 0 |x| 9 × 109 RAD 0 |x| 157079632.7 GRA 0 |x| 1 × 1010 tanx DEG Same as sinx, except when |x| = (2n–1) × 90. RAD Same as sinx, except when |x| = (2n–1) × π/2. GRA Same as sinx, except when |x| = (2n–1) × 100. sin–1 x 0 |x| 1 cos–1 x tan–1 x 0 |x| 9.999999999 × 1099 sinhx 0 |x| 230.2585092 coshx sinh–1 x 0 |x| 4.999999999 × 1099 cosh–1 x 1 x 4.999999999 × 1099 tanhx 0 |x| 9.999999999 × 1099 tanh–1 x 0 |x| 9.999999999 × 10–1
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E-39 logx/lnx 0 x 9.999999999 × 1099 10x –9.999999999 × 1099 x 99.99999999 ex –9.999999999 × 1099 x 230.2585092 ' x 0 x 1 × 10100 x2 |x| 1 × 1050 x–1 |x| 1 × 10100 ; x G 0 3 ' x |x| 1 × 10100 x! 0 x 69 (x is an integer) nPr 0 n 1 × 1010 , 0 r n (n, r are integers) 1 {n!/(n–r)!} 1 × 10100 nCr 0 n 1 × 1010 , 0 r n (n, r are integers) 1 n!/r! 1 × 10100 or 1 n!/(n–r)! 1 × 10100 Pol(x, y) |x|, |y| 9.999999999 × 1099 x2 +y2 9.999999999 × 1099 Rec(r, ) 0 r 9.999999999 × 1099 : Same as sinx °’ ” |a|, b, c 1 × 10100 ; 0 b, c The display seconds value is subject to an error of 1 at the second decimal place. |x| 1 × 10100 Decimal ↔ Sexagesimal Conversions 0°0´0˝ |x| 9999999°59´59˝ xy x 0: –1 × 10100 ylogx 100 x = 0: y 0 x 0: y = n, m 2n+1 (m, n are integers) However: –1 × 10100 ylog |x| 100 x ' y y 0: x G 0, –1 × 10100 1/x logy 100 y = 0: x 0 y 0: x = 2n+1, 2n+1 m (m G 0; m, n are integers) However: –1 × 10100 1/x log |y| 100 ab /c Total of integer, numerator, and denominator must be 10 digits or less (including division marks). RanInt#(a, b) a b; |a|, |b| 1 × 1010 ; b – a 1 × 1010 • Precision is basically the same as that described under “Calculation Range and Precision”, above. • xy , x ' y, 3 ', x!, nPr, nCr type functions require consecutive internal calculation, which can cause accumulation of errors that occur with each calculation. • Error is cumulative and tends to be large in the vicinity of a function’s singular point and inflection point. • The range for calculation results that can be displayed in π form when using Natural Display is |x| 106 . Note, however, that internal calculation error can make it impossible to display some calculation results in π form. It also can cause calculation results that should be in decimal form to appear in π form.
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E-40 Errors The calculator willdisplay an error message whenever an error occurs for any reason during a calculation. There are two ways to exit an error message display: Pressing d or e to display the location of the error, or pressing A to clear the message and calculation. Displaying the Location of an Error While an error message is displayed, press d or e to return to the calculation screen. The cursor will be positioned at the location where the error occurred, ready for input. Make the necessary corrections to the calculation and execute it again. When you input 14 ÷ 0 × 2 = by mistake instead of 14 ÷ 10 × 2 = B 14/02= e (or d) d1= Clearing the Error Message While an error message is displayed, press A to return to the calculation screen. Note that this also clears the calculation that contained the error. Error Messages Math ERROR Cause: • The intermediate or final result of the calculation you are performing exceeds the allowable calculation range. • Your input exceeds the allowable input range (particularly when using functions). • The calculation you are performing contains an illegal mathematical operation (such as division by zero). Action: • Check the input values, reduce the number of digits, and try again. • When using independent memory or a variable as the argument of a function, make sure that the memory or variable value is within the allowable range for the function. Stack ERROR Cause: • The calculation you are performing has caused the capacity of the numeric stack or the command stack to be exceeded. • The calculation you are performing has caused the capacity of the matrix or vector stack to be exceeded. Action: • Simplify the calculation expression so it does not exceed the capacity of the stack. • Try splitting the calculation into two or more parts. Syntax ERROR Cause: There is a problem with the format of the calculation you are performing. Math Math Math Math Math Math
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E-41 Action: Make necessarycorrections. Argument ERROR Cause: There is a problem with the argument of the calculation you are performing. Action: Make necessary corrections. Dimension ERROR (MATRIX and VECTOR Modes only) Cause: • The matrix or vector you are trying to use in a calculation was input without specifying its dimension. • You are trying to perform a calculation with matrices or vectors whose dimensions do not allow that type of calculation. Action: • Specify the dimension of the matrix or vector and then perform the calculation again. • Check the dimensions specified for the matrices or vectors to see if they are compatible with the calculation. Variable ERROR (SOLVE feature only) Cause: • You did not specify a solution variable, and there is no X variable in the equation you input. • The solution variable that you specified is not included in the equation you input. Action: • The equation you input must include an X variable when you do not specify the solution variable. • Specify a variable that is included in the equation you input as the solution variable. Can’t Solve Error (SOLVE feature only) Cause: The calculator could not obtain a solution. Action: • Check for errors in the equation that you input. • Input a value for the solution variable that is close to the expected solution and try again. Insufficient MEM Error Cause: The configuration of TABLE Mode parameters caused more than 30 X-values to be generated for a table. Action: Narrow the table calculation range by changing the Start, End, and Step values, and try again. Time Out Error Cause: The current differential or integration calculation ends without the ending condition being fulfilled. Action: Try increasing the tol value. Note that this also decreases solution precision. Before Assuming Malfunction of the Calculator... Perform the following steps whenever an error occurs during a calculation or when calculation results are not what you expected. If one step does not correct the problem, move on to the next step. Note that you should make separate copies of important data before performing these steps. 1. Check the calculation expression to make sure that it does not contain any errors. 2. Make sure that you are using the correct mode for the type of calculation you are trying to perform. 3. If the above steps do not correct your problem, press the O key. This will cause the calculator to perform a routine that checks whether calculation functions are operating correctly. If the calculator discovers any abnormality, it automatically initializes the calculation mode and clears memory contents.
43.
E-42 For details aboutinitialized settings, see “Configuring the Calculator Setup”. 4. Initialize all modes and settings by performing the following operation: 19(CLR)1(Setup)=(Yes). Replacing the Battery A low battery is indicated by a dim display, even if contrast is adjusted, or by failure of figures to appear on the display immediately after you turn on the calculator. If this happens, replace the battery with a new one. Important: Removing the battery will cause all of the calculator’s memory contents to be deleted. 1. Press 1A(OFF) to turn off the calculator. • To ensure that you do not accidentally turn on power while replacing the battery, slide the hard case onto the front of the calculator (fx-991ES PLUS). 2. Remove the cover as shown in the illustration and replace the battery, taking care that its plus (+) and minus (–) ends are facing correctly. Screw Screw Screw fx-570ES PLUS fx-991ES PLUS 3. Replace the cover. 4. Initialize the calculator: O19(CLR)3(All)=(Yes) • Do not skip the above step! Specifications Power Requirements: fx-570ES PLUS: AAA-size battery R03 (UM-4) × 1 fx-991ES PLUS: Built-in solar cell; button battery LR44 (GPA76) × 1 Approximate Battery Life: fx-570ES PLUS: 17,000 hours (continuous display of flashing cursor) fx-991ES PLUS: 3 years (based on one hour of operation per day) Power Consumption: 0.0002 W (fx-570ES PLUS) Operating Temperature: 0°C to 40°C (32°F to 104°F) Dimensions: fx-570ES PLUS: 13.8 (H) × 80 (W) × 162 (D) mm 1 /2 (H) × 31 /8 (W) × 63 /8 (D) fx-991ES PLUS: 11.1 (H) × 80 (W) × 162 (D) mm 3 /8 (H) × 31 /8 (W) × 63 /8 (D)
44.
E-43 Approximate Weight: fx-570ES PLUS:100 g (3.5 oz) including the battery fx-991ES PLUS: 95 g (3.4 oz) including the battery Frequently Asked Questions k How can I perform input and display results the same way I did on a model that does not have Natural Textbook Display? Perform the following key operation: 1N(SETUP)2(LineIO). See “Configuring the Calculator Setup” on page E-5 for more information. k How can I change a fraction form result to decimal form? How can I change a fraction form result produced by a division operation to decimal form? See “Toggling Calculation Results” on page E-9 for the procedure. k What is the difference between Ans memory, independent memory, and variable memory? Each of these types of memory acts like “containers” for temporary storage of a single value. Ans Memory: Stores the result of the last calculation performed. Use this memory to carry the result of one calculation on to the next. Independent Memory: Use this memory to totalize the results of multiple calculations. Variables: This memory is helpful when you need to uses the same value multiple times in one or more calculations. k What is the key operation to take me from the STAT Mode or TABLE Mode to a mode where I can perform arithmetic calculations? Press N1(COMP). k How can I return the calculator to its initial default settings? Perform the following operation: 19(CLR)1(Setup)=(Yes) k When I execute a function calculation,why do I get a calculation result that is completely different from older CASIO calculator models? With a Natural Textbook Display model, the argument of a function that uses parentheses must be followed by a closing parenthesis. Failing to press )after the argument to close the parentheses may cause unwanted values or expressions to be included as part of the argument. Example: (sin 30) + 15 v Older (S-VPAM) Model: s30+15= 15.5 Natural Textbook Display Model: b s30)+15= 15.5 Failure to press ) here as shown below will result in calculation of sin 45. s30+15= 0.7071067812
45.
Manufacturer: CASIO COMPUTER CO.,LTD. 6-2, Hon-machi 1-chome Shibuya-ku, Tokyo 151-8543, Japan Responsible within the European Union: CASIO EUROPE GmbH Casio-Platz 1 22848 Norderstedt, Germany This mark applies in EU countries only.
46.
CASIO COMPUTER CO.,LTD. 6-2, Hon-machi 1-chome Shibuya-ku, Tokyo 151-8543, Japan SA0907-B |
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What is the Taylor Series?
The Taylor series is a mathematical representation of a function as an infinite sum of terms. It allows us to approximate a function using its derivatives at a particular point. The series is named after the mathematician Brook Taylor, who introduced it in the 18th century.
The Taylor series expansion of a function f(x) around a point a is given by:
f(x) = f(a) + f'(a)(x-a)/1! + f''(a)(x-a)²/2! + f'''(a)(x-a)³/3! + ...
Each term in the series corresponds to a derivative of the function evaluated at the point a, multiplied by powers of the difference between x and a, divided by the factorial of the order of the derivative. The terms capture the local behavior of the function at the point a. The series is often truncated after a certain number of terms to create an approximation of the function. The more terms included, the closer the approximation is to the original function.
The Taylor series expansion is valuable in calculus and mathematical analysis. It provides a way to represent functions that may be difficult to work with directly. It allows for the estimation of function values beyond the range of known values and aids in the understanding of the behavior of functions near a particular point. In quantitative finance, the Taylor series is used in various ways to approximate and analyze financial functions and models, such as:
Option Pricing Models: The Taylor series expansion is employed to approximate option pricing models, such as the Black-Scholes model. By expanding the model's equations using the Taylor series, it is possible to derive simpler approximations or closed-form solutions for pricing options. These approximations can help in quickly estimating option prices and Greeks (sensitivity measures) without relying on complex numerical methods.
Numerical Methods:The Taylor series is utilized in numerical methods to approximate financial derivatives, such as option sensitivities (e.g., delta, gamma, vega). By approximating the derivative using the Taylor series expansion, numerical techniques like finite difference methods can be employed to calculate sensitivities accurately and efficiently.
Risk Management Models:The Taylor series is incorporated into risk management models, such as risk factor models or stress testing frameworks. By expanding the models using the Taylor series, the impact of changes in risk factors on portfolio risk can be analyzed. This enables the assessment of potential losses under different scenarios or shocks.
It's important to note that the Taylor series approximations are most effective for small deviations from the expansion point and may introduce errors as the deviation increases. Careful consideration and validation are necessary when employing Taylor series approximations in quantitative finance to ensure their accuracy and reliability.
The Taylor series is covered in more detail in module 1 of the CQF program.
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A Comprehensive Guide to Correlation Matrix, Missing Data, and Multicollinearity.
“Understanding Correlation Matrix, Handling Missing Data, and Addressing Multicollinearity in Data Analysis”
Taran Kaur
9 min readFeb 16, 2025
In statistical analysis, a correlation matrix is an essential tool used to understand the relationships between multiple variables. It is widely used in data exploration, regression analysis, and predictive modeling. In this guide, we will cover how to read and interpret a correlation matrix, identify and handle missing data, and understand the issues surrounding multicollinearity.
Understanding the Correlation Matrix
A correlation matrix is a table that displays the correlation coefficients between several variables. The correlation coefficient ranges from -1 to +1:
+1 indicates a perfect positive relationship: as one variable increases, the other increases proportionally.
-1 indicates a perfect negative relationship: as one variable increases, the other decreases in perfect proportion.
0 indicates no linear relationship between the variables.
Example of a Correlation Matrix:
The diagonal elements (e.g., 1.00 for X1, X2, and X3) represent the correlation of each variable with itself.
The off-diagonal elements show the correlation between pairs of variables (e.g., X1 and X2 have a correlation of 0.85, indicating a strong positive relationship).
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How to Read a Correlation Matrix
To interpret the correlation matrix effectively:
Magnitude of the Correlation Coefficient:
A strong correlation (above ±0.7) indicates that the variables move in the same or opposite direction.
A moderate correlation (between ±0.4 and ±0.7) suggests a weaker relationship.
A weak or no correlation (between ±0.2 and ±0.4) means the variables are only loosely related or unrelated.
Sign of the Correlation:
Positive Correlation: As one variable increases, the other tends to increase.
Negative Correlation: As one variable increases, the other tends to decrease.
Symmetry of the Matrix:
The matrix is symmetric: the correlation between X1 and X2 is the same as the correlation between X2 and X1. This redundancy can help confirm the calculations.
Why Is the Correlation Matrix So Important?
Understanding Relationships Between Variables:
A correlation matrix provides a quick, visual way to spot relationships between variables. For example, in financial analysis, a correlation matrix of stock returns can help identify which stocks move in the same direction (positive correlation) or in opposite directions (negative correlation).
In customer behavior analysis, correlations between various purchasing behaviors, like frequency of visits and total spending, can help identify key drivers of business.
Guiding Feature Selection in Machine Learning:
Before building machine learning models, identifying highly correlated features through the correlation matrix can inform the feature selection process. Variables that are highly correlated might add redundant information, which can be addressed to improve model performance.
For instance, in predictive models, removing variables with high correlations can improve the accuracy and stability of the model by reducing overfitting and avoiding multicollinearity.
Detecting Multicollinearity:
One of the most crucial tasks when dealing with regression models is checking for multicollinearity. The correlation matrix allows us to see if two or more independent variables are highly correlated with one another, which can interfere with the model’s ability to estimate the relationship between each independent variable and the dependent variable.
Multicollinearity leads to inflated standard errors and unstable coefficients in regression models. Thus, using the correlation matrix as a diagnostic tool is essential to ensure model validity.
Building and Refining Hypotheses:
By showing how variables relate to each other, the correlation matrix can provide insights for hypothesis testing. For example, researchers might hypothesize that a variable X has a positive effect on Y. The correlation matrix can either confirm or challenge this hypothesis, guiding further analysis.
Identifying and Handling Missing Data
Missing data can introduce significant challenges in data analysis. The way missing data is handled depends on the nature of the missingness, which can be classified into three main types:
MCAR (Missing Completely at Random)
MAR (Missing at Random)
MNAR (Missing Not at Random)
Each type has different implications for how missing data should be handled and how it affects the correlation matrix.
MCAR (Missing Completely at Random)
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MCAR occurs when the probability of missing data is unrelated to any other variable, whether observed or unobserved. The missingness is purely random.
Impact on Correlation Matrix: Since the missingness is not related to any variable, the correlation between variables will remain unbiased. The relationships among variables in the matrix are unaffected by missing data.
Detection: MCAR is difficult to detect directly from the correlation matrix, as the missingness does not create any patterns that influence the correlation coefficients. A statistical test such as Little’s MCAR test can formally check if data is MCAR.
Numerical Impact (Examples):
If you have a dataset where the correlation between X1 and X2 is 0.85, and the missing values are MCAR, after handling the missing data appropriately (like imputation or deletion), the correlation will likely still be around 0.85.
Similarly, if the correlation between X2 and X3 is -0.40, the removal or imputation of missing data would not drastically affect this correlation.
Thus, in an MCAR scenario, you can confidently rely on the correlation matrix as a true reflection of the relationships between variables.
Actions:
Imputation: You can use simple imputation techniques, such as replacing missing values with the mean or median, without introducing significant bias into the correlation matrix.
Deletion: Listwise deletion (removing rows with missing values) won’t distort the correlations, as the missingness does not depend on any variable.
2. MAR (Missing at Random)
MAR means that the missingness is related to observed data but not to the missing values themselves. For instance, a certain group of people may be more likely to have missing data based on their income or education level.
Impact on Correlation Matrix: MAR can introduce bias if the missing data is related to an observed variable that is part of the correlation matrix. For example, if income data is missing more frequently for lower-income individuals, this could distort the relationship between income and education.
Distorted Relationships: Variables whose missingness is associated with other observed variables may lead to overestimated or underestimated correlations. For example, if two variables X1 and X2 are related to missing data patterns, imputing or ignoring the missing data may result in incorrect conclusions about the relationship between X1 and X2.
Detection: To identify MAR, look for patterns in missing data that are related to other observed variables. This can be observed by inspecting the correlations or using statistical tests to check for relationships between missingness and observed data.
Numerical Impact (Examples):
Suppose the correlation between X1 (income) and X2 (education level) is 0.65. However, if the missing data for X2 is systematically related to X1 (e.g., missing education data for individuals with lower income), this could lead to an overestimation of the correlation between X1 and X2 after imputation.
On the other hand, if the missing data is related to a third variable, say X3 (age), the relationship between X1 and X2 might become less pronounced, showing a lower correlation.
Thus, MAR creates bias in the correlation matrix, making it less trustworthy unless the missingness pattern is modeled appropriately.
Actions:
Imputation: Use more sophisticated imputation techniques, such as regression imputation or multiple imputation, that account for the relationships between the observed variables.
Sensitivity Analysis: Perform sensitivity analysis to understand how different methods of dealing with MAR might impact the correlations.
Handling MAR: For MAR, sophisticated imputation methods are recommended. Techniques like multiple imputation or regression imputation, where missing values are predicted based on other observed data, can help reduce bias and preserve the integrity of the correlation matrix.
3. MNAR (Missing Not at Random)
MNAR occurs when the probability of missingness is related to the unobserved value itself. For instance, if higher-income individuals are less likely to report their income, this introduces a systematic bias based on the missing values.
Impact on Correlation Matrix: MNAR is the most problematic type of missingness as it distorts correlations between variables. The missing data directly influences the relationship between variables, making the correlation matrix biased and unreliable.
Significant Distortion of Relationships: Because the missing data is directly related to the unobserved values, the correlations between variables can be heavily biased. This leads to distorted correlation coefficients that do not reflect the true relationships between variables.
Underestimation or Overestimation of Correlations: In the case of MNAR, correlations can either be underestimated or overestimated depending on the nature of the missingness. For example, if high-income individuals tend to not report their income (MNAR), this will distort the correlation between income and education, as the correlation may appear weaker than it actually is, due to the missing data.
Detection: Detecting MNAR is challenging. One potential approach is to investigate the relationship between missingness and the missing data itself, but this requires domain knowledge or external information about the data collection process.
Numerical Impact (Examples):
Imagine that the correlation between X1 (income) and X2 (education) is 0.75 in the complete dataset. However, if individuals with extremely high or low incomes are less likely to report their income, the correlation between X1 and X2 might appear artificially low. This is because missing data systematically ignores high-income individuals, thus distorting the relationship.
In a dataset with missing values in X1 and X2 that are dependent on X1 itself, correlations may also appear distorted in the other direction, potentially showing an exaggerated correlation between two variables when high values are missed for one but not the other.
Thus, MNAR can cause significant bias, as it involves missingness that directly correlates with the unobserved data, distorting the integrity of the correlation matrix.
Actions:
Model Missingness: To handle MNAR data, it’s crucial to incorporate the missingness mechanism into the analysis itself using models like pattern mixture models or selection models that attempt to model the missingness process.
Robust Imputation: MNAR data often requires complex imputation techniques that factor in the specific missingness patterns, which may involve external knowledge or domain-specific models.
In summary, the impact of missing data on the correlation matrix is substantial, and understanding the type of missingness — MCAR, MAR, or MNAR — is crucial for accurate interpretation of the correlations. MCAR typically has little to no effect on the correlation matrix, while MAR and MNAR can introduce significant bias that skews the relationships between variables. To deal with missing data effectively, it is essential to apply appropriate techniques for each type of missingness, such as sophisticated imputation methods or models that account for missingness patterns.
Multicollinearity and Its Impact on the Correlation Matrix
Multicollinearity refers to the situation where two or more independent variables in a regression model are highly correlated with each other. This can cause issues in regression analysis by making it difficult to estimate the effect of each variable on the dependent variable.
Signs of Multicollinearity in the Correlation Matrix: Multicollinearity is often indicated by very high correlations (above 0.9) between two or more variables. If, for example, you observe a correlation of 0.95 between two variables, this suggests that multicollinearity may be present.
Impact on the Correlation Matrix: High correlations in the correlation matrix suggest that the variables may provide redundant information. This makes it difficult to understand the individual effect of each variable, as changes in one variable are likely associated with changes in another.
Detecting Multicollinearity:
High Pairwise Correlations: Look for pairs of variables that have very high correlations, typically above 0.9. If these variables are included in a regression model, multicollinearity may cause instability in the estimated coefficients.
Variance Inflation Factor (VIF): While not directly available from the correlation matrix, calculating VIF for each variable can help identify multicollinearity. A VIF above 10 suggests high multicollinearity.
Handling Multicollinearity:
Remove Redundant Variables: If two variables are highly correlated, consider removing one from the analysis to reduce redundancy.
Principal Component Analysis (PCA): PCA can help by combining correlated variables into a smaller set of uncorrelated components.
Regularization: Techniques like Ridge and Lasso regression apply penalties to reduce the influence of highly correlated variables, helping to mitigate multicollinearity.
Conclusion: The Correlation Matrix as a Diagnostic Tool
The correlation matrix is a powerful diagnostic tool for understanding relationships between variables, detecting multicollinearity, and guiding feature selection in statistical and machine learning models. However, it is crucial to interpret the matrix carefully, especially in the presence of missing data. Depending on the nature of the missingness (MCAR, MAR, or MNAR), the correlations between variables can be biased, distorting the relationships and leading to misleading conclusions.
By understanding the strengths and limitations of the correlation matrix, data analysts can use it more effectively to explore their datasets, detect potential issues, and make more informed decisions about the next steps in the analysis process. When used with awareness of its nuances, the correlation matrix is an indispensable tool for accurate and insightful data analysis.
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IgA nephropathy
Last updated: June 27, 2024
Summary
IgA nephropathy (IgAN), also known as Berger disease, is the most common primary glomerulonephritis worldwide. It most frequently affects male individuals in the second to third decades of life. Clinical manifestations are usually triggered by upper respiratory tract or gastrointestinal infections and include gross hematuria and flank pain; however, most individuals with IgAN are asymptomatic. Rarely, IgAN may manifest as rapidly progressive glomerulonephritis (RPGN). Urinalysis usually shows persistent microhematuria and proteinuria, while more severe IgAN may manifest with recurrent episodes of nephritic syndrome. Kidney biopsy is required for definitive diagnosis and is usually indicated in patients with severe or progressive kidney disease. Treatment consists of measures to slow the progression of the disease (e.g., ACE inhibitors); immunosuppressive therapy is reserved for more severe IgAN. Even with treatment, up to 40% of patients progress to end-stage renal disease (ESRD) within 20 years.
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Epidemiology
IgA nephropathy is the most common primary glomerulonephritis in adults.
Peak incidence: second to third decades of life
Sex: ♂ > ♀ (2:1)
Ethnicity: more common in the Asian population (worldwide)
Epidemiological data refers to the US, unless otherwise specified.
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Pathophysiology
The cause is still not entirely understood.
Most likely mechanism: an increased number of defective, circulating IgA antibodies are synthesized (often triggered by mucosal infections, i.e., upper respiratory tract and gastrointestinal infections) → IgA antibodies form immune complexes that deposit in the renal mesangium → mesangial cell and complement system activation → glomerulonephritis (type III hypersensitivity reaction)
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Clinical features
The course and presentation of IgAN are highly variable.
Asymptomatic with urinalysis abnormalities, e.g., microhematuria, proteinuria (most common presentation)
Symptomatic episodes, usually during or immediately following a respiratory or gastrointestinal infection or following strenuous exercise
Gross or microscopic hematuria
Flank pain
Low-grade fever
Nephritic syndrome (including hypertension)
RPGN and/or nephrotic syndrome (∼ 5% of individuals with IgAN)
ESRD
Manifestations of IgAN usually begin < 5 days after the onset of URTI or GI infection symptoms.
IgAN and IgA vasculitis are both IgA-mediated vasculitides triggered by a mucosal infection. IgA vasculitis most commonly occurs in children < 10 years of age and affects multiple organs (palpable purpura, abdominal pain, arthralgia). IgAN is limited to the kidneys and typically affects adults.
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Diagnosis
General principles
Suspect IgAN in patients with:
Suggestive clinical features, e.g., gross hematuria during or following a URTI or gastrointestinal infection
Incidental hematuria and proteinuria
Declining kidney function (especially in patients < 40 years of age)
Consult nephrology for all patients with suspected IgAN.
A renal biopsy is needed for a definitive diagnosis of IgAN.
Additional laboratory studies (e.g., serum IgA, C3 complement level) can help exclude differential diagnoses and secondary causes.
IgAN is the most common cause of kidney failure in individuals < 40 years of age.
Urine studies
Urinalysis: nephritic sediment
Hematuria: episodic gross or microscopic
Proteinuria: usually accompanies hematuria
Red blood cell casts may be present.
24-hour urine profile: proteinuria
Ranges from mild (< 1 g/24 h) to severe (> 3 g/24 h)
More severe proteinuria is associated with an increased likelihood of IgAN progression.
Renal biopsy
Indications
The decision to perform a renal biopsy is often based on regional and institutional practices.
Usually warranted if there are signs of severe or progressive kidney disease, e.g.:
↑ Serum creatinine
Proteinuria > 0.5–1 g/24 h
Secondary hypertension
Findings
Light microscopy: mesangial proliferation and matrix expansion
Immunofluorescent microscopy: mesangial IgA deposits with C3 and occasionally IgM and/or IgG
Electron microscopy: mesangial immune complex deposits
Histological classification using the revised Oxford classification (MEST-C score) helps determine prognosis.
If findings would not impact management decisions, a biopsy may not be indicated in patients with nonsevere kidney disease.
The renal manifestation of IgA vasculitis is histologically indistinguishable from IgAN.
Additional laboratory studies
The following studies may support the diagnosis of IgAN but are not required in the diagnostic workup.
BMP: Creatinine may be elevated.
↑ IgA in serum: in 50% of patients; not specific to IgAN
C3 level: typically normal
Consider further studies to assess for secondary causes of IgAN (e.g., IgAN associated with chronic liver disease, IgA vasculitis) if clinically suspected.
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Differential diagnoses
Poststreptococcal glomerulonephritis
Associated with low complement levels
Clinical features take longer to appear (typically 10–20 days) following an infection
IgA vasculitis nephritis
Lupus nephritis
Membranoproliferative glomerulonephritis
The differential diagnoses listed here are not exhaustive.
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Treatment
General principles
Consult nephrology for all patients.
Management consists mainly of supportive measures, i.e.:
Management of proteinuria and hypertension
ASCVD risk reduction
Lifestyle modifications
Use of immunosuppressive therapy for severe or progressive kidney disease is controversial.
Long-term monitoring of blood pressure, urine protein, and GFR is required for all patients.
Even with treatment, 20–40% of patients with IgAN develop ESRD within 20 years.
Supportive measures
Indicated for all patients to slow disease progression.
Management of hypertension and proteinuria > 0.5 g/24 h
First line: ACE inhibitors or angiotensin II receptor blockers at maximally tolerated doses
SGLT-2 inhibitors may be added to RAASi.
Dual endothelin angiotensin receptor antagonists (e.g., sparsentan) may be considered if other agents are not effective.
See “Antihypertensive treatment by comorbidities.”
ASCVD risk reduction
Management of hyperlipidemia
Management of ASCVD risk factors in chronic kidney disease
Lifestyle modifications
Obesity management
Tobacco cessation
Exercise
Dietary sodium restriction
Immunosuppressive therapy
A nephrology consultation is required before starting immunosuppressive therapy.
Glucocorticoids may be considered for severe or progressive disease (i.e., proteinuria > 1 g/24 h despite supportive measures for ≥ 3 months).
A short course (≤ 6 months) is recommended.
Implement preventive measures for adverse effects of glucocorticoids (e.g., infections, osteoporosis).
Consider PCP prophylaxis.
Cyclophosphamide: for rapidly progressive IgAN, in addition to glucocorticoids
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Prognosis
Symptoms resolve spontaneously in 5–30% of patients.
20–40% of patients with IgAN develop ESRD within 20 years.
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10123 | https://courses.lumenlearning.com/ntcc-collegealgebracorequisite/chapter/methods-for-plotting-linear-equations/ | Writing Equations of Lines | College Algebra Corequisite
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College Algebra Corequisite
Module 3: The Rectangular Coordinate System and Equations of Lines
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Writing Equations of Lines
Learning Outcomes
Use slope-intercept form to plot and write equations of lines.
Use point-slope form to write the equation of a line.
Write the equation of a line in standard form.
Recognize vertical and horizontal lines from their graphs and equations.
Slope-Intercept Form
Perhaps the most familiar form of a linear equation is slope-intercept form written as y=m x+b y=m x+b, where m=slope m=slope and b=y-intercept b=y-intercept. Let us begin with the slope.
The Slope of a Line
The slope of a line refers to the ratio of the vertical change in y over the horizontal change in x between any two points on a line. It indicates the direction in which a line slants as well as its steepness. Slope is sometimes described as rise over run.
m=y 2−y 1 x 2−x 1 m=y 2−y 1 x 2−x 1
If the slope is positive, the line slants upward to the right. If the slope is negative, the line slants downward to the right. As the slope increases, the line becomes steeper. Some examples are shown below. The lines indicate the following slopes: m=−3 m=−3, m=2 m=2, and m=1 3 m=1 3.
A General Note: The Slope of a Line
The slope of a line, m, represents the change in y over the change in x. Given two points, (x 1,y 1)(x 1,y 1) and (x 2,y 2)(x 2,y 2), the following formula determines the slope of a line containing these points:
m=y 2−y 1 x 2−x 1 m=y 2−y 1 x 2−x 1
Example: Finding the Slope of a Line Given Two Points
Find the slope of a line that passes through the points (2,−1)(2,−1) and (−5,3)(−5,3).
Show Solution
We substitute the y- values and the x- values into the formula.
m=3−(−1)−5−2=4−7=−4 7 m=3−(−1)−5−2=4−7=−4 7
The slope is −4 7−4 7.
Analysis of the Solution
It does not matter which point is called (x 1,y 1)(x 1,y 1) or (x 2,y 2)(x 2,y 2). As long as we are consistent with the order of the y terms and the order of the x terms in the numerator and denominator, the calculation will yield the same result.
Try It
Find the slope of the line that passes through the points (−2,6)(−2,6) and (1,4)(1,4).
Show Solution
slope=m=−2 3=−2 3=m=−2 3=−2 3
Example: Identifying the Slope and y- intercept of a Line Given an Equation
Identify the slope and y- intercept given the equation y=−3 4 x−4 y=−3 4 x−4.
Show Solution
As the line is in y=m x+b y=m x+b form, the given line has a slope of m=−3 4 m=−3 4. The y- intercept is b=−4 b=−4.
Analysis of the Solution
The y-intercept is the point at which the line crosses the y- axis. On the y- axis, x=0 x=0. We can always identify the y- intercept when the line is in slope-intercept form, as it will always equal b. Or, just substitute x=0 x=0 and solve for y.
Recall operations on fractions
Fractions are commonly encountered when writing equations for lines using the point-slope formula. Being able to multiply or add and subtract fractions will enable you to avoid decimal approximations for numbers like 1 3 1 3 or 5 7 5 7.
Adding or subtracting fractions:a b±c d=a d±b c b d a b±c d=a d±b c b d
Multiplying fractions:a b⋅c d=a c b d a b⋅c d=a c b d
Don’t forget to simplify fractions in your final answer!
The Point-Slope Formula
Given the slope and one point on a line, we can find the equation of the line using point-slope form.
y−y 1=m(x−x 1)y−y 1=m(x−x 1)
This is an important formula, as it will be used in other areas of College Algebra and often in Calculus to find the equation of a tangent line. We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form.
A General Note: The Point-Slope Formula
Given one point and the slope, using point-slope form will lead to the equation of a line:
y−y 1=m(x−x 1)y−y 1=m(x−x 1)
Example: Finding the Equation of a Line Given the Slope and One Point
Write the equation of the line with slope m=−3 m=−3 and passing through the point (4,8)(4,8). Write the final equation in slope-intercept form.
Show Solution
Using point-slope form, substitute −3−3 for m and the point (4,8)(4,8) for (x 1,y 1)(x 1,y 1).
y−y 1=m(x−x 1)y−8=−3(x−4)y−8=−3 x+12 y=−3 x+20 y−y 1=m(x−x 1)y−8=−3(x−4)y−8=−3 x+12 y=−3 x+20
Analysis of the Solution
Note that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained.
Try It
Given m=4 m=4, find the equation of the line in slope-intercept form passing through the point (2,5)(2,5).
Show Solution
y=4 x−3 y=4 x−3
Example: Finding the Equation of a Line Passing Through Two Given Points
Find the equation of the line passing through the points (3,4)(3,4) and (0,−3)(0,−3). Write the final equation in slope-intercept form.
Show Solution
First, we calculate the slope using the slope formula and two points.
m=−3−4 0−3=−7−3=7 3 m=−3−4 0−3=−7−3=7 3
Next, we use point-slope form with the slope of 7 3 7 3 and either point. Let’s pick the point (3,4)(3,4) for (x 1,y 1)(x 1,y 1).
y−4=7 3(x−3)y−4=7 3 x−7 y=7 3 x−3 y−4=7 3(x−3)y−4=7 3 x−7 y=7 3 x−3
In slope-intercept form, the equation is written as y=7 3 x−3 y=7 3 x−3.
Analysis of the Solution
To prove that either point can be used, let us use the second point (0,−3)(0,−3) and see if we get the same equation.
y−(−3)=7 3(x−0)y+3=7 3 x y=7 3 x−3 y−(−3)=7 3(x−0)y+3=7 3 x y=7 3 x−3
We see that the same line will be obtained using either point. This makes sense because we used both points to calculate the slope.
Standard Form of a Line
Another way that we can represent the equation of a line is in standard form. Standard form is given as
A x+B y=C A x+B y=C
where A A, B B, and C C are integers. The x and y- terms are on one side of the equal sign and the constant term is on the other side.
Example: Finding the Equation of a Line and Writing It in Standard Form
Find the equation of the line with m=−6 m=−6 and passing through the point (1 4,−2)(1 4,−2). Write the equation in standard form.
Show Solution
We begin by using point-slope form.
y−(−2)=−6(x−1 4)y+2=−6 x+3 2 y−(−2)=−6(x−1 4)y+2=−6 x+3 2
From here, we multiply through by 2 as no fractions are permitted in standard form. Then we move both variables to the left aside of the equal sign and move the constants to the right.
2(y+2)=(−6 x+3 2)2 2 y+4=−12 x+3 12 x+2 y=−1 2(y+2)=(−6 x+3 2)2 2 y+4=−12 x+3 12 x+2 y=−1
This equation is now written in standard form.
Try It
Find the equation of the line in standard form with slope m=−1 3 m=−1 3 which passes through the point (1,1 3)(1,1 3).
Show Solution
x+3 y=2 x+3 y=2
Vertical and Horizontal Lines
Most of the lines we have worked with so far have been slanted, or oblique. In other words, they were neither horizontal nor vertical lines. The equations of vertical and horizontal lines do not require any of the preceding formulas, although we can use the formulas to prove that the equations are correct. The equation of a vertical line is given as
x=c x=c
where c is a constant. The slope of a vertical line is undefined, and regardless of the y- value of any point on the line, the x- coordinate of the point will be c.
Suppose that we want to find the equation of a line containing the following points: (−3,−5),(−3,1),(−3,3)(−3,−5),(−3,1),(−3,3), and (−3,5)(−3,5). First, we will find the slope.
m=5−3−3−(−3)=2 0 m=5−3−3−(−3)=2 0
Zero in the denominator means that the slope is undefined and, therefore, we cannot use point-slope form. However, we can plot the points. Notice that all of the x- coordinates are the same and we find a vertical line through x=−3 x=−3.
The equation of a horizontal line is given as
y=c y=c
where c is a constant. The slope of a horizontal line is zero, and for any x- value of a point on the line, the y- coordinate will be c.
Suppose we want to find the equation of a line that contains the following set of points: (−2,−2),(0,−2),(3,−2)(−2,−2),(0,−2),(3,−2), and (5,−2)(5,−2). We can use point-slope form. First, we find the slope using any two points on the line.
m=−2−(−2)0−(−2)=0 2=0 m=−2−(−2)0−(−2)=0 2=0
Use any point for (x 1,y 1)(x 1,y 1) in the formula, or use the y-intercept.
y−(−2)=0(x−3)y+2=0 y=−2 y−(−2)=0(x−3)y+2=0 y=−2
The graph is a horizontal line through y=−2 y=−2. Notice that all of the y- coordinates are the same.
The line x = −3 is a vertical line. The line y = −2 is a horizontal line.
try it
Use an online graphing calculator to graph the following:
A horizontal line that passes through the point (-5,2)
A vertical line that passes through the point (3,3)
Example: Finding the Equation of a Line Passing Through the Given Points
Find the equation of the line passing through the given points: (1,−3)(1,−3) and (1,4)(1,4).
Show Solution
The x- coordinate of both points is 1. Therefore, we have a vertical line, x=1 x=1.
Try It
Find the equation of the line passing through (−5,2)(−5,2) and (2,2)(2,2).
Show Solution
Horizontal line: y=2 y=2
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10124 | https://en.wikibooks.org/wiki/Mathematics_for_Economics/Hyperbolas | Mathematics for Economics/Hyperbolas - Wikibooks, open books for an open world
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Beginning
1 Introduction
2 Revised percentages
3 Metric geometry
4 References
Mathematics for Economics/Hyperbolas
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<Mathematics for Economics
Hyperbolas in Economics
Rectangular hyperbolas and their radii mark out an arena of economic data
Introduction
[edit | edit source]
The approach to mathematical economics in this chapter is geometric. Direct and inverse proportion correspond to constants that determine rays and hyperbola branches in the first quadrant. Entry into this environment of mathematics is illustrated with economic topics of wages, partitioning, stocks and market dynamics, mentioned superficially. Then students are expected to come to grips with hyperbolic geometry.
It was W. S. Jevons that clarified the study of mean values in a ratio scale in 1863. He wrote, "To take the geometrical mean of two ratios we must multiply them together and extract the square root of the product." He used historical rises of prices of 39 commodities to estimate the depreciation of the gold standard in the context of French trade and New World mines.
w:Johann von Thunen had brought hyperbolas into economics in 1826 when he used geometric mean of two values to determine a third. The extreme minimum wage, allowing the worker a bare survival, was much less than the value produced when equipped with the employer’s tools, evidence of capitalization. Von Thunen solved the problem of deciding on the wage paid, by multiplying the two values, then extracting the square root. This algebraic procedure has been called taking the geometric mean of the two original values, and von Thunen called it the natural wage. When a natural wage is set, the plot of two original values giving that result is a hyperbola.
The equi-partition function p = S/n for n receivers, where S is the total supply S=∑1 n p.{\displaystyle S=\sum {1}^{n}p.} When supply _S is rationed equally, then the equi-partition function provides the rationed quantity p. With n a natural number, as in social partitioning, the points (1, S), (2, S/2), ... (k, S/k) lie in the Cartesian plane. The points are connected with the smooth curve y = p(n) = S/n when n takes real values between the integers.
A company has x thousand shares held publicly, and today the shares are selling at k$. In terms of market capitalization, the product $ x k 000 represents the company value in the market for shares. As k varies from day to day, so company value is in flux. Sometimes management finds k intolerably low and company funds are used to repurchase some shares, thus reducing x. The coordinates (x, k) where market capitalization is constant form a hyperbola.
Supply and demand are spoken of as inversely related. Price of an item is a commercial factor, with price and demand related as follows: high price suppresses demand and cheap price augments demand. Supply is known by the vendor, and vendor sets the price, which in case of scarcity can be very high.
The points (x, y) such that x>0 and y>0 is called the first quadrant Q of the Cartesian plane. Curves similar to a hyperbola branch in Q can be seen The Theory of Politcal Economy by Jevons.
Economic behavior observed in Q can be informed of differential geometry associated with it: that Q is a two-dimensional manifold with (0,0) as the only boundary point. In fact, Q can be read as a copy of Lobachevski's hyperbolic plane when the radial and hyperbolic angle coordinates are introduced.
The radial coordinate determining a hyperbola is wage level in the first example, the supply S in the second example, and company value in the third. In retail, total sales receipts forms the radial coordinate. The hyperbolic angle is an area of the corresponding hyperbolic sector to the unit radius. Whereas the conventional circular angle is less than 360 degrees or 2 π radians, the hyperbolic angle is unbounded. Thus, a zero point is required to anchor its measurement. According to tradition, the point (1,1) on the hyperbola xy= 1 is the zero point. The hyperbolic angle then determined by (x, 1/x) is log x, the natural logarithm with base e = 2.718...
For hyperbolic metric geometry, the hyperbolic angle determines a horizontal axis, with vertical given by the square root of the radial coordinate in Q. These new coordinates in the upper half-plane H are given a metric, a positive distance function satisfying the triangle inequality. Tracking distances via H for movement in Q, over days, weeks, months or years, provides consistent measurement for long and short term perspective.
One of the measures of maturity in an individual or community is willingness to delay rewards until work is done. The psychological subject is delayed gratification. The study in economics is called discounting, a factor of consumer behavior where payoff in the future is tested against a lesser payoff earlier. An encouraging result of economic studies shows people prefer hyperbolic discounting over exponential discounting, which is encouraging since reasonable people use this pattern of deferred gratification, a sign of the proverbial economic man, something sometimes doubted in other contexts.
Revised percentages
[edit | edit source]
For hyperbola xy=1, when b/a=e, then the angle is 1.0 wings
The time value of money is poignantly shown by inflation of prices for the necessities of life. As business costs rise, prices of goods and services reflect the monetary tide. More objectively, economists may express the change as a de facto currency devaluation. Wing units of hyperbolic angle follow the units of natural logarithm, so that the angle between (1,1) and (e,1/e) is one unit. One percent change occurs between 1 and 1.01, and log e (1.01) = .0099503 wings or 9.9503 milliwings. The first ten percentage points have these angular values:
1%: 9.9503 milliwings 6%: 58.269 milliwings
2%: 19.8026 " 7%: 67.659 "
3%: 29.558 " 8%: 76.691 "
4%: 39.2207 " 9%: 86.178 "
5%: 48.790 " 10%: 95.310 "
Increments from one percentile to the next decrease; the first percent has the largest angle. In contrast, for mechanical velocities measured in nanowings, each marginal increase is virtually the same as the previous step. (Compare Kinematics/Transformations) Naturally, the milliwings used here for economics are a million times the magnitude of the nanowings of mechanics.
The laws of economy apply on macro and micro scales as far a population goes. Thus the quadrant has an infinite density near the origin, and the level curves xy = constant express a reference but are not paths of least distance in Q, as developed below. The hyperbolic angle parametrization of Q complements the level curves and provides a portal to the higher geometry of Q: the rays from (0,0) represent constant slopes as well as constant hyperbolic angles. The half plane (u,v) with positive v, takes the geometric mean, square root of xy, to determine v. The hyperbolic angle of a point in Q corresponds to u in the half plane: u = log √(x/y).
The half plane is a standard model of hyperbolic geometry, often taken together with the disk model and complex numbers, where a linear fractional transformation makes the connection. The geodesics, or shortest-distance curves in the half plane are semicircles with center on the u-axis, or half lines { (x 0, y): y > 0 }, x 0 fixed. A point (u,v) in the half plane corresponds to a point in Q given by x = v exp(u) and y = v exp(−u). When the geodesics of the half plane are plotted in Q, an economic scenario of growth and decline is suggested. For instance, the production of a particular product increases as it finds a market, rises to its summit of favorability, then falls off as superseded by its market. Similarly, a corporation starts small, grows with success, as can be traced with historic points in Q. But note that the rays in Q converging at the origin correspond to vertical lines in the half plane that each have a different foot on the u-axis. This feature of the half plane's expression of phenomena in Q allows compatibility of macro and micro market analysis.
Metric geometry
[edit | edit source]
Hyperbolic line segments in the half-plane, with the absolute in red.
In a tradition begun by Arthur Cayley, the boundary of the half-plane is called the absolute: {(u,v):v>0,u∈R}.{\displaystyle {(u,\ v):v>0,\ \ u\in \mathbb {R} }.} Semicircles used for geodesics have center on the absolute.
To take meaningful measurements in Q, reference to the half-plane brings a metric that is respected by motions. These are hyperbolic motions corresponding to whichever model of the hyperbolic plane is at hand, in this case the half-plane (and by bijective correspondence, a metric in Q).
The first motion is magnification or contraction about (0,0): (u,v) to (su, sv), where s>0. The second motion is translation along the absolute: (u, v) to (u + t, v). These motions make any semicircle with center on the absolute equivalent to the unit semicircle u u + v v = 1. Indeed, the center can be moved to (0,0) by translation, and the radius normalized by magnification or contraction.
A vertical interval from (u, a) to (u, b), where b>a, is taken to the interval from (su, sa) to (su, sb). The distance formula, or metric, for this interval is log b – log a = log (b/a), which is invariant under the motion. The second type of motion moves the half-plane to the left or right: (u, v) to (u + t, v). Looking a Q, such a motion corresponds to a hyperbolic rotation, a type of linear transformation that preserves difference of hyperbolic angles in Q.
The third hyperbolic motion is an involution and involves inversion in the unit semicircle in the half-plane. Under this inversion the ray (1, tan θ), 0 < θ < π/2, which is tangent to the unit circle, is taken to the semicircle of radius ½ with center at (½ , 0). Since 1+tan 2θ=sec 2θ,{\displaystyle 1+\tan ^{2}\theta =\sec ^{2}\theta ,} the hypotenuse of the right triangle on [0,1] to (1, tan θ) has length sec θ, so the reciprocal is cos θ, leading to the said semicircle. Now any interval on this semicircle corresponds to an interval on the vertical ray, hence a distance is defined on this semicircle. Invoke the second motion to bring the center to (0, 0), and the first motion to double the radius, resulting in the unit semicircle. Thus, any interval on the unit semicircle has a distance by correspondence with the one centered at (½ , 0), radius ½ . Apparently any semicircle with center on the absolute has a distance formula to apply to intervals.
This hyperbolic geometry features in differential geometry as a surface of negative curvature. It is a saddle surface as every point has an expansive feature not found in Euclidean geometry. The treatment given here with Cartesian geometry and standard trigonometry provides accessibility not requiring complex numbers.
References
[edit | edit source]
↑W. S. Jevons (1863) A Serious Fall in the Value of Gold, page 7
↑H. L. Moore (1895) Von Thunen's Theory of the Natural Wage, page 14
↑W. S. Jevons (1957) The Theory of Political Economy, 5th edition, pages 31, 49, 144, and 173
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MathGraphing Linear Equations in One and Two Variables
Graphing Linear Equations in One and Two Variables
Apr 9, 2022
That being said, linear equations can also be used to locate coordinate points in a graph. But, what are linear equations? How can one solve them to locate coordinate points on a graph? Keep reading the headings below to find out the answers to various questions related to graphing linear equations.
A graph contains a grid network of the x-axis and y-axis which is used to locate coordinate points. The value of a coordinate includes a pair in which the first value belongs to the X-axis and the second value belongs to the Y-axis. However, there are other ways of locating a coordinate point in a graph.
What is a Graph?
Have you seen a graph? It looks like a plane divided into several sections made by the intersection of horizontal and vertical lines. In mathematics, a graph is defined as a pictorial representation of data or a diagram that indicates values in an organized structure. A graph is plotted only if both the values of the x-axis and the y-axis are known.
The point on a graph where the lines of these axes intersect is known as a coordinate point. This point is the result of solving linear equations.
A graph is usually drawn on a coordinate plane. It is important to know that a coordinate plane consists of grids made by the intersecting lines of the X-axis and Y-axis.
An x-axis and a y-axis also represent a graph. Along with this, a graph also has a number line or scale, which is used to locate the precise point in a coordinate grid.
One can take advantage of a graph to represent the data of various things. A graph can define the number of supplies, requirements, and other parameters used in day-to-day life. This way, it is easy to represent a given data in a systematic order.
What is a linear equation?
A linear equation is an equation that has 1 as the highest exponential degree. It means that all the variables present in a linear equation have an exponential value of 1.
Linear equations always form a straight line on a graph. It means a straight line can be drawn by solving a linear equation – along with one of the quadrants in a coordinate grid.
A linear equation is expressed as an algebraic equation. It means each component of the equation has 1 as its exponent. Apart from this, a linear equation always forms a straight line when graphed on a coordinate grid. A straight line is formed with a linear equation, whether it is a value on the x scale or the y scale.
Types of linear equations
There are two types of linear equations –
Linear equations in one variable
Linear equations in two variables
The formula for a linear equation
The formula for a linear equation is used to express a linear equation. However, this can also be done in various ways. It means one can define a linear equation in the standard form, point-slope form, and slope-intercept form.
However, the results can vary based on the variables if the standard form is used to express a linear equation. It is also important to know that no matter what the form is, the exponential degree of all the variables present in a linear equation is always 1.
A linear equation in one variable has a standard form which is expressed by-
Ax + B = 0.
Here,
A = coefficient
x = variable
B = constant
Linear equation in two variables has a standard form which is expressed by-
Ax + By = C
Here,
A and B = coefficient
x and y = variables
C = constant
How to solve a linear equation?
To graphing linear equations on a coordinate plane, one should also know how to solve them. The solution comprises simple steps to locate a point on a graph.
An equation is different from other mathematical problems. It means that any creation has values on both sides to make it equal. It suggests that if we subtract a particular number from both sides of an equation, the equation remains the same and has the same value. The same goes for division, multiplication, and addition.
For example: 4x – 3 = 13
In this equation, one can solve it by moving constant values to one side and the variable values to the other side. By doing this, one can calculate the value of the unknown variable in the given equation.
Solution:
4x – 3 = 13
4x = 13 + 3
4x = 16
x = 16/4
x = 4
From the above solution, it is clear that one should perform mathematical operations in such a way that the left-hand side is equal to the right-hand side. It is done so that the balance of the equation is not disturbed. To understand this balance, refer to the example below.
For example: in the linear equation 4x – 3 = 13, if the addition of +3 is made on both sides, then the resulting linear equation will be,
4x – 3 + 3 = 13 + 3
4x = 16
x = 16/4
x = 4
From the above calculation, we understand that even if there is an addition on both sides of an equation, the result does not change. It is a convenient way to graphing linear equations.
What do you understand by Graphing linear equations?
Since it is clear that a Graphing linear equations forms a straight line on the graph, this line can either be parallel to the x-axis or the y-axis.
To understand this, one can suppose that there is a linear equation with a variable that has a value from the x-axis, then to graph a linear equation; a straight line is drawn which is parallel to the y-axis.
In the same way, if a linear equation has a variable with a value from the y-axis, then a straight line is drawn, which is parallel to the x-axis.
How to plot a Graphing linear equations in one variable?
It is a simple task when it comes to plotting a graph using a linear equation. However, there are different problem-solving methods when the linear equation involves one variable and two variables.
Read the following method to understand how to solve a linear equation in one variable and plot it on a graph.
For the equation: 2x – 5 = 9
The solution is,
2x – 5 = 9
2x = 9 + 5
2x = 14
x = 14/2
x = 7
And since this is a positive value on the x-axis, a straight line can be drawn for the same.
How do plot Graphing linear equations in two variables?
As the name suggests, these types of linear equations have two variables. Hence, the calculation differs from that of one variable.
Follow these steps to plot a graphing linear equations in two variables.
For the equation: x – 3y = 3, the steps will be –
The first thing is to convert the given equation into y = mx + b form.
It will give: y = x/3 – 1
After getting this, replace the value of acts with other numbers. for example, 0, 1, 2…
By substituting these values for x in the equation: y = x/3 – 1, one gets different values for y to create coordinates.
When 0 is taken as a value for x, then the resulting equation will be, y = -1
Likewise, if the value for x is taken as 3, the resulting equation will be, y = 0.
In the same way, if the value of x is substituted as 6, the resulting linear equation will be, y = 1.
As these resulting equations satisfy the original equation of y = x/3 – 1, a graph can be plotted against it.
Some characteristics of linear equations in one variable.
The variable of a linear equation and its value is called the root of the linear equation.
The variable of a linear equation is also called the solution
The resulting solution of a linear equation remains the same if a particular number is multiplied, added, divided, or subtracted from both sides (LHS = RHS) of the equation.
It does not matter whether the linear equation is in one variable or two variables; it always forms a straight line when graphed.
Some characteristics of graphing linear equations in two variables
In the coordinate grid, every point on the scale is a solution to the linear equation.
The solution of a linear equation in two variables will always form a point on a line in the coordinate grid.
Example:
Solve Equation x+2y=7
By first setting x=0x=0 and subsequently y=0y=0, you can find two solutions that correspond to the graph’s xx -intercepts and yy -intercepts, respectively.
If x=0x=0, we get:
0+2y=7y=3.50+2y=7y=3.5
If y=0y=0, we get:
x+2(0)=7x=7x+2(0)=7x=7
Therefore, the two points are (7,0)(7,0) and (0,3.5)(0,3.5).
Draw the line that connects these two spots on a plot.
Summary
The above explanation explains Graphing linear equations and the types of linear equations. Not just this, but a simple solution to calculate linear equations in one variable and linear equations in two variables is thoroughly explained. A linear equation is used to represent data on a graph. Hence, learning how to graph it can enable an understanding of various mathematical concepts.
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10126 | https://www.quora.com/How-many-protons-neutrons-and-electrons-are-in-an-isotope-of-carbon-called-Carbon-14 | How many protons, neutrons, and electrons are in an isotope of carbon called Carbon-14? - Quora
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Chemistry
Carbon-14
Atomic Structure Question
Protons
Isotope
Electrons
Structure of an Atom
Carbon Isotopes
Neutrons
5
How many protons, neutrons, and electrons are in an isotope of carbon called Carbon-14?
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Sep 7
Carbon-14 (written 14C) has:
Protons: 6
Carbon's atomic number is 6, so every carbon isotope has 6 protons.
Neutrons: 8
Neutrons = mass number − protons = 14 − 6 = 8.
Electrons: 6 (when neutral)
A neutral 14C atom has 6 electrons equal to the number of protons.
Upvote ·
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More answers below
How many neutrons are in carbon-14?
Carbon-12 is a common isotope of carbon. Carbon-14 is an isotope used in radioactive dating. How many neutrons are in carbon-14? How many protons are in carbon-14? How many protons in carbon-12?
How many protons, neutrons, and electrons does carbon have?
How many electrons does carbon 14 have?
How many protons does a carbon atom have?
Eric Pressman
B.S. in Secondary Science Education&Chemistry, Florida State University (Graduated 1999) · Author has 70 answers and 78.4K answer views
·5y
Originally Answered: How many protons, neutrons, and electrons are in an isotope of carbon called Carbon-14? ·
All Carbon atoms have 6 protons. Since it’s neutral it also has 6 electrons. It has 8 neutrons. The number 14 is the mass number the number of protons plus neutrons. Eric Pressman Miami
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All Carbon atoms have 6 protons. Since it’s neutral it also has 6 electrons. It has 8 neutrons. The number 14 is the mass number the number of protons plus neutrons. Eric Pressman Miami
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Tod Karlson
Former Day SRO/Surveillance Coordinator for Operations at BEA/INL-Advanced Test Reactor (ATR) (2007–2023) · Author has 12.7K answers and 2.2M answer views
·Updated 3y
Originally Answered: How many protons, neutrons, and electrons are in an isotope of carbon called Carbon-14? ·
All Carbon isotopes have 6 protons (+) which have a positive charge, therefore it will have 6 electrons (-) which have a negative charge. Carbon-14 is called that because it has 14 nucleons (protons & neutrons), so a little math tells you that Carbon-14 has 8 neutrons.
Carbon-12 is the more abundant (99%) of the 2 stable Carbon isotopes, the other being C-13 (1%). C-14 (also called “radiocarbon”) is radioactive with a half-life of about 5700 years. Radiocarbon dating is used to calculate the age of materials based on the decay of trace amounts of C-14 that was consumed by once-living creatures/
Continue Reading
All Carbon isotopes have 6 protons (+) which have a positive charge, therefore it will have 6 electrons (-) which have a negative charge. Carbon-14 is called that because it has 14 nucleons (protons & neutrons), so a little math tells you that Carbon-14 has 8 neutrons.
Carbon-12 is the more abundant (99%) of the 2 stable Carbon isotopes, the other being C-13 (1%). C-14 (also called “radiocarbon”) is radioactive with a half-life of about 5700 years. Radiocarbon dating is used to calculate the age of materials based on the decay of trace amounts of C-14 that was consumed by once-living creatures/plants.
Upvote ·
Jared Smith
Radiological Engineer (2009–present) · Author has 962 answers and 266.4K answer views
·Jan 8
Originally Answered: How many protons, neutrons, and electrons are in a carbon-14 atom? Does the ionization state affect the number of these particles? ·
All Carbon has 6 protons, the # protons determines what element it is.
14 is the isotope mass number of C-14 which is one of the radioactive isotopes of Carbon. For C-14, to find the # neutrons of any isotope take the mass number (14) and subtract the # protons (6) = 8 neutrons.
For an element to be neutrally charged, the # of negative charged electrons must equal the # positive charged protons (thus 6 electrons).
The ionization state of any element (or isotope) is the number of electrons stripped or added to the electron shell. Thus the ionization state only affects the number of electrons.
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Related questions
More answers below
How many electrons, protons, and neutrons are there in an atom of carbon 12?
What is the number of protons, neutrons, and electrons in 1 gram of carbon 14 atoms?
How many more neutrons are in an isotope of carbon-14 than in a standard carbon atom?
Could 666 in the Bible be referring to carbon (6 electrons, 6 protons, 6 neutrons)?
How many neutrons does an atom of carbon-14 contain?
Shree Chemolution
Subject Matter Expert (2018–present)
·4y
Originally Answered: An atom of carbon-14 contains what number of protons, neutrons, and electrons? ·
Carbon's atomic number is 6
So no of proton=6
No of protons= no of electrons =6
No of neutrons = 14–6=8
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Meave Gilchrist
B.S. in Bachelor of Science in Biology&Chemistry, University of Texas Pan American (Graduated 1985) · Author has 6.6K answers and 4.7M answer views
·5y
Originally Answered: How many protons, neutrons, and electrons are in an isotope of carbon called Carbon-14? ·
Isotopes are named for their mass number (A). The mass number is the sum of protons (p⁺) and neutrons (n⁰). The atomic number (Z) of an element is the number of protons in the atomic nuclei of its atoms, and is the same in all atoms of the element. In a neutral atom, the number of electrons (e⁻) equals the number of protons.
For Carbon-14
Z = number of protons (p⁺) = 6
number of electrons (e⁻) = number of protons = 6
A = p⁺ + n⁰ = 14
N = number of neutrons (n⁰) = A - Z = 14 - 6 = 8
Upvote ·
Vvprinceeac Vvprinceeac
Dean at STC College, Pollachi, Tamil Nadu, India (2011–present) · Author has 131 answers and 77.2K answer views
·4y
Originally Answered: An atom of carbon-14 contains what number of protons, neutrons, and electrons? ·
Atomic no. of C is 6 .Its mass no. is 14
Hence it contains 6 protons, 6 electrons and 8 neutrons.
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Steven Moss
BS in Physics, The Cooper Union for the Advancement of Science and Art · Author has 569 answers and 87K answer views
·1y
Originally Answered: How many protons, neutrons, and electrons are in a carbon-14 atom? Does the ionization state affect the number of these particles? ·
Carbon has an Atomic Number = 6 (means 6 Protons)
Carbon 14 has an Atomic Weight of 14 (means # of Protons + # of Neutrons = 14)
Carbon 14 has 14–6 =8 Neutrons
of Electrons = # of Protons in Neutral State or Base State. Ionization reduces the number of electrons associated with Atom.
Upvote ·
Brandon Cammon
Quora user at Quora (product) (2018–present) · Author has 2.6K answers and 967.1K answer views
·4y
Originally Answered: How many protons, neutrons, and electrons are in an isotope of carbon called Carbon-14? ·
7 protons
14 neutrons
Amount of electrons depends on if the atom is a negative or positive ion
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Guy Clentsmith
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Malcolm Sargeant
, Degree level applied chemistry + 20yr experience in corrosion prevention and water treatment · Author has 26.5K answers and 19.7M answer views
·Updated 4y
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Carbon-12 is so named because it has six protons and six neutrons. Carbon-14, which is used to determine the numerical age of dead organic material, is an isotope of Carbon-12. Carbon-14, therefore, must have how many protons?
12 C 12 C, 13 C 13 C, 14 C 14 C ALL have SIX PROTONS…
The identity of an atom, ANY ATOM, is determined by its Z the atomic number Z the atomic number, the which is SIMPLY the number of MASSIVE, positively-charged particles in its nuclear core (and we otherwise call these particles PROTONS). And so, Z=1 Z=1, the element is hydrogen; and in fact, most hydrogen nuclei possess the ONE PROTON only to give the protium isotope, i.e. 1 H.1 H.Z=2 Z=2, the element is helium; ….. Z=74 Z=74, the element is tungsten … And there about a 100 or so chemical elements, each characterized by a specific, and INTEGRAL Z the atomic number=1 Z the atomic number=1
Continue Reading
12 C 12 C, 13 C 13 C, 14 C 14 C ALL have SIX PROTONS…
The identity of an atom, ANY ATOM, is determined by its Z the atomic number Z the atomic number, the which is SIMPLY the number of MASSIVE, positively-charged particles in its nuclear core (and we otherwise call these particles PROTONS). And so, Z=1 Z=1, the element is hydrogen; and in fact, most hydrogen nuclei possess the ONE PROTON only to give the protium isotope, i.e. 1 H.1 H.Z=2 Z=2, the element is helium; ….. Z=74 Z=74, the element is tungsten … And there about a 100 or so chemical elements, each characterized by a specific, and INTEGRAL Z the atomic number=1 Z the atomic number=1 , 2 2, 3 3, 4 4, …up to one hundred or so. What do I mean by integral integral?
Within the nuclear core of many atoms (in fact ALL atoms save for 1 H 1 H), are also NEUTRONS, the which are MASSIVE, NEUTRALLY-charged particles. Interactions between protons, and neutrons, at IMPOSSIBLY short nuclear ranges, give rise to the strong nuclear force strong nuclear force, an attractive force that at such ranges is STRONGER than the electrostatic force of repulsion that exists between like-charged particles. And neutrons give rise to the phenomenon of ISOTOPES.
We take the hydrogen atom … and we know, BY DEFINITION of the hydrogen element, that Z the atomic number=1 Z the atomic number=1. And so there is a SINGLE, characterizing proton in its nucleus. But a few nuclei MAY have an extra neutron or two in it nucleus. One neutron, we got 2 H 2 H, the deuterium isotope, two neutrons, 3 H 3 H, the tritium isotope…. With respect to the rest of the Periodic Table, most of the heavier elements have an envelope of isotopes, whose weighted average gives the atomic mass we see printed on the Periodic Table…
In your problem (finally I got there!) you got 14 C 14 C … and you know that there are thus 14 massive particles …. but should you look at the Periodic Table you find that for carbon, Z the atomic number=6 Z the atomic number=6, for EVERY carbon isotope, i.e. this number defines the nuclide as carbon …. And thus you know the number of protons …. and how many neutrons you got for say the 12 C 12 C, 13 C 13 C, 14 C 14 C isotopes…? 37, 18, …. two and a half? And how many electrons in the neutral atom?
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9 2
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Gerry St-Aubin
Former High School Science Teacher at Ottawa-Carleton District School Board (OCDSB) (1987–2017) · Author has 1.2K answers and 2.5M answer views
·5y
Related
Why is carbon-12 considered an isotope? If a normal carbon atom has 6 protons and six neutrons, why would it be an isotope?
All atoms of all elements are isotopes. There is no “normal" atom and all the rest are isotopes.
C-12 happens to be the most abundant naturally occurring isotope of carbon, but that does not make it any more “normal" than C-14. Also, just because an atom has equal numbers of protons and neutrons does not make it “normal" either. You will find that for the first 20 elements a 1:1 ratio between protons and neutrons commonly results in a stable isotope, but as atomic nuclei increase in size (beyond element 20) the nucleus requires more neutrons than protons to remain stable. So, by your definition
Continue Reading
All atoms of all elements are isotopes. There is no “normal" atom and all the rest are isotopes.
C-12 happens to be the most abundant naturally occurring isotope of carbon, but that does not make it any more “normal" than C-14. Also, just because an atom has equal numbers of protons and neutrons does not make it “normal" either. You will find that for the first 20 elements a 1:1 ratio between protons and neutrons commonly results in a stable isotope, but as atomic nuclei increase in size (beyond element 20) the nucleus requires more neutrons than protons to remain stable. So, by your definition, if equal numbers of protons and neutrons make the “normal" atom, then the most abundant atoms of many of the heavier elements would not be ”normal".
When elements are created in stars, nuclei are formed when different groups of neutrons and protons are fused together. The element that is formed will depend only on the number of protons that are found in the resulting nucleus (atomic number). So, if a nucleus is formed that contains 7 protons and 8 neutrons and another nucleus contains 7 protons and 7 neutrons, they will both be nitrogen nuclei and when they each gain 7 electrons they will be nitrogen atoms. One will be the isotope called N-14, the other isotope N-15. Both are perfectly “normal".
Because nuclei of the same element can differ in the number of neutrons they contain, a term was needed to talk about these slightly different atoms of the same element, the term “Isotope” was introduced. Depending on the stability of each nucleus (isotope) some isotopes will be more abundant in nature than others, but that does not make them any more “normal" than any other isotope of that element.
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Stephen Frantz
Former Director of the Reed College Nuclear Reactor · Author has 7.4K answers and 6.4M answer views
·7y
Related
How many protons, neutrons, and electrons does carbon have?
Here is the generic procedure.
Step 1: For the number of protons, it is equal to the atomic number. You can look up the atomic number for any element on the Periodic Table of the Elements, or any Table of the Elements.
If you know the element’s name, the Table will tell you the atomic number, which is the number of protons.
If you know the number of protons, which is also the atomic number, you can look up the element’s name.
BTW, the number of electrons is equal to the number of protons in an unionized atom.
Step 2: For the number of neutrons, you need the Atomic Mass. The Atomic Mass is (roughly)
Continue Reading
Here is the generic procedure.
Step 1: For the number of protons, it is equal to the atomic number. You can look up the atomic number for any element on the Periodic Table of the Elements, or any Table of the Elements.
If you know the element’s name, the Table will tell you the atomic number, which is the number of protons.
If you know the number of protons, which is also the atomic number, you can look up the element’s name.
BTW, the number of electrons is equal to the number of protons in an unionized atom.
Step 2: For the number of neutrons, you need the Atomic Mass. The Atomic Mass is (roughly) the number of protons plus the number of neutrons. So you need the Atomic Number (from Step 1), and the Atomic Mass. You can usually get the Atomic Mass from the same Table of the Elements.
The number of neutrons is equal to the Atomic Mass minus the Atomic Number.
So for your question, the Periodic Table tells us that carbon has an Atomic Number of 6, so there are 6 protons and 6 electrons. The Periodic Table tells us that carbon has an Atomic Mass of ≈12. So there are 12 - 6 = 6 neutrons.
Upvote ·
99 11
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Stephen Frantz
Christian Pastor, Nuclear Physicist · Author has 7.4K answers and 6.4M answer views
·9y
Related
How do you work out the number of protons, neutrons and electrons in isotopes?
The number of protons defines the element. If you know the element you automatically know the number of protons. The number of protons is the same as the atomic number. A list of elements or a periodic table (which is sorted by atomic number) will tell you the number of protons. Ex: All oxygen atoms have 8 protons.
The number of electrons will match the number of protons in an atom unless it has been ionized. Normally the assumption is that the number of electrons equals the number of protons.
The number of neutrons can vary in an element. These variations in neutron number are called isotopes o
Continue Reading
The number of protons defines the element. If you know the element you automatically know the number of protons. The number of protons is the same as the atomic number. A list of elements or a periodic table (which is sorted by atomic number) will tell you the number of protons. Ex: All oxygen atoms have 8 protons.
The number of electrons will match the number of protons in an atom unless it has been ionized. Normally the assumption is that the number of electrons equals the number of protons.
The number of neutrons can vary in an element. These variations in neutron number are called isotopes of the element. The number of protons plus the number of neutrons equals the atomic mass. The atomic mass is usually given as the number after the dash following an element name. Ex: oxygen-18. Sometimes the atomic mass is a superscript before the elemental symbol. Ex: 18O. To get the number of neutrons, simply subtract the number of protons from the atomic mass.
So for oxygen-17, there are 8 protons because it is oxygen, there are probably 8 elections if it is uncharged, and there are 17–8 = 9 neutrons.
Similarly, oxygen-18 has 8 protons and 8 electrons, and 10 neutrons.
Note: isotopes describe the various forms of an element with different number of neutrons.
Nuclide is the generic term for a particular combination of protons and neutrons.
Ex: Oxygen-18 is an isotope of oxygen. Nitrogen-16, oxygen-18, and hydrogen-3 are nuclides.
Upvote ·
9 8
Related questions
How many neutrons are in carbon-14?
Carbon-12 is a common isotope of carbon. Carbon-14 is an isotope used in radioactive dating. How many neutrons are in carbon-14? How many protons are in carbon-14? How many protons in carbon-12?
How many protons, neutrons, and electrons does carbon have?
How many electrons does carbon 14 have?
How many protons does a carbon atom have?
How many electrons, protons, and neutrons are there in an atom of carbon 12?
What is the number of protons, neutrons, and electrons in 1 gram of carbon 14 atoms?
How many more neutrons are in an isotope of carbon-14 than in a standard carbon atom?
Could 666 in the Bible be referring to carbon (6 electrons, 6 protons, 6 neutrons)?
How many neutrons does an atom of carbon-14 contain?
An atom of carbon-14 contains what number of protons, neutrons, and electrons?
What is the number of neutrons present in carbon?
Which isotope has 15 protons, 16 neutrons, and 15 electrons?
How many neutrons are present in isotopes of carbon?
If carbon 12 can capture a neutron and become carbon 13, whats preventing the carbon 13 from capturing a neutron and becoming carbon 14?
Related questions
How many neutrons are in carbon-14?
Carbon-12 is a common isotope of carbon. Carbon-14 is an isotope used in radioactive dating. How many neutrons are in carbon-14? How many protons are in carbon-14? How many protons in carbon-12?
How many protons, neutrons, and electrons does carbon have?
How many electrons does carbon 14 have?
How many protons does a carbon atom have?
How many electrons, protons, and neutrons are there in an atom of carbon 12?
What is the number of protons, neutrons, and electrons in 1 gram of carbon 14 atoms?
How many more neutrons are in an isotope of carbon-14 than in a standard carbon atom?
Could 666 in the Bible be referring to carbon (6 electrons, 6 protons, 6 neutrons)?
How many neutrons does an atom of carbon-14 contain?
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排列组合cn和an公式？
```
排列的公式：A（n，m）=n×（n-1）...（n-m+1）=n!/（n-m）!（n为下标，m为上标,以下同）。
例如：A（4，2）=4!/2!=43=12。（考虑顺序，不考虑顺序则为6）
组合的公式：C（n，m）=P（n，m）/P（m，m） =n!/m!（n-m）!。
例如：C（4，2）=4!/（2!2!）=43/(21)=6。
```
1
2
3
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作者：浣熊数学
链接：
来源：知乎
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
排列组合问题简单起来可以非常简单，比如：一个“田”字里有多少个正方形？难起来也可以非常难，中国的高考、高中数学联赛和美国的 AMC、AIME 都会重点考察这个板块。
很多同学一遇到排列组合公式 P 呀 C 呀什么的就不清楚，这很正常，因为初学者在不一一列举的情况下，很难直观地想清楚哪些算重了，哪些算漏了。我自己作为学生刚接触这个的时候也是这样，每次一遇到排列组合题就很慌，后来发现，学习的关键是：你先得非常明确一些基本模型，这些基本模型往往只用很小的数字就能说明，想清楚后再做一些数字大的问题就轻松了。
一：P 的由来
二：C 的由来
三：5 个组合数的公式直观解释
四：10 个常见题型和方法
一：P 的由来
所谓排列组合，排列在组合之前，咱们要聊的第一个概念是“排列”，排列的英文是 Permutation 或者 Arrangement，因此在数学符号中，用 P 或者 A 表示都可以，二者意思完全一样。
我们常见的 P 右边会跟两个数字（或字母），右下角的数字 n 表示总数，右上角的数字 m 表示抽出的个数。整个符号的意思是“从 n 个人中，有顺序地抽出 m 个人的抽法数”，可以读作“P n 抽 m”。那么，到底什么叫做有顺序的？我们来举个数字很小的例子：
咱们可以用乘法原理：选第一名有 3 种可能性，选第二名有 2 中可能性，因为第一名那个人不可能同时又是第二名了，将这两步相乘起来。（如果你不太理解乘法原理，可以看看下图直观列举的表示。）
作者：浣熊数学
链接：
来源：知乎
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
二：C 的由来
咱们聊的第二个概念是“组合”，它比排列更常用，组合的英文是 Combination，因此在数学符号中用 C 表示，美国和英国教材中，也常用“长括号”表示组合数。
我们常见的 C 右边会跟两个数字（或字母），右下角的数字 n 表示总数，右上角的数字 m 表示抽出的个数。整个符号的意思是“从 n 个人中，不计顺序地抽出 m 个人的抽法数”，可以读作“C n 抽 m”。那么，到底什么叫做不计顺序的？我们也来举个例子：
哈哈，这就可以用到之前排列数的结论了！就让刚才的第一名和第二名去参加会议。但是，对于参加会议来说，谁是第一谁是第二不重要呀！因此我把原图的红色和蓝色都涂成了黑色，以示无区别。（如下图）
至此，第二步中，第一种和第三种都是 A、B 的组合，完全一样，就会有一些算重的，至于有多少个算重，取决于抽出个数 m 的全排列种数，即 m 的阶乘。（如果你不太理解哪些算重了，可以仔细看看下图中箭头所指的对应关系）
于是，组合数公式就是在排列数公式上除以一个 m！。但实际计算中，往往不用阶乘。我的记法是：从大的数字开始往小乘，乘“小的数字那么多”个，再除以“小的数字开始往小乘，乘小的数字那么多个”。
三：组合数的公式直观解释
组合公式Ⅰ：
这个公式课内和竞赛都会常常用到。我在刚学的时候把它联想成“做值日”问题，四个同学中，选三名同学做值日就相当于选一名同学放学直接回家。
比如，班里有 A、B、C、D 四个同学，每天要选出三个同学做值日，有几种选法？这个问题对于学过排列组合的同学自然非常简单了，就是 C 4 抽 3，但是，假如问一个没学过排列组合的人，他会怎么想呢？如果想 ABC，ACD……这种就会比较难想，不如去想它的反面：选Ａ、B、C 或 D 放学直接回家，总共就四种。这就能直观的理解这个公式了。
这个公式对于运算 C 10 抽 8 这样的组合数时非常有用，直接转化成 C 10 抽 2 来计算。
组合公式Ⅱ：
这个公式课内会提到，但不要求熟练掌握，竞赛会常用。可以把它联想成“约妹子看电影”问题，看看在四个妹子中，想约两个妹子有几种约法。
如果四个人都是普通朋友，看作是相同的 A、B、C、D，那自然有 C 4 抽 2 =6 种约法。下面我们来点刺激的：假如这四个人中有一个是你女朋友，她最特殊，你会先问她来不来：
①如果她来，但你还想一共约两个妹子（手动滑稽），那么就需要在其他三个妹子中再约一个，有 C 3 抽 1 种方法；
②如果她不来，那你就需要在其他三个妹子中再约两个，有 C 3 抽 2 种方法。
两类相加，表示的意义就是从 4 个妹子中约两个妹子的情况总数，即公式成立。
这个公式对于处理两个组合数相加问题非常有用，落实在计算上，我把它总结成口诀：上面的数字取大的，底下的数字加一。
组合公式Ⅲ
这个公式课内和竞赛都会常常用到。我把它叫做"抓兔子"问题，想象一个笼子里有两只兔子，抓出来的话有几种抓法？
第一种方法是我去笼子里抓，我在抓的时候就想好是抓 1 只还是抓 2 只，或是抓 0 只（即不抓）。由于先想好了这一点，就会有 C 2 抽 1 和 C 2 抽 2 这些组合数，分别表示按“抓一只”、“抓两只” 分类，每类的情况数；
第二种情况是我把笼子打开，让每只兔子自己选择跳出来或是不跳出来（2 种可能性），每只兔子都是独立的个体，所以可以用乘法原理，总共的情况数是 n 个 2 相乘，即 2 的 n 次方。
两种方法都表示“兔子出来的情况数”，因此一样，即公式得以解释。
这个公式对于处理一系列“底下相同的”组合数相加的问题非常好用，大大节省计算量。而且它与集合、二项式定理等中学数学知识紧密相连，需深入理解。
组合公式Ⅳ
这个公式一般在竞赛中会出现。我把它叫做"火车头"问题：抽出的一些元素，总有一个打头的，称为火车头，它也是火车的一节，只不过是特殊的一节。
具体来讲，比如说你要在 A、B、C、D、E 这 5 个小球中抽取 3 个小球，咱们可以按“哪个小球是第一个”分类
第一类：A 为火车头，那么还需在后面四个小球中抽取两个小球；
第二类：B 为火车头，那么还需在后面三个小球中抽取两个小球；
第三类：C 为火车头，那么还需在后面两个小球中抽取两个小球。
至于 D 或 E 开头的，就不足“三节车厢”了，故不计算。我们把之前说的三类加起来，就直观地理解了这个公式。
这个公式对于处理一系列“上面相同的”组合数相加的问题非常好用，大大节省计算量。记忆方法是：和为上面下面都加一。
组合公式Ⅴ
这个公式是一个相加和相乘结合的公式，看似复杂，但并不难理解。我对它的理解是：可以想象成班里选几名学生，分男女选和不分男女选情况数一样。
比如说，咱们假设班里有 7 名学生，4 男 3 女。如果选出三个人参加竞赛有几种选法？首先容易想到的是 C 7 抽 3 =35。没错，不过咱们还有一个思路，就是按“男女各多少人”分类讨论。
第一类：0 男 3 女，分别抽取，再乘起来。
第二类：1 男 2 女，分别抽取，再乘起来。
第三类：2 男 1 女，分别抽取，再乘起来。
第四类：3 男 0 女，分别抽取，再乘起来。
这四类是互不重叠的，可用加法原理将其相加。原公式就得以直观理解。
hihu.com/search?q=加法原理&search_source=Entity&hybrid_search_source=Entity&hybrid_search_extra={“sourceType”%3A"answer"%2C"sourceId"%3A610713978})将其相加。原公式就得以直观理解。
上面 5 个公式都可以代数证明，也可按照我举得例子通俗理解，如果这二者你都很清楚，那排列组合就能融会贯通啦。
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基本原理
1.加法原理
集合S=s1∪s2∪s3..sm且si∩sj=∅(i
则s=s1+s2+s3..sm
2.乘法原理
s(a,b)
取a有p种，取b有q种
s=pq
公式
1.对于集合s（没有重复元素）:
从s中n个元素中取r个排列的方案数p(n,r)=n!/(n-r)!
从s中n个元素中取r个组合的方案数c(n,r)=n!
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2023 AMC 12A Problems/Problem 9
The following problem is from both the 2023 AMC 10A #11 and 2023 AMC 12A #9, so both problems redirect to this page.
Contents
[hide]
1 Problem
2 Solution 1
3 Solution 2 (Area)
4 Solution 3
5 Solution 4
6 Solution 5
7 Solution 6
8 Solution 7 (Manipulation)
9 Video Solution by Little Fermat
10 Video Solution by Math-X (First understand the problem!!!)
11 Video Solution by Power Solve (easy to digest!)
12 Video Solution (⚡Under 5 min⚡)
13 Video Solution by CosineMethod [🔥Fast and Easy🔥]
14 Video Solution
15 See Also
Problem
A square of area is inscribed in a square of area , creating four congruent triangles, as shown below. What is the ratio of the shorter leg to the longer leg in the shaded right triangle?
Solution 1
The side lengths of the inner square and outer square are and respectively. Let the shorter side of our triangle be , thus the longer leg is . Hence, by the Pythagorean Theorem, we have
By the quadratic formula, we find that , so the answer is
~semisteve ~SirAppel ~ItsMeNoobieboy
Solution 2 (Area)
Looking at the diagram, we know that the square inscribed in the square with area has area . Thus, the area outside of the small square is This area is composed of congruent triangles, so we know that each triangle has an area of .
From solution , the base (short side of the triangle) has a length and the height , which means that .
We can turn this into a quadratic equation: .
By using the quadratic formula, we get .Therefore, the answer is
~ghfhgvghj10 (If I made any mistakes, feel free to make minor edits)
(Clarity & formatting edits by Technodoggo)
Solution 3
Let be the ratio of the shorter leg to the longer leg, and be the length of longer leg. The length of the shorter leg will be .
Because the sum of two legs is the side length of the outside square, we have , which means . Using the Pythagorean Theorem for the shaded right triangle, we also have . Solving both equations, we get . Using to substitute in the second equation, we get . Hence, . By using the quadratic formula, we get . Because be the ratio of the shorter leg to the longer leg, it is always less than . Therefore, the answer is .
~sqroot
Solution 4
The side length of the bigger square is equal to , while the side length of the smaller square is . Let be the shorter leg and be the longer one. Clearly, , and . Using Vieta's to build a quadratic, we get Solving, we get and . Thus, we find .
~vadava_lx
Solution 5
Let be the angle opposite the smaller leg. We want to find .
The area of the triangle is which implies or . Therefore
Solution 6
Allow a and b to be the sides of a triangle. WLOG, suppose We want to find . Notice that the area of a triangle is , which results in . Thus, . However, the square of the hypotenuse of this triangle is , but also . We can write as , and then plug it in. We get , so . Applying the quadratic formula, , or . However, since and must both be solutions of the quadratic, since both equations were cyclic. Since , then , and . To find , we can simply find the square root of . This is , so the answer is . - Sepehr2010
Solution 7 (Manipulation)
Let be the length of the shorter leg and be the longer leg. By the Pythagorean theorem, we can derive that . Using area we can also derive that . as given in the diagram, we can find that because . This means that and . Adding the equations gives and when is plugged in . Rationalizing the denominators gives us .
Video Solution by Little Fermat
~little-fermat
Video Solution by Math-X (First understand the problem!!!)
~Math-X
Video Solution by Power Solve (easy to digest!)
Video Solution (⚡Under 5 min⚡)
~Education, the Study of Everything
Video Solution by CosineMethod [🔥Fast and Easy🔥]
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 10Followed by
Problem 12
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25
All AMC 10 Problems and Solutions
2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 8Followed by
Problem 10
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25
All AMC 12 Problems and Solutions
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
Retrieved from "
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10129 | https://abakus-center.com/blog/the-complete-explanation-of-the-divisibility-rule-for-7 | The complete explanation of the divisibility rule for 7
Mathematics has a set of guidelines to simplify number operations, and divisibility rules are among the most practical tools in the field. One such rule, the divisibility rule of 7, helps determine whether a number is divisible by 7 without performing complex division. This guide provides a deep dive into understanding, applying, and mastering the divisibility rule of 7 for improved mathematical proficiency.
What is the divisibility rule of 7?
The divisibility rule of 7 states that a number is divisible by 7 if a specific operation on its digits results in a number that is either zero or divisible by 7. The steps to apply this rule are straightforward and involve subtracting twice the last digit from the rest of the number.
Steps to apply the rule
Separate the last digit: Take the number and separate its last digit.
Double the last digit: Multiply the separated digit by 2.
Subtract the result: Subtract this doubled value from the remaining part of the number.
Check divisibility: If the resulting number is divisible by 7, then the original number is also divisible by 7.
For example, consider the number 259:
Separate the last digit: 9.
Double it: 9 × 2 = 18.
Subtract: 25 − 18 = 7.
Check divisibility: Since 7 is divisible by 7, 259 is also divisible by 7.
Why does the rule work?
The rule works due to the modular arithmetic properties of numbers. By subtracting twice the last digit, we retain the mathematical equivalence of the number modulo 7. This makes it easier to test divisibility without needing direct division.
Practical applications of the rule
The divisibility rule of 7 is widely used in various contexts:
Quick Mental Calculations: Enables faster verification of divisibility in exams or mental math tasks.
Error Detection in Arithmetic: Helps identify mistakes in large manual calculations.
Cryptography and Algorithms: Plays a role in modular arithmetic used in encryption algorithms.
Examples to master the divisibility rule of 7
Example 1: small numbers
Number: 133
Last digit: 3
Double it: 3 × 2 = 6
Subtract: 13 − 6 = 7
Divisibility: Since 7 is divisible by 7, 133 is also divisible by 7.
Example 2: large numbers
Number: 5,674
Last digit: 4
Double it: 4 × 2 = 8
Subtract: 567 − 8 = 559
Repeat for 559:
Last digit: 9
Double it: 9 × 2 = 18
Subtract: 55 − 18 = 37
37 is not divisible by 7.
Therefore, 5,674 is not divisible by 7.
How to teach the divisibility rule of 7
Teaching this rule can be engaging with the following strategies:
Visual Aids: Use diagrams and flowcharts to illustrate the steps.
Interactive Exercises: Present students with numbers to test in real-time.
Group Challenges: Encourage students to identify divisible numbers from a given list.
Common misconceptions about the rule
Only Works for Certain Numbers: Some believe the rule is unreliable for larger numbers, but this is untrue when applied correctly.
Confusion with Other Rules: It is often confused with the rules for 3 or 9, which involve summing digits.
FAQs about the divisibility rule of 7
Can this rule be used for decimals?
No, the rule applies only to whole numbers.
Is there a shortcut for very large numbers?
Yes, the rule can be applied recursively to simplify calculations.
Why subtract twice the last digit?
This transformation maintains divisibility due to modular arithmetic properties.
Are there similar rules for other numbers?
Yes, divisibility rules exist for 2, 3, 5, 9, 11, and others.
How can this rule improve mental math skills?
It allows quicker verification of divisibility, reducing reliance on calculators.
How can I verify if a number divisible by 7 is correct if the result is negative?
Even if the result after applying the rule is negative, check its absolute value. If the absolute value is divisible by 7, then the original number is also divisible by 7.
Can this rule be applied to numbers written in non-decimal numeral systems?
Yes, but the numbers need to be converted to the decimal system first. The rule works based on modular arithmetic properties specific to the decimal system.
What should I do if the remainder after subtraction is larger than the initial number?
Keep applying the rule repeatedly until you get a smaller number that's easier to work with. This approach is especially useful for large numbers.
Why is the last digit multiplied by 2 specifically, and not any other number?
This is due to the unique properties of the number 10 modulo 7. Multiplying the last digit by 2 ensures the transformation remains valid for divisibility checks.
Are there alternative methods to check divisibility by 7 without using this formula?
Yes, you can manually divide the number by 7 or use remainders from numbers close to the original. However, this rule significantly simplifies and speeds up the process.
By mastering the divisibility rule of 7, we can streamline complex calculations and build confidence in numerical operations. This foundational tool continues to benefit students, educators, and professionals alike.
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10130 | https://www.doubtnut.com/qna/141178649 | The Van't Hoff factor ( i) for 2.0 m aqueous glucose solution is
0.2
0.4
0.6
1.0
More from this Exercise
The correct Answer is:D
Glucose does not undergo dissociation or association. Hence Van't Hoff factor ( i) =1.
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10m
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13m
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20m
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16m
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8m
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16m
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20m
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16m
21. Nuclear Chemistry
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10m
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9m
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7m
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7m
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9m
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7m
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10m
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3m
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4m
+ Band of Stability: Overview
14m
+ Measuring Radioactivity
7m
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12m
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16m
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18m
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14m
22. Organic Chemistry
5h 7m
Introduction to Organic Chemistry
8m
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8m
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10m
+ Skeletal Formula
6m
+ Spatial Orientation of Bonds
3m
+ Intro to Hydrocarbons
16m
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11m
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15m
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11m
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4m
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9m
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13m
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6m
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8m
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11m
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11m
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9m
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5m
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5m
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4m
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8m
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5m
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5m
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7m
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7m
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4m
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4m
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3m
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7m
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4m
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9m
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8m
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7m
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6m
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4m
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3m
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4m
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3m
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4m
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10m
23. Chemistry of the Nonmetals
2h 39m
Main Group Elements: Bonding Types
4m
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7m
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11m
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7m
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16m
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20m
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4m
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11m
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8m
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7m
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7m
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12m
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12m
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6m
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3m
24. Transition Metals and Coordination Compounds
3h 16m
Atomic Radius & Density of Transition Metals
11m
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7m
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11m
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22m
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14m
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4m
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10m
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5m
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Chemical Quantities & Aqueous Reactions
Dilutions
Chemical Quantities & Aqueous Reactions
Dilutions: Videos & Practice Problems
Video Lessons
Practice
Worksheet
Topic summary
A stock solution is a concentrated solution that can be diluted by adding more solvent, typically water, to achieve a lower concentration. The dilution process can be expressed with the equation Math input error, where m1 and v1 are the molarity and volume before dilution, and m2 and v2 are after dilution. This highlights the transition from a higher to a lower molarity solution.
In Dilutions, a solvent (usually water) is added to a concentrated solution.
Concentrated & Diluted Solutions
1
concept
Solution Dilution Process
Video duration:
1m
Play a video:
Solution Dilution Process Video Summary
A stock solution is a concentrated solution that can be diluted for various laboratory applications. Dilution involves adding more solvent, typically water, to a solution to decrease its concentration. For instance, when a dark purple solution, which indicates a high concentration, is gradually mixed with water, the color lightens to a fuchsia hue. This visual change signifies that the solution has become less concentrated. In summary, dilutions are achieved by adding water to the original solution, resulting in a diluted solution with a lower concentration.
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2
example
Ranking Solutions Example
Video duration:
1m
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Ranking Solutions Example Video Summary
In this scenario, we are tasked with arranging solutions based on their molarity, which is defined as the number of moles of solute per liter of solution. Molarity (M) can be calculated using the formula:
M = \frac{n}{V}
where n is the number of moles of solute and V is the volume of the solution in liters.
Let's analyze the provided solutions:
For solution A, there are 5 spheres representing 5 moles of solute in 1 liter of solution. Thus, the molarity is:
M_A = \frac{5 \text{ moles}}{1 \text{ L}} = 5 \text{ M}
For solution B, there are 3 spheres, indicating 3 moles of solute in 2 liters of solution. Therefore, the molarity is:
M_B = \frac{3 \text{ moles}}{2 \text{ L}} = 1.5 \text{ M}
For solution C, there are 6 spheres, which means 6 moles of solute in 3 liters of solution. The molarity is calculated as:
M_C = \frac{6 \text{ moles}}{3 \text{ L}} = 2 \text{ M}
Now, to arrange the solutions from least concentrated to most concentrated based on their molarity, we find:
Solution B: 1.5 M
Solution C: 2 M
Solution A: 5 M
Thus, the order from least concentrated to most concentrated is B, C, and A.
3
concept
Dilution Equation
Video duration:
58s
Play a video:
Dilution Equation Video Summary
Understanding dilution is essential in chemistry, as it allows us to create solutions with lower concentrations from more concentrated ones. The process of dilution can be quantitatively described using the equation:
In this equation, and represent the molarity and volume of the solution before dilution, while and represent the molarity and volume after dilution. It is important to note that , the molarity of the concentrated solution, is always greater than , the molarity of the diluted solution.
The final volume after dilution, , is determined by the initial volume plus the volume of solvent added. This relationship can be expressed as:
By applying these principles, one can effectively prepare solutions with desired concentrations, which is a fundamental skill in various scientific applications.
4
example
Dilution Calculation Example
Video duration:
2m
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Dilution Calculation Example Video Summary
To determine the volume of a concentrated solution needed to prepare a diluted solution, we can apply the concept of dilution, which involves a single compound with two different molarities. In this case, we are working with hydrobromic acid (HBr) and using the dilution equation:
M1V1 = M2V2
Here, M1 represents the molarity of the concentrated solution, while M2 is the molarity of the diluted solution. V1 is the volume of the concentrated solution we need to find, and V2 is the volume of the diluted solution.
In this example, we have:
M1 = 5.2 M (the concentrated solution)
M2 = 2.7 M (the diluted solution)
V2 = 3.5 L (the volume of the diluted solution)
Since we are looking for V1, we can rearrange the equation:
V1 = (M2V2) / M1
Substituting the known values:
V1 = (2.7 M × 3.5 L) / 5.2 M
Calculating this gives:
V1 = 1.8173 L
To convert this volume into milliliters, we use the conversion factor where 1 L = 1000 mL:
V1 = 1.8173 L × 1000 mL/L = 1817.3 mL
Considering significant figures, since the values 5.2, 3.5, and 2.7 all have two significant figures, we round our final answer to:
V1 = 1800 mL
This example illustrates that when dealing with a single compound and two different molarities, we can effectively use the dilution formula to find the required volume of the concentrated solution.
5
Problem
To what final volume would 100 mL of 5.0 M KCl have to be diluted in order to make a solution that is 0.54 M KCl?
72 mL
289 mL
330 mL
930 mL
1400 mL
Problem
If 880 mL of water is added to 125.0 mL of a 0.770 M HBrO4 solution what is the resulting molarity?
A
0.096 M
B
0.136 M
C
0.257 M
D
0.892 M
E
1.76 M
Problem
A student prepared a stock solution by dissolving 25.00 g of NaOH in enough water to make 150.0 mL solution. The student took 20.0 mL of the stock solution and diluted it with enough water to make 250.0 mL solution. Finally taking 75.0 mL of that solution and dissolving it in water to make 500 mL solution. What is the concentration of NaOH for this final solution? (MW of NaOH:40.00 g/mol).
A
0.0500 M
B
0.025 M
C
0.005 M
D
0.500 M
E
0.0100 M
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Here’s what students ask on this topic:
The purpose of performing a dilution in chemistry is to decrease the concentration of a solution by adding more solvent, typically water. This is often done to prepare solutions with specific, lower concentrations for use in experiments or analytical procedures. Dilutions are essential for ensuring accurate and reproducible results in the laboratory, as working with overly concentrated solutions can lead to errors or undesired reactions. By diluting a stock solution, chemists can achieve the desired molarity for their experiments while maintaining precision and control over the solution's properties.
The final concentration of a solution after dilution can be calculated using the dilution equation:
Math input error
Here, and are the molarity and volume of the concentrated solution before dilution, while and are the molarity and volume after dilution. Rearrange the equation to solve for the unknown, such as , if needed. This equation ensures that the number of moles of solute remains constant before and after dilution.
A stock solution is a concentrated solution that contains a high amount of solute per unit volume. It is typically prepared in advance and used as a starting point for creating solutions of lower concentrations. A diluted solution, on the other hand, is created by adding a solvent, usually water, to the stock solution to decrease its concentration. The diluted solution has a lower molarity compared to the stock solution, as the solute is spread out over a larger volume. This process allows chemists to prepare solutions with precise concentrations for specific experimental needs.
During dilution, the color of a solution typically becomes lighter or less intense. This change occurs because the concentration of the solute, which is responsible for the solution's color, decreases as more solvent is added. For example, a dark purple solution may turn into a lighter purple or fuchsia as it is diluted. The color change visually indicates that the solution's concentration has been reduced, although the exact appearance depends on the solute's properties and concentration.
The equation Math input error represents the relationship between the concentration and volume of a solution before and after dilution. Here, and are the molarity and volume of the concentrated solution, while and are the molarity and volume after dilution. This equation ensures that the number of moles of solute remains constant during the dilution process, as the solute is not removed, only spread out over a larger volume.
The volume of solvent added during a dilution can be determined by subtracting the initial volume of the solution () from the final volume of the solution (). The formula is:
Math input error
Here, is the volume of solvent added. This calculation is useful for determining how much solvent is needed to achieve the desired final concentration and volume of the solution.
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ポイントがね ツボがあるんですか [音楽] [音楽] [音楽] 2人とも間違い ない 一旦戻してください いい 今回のこの平方根の考え方使うっていうの はポイントがあるんですねやっぱりいかに ここの形に持っていけるかです 例えば1番1番は2x2乗とX2がついて んだねだからこの形をちょっと違うんだよ どうするかって言うと 2で割ってあげると2乗=なんとかって形 になるこの形になれば平方根の考え2乗を 取ると プラスマイナス5って答えが出てくるんだ ね 他にもね基本的には全部一緒です 2番はですねちょっと特殊な考え方をし てるんですけどちょっと変えていいですか ちょっと変えていいですか 彼女はねここが2乗になってるここも2乗 になってるって気づいたんですねで本当は この後ちょっと移行しなきゃなんですけど ね要するにこの後多分この間にねこの式に 直すっていうのが多分入ってるんじゃない かなって思います 2乗=2乗ってなってるから2乗取るとね こういう形になる -1やっぱこの形2乗=なんとかあって形 に持っていくのがね大切です以上取りまし た平方根ですねで2で割りましたで移行し てあげてここここで終わらないようにね 計算してください プラスマイナスに プラスの1/2になるそうすると答えは -1/2と3/2 の問題かわかりました やっぱこの形二条イコールに持っていくの が大切なんだけどそこまでね 展開とかもしなきゃいけない 移行なんかもしなきゃいけないっていう 問題ですね 結局 展開して移行すると2乗=24となるもん で答えはプラスマイナス で最大のポイント4番です4番わかりまし た 説明してもらって とまずこの式をAX +ax2+bx+c=0の形に直しますで この 式を この形に表したいので両辺に4足しますそ したらx2乗+2x+1=4でこの式は x +1を2乗したやつに変形できるので こうなりますそしてこの2乗を外すと X+1=プラスマイナスとなってX=1と -3になります 持っていく方は人それぞれでいいかなって 思いますポイントはやっぱり一緒ですね ここに持ってきたいんですよ で今回そこに持っていくためにポイントが あるんだね 結局このX+1の2乗1ってどこから来た 石どこから来た そうだねここの1っていうのは ここと関連してるね もっと直線的どういう関係ですか 何のことやらですね ここの数字の半分がここに来てるんだね だってこのカッコの2乗を 展開してあげるとここも数字の2倍がここ に来るんだよねXの係数って考えるとここ のちょうど半分なんです ここの半分 でこの半分の1が何かするとここの数字に なりますね ここの2乗が来るんですねちょっと見 づらくてごめんなさい Xの係数の半分がここになるその2乗が ここに来るこの1ってのが 元の式のここに 付け加えてあげるとカッコの2乗って形が 作れるんですね ここがポイントです 多分裏側の問題こんな問題ばかりですもう ちょっとここ練習しましょう 多分手こずってる人結構いたんじゃないか なこの2乗っていうの作り方をねこれから やっていきましょう ちなみにこの2乗を作り出すこと なんて言いますか 高校でまたやるんですけどその考えも使っ てね解かなきゃいけない問題も出てくると 覚えておいてください 残り時間 こっち今日チャイムなるのかな そうで危険だね終わりますか 次回もうちょっとね 解く時間あげます裏の問題終わりましょう |
10133 | https://www.gamdan.com/blog/interferometry-fringes-and-phases | INTERFEROMETRY - FRINGES AND PHASES — Gamdan Optics
INTERFEROMETRY - FRINGES AND PHASES
INTERFEROMETRY - FRINGES AND PHASES
PHOTONS
Photons are the basic units of light. They display the properties of both waves and particles. If you find this confusing, think how photons must feel! It is the wave-like property that leads to interference. The waves are sinusoidal in character, continuous and alternating in positive or negative sign. Interference can be additive or subtractive, depending on whether the photons are in phase or out of phase.
If two are exactly in phase, they will have twice the intensity of a single photon. But when they are exactly 180 degrees out of phase they will cancel each other, resulting in no light appearing at a detector. Of course, single photons are rare. Under most circumstances there will be groups or bundles of light consisting of countless photons.
What are Fringes?
Fringes are also called light bands. They are alternating lines or regions of dark and light formed by imaging the interference pattern from a reference surface and an optic being tested. They are analogous to contour lines on a topographical map. Each dark fringe is a locus of points for an equal optical path difference between that highly accurate reference and a (generally) distorted test wavefront. These dark points are the result of destructive interference.
Dark fringes are most obvious to the eye, which is sensitive to contrast but not so much to intensity. Hence it appears that there are wide bands of light between dark fringes, because of partial destructive interference between the brightest bands where photons are exactly in phase.
It is uncommon that wavefronts interfering will be exactly the same intensity, though a minor mismatch will still be perceived as dark and light lines. Detectors used for imaging interference patterns are able to discern the slight variation in intensity between the darkest and brightest areas.
So, What is Interference?
Interference is the property of electromagnetic waves that combine to form a resultant wave for which the intensity is reinforced or cancelled. In most cases, the intensity is neither doubled nor totally cancelled, because the waves combining are seldom exactly the same intensity. In general, visible electromagnetic radiation is measured, but infrared or ultraviolet light can be measured with a suitable detector.
When two perfect planar (plano, in optical jargon) waves interfere, they will combine to form a darker or lighter wave, depending on the distance each travels. An optical path length difference will determine if the interference is additive or subtractive. A change of distance difference travelled by light in a purely piston type displacement will change the intensity. If a slight angle or tilt is introduced between the two, alternating light and dark bands, fringes, will be observed. The number of these fringes is directly proportional to the amount of tilt introduced. There are techniques to measure the wedge angle of a flat optic by the fringes produced by that wedge.
Wavefront Measurement by Fringes
Before light sources with considerable coherence length were available (think lasers), opticians formed fringe patterns using a test plate, also known as an optical flat. Spherical test plates also exist, to check the surface of lenses, or curved mirrors. These test plates could be placed in close proximity to each other, so common light sources could be used.
If the mismatch between the surfaces has spherical power, circular fringes will form. For greater mismatch, more fringes form. This leads to some confusion when testing flat surfaces. If the mismatch between the test flat and the optic is small (less than a half wave length of the light being utilized), no round fringes will form. But if there is tilt (or wedge), many fringes can form. It is essentially the straightness of the pattern that determines flatness, not the fringe count.
When two wavefronts interfere, if they are perfect matches but with some tilt introduced, a particular fringe pattern is formed. A perfect pattern has fringes that are straight, parallel, and evenly spaced. Straightness and parallelism are readily evaluated by eye. But spacing difference is not so easily detected. For this reason, it is common to adjust the tilt and observe orthogonal fringes, which will show curvature in the plane of the spacing difference.
A dark region forms when the path length between the reference and test reflection are out of phase by 180 degrees, a half wave. Each additional dark region (typically a fringe) is formed for every increase of path length by a half wave. Light travels to the test surface and back on itself, so a half wave of separation results in a full wave of path length.
When the difference between surfaces (thus reflected wavefronts) is less than a half wave and irregular, and there is no wedge between them, the observer or light detector will see a mottled pattern reminiscent of the shadows of the moon. If wedge is introduced, the fringe lines will be somewhat jagged. But most often optics are polished by a smooth motion, which will produce straight or curving fringes without abrupt features.
Laser interferometers allow producing fringe patterns without requiring that the surfaces be close together. That is a consequence of the relatively long coherence length of most lasers.
Fringe Pattern Conversion to Waves
When multiple circular fringes are detected, counting from a null fringe to the furthest outlying fringe yields double the error in waves, so the rule of thumb is a fringe equals a half wave. If the error is less than a full fringe, some wedge may be introduced. In that case, the departure from the perfect pattern (straight, parallel, equally spaced) must be determined. Usually a photograph or similar output would be produced. Manual processes to analyze the patterns are tedious but yield results. Those may be limited in accuracy or subjective.
A number of automated fringe analyzing methods exist. Some are programmed into the interferometers. Others are separate instruments. These would typically determine points along the fringe centers, and calculate the difference from the ideal pattern. If enough points were measured, the result might be expressed in peak-to-valley, as well as cylinder, astigmatism, and local slope.
Intensity Measurement
More advanced interferometers detect intensity patterns, using a great many pixels. These can measure smaller errors than fringe measuring interferometers. They also measure many more points, approximating a continuous pattern. To calculate the sign of a wavefront (plus or minus, equating to an expanding or converging wavefront, caused by a convex or concave acting optic), requires a series of “snapshots” of the intensity pattern. Usually there are three or more discrete values. The large quantity of points also lends itself to special values, beyond peak-to-valley or other first and second order values. Slope and PSD (power spectral density) expressions can also be calculated.
There are a few different methods of measuring intensity variation and converting them to wavefront error expressions. The amplitude and sign are essential values. To distill the intensity patterns, a series of at least three measurements is made. Each is recorded at a predetermined difference in phase. The phase difference may be incurred by a number of methods. Moving the reference flat of the interferometer is common. Two other techniques actually take advantage of the wave behavior of a laser source. One hops between two close frequencies of the laser. Another shifts rapidly between two polarization states of the laser beam. Both of those methods remove the need to move part of the measurement cavity. They also can be very quickly modulated. In either case the measurement assembly may be more rigid, and the influence of vibration or air currents is minimized.
Summary
It is worth noting that neither fringe nor intensity measurements are measurements of wavefront! Each type of interference pattern is a “map” of intensity variation. Those darker to lighter discriminations are converted by some algorithm built into the measuring device, the interferometer. There is a one-to-one correlation of brightness to phase difference. It is that connection that allows calculating the items of interest, the wavefront values.
What sets GAMDAN Optics apart is our innovation. We work with engineers and scientists to find custom solutions for your material needs. Contact GAMDAN today about your specific material needs.
DENNIS J. GARRITY, AUTHOR
Dennis is an engineer with over 45 years of experience in fabrication, testing, and material evaluation for high precision optics, with extensive hands-on experience. More on the author can be found here.
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QualityDennis Garrityinterferometry, phase matching, fringes, wavefronts, Interferometer, Nonlinear crystal innovation
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10134 | https://www.ck12.org/flexi/cbse-math/complex-numbers/what-are-the-properties-of-the-cube-roots-of-unity/ | What are the properties of the cube roots of unity?
Flexi Says:
The cube roots of unity are the roots of the equation . They are complex numbers and have some interesting properties. Let's denote the cube roots of unity as , , and . Here are some of their properties:
Sum of the roots:
The sum of the cube roots of unity is zero. That is, .
Product of the roots:
The product of the cube roots of unity is one. That is, .
Squares:
The square of each cube root of unity is the other non-trivial cube root of unity. That is, and .
Powers:
Any power of a cube root of unity that is a multiple of 3 is 1. That is, and .
Conjugates:
The conjugate of each non-trivial cube root of unity is the other non-trivial cube root of unity. That is, if , then , and vice versa.
Modulus:
The modulus of each non-trivial cube root of unity is 1. That is, .
Related questions:
What are the basic properties of the modulus of complex numbers?How to divide complex numbers?What do we mean by imaginary part of a complex number?
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10135 | https://es.scribd.com/document/473395583/Ejercicios-Torsion-R1-RESUELTOS | Ejercicios Torsión R1 RESUELTOS | PDF | Esfuerzo de torsión | Rotación
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Ejercicios Torsión R1 RESUELTOS
El documento presenta 9 ejercicios sobre esfuerzos cortantes causados por pares de torsión en ejes cilíndricos. Los ejercicios calculan valores como el esfuerzo cortante máximo, el esfuerzo …
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10136 | https://pubmed.ncbi.nlm.nih.gov/33238161/ | Distinct Structures and Dynamics of Chromatosomes with Different Human Linker Histone Isoforms - PubMed
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. 2021 Jan 7;81(1):166-182.e6.
doi: 10.1016/j.molcel.2020.10.038. Epub 2020 Nov 24.
Distinct Structures and Dynamics of Chromatosomes with Different Human Linker Histone Isoforms
Bing-Rui Zhou1,Hanqiao Feng1,Seyit Kale2,Tara Fox3,Htet Khant3,Natalia de Val3,Rodolfo Ghirlando4,Anna R Panchenko5,Yawen Bai6
Affiliations Expand
Affiliations
1 Laboratory of Biochemistry and Molecular Biology, National Cancer Institute, National Institutes of Health, Bethesda, MD 20892, USA.
2 Izmir Biomedicine and Genome Center, Dokuz Eylul University Health Campus, Balcova, Izmir 35330, Turkey.
3 Center of Macromolecular Microscopy, National Cancer Institute, Cancer Research Technology Program, Frederick National Laboratory for Cancer Research, Leidos Biomedical Research Inc., Frederick, MD 21701, USA.
4 Laboratory of Molecular Biology, National Institute of Diabetes, Digestive and Kidney Diseases, National Institutes of Health, Bethesda, MD 20892, USA.
5 Department of Pathology and Molecular Medicine, School of Medicine, Queen's University, Kingston, ON K7L 3N6, Canada.
6 Laboratory of Biochemistry and Molecular Biology, National Cancer Institute, National Institutes of Health, Bethesda, MD 20892, USA. Electronic address: baiyaw@mail.nih.gov.
PMID: 33238161
PMCID: PMC7796963
DOI: 10.1016/j.molcel.2020.10.038
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Distinct Structures and Dynamics of Chromatosomes with Different Human Linker Histone Isoforms
Bing-Rui Zhou et al. Mol Cell.2021.
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. 2021 Jan 7;81(1):166-182.e6.
doi: 10.1016/j.molcel.2020.10.038. Epub 2020 Nov 24.
Authors
Bing-Rui Zhou1,Hanqiao Feng1,Seyit Kale2,Tara Fox3,Htet Khant3,Natalia de Val3,Rodolfo Ghirlando4,Anna R Panchenko5,Yawen Bai6
Affiliations
1 Laboratory of Biochemistry and Molecular Biology, National Cancer Institute, National Institutes of Health, Bethesda, MD 20892, USA.
2 Izmir Biomedicine and Genome Center, Dokuz Eylul University Health Campus, Balcova, Izmir 35330, Turkey.
3 Center of Macromolecular Microscopy, National Cancer Institute, Cancer Research Technology Program, Frederick National Laboratory for Cancer Research, Leidos Biomedical Research Inc., Frederick, MD 21701, USA.
4 Laboratory of Molecular Biology, National Institute of Diabetes, Digestive and Kidney Diseases, National Institutes of Health, Bethesda, MD 20892, USA.
5 Department of Pathology and Molecular Medicine, School of Medicine, Queen's University, Kingston, ON K7L 3N6, Canada.
6 Laboratory of Biochemistry and Molecular Biology, National Cancer Institute, National Institutes of Health, Bethesda, MD 20892, USA. Electronic address: baiyaw@mail.nih.gov.
PMID: 33238161
PMCID: PMC7796963
DOI: 10.1016/j.molcel.2020.10.038
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Abstract
The repeating structural unit of metazoan chromatin is the chromatosome, a nucleosome bound to a linker histone, H1. There are 11 human H1 isoforms with diverse cellular functions, but how they interact with the nucleosome remains elusive. Here, we determined the cryoelectron microscopy (cryo-EM) structures of chromatosomes containing 197 bp DNA and three different human H1 isoforms, respectively. The globular domains of all three H1 isoforms bound to the nucleosome dyad. However, the flanking/linker DNAs displayed substantial distinct dynamic conformations. Nuclear magnetic resonance (NMR) and H1 tail-swapping cryo-EM experiments revealed that the C-terminal tails of the H1 isoforms mainly controlled the flanking DNA orientations. We also observed partial ordering of the core histone H2A C-terminal and H3 N-terminal tails in the chromatosomes. Our results provide insights into the structures and dynamics of the chromatosomes and have implications for the structure and function of chromatin.
Keywords: Cryo-EM; NMR; chromatin structure; chromatosome; chromatosome dynamics; chromatosome structure; linker histone isoform; linker histone tail; nucleosome; single-chain antibody.
Published by Elsevier Inc.
PubMed Disclaimer
Conflict of interest statement
Declaration of Interests The authors declare no competing interests.
Figures
Figure 1.. Overall Density Maps and Structural…
Figure 1.. Overall Density Maps and Structural Models
(A) Top views of the cryo-EM reconstructions…
Figure 1.. Overall Density Maps and Structural Models
(A) Top views of the cryo-EM reconstructions of the H1.0, H1.4, and H1.10 chromatosomes and the free nucleosome. (B) Top views of the low-pass-filtered (6 Å) density maps (transparent surfaces) as in (A) and corresponding atomic structural models. (C) Side views as in (A). (D) Side views as in (B). See also Figures S1 and S2.
Figure 2.. Interactions between the Globular Domains…
Figure 2.. Interactions between the Globular Domains and DNA
(A) Sequences of the globular domains…
Figure 2.. Interactions between the Globular Domains and DNA
(A) Sequences of the globular domains of H1.0, H1.4, H1.10, and linker DNA. The diagram on the top shows the secondary structures. The residue Gln47 in H1.0 and corresponding Ala residues in H1.4 and H1.10, which are highlighted, interact with dyad DNA. (B) Density maps (transparent surfaces) and cartoon structure models of the globular domains. (C) Densities of the amino acid side chains of the globular domains that interact with the nucleosomal and linker DNAs. (D) Illustration of the difference of the orientation of the globular domain of H1.0 relative to those of H1.4 and H1.10 in the chromatosomes and the residues that are likely responsible for the difference. Structures of the chromatosomes were aligned on core histones in the top panel. The bottom panel showed the alignment of the globular domain structures alone. (E) Illustration of DNA orientation determination in the H1.4 chromatosome by the fitting of cryo-EM densities with DC-95:DG-103 and DC-99:DG-99 pairs in one direction (carbon colored in green), which do not fit when the orientation of the DNA is reversed (carbon colored in gold) (left). In contrast, in the case of free nucleosome, the corresponding cryo-EM densities represent the average of the two positions from opposite DNA orientations (right). (F) The AT-rich base pair region (pink color) in the linker/flanking DNA is bound by the α3 helix through residues Arg78 of H1.4. See also Figures S1 and S2.
Figure 3.. MD of the Chromatosomes
(A)…
Figure 3.. MD of the Chromatosomes
(A) Flexibility of DNA in the H1.0 (red), H1.4…
Figure 3.. MD of the Chromatosomes
(A) Flexibility of DNA in the H1.0 (red), H1.4 (blue), H1.10 (green) chromatosomes, and free nucleosome (black) were illustrated via phosphate atom RMSFs over full molecular dynamics (MD) trajectories. Base pairs that contact globular domains are shown by DNA region: linker DNA 1 (LD1)/linker-L1 (magenta), dyad (orange), and linker DNA 2 (LD2)/linker-α3 (cyan). All panels follow this color convention. The dyad is numbered as 0. (B) Mean and standard deviations of DNA RMSF averaged over base pairs. (C) Distribution of distances between the two terminal base pairs of DNA. (D) Cartoon illustrating the definition of linker strand opening angles, α LD1 and α LD2 for in-nucleosomal-plane motion (top), and β LD1 and β LD2 for out-of-nucleosomal-plane motion. (E) Distributions of differences of linker strand angles, α and β for LD1/linker-L1. Higher values point to more compact structures. (F) Distributions of differences of linker strand angles, α and β for LD2/ linker-α3. Higher values point to more compact structures (G) The number of heavy-atom contacts between the globular domain and DNA averaged over MD trajectories. Residues conserved across all three H1 variants are annotated with red asterisks. Residues in contacts with LD1/linker-L1, dyad region, and LD2/linker-α3 are shown in purple, orange, and cyan, respectively. (H) Average numbers of heavy-atom contacts for three DNA regions. See also Figure S3.
Figure 4.. Multiple Conformations and Interactions of…
Figure 4.. Multiple Conformations and Interactions of Linker DNA
(A) 3D classification of the H1.0…
Figure 4.. Multiple Conformations and Interactions of Linker DNA
(A) 3D classification of the H1.0 chromatosome. The blue circle highlights the observation of densities between the two linker DNA. (B) 3D classification of the H1.4 chromatosome. The blue circles highlight the observation of densities between the two linker DNA. (C) The major class of the H1.10 chromatosome. (D) 3D classification of the free nucleosome. See also Figure S4.
Figure 5.. Linker Histone Tails Control Linker…
Figure 5.. Linker Histone Tails Control Linker DNA Orientation
(A) Amino acid sequence alignment of…
Figure 5.. Linker Histone Tails Control Linker DNA Orientation
(A) Amino acid sequence alignment of the C-terminal tails of H1.0, H1.4, and H1.10. The S/TPKK motifs are highlighted in red with a yellow background. (B) 1 H-15 N HSQC spectra of the H1 isoforms in the chromatosomes and the free form. (C) Deviation of Cα chemical shifts from random coil values (ΔΔCα). (D) Cryo-EM reconstruction of the gH1.10-ncH1.4 chromatosome. Low-pass-filtered (6 Å) cryo-EM density is shown with a transparent gray surface. (E) Cryo-EM density fitted with the globular domain structural model. Density of the amino acid side chains of the globular domain that interact with the DNA. (F) Cartoons of chromatosomes showing differences in linker DNA and H1 tails. See also Figure S5.
Figure 6.. Effects of Linker Histone Binding…
Figure 6.. Effects of Linker Histone Binding on the Conformation of H3 Tails
(A) Comparison…
Figure 6.. Effects of Linker Histone Binding on the Conformation of H3 Tails
(A) Comparison of cryo-EM densities of the free nucleosome and the H1.4 chromatosome. The 147-bp nucleosome density map is from EMD-8938. All maps were low-pass filtered to 6 Å. The density maps were plotted at same intensity levels for core histones. H3 model and densities are zone colored in light blue. Extra densities between the two DNA gyres are colored in light blue, which is connected the H3 N-helix density. Numbers in the circles show the super-helical locations. (B) 1 H-15 N spectra of H3 tails in the chromatosomes and free nucleosome. (C) Chemical shift perturbations (upper panel) and NMR peak intensity changes (lower panel) for the residues in the H3 N-tails upon addition of H1. See also Figure S6.
Figure 7.. Effects of Linker Histone Binding…
Figure 7.. Effects of Linker Histone Binding on the Conformation of H2A Tails
(A) Comparison…
Figure 7.. Effects of Linker Histone Binding on the Conformation of H2A Tails
(A) Comparison of the cryo-EM densities of the free nucleosome and the H1.4 chromatosome. All maps are low-pass filtered to 6 Å. The density maps were plotted at the same intensity levels for core histones. H2A model and density are zone colored in orange. Extra density is also colored in orange, which is closed to one of H2A C-terminal structured region. (B) 1 H-15 N spectra of H2A tails in the chromatosomes and free nucleosome. The dashed lines separate the N-terminal (left) and C-terminal (right) tail regions. (C) Chemical shift perturbations (upper panel) and NMR peak intensity changes (lower panel) for the residues in the H2A tails upon addition of H1. See also Figure S6.
All figures (7)
See this image and copyright information in PMC
References
Adams PD, Afonine PV, Bunkóczi G, Chen VB, Davis IW, Echols N, Headd JJ, Hung LW, Kapral GJ, Grosse-Kunstleve RW, et al. (2010). PHENIX: a comprehensive Python-based system for macromolecular structure solution. Acta Crystallogr. D Biol. Crystallogr 66, 213–221. - PMC - PubMed
Allan J, Hartman PG, Crane-Robinson C, and Aviles FX (1980). The structure of histone H1 and its location in chromatin. Nature 288, 675–679. - PubMed
Allan J, Mitchell T, Harborne N, Bohm L, and Crane-Robinson C (1986). Roles of H1 domains in determining higher order chromatin structure and H1 location. J. Mol. Biol 187, 591–601. - PubMed
Arents G, Burlingame RW, Wang BC, Love WE, and Moudrianakis EN (1991). The nucleosomal core histone octamer at 3.1 A resolution: a tripartite protein assembly and a left-handed superhelix. Proc. Natl. Acad. Sci. USA 88, 10148–10152. - PMC - PubMed
Arimura Y, Ikura M, Fujita R, Noda M, Kobayashi W, Horikoshi N, Sun J, Shi L, Kusakabe M, Harata M, et al. (2018). Cancer-associated mutations of histones H2B, H3.1 and H2A.Z.1 affect the structure and stability of the nucleosome. Nucleic Acids Res. 46, 10007–10018. - PMC - PubMed
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10137 | https://lawnlinguistics.com/2018/12/16/corpora-and-the-second-amendment-bear/ | Corpora and the Second Amendment: “bear” | LAWnLinguistics
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Corpora and the Second Amendment
← Corpora and the Second Amendment: “keep” (part 2)
Corpora and the Second Amendment: “arms” →
Corpora and the Second Amendment: “bear”
Posted onDecember 16, 2018|3 Comments
An introduction and guide to my series of posts “Corpora and the Second Amendment” is available here. The corpus data that is discussed can be downloaded here. That link will take you to a shared folder in Dropbox. Important: Use the “Download” button at the top right of the screen.
Starting with this post, I’m (finally) getting to the meat of what I’ve called “the coming corpus-based reexamination of the Second Amendment.” The plan, as I’ve said before, is to more or less mirror the structure of the Supreme Court’s analysis of keep and bear arms. This post will focus on bear, and subsequent posts will focus separately on arms, bear arms, and keep and bear arms. (I won’t be discussing keep arms because I have nothing to say about it.)
In discussing the meaning of the verb bear, Justice Scalia’s majority opinion in District of Columbia v. Heller said, “At the time of the founding, as now, to ‘bear’ meant to ‘carry.’’’ That statement was backed up by citations to distinguished lexicographic authority—Samuel Johnson, Noah Webster, Thomas Sheridan, and the OED—but evidence that was not readily available when Heller was decided shows that Scalia’s statement was very much an oversimplification. Although bear was sometimes used in the way that Scalia described, it was not synonymous with carry and its overall pattern of use was quite different.
The new evidence I’m referring to comes from two sources. One is the book Borrowed Words: A History of Loanwords in English by Philip Durkin. Durkin is one of the two Deputy Chief Editors of the OED, the OED’s principal etymologyist, the author of the Oxford Guide to Etymology, and the editor of the Oxford Handbook of Lexicography. Borrowed Words wasn’t published until after Heller was decided, so the information I’m about to discuss wasn’t available to Scalia.
One of the things Durkin does in Borrowed Words is to try to determine the point when bear was replaced by carry as the verb primarily used in English to express the meaning that Scalia is talking about. Durkin reports that carry came into Middle English (from French) no later than about 1375. And although he notes that it is “very difficult to say when it became more common than bear in the sense of ‘to bear by bodily effort,’” he finds it “very likely that carry was the basic word in this meaning by the seventeenth century (at least in the ancestor of modern standard English).”
The other source of evidence that bear was not synonymous with carry is the data from the BYU Law Corpora, which very much bears Durkin’s conclusion out. When Justice Scalia said in Heller that in 1789 bear meant carry, he was talking about the carrying of tangible objects. But although the corpus data provides extensive evidence of carry being used to denote that kind of carrying, there was much less evidence of bear being used that way. Overall, the differences in how the two words were used greatly outnumbered the similarities.
Although my discussion of keep was based entirely on data from COFEA (the Corpus of Founding Era American English), starting with this post I will be relying on both COFEA and COEME (the Corpus of Early Modern English). Because COEME includes texts from a period of more than 300 years, I’ve limited my searches in COEME to the period 1760-1799, which is the period covered by COFEA. For more information about the data, and in particular about the issue of duplication within each corpus and between the two corpora, see the Introduction pages of the spreadsheets in which the data is set out, which can be downloadedhere.
(carry or bear)+ nouns denoting tangible objects. Since what we’re concerned with is the carrying of tangible objects, I started by trying to find out which nouns acted most frequently as each verb’s direct object—a task that I carried out by running searches that would return a list of the nouns that occurred within two words to the right of each of the verbs. Looking for nouns within a two-word span after the verb made it possible to find not only plurals and noncount nouns, which can appear immediately to the right of the verb (carry objects, carry stuff), but also count nouns, which can’t (carry an object but not carry object (the asterisk before “carry object” marks it as ungrammatical)). I initially limited the search to the 100 most frequent collocates, but later expanded to the top 200.
With respect to carry, many of the nouns in the search results—including many that were at the top of the list in terms of frequency—denoted tangible objects. These included arms, gun(s),provisions,goods,letters, baggage,supplies,mail,boards,load(s),corn,flag, torches,coal,prizes, packs,basket,bag,parcels,flour,timber,pistols,luggage,bundle,firearms,banners, and bedding.
For bear, on the other hand, nouns like those were much less in evidence. Of the few that I found, the most frequent one by far was arms, which I’ll defer discussion of until the post that deals specifically with bear arms. The other nouns in this category fall into two groups. In the first group, which consisted mainly of torches,flags,prize, cross, and sword, the nouns were used in phrases denoting the kind of physical carrying that Justice Scalia had in mind. The most frequent of these uses was (perhaps unsurprisingly) bear + cross—primarily referring to Christ, but also used metaphorically.
[Update: I want to clarify the statement that the most frequent tangible-object noun used with bear was arms. As discussed in a later post, arms was often used to mean ‘weapons,’ but it was also frequently used to convey a variety of figurative meanings, and when used that way it did not denote tangible objects. Therefore, when I refer in this post to arms as denoting a kind of tangible object (namely, ‘weapons’), that should be understood to mean only that arms was often used that way, not that it was always used that way. In particular, it should not be understood as as a statement about how arms was used when it was part of the phrase bear arms.]
The second group consisted of infrequently-occurring nouns having to do with agriculture and horticulture, such as fruit, wheat, wood (as part of a living plant or tree), acorns, corn, and so on. As far as I could tell, none of those nouns occurred in phrases denoting the same kind of carrying. Rather, the nouns appeared in phrases in which bear was used primarily in the sense of ‘bringing forth,’ as in land bearing wheat, and plants bearing fruit, corn, and acorns. (Bear fruit was also used metaphorically to mean essentially ‘yield results.’) And in a few instances, such as the one below, bearing wood was used as a noun phrase denoting the fruit-bearing portion of a tree or plant.
The mode of bearing in peaches, nectarines, and apricots, is on the last year’s wood , which makes it necessary to shorten, in order to a certain supply of bearing wood for the next year; and to contrive to have succession wood in every part of the tree, is one of the chief arts of the pruner.
Another difference between carry and bear was that the data for carry—but not for bear— included uses in which it was various categories of human beings that were described as being carried from one place to another. While we don’t usually think of humans as objects, that’s only because doing so would feel dehumanizing. So it’s certainly relevant to the current inquiry that the corpus data included statements in which the words captives, passengers, and negroes were used as direct objects of carry, as they did in these examples:
When a corsair takes a prize he carries the captives directly to the palace of the dey, where the European consuls assemble
a heavy penalty upon any of their citizens who ſhall be engaged in the buſineſs of carrying Negroes from Africa to any part of the world whatever
Here We saw two or 3 Passage Waggons—a Vehicle with four Wheels contrived to carry many Passengers and much Baggage.
I have ingaged Capt. Saml. Crozier a very sober usefull Man & who has been in the habit of carrying Passengers to conduct you to Lisbon & he will tomorrow commence on putting the Schooner in compleat order for your reception
Note that in some of these uses, carry was used in a sense that is now largely obsolete, in which it means to take, accompany, or escort someone or something somewhere.
There were also concordance lines, such as the following, that followed a slightly different grammatical pattern, but that nevertheless referred to the carrying of people:
On the 13th of November, a Sweedish ship sailed from Marscilles, bound to Philadelphia, and carried as passengers all the Americans, (late prisoners in Algiers) except those who tarried on board the Fortune.
They came by water in 30 canoes, slew seven Indians and wounded two Sagamores who lived near Boston, and carried away captives one of their wives with divers other Indians
The officers and principal inhabitants among the Swedes, were carried prisoners to New Amsterdam; and thence to Holland
There were no results for bear that fit either of these patterns. Although bear did occur with direct objects denoting various categories of people, in all of those cases the verb was used in a different sense. These nouns included children, babes, sons,infants, citizens, and subjects, and they were modified by phrases such as new born, first born, free born, and natural born.
This isn’t to say that bear was never used to refer to the transportation of human beings from one place to another. There was, for example, this bit of verse:
She, deemed Circassian, on her infant day, Was from thy native frontier borne away.
But from what I saw in the data, examples like this were few and far between.
Among the most frequent noun collocates of carry were home, trade, and places. These collocations differed from those I’ve been discussing, in that the nouns did not act as the direct object of carry, but the phrases in question nevertheless had to do with the carrying of tangible objects (including people).
Home was used to denote the place where something or someone was carried to:
We should think that no cotton can be carried home from Bombay this year, and even the year following.
Our captain, you may be sure, was in no haste to carry her home, being fallen most desperately in love with her
Trade occurred in the noun phrase the carrying trade, which the OED describes as denoting “the trade or business of carrying goods, esp. over sea between different countries.” Similarly, places occurred in the noun phrase carrying places, which denoted “a place where goods, etc. have to be carried overland in inland navigation” (OED). I saw nothing comparable in the data for bear.
Also found in the data for carry were uses in which the literal, physical sense of the word was metaphorically extended to include the carrying of information, as in carry messages, carry intelligence, carry dispatches, carry tidings, carry news, and carry complaints. Once again, I found no analogous uses of bear.
(carry or bear)+ (burden(s) or weight).Perhaps the most significant overlap between the results for carry and those for bear(putting aside carry/bear arms) is seen in the pairs carry+burden(s)&bear+burden(s), on the one hand, and carry+weight&bear+weight on the other. Both of the nouns in these pairs can be used literally, to denote physical burdens and weights, and carrying or bearing a physical burden or weight necessarily amounts to carrying or bearing a tangible object. But both words can also be used metaphorically, and to speak of “carrying” or “bearing” a metaphorical burden amounts to using the verb metaphorically as well. It therefore makes sense to compare the pattern of usage associated with carry+burden(s) against the pattern associated with bear+burden(s), and to similarly compare the pattern associated with carry+weight against the one associated with bear+weight.
In each case, the uses were divided between literal and metaphorical. But there were some significant differences between the patterns of usage for each word—differences that were both quantitative and qualitative.
The biggest quantitative difference was seen in the data for burden(s), with the literal use being 100 times as frequent for carry burden(s) as for bear burden(s). The literal-to-metaphoric ratio for carry was roughly 10:1, while for bear it was the inverse, 1:10.
(These figures are based on the combined results from both COFEA and COEME, after eliminating duplicates. I should note that there’s roughly four times as much data for bear burden(s) as there is for carry burden(s), and that the total dataset for carry burdens is fairly small in absolute terms. There are 349 concordance lines for bear, and 85 for carry. Therefore, the data for bear burden(s) is presumably more reliable than the data for carry burden(s). But even so, a ~~10,000%~~ 1,000% difference in frequency seems pretty substantial.)
However, when we turn to bear weight and carry weight, the data paints a more complicated picture. The ratios of literal uses to metaphoric uses for these two phrases doesn’t show anything like the disparity that we saw with respect to bear burden(s) and carry burden(s). For bear weight the ratio was roughly 0.8:1, while for carry weight it was about 1.2 : 1. Thus, literal uses were only 1.5 times as frequent for carry as they were for bear.
When viewed in isolation, this could be seen as evidence that bear weight was used fairly often to denote the kind of physical carrying that the Supreme Court in Heller had in mind. But when the data is examined more closely, it points toward a different conclusion.
Here’s why. So far, I’ve been looking only at how the words burden and weight were used in each concordance line—whether they were used literally, to denote a physical burden or weight, or instead metaphorically. To the extent that the burden or weight was metaphorical rather than literal, the act of bearing or carrying the weight must similarly be metaphorical. That by itself means that the act couldn’t have constituted “carrying” in the sense relevant here. But that leaves open the question whether the acts denoted by literal uses of bear weight similarly constituted the relevant kind of carrying. And the data suggests that for the most part, they did not.
In most of the instances of bear weight in which weight was used literally, bear was used in a way that denoted what I’ll call “immobile support”—the weight was physically supported, without any horizontal movement, as in these examples:
If the board were one inch thick, and 12 inches deep, and if it were capable of bearing a weight of 12 when the pressure was applied perpendicular to the thin way of the board,
into the woods, by walking along shore upon the ice, which still covered the sea, and had strength sufficient to bear any weight.
on which grow a great number of oak trees, almost all the branches of which, able to bear the weight, are, in the proper season of the year, loaded with eagle’s nests
myself to clamber out at the top, which I had nearly reached when a stone on which my whole weight bore, dislodg’d and down falls the stone, and myself to the bottom again.
In contrast, most of the examples of carry weight highlighted the movement of the weighty object from one place to another, with the supporting of the object’s weight being a part of that action rather than the focus of independent attention. For example:
The camels usually carry 800lb. weight upon their backs, which is not taken off during the whole journey, for they naturally kneel
of the baggage and flour to be on pack horses as most convenient but horse Waggons might be had to carry about 1800 Weight and roads for them easily made,
from place to place by Elephants, and if their necks and trunks have no more room for burthens, they will carry an additional weight in their mouth
It is proposed the frigates of thirty six guns shall carry the same weight of metal of the forty four gun ships, but only to have twenty eight guns upon one
Out of the 60 literal uses of bear weight, in only 4 of them (6.7% of the total) was bear used in this sense of ‘carry.’ Since there were 70 metaphoric uses, those uses were more than 17 times as frequent as the uses that denoted carrying. And while that’s only about one-sixth of the difference in frequency that is seen with respect to bear burden(s) and carry burden(s), it is a big difference from what the raw numbers seem to suggest.
We can see another interesting difference between between bear weight and carry weight when we compare the metaphoric uses of each phrase. In almost all the instances of bear weight, the phrase denotes being metaphorically burdened or weighed down by something, as in these examples:
thus the queen discovers herself to be the firm support of the state, when, after having for a long time borne its weight, she is not even bowed down under its fall.
I’ll support thee; for in addition to the oppression of our common grief, thou, sweet girl, must bear the agonizing weight of disappointed love.
Her father in stern accents replied, do you think Madam, you are so well able to bear the weight of mine anger?
only the little finger of the Episcopalian hierarchy, we should think the burthen comparatively light, though we were called to bear the weight of the loins of the Presbyterians of New-England.
Note the semantic similarity between these examples and the examples of literal uses of bear weight, which I described as denoting “immobile support.” What we see in the metaphoric uses is something that is closely analogous to that. (Which is not surprising, given the parallels between metaphor and analogy.) And a bit later we’ll look at some additional uses of bear whose semantics are similar.
Just as most of the metaphoric uses of bear weight are semantically similar to its literal uses, the metaphoric uses of carry weight are semantically similar to its literal uses. And by the same token, they differ semantically from the metaphoric uses of bear weight in much the same way that the literal uses of carry weight differ from the literal uses of bear weight.
Examples typifying the metaphoric uses of carry weight include these:
I own that in my mind your observations upon that point carry much weight
his Objection, at first Look, seems to carry Weight with it
Let us not look on it as superstitious to suppose that such wishes may still carry weight with them
You will consider yourselves as the leaders of this people; that your example in your public department will ever carry much weight and energy in it
Whereas to bear weight (whether literally or metaphorically) is to be subjected to the force of the weight, these examples show that to metaphorically carry weight is to convey or direct the metaphoric force of that weight onto someone or something else. This parallels what we saw in the literal uses of carry weight, which denoted the movement of the weight from one place to another. And it similarly resembles what we saw in a different set of metaphorical uses of carry, namely those having to do with conveying information: carry messages, carry intelligence, carry dispatches, carry tidings, carry news, and carry complaints. In fact, the metaphoric uses of carry weight can be regarded as an abstract extension of those comparatively more concrete metaphoric uses: rather than denoting the conveyance of specific information or types of information, they denote the conveyance of the persuasive force of information or communication, or of communicatively significant actions or institutions.
So to sum up this part of the discussion: What we can see in the data on bear/carry + burden(s)/weight are patterns of usage along two dimensions. The dimension that’s of primary interest here is the one along which the usage of each word contrasts with that of the other. And even though we’ve been looking at only two nouns, we’ve seen several different kinds of contrast.
The second dimension is the dimension along which we can look at the use of each verb separately, comparing the differences and similarities in how it is used, depending on which noun it combines with. With respect to each verb, there has been an underlying semantic commonality having to do with the flow of energy between the participants in a situation. To carry something, whether a physical object or an abstract entity, is to direct physical or abstract force to the object or entity, and thereby out into the world. To bear something, on the other hand, is to experience force that is imposed by the object. In other words, carrying is a form of acting on something, while bearing is a form of being acted upon.
This isn’t to say that these patterns are followed in every use of bear and carry. They aren’t. But the uses that don’t fit the patterns are merely exceptions; they aren’t so widespread as to prevent the patterns from emerging in the first place.
bear + other nouns
Apart from talking about bear burden(s) and bear weight, I’ve haven’t said much about how bear is used in the corpus data; I’ve talked more about the fact that it hasn’t been used in the ways that we’ve seen carry be used. So I want to talk briefly about what the data for bear does show.
The data can be broken down into several categories.
Language from legal documents. Most of the uses in this category constitute legal boilerplate: bear date, bear name, bear witness, bear testimony, bear test(e). The one use that’s not part of a purely formulaic recital is bear interest. These uses have an abstract similarity to some uses of carry, and in several of them it would be possible to replace the bear with carry with little loss of meaning. Even then, however, bear is used in an abstract sense that doesn’t denote physical carrying.
Birth-related. This category includes uses such as new born babes, born a citizen, natural born citizen, and born in Ireland. There is also an agricultural variation on this theme: bear fruit(in the literal sense), bear clusters of berries, and so on. And also bear fruit in the metaphoric sense.
Compass bearings. This category consists of uses such as At noon Cape Egmont bore about N.E. and in this direction, at about four leagues from the shore, we had forty fathom of water.
Relationship. In the uses in this group, bear is part of an expression asserting that something has a specified relation to something else. The basic template is X bears analogy (resemblance, proportion,/comparison) to Y.
Withstanding trouble and woe. This category is all about suffering through bad conditions and events: bear adversity, bear affliction, bear drought, bear famine, fatigue, hardships, hunger ill-will injuries insults misfortunes reproof trouble….
Is it not composed of a mixture of people from different countries; some more, some less capable of bearing fatigue and hardship?
Prosperity is the touchstone of virtue; for it is less difficult to bear misfortunes than to remain uncorrupted by pleasure.
God , thou knowest that I have borne my sufferings meekly: I have wept for myself, but never cursed my foes
friend, be you assured, for your consolation, that we the American captives, in this city of bondage, will bear our sufferings with fortitude and resignation
These examples may have a familiar ring to them, because they all instantiate the semantic pattern in which bearing X means essentially ‘experiencing X’ or ‘undergoing X.’ And with that in mind, it may not be a surprise when I tell you that there is a corresponding set of uses of carry in which the verb’s direct object similarly denotes bad stuff, but in which the direction of causation is reversed, such as carry terror, carry desolation, carry conquest, carry havoc, and carry devastation. To carry these bad things is not to experience them but to inflict them on others:
the arms of Scanderbeg and the scimitars of his illustrious soldiers, who were an army of heroes, carried death at every stroke, and gave decisive victory
remember, that Great-Britain, during last war, did at one time carry conquest through every quarter of the globe
Ye who worship the mammon of unrighteousness, and sacrifice nations for gain; who have carried desolation to the utmost bounds of the earth
the French partisans at Avignon and citizens of the neighbouring departments, have run over the Comtat, carried devastation with them wherever they met opposition and forced the inhabitants to declare in favor of France
And in conclusion…
Having gone through the corpus data, it seems to me indisputable that at the time of the framing of the Constitution, bear was in general not synonymous with carry. Although it was sometimes used to denote the kind of physical carrying that the Supreme Court in Heller had in mind, those uses were infrequent and were exceptions to the general pattern of usage.
Thus, when Justice Scalia wrote that “at the time of the founding, as now, to ‘bear’ meant to ‘carry,’’’ his statement had a kernel of truth but didn’t accurately reflect how the word was ordinarily used.
However, that doesn’t resolve the question of how bear arms as used in the Second Amendment was likely to have been understood by the American public of 1789. As I’ve said, arms was one of the most common nouns that acted as the direct object of bear. That means that bear arms is potentially the biggest exception to the general rule that bear didn’t mean carry. I haven’t addressed that issue in this post, and am instead saving it for a post of its own. But before I get there, I’m going to talk about arms. Because it turns out that Heller’s short treatment of that word—concluding simply that it meant ‘weapons’—was at least as much of an oversimplification as its discussion of bear.
Further reading.If you’re interested in learning more about the semantics of bear, you’ll find a detailed discussion, based on an informal corpus of texts from the 1960s and 1970s, in On Monosemy: A Study in Linguistic Semantics, by Charles Ruhl (publisher, Google Books, Amazon).
Hat tip to Dennis Baron, who was the first person to look up bear arms in the COFEA and COEME; see his comment to my Language Log post announcing the opening of the two corpora for beta testing (scroll to the bottom), and his subsequent op-ed in the Washington Post. It was Baron’s op-ed that inspired me to undertake this look at the Second Amendment.
Cross-posted on Language Log.
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The coming corpus-based reexamination of the Second AmendmentMay 28, 2018 In ""bear arms""
Corpora and the Second Amendment: Preliminaries and caveatsJune 7, 2018 In ""bear arms""
“bear arms” in the BYU Law corporaMay 8, 2018 In ""bear arms""
This entry was posted in "bear", Corpus linguistics & constitutional interpretation, District of Columbia v. Heller, Law and corpus linguistics, Law and linguistics, Scalia, Second Amendment. Bookmark the permalink.
← Corpora and the Second Amendment: “keep” (part 2)
Corpora and the Second Amendment: “arms” →
3 responses to “Corpora and the Second Amendment: “bear””
DENNIS BARON | December 16, 2018 at 5:35 pm | Neal, 4 out of 60 is 6.7%, not 15%
weboflanguage | December 16, 2018 at 5:36 pm | Neal, 4 tokens out of 60 is indeed 1/15, but that’s 6.7%, not 15%
Neal Goldfarb | December 16, 2018 at 9:28 pm | In the immortal words of Barbie, math is hard. But thank you for the heads-up; my boneheaded error has been corrected.
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10138 | https://www.effortlessmath.com/math-topics/how-to-understand-co-function-even-odd-and-periodicity-identities-in-trigonometry/?srsltid=AfmBOopuo39bfSZFTo5YQ_zTG7PlyNm0ZLzRSGioagVf1xKsgDlEB4Iz | How to Understand Co-Function, Even-Odd, and Periodicity Identities in Trigonometry - Effortless Math: We Help Students Learn to LOVE Mathematics
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How to Understand Co-Function, Even-Odd, and Periodicity Identities in Trigonometry
Trigonometry, a branch of mathematics that deals with the relationships between the sides and angles of triangles, holds profound significance in our daily lives. Among its varied concepts, the Co-Function, Even-Odd, and Periodicity Identities stand as crucial pillars. These concepts are key to comprehending the depth of trigonometric functions, their relationships, and their behavior in mathematics. In this article, we aim to simplify and explain these identities in detail.
1. The Intricacies of Co-Function Identities
In trigonometry, two functions are said to be co-functions if their values complement each other for complementary angles. More simply put, an angle and its complement (the amount needed to make it 90°90°) have the same trigonometric values, albeit associated with different functions.
The co-function identities can be represented as:
𝑆 𝑖 𝑛(90°–θ)=𝑐 𝑜 𝑠(θ)S i n(90°–θ)=c o s(θ)
𝐶 𝑜 𝑠(90°–θ)=𝑠 𝑖 𝑛(θ)C o s(90°–θ)=s i n(θ)
𝑇 𝑎 𝑛(90°–θ)=𝑐 𝑜 𝑡(θ)T a n(90°–θ)=c o t(θ)
𝐶 𝑜 𝑡(90°–θ)=𝑡 𝑎 𝑛(θ)C o t(90°–θ)=t a n(θ)
𝑆 𝑒 𝑐(90°–θ)=𝑐 𝑠 𝑐(θ)S e c(90°–θ)=c s c(θ)
𝐶 𝑠 𝑐(90°–θ)=𝑠 𝑒 𝑐(θ)C s c(90°–θ)=s e c(θ)
These identities reveal a profound symmetry in the structure of trigonometric functions and are key to solving various mathematical problems.
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2. Grasping the Concept of Even and Odd Identities
In mathematics, functions are classified as even, odd, or neither. This classification is based on their symmetry about the 𝑦 y-axis (even) or origin (odd). Trigonometric functions also abide by this classification.
An even function satisfies 𝑓(𝑥)=𝑓(−𝑥)f(x)=f(−x). In trigonometry, 𝑐 𝑜 𝑠 𝑖 𝑛 𝑒(𝑐 𝑜 𝑠)c o s i n e(c o s) and 𝑠 𝑒 𝑐 𝑎 𝑛 𝑡(𝑠 𝑒 𝑐)s e c a n t(s e c) are even functions.
An odd function, on the other hand, satisfies 𝑓(𝑥)=−𝑓(−𝑥)f(x)=−f(−x). 𝑆 𝑖 𝑛 𝑒(𝑠 𝑖 𝑛)S i n e(s i n), 𝑡 𝑎 𝑛 𝑔 𝑒 𝑛 𝑡(𝑡 𝑎 𝑛)t a n g e n t(t a n), 𝑐 𝑜 𝑠 𝑒 𝑐 𝑎 𝑛 𝑡(𝑐 𝑠 𝑐)c o s e c a n t(c s c), and 𝑐 𝑜 𝑡 𝑎 𝑛 𝑔 𝑒 𝑛 𝑡(𝑐 𝑜 𝑡)c o t a n g e n t(c o t) are examples of odd functions in trigonometry.
Understanding these identities assists in simplifying trigonometric expressions and solving equations.
3. Understanding the Significance of Periodicity Identities
In trigonometry, the concept of periodicity refers to the behavior of trigonometric functions over specific intervals, after which the functions repeat their values. This ‘interval’ is known as the period of the function.
The primary trigonometric functions – 𝑠 𝑖 𝑛 𝑒,𝑐 𝑜 𝑠 𝑖 𝑛 𝑒,𝑠 𝑒 𝑐 𝑎 𝑛 𝑡,s i n e,c o s i n e,s e c a n t, and 𝑐 𝑜 𝑠 𝑒 𝑐 𝑎 𝑛 𝑡 c o s e c a n t – have a period of 2 π 2 π, while 𝑡 𝑎 𝑛 𝑔 𝑒 𝑛 𝑡 t a n g e n t and 𝑐 𝑜 𝑡 𝑎 𝑛 𝑔 𝑒 𝑛 𝑡 c o t a n g e n t have a period of π π. These periodic behaviors are represented as follows:
𝑆 𝑖 𝑛(𝑥+2 π)=𝑠 𝑖 𝑛(𝑥)S i n(x+2 π)=s i n(x)
𝐶 𝑜 𝑠(𝑥+2 π)=𝑐 𝑜 𝑠(𝑥)C o s(x+2 π)=c o s(x)
𝑇 𝑎 𝑛(𝑥+π)=𝑡 𝑎 𝑛(𝑥)T a n(x+π)=t a n(x)
𝐶 𝑜 𝑡(𝑥+π)=𝑐 𝑜 𝑡(𝑥)C o t(x+π)=c o t(x)
𝑆 𝑒 𝑐(𝑥+2 π)=𝑠 𝑒 𝑐(𝑥)S e c(x+2 π)=s e c(x)
𝐶 𝑠 𝑐(𝑥+2 π)=𝑐 𝑠 𝑐(𝑥)C s c(x+2 π)=c s c(x)
Mastering the concept of periodicity identities is crucial for mathematical analysis and predicting patterns in a wide array of scientific fields, including physics and engineering.
4. Leveraging These Identities in Real-World Applications
The principles of co-function, even-odd, and periodicity identities are not just abstract mathematical concepts. They have wide-reaching applications in various fields such as physics, engineering, computer science, and even in economics and biology.
For example, periodicity identities are vital in understanding wave behaviors in physics, while the concept of even and odd functions helps computer scientists optimize algorithms. Co-function identities, meanwhile, are extensively used in geometry and navigation.
By understanding and applying these identities, we can simplify complex mathematical problems, make precise calculations, and foster advancements in a myriad of scientific disciplines.
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PH S 1150: Basic Astronomy
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Sat, 21 Aug 2021 22:00:03 GMT
7.2: Force, Mass, and Weight
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Contents
1. Home
2. Campus Bookshelves
3. Chicago State University
4. PH S 1150: Basic Astronomy
5. 7: Classical Physics- Gravity and Energy
6. 7.2: Force, Mass, and Weight
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PH S 1150: Basic Astronomy
Front Matter
0: Review of Mathematics
1: Size and Scope
2: Light
3: Telescopes
4: Moving Through Space
5: Moving Through Time
6: Measuring Cosmic Distances
7: Classical Physics- Gravity and Energy
8: Dark Matter
9: Special Relativity
10: General Relativity
11: Black Holes
12: Gravitational Lenses
13: The Expansion of the Universe
14: The Growth of Structure
15: The Cosmic Microwave Background
16: The Early Universe
17: Dark Energy and the Fate of the Universe
18: Knowls
Back Matter
7.2: Force, Mass, and Weight
Last updated Aug 21, 2021
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7.1: Gravity on Earth
7.3: Gravity Is a Universal Force
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Page ID 47964
Kim Coble, Kevin McLin, & Lynn Cominsky
San Francisco State University, Chico State University, & Sonoma State University
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Mass and Weight
2. Net Force on an Object
1. Questions
MASS AND ACCELERATION
Questions
ACCELERATION AND MASS
Questions
Definition: Newton’s first law of motion
Definition: Newton’s second law of motion
Going Further 7.1: Newton’s Third Law
GOING FURTHER: GRAVITY AND INERTIA
Gravitational Force on a Mass
Questions
Lunar Lander
Mass and Weight
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To understand gravity, we will have to learn a little bit more about what causes changes in the motion of objects. Clearly, gravity causes such changes, but it is not the only effect that can do so. For instance, we can push on a cart, or pull on it with a cord. In either case, the cart’s motion can change. Such a pull or push is called a force by physicists. In physics, force has a very precise mathematical definition:
F→tot=ma→
In this expression, F is force, m is mass, and a is acceleration, and the equation says that total force is the product of mass and acceleration.
There are several things that are important to understand about this equation. The first is the little arrows above the F→ and the a→. They are there to remind us that both force and acceleration have a direction associated with them. Pushing on the back bumper of a car has very different results than pushing on the front bumper - one makes the car accelerate forward, the other makes it accelerate backwards. Quantities that have a size and a direction are called vectors , and the arrows remind us that both force and acceleration are vectors. We will often write the equation without the arrows:
F=m a.
In this case, it means we are not concerned with the direction of the force and acceleration, or sometimes it will mean that their direction is obvious from the context of the problem we are considering. Since the only two vectors in this equation, the F and the a, are on opposite sides of the equals sign, they must point in the same direction.
The second thing we have to consider in this equation is the “tot” subscript on the F. That is there to remind us that we are referring to the total, or net force acting on an object. So, for example, if we push equally hard on the back bumper and on the front bumper of a car in such a way that the two forces are the same size but pointed in opposite directions, then the total force on the car will be zero. In this case, where the total force is zero, the equation says that the acceleration must be zero as well. That means that the car’s velocity will remain constant. This is because acceleration is defined to be the change of an object’s velocity with respect to time.
On the other hand, if both forces are pointed in the same direction, the total force will be the sum of the two. Then the acceleration will be larger than if only one force was acting because the two forces together, acting in the same direction, are larger than either one alone(Figure 7.2).
Figure 7.2: Net force and acceleration. The acceleration of an object points in the same direction as the total, or net, force (F tot) acting on that object. The strength of the force, proportional here to the length of the arrow, is equal to the object’s mass multiplied by its acceleration. Note that in this image, the net force points in the direction of F A because F A is larger than the oppositely directed F B. Therefore, the acceleration also points in the direction of F A. Credit: NASA/SSU/Aurore Simonnet
The SI unit of force is the newton (N), named for Isaac Newton, whose discoveries are the subject of this chapter. In SI units, 1 newton = 1 kg m/s 2. The more familiar unit of force in the USA is the pound (lb); 1 N = 0.2248 lb. However, we will use SI units in these modules.
Net Force on an Object
In this activity, you will use what scientists call a freebody diagram to see how forces acting on an object in different directions can add together to create a total force on the object.
Play Activity
Worked Example:
Consider how two forces add together in one dimension. Assume the first force we are dealing with pushes an object to the right with a force of 3 N. The second force pushes the object to the left with a force of 4 N. What is the total force on the object?
The results of using the Freebody Diagram tool for this situation are shown in Figure A.7.1.
The first force pushes 3 N to the right, or 3 N in the positive x direction. There is no push in the y direction. To enter the first force, type 3 in the “x value” box, and 0 in the “y value” box. Now click the “Show” button to enter this force. Notice that the force now appears on the diagram to the right.
For the second force, there is a push of 4 in the negative x direction, and no push in the y direction. To enter this force, type -4 in the “x value” box, and 0 in the “y value box”. Now click the “Show” button to enter this force.
To add the two forces, click the “Add” button. Toward the bottom of the Input Panel, you should now see the Total Force listed in green. On the right, you can see the two input forces (in red) and the total force (in green) on the diagram.
The total force is -1 (x-direction) + 0 (y-direction) N. You can see that the total force points toward the left.
Once you are ready, click “Clear” before going on to the next example.
Figure A.7.1: Net force for 2 forces in one dimension. Output from the Freebody Diagram tool for 3(x-direction)N + -4(x-direction) N. The red arrows show the individual forces, while the green arrow shows the sum of the two forces. Credit: NASA/SSU
Questions
Now do a few examples on your own, either numerically, or using the interactive Freebody Diagram tool.
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Another important thing to understand in the equation is the meaning of the symbol m, the mass. Mass is a tricky concept to understand at first, but hopefully by the end of this chapter you will understand it. Mass, as defined by the equation above, is a measure of how resistant an object is to changes in its motion. First, we will do a couple of examples to demonstrate how this works. The next thing to be clear about is that mass is not the same as weight. The difference will be addressed at the end of this section.
MASS AND ACCELERATION
Worked Examples:
Assume that we want to accelerate a 5 kg object at a rate of 2 m/s 2. We can use the force equation to compute the force we must apply to the object.
Given: m = 5 kg, a = 2 m/s 2
Find: F
Concept: F = ma
Solution: F = ( 5 kg ) ( 2 m/s 2 ) = 10 N
We would therefore have to exert a 10 N force on the object.
We can also use the force relation to deduce the mass of an object. If we imagine that when we apply a 100 N force to an object it accelerates at 2 m/s 2, we can calculate its mass:
Given: F = 100 N, a = 2 m/s 2
Find: m
Concept: F = ma, or, rearranging, m = F/a
Solution: m = F/a = ( 100 N ) / ( 2 m/s 2 ) = 50 kg
This defines the mass of an object using its inertia, or its resistance to change in motion. This definition of mass is sometimes referred to as inertial mass. We can also define mass in terms of gravity, as we do below. While philosophically different, the two definitions give the same numerical answer in all cases studied to the best of our ability to measure.
Questions
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So the more massive an object is, the more difficult it is to change its velocity. This includes both its direction of motion or its speed. We are all familiar with this from our experiences in the world: It is much less dangerous to run into a person than a freight train. While freight trains do tend to move faster than people, that is not the most vital detail to consider. Even if a train is moving only at walking speed you would probably prefer colliding with a person than with a train. This is because a train has a lot more mass - and thus inertia - than a person has.
ACCELERATION AND MASS
We can use the Freebody Diagram again, this time with mass as well as force, to determine both the direction and strength of an object’s acceleration.
Play Activity
Worked Examples:
Imagine you and your friend are fighting over a textbook that you both really, really want to read. The textbook has a mass of 5 kg. You pull on the book with a force of 6 N in the x direction, and your friend pulls on the book with a force of 5 N in the negative x direction.
Using the Freebody Diagram tool that allows you to enter the mass, you should get something that looks like Figure A.7.2 below.
Figure A.7.2: Net force
and acceleration in one dimension. Output from Freebody Diagram for forces of 6 N (x-direction) and -5 N (x-direction) and a mass of 5 kg. The red arrows show the individual forces, while the green arrow shows the net force. The direction of the acceleration is the same as the direction of the net force. Credit: NASA/SSU
What is the total force on the textbook (strength and direction)?
We see from the tool that the net force is +6 N – 5 N = +1 N in the x-direction (to the right).
What is the total acceleration of the textbook (strength and direction)?
We see from the tool that the acceleration of the textbook is 0.2 m/s 2 in the +x direction. Notice that the direction of the acceleration is the same direction as the net force. The tool calculates the strength of the acceleration from the equation F = ma, or a = F/m. In this case, a = 1 N / 5 kg = 0.2 m/s 2.
Will the textbook accelerate toward you or your friend?
It will move toward me.
Questions
Once you decide the matter of the textbook, you and your friend realize that there is a study guide full of answers to all the homework questions in the text, and you begin to fight over that, too! The study guide has a mass of 2 kg. Again, you pull on the guide with a force of 6 newtons in the x direction. Your friend pulls on the study guide with a force of 5 newtons in the negative x direction.
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In the activity above, you can see that when two friends are pulling on a fairly massive textbook, the book does not accelerate much. However, when the friends pull with the same force on a less massive study guide, the study guide accelerates more. This demonstrates that under the influence of a force, an object’s acceleration is inversely proportional to its mass. In other words, an object with a large mass will experience a smaller acceleration due to a given net force than an object with a smaller mass. The object with a smaller mass will experience a larger acceleration due to a particular net force than the object with the larger mass.
Remember that one of the definitions of mass we discussed above refers to the inertial mass because it is based on the inertia of an object, or in other words, on its tendency to resist changes in its motion. The equation linking force, mass, and acceleration was first deduced by Isaac Newton, and it is called Newton’s second law of motion. Along with his first and third laws of motion, it constitutes the basis of classical mechanics, a branch of physics that describes the motion of bodies through space. We have already encountered Newton’s first law, borrowed from Galileo:
Definition: Newton’s first law of motion
Objects in motion remain at a constant velocity (straight line, constant speed) unless acted upon by a net force.
We have now encountered Newton’s second law in our discussion of force, mass, and acceleration:
Definition: Newton’s second law of motion
The acceleration of an object is directly proportional to the total force on the object, and inversely proportional to its mass. In physics, we write this using the equation F=ma, which we introduced and have worked with in its full vector form already.
For more about Newton’s laws see Going Further 7.1: Newton’s Third Law.
Going Further 7.1: Newton’s Third Law
Our daily experiences might lead us to think that forces are always applied by one object on another; for example, a horse pulls a carriage, a person pushes a grocery cart, or a hammer hits a nail. It took Sir Isaac Newton to realize that things are not so simple, and not so one-sided. True, if a hammer strikes a nail, the hammer exerts a force on the nail (thereby driving it into a board). Yet, the nail must also exert a force on the hammer since the hammer’s state of motion is changed, and according to the first law, this requires a net (outside) force. This is the essence of Newton’s third law: For every action there is an equal and opposite reaction. However, it is important to understand that the action and the reaction are acting on different objects.
Try this: Press the side of your hand against the edge of a table. Notice how your hand becomes distorted. Clearly, a force is being exerted on it. You can see the edge of the table pressing into your hand and feel the table exerting a force on your hand. Now press harder. The harder you press, the harder the table pushes back on your hand. Remember this important point: You can only feel the forces being exerted on you, not the forces you exert on something else. So, it is the force the table is exerting on you that you see and feel in your hand.
Figure B.7.1. Action–reaction pair illustrating Newton’s third law. Newton’s third law is illustrated in this figure, which shows that the action of the hand pushing on the table is equal in strength and opposite in direction to the reaction of the table pushing on the hand. Credit: NASA/SSU/Aurore Simonnet
Action–reaction pairs like the force of the hand on the table and the force of the table on hand are all around us. If you are reading this while sitting in a chair, you can feel the force the chair is exerting upward on you while you exert a force downward on it. And if you are reading on a computer monitor that sits on the table in front of you, then there is a balanced action–reaction pair between the monitor and the table, as well as between the table and the floor, and so on for all the objects you see around you. Each of these is an action–reaction pair. Such action–reaction pairs are the domain of Newton’s third law.
Not every equal and opposite pair of forces is a third law pair, however. In the third law, two objects are involved with the two forces. For example, a hand and a table, a table and a hand. We can have two forces of equal strength and opposite direction due to Newton’s second law, but these forces act on a single object. For example, if you hold a ball still in your hand, there is an upward force due to your hand on the ball that is exactly equal to the downward force of gravity from Earth on the ball. That is why the ball does not move. These two forces are acting on a single object, the ball, whose net force and acceleration are zero by Newton’s second law.
Newton's third law is also related to the concept of conservation of momentum. Momentum is defined as an object's mass times its velocity; because velocity is a vector, momentum is also a vector. Momentum is also related to force: the change in momentum over time is equal to the net force.
So, for two objects that are a third law pair, the change in momentum of one object will be equal in strength and opposite in direction to the change in momentum of the other. This principle is important in collisions. For example, if two billiard balls collide, they will bounce off of each other and travel in opposite directions. It is also important for propulsion. For example, a squid is a sea creature that takes in water; when it wants to move, it squirts water out of its body in one direction, and it moves in the opposite direction. As another example, if you are standing on a frozen pond having a snowball fight in the winter, if you throw a snowball, the snowball will move forward, but you will also slide backward a little. The velocity of the snowball will be greater than your velocity because you have a greater mass (unless you make a really big snowball).
We will use Newton’s laws in our study of objects moving under the influence of gravity. Combined, they provide a basic framework for understanding how objects move, why bridges are able to stand, how water flows in a stream and many other physical phenomena that fill our everyday experience.
What does all of this have to do with gravity? Well, as was noted in Section 7.1, gravity causes an acceleration, g, and this acceleration is constant for all objects on Earth, regardless of their mass. If mass and acceleration are involved, then according to Newton’s second law, there must be a total, or net, force acting.
In fact, we can use the second law to measure the force exerted by gravity on objects of different masses. We have a special name for the force of gravity acting on a mass; we call it the weight of the object. Substituting the acceleration due to gravity into Newton’s second law, we get an expression for the force of gravity on Earth, or the weight of an object:
F g=mg
Again, F g is the gravitational force, or weight, m is the mass, and g is the acceleration due to gravity, 9.8 m/s 2. Now the relationship between mass and weight becomes clear: weight is a force, whereas mass is how much “stuff” there is. The weight and the mass are related by a factor of g, but are not the same.
For more information on why the value of the mass is the same, whether the force is gravity or some other force, see Going Further 7.2: Gravity and Inertia.
GOING FURTHER: GRAVITY AND INERTIA
Gravitational Force on a Mass
Worked Example:
If we consider a 5 kg mass, we can use Newton’s second law to find the force that gravity exerts upon it.
Given: m = 5 kg, g = 9.8 m/s 2
Find: F g
Concept: F g = mg
Solution: F g = mg = ( 5 kg ) ( 9.8 m/s 2 ) = 49 N
This is equivalent to just over two pounds.
Questions
ADAPT 7.2.1
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In the previous example we used Newton’s second law to compute the force acting on an object falling in Earth’s gravity. This force causes objects to accelerate at one g(9.8 m/s 2). But what if an object is not falling? Does a force still act on it? Yes! But in that case the acceleration is zero, so how can there be a force? Here we have to think carefully about what the second law says: it says that acceleration is caused by a total, or net force. A zero acceleration means only that there is no net force acting on the body, not that no forces at all act on it. A picture might help clarify this (Figure 7.3).
Figure 7.3: Apple sitting on a table. An apple sitting on a table feels the force of gravity, Fg, pulling it downward, but it is not accelerating. If the acceleration of the apple is zero, the net force must be zero. Therefore, there must be a force of equal strength but opposite direction to the gravitational force. This opposing force that is pushing up on the apple is F T, from the table to the apple, which keeps the apple from falling. The net force on the apple is zero and the apple remains stationary. Credit: NASA/SSU/Aurore Simonnet
Imagine that we have the situation shown in Figure 7.3: An apple sits on a table. In the numerical activity, we saw that the force of gravity is acting on the apple, which should cause it to accelerate downward. However, in this case the apple is not accelerating. Why not? Because the table is in the way. If we apply Newton’s second law to this case, we can deduce that the table must be applying an upward force equal in magnitude to the force of gravity. How do we know that the net force is zero? Since the apple does not accelerate, we know the net force is zero. We can draw the apple and the forces acting on it, and since only the apple’s weight and the upward force from the table, F T, are present, they must be exactly equal in strength and opposite in direction.
Let’s repeat that again for emphasis: The apple does not remain stationary because there are no forces acting on it. In fact, we have just identified two forces that are acting on the apple. However, one force is from gravity and points downward. The other force is from the table and points upward. The forces are equal in strength and point in opposite directions, so they cancel each other out. The total force (or net force) on the apple is zero.
Lunar Lander
We can put what we have learned so far about Newton’s laws and gravity into practice by using the Lunar Lander to simulate landing a spacecraft on the surface of the Moon. To successfully land the spacecraft, you will have to slow it down so that both its horizontal and vertical velocities are very small. You will also have to land the spacecraft upright. Read the Lunar Lander instructions to learn which keys on your keyboard you can use to fire the lander’s engine and to rotate the lander. Watch your fuel level!
Some things to think about as you try this activity:
When you fire the lander’s engine, you are giving the lander a “push”, or applying a force to it. The direction of the force depends on which direction the lander is pointing. When you stop firing the engine, you stop applying a force to the lander.
The lander starts with a horizontal velocity . You will have to slow the lander down in the horizontal direction.
The Moon exerts a gravitational force on the lander, causing it to accelerate downward. You will have to slow the lander down in the vertical direction, too.
Play Activity
This page titled 7.2: Force, Mass, and Weight is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Kim Coble, Kevin McLin, & Lynn Cominsky.
7.2: Force, Mass, and Weight by Kim Coble, Kevin McLin, & Lynn Cominsky is licensed CC BY-NC-SA 4.0.
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IMO General Regulations §6.6 Contributing Countries The Organising Committee and the Problem Selection Committee of IMO 2024 thank the following 63 countries for contributing 229 problem proposals: Algeria, Australia, Azerbaijan, Bangladesh, Belarus, Brazil, Bulgaria, Canada, China, Colombia, Croatia, Cyprus, Czech Republic, Denmark, Dominican Republic, Ecuador, Estonia, France, Georgia, Germany, Ghana, Greece, Hong Kong, India, Indonesia, Ireland, Iran, Israel, Japan, Kazakhstan, Kosovo, Latvia, Lithuania, Luxembourg, Malaysia, Mexico, Moldova, Netherlands, New Zealand, Norway, Peru, Poland, Portugal, Romania, Senegal, Serbia, Singapore, Slovakia, Slovenia, South Africa, South Korea, Spain, Sweden, Switzerland, Syria, Taiwan, Thailand, Tunisia, Türkiye, Uganda, Ukraine, U.S.A., Uzbekistan.
Problem Selection Committee Aron Thomas, Yinghua Ai, Andrew Ng, Géza Kós, Ivan Guo, Alice Carlotti, James Aaronson, Sam Bealing, Adrian Agisilaou, James Cranch, Joseph Myers (chair), Harvey Yau, Maria-Romina Ivan, Michael Ren, Elisa Lorenzo García Shortlisted problems 3 Problems Algebra A1.
Determine all real numbers α such that the number tαu t2αu ¨ ¨ ¨ tnαu is a multiple of n for every positive integer n. (Here tzu denotes the greatest integer less than or equal to z.) (Colombia) A2.
Let n be a positive integer. Find the minimum possible value of S “ 20x2 0 21x2 1 ¨ ¨ ¨ 2nx2 n, where x0, x1, . . . , xn are nonnegative integers such that x0 x1 ¨ ¨ ¨ xn “ n.
(China) A3.
Decide whether for every sequence panq of positive real numbers, 3a1 3a2 ¨ ¨ ¨ 3an p2a1 2a2 ¨ ¨ ¨ 2anq2 ă 1 2024 is true for at least one positive integer n.
(China) A4.
Let Zą0 be the set of all positive integers. Determine all subsets S of t20, 21, 22, . . . u for which there exists a function f : Zą0 Ñ Zą0 such that S “ tfpa bq ´ fpaq ´ fpbq | a, b P Zą0u.
(Thailand) A5.
Find all periodic sequences a1, a2, . . .
of real numbers such that the following conditions hold for all n ě 1: an2 a2 n “ an a2 n1 and |an1 ´ an| ď 1.
(Kosovo) A6.
Let a0, a1, a2, . . . be an infinite strictly increasing sequence of positive integers such that for each n ě 1 we have an P !an´1 an1 2 , ?an´1 ¨ an1 ) .
Let b1, b2, . . . be an infinite sequence of letters defined as bn “ # A, if an “ 1 2pan´1 an1q; G, otherwise.
Prove that there exist positive integers n0 and d such that for all n ě n0 we have bnd “ bn.
(Czech Republic) A7.
Let Q be the set of rational numbers. Let f : Q Ñ Q be a function such that the following property holds: for all x, y P Q, fpx fpyqq “ fpxq y or fpfpxq yq “ x fpyq.
Determine the maximum possible number of elements of tfpxq fp´xq | x P Qu.
(Japan) 4 Bath, United Kingdom, 10th–22nd July 2024 A8.
Let p ‰ q be coprime positive integers. Determine all infinite sequences a1, a2, . . . of positive integers such that the following conditions hold for all n ě 1: maxpan, an1, . . . , anpq ´ minpan, an1, . . . , anpq “ p and maxpan, an1, . . . , anqq ´ minpan, an1, . . . , anqq “ q.
(Japan) Shortlisted problems 5 Combinatorics C1.
Let n be a positive integer. A class of n students run n races, in each of which they are ranked with no draws. A student is eligible for a rating pa, bq for positive integers a and b if they come in the top b places in at least a of the races. Their final score is the maximum possible value of a ´ b across all ratings for which they are eligible.
Find the maximum possible sum of all the scores of the n students.
(Australia) C2.
Let n be a positive integer. The integers 1, 2, 3, . . . , n2 are to be written in the cells of an n ˆ n board such that each integer is written in exactly one cell and each cell contains exactly one integer. For every integer d with d | n, the d-division of the board is the division of the board into pn{dq2 nonoverlapping sub-boards, each of size d ˆ d, such that each cell is contained in exactly one d ˆ d sub-board.
We say that n is a cool number if the integers can be written on the n ˆ n board such that, for each integer d with d | n and 1 ă d ă n, in the d-division of the board, the sum of the integers written in each d ˆ d sub-board is not a multiple of d.
Determine all even cool numbers.
(Türkiye) C3.
Let n be a positive integer. There are 2n knights sitting at a round table. They consist of n pairs of partners, each pair of which wishes to shake hands. A pair can shake hands only when next to each other. Every minute, one pair of adjacent knights swaps places.
Find the minimum number of exchanges of adjacent knights such that, regardless of the initial arrangement, every knight can meet her partner and shake hands at some time.
(Belarus) C4.
On a board with 2024 rows and 2023 columns, Turbo the snail tries to move from the first row to the last row. On each attempt, he chooses to start on any cell in the first row, then moves one step at a time to an adjacent cell sharing a common side. He wins if he reaches any cell in the last row. However, there are 2022 predetermined, hidden monsters in 2022 of the cells, one in each row except the first and last rows, such that no two monsters share the same column. If Turbo unfortunately reaches a cell with a monster, his attempt ends and he is transported back to the first row to start a new attempt. The monsters do not move.
Suppose Turbo is allowed to take n attempts. Determine the minimum value of n for which he has a strategy that guarantees reaching the last row, regardless of the locations of the monsters.
(Hong Kong) C5.
Let N be a positive integer. Geoff and Ceri play a game in which they start by writing the numbers 1, 2, . . . , N on a board. They then take turns to make a move, starting with Geoff. Each move consists of choosing a pair of integers pk, nq, where k ě 0 and n is one of the integers on the board, and then erasing every integer s on the board such that 2k | n ´ s. The game continues until the board is empty. The player who erases the last integer on the board loses.
Determine all values of N for which Geoff can ensure that he wins, no matter how Ceri plays.
(Indonesia) 6 Bath, United Kingdom, 10th–22nd July 2024 C6.
Let n and T be positive integers. James has 4n marbles with weights 1, 2, . . . , 4n.
He places them on a balance scale, so that both sides have equal weight. Andrew may move a marble from one side of the scale to the other, so that the absolute difference in weights of the two sides remains at most T.
Find, in terms of n, the minimum positive integer T such that Andrew may make a sequence of moves such that each marble ends up on the opposite side of the scale, regardless of how James initially placed the marbles.
(Ghana) C7.
Let N be a positive integer and let a1, a2, . . .
be an infinite sequence of positive integers. Suppose that, for each n ą N, an is equal to the number of times an´1 appears in the list a1, a2, . . . , an´1.
Prove that at least one of the sequences a1, a3, a5, . . .
and a2, a4, a6, . . .
is eventually periodic.
(Australia) C8.
Let n be a positive integer. Given an n ˆ n board, the unit cell in the top left corner is initially coloured black, and the other cells are coloured white. We then apply a series of colouring operations to the board. In each operation, we choose a 2 ˆ 2 square with exactly one cell coloured black and we colour the remaining three cells of that 2 ˆ 2 square black.
Determine all values of n such that we can colour the whole board black.
(Peru) Shortlisted problems 7 Geometry G1.
Let ABCD be a cyclic quadrilateral such that AC ă BD ă AD and =DBA ă 90˝.
Point E lies on the line through D parallel to AB such that E and C lie on opposite sides of line AD, and AC “ DE. Point F lies on the line through A parallel to CD such that F and C lie on opposite sides of line AD, and BD “ AF.
Prove that the perpendicular bisectors of segments BC and EF intersect on the circumcircle of ABCD.
(Ukraine) G2.
Let ABC be a triangle with AB ă AC ă BC, incentre I and incircle ω. Let X be the point in the interior of side BC such that the line through X parallel to AC is tangent to ω.
Similarly, let Y be the point in the interior of side BC such that the line through Y parallel to AB is tangent to ω. Let AI intersect the circumcircle of triangle ABC again at P ‰ A. Let K and L be the midpoints of AB and AC, respectively.
Prove that =KIL =Y PX “ 180˝.
(Poland) G3.
Let ABCDE be a convex pentagon and let M be the midpoint of AB. Suppose that segment AB is tangent to the circumcircle of triangle CME at M and that D lies on the circumcircles of triangles AME and BMC. Lines AD and ME intersect at K, and lines BD and MC intersect at L. Points P and Q lie on line EC so that =PDC “ =EDQ “ =ADB.
Prove that lines KP, LQ, and MD are concurrent.
(Belarus) G4.
Let ABCD be a quadrilateral with AB parallel to CD and AB ă CD. Lines AD and BC intersect at a point P. Point X ‰ C on the circumcircle of triangle ABC is such that PC “ PX. Point Y ‰ D on the circumcircle of triangle ABD is such that PD “ PY .
Lines AX and BY intersect at Q.
Prove that PQ is parallel to AB.
(Ukraine) G5.
Let ABC be a triangle with incentre I, and let Ωbe the circumcircle of triangle BIC.
Let K be a point in the interior of segment BC such that =BAK ă =KAC. The angle bisector of =BKA intersects Ωat points W and X such that A and W lie on the same side of BC, and the angle bisector of =CKA intersects Ωat points Y and Z such that A and Y lie on the same side of BC.
Prove that =WAY “ =ZAX.
(Uzbekistan) G6.
Let ABC be an acute triangle with AB ă AC, and let Γ be the circumcircle of ABC.
Points X and Y lie on Γ so that XY and BC intersect on the external angle bisector of =BAC.
Suppose that the tangents to Γ at X and Y intersect at a point T on the same side of BC as A, and that TX and TY intersect BC at U and V , respectively. Let J be the centre of the excircle of triangle TUV opposite the vertex T.
Prove that AJ bisects =BAC.
(Poland) 8 Bath, United Kingdom, 10th–22nd July 2024 G7.
Let ABC be a triangle with incentre I such that AB ă AC ă BC. The second intersections of AI, BI, and CI with the circumcircle of triangle ABC are MA, MB, and MC, respectively. Lines AI and BC intersect at D and lines BMC and CMB intersect at X. Suppose the circumcircles of triangles XMBMC and XBC intersect again at S ‰ X. Lines BX and CX intersect the circumcircle of triangle SXMA again at P ‰ X and Q ‰ X, respectively.
Prove that the circumcentre of triangle SID lies on PQ.
(Thailand) G8.
Let ABC be a triangle with AB ă AC ă BC, and let D be a point in the interior of segment BC. Let E be a point on the circumcircle of triangle ABC such that A and E lie on opposite sides of line BC and =BAD “ =EAC. Let I, IB, IC, JB, and JC be the incentres of triangles ABC, ABD, ADC, ABE, and AEC, respectively.
Prove that IB, IC, JB, and JC are concyclic if and only if AI, IBJC, and JBIC concur.
(Canada) Shortlisted problems 9 Number Theory N1.
Find all positive integers n with the following property: for all positive divisors d of n, we have that d 1 | n or d 1 is prime.
(Ghana) N2.
Determine all finite, nonempty sets S of positive integers such that for every a, b P S there exists c P S with a | b 2c.
(Netherlands) N3.
Determine all sequences a1, a2, . . .
of positive integers such that, for any pair of positive integers m ď n, the arithmetic and geometric means am am1 ¨ ¨ ¨ an n ´ m 1 and pamam1 ¨ ¨ ¨ anq 1 n´m1 are both integers.
(Singapore) N4.
Determine all positive integers a and b such that there exists a positive integer g such that gcdpan b, bn aq “ g for all sufficiently large n.
(Indonesia) N5.
Let S be a finite nonempty set of prime numbers. Let 1 “ b1 ă b2 ă ¨ ¨ ¨ be the sequence of all positive integers whose prime divisors all belong to S. Prove that, for all but finitely many positive integers n, there exist positive integers a1, a2, . . . , an such that a1 b1 a2 b2 ¨ ¨ ¨ an bn “ R 1 b1 1 b2 ¨ ¨ ¨ 1 bn V .
(Croatia) N6.
Let n be a positive integer.
We say that a polynomial P with integer coeffi-cients is n-good if there exists a polynomial Q of degree 2 with integer coefficients such that QpkqpPpkq Qpkqq is never divisible by n for any integer k.
Determine all integers n such that every polynomial with integer coefficients is an n-good polynomial.
(France) N7.
Let Zą0 denote the set of positive integers. Let f : Zą0 Ñ Zą0 be a function satisfying the following property: for m, n P Zą0, the equation fpmnq2 “ fpm2qfpfpnqqfpmfpnqq holds if and only if m and n are coprime.
For each positive integer n, determine all the possible values of fpnq.
(Japan) 10 Bath, United Kingdom, 10th–22nd July 2024 Shortlisted problems – solutions 11 Solutions Algebra A1.
Determine all real numbers α such that the number tαu t2αu ¨ ¨ ¨ tnαu is a multiple of n for every positive integer n. (Here tzu denotes the greatest integer less than or equal to z.) (Colombia) Answer: All even integers satisfy the condition of the problem and no other real number α does so.
Solution 1.
First we will show that even integers satisfy the condition. If α “ 2m where m is an integer then tαu t2αu ¨ ¨ ¨ tnαu “ 2m 4m ¨ ¨ ¨ 2mn “ mnpn 1q which is a multiple of n.
Now we will show that they are the only real numbers satisfying the conditions of the problem. Let α “ k ϵ where k is an integer and 0 ď ϵ ă 1. Then the number tαu t2αu ¨ ¨ ¨ tnαu “ k tϵu 2k t2ϵu ¨ ¨ ¨ nk tnϵu “ knpn 1q 2 tϵu t2ϵu ¨ ¨ ¨ tnϵu has to be a multiple of n. We consider two cases based on the parity of k.
Case 1: k is even.
Then knpn1q 2 is always a multiple of n. Thus tϵu t2ϵu ¨ ¨ ¨ tnϵu also has to be a multiple of n.
We will prove that tnϵu “ 0 for every positive integer n by strong induction. The base case n “ 1 follows from the fact that 0 ď ϵ ă 1. Let us suppose that tmϵu “ 0 for every 1 ď m ă n.
Then the number tϵu t2ϵu ¨ ¨ ¨ tnϵu “ tnϵu has to be a multiple of n. As 0 ď ϵ ă 1 then 0 ď nϵ ă n, which means that the number tnϵu has to be equal to 0.
The equality tnϵu “ 0 implies 0 ď ϵ ă 1{n. Since this has to happen for all n, we conclude that ϵ “ 0 and then α is an even integer.
12 Bath, United Kingdom, 10th–22nd July 2024 Case 2: k is odd.
We will prove that tnϵu “ n ´ 1 for every natural number n by strong induction. The base case n “ 1 again follows from the fact that 0 ď ϵ ă 1. Let us suppose that tmϵu “ m ´ 1 for every 1 ď m ă n. We need the number knpn 1q 2 tϵu t2ϵu ¨ ¨ ¨ tnϵu “ knpn 1q 2 0 1 ¨ ¨ ¨ pn ´ 2q tnϵu “ knpn 1q 2 pn ´ 2qpn ´ 1q 2 tnϵu “ k 1 2 n2 k ´ 3 2 n 1 tnϵu to be a multiple of n. As k is odd, we need 1 tnϵu to be a multiple of n. Again, as 0 ď ϵ ă 1 then 0 ď nϵ ă n, so tnϵu “ n ´ 1 as we wanted.
This implies that 1 ´ 1 n ď ϵ ă 1 for all n which is absurd. So there are no other solutions in this case.
Solution 2. As in Solution 1 we check that for even integers the condition is satisfied. Then, without loss of generality we can assume 0 ď α ă 2. We set Sn “ tαu t2αu ¨ ¨ ¨ tnαu.
Notice that Sn ” 0 pmod nq (1) Sn ” Sn ´ Sn´1 “ tnαu pmod n ´ 1q (2) Since gcdpn, n ´ 1q “ 1, (1) and (2) imply that Sn ” ntnαu pmod npn ´ 1qq.
(3) In addition, 0 ď ntnαu ´ Sn “ n ÿ k“1 ´ tnαu ´ tkαu ¯ ă n ÿ k“1 ´ nα ´ kα 1 ¯ “ npn ´ 1q 2 α n.
(4) For n large enough, the RHS of (4) is less than npn ´ 1q. Then (3) forces 0 “ Sn ´ ntnαu “ n ÿ k“1 ´ tnαu ´ tkαu ¯ (5) for n large enough.
Since tnαu ´ tkαu ě 0 for 1 ď k ď n, we get from (5) that, for all n large enough, all these inequalities are equalities. In particular tαu “ tnαu for all n large enough, which is absurd unless α “ 0.
Comment. An alternative ending to the previous solution is as follows.
By definition we have Sn ď α npn1q 2 , on the other hand (5) implies Sn ě αn2 ´ n for all n large enough, so α “ 0.
Shortlisted problems – solutions 13 Solution 3. As in other solutions, without loss of generality we may assume that 0 ď α ă 2.
Even integers satisfy the condition, so we assume 0 ă α ă 2 and we will derive a contradiction.
By induction on n, we will simultaneously show that tαu t2αu ¨ ¨ ¨ tnαu “ n2, (6) and 2n ´ 1 n ď α ă 2.
(7) The base case is n “ 1: If α ă 1, consider m “ P 1 α T ą 1, then tαu t2αu ¨ ¨ ¨ tmαu “ 1 is not a multiple of m, so we deduce (7). Hence, tαu “ 1 and (6) follows.
For the induction step: assume the induction hypothesis to be true for n, then by (7) 2n 1 ´ 1 n ď pn 1qα ă 2n 2.
Hence, n2 2n ď tαu t2αu ¨ ¨ ¨ tnαu tpn 1qαu “ n2 tpn 1qαu ă n2 2n 2.
So, necessarily tpn 1qαu “ 2n 1 and tαu t2αu ¨ ¨ ¨ tnαu tpn 1qαu “ pn 1q2 in order to obtain a multiple of n 1. These two equalities give (6) and (7) respectively.
Finally, we notice that condition (7) being true for all n gives a contradiction.
Solution 4.
As in other solutions without loss of generality we will assume that 0 ă α ă 2 and derive a contradiction. For each n, we define bn “ tαu t2αu ¨ ¨ ¨ tnαu n , which is a nonnegative integer by the problem condition and our assumption. Note that tpn 1qαu ě tαu , t2αu , . . . , tnαu and tpn 1qαu ą tαu for all n ą 1 α. It follows that bn1 ą bn ù ñ bn1 ě bn 1 for n ą 1 α. Thus, for all such n, bn ě n C where C is a fixed integer. On the other hand, the definition of bn gives bn “ tαu t2αu ¨ ¨ ¨ tnαu n ď α 2α ¨ ¨ ¨ nα n “ α 2 pn 1q, which is a contradiction for sufficiently large n.
14 Bath, United Kingdom, 10th–22nd July 2024 A2.
Let n be a positive integer. Find the minimum possible value of S “ 20x2 0 21x2 1 ¨ ¨ ¨ 2nx2 n, where x0, x1, . . . , xn are nonnegative integers such that x0 x1 ¨ ¨ ¨ xn “ n.
(China) Answer: The minimum value is npn1q 2 .
Solution 1.
For a fixed n, let fpnq denote the minimum possible value of S. Consider the following variant: among all infinite sequences of nonnegative integers x0, x1, . . . , only finitely many of which are nonzero, satisfying x0 x1 ¨ ¨ ¨ “ n, let gpnq denote the minimum possible value of T “ 20x2 0 21x2 1 22x2 2 ¨ ¨ ¨ .
It is clear that gpnq ď fpnq. Conversely, it is easy to see that if a sequence x0, x1, . . . achieves the minimum of gpnq, then x0 ě x1 ě ¨ ¨ ¨ and thus xn1 “ xn2 “ ¨ ¨ ¨ “ 0. In particular, fpnq “ gpnq.
Now, we hope to get an inductive formula for gpnq.
Note that, in order to minimise T for n ě 1, we must have x0 ě 1 since the sequence pxiq is nonincreasing. Note that the minimal value of 21x2 1 22x2 2 ¨ ¨ ¨ “ 2p20x2 1 21x2 2 ¨ ¨ ¨ q over all infinite sequences of nonnegative integers with x1 x2 ¨ ¨ ¨ “ m is exactly 2gpmq. As a result, for n ě 1 we have gpnq “ min x0Pt1,2,...,nu x2 0 2gpn ´ x0q ˘ .
We now prove gpnq “ npn1q 2 by induction. It is clear that gp0q “ 0. Assume that this has been proved for n “ 0, 1, . . . , N ´ 1. Then, x2 0 2gpN ´ x0q “ x2 0 pN ´ x0qpN ´ x0 1q (1) “ 2x2 0 ´ p2N 1qx0 NpN 1q “ 1 2 “ p2x0 ´ Nqp2x0 ´ N ´ 1q N 2 N ‰ .
The product of two consecutive integers p2x0 ´ Nqp2x0 ´ N ´ 1q is always nonnegative, and it is zero precisely when 2x0 is the even number in tN, N 1u. Thus the minimum of the final expression in equation (1) is 1 2pN 2 Nq, so gpNq “ NpN1q 2 , completing the inductive proof.
Shortlisted problems – solutions 15 Solution 2. Consider the following table of numbers, where the row and column indices start from 0, and ai,j “ 2ip2j 1q for i, j ě 0.
j “ 0 1 2 3 4 5 ¨ ¨ ¨ i “ 0 1 3 5 7 9 11 1 2 6 10 14 18 22 2 4 12 20 28 36 44 3 8 24 40 56 72 88 4 16 48 80 112 144 176 .
.
.
Every number can be written uniquely as a product of a power of 2 and an odd number so every positive integer appears exactly once in the table above. It is easy to see that numbers in each row and each column are strictly increasing. Since the sum of the first x odd positive integers is x2, the sum of the first xk numbers in the kth row is 2kx2 k, the kth term appearing in S.
Thus, the sum S can be interpreted as the result of taking a total of n numbers from the first n rows of the table such that we take the leftmost xk numbers from row k (where řn k“1 xk “ n), and then computing the sum of these n numbers. In particular, the minimum possible value of S is the same as the sum of the smallest n numbers in this table, since every row and every column of the table is strictly increasing.
Moreover, the smallest n numbers, namely 1, 2, . . . , n, appear in the first n rows, so the minimum of S is 1 2 ¨ ¨ ¨ n “ npn 1q 2 .
Comment. As can be seen from the table in Solution 2, the equality case of the problem is given by xi “ Z n 2i1 1 2 ^ .
So xi is the result of rounding n 2i1 to the nearest integer. This also gives a proof of the identity n “ 8 ÿ i“0 Z n 2i1 1 2 ^ , which can be separately proven by induction on n: when n is incremented by 1, exactly one term on the right hand side, namely the one corresponding to i “ ν2pnq, increases by 1 while the others remain the same.
Comment. If the condition that the xi are nonnegative integers is relaxed to the xi being nonnegative reals, the problem can be solved by an application of the Cauchy-Schwarz inequality: p20 2´1 ¨ ¨ ¨ 2´nqp20x2 0 21x2 1 ¨ ¨ ¨ 2nx2 nq ě px0 ¨ ¨ ¨ xnq2 “ n2 ù ñ 20x2 0 21x2 1 ¨ ¨ ¨ 2nx2 n ě n2 2 ´ 2´n .
The equality case for this relaxed problem is given by xi “ 2´in 2 ´ 2´n « Z n 2i1 1 2 ^ .
In fact, when the terms in the optimal sequence for the real case are all rounded to the nearest integer, we obtain the optimal sequence for the original problem. While thinking about the real case may guide one towards the equality case of the original problem, it does not seem like it can be easily continued into a full solution.
16 Bath, United Kingdom, 10th–22nd July 2024 A3.
Decide whether for every sequence panq of positive real numbers, 3a1 3a2 ¨ ¨ ¨ 3an p2a1 2a2 ¨ ¨ ¨ 2anq2 ă 1 2024 (1) is true for at least one positive integer n.
(China) Comment. The question can be asked in several forms, as follows: (i) students could be asked, as above, to show the existence of such an n; (ii) students could be asked to show that this happens for all sufficiently large n; (iii) students could be given a concrete positive integer N and asked to show it for all n ą N.
The solutions below provide varying bounds for N.
Answer: The answer is “yes”: there is always such an n.
Common remarks. We write ε “ 1 2024.
Solution 1. For every positive integer n, let Mn “ maxpa1, a2, . . . , anq. We first prove that 3a1 3a2 ¨ ¨ ¨ 3an p2a1 2a2 ¨ ¨ ¨ 2anq2 ď ˆ3 4 ˙Mn .
(2) For i “ 1, 2, . . . , n, from 3 2 ˘ai ď 3 2 ˘Mn we can obtain 3ai ď 3 4 ˘Mn ¨ 2Mn ¨ 2ai. By summing up over all i, n ÿ i“1 3ai ď ˆ3 4 ˙Mn ¨ 2Mn ¨ n ÿ i“1 2ai ď ˆ3 4 ˙Mn ¨ ˆ n ÿ i“1 2ai ˙2 , which is equivalent to (2).
Now let µ “ log4{3 1 ε, so that µ is the positive real number with 3 4 ˘µ “ ε. If there is an index n such that an ą µ, then Mn ě an ą µ, and hence 3a1 3a2 ¨ ¨ ¨ 3an p2a1 2a2 ¨ ¨ ¨ 2anq2 ď ˆ3 4 ˙Mn ă ˆ3 4 ˙µ “ ε.
Otherwise we have 0 ă ai ď µ for all positive integers i, so 3a1 3a2 ¨ ¨ ¨ 3an p2a1 2a2 ¨ ¨ ¨ 2anq2 ď n ¨ 3µ pn ¨ 1q2 “ 3µ n .
If n ą X3µ ε \ , this is less than ε.
Comment. It is also possible to prove (2) by induction on n. The base case n “ 1 is clear. For the induction step, after ordering a1, a2, . . . , an in increasing order as b1 ď b2 ď ¨ ¨ ¨ ď bn, it suffices, for example, to verify that 3b1 3b2 ¨ ¨ ¨ 3bn p2b1 2b2 ¨ ¨ ¨ 2bnq2 ď 3b1 3b2 ¨ ¨ ¨ 3bn p2b1 2b2 ¨ ¨ ¨ 2bnqp2b2 ¨ ¨ ¨ 2bnq ď 3b2 ¨ ¨ ¨ 3bn p2b2 ¨ ¨ ¨ 2bnq2 .
The second inequality is equivalent to 3b1 n ř i“2 2bi ď 2b1 n ř i“2 3bi, which follows from 3 2 ˘b1 ď 3 2 ˘bi.
Shortlisted problems – solutions 17 Solution 2. We will combine two upper bounds.
First, start with the trivial estimate 3a1 ¨ ¨ ¨ 3an p2a1 ¨ ¨ ¨ 2anq2 ď 3a1 ¨ ¨ ¨ 3an 4a1 ¨ ¨ ¨ 4an .
By applying Jensen’s inequality to the convex function xlog3 4 we get 4a1 ¨ ¨ ¨ 4an n “ 3a1˘log3 4 ¨ ¨ ¨ 3an˘log3 4 n ě ˆ3a1 ¨ ¨ ¨ 3an n ˙log3 4 , so 3a1 ¨ ¨ ¨ 3an p2a1 ¨ ¨ ¨ 2anq2 ď 3a1 ¨ ¨ ¨ 3an 4a1 ¨ ¨ ¨ 4an ď ˆ n 3a1 ¨ ¨ ¨ 3an ˙log3 4´1 .
Hence, (1) holds true whenever 3a1 ¨ ¨ ¨ 3an ą ˆ1 ε ˙ 1 log3 4´1 ¨ n.
(3) Second, trivially 3a1 ¨ ¨ ¨ 3an p2a1 ¨ ¨ ¨ 2anq2 ď 3a1 ¨ ¨ ¨ 3an n2 , so (1) is satisfied if 3a1 ¨ ¨ ¨ 3an ă ε ¨ n2.
(4) If n ą 1 ε ˘1 1 log3 4´1 then 1 ε ˘ 1 log3 4´1 ¨ n ă ε ¨ n2, and therefore at least one of (3) and (4) is satisfied.
Solution 3. Define C “ log4{3 2 ε, so that if ai ą C then 3ai ă ε 2 ¨ 4ai. We divide the sequence into “small” and “large” terms by how they compare to C: let Sn “ ti ď n | ai ď Cu and Ln “ ti ď n | ai ą Cu.
Then (1) is equivalent to ř iPSn 3aiř iPSn 2ai ř iPLn 2ai˘2 ř iPLn 3ai ř iPSn 2ai ř iPLn 2ai˘2 ă ε 2 ε 2.
If Ln is nonempty, we have ř iPLn 3aiř iPSn 2ai ř iPLn 2ai˘2 ă ε 2 ¨ ř iPLn 4aiř iPLn 2ai˘2 ď ε 2, and this also trivially holds when Ln is empty (in which case the LHS is zero).
Now suppose that n ě 2 ε 3 2 ˘C. Note that 3ai ď 3 2 ˘C 2ai for i P Sn, so we have ř iPSn 3ai ř iPSn 2ai ř iPLn 2ai˘2 ď 3 2 ˘C ř iPSn 2aiř iPSn 2ai ř iPLn 2ai˘2 ď 3 2 ˘C ř iPSn 2ai ř iPLn 2ai ă 3 2 ˘C n ď ε 2, so we have (1).
18 Bath, United Kingdom, 10th–22nd July 2024 Solution 4.
For every index i “ 1, 2, . . . , n, apply the weighted AM-GM inequality to numbers 2ai and pn ´ 1q with weights log2 3 2 « 0.585 and log2 4 3 « 0.415 as 2a1 2a2 ¨ ¨ ¨ 2an ě 2ai pn ´ 1q ą log2 3 2 ¨ 2ai log2 4 3 ¨ pn ´ 1q ě 2ai˘log2 3 2 ¨ pn ´ 1qlog2 4 3 “ ˆ3 2 ˙ai ¨ pn ´ 1qlog2 4 3 ą ˆ3 2 ˙ai ¨ pn ´ 1q2{5.
By summing up for i “ 1, 2, . . . , n, p2a1 ¨ ¨ ¨ 2anq2 “ n ÿ i“1 2ai2a1 2a2 ¨ ¨ ¨ 2an˘ ą pn ´ 1q2{5 n ÿ i“1 3ai so 3a1 3a2 ¨ ¨ ¨ 3an p2a1 2a2 ¨ ¨ ¨ 2anq2 ă 1 pn ´ 1q2{5.
If n ě 1 ε ˘5{2 1 then 1 pn´1q2{5 ă ε.
Shortlisted problems – solutions 19 A4.
Let Zą0 be the set of all positive integers. Determine all subsets S of t20, 21, 22, . . . u for which there exists a function f : Zą0 Ñ Zą0 such that S “ tfpa bq ´ fpaq ´ fpbq | a, b P Zą0u.
(Thailand) Answer: S can be any subset of size 1 or 2.
Common remarks.
For this problem, it is convenient to use notation such as ta, b, cu for multisets rather than sets, and the subset relation is likewise that for multisets. Both solutions use the following property of powers of 2: if 2a 2b “ 2c 2d, then ta, bu and tc, du are the same multiset. Define epa, bq “ log2pfpa bq ´ fpaq ´ fpbqq “ epb, aq. Thus, fpa bq “ 2epa,bq fpaq fpbq.
Solution 1. Clearly S must be nonempty. We start with constructions when 1 ď |S| ď 2.
• If S “ t2ku, then take fpxq “ cx ´ 2k for any integer c ą 2k.
• If S “ t2k, 2ℓu where k ą ℓ, then take fpxq “ p2k ´ 2ℓqtαxu ´ 2ℓ, where α ą 2 is not an integer. This works because tαpx yqu ´ ptαxu tαyuq P t0, 1u for all x and y, and takes both values; the lower bound on α ensures the values of f are positive.
Observe that, inductively, fpnq “ 2ep1,1q 2ep2,1q ¨ ¨ ¨ 2epn´1,1q nfp1q.
Lemma 1. For any positive integers n and k, tep1, 1q, ep2, 1q, . . . , epk ´ 1, 1qu Ă tepn, 1q, epn 1, 1q, . . . , epn k ´ 1, 1qu.
Proof. We work by induction on k; in the case k “ 1, the first multiset is empty, which provides our base case.
For the induction step, suppose k ě 2 and we know that tep1, 1q, ep2, 1q, . . . , epk ´ 2, 1qu Ă tepn, 1q, epn 1, 1q, . . . , epn k ´ 2, 1qu.
By definition, fpn kq ´ fpnq ´ fpkq “ 2epn,kq, and using the first observation we see that fpnkq´fpnq´fpkq “ 2epn,1q 2epn1,1q ¨ ¨ ¨ 2epnk´1,1q˘ ´ 2ep1,1q 2ep2,1q ¨ ¨ ¨ 2epk´1,1q˘ .
From the induction hypothesis, we may write tepn, 1q, epn 1, 1q, . . . , epn k ´ 2, 1qu “ tep1, 1q, ep2, 1q, . . . , epk ´ 2, 1qu Y tau for some a. Thus 2epn,kq “ 2a 2epnk´1,1q ´ 2epk´1,1q.
So tepn, kq, epk´1, 1qu “ ta, epnk´1, 1qu. Thus epk´1, 1q “ a or epk´1, 1q “ epnk´1, 1q, and in either case we have our result.
l Lemma 2. The sequence ep1, 1q, ep2, 1q, ep3, 1q, . . . takes at most two different values.
20 Bath, United Kingdom, 10th–22nd July 2024 Proof. Suppose for a contradiction that k ě 2 is the least index with epk, 1q ‰ ep1, 1q, and that some ℓą k has epℓ, 1q R tepk, 1q, ep1, 1qu. By Lemma 1, any block of k consecutive values of the sequence has at least k ´ 1 values equal to ep1, 1q. This forces epℓ´ 1, 1q “ epℓ´ 2, 1q “ ¨ ¨ ¨ “ epℓ´ pk ´ 1q, 1q “ ep1, 1q and epℓ1, 1q “ epℓ 2, 1q “ ¨ ¨ ¨ “ epℓpk ´ 1q, 1q “ ep1, 1q.
But then the block epℓ´1, 1q, epℓ, 1q, epℓ1, 1q, epℓ2, 1q, . . . , epℓpk ´1q, 1q has length k 1 and does not contain epk, 1q, a contradiction.
l Finally, for any a and b we have fpa bq ´ fpaq ´ fpbq “ 2epa,1q 2epa1,1q ¨ ¨ ¨ 2epab´1,1q˘ ´ 2ep1,1q 2ep2,1q ¨ ¨ ¨ 2epb´1,1q˘ “ 2epi,1q for some a ď i ď a b ´ 1. So |S| ď 2.
Comment. In the construction of functions, α ą 2 is only necessary if k “ ℓ1, to make sure fp1q ‰ 0.
Otherwise, any nonintegral α ą 1 suffices.
Solution 2. Subsets of size 1 or 2 can be achieved as in Solution 1, and S must be nonempty.
We consider such a set S with |S| ě 3 and a corresponding function f in order to achieve a contradiction. We will relate the epa, bq to values of epc, 1q with c1 ă ab, leading to a proof of Lemma 2 from Solution 1 that does not depend on Lemma 1 from that solution.
Suppose a ą 1. We have fpa bq ´ fpaq ´ fpbq “ 2epa,bq and also fpaq ´ fpa ´ 1q ´ fp1q “ 2epa´1,1q, so fpa bq ´ fpa ´ 1q ´ fp1q ´ fpbq “ 2epa,bq 2epa´1,1q.
Similarly, fpa bq ´ fpa ´ 1q ´ fpb 1q “ 2epa´1,b1q and fpb 1q ´ fp1q ´ fpbq “ 2epb,1q, so fpa bq ´ fpa ´ 1q ´ fp1q ´ fpbq “ 2epa´1,b1q 2epb,1q.
Thus either epa, bq “ epa ´ 1, b 1q and epa ´ 1, 1q “ epb, 1q or epa, bq “ epb, 1q and epa ´ 1, b 1q “ epa ´ 1, 1q.
For n ě 4, we consider these possibilities as pa, bq ranges over all pairs with a b “ n. If the first case holds for every such pair (that is, if epc, 1q “ epd, 1q for all c d “ n ´ 1), then all the epa, bq for ab “ n are equal (and the above equations do not constrain whether or not the value is the same as any epc, 1q with c 1 ă n). Otherwise, the values of epa, bq with a b “ n are fully determined by the values of epc, 1q for which epc, 1q ‰ epn ´ 1 ´ c, 1q, and are not all equal.
Specifically, if epc, 1q “ j and epn´1´c, 1q “ k with j ‰ k, we have epc, n´cq “ j “ epn´c, cq and epc 1, n ´ c ´ 1q “ k “ epn ´ c ´ 1, c 1q. Every other value of epa, bq with a b “ n is then determined by the rule that epa, bq “ epa ´ 1, b 1q if epa ´ 1, 1q “ epb, 1q: if we have epc, 1q ‰ epn ´ 1 ´ c, 1q, and epc1, 1q ‰ epn ´ 1 ´ c1, 1q, but epd, 1q “ epn ´ 1 ´ d, 1q for all c ă d ă c1, then if c ă c1 ´ 1 we have epc1 ´ 1, n ´ pc1 ´ 1qq “ epc1, n ´ c1q, then if c ă c1 ´ 2 we have epc1 ´ 2, n ´ pc1 ´ 2qq “ epc1 ´ 1, n ´ pc1 ´ 1qq “ epc1, n ´ c1q, and so on until epc 1, n ´ c ´ 1q “ epc1, n ´ c1q (yielding a contradiction if epn ´ c ´ 1, 1q ‰ epc1, 1q; such a contradiction also arises trivially if c 1 “ c1 and epn ´ c ´ 1, 1q ‰ epc1, 1q). If c is the least integer such that epc, 1q ‰ epn ´ 1 ´ c, 1q, the values of epa, bq with a ă c are similarly determined to be equal to epc, n ´ cq (and likewise for a ą n ´ c).
Shortlisted problems – solutions 21 In other words, if we list the values in ascending order of a from 1 to n ´ 1, any gaps between the pairs of adjacent values determined when epc, 1q ‰ epn ´ 1 ´ c, 1q are filled with copies of the previously determined adjacent values, and if the values on either side of such a gap are different, we have a contradiction (including in the degenerate cases where the pairs are adjacent or overlap, if c 1 “ c1). Note in particular that every value of epa, bq is a value of epc, 1q for some c with c 1 ď a b.
If |S| ě 3, that means that epc, 1q takes at least three different values. Let m be such that epm, 1q does not equal any epc, 1q for c ă m, and there are exactly two different values of epc, 1q for c ă m (and thus m ě 3).
Because epm, 1q does not equal any epc, 1q for c ă m, we have that all epa, bq for ab “ m1 are equal, and epc, 1q “ epd, 1q for all c d “ m. We now consider the values of epa, bq for ab “ m2 determined by the above rules. Since epm, 1q ‰ ep1, 1q, we have ep1, m1q “ ep1, 1q and ep2, mq “ epm, 1q. If there were any other epd, 1q ‰ epm 1 ´ d, 1q, consider the one with minimal d ą 1; because epm, 1q ‰ epd, 1q, we arrive at a contradiction. So every epc, 1q “ epd, 1q for c d “ m 1 except for epm, 1q ‰ ep1, 1q. But these equalities form a path connecting all epc, 1q for c ă m: ep1, 1q “ epm ´ 1, 1q “ ep2, 1q “ epm ´ 2, 1q “ ep3, 1q “ ¨ ¨ ¨ which contradicts the assumption we made that there were exactly two different values of epc, 1q for c ă m.
Solution 3. Constructions for 1 ď |S| ď 2 are shown in Solution 1, and S must be nonempty.
We suppose |S| ě 3 to derive a contradiction.
Claim 1. epa, bq, epb, cq, and epa, cq can take at most two different values.
Proof. By expanding fpa b cq in three different ways, we get 2epa,bq 2epc,abq “ 2epb,cq 2epa,bcq “ 2epa,cq 2epb,acq.
The result follows from the equality of the three multisets of exponents.
l For Claims 2 to 4, we fix k and let N be the smallest integer such that epa, N ´ a 1q “ k for some a ď N.
Claim 2. For any b with b ď N, we must have epb, N ´ b 1q “ k.
Proof. Suppose that epa, N ´ a 1q “ k and a ă b. Expanding fpa pb ´ aq N ´ b 1q in two different ways, we see that 2epa,b´aq 2epb,N´b1q “ 2epN´b1,b´aq 2epN´a1,aq.
By the minimality of N, we must have epb, N ´ b 1q “ epN ´ a 1, aq. The case of a ą b follows by replacing a and b with N ´ a 1 and N ´ b 1.
l Claim 3. epa, 1q “ epN ´ a 1, 1q for any a satisfying 1 ă a ă N.
Proof. By Claim 2, epa, N´a1q “ k. Then by Claim 1, epa, N´a1q, epa, 1q, and epN´a1, 1q can take at most two different values. But by the minimality of N, we must have epa, N´a1q ‰ epa, 1q “ epN ´ a 1, 1q.
l Claim 4. epa, 1q “ epN ´ a, 1q for any a satisfying 1 ď a ă N.
Proof. By Claim 2, epa, N ´ a 1q “ epa 1, N ´ aq “ k. Expanding fp1 a pN ´ aqq in two different ways, we see that 2ep1,aq 2epa1,N´aq “ 2ep1,N´aq 2epN´a1,aq.
Therefore ep1, aq “ ep1, N ´ aq, as required.
l If |S| ě 3, then there exist 1 ă Nk ă Nℓwhere Nk and Nℓare the minimal values corre-sponding to k and ℓ. But Claims 3 and 4 imply that epa, 1q is constant for all 1 ď a ă Nℓ, which is a contradiction.
22 Bath, United Kingdom, 10th–22nd July 2024 A5.
Find all periodic sequences a1, a2, . . .
of real numbers such that the following conditions hold for all n ě 1: an2 a2 n “ an a2 n1 and |an1 ´ an| ď 1.
(Kosovo) Answer: The sequences satisfying the conditions of the problem are: c, ´c, c, ´c, . . . , d, d, d, d, . . . , where c P r´ 1 2, 1 2s and d is any real number.
Solution 1. We rewrite the first condition as an2 an1 “ pan1 anqpan1 ´ an 1q.
(1) If there exists a positive integer m such that am1 am “ 0, then from equation (1) we have an1 an “ 0 for all positive integers n ě m. By the fact that the sequence pai1 aiq is periodic, we get ai1 ai “ 0 for every positive integer i. Thus the sequence paiq is of the form c, ´c, c, ´c, . . . for some |c| ď 1 2.
Now suppose that an1 an ‰ 0 for every positive integer n. Let T be the period of the sequence. From equation (1) we have 1 “ T ź i“1 ai2 ai1 ai1 ai “ T ź i“1 pai1 ´ ai 1q.
Combining with the second condition |ai1 ´ ai| ď 1, we have ai1 ´ ai 1 ą 0. Using the AM-GM inequality we get 1 “ T ź i“1 pai1 ´ ai 1q ď ˜řT i“1pai1 ´ ai 1q T ¸T “ 1.
So the equality holds, and thus we get a2 ´ a1 “ a3 ´ a2 “ ¨ ¨ ¨ “ aT1 ´ aT, which means that paiq is a constant sequence.
So all sequences satisfying the conditions of the problem are those listed above.
Shortlisted problems – solutions 23 Solution 2. Define sn “ an1 an and dn “ an1 ´ an, so the original sequence is periodic if and only if both these sequences are periodic. Rearranging the given conditions yields sn1 “ snp1 dnq and dn1 “ dnpsn ´ 1q, with |dn| ď 1 for all n.
If sn “ 0 for some n then si “ 0 for all i ě n, and for the sequence to be periodic we must have all si “ 0 and the sequence c, ´c, c, ´c, . . . , for some |c| ď 1 2. Similarly, if dn “ 0 for some n and the sequence is periodic, then all di “ 0 and the sequence is c, c, c, . . . .
We claim those are the only periodic sequences, so suppose for contradiction that we have a periodic sequence where no si or di is 0. Under this hypothesis, we will prove that psnq, pdnq have the following properties.
1. All sn are positive numbers. As |dn| ď 1 and sn1 “ snp1dnq ‰ 0, it follows that dn ą ´1 and that all sn have the same sign (all positive or all negative). If all sn are negative, then |dn1| “ |dnpsn ´ 1q| ą |dn|, so |di| is a strictly increasing sequence, contradicting periodicity.
2. Whenever dn ą 0 we have 0 ă sn ă 1. Suppose for contradiction that we have dn ą 0 and sn ě 1 for some n. Since dn1 ‰ 0 we have sn ą 1, and then dn1 ą 0, sn1 ą sn ą 1.
Inductively, all di ą 0 for i ě n, and si1 ą si for i ě n, contradicting periodicity.
Now we can get the desired contraction as follows. Suppose that the period of paiq is T, then řT i“1 di “ aT1´a1 “ 0, hence there is an n such that dn ą 0. By property 2 we get 0 ă sn ă 1, and in particular sn ă 2. Suppose that we have si ă 2. If di ă 0, then si1 “ sip1diq ă si ă 2; if di ą 0, then by property 2 we have 0 ă si ă 1, and then si1 “ sip1 diq ď 2si ă 2. In both cases we get si1 ă 2, and then by induction we get sk ă 2 for all k ě n. But then we have |dk1| “ |dkpsk ´ 1q| ă |dk|, which contradicts the fact that pdkq is periodic.
So the only periodic sequences are the two listed above.
Solution 3.
Note that if an1 “ ´an for any n, then an2 “ an “ ´an1, yielding the first answer by periodicity. Also, if an1 “ an for any n, then an2 “ an “ an1, yielding the second answer by periodicity. If an2 “ an for any n, then a2 n “ a2 n1 so an1 “ ˘an and one of those two cases applies. Henceforth, we will assume that the sequence is neither one of the answers and an ‰ an1, ´an1, an2 for all n for the rest of the solution. Note that the recursion rearranges to an2 ´ an1 “ pa2 n1 ´ an1q ´ pa2 n ´ anq “ pan1 ´ anqpan1 an ´ 1q.
(2) Claim 1. We have that an ď 1 2 for all n.
Proof. First, we cannot have an ą 1 2 for all n. Otherwise, an1 an ´ 1 is positive for all n, so (2) implies that an2 ´ an1 has the same sign as an1 ´ an for all n. This would mean that the sequence is monotonic, contradicting periodicity.
On the other hand, if an1 ď 1 2 and an2 ą 1 2, then a2 n1 “ an2 a2 n ´ an ě an2 ´ 1 4 ą 1 4 ù ñ |an1| ą 1 2, where we use the fact that t2 ´ t ě ´ 1 4 for all t P R. As an1 ď 1 2, this means that an1 ă ´ 1 2 so |an1 ´ an2| ą 1, a contradiction.
l The identity (2) now implies that an2 ´ an1 and an1 ´ an are of opposite signs for all n, so that an ă an1 ð ñ an1 ą an2.
Claim 2.
We have that an ą 0 ð ñ an1 ď 0: that is, the signs of the sequence are alternating.
24 Bath, United Kingdom, 10th–22nd July 2024 Proof. First, it cannot be the case that an ą 0 for all n. Indeed, then we would have from Claim 1 that |an1 an ´ 1| ă 1 for all n, which by (2) means that |an1 ´ an| is strictly decreasing in n, a contradiction of the sequence’s periodicity. It also cannot be the case that an ď 0 for all n, as then we would have that |an1 an ´ 1| ą 1 for all n (noting that by the nonconstant assumption we will never have an “ an1 “ 0) so |an1 ´ an| is strictly increasing in n.
Hence, if the signs of an are not alternating, then by periodicity there exists n with an ą 0 and an1, an2 ď 0 or an, an1 ą 0 and an2 ď 0. In either scenario, we have that a2 n ´ an “ a2 n1 ´ an2 ě a2 n1 ě 0 ù ñ an ą 1 as an is positive and an2 is nonpositive.
In the former case, we have that an ´ an1 ą 1, a contradiction. In the latter case, as an1 ą an2, we must have that an ă an1. But then we have that an1 ´an2 ą an ´an2 ą 1, a contradiction.
l Note now that we cannot have an2 ą ´an1 ą an for any n, as we would then have a2 n1 ´ a2 n “ an2 ´ an ą ´an1 ´ an ą 0 ù ñ an ´ an1 ą 1, a contradiction. Similarly, we cannot have an ą ´an1 ą an2 for any n, as we would then have a2 n ´ a2 n1 “ an ´ an2 ą an an1 ą 0 ù ñ an ´ an1 ą 1.
Having ruled out these scenarios, we may conclude that |an1| is not between |an| and |an2| for any n.
Let k be an index such that |ak| is maximal. Note that we cannot have |ak´2| “ |ak|, as that would imply that ak´2 “ ak by Claim 2. We also cannot have |ak´2| ď |ak´1|, as that would imply that |ak´1| is between |ak´2| and |ak|. Hence, we must have that |ak´1| ă |ak´2| ă |ak|. As |ak| is maximal, we cannot have ak “ 0. If ak ą 0, then we have that ak´ak´2 “ a2 k´1´a2 k´2 ă 0, a contradiction. If ak ă 0, then we have that ak ´ ak2 “ a2 k ´ a2 k1 ą 0, a contradiction.
Shortlisted problems – solutions 25 A6.
Let a0, a1, a2, . . . be an infinite strictly increasing sequence of positive integers such that for each n ě 1 we have an P !an´1 an1 2 , ?an´1 ¨ an1 ) .
Let b1, b2, . . . be an infinite sequence of letters defined as bn “ # A, if an “ 1 2pan´1 an1q; G, otherwise.
Prove that there exist positive integers n0 and d such that for all n ě n0 we have bnd “ bn.
(Czech Republic) Common remarks. In fact, all known proofs proceed by showing that the eventual period of the sequence pbnq always consists of some number of occurrences of G (possibly zero) followed by an A.
Such sequences of any period p ě 1 exist. Indeed, consider the sequence . . . , kp, kp´1pk 1q, . . . , kpk 1qp´1, pk 1qp, pk 1qp´1pk 2q, . . . .
The Tournament of the Towns, in Spring 2009 (Junior A-Level Paper, problem 4), considered sequences satisfying exactly this fairly natural criterion.
However, it asked a vastly easier question about them: in the language of this problem, it asked whether every such sequence had pbnq eventually constant. The answer to that problem is “no”, as heavily hinted by the statement of this problem. Thus, at least so far as the Problem Selection Committee knows, this is a novel problem about a family of sequences which has been previously considered.
Solution 1.
We will show that the eventual period of sequence pbnq consists of any fixed number of occurrences of G (possibly zero) followed by a single A.
We look at the ratios of consecutive terms of the sequence panq. Let C and D be coprime positive integers such that a1{a0 “ pC Dq{C. If bn “ G then an{an´1 “ an1{an. If bn “ A and an{an´1 “ pC kDq{pC pk ´ 1qDq for some positive integer k then an1 an “ 2an ´ an´1 an “ C pk 1qD C kD .
Thus, by induction, there is a sequence of positive integers pknq for n ě 1 which satisfies an{an´1 “ pC knDq{pC pkn ´ 1qDq for all positive integers n. Moreover, we have k1 “ 1 and kn1 “ # kn, if bn “ G; kn 1, if bn “ A.
If there are only finitely many values of n such that bn “ A then the problem statement obviously holds (we can choose d “ 1). Thus, we may assume that bn “ A for infinitely many n. This means that the sequence pknq attains all positive integer values. Given a value q ě 1, denote by mq the last index where value q occurs, that is, the index such that kmq “ q and kmq1 “ q 1.
Our aim is to prove that the sequence of differences pmq1 ´mqq is eventually constant. We first show that it is bounded above. To that end, fix t ě 1 (we will choose a suitably large t later on) and consider a sequence sptq0, sptq1, . . . defined for q ě 1 by sptqq “ amq{pC qDqt.
26 Bath, United Kingdom, 10th–22nd July 2024 We note two properties of sptqq. First, simple algebra gives sptqq1 “ amq1 pC pq 1qDqt “ amq pC pq 1qDqt ˆC pq 1qD C qD ˙mq1´mq “ amq pC qDqt ˆC pq 1qD C qD ˙mq1´mq´t “ sptqq ˆC pq 1qD C qD ˙mq1´mq´t .
It follows that sptqq ą sptqq1 sptqq “ sptqq1 sptqq ă sptqq1 , / .
/ -if and only if $ ’ & ’ % mq1 ´ mq ă t, mq1 ´ mq “ t, mq1 ´ mq ą t.
Second, suppose that mq1 ´ mq ě t for some positive integer q. We claim that in that case sptqq is a positive integer. Indeed, we have amqt “ amq ˆC pq 1qD C qD ˙t , because kmq1 “ kmq2 “ ¨ ¨ ¨ “ kmqt “ q 1. Since C pq 1qD and C qD are coprime we have that sptqq “ amq pC qDqt is an integer.
We choose T ě 1 such that spTq1 ă 1 (which exists since C D ą 1). Then, by induction we can show that spTqq ă 1 for all q. Indeed, since spTqq ă 1, it is not a positive integer; this means that mq1 ´ mq ă T by the second property above. Hence by the first property above we have spTqq1 ă spTqq ă 1, as needed.
This means that mq1 ´ mq ă T for all q. Thus there is a largest integer T 1 ď T with the property that an equality mq1 ´ mq “ T 1 holds for infinitely many values of q.
Therefore, for all sufficiently large values of q we have the inequality mq1 ´ mq ď T 1, which by the first property implies that the sequence spT 1q is decreasing from some point on.
Moreover, we know that the sequence attains infinitely many integer values since there are infinitely many values of q for which we have the equality mq1 ´ mq “ T 1. As a consequence, the sequence spT 1q is constant from some sufficiently large index Q onwards.
This in turn means that the equality mq1 ´ mq “ T 1 holds for all q ě Q. Note that bn “ A is equivalent to the fact that n “ mq for some integer q. Thus, the sequence pbnq is periodic for n ě Q with period T 1, and the proof is complete.
Solution 2. First, observe that the statement holds immediately if bn “ G for all n; otherwise, there must be some n for which bn “ A. Without loss of generality, we may assume that n “ 1, as we can translate the sequence without affecting the statement.
We define an arithmetic sequence ppnq by taking p0 “ a0{ gcdpa0, a1q and p1 “ a1{ gcdpa0, a1q.
Note that p0 ă p1, and hence that ppnq is an increasing sequence of positive integers, and also that p2 “ a2{ gcdpa0, a1q.
We also define a sequence of positive integers dn “ an ´ an´1 and a sequence of positive rational numbers qn “ an{an´1.
Then the following facts are immediate consequences of the definitions: • if bn “ G, then qn1 “ qn and dn1 “ dnqn; • if bn “ A, then dn1 “ dn; Shortlisted problems – solutions 27 • q1 “ p1{p0; • if bn “ A and qn “ pi{pi´1, then qn1 “ pi1{pi.
Now, let ki be the number of integers n for which bn “ G and qn “ pi{pi´1. If some ki is infinite then bn is eventually always G; otherwise, all values of ki are nonnegative integers.
The sequence of values for dn can be written as d0, d0 p1 p0 , . . . , d0 ˆp1 p0 ˙k1 , d0 ˆp1 p0 ˙k1 p2 p1 , . . . , d0 ˆp1 p0 ˙k1 ˆp2 p1 ˙k2 , . . .
and in particular all terms in this sequence are positive integers. Furthermore, pi and pi1 are coprime for all i, so the following sequence consists entirely of positive integers: u0 “ d0p´k1 0 , u1 “ d0p´k1 0 pk1´k2 1 , u2 “ d0p´k1 0 pk1´k2 1 pk2´k3 2 , .
.
. .
We will prove that ki is eventually constant, which implies that the sequence of bn is eventu-ally periodic with period consisting of k copies of G followed by an A (where k is that constant value).
Observe that either ki is unbounded, or is bounded with eventual maximum k for some constant k. In the second case, let r0 be minimal such that kr0 “ k; in the first case let r0 “ 0.
We will construct an infinite sequence of integers as follows: • If kri1 ě kri, then ri1 “ ri 1 • If kri1 ă kri, then ri1 is the minimal positive integer greater than ri such that kri1 ě kri.
Observe that in the second case, such an ri1 must exist by our construction of r0.
We claim that uri1 ď uri with equality only if kri1 “ kri (so ri1 “ ri 1). Indeed, if kri1 ě kri then uri1 “ uri1 “ urip kri´kri1 ri ď uri, with equality if and only if kri “ kri1.
Otherwise, we have uri1 uri “ p kri´kri1 ri p kri1´kri2 ri1 ¨ ¨ ¨ p kri1´1´kri1 ri1´1 , so we just need to show that the right hand side is strictly less than 1. But this follows because p kri´kri1 ri p kri1´kri2 ri1 ¨ ¨ ¨ p kri1´1´kri1 ri1´1 ă p kri´kri2 ri1 p kri2´kri3 ri2 ¨ ¨ ¨ p kri1´1´kri1 ri1´1 ă p kri´kri3 ri2 p kri3´kri4 ri3 ¨ ¨ ¨ p kri1´1´kri1 ri1´1 .
.
.
ă p kri´kri1 ri1´1 ď 1, where each inequality besides the last follows from the fact that pj ă pj1 and kri ą kj for j ă ri1, and the last follows from the fact that kri ď kri1.
Finally, the sequence uri is an infinite nonincreasing sequence of positive integers so must eventually be constant, yielding the claim.
Comment. The two solutions above differ in approach, but have some overlap in the structure they reveal. Indeed, the C nD of Solution 1 is the pn of Solution 2, while the mr1 ´ mr of Solution 1 turns out to be equal to the kr of Solution 2.
28 Bath, United Kingdom, 10th–22nd July 2024 A7.
Let Q be the set of rational numbers. Let f : Q Ñ Q be a function such that the following property holds: for all x, y P Q, fpx fpyqq “ fpxq y or fpfpxq yq “ x fpyq.
Determine the maximum possible number of elements of tfpxq fp´xq | x P Qu.
(Japan) Answer: 2 is the maximum number of elements.
Common remarks.
Suppose that f is a function satisfying the condition of the problem.
We will use the following throughout all solutions.
• a „ b if either fpaq “ b or fpbq “ a, • a Ñ b if fpaq “ b, • Ppx, yq to denote the proposition that either fpx fpyqq “ fpxq y or fpfpxq yq “ x fpyq, • gpxq “ fpxq fp´xq.
With this, the condition Ppx, yq could be rephrased as saying that x fpyq „ fpxq y, and we are asked to determine the maximum possible number of elements of tgpxq | x P Qu.
Solution 1. We begin by providing an example of a function f for which there are two values of gpxq. We take the function fpxq “ txu ´ txu, where txu denotes the floor of x (that is, the largest integer less than or equal to x) and txu “ x ´ txu denotes the fractional part of x.
First, we show that f satisfies Ppx, yq. Given x, y P Q, we have fpxq y “ txu ´ txu tyu tyu “ ptxu tyuq ptyu ´ txuq; x fpyq “ txu txu tyu ´ tyu “ ptxu tyuq ptxu ´ tyuq.
If txu ă tyu, then we have that the fractional part of fpxq y is tyu ´ txu and the floor is txu tyu, so fpxq y Ñ x fpyq. Likewise, if txu ą tyu, then x fpyq Ñ fpxq y. Finally, if txu “ tyu, then fpxq y “ x fpyq “ txu tyu is an integer. In all cases, the relation P is satisfied.
Finally, we observe that if x is an integer then gpxq “ 0, and if x is not an integer then gpxq “ ´2, so there are two values for gpxq as required.
Now, we prove that there cannot be more than two values of gpxq. Ppx, xq tells us that x fpxq „ x fpxq, or in other words, for all x, fpx fpxqq “ x fpxq.
(1) We begin with the following lemma.
Lemma 1. f is a bijection, and satisfies fp´fp´xqq “ x.
(2) Shortlisted problems – solutions 29 Proof. We first prove that f is injective. Suppose that fpx1q “ fpx2q; then Ppx1, x2q tells us that fpx1qx2 „ fpx2qx1. Without loss of generality, suppose that fpx1qx2 Ñ fpx2qx1.
But fpx1q “ fpx2q, so fpfpx1q x2q “ fpfpx2q x2q “ fpx2q x2 by (1). Therefore, fpx2q x1 “ fpx2q x2, as required.
Now, (1) with x “ 0 tells us that fpfp0qq “ fp0q and so by injectivity fp0q “ 0.
Applying Ppx, ´fpxqq tells us that 0 „ x fp´fpxqq, so either 0 “ fp0q “ x fp´fpxqq or fpx fp´fpxqqq “ 0 which implies that x fp´fpxqq “ 0 by injectivity. Either way, we deduce that x “ ´fp´fpxqq, or x “ fp´fp´xqq by replacing x with ´x.
Finally, note that bijectivity follows immediately from (2).
l Since f is bijective, it has an inverse, which we denote f ´1. Rearranging (2) (after replacing x with ´x) gives that fp´xq “ ´f ´1pxq. We have gpxq “ fpxq fp´xq “ fpxq ´ f ´1pxq.
Suppose gpxq “ u and gpyq “ v, where u ‰ v are both nonzero. Define x1 “ f ´1pxq and y1 “ f ´1pyq; by definition, we have x1 Ñ x Ñ x1 u y1 Ñ y Ñ y1 v.
Putting in Ppx1, yq gives xy „ x1 y1v, and putting in Ppx, y1q gives xy „ x1y1 u.
These are not equal since u ‰ v, and x y may have only one incoming and outgoing arrow because f is a bijection, so we must have either x1 y1 u Ñ x y Ñ x1 y1 v or the same with the arrows reversed. Swapping px, uq and py, vq if necessary, we may assume without loss of generality that this is the correct direction for the arrows.
Also, we have ´x1 ´ u Ñ ´x Ñ ´x1 by Lemma 1. Putting in Ppx y, ´x1 ´ uq gives y „ y1 v ´ u, and so y1 v ´ u must be either y1 v or y1. This means u must be either 0 or v, and this contradicts our assumption about u and v.
Comment. Lemma 1 can also be proven as follows. We start by proving that f must be surjective.
Suppose not; then, there must be some t which does not appear in the output of f. Ppx, t ´ fpxqq tells us that t „ x fpt ´ fpxqq, and so by assumption fptq “ x fpt ´ fpxqq for all x. But setting x “ fptq ´ t gives t “ fpt ´ fpfptq ´ tqq, contradicting our assumption about t.
Now, choose some t such that fptq “ 0; such a t must exist by surjectivity. Ppt, tq tells us that fptq “ t, or in other words t “ 0 and fp0q “ 0. The remainder of the proof is the same as the proof given in Solution 1.
Solution 2. We again start with Lemma 1, and note fp0q “ 0 as in the proof of that lemma.
Ppx, ´fpyqq gives xfp´fpyqq „ fpxq´fpyq, and using (2) this becomes x´y „ fpxq´fpyq.
In other words, either fpx ´ yq “ fpxq ´ fpyq or x ´ y “ fpfpxq ´ fpyqq. In the latter case, we deduce that fp´px ´ yqq “ fp´fpfpxq ´ fpyqqq fpy ´ xq “ fp´fpfpxq ´ fpyqqq “ fpyq ´ fpxq.
Thus, fpyq ´ fpxq is equal to either fpy ´ xq or ´fpx ´ yq. Replacing y with x d, we deduce that fpx dq ´ fpxq P tfpdq, ´fp´dqu.
Now, we prove the following claim.
Claim. For any n P Zą0 and d P Q, we have that either gpdq “ 0 or gpdq “ ˘gpd{nq.
In particular, if gpd{nq “ 0 then gpdq “ 0.
30 Bath, United Kingdom, 10th–22nd July 2024 Proof. We first prove that if gpd{nq “ 0 then gpdq “ 0. Suppose that gpd{nq “ 0. Then fpd{nq “ ´fp´d{nq and so fpx d{nq ´ fpxq “ fpd{nq for any x. Applying this repeatedly, we deduce that fpx dq ´ fpxq “ nfpd{nq for any x. Applying this with x “ 0 and x “ ´d and adding gives fpdq fp´dq “ 0, so gpdq “ 0, and in particular the claim is true whenever gpdq “ 0.
Now, select n P Zą0 and d P Q such that gpdq ‰ 0, and observe that we must have gpd{nq ‰ 0. Observe that for any k P Z we have that fpkd{nq ´ fppk ´ 1qd{nq P tfpd{nq, ´fp´d{nqu.
Let Ai be the number of k P Z with i ´ n ă k ď i such that this difference equals fpd{nq.
We deduce that for any i P Z, fpid{nq ´ fpid{n ´ dq “ ÿ i´năkďi fpkd{nq ´ fppk ´ 1qd{nq “ Aifpd{nq ´ pn ´ Aiqfp´d{nq “ ´nfp´d{nq Aigpd{nq.
Since gpd{nq is nonzero, this is a nonconstant linear function of Ai. However, there are only two possible values for fpid{nq ´ fpid{n ´ dq, so there must be at most two possible values for Ai as i varies. And since Ai1 ´ Ai P t´1, 0, 1u, those two values must differ by 1 (if there are two values).
Now, we have fpdq ´ fp0q “ ´nfp´d{nq Angpd{nq, and fp0q ´ fp´dq “ ´nfp´d{nq A0gpd{nq.
Subtracting these (using the fact that fp0q “ 0) we obtain fpdq fp´dq “ pAn ´ A0qgpd{nq “ ˘gpd{nq, where the last line follows from the fact that gpdq is nonzero.
l It immediately follows that there can only be one nonzero number of the form gpxq up to sign; to see why, if gpdq and gpd1q are both nonzero, then for some n, n1 P Zą0 we have d{n “ d1{n1. But gpdq “ ˘gpd{nq “ ˘gpd1q.
Finally, suppose that for some d, d1 we have gpdq “ c and gpd1q “ ´c for some nonzero c.
So we have fpdq fp´dq ´ fpd1q ´ fp´d1q “ 2c which rearranges to become pfpdq ´ fpd1qq ´ pfp´d1q ´ fp´dqq “ 2c.
Each of the bracketed terms must be equal to either fpd ´ d1q or ´fpd1 ´ dq. However, they cannot be equal since c is nonzero, so gpd ´ d1q “ fpd ´ d1q fpd1 ´ dq “ ˘2c. This contradicts the assertion that gp´xq “ ˘c for all x.
Solution 3. As in Solution 1, we start by establishing Lemma 1 as above, and write f ´1pxq “ ´fp´xq for the inverse of f, and gpxq “ fpxq ´ f ´1pxq.
We now prove the following.
Lemma 2. If gpxq ‰ gpyq, then gpx yq “ ˘pgpxq ´ gpyqq.
Shortlisted problems – solutions 31 Proof. Assume x and y are such that gpxq ‰ gpyq.
Applying Ppx, f ´1pyqq gives x y „ fpxq f ´1pyq, and applying Ppf ´1pxq, yq gives x y „ f ´1pxq fpyq.
Observe that pfpxq f ´1pyqq ´ pf ´1pxq fpyqq “ pfpxq ´ f ´1pxqq ´ pfpyq ´ f ´1pyqq “ gpxq ´ gpyq.
By assumption, gpxq ‰ gpyq, and so fpxq f ´1pyq ‰ f ´1pxq fpyq. Since f is bijective, this means that these two values must be fpx yq and f ´1px yq in some order, and so gpx yq “ fpx yq ´ f ´1px yq must be their difference up to sign, which is either gpxq ´ gpyq or gpyq ´ gpxq.
l Claim. If x and q are rational numbers such that gpqq “ 0 and n is an integer, then gpxnqq “ gpxq.
Proof. If gpbq “ 0 and gpaq ‰ gpa bq, then the lemma tells us that gpbq “ ˘pgpa bq ´ gpaqq, which contradicts our assumptions. Therefore, gpaq “ gpa bq whenever gpbq “ 0.
A simple induction then gives that gpnbq “ 0 for any positive integer n, and gpnbq “ 0 for negative n as gpxq “ gp´xq. The claim follows immediately.
l 32 Bath, United Kingdom, 10th–22nd July 2024 Lemma 3. There cannot be both positive and negative elements in the range of g.
Proof. Suppose that gpxq ą 0 and gpyq ă 0. Let S be the set of numbers of the form mx ny for integers m, n.
We first show that gpSq has infinitely many elements.
Indeed, suppose gpSq is finite, and let a P S maximise g and b P S maximise ´g.
Then a b P S, and gpa bq “ gpaq ´ gpbq or gpbq ´ gpaq. In the first case gpa bq ą gpaq and in the second case gpa bq ă gpbq; in either case we get a contradiction.
Now, we show that there must exist some nonzero rational number q with gpqq “ 0. Indeed, suppose first that afpaq “ 0 for all a. Then gpaq “ fpaqfp´aq “ 0 for all a, and so g takes no nonzero value. Otherwise, there is some a with afpaq ‰ 0, and so (1) yields that fpqq “ 0 for q “ a fpaq ‰ 0. Noting that fp´qq “ 0 from Lemma 1 tells us that gpqq “ 0, as required.
Now, there must exist integers s and s1 such that xs “ qs1 and integers t and t1 such that yt “ qt1. The claim above gives that the value of gpmx nyq depends only on the values of m mod s and n mod t, so gpmx nyq can only take finitely many values.
l Finally, suppose that gpxq “ u and gpyq “ v where u ‰ v have the same sign. Assume u, v ą 0 (the other case is similar) and assume u ą v without loss of generality.
Ppf ´1pxq, f ´1pyqq gives x ´ y „ f ´1pxq ´ f ´1pyq “ fpxq ´ fpyq ´ pu ´ vq, and Ppx, yq gives x ´ y „ fpxq ´ fpyq. u ´ v is nonzero, so fpx ´ yq and f ´1px ´ yq must be fpxq ´ fpyq ´ pu ´ vq and fpxq ´ fpyq in some order, and since gpx ´ yq must be nonnegative, we have fpxq ´ fpyq ´ pu ´ vq Ñ x ´ y Ñ fpxq ´ fpyq.
Then, Ppx ´ y, f ´1pyqq tells us that px ´ yq y „ pfpxq ´ fpyqq pfpyq ´ vq, so x „ fpxq ´ v, contradicting either v ‰ u or v ą 0.
Comment. Lemma 2 also follows from fpx dq ´ fpxq P tfpdq, ´fp´dqu as proven in Solution 2.
Indeed, we also have fp´xq ´ fp´x ´ dq P tfpdq, ´fp´dqu, and then subtracting the second from the first we get gpxdq´gpxq P tgpdq, ´gpdq, 0u. Replacing xd and x with x and ´y gives the statement of Lemma 2.
Comment. It is possible to prove using Lemma 2 that g must have image of the form t0, c, 2cu if it has size greater than 2. Indeed, if gpxq “ c and gpyq “ d with 0 ă c ă d, then gpx yq “ d ´ c as it must be nonnegative, and gpyq “ gppx yq p´xqq “ |d ´ 2c| provided that d ‰ 2c.
However, it is not possible to rule out t0, c, 2cu based entirely on the conclusion of Lemma 2; indeed, the function given by gpxq “ $ ’ & ’ % 0, if x “ 2n for n P Z; 2, if x “ 2n 1 for n P Z; 1, if x R Z.
satisfies the conclusion of Lemma 2 (even though there is no function f giving this choice of g).
Note. Solution 1 actually implies that the result also holds over R.
The proposal was originally submitted and evaluated over Q as it is presented here, and the Problem Selection Committee believes that this form is more suitable for the competition because it allows for more varied and interesting approaches once Lemma 1 has been established. Even the variant here defined over Q was found to be fairly challenging.
Shortlisted problems – solutions 33 A8.
Let p ‰ q be coprime positive integers. Determine all infinite sequences a1, a2, . . . of positive integers such that the following conditions hold for all n ě 1: maxpan, an1, . . . , anpq ´ minpan, an1, . . . , anpq “ p and maxpan, an1, . . . , anqq ´ minpan, an1, . . . , anqq “ q.
(Japan) Answer: The only such sequences are an “ n C, where C is a nonnegative integer.
Common remarks.
• Denote by ari,js the subsequence ai, ai1, . . . , aj.
• Without loss of generality, in each solution we suppose p ă q. It can be convenient to treat the case where p “ 1 separately.
• The problem can also be posed for sequences of arbitrary integers (rather than positive).
Refer to the comment after Solution 1 for a proof.
Solution 1. Let k “ r q ps. Note that k ě 2.
Lemma 1. If i, j and m are positive integers such that |i ´ j| ď mp then |ai ´ aj| ď mp.
Proof. By the given condition, if |i ´ j| ď p then |ai ´ aj| ď p. So the lemma follows from induction on m and the triangle inequality.
l Lemma 2. For a fixed n, suppose that ai is minimal over i ě n. Then i ď n p ´ 1.
Proof. Suppose for contradiction that i ě n p. Then minpari´p,iq´psq “ ai. Since q ´ p ď pk ´ 1qp, it follows from Lemma 1 that maxpari´p,iq´psq ď ai pk ´ 1qp ă ai q, which is a contradiction.
l Lemma 3. For a fixed n ą q, suppose that ai is maximal over i ď n. Then i ě n ´ p 1.
Proof. Suppose aj is minimal over j ě n ´ q. Then by Lemma 2, j ď n ´ q p ´ 1. So minparn´q,nsq “ aj and ai ě maxparn´q,nsq, which implies that ai ě aj q.
Lemma 2 also implies that if j ě n then aj ě minparn,npsq. So if i ă j, then we have aj ě ai ´ p, which contradicts ai ě aj q. Hence we must have i ą j.
The above inequality also gives |ai ´ aj| ě q ą pk ´ 1qp, so by Lemma 1 it follows that |i ´ j| ą pk ´ 1qp. Therefore i ą j pk ´ 1qp ě n ´ q pk ´ 1qp ě n ´ p 1.
l Let bn be the minimal value of ai for i ě n. By Lemma 2, bnp ą bn for all n. Hence bn “ minparn,npsq “ minparn,nqsq. Let cn be the maximal value of ai for i ď n. By Lemma 3, cn´p ą cn for all n ą q. Hence cn “ maxparn´p,nsq “ maxparn´q,nsq for n ą q.
So if n ą q then bn “ cnp ´ p “ cnq ´ q. So for n ą q we get bnq´p p “ cnq “ bn q, and hence bnq´p “ bn q ´ p.
Next note that bnp ď anp ď bnp. So bnp´bn ď p for all n ą q, and iterating this pq´pq times gives bnppq´pq´bn ď ppq´pq. But using bnq´p “ bnq´p gives bnppq´pq´bn “ ppq´pq.
Since equality occurs, we must have bnp “ bn p.
So for n ą q, bnp “ bn p and bnq´p “ bn q ´ p. Since p and q ´ p are coprime, bn1 “ bn 1 for all n ą q. The only way for bn and bn1 to be different is if bn “ an, so we deduce that an1 “ an 1 and there is a constant C such that an “ n C for all n ą q.
Finally, suppose an “ n C for all n ě N.
Then p “ maxpaN´1, N C p ´ 1q ´ minpaN´1, N Cq. So aN´1 “ N C p or N C ´ 1. Similarly, aN´1 “ N C q or N C ´ 1. Hence aN´1 “ N C ´ 1. So, by induction, we have an “ n C for all positive integers n. Since a1 ě 1, C is a nonnegative integer.
It is trivial to check that an “ n C satisfies the given condition.
34 Bath, United Kingdom, 10th–22nd July 2024 Comment. Here is a variant of Solution 1. Proceed up to proving bn ´ n is eventually periodic with period q ´ p. Then there is some minimal value of bn ´ n. Suppose n attains this minimal value. Since bnp ´ n ´ p ď bn ´ n, n p also attains this minimal value. And since p and q ´ p are coprime, all n ě q must attain this minimal value. Hence bn1 “ bn 1 for all n ě q. Finish as above.
Comment. It is also possible to solve the problem using a weaker version of Lemma 2 and without Lemma 3. For example, the following lemma plays a similar role.
Lemma 2’. Let b1 n “ minparn,npsq. Then b1 n ă b1 np.
Comment. To solve the problem for sequences an of arbitrary integers, we will use the following lemma.
Lemma 4. The sequence an is either bounded above or bounded below.
Proof. Suppose that an is unbounded above and below. Then there is some i such that ai ă a1 ´ p.
There is also some j such that aj ą maxpar1,isq q. Now let al be minimal amongst ar1,js. Since al ď ai, al ă a1 ´ p and al ă aj ´ kp. By Lemma 1, 1 p ă l ă j ´ kp. So minparl´p,lq´psq “ al. By Lemma 1 again, maxparl´p,lq´psq ď al pk ´ 1qp ă al q, which is a contradiction.
l From there, the solution above can be adapted to prove that an “ n C for all n or an “ ´n C for all n, where C can be any constant integer.
Solution 2.
For n, x ě 1, let the x-width of n be maxparn,nxsq ´ minparn,nxsq. We call a positive integer x good if the x-width of n is less than or equal to x for all sufficiently large n, and we call x very good if the x-width of n is equal to x for sufficiently large n.
Lemma 1. If p1 is good and q1 is very good with p1 ă q1 ă 2p1, then 2p1 ´ q1 is also good.
Proof. Note that 0 ă q1 ´ p1 ă p1 ă q1. Let n be a sufficiently large positive integer. Then for k P rn q1 ´ p1, n p1s, we have ak ě maxparn,np1sq ´ p1 and ak ě maxparnq1´p1,nq1sq ´ p1 since p1 is good, which shows ak ě maxparn,nq1sq ´ p1. Similarly we get ak ď minparn,nq1sq p1.
Therefore, for all k P rnq1 ´p1, np1s we have ak P rmaxparn,nq1sq´p1, minparn,nq1sqp1s.
Thus, the p2p1 ´ q1q-width of n q1 ´ p1 is at most pminparn,nq1sq p1q ´ pmaxparn,nq1sq ´ p1q “ 2p1 ´ q1. The lemma follows.
l Lemma 2.
Let p1 be a good number and q1 a very good number with p1 ă q1. For sufficiently large n, take s, t P rn, n q1s such that minparn,nq1sq “ as and maxparn,nq1sq “ at. Then s P rn, n p1s and t P rn q1 ´ p1, n q1s.
Proof. Lemma 2 and Lemma 3 from Solution 1 hold with p and q replaced by p1 and q1 by similar arguments. We can deduce the statement about s from Lemma 2 of Solution 1. We can deduce the statement about t from Lemma 3 of Solution 1.
l Lemma 3. If p1 is good and q1 is very good with 2p1 ă q1, then there exists a positive integer r such that for all sufficiently large n, we have anr ´ an ě r.
Proof. Let r “ q1 ´ 2p1, and let s and t be as defined in Lemma 2. Then consider the identity pat ´ anq1´p1q panp1r ´ anp1q panp1 ´ asq “ at ´ as “ q1.
By Lemma 2, we have s P rn, n p1s and t P rn q1 ´ p1, n q1s, so anp1 ´ s ď p1 and at ´ anq1´p1 ď p1. Combining these, we get anp1r ´ anp1 ě q1 ´ 2p1 “ r. This proves that anr ´ an ě r for sufficiently large n.
l Lemma 4. Suppose pp, qq ‰ p1, 2q. Then there exists a good number p1 such that 2p1 ă q.
Proof. Let p1 be the smallest good positive integer. Note that p is good, so p1 exists and is less than q.
Suppose for contradiction that 2p1 ě q. If 2p1 ą q, then by Lemma 1, 2p1 ´ q is a good number strictly less than p1, which contradicts minimality of p1. If 2p1 “ q, then p1 ă p ă 2p1.
So we can apply Lemma 1 with q0 “ p to get that 2p1 ´ p is a good number that is strictly less than p1, which again contradicts minimality.
l Shortlisted problems – solutions 35 If pp, qq “ p1, 2q then the problem is easily solved. Otherwise, Lemmas 3 and 4 combined give us some r ą 0 such that anr ´ an ě r for n sufficiently large.
By iterating, we get anpr ´ an ě pr for all sufficiently large n, and hence it follows that anp ´ an “ p. Similarly we get anq ´ an “ q. As p and q are coprime, we deduce that an1 ´ an “ 1 for sufficiently large n. Thus we get an “ n C for sufficiently large n, and we can conclude by the same argument as Solution 1.
36 Bath, United Kingdom, 10th–22nd July 2024 Combinatorics C1.
Let n be a positive integer. A class of n students run n races, in each of which they are ranked with no draws. A student is eligible for a rating pa, bq for positive integers a and b if they come in the top b places in at least a of the races. Their final score is the maximum possible value of a ´ b across all ratings for which they are eligible.
Find the maximum possible sum of all the scores of the n students.
(Australia) Answer: The maximum possible sum is npn´1q 2 .
Solution 1. The answer can be achieved by the students finishing in the same order in every race. To show that this is the maximum, we will apply a series of modifications to the results of the races, each of which does not decrease the total score, such that after k such modifications the first k positions are the same in every race. Say that a student is scored on the bth place if their score is a ´ b because they came in the top b places in a of the races and b is minimal with this property for that student.
Supposing that the first k ´ 1 positions are the same in every race, look at the students scored on the kth place. If there are no such students, let ℓą k be minimal such that some student S is scored on the ℓth place. Then, in every race where S appears in any place from the kth through the ℓth inclusive (of which there must be at least ℓ, otherwise S would achieve a higher rating of 0 based on the nth place), reorder the students in places k through ℓso that S finishes in the kth place instead (and otherwise the ordering of those students is arbitrary).
Now S is scored on the kth place, their score has gone up by ℓ´ k and no other scores have gone down (some might have gone up as well).
Now we know that the first k ´ 1 positions are the same in every race and at least one student is scored on the kth place. Pick one such student S. In each race where S finishes behind the kth place, swap them with the student T who finishes in the kth place, leaving the positions of all other students unchanged. Each such swap increases the score of S by 1 and decreases the score of T by at most 1, so such swaps do not decrease the total score. At the end of this process, the first k positions are the same in every race and the total score has not decreased.
Repeating this n times yields the required result.
Comment. The following simpler approach to modifying results of races is tempting: find pairs of students S and T who are scored on places k and ℓrespectively, where k ă ℓ, but where S finishes after T in some race, and swap the positions of those two students in that race so they finish in the same order as the places they are scored on. However, such a swap can decrease the total score; for example, suppose that k “ 1 and ℓ“ 4, and in some race S finishes 6th and T finishes 3rd; then swapping those students reduces the number of races contributing to T’s score without increasing the number contributing to S’s score.
Solution 2. The answer can be achieved by having the same ranking for all n races.
Note that taking a “ b “ n shows each student has a nonnegative score. Consider a student who has race ranks r1, r2, . . . , rn and a final score of s. We first prove that ÿ i ri ď npn ´ sq.
Without loss of generality, suppose that r1 ď r2 ď ¨ ¨ ¨ ď rn. There must exist some k with s 1 ď k ď n and k ´ rk “ s. In order to maximise ř i ri while retaining the score of s, we can Shortlisted problems – solutions 37 replace each of r1, . . . , rk´1 by rk, and replace each of rk1, . . . , rn by n. Then the sum is ÿ i ri ď krk pn ´ kqn “ n2 ´ kpn ´ rkq “ n2 ´ kpn s ´ kq ď n2 ´ sn.
(1) The final inequality follows from the fact that given s 1 ď k ď n, the quantity kpn s ´ kq is minimised when k “ n.
The sum of ranks of all students across all races is n2pn1q 2 . If the total of all student scores is t, then (1) implies n2pn 1q 2 ď n3 ´ tn.
This rearranges to t ď npn´1q 2 , as required.
Solution 3.
In each race, assign the student in the kth place a weight of 1 ´ k n. If a student finishes in the top b places in at least a of the races, the total of their weights is at least a 1 ´ b n ˘ “ a ´ b a n ˘ ě a ´ b. Thus the sum of a student’s weights across all races is at least their score, and so the sum of all weights for all students across all races is at least the sum of all the scores of all students. The sum of weights in each race is n´1 2 , so the sum of all weights across all races is npn´1q 2 . Equality is achieved if and only if, for each student, the values of b and a determining that student’s score have a “ n and they finish in exactly the bth place in all n races; that is, if the students are ranked the same in every race.
Solution 4.
Given a positive integer bpSq for each student S, define abpSq to be the number of races in which S finished in the top bpSq places, and define scorebpSq “ abpSq ´ bpSq; for a race r, let IbpS, rq be 1 if S finished in the top bpSq places in race r and 0 otherwise, so abpSq “ ÿ r IbpS, rq.
Then the problem asks for the maximum across all possible results of the races of max b ÿ S scorebpSq “ max b ˜ÿ r ÿ S IbpS, rq ´ ÿ S bpSq ¸ .
Given b, the sum ř S IbpS, rq is maximised (not necessarily uniquely) for some choice of the rankings in race r, which is the same choice for every race. So the maximum possible sum of the scores of all the students occurs when all students are ranked the same in all races, which yields the given answer.
38 Bath, United Kingdom, 10th–22nd July 2024 C2.
Let n be a positive integer. The integers 1, 2, 3, . . . , n2 are to be written in the cells of an n ˆ n board such that each integer is written in exactly one cell and each cell contains exactly one integer. For every integer d with d | n, the d-division of the board is the division of the board into pn{dq2 nonoverlapping sub-boards, each of size d ˆ d, such that each cell is contained in exactly one d ˆ d sub-board.
We say that n is a cool number if the integers can be written on the n ˆ n board such that, for each integer d with d | n and 1 ă d ă n, in the d-division of the board, the sum of the integers written in each d ˆ d sub-board is not a multiple of d.
Determine all even cool numbers.
(Türkiye) Answer: The even cool numbers are n “ 2k where k is a positive integer.
Solution. We first show by induction that n “ 2k is a cool number. The base case of n “ 2 is trivial as there is no such d.
For induction, assume that 2k is a cool number. We construct a numbering of a 2k1 ˆ 2k1 board that satisfies the conditions.
Take the 2k1 ˆ2k1 board and divide it into four 2k ˆ2k sub-boards. By assumption, there is some numbering P of a 2k ˆ2k board that satisfies the required condition; we write down the numbering P in each sub-board. Next, add 22k to every number in the second sub-board, add 2 ˆ 22k to every number in the third sub-board, and add 3 ˆ 22k to every number in the fourth sub-board. Then the numbers in the cells of the 2k1 ˆ2k1 board are the numbers 1 to 22pk1q.
Now locate 22k from the first sub-board, and swap it with 22k 2k´1 from the second sub-board. Locate 3ˆ22k from the third sub-board, and swap it with 3ˆ22k 2k´1 from the fourth sub-board.
We claim that this numbering of the 2k1 ˆ2k1 board satisfies the required conditions. For any d “ 2i where i ă k, consider any 2i ˆ 2i sub-board. The sum of its cells modulo 2i is not changed in the addition step or the swapping step, so the sum is congruent modulo 2i to the sum of the corresponding 2i ˆ 2i sub-board in P, which is nonzero, as required.
In the case of d “ 2k, we can directly evaluate the sum of the pb 1qth sub-board for b P t0, 1, 2, 3u. The sum is given by 22k´1p1 22kq b24k p´1qb2k´1 ” 2k´1 pmod 2kq.
Therefore all sub-boards satisfy the required conditions and so 2k1 is a cool number, completing the induction.
It remains to show that no other even number is a cool number. Let n “ 2sm where s is a positive integer and m is an odd integer greater than 1. For the sake of contradiction, suppose that there is a numbering of the n ˆ n board satisfying the required conditions.
Claim. In the 2i-division of the board, where 1 ď i ď s, the sum of numbers in each 2i ˆ 2i sub-board is congruent to 2i´1 modulo 2i.
Proof. We prove the claim by induction on i. The base case of i “ 1 holds as the sum of numbers in each 2 ˆ 2 sub-board must be odd. Next, suppose the claim is true for 2i. In the 2i1-division, each 2i1 ˆ 2i1 sub-board is made up of four 2i ˆ 2i sub-boards, each with a sum congruent to 2i´1 modulo 2i. Hence the sum of each 2i1 ˆ 2i1 sub-board is a multiple of 2i.
It cannot be a multiple of 2i1 because of the conditions, which means it must be congruent to 2i modulo 2i1. This proves the claim.
l Shortlisted problems – solutions 39 Back to the problem, since m is odd, summing up the m2 sums of 2s ˆ 2s sub-boards gives 2s´1m2 ” 2s´1 pmod 2sq.
However, the sum of the numbers from 1 to n2 is n2pn2 1q 2 “ 22s´1m2p22sm2 1q ” 0 pmod 2sq.
This is a contradiction. Therefore n is not a cool number.
Comment. In the case of odd n, similar arguments show that prime powers are cool numbers.
If the definition of cool numbers additionally requires that all d ˆ d sub-boards in the d-division have the same nonzero residue modulo d, then the cool numbers are precisely the prime powers.
40 Bath, United Kingdom, 10th–22nd July 2024 C3.
Let n be a positive integer. There are 2n knights sitting at a round table. They consist of n pairs of partners, each pair of which wishes to shake hands. A pair can shake hands only when next to each other. Every minute, one pair of adjacent knights swaps places.
Find the minimum number of exchanges of adjacent knights such that, regardless of the initial arrangement, every knight can meet her partner and shake hands at some time.
(Belarus) Answer: The minimum number of exchanges is npn´1q 2 .
Common remarks.
The solution is divided into three lemmas. We provide multiple proofs of each lemma.
Solution. Join each pair of knights with a chord across the table. We’ll refer to these chords as chains.
First we show that npn ´ 1q{2 exchanges are required for some arrangements.
Lemma 1. If each knight is initially sitting directly opposite her partner, then at least npn´1q{2 exchanges are required for all knights to meet and shake hands with their partners.
Proof 1.
In this arrangement any two chains are initially intersecting. For two knights to be adjacent to each other, it is necessary that their chain does not cross any other chain, and thus every pair of chains must be uncrossed at some time. Each exchange of adjacent knights can only uncross a single pair of intersecting chains, and thus the number of exchanges required is at least the number of pairs of chains, which is npn ´ 1q{2.
l Proof 2. In this arrangement the two knights in each pair are initially separated by n ´ 1 seats in either direction around the table, and so each pair must move a total of at least n ´ 1 steps so as to be adjacent. There are n pairs, and each exchange moves two knights by a single step.
Hence at least npn ´ 1q{2 moves are required.
l We will now prove that npn´1q{2 exchanges is sufficient in all cases. We’ll prove a stronger version of this bound than is required, namely that every knight can shake hands with her partner at the end, after all exchanges have finished.
Begin by adding a pillar at the centre of the table. For each chain that passes through the centre of the table, we arbitrarily choose one side of the chain and say that the pillar lies on that side of the chain. While the pillar may lie on a chain, we will never move a knight if that causes the pillar to cross to the other side of a chain. Say that a chain passes in front of a knight if it passes between that knight and the pillar, and define the length of a chain to be the number of knights it passes in front of. Then each chain has a length between 0 and n ´ 1 inclusive.
Say that a chain C encloses another chain C1 if C and C1 do not cross, and C passes between C1 and the pillar. Say that two chains are intersecting if they cross on the table; enclosing if one chain encloses the other; and disjoint otherwise. Let k, l and m denote respectively the number of enclosing, intersecting and disjoint pairs of chains. Then we have k l m “ npn ´ 1q 2 .
Shortlisted problems – solutions 41 Lemma 2.
2k l exchanges are sufficient to reach a position with all pairs of knights sitting adjacent to each other.
Proof 1. We proceed by induction on 2k l.
If every chain has length 0, then every pair of knights is adjacent and the statement is trivial.
Otherwise, let A and B be a pair of knights whose chain C0 has length q ě 1. Let S0 “ A, and let S1, . . . , Sq be the knights which C0 passes in front of, sitting in that order from A to B.
We know that C0 passes in front of S1, and there are three cases for the chain C1 for knight S1.
If C1 passes in front of S0 then C0 and C1 are intersecting, and we can make them disjoint by exchanging the positions of S0 and S1. This reduces the sum 2k l by 1.
If C1 passes in front of neither S0 nor B then C1 is enclosed by C0, and we can swap S0 and S1 to make C0 and C1 an intersecting pair. This increases l by 1 and decreases k by 1, and hence reduces the sum 2k l by 1.
If this C1 passes in front of B then we cannot immediately find a beneficial exchange.
In the third case, we look instead at the knights Si and Si1, for each i in turn. Each time, we will either find a beneficial exchange, or find that the chain Ci1 for knight Si1 passes in front of B. Eventually we will either find a beneficial exchange in one of the first two cases above, or we will find that the chain Cq for Sq passes in front of B, in which case Cq and C0 are intersecting and we can make Cq and C0 disjoint by swapping Sq and B.
Also note that the only times a chain is increased in length is when it is enclosed by another chain. But this cannot happen for a chain containing the pillar, so no chains ever cross the pillar.
l Proof 2.
We begin by ignoring the seats, and let each knight walk freely to a predetermined destination. Each pair of knights will walk around the table to one of the two points on the circumference midway between their initial locations, such that the chain between them passes between the pillar and the destination. If more than one pair of knights would have the same destination point, then we make small adjustments to the destination points so that each pair has a distinct destination point.
We then imagine each knight walking at a constant speed (which may be different for each knight). They all start and stop walking at the same time. We want to count how many times two knights pass (either in opposite directions, or in the same direction but at different speeds).
For any two pairs of knights, the number of passes depends on the relation between their two chains.
If their two chains are intersecting then there will be one pass, involving the two knights for whom the other chain passes between them and the pillar.
If their two chains are enclosing then there will be two passes, with one of the knights with the enclosing chain passing both of the knights with the shorter enclosed chain.
If their two chains are disjoint then there will be no passes.
The number of passes is therefore 2k l. If multiple pairs of knights would pass at the same time, we can slightly adjust the walking speeds so that the passes happen at distinct times.
We can then convert this sequence of passes into a sequence of seat exchanges in the original problem, which shows that 2k l exchanges is sufficient.
l 42 Bath, United Kingdom, 10th–22nd July 2024 Lemma 3. k ď m.
Proof 1. We proceed by induction on n. The base case n “ 2 is clear.
Consider a chain C of greatest length, and suppose it joins knights A and B. Let x be the number of chains that intersect C, and let y be the number of chains that are enclosed by C.
Note that no chain can enclose C. Then C passes in front of one knight from each pair whose chain intersects C, and both knights in any pair whose chain is enclosed by C. Thus the length of C is x 2y ď n ´ 1. The number of chains that form a disjoint pair with C is then n ´ 1 ´ x ´ y ě px 2yq ´ x ´ y “ y.
Now we can remove A and B and use the induction hypothesis. We need to show that the length of each remaining chain is at most n ´ 2 so the chains remain valid. No chain increases in length after removing A and B. If any chain C had length n ´ 1, then the chain between A and B also had length n ´ 1. Then C must have passed in front of exactly one of A or B, and so has length n ´ 2 after removing A and B.
l Proof 2. Let kC denote the number of chains C1 such that C encloses C1.
Note that if C encloses C1, then kC1 ă kC.
First we will show that there at least kC chains that are disjoint from C. Let x be the length of C, let S be the set of x knights that C passes in front of, and let T be the set of x knights sitting directly opposite them. None of the knights in T can have a chain that encloses or is enclosed by C, and if any knight in T has a chain that intersects C, then her partner must be a knight in S. So we have that 2kC “ number of knights in S whose chain is enclosed by C “ x ´ number of knights in S whose chain intersects C ď x ´ number of knights in T whose chain intersects C ď number of knights in T whose chain is disjoint from C ď 2 ˆ number of chains that are disjoint from C.
Now let mC denote the number of chains C1 with C and C1 disjoint, and kC1 ă kC. We will show that mC ě kC.
Let R be a set of kC chains that are disjoint from C, such that ř C1PR kC1 is minimal. If every chain C1 P R has kC1 ă kC, then we are done. Otherwise, let consider a chain C1 with kC1 ě kC. There are then at least kC chains C2 for which the chain C1 passes between C2 and the pillar. Each of these chains must have kC2 ă kC1, and at least one of them is not in R (otherwise R would contain C1 and at least kC other chains), so we can swap this chain with C1 to obtain a set R1 with ř C1PR1 kC1 ă ř C1PR kC1. But this contradicts the minimality of R.
We finish by summing these inequalities over all chains C: k “ ÿ C kC ď ÿ C mC ď m.
l By Lemma 3, we have that 2k l ď kl m “ npn´1q{2. Combining this with Lemma 2, we have that npn ´ 1q{2 exchanges is enough to reach an arrangement where every knight is sitting next to her partner, as desired.
Comment. Either proof of Lemma 3 can be adapted to show that the configuration in Lemma 1 is the only one which achieves the bound.
Shortlisted problems – solutions 43 C4.
On a board with 2024 rows and 2023 columns, Turbo the snail tries to move from the first row to the last row. On each attempt, he chooses to start on any cell in the first row, then moves one step at a time to an adjacent cell sharing a common side. He wins if he reaches any cell in the last row. However, there are 2022 predetermined, hidden monsters in 2022 of the cells, one in each row except the first and last rows, such that no two monsters share the same column. If Turbo unfortunately reaches a cell with a monster, his attempt ends and he is transported back to the first row to start a new attempt. The monsters do not move.
Suppose Turbo is allowed to take n attempts. Determine the minimum value of n for which he has a strategy that guarantees reaching the last row, regardless of the locations of the monsters.
(Hong Kong) Comment. One of the main difficulties of solving this question is in determining the correct expression for n. Students may spend a long time attempting to prove bounds for the wrong value for n before finding better strategies.
Students may incorrectly assume that Turbo is not allowed to backtrack to squares he has already visited within a single attempt. Fortunately, making this assumption does not change the answer to the problem, though it may make it slightly harder to find a winning strategy.
Answer: The answer is n “ 3.
Solution. First we demonstrate that there is no winning strategy if Turbo has 2 attempts.
Suppose that p2, iq is the first cell in the second row that Turbo reaches on his first attempt.
There can be a monster in this cell, in which case Turbo must return to the first row immediately, and he cannot have reached any other cells past the first row.
Next, suppose that p3, jq is the first cell in the third row that Turbo reaches on his second attempt. Turbo must have moved to this cell from p2, jq, so we know j ‰ i. So it is possible that there is a monster on p3, jq, in which case Turbo also fails on his second attempt. Therefore Turbo cannot guarantee to reach the last row in 2 attempts.
Next, we exhibit a strategy for n “ 3. On the first attempt, Turbo travels along the path p1, 1q Ñ p2, 1q Ñ p2, 2q Ñ ¨ ¨ ¨ Ñ p2, 2023q.
This path meets every cell in the second row, so Turbo will find the monster in row 2 and his attempt will end.
If the monster in the second row is not on the edge of the board (that is, it is in cell p2, iq with 2 ď i ď 2022), then Turbo takes the following two paths in his second and third attempts: p1, i ´ 1q Ñ p2, i ´ 1q Ñ p3, i ´ 1q Ñ p3, iq Ñ p4, iq Ñ ¨ ¨ ¨ Ñ p2024, iq.
p1, i 1q Ñ p2, i 1q Ñ p3, i 1q Ñ p3, iq Ñ p4, iq Ñ ¨ ¨ ¨ Ñ p2024, iq.
The only cells that may contain monsters in either of these paths are p3, i ´ 1q and p3, i 1q.
At most one of these can contain a monster, so at least one of the two paths will be successful.
44 Bath, United Kingdom, 10th–22nd July 2024 Figure 1: Turbo’s first attempt, and his second and third attempts in the case where the monster on the second row is not on the edge. The cross indicates the location of a monster, and the shaded cells are cells guaranteed to not contain a monster.
If the monster in the second row is on the edge of the board, without loss of generality we may assume it is in p2, 1q. Then, on the second attempt, Turbo takes the following path: p1, 2q Ñ p2, 2q Ñ p2, 3q Ñ p3, 3q Ñ ¨ ¨ ¨ Ñ p2022, 2023q Ñ p2023, 2023q Ñ p2024, 2023q.
Figure 2: Turbo’s second and third attempts in the case where the monster on the second row is on the edge. The light gray cells on the right diagram indicate cells that were visited on the previous attempt. Note that not all safe cells have been shaded.
If there are no monsters on this path, then Turbo wins. Otherwise, let pi, jq be the first cell on which Turbo encounters a monster. We have that j “ i or j “ i 1. Then, on the third attempt, Turbo takes the following path: p1, 2q Ñ p2, 2q Ñ p2, 3q Ñ p3, 3q Ñ ¨ ¨ ¨ Ñ pi ´ 2, i ´ 1q Ñ pi ´ 1, i ´ 1q Ñ pi, i ´ 1q Ñ pi, i ´ 2q Ñ ¨ ¨ ¨ Ñ pi, 2q Ñ pi, 1q Ñ pi 1, 1q Ñ ¨ ¨ ¨ Ñ p2023, 1q Ñ p2024, 1q.
Shortlisted problems – solutions 45 Now note that • The cells from p1, 2q to pi ´ 1, i ´ 1q do not contain monsters because they were reached earlier than pi, jq on the previous attempt.
• The cells pi, kq for 1 ď k ď i ´ 1 do not contain monsters because there is only one monster in row i, and it lies in pi, iq or pi, i 1q.
• The cells pk, 1q for i ď k ď 2024 do not contain monsters because there is at most one monster in column 1, and it lies in p2, 1q.
Therefore Turbo will win on the third attempt.
Comment. A small variation on Turbo’s strategy when the monster on the second row is on the edge is possible. On the second attempt, Turbo can instead take the path p1, 2023q Ñ p2, 2023q Ñ p2, 2022q Ñ ¨ ¨ ¨ Ñ p2, 3q Ñ p2, 2q Ñ p2, 3q Ñ ¨ ¨ ¨ Ñ p2, 2023q Ñ p3, 2023q Ñ p3, 2022q Ñ ¨ ¨ ¨ Ñ p3, 4q Ñ p3, 3q Ñ p3, 4q Ñ ¨ ¨ ¨ Ñ p3, 2023q Ñ ¨ ¨ ¨ Ñ p2022, 2023q Ñ p2022, 2022q Ñ p2022, 2023q Ñ p2023, 2023q Ñ p2024, 2023q.
If there is a monster on this path, say in cell pi, jq, then on the third attempt Turbo can travel straight down to the cell just left of the monster instead of following the path traced out in the second attempt.
p1, j ´ 1q Ñ p2, j ´ 1q Ñ ¨ ¨ ¨ Ñ pi ´ 1, j ´ 1q Ñ pi, j ´ 1q Ñ pi, j ´ 2q Ñ ¨ ¨ ¨ Ñ pi, 2q Ñ pi, 1q Ñ pi 1, 1q Ñ ¨ ¨ ¨ Ñ p2023, 1q Ñ p2024, 1q.
Figure 3: Alternative strategy for Turbo’s second and third attempts.
46 Bath, United Kingdom, 10th–22nd July 2024 C5.
Let N be a positive integer. Geoff and Ceri play a game in which they start by writing the numbers 1, 2, . . . , N on a board. They then take turns to make a move, starting with Geoff. Each move consists of choosing a pair of integers pk, nq, where k ě 0 and n is one of the integers on the board, and then erasing every integer s on the board such that 2k | n ´ s. The game continues until the board is empty. The player who erases the last integer on the board loses.
Determine all values of N for which Geoff can ensure that he wins, no matter how Ceri plays.
(Indonesia) Answer: The answer is that Geoff wins when N is of the form 2n for n odd or of the form t2n for n even and t ą 1 odd.
Common remarks.
We will say that a set S wins if the current player wins given S as the current set of integers on the board. Otherwise, we will say that S loses.
We will let JpS, T q “ p2S´1qYp2T q. Note that every subset of Z can be written as JpS, T q for some unique pair pS, T q of subsets of Z.
We will let rns denote the set t1, 2, . . . , nu.
Solution.
Lemma 1.
For any set S, S wins if and only if JpS, Hq wins. Similarly, S wins if and only if JpH, Sq wins.
Proof. Let pk, mq be a move on S, and let T be the result of applying the move. Then we can reduce JpS, Hq to JpT , Hq by applying the move pk 1, 2m ´ 1q.
Conversely, let pk, mq be a move on JpS, Hq. We can express the result of this move as JpT , Hq for some T . Then we can reduce S to T by applying the move pmaxpk´1, 0q, pk1q{2q.
This gives us a natural bijection between games starting with S and games starting with JpS, Hq and thus proves the first part of the lemma. The second part follows by a similar argument.
l Lemma 2. If S and T are nonempty and at least one of them loses, then JpS, T q wins.
Proof. If S is losing, then we can delete JpH, T q using the move p1, tq for some t P JpH, T q, which leaves the losing set JpS, Hq. Similarly, if T is losing, then we can delete JpS, Hq using the move p1, sq for some s P JpS, Hq, leaving the losing set JpH, T q.
l Lemma 3. If S is nonempty and wins, then JpS, Sq loses.
Proof. From this position, we can convert any sequence of moves into another valid sequence of moves by replacing pk, 2n´1q with pk, 2nq, and vice versa. Thus we may assume that the initial move pk, mq has m odd. We want to show that any such move results in a winning position for the other player.
The move p0, mq loses immediately. Otherwise, the move results in the set JpT , Sq for some set T . There are three cases.
If T is empty then the other player gets the winning set JpH, Sq.
If T is losing then the other player can choose the move p1, sq for some s P JpH, Sq, which leaves the losing set JpT , Hq.
If T is nonempty winning then the other player can choose the move pk, m 1q, which results in the position JpT , T q. We can then proceed by induction on |S| to show that this is a losing set.
l Shortlisted problems – solutions 47 Lemma 4. r2ns wins if and only if rns loses.
Proof. Note that r2ns “ Jprns, rnsq. The result then follows directly from the previous two lemmas.
l Lemma 5. For any integer n ě 1, r2n 1s wins.
Proof. By Lemma 4, either rns or r2ns loses. If rns loses, then by Lemma 2 we have that r2n 1s “ Jprn 1s, rnsq wins. Otherwise, r2ns loses, and therefore r2n 1s wins by choosing the move pk, 2n 1q for sufficiently large k so that only 2n 1 is eliminated.
l It remains to verify the original answer. We have two cases to consider: • Suppose N “ 2n for some n. For N “ 1, every move is an instant loss for Geoff. Then by Lemma 4, Geoff wins for N “ 2n if and only if Geoff loses for N “ 2n´1, and thus by induction we have that Geoff wins for N “ 2n if and only if n is odd.
• Otherwise, N “ t2n, for some n and some t ą 1 with t odd. By Lemma 5, Geoff wins when n “ 0. Then by Lemma 4, Geoff wins for N “ t2n if and only if Geoff loses for N “ t2n´1, and thus by induction on n we have that Geoff wins for N “ t2n if and only if n is even.
Comment. We can represent this game as a game on partial binary trees. This representation could be common in rough working, as it facilitates exploration of small cases. If two sets produce trees which are topologically equivalent, then this equivalence leads to a natural bijection between games starting with the two sets. Such equivalences lead to a significant reduction in the number of distinct cases that need to be considered when exploring the game for small N.
The construction is as follows. First we begin by considering an infinite binary tree. For each positive integer n, we consider the binary representation of n ´ 1, starting with the least significant bit and ending with an infinite sequence of leading zeroes. We map this sequence of bits to a path on the binary tree by starting at the root, and then repeatedly choosing the left child if the bit is 0 and the right child if the bit is 1. We can then truncate each path after reaching a sufficient depth to distinguish the path from all other paths in the tree.
‚ ‚ ‚ ‚ ‚ ‚ ‚ ‚ ‚ ‚ ‚ ‚ ‚ ‚ ‚ ‚ ‚ ‚ ‚ 1 2 3 4 5 6 7 8 9 10 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 Valid moves in this representation of the game consist of selecting a node with two children, and removing either the left child or the right child (and its descendants). Selecting and removing the entire graph is also an allowed move (which loses instantly).
Two trees have equivalent games if they’re topologically identical. This equivalence includes swap-ping the left and right children of any single node, or removing a node with a single child by merging the edges above and below it (and decreasing the depth of its children by one).
48 Bath, United Kingdom, 10th–22nd July 2024 Comment. We can also analyse this game using Grundy values (also known as nim-values or nimbers).
This requires a slight modification to the rules, wherein any move that would erase all integers on the board is disallowed, and the first player who cannot move loses.
This is clearly equivalent to the original game.
Let gpSq denote the Grundy value of the game starting with the set S. Note that the bijection in Lemma 1 shows that gpSq “ gpJpS, Hqq “ gpJpH, Sqq.
For any set V , let mexpV q denote the least nonnegative element that is not an element of V . For nonnegative integers x and y, define jpx, yq recursively as jpx, yq “ mexptx, yu Y tjpw, yq | w ă xu Y tjpx, zq | z ă yuq.
The values of jpx, yq for small x and y are: 5 6 7 8 9 1 0 4 5 3 6 2 0 1 3 4 5 1 0 2 9 2 3 4 0 1 6 8 1 2 0 4 5 3 7 0 1 2 3 4 5 6 — 0 1 2 3 4 5 We can show that gpJpS, T qq “ jpgpSq, gpT qq for any nonempty sets S and T . The remainder of the proof follows a similar structure to the given solution.
This page is intentionally left blank 50 Bath, United Kingdom, 10th–22nd July 2024 C6.
Let n and T be positive integers. James has 4n marbles with weights 1, 2, . . . , 4n.
He places them on a balance scale, so that both sides have equal weight. Andrew may move a marble from one side of the scale to the other, so that the absolute difference in weights of the two sides remains at most T.
Find, in terms of n, the minimum positive integer T such that Andrew may make a sequence of moves such that each marble ends up on the opposite side of the scale, regardless of how James initially placed the marbles.
(Ghana) Answer: The minimum value of T is 4n.
Solution 1. We must have T ě 4n, as otherwise we can never move the marble of weight 4n.
We will show that T “ 4n by showing that, for any initial configuration, there is a sequence of moves, never increasing the absolute value of the difference above 4n, that results in every marble ending up on the opposite side of the scale. Because moves are reversible, it suffices to do the following: exhibit at least one configuration C for which this can be achieved, and show that any initial configuration can reach such a configuration C by some sequence of moves.
Consider partitioning the weights into pairs pt, 4n 1 ´ tq. Suppose that each side of the balance contains n of those pairs. If one side of the balance contains the pair pt, 4n 1 ´ tq for 1 ď t ă 2n and the other side contains p2n, 2n 1q, then the following sequence of moves swaps those pairs between the sides without ever increasing the absolute value of the difference above 4n.
t, 4n 1 ´ t | 2n, 2n 1 (1) t, 2n, 4n 1 ´ t | 2n 1 (2) t, 2n | 2n 1, 4n 1 ´ t (3) t, 2n, 2n 1 | 4n 1 ´ t (4) 2n, 2n 1 | t, 4n 1 ´ t (5) Applying this sequence twice swaps any two pairs pt, 4n 1 ´ tq and pt1, 4n 1 ´ t1q between the sides. So we can achieve an arbitrary exchange of pairs between the sides, and C can be any configuration where each side of the balance contains n of those pairs.
We now show that any initial configuration can reach one where each side has n of those pairs. Consider a configuration where one side has total weight A ´ s and the other has total weight A s, for some 0 ď s ď 2n, and where some pair is split between the two sides. (If no pair is split between the two sides, they must have equal weights and we are done.) Valid moves include moving any weight w with 1 ď w ď 2n s from the A s side to the A ´ s side, and moving any weight w with 1 ď w ď 2n ´ s from the A ´ s side to the A s side. Suppose the pair pt, 4n 1 ´ tq, with t ď 2n, is split between the sides. If t is on the A s side, or on the A ´ s side and t ď 2n ´ s, it can be moved to the other side. Otherwise, t is on the A ´ s side and t ě 2n ´ s 1, so 4n 1 ´ t ď 2n s is on the A s side and can be moved to the other side. So we can unite the two weights from that pair without splitting any other pair, and repeating this we reach a configuration where no pair is split between the sides.
Solution 2.
As in Solution 1, T ě 4n. Let δ be the weight of the left side minus the weight of the right side. A configuration is called legal if |δ| ď 4n, and a move is legal if it results in a legal configuration. We will show that if δ “ 0 then there is a sequence of legal moves after which every marble is on the opposite side.
We treat the n “ 1 case separately. The initial configuration has marbles 1, 4 on one side and 2, 3 on the other. So moving marbles 2, 4, 3, 1 in that order is legal and every marble ends on the opposite side. Now assume n ě 2.
Shortlisted problems – solutions 51 Marbles of weight at most 2n are called small. We will make use of the following lemmas: Lemma 1.
If a pair of legal configurations differ only in the locations of small marbles then there is a sequence of legal moves to get from one to the other.
Proof. At first we only move marbles in the wrong position if they are not on the lighter side.
(In the case of a tie, neither side is lighter.) Such a move is always legal. Since this reduces the number of marbles in the wrong position, eventually it will no longer be possible to perform such a move.
Then the only marbles in the wrong position are on the lighter side. So moving one marble in the wrong position at a time will always increase |δ|, and |δ| ď 4n at the end. Hence every move is legal.
l Lemma 2. Let k P N. A positive integer can be expressed as a sum of distinct positive integers up to k if and only if it is at most kpk 1q{2.
Proof. The maximum possible sum of distinct positive integers up to k is kpk 1q{2. For the other direction we use induction on k. The case k “ 1 is trivial. Assume the statement is true for k ´ 1. For positive integers up to k we only need a single term. For larger integers, including k in the expression means we are done by the inductive hypothesis.
l Also note that np2n 1q ě 4n for n ě 2.
Let 2n ă m ď 4n. Marbles of weight greater than m are called big and marbles from 2n1 to m are called medium.
Suppose all big marbles are on the correct side (that is, opposite where they started), m is on the incorrect side and the configuration is legal. Then the following steps give a sequence of legal moves after which m is on the correct side and the big marbles were never moved.
Assume m is on the left. In Step 2, we rearrange the small marbles so we can move m. But this is only possible if the weight of big and medium marbles on the right is not too large. So we may need to move some medium marbles from the right first, which we do in Step 1.
Step 1 Skip to Step 2 if the total weight of medium and big marbles on the right side is at most np4n 1q 2n ´ m. Since the big marbles are in the correct position and m is in the incorrect position, the big marbles on the right can weigh at most np4n 1q ´ m. So there must be a medium marble m1 ă m on the right.
From the first assumption, it is legal to move all small marbles to the left. Then by Lemma 2 we can move some of the small marbles to the right so the right side has weight exactly np4n1q2n. Then moving m1 is legal. Repeat this step. Since the total weight of medium marbles on the right decreases, this step will occur a bounded number of times.
Step 2 Let the total weight of the right side be np4n1q2n´mx and the weight of small marbles on the right side be y. Note that y ě x. If x ď 0 then moving m is legal.
Otherwise, by Lemma 2 there is a set of small marbles of weight y ´ x. By Lemma 1, there is a sequence of legal moves of small marbles such that the right side has weight exactly np4n 1q 2n ´ m. Now moving m is legal.
Applying the process above for m “ 4n, 4n ´ 1, . . . , 2n 1 will move all nonsmall marbles to the opposite side. Then Lemma 1 completes the proof.
52 Bath, United Kingdom, 10th–22nd July 2024 C7.
Let N be a positive integer and let a1, a2, . . .
be an infinite sequence of positive integers. Suppose that, for each n ą N, an is equal to the number of times an´1 appears in the list a1, a2, . . . , an´1.
Prove that at least one of the sequences a1, a3, a5, . . .
and a2, a4, a6, . . .
is eventually periodic.
(Australia) Solution 1.
Let M ą maxpa1, . . . , aNq. We first prove that some integer appears infinitely many times. If not, then the sequence contains arbitrarily large integers. The first time each integer larger than M appears, it is followed by a 1. So 1 appears infinitely many times, which is a contradiction.
Now we prove that every integer x ě M appears at most M ´ 1 times. If not, consider the first time that any x ě M appears for the M th time. Up to this point, each appearance of x is preceded by an integer which has appeared x ě M times. So there must have been at least M numbers that have already appeared at least M times before x does, which is a contradiction.
Thus there are only finitely many numbers that appear infinitely many times. Let the largest of these be k. Since k appears infinitely many times there must be infinitely many integers greater than M which appear at least k times in the sequence, so each integer 1, 2, . . . , k ´ 1 also appears infinitely many times. Since k 1 doesn’t appear infinitely often there must only be finitely many numbers which appear more than k times. Let the largest such number be l ě k. From here on we call an integer x big if x ą l, medium if l ě x ą k and small if x ď k.
To summarise, each small number appears infinitely many times in the sequence, while each big number appears at most k times in the sequence.
Choose a large enough N 1 ą N such that aN1 is small, and in a1, . . . , aN1: • every medium number has already made all of its appearances; • every small number has made more than maxpk, Nq appearances.
Since every small number has appeared more than k times, past this point each small number must be followed by a big number. Also, by definition each big number appears at most k times, so it must be followed by a small number. Hence the sequence alternates between big and small numbers after aN1.
Lemma 1. Let g be a big number that appears after aN1. If g is followed by the small number h, then h equals the amount of small numbers which have appeared at least g times before that point.
Proof. By the definition of N 1, the small number immediately preceding g has appeared more than maxpk, Nq times, so g ą maxpk, Nq. And since g ą N, the gth appearance of every small number must occur after aN and hence is followed by g. Since there are k small numbers and g appears at most k times, g must appear exactly k times, always following a small number after aN. Hence on the hth appearance of g, exactly h small numbers have appeared at least g times before that point.
l Denote by ari,js the subsequence ai, ai1, . . . , aj.
Lemma 2. Suppose that i and j satisfy the following conditions: (a) j ą i ą N 1 2, (b) ai is small and ai “ aj, (c) no small value appears more than once in ari,j´1s.
Then ai´2 is equal to some small number in ari,j´1s.
Shortlisted problems – solutions 53 Proof. Let I be the set of small numbers that appear at least ai´1 times in ar1,i´1s. By Lemma 1, ai “ |I|. Similarly, let J be the set of small numbers that appear at least aj´1 times in ar1,j´1s.
Then by Lemma 1, aj “ |J | and hence by (b), |I| “ |J |. Also by definition, ai´2 P I and aj´2 P J .
Suppose the small number aj´2 is not in I. This means aj´2 has appeared less than ai´1 times in ar1,i´1s. By (c), aj´2 has appeared at most ai´1 times in ar1,j´1s, hence aj´1 ď ai´1.
Combining with ar1,i´1s Ă ar1,j´1s, this implies I Ď J . But since aj´2 P J z I, this contradicts |I| “ |J |. So aj´2 P I, which means it has appeared at least ai´1 times in ar1,i´1s and one more time in ari,j´1s. Therefore aj´1 ą ai´1.
By (c), any small number appearing at least aj´1 times in ar1,j´1s has also appeared aj´1´1 ě ai´1 times in ar1,i´1s. So J Ď I and hence I “ J . Therefore, ai´2 P J , so it must appear at least aj´1 ´ ai´1 “ 1 more time in ari,j´1s.
l For each small number an with n ą N 1 2, let pn be the smallest number such that anpn “ ai is also small for some i with n ď i ă n pn. In other words, anpn “ ai is the first small number to occur twice after an´1. If i ą n, Lemma 2 (with j “ n pn) implies that ai´2 appears again before anpn, contradicting the minimality of pn. So i “ n. Lemma 2 also implies that pn ě pn´2. So pn, pn2, pn4, . . . is a nondecreasing sequence bounded above by 2k (as there are only k small numbers). Therefore, pn, pn2, pn4, . . . is eventually constant and the subsequence of small numbers is eventually periodic with period at most k.
Note. Since every small number appears infinitely often, Solution 1 additionally proves that the sequence of small numbers has period k. The repeating part of the sequence of small numbers is thus a permutation of the integers from 1 to k. It can be shown that every permutation of the integers from 1 to k is attainable in this way.
54 Bath, United Kingdom, 10th–22nd July 2024 Solution 2. We follow Solution 1 until after Lemma 1. For each n ą N 1 we keep track of how many times each of 1, 2, . . . , k has appeared in a1, . . . , an. We will record this information in an updating pk 1q-tuple pb1, b2, . . . , bk; jq where each bi records the number of times i has appeared. The final element j of the pk 1q-tuple, also called the active element, represents the latest small number that has appeared in a1, . . . , an.
As n increases, the value of pb1, b2, . . . , bk; jq is updated whenever an is small. The pk 1q-tuple updates deterministically based on its previous value. In particular, when an “ j is small, the active element is updated to j and we increment bj by 1. The next big number is an1 “ bj.
By Lemma 1, the next value of the active element, or the next small number an2, is given by the number of b terms greater than or equal to the newly updated bj, or |ti | 1 ď i ď k, bi ě bju|.
(1) Each sufficiently large integer which appears i1 times must also appear i times, with both of these appearances occurring after the initial block of N. So there exists a global constant C such that bi1 ´bi ď C. Suppose that for some r, br1 ´br is unbounded from below. Since the value of br1 ´ br changes by at most 1 when it is updated, there must be some update where br1 ´ br decreases and br1 ´ br ă ´pk ´ 1qC. Combining with the fact that bi ´ bi´1 ď C for all i, we see that at this particular point, by the triangle inequality minpb1, . . . , brq ą maxpbr1, . . . , bkq.
(2) Since br1 ´ br just decreased, the new active element is r. From this point on, if the new active element is at most r, by (1) and (2), the next element to increase is once again from b1, . . . , br. Thus only b1, . . . , br will increase from this point onwards, and bk will no longer increase, contradicting the fact that k must appear infinitely often in the sequence. Therefore |br1 ´ br| is bounded.
Since |br1 ´ br| is bounded, it follows that each of |bi ´ b1| is bounded for i “ 1, . . . , k.
This means that there are only finitely many different states for pb1 ´ b1, b2 ´ b1, . . . , bk ´ b1; jq.
Since the next active element is completely determined by the relative sizes of b1, b2, . . . , bk to each other, and the update of b terms depends on the active element, the active element must be eventually periodic. Therefore the small numbers subsequence, which is either a1, a3, a5, . . .
or a2, a4, a6, . . . , must be eventually periodic.
Shortlisted problems – solutions 55 C8.
Let n be a positive integer. Given an n ˆ n board, the unit cell in the top left corner is initially coloured black, and the other cells are coloured white. We then apply a series of colouring operations to the board. In each operation, we choose a 2 ˆ 2 square with exactly one cell coloured black and we colour the remaining three cells of that 2 ˆ 2 square black.
Determine all values of n such that we can colour the whole board black.
(Peru) Answer: The answer is n “ 2k where k is a nonnegative integer.
Solution 1.
We first prove by induction that it is possible the colour the whole board black for n “ 2k. The base case of k “ 1 is trivial. Assume the result holds for k “ m and consider the case of k “ m 1. Divide the 2m1 ˆ 2m1 board into four 2m ˆ 2m sub-boards. Colour the top left 2m ˆ2m sub-board using the inductive hypothesis. Next, colour the centre 2ˆ2 square with a single operation. Finally, each of the remaining 2m ˆ 2m sub-board can be completely coloured using the inductive hypothesis, starting from the black square closest to the centre.
This concludes the induction.
Now we prove that if such a colouring is possible for n then n must be a power of 2.
Suppose it is possible to colour an n ˆ n board where n ą 1. Identify the top left corner of the board by p0, 0q and the bottom right corner by pn, nq. Whenever an operation takes place in a 2 ˆ 2 square centred on pi, jq, we immediately draw an “X”, joining the four cells’ centres (see Figure 4). Also, identify this X by pi, jq. The first operation implies there’s an X at p1, 1q.
Since the whole board is eventually coloured, every cell centre must be connected to at least one X. The collection of all Xs forms a graph G.
Figure 4: L-trominoes placements corresponding to colouring operations (left) and the corre-sponding X diagram (right).
Claim 1. The graph G is a tree.
Proof. Since every operation requires a pre-existing black cell, each newly drawn X apart from the first must connect to an existing X. So all Xs are connected to the first X and G must be connected. Now, suppose G has a cycle. Consider the newest X involved in the cycle, it must connect to previous Xs at at least two points. But this implies the corresponding operation will colour at most two cells, which is a contradiction.
l 56 Bath, United Kingdom, 10th–22nd July 2024 Note that in the following arguments, Claims 2 to 4 only require the condition that G is a tree and every cell is connected to G.
Claim 2. If there’s an X at pi, jq, then 1 ď i, j ď n ´ 1 and i ” j pmod 2q.
Proof. The inequalities 1 ď i, j ď n ´ 1 are clear. Call an X at pi, jq good if i ” j pmod 2q, or bad if i ı j pmod 2q. The first X at pi, jq “ p1, 1q is good. Suppose some Xs are bad. Since G is connected, there must exist a good X connecting to a bad X. But this can only occur if they connect at two points, creating a cycle. This is a contradiction, thus all Xs are good.
l Call an X at pi, jq odd if i ” j ” 1 pmod 2q, even if i ” j ” 0 pmod 2q.
Claim 3. The integer n must be even. Furthermore, there must be 4pn{2´1q odd Xs connecting the cells on the perimeter of the board as shown in Figure 5.
Proof. If n is odd, the four corners of the bottom left cell are pn, 0q, pn ´ 1, 0q, pn ´ 1, 1q and pn, 1q, none of which satisfies the conditions of Claim 2. So the bottom left cell cannot connect to any X. If n is even, each cell on the edge of the board has exactly one corner satisfying the conditions of Claim 2, so the X connecting it is uniquely determined. Therefore the cells on the perimeter of the board are connected to Xs according to Figure 5.
l Figure 5: Highlighting the permitted points for Xs (left) and Xs on the perimeter (right).
Divide the nˆn board into n2{4 blocks of 2ˆ2 squares. Call each of these blocks a big-cell.
We say a big-cell is filled if it contains an odd X on its interior, empty otherwise. By Claim 3, each big-cell on the perimeter must be filled.
Claim 4. Every big-cell is filled.
Proof. Recall that Xs can only be at pi, jq with i ” j pmod 2q. Suppose a big-cell centred at pi, jq is empty. Then in order for its four cells to be coloured, there must be four even Xs on pi´1, j´1q, pi1, j´1q, pi´1, j1q and pi1, j1q, “surrounding” the big-cell (see Figure 6).
By Claim 3, no empty big-cell can be on the perimeter. So if there exist some empty big-cells, the boundary between empty and filled big-cells must consist of a number of closed loops.
Each closed loop is made up of several line segments of length 2, each of which separates a filled big-cell from an empty big-cell.
Since every empty big-cell is surrounded by even Xs and every filled big-cell contains an odd X, the two end points of each such line segment must be connected by Xs. Since these line segments form at least one closed loop, it implies the existence of a cycle made up of Xs (see Figure 6). This is a contradiction, thus no big-cell can be empty.
l Shortlisted problems – solutions 57 Figure 6: An empty big-cell surrounded by even Xs (left) and the boundary between empty and filled Xs creating a cycle (right).
Therefore every big-cell is filled by an odd X, and the connections between them are provided by even Xs. We can now reduce the n ˆ n problem to an n{2 ˆ n{2 problem in the following way. Perform a dilation of the board by a factor of 1{2 with respect to p0, 0q. Each big-cell is shrunk to a regular cell. For the Xs, replace each odd X at pi, jq by the point pi{2, j{2q, and replace each even X at pi, jq by an X at pi{2, j{2q.
We claim the new resulting graph of Xs is a tree that connects all cells of an n{2ˆn{2 board.
First, two connected Xs in the original n ˆ n board are still connected after their replacements (noting that some Xs have been replaced by single points). For each cell in the n{2ˆn{2 board, its centre corresponds to an odd X from a filled big-cell in the original n ˆ n board, so it must be connected to the graph. Finally, suppose there exists a cycle in the new graph. The cycle consists of Xs that correspond to even Xs in the original graph connecting big-cells, forming a cycle of big-cells. Since in every big-cell, the four unit squares were connected by an odd X, this implies the existence of a cycle in the original graph, which is a contradiction.
Thus the new graph of Xs must be a tree that connects all cells of an n{2ˆn{2 board, which are the required conditions for Claims 2 to 4. Hence we can repeat our argument, halving the dimensions of the board each time, until we reach the base case of a 1 ˆ 1 board (where the tree is a single point). Therefore n must be a power of 2, completing the solution.
58 Bath, United Kingdom, 10th–22nd July 2024 Solution 2.
As in Solution 1, it is possible the colour the whole board black for n “ 2k.
The colouring operation is equivalent to the placement of L-trominoes. For each L-tromino we place on the board, we draw an arrow and a node as shown in Figure 7. We also draw a node in the top left corner of the board.
Figure 7: Tromino with corresponding arrow and node drawn.
Claim 1. The arrows and nodes form a directed tree rooted at the top left corner.
Proof. The proof is similar to the proof of Claim 1 in Solution 1, with the additional note that the directions of the arrows inherit the order of the colouring operations, so they must be pointing away from the top left node.
l Note that since all edges of the tree are diagonal, the nodes can only lie on points pi, jq with i j ” 0 pmod 2q. This implies that we can only place down L-trominoes of one particular parity: that is, with the centre of the L-tromino on a point with i j ” 0 pmod 2q. In the remainder of the proof, we will implicitly use this parity property when determining possible positions of L-trominoes.
Next, we show that certain configurations of edges of the tree are impossible.
Claim 2. There cannot be two edges in a “parallel” configuration (see Figure 8).
Proof. In such a configuration, the two edges can either be directed in the same direction or opposite directions. If they point in the same direction (see Figure 8), then the L-trominoes corresponding to the two edges overlap.
Figure 8: Parallel configuration (left) and two parallel edges, case 1 (right).
If they point in opposite directions, then we get the diagram in Figure 9.
The cells marked p‹q must lie inside the n ˆ n board, so they must be covered by L-trominoes. There is only one possible way to cover these with a L-tromino of the right parity. But this makes the arrows form a cycle, which cannot happen. So we have a contradiction.
l ‹ ‹ Figure 9: Two parallel edges, case 2.
Shortlisted problems – solutions 59 Claim 3. There cannot be three edges in a “zigzag” configuration, shown in Figure 10.
Figure 10: Zigzag configuration.
Proof. Assume for contradiction that there is a zigzag. Then take the zigzag with maximal distance from the root of the tree (measured by distance along the graph from the root to the middle edge of the zigzag).
We may assume without loss of generality that the middle edge is directed down-right. Then the right edge must be directed up-right, since no two arrows can point to the same node. Next, we draw in the corresponding L-trominoes, and consider the cell marked p‹q. There are two possible ways to cover it with an L-tromino, because of the parity of L-tromino centres.
We could choose the centre of the L-tromino to be the top right corner of the cell (see Figure 11). This immediately gives another zigzag.
‹ Figure 11: Zigzag configuration, case 1.
The other possibility is if we choose the centre of the L-tromino to be the bottom left corner of the cell (see Figure 12). Then we need to cover the cell marked p‹‹q with an L-tromino. If ‹ ‹‹ Figure 12: Zigzag configuration, case 2.
we placed the centre of the L-tromino on the top left corner of the cell, this would give two parallel edges, contradicting Claim 2. So we must place the centre of the L-tromino on the bottom right corner of the cell, which gives a zigzag.
In each case, we get another zigzag further away from the root of the tree, which contradicts our assumption of maximality. So there cannot be any zigzags.
l 60 Bath, United Kingdom, 10th–22nd July 2024 We now colour the nodes of the tree. Colour the root node yellow. For all other nodes, we colour it white if it has an arrow coming out of it in a different direction to the arrow going in, and black otherwise.
Claim 4. Any child of a black node is white.
‹ Figure 13: Black node configuration.
Proof. Suppose we have a black node with a child. Then the arrow exiting the black node must be in the same direction as the arrow entering it by the definition of our colouring, giving the left diagram of Figure 13.
The cell marked p‹q must be covered by an L-tromino. If the centre of this L-tromino is the bottom left corner, then this would give an arrow leaving the black node in a different direction, which cannot happen. So the centre of the L-tromino must instead be the top right corner, which gives an arrow leaving the upper node in a different direction. Thus the upper node must be white.
l Claim 5. Every white node has three children, all of which are black.
‹ ‹‹ Figure 14: White node configuration.
Proof. Refer to Figure 14. Suppose we have a white node, as in the leftmost diagram. The cell marked p‹q must be covered by an L-tromino. If the centre of this L-tromino is the bottom right corner of the cell, then this would form a zigzag, which by Claim 3 is not allowed. So the centre must be the top left corner.
Next, the cell marked p‹‹q must be covered by an L-tromino. If the centre of this L-tromino is the top right corner, this would form a zigzag, so the centre must be the bottom left corner instead. Thus we have shown that any white node has three children.
Finally, note that if any of the child nodes had three children of their own, then this would give parallel edges in the diagram, which contradicts Claim 2. Therefore the child nodes of the white node must all be black.
l We now know that the node colours alternate between black and white as you go down the tree, so all white nodes lie on points with coordinates p2i, 2jq, and all black nodes lie on points with coordinates p2i 1, 2j 1q.
Now (assuming n ą 1) we will construct a new board whose cells are 2 ˆ 2 squares of our current board. We replace the root node and its child with a single big cell and a big root node, Shortlisted problems – solutions 61 Figure 15: Replacing with larger cells and L-trominoes.
and we replace each white node and its three children with a big L-tromino, big arrow and big node as shown in Figure 15.
Every black node is the child of the root node or a white node, so every L-tromino is involved in exactly one replacement. Also, the parent of any white node is a black node, whose parent, in turn, is a white node or the root. So the starting point of every big arrow will be on a big node. Therefore we obtain an L-tromino tiling forming a tree.
This shows for n ą 1 that if an nˆn board can be tiled by L-trominoes forming a tree, then n is even, and an n{2ˆn{2 board can also be tiled by L-trominoes forming a tree. Since a 1ˆ1 board can trivially be tiled, we conclude that the only values of n for which an n ˆ n board can be tiled are n “ 2k.
62 Bath, United Kingdom, 10th–22nd July 2024 Geometry G1.
Let ABCD be a cyclic quadrilateral such that AC ă BD ă AD and =DBA ă 90˝.
Point E lies on the line through D parallel to AB such that E and C lie on opposite sides of line AD, and AC “ DE. Point F lies on the line through A parallel to CD such that F and C lie on opposite sides of line AD, and BD “ AF.
Prove that the perpendicular bisectors of segments BC and EF intersect on the circumcircle of ABCD.
(Ukraine) Solution 1. Let T be the midpoint of arc Ŕ BAC and let lines BA and CD intersect EF at K and L, respectively. Note that T lies on the perpendicular bisector of segment BC.
A B C D E F T K L Since ABCD is cyclic, BD sin =BAD “ AC sin =ADC. From parallel lines we have =DAF “ =ADC and =BAD “ =EDA. Hence, AF ¨ sin =DAF “ BD ¨ sin =ADC “ AC ¨ sin =BAD “ DE ¨ sin =EDA.
So F and E are equidistant from the line AD, meaning that EF is parallel to AD.
We have that KADE and FADL are parallelograms, hence we get KA “ DE “ AC and DL “ AF “ BD. Also, KE “ AD “ FL so it suffices to prove the perpendicular bisector of KL passes through T.
Triangle AKC is isosceles so =BTC “ =BAC “ 2=BKC. Likewise, =BTC “ 2=BLC.
Since T, K, and L all lie on the same side of BC and T lies on the perpendicular bisector of BC, T is the centre of circle BKLC. The result follows.
Shortlisted problems – solutions 63 Solution 2. Let AF and DE meet ω at X and Y , respectively, and let T be as in Solution 1.
As BD ă AD, DY ∥AB and =BAY “ =DBA ă 90˝, we have DY ă AB and Y lies on the opposite side of line AD to C. Also from BD ă AD, we have B, C, and D all lie on the same side of the perpendicular bisector of AB which shows AC ą AB. Combining these, we get DY ă AB ă AC “ DE and, as Y and E both lie on the same side of line AD, Y lies in the interior of segment DE. Similarly, X lies in the interior of segment DF.
Since AB is parallel to DY , we have Y A “ BD “ FA. Likewise XD “ AC “ ED.
A B C D E F X Y T Claim 1. T is the midpoint of arc Ŋ XY .
Proof. From AX ∥CD and AB ∥DY we have =CAX “ =AXD “ =AY D “ =Y DB.
Since T is the midpoint of arc Ŕ BAC , we have =BAT “ =TDC, so =TAX “ =CAX =BAC ´ =BAT “ =Y DB =BDC ´ =TDC “ =Y DT.
l Recall from above we have AB ă AC and analogously, DC ă DB, which shows that X, Y and T all lie on the same side of line AD. In particular, T and A lie on opposite sides of XY so T lies on the internal angle bisector of =XAY . Since AF “ AY , we have △ATF – △ATY , giving TF “ TY .
Likewise, TE “ TX, so TE “ TF, meaning that T lies on the perpendicular bisector of segment EF as required.
Comment. The statement remains true without the length and angle conditions on cyclic quadrilateral ABCD however additional care is required to consider different cases based on the ordering of points on lines DE and AF. It is also possible for T to be on the external angle bisector of =XAY .
64 Bath, United Kingdom, 10th–22nd July 2024 Solution 3. From AF “ DB, AC “ DE and =pAC, AFq “ =pAC, CDq “ =pAB, BDq “ =pDE, DBq, triangles ACF and DEB are congruent, so CF “ BE.
Let P “ BE X CF. Since =pCP, BPq “ =pCF, BEq “ =pAF, DBq “ =pDC, DBq, we have that P lies on circle ABCD.
A B C D E F X Y T P Finally, let T be the Miquel point of the quadrilateral BCFE so T lies on circles EFP and ABCD. Note that T is the centre of spiral similarity taking segments BE to CF and since BE “ CF, this is in fact just a rotation, so TB “ TC and TE “ TF; that is, the perpendicular bisectors of BC and EF meet at T, on circle ABCD.
Shortlisted problems – solutions 65 G2.
Let ABC be a triangle with AB ă AC ă BC, incentre I and incircle ω. Let X be the point in the interior of side BC such that the line through X parallel to AC is tangent to ω.
Similarly, let Y be the point in the interior of side BC such that the line through Y parallel to AB is tangent to ω. Let AI intersect the circumcircle of triangle ABC again at P ‰ A. Let K and L be the midpoints of AB and AC, respectively.
Prove that =KIL =Y PX “ 180˝.
(Poland) Solution 1.
Let A1 be the reflection of A in I, then A1 lies on the angle bisector AP. Lines A1X and A1Y are the reflections of AC and AB in I, respectively, and so they are the tangents to ω from X and Y . As is well-known, PB “ PC “ PI, and since =BAP “ =PAC ą 30˝, PB “ PC is greater than the circumradius. Hence PI ą 1 2AP ą AI; we conclude that A1 lies in the interior of segment AP.
A B C I K L P X Y A1 We have =APB “ =ACB in the circumcircle and =ACB “ =A1XC because A1X ∥AC.
Hence, =APB “ =A1XC, and so quadrilateral BPA1X is cyclic. Similarly, it follows that CY A1P is cyclic.
Now we are ready to transform =KIL =Y PX to the sum of angles in triangle A1CB. By a homothety of factor 2 at A we have =KIL “ =CA1B. In circles BPA1X and CY A1P we have =APX “ =A1BC and =Y PA “ =BCA1, therefore =KIL =Y PX “ =CA1B =Y PA =APX ˘ “ =CA1B =BCA1 =A1BC “ 180˝.
66 Bath, United Kingdom, 10th–22nd July 2024 Comment. The constraint AB ă AC ă BC was added by the Problem Selection Committee in order to reduce case-sensitivity. Without that, there would be two more possible configurations according to the possible orders of points A, P and A1, as shown in the pictures below. The solution for these cases is broadly the same, but some extra care is required in the degenerate case when A1 coincides with P and line AP is a common tangent to circles BPX and CPY .
A B C I K L A1 “ P X Y A B C I K L P X Y A1 Shortlisted problems – solutions 67 Solution 2. Let BC “ a, AC “ b, AB “ c and s “ abc 2 , and let the radii of the incircle, B-excircle and C-excircle be r, rb and rc, respectively. Let the incircle be tangent to AC and AB at B0 and C0, respectively; let the B-excircle be tangent to AC at B1, and let the C-excircle be tangent to AB at C1. As is well-known, AB1 “ s ´ c and areap△ABCq “ rs “ rcps ´ cq.
Let the line through X, parallel to AC be tangent to the incircle at E, and the line through Y , parallel to AB be tangent to the incircle at D. Finally, let AP meet BB1 at F.
A B C P I D B0 K B1 C0 L C1 E F X Y It is well-known that points B, E, and B1 are collinear by the homothety between the incircle and the B-excircle, and BE ∥IK because IK is a midline in triangle B0EB1.
Similarly, it follows that C, D, and C1 are collinear and CD ∥IL.
Hence, the problem reduces to proving =Y PA “ =CBE (and its symmetric counterpart =APX “ =DCB with respect to the vertex C), so it suffices to prove that FY PB is cyclic. Since ACPB is cyclic, that is equivalent to FY ∥B1C and BF FB1 “ BY Y C .
By the angle bisector theorem we have BF FB1 “ AB AB1 “ c s ´ c.
The homothety at C that maps the incircle to the C-excircle sends Y to B, so BC Y C “ rc r “ s s ´ c.
So, BY Y C “ BC Y C ´ 1 “ s s ´ c ´ 1 “ c s ´ c “ BF FB1 , which completes the solution.
68 Bath, United Kingdom, 10th–22nd July 2024 G3.
Let ABCDE be a convex pentagon and let M be the midpoint of AB. Suppose that segment AB is tangent to the circumcircle of triangle CME at M and that D lies on the circumcircles of triangles AME and BMC. Lines AD and ME intersect at K, and lines BD and MC intersect at L. Points P and Q lie on line EC so that =PDC “ =EDQ “ =ADB.
Prove that lines KP, LQ, and MD are concurrent.
(Belarus) Common remarks. Each of solutions we present consists of three separate parts: (a) proving KP ∥MC and LQ ∥ME; (b) proving KL ∥AB and, optionally, showing that points C, E, K, and L are concyclic; (c) completing the solution either using homotheties or the parallelogram enclosed by lines KP, MK, ML and LQ, or radical axes between three circles.
Solution 1.
(a) Notice that the condition =PDC “ =ADB is equivalent to =ADP “ =BDC, and =EDQ “ =ADB is equivalent to =EDA “ =QDB. From line AB being tangent to circle CME, and circles AMDE and CDME we read =ECM “ =EMA “ =EDA “ =QDB and =MEC “ =BMC “ =BDC “ =ADP.
Using =ADP “ =MEC, the points D, E, K, and P are concyclic, which gives that =EPK “ =EDA “ =ECM. From that, we can see that KP ∥MC. It can be shown similarly that C, D, Q, and L are concyclic, =LQC “ =MEC and therefore LQ ∥ME.
R S E C D A B P Q M K L (b) Let rays DA and DB intersect circle CDE at R and S, respectively. We now observe that =SEC “ =SDC “ =MEC, so points E, M, and S are collinear. We similarly obtain that C, M, and R are collinear.
From =SRC “ =SEC “ =BMC we can see that RS ∥AB. Since M bisects AB, it follows that KL ∥RS.
(c) Consider the homothety at D that sends RS to KL. Because KP ∥RC and LQ ∥SE, that homothety sends the concurrent lines DM, RC, and SE to DM, KP, and LQ, so these lines are also concurrent.
Shortlisted problems – solutions 69 Solution 2.
(a) As in Solution 1, we show the following: =ECM “ =EMA “ =EDA “ =EPK; =MEC “ =BMC “ =BDC “ =LQC; the points C, D, Q, and L are concyclic; the points D, E, K, and P are concyclic; KP ∥MC; and LQ ∥ME.
(b) Notice that triangles EKP and EMC are homothetic at E, so their circumcircles CME and DEKP are tangent to each other at E. Similarly, circle CDQL is tangent to circle CME at C.
Suppose that the tangents to circle CME at C and E intersect at point X. (The case when CE is a diameter in circle CME can be considered as a limit case.) Moreover, let EX and CX intersect circles DEAM and BCDM again at A1 ‰ E and B1 ‰ C, respectively.
X E C D A B P Q M K L A1 B1 We have XE “ XC because they are the tangents from X to circle CME. Moreover, in circle DEAM, chords AM and A1E are tangent to circle CME, so A1E “ AM. Similarly, we have B1C “ BM, hence A1E “ AM “ BM “ B1C. We conclude XA1 “ XB1, so the powers of X with respect to circles DEAM and BCDM are equal. Therefore, X lies on the radical axis of these two circles, which is DM.
Now notice that by XC “ XE, point X has equal powers to circles CDQL and DEKP, so DX is the radical axis of these circles. Point M lies on DX, so ME ¨ MK “ MC ¨ ML; we conclude that C, E, K, and L are concyclic. Hence, by =MKL “ =ECM “ =KMA we have KL ∥AB.
(c) As =EPK “ =EMA “ =QLK, we have that KLQP is cyclic. The radical axes between circles DEKP, CDQL and KLQP are DM, KP and LQ, so they are concurrent at the radical centre of the three circles.
70 Bath, United Kingdom, 10th–22nd July 2024 Solution 3.
(b) We present another proof that KL ∥AB.
Let AD X LQ “ I, BD X KP “ H, AB X LQ “ U and AB X KP “ V . Since =DHP “ =DLM “ 180˝ ´ =CLD “ 180˝ ´ =CQD “ =DQE, point H lies on circle DPQ. Similarly, we obtain that point I lies on this circle. Hence, =LIH “ =QDB “ =EDA “ =EMA, and LQ ∥ME implies that HI ∥AB.
U V E C D A B H I P Q M K L Let AM “ BM “ d, then we have BU IH “ BL LH “ BM MV “ d d AV and AV IH “ AK KI “ AM MU “ d d BU .
Hence, BU ¨ pd AV q “ AV ¨ pd BUq, so BU “ AV . Therefore, △MLU – △V KM which implies KL ∥AB ∥HI.
(c) Lines MK, ML, KP and LQ enclose a parallelogram.
Line DM passes through the midpoint of KL, which the centre of the parallelogram, and passes through the vertex M.
Therefore, DM passes through the opposite vertex, which is the intersection of KP and LQ.
Shortlisted problems – solutions 71 G4.
Let ABCD be a quadrilateral with AB parallel to CD and AB ă CD. Lines AD and BC intersect at a point P. Point X ‰ C on the circumcircle of triangle ABC is such that PC “ PX. Point Y ‰ D on the circumcircle of triangle ABD is such that PD “ PY .
Lines AX and BY intersect at Q.
Prove that PQ is parallel to AB.
(Ukraine) Solution 1. Let M and N be the midpoints of AD and BC, respectively and let the perpen-dicular bisector of AB intersect the line through P parallel to AB at R.
Lemma. Triangles QAB and RNM are similar.
Proof. Let O be the circumcentre of triangle ABC, and let S be the midpoint of CX. Since N, S, and R are the respective perpendicular feet from O to BC, CX, and PR, we have that quadrilaterals PRNO and CNSO are cyclic. Furthermore, P, S, and O are collinear as PC “ PX. Since ABCX is also cyclic, we have that =QAB “ =XCB “ =PON “ 180˝ ´ =NRP “ =MNR.
Analogously, we have that =ABQ “ =RMN, so triangles QAB and RNM are similar.
l A B C D P X Y Q M N O R S Let dpZ, ℓq denote the perpendicular distance from the point Z to the line ℓ. Using that PR ∥AB along with the similarities QAB „ RNM and PAB „ PMN, we have that dpQ, ABq AB “ dpR, MNq MN “ dpP, MNq MN “ dpP, ABq AB , which implies that PQ ∥AB.
72 Bath, United Kingdom, 10th–22nd July 2024 Solution 2. Let BD and AC intersect at T and let the line through P parallel to AB intersect BD at V . Next, let Q1 be the foot of the perpendicular from T to PV . Finally, let Q1A intersect circle ABC again at X1 and Q1B intersect circle ABD again at Y 1.
A B C D P X1 Y 1 Q1 T V L Claim. PQ1 bisects =BQ1D externally.
Proof. Let PT intersect CD at L. Let 8CD be the point at infinity on line CD. From the standard Ceva-Menelaus configuration we have pD, C; L, 8CDq is harmonic. Hence projecting through P we have ´1 “ pD, C; L, 8CDq “ pD, B; T, V q.
As pD, B; T, V q is harmonic, and also =V Q1T “ 90˝ (by construction), the claim follows.
l Now as =Q1PD “ =BAD “ 180˝ ´ =DY 1B “ 180˝ ´ =DY 1Q1 we have Q1PDY 1 cyclic.
By the claim, we have that P is the midpoint of arc Ŕ DQ1Y 1 , so PD “ PY 1.
Since Y is the unique point not equal to D on circle ABD satisfying PD “ PY , we have Y 1 “ Y .
Likewise X1 “ X so Q1 “ Q and we are done.
Shortlisted problems – solutions 73 Solution 3. Let AX intersect circle PCX for the second time at Q1. Then =AQ1P “ =XQ1P “ =XCP “ =XCB “ 180˝ ´ =BAX “ =Q1AB so PQ1 is parallel to AB. Hence, it suffices to show that Q1 is equal to Q. To do so, we aim to show the common chord of circles PCX and PDY is parallel to AB, since then by symmetry Q1 is also the second intersection of BY and circle PDY .
A B C D P X Y Q OX OY OD OC Let the centres of circles PCX and PDY be OX and OY , respectively. Let the centres of circles ABC and ABD be OC and OD, respectively.
Note P, OX, and OC are collinear since they all lie on the perpendicular bisector of CX.
Likewise P, OY , and OD are collinear on the perpendicular bisector of DY . By considering the projections of OX and OC onto BC, and OY and OD onto AD, we have POX POC “ PC 2 PBPC 2 “ PD 2 PAPD 2 “ POY POD .
Hence OXOY is parallel to OCOD, which is perpendicular to AB as desired.
74 Bath, United Kingdom, 10th–22nd July 2024 G5.
Let ABC be a triangle with incentre I, and let Ωbe the circumcircle of triangle BIC.
Let K be a point in the interior of segment BC such that =BAK ă =KAC. The angle bisector of =BKA intersects Ωat points W and X such that A and W lie on the same side of BC, and the angle bisector of =CKA intersects Ωat points Y and Z such that A and Y lie on the same side of BC.
Prove that =WAY “ =ZAX.
(Uzbekistan) Common remarks.
The key step in each solution is to prove that =ZAK “ =IAY and =WAK “ =IAX. The problem is implied by these equalities, as we then have that =WAY “ =WAK =KAI =IAY “ =IAX =KAI =ZAK “ =ZAX.
B C A K I W X Y Z We now present several proofs that =ZAK “ =IAY , with =WAK “ =IAX following in an analogous manner.
Solution 1.
Let Γ be circle ABC and ω be circle AY Z. Let O, M, and S be the centres of Γ, Ω, and ω, respectively. Let AK intersect Γ again at P, and let the angle bisector of =ZAY intersect ω again at N.
Ω Γ ω B C A K I Y Z O S M P N Shortlisted problems – solutions 75 By power of a point from K to Γ and Ω, we have that KA¨KP “ KB ¨KC “ KY ¨KZ, so P also lies on ω. The pairwise common chords of Γ, Ω, and ω are then AP K OS, BC K OM, and Y Z K MS, so we have that =OMS “ =CKY “ =Y KA “ =MSO. As M lies on Γ and OM “ OS, S also lies on Γ. Note that N lies on MS as NY “ NZ, so =PAN “ 1 2=PSN “ 1 2=PSM “ 1 2=PAM.
Thus, AN bisects =PAM in addition to =ZAY , which means that =ZAK “ =IAY as K lies on AP and I lies on AM.
Solution 2.
Define M and P as in Solution 1, and recall that AY PZ is cyclic. Let Q be the second intersection of the line parallel to BC through P with circle ABC and let J be the incentre of triangle APQ.
B C A K I Y Z M P Q J Since PQ is parallel to BC and =BAP ă =PAC, the angle bisector of =APQ is parallel to the angle bisector of =AKC. Hence, PJ is parallel to Y Z. As M is the midpoint of Ŋ PQ on circle APQ, we have that MP “ MJ. Then since segments Y Z and PJ are parallel and have a common point M on their perpendicular bisectors, PJY Z is cyclic with JY “ PZ. It follows that J also lies on circle AY PZ and that =ZAP “ =JAY “ =IAY .
Comment. The proof of the analogous case of =WAK “ =IAX is slightly different. In this case, J should be defined as the A-excentre of APQ so that PJ is the external bisector of =APQ and PJ ∥WX. The proof is otherwise exactly the same.
76 Bath, United Kingdom, 10th–22nd July 2024 Solution 3. As in the previous solutions, let M be the centre of Ω. Let L be the intersection of AM and BC, and let L1 be the reflection of L over Y Z. Let the circle MY Z intersect AM again at T.
B C A K I Y Z M L L1 T Note that as M is the midpoint of Ŋ BC on circle ABC and L is the foot of the bisector of =BAC, we have that MA ¨ ML “ MI2 “ MY 2. It follows by power of a point that MY is tangent to circle ALY , so =LAY “ =LY M. Using directed angles, we then have that >AY T “ >MTY ´ >MAY “ >MZY ´ >LY M “ >ZY M ´ >LY M “ >ZY L “ >L1Y Z, where we use the fact that MY “ MZ and that L and L1 are symmetric about Y Z. Thus, Y T and Y L1 are isogonal in =AY Z. Analogously, we have that ZT and ZL1 are isogonal in =Y ZA.
This means that T and L1 are isogonal conjugates in triangle AY Z, which allows us to conclude that =ZAK “ =IAY since L1 lies on AK and T lies on AI.
Comment. Owing to the condition =BAK ă =KAC, points L1 and T lie inside triangle AY Z.
However, if one tries to write down the same proof for =WAK “ =IAX, the analogues L1 1 and T1 of L1 and T would lie outside triangle AWX. Thus, the solution has been written using directed angles so that it applies directly to this case as well. It is also possible that L1 1 lies on circle AWX and T1 is a point at infinity. In this case, it is straightforward to interpret the directed angle chase to prove the isogonality, and the isogonality also follows from this scenario being a limit case of other configurations.
Note. The original proposal remarks that this problem is a special case of a more general property: A convex quadrilateral ABCD is inscribed in a circle ω.
The bisectors between AC and BD intersect ω at four points, forming a convex quadrilateral PQRS. Then the conditions XA ¨ XC “ XB ¨ XD and >pXP, XQq “ >pXS, XRq on point X are equivalent.
The Problem Selection Committee believes that the proof of this generalisation is beyond the scope of the competition and considers the original problem to be more suitable.
This page is intentionally left blank 78 Bath, United Kingdom, 10th–22nd July 2024 G6.
Let ABC be an acute triangle with AB ă AC, and let Γ be the circumcircle of ABC.
Points X and Y lie on Γ so that XY and BC intersect on the external angle bisector of =BAC.
Suppose that the tangents to Γ at X and Y intersect at a point T on the same side of BC as A, and that TX and TY intersect BC at U and V , respectively. Let J be the centre of the excircle of triangle TUV opposite the vertex T.
Prove that AJ bisects =BAC.
(Poland) Solution 1. Let N be the midpoint of Ŕ BAC on Γ, and let NX and NY intersect BC at W and Z, respectively.
Claim. Quadrilateral WXY Z is cyclic, and its circumcentre is J.
Proof. As N is the midpoint of Ŕ BAC , W and Z lie on BC, and X and Y are the second intersections of NW and NZ with Γ, we have that WXY Z is cyclic.
Let the parallel to BC through N intersect TU and TV at U 1 and V 1, respectively. Then U 1 is the intersection of the tangents to Γ at N and X, so U 1N “ U 1X. As NU 1 ∥BC, U 1NX is similar to UWX, so UW “ UX as well. Hence, the perpendicular bisector of WX is the internal bisector of =XUW, which is the external bisector of =V UT. Analogously, the perpendicular bisector of Y Z is the external bisector of =TV U. This means that the circumcentre of WXY Z is the intersection of the external bisectors of =V UT and =TV U, which is J.
l B C A X N L Y U V T W Z U 1 V 1 J Let AN intersect BC at L, so XY passes through L as well. By power of a point from L to Γ and circle WXY Z, we have that LA ¨ LN “ LX ¨ LY “ LW ¨ LZ, so WANZ is also cyclic.
Thus, A is the Miquel point of quadrilateral WXY Z. As WXY Z is cyclic with circumcentre J and its opposite sides WX and Y Z intersect at N, we have that AN K AJ. Since AN is the external bisector of =BAC, this implies that AJ is the internal bisector of =BAC.
Shortlisted problems – solutions 79 Solution 2. Let the internal and external angle bisectors of =BAC intersect BC at K and L, respectively. Let AK intersect circle ABC again at M, and let D be the intersection of the tangents to Γ at B and C. Let Ωbe the T-excircle of TUV , and let ω be the incircle of DBC.
Claim. The points T, K, and D are collinear.
Proof. Note that BC and XY are the polars of T and D with respect to Γ. By La Hire’s Theorem, TD is the polar of L with respect to Γ. As pB, C; K, Lq “ ´1, K also lies on the polar of L, thus proving the collinearity.
l Claim. The incentre of DBC is M.
Proof. We have that =MBC “ =MAC “ 1 2=BAC “ 1 2=DBC, so BM bisects =DBC.
Similarly, CM bisects =BCD, so M is the incentre of DBC.
l Γ ω Ω B C A X L Y U V T J K M D Claim. The intersection of the common external tangents of Ωand ω is K.
Proof. Let K1 be the intersection of the common external tangents of Ωand ω. As Ωand ω are both tangent to BC and lie on the same side of BC opposite to A, K1 lies on BC. As T is the intersection of the common external tangents of Γ and Ωand D is the intersection of the common external tangents of Γ and ω, by Monge’s theorem K1 lies on TD. As K1 lies on both BC and TD, it is the same point as K.
l Hence, K is collinear with the centres of Ωand ω, which are M and J, respectively. As K and M both lie on the bisector of =BAC, so does J.
Note. It can be shown that circles AUV and ABC are tangent and that the tangents from U and V to circle ABC different from TU and TV intersect at a point W on line TK. Reframing the problem in terms of quadrilateral TUWV using these properties, we obtain the following problem: Let ABCD be a convex quadrilateral with an incircle ω, and let AC and BD intersect at P. Point E lies on ω such that the circumcircle of ACE is tangent to ω. Prove that if B and E lie on the same side of line AC, then the centre of the excircle of triangle ABC opposite the vertex B lies on line EP.
While this is an appealing statement, the Problem Selection Committee is uncertain about its difficulty and whether it has solutions that do not proceed by reducing to the original problem. Thus, it is believed that the original statement is more suitable for the competition.
80 Bath, United Kingdom, 10th–22nd July 2024 G7.
Let ABC be a triangle with incentre I such that AB ă AC ă BC. The second intersections of AI, BI, and CI with the circumcircle of triangle ABC are MA, MB, and MC, respectively. Lines AI and BC intersect at D and lines BMC and CMB intersect at X. Suppose the circumcircles of triangles XMBMC and XBC intersect again at S ‰ X. Lines BX and CX intersect the circumcircle of triangle SXMA again at P ‰ X and Q ‰ X, respectively.
Prove that the circumcentre of triangle SID lies on PQ.
(Thailand) Solution 1.
A B C O I MA MB MC D X S P Q Let O be the circumcentre of △ABC. First we note from standard properties of the Miquel point S we have: • △SMCMB „ △SBC „ △SPQ; p˚q • I and S are inverses with respect to circle ABC; • =OSX “ 90˝.
Claim 1. =MAPB “ =CDA.
Proof. From the above we have △OMAI „ △OSMA and =MAPB “ =MAPX “ =MASX “ 90˝=MASO “ 90˝=OMAI “ =MABA “ =CDA. l Shortlisted problems – solutions 81 Claim 2.
MCB BP “ MBC CQ “ AI ID.
Proof. Observe that =PMCMA “ =BMCMA “ =DAC and =MCMAB “ =ICD. Combining these with Claim 1 shows MCPMAB „ ADCI. Therefore, MCB BP “ AI ID. Similarly, MBC CQ “ AI ID.
l Claim 3.
DP DQ “ IB IC .
Proof. Firstly, observe that =ICB “ =AMBMC and =CBI “ =MBMCA which gives that △IBC „ △AMCMB. This, combined with Claim 2, is enough to show △DPQ „ △IBC by linearity, proving the claim.
l Claim 4.
IP IQ “ IB IC .
Proof. Combining △IBMC „ △ICMB with Claim 2 shows IBMCP „ ICMBQ giving the result.
l Finally, we have that SP SQ “ SB SC “ BMC CMB “ IB IC from p˚q and △IBMC „ △ICMB. Putting this together with Claims 3 and 4, we have that IB IC “ DP DQ “ IP IQ “ SP SQ, which shows that circle SID is an Apollonius circle with respect to P and Q, giving the desired conclusion.
Comment. The condition AB ă AC ensures S ‰ X. We also need to avoid the case =BAC “ 60˝ as then BMC ∥CMB.
82 Bath, United Kingdom, 10th–22nd July 2024 Solution 2.
We use Claim 1 from Solution 1. We will show that P and Q are inverses in circle SID which implies the result. Perform an inversion in circle BIC and denote the inverse of a point ‚ by ‚1.
A B C O I J MA MB MC D Y X S P Q P 1 Q1 P1 Q1 A1 Y 1 Claim 1. S1 “ J where J is the reflection of I across BC.
Proof. We have that S and I are inverses in circle ABC. Inverting this assertion in circle BIC shows that S1 and I are inverses with respect to line BC, which is just a reflection in line BC.
l Let Y “ MBMC X BC. From =IMCMB “ =MBMCA and =AMBMC “ =MCMBI, we see that A and I are reflections in line MBMC so Y A “ Y I. We have that circle SID maps to circle AIJ which, from the previous comment, has centre Y . Inverting the conclusion that P and Q are inverses with respect to circle SID in circle BIC, it suffices to show P 1 and Q1 are inverses with respect to circle AIJ or equivalently, that Y P 1 ¨ Y Q1 “ Y A2.
Claim 2. Circle XSMA maps to line Y J under the inversion in circle BIC.
Proof. Since circle BIC has centre MA, the inverse of this circle is a line. By Claim 1, this line passes through J hence it suffices to prove that circle XSMA passes through Y 1. From inverting line BC in circle BIC, we have that BCMAY 1 is cyclic so Y S ¨ Y X “ Y B ¨ Y C “ Y Y 1 ¨ Y MA.
where we have used that Y , S and X are collinear by a standard property of the Miquel point.
Hence Y 1 lies on circle XSMA as required.
l Shortlisted problems – solutions 83 Let A1 be the reflection of A in the perpendicular bisector of BC. Using Claim 1 from Solution 1, =P 1BMA “ =MAPB “ =CDA “ 180˝ ´ =ACMA “ 180˝ ´ =MABA1.
Hence, P 1, B, and A1 are collinear. Similarly Q1, C, and A1 are collinear. Let P1 and Q1 be the reflections of P 1 and Q1 across BC. As P 1 and Q1 lie on line Y J, it follows that P1 and Q1 lie on line Y I. Also from the previous collinearities, we get BP1 ∥AC and CQ1 ∥AB.
We have now reduced the problem to the following: Claim 3 (Inverted Problem). Let ABC be a triangle with incentre I. Let Y be the point on BC such that Y A “ Y I. Let P1 and Q1 be points on Y I such that BP1 ∥AC and CQ1 ∥AB.
Then Y A2 “ Y P1 ¨ Y Q1.
A B C I Y P1 Q1 E F Proof. Let Y I intersect AB and AC at E and F, respectively. From the parallel lines, we get that △BEP1 and △CQ1F are homothetic with centre Y . Thus we have Y E Y P1 “ Y Q1 Y F ù ñ Y P1 ¨ Y Q1 “ Y E ¨ Y F.
Moreover, AI bisects =EAF and Y A “ Y I so the circle centred at Y with radius Y A is the Apollonius circle of △AEF with respect to the feet of the internal and external angle bisectors at A. This gives Y E ¨ Y F “ Y A2. Combining these results proves the claim.
l 84 Bath, United Kingdom, 10th–22nd July 2024 Solution 3.
As in Solution 1, let O be the circumcentre of △ABC.
Let XI intersect circle XSMA again at Z ‰ X and let Y “ BC X MBMC. Let X˚ be the inverse of X in circle ABC. We will use the properties of Miquel point S noted at the top of Solution 1 and in addition, that S lies on line XY .
A B C O I MA MB MC D Y X S Q Z X˚ K P Claim 1. Y SAD is cyclic.
Proof. From OMAKBC and Y SKOS we have =DY S “ 180˝ ´ =SOMA.
From inverting collinear points A, I and MA in circle ABC we get ASMAO is cyclic which gives =SOMA “ =SAMA “ =SAD ù ñ =SAD =DY S “ 180˝ proving the claim.
l Claim 2. X˚ lies on circle BIC which has centre MA.
Proof. This follows immediately from inverting circle SBCX in circle ABC.
l Claim 3. Z lies on circle SID.
Proof. We have that =IZS “ =XMAS “ =OMAS ´ =OMAX “ =MAIO ´ =MAX˚O “ =DIO ´ =MAX˚O where in the penultimate step we inverted in circle ABC to get the angle equalities.
Shortlisted problems – solutions 85 From Brocard’s Theorem applied to cyclic quadrilateral BMCMBC, we get Y , I, and X˚ collinear and =Y X˚O “ 90˝. This gives that =MAX˚O “ 90˝ ´ =IX˚MA “ 90˝ ´ =MAIX˚ “ 90˝ ´ =AIY, where the second equality is by Claim 2. We have that A and I are reflections in line MBMC.
Hence, 90˝ ´ =AIY “ 90˝ ´ =Y AD “ 90˝ ´ =Y SD “ =DSO where the second step is by Claim 1, and in the last step we are using OSKY S. Putting these together, =IZS “ =DIO ´ =DSO “ =IDS, proving the claim.
l Let the tangents from S and Z to circle XSMA intersect at K. Observe from the standard Ceva-Menelaus configuration, ´1 “ pXY, XI; XB, XCq X “ pS, Z; P, Qq .
This shows that K lies on line PQ. We then have =ZKS “ 180˝ ´ 2=SXZ “ 2 p90˝ ´ =SXIq “ 2 p180˝ ´ =SIZq , where we are using =ISX “ 90˝. As K lies on the perpendicular bisector of SZ, this is enough to show that K is the centre of circle SIDZ completing the proof.
86 Bath, United Kingdom, 10th–22nd July 2024 Solution 4.
Solution 1 solves the problem by establishing SP SQ “ IP IQ “ DP DQ, which implies that circle SID is an Apollonius circle with respect to P and Q. We demonstrate an alternate approach that only requires us to show two of the ratios SP SQ, IP IQ, and DP DQ to be equal. This can arise from missing some of the observations in Solution 1, for example not proving Claim 3.
Claim. Given we have shown two of the ratios listed above to be equal, it suffices to show that circle SID is orthogonal to circle SXMA, which the same circle as SPQ.
Proof. Supposing we have shown the orthogonality, if SP SQ “ IP IQ or SP SQ “ DP DQ, then we immedi-ately have that circle SID is an Apollonius circle with respect to P and Q. If IP IQ “ DP DQ and S does not lie on the Apollonius circle C defined by this common ratio, then I and D lie on two distinct circles orthogonal to circle SPQ, namely circle SID and C. This implies that I and D are inverses with respect to circle SPQ, which is a contradiction as both I and D lie inside circle SPQ.
l Throughout this solution, we will use the properties of S from the beginning of Solution 1.
Define O and Y as in previous solutions, and let E be the second intersection of circles SOMA and SMBMC.
A B C O I MA MB MC Y S E M 1 A B1 C1 Lemma. We have that OE K AY .
Proof. Let M 1 A, B1, and C1 be the respective reflections of MA, B, and C over line MBMC.
As noted in Solution 3, A and I are reflections across MBMC.
Because MA is the centre of circle BIC, it follows that M 1 A is the centre of circle AB1C1. On the other hand, Y lies on MBMC, so we have that Y B ¨ Y C “ Y B1 ¨ Y C1.
Thus, Y lies on the radical axis of circles ABC and AB1C1, so OM 1 A K AY .
Finally, note that the inverses of circles SOMA and SMBMC in circle ABC are line IMA and circle IMBMC respectively, so E and M 1 A are inverses in circle ABC. Thus, E lies on OM 1 A and the lemma follows.
l Shortlisted problems – solutions 87 Let T denote the composition of an inversion at S with radius ?
SI ¨ SO with a reflection across line SI. By standard properties of the Miquel point, T swaps X and Y and any points Z1 and Z2 on circle ABC with I P Z1Z2. Hence, T swaps the pairs pA, MAq, pB, MBq, pC, MCq, pO, Iq, and pX, Y q. As D “ AI X BC and E is the intersection of circles SOMA and SMBMC, we have that T pDq “ E. Thus, T maps circles SID and SXMA to lines OE and AY , so by the Lemma, circles SID and SXMA are orthogonal, as required.
A B C O I MA MB MC D Y X S E M 1 A 88 Bath, United Kingdom, 10th–22nd July 2024 G8.
Let ABC be a triangle with AB ă AC ă BC, and let D be a point in the interior of segment BC. Let E be a point on the circumcircle of triangle ABC such that A and E lie on opposite sides of line BC and =BAD “ =EAC. Let I, IB, IC, JB, and JC be the incentres of triangles ABC, ABD, ADC, ABE, and AEC, respectively.
Prove that IB, IC, JB, and JC are concyclic if and only if AI, IBJC, and JBIC concur.
(Canada) Solution 1.
Let X be the intersection of IBJC and JBIC. We will prove that, provided that AB ă AC ă BC, the following two conditions are equivalent: (1) AX bisects =BAC; (2) IB, IC, JB, and JC are concyclic.
Let circles AIB and AIC meet BC again at P and Q, respectively. Note that AB “ BQ and AC “ CP because the centres of circles AIB and AIC lie on CI and BI, respectively.
Thus, B, P, Q, and C are collinear in this order as BQPC “ ABAC ą BC by the triangle inequality.
Claim 1. Points P, JB, and IC are collinear, and points Q, IB, and JC are collinear.
Proof. We have that =AJBB “ 90˝ 1 2=AEB “ 90˝ 1 2=ACB “ =AIB “ =APB, so ABJBP is cyclic. As A is the centre of spiral similarity between ABE and ADC, it is also the centre of spiral similarity between ABJB and ADIC. Hence, A is the Miquel point of self-intersecting quadrilateral BDICJB, so P lies on JBIC. Analogously, we have that Q lies on IBJC.
l B C D A I E P Q IB IC JB JC X Throughout the rest of the solution, we will use directed angles.
Shortlisted problems – solutions 89 Proof of p1q ù ñ p2q. We assume that p1q holds.
Claim 1 and the similarities ABDIB „ AECJC and ABEJB „ ADCIC tell us that >IBXIC “ >JCQC >BPJB “ >JCAC >BAJB “ >IBAD >DAIC “ >IBAIC, so AIBXIC is cyclic. Also, as X P AI, we have that >IBAX “ >BAI ´ >BAIB “ >IBAIC ´ >IBAD “ >DAIC.
Using these, we have that >IBICP “ >IBAX “ >DAIC “ >BAJB “ >BPJB, so IBIC ∥BC. Hence, >IBICJB “ >BPJB “ >BIJB “ >IBIJB, so IIBJBIC is cyclic. Analogously, we have that IICJCIB is cyclic, so IBJBJCIC is cyclic, thus proving p2q.
l Proof of p2q ù ñ p1q. We assume that p2q holds.
Claim 2. Circles IBC, IJBIC, and IIBJC are tangent at I.
Proof. Using the cyclic quadrilateral BIJBP, we have that >IBC “ >IBP “ >IJBP “ >IJBIC.
As C, IC, and I are collinear, the tangents to circles IJBIC and IBC at I coincide, so circles IJBIC and IBC are tangent at I. Analogously, circles IIBJC and IBC are tangent at I, so all three circles are tangent at I.
l Claim 3. Point I lies on circle IBJBJCIC.
Proof. Suppose that I does not lie on circle IBJBJCIC. Then the circles IIBJC, IJBIC, and IBJBJCIC are distinct. We apply the radical axis theorem to these three circles. By Claim 2, the radical axis of circles IIBJC and IJBIC is the tangent to circle IBC at I. As IBJC and JBIC intersect at X, IX must be tangent to circle IBC.
However, by Claim 1 we have that X is the intersection of PIC and QIB. As D lies on the interior of segment BC, IB lies on the interior of segment BI and IC lies on the interior of segment CI. Hence, IB, P, Q, and IC all lie on the perimeter of triangle IBC in this order, so X must be in the interior of triangle IBC. This means that IX cannot be tangent to circle BIC, so I must lie on circle IBJBJCIC.
l By Claims 2 and 3, circles IIBIC and IBC are tangent, so IBIC ∥BC. Since IBJBJCIC is cyclic, we have that >PJBJC “ >ICJBJC “ >ICIBJC “ >PQIB “ >PQJC, so PJBJCQ is cyclic. By the radical axis theorem on circles AIPJB, AIQJC, and PJBJCQ, we have that AI, IBJC, and JBIC concur at X, thus proving p1q.
l Solution 2. Let X be the intersection of IBJC and JBIC. As in Solution 1, we will prove that conditions p1q and p2q are equivalent. To do so, we introduce the new condition: p3q IBIC ∥BC and show that p3q is equivalent to both p1q and p2q, provided that AB ă AC ă BC.
Note that ABD „ AEC and ABE „ ADC, where „ denotes positive similarity. We will make use of the following fact.
90 Bath, United Kingdom, 10th–22nd July 2024 Fact. For points P, P1, P2, P3, and P4, the positive similarities PP1P2 „ PP3P4 and PP1P3 „ PP2P4 are equivalent.
Proof of p1q ð ñ p3q.
Let AIB and AIC meet BC at S and T, respectively. Let AJB meet BE at K, AJC meet CE at L, and KT and SL meet at Y .
B C D A E IB IC JB JC X S T K L Y Claim 1. Line AY bisects =BAC.
Proof. Let Y 1 be the intersection of KT and the bisector of =BAC. As =BAK “ 1 2=BAE “ 1 2=DAC “ =TAC, AY 1 also bisects =KAT. Hence, Y 1 is the foot of the bisector of =KAT in triangle AKT.
Using the Fact, we have that ABE „ ADC ù ñ ABEK „ ADCT ù ñ ABD „ AKT „ AEC ù ñ ABDS „ AKTY 1 „ AECL ù ñ ABEK „ ASLY 1 „ ADCT.
As K lies on BE, we have that Y 1 lies on SL, so Y “ Y 1 and AY bisects =BAC.
l We show that X lies on AY if and only if IBIC ∥BC, which implies the equivalence of p1q and p3q by Claim 1. Let AY meet IBJC and JBIC at X1 and X2, respectively. As ABD and AEC are similar, we have that AIB AS “ AJC AL , so IBJC ∥SL. Analogously, we have that JBIC ∥KT. Hence, X1 and X2 coincide with X if and only if AIB AS “ AX1 AY “ AX2 AY “ AIC AT , which is equivalent to IBIC ∥BC.
l Shortlisted problems – solutions 91 Proof of p2q ð ñ p3q. Let AJB and AJC meet circle ABC at M and N, respectively, and let I1 B and I1 C be the A-excentres of ABD and ADC, respectively.
B C D A IB IC I1 B I1 C Z Claim 2. Lines IBIC, I1 BI1 C, and BC are concurrent or pairwise parallel.
Proof. We work in the projective plane. Let IBIC and I1 BI1 C meet BC at Z and Z1, respec-tively. Note that Z is the intersection of the external common tangents of the incircles of ABD and ADC and AD is a common internal tangent of the incircles of ABD and ADC, so pAD, AZ; AIB, AICq “ ´1. Applying the same argument to the A-excircles of ABD and ADC gives pAD, AZ1; AI1 B, AI1 Cq “ ´1, which means that Z “ Z1. Thus, IBIC, I1 BI1 C, and BC concur, possibly at infinity.
l B C D A E IB IC JB JC M N I1 B I1 C 92 Bath, United Kingdom, 10th–22nd July 2024 Claim 3. Lines JBIC and CM are parallel, and lines IBJC and BN are parallel.
Proof. Using the Fact, we have that ABE „ ADC ù ñ ABEJB „ ADCIC ù ñ AJBIC „ ABD.
Thus, =pBD, JBICq “ =BAJB “ =BCM, so JBIC ∥CM. Similarly, we have that IBJC ∥BN.
l Claim 4. The centre of spiral similarity between JBJC and I1 BI1 C is A.
Proof. As IB and I1 B are respectively the incentre and A-excentre of triangle ABD, we have that ABI1 B „ AIBD. Using the similarity ABD „ AEC, this means that ABI1 B „ AJCC, so AB¨AC “ AI1 B ¨AJC and =BAI1 B “ =JCAC. Similarly, we have that AB¨AC “ AJB ¨AI1 C and =BAJB “ =I1 CAC. Together, these imply that AI1 B¨AJC “ AJB¨AI1 C and =JBAJC “ =I1 BAI1 C, so AJBJC „ AI1 BI1 C.
l We proceed using directed angles. By Claim 3, we have that IBJBJCIC is cyclic if and only if >IBICJB “ >IBJCJB ð ñ >IBICJB >MCB “ >IBJCJB >MNB ð ñ >pIBIC, BCq “ >pMN, JBJCq.
By Claim 4, we have that >pJBJC, I1 BI1 Cq “ >JBAI1 B “ >BAIB >MAB “ >EAJC >MAB “ >NAC >MAB “ >pMN, BCq, which is equivalent to >pBC, I1 BI1 Cq “ >pMN, JBJCq. Thus, IBJBJCIC is cyclic if and only if >pIBIC, BCq “ >pBC, I1 BI1 Cq.
(˚) Suppose that IBIC is parallel to BC. By Claim 2, I1 BI1 C is also parallel to BC, so we have that >pIBIC, BCq “ >pBC, I1 BI1 Cq “ 0˝. Thus, p˚q is satisfied, so IBJBJCIC is cyclic.
B C D A IB IC I1 B I1 C Z I Shortlisted problems – solutions 93 Suppose now that IBIC is not parallel to BC while IBJBJCIC is cyclic. By Claim 2, IBIC, I1 BI1 C, and BC concur at a point Z. As IB and IC lie on segments BI and CI, Z must lie outside segment BC. Since A is the intersection of the common external tangents of the incircle and A-excircle of ABD and ZD is a common internal tangent of the incircle and A-excircle of ABD, we have that pZA, ZD; ZIB, ZI1 Bq “ ´1. By p˚q, ZD bisects =IBZI1 B, so =AZD “ 90˝: that is, Z is the foot from A to BC. But this implies that =ABC or =BCA is obtuse, contradicting the fact that AB ă AC ă BC.
l Comment. While we have written the solution using harmonic bundles for the sake of brevity, there are ways to prove Claim 2 and obtain the final contradiction without the use of projective geometry.
Claim 2 can be proven using an application of Menelaus’s theorem, and the final contradiction can be obtained using the fact that an excircle of a triangle is always larger than its incircle.
Solution 3. Let ωB and ωC denote circles AIB and AIC, respectively. Introduce P, Q and X as in Solution 1 and recall from Claim 1 in Solution 1 that P, JB and IC are collinear with JB lying on ωB. From this, we can define JB and IC in terms of X by IC “ XP X CI and JB ‰ P as the second intersection of line XP with ωB. Similarly, we can define IB “ XQX BI and JC ‰ Q as the second intersection of line XQ with ωC. Note that this now detaches the definitions of points IB, IC, JB, and JC from points D and E.
Let ℓbe a line passing through I. We now allow X to vary along ℓwhile fixing △ABC and points I, P, and Q. We use the definitions from above to construct IB, IC, JB, and JC. We will classify all cases where these four points are concyclic. Throughout the rest of the solution we use directed angles and directed lengths.
For nondegeneracy reasons, we exclude cases where X “ I and X lies on line BC, which means that IB, JB ‰ B and IC, JC ‰ C. We also exclude the cases where ℓis tangent to either ωB or ωC. Similar results hold in these cases and they can be treated as limit cases.
Γ ωB ωC ℓ A B C X I P Q IB IC JC G H U V JB 94 Bath, United Kingdom, 10th–22nd July 2024 Claim 1. Line IBJB passes through a fixed point on ωB, and line ICJC passes through a fixed point on ωC as X varies on ℓ.
Proof. Let U ‰ JB be the second intersection of IBJB with ωB. We have by the law of sines that sin >IJBU sin >UJBB “ sin >IJBIB sin >IBJBB “ sin >JBIIB sin >JBBIB ¨ IIB IBB “ sin >JBIB sin >JBBI ¨ IIB IBB “ sin >XPQ sin >XPI ¨ IIB IBB .
We also have IIB IBB “ sin >IQIB sin >IBQB ¨ |IQ| |BQ| “ sin >IQX sin >XQP ¨ |IQ| |BQ|.
Combining these and applying Ceva’s Theorem in △PIQ with point X, we get sin >IJBU sin >UJBB “ sin >XPQ sin >XPI ¨ sin >IQX sin >XQP ¨ |IQ| |BQ| “ sin >XIQ sin >XIP ¨ |IQ| |BQ| “ sin >pℓ, IQq sin >pℓ, IPq ¨ |IQ| |BQ|, which is independent of the choice of X on ℓ. As >IJBU >UJBB “ >IJBB “ >IAB is fixed, this is enough to show point U is fixed on ωB.
Similarly, if we define V ‰ JC to be the second intersection of ICJC with ωC, we get that V is fixed on ωC.
l Let G ‰ X and H ‰ X be the second intersections of ℓwith ωB and ωC, respectively which exist as we are assuming ℓis not tangent to either ωB or ωC.
Claim 2. Points U, G and Q are collinear and points V , H and P are collinear.
Proof. Taking X “ G, we have JB “ G and IB “ XQXBI. Both of these points lie on line QG which, by Claim 1, shows that U, G, Q are collinear. Similarly, V , H, P are collinear.
l Claim 3.
Points IB, IC, JB, JC are concyclic if and only if points P, Q, G, H are concyclic.
In particular, this depends only on ℓ, not on the choice of X on ℓ.
Proof. We have that >ICJBIB “ >PJBU “ >PGU “ >PGQ >ICJCIB “ >V JCQ “ >V HQ “ >PHQ.
Thus >ICJBIB “ >ICJCIB ð ñ >PGQ “ >PHQ which proves the claim.
l Claim 4. P, Q, G, H are concyclic if and only if ℓP tIA, IP, IQ, tu where t is the tangent to circle BIC at I.
Proof. When ℓ“ IA, we have G “ H “ A so the cyclic condition from Claim 3 holds. Similarly, when ℓ“ IP or ℓ“ IQ, G “ P or H “ Q, respectively, so again the cyclic condition holds.
Now, consider the case where ℓR tIA, IP, IQu. In this case it is straightforward to see that the four points P, Q, G, and H are distinct. We then have that >QPG “ >BPG “ >BIG, so PQGH concyclic ð ñ >QHG “ >QPG ð ñ >QHG “ >BIG ð ñ QH ∥BI.
We also have that >CQH “ >CIH, so ℓtangent to circle BIC ð ñ >CIH “ >CBI ð ñ >CQH “ >CBI ð ñ QH ∥BI.
Hence, in this case P, Q, G, H are concyclic if and only if ℓis tangent to circle BIC, as claimed.
l Shortlisted problems – solutions 95 We now revert to using points D and E to define points IB, IC, JB, JC, and X, returning to the original set-up.
Claim 5.
Let Γ be the circle passing through P and Q that is tangent to IP and IQ, which exists as IP “ IQ “ IA. Then X lies on Γ. Furthermore, X lies on the same side of BC as A and does not lie on line BC.
Proof. We have that >XPI “ >JBPI “ >JBAI “ >BAI ´ >BAJB “ >JBAJC ´ >JBAE “ >EAJC “ >JCAC “ >JCQC “ >XQP, so circle XPQ is tangent to IP. Similarly, circle XPQ is tangent to IQ, so X lies on Γ.
As D lies in the interior of segment BC, IC lies in the interior of segment CI. Since X is the second intersection of PIC with Γ and IP is tangent to Γ, X lies in the interior of Ŋ PQ on Γ on the same side of BC as A. This implies the second part of the claim.
l By Claim 5, we cannot have ℓP tIP, IQu in the original problem. Furthermore, as shown in Claim 2 of Solution 1, we have that X lies inside triangle IBC, which means that ℓ‰ t.
Thus, the only remaining possibility in Claim 4 is ℓ“ AI. We then have IBICJBJC concyclic Claim 3 ð ù ù ù ù ñ PQGH concyclic Claim 4 ð ù ù ù ù ñ X lies on AI, finishing the problem.
Comment. The condition AB ă AC ă BC is used in an essential way in the solutions. In Solution 1, it is used in the proof of Claim 3 to ensure that X lies in the interior of triangle IBC. In Solution 2, it is used in the final step to ensure that =ABC and =BCA cannot be obtuse. In Solution 3, it is used to exclude the case ℓ“ t. If the condition is removed, then the problem is no longer true: whenever =ABC or =BCA is obtuse, there exists a choice of D on BC such that IBJBJCIC is cyclic but AI, IBJC, and JBIC do not concur. This counterexample configuration can be constructed using Solution 3 by letting X be the intersection of t with Γ that lies on the same side of BC as A and constructing IB, IC, JB, and JC as described in the solution, from which we can reconstruct D.
Conversely, the problem holds whenever =ABC and =BCA are both not obtuse, as can be seen from Solution 2. This is thus the weakest possible condition on triangle ABC that is necessary for the problem to be true.
B C D A I IB IC JB X Z P Q Γ JC When X lies on the tangent to circle IBC at I, there is no contradiction in the proof of Claim 3 in Solution 1: circles IIBJC and IJBIC are distinct, and X is the radical centre of circles IIBJC, IJBIC, and IBJBICJC. There is also no contradiction in the final step of Solution 2, and indeed IBIC and BC intersect at the foot of the altitude from A to BC.
There are no configuration issues with the direction p1q ù ñ p2q. This implication holds without any constraint on triangle ABC, and the proofs in Solutions 1 and 2 apply without any modification.
96 Bath, United Kingdom, 10th–22nd July 2024 Number Theory N1.
Find all positive integers n with the following property: for all positive divisors d of n, we have that d 1 | n or d 1 is prime.
(Ghana) Answer: n P t1, 2, 4, 12u.
Solution 1. It is easy to verify that n “ 1, 2, 4, 12 all work. We must show they are the only possibilities. We write n “ 2km, where k is a nonnegative integer and m is odd. Since m | n, either m 1 is prime or m 1 | n.
In the former case, since m 1 is even it must be 2, so n “ 2k.
If k ě 3, we get a contradiction, since 8 | n but 9 ∤n. Hence k ď 2, so n P t1, 2, 4u.
In the latter case, we have m 1 | 2km and m 1 coprime to m, and hence m 1 | 2k. This means that m 1 “ 2j with 2 ď j ď k (since j “ 1 gives m “ 1, which was considered earlier).
Then we have 2k 1 ∤n: since 2k1 is odd, it would have to divide m but is larger than m.
Hence, by the condition of the problem, 2k 1 is prime. If k “ 2, j must be 2 as well, and this gives the solution n “ 12. Also, 2k´1 1 ∤n for k ą 2: since it is odd, it would have to divide m. However, we have no solutions to 2k´1 1 | 2j ´ 1 with j ď k: the left-hand side is greater than the right unless j “ k, when the left-hand side is just over half the right-hand side.
Since we have 2k | n and 2k 1 ∤n, and 2k´1 | n and 2k´1 1 ∤n, we must have 2k 1 and 2k´1 1 both prime. However, 2a 1 is a multiple of three if a is odd, so we must have 2k 1 “ 3 (impossible as this gives k “ 1) or 2k´1 1 “ 3, which gives j “ k “ 2, whence n “ 12.
Solution 2. We proceed as in Solution 1 as far as determining that n “ 2kp2j ´1q with j ď k.
Now, we have 2j | n but 2j 1 ∤n, as it is odd and does not divide 2j ´ 1. Thus 2j 1 is prime. The theory of Fermat primes tells us we must have j “ 2h with h ą 0.
Then 22h ´ 1 is congruent to 3 or 6 (modulo 9) depending on whether h is odd or even, respectively. In particular it is not divisible by 9, so n “ 2kp22h ´ 1q is not divisible by 9; so we must have k ď 2, since if k ě 3 then 8 | n but 9 ∤n with 9 not prime.
Solution 3. Let p be the smallest integer not dividing n. Since p ´ 1 is a divisor of n, p must be a prime. Let 1 ď r ď p ´ 1 be the remainder of n modulo p. Since p ´ r ă p, we have p ´ r | n, so we may consider the divisor d “ n p´r.
Since p | n ´ r, we have p | n p ´ r, whence p | d 1. Thus d 1 ∤n; so it must be prime.
On the other hand, this prime is divisible by p, so we conclude d 1 “ p, which means that n “ pp ´ 1qpp ´ rq.
Then from p ´ 2, p ´ 3 | n we get pp ´ 2qpp ´ 3q | 2pp ´ rq, from which we find pp ´ 2qpp ´ 3q ď 2pp ´ rq ď 2pp ´ 1q.
Solving this quadratic inequality gives p ď 5, which means that n P t1, 2, 4, 8, 12, 16u. Of this set, n “ 8 and n “ 16 are not solutions.
Solution 4. We suppose that n is not 1 or 2.
Since n | n and n 1 ∤n, we know that n 1 is prime. Thus it is odd, so 2 | n; as n ą 2, we have n 2 | n and n 2 1 ∤n, so n 2 1 is prime. Thus it is also odd, so 4 | n.
We must then have n 4 1 | n or n 4 1 prime.
Shortlisted problems – solutions 97 In the former case, n 4 | 4pn 4 1q ´ n, so n 4 1 | 4. This means that n “ 4 or n “ 12.
In the latter case, n 4 1 must be odd if n ‰ 4. Thus we have n “ 8m where 2m 1, 4m 1, 8m 1 are all prime; n “ 8 does not work, so 3 | m (otherwise one of those numbers would be divisible by 3). Thus 24 | n, so 25 | n as 25 is not prime.
Now suppose that p is the least positive integer not dividing n: as in Solution 3 we know that p is prime, and what we have done so far shows that p ě 7. If p2 ´ 1 “ pp ´ 1qpp 1q is the product of coprime integers less than p, it divides n, and p2 is not prime so also divides n (a contradiction); p ´ 1 and p 1 are even and have no common factor higher than 2, so all odd prime power divisors of their product are less than p and the only case where p2 ´ 1 is not a product of coprime integers less than p is when one of p ´ 1 and p 1 is a power of 2, say 2m (with m ě 3). If p “ 2m ´ 1, then 3p ´ 1 “ 4p3 ˆ 2m´2 ´ 1q and 3 ˆ 2m´2 ´ 1 is an odd integer less than p, so 3p ´ 1 | n and so 3p | n. Finally, if p “ 2m 1, then m is even and 2p ´ 1 “ 2m1 1 is a multiple of 3; the only case where it is a power of 3 is when m “ 2, but we have m ě 3, so 2p ´ 1 is a product of coprime integers less than p and again we have a contradiction.
Solution 5.
As in Solution 4, we deduce that if n ą 2 then n must be even. We write n “ 2 ¨ 3k ¨ r, where k is a nonnegative integer and 3 ∤r.
Since r and 2r are both different and nonzero modulo 3, one of them must be congruent to 2 modulo 3. We’ll say that it is ar, where a P t1, 2u.
Since ar | n, we must have that ar 1 is either prime or a factor of n. In the first case, ar 1 “ 3 because 3 | ar 1, and so n “ 2 ¨ 3k ¨ r, where r “ 2{a is 1 or 2. Noting that we must have k ď 1 (else 9 | n but 10 ∤n), we can examine cases to deduce that n P t2, 4, 12u are the only possibilities.
Otherwise, ar 1 | n. Since ar 1 is coprime to r, we must in fact have that ar 1 | 2 ¨ 3k, and since 3 | ar 1 by assumption we deduce that k ě 1. In particular, 3k 1 is an even number that is at least 4, so is not prime and must divide n. As it is coprime to 3, we must in fact have 3k 1 | 2r.
Let q1 and q2 be such that q1par1q “ 2¨3k and q2p3k 1q “ 2r. We have that q1ar ă 2¨3k and q23k ă 2r, and multiplying these together gives q1q2a ă 4.
If a “ 2 then q1 “ q2 “ 1, so 2r 1 “ 2 ¨ 3k, which is not possible (considering both sides modulo 2).
If a “ 1 then r must be equivalent to 2 modulo 3, so q2p3k 1q “ 2r gives that q2 is equivalent to 1 modulo 3, whence q2 “ 1. So we deduce that 2r “ 3k 1. Thus, we deduce that q1p3k 3q “ 4 ¨ 3k, which rearranges to give 3k´1p4 ´ q1q “ q1, whence 3k´1 ď q1 ă 4 and so k ď 2. We can examine cases to deduce that n “ 12 is the only possibility.
98 Bath, United Kingdom, 10th–22nd July 2024 N2.
Determine all finite, nonempty sets S of positive integers such that for every a, b P S there exists c P S with a | b 2c.
(Netherlands) Answer: The possible sets are S “ ttu and S “ tt, 3tu for any positive integer t.
Solution 1.
Without loss of generality, we may divide all elements of S by any common factor, after which they cannot all be even. As a ∤b 2c for a even and b odd, the elements of S are all odd.
We now divide into three cases: Case 1: |S| “ 1.
The set S “ ttu clearly works.
Case 2: |S| “ 2.
Say S “ tr, su with r ă s, so either s | r 2r or s | r 2s, and in either case s | 3r. We cannot have s “ 3r{2 as we assumed that r is odd, so s “ 3r and S “ tr, 3ru, which clearly works by examining cases for a and b.
Case 3: |S| ě 3.
If all elements of S are odd then for any b, c P S, b 2c ı b pmod 4q. If a | b 2c with a ” b pmod 4q, this means there exists k with b 2c “ ka and k ” 3 pmod 4q, so k ě 3. If a is the greatest element of S and b ă a, we have b 2c ă 3a, a contradiction. Thus when a is the greatest element, no b ă a has b ” a pmod 4q (and thus all elements other than the greatest are congruent modulo 4).
Let d and e be the largest and second largest element of S respectively. Let f ‰ d, e be any other element of S. There is some c P S with e | f2c, and e ı f 2c pmod 4q, so f2c ě 3e, so c ą e. Since e is the second largest element of S, c “ d, so e | f 2d, and this holds for all f P S with f ă e, but can only hold for at most one such f. So |S| ď 3.
Hence the elements of S are d ą e ą f, and by the discussion above without loss of generality we may suppose these elements are all odd, e ” f pmod 4q and d ı e pmod 4q. We have above that e | f 2d. Furthermore, there exists some c P S with d | f 2c, and c ‰ d as d ą f so d ∤f, so c ď e; as f 2e ă 3e, we have e ą d{3. Since f 2c is odd and f 2c ă 3d, we have f 2c “ d.
Subcase 3.1: c “ f.
Here d “ 3f and e | f 2d “ 7f. As e ą f and e ” f pmod 4q, we have e “ 7f{3 and the elements are some multiples of t3, 7, 9u. But a “ 7 and b “ 9 have no corresponding value of c.
Subcase 3.2: c “ e.
Here d “ f 2e and e | f 2d “ 3f 4e so e | 3f. But this is not possible with e ą f and e ” f pmod 4q.
Solution 2.
As in Solution 1, we reduce to the case where all elements of S are odd. Since all one-element sets satisfy the given conditions, we show that if |S| ě 2, then |S| “ 2 and S “ tt, 3tu for some positive integer t.
Let d be the largest element. For any e P S with e ‰ d there must be a f P S such that d | e 2f. This implies 2f ” ´e pmod dq, hence 2f ” d ´ e pmod dq. Now d ´ e is even (because all elements in S are odd) and d is odd, so d´e 2 is an integer and we have f ” d´e 2 pmod dq. Further, 0 ă d´e 2 ă d, while we must also have 0 ă f ď d, so f “ d´e 2 . We conclude that for any e P S with e ‰ d the integer d´e 2 is also in S and not equal to d.
Denote by e1 ă e2 ă ¨ ¨ ¨ ă ek ă d the elements of S, where k ě 1. Then d´e1 2 ą d´e2 2 ą ¨ ¨ ¨ ą d´ek 2 are also elements of S, none of them equal to d. Hence we must have e1 “ d´ek 2 and Shortlisted problems – solutions 99 ek “ d´e1 2 , so 2e1ek “ d “ 2eke1. We conclude e1 “ ek, so k “ 1, and also d “ 2eke1 “ 3e1.
Hence S “ te1, 3e1u for some positive integer e1.
Solution 3.
As in Solution 1, we reduce to the case where all elements of S are odd. Since all one-element sets satisfy the given conditions, we show that if |S| ě 2, then |S| “ 2 and S “ tt, 3tu for some positive integer t.
Let d be the largest element, and let e P S be any other element. We will say that x P S pmod dq if the unique element y in t1, . . . , du such that x ” y pmod dq is an element of S. Note that by the choice of d being the largest element, if x ‰ d, then x ı 0 pmod dq. The given condition implies that if b P S, then ´ b 2 P S pmod dq. Repeating this gives ´ b 2 P S ñ b 4 P S pmod dq, and by iterating, we have b P S ñ b p´2qk P S pmod dq for all k. Since d is odd, there is some g such that p´2qg ” 1 pmod dq, so by setting k “ g ´ 1, we get that for all d ‰ e P S, ´2e P S pmod dq.
Now, if e ą d 2, then ´2e P S pmod dq and d ´ 2e ă 0, so 2d ´ 2e P S, contradicting the lack of even elements. Then e ă d 2 for any e P S z tdu, so we have e P S ñ d ´ 2e P S. Since d ´ 2e ‰ d, we must have d ´ 2e ă d 2, which rearranges to e ą d 4.
Let λ P p0, 1q be a positive real number and suppose we have proved that e ą λd for any e P S z tdu. Then d ´ 2e ą λd, which rearranges to e ă p1´λqd 2 . Then d ´ 2e ă p1´λqd 2 , which rearranges to e ą p1λqd 4 . Defining λ0 “ 1 4 and λi “ 1λi´1 4 for i ě 1, we have shown that for all e P S z tdu and all λi, e ą λid. Now note that the sequence λi is increasing and bounded above by 1 3, so it converges to some limit ℓ, which satisfies ℓ“ 1ℓ 4 , so ℓ“ 1 3. Hence e ě d 3, but then d ´ 2e ě d 3 implies e ď d 3, so e must be d 3, and we are done.
Comment. We can finish Solution 3 alternatively as follows: after showing that if e P S z tdu then d ´ 2e P S z tdu, note that pd ´ 2eq ´ d 3 “ 2d 3 ´ 2e “ ´2 ˆ e ´ d 3 ˙ .
So consider e P S ztdu maximising |e´ d 3|. If e ‰ d 3, them the above shows that |pd´2eq´ d 3| ą |e´ d 3|, which is a contradiction. Thus S z tdu is empty or equal to t d 3u, which completes the proof.
100 Bath, United Kingdom, 10th–22nd July 2024 N3.
Determine all sequences a1, a2, . . .
of positive integers such that, for any pair of positive integers m ď n, the arithmetic and geometric means am am1 ¨ ¨ ¨ an n ´ m 1 and pamam1 ¨ ¨ ¨ anq 1 n´m1 are both integers.
(Singapore) Answer: The only such sequences are the constant sequences (which clearly work).
Solution 1.
We say that an integer sequence b1, b2, . . .
is good if for any pair of positive integers m ď n, the arithmetic mean bmbm1¨¨¨bn n´m1 is an integer. Then the condition in the question is equivalent to saying that the sequences paiq and pνppaiqq for all primes p are good.
Claim 1. If pbiq is a good sequence, then n ´ m | bn ´ bm for all pairs of integers m, n.
Proof. This follows from n ´ m dividing bm bm1 ¨ ¨ ¨ bn´1 and bm1 bm2 ¨ ¨ ¨ bn, and then taking the difference.
l Claim 2.
If pbiq is a good sequence where some integer b occurs infinitely many times, then pbiq is constant.
Proof. Say bn1, bn2, bn3, . . . are equal to b. Then for all m, we have that b ´ bm “ bnj ´ bm is divisible by infinitely many different integers nj ´m, so it must be zero. Therefore the sequence is constant.
l Now, for a given prime p, we look at the sequence pνppaiqq. Let k “ νppa1q. Then Claim 1 tells us that a1 ” anpk11 pmod pk1q for all n, which implies that νppanpk11q “ k for all n.
We now have that k appears infinitely many times in this good sequence, so by Claim 2, the sequence pνppaiqq is constant. This holds for all primes p, so paiq must in fact be constant.
Solution 2. As in Claim 1 of Solution 1, we have that air ” ai pmod rq, which tells us that the sequence ai is periodic modulo p with period p. Also, by a similar argument, we have that air{ai is the rth power of a rational number.
Now suppose that for some i ı j pmod pq we have ai, aj ı 0 pmod pq. As p and p ´ 1 are coprime, we can find some i1 ” i pmod pq, j1 ” j pmod pq such that p ´ 1 | i1 ´ j1. Then ai1{aj1 is a perfect pp ´ 1qth power, so ai1 “ tup´1, aj1 “ tvp´1 for some positive integers t, u, v not divisible by p. By Fermat’s little theorem, up´1 and vp´1 must be 1 modulo p. So we must have ai ” ai1 ” t ” aj1 ” aj pmod pq.
Thus all values of ai that are not divisible by p are congruent modulo p.
For the sum of p consecutive values to be divisible by p, this means that all the ai are congruent modulo p. Since this is true for all primes p, the sequence must therefore be constant.
Shortlisted problems – solutions 101 Solution 3.
Fix an arbitrary index m. First, we show that am divides an for sufficiently large n. Let n be sufficiently large that n ą νppamq m for every prime p. By Claim 1 of Solution 1, we have νppamq ” νppanq pmod n ´ mq.
Since νppamq ă n´m, it follows that νppamq ď νppanq. This holds for every prime p, so am | an.
Next, suppose that there is some index k such that am does not divide ak. By the previous, there is a maximal such k. Then ak1, ak2, . . . are all divisible by am. But now applying the first condition gives am | ak ak1 ¨ ¨ ¨ akam´1, so am divides ak, a contradiction. Therefore every term an is divisible by am.
As m was arbitrary, we now have am | an and vice versa for all m, n. So the sequence must be constant.
102 Bath, United Kingdom, 10th–22nd July 2024 N4.
Determine all positive integers a and b such that there exists a positive integer g such that gcdpan b, bn aq “ g for all sufficiently large n.
(Indonesia) Answer: The only solution is pa, bq “ p1, 1q.
Solution 1. It is clear that we may take g “ 2 for pa, bq “ p1, 1q. Supposing that pa, bq satisfies the conditions in the problem, let N be a positive integer such that gcdpan b, bn aq “ g for all n ě N.
Lemma. We have that g “ gcdpa, bq or g “ 2 gcdpa, bq.
Proof. Note that both aN b and aN1 b are divisible by g. Hence apaN bq ´ paN1 bq “ ab ´ b “ apb ´ 1q is divisible by g. Analogously, bpa ´ 1q is divisible by g. Their difference a ´ b is then divisible by g, so g also divides apb ´ 1q apa ´ bq “ a2 ´ a. All powers of a are then congruent modulo g, so a b ” aN b ” 0 pmod gq. Then 2a “ pa bq pa ´ bq and 2b “ pa bq ´ pa ´ bq are both divisible by g, so g | 2 gcdpa, bq. On the other hand, it is clear that gcdpa, bq | g, thus proving the Lemma.
l Let d “ gcdpa, bq, and write a “ dx and b “ dy for coprime positive integers x and y. We have that gcd ppdxqn dy, pdyqn dxq “ d gcd dn´1xn y, dn´1yn x ˘ , so the Lemma tells us that gcd dn´1xn y, dn´1yn x ˘ ď 2 for all n ě N. Defining K “ d2xy 1, note that K is coprime to each of d, x, and y. By Euler’s theorem, for n ” ´1 pmod φpKqq we have that dn´1xn y ” d´2x´1 y ” d´2x´1p1 d2xyq ” 0 pmod Kq, so K | dn´1xn y. Analogously, we have that K | dn´1yn x. Taking such an n which also satisfies n ě N gives us that K | gcdpdn´1xn y, dn´1yn xq ď 2.
This is only possible when d “ x “ y “ 1, which yields the only solution pa, bq “ p1, 1q.
Solution 2. After proving the Lemma, one can finish the solution as follows.
For any prime factor p of ab 1, p is coprime to a and b. Take an n ě N such that n ” ´1 pmod p ´ 1q. By Fermat’s little theorem, we have that an b ” a´1 b “ a´1p1 abq ” 0 pmod pq, bn a ” b´1 a “ b´1p1 abq ” 0 pmod pq, then p divides g. By the Lemma, we have that p | 2 gcdpa, bq, and thus p “ 2. Therefore, ab1 is a power of 2, and a and b are both odd numbers.
If pa, bq ‰ p1, 1q, then ab 1 is divisible by 4, hence ta, bu “ t´1, 1u pmod 4q. For odd n ě N, we have that an b ” bn a ” p´1q 1 “ 0 pmod 4q, then 4 | g. But by the Lemma, we have that ν2pgq ď ν2p2 gcdpa, bqq “ 1, which is a contradiction.
So the only solution to the problem is pa, bq “ p1, 1q.
Shortlisted problems – solutions 103 N5.
Let S be a finite nonempty set of prime numbers. Let 1 “ b1 ă b2 ă ¨ ¨ ¨ be the sequence of all positive integers whose prime divisors all belong to S. Prove that, for all but finitely many positive integers n, there exist positive integers a1, a2, . . . , an such that a1 b1 a2 b2 ¨ ¨ ¨ an bn “ R 1 b1 1 b2 ¨ ¨ ¨ 1 bn V .
(Croatia) Solution 1.
If S has only one element p, then bi “ pi´1 and we can easily find a1, . . . , an with 2 “ Qřn´1 i“0 1 pi U “ řn´1 i“0 ai pi´1 by taking a1 “ a2 “ ¨ ¨ ¨ “ an´1 “ 1 and choosing an “ pn´1 ´ pp p2 ... pn´2q.
More generally, observe that the sum of 1 bi over all i is ÿ i 1 bi “ ź i ˆ 1 1 pi 1 p2 i . . .
˙ “ ź pPS p p ´ 1.
In particular, if n is large enough, then S n ÿ j“1 1 bj W “ Sź pPS p p ´ 1 W .
For the remainder of the proof, we will only consider n large enough that this equality holds.
Next, we handle the special case S “ t2, 3u, for which this product is 3. Start by setting ai “ # 1, if 2bi ď bn; 2, if 2bi ą bn.
Then, ÿ iďn ν3pbiq“t ai bi “ # 2 3t, if bn ě 3t; 0, otherwise.
As a result, ÿ iďn ai bi “ ÿ tě0 3tďbn 2 3t “ 3 ´ 1 3T where T is the largest t ě 0 with 3t ď bn. Thus, increasing aj by one (where bj “ 3T) gives a sequence of ai that works.
Otherwise, we may assume that |S| ą 1 and S ‰ t2, 3u, which means that the product ś pPS p p´1 is not an integer. Indeed, • if |S| ą 2 then 2 divides the denominator at least twice and so divides the denominator of the overall fraction; • if |S| “ 2 and 2 R S then 2 divides the denominator and not the numerator; 104 Bath, United Kingdom, 10th–22nd July 2024 • if S “ t2, pu then the product is 2p{pp ´ 1q which is not an integer for p ą 3.
It follows that for some fixed α ą 0, we have that Sź pPS p p ´ 1 W “ ź pPS p p ´ 1 α, from which it follows that S n ÿ i“1 1 bi W ´ n ÿ i“1 1 bi ą α.
It will now suffice to prove the following claim.
Claim. Suppose that n is large enough, and let ep be the largest nonnegative integer such that pep ď bn. Let M “ ś pPS pep. If u is a positive integer such that u{M ą α, then there exist nonnegative integers ai such that ÿ i ai bi “ u M .
The problem statement follows after replacing ai with ai 1 for each i.
To prove this, choose some constant c such that ř pPS p´c ă α, and suppose n is large enough that pc ă bn for each p P S; in particular, pc | M with M defined as above.
For each p P S, let ip be such that bip “ pep and choose the smallest nonnegative integer aip satisfying pep´c ˇ ˇ ˇ aip ˆ M pep ˙ ´ u.
Such an aip must exist and be at most pep´c; indeed, M pep is an integer coprime to p, so we can take aip to be equal to u times its multiplicative inverse modulo pep´c. The sum of the contributions to the sum from the aip is at most ÿ pPS pep´c pep “ ÿ pPS p´c ă α.
So, we have u M “ ÿ pPS aip pep r ś pPS pc, where r is an integer because of our choice of aip and r is nonnegative because of the bound on u. Simply choose ai “ r where bi “ ś pPS pc to complete the proof.
Solution 2. We reduce to the claim as in Solution 1, and provide an alternative approach for constructing the ai.
Let p0 P S be the smallest prime in S. Let z0 “ u{M. We construct a sequence z0, z1, z2, . . . and values of ai by the following iterative process: to construct zj1, • select the largest prime p P S dividing the denominator of zj, and let µ be the number of times p divides the denominator of zj; • choose the largest ν such that pν 0pµ ď bn, and let i ď n be such that bi “ pν 0pµ; • choose 0 ď ai ă p such that the denominator of zk ´ ai{bi has at most µ ´ 1 factors of p, and let zk1 “ zk ´ ai{bi; • continue until p0 is the only prime dividing the denominator of zk.
Shortlisted problems – solutions 105 Note that we can always choose ai in step 3; by construction, zkbi has no factors of p in its denominator, so must be realised as an element of Zp.
Each time we do this, bi ą M{p0 by construction, so ai bi ă pp0 M ď p0p1 M , where p1 is the largest prime in S. And the number of times we do this operation is at most ÿ pPS pąp0 ep ď |S| log2pMq, so the sum of the ai{bi we have assigned is at most |S|p0p1 log2pMq{M.
Choose n large enough that log2pMq{M ă α; after subtracting the above choices of ai{bi from u{M, we have a quantity of the form r{p ep0 0 , where r is an integer by construction and r is positive by the above bounds. Simply set ai “ r where bi “ p ep0 0 to complete the proof.
Solution 3.
As in Solution 1, we may handle |S| “ 1 and S “ t2, 3u separately; otherwise, we can define α as we did in that solution. Also define ep to be the largest nonnegative integer such that pep ď bn as we did in Solution 1.
We will show that, for n sufficiently large, we may choose some j ď n, and positive inte-gers ai, such that ÿ i‰j ai bi ´ ÿ i‰j 1 bi ă α.
and all ai bi are integer multiples of 1 bj . We then set aj to be the least positive integer such that the sum on the left is an integer, which will obviously have the required value.
Concretely, choose j such that bj “ ś pPS ptep{|S|u, which is less than bn by construction.
For i ‰ j, set ai “ bi{ gcd pbi, bjq. We have ÿ i‰j ai bi ´ ÿ i‰j 1 bi ă ÿ i‰j aią1 ai bi .
If ai ą 1, then there must be some p P S for which ptep{|S|u1 | bi, and so ai bi “ 1 gcd pbi, bjq ď 1 ptep{|S|u ă p b1{|S| n , where the last inequality follows from the fact that pep1 ą bn.
Now n ď ś pPSplogppbnq 1q ď p2 log bnq|S|, so ÿ i‰j aią1 ai bi ď p2 log bnq|S| b1{|S| n , and so we can choose n large enough that this quantity is less than α, as required.
106 Bath, United Kingdom, 10th–22nd July 2024 N6.
Let n be a positive integer.
We say that a polynomial P with integer coeffi-cients is n-good if there exists a polynomial Q of degree 2 with integer coefficients such that QpkqpPpkq Qpkqq is never divisible by n for any integer k.
Determine all integers n such that every polynomial with integer coefficients is an n-good polynomial.
(France) Answer: The set of such n is any n ą 2.
Solution 1. First, observe that no polynomial is 1-good (because QpXqpPpXqQpXqq always has roots modulo 1) and the polynomial PpXq “ 1 is not 2-good (because QpXqpQpXq 1q is always divisible by 2).
Now, if P is d-good with some Q, then Q ¨ pP Qq has no roots mod d. Therefore, it certainly has no roots mod n for d | n, so P must be n-good. Consequently, it suffices to show that all polynomials are n-good whenever n is an odd prime, or n “ 4.
We start by handling the case n “ 4. We will construct a Q such that QpXq is never divisible by 4 and QpXq PpXq is always odd; this will clearly show that P is 4-good. Note that any function modulo 2 must be either constant or linear – in other words, there are a, b P t0, 1u such that PpXq “ aX b mod 2 for all X. If a “ 0 then set QpXq “ 4X2 b 1, and if a “ 1 then set QpXq “ X2 b 1; in all cases, Q will satisfy the required properties.
It remains to prove that any polynomial is p-good, where p is an odd prime. We will prove that for any function f defined mod p, there is a quadratic Q with no roots mod p such that Qpxq ‰ fpxq mod p for all x; the statement about P then follows with f replaced by ´P. For the remainder of the proof, we will consider all equalities modulo p.
Suppose that a function f not satisfying the above exists; in other words, f has the property that for any quadratic Q with no roots mod p, there is some x such that Qpxq “ fpxq. Without loss of generality, we may assume that f has no roots mod p. To see why, suppose that fpuq “ 0 for some u, and let g be the function such that gpxq “ fpxq for x ‰ u and gpuq “ 1. For any Q with no roots, we know that there is some x ‰ u such that Ppxq “ fpxq, and so Ppxq “ gpxq for that choice of x. In particular, g is also not p-good.
Now, suppose first that there is some nonzero t such that t is not in the image of f. Then we may take QpXq “ pX2 t; this quadratic is never equal to f and is never zero. Thus, f must be surjective onto the nonzero residues mod p. There are p choices for X and p ´ 1 nonzero residues mod p, so there must be some x1 ‰ x2 mod p such that fpx1q “ fpx2q, and f is a bijection from the set of residues mod p not equal to x2 to the set of nonzero residues mod p.
Now, note that we may choose any b and c with b nonzero and replace fpXq with gpXq “ fpbXcq; if there were some Q with no roots such that Qpxq ‰ gpxq for all x, then QpX{b´c{bq would work for f. Choose b and c such that bx1 c “ 1 and bx2 c “ ´1; such b and c must exist (we may take b “ 2{px1 ´ x2q and c “ px1 x2q{px2 ´ x1q). Renaming g to f, we see that we may assume fp1q “ fp´1q.
Let r1 be a quadratic nonresidue mod p. Choose y ‰ 0 such that fpyq “ p1 ´ r1qfp0q, which must exist as the right hand side is nonzero and 1 ´ r1 is not equal to 1. Choose r “ y2{r1, which is a quadratic nonresidue.
Consider ϕpXq “ fpXq{pX2 ´ rq. By definition, ϕp1q “ ϕp´1q and ϕp0q “ ϕpyq, so there are no more than p ´ 2 values in the image of ϕ. Choose some nonzero a not in the image of ϕ, so fpXq{pX2 ´ rq is never equal to a. The quadratic QpXq “ apX2 ´ rq is never zero and also never equal to fpXq, which completes the proof.
Shortlisted problems – solutions 107 Comment. In fact, there is no need to pass from polynomials P to functions f, as any function mod p is a polynomial. Concretely, instead of passing from f to g, we would have instead replaced PpXq with PpXq 1 ´ pX ´ uqp´1, which is a polynomial that is unchanged except at X “ u.
Solution 2.
Given f a function mod p such that f is surjective onto the nonzero ele-ments of Z{pZ and fp1q “ fp´1q, we provide an alternative approach to construct a nonzero quadratic QpXq such that QpXq ‰ fpXq. Let r be the smallest quadratic nonresidue mod p (so r ´ 1 is a square) and let a vary over the nonzero elements mod p; we will show that it is possible to choose QapXq “ apX2 ´ rq for some choice of a. Note that any quadratic of this form will be nowhere zero.
Suppose that no such Qa works. Then, for each a, there exists x such that apx2 ´rq “ fpxq.
We may assume that x ‰ ´1, as if the equality holds for x “ ´1 then it also holds for x “ 1.
However, apx2 ´ rq “ fpxq implies a “ fpxq{px2 ´ rq, so fpxq{px2 ´ rq must be a surjection from tx ‰ ´1u to the set of nonzero a, and so this is a bijection. In particular, for each a, there exists a unique xa such that fpxaq “ apx2 a ´ rq.
We now have ź t‰0 t “ ź a‰0 fpxaq “ ź a‰0 a ź a‰0 px2 a ´ rq “ ź a‰0 a ź x‰´1 px2 ´ rq where the first equality follows because f is surjective onto the nonzero residues mod p, and the second equality follows from the definition of xa. The two products cancel, which means that ś x‰´1px2 ´ rq “ 1.
However, we also get ź x‰´1 px2 ´ rq “ p´rqp1 ´ rq ˜pp´1q{2 ź x“2 px2 ´ rq ¸2 .
However, this is a contradiction as ´rp1 ´ rq “ rpr ´ 1q, which is not a quadratic residue (by our choice of r).
Comment. By Wilson’s theorem, we know that the product of the nonzero elements mod p is ´1; however, this fact was not necessary for the solution so we chose to present the solution without needing to state it.
Comment. One can in fact show that ź x‰´1 px2 ´ rq “ ´4r 1 ´ r.
To do this, note that the polynomial X p´1 2 ´ 1 has the p´1 2 quadratic residues as roots, so we have ź s quad. res.
pX ´ sq “ X p´1 2 ´ 1 and so ź x‰0 pX ´ x2q “ pX p´1 2 ´ 1q2.
Since r is a quadratic nonresidue, by Euler’s criterion r p´1 2 “ ´1, and the result follows.
Therefore, one can replace the condition that r is the smallest quadratic nonresidue with the condition that r is a quadratic nonresidue not equal to ´1 3 (which is possible for all p ě 3).
108 Bath, United Kingdom, 10th–22nd July 2024 Solution 3. As in Solution 1, we will reduce to the case of p being an odd prime and f being a function mod p with no roots which is surjective onto the set of nonzero residues mod p, although we make no assumption about the values of x1 and x2 with fpx1q “ fpx2q.
We will again consider quadratics of the form Qa,b,cpXq “ aRpbX cq, where RpXq “ X2´r for an arbitrary fixed quadratic nonresidue r, a and b are nonzero mod p, and c is any residue mod p.
For each fixed b and c, there must be n pairs pa, xq such that aRpbx cq “ fpxq, because there must be exactly one value of a for each x. If any a appears in no such pair then we are done, so assume otherwise. In other words, there must be exactly one a such that there are two such x, and for all other a there is only one such x.
Thus, for each pb, cq, there is exactly one unordered pair tx1, x2u such that for some a we have fpxiq “ aRpbxi cq; in other words, there is exactly one unordered pair tx1, x2u such that fpx1q{Rpbx1 cq “ fpx2q{Rpbx2 cq.
Now, we show that for each unordered pair tx1, x2u there must be at least one pair pb, cq such that fpx1q{Rpbx1 cq “ fpx2q{Rpbx2 cq. Indeed, let t “ fpx1q{fpx2q. There must be some x1 1, x1 2 such that Rpx1 1q{Rpx1 2q “ t; this is because RpXq and tRpXq both take p1 2 nonzero values mod p, so the intersection must be nonempty by the pigeonhole principle. Choosing b and c such that bx1 c “ x1 1 and bx2 c “ x1 2 gives the claim.
Note further that if pb, cq and tx1, x2u satisfy the relation, then the same is true for p´b, ´cq and tx1, x2u because Rpbx cq “ Rp´bx ´ cq. Since b is nonzero, this means that each pair tx1, x2u corresponds to at least two pairs pb, cq. However, since there are ppp ´ 1q pairs pb, cq with b nonzero and ppp´1q{2 unordered pairs tx1, x2u, each tx1, x2u must correspond to exactly two pairs pb, cq and p´b, ´cq for some pb, cq.
Now, since the image of f has only p ´ 1 elements, there must be some x1, x2 such that fpx1q “ fpx2q. Choose any b, c such that bx1 c “ ´pbx2 cq, so Rpbx1 cq “ Rpbx2 cq and so fpx1q{Rpbx1 cq “ fpx2q{Rpbx2 cq. There is such a pair b, c for any nonzero b, so there are at least p ´ 1 such pairs, and this quantity is greater than 2 for p ě 5.
Finally, for the special case that p “ 3, we observe that there must be at least one allowed value for Qpxq for each x, so there must exist such a quadratic Q by Lagrange interpolation.
Comment. We may also handle the case p “ 3 as follows. Recall that we may assume f is nonzero and surjective onto t1, 2u mod 3, so the image of f must be p1, 1, 2q or p1, 2, 2q in some order. Without loss of generality fp1q “ fp2q, so we either have pfp0q, fp1q, fp2qq “ p1, 2, 2q or p2, 1, 1q. In the first case, take QpXq “ 2X2 2, and in the second case take QpXq “ X2 1.
In some sense, this is equivalent to the Lagrange interpolation approach, as in each case the polynomial QpXq can be determined by Lagrange interpolation.
Solution 4.
Again, we reduce to the case of p being an odd prime and f being a function mod p; we will show that there is a quadratic which is nowhere zero such that Qpxq “ fpxq has no root. We can handle the case of p “ 3 separately as in Solution 3, so assume that p ě 5.
We will prove the following more general statement: let p ě 5 be a prime and let A1, A2, . . . , Ap be subsets of Z{pZ with |Ai| “ 2 for all i. Then there exists a polynomial Q P Z{pZrXs of degree at most 2 such that Qpiq R Ai for all i. Indeed, applying this statement to the sets Ai “ t0, fpiqu (and adding pX2 if necessary) produces a quadratic Q satisfying the desired property.
Choose the coefficients of Q uniformly at random from Z{pZ, and let T be the random variable denoting the number of i for which Qpiq P Ai. Observe that for k ď 3, we have E „ˆT k ˙ȷ “ 2k ˆp k ˙ p´k.
To see why, let k ď 3. If S Ď Z{pZ has size k and paiqiPS is a k-tuple, the probability that Qpiq “ ai on S is equal to p´k; for k “ 3 this follows by Lagrange interpolation, and for k ă 3 Shortlisted problems – solutions 109 it follows from the k “ 3 case by summing. The expectation is therefore equal to the number of S Ď Z{pZ of size k times the probability that Qpiq P Ai for each i P S, which is equal to the right hand side as each Ai has size 2.
Now, observe that we have the identity pt ´ 1qpt ´ 3qpt ´ 4q “ ´12 12 t 1 ˘ ´ 10t 2 ˘ 6t 3 ˘ , so ErpT ´ 1qpT ´ 3qpT ´ 4qs “ ´12 12E „ˆT 1 ˙ȷ ´ 10E „ˆT 2 ˙ȷ 6E „ˆT 3 ˙ȷ “ ´12 12 ¨ 2 ´ 10 ¨ 2 ˆ 1 ´ 1 p ˙ 6 ¨ 4 3 ˆ 1 ´ 1 p ˙ ˆ 1 ´ 2 p ˙ “ ´4 p 16 p2 .
This is negative for p ě 5. Because pt ´ 1qpt ´ 3qpt ´ 4q ě 0 for all integers t ą 0, it then follows that T “ 0 with positive probability, which implies that there must exist some Q with Qpiq R Ai for all i, as desired.
Comment. We do not have much freedom to choose a different polynomial in place of RpTq “ pT ´ 1qpT ´ 3qpT ´ 4q in this argument. Indeed, it can be shown (by comparing coefficients ofT K ˘ ) that if R has degree at most 3, then the expected value of RpTq tends to 1 3pRp4q ` 2Rp1qq as p tends to infinity, so R must have both 1 and 4 as roots.
In particular, R must be of the form RpTq “ pT ´ 1qpT ´ 4qpT ´ dq for some d ě 3, and if d ă 4 then the argument works for any p with p ą 4{p4 ´ dq.
110 Bath, United Kingdom, 10th–22nd July 2024 N7.
Let Zą0 denote the set of positive integers. Let f : Zą0 Ñ Zą0 be a function satisfying the following property: for m, n P Zą0, the equation fpmnq2 “ fpm2qfpfpnqqfpmfpnqq holds if and only if m and n are coprime.
For each positive integer n, determine all the possible values of fpnq.
(Japan) Answer: All numbers with the same set of prime factors as n.
Common remarks. We refer to the given property as Ppm, nq. We use the notation radpnq for the radical of n: the product of the distinct primes dividing n.
Solution 1. We start with a series of straightforward deductions: • From Pp1, 1q, we have fp1q2 “ fp1qfpfp1qq2, so fp1q “ fpfp1qq2.
• From Pp1, fp1qq, we have fpfp1qq2 “ fp1qfpfpfp1qqqfpfpfp1qqq, so fpfpfp1qqq “ 1.
• From Pp1, fpfp1qqq, we have fpfpfp1qqq2 “ fp1qfpfpfpfp1qqqqfpfpfpfp1qqqq, which sim-plifies to 1 “ fp1q3, so fp1q “ 1.
• From Pp1, nq we deduce fpnq “ fpfpnqq for all n.
• From Ppm, 1q we deduce fpmq “ fpm2q for all m.
• Simplifying Ppm, nq, we have that fpmnq2 “ fpmqfpnqfpmfpnqq if and only if m and n are coprime; refer to this as Qpm, nq.
• From Qpm, fpnqq, we have that fpmfpnqq “ fpmqfpnq if and only if m and fpnq are coprime; refer to this as Rpm, nq.
Claim. If fpaq “ 1, then a “ 1.
Proof. If a ‰ 1, then Qpa, aq gives fpaq2 ‰ fpaq2fpafpaqq. If fpaq “ 1, then both sides simplify to 1, a contradiction.
l Claim. If n ‰ 1 then gcdpn, fpnqq ‰ 1.
Proof. If gcdpn, fpnqq “ 1, then Qpfpnq, nq gives fpnfpnqq2 “ fpnq3, and Qpn, fpnqq gives fpnfpnqq2 “ fpnq2fpnfpnqq, which together yield fpnq “ 1 for a contradiction.
l Claim. For all n we have radpnq | fpnq.
Proof. For any prime p | n, write n “ pvn1 with p ∤n1.
From Qppv, n1q we have fpnq2 “ fppvqfpn1qfppvfpn1qq. Since gcdppv, fppvqq ‰ 1, it follows that p | fppvq, so p | fpnq, and thus radpnq | fpnq.
l Claim. If n is coprime to fpkq, then fpnq is coprime to fpkq.
Proof. From Qpfpkq, nq we have fpnfpkqq2 “ fpkqfpnqfpfpkqfpnqq; applying Rpn, kq to the LHS, we conclude that fpkqfpnq “ fpfpkqfpnqq. Applying Rpfpnq, kq we deduce that fpnq is coprime to fpkq, as required.
l Claim. If p is prime then fppq is a power of p.
Shortlisted problems – solutions 111 Proof. Suppose otherwise. We know that p | fppq; let q ‰ p be another prime with q | fppq.
If, for some positive integer N, we have p ∤fpNq, then fppq is coprime to fpNq, so q ∤fpNq, so q ∤N; thus, if q | N, then p | fpNq (and in particular, p | fpqq, by taking N “ q).
Similarly, if q ∤fpNq then fpqq is coprime to fpNq; as p | fpqq, this means p ∤fpNq, so p ∤N.
So if p | N, then q | fpNq.
Together with radpnq | fpnq, this means that for any n not coprime to pq, we have pq | fpnq.
Let m “ mintνppfpxqq | x is not coprime to pqu, and let X be a positive integer not coprime to pq such that νppfpXqq “ m. The argument above shows m ě 1. We can write fpXq “ pmqyX1, where y ě 1, p ∤X1 and q ∤X1. Since fpfpXqq “ fpXq we have fppmqyX1q “ pmqyX1.
Applying Qppm, qyX1q gives ppmqyX1q2 “ fppmqfpqyX1qfppmfpqyX1qq. The RHS is divisible by p3m but the LHS is only divisible by p2m, yielding a contradiction.
l Claim. For any integer n, radpfpnqq “ radpnq.
Proof. We already have that radpnq | fpnq, so it remains only to show that no other primes divide fpnq. If p is prime and p ∤n, the previous Claim shows that n is coprime to fppq, and thus fpnq is coprime to fppq; that is, p ∤fpnq. So exactly the same primes divide fpnq as divide n.
l It remains only to exhibit functions that show all values of fpnq with radpfpnqq “ radpnq are possible. Given eppq ě 1 for each prime p, take fpnq “ ź p|n peppq and we verify by examining exponents of each prime that this satisfies the conditions of the problem.
Comment. A quicker but less straightforward proof that fp1q “ 1 is to let fpnq “ M be the least value that f takes; then Pp1, nq gives M2 “ fpnq2 “ fp1qfpfpnqq2 ě M3 so M “ 1 and fp1q “ 1.
Solution 2.
As in Solution 1, we see that there are indeed functions f satisfying the given condition and producing all the given values of fpnq, and we follow Solution 1 to show the following facts: • fp1q “ 1.
• fpmq “ fpm2q for all m.
• fpnq “ fpfpnqq for all n.
• fpmnq2 “ fpmqfpnqfpmfpnqq if and only if m and n are coprime; refer to this as Qpm, nq.
Taking Qpm, nq together with Qpn, mq gives that fpmfpnqq “ fpnfpmqq if m and n are coprime.
Suppose now that m is coprime to both n and fpnq. We have fpmnq2 “ fpmqfpnqfpmfpnqq and squaring both sides gives fpmnq4 “ fpmq2fpnq2fpmfpnqq2 “ fpmq2fpnq2fpmqfpfpnqqfpmfpfpnqqq “ fpmq3fpnq3fpmfpnqq.
Thus fpmfpnqq “ fpmqfpnq, so fpmnq2 “ fpmq2fpnq2, so fpmnq “ fpmqfpnq “ fpmfpnqq “ fpnfpmqq.
If m is coprime to both n and fpnq but however n is not coprime to fpmq, we have fpnfpmqq2 ‰ fpnqfpfpmqqfpnfpfpmqqq “ fpnqfpmqfpnfpmqq “ fpnfpmqq2, a contradiction. Thus, given that m and n are coprime, we know that m is coprime to fpnq if and only if n is coprime to fpmq. In particular, if p and q are different primes, then p | fpqq if and only if q | fppq, and likewise, for any positive integer k, p | fpqkq if and only if q | fppq.
More generally, if p ∤n, then p | fpnq if and only if n is not coprime to fppq.
Now form a graph whose vertices are the primes, and where there is an edge between primes p ‰ q if and only if p | fpqq (and so q | fppq); every vertex has finite degree. For any integer n, the primes dividing fpnq are all the primes that are neighbours of any prime q | n, together possibly with some further primes p | n.
If p and q are different primes, we have fppfpqqq “ fpqfppqq. The LHS is divisible by all primes that (in the graph) are neighbours of p or neighbours of neighbours of q, and possibly also by p and by some primes that are neighbours of q, and a corresponding statement with p and q swapped applies to the RHS. Thus any prime that is a neighbour of a neighbour of q must be one of: p, q, distance 1 from q, or distance 1 or 2 from p. For any prime r that is distance 2 from q, there are only finitely many primes p that it is distance 2 or less from, so by choosing a suitable prime p (depending on q) we conclude that every prime that is a neighbour of a neighbour of q is actually q itself or a neighbour of q.
So the connected components of the graph are (finite) complete graphs. If m is divisible only by primes in one component, and n is divisible only by primes in another component, then fpmnq “ fpmqfpnq. If n is divisible by more than one prime from a component, considering the expression for fpmnq2 as applied with successive prime power divisors of n shows that fpnq is divisible by all the primes in that component. However, while fppkq is divisible by all the primes in the component of p except possibly for p itself, we do not yet know that p | fppkq.
We now consider cases for the order of a component.
For any prime p, we cannot have fppkq “ 1, because Qppk, pkq gives fpp2kq2 ‰ fppkqfppkqfppkfppkqq, and simplifying using fpm2q “ fpmq results in 1 ‰ 1. So for a component of order 1, fppkq is a positive power of p, so has the same set of prime factors as p, as required.
Now consider a component of order at least 2. Since fpfpnqq “ fpnq, if the component has order at least 3, then for any n ‰ 1 whose prime divisors are in that component, fpnq is divisible by all the primes in that component. If the component has order 2, we saw above that this is true except possibly for n “ pk. However, if the primes in the component are p and q, and fppkq “ qℓ, then fpqℓq “ fpfppkqq “ fppkq “ qℓ, which contradicts p | fpqℓq. So for any component of order at least 2, and any n ‰ 1 whose prime divisors are in that component, fpnq is divisible by all the primes in that component.
In a component of order at least 2, let m be the product of all the primes in that component, and let t be maximal such that mt | fpnq for all n ‰ 1 whose prime divisors are in that component; we have seen that t ě 1. If m and n are coprime numbers greater than 1, all of whose prime divisors are in that component, then Qpm, nq tells us that m3t{2 | fpmnq. For any n1 ‰ 1, all of whose prime divisors are in that component, fpn1q is divisible by all the primes in that component, so can be expressed as such a product, so m3t{2 | fpfpn1qq “ fpn1q. But this means t ě 3t{2, a contradiction, so all components have order 1, and we are done.
Supported by: The activities of the Problem Selection Committee were supported by the University of Cambridge |
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Recurrence relation for Pell's equation $x^2-2y^2=1$
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I am wondering how to find the recurrence relation for solutions for $x$ in the Pell's equation $x^2-2y^2=1$.
I know the formula for the general term. It is $$\frac{(3+2\sqrt2)^n+(3-2\sqrt2)^n}{2}$$ for $x_n$, the $n^{th}$ smallest solution for $x$. Any help would be appreciated! Thanks!
I got a feeling that the recurrence formula is $x_n=6x_{n-1}-x_{n-2}$, but I wonder how to prove this relation true/false and how to derive/generate the recurrence relation. Note that $x_{-1}=1$ and this recurrence formula applies to all nonnegative integral $n$.
combinatorics
number-theory
recurrence-relations
pell-type-equations
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edited Sep 27, 2018 at 7:07
KaiKai
asked Sep 27, 2018 at 6:56
KaiKai
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$\begingroup$ For adding more details, it is recommended that you edit your question instead of creating a comment. Thanks. $\endgroup$
GoodDeeds
– GoodDeeds
2018-09-27 07:03:24 +00:00
Commented Sep 27, 2018 at 7:03
$\begingroup$ Thanks. I will edit the question. $\endgroup$
Kai
– Kai
2018-09-27 07:04:33 +00:00
Commented Sep 27, 2018 at 7:04
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$\begingroup$ I'm not familiar with this particular technique, but it's easy to see that $x_n = 6x_{n-1} - x_{n-2}$ satisfies the formula given. It's also not difficult to see that it is the only recurrence relation of the form $x_n = ax_{n-1} + bx_{n-2}$ that the formula satisfies. If you are looking for such a recurrence relation, you've found it! $\endgroup$
Theo Bendit
– Theo Bendit
2018-09-27 07:17:21 +00:00
Commented Sep 27, 2018 at 7:17
$\begingroup$ I haven't learned generating functions, so I don't quite understand why it works, and why it is the only formula that work. I will try to figure that out. $\endgroup$
Kai
– Kai
2018-09-27 07:25:13 +00:00
Commented Sep 27, 2018 at 7:25
$\begingroup$ Have you tried to expand the formula $x_{n+1}=6 x_n - 1 x_{n-1}$ using the fractional/exponential expression that you already have, simplify and identify the correctness of the recursion-formula? $\endgroup$
Gottfried Helms
– Gottfried Helms
2018-09-27 07:30:54 +00:00
Commented Sep 27, 2018 at 7:30
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If $u_n=A\alpha^n+B\beta^n$ then $\alpha$ and $\beta$ are the roots of the quadratic equation $$(x-\alpha)(x-\beta)=x^2-(\alpha+\beta)x+\alpha\beta=0$$
Let the quadratic be $p(x)=x^2-px+q$, then consider $$0=A\alpha^np(\alpha)+B\beta^np(\beta)=$$$$=(A\alpha^{n+2}+B\beta^{n+2})-p(A\alpha^{n+1}+B\beta^{n+1})+q(A\alpha^{n}+B\beta^{n})=u_{n+2}-pu_{n+1}+qu_n$$from which $$u_{n+2}=pu_{n+1}-qu_n$$Where $p=\alpha+\beta$ and $q=\alpha\beta$
You simply need to identify $\alpha$ and $\beta$ - the surrounding constants drop out in the arithmetic.
Once two successive terms are known, the constants $A$ and $B$ are determined, and there is a unique recurrence of this kind because two consecutive terms determine the rest through the recurrence.
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answered Sep 27, 2018 at 7:25
Mark BennetMark Bennet
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$\begingroup$ Note also that you can show that the recurrence is true simply by showing that it works for the general term you have. This answer shows you how to derive it. $\endgroup$
Mark Bennet
– Mark Bennet
2018-09-27 07:39:34 +00:00
Commented Sep 27, 2018 at 7:39
$\begingroup$ I don't understand the first sentence: "If $u_n = ...$ then $\alpha$ and $\beta$ are the roots [...]." Is $u_n = ...$ a definition? If so, what are $\alpha$ and $\beta$? If it's not a definition, what is $u_n$, and how are $\alpha$ and $\beta$ defined? $\endgroup$
Anakhand
– Anakhand
2023-09-19 10:20:45 +00:00
Commented Sep 19, 2023 at 10:20
$\begingroup$ I think what confused me is the "If ... then ...", which seems to say that $u_n$ being defined like that implies $\alpha$ and $\beta$ are roots of the quadratic, but the latter fact is independent of the former. $\endgroup$
Anakhand
– Anakhand
2023-09-19 10:30:59 +00:00
Commented Sep 19, 2023 at 10:30
$\begingroup$ @Anakhand $\alpha$ and $\beta$ are automatically the roots of that quadratic, it is true and perhaps if ... then ... is not logically required. "Consider ... [the quadratic] ..." might have been a more elegant way to put it. I was trying to emphasise the connection between the roots of polynomials and the solutions of linear recurrences (which is true for three or more roots too - this is a toy example). The connection is not as well known as it should be. $\endgroup$
Mark Bennet
– Mark Bennet
2023-09-20 18:20:52 +00:00
Commented Sep 20, 2023 at 18:20
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$\begingroup$
The fundamental solution is $9-8 = 1,$ meaning $3^2 - 2 \cdot 2^2 = 1.$ As a result, we have the matrix $$ A = \left( \begin{array}{cc} 3 & 4 \ 2 &3 \end{array} \right) $$ which solves the automorphism relation, $A^T H A = H,$ where $$ H = \left( \begin{array}{cc} 1 & 0 \ 0 & -2 \end{array} \right) $$ That is $$ (3x+4y)^2 - 2 (2x+3y)^2 = x^2 - 2 y^2. $$ Next, $$ A^2 - 6 A + I = 0. $$ Since $$ A \left( \begin{array}{c} x_n \ y_n \end{array} \right) = \left( \begin{array}{c} x_{n+1} \ y_{n+1} \end{array} \right) $$ and $$ A^2 \left( \begin{array}{c} x_n \ y_n \end{array} \right) = \left( \begin{array}{c} x_{n+2} \ y_{n+2} \end{array} \right) $$ we find $$ x_{n+2} - 6 x_{n+1} + x_n = 0 $$ $$ y_{n+2} - 6 y_{n+1} + y_n = 0 $$ This is just Cayley-Hamilton.
Caution: This is for $x^2 - 2 y^2 = 1.$ If we change the problem to $x^2 - 2 y^2 = 119 = 7 \cdot 17,$ the recurrence still holds, except that there are now four such families, each using the same recursion:
$$ (11,1) \; \; \; \; (37,25) \; \; \; \; (211,149) ... $$ $$ (13,5) \; \; \; \; (59,41) \; \; \; \; (341,241) ... $$ $$ (19,11) \; \; \; \; (101,71) \; \; \; \; (587,415) ... $$ $$ (29,19) \; \; \; \; (163,115) \; \; \; \; (949,671) ... $$
If you don't mind negative values for $x,y$ you can combine the above four into two families going both forth and back...
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answered Sep 27, 2018 at 18:38
Will JagyWill Jagy
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$\endgroup$
2
$\begingroup$ Is this group theory? $\endgroup$
Kai
– Kai
2018-09-28 03:24:58 +00:00
Commented Sep 28, 2018 at 3:24
$\begingroup$ @Kai the only group theory you need for this is how to multiply 2 by 2 integer matrices of determinant $1$ $\endgroup$
Will Jagy
– Will Jagy
2018-09-28 03:28:40 +00:00
Commented Sep 28, 2018 at 3:28
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-2
$\begingroup$
You can find the general recurrence relationship here:
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edited Sep 28, 2018 at 5:08
answered Sep 28, 2018 at 4:56
Nilotpal SinhaNilotpal Sinha
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$\endgroup$
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10142 | https://dlmf.nist.gov/28.23 | DLMF: §28.23 Expansions in Series of Bessel Functions ‣ Modified Mathieu Functions ‣ Chapter 28 Mathieu Functions and Hill’s Equation
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28 Mathieu Functions and Hill’s EquationModified Mathieu Functions28.22 Connection Formulas28.24 Expansions in Series of Cross-Products of Bessel Functions or Modified Bessel Functions
§28.23 Expansions in Series of Bessel Functions
ⓘ
Keywords:Bessel functions, expansions in series of, expansions in series of Bessel functions, modified Mathieu functions, radial Mathieu functionsReferenced by:§28.24Permalink: also:Annotations for Ch.28
We use the following notations:
28.23.1 𝒞 μ(1)=J μ,
𝒞 μ(2)=Y μ,
𝒞 μ(3)=H μ(1),
𝒞 μ(4)=H μ(2);
ⓘ
Defines:𝒞 μ(j): cylinder functions (locally)Symbols:J ν(z): Bessel function of the first kind, Y ν(z): Bessel function of the second kind, H ν(1)(z): Bessel function of the third kind (or Hankel function), H ν(2)(z): Bessel function of the third kind (or Hankel function) and j: integerA&S Ref:20.4.7 (in different notation)Referenced by:§28.28(ii)Permalink: TeX, TeX, TeX, pMML, pMML, pMML, pMML, png, png, png, pngSee also:Annotations for §28.23 and Ch.28
compare §10.2(ii). For the coefficients c n ν(q) see §28.14. For A n m(q) and B n m(q) see §28.4.
28.23.2 me ν(0,h 2)M ν(j)(z,h)=∑n=−∞∞(−1)nc 2n ν(h 2)𝒞 ν+2n(j)(2hcoshz),
ⓘ
Symbols:me n(z,q): Mathieu function, coshz: hyperbolic cosine function, M ν(j)(z,h): modified Mathieu function, h: parameter, n: integer, j: integer, z: complex variable, ν: complex parameter, c 2m(q): coefficients and 𝒞 μ(j): cylinder functionsPermalink: pMML, pngSee also:Annotations for §28.23 and Ch.28
28.23.3 me ν′(0,h 2)M ν(j)(z,h)=itanhz∑n=−∞∞(−1)n(ν+2n)c 2n ν(h 2)𝒞 ν+2n(j)(2hcoshz),
ⓘ
Symbols:me n(z,q): Mathieu function, coshz: hyperbolic cosine function, tanhz: hyperbolic tangent function, i: imaginary unit, M ν(j)(z,h): modified Mathieu function, h: parameter, n: integer, j: integer, z: complex variable, ν: complex parameter, c 2m(q): coefficients and 𝒞 μ(j): cylinder functionsPermalink: pMML, pngSee also:Annotations for §28.23 and Ch.28
valid for all z when j=1, and for ℜz>0 and |coshz|>1 when j=2,3,4.
28.23.4 me ν(1 2π,h 2)M ν(j)(z,h)=e iνπ/2∑n=−∞∞c 2n ν(h 2)𝒞 ν+2n(j)(2hsinhz),
ⓘ
Symbols:me n(z,q): Mathieu function, π: the ratio of the circumference of a circle to its diameter, e: base of natural logarithm, sinhz: hyperbolic sine function, i: imaginary unit, M ν(j)(z,h): modified Mathieu function, h: parameter, n: integer, j: integer, z: complex variable, ν: complex parameter, c 2m(q): coefficients and 𝒞 μ(j): cylinder functionsPermalink: pMML, pngSee also:Annotations for §28.23 and Ch.28
28.23.5 me ν′(1 2π,h 2)M ν(j)(z,h)=ie iνπ/2cothz∑n=−∞∞(ν+2n)c 2n ν(h 2)𝒞 ν+2n(j)(2hsinhz),
ⓘ
Symbols:me n(z,q): Mathieu function, π: the ratio of the circumference of a circle to its diameter, e: base of natural logarithm, cothz: hyperbolic cotangent function, sinhz: hyperbolic sine function, i: imaginary unit, M ν(j)(z,h): modified Mathieu function, h: parameter, n: integer, j: integer, z: complex variable, ν: complex parameter, c 2m(q): coefficients and 𝒞 μ(j): cylinder functionsPermalink: pMML, pngSee also:Annotations for §28.23 and Ch.28
valid for all z when j=1, and for ℜz>0 and |sinhz|>1 when j=2,3,4.
In the case when ν is an integer
28.23.6 Mc 2m(j)(z,h)=(−1)m(ce 2m(0,h 2))−1∑ℓ=0∞(−1)ℓA 2ℓ 2m(h 2)𝒞 2ℓ(j)(2hcoshz),
ⓘ
Symbols:ce n(z,q): Mathieu function, coshz: hyperbolic cosine function, Mc n(j)(z,h): radial Mathieu function, m: integer, h: parameter, j: integer, z: complex variable, 𝒞 μ(j): cylinder functions and A m(q): Fourier coefficientA&S Ref:20.6.12 Referenced by:§28.23Permalink: pMML, pngSee also:Annotations for §28.23 and Ch.28
28.23.7 Mc 2m(j)(z,h)=(−1)m(ce 2m(1 2π,h 2))−1∑ℓ=0∞A 2ℓ 2m(h 2)𝒞 2ℓ(j)(2hsinhz),
ⓘ
Symbols:ce n(z,q): Mathieu function, π: the ratio of the circumference of a circle to its diameter, sinhz: hyperbolic sine function, Mc n(j)(z,h): radial Mathieu function, m: integer, h: parameter, j: integer, z: complex variable, 𝒞 μ(j): cylinder functions and A m(q): Fourier coefficientPermalink: pMML, pngSee also:Annotations for §28.23 and Ch.28
28.23.8 Mc 2m+1(j)(z,h)=(−1)m(ce 2m+1(0,h 2))−1∑ℓ=0∞(−1)ℓA 2ℓ+1 2m+1(h 2)𝒞 2ℓ+1(j)(2hcoshz),
ⓘ
Symbols:ce n(z,q): Mathieu function, coshz: hyperbolic cosine function, Mc n(j)(z,h): radial Mathieu function, m: integer, h: parameter, j: integer, z: complex variable, 𝒞 μ(j): cylinder functions and A m(q): Fourier coefficientPermalink: pMML, pngSee also:Annotations for §28.23 and Ch.28
28.23.9 Mc 2m+1(j)(z,h)=(−1)m+1(ce 2m+1′(1 2π,h 2))−1×cothz×∑ℓ=0∞(2ℓ+1)A 2ℓ+1 2m+1(h 2)×𝒞 2ℓ+1(j)(2hsinhz),
ⓘ
Symbols:ce n(z,q): Mathieu function, π: the ratio of the circumference of a circle to its diameter, cothz: hyperbolic cotangent function, sinhz: hyperbolic sine function, Mc n(j)(z,h): radial Mathieu function, m: integer, h: parameter, j: integer, z: complex variable, 𝒞 μ(j): cylinder functions and A m(q): Fourier coefficientPermalink: pMML, pngSee also:Annotations for §28.23 and Ch.28
28.23.10 Ms 2m+1(j)(z,h)=(−1)m(se 2m+1′(0,h 2))−1tanhz×∑ℓ=0∞(−1)ℓ(2ℓ+1)B 2ℓ+1 2m+1(h 2)×𝒞 2ℓ+1(j)(2hcoshz),
ⓘ
Symbols:se n(z,q): Mathieu function, coshz: hyperbolic cosine function, tanhz: hyperbolic tangent function, Ms n(j)(z,h): radial Mathieu function, m: integer, h: parameter, j: integer, z: complex variable, 𝒞 μ(j): cylinder functions and B m(q): Fourier coefficientPermalink: pMML, pngSee also:Annotations for §28.23 and Ch.28
28.23.11 Ms 2m+1(j)(z,h)=(−1)m(se 2m+1(1 2π,h 2))−1∑ℓ=0∞B 2ℓ+1 2m+1(h 2)𝒞 2ℓ+1(j)(2hsinhz),
ⓘ
Symbols:se n(z,q): Mathieu function, π: the ratio of the circumference of a circle to its diameter, sinhz: hyperbolic sine function, Ms n(j)(z,h): radial Mathieu function, m: integer, h: parameter, j: integer, z: complex variable, 𝒞 μ(j): cylinder functions and B m(q): Fourier coefficientA&S Ref:20.6.13 Permalink: pMML, pngSee also:Annotations for §28.23 and Ch.28
28.23.12 Ms 2m+2(j)(z,h)=(−1)m(se 2m+2′(0,h 2))−1tanhz×∑ℓ=0∞(−1)ℓ(2ℓ+2)B 2ℓ+2 2m+2(h 2)×𝒞 2ℓ+2(j)(2hcoshz),
ⓘ
Symbols:se n(z,q): Mathieu function, coshz: hyperbolic cosine function, tanhz: hyperbolic tangent function, Ms n(j)(z,h): radial Mathieu function, m: integer, h: parameter, j: integer, z: complex variable, 𝒞 μ(j): cylinder functions and B m(q): Fourier coefficientPermalink: pMML, pngSee also:Annotations for §28.23 and Ch.28
28.23.13 Ms 2m+2(j)(z,h)=(−1)m+1(se 2m+2′(1 2π,h 2))−1×cothz×∑ℓ=0∞(2ℓ+2)B 2ℓ+2 2m+2(h 2)×𝒞 2ℓ+2(j)(2hsinhz).
ⓘ
Symbols:se n(z,q): Mathieu function, π: the ratio of the circumference of a circle to its diameter, cothz: hyperbolic cotangent function, sinhz: hyperbolic sine function, Ms n(j)(z,h): radial Mathieu function, m: integer, h: parameter, j: integer, z: complex variable, 𝒞 μ(j): cylinder functions and B m(q): Fourier coefficientReferenced by:§28.23Permalink: pMML, pngSee also:Annotations for §28.23 and Ch.28
When j=1, each of the series (28.23.6)–(28.23.13) converges for all z. When j=2,3,4 the series in the even-numbered equations converge for ℜz>0 and |coshz|>1, and the series in the odd-numbered equations converge for ℜz>0 and |sinhz|>1.
For proofs and generalizations, see Meixner and Schäfke (1954, §§2.62 and 2.64).
28.22 Connection Formulas28.24 Expansions in Series of Cross-Products of Bessel Functions or Modified Bessel Functions
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10143 | https://www.puzzlestools.com/post/the-role-of-puzzles-in-developing-early-math-skills | The Role of Puzzles in Developing Early Math Skills
March 12, 2025
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Types of Mathematical Children’s Puzzles
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10144 | https://www.osti.gov/servlets/purl/2551737 | CHAPTER
Natural transformation as a tool in Acinetobacter baylyi : Evolution by amplification of gene copy number
7
Isabel Pardo a , Stacy R. Bedore b , Melissa P. Tumen-Velasquez b,† ,Chantel V. Duscent-Maitland b , Alyssa C. Baugh b ,Suvi Santala c , and Ellen L. Neidle b,
aCentro de Investigaciones Biolo ´gicas Margarita Salas (CIB), Spanish National Research Council (CSIC), Madrid, Spain
bDepartment of Microbiology, University of Georgia, Athens, GA, United States
cFaculty of Engineering and Natural Sciences, Tampere University, Tampere, Finland
Corresponding author: e-mail address: eneidle @uga.edu
Abbreviations
Ab R antibiotic resistance
ALE adaptive laboratory evolution
Cit citrate
cPCR colony PCR
EASy Evolution by Amplification and Synthetic biology
GDA gene duplication and amplification
gDNA genomic DNA
GOI gene(s) of interest
Gua guaiacol
Km kanamycin
qPCR quantitative PCR
SBF synthetic bridging fragment
Sm streptomycin
Sp spectinomycin
TPA terephthalic acid †Current affiliation: Biosciences Division, Oak Ridge National Laboratory, Oak Ridge, TN, United States.
Methods in Microbiology, Volume 52, ISSN 0580-9517,
Copyright © 2023 Elsevier Ltd. All rights reserved. 183 1 Introduction
Adaptive laboratory evolution (ALE) is a widespread and powerful tool to engineer microorganisms with enhanced tolerance, increased production capabilities, or new phenotypes (Dragosits & Mattanovich, 2013). By growing the cells under an appro-priate selective pressure for multiple generations, the desired property can be attained due to the accumulation of beneficial mutations that lead to improved ac-tivities (i.e., optimization) or novel functions (i.e., innovation) (Barrick & Lenski, 2013). However, in the timescales typically employed in experimental evolution, the low frequency of spontaneous, beneficial mutations may impose constraints that are difficult to overcome. Specifically, in the early stages of selection, gene/protein expression may be insufficient to sustain growth. Low expression could be problem-atic if selection depends on the weak side activity of an enzyme, if toxic inter-mediates accumulate, and/or if non-native proteins are unstable. In nature, gene duplication and amplification (GDA) facilitate adaptation to new or changing envi-ronments by increasing expression via transient changes in gene dosage. This ampli-fication is a reversible mechanism that allows the reduction of gene copy number once selective pressure is removed or other beneficial mutations are propagated throughout the population (Andersson & Hughes, 2009; Elliott, Cuff, & Neidle, 2013; Seaton et al., 2012). Although gene duplications tend to occur significantly more frequently than point mutations (Anderson & Roth, 1981; Seaton et al., 2012), the probability remains low that a sufficient number of cells in the starting population of a culture will randomly have a duplication in the correct location to be selectively beneficial in ALE (Herrmann et al., 2022). However, if a tandem duplication encompassing the region of interest does exist, it provides long stretches of identical sequence that allow fur-ther amplification mediated by homologous recombination. Gene copy number then fluctuates stochastically, and if increased dosage is advantageous, cells will be se-lected that, on average, have the same level of increased expression (Reams & Neidle, 2003) (Fig. 1A). A paradigmatic example of the relevance of GDA in exper-imental evolution is the emergence of the aerobic citrate-utilizing (Cit +) phenotype in Escherichia coli during a long-term evolution experiment (Blount, Barrick, Davidson, & Lenski, 2012). This event, which took place after >31,000 generations (15 years), was due to the GDA of citT , which encodes a citrate: succinate antiporter. Later works purposefully evolving E. coli for citrate utilization found that Cit + clones obtained after at least 30 days of selection also possessed multiple copies of citT
(Van Hofwegen, Hovde, & Minnich, 2016). To induce GDA in the laboratory at a precise position, the locus of interest may be artificially duplicated using DNA manipulation techniques (Stoudenmire et al., 2017) (Fig. 1B). Recently, we developed a method named Evolution by Amplifica-tion and Synthetic Biology (EASy) in A. baylyi ADP1 (Tumen-Velasquez et al., 2018). A schematic overview of this method is shown in Fig. 2. EASy takes advan-tage of the exceptional natural competence and efficient homologous recombination of A. baylyi . In a first step, the gene(s) of interest (GOI) and an antibiotic resistance (Ab R ) selection marker are integrated in the chromosome of A. baylyi . Next,
184 CHAPTER 7 Evolution by amplification of gene copy number a specific region of the chromosome, encompassing the GOI and the Ab R marker, is amplified by transforming cells with a synthetic DNA bridging fragment (SBF). The SBF provides sufficient sequence identity for initial duplication by homologous recombination, and it precisely delimits the chromosomal segment that is duplicated, i.e., the amplicon. By including the Ab R marker within the amplicon, mutants with multiple copies can be easily selected in the presence of high antibiotic concentrations. This step bypasses a potential bottleneck in the emergence of amplification mutants when direct selection for the desired phenotype is too demanding to the cells. After-wards, the amplification mutants can be transferred to the selective conditions for ALE for the emergence of the desired phenotype. At this point, high antibiotic concentra-tions are removed from the medium. Thus, the only selective pressure driving GDA is the target condition for ALE (e.g., growth on a non-native substrate). Since chromo-somal duplications are inherently unstable, beneficial mutations that result in a growth advantage will be selected during ALE and lead to a decrease in the average copy number of the amplicon in the evolving population. These beneficial mutations may be found in the GOI and/or in other regions of the chromosome. We successfully used EASy to enable the catabolism of non-native aromatic carbon sources in A. baylyi , namely guaiacol and terephthalic acid (TPA) (Pardo et al., 2020; Tumen-Velasquez et al., 2018). Similarly, after the functional replace-ment of a native aromatic catabolic pathway with a non-native pathway, EASy was used to increase thermotolerance (Bedore, 2021). A. baylyi can utilize a broad range of aromatic compounds through the β-ketoadipate pathway, by which many different substrates are converted by peripheral routes into either catechol or protocatechuic acid (Fuchs, Boll, & Heider, 2011; Seaton & Neidle, 2018). The aromatic rings of these compounds are then cleaved by intradiol dioxygenases before metabolites
FIG. 1
Overview of natural and EASy-mediated gene duplication and amplification (GDA). (A) In natural GDA, the infrequent duplication of the gene of interest (GOI), by homologous or illegitimate recombination, in a few individuals in a population provides a stretch of sequence identity for subsequent homologous recombination. With increasing selective pressure, mutants with higher gene dosage are selected. (B) With EASy, the initial duplication of a targeted chromosomal segment is induced by transforming with a synthetic bridging fragment (SBF). High antibiotic concentrations allow the rapid selection of mutants with multiple copies of the antibiotic selection marker (Ab R), which is present in the amplicon delimited by the SBF. Then, growth under selective conditions without antibiotic drives further GDA of the GOI.
185 1 Introduction FIG. 2
Overview of the EASy method. (A) The gene of interest (GOI) and the antibiotic selection marker (Ab R ), flanked by sequences identical to the upstream (U) and downstream (D) regions of the targeted locus, are integrated in the Acinetobacter baylyi chromosome by homologous recombination. Alternatively, a native gene or chromosomal region can be targeted for amplification by simply introducing the Ab R in an adjacent position. (B) Mutants are transformed with a synthetic bridging fragment (SBF), which delimits gene duplication and amplification to the desired chromosomal segment. (C) The amplification mutants that have incorporated the SBF are selected in the presence of high antibiotic concentrations. (D) The amplification mutants are subjected to adaptive laboratory evolution (ALE) under the desired selective conditions, in which antibiotic selection is removed. During ALE, the tandem array of amplicon copies may expand to sustain growth under the selective conditions. Beneficial mutations (shown as stars) may appear in the gene of interest or elsewhere in the chromosome. As the beneficial mutations accumulate, the copy number of the amplicon is reduced. Finally, alleles holding beneficial mutations are selected and fixed throughout the population.
186 CHAPTER 7 Evolution by amplification of gene copy number are channelled into central metabolism. To enable A. baylyi to utilize guaiacol, we introduced into its chromosome the gcoAB genes from Amycolatopsis sp. ATCC 39116, which encode a guaiacol O-demethylase for the conversion of guaiacol to catechol (Fig. 3A). EASy-mediated GDA of gcoAB gave rise to mutants capable of growing on guaiacol as the sole carbon source (Gua + phenotype). After 1000 generations of ALE, Gua + mutants with a single copy of gcoAB were obtained. These clones presented either point mutations in the heterologous gcoAB genes or transla-tional fusions between gcoAB and catA (encoding the catechol dioxygenase). Other mutations in different loci were likewise selected (Tumen-Velasquez et al., 2018). In the second example, EASy was performed on strains harbouring the TPA catabolic genes tph from Comamonas sp. E6, which encode the necessary activities for the conversion of TPA to protocatechuic acid. To allow uptake of TPA into the cell, transport genes ( tpiAB ) were also integrated into the chromosome of A. baylyi. These genes, encoding the transmembrane proteins of the tripartite TPA transporter from
Comamonas sp. E6, were maintained in single copy to avoid potential toxic effects resulting from overexpression (Fig. 3B). In these studies, mutants with a single copy of the tph genes were not obtained. Moreover, no mutations in the tph genes were identified in eight evolved strains. Nonetheless, EASy allowed the identification of native A. baylyi transporters that had evolved for improved uptake of TPA. We further demonstrated that the heterologous Comamonas transporter is not necessary for A. baylyi to grow on TPA. This outcome would not have been possible without prior amplification of the foreign catabolic genes, a step that was necessary for the emergence of the Tpa + phenotype (Pardo et al., 2020). In this chapter, we provide some tips and best practices for performing EASy experiments with A. baylyi , particularly focused on the evolution of heterologous cat-abolic pathways. However, we note that this method is amenable to many variations, and we have successfully applied it in separate laboratories using slightly different approaches. We anticipate that EASy can be adapted for a range of applications.
FIG. 3
Schematic representation of the chromosomal integrations in Acinetobacter baylyi for EASy of (A) guaiacol and (B) TPA utilization. The native A. baylyi genes are shown in grey. The foreign genes for guaiacol and TPA utilization are shown in pink (catabolic genes) and blue (transport genes). Promoters driving the expression of the heterologous genes are indicated with angled arrows. The ΩKm Rcassette used for antibiotic selection contains a kanamycin resistance gene (green) flanked by transcription and translation terminating signals, respectively shown as “T”s and black octagons. The open-reading frame disruption caused by insertion of gcoAB in panel (A) is indicated with dashed lines. The EASy amplicons are bound by purple boxes.
187 1 Introduction 2 General considerations 2.1 Design of the amplicon and synthetic bridging fragment
To design an EASy experiment, the first step is to determine the precise chromo-somal region to be amplified, i.e., the amplicon. In most experiments, this design involves the chromosomal insertion of target DNA that will be amplified, although in some experiments, we have amplified a region of the wild-type chromosome (Stoudenmire et al., 2017). As depicted in Fig. 1A, the endpoint of the amplicon is created by a novel junction that joins two chromosomal regions that are not nor-mally adjacent. This novel junction is engineered by the choice of DNA segments connected on the SBF (Fig. 1B). When used as donor DNA, this SBF promotes homologous recombination with chromosomal DNA to duplicate the desired region. Also essential to the experimental design is inclusion of an Ab R marker in the chro-mosome within the intended amplicon. This marker provides a means to select and maintain increased gene dosage after transforming with the SBF. The simplest way to construct the parent strain is to insert the GOI (or larger region of interest) into the chromosome together with the Ab R marker (Fig. 2). We have used
Ω fragments encoding either resistance to kanamycin (Km R) or streptomycin/ spectinomycin (Sm R/Sp R) (Eraso & Kaplan, 1994; Prentki & Krisch, 1984). In these fragments, the Ab R marker is flanked by transcription and translation termination signals (Fig. 3). This configuration ensures that the insertion of the Ab R marker does not interfere with the expression of upstream and downstream genes, independent of the orientation. However, any DNA construction that isolates the expression of an Ab R marker from the rest of the amplicon is expected to work. An important consideration when designing the amplicon is the potential toxicity caused by overexpression of certain genes. For example, when engineering TPA transport and catabolism in A. baylyi , we found that high expression levels of genes encoding transmembrane proteins were lethal to the cells, even when in single copy in the chromosome (Pardo et al., 2020). Therefore, we decided to insert them as a separate transcription unit that was not included in the amplicon (Fig. 3B). The architecture of the amplicon can also be tailored to different gene expression strat-egies. For EASy of guaiacol utilization, the gcoAB genes were inserted as transcrip-tional fusions directly downstream of catA , so that they were under control of the
cat promoter (Fig. 3A). On the other hand, the tph genes were placed under control of the tac promoter (de Boer, Comstock, & Vasser, 1983), a strong constitutive pro-moter in A. baylyi , so that they were independently transcribed during EASy TPA consumption (Fig. 3B). Synthetic promoter sequences, of varying strength, for use in strain ADP1 have recently been characterized (Biggs et al., 2020). Once the desired amplicon is planned, the SBF is designed accordingly to delimit GDA to the chromosomal segment of interest. For this purpose, the last 1000 bp of the 3 0-end of the amplicon are fused to the first 1000 bp of the 5 0-end, in tandem. This length provides a sufficiently long stretch of sequence identity to enable homologous recombination in A. baylyi . Again, this step is amenable to variations. The SBF can be designed to limit the amplicon to the foreign DNA sequence (Fig. 4, amplicon 1), or to include the native upstream and downstream sequences of the insertion locus (Fig. 4, amplicon 2). We have successfully used both approaches.
188 CHAPTER 7 Evolution by amplification of gene copy number FIG. 4
Workflow for the construction of the synthetic bridging fragment (SBF) for EASy. (A) In a first step, the target chromosomal segment that will be amplified is determined (i.e., amplicon, bound by a dotted line). In the figure, amplicon design 1 (left side) is limited to the gene of interest (GOI) and the antibiotic selection marker (Ab R ), whereas amplicon design 2 (right side) includes the upstream (U) and downstream (D) chromosomal regions of the targeted insertion locus. Once the amplicon is designed, the first and last 1000 bp of the target amplicon are amplified by PCR. The forward primer to amplify the 5 0-end of the amplicon (primer 1) is designed to have a 20 bp overhang identical to the 3 0-end of the target amplicon. Likewise, the reverse primer to amplify the 3 0-end of the target amplicon has a 20 bp overhang that is the reverse complement of the 5 0-end of the amplicon (primer 4). (B) The full-length SBF is assembled in 15 cycles of a primer-less PCR by sequence overlap extension, using the PCR products form the first round of PCR as self-priming templates (fragments X and Y in panel A). Then, primers 2 and 3 are added to amplify the full-length SBF in another 15 cycles of PCR. (C) The SBF is transformed into the recipient strain to precisely delimit GDA to the target chromosomal region, with the integrated SBF forming part of the novel junction (bound by dashed lines). Depending on the design of the SBF, the resulting amplicon will be different (indicated with brackets).
189 2 General considerations 2.2 Chromosomal integration by natural transformation
Having designed the DNA cassette that will be integrated into the chromosome of
A. baylyi , it needs to be flanked by upstream and downstream sequences identical to the chromosomal locus targeted for insertion (Figs. 2 and 6). These regions of iden-tity can be as small as 300 bp (de Berardinis et al., 2008), although longer stretches of identical sequences increase the efficiency of homologous recombination and facil-itate the integration of larger DNA fragments. The construction of the DNA cassette flanked by the homology regions can be performed using any DNA assembly tech-nique (e.g., restriction cloning, sequence overlap extension PCR, Gibson assembly, Golden Gate assembly, etc.). Since linear DNA fragments are used for the integration into the chromosome, the assembly products can be directly transformed into
A. baylyi or sub-cloned into a plasmid for sequence verification and maintenance. If the latter option is used, a simple digestion with a restriction enzyme to linearize the plasmid backbone needs to be performed before transformation. In all cases, the genotype of the resulting transformant should be confirmed. As noted above, A. baylyi has a remarkable competency for natural transforma-tion, which allows using many different variations of the same two-step protocol: (1) incubation of A. baylyi with transforming DNA; and (2) selection or screening to identify desired transformants (Biggs et al., 2020; Metzgar et al., 2004; Santala, Karp, & Santala, 2016; Sua ´rez et al., 2020; Sua ´rez, Renda, Dasgupta, & Barrick, 2017). The incubation step can be done in liquid or solid medium, which can be either minimal or rich. The same protocol can be used for replicative plasmids or linear DNA fragments. For chromosomal integrations, unpurified linear DNA fragments can be added directly from assembly or digestion reactions, and even genomic DNA from lysed cells can be used. Bedore et al. (2023) describe in detail the pro-tocols that we routinely use in the laboratory, although these may be entirely custom-ized. Any one of the transformation protocols given can be used throughout the different steps of EASy.
2.3 Selection of amplification mutants
As noted in the introductory section, the SBF provides sufficient sequence identity for initial duplication of the target chromosomal segment. Further amplification of the duplicated region can occur and be maintained if selectively advantageous. Therefore, following transformation with the SBF, amplification mutants are selected in the presence of high antibiotic concentration. This selection is used to maintain a tandem array of multiple copies of the amplicon. Under the conditions described in the Procedures section, we use a concentration of 1 g/L Km or Sm/Sp (using a com-bination of both Sm and Sp) which inhibits growth of mutants carrying a single copy of the Ab R gene. Nevertheless, we have found that this concentration may need to be optimized when using resistance genes from different sources or growing on different media (unpublished data). In particular, the ΩKm R fragment we typically use (posi-tions 10,499 –12,705 in GenBank MN266288.1) harbours the nptII gene from Tn 5,
190 CHAPTER 7 Evolution by amplification of gene copy number whereas the ΩSm/Sp R fragment (GenBank M60473.1) harbours the aadA gene from plasmid R100 (Fellay, Frey, & Krisch, 1987; Prentki & Krisch, 1984). Researchers implementing EASy in their laboratory should first optimize the selection conditions that only permit growth of the amplification mutants. Spontaneous drug-resistant mutants are not observed. Typically, 5 –10 copies of the drug marker are selected by these antibiotic concentrations. Once colonies are obtained that are resistant to high antibiotic concentrations, GDA can be checked by colony PCR (cPCR) or quantitative real-time PCR (qPCR). In the first approach, a pair of primers that specifically amplify the novel junction created by the SBF can be used (e.g., primers used for PCR 3 in Fig. 4). Although this method is not able to quantify copy number, it is a simple and fast way to screen possible amplification mutants. To quantify gene copy number, qPCR is used instead. Primers targeting any region of the amplicon can be employed. However, primers that specifically bind the Ab R marker have the advantage that they can be re-used in different EASy experiments to give reproducible estimations of the gene copy number. In previous work, we have shown that the estimated copy number is virtually the same using primers binding the Ab R marker or the GOI (Pardo et al., 2020). Clones confirmed to have multiple copies of the amplicon can then be trans-ferred to the selective medium for ALE. The antibiotic is omitted at this point so that gene dosage of the amplicon can vary and be selected based on the desired phenotype. In a variation of this approach, two different chromosomal regions of the same parent strain can be amplified. This approach was recently demonstrated using the
ΩSm/Sp R marker to select amplification in one chromosomal region and the ΩKm Rmarker in a different region (unpublished data). In this case, the two distinct regions were manipulated sequentially in the construction of a synthetic metabolic pathway. The goal is to build complex pathways in a modular fashion. This approach is designed to alleviate problems caused by imbalances between the synthesis and degradation of metabolic intermediates that might be toxic. After moving the doubly-amplified populations to selective growth medium, without antibiotics, the new pathway may evolve to reduce the accumulation of toxic metabolites via variation in the gene dosage of different parts of the pathway.
2.4 Interpreting gene copy number estimations and obtaining single-copy mutants
GDA is a highly dynamic process, where gene copy number rapidly fluctuates, and copy number decreases in the absence of selection. Therefore, genomic DNA extracted from cells growing under the selective conditions should be used for an accurate estimation of the gene copy number by qPCR. It is important to note that this number represents the average for all the cells in a population. On occasions, the gene copy number for an evolving population will stabilize at 2 –4 copies, but a significant proportion of individual cells will carry a single copy of the amplicon. In Section 4.4, we provide two approaches to isolate single-copy mutants. The first
191 2 General considerations approach is based on the screening of isolated colonies from the evolving popula-tion for the absence of the SBF, indicative of a single copy of the amplicon (Section 4.4.1). The second approach consists of replacing the tandem repeats of the amplicon with a single copy of the GOI (Section 4.4.2). An overview of this al-lelic replacement method is shown in Fig. 5. In a first step, the tandem amplicon array in an evolved isolate is replaced with a selectable/counter-selectable marker, such as the sacB -Km R cassette (Jones & Williams, 2003). The counter-selectable marker al-lows the isolation of mutants in a second transformation step, in which a single copy of the non-evolved amplicon is re-introduced. It should be noted that if the original GDA was selected with one Ab R marker (such as Km R ) a different marker (such as Sm/Sp R ) should be used in combination with the sacB counter-selectable marker. Once mutants have been isolated in which the evolved amplicons have been replaced with a single copy of the original version of this DNA, further analyses are possible (Fig. 5B). If these mutants grow in the selective medium, then the mu-tations accrued throughout the chromosome are sufficient to sustain growth with a single copy of the non-evolved GOI. If such mutants do not grow in the selective medium, they can be transformed with individual copies of the evolved GOI alleles, amplified from the multi-copy population by PCR. Beneficial mutations introduced by allelic replacement can be selected and identified (Fig. 5C). We note that recombina-tion between different alleles of the GOI may occur during PCR or transformation. Therefore, mutations that were present in separate copies of the gene may be combined in the same allele. Likewise, two mutations that are sufficiently distant in the same copy of the GOI may be segregated. By plating cells transformed with the evolved alleles on solid ALE selective medium, faster-growing clones that acquire a single copy of an allele with beneficial mutations can be identified and isolated for sequenc-ing. However, given enough time, spontaneous amplification mutants may also arise. The novel junction may occur anywhere in the chromosome—even several kbp apart from the GOI. In this case, screening for the absence of the SBF does not assure that mutants with the desired phenotype will present a single copy of the GOI. Similarly, a PCR using primers that bind the upstream and downstream regions flanking the inser-tion locus of the amplicon may still give a band resembling that of the single-copy parent strain before GDA. Therefore, the presence of the GOI in single copy needs to be confirmed by qPCR or whole-genome sequencing. Any method of whole-genome sequencing should be sufficient for such purposes. The final goal of EASy is to isolate beneficial mutations that improve cell fitness under the selective conditions. Ideally, the epistatic effect of the mutations accrued during ALE allows growth of evolved mutants with a single copy of the amplicon. These mutations are identified with whole-genome sequencing of isolated evolved clones and mapping to the reference genome sequence with appropriate soft-ware (Deatherage & Barrick, 2014). The latest annotated genome sequence for
A. baylyi APD1 is deposited in GenBank under accession number NC_005966. It is possible, however, that evolved mutants with a single copy of the amplicon cannot be obtained, despite following the protocols mentioned above. For example, it may not be possible to remove and reintroduce amplicon copies as illustrated in Fig. 5. One issue that may arise with ALE experiments is a reduction in the transformability
192 CHAPTER 7 Evolution by amplification of gene copy number of ADP1. This problem can be avoided using a transposon-free strain, ADP1-ISx (Sua ´rez et al., 2017). The choice of parent strain may depend on the goal of the ex-periment. In some cases, we prefer to use ADP1, in which the insertion sequences are intact, because transposition generates beneficial genomic mutations during ALE. Regardless of whether a single copy of the GOI is obtained, valuable information can be gained from whole-genome sequencing data, as we discuss in the Section 5.
FIG. 5
Overview of the allelic replacement strategy to obtain single-copy mutants from EASy experiments. (A) An evolved isolate from an EASy experiment carries four copies of the amplicon, which encompasses the gene of interest (GOI) and a kanamycin selection marker (Km R ). One of the copies of the gene of interest presents a beneficial mutation (white star). A second beneficial mutation has been selected elsewhere in the chromosome (purple star). The multi-copy amplicon array is replaced by a DNA cassette carrying a streptomycin/ spectinomycin selection marker (Sm R ) and a sacB counter-selection marker, flanked by sequences identical to upstream (U) and downstream (D) chromosomal regions for homologous recombination. Transformants are selected on Sm/Sp plates. (B) Mutants are then transformed with the parent, non-evolved amplicon. Transformants are selected on medium with sucrose. (C) Mutants obtained in step B are then transformed with single copies of the amplicon derived from the initial EASy isolate. Transformants that incorporate mutated alleles that provide a growth benefit are selected on the same medium used for ALE.
193 2 General considerations 3 Material and equipment 3.1 Strains and culture media
• A. baylyi ADP1 (from culture collections ATCC 33305 or DSM 24193). A transposon-free derivative of ADP1 is also available, ADP1-ISx (Sua ´rez et al., 2017). • Chemically-competent E. coli cells to be used as hosts for plasmid construction and maintenance (any host strain). Typical strains used for this purpose include E. coli strains DH5 α and XL1-Blue (Agilent Technologies). • Metals 44 solution (per litre, 2.5 g EDTA; 10.95 g ZnSO 47H 2 O; 5 g FeSO 47H 2 O; 1.54 g MnSO 47H 2 O; 392 mg CuSO 45H 2 O; 250 mg Co(NO 3 ) 26H 2 O; and 177 mg Na 2 B 4 O710H 2 O) • Concentrated base solution (per litre, 20 g nitriloacetic acid dissolved in 600 mL H 2O with 14.6 g KOH; 28.9 g MgSO 4 ; 6.67 g CaCl 22H 2 O; 18.5 mg Mo 7 O24 4H 2 O; 198 mg FeSO 47H 2 O; and 100 mL of Metals 44 solution) • Minimal medium for Acinetobacter (per litre, 25 mL 0.5 M KH 2 PO 4 ; 25 mL 0.5 M Na 2 HPO 4 ; 10 mL 10% (NH 4 )2 SO 4 ; 1 mL concentrated base) supplemented with an appropriate carbon source (e.g., 20 mM pyruvate, succinate, or acetate). Although organic acids tend to be preferred to sugars, glucose (25 –50 mM) can be used. A longer lag phase will be observed with glucose. In addition, casein amino acids are good carbon sources and can be added in concentration 0.2% (with or without other carbon sources). For plates, 1.5% agar is used. • YTS agar plates for sucrose counterselection (autoclave 3 g yeast extract, 6 g tryptone, 11g agar in 300 mL water; then add 300 mL filter-sterilized 50% sucrose)
3.2 Reagents for DNA manipulation
• Custom PCR primers • Cloning vector, such as pUC18 or pUC19 (Yanisch-Perron, Vieira, & Messing, 1985) • Plasmid carrying the ΩKm R or ΩSm/Sp R fragment, e.g., pHP45 Ω-Km (Fellay et al., 1987), pUI1637, pUI1638 (Eraso & Kaplan, 1994) • High-fidelity DNA polymerase and additional components (e.g., NEB Phusion) • Taq polymerase and reagents for colony PCR (e.g., Bioline MyTaq HS Red Mix, NEB LongAmp Taq DNA polymerase) • NEBuilder HiFi DNA assembly master mix (NEB) • Restriction enzymes • DNA ligase (e.g., NEB Quick ligation kit, T4 DNA ligase) • TaqMan probes • TaqMan Gene Expression Master Mix (Thermo Scientific) • Agarose • TAE (Tris Acetate EDTA) buffer, final working solution for electrophoresis running buffer (1 ): 40 mM Tris, 20 mM acetic acid, and 0.4 mM EDTA • Plasmid purification kit (e.g., Zymo Research ZR Plasmid miniprep kit)
194 CHAPTER 7 Evolution by amplification of gene copy number • PCR purification kit (e.g., Zymo Research DNA Clean & Concentrator kit) • Genomic DNA (gDNA) extraction kit (e.g., Zymo Research Quick-DNA Miniprep Plus kit)
3.3 Equipment
• Orbital shaker • Incubator • DNA electrophoresis chamber • UV or blue-light transilluminator • NanoDrop or comparable DNA quantification instrument • Thermocycler • Real-time PCR cycler • Spectrophotometer or device for following cell growth
4 Experimental procedures 4.1 Construction of the amplicon and chromosomal integration
As indicated in Section 2, different DNA manipulation techniques can be used to assemble the cassette that will be integrated into the A. baylyi chromosome. Here, we describe a procedure using NEBuilder HiFi DNA assembly for plasmid cloning as described in Pardo et al., 2020 (Fig. 6):
Design PCR primers for the amplification of the gene of interest (including 5 0 and 30 untranslated regions) and the upstream and downstream homology regions (1 kbp each) of the targeted insertion locus. These oligos should have 15 –20 nt overhangs complementary to each other and the cloning vector for subsequent assembly. Additionally, the reverse primer amplifying the gene of interest and the forward primer amplifying the downstream homology region should include a restriction site for the insertion of the ΩKm R fragment in Step 7.
Amplify the upstream and downstream homology regions using A. baylyi
genomic DNA (gDNA) or whole cells as template with a high-fidelity DNA polymerase (e.g., Phusion) following the manufacturer’s indications. If whole cells are used, include an initial denaturation step at 95 °C for 5 min in the PCR program. Follow the same protocol to amplify the gene of interest from gDNA of a heterologous host, plasmid, or synthetic DNA.
Linearize the cloning vector (e.g., pUC19) by PCR or digestion with restriction enzymes.
Purify the PCR and/or digestion products and measure DNA concentration with NanoDrop (or comparable instrument for quantification).
Prepare NEBuilder HiFi DNA assembly reaction including the upstream and downstream homology regions, the gene of interest, and the linearized cloning vector following the manufacturer’s indications.
195 4 Experimental procedures 6. Transform an aliquot of the assembly reactions into chemically-competent
E. coli and select on LB agar with the appropriate antibiotic.
Purify plasmids by miniprep and verify correct assembly by sequencing or other methods.
Digest the plasmid carrying the ΩKm R fragment (e.g., pUI1637) with a restriction enzyme whose target sequence is present on the inverted flanking polylinkers. Purify the ΩKm R fragment in an agarose gel (2.4 kbp). Digest the plasmid carrying the gene of interest (recipient) with the same enzyme.
Prepare a T4 ligase reaction with the recipient plasmid and the ΩKm R fragment.
Transform an aliquot of the ligation reactions into chemically-competent
E. coli cells and select on LB agar +50 mg/L Km.
Purify plasmids by miniprep. Optionally, the orientation of the ΩKm R fragment can be checked by PCR or diagnostic restriction digestion.
Digest the resulting plasmid with a restriction enzyme cutting in the vector backbone for transformation into A. baylyi . No purification of the DNA or heat inactivation of the enzyme is necessary.
Transform DNA into A. baylyi using any of the protocols provided in Bedore et al. (2023).
FIG. 6
Workflow for the construction of the amplicon in a replicative plasmid for subsequent integration in Acinetobacter baylyi . The gene of interest (GOI) and the flanking upstream (U) and downstream (D) chromosomal regions for homologous recombination are amplified by PCR, using primers with appropriate 15–20 nt overhangs for NEBuilder HiFi DNA assembly into a cloning vector. The resulting plasmid (recipient) and the donor plasmid carrying the ΩKm Rfragment are digested with the same restriction enzyme. Then, the ΩKm Rfragment is ligated into the recipient plasmid between the GOI and the downstream homology region.
196 CHAPTER 7 Evolution by amplification of gene copy number 4.2 Construction of the SBF and amplification of gene copy number
Once the desired DNA cassette has been integrated into the A. baylyi chromosome, the resulting mutant (i.e., parent strain) is transformed with the SBF to facilitate gene duplication. Here, we describe the construction of the SBF using sequence overlap extension PCR (Fig. 4), although other DNA assembly techniques can be used (or can be commercially synthesized). For future experiments, it is convenient to construct a plasmid carrying the SBF that can be stored and used as needed.
Amplify the first 1000 bp of the target amplicon using a forward primer with a 15 –20 bp overhang identical to the 3 0-end of the amplicon, and a standard reverse primer. Use a high-fidelity DNA polymerase and the plasmid or mutant strain carrying the DNA cassette constructed in Section 4.1 as template.
Similarly, amplify the last 1000 bp of the target amplicon using a standard forward primer and a reverse primer with 15 –20 bp overhang that is the reverse complement of the 5 0-end.
Purify the PCR and/or digestion products and measure DNA concentration.
Prepare a primer-less PCR mix using 30 ng of each of the of the purified DNA fragments and amplify for 15 cycles. The 30 –40 bp stretch of sequence identity generated by the overlapping primers should have a melting temperature of 55 –65 °C. Then, add the standard forward and reverse primers used in steps 1 and 2, respectively (i.e., primers 3 and 4 in Fig. 4). Amplify the full-length SBF for another 15 cycles.
Check correct assembly of the full-length SBF in an agarose gel (expected size of
2 kbp). If needed, purify the correct band by excision from the agarose gel.
Transform the SBF fragment into the parent A. baylyi strain containing a single copy of the region targeted for amplification. For selection of amplification mutants, plate cells on minimal medium containing 1 g/L Km.
We recommend streak purification of colonies at least three times on 1 g/L Km plates to rule out false positive clones. Often, colonies grow poorly on the first transfer.
Confirm integration of the SBF in the high-antibiotic resistant colonies by PCR, using the same primers used for amplification of the SBF in step 4. Amplification mutants should be maintained under selective pressure, such as in the presence of high antibiotic concentration, to prevent undesired decrease of the amplicon copy.
4.3 Adaptive laboratory evolution and monitoring of gene copy number over time
Selection of the amplification mutants in high antibiotic concentrations ensures obtaining colonies in 1 –2 days, which can be quickly checked for increased gene dos-age by colony PCR or qPCR. The confirmed amplification clones can then be trans-ferred to the desired solid selective medium without antibiotic, e.g., minimal medium
197 4 Experimental procedures with a non-native carbon source. In this way, the only selective pressure is the target condition for adaptive laboratory evolution (ALE). In our experience, it can take up to 2 weeks for colonies to appear on the selective agar plates, although growth will be faster when re-streaked on a second selective plate. These clones can then be used to initiate an ALE experiment, which we describe below following a serial passage method. Generally, as in any ALE experiment, it is recommended to start with a moderate selective pressure and gradually increase it over time, e.g., substrate con-centration, temperature, dilution, transfer rate. Optionally, the changes of gene copy number over time can be monitored by qPCR. Here we describe a method using fluorescent probes, e.g., 6FAM-MGBNFQ labelled TaqMan ™ probes (Thermo Scientific). Fluorescent DNA-binding dyes such as SYBR Green can also be used, but we find reduced background noise when using probes. The qPCR protocol described below specifies the DNA concentrations that we normally use and that give suitable amplification curves to estimate gene copy number. We note that we have been able to detect copy numbers above 100 in an EASy experiment.
4.3.1 Adaptive laboratory evolution by serial transfer
Inoculate sterile test tubes containing at least 2 mL liquid selective medium with the amplification clone(s). Use a heavy inoculum for this first transfer to liquid medium. Incubate under the appropriate conditions.
Once stationary phase is reached, transfer an aliquot of the culture to fresh medium, diluting cells at least 100-fold. Repeat this operation as many times as needed until the desired phenotype is reached or no further improvements are observed.
While performing the serial transfers, periodically remove an aliquot of the culture for preservation of the cells as glycerol stocks and extraction of genomic DNA for the quantitation of the average gene copy number of the evolving population (e.g., once a week).
Extract genomic DNA with a molecular biology kit (e.g., Quick-DNA Miniprep Plus kit from Zymo Research) following the manufacturer’s instructions and measure DNA concentration with NanoDrop. Purified DNA can be stored at 20 °C or 80 °C for long-term storage.
4.3.2 Gene copy number analysis by quantitative PCR
Design primer pairs and probes targeting the amplicon and a reference gene that is present in single copy in the chromosome using an appropriate software. In Table 1, we provide the sequences used for qPCR of the Tn 5 nptII gene present in the ΩKm R cassette with rpoA as reference gene.
Prepare separate master mixes (without DNA) for quantitation of the nptII (Km R )and rpoA genes with the TaqMan ™ Gene Expression Master Mix. The final reaction mixes should contain each primer and probe at concentration of 0.2 μM.
Dispense the master mixes into separate wells in a real-time PCR 96-well plate, with at least 3 technical replicates per sample and target. Add DNA to a final
198 CHAPTER 7 Evolution by amplification of gene copy number concentration of 0.05 ng/ μL to each well such that the final reaction volume is 20 μL (1 ng DNA per well). Genomic DNA from the parent strain containing a single copy of the amplicon should be included as control.
Using the genomic DNA from the parent strain as template, prepare a calibration curve with increasing amounts of DNA. We typically use a 5-point curve from 0.02 to 12.5 ng gDNA per well, prepared as 5-fold serial dilutions.
Cover the plate with an optically-clear adhesive film appropriate for real-time PCR and run a program with the following cycles: 50 °C for 2 min; 95 °Cfor 10 min; 40 cycles of 95 °C for 15 s and 60 °C for 1 min.
Using appropriate software, quantify the amount of nptII and rpoA for each sample. The nptII :rpoA ratio is considered to be equivalent to the relative gene copy number of the amplicon. The ratio for the parent strain should be equal to one.
4.4 Obtaining single-copy mutants from EASy
The ultimate goal of EASy is to obtain mutants with a single copy of the amplicon thanks to the accumulation of beneficial mutations. These mutations may appear in the GOI and/or in other chromosomal loci that provide a growth benefit to the cell when combined. On occasions, a single copy of the amplicon will not be reached following ALE. Nevertheless, single-copy mutants can sometimes be obtained by screening or by allelic replacement.
4.4.1 Isolation and screening by colony PCR
The gene copy number that results from qPCR analysis is an average of the cells that make up the population. Therefore, in a population with a copy number of 4 or less some cells will present a single copy of the amplicon. These can be isolated using the following procedure:
Dilute and plate cells from the evolving population on the same selective medium used for ALE with agar. To obtain well isolated colonies, it is generally sufficient to plate 100 μL of a 10 6 dilution from a saturated culture (OD 600 2).
Table 1 Primer and probe sequences for qPCR.
Target Forward primer Reverse primer Probe
nptII agcgttggctaccc gtgata ggaagcggtcagcccatt tgaagagcttggcggc
rpoA bgctcgacgccttc tatttcaa tttacgtcgcattctatt gtcttctt tcaaccacagcagcgc caggc
aSequence within the ΩKm Rcassette, used as a measure of amplicon gene dosage. bSequence within a housekeeping gene, used as a measure of a single-copy chromosomal gene.
199 4 Experimental procedures 2. Screen isolates for the absence of the SBF by colony PCR (cPCR), using the same pair of primers as in Section 4.2, step 7. Only colonies that still possess multiple copies of the amplicon should give a product of the expected size. However, since the qPCR-determined copy number represents the average in a population, colonies need to be screened to identify those with a single copy of the region of interest. If the average copy number is 2 or higher, we recommend screening at least 100 colonies.
Confirm the phenotype of the SBF-negative isolates by streak purification (at least three times) on selective medium to ensure a pure clonal population. We also recommend genotypic and phenotypic confirmation of the presence of the Ab R gene that was initially present in the parent strain prior to amplification and evolution (Section 4.1).
Extract genomic DNA from the selected isolates and repeat the PCR to check for the presence of the SBF. Simultaneously, perform additional PCRs using primers that can distinguish alternative conformations. For example, with a primer pair, each of which binds outside of the amplicon, only clones that harbour a single copy of the amplicon should give an appropriately sized product. However, experimental design may need to be tailored to the specific situation.
Clones that are confirmed to have a single copy of the amplicon can be further analysed by Sanger sequencing of the GOI or whole-genome sequencing to identify beneficial mutations.
4.4.2 Allelic replacement in evolved populations
This method consists of a series of transformations to replace the tandem amplicon array from an isolated clone of the evolved population with a single copy of the amplicon. This method facilitates analysis of the fitness contribution of mutations in the GOI (or in other regions of the chromosome). However, the success of this method may be limited by the known loss of natural competency of A. baylyi during ALE (Renda, Dasgupta, Leon, & Barrick, 2015).
Transform an isolated clone from the evolved population with a DNA cassette carrying a Sm/Sp R selectable marker and a sacB counter-selectable marker. This DNA cassette should be flanked by at least 1000 bp of sequence identical to the upstream and downstream regions of where the tandem array of the amplicon is inserted. The same DNA assembly strategy as the one used for the initial integration of the amplicon can be used (Section 4.1).
Select mutants on agar plates containing 25 mg/L Sm/Sp. Lower transformation efficiency may occur due to the loss of natural competency during ALE.
Confirm Sm/Sp R colonies for correct allelic replacement by cPCR. Also confirm sensitivity to Km and the inability to grow on the ALE selective medium.
Transform mutants from step 3 with a single copy of the non-evolved amplicon, as in Section 4.1. Select for sucrose-resistant transformants on YTS. Since sucrose selection can have high background growth, re-streak colonies on fresh YTS plates at least three times before confirming by cPCR.
200 CHAPTER 7 Evolution by amplification of gene copy number 5. Amplify individual copies of the amplicon from the evolved isolate by PCR, using primers that bind the ends of the amplicon. Alternatively, amplify only the GOI.
Transform cells containing a single copy of the non-evolved allele with the PCR products obtained in step 5. These PCR products can be integrated by direct recombination with the allele present in the chromosome. Include a no DNA control.
Select transformants on plates containing the selective medium used for ALE. Colonies growing faster than the control are expected to have acquired evolved alleles with beneficial mutations.
Confirm that mutants growing on the selective medium present a single copy of the gene of interest by qPCR.
5 Summary and concluding remarks
In this chapter, we provide general guidelines to engineer new phenotypes in
A. baylyi via controlled GDA followed by ALE. Although we focus on expansion of the catabolic capabilities of A. baylyi , the EASy method has potential for use in any application in which the desired phenotype can be linked to improved growth. For example, EASy might facilitate the isolation of mutations that increase detoxi-fying activities or that modify the substrate preference of promiscuous enzymes. The principle of EASy is to compensate for initial low gene expression or enzymatic activities with increased gene dosage. Indeed, the single-copy parent strains used in our EASy experiments were unable to grow under selective conditions, making it impossible to use direct selection in ALE without increased gene dosage. Further-more, we did not observe spontaneous GDA in the parent strains. With EASy, one can target the amplification of the GOI by transforming with a SBF. Since direct selection for the desired phenotype may be too demanding to the cells, GDA is first induced in the presence of high antibiotic concentrations. Growth under these conditions is achieved by introducing an Ab R marker in the target chromosomal segment, i.e., the amplicon. With this strategy, multi-copy mutants can be obtained within 1 –2 days to be transferred to ALE conditions. The EASy method is highly tuneable and can be adapted to various standard DNA manipulation techniques. This flexibility is enabled by the natural compe-tence and genetic malleability of A. baylyi . These same host properties can be exploited to screen beneficial mutations acquired from ALE by reverse engineering (Bedore et al., 2023; Luo et al., 2022). However, EASy presents some limitations. Importantly, there is no “counter-selective” pressure to drive the decrease of gene copy number, other than the inherent instability of the tandem amplicon array. Therefore, it is possible that a single-copy of the amplicon is not fixed in the evolving population after several generations of ALE. In some cases, when the average copy number is 2 –4, it is possible to obtain single-copy mutants, as
201 5 Summary and concluding remarks we have described in Section 4.4.2. In others, average gene copy numbers above 10 make this task more difficult, as we observed for EASy of TPA utilization. While selection for high-level gene dosage using antibiotic selection is conve-nient, there are situations where it may be preferable to use direct selection after transformation with the SBF. Our initial EASy studies to achieve growth on guaiacol and TPA did not allow direct selection. However, recent studies using selection for growth without antibiotic markers have been more successful (unpublished results). By monitoring growth in liquid cultures, the positive effect of gene amplification on growth has been observed without first using antibiotic selection. While this approach is still under development, one benefit could be that gene dosage will more closely match that needed for the target selection. Thus, if selection initially demands low-level amplification, the problem of incomplete allele segregation may be reduced. Regardless of problems that result if a haploid state is not obtained at the end of EASy experiments, valuable information can still be obtained from the whole-genome sequence data of isolates that maintain multiple amplicons. Beneficial mu-tations that are selected elsewhere in the chromosome can be identified and reverse engineered into the single-copy parent strain to evaluate their phenotypic effect. For example, during EASy of TPA utilization, we identified mutations that improved TPA uptake in A. baylyi . When these mutations were engineered into the single-copy parent strain, we observed spontaneous GDA of the tph catabolic genes that led to a Tpa + phenotype, in contrast to the unmodified parent strain. This result proved that uptake of TPA was the first limiting step to obtain TPA-utilizing mutants. The EASy-mediated or spontaneous GDA can be identified with whole-genome se-quencing by an increased coverage of certain regions of the chromosome, usually delimited by short sequence stretches with a high density of polymorphisms— indicative of novel junctions. For the analysis of whole-genome sequencing data, we refer the reader to the open-source breseq computational pipeline, which we have found is optimal for the detection of novel junctions formed during GDA (Deatherage & Barrick, 2014). In conclusion, EASy is a powerful tool to engineer improved cell catalysts through laboratory evolution. Additionally, EASy can be applied in more fundamen-tal research addressing gene regulation or the dynamics of GDA in a microbial population. In all, the simplicity and versatility of the EASy method in combination with the natural competence of A. baylyi make this tool accessible to most microbi-ology laboratories.
Acknowledgements
I. Pardo wishes to thank the Spanish National Research Council, Reina Sofı ´a Foundation, and Primafrı ´o Foundation for funding under agreement no. 20210510. Research at the University of Georgia in the United States, described in this chapter, was funded by grants from the National Science Foundation (MCB2225858) and the U.S. Department of Energy, Office of Science, Office of Biological and Environmental Research, Genomic Science Program (DE-SC0022220). S. Santala would like to thank the Novo Nordisk Foundation (grant NNF21OC0067758) and the Academy of Finland (grant no. 334822 and 347204).
202 CHAPTER 7 Evolution by amplification of gene copy number References
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10145 | https://pmc.ncbi.nlm.nih.gov/articles/PMC4711186/ | Taxonomy, Physiology, and Natural Products of Actinobacteria - PMC
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Taxonomy, Physiology, and Natural Products of Actinobacteria
Essaid Ait Barka
Essaid Ait Barka
a Laboratoire de Stress, Défenses et Reproduction des Plantes, Unité de Recherche Vignes et Vins de Champagne, UFR Sciences, UPRES EA 4707, Université de Reims Champagne-Ardenne, Reims, France
Find articles by Essaid Ait Barka
a,✉, Parul Vatsa
Parul Vatsa
a Laboratoire de Stress, Défenses et Reproduction des Plantes, Unité de Recherche Vignes et Vins de Champagne, UFR Sciences, UPRES EA 4707, Université de Reims Champagne-Ardenne, Reims, France
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a, Lisa Sanchez
Lisa Sanchez
a Laboratoire de Stress, Défenses et Reproduction des Plantes, Unité de Recherche Vignes et Vins de Champagne, UFR Sciences, UPRES EA 4707, Université de Reims Champagne-Ardenne, Reims, France
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a, Nathalie Gaveau-Vaillant
Nathalie Gaveau-Vaillant
a Laboratoire de Stress, Défenses et Reproduction des Plantes, Unité de Recherche Vignes et Vins de Champagne, UFR Sciences, UPRES EA 4707, Université de Reims Champagne-Ardenne, Reims, France
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a, Cedric Jacquard
Cedric Jacquard
a Laboratoire de Stress, Défenses et Reproduction des Plantes, Unité de Recherche Vignes et Vins de Champagne, UFR Sciences, UPRES EA 4707, Université de Reims Champagne-Ardenne, Reims, France
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a, Hans-Peter Klenk
Hans-Peter Klenk
b School of Biology, Newcastle University, Newcastle upon Tyne, United Kingdom
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b, Christophe Clément
Christophe Clément
a Laboratoire de Stress, Défenses et Reproduction des Plantes, Unité de Recherche Vignes et Vins de Champagne, UFR Sciences, UPRES EA 4707, Université de Reims Champagne-Ardenne, Reims, France
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a, Yder Ouhdouch
Yder Ouhdouch
c Faculté de Sciences Semlalia, Université Cadi Ayyad, Laboratoire de Biologie et de Biotechnologie des Microorganismes, Marrakesh, Morocco
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c, Gilles P van Wezel
Gilles P van Wezel
d Molecular Biotechnology, Institute of Biology, Sylvius Laboratories, Leiden University, Leiden, The Netherlands
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a Laboratoire de Stress, Défenses et Reproduction des Plantes, Unité de Recherche Vignes et Vins de Champagne, UFR Sciences, UPRES EA 4707, Université de Reims Champagne-Ardenne, Reims, France
b School of Biology, Newcastle University, Newcastle upon Tyne, United Kingdom
c Faculté de Sciences Semlalia, Université Cadi Ayyad, Laboratoire de Biologie et de Biotechnologie des Microorganismes, Marrakesh, Morocco
d Molecular Biotechnology, Institute of Biology, Sylvius Laboratories, Leiden University, Leiden, The Netherlands
✉
Address correspondence to Essaid Ait Barka, ea.barka@univ-reims.fr, or Gilles P. van Wezel, g.wezel@biology.leidenuniv.nl.
Citation Barka EA, Vatsa P, Sanchez L, Gaveau-Vaillant N, Jacquard C, Klenk H-P, Clément C, Ouhdouch Y, van Wezel GP. 2016. Taxonomy, physiology, and natural products of Actinobacteria. Microbiol Mol Biol Rev 80:1–43. doi:10.1128/MMBR.00019-15.
✉
Corresponding author.
Collection date 2016 Mar.
Copyright © 2015, American Society for Microbiology. All Rights Reserved.
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PMCID: PMC4711186 PMID: 26609051
This article has been corrected. See Microbiol Mol Biol Rev. 2016 Nov 9;80(4):iii.
SUMMARY
Actinobacteria are Gram-positive bacteria with high G+C DNA content that constitute one of the largest bacterial phyla, and they are ubiquitously distributed in both aquatic and terrestrial ecosystems. Many Actinobacteria have a mycelial lifestyle and undergo complex morphological differentiation. They also have an extensive secondary metabolism and produce about two-thirds of all naturally derived antibiotics in current clinical use, as well as many anticancer, anthelmintic, and antifungal compounds. Consequently, these bacteria are of major importance for biotechnology, medicine, and agriculture. Actinobacteria play diverse roles in their associations with various higher organisms, since their members have adopted different lifestyles, and the phylum includes pathogens (notably, species of Corynebacterium, Mycobacterium, Nocardia, Propionibacterium, and Tropheryma), soil inhabitants (e.g., Micromonospora and Streptomyces species), plant commensals (e.g., Frankia spp.), and gastrointestinal commensals (Bifidobacterium spp.). Actinobacteria also play an important role as symbionts and as pathogens in plant-associated microbial communities. This review presents an update on the biology of this important bacterial phylum.
INTRODUCTION
The phylum Actinobacteria is one of the largest taxonomic units among the major lineages currently recognized within the Bacteria domain (1). The actinobacterial genomes sequenced to date belong to organisms relevant to human and veterinary medicine, biotechnology, and ecology, and their observed genomic heterogeneity is assumed to reflect their biodiversity (2). The majority of the Actinobacteria are free-living organisms that are widely distributed in both terrestrial and aquatic (including marine) ecosystems (3). Actinobacteria are Gram-positive filamentous bacteria with a high guanine-plus-cytosine (G+C) content in their genomes. They grow by a combination of tip extension and branching of the hyphae. This is what gave them their name, which derives from the Greek words for ray (aktis or aktin) and fungi (mukēs). Traditionally, actinomycetes were considered transitional forms between fungi and bacteria. Indeed, like filamentous fungi, many Actinobacteria produce a mycelium, and many of these mycelial actinomycetes reproduce by sporulation. However, the comparison to fungi is only superficial: like all bacteria, actinomycetes' cells are thin with a chromosome that is organized in a prokaryotic nucleoid and a peptidoglycan cell wall; furthermore, the cells are susceptible to antibacterial agents (Fig. 1). Physiologically and ecologically, most Actinobacteria are aerobic, but there are exceptions. Further, they can be heterotrophic or chemoautotrophic, but most are chemoheterotrophic and able to use a wide variety of nutritional sources, including various complex polysaccharides (4, 5). Actinobacteria may be inhabitants of soil or aquatic environments (e.g., Streptomyces, Micromonospora, Rhodococcus, and Salinispora species), plant symbionts (e.g., Frankia spp.), plant or animal pathogens (e.g., Corynebacterium, Mycobacterium, or Nocardia species), or gastrointestinal commensals (e.g., Bifidobacterium spp.).
FIG 1.
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Schematic representation of the life cycle of sporulating actinomycetes.
BIOLOGY OF ACTINOBACTERIA
Most of the Actinobacteria (the streptomycetes in particular) are saprophytic, soil-dwelling organisms that spend the majority of their life cycles as semidormant spores, especially under nutrient-limited conditions (6). However, the phylum has adapted to a wide range of ecological environments: actinomycetes are also present in soils, fresh and salt water, and the air. They are more abundant in soils than other media, especially in alkaline soils and soils rich in organic matter, where they constitute an important part of the microbial population. Actinobacteria can be found both on the soil surface and at depths of more than 2 m below ground (7).
The population density of Actinobacteria depends on their habitat and the prevailing climate conditions. They are typically present at densities on the order of 10 6 to 10 9 cells per gram of soil (7); soil populations are dominated by the genus Streptomyces, which accounts for over 95% of the Actinomycetales strains isolated from soil (8). Other factors, such as temperature, pH, and soil moisture, also influence the growth of Actinobacteria. Like other soil bacteria, Actinobacteria are mostly mesophilic, with optimal growth at temperatures between 25 and 30°C. However, thermophilic Actinobacteria can grow at temperatures ranging from 50 to 60°C (9). Vegetative growth of Actinobacteria in the soil is favored by low humidity, especially when the spores are submerged in water. In dry soils where the moisture tension is greater, growth is very limited and may be halted. Most Actinobacteria grow in soils with a neutral pH. They grow best at a pH between 6 and 9, with maximum growth around neutrality. However, a few strains of Streptomyces have been isolated from acidic soils (pH 3.5) (10). The first study on the effect of climate on the distribution of Actinobacteria was done by Hiltner and Strömer (11), who showed that these bacteria account for 20% of the microbial flora of the soil in spring and more than 30% in the autumn because of the large amounts of crop residues available at this time of year. However, during the winter, frost reduces their relative abundance to only 13%.
Taxonomy of Actinobacteria
Actinobacteria represent one of the largest taxonomic units among the 18 major lineages currently recognized within the Bacteria domain, including 5 subclasses, 6 orders, and 14 suborders (1). The genera of this phylum exhibit enormous diversity in terms of their morphology, physiology, and metabolic capabilities. The taxonomy of Actinobacteria has evolved significantly over time with the accumulation of knowledge. The order Actinomycetales, established by Buchanan in 1917 (12), belongs to this group of prokaryotic organisms.
The phylum Actinobacteria is delineated on the basis of its branching position in 16S rRNA gene trees. However, rRNA sequences do not discriminate well between closely related species or even genera, which can create ambiguity. For instance, the taxonomic status of the genus Kitasatospora (13) within the family Streptomycetaceae has been disputed for many years (1, 14, 15), although a recent detailed genetic analysis provided strong evidence that it should be regarded as a separate genus (16). A similar close relationship exists between Micromonospora, Verrucosispora, and Salinispora. Additional genetic markers have therefore been used to discriminate between closely related genera, including rpoB and, most recently, ssgB, which is particularly useful for discriminating between closely related genera (17). Moreover, the massive recent increase in the availability of genome sequence information has provided detailed insights into genome evolution and made it possible to identify genes specific to organisms at the level of genera and family (18).
An updated taxonomy of the phylum Actinobacteria that is based on 16S rRNA trees was recently reported (1). That update eliminated the taxonomic ranks of subclasses and suborders, elevating the former subclasses and suborders to the ranks of classes and orders, respectively (19). The phylum is thus divided into six classes: Actinobacteria, Acidimicrobiia, Coriobacteriia, Nitriliruptoria, Rubrobacteria, and Thermoleophilia.
The class Actinobacteria contains 16 orders, including both of the previously proposed orders, Actinomycetales and Bifidobacteriales (20). The order Actinomycetales is now restricted to the members of the family Actinomycetaceae, and the other suborders that were previously part of this order are now designated distinct orders (19). Consequently, 43 of the 53 families within the phylum Actinobacteria are assigned to a single class, Actinobacteria, whereas the other five classes together contain only 10 families (21).
Morphological classification.
The main characteristics used to delineate the taxonomy of Actinobacteria at the genus and species levels are microscopic morphology and chemotaxonomy. The latter of these characteristics primarily relates to the composition of the cell wall and the whole-cell sugar distribution, although phospholipid composition and menaquinone type may also be considered for fine-tuning purposes (22).
Mycelial fragmentation can be regarded as a special form of vegetative reproduction. However, the Actinobacteria with primarily mycelial lifestyles usually reproduce by forming asexual spores. Actinobacteria exhibit a wide variety of morphologies, differing mainly with respect to the presence or absence of a substrate mycelium or aerial mycelium, the color of the mycelium, the production of diffusible melanoid pigments, and the structure and appearance of their spores (Fig. 1).
(i) Mycelial morphology.
Except for Sporichthya sp., which produces aerial hyphae that are initiated upright on the surface of the medium by holdfasts, Actinobacteria form a substrate mycelium in both submerged and solid-grown cultures. However, on solid surfaces, many differentiate to form aerial hyphae, whose main purpose is to produce reproductive spores (23, 24). The substrate mycelium develops from outgrowth of a germinating spore. The branching substrate mycelium is often monopodial, but in some rare cases, Actinobacteria, such as Thermoactinomyces, exhibit dichotomous branching (25). On the other hand, members of the Micromonosporaceae family produce an extensive substrate mycelium with an absent or rudimentary aerial mycelium.
Actinobacteria exhibit a wide variety of morphologies, including coccoid (Micrococcus) and rod-coccoid (Arthrobacter), as well as fragmenting hyphal forms (Nocardia spp.) and also forms with permanent and highly differentiated branched mycelia (e.g., Streptomyces spp., Frankia) (26). Rhodococci form elongated filaments on the substrate and do not produce a true mycelium (27), while corynebacteria do not produce mycelia at all. However, as in other Actinobacteria, the filaments grow at the apex instead of by lateral wall extension (28, 29). Actinobacteria belonging to the genus Oerskovia are characterized by the formation of branched substrate hyphae that break up into flagellated motile elements (30). Further, mycobacteria and rhodococci do not usually form aerial hyphae, although some exceptions exist (31).
(ii) Spore chain morphology.
Spores are extremely important in the taxonomy of Actinobacteria (32). The initial steps of sporulation in several oligosporic Actinobacteria can be regarded as budding processes, because they satisfy the main criteria used to define budding in other bacteria (Fig. 2). Spores may be formed on the substrate and/or the aerial mycelium as single cells or in chains of different lengths. In other cases, spores may be harbored in special vesicles (sporangia) and endowed with flagella.
FIG 2.
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Schematic drawings of the different types of spore chains produced by actinomycetes.
Thus, in the genera Micromonospora, Micropolyspora, and Thermoactinomycètes, spore formation occurs directly on the substrate mycelium (33), whereas in Streptomyces the spores grow out from the aerial mycelium. The Actinoplanes and Actinosynnema groups are characterized by motile spores, while Thermoactinomyces has unique heat-resistant endospores (33). Some other Actinobacteria genera have sclerotia (Chainia), synnemas (Actinosynnem a), vesicles that contain spores (Frankia), or vesicles that are devoid of spores (Intrasporangium). Other genera, such as Actinoplanes, Ampulariella, Planomonospora, Planobispora, Dactylosporangium, and Streptosporangium, are classified based on their sporangial morphology. Figure 2 illustrates the different types of spores that can be found in actinomycetal genera. Finally, the morphology of the spores themselves can also be used to characterize species: they may have smooth, warty, spiny, hairy, or rugose surfaces (34).
(iii) Spore chain length.
The number of spores per spore chain varies widely from genus to genus. The genera Micromonospora, Salinispora, Thermomonospora, Saccharomonospora, and Promicromonospora produce isolated spores, while Microbispora produces spores in longitudinal pairs. Members of the genera Actinomadura, Saccharopolyspora, Sporicthya, and some Nocardia spp. have short spore chains, while members of the genera Streptomyces, Nocardioides, Kitasatospora, Streptoverticillium, and some Nocardia spp. produce very long chains of up to 100 spores. In contrast, Frankia species produce sporangia, which are essentially bags of spores. Streptomycetes' spore chains can be classified as being straight to flexuous (Rectus-Flexibilis), open loops (Relinaculam-Apertum), open or closed spirals (spira), or verticillate (35).
(iv) Melanoid pigments.
Melanins are polymers with diverse molecular structures that typically appear black or brown and are formed by the oxidative polymerization of phenolic and indolic compounds. They are produced by a broad range of organisms, ranging from bacteria to humans. Actinobacteria have long been known to produce pigments, which may be red, yellow, orange, pink, brownish, distinct brown, greenish brown, blue, or black, depending on the strain, the medium used, and the age of the culture (4).
Generally referred to as melanins, or melanoid pigments, these brown-black metabolic polymers are important not only because of their usefulness in taxonomic studies but also because of their similarity to soil humic substances (36, 37). Melanins are not essential for the organisms' growth and development, but they play a crucial role in improving their survival and competitiveness.
Chemotaxonomic classification.
Chemotaxonomy is the use of the distribution of chemical components to group organisms according to the similarities of their cellular chemistries (38, 39). The most commonly used chemical components in such systematics are cell wall amino acids, lipids, proteins, menaquinones, muramic acid types, sugars, and the base composition of DNA (40, 41). Chemotaxonomic classification and identification can also be performed on the basis of information derived from whole-organism chemical fingerprinting techniques. Below, we discuss chemotaxonomic markers that have been reported to be of particular value for the classification and identification of actinomycetes (1).
Analysis of the cell wall composition of Actinobacteria is taxonomically valuable because it differs between suborders (42). In particular, information on the chemical architecture of the peptidoglycan in the cell wall is valuable for classifying actinomycetes because it facilitates discrimination between groups of Actinobacteria above the genus level. Multiple discriminatory characteristics relating to the structure and composition of their peptidoglycans have been identified (43), including the identity of the amino acid in position 3 of the tetrapeptide side chain, the presence or absence of glycine in interpeptide bridges, and the peptidoglycan's sugar content (43). The presence or absence of specific optical isomers of the chiral nonproteinogenic amino acid 2,6-diaminopimelic acid (DAP) is another chemotaxonomically important characteristic of the cell walls of Gram-positive bacteria: the peptidoglycan of Actinobacteria may contain either ll- or dl-(meso)-DAP, depending on the genus. By considering DAP isomerism and the presence/absence of other amino acids and (amino)sugars, Lechevalier and Lechevalier (44) identified nine distinct actinobacterial cell wall chemotypes (Table 1). However, it is important to realize that while DAP analysis and other chemotaxonomic methods are extremely important in the taxonomy of Actinobacteria, diverse groups share the same DAP profile. For example, the genera Streptomyces, Streptoverticillium, Arachnia, and Nocardioides share the same chemotype (chemotype I), even though their different morphologies indicate that they belong to different families. Therefore, when assessing the phenotypic diversity of Actinobacteria, DAP profiling should be used in combination with other phenotypic or genotypic criteria (45). To this end, a system for classifying Actinobacteria based on both morphological and chemical characteristics has been proposed (4).
TABLE 1.
Different types of cell wall components in Actinomycetesa
| Cell wall type | Major parietal constituent(s) | Genera |
:---
| I | ll-DAP, glycine, no sugar | Arachnia, Nocardioides, Pimelobacter, Streptomyces |
| II | meso-DAP, glycine, arabinose, xylose | Actinomyces, Actinoplanes, Ampulariella, Catellatosporia, Dactylosporangium, Glycomyces, Micromonospora, Pilimelia |
| III | meso-DAP, madurose (3-O-methyl-d-galactose) | Actinocorallia, Actinomadura, Dermatophylus, Frankia, Geodermatophilus, Kitasatospora, Maduromycetes, Microbispora, Microtetraspora, Nonomuraea, Planobispora, Planomonospora, Planotetraspora, some Frankia spp., Spirillosporia, Streptosporangium, Thermoactinomyces, Thermomonospora |
| IV | meso-DAP, arabinose, galactose | Micropolyspora, Nocardioforms |
| V | Deprived of DAP; possesses lysine and ornithine | Actinomyces |
| VI | Deprived of DAP; variable presence of aspartic acid, galactose | Arcanobacterium, Actinomyces, Microbacterium, Oerskovia, Promicromonospora |
| VII | Deprived of DAP; diaminobutyric acid, glycine, with lysine variable | Agromyces, Clavibacter |
| VIII | Deprived of DAP; ornithine | Aureobacterium, Curtobacterium, Cellulomonas |
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a
Information summarized in this table was obtained from references 14, 45, 61, and 602.
Cellular fatty acid patterns are also very useful chemotaxonomic indicators for the identification of specific Actinobacteria genera (46). Bacterial fatty acids range in chain length from two (C 2) to over 90 (C 90) carbon atoms, but only those in the range of C 10 to C 24 are of particular taxonomic value (47). Three major types of fatty acid profiles have been identified in Actinobacteria (46).
Several types of isoprenoid quinones have been characterized in bacteria (48), of which menaquinones are most commonly found in actinomycete cell envelopes (46–49). Menaquinone analysis has provided valuable information for the classification of Actinomadura, Microtetraspora, and Streptomyces strains (46, 50–52). In addition, cyclic menaquinones are characteristic of members of the genus Nocardia (53, 54), while fully saturated cyclic menaquinones have been reported for Pyrobaculum organotrophum (54).
Different types of phospholipids are discontinuously distributed in actinomycetes' cytoplasmic membranes, providing useful information for the classification and identification of actinomycete genera (41, 55). Actinobacteria have been classified into five phospholipid groups based on semiquantitative analyses of major phospholipid markers found in whole-organism extracts (56–58). This classification system was used in the identification of Aeromicrobium (59) and Dietzia (60). Importantly, it has been reported that members of the same Actinobacteria genus have the same phospholipid type.
Finally, sugar composition analysis is also important in chemotaxonomy. At the suprageneric level, neutral sugars (the major constituents of actinomycete cell envelopes) are useful taxonomic markers (Table 2). On the basis of the discontinuous distribution of major diagnostic sugars, Actinomycetes can be divided into five groups. Group A comprises those species whose cell walls contain arabinose and galactose; group B cell walls contain madurose (3-O-methyl-d-galactose); group C consists of those with no diagnostic sugars; group D cell walls contain arabinose and xylose; group E cell walls contain galactose and rhamnose (22, 61). In addition, the presence of 3′-O-methyl-rhamnose in Catellatospora (62) and of tyvelose in Agromyces (63) has been valuable for the classification of some actinomycete taxa.
TABLE 2.
Taxonomic markers used as characteristics to differentiate the genera of Actinomycetes
| Amino acid present | Sugar(s) | Morphological characteristics | Genus |
:--- :--- |
| No diaminopimelic acid | Xylose, madurose | Only substrate mycelium, breaks into motile elements | Oerskovia |
| Sterile aerial mycelium, breaks into nonmotile elements | Promicromonospora |
| | | Sporangia with motile spores | Actinoplanes |
| | | Short chains of conidia on aerial mycelium | Actinomadura |
| l-Diaminopimelic acid | Xylose, madurose | Both aerial and substrate mycelia that break up into rods and coccoid elements | Nocardioides |
| | | Only substrate mycelium, bearing terminal or subterminal vesicles | Intrasporangium |
| | | Aerial mycelium with long chains of spores | Streptomyces, Kitasatospora |
| | | Sclerotia | Streptomyces |
| | | Very short chains of large conidia on the vegetative and aerial mycelia | Streptomyces |
| | | Whorls of small chains of spores | Streptoverticillium |
| | | No aerial mycelium, sporangia on the vegetative mycelium | Kineosporia |
| meso-Diaminopimelic acid | Xylose, arabinose | Conidia isolated on the vegetative mycelium | Micromonospora |
| No sporangia, short chains of conidia | Cattellatospora |
| | | Chains of conidia on the aerial mycelium | Glycomyces |
| | | Dactyloid oligosporic sporangia, motile spores | Dactylosporangium |
| | | Sporangia with spherical and motile spores formed on the surfaced of colonies | Actinoplanes |
| | | Sporangia with rod-shaped spores, motility via polar flagella | Ampullariella |
| | | Sporangia with lateral flagellated spores | Pilimelia |
| | | Multilocular sporangia, spores are nonmotile | Frankia |
| | Madurose | Short chains of conidia on the aerial mycelium | Actinomadura |
| | | Chains of conidia with spores | Microbispora |
| | | Chains of conidia with 2 to 6 spores | Microtetraspora |
| | | Sporangia with 2 motile spores | Planobispora |
| | | Sporangia with 1 motile spore | Planomonospora |
| | | Mycelium with spherical sporangia containing many rod-shaped, motile spores | Spirillospora |
| | Fructose | Multilocular sporangia | Frankia |
| | | Sporangia with motile spores | Actinoplanes |
| | Rhamnose, galactose | Both substrate and aerial mycelia that break into nonmotile elements | Saccharothrix |
| | Rhamnose, galactose, mannose | Same as Streptomyces | Streptoalloteichus |
| | Galactose | Same as Streptomyces | Kitasatospora |
| | Arabinose, galactose | Presence of nocardiomycolic acid (NMA) in whole cells; both substrate and aerial mycelia fragment into rods and coccoid elements | Nocardia |
| | | Presence of NMA; rods and extensively branched substrate mycelium that fragments into irregular rods and cocci | Rhodococcus |
| | | Presence of NMA; straight to slightly curved rods occur singly, in pairs, or in masses; cells are nonmotile, non-spore forming, and do not produce aerial hyphae | Tsukamurella |
| | | Presence of NMA; paired spores borne in longitudinal pairs on vegetative hyphae; aerial mycelium is sparse | Actinobispora |
| | | No NMA, spores are long, cylindrical on aerial mycelium, formed by budding | Pseudonocardia |
| | | No NMA; long chains of conidia on aerial mycelium | Saccharomonospora |
| | | No NMA; aerial mycelium bearing long chains of conidia; halophilic | Actinopolyspora |
| | | No NMA; substrate mycelium tends to break into nonmotile elements; aerial hyphae may form and may also segment | Amycolata, Amycolatopsis |
| | | No NMA; aerial mycelium bearing curled hyphae embedded in amorphous matrix | Kibdelosporangium |
| | | No NMA; both aerial and substrate mycelia bearing long chains of motile spores | Aktinokineospora |
| | | No NMA; aerial mycelium tends to fragment into rods and cocci, short chains of spores | Pseudoamycolata |
| | | Spores formed are not heat resistant | Thermomonospora |
| | | Long chains of spores on aerial mycelium | Nocardiopsis |
| | | Columnar hyphal structures called synnemata bearing chains of conidia capable of forming flagella | Actinosynnema |
| | | Multilocular sporangia containing motile spores | Geodermatophilus |
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Molecular Classification
More recently, the morphological and chemical classification of actinomycetes have been challenged by molecular taxonomic data, much of which were obtained thanks to the rapid advancement of genome sequencing. Notably, some organisms that were inappropriately placed in certain taxonomic groups have recently been reclassified on the basis of molecular analyses (20). A recent example is the final definition of Kitasatospora as a separate genus within the Streptomycetaceae (17); genome sequencing resolved a long-running debate about this group's relationship with the genus Streptomyces and conclusively demonstrated that it is in fact a separate genus (15, 16, 64, 65).
At present, a new species cannot be claimed without genetic analysis based on sequencing the 16S rRNA gene and DNA-DNA hybridization, and even genome sequencing is becoming routine. Molecular and chemical composition criteria have been used to group the order Actinomycetales into 14 suborders: Actinomycineae, Actinopolysporineae, Catenulisporineae, Corynebacterineae, Frankineae, Glycomycineae, Jiangellineae, Kineosporineae, Micrococcineae, Micromonosporineae, Propionibacterineae, Pseudonocardineae, Streptomycineae, and Streptosporangineae (66). Moreover, sequencing of 16S rRNA genes has led to the recognition of 39 families and 130 genera (Fig. 3). All groups previously assigned to the taxonomic rank of “order” were recovered as being strictly monophyletic based on these molecular and chemical criteria, but some paraphyletic groups were found within the rank “suborder.” This might be because the classification was mainly based on 16S rRNA gene trees, which were generated without bootstrap support and may thus include misleading results. The features of some of these genera are summarized below.
FIG 3.
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A genome-based phylogenetic tree based on 97 genome sequences of the phylum Actinobacteria. Type strain genome projects were selected as previously described (676), provided that they yielded at most 25 contigs. Phylogenetic reconstruction, including the assessment of branch support, was done using amino acid sequences according to the methods described by Meier-Kolthoff et al. (677, 678). The tree was visualized by using ITOL (679). Branch support values below 60% are not shown, but the tree generally reveals high support throughout.
The genus Tropheryma.
The most-studied member of the genus Tropheryma is T. whipplei, the causative agent of Whipple's disease, which is characterized by intestinal malabsorption leading to cachexia and death. T. whipplei isolates are typically found in human intracellular niches, such as inside intestinal macrophages and circulating monocytes (67, 68). It has a condensed genome of only 925,938 bp, with a G+C content of only 46% (69, 70), whereas other actinomycete genomes have much larger genomes (up to 10 MBp) and higher G+C contents. T. whipplei has a tropism for myeloid cells, particularly macrophages, although it can be found in various cell types. Further, genome sequencing revealed a lack of key biosynthetic pathways and a lower capacity for energy metabolism. Its small genome and lack of metabolic capabilities suggest that T. whipplei has a host-restricted lifestyle (69). Recent findings have shown that T. whipplei survives phagocyte killing and replicates in macrophages by interfering with innate immune activation (71).
The genus Propionibacterium.
The genus Propionibacterium includes various species belonging to the human cutaneous propionibacteria, including P. acnes, P. avidum, P. granulosum, P. innocuum, and P. propionibacterium. Propionibacterium acnes is a non-spore-forming, anaerobic, pleomorphic rod whose end products of fermentation include propionic acid. The bacterium is omnipresent on human skin, predominantly within sebaceous follicles, where it is generally a harmless commensal. Nonetheless, P. acnes may be an opportunistic pathogen (72). Indeed, the bacterium has been isolated from sites of infection and inflammation in patients suffering from acne and other diverse conditions, including corneal ulcers, synovitis, hyperostosis, endocarditis, pulmonary angitis, and endophthalmitis (73, 74). Recently, Campisano et al. (75) reported a unique example of horizontal interkingdom transfer of P. acnes to the domesticated grapevine, Vitis vinifera L.
The genus Micromonospora.
Micromonospora species are widely distributed in nature, living in different environments. They have long been known as a significant source of secondary metabolites for medicine, and it was recently demonstrated that Micromonospora species may also influence plant growth and development (76); Micromonospora strains have been identified as natural endophytes of legume nodules, although the precise nature and mechanism of their effects on plant development and productivity are currently unclear. While the genus exhibits considerable physiological and biochemical diversity, Micromonospora constitutes a well-defined group in terms of morphology, phylogeny, and chemotaxonomy. Its colonies can be a variety of colors, including white, orange, rose, or brown. However, species of the genus Micromonospora are not always easy to differentiate on the basis of morphology alone. Consequently, phylogenies and species identifications are now more commonly derived by analyzing the sequence of the 16S rRNA gene or gyrB (the gene encoding DNA topoisomerase). The genus Micromonospora consists primarily of soil actinobacteria, which account for 32 of its species, according to the latest version of Bergey's manual (77), although 50 soil actinobacteria in this genus have been validly described as of the time of writing. Most of these species were isolated from alkaline or neutral soils and to a lesser extent from aquatic environments. The spore population of M. echinospora is known to be heterogeneous with respect to its heat response characteristics, suggesting that routine heat activation could be utilized to eliminate the natural variability that exists within populations of this species and its relatives (78). Further, analysis of the genome of M. lupini Lupac 08 revealed a diverse array of genes that may help the bacterium to survive in the soil or in plant tissues. However, despite having many genes that encode putative plant material-degrading enzymes, this bacterium is not regarded as a plant pathogen (79). In addition, genome comparisons showed that M. lupini Lupac 08 is metabolically closely related to Frankia sp. strains ACN14a, CcI3, and EAN1pec. These results suggest that the Micromonospora genus has undergone a previously unidentified process of adaptation from a purely terrestrial to a facultative endophytic lifestyle.
The genus has also been reported to produce a large number of antibiotics (80) and is second only to Streptomyces in this respect, synthesizing up to 500 different molecules with various properties (77). Micromonospora species can produce hydrolytic enzymes, which allows them to play an active role in the degradation of organic matter in their natural habitats. Marine Micromonospora species have recently been reviewed with respect to their broad distribution and their potential use as probiotics (76, 81). Like other endophytic actinobacteria, Micromonospora can suppress a number of pathogens both in vitro and in planta by activating key genes in the systemic acquired resistance (SAR) or jasmonate/ethylene (JA/ET) pathways (76). Unfortunately, there have been few genomic studies on Micromonospora species, and there is a lack of tools for their genetic analysis despite their acknowledged capacity for secondary metabolite production (76).
The genus Salinispora.
Salinispora belongs to the Micromonosporaceae and is the first Actinobacteria genus known to require seawater for growth (82). The genus is widely distributed in tropical and subtropical marine sediments (83) and includes three distinct but closely related clades corresponding to the species S. arenicola, S. pacifica, and S. tropica. Like their terrestrial actinomycete counterparts, Salinispora spp. produce numerous secondary metabolites with diverse potential pharmaceutical applications. For instance, salinosporamide A, isolated from S. tropica, is currently in phase 1 clinical trials in patients with multiple myeloma, lymphomas, leukemia, and solid tumors (84).
Although the three currently known species of Salinispora cooccur at six widely separated and distinct locations (82), only strains of S. tropica isolated from the Caribbean produce the potent anticancer compound salinosporamide A (85). In addition to its production of various secondary metabolites, this genus has attracted major interest for the novel phenomenon of species-specific secondary metabolite production (86, 87). Although it is clear that many of the genes for secondary metabolite production in the Salinispora genome were acquired via horizontal gene transfer, the ecological and evolutionary significance of these mechanisms remain unclear (86).
The genus Mycobacterium.
The relatively simple morphology of mycobacteria partly explains why it is sometimes overlooked when considering criteria for classifying actinomycetes (88, 89). With the genera Corynebacterium and Nocardia, Mycobacterium forms a monophyletic taxon within the Actinobacteria, the so-called CMN group (90). This group shares an unusual waxy cell envelope that contains mycolic acids, meaning these bacteria are unusual in being acid fast and alcohol fast. The mycobacterial cell wall contains various polysaccharide polymers, including arabinogalactan, lipomannan, lipoarabinomannan, and phosphatidylinositol mannosides (91, 92). Representatives of the genus Mycobacterium have been the subjects of three major 16S rRNA sequencing studies (93–95). Mycobacteria are generally free-living saprophytes (96), and they are the causative agents of a broad spectrum of human diseases. Mycobacterial diseases are very often associated with immunocompromised patients, especially those with AIDS. In addition, M. bovis and M. tuberculosis, isolated initially from infected animals, are most likely obligate parasites of humans (97). Both species can survive within macrophages and cause pulmonary disease, although organs other than lungs may be affected. M. leprae, which causes leprosy, lives in Schwann cells and macrophages; infection with this species results in a chronic granulomatous disease of the skin and peripheral nerves (98). Interestingly, the pathogenic M. ulcerans, which is the third most common causative agent of mycobacterial disease, has also been isolated as a soil inhabitant in symbiosis with roots of certain plants living in tropical rain forests and similar environments (99, 100). Mycobacterium marinum was initially identified as a causative organism of tuberculosis in fish in 1926 (101) and was subsequently shown to also cause skin disease in humans (102). M. marinum is a nontuberculosis mycobacterium that is a causative agent of human skin infections acquired through aquatic sources. Most cases of M. marinum infection are reported to have occurred after exposure to contaminated aquarium water or contact with fish and shellfish (103).
The genus Nocardia.
The genus Nocardia is a ubiquitous group of environmental bacteria that is most widely known as the causative agent of opportunistic infection in immunocompromised hosts. It forms a distinct clade that is associated with the genus Rhodococcus. Both the Nocardia and Rhodococcus genera belong to the family Nocardiaceae, which is a suborder of the “aerobic actinomycetes.” Nocardia species are ubiquitous soilborne aerobic actinomycetes, with more than 80 different species identified, of which at least 33 are pathogenic (104). Nocardia infections are mainly induced through inhalation or percutaneous inoculation from environmental sources (105), but nosocomial transmission has also been reported. The pathogen can spread to the brain, kidneys, joints, bones, soft tissues, and eyes, causing disseminated nocardiosis in humans and animals (106). Although Nocardia species are rare, they now account for 1 to 2% of all reported brain abscesses. However, the mortality rate for brain abscesses associated with Nocardia infection is substantially higher (31%) than that for brain abscesses in general (<10%) (107).
Moreover, Nocardia species produce industrially important bioactive molecules, such as antibiotics and enzymes (108, 109). Within the Nocardia clade, two sublines distinguishable by nucleotide differences in helix 37-1 are recognized; one consists of Nocardia asteroides and allied taxa, while the second consists of Nocardia otitidiscaviarum and related species. N. asteroides, the causal agent for most clinical human nocardial infections, was reorganized into multiple species on the basis of drug susceptibility patterns: Nocardia abscessus, the Nocardia brevicatena-Nocardia paucivorans complex, the Nocardia nova complex, the Nocardia transvalensis complex, Nocardia farcinica, and N. asteroides (104). Recently, Nocardia cyriacigeorgica was differentiated from N. asteroides (110).
In the last 2 decades, Nocardia infections have become regarded as an emerging disease among humans and domestic animals worldwide because of improved methods for pathogen isolation and molecular identification and a growing immunocompromised population (111). Nocardia species are recognized as opportunistic pathogens (112) and are known to compromise immune function. Moreover, they have been associated with organ and bone marrow transplants (113), long-term steroid use, connective tissue diseases, human immunodeficiency virus (HIV) infections, chronic obstructive pulmonary disease, alcoholism, cirrhosis, systemic vasculitis, ulcerative colitis, and renal failure (114).
In companion animals, Nocardia infections are usually reported as coinfections with immunosuppressive infectious diseases such as distemper in dogs and leukemia and immunodeficiency in cats (115).
The genus Corynebacterium.
The genus Corynebacterium was initially defined in 1896 to accommodate mainly pathogenic species exhibiting morphological similarity to the diphtheroid bacillus (116). Therefore, the genus comprised, for several decades, an extremely diverse collection of morphologically similar Gram-positive microorganisms, including nonpathogenic soil bacteria (117). Following chemotaxonomic studies and 16S rRNA sequence analysis, there are currently almost 70 recognized Corynebacterium species. Some well-known representatives include C. glutamicum, which (like the thermostable C. efficiens) is widely used in industry for the production of amino acids such as l-glutamic acid and l-lysine for human and animal nutrition, respectively (118). Several genome sequences of Corynebacterium species have been reported, including those of C. ulcerans (119), C. kutscheri (120), C. kroppenstedtii (121), and C. argentoratense (122), providing important new insights into the genomic architecture of the genus. A prophage, CGP3, that integrates into the genome of C. glutamicum and encodes an actin-like protein, AlpC, was recently described (123). CGP3 appears to be inactive in terms of cell lysis and virion production and is therefore referred to as a cryptic prophage, which likely became trapped in the genome in the course of evolution (123). This suggests that bacterial phages use an actin-based transport system similar to that found in vertebrate viruses, such as the herpesvirus. Among the known pathogenic members of Corynebacterium are C. diphtheria, which is a notorious strictly human-adapted species and the causative agent of the acute, communicable disease diphtheria, which is characterized by local growth of the bacterium in the pharynx along with the formation of an inflammatory pseudomembrane (124). The virulence factor in diphtheria is an exotoxin that targets host protein synthesis (125). Another important Corynebacterium pathogen is C. ulcerans, which is increasingly acknowledged as an emerging pathogen in various countries; infections with this species can mimic diphtheria because it harbors lysogenic-β-corynephages that carry the the diphtheria toxin (DT) gene, which is responsible for most of the systemic symptoms of diphtheria (126). C. ulcerans also induces clinical symptoms in the lower respiratory tract, including pneumonia (127) and pulmonary granulomatous nodules (128). However, its pathogenicity does not necessarily depend on the production of DT (129). A final important pathogen in this genus is C. jeikeium, which was initially isolated from human blood cultures and is associated with bacterial endocarditis contracted following cardiac surgery (130). It was subsequently shown to be a natural inhabitant of human skin and has been implicated in a variety of nosocomial infections (131).
The genus Gordonia.
Initially proposed by Tsukamura (132), this genus has been isolated from the sputum of patients with pulmonary disease and also from soil samples. There are currently 29 validly described species in this genus (1). Bacteria of this genus are aerobic and catalase positive, forming rods and cocci. The gordonae are widely distributed and are common in soil, but some strains have been linked with foams found in activated sludge at sewage treatment plants. Three species originally assigned to Rhodococcus, namely, R. bronchialis (132), R. rubropertinctus (133), and R. terrae (132), have more recently been reaffiliated to the genus Gordona as Gordona bronchialis (132), Gordona rubropertincta (133), and Gordona terrae (132). The original spelling Gordona (sic) was corrected to Gordonia by Stackebrandt et al. (134).
The genus Rhodococcus.
The genus Rhodococcus is a heterogeneous group of microorganisms whose members are more closely related to those of the genus Nocardia than to those of the genus Mycobacterium. Rhodococcus species include symbionts (Rhodococcus rhodnii) and pathogens to animals (e.g., R. equi), plants (Rhodococcus fascians), and humans (e.g., R. equi, R. rhodochrous, and R. erythropolis) (135). Rhodococcus equi is the Rhodococcus species that is most likely to act as a pulmonary pathogen in young horses and HIV-infected humans (136).
The Rhodococcus genus has had a long and confused taxonomic pedigree (137, 138). However, many of the early uncertainties have been resolved satisfactorily through the application of chemotaxonomic and phylogenetic character analyses. In the last edition of Bergey's Manual of Systematic Bacteriology, rhodococci were assigned to two aggregate groups based primarily on chemical and serological properties (21). Key diagnostic characteristics for rhodococci are the presence of tuberculostearic acid, mycolic acids with lengths of between 34 and 64 carbon atoms, and with the major menaquinone type being dihydrogenated menaquinones that possess eight isoprenoid units but which lack the cyclic element that is the characteristic motif of the Nocardia genus (135).
Rhodococci are aerobic, Gram-positive, catalase-positive, partially acid-fast, nonmotile actinomycetes that can grow as rods but also as extensively branched substrate hyphae. Some strains produce sparse, aerial hyphae that may be branched or form aerial synnemata, which consist of unbranched filaments that coalesce and project upwards (53). Rhodococci are very important organisms with remarkably catabolic versatility, because they carry genes encoding enzymes that can degrade an impressive array of xenobiotic and organic compounds (139). In addition to their bioremediation potential, they produce metabolites of industrial potential, such as carotenoids, biosurfactants, and bioflocculation agents (140). Some species, such as Rhodococcus rhodochrous, also synthesize commercially valuable products, such as acrylamide (135).
The nomenclature of Rhodococcus equi remains controversial. In a commentary on the nomenclature of this equine pathogen, Goodfellow et al. (141) noted that the taxon is regrettably left without a valid name, because Rhodococcus itself is an illegitimate name and, according to the nomenclature code, should not be used. “Prescottella equi” was suggested as a new name for the taxon that would provide nomenclatural stability; consequently, clinicians and scientists working on this taxon should adopt the name “P. equi.”
The genus Leifsonia.
Evtushenko et al. (142) introduced the genus Leifsonia to accommodate Gram-positive, non-spore-forming, irregular rod- or filament-shaped, motile, mesophilic, catalase-positive bacteria containing dl-2,4-diaminobutyric acid in their peptidoglycan layer. Currently, the genus harbors 12 species and two subspecies, with Leifsonia aquatica as the type species. Members of the genus Leifsonia have been isolated from different ecological niches, including plants (L. poae and L. xyli), soil (L. naganoensis and L. shinshuensis), distilled water (e.g., L. aquatica), Himalayan glaciers, and Antarctic ponds (L. rubra and L. aurea) (142–146).
Leifsonia xyli comprises two subspecies: L. xyli subsp. cynodontis, a pathogen that causes stunting in Bermuda grass (Cynodon dactylon), and L. xyli subsp. xyli (142). Information on the biology and pathogenicity of L. xyli subsp. xyli is limited. Like the gammaproteobacterium Xylella fastidiosa, L. xyli subsp. xyli belongs to a unique group of xylem-limited and fastidious bacterial pathogens and is the causative agent of ratoon stunting disease, the main sugarcane disease worldwide (147).
The genus Bifidobacterium.
Bifidobacteria, first isolated by Tissier (148), are the only family of bacteria in the order Bifidobacteriales. The Bifidobacteriaceae family contains the type genus Bifidobacterium (149), and members of the family Bifidobacteriaceae have different shapes, including curved, short, and bifurcated Y shapes. They were initially classified as Bacillus bifidus communis. The cells have no capsule and they are non-spore-forming, nonmotile, and nonfilamentous bacteria. The genus encompasses bacteria with health-promoting or probiotic properties, such as antimicrobial activity against pathogens that is mediated through the process of competitive exclusion (150), and also bile salt hydrolase activity, immune modulation, and the ability to adhere to mucus or the intestinal epithelium (151). For commercial exploitation, bifidobacterial strains are typically selected for fast growth, antibacterial activity, good adhesion properties, and utilization of prebiotic substrates (151). Among the many probiotic features that have been attributed to bifidobacteria are (i) the induction of immunoglobulin production, (ii) improvement of a food's nutritional value by assimilation of substrates not metabolized by the host, (iii) anticarcinogenic activity, and (iv) folic acid synthesis (152–154). Some bifidobacteria produce antimicrobials (155) and notably, also bacteriocins (156, 157).
The genus Gardnerella.
Classification for the genus Gardnerella is controversial: the genus has often been described as Gram variable but has a Gram-positive wall type (158). Gardnerella vaginalis is a facultative anaerobic bacterium and the only species of this genus belonging to the Bifidobacteriaceae family (159). G. vaginalis is strongly associated with bacterial vaginosis, a disease characterized by malodorous vaginal discharge, but it also occurs frequently in the vaginal microbiota of healthy individuals (160). G. vaginalis-associated vaginosis is a risk factor for poor obstetric and gynecologic outcomes, as well as the acquisition of some sexually transmitted diseases. In addition, clinical studies have demonstrated a relationship between G. vaginalis and preterm delivery (161). The issue of G. vaginalis commensalism is still ambiguous, as the vaginal bacterial community is dynamic and tends to change over the menstrual cycle, leading to a transient dominance of G. vaginalis even in healthy women (162).
The genus Streptomyces.
The various mycelial genera of Actinobacteria harbor some of the most complex known bacteria (163), such as Streptomyces, Thermobifida, and Frankia. Of the three genera, Streptomyces has received particular attention for three main reasons. First, streptomycetes are abundant and important in the soil, where they play major roles in the cycling of carbon trapped in insoluble organic debris, particularly from plants and fungi. This action is enabled by the production of diverse hydrolytic exoenzymes. Second, the genus exhibits a fairly wide phylogenetic spread (164). Third, streptomycetes are among Nature's most competent chemists and produce a stunning multitude and diversity of bioactive secondary metabolites; consequently, they are of great interest in medicine and industry (165). Streptomycetes are the only morphologically complex Actinobacteria whose development has been considered in detail. For more details on this genus, which serves as a model system for bacterial antibiotic production, see the section on “Physiology and Antibiotic Production of Streptomyces,” below.
The genus Frankia.
Frankia is the only nitrogen-fixing actinobacterium and can be distinguished by its ability to enter into symbiotic associations with diverse woody angiosperms known collectively as actinorhizal plants. The most notable plant genera in this group are Alnus, Casuarina, and Elaeaginus, and their symbiosis with Frankia enables them to grow well in nitrogen-poor soils (166, 167). Like Streptomyces, the DNA of Frankia has a particularly high G+C content of 72 to 73% (2). Frankia can form three different cell types, growing as mycelia or as multilocular sporangia. Under nitrogen-limited and aerobic conditions, Frankia develops so-called vesicles at the tips of hyphae or at the ends of short side hyphae (168). For a long time, Frankia spp. were believed to be the only bacteria within the Actinobacteria able to fix atmospheric nitrogen. However, Gtari et al. (169) recently reviewed the sparse physiological and biochemical studies conducted on Actinobacteria over the last 50 years and concluded that nitrogen fixation within this group is unlikely to be restricted to frankiae.
The genus Thermobifida.
The genus Thermobifida, established by Zhang et al. (170), was originally assigned to the highly heterogeneous genus Thermomonospora. A phylogenetic analysis based on 16S rRNA sequences prompted the reclassification of Thermobifida alba and Thermobifida fusca, which were previously classified as Thermomonospora species (33, 137). Later, Thermobifida cellulolytica was added to this genus (171). More recently, Thermobifida halotolerans sp. nov. was proposed as representative of a novel species of Thermobifida (172). Members of the genus Thermobifida are Gram-positive, non-acid-fast, chemo-organotrophic aerobic organisms that form an extensively branched substrate mycelium. Thermobifida species are moderately thermophilic, growing optimally at 55°C, and act as major degraders of plant cell walls in heated organic materials, such as compost heaps, rotting hay, manure piles, or mushroom growth medium.
PHYSIOLOGY AND ANTIBIOTIC PRODUCTION OF STREPTOMYCES
The Streptomyces Life Cycle
Streptomycetes play key roles in soil ecology because of their ability to scavenge nutrients and, in particular, to hydrolyze a wide range of polysaccharides (cellulose, chitin, xylan, and agar) and other natural macromolecules (173). The life cycle of the multicellular mycelial Streptomyces starts with the germination of a spore that grows out to form vegetative hyphae, after which a process of hyphal growth and branching results in an intricately branched vegetative mycelium (174). A prominent feature of the vegetative hyphae of Streptomyces is that they grow by tip extension (28). This in contrast to unicellular bacteria, like Bacillus subtilis and Escherichia coli, where cell elongation is achieved by incorporation of new cell wall material in the lateral wall (175). Exponential growth of the vegetative hyphae is achieved by a combination of tip growth and branching. The fact that cell division during vegetative growth does not lead to cell fission but rather to cross-walls that separate the hyphae into connected compartments (176) makes streptomycetes a rare example of a multicellular bacterium, with each compartment containing multiple copies of the chromosome (177, 178). The spacing of the vegetative cross-walls varies significantly, both between different Streptomyces species and within individual species between different growth conditions and mycelial ages.
Under adverse conditions, such as nutrient depletion, the vegetative mycelium differentiates to form erected sporogenic structures called aerial hyphae. This is also the moment in the life cycle when most antibiotics are produced (179, 180). Streptomyces and other filamentous microorganisms are sessile; when nutrient depletion occurs, the vegetative or substrate mycelium is autolytically degraded by a programmed cell death (PCD)-like mechanism to acquire the building blocks needed to erect a second mass of (aerial) mycelium (181–183). PCD results in the accumulation of amino acids, aminosugars, nucleotides, and lipids around the lysing substrate mycelium (184–186), which inevitably attract motile competing microbes in the habitat; it is logical to assume that antibiotics are produced at this time to protect the pool of nutrients. One well-studied system revolves around the PCD-responsive nutrient sensory regulator DasR, which controls early development and antibiotic production and responds to the accumulation of cell wall-derived N-acetylglucosamine (186, 187). The role of DasR as a regulator of antibiotic production is discussed in more detail in the section on controlling antibiotic production. A cascade of extracellular proteases and protease inhibitors also plays a well-established role in PCD and development in streptomycetes, as reviewed elsewhere (173, 188).
Two rounds of PCD occur during the Streptomyces life cycle (189). After spore germination, a compartmentalized mycelium grows out and then undergoes a first round of PCD that affects the material formed during early vegetative growth. This is then followed by a second round of PCD that is initiated during the onset of development (189). At this stage, the vegetative or substrate hyphae are lysed so as to provide nutrients for the next round of biomass formation, i.e., the growth of the aerial mycelium. The aerial hyphae give the colonies their characteristic fluffy appearance and eventually differentiate to form chains of unigenomic spores (23). Genes that are required for the formation of aerial hyphae are referred to as bld genes, in reference to the bald (“hairless”) phenotype of mutants lacking the fluffy aerial hyphae (190), while mutants whose development is blocked at a stage prior to sporulation are called whi (white), due to their failure to produce the gray spore pigment (174, 191).
Genes that are required for aerial growth or for sporulation were originally identified by screening for mutants after random mutagenesis by using UV irradiation or treatment with chemical mutagens, or by transposon-mediated mutagenesis, resulting in a collection of bld and whi mutants that were subsequently classified on the basis of their morphology (174, 190–195). Several new classes of developmental genes have been identified on the basis of physiological criteria, such as the acceleration of aerial mycelium formation in S. lividans (ram genes, for r apid a erial m ycelium ), complementation of mutants of S. griseus with disturbed sporulation (ssgA-like genes, for s porulation of S treptomyces g riseus ), or disruptions in sugar metabolism (186, 198, 199).
Most bld and whi genes that have been identified to date have a (predicted) regulatory function at the transcriptional or translational level, with many encoding predicted transcription factors. Some of the best-studied examples are bldD, a highly pleiotropic transcription factor that controls hundreds of development-related genes (200–202), the RNA polymerase σ factors bldN (203) and whiG (204, 205), which control early events during sporulation (although bldN is also strongly transcribed during aerial growth), and whiH, which controls the onset of sporulation-specific cell division (206, 207). There is also extensive control at the translational level. A wonderful example is bldA, which specifies a tRNA molecule responsible for the translation of the rare leucine codon UUA (208, 209). Deletion of bldA has a pleiotropic effect on gene expression in streptomycetes (210, 211). A major target of bldA-mediated translational control is bldH (adpA), which encodes an important global regulator of development and antibiotic production (212–215). Transcription of adpA is activated in response to the γ-butyrolactone A-factor in S. griseus and to the related molecule SCB1 in S. coelicolor (216–220). An interesting feedback loop exists whereby the translation of the adpA mRNA depends on BldA (221, 222), while AdpA in turn controls bldA transcription (223).
Recently, it was elegantly shown by the group of Mark Buttner that the activity of BldD, which represses many developmental genes during vegetative growth, is controlled posttranslationally by the signaling molecule cyclic-di-GMP (CDG) (224). Binding of tetrameric CDG to BldD brings together the DNA binding domains of the BldD dimer, thus enabling the protein to bind to its target sites (224). An example of metabolic control is presented by the pleiotropic nutrient sensory regulator DasR, which is essential for development and pleiotropically represses antibiotic production (see below). DNA binding by DasR is controlled by the binding of GlcNAc-related metabolites as ligands (186, 225). An overview of key developmental events and regulatory networks in streptomycetes is presented in Fig. 4. An extensive overview of the very complex and intriguing regulatory networks that control the onset of sporulation is beyond the scope of this review; we refer the reader to the excellent previously published reviews of this field for further information (23, 173, 188, 226, 227).
FIG 4.
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Major events during development of Streptomyces. Nutrient stress is a major trigger of development, leading to the accumulation of ppGpp, resulting in cessation of early growth and repression of the nutrient sensory DasR protein by cell wall-derived metabolites following PCD of the substrate mycelium. Bld proteins and environmental signals control the procession toward aerial growth and antibiotic production. The developmental master regulator BldD (when bound to tetrameric cyclic-di-GMP) represses the transcription of genes for many key developmental regulatory proteins, including WhiB, WhiG, SsgA, and SsgB, as well as FtsZ. Chaplins and SapB provide a supportive hydrophobic layer to allow aerial hyphae to become erect and break through the moist soil surface. White proteins control aerial growth, whereby WhiAB and SsgB likely play a role in growth cessation. Eventually, FtsZ accumulates and localizes to septum sites in an SsgAB-dependent manner. Ladders of FtsZ are formed, which subsequently delimit the spore compartments. Chromosome condensation and segregation are followed by septum closure and spore maturation. The onset of antibiotic production typically correlates temporally to the transition from vegetative to aerial growth. Solid black arrows represent major transitions in development. Dark dotted lines indicate transcriptional control (arrows for activation, ovals for repression).
Environmental Control of Aerial Hypha Formation
In addition to being defective in aerial hypha formation, early developmental (bld) mutants also exhibit disrupted antibiotic production. This underlines the connection between development and secondary metabolism (see below). Most bld mutants fail to produce antibiotics, although some, in particular bldF, are antibiotic overproducers. By definition, all of the nonessential genes that are required for aerial hypha formation are bld genes. Extracellular complementation experiments where bld mutants were grown in close proximity to one another without physical contact suggested the existence of a hierarchical relationship between at least some of the bld genes (228–231). Aerial hypha formation could be restored from one bld mutant to another, which is consistent with the idea of a signaling cascade that generates a signal that ultimately leads to the onset of development. However, these experiments were almost exclusively performed on a single reference medium, namely, nutrient-rich R2YE agar plates with glucose, and many bld mutants have a conditional bald phenotype—in other words, they are able to produce at least some aerial hyphae and spores on minimal media with nonrepressive carbon sources, such as mannitol (192, 198, 227). A logical assumption is that this is the result of carbon catabolite repression (CCR), whereby favorable carbon sources such as glucose signal the presence of abundant food, thus favoring growth over development and antibiotic production (232, 233).
In streptomycetes, CCR largely depends on the glycolytic enzyme glucose kinase, and deletion of the glkA gene encoding glucose kinase therefore abolishes CCR (232, 234, 235). Suggestively, deleting glkA in bldA mutants of S. coelicolor restores their ability to sporulate on glucose-containing media (236). Conversely, mutants that lack the bldB gene (which encodes a small 99-amino-acid [aa] protein) are defective in CCR, although the mode of action of BldB is as yet unclear (192, 237). It should be noted that Glk-independent pathways of CCR that affect development and antibiotic production also exist, adding further complexity to the picture (238). Other bld genes relevant to sugar metabolism are ptsH, ptsI, and crr, which encode the global components HPr, enzyme I (EI), and enzyme IIA (EIIA Crr), respectively, of the phospohoenolpyruvate (PEP)-dependent phosphotransferase system (PTS), which transports sugars such as N-acetylglucosamine and fructose in S. coelicolor (239, 240). Other examples are dasABC, which encodes a chitobiose sugar transporter (199, 241), and the pleiotropic sugar regulators atrA (242) and dasR (186, 187). Perhaps surprisingly, the nonsporulating phenotype of the das and pts transport mutants is independent of the carbon source and thus probably also of the transport activity (186, 199, 241). These examples highlight the important and complex connections between carbon utilization and development in streptomycetes.
Metals also play a key role in the onset of development. Streptomyces lividans requires a large amount of copper for proper aerial growth (243). This defect can be rescued by enhanced expression of the ram cluster (196), which ultimately leads to the production of the surfactant SapB (see the next section). Recently, Sébastien Rigali and colleagues showed that development can be restored to bldJ and bldK mutants by supplementing R2YE agar with iron (244). The bldK gene cluster encodes an oligopeptide transporter (228, 245), while the function of bldJ is unknown (228). Interestingly, mass spectrometric analysis revealed that all of the bld mutants that were tested showed either severely reduced (for bldA, bldJ, and ptsH mutants) or enhanced (for bldF, bldK, crr, and ptsI mutants) production of the iron-binding siderophore desferrioxamine (244). The same paper also mentioned unpublished data suggesting that deregulated desferrioxamine production occurs in bldB, bldC, bldD, dasA, and adpA mutants when grown on R2YE agar plates and may thus be a more general feature of many bld mutants (244). Further complications arise from the fact that both bldJ mutants and mutants lacking the citA gene for citrate synthase (which also fail to develop on R2YE under “standard” conditions), do successfully develop on R2YE when the medium is strongly buffered (246). Besides shedding more light on the nature of bld mutations, these experiments also show that environmental factors, such as metal availability, pH, and carbon and nitrogen sources, have profound effects on the onset of development, and so the composition of the medium should be considered very carefully when planning experiments to study development in streptomycetes.
Facilitating Aerial Growth: the Roles of Chaplins, Rodlins, and SapB
Aerial hyphae differ substantially from vegetative hyphae. One major difference is that aerial hyphae of wild-type cells typically do not branch extensively, are nearly twice as wide as vegetative hyphae, and undergo rapid growth. The aerial hyphae are also surrounded by a sheath that later becomes part of the spore coat (247–251). This sheath is hydrophobic on the air-facing side, allowing the aerial hyphae to break through the moist soil-air surface with the assistance of the turgor pressure generated by the hyphae (252, 253), similarly to what has been proposed for fungi (254). Chaplins are potent surfactants, reducing the surface tension from 72 to 24 mJ m−2 (247, 255). An important function of the sheath may also be to create a channel along the outer hyphal wall that can facilitate nutrient transport, as proposed by Keith Chater (173, 256). This argument suggests that nutrients and other metabolites might diffuse from the vegetative hyphae in the basal part of the colony up to the growing tips of the aerial hyphae (173, 256), which would be an attractive alternative to transport through hyphae and could potentially resolve the longstanding debate about how nutrients are transported efficiently across long distances over the cross-walls (188).
The sheath consists of a number of hydrophobic proteins, in particular chaplins and rodlins (230, 247, 249, 257–260). Together, these proteins form the so-called rodlet layer, which decorates the spores with a seemingly random pattern of small lines running in all directions; high-resolution electron microscopy shows that these lines consist of small protein assemblies (Fig. 5). Typical rodlet layer formation depends on the rodlin proteins RdlA and RdlB; deletion of the rdl genes instead results in decoration with fine lines consisting of the chaplins (248). S. coelicolor contains eight chaplins, three large ones (ChpABC) with two chaplin domains and a sortase domain and five smaller chaplins (ChpDEFGH) bearing a single chaplin domain (247, 249). The chaplins assemble on the hyphal surface into an amphipathic protein layer that consists of amyloid-like fibrils. Of the chaplins, the vegetatively expressed ChpC, ChpE, and ChpH proteins are sufficient for sporulation (261). ChpE and ChpH are secreted into the surrounding medium to reduce the surface tension so as to enable the hyphae to grow into the air (247, 255).
FIG 5.
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Scanning electron micrograph of the surface layer of mature spores, revealing a distinctive rodlet layer. This layer consists of hydrophobic chaplin (Chp) and rodlin (Rdl) proteins. Bar, 100 nm.
Closer analysis of the ChpH protein showed that it has two amyloidogenic domains at its N and C termini, which are both required for aerial hypha formation, while only the C-terminal domain is required for assembly of the rodlet ultrastructure (262). In addition to the chaplins, there are two rodlin proteins that also contribute to the development of the sheath's rodlet ultrastructure and the spore surface (258). Suggestively, most of the genes for the chaplins and the two rodlin genes lie in close proximity on the genome (263). However, the rodlins are not required for sporulation, even though they are needed for the sheath's development into paired rodlet structures (258). It is clear that these hydrophobic structural proteins play key roles in the aerial development of streptomycetes, but the precise role of each of the individual components remains to be resolved.
In addition to ChpE and ChpH, the onset of aerial growth requires the extracellular accumulation of yet another hydrophobic surfactant, SapB (230, 264). SapB-type proteins are widespread in streptomycetes; well-studied examples include AmfS in S. griseus and SapT in S. tendae (265, 266). In S. coelicolor, SapB is encoded by the ramS gene in the ramCSAB gene cluster (196, 267), which is controlled by the orphan response regulatory gene ramR (268–270). In turn, at least in S. griseus, transcription of the amf operon, and thus of amfS, depends on AdpA and therefore ultimately also on BldA (215). RamS is produced as a 42-aa propeptide that is subsequently modified and exported in a way very similar to that for lantibiotics, although the way the propeptide is processed is yet unknown. During modification by RamC, four dehydroalanine residues and two lanthionine bridges are introduced (271). Thus, a highly modified 21-aa molecule of 2,027 Da is produced (271), with all the structural and genetic features of type II lantibiotics (259, 272). Despite the exciting insights that have been obtained into the biological role of SapB so far, the precise mechanism by which it controls the developmental growth of streptomycetes awaits further elucidation, as do the transcriptional and posttranscriptional control mechanisms underlying its biosynthesis (273).
As a final comment on this topic, it is important to note that even the extracellular addition of the fungal hydrophobin SC3 (obtained from the basidiomycete Schizophyllum commune) restores aerial growth to several bld mutants of S. coelicolor that are deficient in the production of chaplins and/or SapB (264, 265). This again underlines the importance of the extracellular accumulation of a hydrophobic layer for the early stages of aerial growth.
From Aerial Hyphae to Spores: Sporulation-Specific Cell Division and the Cytoskeleton
Like vegetative hyphae, aerial hyphae grow by tip extension. Once sufficient aerial biomass is generated, a signal is transmitted that results in growth cessation, followed by the onset of sporulation. The signal for growth cessation is not yet known but likely relates to the Whi regulatory proteins WhiA and WhiB, as well as the cell division activator SsgB. Mutations of whiA and whiB produce identical phenotypes, with hypercoiling and very long aerial hyphae that fail to initiate cell division (23, 174, 274), while ssgB mutations produce a large colony phenotype, forming an extremely large aerial biomass (275).
The landmark event in the onset of sporulation is the initiation of sporulation-specific cell division, which is notable because the process of cell division is completely different between vegetative and aerial hyphae. Wonderful movies of septum formation during early growth of the hyphae of S. coelicolor show how irregular the placement of septa is in vegetative hyphae, with cross-walls dividing the vegetative hyphae into multigenomic compartments (276). In contrast, during sporulation-specific cell division in aerial hyphae, many septa are formed almost simultaneously and in a highly symmetrical fashion, followed by the formation of spore compartments and cell fission, resulting in chains of spores that each contain a single copy of the chromosome (reviewed in references 277 and 278). Most bacteria divide by binary fission, whereby a single mother cell symmetrically divides into identical daughter cells. This process involves the formation of a cytokinetic ring structure, of which the scaffold is formed by the polymerization of thousands of copies of the tubulin homolog FtsZ at division sites (279–283). However, in streptomycetes, the long aerial hyphae differentiate into chains of spores after a uniquely coordinated cell division event. Distinctive ladders of FtsZ are thereby produced that consist of up to 100 septa, and this eventually leads to the production of chains of haploid spores (284–286). Sufficient accumulation of FtsZ is required to support sporulation, and developmental ftsZ transcription is largely dependent on the “early” whi regulatory genes whiA, whiB, whiG, whiH, whiI, and whiJ (287). Consistent with the notion that the control of ftsZ transcription may be a key event, at least in S. coelicolor, the nonsporulating phenotype of many of these early whi mutants could be overruled by constitutive expression of ftsZ during development (288). This also suggests that no other genes that are required for sporulation completely depend on these whi genes, at least not when FtsZ is overexpressed.
A unique feature of Streptomyces biology is that cell division is not required for growth and ftsZ null mutants are viable (289). While the ftsZ mutant also fails to make cross-walls, most of the other cell division mutants are only defective in sporulation-specific cell division (278, 290–293). This illustrates a major difference between vegetative and aerial cell division.
While sporulation-specific cell division is mechanistically very similar to that in bacteria that divide by binary fission, the way septum site localization is controlled is completely different, involving actinomycete-specific proteins (177, 277, 278). In unicellular bacteria, the positioning and timing of the formation of a septum involves the action of negative-control systems such as Min, which prevents Z-ring assembly at the cell poles (294, 295), and nucleoid occlusion, which prevents DNA damage by blocking Z-ring formation over nonsegregated chromosomes (296). The Z-ring is tethered to the membrane by dedicated anchoring proteins, such as FtsA and ZipA in Escherichia coli (297, 298). All of these systems are absent in streptomycetes. It is therefore unclear how streptomycetes avoid DNA damage in the multinucleoid hyphae. Elegant work on DNA partitioning revealed the important role of the ParAB proteins in DNA segregation during growth and development (299–301). FtsK helps to avoid “guillotining” of the DNA by pumping chromosomes into the spore compartments prior to septum closure, and ftsK mutants frequently generate spores with incomplete chromosomes (302–304). Other proteins that should be considered are SmeA and SffA, which play key roles in DNA translocation during sporulation (302), and also the DNA-packaging proteins HupS (305), sIHF (306, 307), Smc (308), and Dps (309).
In terms of septum site localization, a key role is played by the SsgA-like proteins (SALPs), which only occur in sporulating actinobacteria (310, 311). SsgA activates sporulation-specific cell division (312, 313), and both ssgA and ssgB are required for sporulation (275, 314, 315). The symmetrical spacing of the many Z-rings is achieved by SsgB, which directly recruits FtsZ and also stimulates its polymerization (316). SsgB localizes to future division sites prior to and independent of FtsZ (316). Thus, cell division is positively controlled in streptomycetes (Fig. 6). The next obvious question is how SsgB itself is localized, especially given that it lacks a membrane domain. Another important cell division protein that controls sporulation-specific cell division in Streptomyces is CrgA, which affects sporulation-specific cell division by influencing Z-ring assembly (317). In contrast to the SALPs, CrgA also occurs in nonsporulating actinomycetes, and it interacts with FtsZ, FtsI, and FtsQ in Mycobacteria (318). The phenotypes of crgA null mutants and overexpressing strains suggest that CrgA affects both cell division and the cytoskeleton, although its precise mode of action is still unknown.
FIG 6.
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Model for the control of sporulation-specific cell division in Streptomyces. When sporulation starts, SsgA localizes dynamically in young aerial hyphae, while SsgB and FtsZ are still diffuse at this stge. At this point, ParA is constrained to the hyphal tip. During early cell division, SsgA and SsgB colocalize temporarily at either side of the aerial hyphae, with ParA extending downward as filaments along the aerial hypha. ParB complexes are then formed over the uncondensed chromosomes, while FtsZ assembles in spiral-like filaments. Subsequently, FtsZ and SsgB colocalize and stay together until FtsZ disperses, whereby SsgB recruits FtsZ and stimulate its polymerization into protofilaments. The way the SsgB-FtsZ complex is tethered to the membrane in the absence of a membrane domain in either protein is unclear, but a likely role is played by the SepG protein (SCO2078 in S. coelicolor) encoded by a gene upstream of divIVA (L. Zhang, J. Willemse, D. Claessen, and G. P. van Wezel, unpublished data). Z-rings are then formed at the sporulation stage, followed by chromosome condensation and segregation and the production of sporulation septa. SsgA eventually marks the future germination sites. The figure was adapted from references 277 and 316.
Streptomycetes probably have a much more elaborate cytoskeleton than most other bacteria, which may be explained by their hyphal rather than planktonic growth (319). Besides the tubulin homolog FtsZ and the actin-like proteins MreB and Mbl (320, 321), a large number of proteins with coiled-coil structural elements occur in these bacteria, and evidence is accumulating regarding their important role in growth, cell shape, and morphogenesis (319, 322–324). The protein FilP forms intermediate filament-like structures that contribute to mechanical stress resistance (322). In addition, the Scy protein, encoded by a gene immediately adjacent to filP, apparently functions as a “molecular assembler” and sequesters DivIVA (323). DivIVA is essential for growth in streptomycetes and localizes to tips to drive apical growth, although the molecular mechanism of this process is still unclear (325, 326). In this way, Scy establishes growth nuclei for apical growth and branching. It also interacts with the chromosome-partitioning protein ParA (327) and the intermediate filament-like protein FilP, which in turn interacts with DivIVA. The apical assembly that drives tip growth was termed the tip organizing complex (TIPOC) by Gabriella Kelemen and colleagues (323, 328, 329). SsgA (330) and the polysaccharide synthase CslA (323, 328, 329) are other proteins that are part of this TIPOC. Clearly, we can only see the tip of the iceberg at present, and future discoveries will undoubtedly shed new light on these processes.
STREPTOMYCETES AS ANTIBIOTIC FACTORIES
Actinomycetes produce approximately two-thirds of all known antibiotics, the majority of which are produced by streptomycetes. Consequently, these microorganisms are very important in the fight against emerging multidrug-resistant pathogens (331–333). Streptomyces coelicolor is a model system for studying (the control of) antibiotic production. Scientists have marveled for decades at the ability of single streptomycetes species to produce a plethora of different antibiotic compounds. For example, those produced by S. coelicolor include actinorhodin (Act ), undecylprodigiosin (Red ), calcium-dependent antibiotic (CDA ), and methylenomycin (Mmy ), the latter of which is carried on a plasmid. However, when the genome sequence of S. coelicolor was published (263), it became apparent that this species' true potential as a producer of natural products had actually been underestimated: over 20 biosynthetic gene clusters for secondary metabolites were identified (338), including one that appears to be for the production of a cryptic polyketide antibiotic (Cpk ). It rapidly became apparent that such “concealment” of antibiotic-producing capabilities is the norm rather than the exception, with some streptomycetes harboring more than 50 different secondary metabolite gene clusters (340–343). It therefore appears that the potential of these organisms for novel drug production is much greater than originally anticipated. This has prompted extensive research in applied genomics into so-called cryptic, silent, or sleeping antibiotics (reviewed in references 344 to 347) and methods for activating their biosynthesis (348–352).
Correlation between Growth and Antibiotic Production
Programmed cell death and the DasR system.
The production of antibiotics (and other secondary metabolites) is temporally correlated to the onset of development in the Streptomyces life cycle (179, 180). This correlation may exist because of the need to defend the colony when it is undergoing PCD. Evidence supporting such a direct link between PCD and antibiotic production was provided by the observation that cell wall-derived N-acetylglucosamine (GlcNAc) acts as a signal for the onset of development and as a global elicitor molecule for antibiotic production (186, 187). In competitive soil habitats, the timing of development is crucial. However, it is not clear how colonies know when to initiate this process, which has such major consequences for the colony. As long as sufficient nutrients are available, growth should prevail over development, while during starvation, sporulation and subsequent spore dispersal are essential for the survival of the progeny. The signals that trigger such events should be unmistakable, and GlcNAc may serve this purpose well. In nature, GlcNAc can be obtained from hydrolysis of the abundant natural polymer chitin by the chitinolytic system, or from hydrolysis of microorganism cell walls. For bacteria, GlcNAc is a favorable C and N source and a major constituent of the cell wall peptidoglycan. Some 13 chitinases and chitosanases have been identified in S. coelicolor (353–355).
Interestingly, under poor nutritional conditions, supplementing with GlcNAc accelerates both the onset of development and antibiotic production, suggesting that under these conditions GlcNAc signals nutrient stress, resulting in accelerated development. Conversely, in rich media, higher concentrations of GlcNAc block development and antibiotic production, thus inducing a response typical of vegetative growth (187). The different growth conditions of minimal and rich media likely resemble conditions of feast or famine in the natural environment (i.e., the soil), with GlcNAc acting as an important signaling molecule that would typically be derived from chitin in nutrient-rich soil during feast periods or from the Streptomyces cell walls during PCD (famine), respectively. The secret of this dual signaling role appears to lie in the nature of the sugar transporters. Monomeric GlcNAc enters the cell via the NagE2 permease (356), which is part of the PEP-dependent phosphotransferase system (PTS) (357, 358), while chitobiose (dimeric GlcNAc), which is the subunit of chitin, enters via the ABC transporters DasABC or NgcEFG (241, 353, 359). Subsequently, internalized GlcNAc is converted by the enzymes NagA and NagB to glucosamine-6-phosphate (GlcN-6-P) (360), a central metabolite that can then enter glycolysis (as fructose-6P) or the pathway toward peptidoglycan synthesis.
GlcNAc-derived GlcN-6-P acts as an allosteric effector of the GntR family regulator DasR (186), a global regulator that controls the GlcNAc regulon (186, 360, 361), and also the production of antibiotics (187) and siderophores (362). GlcNAc-dependent nutritional signaling is most likely mediated through changes in the intracellular level of GlcN-6-P, which binds as a ligand to the GntR family regulator DasR, leading to derepression of DasR-mediated control of antibiotic production (187). As shown by genome-wide transcription and ChIP-on-chip analysis, all pathway-specific activator genes for antibiotic biosynthetic gene clusters are controlled by DasR (363). Thus, antibiotic biosynthesis and secretion is induced by adding GlcNAc to minimal medium with a poor carbon source. As mentioned above, these conditions activated the cpk gene cluster (187), which encodes genes that for the cryptic polyketide Cpk and was more recently established as coelimycin P1 (364). This suggests that similar conditions could be used to activate other cryptic antibiotic gene clusters which are expressed poorly (or not at all) under normal growth conditions.
It was recently shown that “hostile” interactions between streptomycetes such as antibiosis and suppression of production by competitors primarily occur under nutrient-limiting conditions; conversely, under nutrient-rich conditions, social interactions are favored (365). These observations suggest that antibiotics are indeed used as weapons in nature. This concept is in contrast with studies suggesting that antibiotics act as signals for intercellular communication (366–368). The latter hypothesis is based on the rationale that the secreted antibiotics may not reach sufficiently high concentrations to cause appreciable growth inhibition and the observation that subinhibitory concentrations of antibiotics induce responses such as biofilm formation (369) or virulence (370) that may benefit the target cells (371). In terms of concentrations in the soil, the same argument can be made for GlcNAc, which only activates antibiotic production at millimolar concentrations on agar plates. Presumably, small molecules reach higher local concentrations, for instance, in close proximity to a producing colony or, in the case of natural products, by binding to dead plant or animal material. In this context, it is worth remembering that cellulose is a preferred column material for the purification of natural products. Information on the interactions between microbes is not just of ecological and evolutionary importance, since it could also be useful in drug discovery efforts. Indeed, data on the interactions between microbes could provide key clues concerning the activation of cryptic biosynthetic pathways (372–376).
Stringent control.
As discussed, nutrient deprivation and the resulting growth cessation are the primary triggers for the onset of antibiotic production. Starvation results in the depletion of amino acids and hence uncharged tRNAs, which then occupy the ribosomal A-site. This in turn induces the production of the small molecules guanosine tetraphosphate and pentaphosphate (377). However, (p)ppGpp, or “magic spot,” as it was originally known, is produced in response to basically all processes that relate to changes in nutrient availability and might affect growth, adaptation, secondary metabolism, survival, persistence, cell division, motility, biofilm formation, development, competence, and virulence (378). The synthesis of (p)ppGpp from GTP and ATP can be carried out by either RelA, which is ribosome associated and activated by the binding of uncharged tRNAs to the ribosome, or by SpoT, which produces (p)ppGpp in response to nutrient starvation (378). The ribosomal protein L11, encoded by rplI (relC), activates RelA and thus ppGpp synthesis (379). Deletion of the relA gene suppresses antibiotic production, underlining the important role of the stringent response in controlling antibiotic production (380). Indeed, under nitrogen-limiting conditions ppGpp causes a dramatic switch in the physiology of streptomycetes, activating the expression of genes involved in morphological and chemical differentiation, including the production of CDA and actinorhodin, and at the same time repressing genes involved in normal growth (381). The exact mechanism by which ppGpp acts upon such a wide range of genes remains to be elucidated. However, the response to fatty acid starvation involves the direct binding of the unacylated (“uncharged”) fatty acid acyl carrier protein (ACP) to SpoT (382), and it has been suggested (180) that perhaps the ACPs of polyketide synthases help to control the stringent response in a similar fashion, providing a possible explanation of the relationship between the stringent response and secondary metabolism.
Morphological control.
A complex relationship exists between morphology and antibiotic production. Streptomycetes display a great many different morphologies in submerged cultures, ranging from fragmented growth to dense clumps, and this can have a major influence on their production levels (24). For example, in Saccharopolyspora erythraea, erythromycin production is favored by clumps with a minimum size of around 90 μm in diameter (383). The connection between mycelial morphology and production is further exemplified by avermectin production by S. avermitilis, which is favored by small dense pellets (384), and by S. coelicolor, in which forced fragmentation by overexpression of the cell division activator protein SsgA abolishes actinorhodin production. However, it is dangerous to generalize, as the same strain shows a 20- to 50-fold increase in undecylprodigiosin production in a fermentor (310, 313, 315), and chloramphenicol production by Streptomyces venezuelae is not hampered by the extremely fragmented growth of the producing organism (385, 386). It was previously suggested that antibiotics that are produced during exponential growth may benefit from fragmented growth, while those produced during the transition or stationary phases are produced much more efficiently by clumps (236).
More insights into the genetic factors that control mycelial growth should allow scientists to significantly improve the yield of natural product formation by actinomycetes. A novel set of genes was recently discovered in Streptomyces that control pellet growth, called mat, for mycelial aggregation (387). The genes were discovered 30 years ago via reverse engineering of a strain of S. lividans that was selected for fragmented growth in a chemostat (388). Deletion of the mat genes (SCO2962 and SCO2963 in S. coelicolor) prevented pellet formation and increased both growth rate and enzyme production in S. lividans (387). The mat genes are probably responsible for the production of a secreted polysaccharide, which presumably glues the hyphae together and thus promotes pellet aggregation.
In silico models have been developed to better understand mycelial growth (389–391), although many were primarily based on physiological and nutritional parameters. Helped by the strong increase in computing power in the modern era, new models were developed recently (392, 393). In particular, a three-dimensional model was developed that includes parameters such as hyphal growth, branching, fragmentation, cross-wall formation, and collision detection, as well as oxygen diffusion (392). For the rational design of actinomycetes as production hosts, it is imperative that we better understand how morphology correlates with production. For example, when and especially where are natural products (and enzymes) secreted? Secretion at apical sites would imply that fragmented growth is favorable because it increases the number of hyphal tips per length unit, as opposed to when production primarily takes place inside mycelial clumps. Interestingly, production of enzymes through the twin arginine translocation (Tat) exporter occurs closely behind the hyphal tips in S. coelicolor (16, 394); in line with this concept, Tat substrates are secreted more efficiently in fragmenting strains of S. coelicolor and S. lividans (313). An extensive review of the industrial implications of the correlation between growth and natural product formation is beyond the scope of this review, so we refer the interested reader elsewhere (24, 395–398).
From global control to the activation of specific gene clusters.
The global regulatory networks ultimately relay information toward the individual biosynthetic gene clusters, acting at the level of pathway-specific control. Transcription of antibiotic biosynthetic gene clusters typically depends on a pathway-specific activator gene inside the cluster. Many of these are controlled in a growth phase-dependent manner, with the SARP family regulators (399) being the best-known examples. Particularly well-known members of this family are ActII-ORF4 and RedD, which activate the production of the pigmented antibiotics actinorhodin and undecylprodigiosin, respectively, in S. coelicolor, and StrR for production of the aminoglycoside streptomycin in S. griseus (400–402). As exemplified by the actinorhodin biosynthetic activator gene actII-ORF4, transcription of which is controlled by over 15 different regulatory proteins, the timing and accumulation of a pathway-specific activator may be very complex (236, 395). However, once activated, there appears to be little additional control downstream, as long as the necessary precursors are present. Indeed, when redD, the pathway-specific activator gene for production of prodigionines in S. coelicolor, is placed under the control of another regulatory element, the result is that control of prodigionin production becomes dictated by the regulatory network controlling that element (275, 314, 315). In other words, placing redD under the control of the promoter for the global nitrogen regulator (glnR) or the sporulation-specific sigma factor (sigF) ensures that production is controlled by nitrogen or produced in aerial hyphae, respectively. This implies that there may be few genetic limitations regarding the production of natural products in time and space once an appropriate activator is expressed. From a production point of view this is an advantage, because restrictions due to growth phase-related control mechanisms can be dealt with by changing the regulatory element. This is particularly important for heterologous expression of biosynthetic gene clusters that are being uncovered at a high rate in actinomycetes in the era of genome sequencing. In particular, the combination of synthetic biology approaches with expression in optimized heterologous Streptomyces production platforms is a promising development (403–408).
ACTINOBACTERIA AS SOURCES OF NATURAL PRODUCTS
Actinobacteria as Sources of Antibiotics
Actinobacteria are of great importance in the field of biotechnology, as producers of a plethora of bioactive secondary metabolites with extensive industrial, medical, and agricultural applications (Table 3 provides examples and corresonding references). In particular, Actinobacteria produce the majority of the naturally occurring antibiotics. The first antibiotics discovered in Actinobacteria were actinomycin from a culture of Streptomyces antibioticus in 1940 (409), streptothricin from Streptomyces lavendulae in 1942 (410), and streptomycin from Streptomyces griseus in 1944 (411), all of which were discovered by Waksman and colleagues. Streptomycetes have been the major source of clinical antibiotics and are responsible for over 80% of all antibiotics of actinobacterial origin (333). That actinomycin, streptomycin, and streptothricin were the first to be found is not surprising, as these molecules occur at much higher frequencies than many other antibiotics. For example, streptothricin is found in some 10% of all streptomycetes isolated randomly from soil and streptomycin is found in 1% and actinomycin in 0.1%, while conversely, erythromycin and vancomycin are found in around 10−5 soil isolates, and daptomycin is found only at a frequency of around 10−7 (412). Major classes of clinical antibiotics produced by actinomycetes are the following: aminoglycosides (neomycin, kanamycin, streptomycin (413–415), angucyclines (auricin; also, antitumor agents like landomycin and moromycin (416), ansamycins (rifamycin, geldanamycin) (417), anthracyclines (primarily antitumor agents, e.g., daunorubicin) (418, 419), β-lactams (cephamycins) (420) and also the important β-lactamase inhibitor clavulanic acid (421, 422), chloramphenicol (423), glutarimides (cycloheximide) (424), glycopeptides (vancomycin, teichoplanin) (425, 426), lipopeptides (daptomycin) (427), lantibiotics (mersacidin, actagardine) (272), macrolides (clarythromycin, erythromycin, tylosin, clarithromycin) (428, 429), oxazolidinones (cycloserine) (430), streptogramins (streptogramin) (431), and tetracyclines (432). The producing capacity of individual actinomycetes can also vary enormously. Some Streptomyces species produce a single antibiotic, while others produce a range of different compounds and compound classes.
TABLE 3.
Examples of bioactive molecules produced by Actinobacteria genera and their activities
| Type of compound and producing species | Bioactive agent(s) | Source or reference |
:---
| Antibacterial agent producers | | |
| Verrucosispora spp. | Abyssomycin | 603 |
| Streptomyces anulatus | Actinomycins | 409 |
| Streptomyces canus | Amphomycin | 604 |
| Micromonospora spp. | Anthracyclin | 605 |
| Streptomyces cattley | Antibiotics and fluorometabolites | 606 |
| Streptomyces canus | Aspartocins | 607 |
| Streptomyces avermitilis | Avermectin | 608 |
| Streptomyces venezuelae | Chloramphenicol | 609 |
| Micromonospora spp. | Clostomicins | 609 |
| Streptomyces griseus | Cycloheximide | 620 |
| Streptomyces orchidaceus | Cycloserine | 610 |
| Streptomyces roseosporus | Daptomycin | 611 |
| Saccharopolyspora erythraea | Erythromycin (Ilotycin) | 612 |
| Micromonospora purpurea | Gentamicin | 613 |
| Streptomyces hygroscopicus | Hygromycin | 614 |
| Streptomyces kanamyceticus | Kanamycin | 615 |
| Streptomyces kitasoensis | Leucomycin | 616 |
| Streptomyces lincolnensis | Lincomycin | 617 |
| Marinispora spp. | Marinomycin | 618 |
| Streptomyces fradiae | Neomycins | 619 |
| Micromonospora spp. | Netamicin | 80 |
| Streptomyces niveus | Novobiocin | 620 |
| Streptomyces antibioticus | Oleandomycin | 621 |
| Streptomyces rimosus | Oxytetracycline | 622 |
| Streptomyces spp. | Pristinamycin | 623 |
| Streptomyces lindensis | Retamycin | 624 |
| Streptomyces mediterranei | Rifamycin | 625 |
| Nocardia lurida | Ristocetin | 621 |
| Streptomyces ambofaciens | Spiramycin | 626 |
| Streptomyces virginiae | Staphylomycin | 627 |
| Streptomyces endus | Stendomycin | 628 |
| Streptomyces lydicus | Streptolydigin | 629 |
| Streptomyces griseus | Streptomycin | 411 |
| Streptomyces lavendulae | Streptothricin | 410 |
| Streptomyces aureofaciens | Tetracycline | 630 |
| Micromonospora spp. | Thiocoraline | 631 |
| Amycolatopsis orientalis | Vancomycin | 632 |
| Antifungal agent producers | | |
| Streptomyces anulatus | Actinomycins | 603 |
| Streptomyces nodosus | Amphotericin B | 633 |
| Streptomyces griseochromogenes | Blasticidin | 634 |
| Streptomyces griseus | Candicidin | 635 |
| Streptomyces spp. | Carboxamycin | 636 |
| Streptomyces venezuelae | Chloramphenicol | 637 |
| Streptomyces padanus | Fungichromin | 638 |
| Streptomyces galbus | Galbonolides | 675 |
| Streptomyces violaceusniger YCED-9 | Guanidylfungin | 564 |
| Streptomyces venezuelae | Jadomycin | 639 |
| Streptomyces kasugaensis | Kasugamycin | 452 |
| Streptomyces spp. | Kitamycin | 640 |
| Streptomyces natalensis | Natamycin | 641 |
| Streptomyces tendae | Nikkomycin | 642 |
| Streptomyces diastatochromogenes | Oligomycin | 643 |
| Streptomyces humidus | Phenylacetate | 644 |
| Streptomyces cacaoi | Polyoxin B | 453 |
| Streptomyces canus | Resistomycin | 645 |
| Streptomyces lavendulae | Streptothricin | 410 |
| Streptomyces canus | Tetracenomycin | 645 |
| Nocardia transvalensis | Transvalencin | 646 |
| Streptomyces hygroscopicus | Validamycin | 647 |
| Bioherbicide/biopesticide producers | | |
| Actinomadura spp. | 2,4-Dihydro-4-(β-d-ribofuranosyl)-1, 2, 4 (3H)-triazol-3-one (herbicide) | 648 |
| Streptomyces hygroscopicus | Herbimycin | 649 |
| Streptomyces avermitilis | Ivermectin (derivative of avermectin) | 650 |
| Streptomyces prasinus | Prasinons | 651 |
| Saccharopolyspora spinosa | Spinosad (neurotoxic insecticides) | 652 |
| Antiparasitic agent producers | | |
| Streptomyces avermitilis | Avermectins | 608 |
| Streptomyces coelicolor | Prodiginine | 653 |
| Streptomyces bottropensis | Trioxacarcin | 654 |
| Antiviral agent producers | | |
| Streptomyces antibioticus | 9-β-d-Arabinofuranosyladénine | 655 |
| Streptomyces hygroscopicus | Hygromycin | 614 |
| Streptomyces spp. | Panosialins | 656 |
| Hypercholesterolemia agent producer | | |
| Streptomyces hygroscopicus | Rapamycin | 674 |
| Antitumor agent producers | | |
| Micromonospora spp. | Anthraquinones | 657 |
| Nocardia asteroides | Asterobactine | 658 |
| Streptomyces spp. | Borrelidine | 659 |
| Micromonospora spp. | Diazepinomicin | 660 |
| Actinomadura spp. | IB-00208 | 661 |
| Micromonospora spp. | LL-E33288 complex | 76 |
| Micromonospora spp. | Lomaiviticins | 76 |
| Micromonospora spp. | Lupinacidins | 657 |
| Thermoactinomyces spp. | Mechercharmycin | 662 |
| Marinospora spp. | Marinomycin | 618 |
| Salinispora tropica | Salinosporamide | 621 |
| Streptomyces peucetius | Doxorubicin (adriamycin) | 664 |
| Streptomyces peucetius | Daunorubicin (daunomycin) | 665 |
| Micromonospora spp. | Tetrocarcin | 76 |
| Micromonospora spp. | Thiocoraline | 666 |
| Immunostimulatory agent producers | | |
| Nocardia rubra | Rubratin | 667 |
| Streptomyces olivoreticuli | Bestatin | 668 |
| Kitasatospora kifunense | FR-900494 | 669 |
| Immunosuppressive agent producers | | |
| Nocardia brasiliensis | Brasilicardin | 670 |
| Streptomyces filipinensis | Hygromycin | 671 |
| Streptomyces filipinensis | Pentalenolactone | 671 |
| Therapeutic enzyme (antitumor) producers | | |
| Streptomyces spp. | l-Asparaginase | 672 |
| Streptomyces olivochromogenes | l-Glutaminase | 673 |
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Besides antibiotics, Actinobacteria also produce a wide variety of other secondary metabolites with activity as herbicides (54), antifungals, antitumor or immunosuppressant drugs, and anthelmintic agents (433). Examples are given below.
Actinobacteria as Sources of Insecticides
Macrotetrolides are active against mites, insects (434–436), coccidia (437), and helminths (438), and they also show immunosuppressive effects (439). They are produced by a variety of Streptomyces species (for a review, see Jizba et al. ). However, with regard to the composition of the macrotetrolide complex, only S. aureus S-3466 (440), which produces a mixture of tetranactin (the most active member of the compound group) with dinactin and trinactin (435, 441), has been utilized for commercial purposes (442). Tetranactin, a cyclic antibiotic produced by Streptomyces aureus with a molecular structure related to cyclosporine, is used as emulsion against carmine mites of fruits and tea. A true success story in terms of anthelmintics is ivermectin (443), which is a dehydro derivative of avermectin produced by Streptomyces avermitilis. After its appearance in the late 1970s, ivermectin was the world's first endectocide, which at the time was a completely novel class of antiparasitic agents, with strong and broad-spectrum activity against both internal and external nematodes and arthropods. Recently, the Nobel Prize for Physiology or Medicine 2015 was awarded to Satoshi Omura and William C. Campbell for their discovery of avermectin, jointly with Youyou Tu for the discovery of the antimalarial drug artemisinin.
Actinobacteria as Sources of Bioherbicide and Bioinsecticide Agents
Mildiomycin, an antifungal metabolite isolated from cultures of Streptoverticillium rimofaciens Niida, is strongly active against several powdery mildews on various crops (444) and inhibits fungal protein biosynthesis (445). The primary sites of action of these antibiotics are at locations where chitin synthesis occurs in the cell wall, there is cation leakage from mitochondria, inositol biosynthesis is occurring, or sites of protein and DNA synthesis. The compounds mentioned above are a few examples of agroactive compounds isolated from Actinobacteria. Validamycin A was commercialized by Takeda for the control of pathogens in rice and other plants and as a tool for damping off diseases in vegetable seedlings. On the other hand, some secreted metabolites are cytotoxic and can include chemical structures such as macrolides, α-pyrones, lactones, indoles, terpenes, and quinones (446). For instance, resistomycin, a quinone-related antibiotic, has a unique structure and exhibits bactericidal and vasoconstrictive activity based on the inhibition of RNA and protein synthesis (447, 448).
The genome sequences of important Actinobacteria species reported to date indicate that as much as 90% of the chemical potential of these organisms remains undiscovered and that the biosynthetic machinery encoded by many of these genetic loci may be activatable under laboratory conditions (449). The predictive models of Watve et al. (450) suggested that over 150,000 bioactive metabolites from members of the genus Streptomyces alone within this order are still waiting to be discovered (328, 451). Molecular techniques such as combinatorial biosynthesis may lead to the discovery of drugs that cannot be found naturally and of biosynthetic components that can be interchanged and modified to produce bioactive products with unique properties.
Actinobacteria as Sources of Antifungal Agents
Kasugamycin is a bactericidal and fungicidal metabolite secreted by Streptomyces kasugaensis (452) that acts as an inhibitor of protein biosynthesis in microorganisms but not in mammals. The systemically active kasugamycin was marketed to control rice blast (Pyricularia oryzae cavara) and bacterial Pseudomonas diseases in several crops. In 1965, Isono et al. (453) isolated the first members of a new class of natural fungicides, polyoxins B and D, from metabolites of Streptomyces cacaoi var. asoensis. These substances act by interfering with fungal cell wall synthesis by inhibiting chitin synthase (454). Polyoxin B is applied against a number of fungal pathogens in fruits, vegetables, and ornamentals, while polyoxin D is used to control the causative agent of rice sheath blight, Rhizoctonia solani (455).
In 1968, the validamycin family was detected by researchers at Takeda Chemical Industries in a greenhouse assay for the treatment of sheath blight disease in rice plants caused by the fungus Rhizoctonia solani. Validamycin A, the major and most active component of the complex, was isolated from Streptomyces hygroscopicus var. limoneus. Within the fungal cell, validamycin is converted to validoxylamine A, a particularly strong inhibitor of trehalose that suppresses the breakdown of intracellular trehalose (456). Trehalose is well known as a storage carbohydrate, and trehalase plays an essential role in the transport of glucose in insects and fungi (457). This mode of action gives validamycin A a favorable biological selectivity, since vertebrates do not depend on the hydrolysis of the disaccharide trehalose for their metabolism (457).
INTERACTIONS BETWEEN ACTINOBACTERIA AND OTHER ORGANISMS
Interactions between Actinobacteria and Invertebrates
Insect-bacterium symbioses are widespread in the environment (458), and antibiotic-producing bacterial symbionts are often recruited to protect the host and/or their resources (459, 460). Many insects (e.g., ants, termites, gall midges, and beetles) have developed a specific association with their microbial communities. These interactions are diverse, ranging from antagonism and commensalism to mutualism, and from obligate to facultative (461).
Interaction with ants.
Microbial communities of many groups of insects have been widely studied (462), and particularly complex associations have been documented between gut bacteria and insects (463, 464). Attine ants have evolved a mutualism with Actinobacteria that produce antibiotics that the ants use as weedkillers to keep their fungal gardens free of other microbes (460, 465). For instance, the ants (genera Atta and Acromyrmex) cut leaves and then masticate them into a fine biomass that is fed to the symbiotic fungus (Leucoagaricus gongylophorus) which, in turn, provides lipid- and carbohydrate-rich hyphae known as gongylidia that will be used by the ants (466).
The ants rely on a similar mutualistic association with members of the Actinobacteria (genus Pseudonocardia) that produce antibiotics that help suppress Escovopsis, which has a devastating effect on the fungus gardens of leaf-cutting ants in the absence of the bacterium (467–470), by significantly reducing the colony fitness or even inducing colony death (471). Other Actinobacteria genera may play a similar role (472, 473). The presence and maintenance of Streptomyces bacteria seems to be of prime importance, as the bacteria appear to be the primary defense against Escovopsis, which ants possess (467, 474).
Interactions with beetles.
Multiple bacterial genera of the Gammaproteobacteria and Actinobacteria classes were found in the larvae, pupae, and adult guts of the bark beetle Dendroctonus rhizophagus. The class of Actinobacteria was represented by Ponticoccus gilvus and Kocuria marina, both of which can degrade carboxymethylcellulose in vitro. Neither Actinobacteria species has ever been reported in other bark beetles, suggesting that these bacteria could be involved in the degradation of cellulosic substrates such as pine bark and phloem, enabling them to serve as a carbon source (475). This postulate is supported by the presence of cellulose-degrading bacteria in the gut of insects that feed on woody tree tissues, such as wood-boring beetles, including Saperda vestita and Agrilus planipennis (476, 477).
Interactions with protozoans.
Mycobacterium ulcerans is responsible for a necrotizing cutaneous infection called Buruli ulcer, which has been reported in more than 30 countries worldwide, mainly in tropical and subtropical climates. However, M. ulcerans can probably not live freely due to its natural fragility, slow-growth development, and inability to withstand exposure to direct sunlight. Further, M. ulcerans is sensitive to several antibiotics, such as streptomycin and rifampin. M. ulcerans therefore rarely occurs as a free-living microorganism, even though Streptomyces griseus and Amycolatopsis rifamycinica, producers of streptomycin and rifampin, thrive under such conditions (478, 479).
To survive, Mycobacterium has adapted to a more protected niche by utilizing free-living amoebae (FLA) as carriers. In their protozoan hosts, “hidden” mycobacteria might find easier opportunities to infect vertebrate end hosts, multiplying within protozoans to escape immune reactions (480). This ability to persist within amoebae has been widely documented (481–483). The internalization of infectious agents inside other parasites represents an evolutionary strategy for survival that may sometimes enhance pathogenesis or transmissibility (480).
Interactions between Actinobacteria and Vertebrates
Actinobacteria, Bacteroidetes, Firmicutes, Fusobacteria, and Proteobacteria dominate the gut microbiota, demonstrating a similar overall composition at the phylum level in various gastrointestinal tract locations, including human gastric fluid, intraoral niches (484, 485), throat (486), distal esophagus (487), stomach mucosa (486), and feces (488). Further, Actinobacteria are prominent among the identified microbiota of the oral cavity but are significantly less abundant in the lower gastrointestinal and genital tracts.
Children with diabetes reportedly exhibit substantially lower numbers of Actinobacteria and Firmicutes compared to healthy children. Further, within the Actinobacteria, the number of Bifidobacterium was significantly lower in children with diabetes (489).
Metagenomic studies of mucosal and fecal samples retrieved from healthy subjects demonstrated the presence of six dominant phylogenetic phyla, including Actinobacteria (490). The role of bifidobacteria in gut ecology illustrates the importance of Actinomyces and other Actinobacteria (491). The phyla Actinobacteria, Bacteroidetes, Firmicutes, Fusobacteria, and Proteobacteria constitute more than 99% of all gut microbiota in dogs and cats. Compared with the case in humans, which harbor 10 18 bacterial cells, Bifidobacteria is less abundant in cats and dogs (492, 493).
Although the aerobic Actinobacteria are infrequently encountered in clinical practice, they are important potential causes of serious human and animal infections. These bacteria have emerged as unusual but important potential human pathogens that cause significant disease, affecting not only immunocompetent hosts but also severely immunocompromised patients. McNeil and Brown (494) reviewed the data on medically important aerobic Actinobacteria and epidemic diseases that these organisms cause in animals and humans. For instance, the genus Nocardia comprises several species that are known to be unusual causes of a wide spectrum of clinical diseases in both humans and animals, infecting cattle, horses, dogs, and swine (494). While the majority of nocardial infections have been attributed to Nocardia asteroides, other pathogenic Nocardia species that have been reported include N. brasiliensis, N. otitidiscaviarum, and N. transvalensis. In a taxonomic revision of the N. asteroides taxon, two new species, N. farcinica and Nocardia nova, were separated (495).
Reports of human infection with Rhodococcus spp. have been rare (494). The disease they cause can have a variable clinical presentation depending upon the host's underlying immune status and possibly upon the site of inoculation and the virulence of the infecting microorganism. Consequently, in severely immunocompromised patients, primary pulmonary Rhodococcus equi infections (pneumonia and lung abscesses) have been reported most frequently (494).
Only a few invasive humans infections, such as mycetoma caused by Streptomyces spp., have been documented to date (496). They can, however, be caused by S. somaliensis and S. sudanensis (497, 498). The majority of invasive Streptomyces infections are associated with bacteremia and lung infections, namely, pneumonia, abscess, and pneumonitis (499). Streptomycetes are infrequent pathogens, although S. somaliensis and S. sudanensis can cause infections.
As for Rhodococcus, most of the infected patients had some underlying immunosuppressive condition, such as HIV infection, cancer, systemic lupus erythematosus (SLE), Crohn's disease, etc. Moreover, S. pelletieri, S. griseus, S. lanatus, and S. albus have been isolated from various patients with lung pathology (496, 500–502). Other opportunistic pathogenic Actinobacteria include Amycolata autotrophica, A. orientalis, Micromonospora spp., N. dassonvillei, and Oerskovia spp. (494).
Interactions between Actinobacteria and Plants
As stated previously, actinomycetes are abundantly present in soils and represent a high proportion of the microbial flora of the rhizosphere (503). Actinobacteria thus unsurprisingly play diverse roles in plant-associated microbial communities. Some genera are viewed predominantly as soil saprophytes with crucial roles in nutrient cycling, while others are endophytes, beneficial symbionts, or even pathogens of plants.
Plant-Actinobacteria deleterious interactions.
(i) Actinobacteria as plant pathogens.
The successful infection of a plant host is a complex multistep process, requiring the pathogen to sense the presence of a suitable host, penetrate and colonize the plant tissue, and survive in the presence of host defense mechanisms. In comparison to other bacteria, actinomycetes play a relatively minor role in plant diseases. However, they represent major pathogens of certain crops in particular areas, and under special conditions affect the quality and the quantity of agricultural products. This may result in huge agricultural losses, especially of potatoes but also of other root crops, such as beet, carrot, parsnip, radish, sweet potato, and turnip (504, 505).
Recent progress in molecular genetics and the understanding of the genomics of plant pathogenicity has been made for the Actinobacterial genera Clavibacter, Streptomyces, Leifsonia, and Rhodococcus. Further, plant-pathogenic Actinobacteria in the genera Streptomyces and Rhodococcus have very wide host ranges, including economically important crops and model plants. For instance, in the genus Streptomyces, plant pathogenicity is rare, with only a dozen or so species possessing this trait of the more than 900 species described. Nevertheless, such species have a significant impact on agricultural economies throughout the world due to their ability to cause important crop diseases, such as potato common scab, which is characterized by lesions that form on the potato tuber surface. The most well-known phytopathogenic Actinobacteria are Streptomyces scabiei, S. acidiscabies, and S. turgidiscabies, which induce devastating scab diseases on a broad spectrum of plants. Streptomyces scabiei, the most ancient of these pathogens, is found worldwide, whereas S. turgidiscabies and S. acidiscabies are emergent pathogens that were first described in Japan and the northeastern United States, respectively (506–508).
S. scabiei and S. turgidiscabies both cause “common” scab in potatoes (Solanum tuberosum); the disease is characterized by gray spores borne in spiral chains and that produce melanin (509). Another disease, the “acid scab,” is associated with S. acidiscabies (510). The typical and acid scab strains differ with respect to pigmentation, spore chain morphology, raffinose utilization, and tolerance of low pH. In addition, S. acidiscabies differs from S. scabiei in phenotype and ecology by having flexuous spore chains, a growth medium-dependent spore mass color ranging from white to salmon-pink, a red or yellow pH-sensitive diffusible pigment, and no melanin.
Russet scab is commonly restricted to nature of the potato's skin and affects the quality of the crop. This disease caused by soilborne streptomycetes different from S. scabies has been reported in Europe and the United States since the beginning of the century (511). The disease is divided into two types. The first type is American russet scab, caused by the genus Streptomyces and species different from S. scabiei (which forms a pigmented mycelium, flexuous spore chains, and no melanin) and from S. acidiscabies (mass spore color, inability to grow at pH 4.5). The second type is European russet (or netted) scab, which is apparently distinct from the American variant (512) with respect to cultivar susceptibility, root attack, and optimum soil temperature.
No specific taxonomic investigations have been carried out on S. ipomoeae, the causal agent of the sweet potato soil rot disease characterized by dwarfed plants with little or no growth and minor discolored leaves, with many plants dying before the end of the season. The organism apparently persists for long periods, even in the absence of the host plant.
In the 1980s, Actinobacteria were reported to plug the xylem vessels of silver, sugar, and Norway maples, leading to early decay and dieback of the tree branches (513, 514). A variety of streptomycetes of different species (S. parvus, S. sparsogenes, Streptomyces sp.) were isolated from the plugs. The isolates were capable of growing within the tree vessels and in vitro in the presence of several phenols. Although sugar maples in the northeastern United States are routinely tapped to collect maple sap for conversion to maple syrup, the mode of penetration of the Actinobacteria into the host is not known. Similarly, a lignocellulose-degrading streptomycete (S. flavovirens) was found to decompose the intact cell walls of the phloem of Douglas firs, and hyphae were found in the cavities deriving from the destruction of the walls of the parenchyma and sclereids (515).
Rhodococcus fascians was first isolated and identified in 1930 as the causal agent of sweet pea fasciations (516). Since then, the symptoms caused by R. fascians in diverse plant species have been described (reviewed by Goethals et al. ). The bacterium infects both monocot and dicot hosts, many of which are economically important (518). Extensive epiphytic growth precedes intercellular invasion through stomata. R. fascians causes various effects in its hosts, including leaf deformation and formation of witches' broom, fasciations, and leafy galls (517, 519). The symptoms are caused by the hyperinduction of shoots through activation of dormant axillary meristems and de novo meristem formation, probably as the result of elaborate manipulation of host hormone balances and pathogen-derived auxins and cytokinins (520, 521). In contrast to Streptomyces and Rhodococcus, plant-pathogenic species in the Actinobacteria genera Clavibacter and Leifsonia are host specific at the species or subspecies level.
Clavibacter michiganensis is another aerobic nonsporulating Gram-positive plant-pathogenic member of the Actinobacteria and is currently the only known species within the genus Clavibacter. C. michiganensis is composed of a number of host-specific subspecies, all of which colonize the xylem. Currently, C. michiganensis is represented by five subspecies: C. michiganensis subsp. insidiosus, C. michiganensis subsp. michiganensis, C. michiganensis subsp. nebraskensis, C. michiganensis subsp. sepedonicus, and C. michiganensis subsp. tesselarius.
C. michiganensis subsp. michiganensis provokes bacterial canker formation in tomatoes, a serious emerging disease of tomatoes wherever tomato plants are grown. This disease has caused substantial economic losses worldwide (522). Another important plant pathogen is C. michiganesis subsp. sepedonicus, which causes a disease in potatoes known as ring rot, due to the way it rots the vascular tissue inside potato tubers. C. michiganensis subsp. nebraskensis causes wilt and blight in maize. C. michiganensis subsp. tesselarius induces leaf freckles and leaf spots in wheat. Wilting and stunting in alfalfa (Medicago sativa) are induced by C. michiganensis subsp. insidiosus (523). The severity of these diseases and the difficulty of controlling the spread of the corresponding pathogens has resulted in the pathogens being classified as quarantine organisms under the European Union Plant Health Legislation, as well as the laws of many other countries.
The genus Leifsonia includes xylem-limited, fastidious, bacterial pathogens. The best-known of these pathogens is L. xyli subsp. xyli, the causative agent of a systemic disease called ratoon stunting of sugarcane. Plant growth inhibition, the hallmark of this disease, may be due to a putative fatty acid desaturase that modifies the carotenoid biosynthesis pathway to produce abscisic acid, a growth inhibitor (524).
(ii) Traits of pathogenicity.
The mechanisms by which a pathogen can induce the development of lesions on the host are still not well known, although recent advances in this field have provided some ideas. An important feature in the interaction of a pathogen and its host is the establishment of an equilibrium that allows both partners to survive in nature. If no resistant or tolerant plant cultivars are available, pathogen infection will inevitably lead to disease outbreaks. Despite the economic importance of plant-pathogenic Streptomyces species, very little is known about the molecular mechanisms used by these organisms to sense the presence of a suitable plant host, colonize the host's tissues, and resist its defense mechanisms.
The virulence factors of C. michiganensis subsp. michiganensis seem to be extracellular enzymes, particularly proteases. Dreier et al. (525) reported the participation of three proteases (Pat-1, ChpC, and ChpG) in the pathogen's initial interactions with the host plant, but their exact target remains unknown. Several potential virulence genes are clustered in the chp-tomA region of the bacterial genome, which may be a pathogenicity island and crucial for successful colonization. It was recently shown that the virulence of Clavibacter michiganensis subsp. michiganensis toward tomato plants can be modulated by three different mechanisms: loss of plasmids accompanied by the loss of the pathogenicity factors celA and pat-1, resulting in reduced virulence or even nonvirulence in a plasmid-free derivative; transfer of plasmids to plasmid-free C. michiganensis subsp. michiganensis derivatives, which restores full virulence; and loss of the pathogenicity island due to stress-activated recA-dependent recombination events, which leads to a low-titer colonizer that may carry the plasmids and the pathogenicity factors necessary for effective colonization but which can be considered a nonvirulent endophyte (526).
Ammonium assimilation and nitrogen control in M. tuberculosis have been studied intensively, with a particular focus on glutamine synthetase I (GSI), an enzyme whose extracellular release was identified as a potentially important determinant of pathogenicity (527).
Phytotoxin production is commonly involved in the pathogenicity of Streptomyces. Some early work on potato scab disease demonstrated that darkening of tuber cells during pathogen colonization was a response to the action of a toxin or enzyme secreted by the scab organism. A key virulence determinant in scab-causing streptomycetes is a family of phytotoxic secondary metabolites called thaxtomins, of which thaxtomin A is the most abundant (528). Thaxtomin production is usually positively correlated with pathogenicity. Further, thaxtomin induces a variety of phenotypic changes in the plant host, including cell hypertrophy, root and shoot stunting, tissue necrosis, alterations in plant Ca 2+ and H+ ion influxes, inhibition of cellulose synthesis, programmed cell death, and production of the antimicrobial plant phytoalexin scopoletin (529–533). In addition, thaxtomin helps S. scabiei, S. turgidiscabies, and S. acidiscabies penetrate plant cell walls by inhibiting cellulose biosynthesis (reviewed by Loria et al. ), which presumably enables the bacterium to secrete proteins onto the host cell membrane. An additional virulence determinant that has been described in plant-pathogenic streptomycetes is a secreted necrogenic protein called Nec1. Thaxtomin and Nec1 were the first virulence determinants to be identified and are thought to contribute to tissue penetration and suppression of plant defense responses, respectively. Thaxtomins are cyclic dipeptides (2,5-diketopiperazines) derived from l-phenylalanine and l-tryptophan, and they contain a 4-nitroindole moiety that is essential for their phytotoxicity (528). The production of thaxtomin A was positively correlated with disease severity (534). Mutants of S. scabiei and S. acidiscabies that were deficient in thaxtomin A biosynthesis did not cause symptoms on potato tubers, establishing the thaxtomins as important pathogenicity determinants (535, 536). Eleven family members have been isolated and characterized and are distinguished by the presence or absence of N-methyl and/or hydroxyl groups. Thaxtomin A, the predominant family member produced by S. scabiei, S. turgidiscabies, and S. acidiscabies, is required for scab disease development (536–538). The S. scabiei coronafacic acid-like biosynthetic cluster contributes to host-pathogen interactions, as demonstrated recently by Bignell et al. (504). In addition, an extracellular esterase from S. scabiei has been characterized, sequenced, and identified as a potential virulence factor whose activity may be regulated by the availability of zinc (539).
Several bacterial and fungal genomes have been reported to encode proteins with distant homology to plant expansins (540). These proteins are required for virulence, have C-terminal expansin-like domains, and are found in many plant-associated microorganisms, including the phytopathogenic Actinobacteria organisms Clavibacter michiganensis subsp. sepedonicus (526, 541). The S. scabiei expansin-like proteins exhibit 66% homology with each other at the amino acid level and are also closely related to putative expansin-like proteins from two nonpathogenic Streptomyces species. It is tempting to speculate that SCAB44951 might be important for plant-microbe interactions, because it can mimic specific plant PR-1 proteins and thus manipulate plant defense responses during infection. Interestingly, the PR-1-type protein identified in S. scabiei is part of the pathogenome and is conserved in S. ipomoeae, further supporting a potential role for this protein in Streptomyces-plant interactions.
Propionibacterium acnes produces abundant porphyrins, which might contribute to skin damage (542). The interaction of porphyrins with oxygen is thought to contribute to keratinocyte damage and consequently to have implications regarding the pathogenesis of progressive macular hypomelanosis (543).
Concanamycins are produced by several Streptomyces species, including S. diastatochromogenes (544), S. neyagawaensis (545), S. graminofaciens (546), and S. scabiei (547, 548). For instance, concanamycins A and B, produced by S. scabiei, exhibit phytotoxic activity (549) and have been proposed (but not proven) to be virulence determinants in that organism. It should be noted that the other concanamycin family members are not produced in S. acidiscabies or S. turgidiscabies (548–550).
Plant-Actinobacteria beneficial interactions.
Actinobacteria are microorganisms capable of colonizing the rhizosphere through their antagonistic and competitive characteristics concerning other soil microorganisms (503). Like other beneficial microorganisms, Actinobacteria can affect plant growth in two general ways, either directly or indirectly. Indirect promotion occurs when they prevent the harmful effects of one or more deleterious microorganisms. This is chiefly done through biocontrol or antagonism toward soil plant pathogens. Specifically, colonization or the biosynthesis of antibiotics (551) and other secondary metabolites can prevent pathogen invasion and establishment. Direct promotion of plant growth occurs when the plant is supplied with a compound that is synthesized by the bacteria, or when the latter otherwise facilitates plant uptake of soil nutrients. Possible contributions of this sort include nitrogen fixation, siderophore synthesis, phytohormone synthesis, and solubilization of minerals to make them available for plant uptake and use (552).
(i) Actinobacteria as biological control agents.
The Actinobacteria are widely recognized for their potential in biocontrol (553–555) because they are important producers of bioactive compounds (556). Over the past 50 years, there have been many studies on the mechanisms by which Actinobacteria might inhibit pathogens in soil, including antibiosis, nutrient competition, production of degradative enzymes, nitrous oxide production, and quorum quenching (557–559). Their adaptability to different environments in the rhizosphere makes them a strong competitor. Some are known for their production of siderophores, which can chelate iron, depriving other organisms of this important micronutrient (560, 561). Siderophore production by S. griseorubiginouse is effective in the fight against Fusarium wilt of banana caused by Fusarium oxysporum f. sp. cubenese (561). Actinobacteria have also been reported to secrete enzymes that degrade the mycelial cell walls of fungal parasites, as described in several studies (556, 562–565). Several chitinolytic enzymes have been identified in some species of Actinobacteria, including S. antibioticus (566), S. aureofaciens (566), S. lividens (567), S. plicatus (568), S. halsteii AJ-7 (569), and S. lydicus WYEC108 (570). A biofungicide containing Streptomyces lydicus WYEC 108 was approved by AG (Natural Industries Inc., TX, USA) and registered in 2004 as Actinovate soluble (EPA registration number 73314-1). The product, which is completely water soluble, is a biofungicide that effectively protects against and controls many common foliar and soilborne diseases.
Actinobacteria are also widely known for their ability to produce antibiotics that allow them to inhibit plant pathogens (556, 571, 572). Trejo-Estrada et al. (564) have shown a correlation between the production of antibiotics in soil Actinobacteria and effectiveness in the fight against plant pathogens. For example, Streptomyces violaceusniger YCED9 produces three antifungal compounds, including nigrecine, geltanamycine, and guanidylfingine, that fight against plant pathogens (564). Similarly, several antibiotics produced by Actinobacteria are currently used in biological control (Table 3).
Millard (573) reported that green manures, or crops grown specifically for biomass to be incorporated into soil, could reduce the infection of potatoes by pathogenic Streptomyces scabiei. Later, in 1926, Sanford (574) noted that Actinomyces scabiei was “very sensitive to the secreted products of many molds and bacteria, some of which prevent its growth,” and suggested that green manures favored the antagonistic bacteria that inhibited the pathogen. Subsequently, Millard and Taylor (575) showed that soil inoculation with a saprophytic (nonpathogenic) Actinomycete isolate could significantly reduce both disease (potato scab) and pathogen populations, concluding that the saprophytic inoculated strain outcompetes the pathogen in soil, thereby reducing plant disease.
Antibiotic-mediated inhibition of pathogens is generally the primary focus in efforts to suppress plant diseases. However, the diversity of secondary metabolites produced by Streptomyces and other species also offers great potential for suppressing fungal, bacterial, oomycete, and nematode pathogens.
(ii) Actinobacteria as plant growth-promoting rhizobacteria.
In attempts to develop commercial biocontrol and plant growth-promoting products using rhizobacteria, it is important to recognize the specific challenges they present. To begin with, the interaction between plant growth-promoting rhizobacteria (PGPR) species and their plant symbionts appears to be specific, even within a crop or cultivar (552, 576). While a rhizobacterium screened for growth promotion may reveal positive effects on one crop, it may have no effect or even retard the growth of another (577, 578). Although rhizobacteria may present unique challenges to our attempts to harness their beneficial attributes, the prospects for improving agriculture by using PGPR for biocontrol seem to be excellent. The first step toward exploiting PGPR species to enhance plant growth will be to better understand the systems that enable them to act as efficient plant growth enhancers.
Since the 1990s, several nitrogen-fixing Actinobacteria have been recognized and found to be associated with plants. Corynebacterium sp. AN1 isolated from the plant forest phyllosphere reduces acetylene and can be regarded as a substitute for nitrogenous fertilizer as a means of promoting maize growth (579). Pseudonocardia dioxanivorans CB1190 isolates, which can grow on 1,4-dioxane as their sole source of carbon and energy, have also been shown to fix dinitrogen (580). Despite the well-documented history of Streptomyces in biocontrol and preliminary evidence of their capacity to enhance plant growth (581), the potential of Streptomyces species as PGPR has not been widely studied. This is surprising because streptomycetes, which generally account for an abundant percentage of the soil microflora, are particularly effective colonizers of plant root systems and are able to endure unfavorable growth conditions by forming spores (582).
Merriman et al. (583) reported the use of Streptomyces griseus (Krainsky) Waksman and Henrici isolates to treat the seeds of barley, oat, wheat, and carrot, in order to increase their growth. The isolate was originally selected for the biological control of the pathogen Rhizoctonia solani. Though the S. griseus isolate did increase the average grain yield, dry foliage weight, tiller number, and advanced head emergence for both wheat and oat relative to controls, the differences were not statistically significant. However, the isolate was more successful as a seed treatment for carrots. Marketable yields were increased over controls by 17% and 15% in two separate field trials. In addition, both trials also provided an improved yield of large- and very-large-grade carrots relative to controls (583). Nearly 20 years later, El-Abyad et al. (584) described the use of three Streptomyces spp. in the control of bacterial, Fusarium, and Verticillium wilts, early blight, and bacterial canker of tomato. In addition, tomato growth was significantly improved by the use of the antagonistic Streptomyces spp. for a seed coating. An increased availability of growth regulators produced by the inoculum was the reason proposed for the improvement in tomato growth, although this has not yet been formally tested (584). While studies conducted by El-Abyad et al. (584) and Merriman et al. (583) reported plant growth enhancement to be a function of the magnitude of inoculation with Streptomyces, but the behavior of the inocula under gnotobiotic conditions and the possible mechanisms of streptomycete-mediated growth promotion should be investigated further.
(iii) Actinobacteria as symbionts.
Streptomyces spp. constitute protective mutualistic symbioses in which the host feeds and protects the bacteria and in return the bacteria provide antibiotics to protect the host, or their resources, from pathogens (585). Other genera of Actinobacteria, namely, Frankia and Micromonospora, form mutualistic symbioses with higher organisms via nitrogen-fixing actinonodules in trees and shrubs (586, 587).
For a long time, diazotrophy in the Actinobacteria was thought to be limited to the genus Frankia. However, molecular studies have increased the number of known nifH-containing Actinobacteria beyond Frankia spp. (587–591). The discovery of these actinobaceria has stimulated further discussion and inquiries on the origin and emergence of diazotrophy among Actinobacteria. Nitrogen-fixing Actinobacteria in the genus Frankia live as soil saprophytes and as endophytic symbionts in over 200 plant species (592). The genus Frankia has a special significance as the nitrogen-fixing partner in a symbiosis with certain nonleguminous plants, most notably of the genera Alnus, Casuarina, and Elaeaginus, permitting these plants to grow well in nitrogen-poor soils. Some species of higher plants are symbiotic, forming endophytic associations with actinomycetes to achieve actinorhizal nitrogen fixation (168). The genus Frankia establishes a symbiotic relationship with many flowering plants. The best-known example is the alder (Alnus), where these Actinobacteria are found in the roots, in nodules where nitrogen gas is allowed to reach the nitrogenase (168).
(iv) Actinobacteria as endophytes.
Endophytic Actinobacteria have been isolated from a wide variety of plants. The most frequently observed species belong to the genera Microbispora, Nocardia, Micromonospora, and Streptomyces, the most abundant (566, 593). Unlike pathogenic streptomycetes, endophytic species persist inside the plant host for long periods of time without causing observable disorder symptoms and lack known virulence determinants shared with phytopathogenic Streptomyces spp. (503). Endophytic streptomycetes may improve the growth of their plant host by producing auxins that promote root growth and development (594, 595). Moreover, endophytic colonization of Pisum sativum with the endophyte S. lydicus improves the frequency of root nodulation by Rhizobium spp., causing enhanced iron and molybdenum assimilation and vigorous plant growth (596).
(v) Actinobacteria as elicitors of plant defense.
In addition to direct toxic effects on other microbes, nitrous oxide production by Streptomyces has been suggested to activate plant defenses, improving the plant's protection against pathogens (597). Recently, Mahmoudi et al. (559) reported that streptomycetes can degrade the signaling compounds that coordinate the expression of genes required for pathogenicity in Pectobacterium carotovorum, suggesting a further mechanism for disease suppression. In addition, the production of chitinases or plant growth-promoting compounds has been reported to contribute to disease suppression by some Streptomyces isolates (598–601).
CONCLUSIONS AND FUTURE PERSPECTIVES
In this review, we have provided a comprehensive overview of the current knowledge concerning the biology of the phylum Actinobacteria and applications of its members in medicine, agriculture, and industry. Distributed in both terrestrial and aquatic ecosystems, its members play a crucial role in the recycling of refractory biomaterials. The diversity of this phylum is large and includes many beneficial but also some pathogenic species. Mycobacterium tuberculosis is carried by 2 billion people in the world and is the causative agent of tuberculosis, killing several million people every year. There are also the scab-causing streptomycetes, which have a broad host range, infect plants, and are known for their ability to cause necrotic scab-like lesions on economically important root and tuber crops, such as potato. On the other hand, the Actinobacteria have numerous clear potential benefits for humans as sources of novel antibiotics, antifungals, anticancer agents, and other secondary metabolites that might be used in medicine or to improve plant growth and resistance to diseases. Actinobacteria are also very promising for biocontrol of pests and as plant growth promoters. With the rapid developments in the fields of genomics, synthetic biology, and ecology and the strong requirement for new antimicrobial compounds to combat antimicrobial resistance, the biology of the Actinobacteria is a highly dynamic research field and we expect to see many new advances in this field in the years to come.
ACKNOWLEDGMENTS
We are grateful to Dennis Claessen for discussions.
This work is supported by the research program “Assessing and reducing environmental risks from plant protection products” funded by the French Ministries in charge of Ecology and Agriculture.
Biographies
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Essaid Ait Barka is a professor of Plant Physiology at the University of Reims. He studied Plant Biology at the Cady Ayad University in Marrakesh (1988) and got his Ph.D. from Reims University (1993), on the plant reaction to low temperatures stress. As a postdoc at Laval University (Canada) and Penn State University (USA), he worked as research professor at NSAC (Nova Scotia, Canada). He is interested in the interactions between plants and pathogenic/nonpathogenic microorganisms, and his current research is directed toward basic and applied aspects of using beneficial microorganisms as microbial inoculants to promote plant growth and provide biological resistance against plant biotic and abiotic stresses. His investigations aim to understand the molecular mechanisms of cross talk between plant defense signal transduction pathways and beneficial microorganisms. He has many partnerships with the private and public sectors and is coinventor of two patents in the field of biocontrol.
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Christophe Clément is a Professor at the University of Reims, France. He received his Ph.D. in Plant Physiology from the University of Reims, France, in 1994. He obtained the head position of the lab in 2001. His research is focused on the physio-molecular analysis of plant responses to both biotic and abiotic stresses.
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Gilles P. van Wezel is a Professor of Molecular Biotechnology, at the Institute of Biology at Leiden University. He studied biochemistry at the Free University in Amsterdam (1987) and performed his Ph.D. investigation in the group of Leendert Bosch in Leiden (1994) on the control of translational genes in Streptomyces. As a postdoc (EU Human Capital Mobility) in the laboratory of Mervyn Bibb at the John Innes Centre in Norwich (United Kingdom), he worked on the control of carbon metabolism in Streptomyces, and the link between sugar utilization and antibiotic production is still a major theme in his research. His current research focuses on the systems biology of growth, cell division, and antibiotic production of Streptomyces. A major application is the activation of silent biosynthetic gene clusters and elucidation of novel molecules with bioactivity against multidrug-resistant pathogens.
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Sum of the digits of two consecutive integers divisible by 17?
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Find the smallest positive integer n n such that the digit sums of n n and n+1 n+1 are both divisible by 17 17 or prove that no such solution exists.
My question was inspired when I couldn't find the correct solution to this similar question on brilliant:
I only have a fairly basic understanding of number theory and was wondering how this difficult problem could be broken down in to smaller parts that are easier to understand, or even a simple approach that I haven't spotted yet.
Note that I have change the number from 10 in the original question to 17 in my question so as to preserve the challenge of the original question.
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edited Nov 12, 2015 at 9:41
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hint. I would think the last digit of n n must be 9 Mirko –Mirko 2015-11-10 23:38:14 +00:00 Commented Nov 10, 2015 at 23:38
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Let d s(n):{0,…,9}+→Z d s(n):{0,…,9}+→Z the digit sum of n∈N n∈N in decimal representation, (n)10(n)10. Let u u be a prefix base 10 digit string.
If there is no carrying happening when increasing n→n+1 n→n+1, the number string is of the form (n)10=u d(n)10=u d, with d∈{0,…8}d∈{0,…8}, and we have:
d s(u d)=d s(u)+d d s((n+1)10)=d s(u)+d+1 d s(u d)=d s(u)+d d s((n+1)10)=d s(u)+d+1
so not both can vanish modulo 17 17 (or any other number greater 1 1).
For a number with carrying happening, we have:
d s(u(10 k−1)10)=d s(u)+9 k d s(u(10 k)10)=d s(u)+1 d s(u(10 k−1)10)=d s(u)+9 k d s(u(10 k)10)=d s(u)+1
So for the lower number we choose the smallest number with d s(u)=16 d s(u)=16. This would be 79 79. But it ends with a 9 9 and we need the carrying process to stop, so it should not end with a 9 9. The next candidate seems to be 88 88.
For the upper number k=2 k=2 already leaves the needed rest one (modulo 17 17). Test:
d s(8899)=16+18=34=2⋅17 d s(8900)=17 d s(8899)=16+18=34=2⋅17 d s(8900)=17
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edited Nov 11, 2015 at 0:27
answered Nov 10, 2015 at 23:49
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10147 | https://math.stackexchange.com/questions/1618417/prove-that-sum-k-0m-binomnk-binomn-km-k-2m-binomnm-for-m | combinatorics - Prove that $\sum_{k=0}^m \binom{n}{k} \binom{n-k}{m-k}= 2^m \binom{n}{m}$ for $m < n $ - Mathematics Stack Exchange
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Prove that ∑m k=0(n k)(n−k m−k)=2 m(n m)∑k=0 m(n k)(n−k m−k)=2 m(n m) for m<n m<n
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Prove that ∑m k=0(n k)(n−k m−k)=2 m(n m)∑k=0 m(n k)(n−k m−k)=2 m(n m) for m<n m<n
In short : I've tried to prove this by induction since I can't really see how to interpret this with a combinatorial argument,below I provide my thinking
Inducting on m m I have proved the base case m=1 m=1,since I have
(n 0)(n 1)+(n 1)(n−1 0)n+n=2(n 1)=2 n(n 0)(n 1)+(n 1)(n−1 0)=2(n 1)n+n=2 n
Assuming it hold for m=j m=j I have ∑j k=0(n k)(n−k j−k)=2 j(n j)∑k=0 j(n k)(n−k j−k)=2 j(n j)
(I don't write the claim for sake of space and your time)
So to complete the induction I have to prove that
2 j+1(n j+1)=2 j(n j)+(n k)(n−k(j+1)−k)2 j+1(n j+1)=2 j(n j)+(n k)(n−k(j+1)−k)
which is rather a beast to simplify...
Can someone help ?
(Any proof is welcome,though I would also appreciate if someone can help me with the induction proof)
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edited Dec 10, 2016 at 11:58
Martin Sleziak
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The left hand side of the equation (the one of the problem) looks also similar to Vandermonde's Identity,I don't know if that is related however.Mr. Y –Mr. Y 2016-01-19 17:13:28 +00:00 Commented Jan 19, 2016 at 17:13
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You have
(n k)(n−k m−k)=n!k!(m−k)!(n−m)!=(n m)(m k),(n k)(n−k m−k)=n!k!(m−k)!(n−m)!=(n m)(m k),
so
∑k=0 m(n k)(n−k m−k)=(n m)∑k=0 m(m k)=2 m(n m).∑k=0 m(n k)(n−k m−k)=(n m)∑k=0 m(m k)=2 m(n m).
PS: You can also interpret this formula combinatoricaly: The right side 2 m(n m)2 m(n m) tells us that we have n n colourless balls, out of which we choose m m many (giving us (n m)(n m)), and paint each of them either black or white (giving us 2 m 2 m). The term (n k)(n−k m−k)(n k)(n−k m−k) on the left side tells us that we choose k k balls to paint black and then m−k m−k balls to paint white; summing this over k=0,1,…,m k=0,1,…,m gives us the same result as before.
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edited Jan 19, 2016 at 17:25
answered Jan 19, 2016 at 17:17
Jendrik StelznerJendrik Stelzner
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Was that so simple to you ?Thank you for this neat answer :)Mr. Y –Mr. Y 2016-01-19 17:18:46 +00:00 Commented Jan 19, 2016 at 17:18
By the way in the last step why can you just sum the (m k)(m k) while leaving (n m)(n m) unchanged ?Doesn't it also change while we sum from k=0 k=0 to m m ?Mr. Y –Mr. Y 2016-01-19 17:25:45 +00:00 Commented Jan 19, 2016 at 17:25
1 The term (n m)(n m) does not depend on k k, which we are summing over; it is a constant which we can pull out of the sum.Jendrik Stelzner –Jendrik Stelzner 2016-01-19 17:26:35 +00:00 Commented Jan 19, 2016 at 17:26
Oh,yeah ....I lost that detail . Thanks also for the combinatorial argument . Btw what do you think about the induction proof (is it worth ?)?Mr. Y –Mr. Y 2016-01-19 17:28:15 +00:00 Commented Jan 19, 2016 at 17:28
Proof by induction seems a bit problematic to me, or at least not so easy as your post suggests: Notice that the summand (n k)(n−k n−m)(n k)(n−k n−m) does depend on m m. So in the induction step m→m+1 m→m+1 we do not only get one more summand on the left hand side of the equation, but the summands so far also change by some varying factor.Jendrik Stelzner –Jendrik Stelzner 2016-01-19 17:34:50 +00:00 Commented Jan 19, 2016 at 17:34
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Another solution is by double counting.
Suppose that we have n balls, and we want to choose m balls among them and color each of them by blue or red. In how many ways we can do that?
Right side: We can simply choose m of n balls ( (n m)(n m) ) and then choose a color for each of these m balls (2 m 2 m).
Left side: We can choose k k balls to be colored by blue ( (n k)(n k) ) and the m-k balls from the remaining balls to be colored by red ( (n−k m−k)(n−k m−k) ) and we must do so for all 1≤k≤m 1≤k≤m.
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answered Jan 19, 2016 at 17:27
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Hint: The following technique can also be useful in more complicated situations.
We use the coefficient of operator [x n][x n] to denote the coefficient of x n x n in a polynomial or series A(x)=∑∞k=0 a k x k A(x)=∑k=0∞a k x k. We can write e.g.
(n k)=x kn(n k)=x kn
We obtain for 0≤m≤n 0≤m≤n
∑k=0 m(n k)(n−k m−k)=∑k=0∞x kny m−kn−k=y mn∑k=0∞y k(1+y)−kx kn=y mn(1+y 1+y)n=y mn=(n m)2 m(1)(2)(3)∑k=0 m(n k)(n−k m−k)(1)=∑k=0∞x kny m−kn−k(2)=y mn∑k=0∞y k(1+y)−kx kn(3)=y mn(1+y 1+y)n=y mn=(n m)2 m
Comment:
In (1) we use the coefficient of operator twice and set the upper limit of the sum to ∞∞ without changing anything, since we add only zero.
In (2) we rearrange the sum by using the linearity of the coefficient of operator and [x n+k]A(x)=[x n]x−k A(x)[x n+k]A(x)=[x n]x−k A(x).
In (3) we apply the substitution rule
A(x)=∑k=0∞a k x k=∑k=0∞x k[y k]A(y)A(x)=∑k=0∞a k x k=∑k=0∞x k[y k]A(y) with a k=[y k]A(y)a k=[y k]A(y).
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edited Jan 21, 2016 at 11:14
answered Jan 21, 2016 at 10:51
Markus ScheuerMarkus Scheuer
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10148 | https://en.wikipedia.org/wiki/Kraft%E2%80%93McMillan_inequality | Kraft–McMillan inequality - Wikipedia
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(Top)
1 Applications and intuitions
2 Formal statement
3 Example: binary trees
4 ProofToggle Proof subsection
4.1 Proof for prefix codes
4.2 Proof of the general case
4.3 Alternative construction for the converse
5 Generalizations
6 Notes
7 References
8 See also
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Kraft–McMillan inequality
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From Wikipedia, the free encyclopedia
Concept in coding theory
In coding theory, the Kraft–McMillan inequality gives a necessary and sufficient condition for the existence of a prefix code (in Leon G. Kraft's version) or a uniquely decodable code (in Brockway McMillan's version) for a given set of codeword lengths. Its applications to prefix codes and trees often find use in computer science and information theory. The prefix code can contain either finitely many or infinitely many codewords.
Kraft's inequality was published in Kraft (1949). However, Kraft's paper discusses only prefix codes, and attributes the analysis leading to the inequality to Raymond Redheffer. The result was independently discovered in McMillan (1956). McMillan proves the result for the general case of uniquely decodable codes, and attributes the version for prefix codes to a spoken observation in 1955 by Joseph Leo Doob.
Applications and intuitions
[edit]
Kraft's inequality limits the lengths of codewords in a prefix code: if one takes an exponential of the length of each valid codeword, the resulting set of values must look like a probability mass function, that is, it must have total measure less than or equal to one. Kraft's inequality can be thought of in terms of a constrained budget to be spent on codewords, with shorter codewords being more expensive. Among the useful properties following from the inequality are the following statements:
If Kraft's inequality holds with strict inequality, the code has some redundancy.
If Kraft's inequality holds with equality, the code in question is a complete code.
If Kraft's inequality does not hold, the code is not uniquely decodable.
For every uniquely decodable code, there exists a prefix code with the same length distribution.
Formal statement
[edit]
Let each source symbol from the alphabet
S={s 1,s 2,…,s n}{\displaystyle S={\,s_{1},s_{2},\ldots ,s_{n}\,}}
be encoded into a uniquely decodable code over an alphabet of size r{\displaystyle r} with codeword lengths
ℓ 1,ℓ 2,…,ℓ n.{\displaystyle \ell {1},\ell {2},\ldots ,\ell _{n}.}
Then
∑i=1 n r−ℓ i⩽1.{\displaystyle \sum {i=1}^{n}r^{-\ell {i}}\leqslant 1.}
Conversely, for a given set of natural numbers ℓ 1,ℓ 2,…,ℓ n{\displaystyle \ell {1},\ell {2},\ldots ,\ell _{n}} satisfying the above inequality, there exists a uniquely decodable code over an alphabet of size r{\displaystyle r} with those codeword lengths.
Example: binary trees
[edit]
9, 14, 19, 67 and 76 are leaf nodes at depths of 3, 3, 3, 3 and 2, respectively.
Any binary tree can be viewed as defining a prefix code for the leaves of the tree. Kraft's inequality states that
∑ℓ∈leaves 2−depth(ℓ)⩽1.{\displaystyle \sum _{\ell \in {\text{leaves}}}2^{-{\text{depth}}(\ell )}\leqslant 1.}
Here the sum is taken over the leaves of the tree, i.e. the nodes without any children. The depth is the distance to the root node. In the tree to the right, this sum is
1 4+4(1 8)=3 4⩽1.{\displaystyle {\frac {1}{4}}+4\left({\frac {1}{8}}\right)={\frac {3}{4}}\leqslant 1.}
Proof
[edit]
Proof for prefix codes
[edit]
Example for binary tree. Red nodes represent a prefix tree. The method for calculating the number of descendant leaf nodes in the full tree is shown.
First, let us show that the Kraft inequality holds whenever the code for S{\displaystyle S} is a prefix code.
Suppose that ℓ 1⩽ℓ 2⩽⋯⩽ℓ n{\displaystyle \ell {1}\leqslant \ell {2}\leqslant \cdots \leqslant \ell {n}}. Let A{\displaystyle A} be the full r{\displaystyle r}-ary tree of depth ℓ n{\displaystyle \ell {n}} (thus, every node of A{\displaystyle A} at level <ℓ n{\displaystyle <\ell {n}} has r{\displaystyle r} children, while the nodes at level ℓ n{\displaystyle \ell {n}} are leaves). Every word of length ℓ⩽ℓ n{\displaystyle \ell \leqslant \ell {n}} over an r{\displaystyle r}-ary alphabet corresponds to a node in this tree at depth ℓ{\displaystyle \ell }. The i{\displaystyle i}th word in the prefix code corresponds to a node v i{\displaystyle v{i}}; let A i{\displaystyle A_{i}} be the set of all leaf nodes (i.e. of nodes at depth ℓ n{\displaystyle \ell {n}}) in the subtree of A{\displaystyle A} rooted at v i{\displaystyle v{i}}. That subtree being of height ℓ n−ℓ i{\displaystyle \ell {n}-\ell {i}}, we have
|A i|=r ℓ n−ℓ i.{\displaystyle |A_{i}|=r^{\ell {n}-\ell {i}}.}
Since the code is a prefix code, those subtrees cannot share any leaves, which means that
A i∩A j=∅,i≠j.{\displaystyle A_{i}\cap A_{j}=\varnothing ,\quad i\neq j.}
Thus, given that the total number of nodes at depth ℓ n{\displaystyle \ell {n}} is r ℓ n{\displaystyle r^{\ell {n}}}, we have
|⋃i=1 n A i|=∑i=1 n|A i|=∑i=1 n r ℓ n−ℓ i⩽r ℓ n{\displaystyle \left|\bigcup {i=1}^{n}A{i}\right|=\sum {i=1}^{n}|A{i}|=\sum {i=1}^{n}r^{\ell {n}-\ell {i}}\leqslant r^{\ell {n}}}
from which the result follows.
Conversely, given any ordered sequence of n{\displaystyle n} natural numbers,
ℓ 1⩽ℓ 2⩽⋯⩽ℓ n{\displaystyle \ell {1}\leqslant \ell {2}\leqslant \cdots \leqslant \ell {n}}
satisfying the Kraft inequality, one can construct a prefix code with codeword lengths equal to each ℓ i{\displaystyle \ell {i}} by choosing a word of length ℓ i{\displaystyle \ell {i}} arbitrarily, then ruling out all words of greater length that have it as a prefix. There again, we shall interpret this in terms of leaf nodes of an r{\displaystyle r}-ary tree of depth ℓ n{\displaystyle \ell {n}}. First choose any node from the full tree at depth ℓ 1{\displaystyle \ell {1}}; it corresponds to the first word of our new code. Since we are building a prefix code, all the descendants of this node (i.e., all words that have this first word as a prefix) become unsuitable for inclusion in the code. We consider the descendants at depth ℓ n{\displaystyle \ell {n}} (i.e., the leaf nodes among the descendants); there are r ℓ n−ℓ 1{\displaystyle r^{\ell {n}-\ell {1}}} such descendant nodes that are removed from consideration. The next iteration picks a (surviving) node at depth ℓ 2{\displaystyle \ell {2}} and removes r ℓ n−ℓ 2{\displaystyle r^{\ell {n}-\ell _{2}}} further leaf nodes, and so on. After n{\displaystyle n} iterations, we have removed a total of
∑i=1 n r ℓ n−ℓ i{\displaystyle \sum {i=1}^{n}r^{\ell {n}-\ell {i}}}
nodes. The question is whether we need to remove more leaf nodes than we actually have available — r ℓ n{\displaystyle r^{\ell {n}}} in all — in the process of building the code. Since the Kraft inequality holds, we have indeed
∑i=1 n r ℓ n−ℓ i⩽r ℓ n{\displaystyle \sum {i=1}^{n}r^{\ell {n}-\ell {i}}\leqslant r^{\ell {n}}}
and thus a prefix code can be built. Note that as the choice of nodes at each step is largely arbitrary, many different suitable prefix codes can be built, in general.
Proof of the general case
[edit]
Now we will prove that the Kraft inequality holds whenever S{\displaystyle S} is a uniquely decodable code. (The converse needs not be proven, since we have already proven it for prefix codes, which is a stronger claim.) The proof is by Jack I. Karush.
We need only prove it when there are finitely many codewords. If there are infinitely many codewords, then any finite subset of it is also uniquely decodable, so it satisfies the Kraft–McMillan inequality. Taking the limit, we have the inequality for the full code.
Denote C=∑i=1 n r−l i{\displaystyle C=\sum {i=1}^{n}r^{-l{i}}}. The idea of the proof is to get an upper bound on C m{\displaystyle C^{m}} for m∈N{\displaystyle m\in \mathbb {N} } and show that it can only hold for all m{\displaystyle m} if C≤1{\displaystyle C\leq 1}. Rewrite C m{\displaystyle C^{m}} as
C m=(∑i=1 n r−l i)m=∑i 1=1 n∑i 2=1 n⋯∑i m=1 n r−(l i 1+l i 2+⋯+l i m){\displaystyle {\begin{aligned}C^{m}&=\left(\sum {i=1}^{n}r^{-l{i}}\right)^{m}\&=\sum {i{1}=1}^{n}\sum {i{2}=1}^{n}\cdots \sum {i{m}=1}^{n}r^{-\left(l_{i_{1}}+l_{i_{2}}+\cdots +l_{i_{m}}\right)}\\end{aligned}}}
Consider all m-powersS m{\displaystyle S^{m}}, in the form of words s i 1 s i 2…s i m{\displaystyle s_{i_{1}}s_{i_{2}}\dots s_{i_{m}}}, where i 1,i 2,…,i m{\displaystyle i_{1},i_{2},\dots ,i_{m}} are indices between 1 and n{\displaystyle n}. Note that, since S was assumed to uniquely decodable, s i 1 s i 2…s i m=s j 1 s j 2…s j m{\displaystyle s_{i_{1}}s_{i_{2}}\dots s_{i_{m}}=s_{j_{1}}s_{j_{2}}\dots s_{j_{m}}} implies i 1=j 1,i 2=j 2,…,i m=j m{\displaystyle i_{1}=j_{1},i_{2}=j_{2},\dots ,i_{m}=j_{m}}. This means that each summand corresponds to exactly one word in S m{\displaystyle S^{m}}. This allows us to rewrite the equation to
C m=∑ℓ=1 m⋅ℓ m a x q ℓ r−ℓ{\displaystyle C^{m}=\sum {\ell =1}^{m\cdot \ell {max}}q_{\ell }\,r^{-\ell }}
where q ℓ{\displaystyle q_{\ell }} is the number of codewords in S m{\displaystyle S^{m}} of length ℓ{\displaystyle \ell } and ℓ m a x{\displaystyle \ell {max}} is the length of the longest codeword in S{\displaystyle S}. For an r{\displaystyle r}-letter alphabet there are only r ℓ{\displaystyle r^{\ell }} possible words of length ℓ{\displaystyle \ell }, so q ℓ≤r ℓ{\displaystyle q{\ell }\leq r^{\ell }}. Using this, we upper bound C m{\displaystyle C^{m}}:
C m=∑ℓ=1 m⋅ℓ m a x q ℓ r−ℓ≤∑ℓ=1 m⋅ℓ m a x r ℓ r−ℓ=m⋅ℓ m a x{\displaystyle {\begin{aligned}C^{m}&=\sum {\ell =1}^{m\cdot \ell {max}}q_{\ell }\,r^{-\ell }\&\leq \sum {\ell =1}^{m\cdot \ell {max}}r^{\ell }\,r^{-\ell }=m\cdot \ell _{max}\end{aligned}}}
Taking the m{\displaystyle m}-th root, we get
C=∑i=1 n r−l i≤(m⋅ℓ m a x)1 m{\displaystyle C=\sum {i=1}^{n}r^{-l{i}}\leq \left(m\cdot \ell {max}\right)^{\frac {1}{m}}}
This bound holds for any m∈N{\displaystyle m\in \mathbb {N} }. The right side is 1 asymptotically, so ∑i=1 n r−l i≤1{\displaystyle \sum {i=1}^{n}r^{-l_{i}}\leq 1} must hold (otherwise the inequality would be broken for a large enough m{\displaystyle m}).
Alternative construction for the converse
[edit]
Given a sequence of n{\displaystyle n} natural numbers,
ℓ 1⩽ℓ 2⩽⋯⩽ℓ n{\displaystyle \ell {1}\leqslant \ell {2}\leqslant \cdots \leqslant \ell {n}}
satisfying the Kraft inequality, we can construct a prefix code as follows. Define the _i th codeword, C i, to be the first ℓ i{\displaystyle \ell {i}} digits after the radix point (e.g. decimal point) in the base _r representation of
∑j=1 i−1 r−ℓ j.{\displaystyle \sum {j=1}^{i-1}r^{-\ell {j}}.}
Note that by Kraft's inequality, this sum is never more than 1. Hence the codewords capture the entire value of the sum. Therefore, for j>i, the first ℓ i{\displaystyle \ell {i}} digits of _C j form a larger number than C i, so the code is prefix free.
Generalizations
[edit]
The following generalization is found in.
Theorem—If C,D{\textstyle C,D} are uniquely decodable, and every codeword in C{\textstyle C} is a concatenation of codewords in D{\textstyle D}, then ∑c∈C r−|c|≤∑c∈D r−|c|{\displaystyle \sum {c\in C}r^{-|c|}\leq \sum {c\in D}r^{-|c|}}
The previous theorem is the special case when D={a 1,…,a r}{\displaystyle D={a_{1},\dots ,a_{r}}}.
Proof
Let Q C(x){\textstyle Q_{C}(x)} be the generating function for the code. That is, Q C(x):=∑c∈C x|c|{\displaystyle Q_{C}(x):=\sum _{c\in C}x^{|c|}}
By a counting argument, the k{\textstyle k}-th coefficient of Q C n{\textstyle Q_{C}^{n}} is the number of strings of length n{\textstyle n} with code length k{\textstyle k}. That is, Q C n(x)=∑k≥0 x k#(strings of length n with C-codes of length k){\displaystyle Q_{C}^{n}(x)=\sum _{k\geq 0}x^{k}#({\text{strings of length }}n{\text{ with }}C{\text{-codes of length }}k)} Similarly,
1 1−Q C(x)=1+Q C(x)+Q C(x)2+⋯=∑k≥0 x k#(strings with C-codes of length k){\displaystyle {\frac {1}{1-Q_{C}(x)}}=1+Q_{C}(x)+Q_{C}(x)^{2}+\cdots =\sum _{k\geq 0}x^{k}#({\text{strings with }}C{\text{-codes of length }}k)}
Since the code is uniquely decodable, any power of Q C{\textstyle Q_{C}} is absolutely bounded by r|x|+r 2|x|2+⋯=r|x|1−r|x|{\textstyle r|x|+r^{2}|x|^{2}+\cdots ={\frac {r|x|}{1-r|x|}}}, so each of Q C,Q C 2,…{\textstyle Q_{C},Q_{C}^{2},\dots } and 1 1−Q C(x){\textstyle {\frac {1}{1-Q_{C}(x)}}} is analytic in the disk |x|<1/r{\textstyle |x|<1/r}.
We claim that for all x∈(0,1/r){\textstyle x\in (0,1/r)}, Q C n≤Q D n+Q D n+1+⋯{\displaystyle Q_{C}^{n}\leq Q_{D}^{n}+Q_{D}^{n+1}+\cdots }
The left side is ∑k≥0 x k#(strings of length n with C-codes of length k){\displaystyle \sum _{k\geq 0}x^{k}#({\text{strings of length }}n{\text{ with }}C{\text{-codes of length }}k)} and the right side is
∑k≥0 x k#(strings of length≥n with D-codes of length k){\displaystyle \sum _{k\geq 0}x^{k}#({\text{strings of length}}\geq n{\text{ with }}D{\text{-codes of length }}k)}
Now, since every codeword in C{\textstyle C} is a concatenation of codewords in D{\textstyle D}, and D{\textstyle D} is uniquely decodable, each string of length n{\textstyle n} with C{\textstyle C}-code c 1…c n{\textstyle c_{1}\dots c_{n}} of length k{\textstyle k} corresponds to a unique string s c 1…s c n{\textstyle s_{c_{1}}\dots s_{c_{n}}} whose D{\textstyle D}-code is c 1…c n{\textstyle c_{1}\dots c_{n}}. The string has length at least n{\textstyle n}.
Therefore, the coefficients on the left are less or equal to the coefficients on the right.
Thus, for all x∈(0,1/r){\textstyle x\in (0,1/r)}, and all n=1,2,…{\textstyle n=1,2,\dots }, we have Q C≤Q D(1−Q D)1/n{\displaystyle Q_{C}\leq {\frac {Q_{D}}{(1-Q_{D})^{1/n}}}} Taking n→∞{\textstyle n\to \infty } limit, we have Q C(x)≤Q D(x){\textstyle Q_{C}(x)\leq Q_{D}(x)} for all x∈(0,1/r){\textstyle x\in (0,1/r)}.
Since Q C(1/r){\textstyle Q_{C}(1/r)} and Q D(1/r){\textstyle Q_{D}(1/r)} both converge, we have Q C(1/r)≤Q D(1/r){\textstyle Q_{C}(1/r)\leq Q_{D}(1/r)} by taking the limit and applying Abel's theorem.
There is a generalization to quantum code.
Notes
[edit]
^Cover, Thomas M.; Thomas, Joy A. (2006), "Data Compression", Elements of Information Theory (2nd ed.), John Wiley & Sons, Inc, pp.108–109, doi:10.1002/047174882X.ch5, ISBN978-0-471-24195-9
^De Rooij, Steven; Grünwald, Peter D. (2011), "LUCKINESS AND REGRET IN MINIMUM DESCRIPTION LENGTH INFERENCE", Philosophy of Statistics (1st ed.), Elsevier, p.875, ISBN978-0-080-93096-1
^Karush, J. (April 1961). "A simple proof of an inequality of McMillan (Corresp.)". IEEE Transactions on Information Theory. 7 (2): 118. doi:10.1109/TIT.1961.1057625. ISSN0018-9448.
^Cover, Thomas M.; Thomas, Joy A. (2006). Elements of information theory (2nd ed.). Hoboken, N.J: Wiley-Interscience. ISBN978-0-471-24195-9.
^Foldes, Stephan (2008-06-21). "On McMillan's theorem about uniquely decipherable codes". arXiv:0806.3277 [math.CO].
^Schumacher, Benjamin; Westmoreland, Michael D. (2001-09-10). "Indeterminate-length quantum coding". Physical Review A. 64 (4) 042304. arXiv:quant-ph/0011014. Bibcode:2001PhRvA..64d2304S. doi:10.1103/PhysRevA.64.042304. S2CID53488312.
References
[edit]
Kraft, Leon G. (1949), A device for quantizing, grouping, and coding amplitude modulated pulses (Thesis), Cambridge, MA: MS Thesis, Electrical Engineering Department, Massachusetts Institute of Technology, hdl:1721.1/12390.
McMillan, Brockway (1956), "Two inequalities implied by unique decipherability", IEEE Trans. Inf. Theory, 2 (4): 115–116, doi:10.1109/TIT.1956.1056818.
See also
[edit]
Chaitin's constant
Canonical Huffman code
Retrieved from "
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Inequalities (mathematics)
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Kraft–McMillan inequality
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10149 | https://www.gauthmath.com/solution/1823638091213894/22-En-el-gr-fico-mostrado-las-esferas-reali-zan-un-MRU-se-mueven-en-v-as-paralel | Question
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10150 | https://www.geeksforgeeks.org/r-language/pearson-correlation-testing-in-r-programming/ | Pearson Correlation Testing in R Programming - GeeksforGeeks
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Pearson Correlation Testing in R Programming
Last Updated : 06 Aug, 2025
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Pearson correlation is a parametric statistical method used to measure the linear relationship between two continuous variables. It indicates both the strength and direction of the relationship and returns a value between -1 and +1. In R Programming Language it is used to analyze the association between two normally distributed variables.
There are mainly two types of correlation:
Parametric Correlation: It measures a linear dependence between two variables (x and y) is known as a parametric correlation test because it depends on the distribution of the data.
Non-Parametric Correlation:They are rank-based correlation coefficients and are known as non-parametric correlation.
Pearson Correlation Formula:
r=Σ(x–m x)(y–m y)Σ(x–m x)2 Σ(y–m y)2\displaystyle r = \frac { \Sigma(x – m_x)(y – m_y) }{\sqrt{\Sigma(x – m_x)^2 \Sigma(y – m_y)^2}} r=Σ(x–m x)2 Σ(y–m y)2Σ(x–m x)(y–m y)
Parameters:
r r r: pearson correlation coefficient
x x xandy y y: two vectors of length n
m x m_x m xandm y m_y m y: corresponds to the means of x and y, respectively.
Implementation of Pearson Correlation Testing
We implement Pearson correlation testing in R using two primary functions:
1. Calculating the Correlation Coefficient Using cor()
We calculate the Pearson correlation coefficient between two numeric vectors using the cor() function.
cor: Computes the correlation coefficient between two numeric vectors.
x, y: Input numeric vectors of the same length.
method: Specifies the correlation method to be used (here, it is "pearson").
cat: Used to concatenate and print values.
R
r
x = c(1, 2, 3, 4, 5, 6, 7)
y = c(1, 3, 6, 2, 7, 4, 5)
result = cor(x, y, method = "pearson")
cat("Pearson correlation coefficient is:", result)
Output:
Pearson correlation coefficient is: 0.5357143
2. Performing Correlation Test Using cor.test()
We perform the Pearson correlation test which returns the coefficient, p-value and confidence interval.
cor.test: Performs a test of association between paired samples.
t: Test statistic used to calculate the p-value.
p-value: Indicates the probability of observing the data under the null hypothesis.
alternative hypothesis: States the direction of the correlation (not equal to zero by default).
sample estimates: Returns the computed correlation coefficient.
R
r
x = c(1, 2, 3, 4, 5, 6, 7)
y = c(1, 3, 6, 2, 7, 4, 5)
result = cor.test(x, y, method = "pearson")
print(result)
Output:
Output
In the output above:
T is the value of the test statistic (T = 1.4186)
p-value is the significance level of the test statistic (p-value = 0.2152).
alternative hypothesis is a character string describing the alternative hypothesis (true correlation is not equal to 0).
sample estimates is the correlation coefficient. For Pearson correlation coefficient it’s named as cor (Cor.coeff = 0.5357).
Implementation for Statistical Significance
We test the statistical significance of correlations using the rcorr function and visualize relationships using ggplot2.
1. Installing and Loading Required Packages
We first install and then load the required packages. We use the built-in mtcars dataset.
install.packages: Installs external packages
library: Loads the installed packages
data: Loads datasets
R
```r
install.packages("ggplot2")
install.packages("Hmisc")
install.packages("corrplot")
library(ggplot2)
library(Hmisc)
library(corrplot)
data("mtcars")
```
2. Pearson Correlation Testing
We use the rcorr function to calculate Pearson correlation and p-values. It requires data in matrix form.
rcorr: Calculates Pearson correlation and significance
as.matrix: Converts data frame to matrix
cor_test$r: Correlation coefficients
cor_test$P: P-values for significance
R
r
cor_test <- rcorr(as.matrix(mtcars[, c("mpg", "wt", "hp", "disp")]), type = "pearson")
cor_test$r
cor_test$P
Output:
Output
3. Scatter Plot with Regression Line
We use ggplot2 to show the correlation between two variables with a regression line.
ggplot: Starts the plot
aes: Sets axes
geom_point: Plots data points
geom_smooth: Adds regression line
labs: Adds title and labels
theme_minimal: Applies a clean theme
R
r
ggplot(mtcars, aes(x = wt, y = mpg)) +
geom_point(color = "blue", size = 2) +
geom_smooth(method = "lm", color = "red", se = FALSE) +
labs(title = "Scatter Plot with Pearson Correlation",
x = "Weight (wt)", y = "Miles Per Gallon (mpg)") +
theme_minimal()
Output:
Output
The scatter plot shows a strong negative correlation between weight and mileage, where heavier cars tend to have lower miles per gallon, as indicated by the downward-sloping red regression line
Pearson Correlation Testing in R Programming
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10151 | https://www.alevelh2chemistry.com/a-level-h2-chemistry-common-errors-in-calculating-empirical-formula/ | Skip to primary navigation
A-Level H2 Chemistry Tuition by 10 Year Series Author
Advanced Chemistry Made Easy
You are here: Home / Mole Calculations / Common Errors in Calculating Empirical Formula
Common Errors in Calculating Empirical Formula
By Sean Chua 3 Comments
One of the most common errors in have observed for both GCE A-Level H2 Chemistry (as well as GCE O-Level Chemistry) students is when they are Calculating Empirical Formulae from Composition by Mass.
Take the following question that was given to my A-Level H2 Chemistry Weekly Tuition Class:
Question:
Calculate the empirical formula of a compound that has the composition: 48.8% carbon, 13.5% hydrogen and 37.7% nitrogen.
After calculations, one of my students came up confidently with the answer of C2H5N.
– which is the Incorrect Answer.
After asking her to present her working to the class, i we realised that she made a mistake when trying to round off final numbers.
Let’s take a look at my suggested answer & then see how she made the mistake – which is a Common Error for many Chemistry students year-after-year.
Suggested Answer:
| | | | |
--- --- |
| Element | C | H | N |
| Mass | 48.8 | 13.5 | 37.7 |
| Ar | 12 | 1 | 14 |
| Mole | 48.8 / 12 = 4.07 | 13.5 / 1 = 13.5 | 37.7 / 14 = 2.69 |
| Molar Ratio | 4.07 / 2.69 = 1.51 | 13.5 / 2.69 = 5.02 | 2.69 / 2.69 = 1 |
| Simplest Ratio(x 2) | 3 | 10 | 2 |
As such, my suggested answer for the empirical formulae will be C3H10N2
However, my student insists argues that when we get the Molar Ratio of 1.51 – it is more than the half way mark of 1.5 and we should round it up to 2. As such, her answer will be C2H5N. Do note that this is Incorrect!
The correct strategy is to try to get rid of the fraction (1.51 ~ 3/2) and in this case, multiplying throughout by a factor of X2 will solve the problem and give us the correct empirical formula of C3H10N2.
Hope the above helps you in clarifying some of your doubts.
PS: Let me know how you find about this post on Common Errors in Chemistry. I would love to hear from you. =)
Related Articles:
Determining Molecular Formula from Empirical Formula
Key Definitions & Formulae in Atoms, Molecules and Stoichiometry
Organic Chemistry: Types of Formulae
Ideal Gas Law: Exam-Based Question on pV = nRT
Calculating Relative Atomic Mass from Isotopic Abundance
Reader Interactions
Comments
jauhar zubair kisambira says
I have also done the same mistake like one of your students as i was trying it out but i hope i will rectify it follwing your suggested answer.However why are you multiplying it the the simplest ratio by two if in our lessons in o leval we were told to round off to the nearest whole number ?……………….lastly but not least, i thank you for your suvice and please provide more if possible.
2. zain says
i got the right answer. thankx for the post.
3. Sean says
When the no. of moles is calculated to be close to an integer (full number), then we can usually round it up or down…
E.g. 1.97 you can safely say it’s 2.E.g. 0.986 you can safely say it’s 1.
because the error is very small.
In fact, even in my GCE O-Level Pure Chemistry Tuition Classes, we teach them the proper way. i.e. any calculated value is not near to an integer, we must do the above.
Hope this is clear to everyone.
PS: If you find this useful, feel free to share it with your friends.
Leave a Reply |
10152 | https://pressbooks.bccampus.ca/introductorygeneralphysics2phys1207/chapter/22-5-force-on-a-moving-charge-in-a-magnetic-field-examples-and-applications/ | Skip to content
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Chapter 6 Magnetism
6.5 Force on a Moving Charge in a Magnetic Field: Examples and Applications – Mass Spectrometers
Summary
Describe the effects of a magnetic field on a moving charge.
Calculate the radius of curvature of the path of a charge that is moving in a magnetic field.
Magnetic force can cause a charged particle to move in a circular or spiral path. Cosmic rays are energetic charged particles in outer space, some of which approach the Earth. They can be forced into spiral paths by the Earth’s magnetic field. Protons in giant accelerators are kept in a circular path by magnetic force. The bubble chamber photograph in Figure 1 shows charged particles moving in such curved paths. The curved paths of charged particles in magnetic fields are the basis of a number of phenomena and can even be used analytically, such as in a mass spectrometer.
So does the magnetic force cause circular motion? Magnetic force is always perpendicular to velocity, so that it does no work on the charged particle. The particle’s kinetic energy and speed thus remain constant. The direction of motion is affected, but not the speed. This is typical of uniform circular motion. The simplest case occurs when a charged particle moves perpendicular to a uniform B-field, such as shown in Figure 2. (If this takes place in a vacuum, the magnetic field is the dominant factor determining the motion.) Here, the magnetic force supplies the centripetal force Fcentripetal = m v2 / r Noting that sin θ = 1 as the angle is 90.0 degrees, we see that F = q v B.
Figure 2. A negatively charged particle moves in the plane of the page in a region where the magnetic field is perpendicular into the page (represented by the small circles with x’s—like the tails of arrows). The magnetic force is perpendicular to the velocity, and so velocity changes in direction but not magnitude. Uniform circular motion results.
Because the magnetic force Fsupplies the centripetal force FC, we have
Solving for ryields . r = m v / q B
Here, r is the radius of curvature of the path of a charged particle with mass m and charge q, moving at a speed vperpendicular to a magnetic field of strength B. If the velocity is not perpendicular to the magnetic field, then vis the component of the velocity perpendicular to the field. The component of the velocity parallel to the field is unaffected, since the magnetic force is zero for motion parallel to the field. This produces a spiral motion rather than a circular one.
Example 1: Calculating the Curvature of the Path of an Electron Moving in a Magnetic Field: A Magnet on a TV Screen
A magnet brought near an old-fashioned TV screen such as in Figure 3 (TV sets with cathode ray tubes instead of LCD screens) severely distorts its picture by altering the path of the electrons that make its phosphors glow. (Don’t try this at home, as it will permanently magnetize and ruin the TV.) To illustrate this, calculate the radius of curvature of the path of an electron having a velocity of 6.00 x107 m/s (corresponding to the accelerating voltage of about 10.0 kV used in some TVs) perpendicular to a magnetic field of strength B = 0.500 T (obtainable with permanent magnets).
Figure 3. Side view showing what happens when a magnet comes in contact with a computer monitor or TV screen. Electrons moving toward the screen spiral about magnetic field lines, maintaining the component of their velocity parallel to the field lines. This distorts the image on the screen.
Strategy
We can find the radius of curvature r directly from the equation r = mv/qB since all other quantities in it are given or known.
Solution
Using known values for the mass and charge of an electron, along with the given values of v and B gives us
or
r = 0.683 mm
Discussion
The small radius indicates a large effect. The electrons in the TV picture tube are made to move in very tight circles, greatly altering their paths and distorting the image.
Figure 4 shows how electrons not moving perpendicular to magnetic field lines follow the field lines. The component of velocity parallel to the lines is unaffected, and so the charges spiral along the field lines. If field strength increases in the direction of motion, the field will exert a force to slow the charges, forming a kind of magnetic mirror, as shown below.
Figure 4. When a charged particle moves along a magnetic field line into a region where the field becomes stronger, the particle experiences a force that reduces the component of velocity parallel to the field. This force slows the motion along the field line and here reverses it, forming a “magnetic mirror.”
The properties of charged particles in magnetic fields are related to such different things as the Aurora Australis or Aurora Borealis and particle accelerators. Charged particles approaching magnetic field lines may get trapped in spiral orbits about the lines rather than crossing them, as seen above. Some cosmic rays, for example, follow the Earth’s magnetic field lines, entering the atmosphere near the magnetic poles and causing the southern or northern lights through their ionization of molecules in the atmosphere. This glow of energized atoms and molecules is seen in Chapter 22 Introduction to Magnetism. Those particles that approach middle latitudes must cross magnetic field lines, and many are prevented from penetrating the atmosphere. Cosmic rays are a component of background radiation; consequently, they give a higher radiation dose at the poles than at the equator.
Figure 5. Energetic electrons and protons, components of cosmic rays, from the Sun and deep outer space often follow the Earth’s magnetic field lines rather than cross them. (Recall that the Earth’s north magnetic pole is really a south pole in terms of a bar magnet.)
Some incoming charged particles become trapped in the Earth’s magnetic field, forming two belts above the atmosphere known as the Van Allen radiation belts after the discoverer James A. Van Allen, an American astrophysicist. (See Figure 6.) Particles trapped in these belts form radiation fields (similar to nuclear radiation) so intense that manned space flights avoid them and satellites with sensitive electronics are kept out of them. In the few minutes it took lunar missions to cross the Van Allen radiation belts, astronauts received radiation doses more than twice the allowed annual exposure for radiation workers. Other planets have similar belts, especially those having strong magnetic fields like Jupiter.
Figure 6. The Van Allen radiation belts are two regions in which energetic charged particles are trapped in the Earth’s magnetic field. One belt lies about 300 km above the Earth’s surface, the other about 16,000 km. Charged particles in these belts migrate along magnetic field lines and are partially reflected away from the poles by the stronger fields there. The charged particles that enter the atmosphere are replenished by the Sun and sources in deep outer space.
Back on Earth, we have devices that employ magnetic fields to contain charged particles. Among them are the giant particle accelerators that have been used to explore the substructure of matter. (See Figure 7.) Magnetic fields not only control the direction of the charged particles, they also are used to focus particles into beams and overcome the repulsion of like charges in these beams.
Figure 7. The Fermilab facility in Illinois has a large particle accelerator (the most powerful in the world until 2008) that employs magnetic fields (magnets seen here in orange) to contain and direct its beam. This and other accelerators have been in use for several decades and have allowed us to discover some of the laws underlying all matter. (credit: ammcrim, Flickr)
Thermonuclear fusion (like that occurring in the Sun) is a hope for a future clean energy source. One of the most promising devices is the tokamak, which uses magnetic fields to contain (or trap) and direct the reactive charged particles. (See Figure 8.) Less exotic, but more immediately practical, amplifiers in microwave ovens use a magnetic field to contain oscillating electrons. These oscillating electrons generate the microwaves sent into the oven.
Figure 8. Tokamaks such as the one shown in the figure are being studied with the goal of economical production of energy by nuclear fusion. Magnetic fields in the doughnut-shaped device contain and direct the reactive charged particles. (credit: David Mellis, Flickr)
Mass spectrometers have a variety of designs, and many use magnetic fields to measure mass. The curvature of a charged particle’s path in the field is related to its mass and is measured to obtain mass information. Historically, such techniques were employed in the first direct observations of electron charge and mass. Today, mass spectrometers (sometimes coupled with gas chromatographs) are used to determine the make-up and sequencing of large biological molecules.
Section Summary
Magnetic force can supply centripetal force and cause a charged particle to move in a circular path of radius
where v is the component of the velocity perpendicular to B for a charged particle with mass m and charge q.
Conceptual Questions
1: How can the motion of a charged particle be used to distinguish between a magnetic and an electric field?
2: High-velocity charged particles can damage biological cells and are a component of radiation exposure in a variety of locations ranging from research facilities to natural background. Describe how you could use a magnetic field to shield yourself.
3: If a cosmic ray proton approaches the Earth from outer space along a line toward the centre of the Earth that lies in the plane of the equator, in what direction will it be deflected by the Earth’s magnetic field? What about an electron? A neutron?
4: What are the signs of the charges on the particles in Figure 9?
Figure 9.
5: Which of the particles in Figure 10 has the greatest velocity, assuming they have identical charges and masses?
Figure 10.
6: Which of the particles in Figure 10 has the greatest mass, assuming all have identical charges and velocities?
7: While operating, a high-precision TV monitor is placed on its side during maintenance. The image on the monitor changes color and blurs slightly. Discuss the possible relation of these effects to the Earth’s magnetic field.
Problem & Exercises
1: A cosmic ray electron moves at 7.50 x 106m/s perpendicular to the Earth’s magnetic field at an altitude where field strength is 1.00 x 10-5 T. What is the radius of the circular path the electron follows?
2: A proton moves at 7.50 x107 m/s perpendicular to a magnetic field. The field causes the proton to travel in a circular path of radius 0.800 m. What is the field strength?
3: (a) Viewers of Star Trek hear of an antimatter drive on the Starship Enterprise. One possibility for such a futuristic energy source is to store antimatter charged particles in a vacuum chamber, circulating in a magnetic field, and then extract them as needed. Antimatter annihilates with normal matter, producing pure energy. What strength magnetic field is needed to hold antiprotons, moving at 5.00 x 107 m/s in a circular path 2.00 m in radius? Antiprotons have the same mass as protons but the opposite (negative) charge. (b) Is this field strength obtainable with today’s technology or is it a futuristic possibility?
4: (a) An oxygen-16 ion with a mass of 2.66 x 10-26 kg travels at 5.00 x 106 m/s perpendicular to a 1.20-T magnetic field, which makes it move in a circular arc with a 0.231-m radius. What positive charge is on the ion? (b) What is the ratio of this charge to the charge of an electron? (c) Discuss why the ratio found in (b) should be an integer.
5: What radius circular path does an electron travel if it moves at the same speed and in the same magnetic field as the proton in Exercise 2. The speed is 7.50 x107 m/s perpendicular to a magnetic field of strength 0.979 T.
6: A velocity selector in a mass spectrometer uses a 0.100-T magnetic field. (a) What electric field strength is needed to select a speed of 4.00 x 106 m/s ? (b) What is the voltage between the plates if they are separated by 1.00 cm?
7: An electron in a TV CRT moves with a speed of 6.00 x 107 m/s , in a direction perpendicular to the Earth’s field, which has a strength of 5.00 x 10-5 T. (a) What strength electric field must be applied perpendicular to the Earth’s field to make the electron moves in a straight line? (b) If this is done between plates separated by 1.00 cm, what is the voltage applied? (Note that TVs are usually surrounded by a ferromagnetic material to shield against external magnetic fields and avoid the need for such a correction.)
8: (a) At what speed will a proton move in a circular path of the same radius as the electron in Chapter 22.5 Exercise 1? (b) What would the radius of the path be if the proton had the same speed as the electron? (c) What would the radius be if the proton had the same kinetic energy as the electron? (d) The same momentum?
9: A mass spectrometer is being used to separate common oxygen-16 from the much rarer oxygen-18, taken from a sample of old glacial ice. (The relative abundance of these oxygen isotopes is related to climatic temperature at the time the ice was deposited.) The ratio of the masses of these two ions is 16 to 18, the mass of oxygen-16 is 2.66 x 10-26 kg, and they are singly charged and travel at 5.00 x 106 m/s in a 1.20-T magnetic field. What is the separation between their paths when they hit a target after traversing a semicircle? Hint – find the radius of the path for each isotope, then subtract that answer to find the separation.
10: (a) Triply charged uranium-235 and uranium-238 ions are being separated in a mass spectrometer. (The much rarer uranium-235 is used as reactor fuel.) The masses of the ions are 3.90 x 10-25 kg and 3.95 x 10-25 kg, respectively, and they travel at 3.00 x 105 m/s in a 0.250-T field. What is the separation between their paths when they hit a target after traversing a semicircle? (b) Discuss whether this distance between their paths seems to be big enough to be practical in the separation of uranium-235 from uranium-238.
Solutions
Problems & Exercises
1: 4.27 m
2: 0.979 T
3: (a) 0.261 T (b) This strength is definitely obtainable with today’s technology. Magnetic field strengths of 0.500 T are obtainable with permanent magnets.
5: 4.36 x 10-4 m
7: (a) 3.00 kV/m (b) 30.0 V
9: 0.173 m |
10153 | https://courses.lumenlearning.com/introstatscorequisite/chapter/translate-verbal-statements-to-mathematical-expressions/ | Translate Verbal Statements to Mathematical Expressions
Learning Outcomes
Express verbal statements as mathematical expressions
Translate word problems into expressions
To solve a word problem, you often need to translate a verbal expression into a mathematical expression. In the following table are some key words which correspond to common mathematical operations and relations.
| | | |
---
| Addition | + | sum, total, together, more than, increased by, … |
| Subtraction | – | difference, less than, decreased by, … |
| Multiplication | | product, double, half, percent of, … |
| Equality | = | is, is the same as, is not different from, … |
Example
| | |
--- |
| Verbal Statement | Mathematical Statement |
| The sum of a number and 5 is 12 | n+5=12 |
| A number decreased by 5 is 12 | n-5=12 |
| Half of a number is 12 | 12n=12 |
| The total of a number and 7 is the same as twice the number | n+7=2n |
In this example video, we show how to translate more words into mathematical expressions.
For more examples of how to translate verbal statements into algebraic statements, watch the following video.
Try It
Translate verbal statements involving inequalities into expressions
Often we are interested in how a quantity compares to a particular value. For example, if more than 50% of voters vote in favor of a proposal, the proposal passes and is adopted. If we let p represent the proportion of voters who will vote in favor of a proposal, we can express the outcome “the proposal passes” as “p > 0.5.” Any value of p which is more than 0.5 satisfies this inequality.
The table below shows the relationships between possible verbal statements about the relationship between a variable x and a constant k, and the corresponding mathematical statement.
| | |
--- |
| Verbal Statement: x is… | Mathematical Statement |
| greater than or equal to k at least k not more than k | x≥k |
| less than or equal to k at most k not more than k | x≤k |
| less than k below k fewer than k | [latex]x |
| greater than k more than k above k | x>k |
| equal to k is k exactly k the same as k | x=k |
| not equal to k not k different from k | x≠k |
EXAMPLE: EXPRESS A VERBAL STATEMENT INVOLVING INEQUALITY AS A MATHEMATICAL STATEMENT
Write the following as a mathematical statement, using x for the variable.
The number of tries is no more than 12.
Show Answer
Here we are asked to use x for the unknown number of tries. The statement compares x to 12. Since the comparison is “no more than,” we use the symbol ≤.
x≤12
EXAMPLE: EXPRESS A VERBAL STATEMENT INVOLVING INEQUALITY AS A MATHEMATICAL STATEMENT
Write the following as a mathematical statement, using x for the variable.
The number of hearts is fewer than 5.
Show Answer
Here we are asked to use x for the unknown number of tries. The statement compares x to 5. Since the comparison is “fewer than,” we use the symbol <.
x<5
Some verbal statements involve compound inequalities. That is, two inequalities must both be satisfied.
EXAMPLE: EXPRESS A VERBAL STATEMENT INVOLVING INEQUALITY AS A MATHEMATICAL STATEMENT
Write the following as a mathematical statement, using x for the variable.
The number of defects is at least 1, but no more than 5.
Show Answer
Here we are asked to use x for the number of defects. The statement compares x to both 1 and 5.
Since the first comparison is “at least,” we use ≥. So we require x≥1.
Since the second comparison is “no more than,” we use ≤. So we require x≤5.
In order for both comparisons to be true, we need x≥1 and x≤5. Since we can interchange the expressions on each side of the inequality as long as we reverse the inequality, x≥1 can also be written 1≤x. We can write a single mathematical statement which describes the values of x which satisfy both conditions.
1≤x≤5
Candela Citations
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Write Algebraic Expressions from Statements: Form ax+b and a(x+b). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution
Write Algebraic Expressions. Provided by: James Sousa (Mathispower4u.com). Located at: License: CC BY: Attribution
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CC licensed content, Original
Write Algebraic Expressions from Statements: Form ax+b and a(x+b). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: License: CC BY: Attribution
Write Algebraic Expressions. Provided by: James Sousa (Mathispower4u.com). Located at: License: CC BY: Attribution
Provided by: Lumen Learning. License: CC BY: Attribution
CC licensed content, Shared previously
Question ID 142722. Authored by: Lumen Learning. License: CC BY: Attribution. License Terms: IMathAS Community License, CC-BY + GPL
CC licensed content, Specific attribution
Prealgebra. Provided by: OpenStax. Located at: License: CC BY: Attribution. License Terms: Access for free at |
10154 | https://periodictable.com/Elements/053/data.html | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | | Technical data for IodineClick any property name to see plots of that property for all the elements. | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | | | | --- | | Overview | | | Name | Iodine | | Symbol | I | | Atomic Number | 53 | | Atomic Weight | 126.90447 | | Density | 4.94 g/cm3 | | Melting Point | 113.7 °C | | Boiling Point | 184.3 °C | | Thermal properties | | | Phase | Solid | | Melting Point | 113.7 °C | | Boiling Point | 184.3 °C | | Absolute Melting Point | 386.85 K | | Absolute Boiling Point | 457.4 K | | Critical Pressure | 11.7 MPa (115.5 Atm) | | Critical Temperature | 819 K | | Heat of Fusion | 7.76 kJ/mol | | Heat of Vaporization | 20.9 kJ/mol | | Specific Heat | 429 J/(kg K)[note] | | Adiabatic Index | N/A | | Neel Point | N/A | | Thermal Conductivity | 0.449 W/(m K) | | Thermal Expansion | N/A | | Bulk physical properties | | | Density | 4.94 g/cm3 | | Density (Liquid) | N/A | | Molar Volume | 0.000025689 | | Brinell Hardness | N/A | | Mohs Hardness | N/A | | Vickers Hardness | N/A | | Bulk Modulus | 7.7 GPa | | Shear Modulus | N/A | | Young Modulus | N/A | | Poisson Ratio | N/A | | Refractive Index | N/A | | Speed of Sound | N/A | | Thermal Conductivity | 0.449 W/(m K) | | Thermal Expansion | N/A | | Reactivity | | | Valence | 7 | | Electronegativity | 2.66 | | ElectronAffinity | 295.2 kJ/mol | | Ionization Energies | | | | 1008.4, 1845.9, 3180 kJ/mol | | | Health and Safety | | | DOT Hazard Class | | | | 8 | | | DOT Numbers | 1759 | | RTECS Number | | | | RTECSNN1575000 | | | NFPA Label | | | | | | --- | | Classifications | | | Alternate Names | None | | Names of Allotropes | Diiodine | | Block | p | | Group | 17 | | Period | 5 | | Series | Halogen | | Electron Configuration | [Kr]5s24d105p5 | | Color | SlateGray | | Discovery | | | | 1811 in France | | | Gas phase | N/A | | CAS Number | CAS7553-56-2 | | CID Number | | | | CID807 | | | RTECS Number | | | | RTECSNN1575000 | | | Electrical properties | | | Electrical Type | Insulator | | Electrical Conductivity | 1×10-7 S/m | | Resistivity | 1×107 m Ω | | Superconducting Point | N/A | | Magnetic properties | | | Magnetic Type | Diamagnetic | | Curie Point | N/A | | Mass Magnetic Susceptibility | -4.5×10-9 m3/Kg | | Molar Magnetic Susceptibility | -1.14×10-9 m3/mol | | Volume Magnetic Susceptibility | -0.0000222 | | Abundances | | | % in Universe | 1×10-7% | | % in Sun | N/A | | % in Meteorites | 0.000025% | | % in Earth's Crust | 0.000049% | | % in Oceans | 6×10-6% | | % in Humans | 0.00002% | | Atomic dimensions and structure | | | Atomic Radius | 115 pm | | Covalent Radius | 139 pm | | Van der Waals Radius | 198 pm | | Crystal Structure | Base Orthorhombic | | Lattice Angles | | | | π/2, π/2, π/2 | | | Lattice Constants | | | | 718.02, 471.02, 981.03 pm | | | Space Group Name | Cmca | | Space Group Number | 64 | | Nuclear Properties | | | Half-Life | Stable | | Lifetime | Stable | | Decay Mode | N/A | | Quantum Numbers | 2P3/2 | | Neutron Cross Section | 6.3 | | Neutron Mass Absorption | 0.0018 | | Known Isotopes | | | | 108I, 109I, 110I, 111I, 112I, 113I, 114I, 115I, 116I, 117I, 118I, 119I, 120I, 121I, 122I, 123I, 124I, 125I, 126I, 127I, 128I, 129I, 130I, 131I, 132I, 133I, 134I, 135I, 136I, 137I, 138I, 139I, 140I, 141I, 142I, 143I, 144I | | | Stable Isotopes | | | | 127I | | | Isotopic Abundances | | | | --- | | 127I | 100% | | | | Notes on the properties of Iodine: Specific Heat: Value given for solid phase. Up to date, curated data provided byMathematica's ElementData function from Wolfram Research, Inc. |
| | | | | --- | | | Click here to buy a book, photographic periodic table poster, card deck, or 3D print based on the images you see here! | |
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10155 | https://www.reddit.com/r/askscience/comments/9oethx/why_does_rna_have_uracil_instead_of_thymine/ | Why does RNA have Uracil instead of Thymine? : r/askscience
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Antoni2000
Why does RNA have Uracil instead of Thymine?
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10156 | https://courses.lumenlearning.com/suny-ap2/chapter/cardiac-muscle-and-electrical-activity/ | Cardiac Muscle and Electrical Activity | Anatomy and Physiology II
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Anatomy and Physiology II
Module 3: The Cardiovascular System: The Heart
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Cardiac Muscle and Electrical Activity
Learning Objectives
By the end of this section, you will be able to:
Describe the structure of cardiac muscle
Identify and describe the components of the conducting system that distributes electrical impulses through the heart
Compare the effect of ion movement on membrane potential of cardiac conductive and contractile cells
Relate characteristics of an electrocardiogram to events in the cardiac cycle
Identify blocks that can interrupt the cardiac cycle
Recall that cardiac muscle shares a few characteristics with both skeletal muscle and smooth muscle, but it has some unique properties of its own. Not the least of these exceptional properties is its ability to initiate an electrical potential at a fixed rate that spreads rapidly from cell to cell to trigger the contractile mechanism. This property is known as autorhythmicity. Neither smooth nor skeletal muscle can do this. Even though cardiac muscle has autorhythmicity, heart rate is modulated by the endocrine and nervous systems.
There are two major types of cardiac muscle cells: myocardial contractile cells and myocardial conducting cells. The myocardial contractile cells constitute the bulk (99 percent) of the cells in the atria and ventricles. Contractile cells conduct impulses and are responsible for contractions that pump blood through the body. The myocardial conducting cells (1 percent of the cells) form the conduction system of the heart. Except for Purkinje cells, they are generally much smaller than the contractile cells and have few of the myofibrils or filaments needed for contraction. Their function is similar in many respects to neurons, although they are specialized muscle cells. Myocardial conduction cells initiate and propagate the action potential (the electrical impulse) that travels throughout the heart and triggers the contractions that propel the blood.
Structure of Cardiac Muscle
Compared to the giant cylinders of skeletal muscle, cardiac muscle cells, or cardiomyocytes, are considerably shorter with much smaller diameters. Cardiac muscle also demonstrates striations, the alternating pattern of dark A bands and light I bands attributed to the precise arrangement of the myofilaments and fibrils that are organized in sarcomeres along the length of the cell. These contractile elements are virtually identical to skeletal muscle. T (transverse) tubules penetrate from the surface plasma membrane, the sarcolemma, to the interior of the cell, allowing the electrical impulse to reach the interior. The T tubules are only found at the Z discs, whereas in skeletal muscle, they are found at the junction of the A and I bands. Therefore, there are one-half as many T tubules in cardiac muscle as in skeletal muscle. In addition, the sarcoplasmic reticulum stores few calcium ions, so most of the calcium ions must come from outside the cells. The result is a slower onset of contraction. Mitochondria are plentiful, providing energy for the contractions of the heart. Typically, cardiomyocytes have a single, central nucleus, but two or more nuclei may be found in some cells.
Cardiac muscle cells branch freely. A junction between two adjoining cells is marked by a critical structure called an intercalated disc, which helps support the synchronized contraction of the muscle. The sarcolemmas from adjacent cells bind together at the intercalated discs. They consist of desmosomes, specialized linking proteoglycans, tight junctions, and large numbers of gap junctions that allow the passage of ions between the cells and help to synchronize the contraction. Intercellular connective tissue also helps to bind the cells together. The importance of strongly binding these cells together is necessitated by the forces exerted by contraction.
Figure 1. (a) Cardiac muscle cells have myofibrils composed of myofilaments arranged in sarcomeres, T tubules to transmit the impulse from the sarcolemma to the interior of the cell, numerous mitochondria for energy, and intercalated discs that are found at the junction of different cardiac muscle cells. (b) A photomicrograph of cardiac muscle cells shows the nuclei and intercalated discs. (c) An intercalated disc connects cardiac muscle cells and consists of desmosomes and gap junctions. LM × 1600. (Micrograph provided by the Regents of the University of Michigan Medical School © 2012)
Cardiac muscle undergoes aerobic respiration patterns, primarily metabolizing lipids and carbohydrates. Myoglobin, lipids, and glycogen are all stored within the cytoplasm. Cardiac muscle cells undergo twitch-type contractions with long refractory periods followed by brief relaxation periods. The relaxation is essential so the heart can fill with blood for the next cycle. The refractory period is very long to prevent the possibility of tetany, a condition in which muscle remains involuntarily contracted. In the heart, tetany is not compatible with life, since it would prevent the heart from pumping blood.
Everyday Connection:Repairand Replacement
Damaged cardiac muscle cells have extremely limited abilities to repair themselves or to replace dead cells via mitosis. Recent evidence indicates that at least some stem cells remain within the heart that continue to divide and at least potentially replace these dead cells. However, newly formed or repaired cells are rarely as functional as the original cells, and cardiac function is reduced. In the event of a heart attack or MI, dead cells are often replaced by patches of scar tissue. Autopsies performed on individuals who had successfully received heart transplants show some proliferation of original cells. If researchers can unlock the mechanism that generates new cells and restore full mitotic capabilities to heart muscle, the prognosis for heart attack survivors will be greatly enhanced. To date, myocardial cells produced within the patient (in situ) by cardiac stem cells seem to be nonfunctional, although those grown in Petri dishes (in vitro) do beat. Perhaps soon this mystery will be solved, and new advances in treatment will be commonplace.
Conduction System of the Heart
If embryonic heart cells are separated into a Petri dish and kept alive, each is capable of generating its own electrical impulse followed by contraction. When two independently beating embryonic cardiac muscle cells are placed together, the cell with the higher inherent rate sets the pace, and the impulse spreads from the faster to the slower cell to trigger a contraction. As more cells are joined together, the fastest cell continues to assume control of the rate. A fully developed adult heart maintains the capability of generating its own electrical impulse, triggered by the fastest cells, as part of the cardiac conduction system. The components of the cardiac conduction system include the sinoatrial node, the atrioventricular node, the atrioventricular bundle, the atrioventricular bundle branches, and the Purkinje cells.
Figure 2. Specialized conducting components of the heart include the sinoatrial node, the internodal pathways, the atrioventricular node, the atrioventricular bundle, the right and left bundle branches, and the Purkinje fibers.
Sinoatrial (SA) Node
Normal cardiac rhythm is established by the sinoatrial (SA) node, a specialized clump of myocardial conducting cells located in the superior and posterior walls of the right atrium in close proximity to the orifice of the superior vena cava. The SA node has the highest inherent rate of depolarization and is known as the pacemaker of the heart. It initiates the sinus rhythm, or normal electrical pattern followed by contraction of the heart.
This impulse spreads from its initiation in the SA node throughout the atria through specialized internodal pathways, to the atrial myocardial contractile cells and the atrioventricular node. The internodal pathways consist of three bands (anterior, middle, and posterior) that lead directly from the SA node to the next node in the conduction system, the atrioventricular node. The impulse takes approximately 50 ms (milliseconds) to travel between these two nodes. The relative importance of this pathway has been debated since the impulse would reach the atrioventricular node simply following the cell-by-cell pathway through the contractile cells of the myocardium in the atria. In addition, there is a specialized pathway called Bachmann’s bundle or the interatrial band that conducts the impulse directly from the right atrium to the left atrium. Regardless of the pathway, as the impulse reaches the atrioventricular septum, the connective tissue of the cardiac skeleton prevents the impulse from spreading into the myocardial cells in the ventricles except at the atrioventricular node. Figure 3 illustrates the initiation of the impulse in the SA node that then spreads the impulse throughout the atria to the atrioventricular node.
Figure 3. (1) The sinoatrial (SA) node and the remainder of the conduction system are at rest. (2) The SA node initiates the action potential, which sweeps across the atria. (3) After reaching the atrioventricular node, there is a delay of approximately 100 ms that allows the atria to complete pumping blood before the impulse is transmitted to the atrioventricular bundle. (4) Following the delay, the impulse travels through the atrioventricular bundle and bundle branches to the Purkinje fibers, and also reaches the right papillary muscle via the moderator band. (5) The impulse spreads to the contractile fibers of the ventricle. (6) Ventricular contraction begins.
The electrical event, the wave of depolarization, is the trigger for muscular contraction. The wave of depolarization begins in the right atrium, and the impulse spreads across the superior portions of both atria and then down through the contractile cells. The contractile cells then begin contraction from the superior to the inferior portions of the atria, efficiently pumping blood into the ventricles.
Atrioventricular (AV) Node
The atrioventricular (AV) node is a second clump of specialized myocardial conductive cells, located in the inferior portion of the right atrium within the atrioventricular septum. The septum prevents the impulse from spreading directly to the ventricles without passing through the AV node. There is a critical pause before the AV node depolarizes and transmits the impulse to the atrioventricular bundle (see image above, step 3). This delay in transmission is partially attributable to the small diameter of the cells of the node, which slow the impulse. Also, conduction between nodal cells is less efficient than between conducting cells. These factors mean that it takes the impulse approximately 100 ms to pass through the node. This pause is critical to heart function, as it allows the atrial cardiomyocytes to complete their contraction that pumps blood into the ventricles before the impulse is transmitted to the cells of the ventricle itself. With extreme stimulation by the SA node, the AV node can transmit impulses maximally at 220 per minute. This establishes the typical maximum heart rate in a healthy young individual. Damaged hearts or those stimulated by drugs can contract at higher rates, but at these rates, the heart can no longer effectively pump blood.
Atrioventricular Bundle (Bundle of His), Bundle Branches, and Purkinje Fibers
Arising from the AV node, the atrioventricular bundle, or bundle of His, proceeds through the interventricular septum before dividing into two atrioventricular bundle branches, commonly called the left and right bundle branches. The left bundle branch has two fascicles. The left bundle branch supplies the left ventricle, and the right bundle branch the right ventricle. Since the left ventricle is much larger than the right, the left bundle branch is also considerably larger than the right. Portions of the right bundle branch are found in the moderator band and supply the right papillary muscles. Because of this connection, each papillary muscle receives the impulse at approximately the same time, so they begin to contract simultaneously just prior to the remainder of the myocardial contractile cells of the ventricles. This is believed to allow tension to develop on the chordae tendineae prior to right ventricular contraction. There is no corresponding moderator band on the left. Both bundle branches descend and reach the apex of the heart where they connect with the Purkinje fibers (see image above, step 4). This passage takes approximately 25 ms.
The Purkinje fibers are additional myocardial conductive fibers that spread the impulse to the myocardial contractile cells in the ventricles. They extend throughout the myocardium from the apex of the heart toward the atrioventricular septum and the base of the heart. The Purkinje fibers have a fast inherent conduction rate, and the electrical impulse reaches all of the ventricular muscle cells in about 75 ms (see image above, step 5). Since the electrical stimulus begins at the apex, the contraction also begins at the apex and travels toward the base of the heart, similar to squeezing a tube of toothpaste from the bottom. This allows the blood to be pumped out of the ventricles and into the aorta and pulmonary trunk. The total time elapsed from the initiation of the impulse in the SA node until depolarization of the ventricles is approximately 225 ms.
Membrane Potentials and Ion Movement in Cardiac Conductive Cells
Action potentials are considerably different between cardiac conductive cells and cardiac contractive cells. While Na+ and K+ play essential roles, Ca 2+ is also critical for both types of cells. Unlike skeletal muscles and neurons, cardiac conductive cells do not have a stable resting potential. Conductive cells contain a series of sodium ion channels that allow a normal and slow influx of sodium ions that causes the membrane potential to rise slowly from an initial value of −60 mV up to about –40 mV. The resulting movement of sodium ions creates spontaneous depolarization (or prepotential depolarization). At this point, calcium ion channels open and Ca 2+ enters the cell, further depolarizing it at a more rapid rate until it reaches a value of approximately +5 mV. At this point, the calcium ion channels close and K+ channels open, allowing outflux of K+ and resulting in repolarization. When the membrane potential reaches approximately −60 mV, the K+ channels close and Na+ channels open, and the prepotential phase begins again. This phenomenon explains the autorhythmicity properties of cardiac muscle (Figure 4).
Figure 4. The prepotential is due to a slow influx of sodium ions until the threshold is reached followed by a rapid depolarization and repolarization. The prepotential accounts for the membrane reaching threshold and initiates the spontaneous depolarization and contraction of the cell. Note the lack of a resting potential.
Membrane Potentials and Ion Movement in Cardiac Contractile Cells
There is a distinctly different electrical pattern involving the contractile cells. In this case, there is a rapid depolarization, followed by a plateau phase and then repolarization. This phenomenon accounts for the long refractory periods required for the cardiac muscle cells to pump blood effectively before they are capable of firing for a second time. These cardiac myocytes normally do not initiate their own electrical potential, although they are capable of doing so, but rather wait for an impulse to reach them.
Contractile cells demonstrate a much more stable resting phase than conductive cells at approximately −80 mV for cells in the atria and −90 mV for cells in the ventricles. Despite this initial difference, the other components of their action potentials are virtually identical. In both cases, when stimulated by an action potential, voltage-gated channels rapidly open, beginning the positive-feedback mechanism of depolarization. This rapid influx of positively charged ions raises the membrane potential to approximately +30 mV, at which point the sodium channels close. The rapid depolarization period typically lasts 3–5 ms. Depolarization is followed by the plateau phase, in which membrane potential declines relatively slowly. This is due in large part to the opening of the slow Ca 2+ channels, allowing Ca 2+ to enter the cell while few K+ channels are open, allowing K+ to exit the cell. The relatively long plateau phase lasts approximately 175 ms. Once the membrane potential reaches approximately zero, the Ca 2+ channels close and K+ channels open, allowing K+ to exit the cell. The repolarization lasts approximately 75 ms. At this point, membrane potential drops until it reaches resting levels once more and the cycle repeats. The entire event lasts between 250 and 300 ms (Figure 5).
Figure 5. (a) Note the long plateau phase due to the influx of calcium ions. The extended refractory period allows the cell to fully contract before another electrical event can occur. (b) The action potential for heart muscle is compared to that of skeletal muscle.
The absolute refractory period for cardiac contractile muscle lasts approximately 200 ms, and the relative refractory period lasts approximately 50 ms, for a total of 250 ms. This extended period is critical, since the heart muscle must contract to pump blood effectively and the contraction must follow the electrical events. Without extended refractory periods, premature contractions would occur in the heart and would not be compatible with life.
Calcium Ions
Calcium ions play two critical roles in the physiology of cardiac muscle. Their influx through slow calcium channels accounts for the prolonged plateau phase and absolute refractory period that enable cardiac muscle to function properly. Calcium ions also combine with the regulatory protein troponin in the troponin-tropomyosin complex; this complex removes the inhibition that prevents the heads of the myosin molecules from forming cross bridges with the active sites on actin that provide the power stroke of contraction. This mechanism is virtually identical to that of skeletal muscle. Approximately 20 percent of the calcium required for contraction is supplied by the influx of Ca 2+ during the plateau phase. The remaining Ca 2+ for contraction is released from storage in the sarcoplasmic reticulum.
Comparative Rates of Conduction System Firing
The pattern of prepotential or spontaneous depolarization, followed by rapid depolarization and repolarization just described, are seen in the SA node and a few other conductive cells in the heart. Since the SA node is the pacemaker, it reaches threshold faster than any other component of the conduction system. It will initiate the impulses spreading to the other conducting cells. The SA node, without nervous or endocrine control, would initiate a heart impulse approximately 80–100 times per minute. Although each component of the conduction system is capable of generating its own impulse, the rate progressively slows as you proceed from the SA node to the Purkinje fibers. Without the SA node, the AV node would generate a heart rate of 40–60 beats per minute. If the AV node were blocked, the atrioventricular bundle would fire at a rate of approximately 30–40 impulses per minute. The bundle branches would have an inherent rate of 20–30 impulses per minute, and the Purkinje fibers would fire at 15–20 impulses per minute. While a few exceptionally trained aerobic athletes demonstrate resting heart rates in the range of 30–40 beats per minute (the lowest recorded figure is 28 beats per minute for Miguel Indurain, a cyclist), for most individuals, rates lower than 50 beats per minute would indicate a condition called bradycardia. Depending upon the specific individual, as rates fall much below this level, the heart would be unable to maintain adequate flow of blood to vital tissues, initially resulting in decreasing loss of function across the systems, unconsciousness, and ultimately death.
Electrocardiogram
Figure 6. In a 12-lead ECG, six electrodes are placed on the chest, and four electrodes are placed on the limbs.
By careful placement of surface electrodes on the body, it is possible to record the complex, compound electrical signal of the heart. This tracing of the electrical signal is the electrocardiogram (ECG), also commonly abbreviated EKG (K coming kardiology, from the German term for cardiology). Careful analysis of the ECG reveals a detailed picture of both normal and abnormal heart function, and is an indispensable clinical diagnostic tool. The standard electrocardiograph (the instrument that generates an ECG) uses 3, 5, or 12 leads. The greater the number of leads an electrocardiograph uses, the more information the ECG provides. The term “lead” may be used to refer to the cable from the electrode to the electrical recorder, but it typically describes the voltage difference between two of the electrodes. The 12-lead electrocardiograph uses 10 electrodes placed in standard locations on the patient’s skin (Figure 6). In continuous ambulatory electrocardiographs, the patient wears a small, portable, battery-operated device known as a Holter monitor, or simply a Holter, that continuously monitors heart electrical activity, typically for a period of 24 hours during the patient’s normal routine.
A normal ECG tracing is presented in Figure 7. Each component, segment, and interval is labeled and corresponds to important electrical events, demonstrating the relationship between these events and contraction in the heart.
There are five prominent points on the ECG: the P wave, the QRS complex, and the T wave. The small P wave represents the depolarization of the atria. The atria begin contracting approximately 25 ms after the start of the P wave. The large QRS complex represents the depolarization of the ventricles, which requires a much stronger electrical signal because of the larger size of the ventricular cardiac muscle. The ventricles begin to contract as the QRS reaches the peak of the R wave. Lastly, the T wave represents the repolarization of the ventricles. The repolarization of the atria occurs during the QRS complex, which masks it on an ECG.
Figure 7. A normal tracing shows the P wave, QRS complex, and T wave. Also indicated are the PR, QT, QRS, and ST intervals, plus the P-R and S-T segments.
The major segments and intervals of an ECG tracing are indicated in the image below. Segments are defined as the regions between two waves. Intervals include one segment plus one or more waves. For example, the PR segment begins at the end of the P wave and ends at the beginning of the QRS complex. The PR interval starts at the beginning of the P wave and ends with the beginning of the QRS complex. The PR interval is more clinically relevant, as it measures the duration from the beginning of atrial depolarization (the P wave) to the initiation of the QRS complex. Since the Q wave may be difficult to view in some tracings, the measurement is often extended to the R that is more easily visible. Should there be a delay in passage of the impulse from the SA node to the AV node, it would be visible in the PR interval. Figure 8 correlates events of heart contraction to the corresponding segments and intervals of an ECG.
Figure 8. This diagram correlates an ECG tracing with the electrical and mechanical events of a heart contraction. Each segment of an ECG tracing corresponds to one event in the cardiac cycle.
Visit this site for a more detailed analysis of ECGs.
Everyday Connection:ECG Abnormalities
Occasionally, an area of the heart other than the SA node will initiate an impulse that will be followed by a premature contraction. Such an area, which may actually be a component of the conduction system or some other contractile cells, is known as an ectopic focus or ectopic pacemaker. An ectopic focus may be stimulated by localized ischemia; exposure to certain drugs, including caffeine, digitalis, or acetylcholine; elevated stimulation by both sympathetic or parasympathetic divisions of the autonomic nervous system; or a number of disease or pathological conditions. Occasional occurances are generally transitory and nonlife threatening, but if the condition becomes chronic, it may lead to either an arrhythmia, a deviation from the normal pattern of impulse conduction and contraction, or to fibrillation, an uncoordinated beating of the heart.
While interpretation of an ECG is possible and extremely valuable after some training, a full understanding of the complexities and intricacies generally requires several years of experience. In general, the size of the electrical variations, the duration of the events, and detailed vector analysis provide the most comprehensive picture of cardiac function. For example, an amplified P wave may indicate enlargement of the atria, an enlarged Q wave may indicate a MI, and an enlarged suppressed or inverted Q wave often indicates enlarged ventricles. T waves often appear flatter when insufficient oxygen is being delivered to the myocardium. An elevation of the ST segment above baseline is often seen in patients with an acute MI, and may appear depressed below the baseline when hypoxia is occurring.
As useful as analyzing these electrical recordings may be, there are limitations. For example, not all areas suffering a MI may be obvious on the ECG. Additionally, it will not reveal the effectiveness of the pumping, which requires further testing, such as an ultrasound test called an echocardiogram or nuclear medicine imaging. It is also possible for there to be pulseless electrical activity, which will show up on an ECG tracing, although there is no corresponding pumping action. Common abnormalities that may be detected by the ECGs are shown in Figure 9.
Figure 9. (a) In a second-degree or partial block, one-half of the P waves are not followed by the QRS complex and T waves while the other half are. (b) In atrial fibrillation, the electrical pattern is abnormal prior to the QRS complex, and the frequency between the QRS complexes has increased. (c) In ventricular tachycardia, the shape of the QRS complex is abnormal. (d) In ventricular fibrillation, there is no normal electrical activity. (e) In a third-degree block, there is no correlation between atrial activity (the P wave) and ventricular activity (the QRS complex).
Visit this sitefor a more complete library of abnormal ECGs.
Everyday Connection:External AutomatedDefibrillators
In the event that the electrical activity of the heart is severely disrupted, cessation of electrical activity or fibrillation may occur. In fibrillation, the heart beats in a wild, uncontrolled manner, which prevents it from being able to pump effectively. Atrial fibrillation (see Figure 10a) is a serious condition, but as long as the ventricles continue to pump blood, the patient’s life may not be in immediate danger. Ventricular fibrillation (see Figure 10b) is a medical emergency that requires life support, because the ventricles are not effectively pumping blood. In a hospital setting, it is often described as “code blue.” If untreated for as little as a few minutes, ventricular fibrillation may lead to brain death. The most common treatment is defibrillation, which uses special paddles to apply a charge to the heart from an external electrical source in an attempt to establish a normal sinus rhythm. A defibrillator effectively stops the heart so that the SA node can trigger a normal conduction cycle. Because of their effectiveness in reestablishing a normal sinus rhythm, external automated defibrillators (EADs) are being placed in areas frequented by large numbers of people, such as schools, restaurants, and airports. These devices contain simple and direct verbal instructions that can be followed by nonmedical personnel in an attempt to save a life.
Figure 10. (a) An external automatic defibrillator can be used by nonmedical personnel to reestablish a normal sinus rhythm in a person with fibrillation. (b) Defibrillator paddles are more commonly used in hospital settings. (credit b: “widerider107”/flickr.com)
A heart block refers to an interruption in the normal conduction pathway. The nomenclature for these is very straightforward. SA nodal blocks occur within the SA node. AV nodal blocks occur within the AV node. Infra-Hisian blocks involve the bundle of His. Bundle branch blocks occur within either the left or right atrioventricular bundle branches. Hemiblocks are partial and occur within one or more fascicles of the atrioventricular bundle branch. Clinically, the most common types are the AV nodal and infra-Hisian blocks.
AV blocks are often described by degrees. A first-degree or partial block indicates a delay in conduction between the SA and AV nodes. This can be recognized on the ECG as an abnormally long PR interval. A second-degree or incomplete block occurs when some impulses from the SA node reach the AV node and continue, while others do not. In this instance, the ECG would reveal some P waves not followed by a QRS complex, while others would appear normal. In the third-degree or complete block, there is no correlation between atrial activity (the P wave) and ventricular activity (the QRS complex). Even in the event of a total SA block, the AV node will assume the role of pacemaker and continue initiating contractions at 40–60 contractions per minute, which is adequate to maintain consciousness.
When arrhythmias become a chronic problem, the heart maintains a junctional rhythm, which originates in the AV node. In order to speed up the heart rate and restore full sinus rhythm, a cardiologist can implant an artificial pacemaker, which delivers electrical impulses to the heart muscle to ensure that the heart continues to contract and pump blood effectively. These artificial pacemakers are programmable by the cardiologists and can either provide stimulation temporarily upon demand or on a continuous basis. Some devices also contain built-in defibrillators.
Cardiac Muscle Metabolism
Normally, cardiac muscle metabolism is entirely aerobic. Oxygen from the lungs is brought to the heart, and every other organ, attached to the hemoglobin molecules within the erythrocytes. Heart cells also store appreciable amounts of oxygen in myoglobin. Normally, these two mechanisms, circulating oxygen and oxygen attached to myoglobin, can supply sufficient oxygen to the heart, even during peak performance.
Fatty acids and glucose from the circulation are broken down within the mitochondria to release energy in the form of ATP. Both fatty acid droplets and glycogen are stored within the sarcoplasm and provide additional nutrient supply. (Seek additional content for more detail about metabolism.)
Chapter Review
The heart is regulated by both neural and endocrine control, yet it is capable of initiating its own action potential followed by muscular contraction. The conductive cells within the heart establish the heart rate and transmit it through the myocardium. The contractile cells contract and propel the blood. The normal path of transmission for the conductive cells is the sinoatrial (SA) node, internodal pathways, atrioventricular (AV) node, atrioventricular (AV) bundle of His, bundle branches, and Purkinje fibers. The action potential for the conductive cells consists of a prepotential phase with a slow influx of Na+ followed by a rapid influx of Ca 2+ and outflux of K+. Contractile cells have an action potential with an extended plateau phase that results in an extended refractory period to allow complete contraction for the heart to pump blood effectively. Recognizable points on the ECG include the P wave that corresponds to atrial depolarization, the QRS complex that corresponds to ventricular depolarization, and the T wave that corresponds to ventricular repolarization.
Critical Thinking Questions
Why is the plateau phase so critical to cardiac muscle function?
How does the delay of the impulse at the atrioventricular node contribute to cardiac function?
How do gap junctions and intercalated disks aid contraction of the heart?
Why do the cardiac muscles cells demonstrate autorhythmicity?
Show Answers
It prevents additional impulses from spreading through the heart prematurely, thereby allowing the muscle sufficient time to contract and pump blood effectively.
It ensures sufficient time for the atrial muscle to contract and pump blood into the ventricles prior to the impulse being conducted into the lower chambers.
Gap junctions within the intercalated disks allow impulses to spread from one cardiac muscle cell to another, allowing sodium, potassium, and calcium ions to flow between adjacent cells, propagating the action potential, and ensuring coordinated contractions.
Without a true resting potential, there is a slow influx of sodium ions through slow channels that produces a prepotential that gradually reaches threshold.
Glossary
artificial pacemaker:medical device that transmits electrical signals to the heart to ensure that it contracts and pumps blood to the body
atrioventricular bundle:(also, bundle of His) group of specialized myocardial conductile cells that transmit the impulse from the AV node through the interventricular septum; form the left and right atrioventricular bundle branches
atrioventricular bundle branches:(also, left or right bundle branches) specialized myocardial conductile cells that arise from the bifurcation of the atrioventricular bundle and pass through the interventricular septum; lead to the Purkinje fibers and also to the right papillary muscle via the moderator band
atrioventricular (AV) node:clump of myocardial cells located in the inferior portion of the right atrium within the atrioventricular septum; receives the impulse from the SA node, pauses, and then transmits it into specialized conducting cells within the interventricular septum
autorhythmicity:ability of cardiac muscle to initiate its own electrical impulse that triggers the mechanical contraction that pumps blood at a fixed pace without nervous or endocrine control
Bachmann’s bundle:(also, interatrial band) group of specialized conducting cells that transmit the impulse directly from the SA node in the right atrium to the left atrium
bundle of His:(also, atrioventricular bundle) group of specialized myocardial conductile cells that transmit the impulse from the AV node through the interventricular septum; form the left and right atrioventricular bundle branches
electrocardiogram (ECG):surface recording of the electrical activity of the heart that can be used for diagnosis of irregular heart function; also abbreviated as EKG
heart block:interruption in the normal conduction pathway
interatrial band:(also, Bachmann’s bundle) group of specialized conducting cells that transmit the impulse directly from the SA node in the right atrium to the left atrium
intercalated disc:physical junction between adjacent cardiac muscle cells; consisting of desmosomes, specialized linking proteoglycans, and gap junctions that allow passage of ions between the two cells
internodal pathways:specialized conductile cells within the atria that transmit the impulse from the SA node throughout the myocardial cells of the atrium and to the AV node
myocardial conducting cells:specialized cells that transmit electrical impulses throughout the heart and trigger contraction by the myocardial contractile cells
myocardial contractile cells:bulk of the cardiac muscle cells in the atria and ventricles that conduct impulses and contract to propel blood
P wave:component of the electrocardiogram that represents the depolarization of the atria
pacemaker:cluster of specialized myocardial cells known as the SA node that initiates the sinus rhythm
prepotential depolarization:(also, spontaneous depolarization) mechanism that accounts for the autorhythmic property of cardiac muscle; the membrane potential increases as sodium ions diffuse through the always-open sodium ion channels and causes the electrical potential to rise
Purkinje fibers:specialized myocardial conduction fibers that arise from the bundle branches and spread the impulse to the myocardial contraction fibers of the ventricles
QRS complex:component of the electrocardiogram that represents the depolarization of the ventricles and includes, as a component, the repolarization of the atria
sinoatrial (SA) node:known as the pacemaker, a specialized clump of myocardial conducting cells located in the superior portion of the right atrium that has the highest inherent rate of depolarization that then spreads throughout the heart
sinus rhythm:normal contractile pattern of the heart
spontaneous depolarization:(also, prepotential depolarization) the mechanism that accounts for the autorhythmic property of cardiac muscle; the membrane potential increases as sodium ions diffuse through the always-open sodium ion channels and causes the electrical potential to rise
T wave:component of the electrocardiogram that represents the repolarization of the ventricles
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UPSC » UPSC CSE Study Materials » Mathematics » A Guide To The Median Properties In Statistics
A Guide To The Median Properties In Statistics
Everything you need to know about A Guide to the Median Properties in Statistics, what is median in statistics, the median in statistics, mean median mode relation, and all other topics related to A Guide to the Median Properties in Statistics
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When the data are arranged from lowest to highest, the median is the value that reflects the middle line. Fifty percent of the scores are either higher or lower than the median. As a result, it is sometimes referred to as the top quarter or positional average. The median is placed based on whether the data set includes an even or odd range of items. Depending on whether there are an even or an odd number of cases, the process for determining the median varies.
Median Characteristics
The qualities of the median were outlined in statistics in the following sections.
· The median is unaffected by any of the information values in the dataset.
· Individual values do not correspond to the median value, which is determined by its location.
· The gap between the median and the other variables will be smaller than any other point.
· There is only one median in each array, and it cannot be modified algebraically. It cannot be measured or mixed. The median remains constant in a grouping strategy.
· The median does not apply to qualitative data; the variables must be linked and arranged before the median can be determined.
· The median of a ratio, interval, or ordinal scale can be calculated.
· When an allocation is skewed, the median is better to be considered than the mean.
What exactly is the Median?
The median is the number in the middle of an organized, ascending, or descending list of numbers, and it may be more revealing of the data set than the average.
The median is the number in the middle of an ordered, ascending, or descending list of numbers, and it may be more revealing of the data set than the average.
When there are exceptions in the series that may impact the average of the values, the median is typically used rather than the mean.
If there are an odd number of integers, the result in the center is the median number, with the same number below and above it.
If the list has an even number of values, find the middle pair, add them together, then divide by two to get the median value.
Recognizing the Median
The median is the number in the middle of a group of numbers that has been sorted. To calculate the median value of a set of numbers, they must first be classified or ranked in value order from lowest to highest or largest to lowest. Although the median can be used to approximate an average or mean, it should not be confused with the true mean.
If there are an odd number of integers, the result in the center is the median number, with the same number below and above it.
If the table has an even number of numbers, find the middle pair, add them together, then divide by two to get the median value.
When there are exceptions in the series that may impact the total of the statistics, the median is typically used rather than the mean. Outliers have a smaller influence on the median of a series than they do on the mean.
Conclusion
We have learned about A Guide to the Median Properties in Statistics, what is median in statistics is, the median in statistics, the mean median mode relation, and all other topics related to A Guide to the Median Properties in Statistics.
The median is the number in the center of a sorted, ascending, or descending list of numbers, and it might be more informative of that data collection than the average. When there are outliers in the series that might affect the average of the numbers, the median is frequently utilized instead of the mean.
Frequently asked questions
Get answers to the most common queries related to the UPSC Examination Preparation.
What is the Relationship Between Mean, Median, and Mode?
Ans. The relationship between mean, median, and modes is critical in statistics and is beneficial when dealing with ...Read full
What is mean?
Ans. Given a collection of values, the mean is calculated by adding all of the entries and multiplying the total by ...Read full
What are the statistical qualities of the median?
Ans. The median is not affected by all of the data values in a dataset. Individual values do not reflect the median ...Read full
Ans. The relationship between mean, median, and modes is critical in statistics and is beneficial when dealing with comparable situations. In the case of a highly skewed distribution, the discrepancy between mean and mode is three times the variance between mean and median.
Ans. Given a collection of values, the mean is calculated by adding all of the entries and multiplying the total by the number of values. In arithmetic notation, you have a collection of values called the populace X1,……, Xn.
Ans. The median is not affected by all of the data values in a dataset. Individual values do not reflect the median value, which is determined by its location.
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Trigonometry Olympiad Problems
This document provides numerous inequalities relating to geometric properties of triangles. It begins by listing several trigonometric identities that hold when the sums of the angles of a t…
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Nov 12
Math Explanations November 16, 2015
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a
2
+
b
2
+
c
2
)
ab
+
bc
+
ca
3
adDownload to read ad-free
a
3
+
b
3
+
c
3
≤
8
s
(
R
2
−
r
2
)in terms of the semiperimeter ”s”;
r
(
r
4
R
)
≥√
3
·
T
in terms of the area ”T”
s
√
3
≤
r
4
R
and
s
2
≥
16
Rr
−
5
r
2
in terms of the semiperimeter ”s”; and 2
R
2
10
Rr
−
r
2
−
2(
R
−
2
r
)
R
2
−
2
Rr
≤
s
2
≤
2
R
2
10
Rr
−
r
2
2(
R
−
2
r
)
R
2
−
2
Rr
also in terms of the semiperimeter.¡ref name=SV/¿rp—p. 206¡ref name=Birsan¿Birsan, Temis-tocle (2015), ”Bounds for elements of a triangle expressed by R, r, and s”, ”Forum Geometricorum”15, 99103. umgeom.fau.e du/F G2015v olume15/FG 201508.pd f ¡/ref¿rp—p. 99 In the lat-ter double inequality, the first part holds with equality if and only if the triangle is isosceles with an angle of at least 60, and the last part holds with equality if and only if the triangle is isosceles with an apex angle of at most 60. Thus both are equalities if and only if the trian gle is equila teral.¡ref name=Birs an/¿r p—Thm. 1 Further,9
r
2
T
≤
1
a
+1
b
+1
c
≤
9
R
4
T.
Blundon’s inequality states that
s
≤
(3
√
3
−
4)
r
2
R
For incircle center ”I”, let ”AI”, ”BI”, and ”CI” extend beyond ”I” to intersect the circumcircle at ”D”, ”E”, and ”F” respectively. Then
AI ID
+
B I IE
+
CI IF
≥
3
.
In terms of the vertex angles we have cos
A
·
cos
B
·
cos
C
≤
r R
√
2
2
===Circumradius and other lengths===For the circumradius ”R” we have 18
R
3
≥
(
a
2
+
b
2
+
c
2
)
R
+
abc
√
3 and 4
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10159 | https://en.wikipedia.org/wiki/Displacement_(geometry) | Displacement (geometry) - Wikipedia
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1 Formulation
2 Rigid body
3 Derivatives
4 Discussion
5 See also
6 References
7 External links
Displacement (geometry)
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From Wikipedia, the free encyclopedia
Vector relating the initial and the final positions of a moving point
For other uses, see Displacement (disambiguation).
| Displacement |
| Displacement versus distance travelled along a path |
| Common symbols | d, s, ∆s , ∆x , ∆y, ∆z |
| SI unit | metre |
| In SI base units | m |
| Dimension | L |
Part of a series on
Classical mechanics
F=d p d t{\displaystyle {\textbf {F}}={\frac {d\mathbf {p} }{dt}}}
Second law of motion
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In geometry and mechanics, a displacement is a vector whose length is the shortest distance from the initial to the final position of a point P undergoing motion. It quantifies both the distance and direction of the net or total motion along a straight line from the initial position to the final position of the point trajectory. A displacement may be identified with the translation that maps the initial position to the final position. Displacement is the shift in location when an object in motion changes from one position to another. For motion over a given interval of time, the displacement divided by the length of the time interval defines the average velocity (a vector), whose magnitude is the average speed (a scalar quantity).
Formulation
[edit]
A displacement may be formulated as a relative position (resulting from the motion), that is, as the final position x f of a point relative to its initial position x i. The corresponding displacement vector can be defined as the difference between the final and initial positions: s=x f−x i=Δ x{\displaystyle s=x_{\textrm {f}}-x_{\textrm {i}}=\Delta {x}}
Rigid body
[edit]
In dealing with the motion of a rigid body, the term displacement may also include the rotations of the body. In this case, the displacement of a particle of the body is called linear displacement (displacement along a line), while the rotation of the body is called angular displacement.
Derivatives
[edit]
See also: Position (geometry) §Derivatives
For a position vector s{\displaystyle \mathbf {s} } that is a function of time t{\displaystyle t}, the derivatives can be computed with respect to t{\displaystyle t}. The first two derivatives are frequently encountered in physics.
Velocityv=d s d t{\displaystyle \mathbf {v} ={\frac {d\mathbf {s} }{dt}}}Accelerationa=d v d t=d 2 s d t 2{\displaystyle \mathbf {a} ={\frac {d\mathbf {v} }{dt}}={\frac {d^{2}\mathbf {s} }{dt^{2}}}}Jerkj=d a d t=d 2 v d t 2=d 3 s d t 3{\displaystyle \mathbf {j} ={\frac {d\mathbf {a} }{dt}}={\frac {d^{2}\mathbf {v} }{dt^{2}}}={\frac {d^{3}\mathbf {s} }{dt^{3}}}}
These common names correspond to terminology used in basic kinematics. By extension, the higher order derivatives can be computed in a similar fashion. Study of these higher order derivatives can improve approximations of the original displacement function. Such higher-order terms are required in order to accurately represent the displacement function as a sum of an infinite series, enabling several analytical techniques in engineering and physics. The fourth order derivative is called jounce.
Discussion
[edit]
In considering motions of objects over time, the instantaneous velocity of the object is the rate of change of the displacement as a function of time. The instantaneous speed, then, is distinct from velocity, or the time rate of change of the distance travelled along a specific path. The velocity may be equivalently defined as the time rate of change of the position vector. If one considers a moving initial position, or equivalently a moving origin (e.g. an initial position or origin which is fixed to a train wagon, which in turn moves on its rail track), the velocity of P (e.g. a point representing the position of a passenger walking on the train) may be referred to as a relative velocity; this is opposed to an absolute velocity, which is computed with respect to a point and coordinate axes which are considered to be at rest (a inertial frame of reference such as, for instance, a point fixed on the floor of the train station and the usual vertical and horizontal directions).
See also
[edit]
Mathematics portal
Physics portal
Affine space
Deformation (mechanics)
Displacement field (mechanics)
Equipollence (geometry)
Motion vector
Position vector
Radial velocity
Screw displacement
References
[edit]
^Tom Henderson. "Describing Motion with Words". The Physics Classroom. Retrieved 2 January 2012.
^Moebs, William; Ling, Samuel J.; Sanny, Jeff (2016-09-19). "3.1 Position, Displacement, and Average Velocity - University Physics Volume 1 | OpenStax". openstax.org. Retrieved 2024-03-11.
^"Angular Displacement, Velocity, Acceleration". NASA Glenn Research Center. National Aeronautics and Space Administration. 13 May 2021. Retrieved 9 November 2023.
^Stewart, James (2001). "§2.8 - The Derivative As A Function". Calculus (2nd ed.). Brooks/Cole. ISBN0-534-37718-1.
External links
[edit]
Media related to Displacement vector at Wikimedia Commons
| show v t e Kinematics |
| ← Integrate … Differentiate → |
| Absement Displacement (Distance) Velocity (Speed) Acceleration Jerk Higher derivatives |
| show v t e Classical mechanicsSI units |
| Linear/translational quantities Angular/rotational quantities Dimensions 1 L L 2 Dimensions 1 θ θ 2 Ttime: t sabsement: A m sTtime: t s 1distance: d, position: r, s, x, displacement marea: A m 21angle: θ, angular displacement: θ radsolid angle: Ω rad 2, sr T−1frequency: f s−1, Hzspeed: v, velocity: v m s−1kinematic viscosity: ν, specific angular momentum:h m 2 s−1 T−1frequency: f, rotational speed: n, rotational velocity: n s−1, Hzangular speed: ω, angular velocity: ω rad s−1 T−2acceleration: a m s−2T−2rotational acceleration s−2angular acceleration: α rad s−2 T−3jerk: j m s−3 T−3angular jerk: ζ rad s−3 Mmass: m kgweighted position: M ⟨x⟩ = ∑ m xmoment of inertia:I kg m 2ML MT−1Mass flow rate: m˙{\displaystyle {\dot {m}}} kg s−1momentum: p, impulse: J kg m s−1, N saction: 𝒮, actergy: ℵ kg m 2 s−1, J sMLT−1angular momentum: L, angular impulse: ΔL kg m rad s−1 MT−2force: F, weight: Fg kg m s−2, Nenergy: E, work: W, Lagrangian: L kg m 2 s−2, JMLT−2torque: τ, moment: M kg m rad s−2, N m MT−3yank: Y kg m s−3, N s−1power: P kg m 2 s−3,WMLT−3rotatum: P kg m rad s−3, N m s−1 |
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Antifungal Agents
Antifungal Agents
There are 200,000 known species of fungi, and estimates of the total size of Kingdom Fungi range to well over a million. Residents of the kingdom are quite diverse and include yeasts, molds, mushrooms, smuts, the pathogens Aspergillus fumigatus and Candida albicans, and the source of penicillin, Penicillium chrysogenum. Fortunately, only ~400 fungi cause disease in animals, and even fewer cause significant human disease. However, fungal infections are becoming more common in patients with compromised immune systems. Fungi are eukaryotes with unique cell walls containing glucans and chitin, and their eradication requires different strategies than those for treatment of bacterial infections. Available agents have effects on the synthesis of membrane and cell-wall components, on membrane permeability, on the synthesis of nucleic acids, and on microtubule/mitotic spindle function (Figure 57–1). Antifungal agents described in this chapter are discussed under two major headings, systemic and topical, although this distinction is somewhat arbitrary. The imidazole, triazole, and polyene antifungal agents may be used either systemically or topically, and many superficial mycoses can be treated either systemically or topically. Table 57–1 summarizes common mycoses and their pharmacotherapy.
Figure 57–1Sites of action of antifungal drugs. Amphotericin B and other polyenes (e.g., nystatin) bind to ergosterol in fungal cell membranes and increase membrane permeability. The imidazoles and triazoles (itraconazole, et al.) inhibit 14-α-sterol demethylase, prevent ergosterol synthesis, and lead to the accumulation of 14-α-methylsterols. The allylamines (e.g., naftifine and terbinafine) inhibit squalene epoxidase and prevent ergosterol synthesis. The echinocandins, such as caspofungin, inhibit the formation of glucans in the fungal cell wall.
Table 57–1
Pharmacotherapy of Mycoses
SYSTEMIC ANTIFUNGAL AGENTS
DRUGS FOR DEEPLY INVASIVE FUNGAL INFECTIONS
AMPHOTERICIN B. Amphotericin B is an amphoteric polyene macrolide with broad spectrum antifungal activity.
MECHANISM OF ACTION. Amphotericin B has useful clinical activity against a broad range of pathogenic fungi and limited activity against the protozoa Leishmania spp. and Naegleria fowleri. The antifungal activity of amphotericin B depends principally on its binding to a sterol moiety, primarily ergosterol in the membrane of sensitive fungi. By virtue of their interaction with these sterols, polyenes appear to form pores or channels that increase the permeability of the membrane, allowing leakage of a variety of small molecules (seeFigure 57–1).
FORMULATIONS.Table 57–2 summarizes the pharmacokinetic properties of the 4 available preparations.
C-AMB (conventional amphotericin B, F UNGIZONE). Amphotericin B is insoluble in water but is formulated for intravenous infusion by complexing it with the bile salt deoxycholate. The complex is marketed as a lyophilized powder for injection. C-AMB forms a colloid in water, with particles largely <0.4 μm in diameter. Filters in intravenous infusion lines that trap particles >0.22 μm in diameter will remove significant amounts of drug. Addition of electrolytes to infusion solutions causes the colloid to aggregate.
ABCD. Amphotericin B colloidal dispersion (A MPHOTEC, A MPHOCIL) contains roughly equimolar amounts of amphotericin B and cholesteryl sulfate formulated for injection. Like C-AMB, ABCD forms a colloidal solution when dispersed in aqueous solution.
L-AMB. Liposomal amphotericin B is a small, unilamellar vesicle formulation of amphotericin B (A MBISOME). The drug is supplied as a lyophilized powder, which is reconstituted with sterile water for injection.
ABLC (A BELCET). Amphotericin B lipid complex is a complex of amphotericin B with lipids (dimyristoylphosphatidylcholine and dimyristoylphosphatidylglycerol).
The 3 lipid formulations collectively appear to reduce the risk of the patient’s serum creatinine doubling during therapy by 58%. However, the cost of the lipid formulations greatly exceeds that of C-AMB, making them unavailable in many countries.
ADME. GI absorption of all amphotericin B formulations is negligible and IV delivery is indicated for systemic use. Amphotericin B in plasma is more than 90% bound to proteins. Pharmacokinetic parameters vary with the formulation. Azotemia, liver failure, or hemodialysis does not have a measurable impact on plasma concentrations. Concentrations of amphotericin B (via C-AMB) in fluids from inflamed pleura, peritoneum, synovium, and aqueous humor are approximately two-thirds of trough concentrations in plasma. Little amphotericin B from any formulation penetrates into cerebrospinal fluid (CSF), vitreous humor, or normal amniotic fluid.
ANTIFUNGAL ACTIVITY; FUNGAL RESISTANCE. Amphotericin B has useful clinical activity against Candida spp., Cryptococcus neoformans, Blastomyces dermatitidis, Histoplasma capsulatum, Sporothrix schenckii, Coccidioides spp., Paracoccidioides braziliensis, Aspergillus spp., Penicillium marneffei, and the agents of mucormycosis. Amphotericin B has limited activity against the protozoa Leishmania spp. and N. fowleri. The drug has no antibacterial activity. Some isolates of Candida lusitaniae have been relatively resistant to amphotericin B. Aspergillus terreus and perhaps Aspergillus nidulans may be more resistant to amphotericin B than other Aspergillus species.
THERAPEUTIC USES. The recommended doses for each formulation are summarized in Table 57–2. Candida esophagitis responds to much lower doses than deeply invasive mycoses. Intrathecal infusion of C-AMB is useful in patients with meningitis caused by Coccidioides. Too little is known about intrathecal administration of lipid formulations to recommend them. C-AMB can be injected into the CSF of the lumbar spine, cisterna magna, or lateral cerebral ventricle. Fever and headache are common reactions that may be decreased by intrathecal administration of 10-15 mg of hydrocortisone. Local injections of amphotericin B into a joint or peritoneal dialysate fluid commonly produce irritation and pain. Intraocular injection following pars plana vitrectomy has been used successfully for fungal endophthalmitis. Intravenous administration of amphotericin B is the treatment of choice for mucormycosis and is used for initial treatment of cryptococcal meningitis, severe or rapidly progressing histoplasmosis, blastomycosis, coccidioidomycosis, and penicilliosis marneffei, as well as in patients not responding to azole therapy of invasive aspergillosis, extracutaneous sporotrichosis, fusariosis, alternariosis, and trichosporonosis. Amphotericin B (C-AMB or L-AMB) is often given to selected patients with profound neutropenia who have fever that does not respond to broad-spectrum antibacterial agents over 5-7 days.
Table 57–2
Pharmocokinetic Parameters for Amphotericin B Formulations after Multiple Administrations in Humans
UNTOWARD EFFECTS. The major acute reactions to intravenous amphotericin B formulations are fever and chills. Infusion-related reactions are worst with ABCD and least with L-AMB. Tachypnea and respiratory stridor or modest hypotension also may occur, but true bronchospasm or anaphylaxis is rare. Patients with pre-existing cardiac or pulmonary disease may tolerate the metabolic demands of the reaction poorly and develop hypoxia or hypotension. The reaction ends spontaneously in 30-45 min; meperidine may shorten it. Pretreatment with oral acetaminophen or use of intravenous glucocorticoids at the start of the infusion decreases reactions.
Azotemia occurs in 80% of patients who receive C-AMB for deep mycoses. Lipid formulations are less nephrotoxic, being much less with ABLC, even less with L-AMB, and minimal with ABCD. Toxicity is dose-dependent and usually transient and increased by concurrent therapy with other nephrotoxic agents, such as aminoglycosides or cyclosporine. Permanent functional impairment is uncommon in adults with normal renal function prior to treatment unless the cumulative dose exceeds 3-4 g. Renal tubular acidosis and renal wasting of K+ and Mg 2+ also may be seen during and for several weeks after therapy, often requiring repletion. Administration of 1 L of normal saline intravenously on the day that C-AMB is to be given has been recommended for adults who are able to tolerate the Na+ load.
Hypochromic, normocytic anemia commonly occurs during treatment with C-AMB. Anemia is less with lipid formulations and usually not seen over the first 2 weeks. The anemia is most likely due to decreased production of erythropoietin and often responds to administration of recombinant erythropoietin. Headache, nausea, vomiting, malaise, weight loss, and phlebitis at peripheral infusion sites are common. Arachnoiditis has been observed as a complication of injecting C-AMB into the CSF.
FLUCYTOSINE
Flucytosine (5-fluorocytosine) has a spectrum of antifungal activity that is considerably more restricted than that of amphotericin B.
MECHANISM OF ACTION. All susceptible fungi are capable of deaminating flucytosine to 5-fluorouracil (5-FU) (Figure 57–2), a potent antimetabolite that is used in cancer chemotherapy (seeChapter 61). Fluorouracil is metabolized first to 5-fluorouracil-ribose monophosphate (5-FUMP) by the enzyme uracil phosphoribosyl transferase (UPRTase). 5-FUMP then is either incorporated into RNA (via synthesis of 5-fluorouridine triphosphate) or metabolized to 5-fluoro-2′-deoxyuridine-5′-monophosphate (5-FdUMP), a potent inhibitor of thymidylate synthetase and thus of DNA synthesis. The selective action of flucytosine is due to the lack of cytosine deaminase in mammalian cells, which prevents metabolism to fluorouracil.
Figure 57–2Action of flucytosine in fungi. Flucytosine (5-fluorocytosine) is transported by cytosine permease into the fungal cell, where it is deaminated to 5-fluorouracil (5-FU). The 5-FU is converted to 5-fluorouracil-ribose monophosphate (5-FUMP) and then is either converted to 5-fluorouridine triphosphate (5-FUTP) and incorporated into RNA or converted to 5-fluoro-2′-deoxyuridine-5′-monophosphate (5-FdUMP), a potent inhibitor of thymidylate synthase. 5-FUDP, 5-fluorouridine-5′-diphosphate; dUMP, deoxyuridine-5′-monophosphate; dTMP, deoxythymidine-5′-monophosphate; PRT, phosphoribosyltransferase.
ANTIFUNGAL ACTIVITY. Flucytosine has clinically useful activity against C. neoformans, Candida spp., and the agents of chromoblastomycosis.
FUNGAL RESISTANCE. Drug resistance arising during therapy (secondary resistance) is an important cause of therapeutic failure when flucytosine is used alone for cryptococcosis and candidiasis. The mechanism for this resistance can be loss of the permease necessary for cytosine transport or decreased activity of either UPRTase or cytosine deaminase (seeFigure 57–2).
ADME. Flucytosine is absorbed rapidly and well from the GI tract and widely distributed in the body. Approximately 80% of a given dose is excreted unchanged in the urine. The t 1/2 of the drug is 3-6 h but may reach 200 h in renal failure. The clearance of flucytosine is approximately equivalent to that of creatinine. Reduction of dosage is necessary in patients with decreased renal function, and concentrations of drug in plasma should be measured periodically (desirable range of peak concentrations, 50 and 100 μg/mL). Flucytosine is cleared by hemodialysis, and patients undergoing such treatment should receive a single dose of 37.5 mg/kg after dialysis; the drug also is removed by peritoneal dialysis.
THERAPEUTIC USES. Flucytosine (A NCOBON) is given orally at 50-150 mg/kg/day, in 4 divided doses at 6-h intervals. Flucytosine is used predominantly in combination with amphotericin B. An all-oral regimen of flucytosine plus fluconazole also has been advocated for therapy of AIDS patients with cryptococcosis, but the combination has substantial GI toxicity with no evidence that flucytosine adds benefit. The addition of flucytosine to ≥6 weeks of therapy with C-AMB runs the risk of substantial bone marrow suppression or colitis if the flucytosine dose is not promptly adjusted downward as amphotericin B–induced azotemia occurs. The guidelines for the treatment of cryptococcal meningoencephalitis recommend addition of flucytosine (100 mg/kg orally in 4 divided doses) for the first 2 weeks of treatment with amphotericin B in AIDS patients.
Untoward Effects. Flucytosine may depress the bone marrow and lead to leukopenia. Other untoward effects include rash, nausea, vomiting, diarrhea, and severe enterocolitis. In ~5% of patients, plasma levels of hepatic enzymes are elevated, but this effect reverses when therapy is stopped. Toxicity is more frequent in patients with AIDS or azotemia and when plasma drug concentrations exceed 100 μ/mL.
IMIDAZOLES AND TRIAZOLES
The azole antifungals include two broad classes, imidazoles and triazoles. Of the drugs now on the market in the U.S., clotrimazole, miconazole, ketoconazole, econazole, butoconazole, oxiconazole, sertaconazole, and sulconazole are imidazoles; terconazole, itraconazole, fluconazole, voriconazole, and posaconazole are triazoles. The topical use of azole antifungals is described in the second section of this chapter.
MECHANISM OF ACTION. The major effect of imidazoles and triazoles on fungi is inhibition of 14-α-sterol demethylase, a microsomal CYP (seeFigure 57–1). Imidazoles and triazoles thus impair the biosynthesis of ergosterol for the cytoplasmic membrane and lead to the accumulation of 14-α-methylsterols. These methylsterols may disrupt the close packing of acyl chains of phospholipids, impairing the functions of certain membrane-bound enzyme systems, thus inhibiting growth of the fungi.
ANTIFUNGAL ACTIVITY. Azoles as a group have clinically useful activity against C. albicans, Candida tropicalis, Candida parapsilosis, Candida glabrata, C. neoformans, B. dermatitidis, H. capsulatum, Coccidioides spp., Paracoccidioides brasiliensis, and ringworm fungi (dermatophytes). Aspergillus spp., Scedosporium apiospermum (Pseudallescheria boydii), Fusarium, and S. schenckii are intermediate in susceptibility. Candida krusei and the agents of mucormycosis are more resistant. These drugs have antiprotozoal effects against Leishmania major. Posaconazole has slightly improved activity in vitro against the agents of mucormycosis.
RESISTANCE. Azole resistance emerges gradually during prolonged azole therapy, causing clinical failure in patients with far-advanced HIV infection and oropharyngeal or esophageal candidiasis. The primary mechanism of resistance in C. albicans is accumulation of mutations in ERG11, the gene coding for the 14-,-sterol demethylase; cross-resistance is conferred to all azoles.
INTERACTION OF AZOLE ANTI-FUNGALS WITH OTHER DRUGS. The azoles interact with hepatic CYPs as substrates and inhibitors (Table 57–3), providing myriad possibilities for the interaction of azoles with many other medications. Thus, azoles can elevate plasma levels of some coadministered drugs (Table 57–4). Other coadministered drugs can decrease plasma concentrations of azole antifungal agents (Table 57–5). As a consequence of myriad interactions, combinations of certain drugs with azole antifungal medications may be contraindicated (Table 57–6).
Table 57–3
Interaction of Azole Antifungal Agents with Hepatic CYPs
Table 57–4
Drugs Exhibiting Elevated Plasma Concentrations When Co-Administered with Azole Anti-Fungal Agents
Table 57–5
Some Agents that Decrease Triazole Concentration
Table 57–6
Some Additional Contraindicated Azole Drug Combinations
KETOCONAZOLE
Ketoconazole, administered orally, has been replaced by itraconazole for the treatment of all mycoses except when the lower cost of ketoconazole outweighs the advantage of itraconazole. Ketoconazole sometimes is used to inhibit excessive production of glucocorticoids in patients with Cushing syndrome (seeChapter 42) and is available for topical use.
ITRACONAZOLE
Itraconazole lacks ketoconazole’s corticosteroid suppression while retaining most of ketoconazole’s pharmacological properties and expanding the antifungal spectrum. This synthetic triazole is an equimolar racemic mixture of 4 diastereoisomers.
ADME. Itraconazole (S PORANOX, others) is available as a capsule and a solution in hydroxypropyl-β-cyclodextrin for oral use. The capsule form of the drug is best absorbed in the fed state, but the oral solution is better absorbed in the fasting state, providing peak plasma concentrations >150% of those obtained with the capsule. Itraconazole is metabolized in the liver; it is both a substrate for and a potent inhibitor of CYP3A4. Itraconazole is present in plasma with an approximately equal concentration of a biologically active metabolite, hydroxy-itraconazole. The native drug and metabolite are >99% bound to plasma proteins. Neither appears in urine or CSF. The t 1/2 of itraconazole at steady state is ~30-40 h. Steady-state levels of itraconazole are not reached for 4 days and those of hydroxy-itraconazole for 7 days; thus, loading doses are recommended when treating deep mycoses. Severe liver disease will increase itraconazole plasma concentrations, but azotemia and hemodialysis have no effect.
DRUG INTERACTIONS.Tables 57–4, 57–5, and 57–6 list select interactions of azoles with other drugs.
THERAPEUTIC USES. Itraconazole is the drug of choice for patients with indolent, nonmeningeal infections due to B. dermatitidis, H. capsulatum, P. brasiliensis, and Coccidioides immitis
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10161 | https://byjus.com/ncert-solutions-class-9-maths/chapter-12-herons-formula/ | NCERT Solutions Class 9 Maths Chapter 12 – Download Free PDF
According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 10.
NCERT Solutions for Class 9 Maths Chapter 12 – Heron’s Formula is provided here. Heron’s formula is a fundamental concept that finds significance in countless areas and is included in the CBSE Syllabus of Class 9 Maths. Therefore, it is imperative to have a clear grasp of the concept. And one of the best ways to do just that is by referring to the NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula. These solutions are designed by knowledgeable teachers with years of experience in accordance with the latest update on the CBSE syllabus 2023-24. The NCERT Solutions for Class 9 aim at equipping the students with detailed and step-wise explanations for all the answers to the questions given in the exercises of this Chapter.
Download Exclusively Curated Chapter Notes for Class 9 Maths Chapter – 12 Heron’s Formula
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Download Most Important Questions for Class 9 Maths Chapter – 12 Heron’s Formula
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Hence, one of the best guides you could adapt to your study needs is NCERT Solutions. Relevant topics are presented in an easy-to-understand format, barring the use of any complicated jargon. Furthermore, its content is updated as per the last CBSE syllabus and its guidelines.
Chapter 12-Heron’s Formula
Chapter 1 Number System
Chapter 2 Polynomials
Chapter 3 Coordinate Geometry
Chapter 4 Linear Equations in Two Variables
Chapter 5 Introduction to Euclids Geometry
Chapter 6 Lines and Angles
Chapter 7 Triangles
Chapter 8 Quadrilaterals
Chapter 9 Areas of Parallelograms and Triangles
Chapter 10 Circles
Chapter 11 Constructions
Chapter 12 Heron’s Formula
Chapter 13 Surface Areas and Volumes
Chapter 14 Statistics
Chapter 15 Introduction to Probability
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List of Exercises in NCERTclass 9 Maths Chapter 12:
Exercise 12.1 Solutions – 6 Questions
Exercise 12.2 Solutions – 9 Questions
Access Answers of NCERT Class 9 Maths Chapter 12 – Heron’s Formula
Exercise: 12.1 (Page No: 202)
1. A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?
Side of the signal board = a
Perimeter of the signal board = 3a = 180 cm
∴ a = 60 cm
Semi perimeter of the signal board (s) = 3a/2
By using Heron’s formula,
Area of the triangular signal board will be =
2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see Fig. 12.9). The advertisements yield an earning of ₹5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?
Solution:
The sides of the triangle ABC are 122 m, 22 m and 120 m respectively.
Now, the perimeter will be (122+22+120) = 264 m
Also, the semi perimeter (s) = 264/2 = 132 m
Using Heron’s formula,
Area of the triangle =
=1320 m2
We know that the rent of advertising per year = ₹ 5000 per m2
∴ The rent of one wall for 3 months = Rs. (1320×5000×3)/12 = Rs. 1650000
3. There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN” (see Fig. 12.10 ). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.
Solution:
It is given that the sides of the wall as 15 m, 11 m and 6 m.
So, the semi perimeter of triangular wall (s) = (15+11+6)/2 m = 16 m
Using Heron’s formula,
Area of the message =
= √[16(16-15)(16-11) (16-6)] m2
= √[16×1×5×10] m2 = √800 m2
= 20√2 m2
4. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42cm.
Solution:
Assume the third side of the triangle to be “x”.
Now, the three sides of the triangle are 18 cm, 10 cm, and “x” cm
It is given that the perimeter of the triangle = 42cm
So, x = 42-(18+10) cm = 14 cm
∴ The semi perimeter of triangle = 42/2 = 21 cm
Using Heron’s formula,
Area of the triangle,
=
= √[21(21-18)(21-10)(21-14)] cm2
= √[21×3×11×7] m2
= 21√11 cm2
5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.
Solution:
The ratio of the sides of the triangle are given as 12 : 17 : 25
Now, let the common ratio between the sides of the triangle be “x”
∴ The sides are 12x, 17x and 25x
It is also given that the perimeter of the triangle = 540 cm
12x+17x+25x = 540 cm
54x = 540cm
So, x = 10
Now, the sides of triangle are 120 cm, 170 cm, 250 cm.
So, the semi perimeter of the triangle (s) = 540/2 = 270 cm
Using Heron’s formula,
Area of the triangle
= 9000 cm2
6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Solution:
First, let the third side be x.
It is given that the length of the equal sides is 12 cm and its perimeter is 30 cm.
So, 30 = 12+12+x
∴ The length of the third side = 6 cm
Thus, the semi perimeter of the isosceles triangle (s) = 30/2 cm = 15 cm
Using Heron’s formula,
Area of the triangle
=
= √[15(15-12)(15-12)(15-6)] cm2
= √[15×3×3×9] cm2
= 9√15 cm2
Exercise: 12.2 (Page No: 206)
1. A park, in the shape of a quadrilateral ABCD, has C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
Solution:
First, construct a quadrilateral ABCD and join BD.
We know that
C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m
The diagram is:
Now, apply Pythagoras theorem in ΔBCD
BD2 = BC2 +CD2
⇒ BD2 = 122+52
⇒ BD2 = 169
⇒ BD = 13 m
Now, the area of ΔBCD = (½ ×12×5) = 30 m2
The semi perimeter of ΔABD
(s) = (perimeter/2)
= (8+9+13)/2 m
= 30/2 m = 15 m
Using Heron’s formula,
Area of ΔABD
= 6√35 m2 = 35.5 m2 (approximately)
∴ The area of quadrilateral ABCD = Area of ΔBCD+Area of ΔABD
= 30 m2+35.5m2 = 65.5 m2
2. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
Solution:
First, construct a diagram with the given parameter.
Now, apply Pythagorean theorem in ΔABC,
AC2 = AB2+BC2
⇒ 52 = 32+42
⇒ 25 = 25
Thus, it can be concluded that ΔABC is a right angled at B.
So, area of ΔBCD = (½ ×3×4) = 6 cm2
The semi perimeter of ΔACD (s) = (perimeter/2) = (5+5+4)/2 cm = 14/2 cm = 7 m
Now, using Heron’s formula,
Area of ΔACD
= 2√21 cm2 = 9.17 cm2 (approximately)
Area of quadrilateral ABCD = Area of ΔABC + Area of ΔACD = 6 cm2 +9.17 cm2 = 15.17 cm2
3. Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find the total area of the paper used.
Solution:
For the triangle I section:
It is an isosceles triangle and the sides are 5 cm, 1 cm and 5 cm
Perimeter = 5+5+1 = 11 cm
So, semi perimeter = 11/2 cm = 5.5 cm
Using Heron’s formula,
Area = √[s(s-a)(s-b)(s-c)]
= √[5.5(5.5- 5)(5.5-5)(5.5-1)] cm2
= √[5.5×0.5×0.5×4.5] cm2
= 0.75√11 cm2
= 0.75 × 3.317cm2
= 2.488cm2 (approx)
For the quadrilateral II section:
This quadrilateral is a rectangle with length and breadth as 6.5 cm and 1 cm respectively.
∴ Area = 6.5×1 cm2=6.5 cm2
For the quadrilateral III section:
It is a trapezoid with 2 sides as 1 cm each and the third side as 2 cm.
Area of the trapezoid = Area of the parallelogram + Area of the equilateral triangle
The perpendicular height of the parallelogram will be
= 0.86 cm
And, the area of the equilateral triangle will be (√3/4×a2) = 0.43
∴ Area of the trapezoid = 0.86+0.43 = 1.3 cm2 (approximately).
For triangle IV and V:
These triangles are 2 congruent right angled triangles having the base as 6 cm and height 1.5 cm
Area triangles IV and V = 2×(½×6×1.5) cm2 = 9 cm2
So, the total area of the paper used = (2.488+6.5+1.3+9) cm2 = 19.3 cm2
4. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Solution:
Given,
It is given that the parallelogram and triangle have equal areas.
The sides of the triangle are given as 26 cm, 28 cm and 30 cm.
So, the perimeter = 26+28+30 = 84 cm
And its semi perimeter = 84/2 cm = 42 cm
Now, by using Heron’s formula, area of the triangle =
= √[42(42-26)(42-28)(42-30)] cm2
= √[42×16×14×12] cm2
= 336 cm2
Now, let the height of parallelogram be h.
As the area of parallelogram = area of the triangle,
28 cm× h = 336 cm2
∴ h = 336/28 cm
So, the height of the parallelogram is 12 cm.
5. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?
Solution:
Draw a rhombus-shaped field first with the vertices as ABCD. The diagonal AC divides the rhombus into two congruent triangles which are having equal areas. The diagram is as follows.
Consider the triangle BCD,
Its semi-perimeter = (48 + 30 + 30)/2 m = 54 m
Using Heron’s formula,
Area of the ΔBCD =
= 432 m2
∴ Area of field = 2 × area of the ΔBCD = (2 × 432) m2 = 864 m2
Thus, the area of the grass field that each cow will be getting = (864/18) m2 = 48 m2
6. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig.12.16), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?
Solution:
For each triangular piece, The semi perimeter will be
s = (50+50+20)/2 cm = 120/2 cm = 60cm
Using Heron’s formula,
Area of the triangular piece
=
= √[60(60-50)(60-50)(60-20)] cm2
= √[60×10×10×40] cm2
= 200√6 cm2
∴ The area of all the triangular pieces = 5 × 200√6 cm2 = 1000√6 cm2
7. A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in Fig. 12.17. How much paper of each shade has been used in it?
Solution:
As the kite is in the shape of a square, its area will be
A = (½)×(diagonal)2
Area of the kite = (½)×32×32 = 512 cm2.
The area of shade I = Area of shade II
512/2 cm2 = 256 cm2
So, the total area of the paper that is required in each shade = 256 cm2
For the triangle section (III),
The sides are given as 6 cm, 6 cm and 8 cm
Now, the semi perimeter of this isosceles triangle = (6+6+8)/2 cm = 10 cm
By using Heron’s formula, the area of the III triangular piece will be
=
= √[10(10-6)(10-6)(10-8)] cm2
= √(10×4 ×4×2) cm2
= 8√5 cm2 = 17.92 cm2 (approx.)
8. A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see Fig. 12.18). Find the cost of polishing the tiles at the rate of 50p per cm2 .
Solution:
The semi perimeter of the each triangular shape = (28+9+35)/2 cm = 36 cm
By using Heron’s formula,
The area of each triangular shape will be
= 36√6 cm2 = 88.2 cm2
Now, the total area of 16 tiles = 16×88.2 cm2 = 1411.2 cm2
It is given that the polishing cost of tiles = 50 paise/cm2
∴ The total polishing cost of the tiles = Rs. (1411.2×0.5) = Rs. 705.6
9. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.
Solution:
First, draw a line segment BE parallel to the line AD. Then, from B, draw a perpendicular on the line segment CD.
Now, it can be seen that the quadrilateral ABED is a parallelogram. So,
AB = ED = 10 m
AD = BE = 13 m
EC = 25-ED = 25-10 = 15 m
Now, consider the triangle BEC,
Its semi perimeter (s) = (13+14+15)/2 = 21 m
By using Heron’s formula,
Area of ΔBEC =
= 84 m2
We also know that the area of ΔBEC = (½)×CE×BF
84 cm2 = (½)×15×BF
BF = (168/15) cm = 11.2 cm
So, the total area of ABED will be BF×DE i.e. 11.2×10 = 112 m2
∴ Area of the field = 84+112 = 196 m2
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| NCERT Exemplar for Class 9 Maths Chapter 12 |
| CBSE Notes for Class 9 Maths Chapter 12 |
Chapter 12 Heron’s Formula belongs to Unit 5: Mensuration. This unit carries a total of 13 marks out of 100. Therefore, this is an important chapter and should be studied thoroughly. The important topics that are covered in this chapter are:
Area of a Triangle – by Heron’s Formula
Application of Heron’s Formula in Finding Areas of Quadrilaterals
Heron’s formula helps us to find the area of a triangle with 3 side lengths. Besides the formula, Heron also contributed in other ways – the most notable one being the inventor of the very first steam engine called the Aeolipile. However, Heron couldn’t find any practical applications for it; instead, it ended up being used as a toy and an object of curiosity for the ancient Greeks.
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Dropped Topics – 12.1 Introduction and 12.3 Application of Heron’s formula in finding areas of quadrilaterals.
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How do you complete the square in quadratic equations?
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The method of completing the square is a way to solve a quadratic equation by rewriting it in a way that makes it easier to take the square root of both sides. To complete the square, we need to add a constant term that turns the expression into a perfect trinomial.
Example: Complete the square of the quadratic expression x 2+10 x.
To complete the square we need a constant term that turns the expression into a perfect square trinomial. Since the middle term in a perfect square trinomial is always 2 times the product of the square roots of the other two terms, we re-write the quadratic expression as:
x 2+2(5)(x)
We see that the constant term that we are seeking is (5)2:
x 2+2(5)(x)+(5)2
So, by adding (5)2 i.e., 25, the given expression can be factored as (x+5)2.
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10163 | https://oeis.org/A000027 | The OEIS is supported by the many generous donors to the OEIS Foundation.
The positive integers. Also called the natural numbers, the whole numbers or the counting numbers, but these terms are ambiguous.
(Formerly M0472 N0173)
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77
(list;
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graph;
refs;
listen;
history;
text;
internal format)
For some authors, the terms "natural numbers" and "counting numbers" include 0, i.e., refer to the nonnegative integers A001477; the term "whole numbers" frequently also designates the whole set of (signed) integers A001057.
a(n) is smallest positive integer which is consistent with sequence being monotonically increasing and satisfying a(a(n)) = n (cf. A007378).
Inverse Euler transform of A000219.
The rectangular array having A000027 as antidiagonals is the dispersion of the complement of the triangular numbers, A000217 (which triangularly form column 1 of this array). The array is also the transpose of A038722. - Clark Kimberling, Apr 05 2003
For nonzero x, define f(n) = floor(nx) - floor(n/x). Then f=A000027 if and only if x=tau or x=-tau. - Clark Kimberling, Jan 09 2005
Numbers of form (2^i)k for odd k (i.e., n = A006519(n)A000265(n)); thus n corresponds uniquely to an ordered pair (i,k) where i=A007814, k=A000265 (with A007814(2n)=A001511(n), A007814(2n+1)=0). - Lekraj Beedassy, Apr 22 2006
If the offset were changed to 0, we would have the following pattern: a(n)=binomial(n,0) + binomial(n,1) for the present sequence (number of regions in 1-space defined by n points), A000124 (number of regions in 2-space defined by n straight lines), A000125 (number of regions in 3-space defined by n planes), A000127 (number of regions in 4-space defined by n hyperplanes), A006261, A008859, A008860, A008861, A008862 and A008863, where the last six sequences are interpreted analogously and in each "... by n ..." clause an offset of 0 has been assumed, resulting in a(0)=1 for all of them, which corresponds to the case of not cutting with a hyperplane at all and therefore having one region. - Peter C. Heinig (algorithms(AT)gmx.de), Oct 19 2006
Define a number of points on a straight line to be in general arrangement when no two points coincide. Then these are the numbers of regions defined by n points in general arrangement on a straight line, when an offset of 0 is assumed. For instance, a(0)=1, since using no point at all leaves one region. The sequence satisfies the recursion a(n) = a(n-1) + 1. This has the following geometrical interpretation: Suppose there are already n-1 points in general arrangement, thus defining the maximal number of regions on a straight line obtainable by n-1 points, and now one more point is added in general arrangement. Then it will coincide with no other point and act as a dividing wall thereby creating one new region in addition to the a(n-1)=(n-1)+1=n regions already there, hence a(n)=a(n-1)+1. Cf. the comments on A000124 for an analogous interpretation. - Peter C. Heinig (algorithms(AT)gmx.de), Oct 19 2006
The sequence a(n)=n (for n=1,2,3) and a(n)=n+1 (for n=4,5,...) gives to the rank (minimal cardinality of a generating set) for the semigroup I_n\S_n, where I_n and S_n denote the symmetric inverse semigroup and symmetric group on [n]. - James East, May 03 2007
The sequence a(n)=n (for n=1,2), a(n)=n+1 (for n=3) and a(n)=n+2 (for n=4,5,...) gives the rank (minimal cardinality of a generating set) for the semigroup PT_n\T_n, where PT_n and T_n denote the partial transformation semigroup and transformation semigroup on [n]. - James East, May 03 2007
"God made the integers; all else is the work of man." This famous quotation is a translation of "Die ganzen Zahlen hat der liebe Gott gemacht, alles andere ist Menschenwerk," spoken by Leopold Kronecker in a lecture at the Berliner Naturforscher-Versammlung in 1886. Possibly the first publication of the statement is in Heinrich Weber's "Leopold Kronecker," Jahresberichte D.M.V. 2 (1893) 5-31. - Clark Kimberling, Jul 07 2007
Binomial transform of A019590, inverse binomial transform of A001792. - Philippe Deléham, Oct 24 2007
Writing A000027 as N, perhaps the simplest one-to-one correspondence between N X N and N is this: f(m,n) = ((m+n)^2 - m - 3n + 2)/2. Its inverse is given by I(k)=(g,h), where g = k - J(J-1)/2, h = J + 1 - g, J = floor((1 + sqrt(8k - 7))/2). Thus I(1)=(1,1), I(2)=(1,2), I(3)=(2,1) and so on; the mapping I fills the first-quadrant lattice by successive antidiagonals. - Clark Kimberling, Sep 11 2008
a(n) is also the mean of the first n odd integers. - Ian Kent, Dec 23 2008
Equals INVERTi transform of A001906, the even-indexed Fibonacci numbers starting (1, 3, 8, 21, 55, ...). - Gary W. Adamson, Jun 05 2009
These are also the 2-rough numbers: positive integers that have no prime factors less than 2. - Michael B. Porter, Oct 08 2009
Totally multiplicative sequence with a(p) = p for prime p. Totally multiplicative sequence with a(p) = a(p-1) + 1 for prime p. - Jaroslav Krizek, Oct 18 2009
Triangle T(k,j) of natural numbers, read by rows, with T(k,j) = binomial(k,2) + j = (k^2-k)/2 + j where 1 <= j <= k. In other words, a(n) = n = binomial(k,2) + j where k is the largest integer such that binomial(k,2) < n and j = n - binomial(k,2). For example, T(4,1)=7, T(4,2)=8, T(4,3)=9, and T(4,4)=10. Note that T(n,n)=A000217(n), the n-th triangular number. - Dennis P. Walsh, Nov 19 2009
Hofstadter-Conway-like sequence (see A004001): a(n) = a(a(n-1)) + a(n-a(n-1)) with a(1) = 1, a(2) = 2. - Jaroslav Krizek, Dec 11 2009
a(n) is also the dimension of the irreducible representations of the Lie algebra sl(2). - Leonid Bedratyuk, Jan 04 2010
Floyd's triangle read by rows. - Paul Muljadi, Jan 25 2010
Number of numbers between k and 2k where k is an integer. - Giovanni Teofilatto, Mar 26 2010
Generated from a(2n) = ra(n), a(2n+1) = a(n) + a(n+1), r = 2; in an infinite set, row 2 of the array shown in A178568. - Gary W. Adamson, May 29 2010
1/n = continued fraction [n]. Let barover[n] = [n,n,n,...] = 1/k. Then k - 1/k = n. Example: [2,2,2,...] = (sqrt(2) - 1) = 1/k, with k = (sqrt(2) + 1). Then 2 = k - 1/k. - Gary W. Adamson, Jul 15 2010
Number of n-digit numbers the binary expansion of which contains one run of 1's. - Vladimir Shevelev, Jul 30 2010
From Clark Kimberling, Jan 29 2011: (Start)
Let T denote the "natural number array A000027":
1 2 4 7 ...
3 5 8 12 ...
6 9 13 18 ...
10 14 19 25 ...
T(n,k) = n+(n+k-2)(n+k-1)/2. See A185787 for a list of sequences based on T, such as rows, columns, diagonals, and sub-arrays. (End)
The Stern polynomial B(n,x) evaluated at x=2. See A125184. - T. D. Noe, Feb 28 2011
The denominator in the Maclaurin series of log(2), which is 1 - 1/2 + 1/3 - 1/4 + .... - Mohammad K. Azarian, Oct 13 2011
As a function of Bernoulli numbers B_n (cf. A027641: (1, -1/2, 1/6, 0, -1/30, 0, 1/42, ...)): let V = a variant of B_n changing the (-1/2) to (1/2). Then triangle A074909 (the beheaded Pascal's triangle) [1, 1/2, 1/6, 0, -1/30, ...] = the vector [1, 2, 3, 4, 5, ...]. - Gary W. Adamson, Mar 05 2012
Number of partitions of 2n+1 into exactly two parts. - Wesley Ivan Hurt, Jul 15 2013
Integers n dividing u(n) = 2u(n-1) - u(n-2); u(0)=0, u(1)=1 (Lucas sequence A001477). - Thomas M. Bridge, Nov 03 2013
For this sequence, the generalized continued fraction a(1)+a(1)/(a(2)+a(2)/(a(3)+a(3)/(a(4)+...))), evaluates to 1/(e-2) = A194807. - Stanislav Sykora, Jan 20 2014
Engel expansion of e-1 (A091131 = 1.71828...). - Jaroslav Krizek, Jan 23 2014
a(n) is the number of permutations of length n simultaneously avoiding 213, 231 and 321 in the classical sense which are breadth-first search reading words of increasing unary-binary trees. For more details, see the entry for permutations avoiding 231 at A245898. - Manda Riehl, Aug 05 2014
a(n) is also the number of permutations simultaneously avoiding 213, 231 and 321 in the classical sense which can be realized as labels on an increasing strict binary tree with 2n-1 nodes. See A245904 for more information on increasing strict binary trees. - Manda Riehl, Aug 07 2014
a(n) = least k such that 2Pi - Sum_{h=1..k} 1/(h^2 - h + 3/16) < 1/n. - Clark Kimberling, Sep 28 2014
a(n) = least k such that Pi^2/6 - Sum_{h=1..k} 1/h^2 < 1/n. - Clark Kimberling, Oct 02 2014
Determinants of the spiral knots S(2,k,(1)). a(k) = det(S(2,k,(1))). These knots are also the torus knots T(2,k). - Ryan Stees, Dec 15 2014
As a function, the restriction of the identity map on the nonnegative integers {0,1,2,3...}, A001477, to the positive integers {1,2,3,...}. - M. F. Hasler, Jan 18 2015
See also A131685(k) = smallest positive number m such that c(i) = m (i^1 + 1) (i^2 + 2) ... (i^k+ k) / k! takes integral values for all i>=0: For k=1, A131685(k)=1, which implies that this is a well defined integer sequence. - Alexander R. Povolotsky, Apr 24 2015
a(n) is the number of compositions of n+2 into n parts avoiding the part 2. - Milan Janjic, Jan 07 2016
Does not satisfy Benford's law [Berger-Hill, 2017] - N. J. A. Sloane, Feb 07 2017
Parametrization for the finite multisubsets of the positive integers, where, for p_j the j-th prime, n = Product_{j} p_j^(e_j) corresponds to the multiset containing e_j copies of j ('Heinz encoding' -- see A056239, A003963, A289506, A289507, A289508, A289509). - Christopher J. Smyth, Jul 31 2017
The arithmetic function v_1(n,1) as defined in A289197. - Robert Price, Aug 22 2017
For n >= 3, a(n)=n is the least area that can be obtained for an irregular octagon drawn in a square of n units side, whose sides are parallel to the axes, with 4 vertices that coincide with the 4 vertices of the square, and the 4 remaining vertices having integer coordinates. See Affaire de Logique link. - Michel Marcus, Apr 28 2018
a(n+1) is the order of rowmotion on a poset defined by a disjoint union of chains of length n. - Nick Mayers, Jun 08 2018
Number of 1's in n-th generation of 1-D Cellular Automata using Rules 50, 58, 114, 122, 178, 186, 206, 220, 238, 242, 250 or 252 in the Wolfram numbering scheme, started with a single 1. - Frank Hollstein, Mar 25 2019
(1, 2, 3, 4, 5, ...) is the fourth INVERT transform of (1, -2, 3, -4, 5, ...). - Gary W. Adamson, Jul 15 2019
REFERENCES
T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 1.
T. M. Apostol, Modular Functions and Dirichlet Series in Number Theory, Springer-Verlag, 1990, page 25.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 22.
W. Fulton and J. Harris, Representation theory: a first course, (1991), page 149. [From Leonid Bedratyuk, Jan 04 2010]
I. S. Gradstein and I. M. Ryshik, Tables of series, products, and integrals, Volume 1, Verlag Harri Deutsch, 1981.
R. E. Schwartz, You Can Count on Monsters: The First 100 numbers and Their Characters, A. K. Peters and MAA, 2010.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
N. J. A. Sloane, Table of n, a(n) for n = 1..500000 [a large file]
Archimedes Laboratory, What's special about this number?
Affaire de Logique, Pick et Pick et Colegram (in French), No. 1051, 18-04-2018.
Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
James Barton, The Numbers
A. Berger and T. P. Hill, What is Benford's Law?, Notices, Amer. Math. Soc., 64:2 (2017), 132-134.
A. Breiland, L. Oesper, and L. Taalman, p-Coloring classes of torus knots, Online Missouri J. Math. Sci., 21 (2009), 120-126.
N. Brothers, S. Evans, L. Taalman, L. Van Wyk, D. Witczak, and C. Yarnall, Spiral knots, Missouri J. of Math. Sci., 22 (2010).
C. K. Caldwell, Prime Curios
Case and Abiessu, interesting number
S. Crandall, notes on interesting digital ephemera
O. Curtis, Interesting Numbers
M. DeLong, M. Russell, and J. Schrock, Colorability and determinants of T(m,n,r,s) twisted torus knots for n equiv. +/-1(mod m), Involve, Vol. 8 (2015), No. 3, 361-384.
Walter Felscher, Historia Matematica Mailing List Archive.
P. Flajolet and R. Sedgewick, Analytic Combinatorics, 2009; see page 371.
Robert R. Forslund, A logical alternative to the existing positional number system, Southwest Journal of Pure and Applied Mathematics, Vol. 1 1995 pp. 27-29.
E. Friedman, What's Special About This Number?
R. K. Guy, Letter to N. J. A. Sloane
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Kival Ngaokrajang, Illustration about relation to many other sequences, when the sequence is considered as a triangular table read by its antidiagonals. Additional illustrations when the sequence is considered as a centered triangular table read by rows.
Mike Keith, All Numbers Are Interesting: A Constructive Approach
Leonardo of Pisa [Leonardo Pisano], Illustration of initial terms, from Liber Abaci [The Book of Calculation], 1202 (photo by David Singmaster).
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See pp. 15, 24.
Robert Munafo, Notable Properties of Specific Numbers.
G. Pfeiffer, Counting Transitive Relations, Journal of Integer Sequences, Vol. 7 (2004), Article 04.3.2.
R. Phillips, Numbers from one to thirty-one
J. Striker, Dynamical Algebraic Combinatorics: Promotion, Rowmotion, and Resonance, Notices of the AMS, June/July 2017, pp. 543-549.
G. Villemin's Almanac of Numbers, NOMBRES en BREF (in French)
Eric Weisstein's World of Mathematics, Natural Number, Positive Integer, Counting Number Composition, Davenport-Schinzel Sequence, Idempotent Number, N, Smarandache Ceil Function, Whole Number, Engel Expansion, and Trinomial Coefficient
Wikipedia, List of numbers, Interesting number paradox, and Floyd's triangle
Robert G. Wilson v, English names for the numbers from 0 to 11159 without spaces or hyphens
Robert G. Wilson v, American English names for the numbers from 0 to 100999 without spaces or hyphens
Index entries for "core" sequences
Index entries for sequences of the a(a(n)) = 2n family
Index entries for sequences that are permutations of the natural numbers
Index entries for related partition-counting sequences
Index entries for linear recurrences with constant coefficients, signature (2,-1).
Index to divisibility sequences
Index entries for sequences related to Benford's law
FORMULA
a(2k+1) = A005408(k), k >= 0, a(2k) = A005843(k), k >= 1.
Multiplicative with a(p^e) = p^e. - David W. Wilson, Aug 01 2001
Another g.f.: Sum_{n>0} phi(n)x^n/(1-x^n) (Apostol).
When seen as an array: T(k, n) = n+1 + (k+n)(k+n+1)/2. Main diagonal is 2n(n+1)+1 (A001844), antidiagonal sums are n(n^2+1)/2 (A006003). - Ralf Stephan, Oct 17 2004
Dirichlet generating function: zeta(s-1). - Franklin T. Adams-Watters, Sep 11 2005
G.f.: x/(1-x)^2. E.g.f.: xexp(x). a(n)=n. a(-n)=-a(n).
Series reversion of g.f. A(x) is xC(-x)^2 where C(x) is the g.f. of A000108. - Michael Somos, Sep 04 2006
G.f. A(x) satisfies 0 = f(A(x), A(x^2)) where f(u, v) = u^2 - v - 4uv. - Michael Somos, Oct 03 2006
Convolution of A000012 (the all-ones sequence) with itself. - Tanya Khovanova, Jun 22 2007
a(n) = 2a(n-1)-a(n-2); a(1)=1, a(2)=2. a(n) = 1+a(n-1). - Philippe Deléham, Nov 03 2008
a(n) = A000720(A000040(n)). - Juri-Stepan Gerasimov, Nov 29 2009
a(n+1) = Sum_{k=0..n} A101950(n,k). - Philippe Deléham, Feb 10 2012
a(n) = Sum_{d | n} phi(d) = Sum_{d | n} A000010(d). - Jaroslav Krizek, Apr 20 2012
G.f.: x Product_{j>=0} (1+x^(2^j))^2 = x (1+2x+x^2) (1+2x^2+x^4) (1+2x^4+x^8) ... = x + 2x^2 + 3x^3 + ... . - Gary W. Adamson, Jun 26 2012
a(n) = det(binomial(i+1,j), 1 <= i,j <= n). - Mircea Merca, Apr 06 2013
E.g.f.: xE(0), where E(k) = 1 + 1/(x - x^3/(x^2 + (k+1)/E(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 03 2013
From Wolfdieter Lang, Oct 09 2013: (Start)
a(n) = Product_{k=1..n-1} 2sin(Pik/n), n > 1.
a(n) = Product_{k=1..n-1} (2sin(Pik/(2n)))^2, n > 1.
These identities are used in the calculation of products of ratios of lengths of certain lines in a regular n-gon. For the first identity see the Gradstein-Ryshik reference, p. 62, 1.392 1., bringing the first factor there to the left hand side and taking the limit x -> 0 (L'Hôpital). The second line follows from the first one. Thanks to Seppo Mustonen who led me to consider n-gon lengths products. (End)
a(n) = Sum_{j=0..k} (-1)^(j-1)jbinomial(n,j)binomial(n-1+k-j,k-j), k>=0. - Mircea Merca, Jan 25 2014
a(n) = A052410(n)^A052409(n). - Reinhard Zumkeller, Apr 06 2014
a(n) = Sum_{k=1..n^2+2n} 1/(sqrt(k)+sqrt(k+1)). - Pierre CAMI, Apr 25 2014
a(n) = floor(1/sin(1/n)) = floor(cot(1/(n+1))) = ceiling(cot(1/n)). - Clark Kimberling, Oct 08 2014
a(n) = floor(1/(log(n+1)-log(n))). - Thomas Ordowski, Oct 10 2014
a(k) = det(S(2,k,1)). - Ryan Stees, Dec 15 2014
a(n) = 1/(1/(n+1) + 1/(n+1)^2 + 1/(n+1)^3 + ...). - Pierre CAMI, Jan 22 2015
a(n) = Sum_{m=0..n-1} Stirling1(n-1,m)Bell(m+1), for n >= 1. This corresponds to Bell(m+1) = Sum_{k=0..m} Stirling2(m, k)(k+1), for m >= 0, from the fact that Stirling2Stirling1 = identity matrix. See A048993, A048994 and A000110. - Wolfdieter Lang, Feb 03 2015
a(n) = Sum_{k=1..2n-1}(-1)^(k+1)k(2n-k). In addition, surprisingly, a(n) = Sum_{k=1..2n-1}(-1)^(k+1)k^2(2n-k)^2. - Charlie Marion, Jan 05 2016
G.f.: x/(1-x)^2 = (x r(x) r(x^3) r(x^9) r(x^27) ...), where r(x) = (1 + x + x^2)^2 = (1 + 2x + 3x^2 + 2x^3 + x^4). - Gary W. Adamson, Jan 11 2017
a(n) = floor(1/(Pi/2-arctan(n))). - Clark Kimberling, Mar 11 2020
a(n) = Sum_{d|n} mu(n/d)sigma(d). - Ridouane Oudra, Oct 03 2020
a(n) = Sum_{k=1..n} phi(gcd(n,k))/phi(n/gcd(n,k)). - Richard L. Ollerton, May 09 2021
a(n) = S(n-1, 2), with the Chebyshev S-polynomials A049310. - Wolfdieter Lang, Mar 09 2023
From Peter Bala, Nov 02 2024: (Start)
For positive integer m, a(n) = (1/m) Sum_{k = 1..2mn-1} (-1)^(k+1) k (2mn - k) = (1/m) Sum_{k = 1..2mn-1} (-1)^(k+1) k^2 (2mn - k)^2 (the case m = 1 is given above).
a(n) = Sum_{k = 0..3n} (-1)^(n+k+1) k binomial(3n+k, 2k). (End)
MAPLE
A000027 := n->n; seq(A000027(n), n=1..100);
MATHEMATICA
Range@ 77 ( Robert G. Wilson v, Mar 31 2015 )
PROG
(Magma) [ n : n in [1..100]];
(PARI) {a(n) = n};
(R) 1:100
(Shell) seq 1 100
(Haskell)
a000027 = id
a000027_list = [1..] -- Reinhard Zumkeller, May 07 2012
(Maxima) makelist(n, n, 1, 30); / Martin Ettl, Nov 07 2012 /
(Python)
def A000027(n): return n # Chai Wah Wu, May 09 2022
(Julia) print([n for n in 1:280]) # Paul Muljadi, Apr 09 2024
(Perl) print join(", ", 1..280) # Paul Muljadi, May 29 2024
A001477 = nonnegative numbers.
Partial sums of A000012.
Cf. A001478, A001906, A007931, A007932, A027641, A074909, A089353 (multisets), A178568, A194807.
Cf. A026081 = integers in reverse alphabetical order in U.S. English, A107322 = English name for number and its reverse have the same number of letters, A119796 = zero through ten in alphabetical order of English reverse spelling, A005589, etc. Cf. A185787 (includes a list of sequences based on the natural number array A000027).
Cf. Boustrophedon transforms: A000737, A231179;
Cf. A038722 (mirrored when seen as triangle), A056011 (boustrophedon).
Cf. A048993, A048994, A000110 (see the Feb 03 2015 formula).
Cf. A289187, A000010, A008683, A000203, A049310.
Sequence in context: A131738 A199969 A303502 A001477 A087156 A254109
Adjacent sequences: A000024 A000025 A000026 A000028 A000029 A000030
core,nonn,easy,mult,tabl
N. J. A. Sloane
Links edited by Daniel Forgues, Oct 07 2009. |
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A set is a collection of well-defined items like numbers, letters, objects, or symbols, and we write them inside curly brackets { }. For example, a set can contain natural numbers like {1, 2, 3}, or even shapes like {circle, square, triangle}. The items in a set are called elements.
A subset is a smaller group formed by selecting some or all elements from a set. If all the elements of one set are also present in another set, then the first set is called a subset of the second. We use the symbol ⊆ to show this relationship.
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For example, if Set P has all odd numbers and Set Q has {1, 5, 7}, then Q is a subset of P, written as Q ⊆ P, because all the elements in Q are also in P. In this case, P is called the superset of Q.
What is a Subset?
A subset is a subgroup of any set. Consider two sets, A and B then A will be a subset of B if and only if all the components of A are present in B. We can also say that A is contained in B.
To understand the subset definition more clearly, consider a set P such that P comprises the names of all the cities of a country. Another set Q includes the names of cities in your region. Here Q will be a subset of P because all the cities in your region would also be cities of your country; hence, Q is a subset of P. There are only a definite number of distinct/unique subsets for any set, therefore the remaining are irrelevant and repetitive.
Subset Example 1: A subset as far as our understanding is a set contained in another set. It is like one can pick ice cream from the following flavours:{mango, chocolate, butterscotch}
You can take any one flavor {mango}, {chocolate}, or {butterscotch},
or any two flavors: {mango chocolate}, {chocolate, butterscotch}, or {mango, butterscotch}.
What is a Subset in Maths?
A Set 1 is supposed to be a subset of Set 2 if all the components of Set 1 are also existing in Set 2. In other words, set1 is included inside Set2.
If Set1={A,B,C} and Set2={A,B,C,D,E,F,G,H,I} then we can say that Set1 is a subset of Set2 as all the elements in set 1 are available in set 2.
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Subset Symbol
In the set theory, a subset is expressed by the symbol ⊆ and addressed as ‘is a subset of’. Applying this symbol we can represent subsets as follows:
P ⊆ Q: which is read as Set P is a subset of Set Q
Note: A subset can be identical to the set i.e, a subset can contain all the elements that are present in the set.
Subset Example 2: Find whether P is a subset of Q.
P = {set of even digits}, Q = {set of whole numbers}
Solution: The set of even numbers can be represented as:
P = {0, 2, 4, 6, 8, 10, 12 …}
Similarly, the set of all whole numbers can be represented as:
Q = {0, 1, 2, 3, 4, 5, 6, 7…}
From the set components of P and Q, we can figure out that the elements of P are present in the set Q. Therefore P is a subset of Q.
Subset Example 3: Determine whether P is a subset of Q.
P = {1, 3, 5, 7}, Q= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14…..}
It can be analyzed from the set elements that P’s elements relate to set Q. Hence P is a subset of Q.
Subset Example 4: Conclude whether X is a subset of Y.
X = {All writing material in a stationary workshop}, Y = {Pencils}
Set X involves pens, sketch pens, markers, pencils, notepads, etc. Whereas set Y only carries pencils. So we cannot state that all X’s elements are present in Y, which is a requirement for X to be a subset of Y. In this particular case, we can state that Y is a subset of X, but X is not a subset of Y.
Subset Example 5: Determine whether A is a subset of B.
A = {Toyota}, B = {All brands of cars}
Set B covers all brands of cars; Maruti Suzuki, Hyundai, Toyota Mahindra, Tata Motors, Mercedes Benz, etc. Moreover, A is a set of Toyota. Then we can say that all elements of A are incorporated into B. Hence, A is a subset of B.
Learn about Number Systems
All Subsets of a Set
The no. of subsets of a set of any set consisting of all likely sets including its components and the null set. Let us learn with an example.
Subset Example 6: Find all the subsets of set P = {2, 4, 6, 8}
Solution: Given, P = {2, 4, 6, 8}
Number of subsets of P are =
{}
{2}, {4}, {6}, {8},
{2, 4}, {2, 6}, {2, 8}, {4, 6},{4, 8}, {6, 8},
{2, 4, 6}, {4, 6, 8}, {2, 6, 8}, {2, 4, 8}
{2, 4, 6, 8}
Types of Subsets
There are 2 types of Subsets:
Proper Subsets
Improper Subsets
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Proper Subset
Any set say “P” is supposed to be a proper subset of Q if there is at least one element in Q, which is not available in set P. That is, a proper subset is one that contains a few components of the original set.
We can say that if P and Q are unequal sets and all elements of P are present in Q, then P is the proper subset of Q.
It is also termed a strict subset.
Proper Subset Symbol: A proper subset is expressed by ⊂ and is addressed as ‘is a proper subset of’. For example P ⊂ Q
Proper Subset Examples
Subset Example 7: Is P a proper subset of Q where P = {1, 3, 7, 8} and Q = {1, 3, 7, 8}?
Solution: The answer would be no, P is not a proper subset of Q as both are identical, and Q does not have any unique element, which is not existing in P.
Subset Example 8: Is X a proper subset of Y when X = {1, 6} and Y = {1, 4, 6, 8}?
Solution: The answer would be yes, X is a proper subset of Y as all the elements of X are present in Y and X is not equal to Y as well.
Learn about Roster Notation
Improper Subset
Suppose two sets, X and Y then X is an improper subset of Y if it includes all the elements of Y. This implies that an improper subset comprises every element of the primary set with the null set. The improper subset symbol is ⊆. For example P ⊆Q
Subset Example 9: If set P = {2, 3, 5}, then determine the number of subsets, proper subsets and improper subsets.
Number of subsets: {2}, {3}, {5}, {2,3}, {3, 5}, {2,5}, {2, 3, 5} and Φ or {}.
Proper Subsets: {}, {2}, {3}, {5}, {2,3}, {3, 5}, {2,5}
This can be denoted as {}, {2}, {3}, {5}, {2,3}, {3, 5}, {2,5} ⊂ P.
Improper Subset: {2, 3, 5}.
This can be denoted as {2, 3, 5} ⊆ P.
Learn about Finite and Infinite Sets
Subset Formula
The set theory symbols were explained by mathematicians to represent the collections of objects. If it is required to select n number of elements from a set including N number of elements, it can be performed in (^NC_n) ways.
Proper Subset Formula
If a set holds “n” elements, then the number of the subset for the given set is (2^n) and the number of proper subsets of the provided subset is calculated by the formula (2^n-1).
Subset Example 10: For a set P with the elements, P = {1, 2}, determine the proper subset.
The proper subset formula is (2^n-1) (where n is the number of elements in the set)
P = {1, 2}
Total number of elements (n) in the set=2
Hence the number of proper subset=(2^2−1) =3
Therefore the total number of proper subsets for the given set is { }, {1}, and {2}.
Subset Example 11: For the given set determine the power set.
Set Y={2,3,6}
Total number of components in the set Y=3
The power set of Y is:
P(Y)={},{2},{3},{6},{2,3},{3,6},{2,6},{2,3,6}
P(Y)=(2^n)
Substituting n=3
P(X)=(2^{3})=8
Learn about Vector Algebra
How to Represent Subsets
We are quite clear with what a subset is, now let us check some of the representations for the same. A subset, like any other set, is addressed with its elements inside curly braces.
Hence, consider two sets, X and Y:
X ⊆ Y: The above notation indicates that X is a subset of Y.
X ⊈ Y: Here the notation indicates that X is not a subset of Y.
X ⊂ Y: If X is a proper subset of Y, then it is expressed by the above notation.
X⊄ Y: If X is not a proper subset of Y, then we address it by the above notation.
Properties of Subsets
Some of the important properties of subsets are as follows:
Every set is said to be a subset of the provided set itself. Whether the set is finite or infinite, a set itself will be taken as the subset of itself.
For example, for a finite set A = {3,6}, all the possible subsets for the given data is:
A ={}, {3}, {6}, {3,6}.
As you can recognize, we have included a subset with identical elements as the initial set to satisfy the property.
We can state, an empty set is regarded as a subset of every set.
For example, take a finite set B = {a, b}, so all the possible subsets of this set are:
A = {}, {a}, {b}, {a, b}
If P is a subset of Q, then we can state that all the elements in P are available in Q.
Consider a set P= {2, 6, 9} and another set as Q = {2, 3, 4, 5, 6, 7, 8, 9}. Here we can say P is a subset of Q as all the elements of P are present in Q.
If set A is a subset of set B then we can assume that B is a superset of A.
There are (2^n) subsets and (2^n-1) proper subsets for a provided set of data.
Subsets of Real Numbers
Real numbers are the numbers we use every day, like fractions, decimals, and whole numbers. They can be written in decimal form and include different groups, called subsets. Let’s look at the main subsets of real numbers:
Rational Numbers These are numbers that can be written as a fraction (p/q), where both p and q are integers and q is not zero. In decimal form, they either end (like 0.25) or repeat (like 0.333...). Examples: 1/2, -5/9, 4, 0.75
Irrational Numbers These numbers cannot be written as a simple fraction. Their decimal forms go on forever without repeating. Examples: √2, π (pi), e
Integers These include positive numbers, negative numbers, and zero, but no fractions or decimals. Examples: -3, -1, 0, 2, 5
Whole Numbers These are all non-negative numbers starting from 0. They include zero and counting numbers but no negatives. Examples: 0, 1, 2, 10, 100
Natural Numbers These are the numbers used for counting, starting from 1. They do not include 0 or any negative number. Examples: 1, 2, 3, 50.
Subsets of Integers
Integers are numbers that include positive numbers, negative numbers, and zero. They do not include fractions or decimal numbers. The symbol used to represent integers is Z.
Here are the common subsets of integers:
Whole Numbers Whole numbers are all non-negative numbers starting from 0. They do not include any negative numbers. Examples: 0, 1, 2, 5, 100
Natural Numbers Natural numbers are also called counting numbers. These are the numbers we usually use to count things. Natural numbers start from 1 and go on. They do not include zero or negative numbers. Examples: 1, 2, 3, 10, 50
Representation of Subsets through Venn Diagrams
To understand the links between different sets, the Venn diagram is the most proper tool to reflect logical connections among certain sets. They are employed abundantly for the design of sets, more importantly for finite sets. A Venn diagram indicates the sets as the area inside a circular target with the elements as points inside the area.
As subsets usually involve two sets, we can easily practice a Venn diagram to illustrate and visualize them.
Subset Example 12: For the set X = {1, 3, 6} and set Y = {1, 3, 6, 9, 12, 15, 18} draw the venn diagram.
The Venn diagram illustration of sets X and Y are as follows:
As we can recognize from the diagram that X, surrounded by a region denoted by its set, is a portion of region Y. Each area has its elements expressed as points inside the region.
Difference between Proper and Improper Subsets
A subset is a set formed using some or all elements of another set.
There are two types of subsets: Proper Subset and Improper Subset.
Here’s how they are different:
| | |
--- |
| Proper Subset | Improper Subset |
| It includes some but not all elements of a set. | It includes all elements of the original set. |
| It is never equal to the original set. | It is always equal to the original set. |
| The total number of proper subsets of a set with n elements is 2ⁿ – 1. | A set has only one improper subset. |
| The symbol used is ⊂ (proper subset). | The symbol used is ⊆ (subset, including improper). |
Example: If A = {1, 2},
Proper subsets of A: {1}, {2}
Improper subset of A: {1, 2} (same as A)
Difference between Proper Subset and Superset
A set P is a proper subset of Q if Q possesses at least one component that is not present in P. It is indicated by the symbol ⊂ whereas Q will be the superset of P if and only if all the elements existing in P are a part of Q which states that Q is greater in size when compared to P. If P denotes the proper subset of Q then Q will be the superset of P. Denoted by the symbol ⊇.
Subset Example 13: For the given two sets X = {2, 4, 6, 8} and Y= {2, 4, 6} check if X is a superset of Y.
For X to be a superset of Y, it requires holding all the elements present in set Y. As we can notice from the given data that X has all the elements that are present in Y. Also, Y is a proper subset of X; hence, X is Y’s superset.
We hope that the above article on Subsets is helpful for your understanding and exam preparations. Stay tuned to the Testbook app for more updates on related topics from Mathematics, and various such subjects. Also, reach out to the test series available to examine your knowledge regarding several exams.
| |
| If you are checking Subsets article, also check related maths articles: |
| Intersection of sets | Set Difference |
| Set Theory Symbols | Cartesian Product of Sets |
| Multiplicative Identity | Union of Sets |
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FAQs For Subsets
What is a subset?
A subset, as the name implies, is a subgroup of any set. Consider two sets, A and B Mathematically speaking, A will be a subset of B if and only if all the components of A are present in B. We can also say that A is contained in B.
What is a proper subset?
Any set say "P" is supposed to be a proper subset of Q if there is at least one element in Q, which is not available in set P. That is, a proper subset is one that contains a few components of the original set.
How to determine the no of subsets of a set?
If a set holds “n” elements, then the number of the subset for the given set is (2^n).
What is an improper subset?
Suppose two sets, X and Y then X is an improper subset of Y if it includes all the elements of Y. This implies that an improper subset comprises every element of the primary set with the null set.
What is the proper subset symbol?
A proper subset is expressed by ⊂ and is addressed as ‘is a proper subset of’. For example: P ⊂
What is the difference between a subset and a proper subset?
A subset can be equal to the original set, but a proper subset must be smaller (not equal).
What is the empty set a subset of?
The empty set (∅) is a subset of every set.
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10165 | https://www.zhihu.com/question/293799396 | 高中几何体斜二测画法为什么会改变平面图形的表面积？ - 知乎
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高中几何体斜二测画法为什么会改变平面图形的表面积？
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高中数学
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高中几何体斜二测画法为什么会改变平面图形的表面积？
如题，最近刚学斜二测画法画直观图，有类型题问: 斜二测画法所作出的图形面积是原平面图形的几倍？当然，斜二测画法y‘轴上的高变为原来的1/2，如果按照平…显示全部
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他所说的表面积减小是指，你看上去用斜二测平面图形表现出来的面积，比你直接正视这个图形表面积小，不是说改变了原物体的面积。
这是正视看到的图
这是斜视看到的图
可以很明显看出，第二个图表现出来的面积比第一个图要小了(仅看红圈围出的面积)，但事实上，你从视觉直观上，并不会觉得第二张图凳子的真实面积比第一张图小，只要像斜二测画法那样规定出变换关系(斜45°，y轴压缩一半)，那样不论是从定性角度还是定量角度，都不会有任何的信息丢失。
而斜二测画法的好处是，它不仅可以将一个需要三个平面投影才能画出的立体图形用一个平面图形表示出来，而且还比较合乎人们的直观认识，图形画出来非常直观容易理解，从这个角度上看，当然是有意义的
最后另附斜二测画法产生的面积比的推导
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卧龙05学长国涛
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高中数学知识：基本立体图形，直观图，几何体的表面积和体积，空间点、直线、平面之间的关系等知识总结。30 赞同 · 5 评论 文章
我们在学习的时候知道空间图形的直观图是通过斜二测画法得到的，首先我们要知道斜二测画法的相关规则。
我们知道在斜二测画法中，横坐标和纵坐标夹角为45度，原图的横坐标保持不变，但是纵坐标变为原来的一半，这个时候我们通过计算得知，原图的面积通过斜二测画法得到的直观图面积比原图要小。
下面图片为斜二测画法的步骤和规则，你可以看看了解一下就可以了解了。
总而言之，你只要知道斜二测画法的步骤和规则，自然也就知道如何通过斜二测画法得到相应几何体的直观图，同时也就知道它的表面积会发生变化了。
其实无论哪门课程的学习都要把基础的知识概念弄明白，这样才能更好地学习并掌握相应的更深层次的学习。学习就是这样，从最简单的知识学起，步步为营，最终就可以不断取得进步。
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袁规炉
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你的理解不对，这种作图是为了立体几何的直观，并没有改变图形表面积，你用向量直接计算就能看出来；发生改变的一般出现在投影上，但是那是因为存在两面夹角。
发布于 2020-06-02 01:01
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10166 | https://math.stackexchange.com/questions/3242528/how-can-i-express-a-median-in-terms-of-barycentric-coordinates | computational geometry - how can I express a median in terms of barycentric coordinates? - Mathematics Stack Exchange
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how can I express a median in terms of barycentric coordinates?
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I am trying to understand relationship between Median and barycentric coordinates.
here is a figure (fig_1) comes from wiki
that post gives a good explanation and I am trying to connect the notation in that post to fig_1.
Vertex of such small triangle at the first median has B.C.(baricentric coordinates) like (g,f,f), where g+f+f=1, at the second median (f,g,f), and (f,f,g) at the third one. We can see that for g=1/3 triangle size is zero (coefficient cf= 0), for g=1 triangle is equal to large triangle (cf = 0).
in this context, is it correct to consider point E as (g,f,f)?
computational-geometry
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edited May 28, 2019 at 9:40
Bernard
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asked May 28, 2019 at 9:32
JJJohnJJJohn
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Every point on the line segment A E A E has barycentric coordinates of the form (g,f,f)(g,f,f) where g+2 f=1 g+2 f=1. In particular, E E corresponds to g=0,f=1/2 g=0,f=1/2.user856 –user856 2019-05-28 10:59:08 +00:00 Commented May 28, 2019 at 10:59
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Yes, a way to understand the situation is as follows. Let x,y,z x,y,z be the normed barycentric coordinates of O O, x+y+z=1 x+y+z=1. By definition:
O=x A+y B+z C.O=x A+y B+z C.
(Understand this either using affix points A,B,C A,B,C, or using a "testing point" P P and rewrite with vectors the relation as P O=x P A+y P B+z P C P O=x P A+y P B+z P C.)
These x,y,z x,y,z are barycentric coordinates, thus coordinates "in the plane", and the argument reduces them, their symmetry to one "on the line". Let us see how. Just rewrite:
O E=x A+(y+z)E,where=y y+z B+z y+z C.O=x A+(y+z)E,where E=y y+z B+z y+z C.
(From the first relation, E E is on O A O A, from the second relation it is on B C B C, so it is indeed the point E E in the picture.)
Since E E is the middle of B C B C, we must have the corresponding coordinates "in the line" equal, so y/(y+z)=z/(y+z)y/(y+z)=z/(y+z), so y=z y=z.
By symmetry to y=z y=z we also get x=y x=y (and x=z x=z), so x=y=z=1/3 x=y=z=1/3.
Note that already knowing the porportion O E:A E O E:A E lets us know the x x.
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answered May 28, 2019 at 11:00
dan_fuleadan_fulea
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thanks for your answer! is $O = xA+yB+zC$ in Cartesian coordinates or in barycentric coordinates?JJJohn –JJJohn 2019-05-28 12:09:13 +00:00 Commented May 28, 2019 at 12:09
2 I will give an answer using examples. See also faculty.evansville.edu/ck6/encyclopedia/ETC.html for more on trilinear and barycentric coordinates. For the intersection of medians we have the (not normalized) coordinates (1:1:1)(1:1:1). To make them sum to one, we normalize to (1/3,1/3,1/3)(1/3,1/3,1/3). Then the relation G=1 3 A+1 3 B+1 3 C G=1 3 A+1 3 B+1 3 C (using G G instead of O O) holds in cartesian coordinates, for instance, if A(1,2)A(1,2), B(2,3)B(2,3), C(3,−2)C(3,−2), then we simply build 1 3(A+B+C)=(1 3(1+2+3),1 3(2+3−2))1 3(A+B+C)=(1 3(1+2+3),1 3(2+3−2)).dan_fulea –dan_fulea 2019-05-28 14:15:05 +00:00 Commented May 28, 2019 at 14:15
1 A very good description is here: web.evanchen.cc/handouts/bary/bary-short.pdfdan_fulea –dan_fulea 2019-05-28 14:15:11 +00:00 Commented May 28, 2019 at 14:15
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10167 | https://pmc.ncbi.nlm.nih.gov/articles/PMC5766074/ | Lymphadenitis associated with cat-scratch disease simulating a neoplasm: Imaging findings with histopathological associations - PMC
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Lymphadenitis associated with cat-scratch disease simulating a neoplasm: Imaging findings with histopathological associations
Ying Chen
Ying Chen
1 Department of Radiology, Second Affiliated Hospital, Zhejiang University School of Medicine, Hangzhou, Zhejiang 310009, P.R. China
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1,, Yan-Biao Fu
Yan-Biao Fu
2 Department of Pathology, Second Affiliated Hospital, Zhejiang University School of Medicine, Hangzhou, Zhejiang 310009, P.R. China
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2,, Xiu-Fang Xu
Xiu-Fang Xu
3 Department of Radiology, Hangzhou Medical College, Hangzhou, Zhejiang 310053, P.R. China
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3, Yao Pan
Yao Pan
1 Department of Radiology, Second Affiliated Hospital, Zhejiang University School of Medicine, Hangzhou, Zhejiang 310009, P.R. China
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1, Chen-Ying Lu
Chen-Ying Lu
1 Department of Radiology, Second Affiliated Hospital, Zhejiang University School of Medicine, Hangzhou, Zhejiang 310009, P.R. China
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1 Department of Radiology, Second Affiliated Hospital, Zhejiang University School of Medicine, Hangzhou, Zhejiang 310009, P.R. China
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1 Department of Radiology, Second Affiliated Hospital, Zhejiang University School of Medicine, Hangzhou, Zhejiang 310009, P.R. China
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1 Department of Radiology, Second Affiliated Hospital, Zhejiang University School of Medicine, Hangzhou, Zhejiang 310009, P.R. China
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1 Department of Radiology, Second Affiliated Hospital, Zhejiang University School of Medicine, Hangzhou, Zhejiang 310009, P.R. China
2 Department of Pathology, Second Affiliated Hospital, Zhejiang University School of Medicine, Hangzhou, Zhejiang 310009, P.R. China
3 Department of Radiology, Hangzhou Medical College, Hangzhou, Zhejiang 310053, P.R. China
✉
Correspondence to: Dr Ri-Sheng Yu, Department of Radiology, Second Affiliated Hospital, Zhejiang University School of Medicine, 88 Jiefang Road, Hangzhou, Zhejiang 310009, P.R. China, E-mail: cjr.yurisheng@vip.163.com
Contributed equally
Received 2016 Aug 31; Accepted 2017 Sep 28; Issue date 2018 Jan.
Copyright: © Chen et al.
This is an open access article distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivs License, which permits use and distribution in any medium, provided the original work is properly cited, the use is non-commercial and no modifications or adaptations are made.
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PMCID: PMC5766074 PMID: 29399138
Abstract
The lymphadenitis associated with cat-scratch disease (CSD) is often confused with neoplasms by a number of radiologists and clinicians, and consequently, unnecessary invasive procedures or surgeries are performed. In the present study, the contrast-enhanced computed tomography (CT) and magnetic resonance imaging (MRI) findings of 10 patients (6 men and 4 women) with clinically and pathologically confirmed lymphadenitis associated with CSD were retrospectively analyzed (CT in 3 patients, MRI in 6 patients, and CT and MRI in 1 patient) at The Second Affiliated Hospital of Zhejiang University School of Medicine (Hangzhou, China) between January 2007 and November 2014. As a result, 17 enlarged lymph nodes were identified in 10 cases. The 5 nodes identified by CT scan exhibited relatively inhomogeneous isodensity to muscle, with patchy low density in the center. All 14 nodes identified by MRI scan exhibited homogeneous or heterogeneous isointensity to muscle or slightly increased intensity compared with that of muscle on T1-weighted images (T1WI), and homogeneous or heterogeneous hyperintensity on fat-suppressed T2WI. Following enhancement, all 17 enlarged lymph nodes associated with CSD demonstrated the following 3 different enhancement patterns: Moderate homogeneous enhancement (n=8), which was associated with histologically identified early disease stage; marked heterogeneous enhancement with no enhancement of the necrotic areas (n=4), and heterogeneous enhancement with progressively ‘spoke-wheel-like’ (defined as radiating enhancement from the center) enhancement of the patchy low-density area (n=1), which was associated with histologically identified intermediate disease stage; and astral low-density/hypointensity with marked enhancement (n=2) or a ‘rose flower’ sign (n=2), which was associated with histologically identified late disease stage. We hypothesized that the CT and MRI results of lymphadenitis in CSD may be associated with the pathological features. It may be suggested that the diagnosis of CSD may be formed when considering the characteristic CT and MRI features of astral low-density/hypointensity with marked enhancement or a ‘rose flower’ sign (defined as marginal petaloid enhancement) in the late disease stage, or the MRI results of homogeneous, moderate enhancement in the early disease stage, or the CT/MRI data of heterogeneous enhancement with non-enhancing area in the center in the intermediate disease stage, in solitary or multiple enlarged lymph nodes associated with general subcutaneous edema in the vicinity of the nodes on CT/MRI and with a history of cat exposure.
Keywords: lymphadenitis, cat-scratch disease, computed tomography, magnetic resonance imaging, pathology
Introduction
Cat-scratch disease (CSD) is an infrequent infection caused by Bartonella henselae, a Gram-negative, facultative intracellular bacillus acquired from exposure to cats or cat fleas (1–3). Bartonella henselae are able to contaminate feline saliva and infect humans through scratching or biting (4), with the lowest incidence occurring in mountainous, arid regions and the highest incidence occurring in the humid, warm climate of the South of the United States (5). As Bartonella species are difficult to culture, the serological study of specific immunoglobulin (Ig) G is generally recommended in order to obtain a CSD diagnosis. A positive IgM test indicates acute infection; however, the production of IgM antibodies only occurs over a very short period of time is rarely detected in CSD (6).
The typical manifestation of CSD is a regional lymphadenitis, which presents in ~90% of cases as a subacute regional lymphadenopathy (7), and it occurs predominantly in children and young adults (8). Although the lymphadenitis usually spontaneously resolves, or may be treated with a short course of antibiotics (9–11), it occasionally progresses to severe life-threatening complications. Atypical CSD may present with a wide range of clinical manifestations, including neuroretinitis, periadenitis, hepatosplenic lesions, osteomyelitis, endocarditis, encephalitis and fevers of unknown origins (12–14), and occurs in 5% of immunocompetent patients. Additionally, the lymphadenitis associated with CSD is often confused with neoplasms by a number of radiologists and clinicians, as they are not familiar with this pseudo-tumor lesion (15). Therefore, the early consideration of this disorder will avoid unnecessary invasive procedures or surgery.
The estimated annual incidence of CSD in the United States is 4.5–9.3 outpatient diagnoses per 100,000 individuals and 0.19–0.86 hospital admissions per 100,000 individuals (16,17). However, a review of the English literature revealed that <20 cases of lymphadenitis associated with CSD on contrast-enhanced computed tomography (CT) and magnetic resonance imaging (MRI) data have been well addressed worldwide (15,18–20) since its first description in 1889 by Parinaud (21), and the majority of these studies were of sporadic cases and small case series. Additionally, no distinguishable early, intermediate or late stage imaging manifestations or pathological features of CSD have been identified.
The present study highlights the predominant radiological and pathological features in 10 cases of regional lymphadenitis associated with CSD, focusing on the association between the early, intermediate and late stages of the imaging results, and the pathological features of each stage, which, to the best of our knowledge, have not been described in-depth previously in the English literature.
Patients and methods
Patients
The present study was approved by the Institutional Research and Ethics Board of The Second Affiliated Hospital of Zhejiang University School of Medicine (Hangzhou, China) and the requirement for patient informed consent was waived. Enhanced CT (n=3), enhanced MRI (n=6), combined enhanced CT and MRI (n=1), clinical (n=10) and pathological (n=10) features of 10 cases of CSD were retrospectively reviewed at The Second Affiliated Hospital of Zhejiang University School of Medicine between January 2007 and November 2014. The differences in the diagnostic techniques used are due to the retrospective nature of the study. Among the 10 patients, there were 6 men and 4 women, with ages ranging from 18 to 57 years (mean, 40.9 years). The duration of symptoms ranged from 7 days to 8 years.
The clinical signs and symptoms included painless nodes (n=5), painful nodes (n=3), painful swelling (n=3) and mild fever (n=2). A total of 2 patients without any complaints were incidentally identified by a routine physical examination. A total of 7 patients had pet cats and 3 patients had pet dogs, but they all initially denied a history of cat or dog scratches. Physical examinations were unremarkable, with the exception of tenderly mobile nodular lesions with a smooth surface and slight tenderness at the elbow (n=6), neck (n=2), groin (n=1) and axilla (n=1). Laboratory tests indicated leukocytosis (11.4×10 9/l and 12.8×10 9/l; normal range, 4.0–10.0×10 9/l) and slightly elevated C-reactive protein (20.5 and 31.8 mg/l; normal range, <10.0 mg/l) in 2 cases. No evidence of peripheral eosinophilia was identified. Blood cultures and viral serological tests were all negative. Tuberculin tests (purified protein derivative test) were all negative. The expression levels of various tumor markers (carcinoembryonic antigen, α-fetoprotein, carbohydrate antigen (CA)19-9, CA125, CA242, neuron-specific enolase, squamous cell carcinoma antigen, prostate-specific antigen and β-human chorionic gonadotrophin) were all negative and no primary malignancy was identified in all 10 patients. No serological or skin tests for CSD were performed.
Radiological examination
CT examinations were performed in 4 patients using 16-row CT scanners (Siemens Somatom Sensation 16; Siemens AG, Munich, Germany) with the following parameters: 6-mm axial sections ranging from the plane of the ilium to the last plane of the inferior pubic ramus for the pelvic cavity in 1 patient; 4-mm axial sections ranging from the plane of the mandible to the last plane of the thyroid cartilage for the neck in 1 patient; 6-mm axial sections ranging from the plane of the clavicle to the last plane of the shoulder blade for the axilla in 1 patient; and 5-mm axial sections ranging from the plane of the middle humerus to the last plane of the middle ulna for the elbow in 1 patient. All patients were examined in a fasting state with plain scanning at first, and then an 80 ml/bolus injection of non-ionic contrast medium (OMNIPAQUE 300; GE Healthcare, Chicago, IL, USA) was administered via the antecubital vein for multiple-phase enhanced scanning.
MRI examinations were performed using a 1.5 T or 3.0 T magnet (SIGNA™; GE Healthcare). The elbows in 6 patients and the neck in 1 patient all underwent spin-echo (SE) T1-weighted imaging (T1WI), turbo SE (TSE) fat-suppression T2WI, and axial, coronal and sagittal contrast-enhanced T1WI (Omniscan 0.1 mmol/kg body weight; GE Healthcare, Chicago, IL, USA) with fat suppression. The imaging parameters of the elbow scans were as follows: Repetition time/echo time (TR/TE) of T1WI, 240–700/3.6–13 msec; and TR/TE of T2WI, 2,840–2,900/72-85 msec. The matrix was 256×256, the number of excitations (NEX) was 1.0–2.0, the field-of-view (FOV) was 140×140-180×200 mm, and slice thickness was 4.0 mm, with a 0.4–0.5 mm interslice gap. The imaging parameters of the neck scans were as follows: T1WI TR/TE, 400–600/8-9.9 msec; T2WI TR/TE, 4,020–4,820/84.8 msec; NEX, 2.0–3.0; 256×256 matrix; FOV, 140×140-180×200 mm; and slice thickness, 4.0 mm, with no interslice gap.
Image analysis
All CT and MRI images were reviewed separately by 2 radiologists who had no knowledge of the clinical and pathological outcomes. Discordance between opinions was resolved by consensus. The CT and MRI evaluation included the site, number, size, density or signal intensity and the enhancement patterns of the loco-regional lesions, and the existence of strands in the surrounding soft tissue.
Pathological analysis
The pathological specimens of lymph node tissues were obtained by resection in 6 patients and by percutaneous needle biopsy in 4 patients. The lymph node tissue samples were fixed with 10% neutral formaldehyde solution for 24 h at 18°C, embedded in paraffin and cut into 3-µm-thick sections. The sections were subsequently stained with routine hematoxylin and eosin for 2 h at 18°C prior to being dehydrated with 95% alcohol 5 min (4 times) and xylene 5 min (twice), and sealed with neutral balsam. Subsequently, the structure of the tissue cells was observed at ×200 magnification using a BX50 Olympus microscope (Olympus Corporation, Tokyo, Japan).
All specimen analyses were confirmed by an experienced pathologist for diagnostic accuracy. The histological staging of the lesions and the association between the enhanced imaging data and the pathological features were evaluated in each case, with a particular focus on the coagulation or liquid necrosis, astral abscess and the presence of fibrosis.
Results
Imaging data
In total, there were 17 enlarged lymph nodes identified in 10 cases, of which 12 nodes were scanned on enhanced MRI, 3 on enhanced CT, and 2 on enhanced CT and MRI. Among them, 13 nodes were located at the elbow, 2 in the neck, 1 in the groin and 1 in the axilla. Multiple lesions were identified in 4 cases, and singular lesions in 6 patients. All the lesions were round, well-delineated nodes or nodules. The size of the enlarged lymph nodes ranged from 0.8–3.0 cm. On the CT scans, all the 5 nodes identified exhibited relatively inhomogeneous isodensity to muscle, with patchy or astral low density in the center. Following enhancement, 2 of the 5 nodes demonstrated marked heterogeneous enhancement, with no enhancement of the patchy low-density central areas, 1 of the 5 nodes displayed heterogeneous enhancement with progressive ‘spoke-wheel-like’ (defined as radiating enhancement from the center) enhancement of the patchy low-density area (Fig. 1), and the other 2 nodes demonstrated heterogeneous enhancement with peripheral flower ring reinforcement of the astral low-density area, also known as the ‘rose flower’ sign.
Figure 1.
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Lymphadenopathy in a 57-year-old man with cat-scratch disease at the intermediate stage in the left neck. The node (arrow) displays heterogeneous enhancement with ‘spoke-wheel-like’ enhancement of the patchy low-density area on an enhanced computed tomography scan following intravenous contrast administration.
On the MRI scans, all the 14 nodes identified were homogeneously or heterogeneously isointense to muscle or with slightly increased intensity compared with that of muscle on T1WI, and homogeneously or heterogeneously hyperintense on fat-suppressed T2WI. Subsequent to enhancement, 8 of the 14 nodes displayed moderately homogeneous enhancement with no necrotic areas (Fig. 2) and 2 of the 14 nodes demonstrated marked heterogeneous enhancement with no enhancement of the necrotic central areas, which demonstrated patchy hypointense on T1WI and patchy hyperintense on T2WI in the center (Fig. 3A-C). Another 2 of the 14 nodes exhibited heterogeneous enhancement with marked enhancement of the astral hypointense area on T1WI and T2WI (Fig. 4A-C), and the other 2 exhibited heterogeneous enhancement with marginal petaloid enhancement of the astral hypointense area, namely the ‘rose flower’ sign (Fig. 5A-C). In addition, all 10 cases demonstrated general subcutaneous edema in the vicinity of the nodes.
Figure 2.
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Multiple lymph node involvement in a 50-year-old man with cat-scratch disease at the early stage in the left elbow. The nodes (arrows) exhibit moderately homogeneous enhancement with no necrotic areas on an enhanced fat-suppressed sagittal T1-weighted imaging scan.
Figure 3.
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Lymphadenopathy in a 51-year-old woman with cat-scratch disease at the intermediate stage in the left elbow. (A) The node (arrow) indicates heterogeneous hyperintensity with patchy hyperintensity (arrow head) on a fat-suppressed coronal T2-weighted imaging scan, with general subcutaneous edema (star) in the vicinity of the node. (B) Following enhancement, the node demonstrates marked heterogeneous enhancement with no enhancement of the necrotic central areas (arrowhead) on an enhanced fat-suppressed axial T1-weighted imaging scan.
Figure 4.
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Lymphadenopathy in an 18-year-old man with cat-scratch disease at the late stage in the left elbow. (A) The node (arrow) exhibits heterogeneous isointensity to muscle with astral hypointensity (arrowhead) on T1WI, and (B) heterogeneous hyperintensity with astral hypointensity (arrowhead) on fat-suppressed T2WI, with general subcutaneous edema (star) in the vicinity of the node. (C) Following enhancement, the node (black arrow) exhibits heterogeneous enhancement with marked enhancement of the astral hypointense area (black arrowhead) on an enhanced fat-suppressed axial T1-weighted imaging scan. (D) The two nodes (white arrow) demonstrate heterogeneous enhancement with marked enhancement of the astral hypointense area (white arrowhead) on an enhanced fat-suppressed coronal T1-weighted imaging scan.
Figure 5.
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Lymphadenopathy in a 37-year-old woman with cat-scratch disease at the late stage in the right elbow. (A) The node (arrow) demonstrates heterogeneous isointensity to muscle with astral hypointensity (arrowhead) on T1WI, and (B) heterogeneous hyperintensity with astral hypointensity (arrowhead) on fat-suppressed T2WI, with general subcutaneous edema (star) in the vicinity of the node. (C) Following enhancement, the node (arrow) demonstrates heterogeneous enhancement with peripheral flower ring reinforcement of the astral hypointense area (arrowhead) on an enhanced axial T1-weighted imaging scan. (D) The two nodes (arrow) exhibit heterogeneous enhancement with peripheral flower ring reinforcement of the astral hypointense area (arrowhead), on an enhanced sagittal T1-weighted imaging scan.
Pathological features
There were 3 stages of CSD histologically, including early stage (n=3), intermediate stage (n=4) and late stage (n=3). In the early stage, there were numerous granulomas with proliferation of histiocytes, lymphoid follicles and neutrophilic aggregation (Fig. 6). In the intermediate stage, there were irregular granulomas with stellate abscesses composed of marked central necrosis surrounded by an inner layer of palisading histiocytes, an intermediate lymphocytic rim, and an outermost zone of fibrosis (Fig. 7). In the late stage, the inflammatory and necrotic regions were replaced or surrounded by the spindle collagen-producing cells, which were composed of variable amounts of fibroblasts or fibrocytes (Fig. 8).
Figure 6.
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Histological microscopy indicates that there are numerous granulomas with proliferation of histiocytes, lymphoid follicles and neutrophilic aggregation in the early stage of cat-scratch disease (hematoxylin and eosin staining; magnification, ×200).
Figure 7.
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Histological microscopy indicates that there are irregular granulomas with stellate abscesses composed of marked central necrosis surrounded by an inner layer of palisading histiocytes (arrow), an intermediate lymphocytic rim (arrowhead) and an outermost zone of fibrosis (star) in the intermediate stage of cat-scratch disease (hematoxylin and eosin staining; magnification, ×100).
Figure 8.
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Histological microscopy demonstrates that the inflammatory and necrotic regions are replaced or surrounded by the spindle collagen-producing cells (star), which are composed of variable amounts of fibroblasts or fibrocytes, in the late stage of cat-scratch disease (hematoxylin and eosin staining; magnification, ×100).
Enhanced imaging data associated with the pathological features
There were 3 types of CT and MRI enhancement patterns in the 17 enlarged lymph nodes associated with CSD. The different enhancement patterns were associated with different pathological features (Tables I–III). According to Tables I–III, the 8 (47.1%) nodes with moderately homogeneous enhancement (all on MRI) belonged to the histologically determined early stage (stage total, 47.1%); the 4 (23.5%) nodes with marked heterogeneous enhancement with no enhancement of the necrotic central areas (2 on CT and 2 on MRI) and heterogeneous enhancement with progressively spoke-wheel-like enhancement of the patchy low-density area in 1 (5.9%) node (on CT) belonged to the histologically determined intermediate stage (stage total, 29.4%); and the 2 (11.8%) nodes with astral low-density/hypointensity with marked enhancement (both on MRI) or the ‘rose flower’ sign in 2 (11.8%) nodes (both on CT/MRI) belonged to the histologically determined late stage (stage total, 23.6%).
Table I.
Imaging data of lymphadenitis associated with cat-scratch disease is associated with pathological data in the early disease stage.
| Case no. | Sex | Age, years | Site | Lesions, n | Examination method | Density or intensity features on unenhanced CT/MRI | Enhanced imaging data on CT/MRI | Pathological data |
:---: :---: :---: :---:
| 1 | Female | 28 | Elbow | 1 | MRI | Homogeneous isointensity to muscle on T1WI and homogeneous hyperintensity on fat-suppressed T2WI | Moderately homogeneous enhancement with no necrotic area | Numerous granulomas with proliferation of histiocytes, lymphoid follicle and neutrophilic aggregation |
| 2 | Female | 42 | Elbow | 1 | MRI | | | |
| 3 (Fig. 2) | Male | 50 | Elbow | 5 | MRI | | | |
| 4 (Fig. 3) | Female | 51 | Elbow | 1 | MRI | | | |
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Table III.
Imaging data of lymphadenitis associated with cat-scratch disease is associated with pathological data in the late disease stage.
| Case no. | Sex | Age, years | Site | Lesion Number | Examination method | Density or intensity features on unenhanced CT/MRI | Enhanced imaging data on CT/MRI | Pathological data |
:---: :---: :---: :---:
| 9 (Fig. 4) | Male | 18 | Elbow | 2 | MRI | Heterogeneous isointensity to muscle with astral hypointense area on T1WI and heterogeneous hyperintensity with astral hypointense area on fat-suppressed T2WI | Heterogeneous enhancement with clear enhancement of the astral hypointense areas | Inflammatory and coagulation-necrosis regions surrounded by fibrinopurulent debris, and stellate fibrosis |
| 10 (Fig. 5) | Female | 37 | Elbow | 2 | MRI | Heterogeneous enhancement with peripheral flower ring reinforcement of the astral hypointense area, namely the ‘rose flower’ sign | | |
| | | | | | CT | Relatively inhomogeneous isodensity to muscle with astral low density in the center | Heterogeneous enhancement with peripheral flower ring reinforcement of the astral low-density area, namely the ‘rose flower’ sign | |
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MRI, magnetic resonance imaging; CT, computed tomography; T1WI, T1-weighted imaging; T2WI, T2-weighted imaging.
Discussion
CSD is an infrequent zoonotic infectious illness characterized by regional lymphadenopathy and usually caused by the Rickettsia-like microorganism termed Bartonella (formerly Rochalimæa) henselae, a gram-negative coccobacillus (18,22). The condition is most commonly diagnosed in individuals between the ages of 5 and 21 years (18,23). In the present study, there was only 1 patient belonging to this age group, which was not compatible with other studies in the literature. The majority of patients with the disease confirmed recent contact with a cat, usually a kitten, however, occasionally it was not identified by the patient initially. In the present study, although 7 patients had pet cats and 3 patients had pet dogs, none of them recalled this characteristic history prior to surgery or biopsy. Therefore, it is extremely important to obtain a detailed disease history prior to diagnoses.
The clinical manifestation of CSD is variable, with the majority of cases exhibiting non-specific symptoms, such as papules or pustules appearing at the site of the scratch between 7 and 12 days after exposure, which usually heal spontaneously in days to weeks. During the next 1–3 weeks, the enlargement of regional lymph nodes occurs. On the basis of the frequency of scratches, the involved lymph nodes are situated at upper extremities (46%), neck (26%), groin (17%) or other locations (24,25). The medial epitrochlear region is the most prevalent location (18,20). In the patients of the present study, 6/10 cases of CSD were located at the elbow, probably due to a high frequency of scratches on the hands. A total of 2 patients exhibited lymphadenopathy in the neck, 1 in the inguinal region and the other in the axilla, results which were in accordance with those of previous studies (24,25). The majority of enlarged lymph nodes often regress over weeks to months, and rarely may persist for up to 2 years (25,26). In the present study, the duration of enlarged lymph nodes ranged from 7 days to 8 years, which was the longest recorded time to persist. Regional lymphadenitis may be accompanied by systemic symptoms, including malaise and mild fever (20). In the cases of the present study, 2/10 exhibited mild fever. Physical examination often reveals a painful swelling or node, which was demonstrated in 3/10 cases.
All the cases of the present study were consistent with the following diagnosis criteria (9,27) of CSD: i) History of animal contact (particularly cats or kittens) with a regional inoculation lesion or scratch mark; ii) laboratory data negative for other infectious causes of lymphadenopathy; and iii) histopathological data demonstrating granulomatous inflammation compatible with CSD.
A literature review revealed that only a small number of radiological data of lymphadenitis in CSD have been described previously (15,18–20), and to the best of our knowledge, the majority of these studies consisted of sporadic cases and small case series. The 10 cases of lymphadenitis in CSD of the present study may be the largest so far. The involved nodes may be single or multiple in nature, and range in size from 1–5 cm (20,28). Single nodal involvement has been identified in 44–85% of patients (18) and multiple nodal involvement at a single site has been observed in 24% of cases (18,20). All the multiple lesions were round, well-delineated nodes or nodes with no fusion change (15,20). Therefore, single nodes were more common compared with multiple nodes (15,18–20). In the present study population, 4 patients presented with multiple node involvement, and 6 with single node involvement, with sizes ranging from 0.8–3.0 cm in all nodes. These findings were similar to the results of previous studies (15,18–20).
The enhanced CT and MRI data of lymphadenitis in CSD were associated with the pathological features, and the different enhancement patterns may reflect different pathological stages of lesions. To the best of our knowledge, the association between radiological data with the pathological features of lymphadenitis in CSD have not been well addressed.
According to previous studies (15,18–20), and the 10 cases of lymphadenitis in CSD in the present study, the most common plain CT images demonstrated nodes that were ovoid, well-delineated and with relatively inhomogeneous isodensity to muscle, with patchy low density in the center. The corresponding unenhanced MRI revealed nodes that were heterogeneously isointense to muscle or with slightly increased intensity compared with that of muscle on T1WI scans, with patchy hypointensity in the center or periphery, and heterogeneous hyperintensity with patches of increased hyperintensity in the center or intermediate signal intensity with patchy hyperintensity in the periphery on fat-suppressed T2WI. Following intravenous contrast administration, the peripheral or central areas demonstrated marked enhancement on the plain CT, whereas on the enhanced CT image, the central or peripheral areas indicated no enhancement at all, or heterogeneous enhancement with progressive ‘spoke-wheel-like’ enhancement of the patchy low-density central area. The corresponding histological analysis demonstrated irregular granulomas with stellate abscesses composed of marked central necrosis, surrounded by an inner layer of palisading histiocytes, an intermediate lymphocytic rim and an outermost zone of fibrosis. This pattern was observed in 5/17 (29.4%) nodes in the present study, and we hypothesized that the non-enhancing areas of the scanning images corresponded to areas of liquid necrosis, and that they belonged to the intermediate disease stage, as identified histologically. Although it was difficult to differentiate CSD at this point from tuberculous granuloma with liquid necrosis by radiological data (29), if associated with the special disease history of cat exposure, it may be diagnosed correctly at a relatively early time-point. It may be that heterogeneous enhancement with progressive ‘spoke-wheel-like’ enhancement of the central patchy low-density area on CT images is the characteristic manifestation of lymphadenitis in CSD, which has not been well addressed previously to the best of our knowledge, and requires additional cases for verification.
The second most common radiological result of lymphadenitis in CSD was of relatively uniform density on CT or homogeneous signal intensity on both T1WI and T2WI, with homogeneous, moderate enhancement following intravenous contrast administration. The corresponding histological analysis revealed that there were numerous granulomas, with proliferation of histiocytes, lymphoid follicles and neutrophilic aggregation. We hypothesized that homogeneous, moderate enhancement corresponded with inflammatory granulomas, and that they were indicative of the early stage of infection without marked necrosis, which was consistent with previous reports (20,30), as they were observed histologically in 8 (47.1%) nodes in the present study. As for the intermediate stage, although it was difficult to differentiate CSD at this point from other chronic granulomatous diseases using radiological data (29), if associated with the special disease history of cat exposure, CSD may also be diagnosed correctly at an early time-point.
The less common radiological result of lymphadenitis in CSD from the present study was exhibited as heterogeneous low-density on the CT scans, heterogeneous isointensity to muscle with astral hypointense areas on T1WI and heterogeneous hyperintensity with an astral hypointense area on fat-suppressed T2WI, which was demonstrated in 4 (23.5%) nodes. Following intravenous contrast administration, there were 2 different enhancement patterns in these nodes: i) Heterogeneous enhancement, with marked enhancement of the astral hypointense areas; and ii) heterogeneous enhancement with a ‘rose flower’ sign enhancement of the astral hypointense area. The corresponding histological analysis demonstrated that the inflammatory and necrotic regions were replaced or surrounded by spindle collagen-producing cells, which were composed of variable amounts of fibroblasts or fibrocytes, and we hypothesized that these 2 different enhancement patterns corresponded to the astral low-density/hypointensity area of the nodes composed of variable amounts of fibroblasts or fibrocytes. In cases where immature fibroblasts were the primary component of the astral low-density/hypointensity area, marked enhancement was observed, and in cases where the mature fibrocytes were the primary component in the center surrounded by the fibroblasts, there was no enhancement in the center surrounded by peripheral flower ring reinforcement, indicating that it belonged to a histologically-determined late disease stage. This stage exhibited marked characteristics in the radiological data, particularly the ‘rose flower’ sign, which had not been described previously in the English language literature, to the best of our knowledge.
Imaging data in CSD usually also demonstrates extensive surrounding edema in the area of lymphatic drainage proximal to the site of inoculation (15). In particular, MRI T2WI images exhibited surrounding edema more clearly compared with CT scans, which was a characteristic of inflammatory lesions, generally not appearing in benign tumors and lymphoma.
For differential diagnosis, the most important factor of cat exposure and imaging features of astral low-density/hypointensity area in enlarged lymph nodes with clear enhancement or the ‘rose flower’ sign associated with the general subcutaneous edema in the vicinity of the nodes may differentiate CSD from soft-tissue hemangioma, neurogenic tumors lymphoma, metastasis and soft-tissue sarcoma, although it is difficult to form the diagnosis of CSD without a history of cat exposure.
The lymphadenitis in CSD often resolves spontaneously within 3 months. Antibiotics are not indicated in the majority of cases, but they may be considered for severe or systemic disease involving the bone marrow, liver, spleen, brain and eyes (31–35), particularly in immunocompromised patients. A total of 10–35% of the infected nodes progress to suppuration, and evacuating the pus is necessary in this condition (36).
In summary, the diagnosis of CSD should be considered in relatively young patients with lymphadenopathy in the upper extremities or head and neck region who also exhibit i) astral low-density/hypointensity areas with marked enhancement or the ‘rose flower’ sign on CT/MRI scans in the late stages of infection, or ii) homogeneous, moderate enhancement on MRI scans in the early stages of infection, or iii) heterogeneous enhancement with non-enhancing areas on CT/MRI scans or heterogeneous enhancement with progressively ‘spoke-wheel-like’ enhancement of the central patchy low-density area on CT scans in the intermediate stages of infection, in single or multiple enlarged lymph nodes associated with general subcutaneous edema in the vicinity of the nodes on CT/MRI. This should be in addition to a history or a renewed anamnesis of exposure to a cat, therefore avoiding unnecessary surgical resection of this pseudotumor. MRI scans were more advantageous compared with CT in displaying the imaging features in CSD, particularly the characteristics of marked enhancement of an astral hypointensity area or the ‘rose flower’ sign.
Table II.
Imaging data of lymphadenitis associated with cat-scratch is associated with pathological data in the intermediate disease stage.
| Case no. | Sex | Age, years | Site | Lesion Number | Examination method | Density or intensity features on unenhanced CT/MRI | Enhanced imaging data on CT/MRI | Pathological data |
:---: :---: :---: :---:
| 4 (Fig. 3) | Female | 51 | Elbow | 1 | MRI | Heterogeneous isointensity to muscle with patchy hypointensity in the center on T1WI and heterogeneous hyperintensity with patches of increased hyperintensity in the center on fat-suppressed T2WI | Clear heterogeneous enhancement with no enhancement of the necrotic central areas | Irregular granulomas with stellate abscesses composed of clear central necrosis surrounded by an inner layer of palisading histiocytes, an intermediate lymphocytic rim, and an outermost zone of fibrosis |
| 5 | Male | 38 | Groin | 1 | CT | Relatively inhomogeneous isodensity to muscle with patchy low density in the center | | |
| 6 | Male | 53 | Axilla | 1 | CT | | | |
| 7 | Male | 35 | Neck | 1 | MRI | Heterogeneous slightly increased intensity compared with that of muscle on T1WI with patchy hypointensity in the center on T1WI and heterogeneous hyperintensity with patchy hyperintensity in the center on fat-suppressed T2WI | | |
| 8 (Fig. 1) | Male | 57 | Neck | 1 | CT | Relatively inhomogeneous isodensity to muscle with patchy low density in the center | Heterogeneous enhancement with progressively spoke-and-wheel enhancement of the patchy low-density area | |
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MRI, magnetic resonance imaging; CT, computed tomography; T1WI, T1-weighted imaging; T2WI, T2-weighted imaging.
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10169 | https://stackoverflow.com/questions/63021828/solving-modular-linear-congruences-for-large-numbers | Skip to main content
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Solving modular linear congruences for large numbers
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I'm looking for a better algorithm than one I found on stackoverflow to handle 4096 byte numbers, i'm hitting a maximum recursion depth.
Code from stackoverlow post, taken from this Stack Overflow post:
```
def linear_congruence(a, b, m):
if b == 0:
return 0
if a < 0:
a = -a
b = -b
b %= m
while a > m:
a -= m
return (m linear_congruence(m, -b, a) + b) // a
```
This works fine for smaller numbers, for example:
```
In : pow_mod(8261, 63, 4033)
63 1 8261 4033
31 195 1728 4033
15 2221 1564 4033
7 1231 2098 4033
3 1518 1601 4033
1 2452 2246 4033
0 2147 3266 4033
Out: 2147
And the linear congruence works:
linear_congruence(8261, 3266, 4033):
2147
```
But i hit maximum recursion depth with larger numbers. Is there a better algorithm or non recursive algorithm of the linear_congruence algorithm i provided?
Based on Eric Postpischil's remark, i wrote the pseudocode from the wikipedia entry and created a very fast linear congruence algorithm utilizing the method from here: .
This works nicely on pows with a powers of 2-1, to get the answer. I'm looking into how offsetting from this changes the answer and hope to incorporate it to work for those answers as well, but for now, i have what i need since i'm working with powers of 2 -1 for y in pow(x, y, z):
```
def fastlinearcongruencex(powx, divmodx, N, withstats=False):
x, y, z = egcditerx(powx, N, withstats)
if x > 1:
powx//=x
divmodx//=x
N//=x
if withstats == True:
print(f"powx = {powx}, divmodx = {divmodx}, N = {N}")
x, y, z = egcditerx(powx, N)
if withstats == True:
print(f"x = {x}, y = {y}, z = {z}")
answer = (ydivmodx)%N
if withstats == True:
print(f"answer = {answer}")
return answer
def egcditerx(a, b, withstats=False):
s = 0
r = b
old_s = 1
old_r = a
while r!= 0:
quotient = old_r // r
old_r, r = r, old_r - quotient r
old_s, s = s, old_s - quotient s
if withstats == True:
print(f"quotient = {quotient}, old_r = {old_r}, r = {r}, old_s = {old_s}, s = {s}")
if b != 0:
bezout_t = quotient = (old_r - old_s a) // b
if withstats == True:
print(f"bezout_t = {bezout_t}")
else:
bezout_t = 0
if withstats == True:
print("Bézout coefficients:", (old_s, bezout_t))
print("greatest common divisor:", old_r)
return old_r, old_s, bezout_t
```
It even works instantaneously on 4096 byte numbers which is great:
```
In : rpowxxxwithbitlength(1009,offset=0, withstats=True, withx=True, withbl=True)
63 1 272 1009
31 272 327 1009
15 152 984 1009
7 236 625 1009
3 186 142 1009
1 178 993 1009
0 179 256 1009
Out: (179, 256, True, 272)
In : fastlinearcongruencex(272,256,1009)
Out: 179
```
Thank you Eric for pointing out what this was, i wrote an extremely fast linear congruence algorithm utilizing egcd and the procedure from the pdf above. If any stackoverflowers need a fast algorithm, please point them to this one. I also learned that congruence is always maintained when the pow(x,y,z) has a y off of the powers of 2-1. I will look into this further to see if an offset change exists to keep the answers intact and will follow up in the future if found.
python
math
linear-programming
modulo
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edited Dec 9, 2023 at 12:53
The_spider
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asked Jul 21, 2020 at 19:32
oppressionslayeroppressionslayer
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6
Do you know in advance that a is relatively prime to m? If not, how do you want to handle cases involving multiple solutions, or no solutions?
– Mark Dickinson
Commented
Jul 21, 2020 at 19:37
If i could get multiple solutions that would be great, i have noticed that it's not 100% so i'm glad to hear there are multiple solutions. I'm only using this for numbers that are pow mod, so i haven't seen a case of no solutions so i'm not sure how i would want to handle those, maybe just get no answer. Right now, if i don't get the right answer, i do the next number above and usually that works.
– oppressionslayer
Commented
Jul 21, 2020 at 19:39
Also, i don't know that a is relatively prime to m. I'm using an algorithim to generate a, and working on a personal educational math project to see if i can formulate numbers that i can predict the last modulus x. i know, that sounds impossible, but it's a fun project, and i'm having fun working on something that sounds impossible :-)
– oppressionslayer
Commented
Jul 21, 2020 at 19:45
3
Rewrite the code as a loop, not a recursive call. This is the Extended Euclidean Algorithm and should not be written with recursion.
– Eric Postpischil
Commented
Jul 21, 2020 at 20:57
The source was most likely this answer.
– beaker
Commented
Jul 21, 2020 at 21:47
|
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If you have Python 3.8 or later, you can do everything you need to with a very small number of lines of code.
First some mathematics: I'm assuming that you want to solve ax = b (mod m) for an integer x, given integers a, b and m. I'm also assuming that m is positive.
The first thing you need to compute is the greatest common divisor g of a and m. There are two cases:
if b is not a multiple of g, then the congruence has no solutions (if ax + my = b for some integers x and y, then any common divisor of a and m must also be a divisor of b)
if b is a multiple of g, then the congruence is exactly equivalent to (a/g)x = (b/g) (mod (m/g)). Now a/g and m/g are relatively prime, so we can compute an inverse to a/g modulo m/g. Multiplying that inverse by b/g gives a solution, and the general solution can be obtained by adding an arbitrary multiple of m/g to that solution.
Python's math module has had a gcd function since Python 3.5, and the built-in pow function can be used to compute modular inverses since Python 3.8.
Putting it all together, here's some code. First a function that finds the general solution, or raises an exception if no solution exists. If it succeeds, it returns two integers. The first gives a particular solution; the second gives the modulus that provides the general solution.
```
def solve_linear_congruence(a, b, m):
""" Describe all solutions to ax = b (mod m), or raise ValueError. """
g = math.gcd(a, m)
if b % g:
raise ValueError("No solutions")
a, b, m = a//g, b//g, m//g
return pow(a, -1, m) b % m, m
```
And then some driver code, to demonstrate how to use the above.
```
def print_solutions(a, b, m):
print(f"Solving the congruence: {a}x = {b} (mod {m})")
try:
x, mx = solve_linear_congruence(a, b, m)
except ValueError:
print("No solutions")
else:
print(f"Particular solution: x = {x}")
print(f"General solution: x = {x} (mod {mx})")
```
Example use:
```
print_solutions(272, 256, 1009)
Solving the congruence: 272x = 256 (mod 1009)
Particular solution: x = 179
General solution: x = 179 (mod 1009)
print_solutions(98, 105, 1001)
Solving the congruence: 98x = 105 (mod 1001)
Particular solution: x = 93
General solution: x = 93 (mod 143)
print_solutions(98, 107, 1001)
Solving the congruence: 98x = 107 (mod 1001)
No solutions
```
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edited Jul 22, 2020 at 16:42
answered Jul 22, 2020 at 16:30
Mark DickinsonMark Dickinson
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1
Thanks! I tried this out in a VM as i don't have 3.8, but that is an awesome addition. This works great
– oppressionslayer
Commented
Jul 23, 2020 at 1:10
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Suppose that for some reason the linear congruence equations you'll be 'attacking' come up 'empty' (no solutions) often enough to be a design criteria for your algorithm.
It turns out that you can use just (with any real overhead) the residue operations to answer that binary question -
There exist solutions XOR There are no solutions
This might have utility in cryptography; see also the abstract,
Introduction of the Residue Number Arithmetic Logic Unit
With Brief Computational Complexity Analysis
Once you determine that a solution exist, you can use back substitution
and the ALU to determine a solution.
Also, you'll have calculated the gcd(a,m) and can construct the coefficients of Bézout's identity
(if you need them).
Following is python program that incorporates the above ideas; it calculates the minimal solution when it exists and prints out Bézout's identity.
```
test_data = [ \
(32,12,82), \
(9,3,23), \
(17,41,73), \
(227,1,2011), \
(25,15,29), \
(2,22,71), \
(7,10,21), \
(124,58,900), \
(46, 12, 240), \
]
for lc in test_data:
LC = lc
back_sub_List = []
while True:
back_sub_List.append(LC)
n_mod_a = LC % LC
if n_mod_a == 0:
break
LC = (n_mod_a, -LC % LC, LC)
gcd_of_a0_n0 = LC
if LC % LC != 0:
print(f"No solution for {back_sub_List}x = {back_sub_List} (mod {back_sub_List})")
else:
k = 0
for LC in back_sub_List[::-1]: # solve with back substitution
a,b,m = LC
k = (b + km) // a # optimize calculation since the remainder is zero?
print(f"The minimal solution for {back_sub_List}x = {back_sub_List} (mod {back_sub_List}) is equal to {k}")
# get bezout
S = [1,0]
T = [0,1]
for LC in back_sub_List:
a,b,n = LC
q = n // a
s = S - q S
S = [S, s]
t = T - q T
T = [T, t]
print(f" Bézout's identity: ({S})({lc}) + ({T})({lc}) = {gcd_of_a0_n0}")
```
PROGRAM OUTPUT
```
The minimal solution for 32x = 12 (mod 82) is equal to 26
Bézout's identity: (-7)(82) + (18)(32) = 2
The minimal solution for 9x = 3 (mod 23) is equal to 8
Bézout's identity: (2)(23) + (-5)(9) = 1
The minimal solution for 17x = 41 (mod 73) is equal to 11
Bézout's identity: (7)(73) + (-30)(17) = 1
The minimal solution for 227x = 1 (mod 2011) is equal to 1320
Bézout's identity: (78)(2011) + (-691)(227) = 1
The minimal solution for 25x = 15 (mod 29) is equal to 18
Bézout's identity: (-6)(29) + (7)(25) = 1
The minimal solution for 2x = 22 (mod 71) is equal to 11
Bézout's identity: (1)(71) + (-35)(2) = 1
No solution for 7x = 10 (mod 21)
Bézout's identity: (0)(21) + (1)(7) = 7
No solution for 124x = 58 (mod 900)
Bézout's identity: (4)(900) + (-29)(124) = 4
The minimal solution for 46x = 12 (mod 240) is equal to 42
Bézout's identity: (-9)(240) + (47)(46) = 2
```
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edited Nov 25, 2020 at 1:29
answered Nov 24, 2020 at 18:35
CopyPasteItCopyPasteIt
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I've recognized that to compute Bézout's coefficients a computer algorithm should use exact integer division. Of interest: In 1993 Tudor Jebelean wrote two papers (1) A Generalization of the Binary GCD Algorithm (2) An Algorithm for Exact Division
– CopyPasteIt
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Nov 25, 2020 at 8:47
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10170 | https://genomely.substack.com/p/similar-by-chance-understanding-convergent | Similar by Chance? Understanding Convergent Evolution
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Similar by Chance? Understanding Convergent Evolution
Convergent evolution and four classic examples thereof
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Convergent evolution is a fascinating and significant concept in the field of evolutionary biology, illustrating how different species can independently evolve similar traits in response to analogous environmental pressures or ecological niches. This phenomenon underscores the adaptability and diversity of life, offering insights into the mechanisms of evolution and the predictability of evolutionary outcomes. In this blog post, we delve into the intricacies of convergent evolution, exploring its principles, examples, and implications for our understanding of life on Earth.
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Understanding Convergent Evolution
Convergent evolution occurs when organisms that are not closely related independently evolve similar features or functions as a result of having to adapt to similar environments or ecological roles. This convergence can be seen in various aspects, including physical structures, behaviors, and even molecular processes. It is a testament to the power of natural selection, shaping organisms in a way that best suits their survival and reproduction in their respective habitats.
Principles of Convergent Evolution
The driving force behind convergent evolution is natural selection, the process by which certain heritable traits become more common in a population because they are associated with increased survival and reproductive success. When different species face similar selective pressures—such as climate, predation, or food availability—they may independently develop analogous adaptations that help them thrive in similar ecological niches.
Convergent evolution is distinct from parallel evolution, where related species evolve similar traits, and divergent evolution, where closely related species evolve different traits. It highlights the creativity of evolution, producing similar solutions to life's challenges across disparate lineages.
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Examples of Convergent Evolution
The natural world is replete with examples of convergent evolution, showcasing the adaptability of life across the globe. Some notable instances include:
Wings in Bats and Birds
Source: evolution.berkeley.edu
Despite their distant common ancestry, both bats and birds have evolved wings that enable flight. This adaptation to a flying lifestyle involves significantly different structures (mammalian limbs vs. avian feathers) but serves the same functional purpose.
Eye Structure in Cephalopods and Vertebrates
Napoli et al. 2022
The complex eyes of cephalopods (like squids and octopuses) and those of vertebrates (including humans) have evolved independently. Both eye types have lenses, retinas, and irises, yet their common ancestors had only simple light-sensitive spots.
Genomely is a reader-supported publication. To receive new posts and support my work, consider becoming a free or paid subscriber. Subscribe
3. Echolocation in Bats and Dolphins
r/fascinating
Echolocation, the ability to navigate and find food using sound waves, has independently evolved in bats and dolphins. These animals emit sounds that bounce off objects, returning echoes that provide information about the surroundings.
Cactus Plants in the Americas and Euphorbias in Africa
Plant convergent evolution
Both cacti and euphorbias have developed thick, fleshy stems for water storage and spines for protection, adaptations to arid environments. Despite their similarities, they belong to very different plant families.
Implications for Evolutionary Biology
Convergent evolution challenges the notion of evolution as a purely random process, suggesting that there are predictable paths that evolution may take in response to similar environmental challenges. It also raises intriguing questions about the limits of adaptation and the potential for predicting evolutionary outcomes.
Furthermore, convergent evolution has practical implications for the study of biodiversity and conservation. Understanding how similar traits evolve in different species can inform conservation strategies, especially in the face of changing climates and habitats.
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Conclusion
Convergent evolution is a testament to the adaptability and diversity of life on Earth. It highlights how similar environmental pressures can shape the evolution of unrelated species, leading to remarkable examples of similarity across the tree of life. By studying these convergent patterns, scientists can gain deeper insights into the principles of natural selection, the adaptability of organisms, and the predictability of evolutionary pathways. As we continue to explore the natural world, the concept of convergent evolution will undoubtedly play a crucial role in enhancing our understanding of the complexity and interconnectedness of life.
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10171 | https://math.libretexts.org/Courses/Fresno_City_College/Math_3A%3A_College_Algebra_-_Fresno_City_College/03%3A_Linear_and_Quadratic_Functions/3.01%3A_Linear_Functions | Skip to main content
3.1: Linear Functions
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: Jul 26, 2022
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3: Linear and Quadratic Functions
3.2: Graphs of Linear Functions
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Just as with the growth of a bamboo plant, there are many situations that involve constant change over time. Consider, for example, the first commercial maglev train in the world, the Shanghai MagLev Train (maglev train in the world, the Shanghai MagLev Train (Figure ). It carries passengers comfortably for a 30-kilometer trip from the airport to the subway station in only eight minutes.
Suppose a maglev train were to travel a long distance, and that the train maintains a constant speed of 83 meters per second for a period of time once it is 250 meters from the station. How can we analyze the train’s distance from the station as a function of time? In this section, we will investigate a kind of function that is useful for this purpose, and use it to investigate real-world situations such as the train’s distance from the station at a given point in time.maglev train were to travel a long distance, and that the train maintains a constant speed of 83 meters per second for a period of time once it is 250 meters from the station. How can we analyze the train’s distance from the station as a function of time? In this section, we will investigate a kind of function that is useful for this purpose, and use it to investigate real-world situations such as the train’s distance from the station at a given point in time.
Representing Linear Functions
The function describing the train’s motion is a linear function, which is defined as a function with a constant rate of change, that is, a polynomial of degree 1. There are several ways to represent a linear function, including word form, function notation, tabular form, and graphical form. We will describe the train’s motion as a function using each method.
Representing a Linear Function in Word Form
Let’s begin by describing the linear function in words. For the train problem we just considered, the following word sentence may be used to describe the function relationship.
The train’s distance from the station is a function of the time during which the train moves at a constant speed plus its original distance from the station when it began moving at constant speed.
The speed is the rate of change. Recall that a rate of change is a measure of how quickly the dependent variable changes with respect to the independent variable. The rate of change for this example is constant, which means that it is the same for each input value. As the time (input) increases by 1 second, the corresponding distance (output) increases by 83 meters. The train began moving at this constant speed at a distance of 250 meters from the station.
Representing a Linear Function in Function Notation
Another approach to representing linear functions is by using function notation. One example of function notation is an equation written in the form known as the slope-intercept form of a line, where xis the input value, is the rate of change, and is the initial value of the dependent variable.
In the example of the train, we might use the notation in which the total distance is a function of the time . The rate, , is 83 meters per second. The initial value of the dependent variable is the original distance from the station, 250 meters. We can write a generalized equation to represent the motion of the train.
Representing a Linear Function in Tabular Form
A third method of representing a linear function is through the use of a table. The relationship between the distance from the station and the time is represented in Figure . From the table, we can see that the distance changes by 83 meters for every 1 second increase in time.
Q/A? Can the input in the previous example be any real number?
No. The input represents time, so while nonnegative rational and irrational numbers are possible, negative real numbers are not possible for this example. The input consists of non-negative real numbers.
Representing a Linear Function in Graphical Form
Another way to represent linear functions is visually, using a graph. We can use the function relationship from above, , to draw a graph, represented in Figure . Notice the graph is a line. When we plot a linear function, the graph is always a line.
The rate of change, which is constant, determines the slant, or slope of the line. The point at which the input value is zero is the vertical intercept, or y-intercept, of the line. We can see from the graph in Figure that the y-intercept in the train example we just saw is and represents the distance of the train from the station when it began moving at a constant speed.
Notice that the graph of the train example is restricted, but this is not always the case. Consider the graph of the line . Ask yourself what numbers can be input to the function, that is, what is the domain of the function? The domain is comprised of all real numbers because any number may be doubled, and then have one added to the product.
Definition: Linear Function
A linear function is a function whose graph is a line. Linear functions can be written in the slope-intercept form of a line
where is the initial or starting value of the function (when input, ), and is the constant rate of change, or slope of the function. The y-intercept is at .
Example : Using a Linear Function to Find the Pressure on a Diver
The pressure, , in pounds per square inch (PSI) on the diver in Figure depends upon her depth below the water surface, , in feet. This relationship may be modeled by the equation, . Restate this function in words.
To restate the function in words, we need to describe each part of the equation. The pressure as a function of depth equals four hundred thirty-four thousandths times depth plus fourteen and six hundred ninety-six thousandths.
Analysis
The initial value, 14.696, is the pressure in PSI on the diver at a depth of 0 feet, which is the surface of the water. The rate of change, or slope, is 0.434 PSI per foot. This tells us that the pressure on the diver increases 0.434 PSI for each foot her depth increases.
Determining whether a Linear Function Is Increasing, Decreasing, or Constant
The linear functions we used in the two previous examples increased over time, but not every linear function does. A linear function may be increasing, decreasing, or constant. For an increasing function, as with the train example, the output values increase as the input values increase. The graph of an increasing function has a positive slope. A line with a positive slope slants upward from left to right as in Figure (a). For a decreasing function, the slope is negative. The output values decrease as the input values increase. A line with a negative slope slants downward from left to right as in Figure (b). If the function is constant, the output values are the same for all input values so the slope is zero. A line with a slope of zero is horizontal as in Figure (c).
Increasing and Decreasing Functions
The slope determines if the function is an increasing linear function, a decreasing linear function, or a constant function.
is an increasing function if .
is an decreasing function if .
is a constant function if .
Example : Deciding whether a Function Is Increasing, Decreasing, or Constant
Some recent studies suggest that a teenager sends an average of 60 texts per day. For each of the following scenarios, find the linear function that describes the relationship between the input value and the output value. Then, determine whether the graph of the function is increasing, decreasing, or constant.
The total number of texts a teen sends is considered a function of time in days. The input is the number of days, and output is the total number of texts sent.
A teen has a limit of 500 texts per month in his or her data plan. The input is the number of days, and output is the total number of texts remaining for the month.
A teen has an unlimited number of texts in his or her data plan for a cost of $50 per month. The input is the number of days, and output is the total cost of texting each month.
Solution
Analyze each function.
The function can be represented as where is the number of days. The slope, 60, is positive so the function is increasing. This makes sense because the total number of texts increases with each day.
The function can be represented as where is the number of days. In this case, the slope is negative so the function is decreasing. This makes sense because the number of texts remaining decreases each day and this function represents the number of texts remaining in the data plan after days.
The cost function can be represented as because the number of days does not affect the total cost. The slope is 0 so the function is constant.
Calculating and Interpreting Slope
In the examples we have seen so far, we have had the slope provided for us. However, we often need to calculate the slope given input and output values. Given two values for the input, and , and two corresponding values for the output, and —which can be represented by a set of points, and —we can calculate the slope , as follows
where is the vertical displacement and is the horizontal displacement. Note in function notation two corresponding values for the output and for the function , and , so we could equivalently write
Figure indicates how the slope of the line between the points, and , is calculated. Recall that the slope measures steepness. The greater the absolute value of the slope, the steeper the line is.
Q & A
Are the units for slope always ?
Yes. Think of the units as the change of output value for each unit of change in input value. An example of slope could be miles per hour or dollars per day. Notice the units appear as a ratio of units for the output per units for the input.
Calculate Slope
The slope, or rate of change, of a function can be calculated according to the following:
where and are input values, and are output values.
Given two points from a linear function, calculate and interpret the slope.
Determine the units for output and input values.
Calculate the change of output values and change of input values.
Interpret the slope as the change in output values per unit of the input value.
Example : Finding the Slope of a Linear Function
If is a linear function, and and are points on the line, find the slope. Is this function increasing or decreasing?
Solution
The coordinate pairs are and . To find the rate of change, we divide the change in output by the change in input.
We could also write the slope as . The function is increasing because .
Analysis
As noted earlier, the order in which we write the points does not matter when we compute the slope of the line as long as the first output value, or y-coordinate, used corresponds with the first input value, or x-coordinate, used.
Try it!
If is a linear function, and and are points on the line, find the slope. Is this function increasing or decreasing?
Answer
: ; decreasing because .
Example : Finding the Population Change from a Linear Function
The population of a city increased from 23,400 to 27,800 between 2008 and 2012. Find the change of population per year if we assume the change was constant from 2008 to 2012.
The rate of change relates the change in population to the change in time. The population increased by people over the four-year time interval. To find the rate of change, divide the change in the number of people by the number of years.
So the population increased by 1,100 people per year.
Analysis
Because we are told that the population increased, we would expect the slope to be positive. This positive slope we calculated is therefore reasonable.
Try it!
The population of a small town increased from 1,442 to 1,868 between 2009 and 2012. Find the change of population per year if we assume the change was constant from 2009 to 2012.
Answer
Writing the Point-Slope Form of a Linear Equation
Up until now, we have been using the slope-intercept form of a linear equation to describe linear functions. Here, we will learn another way to write a linear function, the point-slope form.
The point-slope form is derived from the slope formula.
Keep in mind that the slope-intercept form and the point-slope form can be used to describe the same function. We can move from one form to another using basic algebra. For example, suppose we are given an equation in point-slope form, . We can convert it to the slope-intercept form as shown.
Therefore, the same line can be described in slope-intercept form as .
Point-Slope Form of a Linear Equation
The point-slope form of a linear equation takes the form
where is the slope, and are the and coordinates of a specific point through which the line passes.
Writing the Equation of a Line Using a Point and the Slope
The point-slope form is particularly useful if we know one point and the slope of a line. Suppose, for example, we are told that a line has a slope of 2 and passes through the point . We know that and that and . We can substitute these values into the general point-slope equation.
If we wanted to then rewrite the equation in slope-intercept form, we apply algebraic techniques.
Both equations, and , describe the same line. See Figure .
Example : Writing Linear Equations Using a Point and the Slope
Write the point-slope form of an equation of a line with a slope of 3 that passes through the point . Then rewrite it in the slope-intercept form.
Solution
Let’s figure out what we know from the given information. The slope is 3, so . We also know one point, so we know and . Now we can substitute these values into the general point-slope equation.
Then we use algebra to find the slope-intercept form.
Try it!
Write the point-slope form of an equation of a line with a slope of –2 that passes through the point . Then rewrite it in the slope-intercept form.
Answer
: ;
Writing the Equation of a Line Using Two Points
The point-slope form of an equation is also useful if we know any two points through which a line passes. Suppose, for example, we know that a line passes through the points and . We can use the coordinates of the two points to find the slope.
Now we can use the slope we found and the coordinates of one of the points to find the equation for the line. Let use for our point.
As before, we can use algebra to rewrite the equation in the slope-intercept form.
Both equations describe the line shown in Figure .
Example : Writing Linear Equations Using Two Points
Write the point-slope form of an equation of a line that passes through the points and . Then rewrite it in the slope-intercept form.
Let’s begin by finding the slope.
So . Next, we substitute the slope and the coordinates for one of the points into the general point-slope equation. We can choose either point, but we will use .
The point-slope equation of the line is . To rewrite the equation in slope-intercept form, we use algebra.
The slope-intercept equation of the line is .
: Write the point-slope form of an equation of a line that passes through the points and . Then rewrite it in the slope-intercept form.
Solution
;
Writing and Interpreting an Equation for a Linear Function
Now that we have written equations for linear functions in both the slope-intercept form and the point-slope form, we can choose which method to use based on the information we are given. That information may be provided in the form of a graph, a point and a slope, two points, and so on. Look at the graph of the function in Figure .
We are not given the slope of the line, but we can choose any two points on the line to find the slope. Let’s choose and . We can use these points to calculate the slope.
Now we can substitute the slope and the coordinates of one of the points into the point-slope form.
If we want to rewrite the equation in the slope-intercept form, we would find
If we wanted to find the slope-intercept form without first writing the point-slope form, we could have recognized that the line crosses the y-axis when the output value is 7. Therefore, . We now have the initial value and the slope so we can substitute and into the slope-intercept form of a line.
So the function is , and the linear equation would be .
Given the graph of a linear function, write an equation to represent the function.
Identify two points on the line.
Use the two points to calculate the slope.
Determine where the line crosses the y-axis to identify the y-intercept by visual inspection.
Substitute the slope and y-intercept into the slope-intercept form of a line equation.
Example : Writing an Equation for a Linear Function
Write an equation for a linear function given a graph of shown inFigure .
Solution
Identify two points on the line, such as and . Use the points to calculate the slope.
Substitute the slope and the coordinates of one of the points into the point-slope form.
We can use algebra to rewrite the equation in the slope-intercept form.
Analysis
This makes sense because we can see from Figure that the line crosses the y-axis at the point , which is the y-intercept, so .
Example : Writing an Equation for a Linear Cost Function
Suppose Ben starts a company in which he incurs a fixed cost of 37.50 per item. Write a linear function where is the cost for items produced in a given month.
Solution
The fixed cost is present every month, 37.50 for Ben. The variable cost, called the marginal cost, is represented by 37.5. The cost Ben incurs is the sum of these two costs, represented by .
Analysis
If Ben produces 100 items in a month, his monthly cost is represented by
So his monthly cost would be $5,000.
Example : Writing an Equation for a Linear Function Given Two Points
If is a linear function, with , and , find an equation for the function in slope-intercept form.
Solution
We can write the given points using coordinates.
We can then use the points to calculate the slope.
Substitute the slope and the coordinates of one of the points into the point-slope form.
We can use algebra to rewrite the equation in the slope-intercept form.
Tri It!
If is a linear function, with , and , find an equation for the function in slope-intercept form.
Answer
Modeling Real-World Problems with Linear Functions
In the real world, problems are not always explicitly stated in terms of a function or represented with a graph. Fortunately, we can analyze the problem by first representing it as a linear function and then interpreting the components of the function. As long as we know, or can figure out, the initial value and the rate of change of a linear function, we can solve many different kinds of real-world problems.
Given a linear function and the initial value and rate of change, evaluate .
Determine the initial value and the rate of change (slope).
Substitute the values into .
Evaluate the function at .
Example : Using a Linear Function to Determine the number of Songs in a Music Collection
Marcus currently has 200 songs in his music collection. Every month, he adds 15 new songs. Write a formula for the number of songs, , in his collection as a function of time, , the number of months. How many songs will he own in a year?
Solution
The initial value for this function is 200 because he currently owns 200 songs, so , which means that .
The number of songs increases by 15 songs per month, so the rate of change is 15 songs per month. Therefore we know that . We can substitute the initial value and the rate of change into the slope-intercept form of a line.
We can write the formula .
With this formula, we can then predict how many songs Marcus will have in 1 year (12 months). In other words, we can evaluate the function at .
Marcus will have 380 songs in 12 months.
Analysis
Notice that is an increasing linear function. As the input (the number of months) increases, the output (number of songs) increases as well.
Example : Using a Linear Function to Calculate Salary Plus Commission
Working as an insurance salesperson, Ilya earns a base salary plus a commission on each new policy. Therefore, Ilya’s weekly income, I,depends on the number of new policies, , he sells during the week. Last week he sold 3 new policies, and earned 920. Find an equation for , and interpret the meaning of the components of the equation.
Solution
The given information gives us two input-output pairs: and . We start by finding the rate of change.
Keeping track of units can help us interpret this quantity. Income increased by 80 per policy. Therefore, Ilya earns a commission of $80 for each policy sold during the week.
We can then solve for the initial value.
The value of is the starting value for the function and represents Ilya’s income when , or when no new policies are sold. We can interpret this as Ilya’s base salary for the week, which does not depend upon the number of policies sold.
We can now write the final equation.
Our final interpretation is that Ilya’s base salary is 80 commission for each policy sold.
Example : Using Tabular Form to Write an Equation for a Linear Function
Table relates the number of rats in a population to time, in weeks. Use the table to write a linear equation.
Table
| w, number of weeks | 0 | 2 | 4 | 6 |
| P(w), number of rats | 1000 | 1080 | 1160 | 1240 |
Solution
We can see from the table that the initial value for the number of rats is 1000, so .
Rather than solving for , we can tell from looking at the table that the population increases by 80 for every 2 weeks that pass. This means that the rate of change is 80 rats per 2 weeks, which can be simplified to 40 rats per week.
If we did not notice the rate of change from the table we could still solve for the slope using any two points from the table. For example, using and
Is the initial value always provided in a table of values like Table ?
No. Sometimes the initial value is provided in a table of values, but sometimes it is not. If you see an input of 0, then the initial value would be the corresponding output. If the initial value is not provided because there is no value of input on the table equal to 0, find the slope, substitute one coordinate pair and the slope into , and solve for .
: A new plant food was introduced to a young tree to test its effect on the height of the tree. Table shows the height of the tree, in feet, months since the measurements began. Write a linear function, , where is the number of months since the start of the experiment.
Table
| | 0 | 2 | 4 | 8 | 12 |
| | 12.5 | 13.5 | 14.5 | 16.5 | 18.5 |
Solution
Key Equations
slope-intercept form of a line:
slope:
point-slope form of a line:
Key Concepts
The ordered pairs given by a linear function represent points on a line.
Linear functions can be represented in words, function notation, tabular form, and graphical form.
The rate of change of a linear function is also known as the slope.
An equation in the slope-intercept form of a line includes the slope and the initial value of the function.
The initial value, or y-intercept, is the output value when the input of a linear function is zero. It is the y-value of the point at which the line crosses the y-axis.
An increasing linear function results in a graph that slants upward from left to right and has a positive slope.
A decreasing linear function results in a graph that slants downward from left to right and has a negative slope.
A constant linear function results in a graph that is a horizontal line.
Analyzing the slope within the context of a problem indicates whether a linear function is increasing, decreasing, or constant.
The slope of a linear function can be calculated by dividing the difference between y-values by the difference in corresponding x-values of any two points on the line.
The slope and initial value can be determined given a graph or any two points on the line.
One type of function notation is the slope-intercept form of an equation.
The point-slope form is useful for finding a linear equation when given the slope of a line and one point.
The point-slope form is also convenient for finding a linear equation when given two points through which a line passes.
The equation for a linear function can be written if the slope and initial value are known.
A linear function can be used to solve real-world problems.
A linear function can be written from tabular form.
Footnotes
1 www.chinahighlights.com/shang...glev-train.htm
2 www.cbsnews.com/8301-501465_1...ay-study-says/
Glossary
decreasing linear function
a function with a negative slope: If , then .
increasing linear function
a function with a positive slope: If , then .
linear function
a function with a constant rate of change that is a polynomial of degree 1, and whose graph is a straight line
point-slope form
the equation for a line that represents a linear function of the form (y−y_1=m(x−x_1)
the ratio of the change in output values to the change in input values; a measure of the steepness of a line
slope-intercept form
the equation for a line that represents a linear function in the form
the value of a function when the input value is zero; also known as initial value
3: Linear and Quadratic Functions
3.2: Graphs of Linear Functions |
10172 | https://www.tutorchase.com/answers/a-level/physics/how-does-a-diverging-lens-correct-myopia | How does a diverging lens correct myopia?
A diverging lens corrects myopia by diverging the incoming light rays before they reach the eye.
Myopia, also known as nearsightedness, is a condition where the eye is unable to focus on distant objects. This is because the eyeball is too long or the cornea is too curved, causing light rays to converge in front of the retina instead of on it. As a result, distant objects appear blurry.
A diverging lens is a concave lens that causes light rays to spread out or diverge. When placed in front of a myopic eye, the diverging lens diverges the incoming light rays even further before they reach the eye. This reduces the overall convergence of the light rays and moves the focal point back to the retina, allowing the eye to focus on distant objects.
The amount of divergence required depends on the severity of the myopia. A stronger diverging lens is needed for a more myopic eye, while a weaker lens is sufficient for a less myopic eye. The power of the lens is measured in dioptres, with a higher negative value indicating a stronger lens.
In summary, a diverging lens corrects myopia by diverging the incoming light rays before they reach the eye, reducing the overall convergence of the light rays and moving the focal point back to the retina.
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10173 | https://tw.rigol.com/tw/Images/DG900_UserGuide_EN_tcm17-4035.pdf | RIGOL User’s Guide DG900 Series Function/Arbitrary Waveform Generator Aug. 2018 RIGOL (SUZHOU) TECHNOLOGIES INC.
RIGOL DG900 User's Guide I Guaranty and Declaration Copyright © 2018 RIGOL (SUZHOU) TECHNOLOGIES INC. All Rights Reserved. Trademark Information RIGOL is a registered trademark of RIGOL (SUZHOU) TECHNOLOGIES INC. Publication Number UGB10100-1110 Software Version 00.00.01 Software upgrade might change or add product features. Please acquire the latest version of the manual from RIGOL website or contact RIGOL to upgrade the software. Notices RIGOL products are covered by P.R.C. and foreign patents, issued and pending. RIGOL reserves the right to modify or change parts of or all the specifications and pricing policies at the company’s sole decision. Information in this publication replaces all previously released materials. Information in this publication is subject to change without notice. RIGOL shall not be liable for either incidental or consequential losses in connection with the furnishing, use, or performance of this manual, as well as any information contained. Any part of this document is forbidden to be copied, photocopied, or rearranged without prior written approval of RIGOL. Product Certification RIGOL guarantees that this product conforms to the national and industrial standards in China as well as the ISO9001:2015 standard and the ISO14001:2015 standard. Other international standard conformance certifications are in progress. Contact Us If you have any problem or requirement when using our products or this manual, please contact RIGOL. E-mail: service@rigol.com Website: www.rigol.com RIGOL II DG900 User's Guide Safety Requirement General Safety Summary Please review the following safety precautions carefully before putting the instrument into operation so as to avoid any personal injury or damage to the instrument and any product connected to it. To prevent potential hazards, please follow the instructions specified in this manual to use the instrument properly. Use the BNC Output Connector Properly. The BNC output connector on the front panel only allows to output the signal but not to input the signal. Use Proper Power Cord. Only the exclusive power cord designed for the instrument and authorized for use within the local country could be used. Ground the Instrument. The instrument is grounded through the Protective Earth lead of the power cord. To avoid electric shock, connect the earth terminal of the power cord to the Protective Earth terminal before connecting any input or output terminals. Connect the Probe Correctly. If a probe is used, the probe ground lead must be connected to earth ground. Do not connect the ground lead to high voltage. Improper way of connection could result in dangerous voltages being present on the connectors, controls or other surfaces of the oscilloscope and probes, which will cause potential hazards for operators. Observe All Terminal Ratings. To avoid fire or shock hazard, observe all ratings and markers on the instrument and check your manual for more information about ratings before connecting the instrument. Use Proper Overvoltage Protection. Ensure that no overvoltage (such as that caused by a bolt of lightning) can reach the product. Otherwise, the operator might be exposed to the danger of an electric shock. Do Not Operate Without Covers. Do not operate the instrument with covers or panels removed. Do Not Insert Objects Into the Air Outlet. Do not insert objects into the air outlet, as doing so may cause damage to the instrument. RIGOL DG900 User's Guide III Use Proper Fuse. Please use the specified fuses. Avoid Circuit or Wire Exposure. Do not touch exposed junctions and components when the unit is powered on. Do Not Operate With Suspected Failures. If you suspect that any damage may occur to the instrument, have it inspected by RIGOL authorized personnel before further operations. Any maintenance, adjustment or replacement especially to circuits or accessories must be performed by RIGOL authorized personnel. Provide Adequate Ventilation. Inadequate ventilation may cause an increase of temperature in the instrument, which would cause damage to the instrument. So please keep the instrument well ventilated and inspect the air outlet and the fan regularly. Do Not Operate in Wet Conditions. To avoid short circuit inside the instrument or electric shock, never operate the instrument in a humid environment. Do Not Operate in an Explosive Atmosphere. To avoid personal injuries or damage to the instrument, never operate the instrument in an explosive atmosphere. Keep Product Surfaces Clean and Dry. To avoid dust or moisture from affecting the performance of the instrument, keep the surfaces of the instrument clean and dry. Prevent Electrostatic Impact. Operate the instrument in an electrostatic discharge protective environment to avoid damage induced by static discharges. Always ground both the internal and external conductors of cables to release static before making connections. Use the Battery Properly. Do not expose the battery (if available) to high temperature or fire. Keep it out of the reach of children. Improper change of a battery (lithium battery) may cause an explosion. Use the RIGOL specified battery only. Handle with Caution. Please handle with care during transportation to avoid damage to keys, knobs, interfaces, and other parts on the panels. RIGOL IV DG900 User's Guide Safety Notices and Symbols Safety Notices in this Manual: WARNING Indicates a potentially hazardous situation or practice which, if not avoided, will result in serious injury or death. CAUTION Indicates a potentially hazardous situation or practice which, if not avoided, could result in damage to the product or loss of important data. Safety Terms on the Product: DANGER It calls attention to an operation, if not correctly performed, could result in injury or hazard immediately. WARNING It calls attention to an operation, if not correctly performed, could result in potential injury or hazard. CAUTION It calls attention to an operation, if not correctly performed, could result in damage to the product or other devices connected to the product. Safety Symbols on the Product: Hazardous Voltage Safety Warning Protective Earth Terminal Chassis Ground Test Ground RIGOL DG900 User's Guide V Care and Cleaning Care Do not store or leave the instrument where it may be exposed to direct sunlight for long periods of time. Cleaning Clean the instrument regularly according to its operating conditions. 1. Disconnect the instrument from all power sources. 2. Clean the external surfaces of the instrument with a soft cloth dampened with mild detergent or water. Avoid having any water or other objects into the chassis via the heat dissipation hole. When cleaning the LCD, take care to avoid scarifying it. CAUTION To avoid damage to the instrument, do not expose it to caustic liquids. WARNING To avoid short-circuit resulting from moisture or personal injuries, ensure that the instrument is completely dry before connecting it to the power supply. Environmental Considerations The following symbol indicates that this product complies with the WEEE Directive 2002/96/EC. Product End-of-Life Handling The equipment may contain substances that could be harmful to the environment or human health. To avoid the release of such substances into the environment and avoid harm to human health, we recommend you to recycle this product appropriately to ensure that most materials are reused or recycled properly. Please contact your local authorities for disposal or recycling information. You can log in to RIGOL official website (www.rigol.com) to download the latest version of the RoHS-WEEE certification file. RIGOL VI DG900 User's Guide DG900 Series Function/Arbitrary Waveform Generator Overview As a multi-functional signal generator, DG900 series function/arbitrary waveform generator integrates many instruments into 1, such as function generator, arbitrary waveform generator, noise generator, pulse generator, harmonic generator, analog/digital modulator, and frequency counter. As a multi-functional and portable instrument, it offers you a new choice in education, R&D, production, measurement, and other industries with its user-friendly touch screen and high performance at an unprecedented price point. Main Features: Unique SiFi II (Signal Fidelity II) technology: generate the arbitrary waveforms point by point; recover the signal without distortion; sample rate accurate and adjustable; jitter of all the output waveforms (including Sine, Pulse, etc.) as low as 200 ps 16 Mpts memory depth per channel for arbitrary waveforms Standard dual-channel with the same performance, equivalent to two independent signal sources High frequency stability: ±1 ppm; low phase noise: -105 dBc/Hz Built-in up to 8 orders harmonics generator Built-in 7 digits/s, 240 MHz bandwidth full featured frequency counter Up to 160 built-in arbitrary waveforms, covering the common signals in engineering application, medical electronics, auto electronics, math processing, and other various fields Sample rate up to 250 MSa/s, vertical resolution 16 bits Arbitrary waveform sequence editing function available; arbitrary waveforms also can be generated through the PC software Various analog and digital modulation functions: AM, FM, PM, ASK, FSK, PSK, and PWM. Standard waveform combine function, capable of outputting specified waveforms combined with the basic waveforms Standard channel tracking function, when enabled, all the parameters of both channels are updated based on users' configurations USB HOST&DEVICE interface (standard); USB-GPIB function supported 4.3'' TFT color touch screen RS232, PRBS, and Dualtone outputs supported RIGOL DG900 User's Guide VII Document Overview Main Topics in this Manual Chapter 1 Quick Start Introduces the appearance and dimensions of DG900, its front/rear panel, and user interface. Chapter 2 Front Panel Operations Introduces the main functions and operation methods of DG900. Chapter 3 Remote Control Introduces how to control DG900 remotely. Chapter 4 Troubleshooting Lists the possible failures and solutions in using DG900. Chapter 5 Appendix Provides the information about the options and accessories list, as well as warranty information of DG900. Format Conventions in this Manual 1. Keys: The keys on the front panel are usually denoted by the format of "Key Name (Bold) + Text Box". For example, Utility denotes the Utility key. 2. Menu Labels: The menu labels are usually denoted by the format of "Menu Word (Bold) + Character Shading". For example, System Setting. 3. Connectors: The connectors on the front or rear panel are usually denoted by the format of "Connector Name (Bold) + Square Brackets (Bold)". For example, [Counter]. 4. Operation Procedures: "" represents the next step of operation. For example, Utility System Setting denotes that first press Utility on the front panel, and then tap the System Setting menu label. Content Conventions in this Manual 1. DG900 series function/arbitrary waveform generator includes the following models: This manual takes DG992 as an example to illustrate the operation methods of the DG900 series. Model No. of Channels Max. Output Frequency DG952 2 50 MHz DG972 2 70 MHz DG992 2 100 MHz 2. DG900 series function/arbitrary waveform generator is equipped with two channels (CH1 and CH2). Unless otherwise specified, this manual takes CH1 as RIGOL VIII DG900 User's Guide an example to introduce the operation methods of the generator. The operation methods of CH2 is the same as that of CH1. Manuals of this Product The manuals of this product mainly include the quick guide, user’s guide, programming guide, data sheet, and etc. For the latest version of this manual, download it from the official website of RIGOL (www.rigol.com). Contents RIGOL DG900 User's Guide IX Contents Guaranty and Declaration ......................................................................... I Safety Requirement ................................................................................ II General Safety Summary ........................................................................... II Safety Notices and Symbols ...................................................................... IV Care and Cleaning .................................................................................... V Environmental Considerations .................................................................... V DG900 Series Function/Arbitrary Waveform Generator Overview ......... VI Document Overview .............................................................................. VII Chapter 1 Quick Start ......................................................................... 1-1 General Inspection ................................................................................ 1-2 Appearance and Dimensions ................................................................... 1-3 Front Panel Overview ............................................................................. 1-4 Rear Panel Overview .............................................................................. 1-8 To Prepare for Use ................................................................................ 1-11 To Connect to AC Power ................................................................. 1-11 Turn-on Checkout .......................................................................... 1-11 To Set the System Language ........................................................... 1-11 User Interface ...................................................................................... 1-12 To Use the Built-in Help System ............................................................. 1-15 Chapter 2 Front Panel Operations ...................................................... 2-1 To Output Basic Waveform ..................................................................... 2-2 To Select Output Channel ................................................................ 2-2 To Select Basic Waveform ................................................................ 2-3 To Set Frequency/Period .................................................................. 2-4 To Set Amplitude/High Level ............................................................ 2-5 To Set Offset/Low Level ................................................................... 2-7 To Set Start Phase .......................................................................... 2-8 To Set Duty Cycle (Square) .............................................................. 2-9 To Set Symmetry (Ramp) ................................................................ 2-10 To Set Pulse Width/Duty Cycle (Pulse) ............................................. 2-11 To Set Rising/Falling Edge (Pulse) .................................................... 2-12 To Enable Channel Output .............................................................. 2-13 Align Phase ................................................................................... 2-14 Example: To Output Sine ................................................................ 2-15 To Output the Arbitrary Waveform .......................................................... 2-17 To Enable Arbitrary Waveforms........................................................ 2-17 To Select the Waveform .................................................................. 2-17 To Set Parameters .......................................................................... 2-22 To Output Harmonic ............................................................................. 2-23 Harmonic Overview ........................................................................ 2-23 RIGOL Contents X DG900 User's Guide To Set Fundamental Waveform Parameters ...................................... 2-24 To Select Harmonic Type ................................................................ 2-24 To Set Harmonic Order ................................................................... 2-24 To Select Harmonic Amplitude ......................................................... 2-25 To Set Harmonic Phase .................................................................. 2-25 Example: To Output Harmonic ........................................................ 2-25 DC ...................................................................................................... 2-28 To Output Dual-tone Waveform ............................................................. 2-29 To Output Advanced Waveform ............................................................. 2-30 PRBS ............................................................................................ 2-30 RS232 .......................................................................................... 2-31 Sequence ...................................................................................... 2-32 Modulation .......................................................................................... 2-34 Amplitude Modulation (AM) ............................................................ 2-34 Frequency Modulation (FM) ............................................................ 2-38 Phase Modulation (PM) .................................................................. 2-41 Amplitude Shift Keying (ASK) .......................................................... 2-44 Frequency Shift Keying (FSK) .......................................................... 2-47 Phase Shift Keying (PSK) ................................................................ 2-50 Pulse Width Modulation (PWM) ....................................................... 2-53 Sweep................................................................................................. 2-56 To Enable the Sweep Function ........................................................ 2-56 Sweep Type .................................................................................. 2-56 Sweep Time .................................................................................. 2-58 Return Time .................................................................................. 2-58 Start Frequency and Stop Frequency ............................................... 2-59 Center Frequency and Frequency Span ............................................ 2-59 Sweep Trigger Source .................................................................... 2-60 Marker Freq .................................................................................. 2-61 Start Hold ..................................................................................... 2-62 Stop Hold ..................................................................................... 2-62 Burst .................................................................................................. 2-63 To Enable the Burst Function .......................................................... 2-63 Burst Type .................................................................................... 2-63 Burst Delay ................................................................................... 2-65 Burst Period .................................................................................. 2-66 Idle Level ...................................................................................... 2-66 Burst Trigger Source ...................................................................... 2-67 Gated Polarity ............................................................................... 2-68 Frequency Counter ............................................................................... 2-69 To Enable the Frequency Counter .................................................... 2-69 To Set the Frequency Counter ......................................................... 2-70 Store and Recall ................................................................................... 2-73 Storage System ............................................................................. 2-73 File Type ....................................................................................... 2-74 Categories .................................................................................... 2-74 Contents RIGOL DG900 User's Guide XI File Operation ................................................................................ 2-75 Seamless Interconnection with Oscilloscope ..................................... 2-79 Channel Setting .................................................................................... 2-81 Output Setting ............................................................................... 2-81 Sync Setting .................................................................................. 2-83 Coupling Setting ............................................................................ 2-85 Waveform Combination .................................................................. 2-89 Common Settings ................................................................................. 2-90 System Setting .............................................................................. 2-90 Interface ....................................................................................... 2-92 System Info ................................................................................... 2-96 Option .......................................................................................... 2-96 Display Setting .............................................................................. 2-96 Print Setting .................................................................................. 2-97 System Utility Function ......................................................................... 2-98 To Restore Preset ........................................................................... 2-98 Channel Copy .............................................................................. 2-102 To Lock the Keyboard ................................................................... 2-103 Chapter 3 Remote Control .................................................................. 3-1 Remote Control via USB ......................................................................... 3-2 Remote Control via LAN ......................................................................... 3-2 Remote Control via GPIB ........................................................................ 3-4 Chapter 4 Troubleshooting ................................................................. 4-1 Chapter 5 Appendix ............................................................................ 5-1 Appendix A: Accessories and Options ...................................................... 5-1 Appendix B: Warranty ............................................................................ 5-2 Index ....................................................................................................... 1 Chapter 1 Quick Guide RIGOL DG900 User's Guide 1-1 Chapter 1 Quick Start This chapter briefly introduces the appearance and dimensions of DG900 series function/arbitrary waveform generator, its front/rear panel, and user interface. Contents in this chapter: General Inspection Appearance and Dimensions Front Panel Overview Rear Panel Overview To Prepare for Use User Interface To Use the Built-in Help System RIGOL Chapter 1 Quick Guide 1-2 DG900 User's Guide General Inspection 1． Inspect the packaging If the packaging has been damaged, do not dispose the damaged packaging or cushioning materials until the shipment has been checked for completeness and has passed both electrical and mechanical tests. The consigner or carrier shall be liable for the damage to the instrument resulting from shipment. RIGOL would not be responsible for free maintenance/rework or replacement of the instrument. 2． Inspect the instrument In case of any mechanical damage, missing parts, or failure in passing the electrical and mechanical tests, contact your RIGOL sales representative. 3． Check the accessories Please check the accessories according to the packing lists. If the accessories are damaged or incomplete, please contact your RIGOL sales representative. Chapter 1 Quick Guide RIGOL DG900 User's Guide 1-3 Appearance and Dimensions Front View Unit: mm Side View Unit: mm RIGOL Chapter 1 Quick Guide 1-4 DG900 User's Guide Front Panel Overview Figure 1-1 Front Panel 1. Power Key Turns on or off the generator. 2. Align Key Performs the phase alignment operation. For details, refer to descriptions in "Align Phase". 3. CH1 Output Connector BNC connector, with 50 Ω nominal output impedance. When Output1 is enabled (the backlight turns on), this connector outputs waveforms according to the current configuration of CH1. 4. Channel Control Area It is used to control the output of CH1. — Press this key to enable the output of CH1, the backlight turns on. At this time, the [CH1] connector outputs the waveforms according to the current configuration of CH1. — Press this key again to disable the output of CH1, and the backlight turns off. It is used to control the output of CH2. — Press this key to enable the output of CH2, the backlight turns on. At this time, the [CH2] connector outputs the waveforms according to the current configuration of CH2. — Press this key again to disable the output of CH2, and the backlight turns off. 1 2 3 4 5 6 7 13 12 11 10 9 8 Chapter 1 Quick Guide RIGOL DG900 User's Guide 1-5 5. CH2 Output Connector BNC connector, with 50 Ω nominal output impedance. When Output2 is enabled (the backlight turns on), this connector outputs waveforms according to the current configuration of CH2. 6. Counter Measurement Signal Input Connector BNC connector, with 1 MΩ input impedance. It is used to receive the signal measured by the counter. 7. Frequency Counter Enables or disables the frequency counter. — Press this key to enable the frequency counter, and the backlight turns on and blinks continuously. — Press this key again to disable the frequency counter, and the backlight turns off. At this time, the frequency counter is disabled. Note: When the frequency counter is enabled, no waveforms will be output from the CH2 connector. When the frequency counter is disabled, waveforms are allowed to be output from the CH2 connector. 8. Arrow Key — It is used to move the cursor to select the digit to be edited when you use the knob to set the parameters (pressing down the knob can enter the editing mode). — In the user interface, it is used to move left or right the cursor. 9. Knob — When you select a menu label in the interface, the knob can be used to move the cursor down (clockwise) or up (counterclockwise). — It can be used to increase (clockwise) or decrease (counterclockwise) the value marked by the cursor when you use the knob to set the parameters (pressing down the knob can enter the editing mode). Press down the knob again to exit the editing mode. — It can be used to select the desired waveform by moving the cursor with the knob when you select the waveform (pressing the right arrow key will locate the cursor to the right of the interface). Press down the knob to select the desired waveform. — When you store or read a file, it can be used to select a storage location or select a file to be read. Press down the knob to unfold the currently selected directory. — It can be used to select the desired parameter by moving the cursor with the knob when you set the common information (pressing the right arrow key will locate the cursor to the right of the interface). Press down the knob CAUTION To avoid damages to the instrument, the input signal voltage cannot exceed ±2.5 V. RIGOL Chapter 1 Quick Guide 1-6 DG900 User's Guide to select the desired parameter. Then, rotate the knob to modify the parameter, and press it down again to confirm your modification. — It is used to select the desired configuration type in the Preset interface. Press down the knob to confirm your selection. At this time, a dialog box is displayed. Use the knob to select the corresponding button, then press down the knob to perform the corresponding operation (note that only when the button turns green, can your operation on the knob be valid). 10. Menu Key Enters the waveform mode selection interface. 11. Home Key Enters the main interface of the instrument. 12. Function Keys Restores the instrument to its preset state. At most 10 states can be preset. Locks or unlocks the front-panel keys and the touch screen. In the unlocked state, press Lock to lock the front-panel keys and the touch screen. At this time, except the Lock key, all other keys on the front panel and the touch screen operation are invalid. Press the Lock key again to unlock the keys and the touch screen. Used for manual trigger. — The default trigger setting for the generator is Internal trigger. In this mode, when you select the sweep or burst mode, the generator outputs the waveforms continuously. At this time, press the Trig key, and instrument automatically switches to the manual trigger mode from auto trigger. — Each time you press the Trig key, one sweep will be triggered manually or one burst will be output. It is used to set the utility function parameters and system parameters. Stores or recalls the instrument state or the user-defined arbitrary waveform data. A non-volatile memory (Disk C) is built in, and a USB storage device (Disk D) can be externally connected. Gets the help information of any front-panel keys and the help information of the current interface. Chapter 1 Quick Guide RIGOL DG900 User's Guide 1-7 Note: When the instrument is in the remote mode, press this key to return to the local mode. 13. LCD 4.3-inch TFT (480×272) color LCD display. The menu label and parameter settings of the current function, system state, prompt messages, and other information can be clearly displayed on the LCD. For details, refer to descriptions in "User Interface". RIGOL Chapter 1 Quick Guide 1-8 DG900 User's Guide Rear Panel Overview Figure 1-2 Rear Panel 1. [10MHz In/Out] BNC female connector, with 50 Ω nominal impedance. Its function is determined by the clock type used by the instrument. 1) When internal clock source is selected, this connector (as 10MHz Out) outputs the 10 MHz clock signal generated by the internal crystal oscillator inside the generator. 2) When external clock source is selected, this connector (as 10MHz In) receives an external 10 MHz clock signal. This connector is usually used to realize synchronization among multiple instruments. For details about the signals mentioned above, please refer to the descriptions in "Clock Source". 2. [CH1/Sync/Ext Mod/Trig/FSK] BNC female connector, with 50 Ω nominal impedance. Its function is determined by the current working mode of CH1. 1) Sync When the output of CH1 is enabled, this connector outputs the corresponding sync signal that matches the current configuration of CH1. For detailed information about the characteristics of the sync signals that correspond to various output signals, refer to the descriptions in "Sync Setting". 2) Ext Mod When AM, FM, PM, or PWM of CH1 is enabled and external modulation source is selected, this connector receives an external modulation signal. Its input impedance is 1000 Ω. For details, refer to descriptions in "Modulation". 1 2 3 4 5 6 Chapter 1 Quick Guide RIGOL DG900 User's Guide 1-9 3) FSK When ASK, FSK, or PSK of CH1 is enabled and external modulation source is selected, this connector receives an external modulation signal whose polarity can be set by users. Its input impedance is 1000 Ω. For details, refer to descriptions in "Modulation". 4) Trig In When Sweep or Burst of CH1 is enabled and external trigger source is selected, this connector receives an external trigger signal whose polarity can be set by users. 5) Trig Out When Burst of CH1 is enabled, and the internal/manual trigger source is selected, this connector outputs a trigger signal with specified edge type. 3. [CH2/Sync/Ext Mod/Trig/FSK] BNC female connector, with 50 Ω nominal impedance. Its function is determined by the current working mode of CH2. 1) Sync When the output of CH2 is enabled, this connector outputs the corresponding sync signal that matches the current configuration of CH2. For detailed information about the characteristics of the sync signals that correspond to various output signals, refer to the descriptions in "Sync Setting". 2) Ext Mod When AM, FM, PM or PWM of CH2 is enabled and external modulation source is selected, this connector receives an external modulation signal. Its input impedance is 1000 Ω. For details, refer to descriptions in "Modulation". 3) FSK When ASK, FSK, or PSK of CH2 is enabled and external modulation source is selected, this connector receives an external modulation signal whose polarity can be set by users. Its input impedance is 1000 Ω. For details, refer to descriptions in "Modulation". 4) Trig In When Sweep or Burst of CH2 is enabled and external trigger source is selected, this connector receives an external trigger signal whose polarity can be set by users. 5) Trig Out When Burst of CH2 is enabled, and the internal/manual trigger source is selected, this connector outputs a trigger signal with specified edge type. 4. USB HOST Supports FAT32 format Flash type USB storage device, RIGOL TMC digital oscilloscope (DS), and the USB-GPIB interface converter. USB storage device: reads the waveform files or state files saved in the USB storage device; or stores the current instrument states or edited waveform data into the USB storage device. Besides, the contents displayed on the RIGOL Chapter 1 Quick Guide 1-10 DG900 User's Guide screen can also be saved to the USB storage device in the format of an image (.Bmp). TMC DS: seamlessly interconnects with the RIGOL DS that meets the TMC standard. Reads and stores the waveform data collected by the DS and rebuilds waveforms without distortion. USB-GPIB interface converter (optional accessory): extends the GPIB interface for RIGOL instruments that integrates the USB HOST interface but not the GPIB interface. 5. USB DEVICE It is used to connect the generator to a computer which can control the generator remotely by using PC software or by programming. 6. AC Power Cord Connector The rated AC power source supported by the signal generator is (100-127 V, 45-440 Hz) or (100-240 V, 45-65 Hz), and its maximum input power shall not exceed 30 W. The specification of the fuse is 250 Vac, T4.0 A. Chapter 1 Quick Guide RIGOL DG900 User's Guide 1-11 To Prepare for Use To Connect to AC Power Please use the power cord provided in the accessories to connect the signal generator to the AC power source, as shown in the figure below. The rated AC power source supported by the signal generator is (100-127 V, 45-440Hz) or (100-240 V, 45-65Hz), and its maximum input power shall not exceed 30 W. When the signal generator is connected to the AC power source via the power cord, the instrument automatically adjusts itself to within the proper voltage range, and you do not need to select the voltage range manually. CAUTION To avoid electric shock, ensure that the instrument is correctly grounded. Turn-on Checkout After connecting the instrument to the power source properly, press on the front panel to start the signal generator. During the start-up, the instrument will undergo the initialization and self-check process. Then, it will enter a default interface. If you still fail to power on the instrument normally, refer to the methods in "Troubleshooting" to resolve the problem. To Set the System Language DG900 arbitrary waveform generator supports multiple languages. You can press Utility System Setting, and then select a desired language from the "Language" drop-down list. RIGOL Chapter 1 Quick Guide 1-12 DG900 User's Guide User Interface The DG900 user interface is shown in the following figure. Figure 1-1 User Interface 1. Channel Output Configuration Status Bar Displays the current output configuration of the channel. Note: Two channels can be enabled simultaneously, but you cannot select both channels at the same time. Output Impedance Type: High impedance: displays HighZ. Load: displays impedance (the default is 50 Ω and the range is from 1 Ω to 10 kΩ). Channel Output State: ON: illuminated in yellow. OFF: grayed out. Selected Waveform: Sine/Square/Ramp/Pulse/Noise/Prbs/Dualtone/Harm /Rs232/DC/Arb/Sequence Modulation Type: AM/FM/PM/ASK/FSK/PSK/PWM Sweep Type: Linear/Log/Step Burst Type: Ncycle/Infinite/Gated 1 2 3 7 6 5 4 8 9 10 11 Chapter 1 Quick Guide RIGOL DG900 User's Guide 1-13 2. Up and Down Scroll Bar Prompts you to move up and down with your fingers on the screen to view and set parameters. 3. Information Setting : opens the Store interface. : opens the Utility interface. : performs the channel copy function. : performs the screen print operation. 4. Right Arrow Prompts you to slide right on the screen to switch to the waveform selection interface. 5. Status Bar : indicates that the front-panel keys and the screen are locked. : indicates that the beeper is disabled. : indicates that the instrument is in programming-controlled mode. : indicates that the instrument has been successfully connected to the local area network by using the network cable. : indicates that a USB storage device is found. 6. Waveform Displays the currently selected waveform of each channel. 7. Interface Label Displays the label of the current interface. 8. Frequency Displays the frequency of the current waveform of each channel. Tap the Freq parameter input field to modify the parameter with the pop-up numeric keypad. You can also use the arrow keys and the knob to modify the parameter. 9. Amplitude Displays the amplitude of the current waveform of each channel. Tap the Ampl parameter input field to modify the parameter with the pop-up numeric keypad. You can also use the arrow keys and the knob to modify the parameter. 10. Offset Displays the DC offset of the current waveform of each channel. Tap the Offset parameter input field to modify the parameter with the pop-up numeric keypad. You can also use the arrow keys and the knob to modify the parameter. RIGOL Chapter 1 Quick Guide 1-14 DG900 User's Guide 11. Phase Displays the phase of the current waveform of each channel. Tap the Phase parameter input field to modify the parameter with the pop-up numeric keypad. You can also use the arrow keys and the knob to modify the parameter. Chapter 1 Quick Guide RIGOL DG900 User's Guide 1-15 To Use the Built-in Help System DG900 series provides the help information for each front-panel function menu and the current display interface. You can view the help information if you have any questions during the operation process. 1. Obtain the help information of the front panel keys To get the help information about any front-panel key or menu softkey, press the Help/Local key first, and then press the desired key for the help information. Then, the corresponding help information is displayed. 2. Obtain the common help topics Press Help/Local on the front panel, and then the following interface is displayed below. Tap "Help" to enter the help interface. At this time, you can tap on the touch screen to move up and down the help items or rotate the knob to scroll up and down the list to select the desired help item. Then, the help information for the item is displayed in the interface. 3. Obtain the descriptions of the data in the interface Press Help/Local on the front panel to enter the interface, as shown above. Tap "Center" to view the descriptions for the data in the center of the current interface. Tap "Top" to view the descriptions for the data in the top part of the current interface. Tap "Bottom" to view the descriptions for the data in the bottom part of the current interface. Tap "Guide" to enter the guide interface. 4. Page up/down operation When the help information is displayed in multiple pages, you can tap to move up and down the touch screen to view the help information. 5. Close the current help information When the help information is displayed in the interface, press Help/Local on the front panel to close the help information currently displayed on the screen. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-1 Chapter 2 Front Panel Operations Introduces the main functions and operation methods of DG900. Contents in this chapter: To Output Basic Waveform To Output the Arbitrary Waveform To Output Harmonic DC To Output Dual-tone Waveform To Output Advanced Waveform Modulation Sweep Burst Frequency Counter Store and Recall Channel Setting Common Settings System Utility Function RIGOL Chapter 2 Front Panel Operations 2-2 DG900 User's Guide To Output Basic Waveform DG900 series can output basic waveforms (including Sine, Square, Ramp, Pulse, and Noise) from one channel or from both two channels at the same time. At start-up, the dual channels are configured to output a sine waveform with 1 kHz frequency and 5 Vpp amplitude by default. Users can configure the instrument to output various basic waveforms. To Select Output Channel At start-up, the instrument displays the CH1 parameter interface by default. You can press Output1 or Output2 (also, you can tap the channel output configuration status bar or ) to switch between CH1 and CH2 as the currently selected channel. The interface of CH1 channel is displayed in red, and that of CH2 is displayed in blue. After the desired channel is selected, you can configure the waveform and parameters of the selected channel. Key Points: CH1 and CH2 cannot be both selected at the same time. You can first select CH1, after configuring the waveform and parameters of CH1, then select CH2 for configuration. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-3 To Select Basic Waveform DG900 can output 5 basic waveforms, including Sine, Square, Ramp, Pulse and Noise. Press Menu on the front panel, then tap the touch screen to select the desired waveform. After selecting the desired waveform, you are automatically directed to the waveform parameter setting interface. To go back to the waveform selection interface, tap to slide right or press the Menu key on the front panel. At start-up, Sine is selected by default. Table 2-1 Basic Waveforms Basic Waveforms Sine Square Ramp Pulse Noise Function Name Sine Square Ramp Pulse Noise Freq/Period √ √ √ √ Ampl/HighL √ √ √ √ √ Offset/LowL √ √ √ √ √ Start Phase √ √ √ √ Align Phase √ √ √ √ Duty √ √ Symmetry √ Width/Duty √ RisEdge √ FallEdge √ RIGOL Chapter 2 Front Panel Operations 2-4 DG900 User's Guide To Set Frequency/Period Frequency is one of the most important parameters of basic waveforms. For different instrument models and waveforms, the setting ranges of frequency are different. The default frequency is 1 kHz. Sine Square Ramp Pulse Noise (-3 dB) DG952 1 μHz to 50 MHz 1 μHz to 15 MHz 1 μHz to 1.5 MHz 1 μHz to 15 MHz 100 MHz Bandwidth DG972 1 μHz to 70 MHz 1 μHz to 20 MHz 1 μHz to 1.5 MHz 1 μHz to 20 MHz 100 MHz Bandwidth DG992 1 μHz to 100 MHz 1 μHz to 25 MHz 1 μHz to 2 MHz 1 μHz to 25 MHz 100 MHz Bandwidth The frequency displayed on the screen is the default value or the frequency previously set. When the instrument function is changed, if this frequency is valid under the new function, the instrument will still use this frequency; otherwise, the instrument would display a prompt message and set the frequency to the frequency upper limit of the new function automatically. Tap the Freq parameter input field to input the desired frequency value with the numeric keypad, then select the unit. Tap "Ok". The frequency units available are MHz, kHz, Hz, mHz, and μHz. Tap the label again to switch to Period setting. The period units available are s, ms, μs, and ns. You can also use the arrow keys and the knob to set the parameter value: press down the knob to enter the editing mode, use the arrow keys to move the cursor to select the digit to be edited, and then rotate the knob to modify the value. Besides, you can use the knob to switch between different parameters. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-5 To Set Amplitude/High Level The amplitude setting range is limited by the "Impedance" and "Freq/Period" settings. By default, it is 5 Vpp. The amplitude displayed on the screen is the default value or the amplitude previously set. When the instrument configuration (e.g. frequency) is changed, if this amplitude is valid, the instrument will still use this amplitude; otherwise, the instrument would display a prompt message and set the amplitude to the amplitude upper limit of the new configuration automatically. You can also use "High Level" or "Low Level" to set the amplitude. Tap the Ampl parameter input field to input the desired frequency value with the numeric keypad, then select the unit. Tap "Ok". The amplitude units available are Vpp, mVpp, Vrms, mVrms, and dBm (invalid in HighZ). Tap the label again to switch to High Level setting. The high level units available are V and mV. You can also use the arrow keys and the knob to set the parameter value: press down the knob to enter the editing mode, use the arrow keys to move the cursor to select the digit to be edited, and then rotate the knob to modify the value. Besides, you can use the knob to switch between different parameters. Key Points: 1． How to convert the amplitude in Vpp to the corresponding value in Vrms? Method: Vpp is the unit for signal peak-peak value and Vrms is the unit for root-mean-square value. The default unit is Vpp. When you set the amplitude, input a decimal point in the numeric keypad, and then select a unit. Tap "Ok" to switch to the unit of the current amplitude. Remarks: For different waveforms, the relation between Vpp and Vrms is different. The relation of the two units is as shown in the figure below (take sine waveform as an example). Vpp=2Vamp Vrms=0.707Vamp Vamp RIGOL Chapter 2 Front Panel Operations 2-6 DG900 User's Guide According to the figure above, the conversion relation between Vpp and Vrms fulfills the following equation: Vrms 2 2 Vpp = For example, if the current amplitude is 5 Vpp, input a decimal point in the numeric keypad, and then select a unit Vrms. Tap "Ok". Then, it can be converted to a value with Vrms as the unit. For sine waveform, the converted value is 1.768 Vrms. 2． How to set the amplitude of the waveform in dBm? Method: 1) Press Output1 or Output2 (also, you can tap the channel output configuration status bar or ) to select the desired channel. 2) Tap the channel output configuration status bar (CH1) or (CH2). In the channel setting interface, tap the OutputSet menu label and select "Off". Then use the numeric keypad to set a proper load value. 3) Select the desired waveform, tap the Ampl menu label, and then input the desired value by using the numeric keypad. Then select the unit "dBm" from the pop-up menu. Remarks: dBm is the unit for signal power absolute value， and the conversion relation between dBm and Vrms fulfills the following equation: 2 1 10lg( ) 0.001 Vrms dBm R W = × Wherein, R represents the channel output impedance value and it must be a specific value, so the unit dBm is not available when the output impedance is "HighZ". For example, if the current output impedance is 50 Ω and the amplitude is 1.768 Vrms (i.g. 5 Vpp), input a decimal point in the numeric keypad and then select "dBm". Tap "Ok" to convert the amplitude value to the corresponding value in dBm. The converted value is 17.9601 dBm. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-7 To Set Offset/Low Level The DC offset setting range is limited by the "Impedance" and "Amplitude/High Level" settings. The default value is 0 Vdc. The DC offset voltage displayed on the screen is the default value or the offset previously set. When the instrument configuration (e.g. impedance) is changed, if this offset is valid, the instrument will still use this offset; otherwise, the instrument would display a prompt message and set the offset to the offset upper limit of the new configuration automatically. Tap the Offset parameter input field to input the desired offset value with the numeric keypad, then select the unit. Tap "Ok". The DC offset voltage units available are Vdc and mVdc. Tap the label again to switch to Low Level setting. The low level should be at least 1 mV smaller than the high level (output impedance: 50 Ω). The low level units available are V and mV. You can also use the arrow keys and the knob to set the parameter value: press down the knob to enter the editing mode, use the arrow keys to move the cursor to select the digit to be edited, and then rotate the knob to modify the value. Besides, you can use the knob to switch between different parameters. RIGOL Chapter 2 Front Panel Operations 2-8 DG900 User's Guide To Set Start Phase The setting range of start phase is from 0° to 360°. The default is 0°. The start phase displayed on the screen is the default value or the phase previously set. When the instrument function is changed, the new function will still use this phase. Tap the Phase parameter input field to input the desired phase value with the numeric keypad, then select the unit "°". Tap "Ok". You can also use the arrow keys and the knob to set the parameter value: press down the knob to enter the editing mode, use the arrow keys to move the cursor to select the digit to be edited, and then rotate the knob to modify the value. Besides, you can use the knob to switch between different parameters. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-9 To Set Duty Cycle (Square) Duty cycle is defined as the percentage that the high level takes up in the whole period (as shown in the figure below). This parameter is only available when Square or Pulse is selected. T t Duty Cycle=t/T100% The available range of the duty cycle is from 0.01% to 99.99% (limited by the current frequency setting). The default is 50%. Tap the Duty parameter input field to input the desired duty cycle value with the numeric keypad, then select the unit "%". Tap "Ok". You can also use the arrow keys and the knob to set the parameter value: press down the knob to enter the editing mode, use the arrow keys to move the cursor to select the digit to be edited, and then rotate the knob to modify the value. Besides, you can use the knob to switch between different parameters. RIGOL Chapter 2 Front Panel Operations 2-10 DG900 User's Guide To Set Symmetry (Ramp) Symmetry is defined as the percentage that the rising period of the ramp takes up in the whole period (as shown in the figure below). This parameter is only available when Ramp is selected. T t Symmetry=t/T100% The setting range of symmetry is from 0% to 100%. The default is 50%. Tap the Symm parameter input field to input the desired symmetry value with the numeric keypad, then select the unit "%". Tap "Ok". You can also use the arrow keys and the knob to set the parameter value: press down the knob to enter the editing mode, use the arrow keys to move the cursor to select the digit to be edited, and then rotate the knob to modify the value. Besides, you can use the knob to switch between different parameters. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-11 To Set Pulse Width/Duty Cycle (Pulse) Pulse width is defined as the time from the 50% threshold of a pulse's rising edge to the 50% threshold of the next falling edge (as shown in the figure below). The settable range of the pulse width is from 16 ns to 999.999982118 ks (limited by "minimum pulse width" and "pulse period"). The default is 500 μs. Pulse Width ≥ Minimum Pulse Width Pulse Width < Pulse Period - 2 × Minimum Pulse Width Pulse duty cycle is defined as the percentage that the pulse width takes up in the whole pulse period. Pulse duty cycle and pulse width are correlative. Modifying either of them (pulse duty cycle or pulse width) will automatically affect the other. The settable range of the pulse duty cycle is from 0.001% to 99.999% (limited by "minimum pulse width" and "pulse period"). The default is 50%. Pulse Duty Cycle ≥ 100 × Minimum Pulse Width ÷ Pulse Period Pulse Duty Cycle < 100 × (1 - 2 × Minimum Pulse Width ÷ Pulse Period) Tap the Width parameter input field to input the desired pulse width value with the numeric keypad, then select the unit. Tap "Ok". The pulse width units available are s, ms, μs, and ns. Tap this menu label again to switch to duty cycle setting. You can also use the arrow keys and the knob to set the parameter value: press down the knob to enter the editing mode, use the arrow keys to move the cursor to select the digit to be edited, and then rotate the knob to modify the value. Besides, you can use the knob to switch between different parameters. Pulse 10% tRise tFall 90% 50% Pulse Period RIGOL Chapter 2 Front Panel Operations 2-12 DG900 User's Guide To Set Rising/Falling Edge (Pulse) The rising edge time is defined as the duration of the pulse amplitude rising from 10% to 90% threshold, while falling edge time is defined as the duration of the pulse amplitude moving down from 90% to 10% threshold (as shown in the figure below). The setting range of rising/falling edge time is limited by the currently specified pulse width limit (as shown in the formula below). DG900 will automatically adjust the edge time to match the specified pulse width if the value currently set exceeds the limit value. Rising/Falling Edge Time ≤ 0.625 × Pulse Width Tap the RisEdge or FallEdge parameter input field. Use the numeric keypad to input the desired value and then select the desired unit from the pop-up menu. The pulse width units available are sec, msec, μsec, and nsec. The rising and falling edge time are independent of each other, and users can set them separately. You can also use the arrow keys and the knob to set the parameter value: press down the knob to enter the editing mode, use the arrow keys to move the cursor to select the digit to be edited, and then rotate the knob to modify the value. Besides, you can use the knob to switch between different parameters. Pulse 10% tRise tFall 90% 50% Pulse Period Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-13 To Enable Channel Output After configuring the parameters of the waveform selected, enable the channel to output the waveforms. Before enabling channel output, you can also tap or and then tap the corresponding menu label in the channel setting interface to set the parameters related with the channel output. For details, refer to descriptions in "Channel Setting". Press Output1 or Output2 on the front panel, and then the backlight turns on. You can also tap the channel output configuration status bar or enable the channel output in the channel setting interface. When the channel status bar is highlighted, (i.g. or ), the configured waveforms are output from the corresponding output connector on the front panel. RIGOL Chapter 2 Front Panel Operations 2-14 DG900 User's Guide Align Phase DG900 series dual-channel function/arbitrary waveform generator enables you to align the phases of the two channels. Press Align on the front panel, then the instrument will re-configure the two channels and enable the generator to output with specified frequency and start phase. For two signals whose frequencies are the same or in multiple relationship, this operation can align their phases. For example, assume a sine waveform (1 kHz, 5 Vpp, 0°) is output from CH1, while another one (1 kHz, 5 Vpp, 180°) from CH2. Use the oscilloscope to acquire the waveforms of the two channels and stably display the waveforms. It can be found that the phase deviation between the two waveforms is no longer 180°. At this point, press Align on the generator and the waveforms shown on the oscilloscope will have a phase deviation of 180° without manual adjustment of the start phase of the generator. Figure 2-1 Before Aligning Phase Figure 2-2 After Aligning Phase CH1 CH2 CH1 CH2 Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-15 Example: To Output Sine This section mainly introduces how to output Sine waveforms (frequency 20 kHz, amplitude 2.5 Vpp, offset 500 mVdc, start phase 90°) from the [CH1] connector. 1. Select the output channel: Press Output1 on the front panel or tap the channel output configuration status bar to select CH1. At this time, the channel is indicated in red in the status bar. 2. Select Sine waveforms: Press Menu on the front panel, and then the waveform selection interface is displayed. Tap Continuous and then select the "Sine" icon to go to the sine waveform parameter setting interface automatically. 3. Set frequency: Tap the Freq parameter input field to input 20 with the pop-up numeric keypad, and then select the unit "kHz". Tap "Ok". 4. Set amplitude: Tap the Ampl parameter input field to input 2.5 with the pop-up numeric keypad, and then select "Vpp" as the unit. Tap "Ok". 5. Set offset voltage: Tap the Offset parameter input field to input 500 with the pop-up numeric keypad, and then select "mVdc" as the unit. Tap "Ok". 6. Set start phase: Tap the Phase parameter input field to input 90 with the pop-up numeric keypad, and then select "°" as the unit. Tap "Ok". 7. Enable channel output: Press Output1 and the backlight turns on. Also, you can tap the channel output configuration status bar to enable the channel output. Then, the channel status is highlighted (i.g. ), and the Sine signal is output from the [CH1] connector based on the current configurations. 8. Observe the output waveform: Connect the [CH1] connector of DG900 to the oscilloscope by using the BNC cable. The waveform is as shown in the figure below. RIGOL Chapter 2 Front Panel Operations 2-16 DG900 User's Guide Figure 2-3 Sine Waveform Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-17 To Output the Arbitrary Waveform DG900 can output built-in waveforms from a single channel or from two channels at the same time. The 160 kinds of built-in arbitrary waveforms are stored in the internal non-volatile memory. To Enable Arbitrary Waveforms Press Menu Continuous "Arb" to enable arbitrary waveform function and open the arbitrary waveform selection interface, as shown in the figure below. Figure 2-4 Arbitrary Waveform Selection Interface To Select the Waveform DG900 allows users to select 160 built-in waveforms and arbitrary waveforms stored in the internal or external memory of the instrument. Key Points: After selecting the desired waveform, press the corresponding channel output control key (Output1 or Output2), and the specified waveform will be output from the channel. For details, refer to descriptions in "To Enable Channel Output". Note that when the frequency of the arbitrary waveforms is greater than 1 MHz, some waveforms will be distorted. RIGOL Chapter 2 Front Panel Operations 2-18 DG900 User's Guide Built-in Waveform DG900 has 160 built-in arbitrary waveforms, as shown in Table 2-2. In the arbitrary waveform selection interface, tap the Engineering, Medical, AutoElec, or Maths menu label to select the corresponding type. Tap or rotate the knob (pressing the right arrow key will locate the cursor to the right of the interface) to select the desired waveform (the selected waveform is highlighted). The waveforms under the Common menu label are most frequently selected ones by users. At most, 8 waveforms can be stored. Table 2-2 160 Built-in Arbitrary Waveforms Type Waveform Remarks Engineering Sinc Sinc function Lorentz Lorentz function Log Logarithm function and the base is 10 GausPul Gauss pulse NegRamp Negative ramp NPulse Negative pulse PPulse Positive pulse SineTra Sine-Tra waveform SineVer Sine-Ver waveform StairDn Stair-down waveform StairUD Stair-up and stair-down waveform StairUp Stair-up waveform Trapezia Trapezia waveform AmpALT Gain oscillation curve AttALT Attenuation oscillation curve RouHalf RoundHalf Wave RousPM RoundsPM Waveform BlaWave Time-velocity curve of explosive vibration DampOsc Time-displacement curve of damped oscillation SwigOsc Kinetic energy- time curve of swing oscillation Dischar Discharge curve of Ni-MH battery Pahcur Current waveform of DC brushless motor Combin Combination function SCR SCR firing profile Bworth Butterworth filter Chshev1 Chebyshev1 filter Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-19 Chshev2 Chebyshev2 filter TV TV signal Voice Voice signal Surge Surge signal Radar Radar signal DualTone Dual-tone signal Ripple Power ripple Quake Analog quake waveform Gamma Gamma signal StepResp Step-response signal BandLim Bandwidth-limited signal CPulse C-Pulse signal CWPulse CW pulse signal GateVibr Gate self-oscillation signal LFMPulse Linear FM pulse signal MCNoise Mechanical construction noise AM Sectioned sine AM signal FM Sectioned sine FM signal PFM Sectioned pulse FM signal PM Sectioned sine PM signal PWM Sectioned PWM signal Medical Cardiac Cardiac signal EOG Electro-oculogram EEG Electroencephalogram EMG Electromyogram Pulgram Pulsilogram ResSpd Speed curve of the respiration ECG1 Electrocardiogram 1 ECG2 Electrocardiogram 2 ECG3 Electrocardiogram 3 ECG4 Electrocardiogram 4 ECG5 Electrocardiogram 5 ECG6 Electrocardiogram 6 ECG7 Electrocardiogram 7 ECG8 Electrocardiogram 8 ECG9 Electrocardiogram 9 ECG10 Electrocardiogram 10 ECG11 Electrocardiogram 11 ECG12 Electrocardiogram 12 ECG13 Electrocardiogram 13 ECG14 Electrocardiogram 14 ECG15 Electrocardiogram 15 RIGOL Chapter 2 Front Panel Operations 2-20 DG900 User's Guide LFPulse Waveform of the low frequency pulse electrotherapy Tens1 Waveform 1 of the nerve stimulation electrotherapy Tens2 Waveform 2 of the nerve stimulation electrotherapy Tens3 Waveform 3 of the nerve stimulation electrotherapy AutoElec Ignition Ignition waveform of the automotive motor SP Automotive starting profile with ringing VR Automotive supply voltage profile for resetting TP1 Automotive transients arising from disconnection TP2A Automotive transients arising from inductance in wiring TP2B Automotive transients arising from the ignition switching off TP3A Automotive transients arising from switching TP3B Automotive transients arising from switching TP4 Automotive working profile during start-up TP5A Automotive transients arising from cut-off of battery power TP5B Automotive transients arising from cut-off of battery power Maths Airy Airy function Besseli Bessel functions of the first kind Bessely Bessel functions of the second kind Cubic Cubic function Dirichlet Dirichlet function Erf Error function Erfc Complementary error function ErfcInv Inverted complementary error function ErfInv Inverted error function ExpFall Exponential fall function ExpRise Exponential rise function HavSin HaverSine function Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-21 Laguerre 4-times Laguerre polynomial Legend 5-times Legend polynomial Versiera Versiera ARB_X2 Square function Gauss Gaussian distribution or normal distribution Weibull Weibull distribution LogNorm Logarithmic normal distribution Laplace Laplace distribution Maxwell Maxwell distribution Rayleigh Rayleigh distribution Cauchy Cauchy distribution CosH Hyperbolic cosine CosInt Integral cosine Cot Cotangent CotHCon Concave hyperbolic cotangent CotHPro Protuberant hyperbolic cotangent CscCon Concave cosecant CscPro Protuberant cosecant CscHCon Concave hyperbolic cosecant CscHPro Protuberant hyperbolic cosecant RecipCon Concave reciprocal RecipPro Protuberant reciprocal SecCon Concave secant SecPro Protuberant secant SecH Hyperbolic secant SinH Hyperbolic sine SinInt Integral sine Sqrt Square root Tan Tangent TanH Hyperbolic tangent AbsSine Absolute value of sine AbsSinH Absolute value of half sine ACos Arc cosine ACosH Arc hyperbolic cosine ACotCon Concave arc cotangent ACotPro Protuberant arc cotangent ACotHCon Concave arc hyperbolic cotangent ACotHPro Protuberant arc hyperbolic cotangent ACscCon Concave arc cosecant ACscPro Protuberant arc cosecant ACscHCon Concave arc hyperbolic cosecant ACscHPro Protuberant arc hyperbolic cosecant ASecCon Concave arc secant RIGOL Chapter 2 Front Panel Operations 2-22 DG900 User's Guide ASecPro Protuberant arc secant ASecH Arc hyperbolic secant ASin Arc Sinc ASinH Arc hyperbolic sine ATan Arc tangent ATanH Arc hyperbolic tangent Bartlett Bartlett window BarWin Modified Bartlett-Hann window Blkman Blackman window BlkmanH Blackman-Harris window BohWin Bohman window Boxcar Rectangular window ChebWin Chebyshev window FlatWin Flat Top weighted window Ham Hamming window Hanning Hanning window Kaiser Kaiser window NutWin Nuttall-defined minimum 4-term Blackman-Harris window ParWin Parzen window TayWin Taylor window Triang Triangle window (Fejer window) TukWin Tukey (tapered cosine) window To Set Parameters After selecting the desired waveform, you are automatically directed to the waveform parameter setting interface. 1. Freq/Period: sets the output frequency/period of the arbitrary waveform. 2. Ampl/HighL: sets the output amplitude/high level of the arbitrary waveform. 3. Offset/LowL: sets the output offset/low level of the arbitrary waveform. 4. Phase: sets the output start phase of the arbitrary waveform. Please refer to "To Output Basic Waveform" to configure the parameters and output for the channel. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-23 To Output Harmonic DG900 can be used as a harmonic generator to output harmonic with specified order, amplitude and phase. It is usually used in the test of harmonic detector device or harmonic filter device. This section introduces how to configure the generator to output harmonic. Harmonic Overview According to Fourier transform, time domain waveform is the superposition of a series of sine waveforms, as shown in the equation below: ......
) 2 sin( ) 2 sin( ) 2 sin( ) ( 3 3 3 2 2 2 1 1 1 + + + + + + = ϕ π ϕ π ϕ π t f A t f A t f A t f Generally, component with 1 f frequency is called fundamental waveform, 1 f is the fundamental waveform frequency, 1 A is the fundamental waveform amplitude, and 1 ϕ is the fundamental waveform phase. The frequencies of the other components (called harmonics) are all integral multiple of the fundamental waveform frequency. Components whose frequencies are odd multiples of the fundamental waveform frequency are called odd harmonics, and components whose frequencies are even multiples of the fundamental waveform frequency are called even harmonics. At most, 8th harmonic can be output by DG900. After selecting CH1 or CH2, press Menu Continuous "Harm" to enter the harmonic setting menu. You can set fundamental waveform parameters, select the type of output harmonic, specify the highest order of harmonic, and set the amplitude and phase of each order of harmonic. After completing the harmonic parameters setting, press Output1, and the backlight turns on. The instrument outputs the specified harmonic from the corresponding output terminal. For details, refer to descriptions in "To Enable Channel Output". RIGOL Chapter 2 Front Panel Operations 2-24 DG900 User's Guide To Set Fundamental Waveform Parameters DG900 allows users to set various fundamental waveform parameters such as frequency, period, amplitude, DC offset voltage, high level, low level, and start phase. It also supports align phase operation. To set the above fundamental waveform parameters, refer to descriptions in "To Output Basic Waveform". To Select Harmonic Type DG900 can output even harmonic, odd harmonic, both harmonics, and user-defined order of harmonic. After entering the harmonic setting interface, tap the Type parameter selection field to select the desired harmonic type. 1. Even: outputs fundamental waveform and even harmonics. 2. Odd: outputs fundamental waveform and odd harmonics. 3. Both: outputs fundamental waveform and all the harmonics in order. 4. User: outputs the user-defined orders of harmonics. The highest order is 8. 8 bits binary data are used to represent the output status of the 8 orders of harmonics respectively. Wherein, 1 represents enabling the output of the corresponding harmonic, and 0 represents disabling the output of the corresponding harmonic. You only need to tap the User parameter input field, and then modify the value of each data bit by using the numeric keypad (note that the leftmost bit representing fundamental waveform is always X and cannot be modified). For example, if you set the 8 bits data to X001 0001, it indicates that the fundamental waveform, 4th and 8th orders of harmonics are output. Note: The actually output harmonics is determined by the currently specified harmonic order and harmonic type. To Set Harmonic Order The highest order of harmonic output from DG900 cannot be greater than the specified value. After entering the harmonic setting interface, tap the Count parameter selection field to input a value with numeric keypad. The range is limited by the maximum output frequency of the instrument and current fundamental waveform frequency. Range: 2 to maximum output frequency ÷ fundamental waveform Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-25 frequency, an integer. The maximum is 8. You can also use the arrow keys and the knob to set the parameter value: press down the knob to enter the editing mode, use the arrow keys to move the cursor to select the digit to be edited, and then rotate the knob to modify the value. Besides, you can use the knob to switch between different parameters. To Select Harmonic Amplitude After entering the harmonic setting interface, tap the Harm Ampl parameter input field to set the amplitude of each order of harmonic. 1) No.: selects the sequence number of the harmonic to be set. 2) Harm Ampl: sets the amplitude of the specified order of harmonic. Use the numeric keypad to input the amplitude value, and then select the desired unit. Tap "Ok". The amplitude units available are Vpp, mVpp, Vrms, mVrms, and dBm (invalid in HighZ). To Set Harmonic Phase After entering the harmonic setting interface, tap the Harm Phase parameter input field to set the phase of each order of harmonic. 1. No.: selects the sequence number of the harmonic to be set. 2. Harm Phase: sets the phase of the specified order of harmonic. Use the numeric keypad to input the phase value, and then select the desired unit "°". Tap "Ok". Example: To Output Harmonic This section introduces how to output 2th order of harmonic (harmonic amplitude: 2 Vpp, harmonic phase: 30°) and 4th orders of harmonic (harmonic amplitude: 1 Vpp, harmonic phase: 50°) from the [CH1] connector. 1. Select the output channel: Press Output1 on the front panel or tap the channel output configuration status bar to select CH1. At this time, the channel is indicated in red in the status bar. RIGOL Chapter 2 Front Panel Operations 2-26 DG900 User's Guide 2. Set fundamental waveform parameters: In this example, various parameters of the fundamental waveform take the default values. You can also refer to descriptions in "To Output Basic Waveform" to set various fundamental waveform parameters such as frequency/period, amplitude/high level, offset/low level, and start phase. 3. Enable harmonic function: Press Menu on the front panel, and then the waveform selection interface is displayed. Tap Continuous and then select the "Harm" icon to go to the harmonic parameter setting interface automatically. 4. Set harmonic count: In the harmonic parameter setting interface, tap the Count parameter input field to input a value 5 with pop-up numeric keypad. Tap "Ok". 5. Select the harmonic type: In the harmonic parameter setting interface, tap the Type parameter selection field to select the desired even harmonic. 6. Set harmonic amplitude: In the harmonic parameter setting menu, set the amplitude of 2th and 4th orders of harmonic. 1) Tap the No. parameter input field to input 2 with the pop-up numeric keypad, and then tap "Ok". 2) Tap the Harm Ampl parameter input field to input 2 with the pop-up numeric keypad, select the unit "Vpp". Then tap "Ok". 3) Refer to Step 1 and Step 2 to set the amplitude of the 4th order harmonic to 1 Vpp. 7. Set harmonic phase: In the harmonic parameter setting menu, set the phases of 2th and 4th orders of harmonic in sequence. 1) Tap the No. parameter input field to input 2 with the pop-up numeric keypad, and then tap "Ok". 2) Tap the Harm Phase parameter input field to input 30 with the pop-up numeric keypad, select the unit "°". Then tap "Ok". 3) Refer to Step 1 and Step 2 to set the phase of the 4th order harmonic to 50°. 8. Enable channel output: Press Output1, and the backlight turns on. Also, you can tap the channel output configuration status bar to enable the channel output. Then, the channel status is highlighted (i.g. ), and the fundamental waveform, 2th and 4th order of harmonics are output from the [CH1] connector based on the current configurations. 9. Observe the output waveform: Connect the [CH1] connector of DG900 to the oscilloscope by using the BNC cable. The waveform is as shown in the figure below. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-27 Figure 2-5 Harmonic RIGOL Chapter 2 Front Panel Operations 2-28 DG900 User's Guide DC DG900 can output DC signal with amplitude range from -10 V to 10 V (HighZ) or from -5 V to 5 V (when the load is 50 Ω). The figure below shows a sketch of the DC signal. Press Menu Continuous "DC" to select the DC signal. Figure 2-6 Sketch of DC signal Tap the Offset parameter input field to input the desired offset value with the numeric keypad, then select the unit. Tap "Ok". You can also use the arrow keys and the knob to set the parameter value: press down the knob to enter the editing mode, use the arrow keys to move the cursor to select the digit to be edited, and then rotate the knob to modify the value. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-29 To Output Dual-tone Waveform Press Menu Continuous "Dualtone" to enter the dual-tone signal parameter setting interface. 1. Freq1/CentFreq: sets the output frequency/period of Signal 1 in the dual-tone signal. 2. Freq2/OffsFreq: sets the output frequency/period of Signal 2 in the dual-tone signal. 3. Ampl/HighL: sets the output amplitude/high level of the dual-tone signal. 4. Offset/LowL: sets the output offset/low level of the dual-tone signal. Please refer to "To Output Basic Waveform" to configure the parameters and output for the channel. RIGOL Chapter 2 Front Panel Operations 2-30 DG900 User's Guide 6 7 8 9 To Output Advanced Waveform DG900 provides three advanced waveforms: PRBS, RS232, and Sequence. Note: When one channel is set to one of the advanced waveforms, the other channel will automatically switch to the set advanced waveform from other waveform. PRBS A PRBS is generated by a linear-feedback shift register (LFSR), as shown below. An LFSR is specified by the number of stages it contains (L) and which stages (taps) feed the exclusive-or (XOR) gates in its feedback network. The PRBS output is taken from the last stage. After selecting proper taps, an L-stage LFSR produces a repetitive PRBS of length 2L-1. The clock frequency of LFSR determines the "bit rate" of PRBS. Press Menu Advanced "PRBS" to open the PRBS parameter setting interface. 1. Bit Rate: sets the output bit rate of the PRBS waveform. 2. Ampl/HighL: sets the output amplitude/high level of the PRBS waveform. 3. Offset/LowL: sets the output offset/low level of the PRBS waveform. 4. Data: selects the data type of the PRBS waveform to PRBS7, PRBS9, PRBS11, resulting in sequences of 127, 511, and 2047 bits in length. By default, it is PRBS7. Please refer to "To Output Basic Waveform" to configure the parameters and output for the channel. 1 2 3 4 5 XOR PRBS Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-31 RS232 In RS232 serial protocol, a character is transmitted as a frame of data. The frame consists of 1 start bit, 5-8 data bits, 0-1 check bit, and 1-2 stop bits. Its format diagram is as shown in the figure below. RS232 uses "Negative Logic", i.g. high level is logic "0" and low level is logic "1". Press Menu Advanced "RS232" to open the RS232 waveform parameter setting interface. 1. Baud Rate: sets the output baud rate of RS232 waveform. 2. Ampl/HighL: sets the output amplitude/high level of RS232 waveform. 3. Offset/LowL: sets the output offset/low level of RS232 waveform. 4. Data Bits: sets the data bits of RS232 waveform to 5, 6, 7, or 8. 5. Stop Bits: sets the stop bits of RS232 waveform to "1", "1.5", or "2". 6. Parity Bit: sets the check bit of RS232 waveform to "Odd", "Even", or "None". 7. Data: sets the data of RS232 waveform. The available value is from 0 to 255. Please refer to "To Output Basic Waveform" to configure the parameters and output for the channel. RIGOL Chapter 2 Front Panel Operations 2-32 DG900 User's Guide Sequence You can self-define the sequence, and the edited waveforms can be stored in the internal or external memory of the instrument in ".SEQ" format. Press Menu Advanced "Sequence" to open the Sequence parameter setting interface. Tap the icon at the upper-right corner of the interface to enter the Sequence editing interface, as shown in the figure below. Figure 2-7 Sequence Editing Interface No.: indicates the serial number represented by "n". The available range of "n" is from 1 to 8. Wave: selects the waveform that corresponds to the current No. Period: sets the number of cycles of the currently selected waveform in "n". Tap "Ok" to confirm the current editing and go back to the Sequence parameter setting interface. You can preview the currently set waveforms in the right section of the interface and set the waveform parameters at the left section of the interface. Tap "Store" to save the currently edited Sequence waveform to the internal non-volatile memory (Disk C) or the external memory (Disk D, when a USB storage device is inserted). 1. Filter: selects the filter type of Sequence. Smooth: The frequency response is relatively wide and flat. The edge time is fast but the sharp transitions will have overshoot and ringing. Step: The sharp transitions are relatively ideal, with a narrower frequency bandwidth and the edge time is relatively slow. Interpolation: An abrupt change may occur among the output points-to-points. The edge time can be adjustable. 2. Samp. Rate: sets the output sample rate of Sequence. 3. Ampl/HighL: sets the output amplitude/high level of Sequence. 4. Offset/LowL: sets the output offset/low level of Sequence. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-33 5. Phase: sets the output start phase of Sequence. 6. Edge Time: sets the edge time of Sequence. Please refer to "To Output Basic Waveform" to configure the parameters and output for the channel. RIGOL Chapter 2 Front Panel Operations 2-34 DG900 User's Guide Modulation DG900 can output modulated waveforms from a single channel or from dual channels at the same time. Modulation is the process of modifying certain parameters (e.g. amplitude, frequency, or phase) of the carrier waveform signal according to the changes of the modulating signal. The carrier waveform can be Sine, Square, Ramp, Arbitrary waveform, or Pulse (only in PWM). The modulating waveform can be from internal or external modulation source. The modulation types supported by DG900 include AM, FM, PM, ASK, FSK, PSK, and PWM. Amplitude Modulation (AM) For amplitude modulation (AM), the amplitude of the carrier waveform varies with the instantaneous voltage of the modulating waveform. To Select AM Modulation Press Menu Modulation "AM" to enable the AM function. When Modulation is enabled, Sweep or Burst will be automatically disabled (if enabled currently). Tap the icon at the upper-right corner of the interface (when Sine is selected as the carrier waveform) to view and set the parameters of the current carrier waveform. Tap the icon at the upper-right corner of the interface again to go back to the current modulation parameter interface. To Select the Carrier Waveform AM carrier waveform could be Sine, Square, Ramp, or Arbitrary waveform. The default is Sine. Press Menu Continuous to select the desired carrier waveform. Pulse, Noise, Harmonic, Dual-tone, and DC could not be used as carrier waveform. To Set Carrier Waveform Parameters The different settings of various parameters (e.g. frequency, amplitude, offset, and start phase) of the carrier waveform will affect the output AM modulated waveform. For different carrier waveform shapes, the ranges of the various parameters are different (the ranges are related to the instrument model that you are using and the currently selected carrier waveform). For all carrier waveform shapes, the default values are 1 kHz frequency, 5 Vpp amplitude, 0 Vdc offset, and 0° start phase. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-35 When the carrier waveform currently selected is Sine, Square, or Ramp, please refer to descriptions in "To Output Basic Waveform" to set the parameters of the carrier waveform currently selected. When the carrier waveform currently selected is arbitrary waveform, please refer to descriptions in "To Output the Arbitrary Waveform" to set the parameters of the carrier waveform currently selected. To Select Modulation Source DG900 can receive modulating waveform from the internal or external modulation source. Tap the Source parameter selection field to select an "Internal" or "External" modulation source. 1. Internal Source When the internal modulation source is selected, tap the Mod.Wave parameter selection field to select Sine, Square, Triangle, UpRamp, DnRamp, Noise, or Arb as modulating waveform. By default, Sine is selected. Square: 50% duty cycle. Triangle: 50% symmetry. UpRamp: 100% symmetry. DnRamp: 0% symmetry. Arb: the arbitrary waveform selected for the current channel. Note: Noise can be used as the modulating waveform but cannot be used as carrier waveform. 2. External Source When the external modulation source is selected, Mod.Wave and Mod.Freq will be grayed out and disabled. The generator receives the external modulating signal from the rear-panel [CH1/Sync/Ext Mod/Trig/FSK] connector. At this time, amplitude of the modulated waveform is controlled by the ±5 V signal level of the connector. For example, if the modulation depth is set to 100%, the output amplitude will be the maximum when the modulating signal is +5 V and the minimum when the modulating signal is -5 V. RIGOL Chapter 2 Front Panel Operations 2-36 DG900 User's Guide Key Points: How to realize the intermodulation between dual channels? The following example takes the output signal of CH2 as the modulating waveform. 1. Connect the CH2 output terminal to the rear-panel [CH1/Sync/Ext Mod/Trig/FSK] connector by using the dual BNC cable. 2. Select CH1, then press Menu Modulation to select the desired modulation type and set the corresponding parameters. After that, select the external modulation source. 3. Select CH2 and set the desired modulating waveform and the corresponding parameters. 4. Press Output1 to enable the output of CH1. To Set Modulating Waveform Frequency When the internal modulation source is selected, tap the Mod.Freq parameter input field, use the numeric keypad to input the modulating waveform frequency. You can also use the arrow keys and the knob to input a desired frequency value. The modulating waveform frequency ranges from 2 mHz to 1 MHz, and the default value is 100 Hz. Note: When the external modulation source is selected, this menu will be grayed out and disabled. Set the Modulation Depth Modulation depth indicates the amplitude variation. It is expressed in percentage. The AM modulation depth ranges from 0% to 120%. Tap the Mod.Depth parameter input field to set the AM modulation depth. At 0% depth, the output amplitude is half of the carrier waveform amplitude. At 100% depth, the output amplitude is equal to carrier waveform amplitude. At >100% depth, the output amplitude of the instrument will not exceed 10 Vpp (into a 50 Ω load). When the external modulation source is selected, the output amplitude of the instrument is controlled by the ±5 V signal level of the rear-panel [CH1/Sync/Ext Mod/Trig/FSK] connector. For example, if the modulation depth is set to 100%, the output amplitude will be the maximum when the modulating signal is +5 V and the minimum when the modulating signal is -5 V. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-37 Carrier Waveform Suppression DG900 supports two types of amplitude modulation: normal amplitude modulation and double sideband suppressed carrier (DSSC) amplitude modulation. In the normal amplitude modulation, the modulated waveform contains carrier waveform components. Because carrier waveform components carry no information, the modulation is less efficient. In order to improve the modulation efficiency, the carrier waveform components are suppressed on the basis of the normal amplitude modulation. At this time, the modulated waveforms all carry information. This method is called double sideband suppressed carrier modulation. By default, DG900 selects the normal amplitude modulation. Tap the DSSC parameter selection field to select "On" to enable the double sideband suppressed carrier modulation. RIGOL Chapter 2 Front Panel Operations 2-38 DG900 User's Guide Frequency Modulation (FM) For frequency modulation (FM), the frequency of the carrier waveform varies with the instantaneous voltage of the modulating waveform. To Select FM Modulation Press Menu Modulation "FM" to enable the FM function. When Modulation is enabled, Sweep or Burst will be automatically disabled (if enabled currently). Tap the icon at the upper-right corner of the interface (when Sine is selected as the carrier waveform) to view and set the parameters of the current carrier waveform. Tap the icon at the upper-right corner of the interface again to go back to the current modulation parameter interface. To Select the Carrier Waveform FM carrier waveform could be Sine, Square, Ramp, or Arbitrary waveform. The default is Sine. Press Menu Continuous to select the desired carrier waveform. Pulse, Noise, Harmonic, Dual-tone, and DC could not be used as carrier waveform. To Set Carrier Waveform Parameters The different settings of various parameters (e.g. frequency, amplitude, offset, and start phase) of the carrier waveform will affect the output FM modulated waveform. For different carrier waveform shapes, the ranges of the various parameters are different (the ranges are related to the instrument model that you are using and the currently selected carrier waveform). For all carrier waveform shapes, the default values are 1 kHz frequency, 5 Vpp amplitude, 0 Vdc offset, and 0° start phase. When the carrier waveform currently selected is Sine, Square, or Ramp, please refer to descriptions in "To Output Basic Waveform" to set the parameters of the carrier waveform currently selected. When the carrier waveform currently selected is arbitrary waveform, please refer to descriptions in "To Output the Arbitrary Waveform" to set the parameters of the carrier waveform currently selected. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-39 To Select Modulation Source DG900 can receive modulating waveform from the internal or external modulation source. Tap the Source parameter selection field to select an "Internal" or "External" modulation source. 1. Internal Source When internal modulation source is selected, tap the Mod.Wave parameter selection field to select Sine, Square, Triangle, UpRamp, DnRamp, Noise, or Arb as modulating waveform. By default, Sine is selected. Square: 50% duty cycle. Triangle: 50% symmetry. UpRamp: 100% symmetry. DnRamp: 0% symmetry. Arb: the arbitrary waveform selected for the current channel. Note: Noise can be used as the modulating waveform but cannot be used as carrier waveform. 2. External Source When the external modulation source is selected, Mod.Wave and Mod.Freq will be grayed out and disabled. The generator receives the external modulating signal from the rear-panel [CH1/Sync/Ext Mod/Trig/FSK] connector. At this time, the frequency deviation of the modulated waveform is controlled by the ±5 V signal level of the connector. For example, if the frequency deviation is set to 1 kHz, +5 V signal level corresponds to a 1 kHz increase in frequency and -5 V signal level corresponds to a 1 kHz decrease in frequency. To Set Modulating Waveform Frequency When the internal modulation source is selected, tap the Mod.Freq parameter input field and use the numeric keypad to set the modulating waveform frequency. You can also use the arrow keys and the knob to input a desired frequency value. The modulating waveform frequency ranges from 2 mHz to 1 MHz, and the default value is 100 Hz. Note: When the external modulation source is selected, this menu will be grayed out and disabled. To Set the Frequency Deviation Frequency deviation indicates the variation in the modulating waveform frequency from the carrier frequency. Tap the Freq.Dev parameter input field to set the FM frequency deviation. RIGOL Chapter 2 Front Panel Operations 2-40 DG900 User's Guide The frequency deviation must be smaller than or equal to the carrier frequency. The sum of frequency deviation and carrier frequency must be smaller than or equal to the the upper limit of the current carrier frequency plus 1 kHz. Note: If Sine is selected currently as the carrier waveform, the carrier amplitude will be limited at 2 Vpp when the sum of the frequency deviation and the carrier frequency is greater than the frequency upper limit of the current carrier. When the external modulation source is selected, the frequency deviation is controlled by the ±5 V signal level on the rear-panel [CH1/Sync/Ext Mod/Trig/FSK] connector. Positive signal level corresponds to frequency increase, and negative signal level corresponds to frequency decrease. Lower signal levels produce less deviation. For example, if the frequency deviation is set to 1 kHz, +5 V signal level corresponds to a 1 kHz increase in frequency and -5 V signal level corresponds to a 1 kHz decrease in frequency. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-41 Phase Modulation (PM) For phase modulation (PM), the phase of the carrier waveform varies with the instantaneous voltage of the modulating waveform. To Select PM Modulation Press Menu Modulation "PM" to enable the PM function. When Modulation is enabled, Sweep or Burst will be automatically disabled (if enabled currently). Tap the icon at the upper-right corner of the interface (when Sine is selected as the carrier waveform) to view and set the parameters of the current carrier waveform. Tap the icon at the upper-right corner of the interface again to go back to the current modulation parameter interface. To Select the Carrier Waveform PM carrier waveform could be Sine, Square, Ramp, or Arbitrary waveform. The default is Sine. Press Menu Continuous to select the desired carrier waveform. Pulse, Noise, Harmonic, Dual-tone, and DC could not be used as carrier waveform. To Set Carrier Waveform Parameters The different settings of various parameters (e.g. frequency, amplitude, and offset) of the carrier waveform will affect the output PM modulated waveform. For different carrier waveform shapes, the ranges of the various parameters are different (the ranges are related to the instrument model that you are using and the currently selected carrier waveform). For all carrier waveform shapes, the default values are 1 kHz frequency, 5 Vpp amplitude, and 0 Vdc offset. When the carrier waveform currently selected is Sine, Square, or Ramp, please refer to descriptions in "To Output Basic Waveform" to set the parameters of the carrier waveform currently selected. When the carrier waveform currently selected is arbitrary waveform, please refer to descriptions in "To Output the Arbitrary Waveform" to set the parameters of the carrier waveform currently selected. Note: After the PM function is enabled, the start phase of the carrier waveform cannot be set. RIGOL Chapter 2 Front Panel Operations 2-42 DG900 User's Guide To Select Modulation Source DG900 can receive modulating waveform from the internal or external modulation source. Tap the Source parameter selection field to select an "Internal" or "External" modulation source. 1. Internal Source When internal modulation source is selected, tap the Mod.Wave parameter selection field to select Sine, Square, Triangle, UpRamp, DnRamp, Noise, or Arb as modulating waveform. By default, Sine is selected. Square: 50% duty cycle. Triangle: 50% symmetry. UpRamp: 100% symmetry. DnRamp: 0% symmetry. Arb: the arbitrary waveform selected for the current channel. Note: Noise can be used as the modulating waveform but cannot be used as carrier waveform. 2. External Source When the external modulation source is selected, Mod.Wave and Mod.Freq will be grayed out and disabled. The generator receives the external modulating signal from the rear-panel [CH1/Sync/Ext Mod/Trig/FSK] connector. At this time, the phase deviation of the modulated waveform is controlled by the ±5 V signal level of the connector. For example, if the phase deviation is set to 180°, +5 V signal level corresponds to a 180º phase variation. The lower external signal levels produce less deviation. To Set Modulating Waveform Frequency When the internal modulation source is selected, tap the Mod.Freq parameter input field and use the numeric keypad to set the modulating waveform frequency. You can also use the arrow keys and the knob to input a desired frequency value. The modulating waveform frequency ranges from 2 mHz to 1 MHz, and the default value is 100 Hz. Note: When the external modulation source is selected, this menu will be grayed out and disabled. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-43 To Set Phase Deviation Phase deviation indicates the deviation of the modulating waveform phase from the carrier waveform phase. Tap the Phas.Dev parameter input field and use the numeric keypad to set the PM phase deviation. You can also use the arrow keys and the knob to input a desired phase value. The setting range of the phase deviation is from 0° to 360°, and the default is 90°. When the external modulation source is selected, the phase deviation is controlled by the ±5 V signal level on the rear-panel [CH1/Sync/Ext Mod/Trig/FSK] connector. For example, if the phase deviation is set to 180°, +5 V signal level corresponds to a 180º phase variation. The lower external signal levels produce less deviation. RIGOL Chapter 2 Front Panel Operations 2-44 DG900 User's Guide Amplitude Shift Keying (ASK) When using ASK (Amplitude Shift Keying), you can configure the generator to "shift" its output amplitude between two preset amplitude values ("carrier amplitude" and "modulating amplitude"). To Select ASK Modulation Press Menu Modulation "ASK" to enable the ASK function. When Modulation is enabled, Sweep or Burst will be automatically disabled (if enabled currently). Tap the icon at the upper-right corner of the interface (when Sine is selected as the carrier waveform) to view and set the parameters of the current carrier waveform. Tap the icon at the upper-right corner of the interface again to go back to the current modulation parameter interface. To Select the Carrier Waveform ASK carrier waveform could be Sine, Square, Ramp, or Arbitrary waveform. The default is Sine. Press Menu Continuous to select the desired carrier waveform. Pulse, Noise, Harmonic, Dual-tone, and DC could not be used as carrier waveform. To Set Carrier Waveform Parameters The different settings of various parameters (e.g. frequency, amplitude, offset, and start phase) of the carrier waveform will affect the output ASK modulated waveform. For different carrier waveform shapes, the ranges of the various parameters are different (the ranges are related to the instrument model that you are using and the currently selected carrier waveform). For all carrier waveform shapes, the default values are 1 kHz frequency, 5 Vpp amplitude, 0 Vdc offset, and 0° start phase. When the carrier waveform currently selected is Sine, Square, or Ramp, please refer to descriptions in "To Output Basic Waveform" to set the parameters of the carrier waveform currently selected. When the carrier waveform currently selected is arbitrary waveform, please refer to descriptions in "To Output the Arbitrary Waveform" to set the parameters of the carrier waveform currently selected. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-45 To Select Modulation Source Tap the Source parameter selection field to select an "Internal" or "External" modulation source. 1. Internal Source When the internal source is selected, the modulating waveform is set as a Square with 50% duty cycle. At this time, the rate at which the output amplitude "shifts" between "carrier amplitude" and "modulating amplitude" is determined by the modulation rate. 2. External Source When the external source is selected, the generator receives the external modulating signal from the rear-panel [CH1/Sync/Ext Mod/Trig/FSK] connector. Note: The [CH1/Sync/Ext Mod/Trig/FSK] connector controlling ASK modulation externally is different from controlling AM/FM/PM modulations externally. While controlling the ASK modulation, you can set the polarity. To Set Modulating Rate When the internal source is selected, tap the Mod.Rate parameter input field and use the numeric keypad to set the rate at which the output amplitude shifts between "carrier amplitude" and "modulating amplitude". You can also use the arrow keys and the knob to input a desired frequency value. The frequency range is from 2 mHz to 1 MHz, and the default is 100 Hz. Note: When the external modulation source is selected, this menu will be grayed out and disabled. To Set the Modulation Amplitude Tap the Mod.Ampl parameter input field and use the numeric keypad to set the modulation amplitude. You can also use the arrow keys and the knob to input a desired amplitude value. The range of amplitude (HighZ) is from 0 to 10 Vpp, and the default is 2 Vpp. RIGOL Chapter 2 Front Panel Operations 2-46 DG900 User's Guide To Set the Modulation Polarity Tap the Polarity parameter selection field to select "Positive" or "Negative" polarity of the modulating waveform to control the output amplitude. In the internal modulation, set the polarity to "Positive", and the generator will output an amplitude whichever is smaller between the carrier amplitude and modulating amplitude when the modulating waveform is a logic low level. It will output an amplitude whichever is greater between the carrier amplitude and modulating amplitude when the modulating waveform is a logic high level. The situation is the opposite when the polarity is set to "Negative". In the external modulation, set the polarity to "Positive", and the generator will output an amplitude whichever is smaller between the carrier amplitude and modulating amplitude when the external input signal is a logic low level. It will output an amplitude whichever is greater between the carrier amplitude and the modulating amplitude when the external input signal is a logic high level. The situation is the opposite when the polarity is set to "Negative". Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-47 Frequency Shift Keying (FSK) When using FSK (Frequency Shift Keying), you can configure the generator to "shift" its output frequency between two preset frequency values ("carrier amplitude" and "hop frequency"). To Select FSK Modulation Press Menu Modulation "FSK" to enable the FSK function. When Modulation is enabled, Sweep or Burst will be automatically disabled (if enabled currently). Tap the icon at the upper-right corner of the interface (when Sine is selected as the carrier waveform) to view and set the parameters of the current carrier waveform. Tap the icon at the upper-right corner of the interface again to go back to the current modulation parameter interface. To Select the Carrier Waveform FSK carrier waveform could be Sine, Square, Ramp, or Arbitrary waveform. The default is Sine. Press Menu Continuous to select the desired carrier waveform. Pulse, Noise, Harmonic, Dual-tone, and DC could not be used as carrier waveform. To Set Carrier Waveform Parameters The different settings of various parameters (e.g. frequency, amplitude, offset, and start phase) of the carrier waveform will affect the output FSK modulated waveform. For different carrier waveform shapes, the ranges of the various parameters are different (the ranges are related to the instrument model that you are using and the currently selected carrier waveform). For all carrier waveform shapes, the default values are 1 kHz frequency, 5 Vpp amplitude, 0 VdC offset, and 0° start phase. When the carrier waveform currently selected is Sine, Square, or Ramp, please refer to descriptions in "To Output Basic Waveform" to set the parameters of the carrier waveform currently selected. When the carrier waveform currently selected is arbitrary waveform, please refer to descriptions in "To Output the Arbitrary Waveform" to set the parameters of the carrier waveform currently selected. RIGOL Chapter 2 Front Panel Operations 2-48 DG900 User's Guide To Select Modulation Source Tap the Source parameter selection field to select an "Internal" or "External" modulation source. 1. Internal Source When the internal source is selected, the modulating waveform is set as a Square with 50% duty cycle. At this time, the rate at which the output frequency "shifts" between "carrier frequency" and "hop frequency" is determined by the modulation rate. 2. External Source When the external source is selected, the generator receives the external modulating signal from the rear-panel [CH1/Sync/Ext Mod/Trig/FSK] connector. Note: The [CH1/Sync/Ext Mod/Trig/FSK] connector controlling ASK modulation externally is different from controlling AM/FM/PM modulations externally. While controlling the FSK modulation, you can set the polarity. To Set Modulating Rate When the internal source is selected, tap the Mod.Rate parameter input field and use the numeric keypad to set the rate at which the output frequency shifts between "carrier frequency" and "hop frequency". You can also use the arrow keys and the knob to input a desired frequency value. The frequency range is from 2 mHz to 1 MHz, and the default is 100 Hz. Note: When the external modulation source is selected, this menu will be grayed out and disabled. To Set Hop Frequency Hop frequency indicates the frequency of the modulating waveform. The range of the hop frequency is determined by the currently selected carrier waveform. Tap the Hop Freq parameter input field and use the numeric keypad to input a desired frequency value. You can also use the arrow keys and the knob to input a desired frequency value. Sine: 1 μHz to 100 MHz Square: 1 μHz to 25 MHz Ramp: 1 μHz to 2 MHz Arb: 1 μHz to 25 MHz Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-49 To Set the Modulation Polarity Tap the Polarity parameter selection field to select "Positive" or "Negative" polarity of the modulating waveform to control the output frequency. In the internal modulation, set the polarity to "Positive", and the generator will output the carrier frequency when the modulating waveform amplitude is a logic low level. It will output the hop frequency when the modulating waveform amplitude is a logic high level. The situation is the opposite when the polarity is set to "Negative". In the external modulation, set the polarity to "Positive", and the generator will output the carrier frequency when the external input signal is a logic low level. It will output the hop frequency when the external input signal is a logic high level. The situation is the opposite when the polarity is set to "Negative". RIGOL Chapter 2 Front Panel Operations 2-50 DG900 User's Guide Phase Shift Keying (PSK) When using PSK (Phase Shift Keying), you can configure the generator to "shift" its output phase between two preset phase values ("carrier phase" and "modulating phase"). To Select PSK Modulation Press Menu Modulation "PSK" to enable the PSK function. When Modulation is enabled, Sweep or Burst will be automatically disabled (if enabled currently). Tap the icon at the upper-right corner of the interface (when Sine is selected as the carrier waveform) to view and set the parameters of the current carrier waveform. Tap the icon at the upper-right corner of the interface again to go back to the current modulation parameter interface. To Select the Carrier Waveform PSK carrier waveform could be Sine, Square, Ramp, or Arbitrary waveform. The default is Sine. Press Menu Continuous to select the desired carrier waveform. Pulse, Noise, Harmonic, Dual-tone, and DC could not be used as carrier waveform. To Set Carrier Waveform Parameters The different settings of various parameters (e.g. frequency, amplitude, offset, and start phase) of the carrier waveform will affect the output PSK modulated waveform. For different carrier waveform shapes, the ranges of the various parameters are different (the ranges are related to the instrument model that you are using and the currently selected carrier waveform). For all carrier waveform shapes, the default values are 1 kHz frequency, 5 Vpp amplitude, 0 VdC offset, and 0° start phase. When the carrier waveform currently selected is Sine, Square, or Ramp, please refer to descriptions in "To Output Basic Waveform" to set the parameters of the carrier waveform currently selected. When the carrier waveform currently selected is arbitrary waveform, please refer to descriptions in "To Output the Arbitrary Waveform" to set the parameters of the carrier waveform currently selected. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-51 To Select Modulation Source Tap the Source parameter selection field to select an "Internal" or "External" modulation source. 1. Internal Source When the internal source is selected, the modulating waveform is set as a Square with 50% duty cycle. At this time, the rate at which the output phase "shifts" between "carrier phase" and "modulating phase" is determined by the modulation rate. 2. External Source When the external source is selected, the generator receives the external modulating signal from the rear-panel [CH1/Sync/Ext Mod/Trig/FSK] connector. Note: The [CH1/Sync/Ext Mod/Trig/FSK] connector controlling PSK modulation externally is different from controlling AM/FM/PM modulations externally. While controlling the PSK modulation, you can set the polarity. To Set Modulating Rate When the internal source is selected, tap the Mod.Rate parameter input field and use the numeric keypad to set the rate at which the output phase shifts between "carrier phase" and "modulating phase". You can also use the arrow keys and the knob to input a desired frequency value. The frequency range is from 2 mHz to 1 MHz, and the default is 100 Hz. Note: When the external modulation source is selected, this menu will be grayed out and disabled. To Set Modulating Phase Modulating phase indicates the phase of the modulating waveform. Tap the Phase parameter input field and use the numeric keypad to set the modulation phase. You can also use the arrow keys and the knob to input a desired phase value. The phase range is from 0° to 360°, and the default is 180°. RIGOL Chapter 2 Front Panel Operations 2-52 DG900 User's Guide To Set Modulation Polarity Tap the Polarity parameter selection field to select "Positive" or "Negative" polarity of the modulating waveform to control the output phase. In the internal modulation, set the polarity to "Positive", and the generator will output the carrier phase when the modulating waveform amplitude is a logic low level. It will output the modulating phase when the modulating waveform amplitude is a logic high level. The situation is the opposite when the polarity is set to "Negative". In the external modulation, set the polarity to "Positive", and the generator will output the carrier phase when the external input signal is a logic low level. It will output the modulation phase when the external input signal is a logic high level. The situation is the opposite when the polarity is set to "Negative". Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-53 Pulse Width Modulation (PWM) For PWM (Pulse Width Modulation), the pulse width of the carrier waveform varies with the instantaneous voltage of the modulating waveform. To Select PWM Modulation PWM can only be used to modulate pulse. To select PWM modulation, first select Pulse, then press Menu Modulation "PWM" to enable PWM function. Tap the icon at the upper-right corner of the interface to view and set the parameters of the current carrier waveform. Tap the icon at the upper-right corner of the interface again to go back to the current modulation parameter interface. If the "Pulse" waveform is not selected, PWM modulation is unavailable. When Modulation is enabled but the modulation type is not PWM, pressing "Pulse" will select PWM automatically. When Modulation is enabled, Sweep or Burst will be automatically disabled (if enabled currently). To Select the Carrier Waveform As mentioned before, the carrier waveform of PWM can only be Pulse. To select Pulse waveform, press Menu Continuous "Pulse". To Set Carrier Waveform Parameters The different settings of various parameters (e.g. frequency, amplitude, offset, pulse width, and duty cycle) of the Pulse waveform will affect the output PWM modulated waveform. For all the pulses, the ranges of the various parameters are different. The default values are 1 kHz frequency, 5 Vpp amplitude, 0 Vdc offset, 500 µs pulse width, and 50% duty cycle. Please refer to descriptions in "To Output Basic Waveform" to set the parameters of the carrier waveform. RIGOL Chapter 2 Front Panel Operations 2-54 DG900 User's Guide To Select Modulation Source Tap the Source parameter selection field to select an "Internal" or "External" modulation source. 1. Internal Source When internal modulation source is selected, tap the Mod.Wave parameter selection field to select Sine, Square, Triangle, UpRamp, DnRamp, Noise, or Arb as modulating waveform. By default, Sine is selected. Square: 50% duty cycle. Triangle: 50% symmetry. UpRamp: 100% symmetry. DnRamp: 0% symmetry. Arb: the arbitrary waveform selected for the current channel. Note: Noise can be used as the modulating waveform but cannot be used as carrier waveform. 2. External Source When the external modulation source is selected, Mod.Wave and Mod.Freq will be grayed out and disabled. The generator receives the external modulating signal from the rear-panel [CH1/Sync/Ext Mod/Trig/FSK] connector. At this time, the width deviation or the duty cycle deviation of the modulated waveform is controlled by the ±5 V signal level on the connector. For example, if the width deviation is set to 10 s, the +5 V signal level corresponds to a 10 s width variation. To Set Modulating Waveform Frequency When the internal modulation source is selected, tap the Mod.Freq parameter input field and use the numeric keypad to set the modulating waveform frequency. You can also use the arrow keys and the knob to input a desired frequency value. The modulating waveform frequency ranges from 2 mHz to 1 MHz, and the default value is 100 Hz. Note: When the external modulation source is selected, this menu will be grayed out and disabled. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-55 To Set Width/Duty Deviation If "Width" is currently selected in the pulse of the specified channel, "Width Dev" is displayed in the interface when the PWM modulation function is enabled; if the "Duty" is currently selected in the pulse of the specified channel, "Duty Dev" is displayed in the interface when the PWM modulation function is enabled. Tap the Width Dev (or Duty Dev) parameter input field and use the numeric keypad to input a desired value. You can also use the arrow keys and the knob to input a desired value. Width deviation represents the variation of the modulated waveform pulse width relative to the original pulse width (the available units are ns, μs, ms, s, and ks). Width deviation cannot exceed the current pulse width. The width deviation is limited by the minimum pulse width and current edge time setting. Duty deviation represents the variation of the modulated waveform pulse duty cycle relative to the original duty cycle (expressed in %). Duty deviation cannot exceed the current duty cycle of the pulse. The duty deviation is limited by the minimum duty cycle and current edge time setting. When the external modulation source is selected, the width deviation is controlled by the ±5 V signal level on the rear-panel [CH1/Sync/Ext Mod/Trig/FSK] connector. For example, if the width deviation is set to 10 s, the +5 V signal level corresponds to a 10 s width variation. RIGOL Chapter 2 Front Panel Operations 2-56 DG900 User's Guide Sweep DG900 can output sweep waveforms from a single channel or from dual channels at the same time. In sweep mode, the generator outputs a signal variably from the start frequency to stop frequency within the specified sweep time. DG900 supports linear, log, and step sweep modes; allows users to set start hold, stop hold, and return time; supports internal, external, or manual trigger source; and capable of generating sweep output for Sine, Square, Ramp, and Arbitrary waveform. To Enable the Sweep Function Press Menu Sweep to select "Linear", "Log", or "Step" to enable the sweep function. The Modulation or Burst function will be disabled automatically (if enabled currently). At this time, the generator will generate the sweep waveform from the corresponding channel (if enabled currently) according to the current configuration. You can also reset the sweep parameters. For details, refer to the following section. Sweep Type DG900 provides Linear, Log, and Step sweep types. Linear Sweep In the linear sweep type, the output frequency of the instrument varies linearly in the way of "several Hertz per second". The variation is controlled by "Start Frequency", "Stop Frequency" and "Sweep Time". Press Menu Sweep "Linear" to enable the linear sweep. A straight line is displayed on the waveform on the screen, indicating that the output frequency varies linearly. Figure 2-8 Linear Sweep Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-57 Log Sweep In Log Sweep type, the output frequency of the instrument varies in a logarithmic fashion, that is, the output frequency changes in the way of "octave per second" or "decade per second". The variation is controlled by "Start Frequency", "Stop Frequency", and "Sweep Time". When Log Sweep is enabled, you can set the following parameter: start frequency start F , stop frequency stop F , and sweep time sweep T . The function prototype of Log Sweep is: T current F P = current F is the instantaneous frequency of the current output. The parameter P and T can be expressed with the above parameter. sweep start stop T F F P / ) / lg( 10 = ) lg( / ) lg( P F t T start + = Wherein, t is the time from the start of the sweep, and its range is from 0 to sweep T . Press Menu Sweep "Log" to enable the log sweep. An exponential function curve is displayed on the screen, indicating that the output frequency changes in a logarithmic mode. Figure 2-9 Log Sweep Step Sweep In Step Sweep type, the output frequency of the instrument "steps" from "Start Frequency" to "Stop Frequency" with a stepwise increase. The duration of the output signal on each frequency point is determined by "Sweep Time" and "Steps". Press Menu Sweep "Step" to enable the step sweep. A step waveform is displayed on the screen, indicating that the output frequency varies with a stepwise RIGOL Chapter 2 Front Panel Operations 2-58 DG900 User's Guide increase. At this point, tap to slide the screen or use the knob to page up or down. Tap the Step parameter input field and use the numeric keypad to set the step. You can also use the arrow keys and the knob to set the step. By default, it is 2, and its settable range is from 2 to 1024. Note: In the "Linear" and "Log" sweep modes, the Step menu label is grayed out and disabled. Figure 2-10 Step Sweep Sweep Time When the sweep function is enabled, tap the Sweep Time parameter input field and use the numeric keypad to modify the sweep time. You can also use the arrow keys and the knob to modify the sweep time. By default, it is 1 s, and its settable range is from 1 ms to 500 s. The generator will restart to sweep and output from the specified "Start Frequency" after the sweep time is modified. Return Time Return time indicates the time that the output signal restores from "End Freq" to "Start Freq" after the signal generator sweeps from "Start Freq" to "End Freq" and the "StopHold" time expires. When the sweep function is enabled, tap the Return Time parameter input field and use the numeric keypad to modify the return time. You can also use the arrow keys and the knob to modify the return time. By default, it is 0 s, and its settable range is from 0 s to 500 s. The generator will restart to sweep and output from the specified "Start Frequency" once the return time is modified. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-59 Start Frequency and Stop Frequency Start frequency and stop frequency are the upper and lower limits of the frequency for frequency sweep. The generator always sweeps from the start frequency to the stop frequency, and then returns back to the start frequency. Start Frequency < Stop Frequency: the generator sweeps from low frequency to high frequency. Start Frequency > Stop Frequency: the generator sweeps from high frequency to low frequency. Start Frequency = Stop Frequency: the generator outputs with a fixed frequency. After the sweep function is enabled, tap the Start/Center menu label to enable the display of "Start". Notice that the "Stop" in Stop/Span menu label will also be highlighted at the same time. Input the desired frequency value by using the numeric keypad, the arrow keys, or the knob. By default, the start frequency is 100 Hz, and the stop frequency is 1 kHz. Different sweep waveforms correspond to different start frequency and stop frequency ranges. After modifying the "Start Freq" or "Stop Freq", the signal generator will restart to output from the specified "Start Freq". Center Frequency and Frequency Span You can also set the frequency boundaries of frequency sweep through center frequency and frequency span. Center Frequency = (|Start Frequency + Stop Frequency|)/2 Frequency Span = Stop Frequency – Start Frequency After the sweep function is enabled, tap the Start/Center menu label to enable the display of "Center". At this time, the "Span" in Stop/Span menu label will also be highlighted at the same time. Input the desired frequency value by using the numeric keypad, the arrow keys, or the knob. By default, the center frequency is 550 Hz, and the frequency span is 900 Hz. Different sweep waveforms correspond to different center frequency and frequency span ranges. Besides, the center frequency and frequency span are mutually affected. Define the minimum frequency of the currently selected waveform min F , the maximum frequency as max F and 2 / ) ( max min F F Fm + = . The available range of the center frequency is from min F to max F . The parameters of different waveforms are as follows: Sine: 1 μHz to 100 MHz Square: 1 μHz to 25 MHz RIGOL Chapter 2 Front Panel Operations 2-60 DG900 User's Guide Ramp: 1 μHz to 2 MHz Arb: 1 μHz to 25 MHz The range of the frequency span is affected by the center frequency. Center frequency < m F : the range of the frequency span is ±2× (center frequency – min F ); Center frequency ≥ m F : the range of the frequency span is ±2× ( max F - center frequency). Take sine as an example. min F is 1 μHz, max F is 100 MHz, and m F is 50 MHz. If the center frequency is 550 Hz, the frequency span range is ±2×(550 Hz–1 μHz) = ±1.099999998 kHz; if the center frequency is 60 MHz, the frequency span range is ±2×(100 MHz – 60 MHz) = ±80 MHz. The generator will restart to sweep from the specified "start frequency" after the "center frequency" or "frequency span" is modified. Tip: In large-scale sweep, the amplitude characteristic of the output signal might change. Sweep Trigger Source The trigger sources of the sweep can be Internal, External, or Manual. When the signal generator receives a trigger signal, a sweep output is generated, and then the signal generator waits for another trigger signal. After the sweep function is enabled, tap the Source to select "Internal", "External", or "Manual". The default is "Internal". 1. Internal Trigger The generator outputs continuous sweep waveforms when the internal trigger is selected. The trigger period is determined by the specified sweep time, return time, start hold, and stop hold time. 2. External Trigger In the External trigger, the signal generator receives the trigger signal input from the rear-panel [CH1/Sync/Ext Mod/Trig/FSK] connector. Once it receives a TTL pulse with a specified polarity, a sweep is initiated. To specify the polarity of the TTL pulse, tap the Trig In to select "RisEdge" or "FallEdge". By default, it is "RisEdge". Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-61 3. Manual Trigger In the Manual trigger, when you press Trig on the front panel for one time, one sweep is immediately launched for the corresponding channel. Key Points: When "External" trigger is selected, tap the Trig In to select "RisEdge" or "FallEdge". The rear-panel [CH1/Sync/Ext Mod/Trig/FSK] connector is used as the input terminal of the external trigger signal. When "RisEdge" is selected, the instrument triggers one sweep at the rising edge of the input signal. When "FallEdge" is selected, the instrument triggers one sweep at the falling edge of the input signal. Marker Freq The sync signal output from the rear-panel [CH1/Sync/Ext Mod/Trig/FSK] connector that corresponds to the channel on the front panel always changes from low level to high level at the start of each sweep. When the "Marker" function is disabled, the sync signal will change to low level at the center frequency point. When the "Marker" function is enabled, the sync signal changes to low level at the specified marker frequency point. After the sweep function is enabled, tap the Marker menu label to select "On". Then tap the Marker Freq menu label, and use the numeric keypad to modify the marker frequency. You can also use the arrow keys and the knob to modify the marker frequency. The default value is 550 Hz, and the available range is limited by "Start Frequency" and "Stop Frequency". The generator will restart to sweep and output from the specified "Start Frequency" after the mark frequency is modified. Key Points: For step sweep (the sweep points determined by the start frequency, stop frequency, and step respectively are f1, f2, ……, fn, fn+1……), if the marker frequency setting is one of the sweep points values, the sync signal is TTL high level at the start of the sweep and will change to low level at the marker frequency point. If the set marker frequency is not equal to the value at the sweep point, the sync signal will change to low level at the sweep point which is closest to this marker frequency. RIGOL Chapter 2 Front Panel Operations 2-62 DG900 User's Guide Start Hold Start hold is the duration that the output signal outputs with the "Start Frequency" after the sweep starts. After the start hold time expires, the generator will continue outputting with varied frequencies according to the current sweep type. When the sweep function is enabled, tap the StartHold menu label and use the numeric keypad to modify the start hold time. You can also use the arrow keys and the knob to modify the start hold time. By default, it is 0 s, and its settable range is from 0 s to 500 s. The generator will restart to sweep and output from the specified "Start Frequency" once the Start Hold time is changed. Stop Hold Stop Hold is the duration that the output signal continues outputting with the "Stop Frequency" after the generator has swept from the "Start Frequency" to the "Stop Frequency". When the sweep function is enabled, tap the StopHold menu label and use the numeric keypad to modify the stop hold time. You can also use the arrow keys and the knob to modify the stop hold time. By default, it is 0 s, and its settable range is from 0 s to 500 s. The generator will restart to sweep and output from the specified "Start Frequency" after the Stop Hold time is modified. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-63 Burst DG900 can output the waveforms with a specified number of cycles (Burst) from a single channel or from dual channels at the same time. DG900 supports control of burst output by internal, manual or external trigger source; supports three kinds of burst types, including N cycle, Infinite, and Gated. The generator can generate burst by using Sine, Square, Ramp, Pulse, Noise (only for gated burst), arbitrary waveform, PRBS, RS232, or Sequence. To Enable the Burst Function Press Menu Burst to select "NCycle", "Infinite", or "Gated" to enable the pulse function. The Modulation or Sweep function will be disabled automatically (if enabled currently). At this time, the generator will output burst waveform from the corresponding channel (if currently turned on) according to the current configuration. You can also reset the burst function menu. Please refer to the following section for details. Burst Type DG900 can output three types of bursts, including N cycle, Infinite and Gated. Table 2-3 Relations among Burst type, Trigger Source, and Carrier Waveforms Type Trigger Source Carrier Waveform N Cycle Internal/External/Manual Sine, Square, Ramp, Pulse, Arbitrary Waveform, PRBS, RS232, Sequence Infinite External/Manual Sine, Square, Ramp, Pulse, Arbitrary Waveform, PRBS, RS232, Sequence Gated External Sine, Square, Ramp, Pulse, Noise, Arbitrary Waveform, PRBS, RS232, Sequence N Cycle In N Cycle mode, the generator will output waveforms with a specified number of cycles after receiving the trigger signal. Press Menu Burst "NCycle" to enable the N Cycle burst. Tap the Cycles menu label, then use the numeric keypad to input the number of cycles. You can also use the arrow keys and the knob to modify the number of cycles. The default is 1, and the range is from 1 to 1,000,000 (external or manual trigger) or from 1 to 500,000 (internal trigger). RIGOL Chapter 2 Front Panel Operations 2-64 DG900 User's Guide Figure 2-11 N Cycle Burst For N cycle burst, "Internal", "External", or "Manual" trigger source could be selected. Besides, you can also set the parameters of "Period" (internal trigger), "Delay", "Trig In" (external trigger) and "Trig Out" (internal and manual trigger). Infinite Burst In Infinite mode, the cycle number of the waveform is set as an infinite value. The generator outputs a continuous waveform after receiving a trigger signal. Press Menu Burst "Infinite" to enable the Infinite burst. The instrument will set the trigger source to "Manual" automatically. A diagram of an infinite cycle burst will appear on the screen. Figure 2-12 Infinite Burst For the Infinite burst, "External" or "Manual" trigger source could be selected. Besides, users can set the parameters of "Delay", "Trig In" (external trigger), and "Trig Out" (manual trigger). Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-65 Gated Burst In Gated Burst Mode, the generator controls the waveform output according to the external signal level input from the rear-panel [CH1/Sync/Ext Mod/Trig/FSK] connector. Press Menu Burst "Gated" to enable the gated burst. Then tap the Polarity menu label to select "Positive" or "Negative". Only when the gated signal is Positive or Negative, can the waveforms be output. Figure 2-13 Gated Burst When the gated signal is "True", the generator outputs a continuous waveform. When the gated signal is "False", the generator first completes the output of the current period and then stops. For Noise waveform, the output will stop immediately once the gated signal becomes "False". Gated Burst could only be triggered by "External" trigger source. Burst Delay Burst delay is only available for N cycle and Infinite burst mode. It is defined as the duration from the time when the generator receives the trigger signal to the time when it starts to output the N Cycle (or Infinite) burst. Press Menu Burst to select "NCycle" or "Infinite", and then tap the Delay menu label and use the numeric keypad to input the desired delay time. You can also use the arrow keys and the knob to input a desired delay time. The range of the delay time is: 0 s ≤delay time ≤ 100 s. The default is 0 s. RIGOL Chapter 2 Front Panel Operations 2-66 DG900 User's Guide Burst Period Burst period is only available for N cycle burst in the internal trigger. It is defined as the time from the start of a burst to the time when the next burst starts. Burst Period ≥ 1 μs + waveform period × number of bursts. Wherein, the waveform period indicates the period of burst function (e.g. Sine and Square). If the burst period currently set is too small, the generator will increase this period automatically to allow the output of the specified number of cycles. Press Menu Burst "NCycle" to enable the N Cycle burst. Tap the Source menu label to select "Internal", and then tap the Period menu label to input the desired period by using the numeric keypad. You can also use the arrow keys and the knob to input a desired period. By default, it is 10 ms, and its available range is from 1 μs to 500 s. Idle Level In burst mode, the generator outputs the carrier waveform with a specified number of cycles and then outputs a level, as shown in the figure below. The level is defined as idle level. Figure 2-14 Definition of Idle Level When Burst is enabled, tap Idle Level to select your desired menu to set the idle level. 1. 1st Point Sets the level at the first point of the carrier waveform as the idle level. 2. Top Sets the level at the top point of the carrier waveform as the idle level. 3. Center Sets the level at the center point of the carrier waveform as the idle level. 4. Bottom Sets the level at the bottom point of the carrier waveform as the idle level. 5. User Sets the level at the specified point of the carrier waveform as the idle level. Idle Level Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-67 Tap the User menu label, and input the desired value by using the numeric keypad. You can also use the arrow keys and the knob to input a desired value. Its range is from 0 to 65535. Burst Trigger Source Burst trigger source can be internal, external, or manual. The generator will generate a burst output when a trigger signal is received, and then it waits for the next trigger signal. After the burst function is enabled, tap the Source to select "Internal", "External", or "Manual". The default is "Internal". 1. Internal Trigger When internal trigger is selected, the generator can only output N cycle burst, and the output burst frequency is determined by the "Burst Period". Tap the Trig Out menu label to select "RisEdge" or "FallEdge". The rear-panel [CH1/Sync/Ext Mod/Trig/FSK] connector will output the trigger signal with a specified edge. If you select "Off", the trigger output is disabled. 2. External Trigger When external trigger is selected, the generator can only output N cycle burst, Infinite burst, or Gated burst. The signal generator receives the trigger signal input from the rear-panel [CH1/Sync/Ext Mod/Trig/FSK] connector. Once it receives a TTL pulse with a specified polarity, a burst output is initiated. To specify the polarity of the TTL pulse, tap Trig In to select "RisEdge" or "FallEdge". By default, it is "RisEdge". 3. Manual Trigger When manual trigger is selected, the generator can output N cycle burst or Infinite burst. When you press Trig on the front panel for one time, one burst output is immediately launched for the corresponding channel (if enabled currently). Tap the Trig Out menu label to select "RisEdge" or "FallEdge". The rear-panel [CH1/Sync/Ext Mod/Trig/FSK] connector will output the trigger signal with a specified edge. If you select "Off", the trigger output is disabled. RIGOL Chapter 2 Front Panel Operations 2-68 DG900 User's Guide Gated Polarity Gated polarity is only available in gated burst mode. The instrument outputs the burst when the gated signal on the rear-panel [CH1/Sync/Ext Mod/Trig/FSK] connector is "High Level" or "Low Level". Press Menu Burst "Gated" to enable the gated burst. Tap the Polarity to select "Positive" or "Negative", and the default is "Positive". Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-69 Frequency Counter DG900 provides a frequency counter which can measure various parameters (e.g. frequency, period, duty cycle, positive pulse width, and negative pulse width) of the external input signal and make statistics of measurement results. When the statistical function is enabled, the instrument calculates the maximum, minimum, average, and standard deviation of the measurement values automatically and displays the variation tendency of the measurement values in "Digital" or "Curve" mode. In addition, CH1 output can work normally while the frequency counter is making measurements. To Enable the Frequency Counter Press Counter on the front panel (the backlight turns on) to enable the frequency counter function and enter the frequency counter setting interface. Figure 2-15 Frequency Counter Parameter Setting Interface If the frequency counter is currently turned on and the screen displays the frequency counter interface, press Counter again to disable the frequency counter function. If the frequency counter is currently enabled and the screen displays interfaces other than the frequency counter interface, press Counter again to switch to the frequency counter interface. Note: When the frequency counter is enabled, the output of CH2 will be disabled. RIGOL Chapter 2 Front Panel Operations 2-70 DG900 User's Guide To Set the Frequency Counter 1. Gate Time Tap the GateTime parameter selection field to select the gate time for the measurement system. The default is "100ms". When "Auto" is selected, the instrument will select an appropriate gate time automatically according to the characteristics of the signal to be measured. 1 ms 1.048 ms 10 ms 8.389 ms 100 ms 134.218 ms 1 s 1.074 s 10 s 8.590 s >10 s >8.590 s 2. To Select the Parameters to be Measured Tap the Meas.Para parameter selection field to select the type of parameter to be measured by the frequency counter. The frequency counter can measure the following parameters: frequency, period, duty cycle, positive pulse width, and negative pulse width. The default is "frequency". 3. Statistical Function Tap the Statistics parameter selection field to enable or disable the statistical function of the frequency counter. When the statistical function is enabled, the instrument calculates the maximum, minimum, average, and standard deviation of the measurement values automatically and displays the variation tendency of the measurement values in "Digital" or "Curve" mode. Slide left the touch screen to switch to the statistics display interface. Figure 2-16 Statistics Display Interface Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-71 (1) To Clear the Statistics Results Tap the "Clear" menu label, and the signal generator clears the current statistics results. (2) Operating Status Tap the Status parameter selection field to select the running status of the frequency counter. When you press Counter on the front panel, the frequency counter enters "Run" state automatically and it will continuously measure the input signal according to the current configuration. Tap the Status menu label to select "Single". After finishing the current measurement, it ethers the "Stop" status. When the frequency counter enters the "Stop" status, each time you press the Single menu label, the frequency counter executes one measurement. Note: When the statistical function is disabled, sliding left in the frequency counter parameter setting interface will switch to the frequency counter parameter measurement interface. Figure 2-17 Frequency Counter Parameter Measurement Interface 4. Sensitivity Sets the trigger sensitivity of the measurement system. Tap the Sensitivity parameter selection field to select "High" or "Low". 5. Trigger Level Sets the trigger level of the measurement system. The system triggers and gets the measurement readings when the input signal reaches the specified trigger level. The default is 0 V, and the range available is from -2.5 V to 2.5 V. Tap the Trig Lev parameter input field to input the desired value by using the numeric keypad and select the unit "V" or "mV" from the pop-up menu. 6. Coupling Sets the coupling mode of the input signal to "AC" or "DC". The default is "DC". RIGOL Chapter 2 Front Panel Operations 2-72 DG900 User's Guide 7. High Frequency Rejection High frequency rejection can be used to filter out the high-frequency components and improve the measurement accuracy in low-frequency signal measurement. Tap the High Freq Rejection parameter selection field to enable or disable the high frequency rejection function. Note: Enable the high frequency rejection when low-frequency signal with less than 200 kHz frequency is measured to filter out the high-frequency noise interference. Disable the high frequency rejection when high-frequency signal with greater than 200 kHz frequency is measured. At this time, the maximum input frequency can be 240 MHz. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-73 Store and Recall DG900 can store the current instrument state and user-defined Sequence waveforms in the internal or external memory. You can recall them when needed. DG900 can also seamlessly interconnect with RIGOL's oscilloscopes that support USB-TMC; rebuild and output the waveform data collected by the oscilloscope losslessly. Press Store on the front panel or tap the icon in the information setting area at the lower-right corner of the user interface to enable the Store/Recall function and open the Store/Recall interface as shown in the figure below. Figure 2-18 Store and Recall Interface Storage System DG900 can store the current instrument state and user-defined Sequence waveforms in the internal or external memory. You can recall them when needed. DG900 provides an internal non-volatile memory (Disk C) and an external memory (Disk D). 1． Disk C: You can store instrument states and Sequence waveform files (created by users or downloaded through remote commands) to Disk C. Also, you can copy the files from Disk C to the USB storage device. 2． Disk D: available when the USB storage device is detected by the rear-panel USB HOST interface. You can store state files or arbitrary waveform files to Disk D. The number of the files that can be stored is determined by the storage capacity of the USB storage device. The Csv files and Bmp files stored in the USB storage device can also be read. Note: DG900 can only identify the file whose filename is named in the filename input interface. If other characters are used to name the file, the name might not be normally displayed in the store and recall interface. RIGOL Chapter 2 Front Panel Operations 2-74 DG900 User's Guide File Type The file types supported by DG900 include State File, Arb File, Csv File, and Bmp File. 1. State File Stores the instrument state in the internal or external memory in ".RSF" format. The state file stored includes the dual-channel selected waveform, frequency, amplitude, offset, duty cycle, symmetry, phase, modulation, sweep, burst parameters, frequency counter parameters, as well as utility function parameters and system parameters under the Utility menu. 2. Arb File Stores the user-defined arbitrary waveform in the internal or external memory in ".RAF" format. For the arbitrary waveform file, the voltage values corresponding to each waveform point are stored in binary data format. The voltage value at each point takes up 2 bytes (16 bits). Therefore, its format of the binary data is from 0x0000 to 0xFFFF. The stored arbitrary waveform files can be read by the DG900 series. Besides, the arbitrary waveform files stored in the USB storage device can be opened and edited by data editing tools (e.g. WinHex or UltraEdit) on the PC. 3. Csv File Reads the data file in Csv format from the external memory. After finishing reading the file, the instrument automatically enters the arb waveform function interface. The arbitrary waveforms that have been read are stored in the RAM. 4. Bmp File Reads the "BMP" format bitmap file from the external memory. It is only available when you self-define splash screen. Categories In the Store and Recall interface, tap the "Arb File" or "State File" under the Categories menu label to view the arb file or state file in the first-level folder of Disk C and D (when inserted with a USB storage device). You can also use the Left arrow key to locate the cursor to the left side of the user interface, and then rotate the knob to select the "Arb File" or the "State File" folder. Press down the knob to enter the current directory. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-75 File Operation In the Disk C or D (when inserted with a USB storage device), you can perform a series of operations on the files, including Save, Read, Copy, Paste, and Delete. Save 1. To Select the Save Path Selects Disk C or D (when inserted with a USB storage device) as the current save path. You can tap Disk C or D to enter the current directory. You can also use the Left arrow key to locate the cursor to the left side of the user interface, and then rotate the knob to select Disk C or D. Press down the knob to enter the current directory. Only the state file and the arb file can be saved locally. If the current file is "Arb File", press Menu Advanced "Sequence". In the Sequence editing interface, tap "Store" to save the file. 2. To Open the Filename Input Interface After selecting the path, tap Save to enter the filename input interface, as shown in the following figure. Figure 2-19 Filename Input Interface (English) 3. To Input a Filename The length of the filename is limited to 7 characters. Input English Letters: Tap in the filename input interface to select English input method (EN is highlighted). Tap to switch between the upper-case and lower-case letters. Filename Input Area RIGOL Chapter 2 Front Panel Operations 2-76 DG900 User's Guide Tap the desired character, and the selected character will be displayed in the filename input area. Use the same method to input all the characters of the filename. You can also tap the delete the unwanted characters in the filename input area. Input Numbers Tap in the filename input interface to switch to the numbers and symbols selection interface, as shown in the following figure. The input method is similar with that of the English input. Figure 2-20 Filename Input Interface (Numbers) Input Chinese Tap in the filename input interface to select Chinese input method (CH is highlighted). Note that if the current interface is number input interface, tap to switch to English input interface. Then, select Chinese input. Figure 2-21 Filename Input Interface (Chinese) Filename Input Area Pinyin Input Area Chinese Character Display Area Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-77 Tap the desired character, and the selected character will be displayed in the Pinyin input area. After inputting the Pinyin for a Chinese character, tap the desired Chinese character in the Chinese character display area (if no proper Chinese character exists, tap or to page down to search for the desired Chinese character). The selected Chinese character will be displayed in the filename input area. Use the same method to input all the Chinese characters of the filename. You can also tap the delete the unwanted characters in the Pinyin input area, and then delete the Chinese character located ahead of the cursor in the filename input area. 4. To Save a File After finishing inputting filename in the filename input interface, press "Ok", and the generator will save the file with the specified filename to the currently selected directory. Tap "Cancel" to cancel the save operation. Read 1. To Select the Read Path Selects Disk C or D (when inserted with a USB storage device) as the current read path. You can tap Disk C or D to enter the current directory. You can also use the Left arrow key to locate the cursor to the left side of the user interface, and then rotate the knob to select Disk C or D. Press down the knob to enter the current directory. 2. To Select the File to be Read In the selected path, tap or use the knob to select the file to be read. For Disk C, only state file and arb file can be read. For Disk D, the state files, arb waveform files, Csv files, and bitmap files can be read. Note: the bitmap file can only be read when you self-define splash screen. 3. Read a file Press Read, and the generator will read the currently selected file and display the corresponding prompt message when the file is read successfully. Copy and Paste 1. To Select a File to be Copied In the selected path, tap or use the knob to select the file to be copied. 2. To Copy a File Tap Copy, the generator will copy the currently selected file. RIGOL Chapter 2 Front Panel Operations 2-78 DG900 User's Guide 3. To Select a Paste Location When copying file from Disk C to Disk D, tap or use the knob to select the Disk D directory. When copying file from Disk D to Disk C, tap or use the knob to select the Disk C directory. 4. To Paste a File Tap Paste, the generator will paste the copied file to the currently selected directory. After that, a corresponding prompt message will be displayed, indicating that the paste operation is completed. Delete 1. To Select a File to be Deleted In the selected path, tap or use the knob to select the file to be deleted. 2. To Delete a File or Folder Tap Delete, the generator will delete the currently selected file. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-79 Seamless Interconnection with Oscilloscope DG900 can also seamlessly interconnect with RIGOL's oscilloscopes that support USB-TMC; rebuild and output the waveform data collected by the oscilloscope losslessly. The connection method between the instruments is as shown in the figure below. Operation Procedures: 1. Make sure the signal under test is displayed stably on the screen of DS1. 2. Connect the USB HOST interface of DG900 with the USB DEVICE interface of DS1 by using the USB cable. Now, DS1 will enter the remote mode automatically. 3. Use DG900 to read the signal under test that is displayed currently on the screen of DS1: Press Store on the front panel of DG900 to enter the store and recall interface. At this time, the Oscilloscope menu label is displayed below the Categories menu label, and the model identifier of DS1 will be displayed under Oscilloscope. Tap or use the knob of DG900 to select the DS1 model identifier to enter the seamless interconnection operation interface. Figure 2-22 Seamless Interconnection Operation Interface All the channels of DS1 and their on/off status are displayed at the left section of the interface. The parameter settings are displayed at the right section of the interface. Tap the first parameter selection field to set the data type of the current channel to be read to "Screen Data" or "Ram Data". Tap the second parameter selection field to set the data position (only available when data type is set to "Ram Data"), and its available choices include "Head Pos", "Trig Pos", DG900 DS1 DS2 USB DEVICE USB HOST Output Terminal Input Terminal USB Cable BNC Cable RIGOL Chapter 2 Front Panel Operations 2-80 DG900 User's Guide and "Tail Pos". Tap or use the knob to select the channel to be read, and then set the parameters of the corresponding channel. Tap Read. At this time, DS1 will enter the "Stop" status automatically, and DG900 will read the arbitrary waveform data (i.g. the waveform data collected by DS1) automatically. After reading the data, the instrument saves them to the internal RAM of the current channel and switches to Sequence mode (if the current mode is not Sequence mode) automatically. 4. Use the BNC cable to connect the output of the current channel of DG900 to the input of DS2. Turn on the channel output and DG900 will output the arbitrary waveform collected by DS1. You can observe the waveform collected by DS1 from DS2 by setting DS2. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-81 Channel Setting The function and setting methods of CH2 are the same as that of CH1. This section takes CH1 as an example to introduce the channel setting method. To set the CH2 channel output parameters, select CH2 (refer to "To Select Output Channel") and then set the channel according to the descriptions in this section. Tap the channel output configuration status bar or under the user interface to enter the channel setting interface. Tap the corresponding menu label to set the parameters related with the channel. Figure 2-23 Channel Setting Interface Output Setting 1. Output On/Off Enable or disable the signal on the [CH1] connector. In the channel setting interface, tap the OutputSet menu label, and then tap the Output parameter selection field to select "On" or "Off" to enable or disable the signal output. By default, it is set to "Off". 2. Output Polarity Set the signal on the [CH1] connector as Normal or Inverted. Waveform is inverted relative to the offset voltage. In the channel setting interface, tap the OutputSet menu label, and then tap the Inverted parameter selection field to select "On" or "Off". By default, it is set to "Off". If "Off", the instrument outputs the normal waveform; if "On", the instrument outputs the inverted waveform. For example, if Inverted is "Off" (normal mode), the first half period of the waveform in the cycle is Positive; if Inverted is "On", the first half period of the waveform in the cycle is Negative, as shown in the figure below. RIGOL Chapter 2 Front Panel Operations 2-82 DG900 User's Guide Note: When the waveform is inverted, the sync signal related to the waveform is not inverted. To set the invert of the sync signal, tap SyncSet Polarity to select "Negative". 3. Output Impedance The output impedance setting affects the output amplitude and DC offset. The instrument has a 50 Ω fixed serial output impedance for the front-panel [CH1] connector. If the actual load is different from the specified value, then the displayed voltage level will not match the voltage level of the components under test. To ensure a correct voltage level, ensure that the load impedance setting must match the actual load. In the channel setting interface, tap the OutputSet menu label, and then tap the HighZ parameter selection field to select "On" or "Off". The default is "On". When "Off", tap the Impedance menu label, and input the desired value by using the numeric keypad. You can also use the arrow keys and the knob to set the impedance. The default is 50 Ω, and the range is from 1 Ω to 10 kΩ. Impedance setting will be displayed on the screen. The generator will adjust the output amplitude and offset voltage automatically once the impedance setting is modified. For example, the current amplitude is 5 Vpp. At this point, modify the output impedance from "50Ω" to "HighZ". Then, the amplitude value displayed on the screen will be doubled to 10 Vpp. If the output impedance is modified from "HighZ" to "50Ω", the amplitude value will be reduced to half of the previous value (i.g. 2.5 Vpp). Note that only the displayed values change with the parameter and the actual output from the generator does not change. 4. Level Limit Enables or disables the level limit function. In the channel setting interface, tap the OutputSet menu label, and then tap the Level Limit parameter selection field to select "On" or "Off". By default, it is set to "Off". When "On", the set high level and low level values will be limited within the range of the "HighL Limit" and "LowL Limit". Tap HighL Limit and LowL Limit respectively, and then use the numeric keypad to set the desired high level limit and low level limit. The settable ranges of the high level limit and low level limit are determined by the currently set amplitude and offset values. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-83 Sync Setting DG900 can output the basic waveforms (except Noise and DC), arb waveforms, harmonics, sweep waveforms, burst waveforms, and the sync signal of the modulated waveforms from a single channel or the dual channels at the same time. The signal will be output from the rear-panel [CH1/Sync/Ext Mod/Trig/FSK] connector. 1. Sync Switch Enables or disables the sync signal on the [CH1/Sync/Ext Mod/Trig/FSK] connector. In the channel setting interface, tap the SyncSet menu label, and then tap the SyncState parameter selection field to select "On" or "Off" to enable or disable the sync signal output. By default, it is "On". That is, send the sync signal to the [CH1/Sync/Ext Mod/Trig/FSK] connector. When disabling the sync signal, the output level on the [CH1/Sync/Ext Mod/Trig/FSK] connector is the logic low level. 2. Sync Polarity Sets the signal on the [CH1/Sync/Ext Mod/Trig/FSK] connector as "Normal" or "Inverted". In the channel setting interface, tap the SyncSet menu label, and then tap the Polarity parameter selection field to select "Positive" or "Negative". Positive: outputs the normal sync signal. Negative: outputs the inverted sync signal. Key Points: Sync signals of various waveforms: 1. Basic Waveforms 1) When the frequency of the basic waveform is less than or equal to 100 kHz: The sync signal is a Square with 50% duty cycle, and the same frequency as that of the basic waveform. When the first waveform point is output, the sync signal is TTL high level. 2) When the frequency of the basic waveform is greater than 100 kHz: Sine and Ramp: the sync signal is a Square with 50% duty cycle. Square and Pulse: the sync signal is a Square with a variable duty cycle. The duty cycle varies with the duty cycle of the output signal. Take 0 V voltage or the DC offset of the basic waveform as the reference, the sync signal is TTL high level when the corresponding value of the output signal is greater than the reference value. When the frequency of the basic waveform is less than or equal to 30 MHz, the sync signal has the same frequency as the basic waveform. When the frequency of the basic waveform is greater than 30 MHz, the frequency of the sync signal RIGOL Chapter 2 Front Panel Operations 2-84 DG900 User's Guide is (the basic waveform frequency÷2n). Wherein, n represents the frequency dividing coefficient and it is 1 when the frequency of the basic waveform is greater than 30 MHz and less than or equal to 60 MHz. 3) Noise: there is no sync signal output. 2. Harmonic For harmonics, the sync signal is a Square with 50% duty cycle, and its frequency is the same as that of the fundamental waveform. When the first waveform point is output, the sync signal is TTL high level. 3. Arbitrary Waveform For arbitrary waveforms, the sync signal is a Square with 50% duty cycle, and its frequency is the same as that of the arbitrary waveform. When the first waveform point is output, the sync signal is TTL high level. 4. Modulated Waveform 1) When internal modulation source is selected, the sync signal is a Square with 50% duty cycle. For AM, FM, PM, and PWM, the frequency of the sync signal is the modulating frequency. For ASK, FSK, and PSK, the frequency of the sync signal is the modulating rate. 2) When external modulation source is selected: The terminal is used to receive the external modulation signal, and there is no sync signal output. 5. Sweep Output Waveform 1) When the internal or manual trigger source is selected: For the sweep with "Marker Freq" off: The sync signal is a Square with 50% duty cycle. The sync signal is TTL high level at the start of the sweep and changes to low level at the mid-point of the sweep. The frequency of the sync signal is the reciprocal of the sum of the specified sweep time, return time, start hold, and stop hold time. For the sweep with "Marker Freq" on: For linear and log sweep, the sync signal is TTL high level at the start of the sweep and changes to low level at the marker frequency. For step sweep (the sweep points determined by the start frequency, stop frequency, and step respectively are f1, f2, ……, fn, fn+1……), if the marker frequency setting is one of the sweep points values, the sync signal is TTL high level at the start of the sweep and will change to low level at the marker frequency point. If the set marker frequency is not equal to the value at the sweep point, the sync signal will change to low level at the sweep point which is closest to Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-85 this marker frequency. 2) When external trigger source is selected: The terminal is used to receive the external trigger signal, and there is no sync signal output. 6. Burst Waveform 1) When the internal or manual trigger source is selected: Infinite burst: the sync signal is the same as that of the basic waveform. N cycle burst: the sync signal is TTL high level at the start of the burst and will change to TTL low level after the specified number of cycles is finished. For the sync signal, the frequency is the reciprocal of the Burst Period and the duty cycle is Carrier PeriodCycles/Burst Period. Gated burst: the terminal is used to receive the external gate signal, and there is no sync signal output. 2) When external trigger source is selected: The terminal is used to receive the external trigger signal, and there is no sync signal output. Coupling Setting DG900 supports frequency, amplitude, and phase couplings. You can set the frequency deviation (frequency ratio), amplitude deviation (amplitude ratio), or phase deviation (phase ratio) of the two channels. When the corresponding coupling functions are enabled, CH1 and CH2 are mutual base sources. When the frequency, amplitude, or phase of one channel (as the base source) is changed, the corresponding parameter of the other channel will be changed automatically and always keeps the specified frequency deviation (ratio), amplitude deviation (ratio), or phase deviation (ratio) relative to the base channel. In the channel setting interface, tap the CoupleSet menu label to enter the channel coupling setting interface. Frequency Coupling 1. To Enable the Coupling Function Tap the Freq Cpl parameter selection field to select "On" or "Off" to enable or disable the frequency coupling function. By default, it is set to "Off". 2. Freq Coupling Mode Tap the FreqCplMode parameter selection field to select "Deviation" or "Ratio". Then, tap the Freq Dev or Freq Ratio parameter selection field to input the RIGOL Chapter 2 Front Panel Operations 2-86 DG900 User's Guide desired value by using the numeric keypad, the arrow keys, or the knob. Freq Deviation: indicates the frequency deviation between CH1 and CH2. The relations among the parameters are as follows: FCH2=FCH1+FDev (take CH1 as the reference source); FCH1=FCH2+FDev (take CH2 as the reference source). Freq Ratio: indicates the ratio of the frequencies of CH1 to that of CH2. The relations among the parameters are as follows: FCH2=FCH1FRatio (take CH1 as the reference source); FCH1=FCH2FRatio (take CH2 as the reference source). Note: Please set this parameter before enabling the frequency coupling function. When the frequency coupling is enabled, this menu is grayed out and disabled. You cannot set the frequency deviation or frequency ratio. Amplitude Coupling 1. To Enable the Coupling Function Tap the Ampl Cpl parameter selection field to select "On" or "Off" to enable or disable the amplitude coupling function. By default, it is set to "Off". 2. Amp Coupling Mode Tap the AmplCplMode parameter selection field to select "Deviation" or "Ratio". Then, tap the Ampl Dev or Ampl Ratio parameter input field to input the desired value by using the numeric keypad. Ampl Deviation: indicates the amplitude deviation between CH1 and CH2. The relations among the parameters are as follows: ACH2=ACH1+ADev (take CH1 as the reference source); ACH1=ACH2+ADev (take CH2 as the reference source). Ampl Ratio: indicates the ratio of the amplitudes of CH1 to that of CH2. The relations among the parameters are as follows: ACH2=ACH1ARatio (take CH1 as the reference source); ACH1=ACH2ARatio (take CH2 as the reference source). Note: Please set this parameter before enabling the amplitude coupling function. When the amplitude coupling is enabled, this menu is grayed out and disabled. You cannot set the amplitude deviation or amplitude ratio. Phase Coupling 1. To Enable the Coupling Function Tap the Phase Cpl parameter selection field to select "On" or "Off" to enable or disable the phase coupling function. By default, it is set to "Off". 2. PhasCplMod Tap the PhasCplMode parameter selection field to select "Deviation" or "Ratio". Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-87 Then, tap the Phase Dev or Phase Ratio parameter input field to input the desired value by using the numeric keypad. Phase Deviation: indicates the phase deviation between CH1 and CH2. The relations among the parameters are as follows: PCH2=PCH1+PDev (take CH1 as the reference source); PCH1=PCH2+PDev (take CH2 as the reference source). Phase Ratio: indicates the ratio of the phases of CH1 to that of CH2. The relations among the parameters are as follows: PCH2=PCH1PRatio (take CH1 as the reference source); PCH1=PCH2PRatio (take CH2 as the reference source). Note: Please set this parameter before enabling the phase coupling function. When the phase coupling is enabled, this menu is grayed out and disabled. You cannot set the phase deviation or phase ratio. Trigger Coupling Tap the Trig Coupling parameter selection field to select "On" or "Off" to enable or disable the trigger coupling function. By default, it is set to "Off". When enabling the output of the two channels at the same time and enabling the trigger coupling, if one channel executes the trigger, then another channel will be triggered automatically. When the corresponding coupling function is enabled, there will be a yellow mark at the left side of the frequency, amplitude, and phase of the two channels. As shown in the following figure, the frequency, amplitude, and phase are in the coupling state. For example, when the frequency, amplitude, and phase deviations are set to "100Hz", "1Vpp", and "10°" respectively, and the frequency, amplitude, phase of CH1 are modified to "200Hz", "2Vpp", and "20°", then the corresponding parameters of CH2 will be automatically modified to "300Hz", "3Vpp", and "30°". Switch to CH2 and modify the frequency, amplitude, phase of CH2 to "200Hz", "2Vpp", and "20°", then, the corresponding parameters of CH1 will be automatically modified to "100Hz", "1Vpp", and "10°". RIGOL Chapter 2 Front Panel Operations 2-88 DG900 User's Guide Figure 2-24 Channel Coupling Key Points: Channel coupling is only available when both the waveforms of the two channels are basic waveforms (Sine, Square, Ramp, or Arbitrary waveform). The generator will adjust the frequency, phase, or amplitude of the channel to avoid parameter over-limit if the following conditions occur: 1. the result of the frequency, amplitude or phase value of CH1 plus (or multiply) the set deviation (or ratio) exceeds the upper limit of frequency, amplitude or phase of CH2; 2. the result of the frequency, amplitude or phase value of CH2 minus (or is divided by) the set deviation (or ratio) is below the lower limit of frequency, amplitude or phase of CH1. When the phase of the one channel is modified, the phase (phase displayed on the interface) of the other channel will be modified accordingly. At this point, aligning phase between the two channels can be realized without executing the Align Phase operation. Track Mode Tap the Track Mode parameter selection field and select the track type to "On", "Inverted", or "Off". On: enables the track function. The instrument copies the various parameters and states (except the channel output state) of CH1 to CH2 automatically. When the parameters or states of CH1 are modified, the corresponding parameters or states of CH2 (except the channel output state) will be adjusted to the same values or states automatically. At this time, the dual channels (if currently enabled) can output the same signal. Inverted: the track function has been enabled. The instrument copies the various parameters and states (except the channel output state) of CH1 to CH2 automatically. When the parameters or states of CH1 are modified, the corresponding parameters or states of CH2 (except the channel output state) will be adjusted to the same values or states automatically. At this time, CH2 Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-89 outputs the inverted signal of the output signal of CH1 (if channels are currently enabled). Off: disables the track function. This is the default state. Note: When the track function is enabled, the coupling function is disabled. Waveform Combination 1. To Enable the Waveform Combination Function In the channel setting interface, tap the CombineSet menu label and then tap the Combine parameter selection field to select "On" to enable the waveform combine function. Then, you can combine the specified waveforms based on the current fundamental waveform. If you select "Off", the waveform combine function is disabled. Note: This function is only valid for the basic waveforms. 2. To Select the Waveform to be Combined In the channel setting interface, tap the CombineSet menu label and then tap the Waveform parameter selection field to select a waveform to be combined to the current basic waveform. 3. Freq Combination Sets the frequency of the waveform to be combined to the current basic waveform. In the channel setting interface, tap the CombineSet menu label and then tap the Freq parameter input field to input the desired value by using the numeric keypad. The range is related to the currently selected basic waveform. 4. Ratio Sets the ratio of the amplitude of the waveform to be combined to that of the current basic waveform. In the channel setting interface, tap the CombineSet menu label and then tap the Ratio parameter input field to input the desired value by using the numeric keypad. RIGOL Chapter 2 Front Panel Operations 2-90 DG900 User's Guide Common Settings Press Utility on the front panel or tap the icon in the information setting area at the lower-right corner of the user interface to open the following operation interface. The interface displays the system parameters of the currently selected channel. Figure 2-25 Utility Interface System Setting To Set the System Language At present, DG900 supports menus in multiple languages. Both Chinese and English are available for the display of the help information, prompt messages, and interface display. Chinese and English input methods are also supported. Press Utility System Setting Language to select the desired language. This setting is stored in non-volatile memory and will not be affected by the "restore to defaults" operation. Power-on Selects "Default" or "Last" as the power-on system state. The default setting is "Default". Last: includes all system parameters and states, except channel output state and clock source. Default: denotes the factory defaults except certain parameters (e.g. Language). Press Utility System Setting Power-on to select the desired configuration type. This setting is stored in non-volatile memory and will not be affected by the "restore to defaults" operation. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-91 Clock Source DG900 provides an internal 10MHz clock source and receives the external clock source from the rear-panel [10MHz In/Out] connector. It can also output clock source from the [10MHz In/Out] connector for other devices. Press Utility System Setting Clk Source to select "Internal" or "External". The default is "Internal". If "External" is selected, the instrument will detect whether a valid external clock signal is input from the rear-panel [10MHz Out/In] connector. If no valid clock source is detected, a prompt message "No valid external clock is detected!" is displayed, and the clock source will be switched to "Internal". You can synchronize two or multiple instruments by setting the clock source. When two instruments are synchronized, the "Align Phase" function cannot be used. It is only applicable in adjusting the phase relations between two output channels of the same instrument and cannot be used to change the phase relations between the output channels of two instruments. Of course, you can change the phase relations between two instruments by modifying the "Start Phase" of each output channel. Sync methods for two or multiple instruments: Synchronization between two instruments: Connect the [10MHz In/Out] connector of Generator A ("Internal" clock) to the [10MHz In/Out] connector of Generator B ("External" clock) and set the output frequencies of A and B as the same value to realize synchronization between two instruments. Synchronization among multiple instruments: Divide the 10MHz clock source of a Generator ("Internal" clock) into multiple channels, and then connect them to the [10MHz In/Out] connectors of other generators ("External" clock) respectively, and finally set the output frequencies of all the generators as the same value to realize synchronization among multiple instruments. Beeper When the beeper in DG900 is enabled, a beep sound is generated when an error occurs during front-panel operation, touch screen operation, or remote operation. Press Utility System Setting Beeper to select "On" or "Off". The default is "On". Decimal Point Press Utility System Setting Decimal to set the display format of the RIGOL Chapter 2 Front Panel Operations 2-92 DG900 User's Guide decimal point in the number parameter. It can be set to a decimal "." or a comma ",". By default, it is a decimal ".". This setting is stored in non-volatile memory and will not be affected by the "restore to defaults" operation. Delimiter Press Utility System Setting Delimiter to set the display format of the delimiter in the number parameter. This setting is stored in non-volatile memory and will not be affected by the "restore to defaults" operation. When the decimal is set to ".", the delimiter can be set to "," "Space", or "None"; when the decimal point is set to ",", the delimiter can be set to ".", "Space", or "None". The decimal point and the delimiter cannot be set to a decimal "." or a comma "," at the same time. System Log Enables or disables the system log function. When enabled, the keys and prompt messages can be recorded, and they will be saved to the internal memory in Syslog.log file format. Press Utility System Setting System Log to select "On" or "Off". The default is "Off". Interface Press Utility Interface to select a different interface type. You can set the GPIB address, LAN interface parameters, or select the type of the device connected to the USB interface. The interface selection is stored in the non-volatile memory. To Set the GPIB Address Each device connected to GPIB interface must have a unique address. Press Utility Interface GPIB, and then use the numeric keypad to set the GPIB address to any value ranging from 0 to 30. The default is "2". The selected address is stored in the non-volatile memory. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-93 To Set the LAN Parameters Press Utility Interface LAN to open the LAN parameters configuration interface as shown in the figure below. You can view the network status and configure the network parameters. Figure 2-26 LAN Setting Interface Network Status Different prompt messages will be displayed based on the current network status. Connected: indicates that the network connection is successful. Disconnect: indicates that the network connection is failed. Adapter not found: indicates that the network is not connected. IP Configuration IP configuration mode can be DHCP, AutoIP, or ManualIP. In different IP configuration modes, the configurations for the IP address and other network parameters are different. 1. DHCP In DHCP mode, the DHCP server in the current network assigns the network parameters (e.g. IP address) for the signal generator. Tap the DHCP parameter selection field to select "On" or "Off" to turn DHCP mode on or off. 2. AutoIP In Auto IP mode, the generator acquires IP address within the range from 169.254.0.1 to 169.254.255.254, and gets the subnet mask 255.255.0.0 automatically according to the current network configuration. Tap the AutoIP parameter selection field to select "On" or "Off" to turn the RIGOL Chapter 2 Front Panel Operations 2-94 DG900 User's Guide AutoIP mode on or off. When DHCP and AutoIP modes are turned on at the same time, the DHCP mode takes precedence. Therefore, to enable this mode, set DHCP to "Off". 3. ManualIP In ManualIP mode, the network parameters of the generator, e.g. IP address, are defined by users. Tap the ManualIP parameter selection field to select "On" or "Off" to turn the ManualIP mode on or off. If all the three IP configuration modes are set to "On", the priority of parameters configuration from high to low is "DHCP", "AutoIP", and "ManualIP". Therefore, to enable this mode, set DHCP and AutoIP to "Off". The format of the IP address is "nnn.nnn.nnn.nnn". The range for the first segment (nnn) of the address is from 1 to 223 (except 127); and the range for the other three segments is from 0 to 255. You are recommended to ask your network administrator for an IP address available. Tap the IP parameter input field in the LAN setting interface, and then input a desired IP address by using the numeric keypad. The setting is stored in the non-volatile memory and will be loaded automatically at next power-on if DHCP and AutoIP are set to "Off". Tip: If all the three IP configuration modes are set to "On", the priority of parameters configuration from high to low is "DHCP", "AutoIP", and "ManualIP". The three IP configuration modes cannot be set to "Off" at the same time. VISA Address VISA (Virtual Instrument Software Architecture), developed by NI (National Instrument), provides an advanced programming interface to communicate with various instruments through their bus lines. VISA enables you to communicate with the instrument in the same way, regardless of the interface type of the instrument (e.g. GPIB, USB, LAN/Ethernet, or RS232). The GPIB, USB, LAN/Ethernet or RS232 instrument which wants to communicate with VISA is called "resource". VISA descriptor is the resource name and it describes the accurate name and location of the VISA resource. If LAN interface is currently used for communication with the instrument, the VISA descriptor is TCPIP0::192.168.001.102::INSTR. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-95 To Set the Subnet Mask In ManualIP mode, the subnet mask can be manually set. The format of the subnet mask is nnn.nnn.nnn.nnn. Wherein, the range of "nnn" is from 0 to 255. You are recommended to ask your network administrator for a subnet mask available. Tap the Sub parameter input field in the LAN setting interface, and then input a desired subnet mask by using the numeric keypad. The setting is stored in non-volatile memory and will be automatically loaded when the generator is powered on the next time if DHCP and AutoIP are set as "Off". To Set Default Gateway In ManualIP mode, the default gateway can be manually set. The format of the default gateway is "nnn.nnn.nnn.nnn". The range for the first segment (nnn) is from 1 to 223 (except 127); and the range for the other three segments is from 0 to 255. You are recommended to ask your network administrator for a gate address available. Tap the Gate parameter input field in the LAN setting interface, and then input a desired gateway address by using the numeric keypad. The setting is stored in non-volatile memory and will be automatically loaded when the generator is powered on the next time if DHCP and AutoIP are set to "Off". To Set DNS In ManualIP mode, the DNS can be manually set. The format of the DNS address is "nnn.nnn.nnn.nnn". The range for the first segment (nnn) of the address is from 1 to 223 (except 127); and the range for the other three segments is from 0 to 255. You are recommended to ask your network administrator for an address available. Tap the DNS parameter input field in the LAN setting interface, and then input a desired address by using the numeric keypad, the arrow keys, and the knob. The setting is stored in the non-volatile memory and will be automatically loaded when the generator is powered on the next time if DHCP and AutoIP are set to "Off". Default Settings Tap Default to restore the network parameters to the default settings. By default, DHCP and AutoIP are enabled and ManualIP is disabled. RIGOL Chapter 2 Front Panel Operations 2-96 DG900 User's Guide Apply Tap Apply to apply the currently set network parameters. System Info Press Utility System Info, and the model, serial number, and software version number of the current instrument are displayed in the system information interface. Option Press Utility Option, and the installation status of the options for the current instrument is displayed in the system information interface. Display Setting Press Utility DispSet to enter the display setting interface. 1. Brightness Setting Tap the Brightness parameter input field and use the numeric keypad to modify the brightness. The range is from 1% to 100%. The default is 50%. This setting is stored in non-volatile memory and will not be affected by the "restore to defaults" operation. 2. Screen Saver Enables or disables the screen saver mode. Tap the Screen Saver parameter selection field to select "On" or "Off" to enable or disable the screen saver function. By default, it is set to "Off". 3. Splash Screen DG900 allows you to self-define the splash screen. You can save the contents displayed on the screen to the USB storage device in BMP format. Insert a USB storage device into the USB HOST interface on the rear panel. Tap "Recall" at the right side of the Splash Screen menu label to enter the store and recall interface. Select the desired file in BMP format and tap Read. Note: The internal memory cannot store the file in BMP format. You need to store the contents in the USB storage device in BMP format first. The size of the image cannot exceed 480272. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-97 Print Setting You can store the contents displayed on the screen in the external USB storage device in picture format. 1. Insert a USB storage device into the USB HOST interface on the rear panel. 2. Press Utility Print Format to select "BMP" or "PNG". 3. Switch the content displayed on the screen to the interface to be printed. Tap at the lower-right corner of the user interface. The system will store the screen in the USB storage device in the image format according to the preset settings. RIGOL Chapter 2 Front Panel Operations 2-98 DG900 User's Guide System Utility Function To Restore Preset Press Preset to restore the preset status interface, as shown in the figure below. Figure 2-27 Preset Status Restore Interface Factory: Tap the Def icon, then a dialog box is displayed, tap "Apply" to restore the system to its factory defaults. User save: Tap any one of the items from "User1" to "User10". Then a dialog box is displayed. Tap "Save" to save the current system status to the internal non-volatile memory as user-defined settings. To use the state again, tap the icon (any one of the items from "User1" to "User10") to quickly recall the setting. You can store up to 10 system states. The defaults are shown in the following table. Note that the item with "" is the factory default value, and its value is related to the user setting during its usage but not affected by the factory default restore operation. Table 2-4 Factory Defaults Parameter Factory Settings Channel Parameter Current Carrier Waveform Sine Output Impedance HighZ Output Load 50 Ω Channel Output Off Output Inverted Off Level Limit Off HighL Limit 0 V LowL Limit 0 V SyncState Off Sync Polarity Negative Freq Cpl Off Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-99 FreqCplMode Deviation Freq Dev 0 μHz Freq Ratio 1 Ampl Cpl Off AmplCplMode Deviation Ampl Dev 0 Vpp Ampl Ratio 1 Phase Cpl Off PhaseCplMode Deviation Phase Dev 0° Phase Ratio 1 Trig Coupling Off Track Mode Off Combine Off Waveform Sine Freq 1 kHz Ratio 10% Basic Waveforms Frequency 1 kHz Amplitude 5 Vpp Amplitude Unit Vpp Offset 0 Vdc Phase 0° Square Duty Cycle 50% Ramp Waveform Symmetry 50% Pulse Duty Cycle 50% Pulse Width 500 μs Pulse Rising Edge 10 ns Pulse Falling Edge 10 ns Harmonic Type Even Harmonic Order 2 Harmonic Phase (7) 0° Harmonic No. 2 Harmonic Amplitude (7) 1.2647 Vpp User X0000000 DC Offset 0 Vdc Built-in Arbitrary Waveforms Sinc Advanced Waveform Amplitude 5 Vpp Offset 0 Vdc PRBS Bit Rate 2 kbps PRBS Data PRBS7 RIGOL Chapter 2 Front Panel Operations 2-100 DG900 User's Guide RS232 Baud Rate 9600 RS232 Data Bits 8 RS232 Stop Bits 1 RS232 Parity Bit None RS232 Data 85 Sequence Filter Interpolation Sequence Sample Rate 1 MSa/s Sequence Phase 0° Modulation AM Modulation Source Internal Mod.Wave Sine Mod.Freq 100 Hz Mod.Depth 100% DSSC Off FM Modulation Source Internal Mod.Wave Sine Mod.Freq 100 Hz Freq.Dev 1 kHz PM Modulation Source Internal Mod.Wave Sine Mod.Freq 100 Hz Phas.Dev 90° ASK Modulation Source Internal Mod.Rate 100 Hz Mod.Ampl 2 Vpp Polarity Positive FSK Modulation Source Internal Mod.Rate 100 Hz Hop Freq 10 kHz Polarity Positive PSK Modulation Source Internal Mod.Rate 100 Hz Phase 180° Polarity Positive PWM Modulation Source Internal Mod.Wave Sine Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-101 Mod.Freq 100 Hz Width Deviation 200 μs Duty Deviation 20% Sweep Sweep Time 1 s Return Time 0 ms Start Freq 100 Hz Stop Freq 1 kHz Center Frequency 550 Hz Frequency Span 900 Hz Start Hold Time 0 ms Stop Hold Time 0 ms Marker Off Marker Freq 550 Hz Trigger Source Internal Trig In RisEdge Step 2 Burst Cycles 1 Period 10 ms Gated Polarity Positive Idle Level 1st Point Trigger Source Internal Trig Out Off Trig In RisEdge Delay 0 ns Interface Focus Items Freq/Period Freq Ampl/HighL Ampl Offset/LowL Offset Duty/Width Duty Start/Center Start Stop/Span Stop Default Channel CH1 Frequency Counter Meas.Para Freq GateTime 100 ms Statistics Off Sensitivity Low Trig Level 0 V RIGOL Chapter 2 Front Panel Operations 2-102 DG900 User's Guide Coupling DC High Freq Rejection Off System Parameter System Setting Language Determined by Factory Delivery Setting Power-on Default Clk Source Internal Beeper On Decimal Dot Delimiter Comma System Log Off DispSet Screen Saver Off Brightness 100 Print Setting Location USB Storage Device Format PNG Interface GPIB 2 DHCP On (default setting in LAN) Auto IP On (default setting in LAN) Manual IP Off (default setting in LAN) Channel Copy DG900 supports states and waveform copy function between its two channels, i.g. copying all parameters and states (except the channel output state), as well as the arbitrary waveform data of one channel to the other one. Modify the channel setting of CH1, and then tap at the lower-right corner of the user interface to copy all parameters and states (except the channel output state), as well as the arbitrary waveform data of CH1 to CH2. Modify the channel setting of CH2, and then tap at the lower-right corner of the user interface to copy all parameters and states (except the channel output state), as well as the arbitrary waveform data of CH2 to CH1. Chapter 2 Front Panel Operations RIGOL DG900 User's Guide 2-103 To Lock the Keyboard You can lock any key or all the keys on the front panel by using the keyboard lock command. 1. Keyboard Lock Command Introduction :SYSTem:KLOCk ,{ON|OFF|0|1} /Locks or unlocks the specified key./ :SYSTem:KLOCk? /Queries whether the specified key is locked./ Wherein, is used to specify the key, and its range is as follows: HOME|MENU|PRESET|STORE|UTILITY|HELP| LOCK|TRIG| /Function keys/ LEFT|RIGHT|KNOB| /Arrow keys and knob/ OUTPUt1|OUTPUt2| /Output control keys/ COUNTER| /Frequency counter key/ ALL /All the keys and the knob on the front panel/ {ON|OFF|0|1} is used to lock or unlock the keys. ON|1 denotes locking the specified key and OFF|0 denotes unlocking the specified key. 2. Lock the Specified Key by Sending the Command through Ultra Sigma Build the communication between the generator and the computer. Run Ultra Sigma and search for the instrument resource. Open SCPI Panel Control and send the command. For the details, refer to "Remote Control". Chapter 3 Remote Control RIGOL DG900 User's Guide 3-1 Chapter 3 Remote Control DG900 can communicate with PC through USB, LAN, and GPIB (with the USB to GPIB interface converter provided by RIGOL) interface. The remote control can be realized by using SCPI (Standard Commands for Programmable Instruments) commands. It can be realized through the following two methods: 1. User-defined programming You can program and control the instrument by using the SCPI commands on the basis of NI-VISA (National Instrument – Virtual Instrument Software Architecture) library. For details about the SCPI commands and programming, refer to DG900 Programming Guide. 2. PC software You can use the RIGOL PC software (Ultra Sigma) to send the SCPI commands to remotely control the instrument. This chapter will illustrate how to use the Ultra Sigma software to remotely control the signal generator (take DG992 as an example) via the various interfaces. Please refer to Ultra Sigma help documentation to install the software and the required components correctly. You can download the latest version of the software from www.rigol.com. When the instrument is in remote control, the front panel keys (except the Power key and Help/Local key) and the touch screen are locked. At this time, you can press Help/Local to exit the remote mode. Contents in this chapter: Remote Control via USB Remote Control via LAN Remote Control via GPIB RIGOL Chapter 3 Remote Control 3-2 DG900 User's Guide Remote Control via USB 1. Connect the device Use the USB cable to connect the rear-panel USB DEVICE interface of the signal generator to the USB HOST interface of the PC. 2. Search for the device resource Start up the Ultra Sigma and the software will automatically search for the generator resources currently connected to the PC. You can also click to search the resources. 3. View the device resource The resources found will appear under the "RIGOL Online Resource" directory, and the model number and USB interface information of the instrument will also be displayed. For example, DG992 (USB0::0x1AB1::0x0642::DG90000000001::INSTR). 4. Communication test Right-click the resource name "DG992 (USB0::0x1AB1::0x0642::DG90000000001::INSTR)" and select "SCPI Panel Control" to turn on the remote command control panel, through which you can send commands and read data. For details about the SCPI commands and programming, refer to DG900 Programming Guide. Remote Control via LAN 1. Connect the device Use the USB-LAN interface converter to extend the LAN interface for the signal generator. Use the network cable to connect the signal generator to your local area network (LAN). 2. Configure network parameters Configure the network parameters of the signal generator according to "To Set the LAN Parameters". 3. Search for the device resource Start up Ultra Sigma and click . The window as shown in figure (a) is displayed. Click and the software searches for the instrument resources currently connected to the LAN and the resources found are displayed at the right section of the window as shown in figure (b). Click to add it. Chapter 3 Remote Control RIGOL DG900 User's Guide 3-3 (a) (b) 4. View the device resource The resources found are shown under the "RIGOL Online Resource" directory. For example, DG992 (TCPIP::172.16.3.82::INSTR). 5. Communication test Right-click the resource name "DG992 (TCPIP::172.16.3.82::INSTR)" and select "SCPI Panel Control" to turn on the remote command control panel, through which you can send commands and read data. RIGOL Chapter 3 Remote Control 3-4 DG900 User's Guide Remote Control via GPIB 1. Connect the device Connect the generator to your computer by using a USB-GPIB interface converter. Note: Please make sure that a GPIB card has been installed to your PC. Connect the USB terminal of the USB-GPIB interface converter to the USB HOST interface on the rear panel of the generator. Connect its GPIB terminal to the GPIB card terminal of the PC. 2. Install the driver of GPIB card Correctly install the driver of the GPIB card which has been connected to the PC. 3. Set the GPIB address Press Utility Interface GPIB set the GPIB address of the generator. 4. Search for the device resource Start up the Ultra Sigma and click to open the GPIB resource search panel as shown in figure (a). Click "Search" and the software will search the GPIB instrument resources connected to the PC. The device resources found will be displayed on the right side of the panel as shown in figure (b). Click "OK" to add the resource. (a) Chapter 3 Remote Control RIGOL DG900 User's Guide 3-5 (b) If the resource cannot be searched automatically: Select the GPIB card address of the PC from the drop-down box of "GPIB0::" and select the GPIB address set in the signal generator from the drop-down box of "0::INSTR". Click "Test" to check whether the GPIB communication works normally. If not, please follow the corresponding prompt messages to solve the problem. 5. View the device resource Click to go back to the main interface of Ultra Sigma. The GPIB resources found will appear under the "RIGOL Online Resource" directory. For example, DG992 (GPIB0::16::INSTR). 6. Communication test Right-click the resource name "DG992 (GPIB0::16::INSTR)", and select "SCPI Panel Control" to open the remote command control panel. Then you can send commands and read data through the panel. Chapter 4 Troubleshooting RIGOL DG900 User's Guide 4-1 Chapter 4 Troubleshooting The commonly encountered failures of DG900 and their solutions are listed below. If the following problems occur, locate and resolve the problems according to the following steps. If the problems still persist, contact RIGOL and provide your instrument information to us. (For instrument information, press Utility System Info.) 1. When I power on the instrument, the instrument stays black and does not display anything. 1) Check whether the power is correctly connected. 2) Check whether the power key is really pressed. 3) Restart the instrument after finishing the above inspections. 4) If the problem still persists, please contact RIGOL. 2. The screen is too dark to see the contents on the screen clearly. 1) Check whether the brightness settings is too low. 2) Press Utility DispSet to enter the display setting menu. Tap the Brightness menu label and use the numeric keypad to adjust the brightness of the screen to an optimal state. You can also use the arrow keys and the knob to adjust the brightness. 3. The generator is locked. 1) Check whether the generator is working in remote mode (in remote mode, is displayed in the status bar of the user interface). Press Help/Local to exit the remote control mode, unlock the front panel and the touch screen. 2) Restarting the generator can also unlock the generator. 4. The settings are correct but no waveform is generated. 1) Check whether the BNC cable is connected to the corresponding channel output terminal ([CH1] or [CH2]) tightly. 2) Check whether the BNC cable has internal damage. 3) Check whether the BNC cable is connected to the test instrument tightly. 4) Check whether the backlight of Output1 or Output2 is turned on. If not, press the corresponding key to illuminate the backlight. 5) After finishing the above inspections, press Utility System Setting Power-on to select "Last" and then restart the instrument. 6) If the problem still persists, please contact RIGOL. 5. The USB storage device cannot be recognized. 1) Check whether the USB storage device can work normally when connected to other instruments or PC. 2) Make sure that the USB storage device is FAT32 format and flash type. The generator doesn’t support hardware USB storage device. RIGOL Chapter 4 Troubleshooting 4-2 DG900 User's Guide 3) After restarting the instrument, insert the USB storage device again to check whether it can work normally. 4) If the USB storage device still cannot work normally, please contact RIGOL. 6. How do you set the amplitude of the waveform in dBm? 1) Select the desired channel. 2) In the channel setting interface, check whether OutputSet HighZ is set to "On". If yes, you cannot set the amplitude of the waveform in dBm at this time. Select "Off" to disable the HighZ, and use the numeric keypad, the arrow keys, and knob to set it to a proper value. 3) Select the desired waveform, tap the Ampl menu label, and then input the desired value by using the numeric keypad. Then select the unit "dBm" from the pop-up menu. 7. Performance verification test is failed. 1) Check whether the generator is within calibration period (1 year). 2) Check whether the generator has been warmed up for at least 30 minutes before the test. 3) Check whether the generator is under the specified temperature. 4) Check whether the test is under strong-magnetism environment. 5) Check whether the power supplies of the generator and test system have a strong interference. 6) Check whether the performance of the test device used meets the requirement. 7) Make sure that the test device used is within the calibration period. 8) Check whether the test devices used meets the required conditions of the manual. 9) Check whether all the connections are tight. 10) Check whether any cable has internal damage. 11) Make sure that the operations conform to settings and processes which are required by the performance verification manual. 12) Check whether the error calculation has faults. 13) Correctly understand the definition of "typical value" for this product: the performance specification of this product under specified conditions. Chapter 5 Appendix RIGOL DG900 User's Guide 5-1 Chapter 5 Appendix Appendix A: Accessories and Options Description Order No. Model DG952 (50MHz, Dual-channel) DG952 DG972 (70MHz, Dual-channel) DG972 DG992 (100MHz, Dual-channel) DG992 Standard Accessories 1 Power Cord conforming to the standard of the destination country - 1 USB Cable CB-USBA-USBB-FF-150 1 BNC Cable CB-BNC-BNC-MM-100 1 Quick Guide - 1 Product Warranty Card - Optional Accessories 40 dB Attenuator RA5040K USB-GPIB Interface Converter USB-GPIB-L RIGOL Chapter 5 Appendix 5-2 DG900 User's Guide Appendix B: Warranty RIGOL (SUZHOU) TECHNOLOGIES INC. (hereinafter referred to as RIGOL) warrants that the product will be free from defects in materials and workmanship within the warranty period. If a product proves defective within the warranty period, RIGOL guarantees free replacement or repair for the defective product. To get repair service, please contact with your nearest RIGOL sales or service office. There is no other warranty, expressed or implied, except such as is expressly set forth herein or other applicable warranty card. There is no implied warranty of merchantability or fitness for a particular purpose. Under no circumstances shall RIGOL be liable for any consequential, indirect, ensuing, or special damages for any breach of warranty in any case. Index RIGOL DG900 User's Guide 1 Index .RAF ...................................... 2-74 .RSF ...................................... 2-74 Align Phase .............................. 2-14 AM .......................................... 2-34 Amplitude .................................. 2-5 Amplitude Coupling ................... 2-86 Arb File .................................... 2-74 ASK ......................................... 2-44 AutoIP ..................................... 2-93 Bmp File .................................. 2-74 Built-in Waveform ..................... 2-18 Burst ....................................... 2-63 Burst Delay .............................. 2-65 Burst Period ............................. 2-66 Carrier Waveform Suppression ... 2-37 Center Frequency ..................... 2-59 Channel Copy ......................... 2-102 Clock Source ............................ 2-91 Copy files ................................. 2-77 Coupling .................................. 2-71 Csv File .................................... 2-74 DC Offset ................................... 2-7 Default Gateway ....................... 2-95 Delete ..................................... 2-78 DHCP....................................... 2-93 DNS ........................................ 2-95 Duty Cycle ................................. 2-9 factory defaults......................... 2-98 FallEdge ................................... 2-12 FM .......................................... 2-38 Freq Dev .................................. 2-39 Frequency .................................. 2-4 Frequency Counter .................... 2-69 Frequency Coupling .................. 2-85 Frequency Span ........................ 2-59 FSK ......................................... 2-47 Gate Time ................................ 2-70 Gated Burst .............................. 2-65 Gated Polarity ........................... 2-68 GPIB address............................ 2-92 Harmonic ................................. 2-23 Harmonic Amplitude .................. 2-25 Harmonic Order ........................ 2-24 Harmonic Phase ........................ 2-25 Harmonic Type .......................... 2-24 High Frequency Rejection ........... 2-72 Idle Level ................................. 2-66 Infinite Burst ............................ 2-64 Linear Sweep ............................ 2-56 Log Sweep ................................ 2-57 Manual configuration ................. 2-94 Marker Freq .............................. 2-61 Modulation Depth ...................... 2-36 Modulation Polarity .................... 2-46 N Cycle .................................... 2-63 Network Status ......................... 2-93 Output Setting .......................... 2-81 Paste a File ............................... 2-78 Phase Coupling ......................... 2-86 Phase Dev ................................ 2-43 PM ........................................... 2-41 Print ......................................... 2-97 PSK .......................................... 2-50 Pulse ........................................ 2-11 PWM ........................................ 2-53 Return Time .............................. 2-58 RisEdge .................................... 2-12 Seamless Interconnection .......... 2-79 Self-define the Splash Screen ..... 2-96 Sensitivity ................................. 2-71 Start Freq ................................. 2-59 Start Hold ................................. 2-62 Start Phase ................................ 2-8 State File .................................. 2-74 Statistical Function .................... 2-70 Step Sweep .............................. 2-57 Stop Freq ................................. 2-59 Stop Hold ................................. 2-62 Subnet mask ............................. 2-94 Sweep ...................................... 2-56 Sweep Time .............................. 2-58 Sweep Trigger Source ................ 2-60 Symmetry ................................. 2-10 Synchronous Output .................. 2-83 Track ........................................ 2-88 USB HOST ................................. 1-9 RIGOL Index 2 DG900 User's Guide VISA Address ........................... 2-94 Waveform Combination ............. 2-89 Width Deviation ........................ 2-55 |
10174 | https://artofproblemsolving.com/community/c989506h1942036_a_treatise_on_functional_divisibility_the_power_of_primes?srsltid=AfmBOopvsKM-Lw52-JCR9zAGIV87g-oF0QVryg797jF-KyXb5zvd9kUo | Just math : A treatise on functional divisibility; the power of primes!
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A treatise on functional divisibility; the power of primes!
by ubermensch, Oct 28, 2019, 2:10 PM
Well, this has been a little while coming... FEs have quietly risen in popularity, and FDs (functional divisibilities) rightly along with it. Although FEs from to are pretty well "documented", I haven't really seen any literature on techniques to attack FDs, and this is partly because of the large reliance on intuition and experience in attacking such problems, and partly because of the comparative recent-ness of these kind of problems. So let's just delve into it, aye?
Let's start our exploration with a nice N1 from 2013:
ISL 2013 N1 wrote:
Let be the set of positive integers. Find all functions such that
for all positive integers and .
Although this can be done without POP (Power of Primes ), this has a pretty clean and straightforward solution using this technique, so bear with me...
Well, the first thing we must realise is that we can extract the most out of a problem with POP if we can make our RHS a function purely of primes. This way we can strongly restrict our RHS, perhaps with bounding, or perhaps purely with divisibility; luckily, this turns out to be just the kind of suitable problem for us to apply the first tool of POP- the simple prime substitution.
Now say that doesn't divide - clearly this would imply which is of course not possible. Thus for some integer (variable) .
This comes out to be a much more powerful result than you might expect- it's technically just applicable for primes, and we don't even have a fixed to play with. Turns out, coupled with a couple of standard substitutions, this ends up almost killing the problem; see for yourself-
Next, naturally would be perhaps to try out the standard substitutions and see if we get something useful out of it-
(which also implies )
(which also implies )
Now, using the last equation and our little prime fact, now gives us:
Here we first glimpse of the true power of playing with primes- there are an infinite number of them! Speaking specifically, we can make them as large as we want, and thus-
As we can make our as big as we want, and for arbitrarily large .
Now, as and for all natural.
Well, so what were the crucial moments while solving this problem? The bounding of is certainly not trivial to think of, but it can certainly be classified as semi-standard. No, there were exactly 2 tipping points:
Noticing how useful the prime substitution can be, and then working on it.
Figuring out how the infinite-ness of primes finally killed the problem, forcing something to become zero and getting the required condition (this wasn't as important in this problem, but this can be the crucial difference between cracking the problem and making a lot of progress in (generally) harder problems).
Ah, so this was quite a good problem, but time to move on (to USAMO 2012 P4!):
USAMO 2012 P4 wrote:
Find all functions (where is the set of positive integers) such that for all positive integers and such that divides for all distinct positive integers .
Remember how forcing something to be zero by using the infinite-ness of primes played out for us last time- this problem has a very similar feel to it, no? It doesn't turn out to be based on primes basically at all, but it hammers the second turning point for our N1 home:
Let's start with the obvious- using , we get or and (from now on, I will use this to mean or ), the first natural substitution was (which didn't turn out to be that useful as a general statement, unfortunately, but set us on the right track), giving us:
This immediately gives us by Wilson's, which, unfortunately doesn't give up too much about the problem; but it never hurts to play around with what we've got, so let's see what happens when we put in small primes- well, (here will mean or ). Thus can either be , or .
Well, when our problem boils down instantly, as you might expect-
If , by and , we get that , for arbitrarily large , will have values .
Hence - thus , for constant or .
Let's take the other case, when . Again, by , there must be infinitely many, and arbitrarily large such that .
Now say that for large enough - thus , and can get as big as we want for all .
Thus our solutions are ,, and .
So what's the lesson here- well, again, what were our turning points- well, one was clearly when we realised , and substituted to get at least some casework for . Another one, and the one which hammered home the importance of making something arbitrarily large while keeping a quantity either invariant or at least bounded, hence forcing our RHS to become zero.
Well, till now we've only dealt with N1s and USAMO P4s- perhaps you aren't convinced, as there were other clean ways to attack these problems too: Watch as we take down a monster of an N6, purely by POP:
ISL 2016 N6 wrote:
Denote by the set of all positive integers. Find all functions such that for all positive integers and , the integer is nonzero and divides .
Proposed by Dorlir Ahmeti, Albania
It'd be quite hard to share the motivations for each step in an N6, so (try the problem first!, then) let's just follow along:
First let's get over with the obvious substitutions-
. Putting gives you .
(Corollary which'll be useful: .
Time to spam our first technique of POP, eh?
For prime:
. Hey this seems pretty useful- it means must be equal to one of .
Well, we've got nothing much else to do, and it pretty much seems like has got to be the only solution, so let's just bash it out using ? (and I promise it gets easier as you go through the cases)
Case 1:
Thus possible.
Case 2:
Thus possible for .
Case 3:
Thus possible.
Case 4:
Thus possible for .
Case 5: possible.
Case 6:
possible for , and possible as for
Aha, so which case did I sneakily not include when listing out these cases which end up not working- well, of course , which works for all . Thus we can conclude for .
Now's the time to apply our second statement- for a fixed , choose some prime such that .
.
As
.
Notice that we can choose arbitrarily large such that becomes arbitrarily large while stays fixed- thus must equal .
Thus the only solution is , and it obviously works, hence we're done.
I'm sure everyone enjoyed the casework, eh?
I think I've shown how useful POP really is... but looking back, what does this technique even use?
Substitute primes to get a strong divisibility condition.
There are an infinite number of primes, which can be as large as want.
This is why I like FDs so much, and even FEs in specific.
As a parting gift, here's a problem to try out POP on- not too hard either, so do try it out:
APMO 2019 P1 wrote:
Let be the set of positive integers. Determine all functions such that is divisible by for all positive integers .
For reference, here's a non POP solution: Solution
.
.
.
.
Substituting
.
Thus . Thus is the only solution.
This has been long enough, and hopefully you've taken away something from this; also do tell me if there are more problems that can be demolished (or at least simplified ) using POP. And yes, I know this was long.
Edit: Thanks to ShaftDraftKiller for informing me of ISL 2004 N3, which is almost a perfect example of both our points of POP, and is rather similar to ISL 2013 N1. Anyhow, here's the problem:
ISL 2004 N3 wrote:
Find all functions satisfying
for any two positive integers and .
Remark. The abbreviation stands for the set of all positive integers:
.
By , we mean (and not ).
Proposed by Mohsen Jamali, Iran
Here's a (non-motivated) solution (hidden for length): POP solution
.
For primes ,
or .
If
For large enough , RHS>0 and , thus for large enough .
(for large enough ):
, for arbitrarily large :
Thus for all naturals .
This post has been edited 4 times. Last edited by ubermensch, Nov 2, 2019, 11:39 AM
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nice note!
isn't it usamo 2012 4 tho, not usamo 2014 4? the latter was a game iirc
You're right, thanks for pointing it out!
This post has been edited 1 time. Last edited by ubermensch, Oct 28, 2019, 2:51 PM
by khina, Oct 28, 2019, 2:44 PM
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Pls also consider adding ISL 2004 N3
by ShaftDraftKiller, Nov 1, 2019, 6:58 PM
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Shaft you’re right, ISL 2004 N3 seems to be almost the perfect problem to demonstrate POP- with the substitution and the “making the LHS arbitrarily large” trick, it ends up working out rather similar to ISL 2013 N1... perhaps I’ll add this as an extra bonus problem when I finish typing up it’s solution at the bottom of the post. Thanks for the great suggestion man!
by ubermensch, Nov 1, 2019, 9:12 PM
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I'll just dump these here...
Balkan MO 2017 P3 wrote:
Let denote the set of positive integers. Find all functions such that
for all
Proposed by Dorlir Ahmeti, Albania
Middle European MO 2016 P4 wrote:
Find all such that divides for all .
Remark: denotes the set of the positive integers.
ISL 2011 N3 wrote:
Let be an odd integer. Determine all functions from the set of integers to itself, such that for all integers and the difference divides
Proposed by Mihai Baluna, Romania
ISL 2007 N5 wrote:
Find all surjective functions such that for every and every prime the number is divisible by if and only if is divisible by .
Author: Mohsen Jamaali and Nima Ahmadi Pour Anari, Iran
Turkey TST 2016 P5 wrote:
Find all functions such that for all holds and .
Iran Round 2 2017 N3 wrote:
Let be a positive integer. Find all functions satisfying the following two conditions:
• For infinitely many prime numbers there exists a positve integer such that .
• For all positive integers and , divides .
Iran TST 2014 #3 P2 wrote:
is there a function such that
the number of divisors of is if and only if the number of divisors of is
Iran TST 2008 P11 wrote:
is a given natural number. Find all functions such that for each the following holds:
Iran TST 2005 P4 wrote:
Find all that there exist and a prime that:
and also if then
by AlastorMoody, Dec 14, 2019, 11:18 AM
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Cool!
I made a (shorter) blog post which admittedly doesn't have solutions in it, but covers the basics:
by AshAuktober, Jan 2, 2025, 3:48 PM
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25P3 lol
by PikaBlaze, Aug 23, 2025, 1:14 AM
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Well... just math, for the most part
ubermensch
Archives
October 2021
It's All over Now, Baby Blue
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My Last INMO
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A nice little N6 (a confidence booster for INMO ;))
November 2019
We-ell, you can only avoid this technique for so long...
Another FE from the APMO- 2011 P5
The $f(x^n)$ Cauchy; easy FEs, but an important generalisation...
My very first P6! (Featuring the infamous LV trick)
Peaks ad infinitum
A cute N3 from 2006- tiny peaks and a neat identity!
An ugly trig bash- my favourite type of geo!
October 2019
Another fun NT- a synthetic divisibility from APMO 2014
A treatise on functional divisibility; the power of primes!
A powerful tool from 1995
An N7 from 2009... The trap of the trivial-looking!
An instructive FE from 2008
Post dedicated to favourite ISL 2008 problem- ISL 2008 N5!
APMO 2019 P5
RMO 2019 P6... Favourite Problem on the test!
Messing around, just saw this blog feature on AOPS, so... why not, eh?
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CAN I HAVE CONTRIBUTOR NOOB
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oh god i saw ur answer on quora but didnt realise it was u
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Wow I had completely forgotten I had subscribed to this blog! Also have fun at imotc
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Awsome blog. Keep posting FE .
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amazing blog !, the way you provide the motivation of your solution is superb Please post more NT
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10175 | https://sciencenotes.org/moles-to-grams-conversion-examples/ | Moles to Grams Conversion Examples
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Moles to Grams Conversion Examples
This entry was posted on August 7, 2022by Anne Helmenstine (updated on February 1, 2023)
Do the moles to gram conversion by multiplying the number of moles by the molar mass of the element or compound.
Converting moles to grams or grams to moles is something you do all the time in chemistry. Chemical equations relate quantities of reactants and products using moles, yet you measure chemicals on scales and balances, which use grams. The conversion is easy, providing you set it up right.
How to Convert Moles to Grams
All you need is a periodic table for looking up the atomic masses of the elements. Here is the conversion factor:
Grams = Moles x Atomic Mass
g = mol x (g/mol)
Here are the steps for converting moles to grams:
Start with the number of moles and the chemical formula of the substance.
Find the molar mass of the substance. Add together the atomic mass of each element multiplied by its subscript. If there is no subscript, use the atomic mass of that element.
Multiply the number of moles by the molar mass. This gives an answer in grams.
Moles to Grams Example #1
For example, find the mass in grams of 3.6 moles of sulfuric acid (H 2 SO 4).
First, look up the atomic masses of the elements:
H = 1.008
S = 32.06
O = 16.00
Now, use the chemical formula and find the molar mass of H 2 SO 4. You multiply the atomic mass of each element by its subscript. Remember, if there is no subscript, it’s the same as multiplying by 1. Add up the masses of each element according to the chemical formula.
molar mass of H 2 SO 4 = (1.008)(2) + 32.06 + (16.00)(4) = 2.016 + 32.06 + 64.00 = 98.076 grams per mole (g/mol)
Finally, multiply the number of moles by the molar mass:
mass of 3.6 moles H 2 SO 4 = 3.6 moles x 98.076 grams/mole = 353.07 grams
Moles to Grams Example #2
Find the mass in grams of 0.700 moles of hydrogen peroxide (H2O2).
This time let’s pay attention to the number of significant figures. You have 3 significant figures in 0.700 moles.
Write down the atomic masses of the elements:
H = 1.008
O = 15.999
Calculate the molar mass of H 2 O 2:
molar mass of H 2 O 2 = (1.008)(2) + (15.999)(2) = 2.016 + 31.998 = 34.014 g/mol
Now, multiply the moles of hydrogen peroxide by its molar mass for the answer in grams:
grams of H 2 O 2 = 0.700 moles x 34.014 grams/mole = 23.810 grams
Grams to Moles Conversion
Converting grams to moles is just as easy. Once again, you use the molar mass of the substance.
Moles = Grams / Atomic Mass
mol = g x mol/g
Start with the number of grams and the chemical formula.
Find the molar mass of the substance.
Divide the mass by the molar mass for an answer in moles.
For example, find the number of moles in 25.0 grams of potassium permanganate (KMnO 4).
Using a periodic table, look up the masses of the elements:
K = 39.01
Mn = 54.94
O = 16.00
Calculate the molar mass of KMnO 4:
molar mass of KMnO 4 = 39.01 + 54.94 + (16.00)(4) = 157.95 g/mol
Divide the mass in grams by the molar mass (g/mol) and get moles:
moles of KMnO4 = 25.0 g / 157.95 g/mol = 0.158 moles
References
Andreas, Birk; et al. (2011). “Determination of the Avogadro Constant by Counting the Atoms in a 28Si Crystal”.Physical Review Letters. 106 (3): 30801. doi:10.1103/PhysRevLett.106.030801
Barański, Andrzej (2012). “The Atomic Mass Unit, the Avogadro Constant, and the Mole: A Way to Understanding”. Journal of Chemical Education. 89 (1): 97–102. doi:10.1021/ed2001957
Cooper, G.; Humphry, S. (2010). “The ontological distinction between units and entities”.Synthese. 187 (2): 393–401. doi:10.1007/s11229-010-9832-1
International Bureau of Weights and Measures (2006). The International System of Units (SI) (8th ed.). ISBN 92-822-2213-6.
“Weights and Measures Act 1985 (c. 72)”. The UK Statute Law Database. Office of Public Sector Information.
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AM-GM Inequality
In algebra, the AM-GM Inequality, also known formally as the Inequality of Arithmetic and Geometric Means or informally as AM-GM, is an inequality that states that any list of nonnegative reals' arithmetic mean is greater than or equal to its geometric mean. Furthermore, the two means are equal if and only if every number in the list is the same.
In symbols, the inequality states that for any real numbers , with equality if and only if .
The AM-GM Inequality is among the most famous inequalities in algebra and has cemented itself as ubiquitous across almost all competitions. Applications exist at introductory, intermediate, and olympiad level problems, with AM-GM being particularly crucial in proof-based contests.
Contents
[hide]
1 Proofs
2 Generalizations
2.1 Weighted AM-GM Inequality
2.2 Mean Inequality Chain
2.3 Power Mean Inequality
3 Problems
3.1 Introductory
3.2 Intermediate
3.3 Olympiad
4 See Also
Proofs
Main article: Proofs of AM-GM
All known proofs of AM-GM use induction or other, more advanced inequalities. Furthermore, they are all more complex than their usage in introductory and most intermediate competitions. AM-GM's most elementary proof utilizes Cauchy Induction, a variant of induction where one proves a result for , uses induction to extend this to all powers of , and then shows that assuming the result for implies it holds for .
Generalizations
The AM-GM Inequality has been generalized into several other inequalities. In addition to those listed, the Minkowski Inequality and Muirhead's Inequality are also generalizations of AM-GM.
Weighted AM-GM Inequality
The Weighted AM-GM Inequality relates the weighted arithmetic and geometric means. It states that for any list of weights such that , with equality if and only if . When , the weighted form is reduced to the AM-GM Inequality. Several proofs of the Weighted AM-GM Inequality can be found in the proofs of AM-GM article.
Mean Inequality Chain
Main article: Mean Inequality Chain
The Mean Inequality Chain, also called the RMS-AM-GM-HM Inequality, relates the root mean square, arithmetic mean, geometric mean, and harmonic mean of a list of nonnegative reals. In particular, it states that with equality if and only if . As with AM-GM, there also exists a weighted version of the Mean Inequality Chain.
Power Mean Inequality
Main article: Power Mean Inequality
The Power Mean Inequality relates all the different power means of a list of nonnegative reals. The power mean is defined as follows: The Power Mean inequality then states that if , then , with equality holding if and only if Plugging into this inequality reduces it to AM-GM, and gives the Mean Inequality Chain. As with AM-GM, there also exists a weighted version of the Power Mean Inequality.
Problems
Introductory
For nonnegative real numbers , demonstrate that if then . (Solution)
Find the maximum of for all positive . (Solution)
Intermediate
Find the minimum value of for .
(Source)
Olympiad
Let , , and be positive real numbers. Prove that
(Source)
See Also
Proofs of AM-GM
Mean Inequality Chain
Power Mean Inequality
Cauchy-Schwarz Inequality
Inequality
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10177 | https://physics.nist.gov/PhysRefData/ASD/levels_form.html | NIST: Atomic Spectra Database - Energy Levels Form
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10178 | https://www.youtube.com/watch?v=8XDxDMcgh0g | The Taylor Series for f(x) = ln(x) at x = 1
MathDoctorBob
66400 subscribers
182 likes
Description
81050 views
Posted: 17 Dec 2010
Calculus: We derive the Taylor series for f(x) = ln(x) at x = 1 and use the 4th Taylor polynomial to estimate ln(.9). We then apply Taylor's Theorem to obtain a bound for the error.
For more videos like this one, please visit the Calculus playlists at this channel.
33 comments
Transcript:
find the tailor series for f ofx equal to natural log of x centered at xal 1 then use that to estimate natural log of 0.9 then use Tailor's theorem to get a Bound for the error of our estimate so our function is going to be natural log of x I'm going to set myself up a tailor Series so what do we do our Center's at one so we're to take powers of x -1 and then we're going to load them up with coefficients we get our kth coefficient by taking the K derivative of f evaluate oneide K factorial so let's turn through a few terms of this okay so first we're going to need a bunch of derivatives so der natural log of x is just going to be 1/x which is x -1 take another derivative the minus1 comes down take one off I get X-2 minus do another we bring the minus two down it becomes a plus two in front that becomes xus 3 as we keep doing that what we're going to notice is the pattern is you're going to alternate minus signs so we'll have a minus one k + one gets the sign correct and then as these exponents come down they're going to accumulate giving us K minus one factorial and then what's left is going to be just X raised to the minus K power so we're going to evaluate all these at one F of one is zero FR Prime of one is one F Prime at one is minus one third derivative at one is going to be two and then we get down to the K derivative at one is going to be minus one to the k + 1 K -1 factorial so now we divide each by its factorial so 0 factorial is equal to 1 1 factorial is equal to 1 2 factorial is equal to 2 1 is 2 so I get minus a half 3 factorial is equal to 3 2 1 which is 6 so we get a third and then when I get down to the K term we're going to take this divide by K factorial but K -1 factorial over K factorial is just equal to 1 over K so our general term is going to be Min -1 to the k + 1 times 1 over K we write out the first few terms of our series it's going to look like X X - 1 - half x - 1^ 2 + 3 x - 1/3 - a/4 x - 1 ra 4th and so on okay I can put it in Clos form as we have here and then we'll note the interval of convergence for this if we go and do the work is going to be 0 to two so we're centered at one and then we'll include zero but we'll throw away the two so that's my tailor series for natural log of x centered at one okay let's do an estimate so what we'll do is for here I've worked out up to the fourth power we have that written down so we'll just use that as our function we'll use to estimate so that's going to be the fourth tailor polinomial okay I put 0.9 in there and so all of our xus 1es will turn into minus. ons and then when I evaluate I get minus 105358 and then that three repeats on forever so that's our estimate now if we want to get a Bound for the error in the estimate we're going to appeal to Tailor's theorem so what will that say that's going to say the remainder for the fourth tailor polinomial evaluated X is going to be given by you take the fifth derivative of your function f there's going to be an X Prime which we evaluate and then we multiply by x - x0 rais the fifth power / 5 factorial this x Prime is going to be somewhere between X and x0 okay in our case x0 is going to be equal to one and X is going to be equal to 0.9 the only problem is we're not told how to find the X Prime so what we do instead is we just look at the entire interval from 0.9 to one and then we're just going to take the maximum of our fifth derivative there that way if I could figure out what this is I don't get the exact formula for the remainder but what I'll get is just an estimate for how bad it can be all right so our X is going to be 0.9 so what happens 5 factorial is 1 over 120 we're going to raise this thing a minus1 5th power but since I have the absolute value on the minus sign goes away now all I got to do is check this maximum so the fifth derivative is going to be given by 24 over X raised to the 5th power and so we get that from our general formula right here now let's plot our points at 0.9 we're going to wind up getting 40. 64 and then at one I'm just going to get a 24 and then we know the general shape of X the 5th that's going to go like this so we can see that the maximum is going to be my 4.64 so what I'll use for this term here is just going to be 41 to keep things sane so I multiply all this through that's going to give me0 5 zeros and then 3 four okay let's take a look at what the actual value is to see how bad the error is our actual value for natural log of 0.9 you go to the calculator and you get this guy minus 10536 and so on so we check against our estimate up here what's our error well you got five zeros and then 22 and then you know that's going to be strictly less than 5 Z 34 so Tailor's theorem is nailing down our actual error |
10179 | https://www.youtube.com/watch?v=0I9fwEsNClM | The Confusing Hierarchy of the Catholic Church
Intrigued Mind
19200 subscribers
1941 likes
Description
138008 views
Posted: 5 Dec 2022
The bishop of Rome, better known as the pope, is at the top of the hierarchy of the Catholic Church. Below him are the cardinals, who pick the next pope. Then there are the archbishops and the bishops. Then the priests, then the deacons, then the non-clergy, the laity. It’s a ladder that can be confusing. In this episode of Intrigued Mind, we are taking a look at the sometimes confusing ladder of authority for the biggest church in the world.
If you’re interested in early access to videos and live chats with the creator of Intrigued Mind, consider subscribing to our Patreon. Your support will greatly help us keep the channel producing more intriguing content.
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234 comments
Transcript:
[Music] even if you're not Catholic you've definitely heard the terms Bishop Cardinal and Priests get thrown around but what is the actual structure of the hierarchy of the Catholic Church who is in charge of what how does someone become the Pope in this episode of intrigued mind we'll be taking a look at the sometimes confusing ladder of Authority for the biggest church in the world if the hierarchy of the Catholic church can seem confusing that's because it kind of is there are all sorts of titles and ranks and you can easily get lost trying to figure out who is in charge of what exactly the Catholic church is one of if not the most influential groups in all of Western history so it's worth taking a look at how exactly it's structured in the ecclesiological meaning of the term hierarchy means the holy ordering of the church which is the body of Christ but generally speaking the term hierarchy refers to who is exercising Authority in the church in the Catholic Church below the pope Authority is mostly held by the bishops priests and deacons service the assistants and co-workers of the Bishops the term Pope was actually used sort of loosely until the 6th Century this is also in the term hierarchy began to be popular the Catholic church is big and a particular area of the church's jurisdiction is known as the diocese there are currently 2 903 diocese and each of these diocese has their own Bishop 144 of those diocese are in the United States diocese are divided up into individual churches which are called parishes each Church Parish has a priest but having more than one isn't uncommon each Church also has a handful of that's how things are ideally supposed to go anyway in reality there are some churches that don't have a priest about one in five Catholic churches do not have a designated priest that serves just that Parish there are also about 3500 churches around the world that are led by a deacon or even a minister who isn't an officially anointed priest or member of the clergy all members of the clergy including deacons priests and Bishops can preach teach baptize witness marriages and conduct funeral liturgies however only priests and Bishops can celebrate the sacraments of the Eucharist better known in the Protestant world as communion priests and Bishops are also the only ones who can hear confession and anoint the sick only a bishop can administer the sacrament of the holy orders meaning that they are the only ones who can make a new person into a bishop priest or deacon at the very top of course is the pope but before we dive into that if you're interested in Early Access to videos and live chats with the creator of intrigued mind consider subscribing to our patreon your support will greatly help us keep the channel Purdue using more intriguing content what most people don't understand about the pope is that he is a bishop Pope is not actually a title but simply an honorific term used for the bishop of Rome that means father whoever the bishop of Rome is that's who is in charge of the Catholic Church there are a few titles given to the bishop of Rome including Supreme pontiff of the universal church and patriarch of the Latin church these aren't exactly as quick and easy to say as Pope however there is also the title servant of the Servants of God which is intended to serve as a reminder that in Christianity leadership is really supposed to be about serving others the pump is addressed as his Holiness the pope is free to resign if he so chooses as was the case with Pope Benedict XVI in 2013. if he does not resign then he serves his Pope until his death the pope lives in Vatican City which is an independent state inside of the city of Rome it was established in its current form in 1929 by Pacs between the church and Italy the pope exercises total civil Authority within the microstate of the Vatican City this particular authority of his doesn't mean as much as you might think since Vatican City comprises only 121 acres and has a total population of 453 people it is the smallest state in the world in terms of area and population this results in all kinds of statistical anomalies that make the pope State an outlier for example it is the stained with the most out of balance male to female ratio despite having no real population or economy Vatican City has an unusually high per capita crime rate due to Petty crimes such as pickpocketing against tourists Vatican City leads the world in per capita wine consumption all of the officials who assist the pope and live in Vatican City are as a whole known as the Roman curia the term Holy See is used to refer to the pope and the curia the pope is assisted in his job by other Bishops who are considered to be the spiritual successors of the original 12 apostles who followed Jesus in the Bible in addition to the Bishops there are Cardinals Cardinals are seen as the princes of the church and are appointed directly by the pope usually the pope chooses various important bishops from all over the world to be Cardinals and form the College of cardinals to advise him Cardinals who are under the age of 80 pick the next Pope when a pope dies or resigns during the transition period where there's no Pope at all the Cardinals are the ones in charge of the church apart from this Cardinals aren't really an integral part of the structure of the Catholic church it's mostly an honorary title when they aren't busy picking the next Pope because of their crucial role in voting in the new pope their name comes from the Latin word cardo which means hinge currently 16 of the 226 Colonels are from the United States the only country with more Cardinals is Italy which has 48 having the Vatican inside of your country probably helps with that a two-third super majority is required in order to elect a new pope there are archbishops as well as Bishops a bishop oversees a diocese which is just a collection of local churches an Archbishop is in charge of an archdiocese which is really just a large diocese an example of an archdiocese would be a major metropolitan area somewhere with a lot of people archbishops have more Authority simply because they oversee more people every Bishop must go and visit the pope every five years and give a report on their particular diocese at least once a year all of a country's Bishops will get together in what's known as an Episcopal conference Bishops have their own authority to run their diocese and aren't just ambassadors of the Pope priests are the clergy members that people are most familiar with they serve under the Bishops and are in charge of the day-to-day Services of the church they preach hear confessions and serve communion the lowest rung of the latter are the deacons deacons are ordained ministers of the church who are intended to focus on things like helping the poor and not things like church leadership they are usually connected to a particular Church where they have certain functions during the service they will sometimes preach sermons and can preside over baptisms weddings and funerals sometimes deacons are people in seminary who are getting ready to be ordained into the priesthood these are called transitional deacons whereas permanent deacons are people who have no intention of becoming priests it's much easier to become a deacon than it is to become any other position in the Catholic church to be ordained as a deacon you must be 25 years old if you're not married if you are married then you have to be at least 35 years old and you have to have the consent of your wife most Catholics are not members of the clergy if you're not in the clergy then you're a part of the laity this term is derived from the Greek Leo stayo which means people of God but just because someone isn't a part of the clergy doesn't mean they don't have any sort of a job at the church lady can serve in different pastoral and administrative roles depending on what the church needs laity also play a role in Services by being acolytes lecters and cantors not everyone who participates in the Liturgy and not everyone who works at the Catholic church is part of the clergy there are lay Catholics who have full-time professional jobs working for the church this is actually fairly widespread especially in North America and Europe while laypeople fill out much of the administrative ranks of the church the ultimate organization and definition of the church's Ministry is left up to the National Bishop conferences these full-time influential church members are known as the lay ecclesial Ministry if it seems a bit complicated that's because it is however hopefully this video has helped clear up a few things about the Catholic Church [Music] for more videos on the most amazing forgotten parts of our history be sure to subscribe to the intriguemind channel like the video and leave your suggestions in the comments below |
10180 | https://www.facebook.com/marrowmed/posts/mcqwithmarrow-imagebaseda-38-year-old-male-was-found-dead-in-a-closed-space-near/3182275045184468/ | Marrow - #mcqwithmarrow #imagebased A 38-year-old male was... | Facebook
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Marrow
July 22, 2020 ·
#mcqwithmarrow#imagebased
A 38-year-old male was found dead in a closed space near his car with the engine turned on. The post mortem lividity is as depicted in the image below, softening of the basal ganglia was noted during the autopsy. What is the cause of death in this case?
A. Histotoxic hypoxia
B. Ischemic hypoxia
C. Anemic hypoxia
D. Hypoxic hypoxia
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Marrow
"The image shows Cherry red postmortem lividity and damage to the basal ganglia is a characteristic postmortem finding in case of carbon monoxide poisoning. Carbon monoxide has a higher affinity to hemoglobin compared to oxygen and hence the amount of hemoglobin available for oxygen transport is markedly reduced.
It is important to note that in the case of anemic hypoxia, O2 content is low but dissolved O2 is within normal limits so it does not stimulate peripheral chemoreceptors.
Lungs show congestion with pink fluid blood, followed by pulmonary edema and bronchopulmonary consolidation.
Pleural and pericardial anoxic haemorrhage, tiny focal necroses in myocardium are late changes (5 days).
Bilateral symmetrical necrosis and cavitation of basal ganglia (esp globus pallidus and putamen)
Spectroscopic examination, Hoppe – Seyler’s test (10% NaOH), kunkel’s (tannic acid) test and adding water (15 ml) in 2 drops of blood can detect CO."
5y
27
Shanmugapriya Sakkaravel
C.. co poisoning... .cherry red colour.. necrosis of putamen and globus pallidus..
5y
2
DrSarthak Roy
Anaemic Hypoxia... Cherry red Carbon Monoxide poisoning
5y
Sukriti Tiwari
C.
CO poisoning which will lead to anemic hypoxia.
5y
Sukanta Singha Roy
C..anaemic hypoxia...CO poisoning..cherry red colour of PM staining...
5y
Leesha Mootha
Co poisoning..cherry red colour in pm lividity...c..
5y
Hammaad Jaafar
Marrow , what is the answer?
5y
View all 2 replies
Bijit Bharali
C : CO poisoning: cherry red postmortem staining - Anemic hypoxia
5y
3
Spirono Lactone
Cherry red colour, basal ganglia softening... CO Poisoning..... Anemic hypoxia
5y
Oli Choudhury
C ..co poisoning,cherry red colour, basal ganglia degeneration..anemic hypoxia
5y |
10181 | https://www.quora.com/In-an-arithmetic-progression-can-the-common-difference-be-zero | In an arithmetic progression, can the common difference be zero? - Quora
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In an arithmetic progression, can the common difference be zero?
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Sep 7
Yes. A common difference of zero is allowed and yields a constant arithmetic progression.
Key points:
Definition: An arithmetic progression (AP) is a sequence a, a + d, a + 2d, ... where a is the first term and d the common difference.
If d = 0, every term equals a: a, a, a, a, ... — a valid AP (often called a constant sequence).
Properties that remain valid for d = 0:
The nth term formula: a_n = a + (n − 1)d gives a_n = a.
The difference between consecutive terms is constant (0).
The average of any two symmetric terms equals the middle term (trivially true since all terms equal a).
Distinction: Some
Continue Reading
Yes. A common difference of zero is allowed and yields a constant arithmetic progression.
Key points:
Definition: An arithmetic progression (AP) is a sequence a, a + d, a + 2d, ... where a is the first term and d the common difference.
If d = 0, every term equals a: a, a, a, a, ... — a valid AP (often called a constant sequence).
Properties that remain valid for d = 0:
The nth term formula: a_n = a + (n − 1)d gives a_n = a.
The difference between consecutive terms is constant (0).
The average of any two symmetric terms equals the middle term (trivially true since all terms equal a).
Distinction: Some contexts (especially when discussing “nondegenerate” APs) explicitly exclude d = 0, but standard definitions do not require excluding it. Always check whether a problem intends to allow the constant case.
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Can the digit "0" can be a common difference in an Arithmetic progression? If yes, then explain kindly, how it will be a progression?
Is the common difference of AP zero?
What are 1 examples of each on arithmetic progression whose common difference is 0, 1, -1?
Why can "n" not be negative in an arithmetic progression?
What is the difference between Fibonacci sequence and arithmetic progression?
Kalpak Patil
Studied at Shakuntala Vidyalaya Titwala
·Updated 3y
Yes it is possible the common difference can be positive negative and zero .
The above pic is from navneet digest std 10th ssc
please upvote !
Continue Reading
Yes it is possible the common difference can be positive negative and zero .
The above pic is from navneet digest std 10th ssc
please upvote !
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Shashank Bajpai
Studied at St. Joseph's College, Allahabad
·4y
Definitely.
The fundamental definition of an Arithmetic Progression (AP) is that the difference between two consecutive terms of an AP should be the same. I.e. to say that the terms should increase or decrease by the same numerical value.
So, yes that numerical value can also be equal to zero 0.
Sum of n terms of an AP can be found using the following formulae —
S = n/2 [2a + (n-1)d]
S = n/2 [a + l]
where, n — no. of terms in the given AP ; a — first term of the said AP ; l — last term of the said AP ; d — common difference between two consecutive terms of the said AP;
So, if d = 0 then, S = n/2 (2a)
Continue Reading
Definitely.
The fundamental definition of an Arithmetic Progression (AP) is that the difference between two consecutive terms of an AP should be the same. I.e. to say that the terms should increase or decrease by the same numerical value.
So, yes that numerical value can also be equal to zero 0.
Sum of n terms of an AP can be found using the following formulae —
S = n/2 [2a + (n-1)d]
S = n/2 [a + l]
where, n — no. of terms in the given AP ; a — first term of the said AP ; l — last term of the said AP ; d — common difference between two consecutive terms of the said AP;
So, if d = 0 then, S = n/2 (2a) = n.a
which holds true for a series in which d=0
Example. Let n = 5, a=2, and d=0 , therefore the given series is — 2, 2, 2, 2, 2
Therefore S = n/2 [2a + (n—1)d] = 5/2 [2(2) + (5–2).0] = 5/2 [4 + 0] = 5x2 = 10
Or using second formula,
S = n/2 [a + l] = 5/2 [2 + 2] = (5/2) x 4 = 10
Hence, we can say that in an Arithmetic Progression the common difference can be zero (0).
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Sourabh Soni
Founder and CEO at Www.fundootutor.com (2020–present) · Author has 1.1K answers and 1.4M answer views
·3y
From this video you will get a knowledge about:- 1. An Arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.
For instance the sequence 5, 7, 9, 11, 13, 15, . .is an arithmetic progression with a common difference of 2.
A finite portion of an arithmetic progression is called a finite arithmetic progression. 4. The sum of a finite arithmetic progression is called an arithmetic series.
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How arithmetic progression is applied in real life?
What is the formula for finding the common difference and sum of an arithmetic progression (AP) series?
What is the definition of a common difference in an arithmetic progression? Can it be negative?
The 4th and 10th terms of an arithmetic progression are 13 and 25 respectively. What are the first term and the common difference of the arithmetic progression? What is its 17th term?
What is the formula for summing an arithmetic progression given its initial term and common difference?
Ashutosh Patnaik
Student at IIT Kharagpur in aerospace dual degree (exp.2027) · Author has 57 answers and 49.8K answer views
·5y
Originally Answered: Can the digit "0" can be a common difference in an Arithmetic progression? If yes, then explain kindly, how it will be a progression? ·
Well to be honest , it is a arithmetic progression . As the condition states that all the terms are equal which indirectly indicates to the fact that the series is sort of a all in one it is hp , gp but as all things in maths come up with few conditions this one has too. The condition is thar none of the terms of the series can be zero which essentially translated to the first term of an ap cannot be zero .
Coming to why it is ap , the basic definition of ap states that the difference of consecutive terms must be equal
Tn : this indicates the nth term
T3 - t2 = t2 - t1
Which is true in the given c
Continue Reading
Well to be honest , it is a arithmetic progression . As the condition states that all the terms are equal which indirectly indicates to the fact that the series is sort of a all in one it is hp , gp but as all things in maths come up with few conditions this one has too. The condition is thar none of the terms of the series can be zero which essentially translated to the first term of an ap cannot be zero .
Coming to why it is ap , the basic definition of ap states that the difference of consecutive terms must be equal
Tn : this indicates the nth term
T3 - t2 = t2 - t1
Which is true in the given condition and according to the condition is equal to 0
Hope that solves your query
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Parth Chopra
Studied at Indian Institute of Technology, Delhi (Graduated 2020) · Author has 346 answers and 517.9K answer views
·8y
Originally Answered: In an arithmetic progression, can the common difference be zero ? ·
Yes it is strange it is an AP why a significant reason is it progresses with zero progress!
WTF you say
I say when with negative common difference you are happy calling it a progression why not zero progression
On better note,
It is not significant to go to fight on these rather to do some development in maths
Acceptance is the Key
Upvote ·
99 10
9 1
Ram Kushwah
Up and coming Most viewed writer · Author has 6.8K answers and 15.7M answer views
·Aug 17
Originally Answered: Can the digit "0" can be a common difference in an Arithmetic progression? If yes, then explain kindly, how it will be a progression? ·
Google says
1.The common difference can be positive, negative or 'zero'.1.The common difference can be positive, negative or 'zero'.
2.The English definition of the word 'progression' has nothing to do with the mathematical definition of arithmetic progression.
3.Thus, a, a, a, a, a..... is a valid A.P. with c.d. 0.(It is also a geometric progression with common ratio=1
Footnotes
Can a arithmetic progression have a common difference of zero & a geometric progression have common ratio one?
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Prasad K
Studying and loves mathematics as a student
·8y
Originally Answered: In an arithmetic progression, can the common difference be zero ? ·
Yes ofcourse. We shouldn't take the literal meaning of progression. Like if the common difference is negative then the series world go on decreasing. But it's still an AP. So yes the constant sequence is also termed as AP.
Upvote ·
9 3
Tejswo Tiwary
Works at BestmeBuy (2020–present)
·Updated 7y
Originally Answered: In an arithmetic progression, can the common difference be zero ? ·
Yes. The common difference in an arithmetic progression can be zero.
Sum of n terms in an AP (Sn)=n/2(2a+(n-1)d)
If the common difference d=0 then we have,
Sn=2an/2 = an
And a×n is a defined number.
Upvote ·
9 4
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Sudhansu Bhushan Mishra
Interested in Mathematics · Author has 616 answers and 1.8M answer views
·8y
Originally Answered: In an arithmetic progression, can the common difference be zero ? ·
Yes it can, but that way, all of your terms of the progression would be the same.
So, to add those terms for an arithmetic progression of n terms just multiply the first term by n.
Upvote ·
9 2
Bill Bell
Graduate studies so long ago that they are mere diaphanous memories. · Author has 4.1K answers and 6.2M answer views
·9y
Originally Answered: In an arithmetic progression, can the common difference be zero ? ·
Yes, the common difference can be zero. I agree that calling it a progression seems odd in this case but it must be admitted that this is just a special case.
Upvote ·
9 1
9 1
Ganesh Ram
M.B.A. in Finance, Anna University, Tamil Nadu, India (Graduated 2014) · Author has 2.1K answers and 527.7K answer views
·5y
Originally Answered: Can the digit "0" can be a common difference in an Arithmetic progression? If yes, then explain kindly, how it will be a progression? ·
If 0 is your common difference, then it is not a Arithmetic Progression. Progression means some common difference should be there between any 2 successive terms. That is how you essentially form a Arithmetic Progression (A.P.) sequence. With 0 as your common difference, the next term also should be the same 1st number, right ? So, there is no A.P. sequence possible with 0 as the common difference.
Upvote ·
9 2
9 1
MATHOPEDIA
Answered by
Rajesh Kumar Chaurasiya
· Author has 241 answers and 173.2K answer views
·Aug 14, 2021
Yes.
Consider the Arithmetic Progression
3,3,3,….3,3,3,….
O r,O r,
0,0,0,….
O r,O r, in general
a,a,a,…..a,a,a,…..
In all these above examples of A.P A.P common difference is 0 0. And these A.P′s A.P′s are known as Constant Arithmetic Progression.
Upvote ·
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URGE
Thread startervoracek
Start dateFeb 7, 2022
V
voracek
Member
Czech
Feb 7, 2022
#1
I would like to ask about the verb URGE. In some textbooks it is written that it can be followed by a gerund or an infinitive without changing their meaning. What is important is the structure of the sentence:
gerund: verb + -ing
infinitive: verb + person (as object) + infinitive with to
_They urge doing the test.
They urge us to do the test._
Could any of your confirm? Thank you for your answers.
owlman5
Senior Member
Colorado
English-US
Feb 7, 2022
#2
voracek said:
_They urge doing the test.
They urge us to do the test._
Could any of your confirm?
Click to expand...
The grammar in both sentences is normal. Both sentences mean roughly the same thing. Of course, you could replace us with some other reference to people: They urge everybody to do the test.
Uncle Jack
Senior Member
Cumbria, UK
British English
Feb 7, 2022
#3
Urge + object + to-infinitive is standard, and is generally better than other options such as using a that-clause.
Urge + gerund is relatively unusual. It only works when the person to carry out the action of the gerund is either obvious from the context or (perhaps more likely) does not matter. I usually associate this use with general statements not aimed at anyone in particular, and "They urge doing the test," for example, does not say who is being urged. Really, it isn't much of an urging if no one is actually being urged to do the thing.
Last edited: Feb 7, 2022
Reactions:owlman5
V
voracek
Member
Czech
Feb 7, 2022
#4
Thank you!
I am thinking of a sentence like this:
The doctors urge getting vaccinated.
Would it be possible to say The doctors urge to get vaccinated?
owlman5
Senior Member
Colorado
English-US
Feb 7, 2022
#5
voracek said:
Would it be possible to say The doctors urge to get vaccinated?
Click to expand...
You really need an object in that sentence, voracek: The doctors urge people/everyone to get vaccinated.
PaulQ
Senior Member
UK
English - England
Feb 7, 2022
#6
voracek said:
What is important is the structure of the sentence:
Click to expand...
They urge {doing the test}. doing the test is a gerund phrase. This is a type of noun phrase.
doing the test is the complement of They urge.
They urge us to do the test. - us is the object of urge _.
to do the test_ is the complement of They urge.
If you use "urge to do something" (in the active voice) you must have an object. -> "subject someoneto do something"
Loob
Senior Member
English UK
Feb 7, 2022
#7
I can't think of a situation in which I'd use They urge [verb]ing, myself.
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10185 | https://testbook.com/question-answer/using-algebraic-identities-simplify-the-following--61a73344cb7ed8272b0aa3d6 | [Solved] Using algebraic identities, simplify the following expressio
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Using algebraic identities, simplify the following expression.
(x 4+x 2+1)(x 2+x+1)
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SSC CHSL 2020 Official Paper 16 (Held On: 4 Aug 2021 Shift 1)
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(x 2- x + 1)
(x 2+ x + 1)
(x 2- 2x + 1)
(x 2+ 2x + 1)
Answer (Detailed Solution Below)
Option 1 : (x 2- x + 1)
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Given:
⇒ ? = (x 4+x 2+1)(x 2+x+1)
Formula:
a 2 - b 2 = (a + b)(a - b)
Calculation:
⇒ ? = (x 4 + 2x 2 + 1 - x 2)/(x 2 + x + 1)
⇒ ? = [(x 2 + 1) - x 2]/(x 2 + x + 1)
⇒ ? = [(x 2 + 1 - x)(x 2 + 1 + x)]/(x 2 + x + 1)
⇒ ? = x 2 - x + 1
∴(x 4+x 2+1)(x 2+x+1) = x 2 - x + 1
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More Identities Questions
Q1.0.728×0.728−0.272×0.272 0.456=?
Q2.If X−1 x=11 and x > 0, what is the value of(x 2−1 x 2)?
Q3.If x4- 12x2+ 1 = 0, then what will be the value of x 4+1 x 4?
Q4.If a3= 1 + 7, 33= 1 + 7 + b, 43= 1 + 7 + c then the value of a + b + c is:
Q5.If x=2+3,what is the value of x 2+1 x 2?
Q6.If a2 + b2 = 90 and ab = 27, then find the possible value of a+b a−b
Q7.If x2+y2−12x+ 18y + 117=0, then the value of x2+y2 is:
Q8.If a + b = 40 and a2+ b2= 116 , find the value of a3 + b3 .
Q9.The sum of the square of two positive numbers a and b (where a > b) is 7 times of their product. Find the difference between their squares.
Q10.If x 2+1 x 2=14 then what is the value of x 3+1 x 3?(x>0)
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Q2.A rectangular field has dimensions where one side is longer than the other. A person decides to walk directly across the field along the diagonal instead of walking along the two adjacent sides. By doing this, the person covers a distance that is shorter than the total walking path by exactly one-third of the length of the longer side. Determine the ratio of the length of the shorter side to the longer side of the field.
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Q5.0.728×0.728−0.272×0.272 0.456=?
Q6.Assertion (A): The system of equations 2× + y=5 and 4× + 2y = 10 has a unique solution Reason (R): A system of linear equations has a unique solution if the ratio of the coefficients of × and y are equal, and the ratio of the constants is also equal.
Q7.Given below are two equations with instructions. 1. 3x2− px + 24 = 0 2. 2y2− ry + 30 = 0 A and B are two roots of the equation 1 while C and D are roots of equation 2. Conditions:- A - B = 2 One root of equation 2 is 60% of the second root. C < D C, D > 0 and A, B > 0 Which of the equations of given options have roots as p and r?
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Q1.Solve the given series below and answer the following question. Series: 125r, M, 120q, (100p + 4), (30q + 42), (8r+ 4.8) Note: A) p and q are consecutive prime numbers where q > p. B) r = p + q C) r is a factor of 60 and greater than 10. Which of the series follows the same pattern as the above series? I) 2000, 1600, 1120, 672, 336, 134.4 II) 1000, 400, 240, 192, 192, 230.4 III) 1200, 1300, 1500, 1800, 2200, 2700
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10186 | https://nationalmaglab.org/magnet-academy/watch-play/interactive-tutorials/magnetic-domains/ | Magnetic Domains - Magnet Academy
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Take a field trip to these high magnetic field landmarks around the globe.
Try This at Home Try This at Home Science can happen anywhere! Try these hands-on science activities at home, school, club, or wherever you might find yourself! They’re easy, fun and can be done with stuff you have around the house. Activity BooksAtomic OrnamentBuild a BacteriaBrain Cerebrum HatCandy DNACandy NeuronCrystal GrowingDIY KaleidoscopeDrawing Magnetic Field LinesFill-in FractalGroovin Ghost Inflating Puffer FishMagnetic HoppingMake a Compass ActivityMakeshift MagnetsMaking ElectromagnetsMaking FerrofluidsMöbius StripNewton's Color WheelPie ChartSee Iron in FoodShapeshifting SlimeSolar System Hats
Plan a Lesson Plan a Lesson Looking for a lesson plan for science class or to incorporate science concepts into other subjects? These lessons plans offer detailed directions for K12 teachers to conduct hands-on lessons and align with next generation science standards. CompassesDemagnetizingElectric MotorsElectromagnetsMagnet Exploration Magnetic PuttyMagnetizing and UnmagnetizingMaking a CircuitPancake ParticlesPlotting Electric Field Lines
Magnetic Domains
Why can some materials be turned into magnets? It’s all thanks to magnetic domains.
In ferromagnetic materials, like iron and nickel, groups of atoms band together in areas called domains. The magnetic strength and orientation, also called the magnetic moments, of the individual atoms in such a domain are aligned with one another and all point in the same direction. It’s those special domains that can turn the material into a magnet.
Every electron is a teeny tiny magnet. They have a north and a south pole and spin around an axis. This spinning results in a very small but extremely significant magnetic field.
In most materials, the magnetic orientation of one electron cancels out the orientation of another. When a material is ferromagnetic, though, those electrons and their magnetic fields join together and their axes align. This makes those little magnetic fields add up, instead of cancelling out.
The tutorial below shows you how these domains respond to an outside magnetic field.
Instructions
Observe the different directions the arrows inside the piece of ferromagnetic material are pointing. Each arrow represents the magnetic orientation of a domain.
Move the bar magnet closer to the ferromagnetic material by using the slider.
Watch how the magnetic orientation of the domains rotate along an axis when they’re effected by the magnetic field of the bar magnet.
Once the magnetic fields are all aligned, hit reset.
See what happens to the domains when you have the magnet in different positions. Reset to try again.
When all the domains inside of the ferromagnetic material are aligned you have created a new permanent magnet. When magnetized, the magnetic moments of the atoms inside of the ferromagnetic material are organized to create a macroscopic magnet, meaning the piece of material has a north and south pole.
There are only four elements in the world that are ferromagnetic at room temperature and can become permanently magnetized: iron, nickel, cobalt and gadolinium. Ferromagnets stay magnetized for a long time, sometimes for millions of years.
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Distances between the circumcenter and each of the three excenters of a triangle
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$\begingroup$
Prove the distances between the circumcenter, $O$, and the three excenters, $I_1$, $I_2$, $I_3$, of a triangle are given by $$\begin{align} |OI_1|^2 &= R(R + 2r_1) \ |OI_2|^2 &= R(R + 2r_2) \ |OI_3|^2 &= R(R + 2r_3) \end{align}$$ where $R$ is the circumradius and $r_1$, $r_2$, $r_3$ are the respective exradii.
I know that the circumcenter is the point of concurrence of perpendicular bisectors of sides; center of circumscribed circle). I also know that excenter is the center of a circle that is tangent to the three lines extended along the sides of a triangle.
geometry
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edited Nov 17, 2013 at 0:22
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angle bisector $AI$ cut the circumcircle of $\triangle ABC$ at $D$$\angle DBI=\angle DBC+\angle IBC=\angle DAB+\angle ABI=\angle BID$ and then $DB=DI$Likewise, $DC=DI$ and then $DB=BI=DC$$I_{A}C$ bisect $\angle BCT$ $\Longrightarrow$ $\angle ICI_{A}=90^{\circ}$thus, $DI=DC=DI_{A}$
the perpendicular bisector of $BC$ cut the circumcircle of $\triangle ABC$ at $M$$\triangle SAI_{A}$ is similar to $\triangle BMD$power of $I_{A}$ with respect to the circumcircle of $\triangle ABC$ is $OI_{A}^{2}-R^{2}$Also, it is $I_{A}D\cdot I_{A}A$
$\triangle SAI_{A}\sim\triangle BMD$ $\Longrightarrow$ $\dfrac{MD}{BD}=\dfrac{I_{A}A}{SI_{A}}$$\Longrightarrow$ $2Rr_{A}=BD\cdot I_{A}A=DI_{A}\cdot I_{A}A$
hence, $2Rr_{A}=DI_{A}\cdot I_{A}A=OI_{A}^{2}-R^{2}$
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answered Nov 18, 2013 at 12:06
chloe_shichloe_shi
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$\begingroup$ what do you mean by power of $I_A$ with respect to the circumcircle and how $\angle DBC + \angle IBC = \angle DAB + \angle ABI$ $\endgroup$
Shiv
– Shiv
2022-05-19 06:23:07 +00:00
Commented May 19, 2022 at 6:23
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Problem 45 Knight Calc - 5th Edition
Textbook Question
A Carnot engine operates between temperatures of 5℃ and 500℃. The output is used to run a Carnot refrigerator operating between -5℃ and 25℃. How many joules of heat energy does the refrigerator exhaust into the room for each joule of heat energy used by the heat engine?
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1
Convert all temperatures from Celsius to Kelvin by adding 273.15 to each value. For the Carnot engine: T_hot = 500 + 273.15 K and T_cold = 5 + 273.15 K. For the Carnot refrigerator: T_high = 25 + 273.15 K and T_low = -5 + 273.15 K.
Calculate the efficiency of the Carnot engine using the formula: η = 1 - (T_cold / T_hot), where T_cold and T_hot are the temperatures of the Carnot engine in Kelvin.
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Determine the coefficient of performance (COP) of the Carnot refrigerator using the formula: COP = T_low / (T_high - T_low), where T_low and T_high are the temperatures of the refrigerator in Kelvin.
Relate the work output of the Carnot engine to the work input of the Carnot refrigerator. For each joule of work produced by the engine, it is used as input work for the refrigerator. Use the relationship: Q_exhaust = COP × Work_input, where Q_exhaust is the heat energy exhausted by the refrigerator into the room.
Combine the efficiency of the Carnot engine and the COP of the refrigerator to find the total heat energy exhausted into the room for each joule of heat energy used by the heat engine. Use the relationship: Q_exhaust = COP × (1 / η).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Carnot Engine
A Carnot engine is an idealized heat engine that operates on the Carnot cycle, which is the most efficient cycle possible between two temperature reservoirs. It converts heat energy from a hot reservoir into work while rejecting some heat to a cold reservoir. The efficiency of a Carnot engine depends solely on the temperatures of the hot and cold reservoirs, given by the formula η = 1 - (T_c/T_h), where T_c and T_h are the absolute temperatures of the cold and hot reservoirs, respectively.
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Carnot Refrigerator
A Carnot refrigerator is an idealized refrigeration cycle that operates between two temperature reservoirs, absorbing heat from a cold reservoir and expelling it to a hot reservoir. Its efficiency is defined by the coefficient of performance (COP), which is the ratio of heat removed from the cold reservoir to the work input. The COP for a Carnot refrigerator is given by COP = T_c / (T_h - T_c), where T_c and T_h are the absolute temperatures of the cold and hot reservoirs, respectively.
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Heat Transfer and Energy Conservation
Heat transfer refers to the movement of thermal energy from one object or system to another due to a temperature difference. In thermodynamic systems, the principle of energy conservation states that energy cannot be created or destroyed, only transformed. In the context of the Carnot engine and refrigerator, the heat energy output from the engine becomes the input for the refrigerator, illustrating the conservation of energy as it moves through the two systems.
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Textbook Question A Carnot refrigerator operates between energy reservoirs at 0℃ and 250℃. A 2.4-cm-diameter, 50-cm-long copper bar connects the two energy reservoirs. At what rate, in W, must work be done on the refrigerator to remove heat from the cold reservoir at the same rate that it arrives through the copper bar? 774 views
Textbook Question An ideal refrigerator utilizes a Carnot cycle operating between 0℃ and 25℃. To turn 10 kg of liquid water at 0℃ into 10 kg of ice at 0℃, (a) how much heat is exhausted into the room and (b) how much energy must be supplied to the refrigerator? 552 views
Textbook Question FIGURE P21.46 shows a Carnot heat engine driving a Carnot refrigerator. Determine Q 2, Q 3 and Q 4. 782 views
Textbook Question A typical coal-fired power plant burns 300 metric tons of coal every hour to generate 750 MW of electricity. 1 metric ton = 1000 kg. The density of coal is 1500 kg/m³ and its heat of combustion is 28 MJ/kg. Assume that all heat is transferred from the fuel to the boiler and that all the work done in spinning the turbine is transformed into electric energy. Suppose the coal is piled up in a 10 m ✕ 10 m room. How tall must the pile be to operate the plant for one day? 804 views
Textbook Question A heat engine exhausts 7600 J of heat while performing 2600 J of useful work. What is the efficiency of this engine? 700 views
Textbook Question A certain power plant puts out 680 MW of electric power. Estimate the heat discharged per second, assuming that the plant has an efficiency of 32%. 452 views
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10189 | https://courses.grainger.illinois.edu/ece110/sp2019/content/labs/Modules/M200_NonIdeal_Instruments.SP19.v1.pdf | Non-Ideal Behavior (DC) Notes: Module 200: Non-Ideal Behavior of Instruments In the core part of Lab 1 you learned how to use the power supply as a source of electrical energy. And you used the multimeters to quantify the behavior of a circuit by measuring the voltage and current associated with each device. You even characterized the resistors used in the circuit you built by measuring the resistance, again, using the multimeter. When using this equipment you did not have to consider how the power was supplied or how the measurements were taken. The assumption was made that these are ideal devices. The plots in Figure 1 illustrate how we think about ideal sources and measuring devices. The ideal power sources maintains a set voltage Vs or current Is no matter what circuit is connected across the terminals. Even if the terminals are shorted (connected together by a wire) the ideal supply maintains the voltage. Ideal measuring devices do not modify the circuit behavior in any way. The ideal voltmeter draws no current into itself but still registers the correct voltage between two points with a circuit. The ideal ammeter accepts any current, creating no voltage drop, and also registers the correct current flowing through the device/circuit tested. The graphs illustrate the ideal behavior as presented in the core lab and in lecture. In the labs you also measure resistance using the Ohmmeter. Is there such a thing as an ideal ohmmeter? How would you know? This module explores the limitations of your bench equipment. Experimenting with the limits you will come to a better understanding of how the multimeter and power supply works. Even though you will rarely be aware of the limitations when using the bench equipment these procedures are meant to give you additional understanding so that when you are using non-ideal components like a battery you will be able to work within the limitations of the device. Let’s start by pushing the limits. You will build very simple circuits choosing resistances chosen to make powering the circuit with a quantifiable voltage and current imprecise. You will build other circuits that tax the ability of the multimeter to make precise measurements. Even the bench equipment is really just a circuit – a very well-designed circuit – but a circuit, just like the one you are testing and it must considered as part of the circuit if you want precise measurements. Non-Ideal Behavior (DC) Notes: Non-Ideal Behavior (DC) Notes: Procedures Getting the voltmeter to give an Improper Value of the Voltage Using the simple circuit below, you will cause the voltmeter to provide an incorrect voltage reading. This is a non-invasive procedure and will not hurt the meter and will not damage the circuit components. Build the circuit below consisting of two resistors in series connected to the power supply. The exact value of the two resistors doesn’t matter, but the resistors should have a resistance ≥ 𝟏𝟏𝟏𝟏𝟏𝟏. The schematic below shows two 24 MΩ resistors connected in series with the power supply. Set the power supply to 5V. Set the multimeter to measure DC voltage. Answer Question 1. The measurement does not give the value you have come to expect because our circuit does not see the voltmeter as an infinite resistance. The figure below shows that a simple model for the digital multimeter (DMM) when measuring DC voltage is a parallel connection of a perfect voltage-measuring device that reads the voltage across a resistor with a large (but not infinite!) resistance. A small amount of current flows through 𝑅𝑅𝐷𝐷𝐷𝐷𝐷𝐷 across which a voltage appears and is measured by our instrument. For that measured voltage to be the same as the voltage across the 24 𝑀𝑀Ω resistor in the absence of the voltmeter, the current through 𝑅𝑅𝐷𝐷𝐷𝐷𝐷𝐷 must be very small compared to the current through the 24 𝑀𝑀Ω resistor that sits in parallel with it. The value of 𝑅𝑅𝐷𝐷𝐷𝐷𝐷𝐷 is approximately 10 𝑀𝑀𝑀𝑀 for our multimeters. This value that is huge compared to the resistances you normally use in the lab and your designs. But the resistances in this particuluar circuit are large. The power supply now sees a single 24 𝑀𝑀Ω resistor in series with a resistance that is the parallel combination of 24 𝑀𝑀Ω and 𝑅𝑅𝐷𝐷𝐷𝐷𝐷𝐷= 10 𝑀𝑀𝑀𝑀. Non-Ideal Behavior (DC) Notes: Answer Question 2. Linear model of DMM as a DC Non-Ideal Behavior (DC) Notes: You can get the ammeter to misread the current as well, but that’s a bit more risky so let’s do a thought experiment. Think about this…when measuring current, you break the circuit and insert the ammeter in series with the “circuitry under test” (the devices for which you want the current to be measured). Below is a schematic of a circuit with a single resistor and the ammeter in DC I mode measuring current. Remember that in the ideal case (ideal ammeter), 𝑅𝑅𝐷𝐷𝐷𝐷𝐷𝐷≈0 Ω. In reality, for our ammeter, 𝑅𝑅𝐷𝐷𝐷𝐷𝐷𝐷 will be in the range from 0.1 Ω to 2 Ω. Answer Question 3. Linear model of DMM as a DC ammeter (DC I) Non-Ideal Behavior (DC) Notes: Getting the Ohmmeter to give an Improper Value of the Resistance Disconnect the Power Supply. Continuing with circuit below. Set the digital multimeter to measure resistance. Answer Question 4. Non-Ideal Behavior (DC) Notes: When the multimeter is put into resistance measuring mode, the resistor, or resistive circuit under test must be disconnected from the rest of the circuit! How does the Ohmmeter measure resistance? As the figure below shows, the Ohmmeter uses the same circuitry as the voltmeter including the internal resistance or 𝑅𝑅𝐷𝐷𝐷𝐷𝐷𝐷= 10 𝑀𝑀Ω. This time, however, the multimeter provides the power (for the voltmeter and ammeter, the power came from the circuit under test). Additional circuitry inside the multimeter injects a known current through the circuit. Answer Questions 5 and 6. Linear model of DMM as an ohmmeter (DC Ω) Non-Ideal Behavior (DC) Notes: Pushing the Power Supply It is more difficult to drive the power supply into exhibiting non-ideal behavior. The power supplies are very well-designed so that the simplest model is an ideal source with a very small series resistance. For the power supply to fail to deliver the correct voltage the circuit being powered must have a very low resistance. Be certain when doing this portion that the small resistor is up to the task. That is, the resistor must have a high-enough power rating! Build the circuit on the left below. USE A 1Ω RESISTOR THAT IS MADE TO CARRYING A LARGE CURRENT. Your TA can help you find a suitable resistor. DO NOT USE THE RESISTORS IN YOUR KIT as they are only ¼ Watt. Be certain that the power supply is not connected until you have set it to 1V. Set the power supply to 1V. When you are ready to answer the next questions, turn it on. Answer Question 7. DC Power Supply Linear model of Non-Ideal Behavior (DC) Notes: Explore More! Points awarded: __ Name: ____ Net ID: ______ Module 200: Non-Ideal Behavior (DC) Probe the voltage across each of the resistors. Record the values that you read from the voltmeter. Calculate the expected voltage using KVL or the voltage divider rule. Are the measured values the same as the calculated values? Using the voltage divider rule estimate the voltage you expect to read on the multimeter now that you are considering the internal resistance of the multimeter when it is set in DC V mode. Is this value closer to values you measured? Assuming 𝑅𝑅𝐷𝐷𝐷𝐷𝐷𝐷= 0.5 Ω what value of 𝑅𝑅 will cause the ammeter to measure a current that it is 10% lower than the value predicted by Ohm’s Law = 𝑉𝑉𝑆𝑆 𝑅𝑅 ? Measure the resistance of one of 24 𝑀𝑀Ω resistors and record the value. Non-Ideal Behavior (DC) Notes: Draw the model of the multimeter across the terminals of 𝑅𝑅2. Use the diagram to explain why you will not measure the value of 𝑅𝑅2 unless you disconnect the resistor being tested from the power source. Again, draw the model of the multimeter across the terminals of R2. Use the diagram to explain why you must isolate the resistor being tested from the rest of the circuit. Using one of the multimeters, measure the voltage across the 1Ω resistor. Is it 1V? It is probably pretty close, but it has likely changed by some amount. Make an estimate of the value of the internal resistance of the power supply from the voltage you measured. |
10190 | https://www.teacherspayteachers.com/Product/Multiply-by-10-100-1000-Worksheets-and-multiples-of-10-100-1000-Worksheets-6154469 | Multiply by 10, 100 & 1000 Worksheets and multiples of 10, 100 & 1000 Worksheets
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These are simple, basic worksheets for students who need extra practice learning to multiply numbers by 10, 100 and 1000. Contains 18 student worksheets and 2 quizzes.
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Multiplying 1-digit numbers by 10, 100, and 1000
Multiplying 2-digit, 3-digit, and 4-digit numbers by 10, 100, and 1000
Multiplying 1-digit numbers by 10, 100, and 1000 with a missing factor
Multiplying 1-digit numbers by multiples of 10
Multiplying two 2-digit numbers that are multiples of 10
Multiplying 1-digit numbers by multiples of 10, 100, and 1000
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10191 | https://math.answers.com/other-math/Why_division_is_not_commutative_property | Why division is not commutative property? - Answers
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Why division is not commutative property?
Anonymous ∙ 14 y ago
Updated: 4/28/2022
It is because a/b does not necessarily equal b/a. If this were true, then a2 = b2, which implies a = +/- b, which does not have to be true.
Likewise, subtraction is not commutative, because a-b does not have to be equal to b-a. Here, subtraction is anticommutative because a-b = -(b-a).
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### Why is there no commutative property for subtraction or division? There is no commutative property in subtraction or division because the order of the numbers cannot be change. This means that when multiplying or adding it does not matter the order of the numbers because the answer comes out the same.
### What is the commutative property in math? The commutative property of a binary operator states that the order of the operands does not affect the result. Thus x ^ y = y ^ x where ^ is the binary operator. Addition and multiplication of numbers are two common operators that are commutative. Subtraction and division are two common ones that are not commutative.
### Which property allows you to regroup addends? The commutative property of addition and the commutative property of multiplication.
### Give an example showing that the commutative property does not hold for division of whole numbers? Here is an example: 4/2 = 2 Commutative property is when you can move numbers around in a problem, and it wouldn't change. This is why it doesn't work in division 2/4 = 1/2 The commutative property applies to only addition and multiplication. It does not apply to division or subtraction. More examples: Addition: 2 + 3 = 3 + 2 = 5 Subtraction: 2 - 3 = -1, 3 - 2 = 1 Division: (see above) Multiplication: 3(5) = 5(3) = 15
### Is 7 divided by x equals x divided by 7 commutative property? 1
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10192 | https://www.sciencedirect.com/science/article/abs/pii/S1040842816301627 | Revisiting IL-6 antagonism in multiple myeloma - ScienceDirect
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Critical Reviews in Oncology/Hematology
Volume 105, September 2016, Pages 1-4
Revisiting IL-6 antagonism in multiple myeloma
Author links open overlay panel Thomas Matthes a, Benoit Manfroi b, Bertrand Huard b
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Abstract
IL-6, a cytokine with broad functions in inflammation and immunity, has been extensively studied for its role on normal antibody-producing plasma cells. In addition, IL-6 is recognized as a proliferative factor for multiple myeloma (MM), a malignant plasma cell tumor developing in the bone marrow. Blocking IL-6 signaling was thus developed into a therapeutic approach for MM already early after its discovery, in 1991. Unfortunately, the first clinical trials did not demonstrate a clear benefit, but despite this apparent failure hopes on IL-6 antagonism are still high and trials ongoing. The cellular source of IL-6 has long been a matter of debate. IL-6 was first recognized as an autocrine factor produced by the malignant plasma cells themselves, but later reports clearly showed that IL-6 was a paracrine factor, produced by the microenvironment, mostly by cells from the myeloid lineage. Recently, we have confirmed that IL-6 originates from myeloid lineage cells, mainly from myeloid precursors. We have also demonstrated that IL-6 amplifies the pool of myeloid cells producing a second key factor for MM, a proliferation inducing ligand (APRIL). These findings form a new rationale for IL-6 inhibition in MM and for new ways to use IL-6 blocking in the clinics.
Introduction
The bone marrow niche constitutes a specialized microenvironment composed of a variety of cellular elements, which play an essential role in the survival, growth and differentiation of normal stem and progenitor cells, but also provide optimal growth conditions for various hematologic malignancies including multiple myeloma (MM). A complex interplay of cytokines, chemokines, proteolytic enzymes and adhesion molecules provides MM plasma cells with survival signals and chemoresistance to current therapies. Among these factors the cytokine IL-6 plays a pivotal role and it has been considered a promising therapeutic target. In this review we will focus on some new findings about the cellular sources of IL-6 and discuss implications on clinical treatment strategies.
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IL-6: a rapid translation from in vitro studies into clinics
In 1986, Nordan and Potter reported on a growth factor produced form peritoneal macrophages and capable of supporting proliferation and survival of plasmocytoma cell lines. In the same year, Hirano et al. described the molecular cloning of this molecule, which they named B-cell stimulating factor-2 (BSF-2) (Hirano et al., 1986), in line with the previously discovered molecule BSF-1, now termed IL-4. BSF-2 was one of several factors detected in T-cell conditioned supernatants able to substitute
IL-6 antagonism in multiple myeloma
In 1991, Klein et al., treated a patient by sequential injections of two murine anti-IL-6 antibodies with a blocking activity directed against recombinant human IL-6. The authors obtained a reduction in MM cell proliferation, thus establishing the proof of concept. In another case report Chari et al. described a patient with relapsed refractory MM after 8 lines of therapy, in whom they were able to obtain a complete, one-year lasting remission by treatment with single agent anti-IL 6
The cellular sources of IL-6 in MM
The cellular sources of IL-6 in MM-infiltrated bone marrow have long been a matter of debate. Early reports, sometimes contradictory, described a paracrine and an autocrine origin (Kawano et al., 1988, Klein et al., 1989). Kawani et al. was the first to describe plasma cells as producers of IL-6, and several independent groups have later confirmed autocrine IL-6 production in MM. IL-6 mRNA in MM cells was also reported (Anderson et al., 1989, Freeman et al., 1989, Hata et al., 1993) and the
Can anti-IL-6 therapeutic strategies be improved in multiple myeloma?
Despite negative results obtained in clinical trials so far, we believe that anti-IL-6 treatments should continued to be tested in MM. In fact, the myeloid precursor origin for IL-6 might explain these failures, since IL-6 antagonism might be redundant in a situation where immature, highly proliferating myeloid cells are already extremely susceptible to standard chemotherapy. In all recent trials with the optimized anti-IL-6 antibody Siltuximab, the antibody was indeed associated to
Conclusion
There have been numerous studies in MM reporting the characterization of IL-6-producing cells from the bone marrow microenvironment (i.e.: osteoblasts, eosinophils, megacaryocytes, stromal cells). We have shown by our recent work that immature myeloid precursors constitute the major producers of IL-6 in BM. These findings may have an impact on the way IL-6 should be used in future MM trails.
Disclosure of fundings
The authors have not received any specific funding for the present article.
Conflict of interest
The authors declare no conflict of interest.
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Preparation and activities of selenium polysaccharide from plant such as Grifola frondosa
2020, Carbohydrate Polymers Citation Excerpt :
Polysaccharide selenide could significantly improve the activity of Lilium brownii polysaccharide. The results showed that Lilium selenopolysaccharide could promote the proliferation of lymphocytes, improve the level of serum antibody, and promote the secretion of IL-2 and IFN-γ by auxiliary T cells, and the expression of IL-6 produced by the cells was significantly higher than that of Lilium polysaccharide, indicating that Lilium selenopolysaccharide could promote the proliferation and differentiation of T cells, and play an important role in regulating immune response and anti-infection of the body (Matthes, Manfroi, & Huard, 2016). The Codonopsis selenium polysaccharide could significantly improve the phagocytosis index of peritoneal macrophages and induce the secretion of TNF-α and IL-6. Show abstract Selenium polysaccharide is an organic selenium compound formed by the combination of polysaccharide and selenium. The biological activity of selenium polysaccharide from plant such as Grifola frondosa is generally higher than that of selenium and polysaccharide, and it is more easily absorbed and utilized by human body. Therefore, selenium polysaccharide has a wide range of applications in immune regulation, anti-tumor, anti-oxidation, anti-aging and so on. Due to the unique pharmacological activity of selenium polysaccharide, the research of selenium polysaccharide has gradually become a hot spot. At present, there are few kinds of selenium polysaccharide, but its structure is very complex. The chemical structure and mechanism of selenium polysaccharide in vivo are not completely clear and need further study. Herein, the preparation and biological activities of selenium polysaccharides were introduced systematically, which provided theoretical basis for further research and application of selenium polysaccharides.
### Rapeseed protein-derived ACE inhibitory peptides LY, RALP and GHS show antioxidant and anti-inflammatory effects on spontaneously hypertensive rats
2019, Journal of Functional Foods Citation Excerpt :
It has a variety of biological effects, such as mediating anti-tumor and regulating immune functions and is also one of the mediators of the inflammatory reaction (Kim, Ko, et al., 2018). IL-6 is mainly produced by macrophages, T cells and B cells, and plays an important role in regulating cell growth and differentiation, the immune response, hematopoietic function and anti-infective function (Matthes, Manfroi, & Huard, 2016). As shown in Fig. 6, the levels of TNF-α and IL-6 in the serum of the SHR group were significantly (p < 0.05) higher than in the control group due to the presence of hypertension. Show abstract Food-derived bioactive peptides have received considerable attention as low toxicity nutrient supplements with multiple physiological activities, such as antioxidant, antihypertensive and anti-inflammatory properties. In this study, a RAW 264.7 cell model and spontaneously hypertensive rats model were employed to investigate the antioxidant and anti-inflammatory activities of the angiotensin I-converting enzyme inhibitory peptides Leu-Tyr (LY), Arg-Ala-Leu-Pro (RALP) and Gly-His-Ser (GHS). The results showed that in vitro, LY, RALP and GHS significantly inhibited the secretion of nitric oxide, interleukin-6 and tumor necrosis factor-α in lipopolysaccharide-stimulated RAW 264.7 macrophages. In vivo, LY, RALP and GHS inhibited the release of nitric oxide and the production of lipid peroxides and reactive oxygen species, and improved cell damage caused by oxidative stress in spontaneously hypertensive rats. Furthermore, LY, RALP, and GHS inhibited proinflammatory cytokines including interleukin-6 and tumor necrosis factor-α in plasma. Thus, LY, RALP and GHS can protect the body from oxidative and inflammatory damage.
### Targeting signaling pathways in multiple myeloma: Pathogenesis and implication for treatments
2018, Cancer Letters Show abstract Multiple myeloma (MM), which is characterized by osteolytic bone lesions, anemia, hypercalcemia, and renal failure, accounts for approximately 10% of all hematologic malignancies. Although the therapeutic landscape of MM has evolved spectacularly over the past decades with 5-year median survival over 50%, most of these patients relapse eventually. The widely recognized therapeutic approaches include chemotherapy, radiation, stem cell transplant, and monoclonal antibody therapy. Former studies have implied that the proliferation, survival, migration and drug resistance of MM cells are in association with the activation of several signaling pathways. In this review, we intended to focus on the major signaling pathways such as PI3K/Akt/mTOR, Ras/Raf/MEK/MAPK, JAK/STAT, NF-κB, Wnt/β-catenin, and RANK/RANKL/OPG, that contribute to the pathogenesis of the MM and the therapeutic approaches developed to target them.
### Interleukin 6: at the interface of human health and disease
2023, Frontiers in Immunology
### Microbiota-driven interleukin-17-producing cells and eosinophils synergize to accelerate multiple myeloma progression
2018, Nature Communications
### Inflammatory and Anti-Inflammatory Equilibrium, Proliferative and Antiproliferative Balance: The Role of Cytokines in Multiple Myeloma
2017, Mediators of Inflammation
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10193 | https://ajronline.org/doi/10.2214/AJR.12.9888 | Skip to main content
Correctly Using Sensitivity, Specificity, and Predictive Values in Clinical Practice: How to Avoid Three Common Pitfalls
Authors: David M. Naeger, Maureen P. Kohi, Emily M. Webb, Andrew Phelps, Karen G. Ordovas, and Thomas B. NewmanAuthor Info & Affiliations
Volume 200, Issue 6
10,07847
Abstract
OBJECTIVE. Radiology is the specialty of imaging-based diagnostic tests. Understanding the science behind evaluating diagnostic test performance is essential for radiologists because we provide care to patients and interact with our colleagues.
CONCLUSION. Here, we review the key terminology used and common pitfalls encountered in the literature and in day-to-day discussions of diagnostic test performance.
Radiology is the specialty of imaging-based diagnostic tests. The science behind evaluating diagnostic tests, so-called “evidence-based diagnosis” , is fundamental to our discipline [2–4]. Here, we review key terminology used to describe the performance of a diagnostic test (imaging and nonimaging), and we review three very common pitfalls encountered when these principles are discussed. Each pitfall is framed in the form of a case.
Case 1
You are covering the emergency department (ED) radiology service for the day and you encounter a minimally displaced scaphoid fracture in a patient who fell on her outstretched hand. You tell the ED physician your findings, and he asks you for the sensitivity of radiographs for such fractures. Having read a recent study, you reply “About 75%.” The ED physician says, “Well, if you're that far away from 100% certain of a fracture, we might need an MRI before I call the surgeon. Can you arrange one?” What do you say?
What Exactly Do Sensitivity and Specificity Tell You?
Sensitivity and specificity describe how a test performs in people with known disease status. In clinical practice, we deal with patients of unknown disease status; this should immediately give us pause about how we use these terms.
Sensitivity expresses how a test performs in people known to have the disease. Highly sensitive tests tend to be positive in patients with disease. This parameter, therefore, depends on the biology of the disease (the chemical or anatomic abnormalities that result from the disease) and characteristics of the test (e.g., how well the machinery or chemical tests detect the abnormalities). For example, the sensitivity of ultrasound for gallstones depends on the underlying biology of gallstones (size and composition), the technology of the ultrasound machine, the technique of the sonographer, and the skill of the reader. Every step along the way affects the overall sensitivity of the test.
Specificity is how the test performs in people who are known to not have disease. Highly specific tests tend to be negative in patients without disease. This parameter also depends on the underlying biology of the disease and characteristics of the test. Even though sensitivity and specificity both involve the test and the underlying biology, they are inherently separate concepts. The specificity of ultrasound for gallstones, for example, involves the biology of healthy people (i.e., how often do patients without gallstones have high-density sludge, polyps, or other stone mimics) and the characteristics of the ultrasound machine, technician, or reader when nothing is truly there. For example, how often does the technologist capture images with artifacts over the lumen mimicking stones? Does the reader tend to over-call gallstones?
By definition, these are separate characteristics of tests. They are determined in different populations (people known to have disease vs not) and rely on different characteristics of the test (how good is it at finding the abnormality vs how likely is it to incorrectly suggest an abnormality).
If you are very familiar with these concepts, you might argue that, although they are separate characteristics, there is a connection between sensitivity and specificity. It is true that extremely sensitive tests tend to have lower specificity and vice versa. This is not a rule, however. Also, for any given test in which a threshold is used to determine positivity, adjusting the threshold almost always improves sensitivity or specificity at the cost of the other.
Let us address an important question: how do we determine the sensitivity and specificity of a test if they can only be calculated in people with known disease status? After all, the point of developing a test is to diagnose the disease. These parameters are determined in research studies that use a reference standard to confirm the true disease status. What happens if the reference standard is imperfect or if the reference standard is a clinical determination that partly relies on the diagnostic test in question? These are great questions that may suggest that the reference standard is not entirely the “reference” and, therefore, a potential source of bias. Sensitivity and specificity calculated with poor reference standards may be inaccurate. As astute readers, we should closely evaluate the reference standard in any study reporting diagnostic test performance.
A common way to recruit patients for a study of a diagnostic test is to use case-control sampling. The word “sampling” can be thought of as a synonym for “recruitment” and specifically refers to the way in which patients are recruited, or sampled, from the underlying population. With case-control sampling, a group of diseased patients (case patients, in whom the disease is confirmed with the reference standard test) is compared with a group of nondiseased patients (control subjects, in whom the lack of disease is also confirmed with the reference standard). The number of control subjects studied is often chosen to achieve a one-toone or other ratio to the number of cases. Both groups undergo the test. Although we can summarize the results of a study such as this in a standard 2 × 2 table (Table 1), it may be helpful to first think of the results summarized in two 2 × 1 tables (Table 2).
TABLE 1: Standard 2 × 2 Table for Diagnostic Test
| | | |
---
| | Disease Positive | Disease Negative |
| Test positive | A | B |
| Test negative | C | D |
| Total | A + C | B + D |
Note—As per convention, true disease status as determined by reference standard is on top. Although many different study designs can be summarized in 2 × 2 table, if population is sampled irrespective of disease state (i.e., cross-sectional sampling), the proportion of study population with disease should mirror the proportion in the population. The proportion with disease (often called prevalence) equals those with disease (A + C) over everyone recruited (A + B + C + D). Thus, disease prevalence equals (A + C) / [(A + C) + (B + D)].
TABLE 2: Standard 2 × 2 Table Broken Down Into Two 2 × 1 Columns
| | | | | |
--- ---
| | Disease Positive | | | Disease Negative |
| Test positive | A | | Test positive | B |
| Test negative | C | Test negative | D |
| Total | A + C | Total | B + D |
Note—When you only know sensitivity and specificity of test, it may be helpful to think of the 2 × 2 table as two columns that can be evaluated separately. Sensitivity (two left columns) can only be determined in population confirmed to have disease. Total disease population tested is equal to A + C. A is number of diseased people in whom test is correct, and B is number in whom test is incorrect; thus, sensitivity equals A / (A + C). Specificity (two right columns) is the proportion of patients with correct test result (i.e., “negative”) of all nondiseased people; thus, specificity equals D / (B + D).
Case 1 Answer: Pitfall 1—Confusing Sensitivity or Specificity for Predictive Values
The conversation in case 1 contains a number of problems. With a positive radiograph, the real question is how likely is it that this positive radiograph reflects a true fracture versus something that looks like a fracture but is not (a false-positive)? If one would like to get an idea for how often this second possibility happens, a false-positive finding, one would ask about specificity (not sensitivity). Highly specific tests rarely result in false-positives (i.e., rarely is positive in those without disease), and low-specificity tests often result in falsely positive findings.
The second, larger, problem is that the ED physician was not trying to ascertain the specificity of the test to get an idea how often it may be falsely positive; he actually wanted a precise measure of “certainty” after receiving a positive test result. Neither specificity nor sensitivity tells us exactly how certain we are of a diagnosis after a diagnostic test result. That question depends on the precise balance of true-positives (actual fractures) versus false-positives (fracture “fake-outs”). With a positive or negative test in hand, the certainty of diagnosis is addressed with the concept of a “predictive value,” as described in the next section.
Case 2
Suppose a new disease was recently discovered in people who live below high-voltage power lines. Apparently, only years and years of exposure cause the disease, but even then, it is very rare. Only 100 cases are known. There is a long presymptomatic phase, but once it is clinically evident, the disease is irreversible and debilitating. The government has encouraged people near power lines to move, but the costs are tremendous and residents are reluctant to leave because the disease is so rare. For years, an accurate diagnostic test for presymptomatic patients was lacking, until the field of radiology comes to the rescue. High-density CSF on head CT proves to be 99% sensitive and 99% specific for the disease. The landmark study describing this finding recruits all 100 clinically symptomatic cases known; 99 have high-density CSF. They also recruit 100 healthy subjects taken from nearby towns who do not live near the power lines; 99 of them do not have high-density CSF. Table 3 is an example of a 2 × 2 table created from these data.
TABLE 3: Patients With and Without Disease, by CSF Status (Case 2)
| CSF Status | Disease Positive | Disease Negative | Total |
--- --- |
| High-density CSF | 99 | 1 | 100 |
| Normal-density CSF | 1 | 99 | 100 |
| Total | 100 | 100 | 200 |
In this scenario, the authors report that the sensitivity, specificity, positive predictive value (PPV), and negative predictive value (NPV) are all 99% (each is 99/100). They state, “This is a ground-breaking finding. A positive test confers a tremendously high risk of disease (99%). Everyone near power lines should be screened and everyone with a positive test should be moved at any cost to avoid almost certain debilitation from this dreaded disease.” Do you agree?
Putting the Two Columns Together: What Assumptions Are Made When Creating a 2 × 2 Table?
How do we move from a world where we only know sensitivity and specificity (which can be graphically depicted as two columns, as in Table 2) to a world where we put them together and can consider predictive values? The most intuitive use of a 2 × 2 table is when the numbers in each box come from a study design that uses cross-sectional sampling. Again, the word “sampling” refers to how patients are recruited, or sampled, from a predefined underlying population. In a study with cross-sectional sampling, a random or consecutive sample of the population is selected irrespective of disease or test status. With a screening test, the underlying population might be quite large (e.g., all people older than 50 years), but with targeted tests, the population may be quite small (e.g., only people with certain unusual symptoms). Regardless, patients are recruited from the underlying population without the investigators knowing their disease or test status. Once the patients are recruited, the authors then have to identify who has the disease using the reference standard and who has a positive test with the test in question. With rare diseases, a very large sample would be needed using this type of sampling, or else there would be very few people with the disease. Table 1 is a standard 2 × 2 table. An example study from the radiology literature is summarized in Table 4; the test in this case, CT colonography, is compared with a reference standard of optical colonoscopy . Note in this example that the disease, having a polyp, is actually quite common, at 61% of the sampled subjects. The important point is, when a study recruits patients irrespective of disease status, the prevalence in the study population should reflect the prevalence in the underlying population.
TABLE 4: Example 2 × 2 Table From Study Using Cross-Sectional Sampling
| Polyps Found on CT Colonography | Polyps Found on Colonoscopy | |
---
| ≤ 1 Polyp | 0 Polyps |
| ≤ 1 polyp | 164 | 33 |
| 0 polyps | 18 | 85 |
| Total | 182 | 118 |
Note—Three hundred patients were recruited who underwent CT colonography (test) and colonoscopy (reference standard). Patients were recruited sequentially before either test was performed (i.e., cross-sectional sample). Sensitivity (90% [164/182]) and specificity (72% [85/118]) were calculated. Disease prevalence in study (61% [182/300]) should mirror prevalence of standard referral population at institution (barring significant biases). Please note that this is simplified summary of one finding in a complex and detailed article.
Predictive Values
With a 2 × 2 table filled with cross-sectional sampling study data, we can now discuss PPVs and NPVs. These terms are test centric, meaning that they provide information about what the test means in patients with positive or negative test results. This is unlike sensitivity and specificity, which reflect the test's performance in people with known disease or nondisease (i.e., these are disease-centric terms). The PPV is the proportion of patients with a true-positive test among all patients with a positive test (which includes both true-and false-positive tests). The NPV is the proportion of patients with a true-negative test among all patients with a negative test (which includes both true- and false-negative tests). Note that the test result (positive or negative) defines the group of people in the denominator, from which we find the proportion of people who test correctly. As when we calculate sensitivity and specificity, the numerator is always the “correct answers.”
In a 2 × 2 table, predictive values are terms that involve looking at the numbers along the rows of the 2 × 2 table, not the columns. This concept is illustrated in Table 5.
TABLE 5: Determining Predictive Values From Standard 2 × 2 Table
| | | | |
--- --- |
| | Disease Positive | Disease Negative | Total |
| Test positiv | A | B | A + B |
| Test negative | C | D | C + D |
| Total | A + C | B + D | (A + C) + (B + D) |
Note—Predictive values are determined by looking at rows of 2 × 2 table. Positive predictive value is proportion of true-positives (A) divided by all positive test results (A + B), or A / (A + B). Negative predictive value is proportion of true-negatives (D) divided by all negative test results (C + D), or D / (C + D). Disease prevalence equals (A + C) / [(A + C) + (B + D)].
Using (and Misusing) Predictive Values
PPVs and NPVs are the holy grail of diagnostic test research. Why? These parameters can be directly applied to patients in clinical settings . For example, the referring physician in case 1 was, in truth, trying to ascertain the PPV to help him determine the next step in his management. When we do not have predictive values readily available, physicians consciously or unconsciously tend to estimate them and apply them clinically. The importance of predictive values (known or estimated) cannot be overstated.
What makes these values so important? When we are using diagnostic tests on clinical patients, we do not know their disease status. Presumably, patients undergoing diagnostic tests have not had the reference standard. If the reference standard were cheap and easy, there would be no reason to use some other diagnostic test at all. With clinical patients, we want to know the likelihood that the patient has the disease after a positive or negative diagnostic test result. This is exactly what the predictive values tell us.
Predictive values can be misused, as occurred in case 2. The misapplication is often not in the math but rather in calculating it with the wrong type of data, most commonly data from studies with case-control sampling (again, “sampling” refers to how the patients were recruited). These types of studies are often easier to perform than studies with cross-sectional sampling and are quite common in the radiology literature. Their ease comes from usually small sample sizes and the guarantee of having a reasonable number of both sick and healthy patients. Cross-sectional sampling may require large sample sizes, particularly if the disease is rare, and can therefore be very costly. The benefit to cross-sectional sampling, however, is it can yield sensitivity, specificity, and predictive values, as we showed already when we reviewed predictive values. Case-control sampling can only directly yield sensitivity and specificity.
What happens when predictive values are inappropriately calculated from a study with case-control sampling? Case-control–sampled studies usually have a disease prevalence in the study that is higher than that in the underlying population (i.e., reality). For example, one-toone case-control recruitment always results in a study disease prevalence of 50%, which is a higher prevalence than observed with most diseases. A PPV inappropriately calculated from a study with case-control sampling (with artificially high disease prevalence) will be artificially high; this is because the higher disease prevalence in the study results in a greater abundance of true-positives. An NPV inappropriately calculated from a study with case-control sampling (with artificially high disease prevalence) is artificially low; this is because the artificially high study disease prevalence results in fewer healthy patients and fewer true-negatives.
Case 2 Answer: Pitfall 2—Inappropriately Calculating Predictive Values From Studies Using Case-Control Sampling
The case 2 study is one with case-control sampling. This is a very rare disease, and all 100 known cases are recruited. With 100 control subjects, we have an apparent disease prevalence in the study of 50%. That is obviously not representative of the population as a whole. As a consequence, the inappropriately calculated predictive values do not represent the predictive values of the test if it were applied to the underlying population in question. However, the calculated sensitivity and specificity are entirely correct and are the main contribution of this study. To illustrate this point, assume that the disease prevalence is actually 0.1% in those who would be tested (i.e., people living under power lines). To get 100 cases, one would have to recruit 100,000 subjects and perform 100.000 head CT examinations. The 2 × 2 table would look like Table 6.
TABLE 6: People Living Under Power Lines, With and Without Disease, by CSF Status (Case 2)
| CSF Status | Disease Positive | Disease Negative | Total |
--- --- |
| High-density CSF | 99 | 9,990 | 10,089 |
| Normal-density CSF | 1 | 89,910 | 89,911 |
| Total | 100 | 99,900 | 100,000 |
The PPV in this case would be 0.009 (99/10,089), so the researchers in case 2 incorrectly overestimated the PPV. If this test were applied to the true population, a person with a positive test result would only have a 0.9% chance of actually having the disease. Although the test is very sensitive (i.e., nearly all diseased individuals will be detected), there are so many nondiseased individuals in the population being tested that the number of false-positive results becomes a real problem. If the government were to adopt screening with this test and move all individuals with a positive test result, 100 people would have to move to prevent 1 person from developing the disease.
You might think the world of research is doomed: studies with case-control sampling give the wrong PPV and NPV, and studies with cross-section sampling of rare diseases must be prohibitively huge. The truth is that there are ways around these problems: sensitivity and specificity can be combined mathematically with the underlying risk of disease (which must be known) to derive predictive values. The mechanics of this are discussed in detail elsewhere [1, 7].
If you are thinking to yourself that this is an overstated problem and that no one would make this mistake, here are two examples from the literature [8, 9]. In the first example, an inappropriately calculated predictive value is reported in the abstract , and in the second, inappropriately calculated predictive values are reported in a table .
Case 3
Suppose a frightening new infection is striking the nation. Approximately 1% of all Americans become infected before a cure could be found. The cure is extraordinarily expensive, as is the only known test for the disease (i.e., the reference standard), which also takes weeks of processing. The medical community begins searching for a less expensive and quicker way to identify those with early treatable disease. The field of radiology comes to the rescue. It is discovered that pericardial calcification detected on chest radio-graph is an excellent sign, with 99% sensitivity and 99% specificity. The landmark study reports these values and also highlights the impressively low false-negative and false-positive rates: both 1%. Your institution immediately begins screening. You get a call from an upset patient who states, “Before I was tested, I had a 1% chance of having this dreaded infection [the prevalence of the disease]. I have a negative x-ray, which has a 1% false-negative rate, so that means I still have a 1% chance of having the disease. What good was this test?” What do you say?
False-Positive and False-Negative Rate Confusion
In a standard 2 × 2 table (Table 1), there is no confusion about who are the false-positives (group C) and false-negatives (group B). However, people like to think in terms of proportions and will often refer to a false-positive or false-negative rate.
The word rate itself is often confusing in medical writing and can have many meanings, usually referring to the frequency with which events happen over a period time. In this context, the word rate simply implies a proportion or percentage. Though we very much dislike it, we will use the word rate here because it is quite commonly used.
Even if we correctly understand the meaning of rate, we are still left with confusion. What is the denominator used to calculate a false-positive or false-negative rate? Take the false-negative rate as an example: the number of false-negatives can be divided by the number of all negative test results (i.e., this would be a test-centric way of thinking about it, and the rate would be equal to 1 – NPV) or the denominator could be all truly disease-positive patients (i.e., this would be a disease-centric way of thinking about it, and the rate would be equal to 1 – sensitivity). In clinical practice, the test-centric value (1 – NPV) is the most useful, because it addresses the question of all the negative test results—that is, how many of those individuals actually have the disease despite having negative test results. The more conventional usage is actually the other, the disease-centric value (1 ? sensitivity), probably because of the ease of calculating it from relatively ubiquitous sensitivity values.
Unfortunately, there are not unique terms to distinguish to which denominator a false-positive or false-negative rate refers. Although convention tends toward the disease-centric definition (with false-positive and false-negative rates implying 1 – specificity and 1 – sensitivity, respectively), many people intentionally or unintentionally use the other definition. One should not make assumptions when a speaker or author uses the phrase “false-negative rate”; rather, we should attempt to determine how the value was calculated. Also, as a writer or speaker yourself, it is important to be clear about which one you mean.
Case 3 Answer: Pitfall 3—False-Negative (and -Positive) Confusion
The article about the outbreak reports a 1% false-negative rate. Given that the sensitivity is reported as 99%, we assume that the calculation was made using the disease-centric understanding (1 – sensitivity). The patient, however, interpreted the rate to mean the more clinically relevant and test-centric understanding (1 – NPV), which would be the proportion of patients with a negative test (like this patient) who actually have the disease despite the negative test. To determine that number, we need to first know the NPV, which is not reported in the scenario, but which we can derive from the information in the problem. We could fill in a theoretic 2 × 2 table assuming we had a large population (say, 10,000 people). We know the prevalence is 1% (so we fill in the lower row with 100 and 9900). We are also told the sensitivity and specificity, so we can fill in the remaining cells, as in Table 7.
TABLE 7: Determining Predictive Values and Chance of Becoming Infected, by Presence or Absence of Pericardial Calcification (Case 3)
| Pericardial Calcification | Disease Positive | Disease Negative | Total |
--- --- |
| Present | 99 | 99 | 198 |
| Absent | 1 | 9801 | 9802 |
| Total | 100 | 9900 | 10,000 |
The NPV is 0.9999 (9801/9802), meaning a negative test confers a 99.99% chance of not having the disease. The risk of actually still having the disease given a negative test is 0.01% (100% ? 99.99%). Even without doing the math, when presented with a situation such as this, we can realize two different meanings of the “false-negative rate” are being used, leading to confusion.
Do you think that this does not happen in practice? Here are two articles reporting the accuracy of a rapid influenza test against a reference standard: the Centers for Disease Control and Prevention website reports a false-positive rate equal to 1 minus PPV, whereas the journal article reports a false-positive rate equal to 1 minus specificity.
Conclusion
We have reviewed the key terms used to describe diagnostic test performance, which is the cornerstone of radiology. We have reviewed three very common pitfalls.
The first pitfall is confusing sensitivity and specificity with predictive values. Sensitivity and specificity describe the performance of a diagnostic test and are estimated in research studies. In daily radiology practice, predictive values are of the greatest utility because they can be applied to individual patients after they are tested (e.g., “What is the significance of this test result?”).
The second pitfall is inappropriately calculating predictive values from studies with case-control sampling. When studies select an arbitrary number, or ratio, of subjects with and without the disease, predictive values cannot be directly calculated.
The third pitfall is confusion regarding false-negative and false-positive “rates.” The proportion of patients with an incorrect test result (often confusingly referred to as a “rate”) can be calculated with two different denominators, either all the people with the same test result (positive or negative) or all the people of the same disease status (with or without the disease). You must dig deeper if you encounter these terms to figure out which is being used; usually “rates” that refer to 1 minus predictive value are more clinically relevant.
Although the basics of epidemiology are taught in medical school, and are even tested on national licensure examinations, we find that these pitfalls are fairly common both in daily discussions and in the literature. Being familiar with evidence-based diagnosis will help us serve our patients and referring clinicians. For interested readers, additional articles are available describing these terms in intuitive ways [6, 12–15], and many textbooks that provide more in-depth discussions are available [1, 7, 16].
Footnote
WEB
This is a Web exclusive article.
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Information & Authors
Information
Published In
American Journal of Roentgenology
Volume 200 | Issue 6 | June 2013
Pages: W566 - W570
PubMed: 23701086
Copyright
© American Roentgen Ray Society.
History
Submitted: August 29, 2012
Accepted: October 23, 2012
First published: May 23, 2013
Keywords
clinical epidemiology
evidence-based diagnosis
predictive values
sensitivity
specificity
Authors
Affiliations
David M. Naeger
Department of Radiology and Biomedical Imaging, University of California, San Francisco, 505 Parnassus Ave, M-391, Box 0628, San Francisco, CA 94143-0628.
View all articles by this author
Maureen P. Kohi
Department of Radiology and Biomedical Imaging, University of California, San Francisco, 505 Parnassus Ave, M-391, Box 0628, San Francisco, CA 94143-0628.
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Emily M. Webb
Department of Radiology and Biomedical Imaging, University of California, San Francisco, 505 Parnassus Ave, M-391, Box 0628, San Francisco, CA 94143-0628.
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Andrew Phelps
Department of Radiology and Biomedical Imaging, University of California, San Francisco, 505 Parnassus Ave, M-391, Box 0628, San Francisco, CA 94143-0628.
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Karen G. Ordovas
Department of Radiology and Biomedical Imaging, University of California, San Francisco, 505 Parnassus Ave, M-391, Box 0628, San Francisco, CA 94143-0628.
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Thomas B. Newman
Department of Epidemiology and Biostatistics, University of California, San Francisco, San Francisco, CA.
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Notes
Address correspondence to D. M. Naeger (david.naeger@radiology.ucsf.edu).
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TABLE 1: Standard 2 × 2 Table for Diagnostic Test
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TABLE 2: Standard 2 × 2 Table Broken Down Into Two 2 × 1 Columns
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TABLE 3: Patients With and Without Disease, by CSF Status (Case 2)
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TABLE 4: Example 2 × 2 Table From Study Using Cross-Sectional Sampling
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TABLE 5: Determining Predictive Values From Standard 2 × 2 Table
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TABLE 6: People Living Under Power Lines, With and Without Disease, by CSF Status (Case 2)
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TABLE 7: Determining Predictive Values and Chance of Becoming Infected, by Presence or Absence of Pericardial Calcification (Case 3)
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References
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10194 | https://www.chemguideforcie.co.uk/2022section27/learning27p1p2.html | chemguide: CIE A level chemistry support: Learning outcome 27.1.2
Chemguide: Support for CIE A level Chemistry
Learning outcome 27: Group 2
27.1: Similarities and trends in the properties of the Group 2 metals, magnesium to barium, and their compounds
Learning outcome 27.1.2
This statement asks for the trend in the solubilities of the Group 2 sulfates and hydroxides, and an explanation for that trend.
Before you go on, you should find and read the statement in your copy of the syllabus.
The Chemguide pages about this are far more complicated than you will need for the sort of questions that CIE ask, and the answers they expect, so I will cover most of what you need to know below.
The explanation given on this page is only a part of the proper explanation. If you go on to do Chemistry at a higher level, you should be aware that this explanation is very over-simplified. It is, however, what CIE expect.
You won't be able to follow this explanation unless you have already covered enthalpies of solution, lattice enthalpies and hydration enthalpies from the energetics part of the syllabus (section 23).
The trends
Sulfates
The sulfates become less soluble as you go down the group.
Magnesium sulfate is soluble in water.
Calcium sulfate is only very slightly soluble in water.
Strontium and barium sulfates are virtually insoluble in water.
You almost certainly know enough simple chemistry to be able to work this trend out.
At some point, you will certainly have met the reaction between magnesium and dilute sulphuric acid to give hydrogen and a solution of magnesium sulfate. You know that magnesium sulfate is soluble.
You will also have come across the test for a sulfate by adding barium chloride (or nitrate) solution to a solution of a sulfate. You get a white precipitate of barium sulfate. So you know that barium sulfate is insoluble.
Sulfates become less soluble as you go down the group.
Hydroxides
The hydroxides become more soluble as you go down the group.
None of them are very soluble, but the solubility increases as you go down the Group. See the beginning of the page about the solubility of the hydroxides (etc).
The explanation for sulfates
The simple explanation is in terms of the changes which occur when an ionic compound dissolves in water. Energy has to be supplied to break up the lattice of ions, and energy is released when these ions form bonds with water molecules.
Breaking up the lattice
To break up an ionic lattice, you need to supply lattice dissociation enthalpy.
The size of the lattice dissociation enthalpy depends on the charges on the ions, and the distances between their centres. All of the Group 2 sulfates consist of 2+ ions attracting 2- ions, and so the only thing that matters is the distance between the ion centres.
As you go down the group, the energy needed to break up the lattice falls as the positive ions get bigger. The bigger the ions, the more distance there is between their centres, and the weaker the forces holding them together.
Releasing energy by forming bonds with water molecules
Energy is released as hydration enthalpy when water molecules cluster around the free metal ions and sulfate ions.
As the positive ions get bigger, the energy released as the ions bond to water molecules falls. Bigger ions aren't so strongly attracted to the water molecules.
The overall effect
Both lattice dissociation enthalpy and hydration enthalpy fall as you go down the group. What matters is how fast they fall relative to each other.
As you go down the group, the lattice dissociation enthalpies don't fall as much as the hydration enthalpies of the positive ions.
The size of the hydration enthalpy of a positive ion is due only to the size of that ion. But that isn't so for lattice dissociation enthalpy.
The lattice dissociation enthalpy is governed by the distance between the centres of the ions, and that is made up of the radius of the large sulfate ion, plus the radius of the smaller positive ion.
The effect of the change in size of the positive ion is being diluted by the presence of the large sulfate ion.
Putting some numbers on this
This is much easier to understand if you have got some numbers to work with. All the values in the table are in kJ per mole.
Note:I have no confidence that the numbers I am going to use are reliable. There seem to be as many different values for these as there are sources of information (more about this at the bottom of the page).
| | lattice enthalpy | hydration enthalpy M 2+ | hydration enthalpy SO 4 2- | overall change |
--- ---
| CaSO 4 | +2653 | -1583 | -1137 | -67 |
| SrSO 4 | +2603 | -1450 | -1137 | +16 |
You can see that the lattice dissociation enthalpy has fallen by only 50 kJ, whereas the hydration enthalpy of the positive ion has fallen by 133 kJ. A bit less heat had to be put in in order to break the lattice, but quite a lot less was given out when the ions bonded to the water.
The net effect is that the overall process becomes less exothermic (or, in this case, actually becomes endothermic).
The explanation for hydroxides
The underlying explanation is still the same. Both lattice dissociation enthalpy and hydration enthalpy fall as you go down the group, and what matters is how fast they fall relative to each other.
Hydroxide ions are much smaller than sulfate ions, and so the size of the positive ion makes up a greater proportion of the distance between the positive and negative ions in the hydroxide case.
In this case, the lattice dissociation enthalpy falls by more than the hydration enthalpy as you go down the group.
As you go down the group, the energy you need to put in falls by more than the energy you get out. That makes the overall process more exothermic as you go from magnesium hydroxide to barium hydroxide.
What do CIE expect you to say?
A straightforward question asking about this would probably ask you to state and explain the trend in solubilities of the sulfates or hydroxides of Group 2 elements. Your answer would need to include:
For sulfates:
Solubility decreases as you go down the group.
The lattice dissociation enthalpy and hydration enthalpy both decrease as you go down the group.
The hydration enthalpy decreases more than the lattice dissociation enthalpy.
Therefore the enthalpy of solution becomes more endothermic (or less exothermic).
For hydroxides:
Solubility increases as you go down the group.
The lattice dissociation enthalpy and hydration enthalpy both decrease as you go down the group.
The lattice dissociation enthalpy decreases more than the hydration enthalpy.
Therefore the enthalpy of solution becomes more exothermic (or less endothermic).
There is a very straightforward question on this in the specimen paper for the syllabus starting in 2022. See paper 4 Q1(b). Don't forget to look at the mark scheme that goes with it.
There is a more complex question involving some calculations about the relative solubilities of magnesium and strontium sulfates and hydroxides in May /June 2010 paper 42 Q2 parts (a) and (b). Again, the mark scheme is essential.
Read this if you are likely to do chemistry beyond A level:I have no confidence whatsoever in this explanation, even though you will find it in lots of books, and all over the web. It looks to me like one of those explanations which "everybody knows", but which falls to pieces as soon as you look at the data. There seem to be a lot of these in inorganic chemistry at this level!
You will find the problems discussed in some detail on the page problems in explaining the solubility of Group 2 compounds. Don't even think about reading this unless your chemistry is really good.
The problem basically is that it is impossible to explain these patterns unless you include entropy in your explanation. Any explanation which doesn't include entropy is at best incomplete, and at worst, wrong.
I haven't been able to find any reliable data for this topic. Different data sources give different values both for lattice energies and hydration energies. Although values for calcium sulfate and strontium sulfate produce the same result whatever source you use (i.e. that strontium sulfate is likely to be less soluble than calcium sulfate), that doesn't hold true if you extend it to barium sulfate. Selecting values to fit your hypothesis, and ignoring others, is just bad science.
If anyone knows where I can get reliable values for the necessary lattice enthalpies and hydration enthalpies for all the Group 2 sulfates, could you let me know via the address on the about the CIE section page. I would also like to know why you think that particular set of values is reliable.
Go to the Section 27 Menu . . .
To return to the list of learning outcomes in Section 27
Go to the CIE Main Menu . . .
To return to the list of all the CIE sections
Go to Chemguide Main Menu . . .
This will take you to the main part of Chemguide.
© Jim Clark 2020 |
10195 | https://artofproblemsolving.com/wiki/index.php/Rearrangement_Inequality?srsltid=AfmBOoqBRNh-uuiRBB_-J-b-6n6xrp1eHNs9xsSZX6BpiQQiGYTatvG_ | Art of Problem Solving
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Rearrangement Inequality
The Rearrangement Inequality states that, if is a permutation of a finiteset (in fact, multiset) of real numbers and is a permutation of another finite set of real numbers, the quantity is maximized when and are similarly sorted (that is, if is greater than or equal to exactly of the other members of , then is also greater than or equal to exactly of the other members of ). Conversely, is minimized when and are oppositely sorted (that is, if is less than or equal to exactly of the other members of , then is also greater than or equal to exactly of the other members ).
Contents
1 Introductory
2 Intermediate
2.1 Proof of the Rearrangement Inequality
3 Uses
4 See Also
5 External Links
Introductory
Consider the following simple application: suppose you are involved in the hold-up of a convenience store. You note, as you are emptying the register, that there are different numbers of each denomination (penny, nickel, dime, quarter, dollar bill, five dollar bill, ten dollar bill and twenty dollar bill) in the register. When would your take be maximized? Certainly, you would hope that there would be the largest number of twenty dollar bills, then the next largest number of tens, etc. Meanwhile, you would find yourself very disappointed if there were more pennies than nickels, more nickels than dimes, and so on. This is a simple application of the rearrangement inequality. It is also an application of the greedy algorithm, so one possible interpretation of the rearrangement inequality is that sometimes, the greedy algorithm works.
Intermediate
Proof of the Rearrangement Inequality
The proof of the Rearrangment Inequality can be handled with proof by contradiction. Only the maximization form is proved here; the minimization proof is virtually identical.
Before we begin the proof properly, it is useful to consider the case where . Without loss of generality, sort and so that and . By hypothesis, . Expanding and taking some terms to the other side of the inequality, we get , as desired.
Now for the general case. Again, without loss of generality, set and ; and suppose the grouping that maximizes the desired sum of products is not the one that pairs with , with , and so on. This means that there is at least one instance where is paired with while is paired with , where and . However, using the technique seen above to prove the inequality for , we can see that the sum of products can only increase if we instead pair with and with (unless both a's or both b's are equal, in which case either we can choose another pair of products or note that the current arrangement is actually identical to the desired one), which contradicts our assumption that the arrangement we had was already the largest one.
Note: The minimization equality can be very easily proved by noting that if we have the set , ordered in decreasing order and the set , ordered in increasing order, then the maximum sum is just . Thus, by negating all values the inequality follows.
Uses
The Rearrangement Inequality has a wide range of uses, from MathCounts level optimization problems to Olympiad level inequality problems. A relatively simple example of its use in solving higher-level problems is found in the proof of Chebyshev's Inequality. It is particularly useful in that it does not require any terms of either sequence to be positive or negative, unlike the power-mean family of inequalities.
See Also
Chebyshev's Inequality
Power Mean Inequality
External Links
The Rearrangement Inequality by Dragos Hrimiuc
Retrieved from "
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Inequalities
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10196 | https://bio.libretexts.org/Bookshelves/Introductory_and_General_Biology/Biology_(Kimball)/18%3A_Evolution/18.06%3A_The_Hardy-Weinberg_Equilibrium | Skip to main content
18.6: The Hardy-Weinberg Equilibrium
Last updated
: Mar 17, 2025
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18.5: Mutation and Evolution
18.7: Polymorphisms
Page ID
: 5926
John W. Kimball
Tufts University & Harvard
( \newcommand{\kernel}{\mathrm{null}\,})
If two individuals mate that are heterozygous (e.g., Bb) for a trait, we find that
25% of their offspring are homozygous for the dominant allele (BB)
50% are heterozygous like their parents (Bb)
25% are homozygous for the recessive allele (bb) and thus, unlike their parents, express the recessive phenotype.
This is what Mendel found when he crossed monohybrids. It occurs because meiosis separates the two alleles of each heterozygous parent so that 50% of the gametes will carry one allele and 50% the other and when the gametes are brought together at random, each B (or b)-carrying egg will have a 1 in 2 probability of being fertilized by a sperm carrying B (or b). (Left table)
| Results of random union of the two gametes produced by two individuals, each heterozygous for a given trait. As a result of meiosis, half the gametes produced by each parent with carry allele B; the other half allele b. | | | Results of random union of the gametes produced by an entire population with a gene pool containing 80% B and 20% b. | | |
--- --- --- |
| | 0.5 B | 0.5 b | | 0.8 B | 0.2 b |
| 0.5 B | 0.25 BB | 0.25 Bb | 0.8 B | 0.64 BB | 0.16 Bb |
| 0.5 b | 0.25 Bb | 0.25 bb | 0.2 b | 0.16 Bb | 0.04 bb |
However, the frequency of two alleles in an entire population of organisms is unlikely to be exactly the same. Let us take as a hypothetical case, a population of hamsters in which 80% of all the gametes in the population carry a dominant allele for black coat (B) and 20% carry the recessive allele for gray coat (b).
Random union of these gametes (right table) will produce a generation:
64% homozygous for BB (0.8 x 0.8 = 0.64)
32% Bb heterozygotes (0.8 x 0.2 x 2 = 0.32)
4% homozygous (bb) for gray coat (0.2 x 0.2 = 0.04)
So 96% of this generation will have black coats; only 4% gray coats.
Will gray coated hamsters eventually disappear? No. Let's see why not.
All the gametes formed by BB hamsters will contain allele B as will one-half the gametes formed by heterozygous (Bb) hamsters.
So, 80% (0.64 + .50.32) of the pool of gametes formed by this generation with contain B.
All the gametes of the gray (bb) hamsters (4%) will contain b but one-half of the gametes of the heterozygous hamsters will as well.
So 20% (0.04 + .50.32) of the gametes will contain b.
So we have duplicated the initial situation exactly. The proportion of allele b in the population has remained the same. The heterozygous hamsters ensure that each generation will contain 4% gray hamsters. Now let us look at an algebraic analysis of the same problem using the expansion of the binomial (p+q)2.
The total number of genes in a population is its gene pool.
Let represent the frequency of one gene in the pool and the frequency of its single allele.
So,
= the fraction of the population homozygous for
= the fraction homozygous for
= the fraction of heterozygotes
In our example, p = 0.8, q = 0.2, and thus
The algebraic method enables us to work backward as well as forward. In fact, because we chose to make B fully dominant, the only way that the frequency of B and b in the gene pool could be known is by determining the frequency of the recessive phenotype (gray) and computing from it the value of q.
q2 = 0.04, so q = 0.2, the frequency of the b allele in the gene pool. Since p + q = 1, p = 0.8 and allele B makes up 80% of the gene pool. Because B is completely dominant over b, we cannot distinguish the Bb hamsters from the BB ones by their phenotype. But substituting in the middle term (2pq) of the expansion gives the percentage of heterozygous hamsters. 2pq = (2)(0.8)(0.2) = 0.32
So, recessive genes do not tend to be lost from a population no matter how small their representation.
Hardy-Weinberg law
So long as certain conditions are met (discussed below), gene frequencies and genotype ratios in a randomly-breeding population remain constant from generation to generation. This is known as the Hardy-Weinberg law.
The Hardy-Weinberg law is named in honor of the two men who first realized the significance of the binomial expansion to population genetics and hence to evolution. Evolution involves changes in the gene pool. A population in Hardy-Weinberg equilibrium shows no change. What the law tells us is that populations are able to maintain a reservoir of variability so that if future conditions require it, the gene pool can change. If recessive alleles were continually tending to disappear, the population would soon become homozygous. Under Hardy-Weinberg conditions, genes that have no present selective value will nonetheless be retained.
When the Hardy-Weinberg Law Fails
To see what forces lead to evolutionary change, we must examine the circumstances in which the Hardy-Weinberg law may fail to apply. There are five:
mutation
gene flow
genetic drift
nonrandom mating
natural selection
Mutation
The frequency of gene B and its allele b will not remain in Hardy-Weinberg equilibrium if the rate of mutation of B -> b (or vice versa) changes. By itself, this type of mutation probably plays only a minor role in evolution; the rates are simply too low. However, gene (and whole genome) duplication - a form of mutation - probably has played a major role in evolution. In any case, evolution absolutely depends on mutations because this is the only way that new alleles are created. After being shuffled in various combinations with the rest of the gene pool, these provide the raw material on which natural selection can act.
Gene Flow
Many species are made up of local populations whose members tend to breed within the group. Each local population can develop a gene pool distinct from that of other local populations. However, members of one population may breed with occasional immigrants from an adjacent population of the same species. This can introduce new genes or alter existing gene frequencies in the residents.
In many plants and some animals, gene flow can occur not only between subpopulations of the same species but also between different (but still related) species. This is called hybridization. If the hybrids later breed with one of the parental types, new genes are passed into the gene pool of that parent population. This process is called introgression. It is simply gene flow between species rather than within them.
Comparison of the genomes of contemporary humans with the genome recovered from Neanderthal remains shows that from 1–3% of our genes were acquired by introgression following mating between members of the two populations tens of thousands of years ago.
Whether within a species or between species, gene flow increases the variability of the gene pool.
Genetic Drift
As we have seen, interbreeding often is limited to the members of local populations. If the population is small, Hardy-Weinberg may be violated. Chance alone may eliminate certain members out of proportion to their numbers in the population. In such cases, the frequency of an allele may begin to drift toward higher or lower values. Ultimately, the allele may represent 100% of the gene pool or, just as likely, disappear from it.
Drift produces evolutionary change, but there is no guarantee that the new population will be more fit than the original one. Evolution by drift is aimless, not adaptive.
Nonrandom Mating
One of the cornerstones of the Hardy-Weinberg equilibrium is that mating in the population must be random. If individuals (usually females) are choosy in their selection of mates, the gene frequencies may become altered. Darwin called this sexual selection.
Nonrandom mating seems to be quite common. Breeding territories, courtship displays, "pecking orders" can all lead to it. In each case certain individuals do not get to make their proportionate contribution to the next generation.
Assortative mating
Humans seldom mate at random preferring phenotypes like themselves (e.g., size, age, ethnicity). This is called assortative mating. Marriage between close relatives is a special case of assortative mating. The closer the kinship, the more alleles shared and the greater the degree of inbreeding. Inbreeding can alter the gene pool. This is because it predisposes to homozygosity. Potentially harmful recessive alleles — invisible in the parents — become exposed to the forces of natural selection in the children.
It turns out that many species - plants as well as animals - have mechanisms be which they avoid inbreeding. Examples:
Link to discussion of self-incompatibility in plants.
Male mice use olfactory cues to discriminate against close relatives when selecting mates. The preference is learned in infancy - an example of imprinting. The distinguishing odors are controlled by the MHC alleles of the mice and are detected by the vomeronasal organ (VNO).
Natural Selection
If individuals having certain genes are better able to produce mature offspring than those without them, the frequency of those genes will increase. This is simply expressing Darwin's natural selection in terms of alterations in the gene pool. (Darwin knew nothing of genes.) Natural selection results from differential mortality and/or differential fecundity.
Mortality Selection
Certain genotypes are less successful than others in surviving through to the end of their reproductive period. The evolutionary impact of mortality selection can be felt anytime from the formation of a new zygote to the end (if there is one) of the organism's period of fertility. Mortality selection is simply another way of describing Darwin's criteria of fitness: survival.
Fecundity Selection
Certain phenotypes (thus genotypes) may make a disproportionate contribution to the gene pool of the next generation by producing a disproportionate number of young. Such fecundity selection is another way of describing another criterion of fitness described by Darwin: family size. In each of these examples of natural selection, certain phenotypes are better able than others to contribute their genes to the next generation. Thus, by Darwin's standards, they are more fit. The outcome is a gradual change in the gene frequencies in that population.
Calculating the Effect of Natural Selection on Gene Frequencies
The effect of natural selection on gene frequencies can be quantified. Let us assume a population containing
36% homozygous dominants (AA)
48% heterozygotes (Aa) and
16% homozygous recessives (aa)
The gene frequencies in this population are and . The heterozygotes are just as successful at reproducing themselves as the homozygous dominants, but the homozygous recessives are only 80% as successful. That is, for every 100 AA (or Aa) individuals that reproduce successfully only 80 of the aa individuals succeed in doing so. The fitness () of the recessive phenotype is thus 80% or 0.8.
Their relative disadvantage can also be expressed as a selection coefficient, , where
In this case,
The change in frequency of the dominant allele () after one generation is expressed by the equation
where and are the initial frequencies of the dominant and recessive alleles respectively. Substituting, we get
So, in one generation, the frequency of allele A rises from its initial value of 0.6 to 0.62 and that of allele a declines from 0.4 to 0.38 ( ).
The new equilibrium produces a population of
38.4% homozygous dominants (an increase of 2.4%) (p2 = 0.384)
47.1% heterozygotes (a decline of 0.9%)(2pq = 0.471) and
14.4% homozygous recessives (a decline of 1.6%)(q2 = 0.144)
If the fitness of the homozygous recessives continues unchanged, the calculations can be reiterated for any number of generations. If you do so, you will find that although the frequency of the recessive genotype declines, the rate at which a is removed from the gene pool declines; that is, the process becomes less efficient at purging allele a. This is because when present in the heterozygote, a is protected from the effects of selection.
18.5: Mutation and Evolution
18.7: Polymorphisms |
10197 | https://www.youtube.com/watch?v=B_yYrgRs8Sk | Rotational Dynamics Pulley Tension Atwood Machine Worked Example | Doc Physics
Doc Schuster
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Posted: 10 Apr 2015
I work through a healthy 2-mass Atwood Machine problem that my AP Physics 1 students just encountered. There's no friction, but the dang pulley has mass. This means that the tension in the rope is different between the two sides of the pulley. Even though it's the same rope.
If some nerdy guy works a problem for you, you should always say thanks, and then run away to work through the steps for yourself. That's how you will learn. DO THIS PROBLEM.
65 comments
Transcript:
so you walk into class right and there's this teacher and he's standing he's looking all cool you hate these teachers who think they're cool and they got haircuts and they were bow ties and all this stuff anyway he says hey bro what is the acceleration of these blocks oh and by the way find each of these tensions this one and this one and find the normal force on the whole apparatus oh and you can't use any other things except for M and R and Baby G so you're coming at it kind of stressed out right but you know what you need to do you need to start analyzing this and there's probably going to be some rotation and they say oh yeah and by the way there was this other thing I noted this is a mass on a uh like a pulley it's like a pulley e Mass it's like a Massy pulley but there's no friction so there's an axis of rotation right here and they tell you that the moment of inertia well the moment of inertia I guess would be because it's a solid um what is it a solid disc you would say that it's 12 m r s and they're giving you this radius here they call that lowercase R but since the mass is actually 3 m this moment of inertia ends up being 3 m R squ okay so this is what we would anticipate and they give that to us also so I'm trying to find all of this stuff and you know what the first step is you just cut off the question and start looking at this thing right fine first question is something about acceleration wasn't it let me see got a little bit disorganized right there I'm thinking that we should start out working in no Violet's no good let's start out working with we're mad at this guy for giving us this problem so let's start out with some aggressive red right here my first step is going to be making some free body diagrams of everything because if I'm ever going to find accelerations then I need to make free body diagrams so my first step is a free body diagram for this little M right here this little M of mine has mg pointing down and there's also a tension pointing up and this other guy is twice as massive so it's uh got a longer Vector right here and I'm going to write 2 mg pointing down and there's a tension pointing up and I immediately become uncomfortable sometimes students want to say that the upward pull on this one is equal to the weight of that one and the upward pull on this one is equal to the weight of that one that's nuts that's going way too far you don't know yet and another thing that's funny about this set situation is those are two tensions and it's the same dang rope but the Rope might be causing this thing to rotate in which case the Rope might be exerting a net torque on this pulley in which case the tension over here might not be the tension over here that's dangerous and I just don't want to assume that they're the same thing so I'm going to call this tension one over here and I'm going to call this tension two over here T1 T2 and that way I won't get confused about periods either if we have to worry about periods so there's one more object that needs a um a uh free body diagram and that's this loopy thing over here and my students took this problem and this is where they struggled on these three free body diagrams a lot of people got those but then they started wondering about what the tensions were this guy here is a mess we'll put a dot at the center and we'll start acknowledging that there are some forces on the sides and these forces might cause rotation so they are torky forces and here's one of them pointing that direction that would be a positive torque and it's called T1 it's just T1 there it is pointing down okay and causing a positive torque that T1 is not the torque but it's causing the torque if I calculated the torque I'd probably need this distance over here which is R that's great and then I've also got a T2 on this side I don't know how long they are but they're probably not the same length because that would be a fantastic coincidence so I've got two forces down wait a second doesn't this thing have itself its own mass of course it does so there's probably also 3 mg pointing down okay and then also there's going to be a normal force pointing up of some unknown value but we kind of get the impression that maybe all of these forces are going to be adding to zero because there's no net force on this guy there might be a net torque on it but the center of mass of that's not moving cuz the whole thing's sitting on a table it'd be cool if you put this on a balloon though and saw what the balloon okay but we're not going to do that our goal is to find accelerations so I'm going to give you some equations with acceleration in them and we're going to do that in see if you can handle it hot pink baby let's go this one gives us a free body diagram we need to Define positive though and uh for the sake of uh consistency with torque I'm going to say that this direction is positive so here x is positive that direction and here x is positive that direction I hope you don't see that as inconsistent because if this guy moves that way then that guy moves that way we're going to have some minus signs here because we already anticipate the acceleration is going to be that direction so uh that'll be okay when we get those minus signs we'll be fine with that so here's my first free body diagram's Newton's Second Law Newton Second Law is where you should go immediately after doing a free body diagram every time Newton's Second Law here says that net force equals mass time Exel ceration and the mass is M and the acceleration is what we don't know and there are two forces it looks like the downward one is positive so I write mg minus T1 there's our Newton's Second Law statement for that guy this one is going to generate a different Newton Second Law because mg is negative so it's going to be T2 minus M2 sorry not M2 but 2 m time Baby G this is getting a a little bit messy it's T2 minus mg but mg is 2 mg because they called it 2 m cuz it's twice as much as M all right great then I say this is equal to 2 m a all right wait a second these M these M's are different well I mean the letter M is different but the masses are different but the acceleration the accelerations are the same so we could plug these two equations into each other and have ourselves a party but the problem is we've already got three variables we've got T1 which we don't know and we've got a which we don't know and we've got T2 which we don't know we're going to need a third equation if you have three unknowns dang it you need three equations and that's this guy over here this guy's statement of Newton's second law is that the net torqus are equal to I Alpha that's Newton's second law for rotations and I'm going to switch to a more Bleak color at this time because a lot of students don't really like rotation yet but you know the moment of inertia I'm just going to leave it here as I but we know what it is and we know Alpha is related to a so instead of alpha I'm going to write a over ARA where this ARA right here is the radius of our spinny thing and then I'm going to go over here and look at the net torqus I've got one positive torque and that's T1 AA and I've got a negative torque and that's t 2 ARA all right T2 is a negative torque because it's causing it to rotate clockwise T1 now another huge mistake that my students made when they did this problem is they said that these forces aren't tensions they said they were equal to mg and they were equal to 2 mg now I want you to think about that for a moment if you know that this tension is equal to mg then it's like you've already gone over to this free body diagram and you said by the way I know that this mass is not accelerating well if the dang mass is not accelerating why are you doing the problem if the mass is not accelerating if it's in equilibrium if you know the two forces on an equal stop just stop write down acceleration equals zero Professor I'll see you Monday now we can't do that because it's a hard problem we've got ourselves three variables right here see these three variables oh nasty I'm going to solve this sucker for T1 minus T2 I'm factoring out an AA and then I'm going to divide both sides by AA I get I a over r² now isn't that nice isn't that nice now what if we even plugged in what i is um um oh that's cool I get three M R square a over R square whoop whoop whoop whoop whoop whoop whoop whoop awesome this is three halves look at this it's three ma 3 mass time acceleration and it's time for us to move into a wee bit of algebra so I'm going to scoot our setup out of the way and come back to it later I can still trace the wounds from that knife attack here brown brown this is the poop phase it's a whole bunch of math I'm going to take this guy and solve it for T1 I'm going to take that guy and solve it for T2 and then I'm going to plug him into this equation that's down here watch me go watch me go T1 equals mg minus M A did I make a mistake remember remember remember remember okay cool and then I'm going to take this guy and solve it for T2 that's even easier T2 equals 2 m a + 2 mg we Mak any mistakes yet I think we're pretty much rocking so far now I'm going to take these two and I'm going to plug them into uh watch I'm going to plug them into right here so I'm going to get mg minus M A minus all of this stuff so that's Min - 2 m a minus 2 mg equals that's T1 minus T2 as solved by this equation and that equation equals it's time for a color transition let's go orange now all that stuff T1 minus T2 supposed to be equal to three Hales mass times acceleration you see the beauty of what we've done here we took these three equations we plugged them in in such a way to eliminate all the tensions and ah it's a soothing Friday afternoon we have eliminated all tensions it's like a lemonade and a hammock this equation right here is like a lemonade and a hammock because look what happens look what happens I'm going to get the purple to kill M M M M M every term has an eminent and they are all gone they are all get it eminently doomed did you get it because there's an eminent if there's an eminent then it's eminently all right let's do it let's do this problem we're trying to solve for a and there are A's all over the place I need to write the equation again it says Gus a - 2 A's - 2 G's equals 3 Hales a well that's pretty cool I could uh I could start combining some things over here I have negative G and I also have - 3 a = 3 a oh it's looking kind of gross maybe the orange maybe it's time to retire orange I don't know let's look at this bold we got a Sharpie Violet right here um I'm going to get that over to there and I'm going to find that this is 3 a + 3 a = negative Baby G well first of all we're going to have a negative a does that make you uncomfortable I don't know it doesn't bother me too too much how many halves is three looks like six halves so we're going to have 9es of a is negative Baby G so a is 2 9ths of G but it's negative we knew it was negative so I'll just multiply by negative one right there the acceleration is 2 9ths of Baby G fascinating fascinating let's go back up to here and say that these guys first of all had some students who did this problem and they ran into accelerations that were equal to Baby G if you run into that then you need to say hold up is it possible that they are in freef Fall did somebody cut this and cut that and everybody says no no that is not possible if you get Baby G for the acceleration chances are you made a mistake somewhere so there are all kinds of places you can check in now what about this answer this says a equals let me write it a little bit nicer in um um um um oh let's just tear all this stuff off should we no let's not do that yet okay a = -2 9 Baby G this equation right here says the acceleration and it is only using M and R and Baby G golly it's only using Baby G that's a cool thing the M's just came into this factor out front and we just found some ratio of Baby G some uh some multiple of BG I mean okay there were some other questions that they had asked us uh remember when the professor came over and said hey bro we're going to find each tension right so that's going to be really easy we've got these two equations for the tension and all I have to do is plug in what a is so over here I know this is mg minus now Ma a but Ma a is oh gosh look it's negative right here so it's going to be plus because I'm subtracting a negative M 2 G / 9 oh my my goodness this is mg plus 2 9th mg that's 11 9th mg cool that's T1 that's the tension over on this side so the tension is more than mg that means if the tension is more than mg then this mass will accelerate upwards oh well maybe it is accelerating upwards all right let's do this other one this one we've got 2 Ma and the a is -2 9 so I'm going to get4 9 mg you agree with that so far okay and then I'm going to add to it 2 mg I get ne oh my gosh I'm going to have to go into nth this is 18 9ths and this sucker is 49ths that I'm taking away that's 14 9ths ew 149 mg oh gosh so um that's less than 2 mg right it's less than 2 mg right here that means that this tension is less than 2 mg which means this mass will be accelerating downwards oh it all comes together it's okay it's going to be okay it's going to be okay oh and there was one last question they said what is the normal force on the apparatus so to do that I need to consider the forces on the apparatus but to do that I need to consider the upward force on the pulley what is this normal Force right here let's look at that normal force by heck just reopen the old wound check this out oh sorry got to line on the camera again I'm looking at this guy right here and I'm saying that the normal force must be equal to T1 plus T2 + 3mg we know what T1 is I hope you'll recall I have it here in this puzzle piece pie um T1 was 11 9th mg it's 11 9th mg and then T2 check it T2 we said was 14 9ths mg plus 14 9th mg plus 3 mg well this is the normal force remember careful careful careful this is the normal force on the pulley so it's not the normal force on the whole apparatus we'll get there really soon though um uh this is the normal force on the pulley and it's 11 oh gosh 25 9ths how many 9ths is that that's 27 9ths holy cow what is this 52 9ths are you kidding me did I make a mistake this is 25 9ths and that's 27 9ths how terribly inconvenient 52 9ths mg okay but that's what it is that's what we find but then we also have to add on the additional mass of this guy which is in equilibrium so I'm going to take oh so I'm going to say FN whole thing this is a final calculation FN of the whole thing is 529 mg plus 4 mg so that is golly another 36 9ths can we handle this another 36 9th that's 88 9th mg now that's really interesting because 88 9ths mg is really close to 99th mg but it's not the same thing if we had 99th mg that would be 10 mg and I want you to go back to the original problem and I want you to notice that the total mass is 10 m let's check let's check see we got M we got 2 m that's 3M here's another 3M that's 6m here's another 4M that's 10m and yet the normal force on the entire thing is less than 10 mg what does that mean it means my dear students that the entire apparatus is accelerating but which way |
10198 | https://www.tmcec.com/wp-content/uploads/2023/06/Steps-in-a-Criminal-Trial-Lesson-4-30-2020.pdf | STEPS IN A CRIMINAL TRIAL Vocabulary: 1. Opening Statement—the attorneys from both sides introduce themselves, inform the jury about the facts, issues and evidence of the case, and ask for the verdict they want. 2. Direct Examination—the questioning of witness by the attorney who called the witness to testify. 3. Cross-Examination—when the opposing attorneys question witnesses. 4. Closing Argument—the attorneys from both sides speak to the jury, emphasizing the strengths of their case and try to persuade them that they have proven their side of the case. 5. Rebuttal—an additional argument given by the prosecution after the defense has presented its closing argument. 6. Verdict—the decision made by a judge or jury during a criminal trial. Teaching Strategy: 1. Have the class brainstorm the various steps in a trial. The teacher should record responses on the chalkboard or overhead. 2. Divide the class into groups of three or four students. Give each group an envelope containing a set of “Steps in a Trial” footsteps (Attachment 1). Ask each group to arrange the strips in the order in which they think each event occurs during a trial in the courtroom. TEKS: SS 3rd– 3.14B; 3.15A, 4th-4.19B, 4.21A, 5th-5.21B; 5.25A, 7th-7.20B; 7.22A, 8th -8.29B; 30A, Govt-19A; 20A Materials Needed: Five or six sets of “Steps in a Trial” footsteps (Handout 1) cut out, “Steps in a Trial” (Transparency 2). Vocabulary: prosecuting, opening statement, defense attorney, direct examination, cross-examination, defendant, rebuttal LEVEL TWO-9 Learning Objectives: Students will 1. Identify the steps in a trial. 2. Describe what is meant by burden of proof. LEVEL TWO-10 3. Debrief the activity by reviewing the correct order of the steps and discussing or clarifying any questions about courtroom procedures. IMPORTANT: Point out that because our legal system assumes the defendant is not guilty until proven guilty in a court of law, the prosecution goes first because the “burden of proof” is always on the prosecution. 4. The correct order of the steps in a trial is as follows: 1. The bailiff calls the case. 2. The judge enters the room and takes his or her seat (the bench). A) Judge calls the case before the court B) Prosecutor reads the charges aloud C) Defendant pleads “not guilty” 3. The prosecution’s attorney makes an opening statement. 4. The defendant’s attorney makes an opening statement. 5. The prosecution’s attorney questions witnesses that will help its side of the case (direct examination). 6. The defendant’s attorney cross-examines witnesses for the prosecution. 7. The defendant’s attorney questions witnesses that will help the defendant’s side of the case (direct examination). 8. The prosecution’s attorney cross-examines witnesses for the defense. 9. The prosecution’s attorney gives closing argument. 10. The defendant’s attorney gives closing argument. 11. The prosecution’s attorney gives a rebuttal (optional). 12. The judge explains to the jury how they are to determine if the defendant is not guilty or guilty (jury instructions). 13. The jury decides the verdict. 14. The decision is announced in court. LEVEL TWO-11 STEPS IN A TRIAL The correct order of the steps in a trial is as follows: 1. Bailiff/Clerk opens the court session. 2. Judge enters room and takes his or her seat (the bench). 3. Prosecuting attorney makes an opening statement. 4. Defense attorney makes an opening statement. 5. Prosecuting attorney questions witnesses that will help its side of the case (direct examination). 6. Defense attorney cross-examines witnesses for the prosecution. 7. Defense attorney questions witnesses who will help the defendant’s side of the case (direct examination). 8. Prosecuting attorney cross-examines witness for the defense. 9. Prosecuting attorney gives closing argument. 10. Defense attorney gives closing argument. 11. Prosecuting attorney gives a rebuttal (optional). 12. Judge explains to the jury how they are to determine if the defendant is guilty or not guilty. 13. The jury decides the verdict. 14. The decision is announced in court. TRANSPARENCY 1 Judge enters room and takes his or her seat (the bench). Bailiff/Clerk opens the court session. STEPS IN A TRIAL LEVEL TWO-12 LEVEL TWO-13 STEPS IN A TRIAL Prosecuting attorney makes an opening statement. Defense attorney makes an opening statement. LEVEL TWO-14 STEPS IN A TRIAL Prosecuting attorney questions witnesses that will help its side of the case (direct examination). Defense attorney cross-examines witnesses for the prosecution. LEVEL TWO-15 STEPS IN A TRIAL Defense attorney questions witnesses who will help the defendant’s side of the case (direct examination). The prosecuting attorney cross-examines witnesses for the defense. LEVEL TWO-16 STEPS IN A TRIAL Prosecuting attorney gives closing argument. Defense attorney gives closing argument. LEVEL TWO-17 STEPS IN A TRIAL Prosecuting attorney gives a rebuttal. (optional) Judge explains to the jury how they are to determine if the defendant is guilty or not guilty. LEVEL TWO-18 STEPS IN A TRIAL The jury decides the verdict. The decision is announced in court. |
10199 | https://www.usgs.gov/centers/national-minerals-information-center/sulfur-statistics-and-information | An official website of the United States government
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